Thermodynamics Solutions (6ed 2134 Pag)

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Chapter 1 INTRODUCTION AND BASIC CONCEPTS Thermodynamics 1-1C Classical thermodynamics is based on experimental observations whereas statistical thermodynamics is based on the average behavior of large groups of particles. 1-2C On a downhill road the potential energy of the bicyclist is being converted to kinetic energy, and thus the bicyclist picks up speed. There is no creation of energy, and thus no violation of the conservation of energy principle. 1-3C There is no truth to his claim. It violates the second law of thermodynamics. 1-4C A car going uphill without the engine running would increase the energy of the car, and thus it would be a violation of the first law of thermodynamics. Therefore, this cannot happen. Using a level meter (a device with an air bubble between two marks of a horizontal water tube) it can shown that the road that looks uphill to the eye is actually downhill.

Mass, Force, and Units 1-5C Pound-mass lbm is the mass unit in English system whereas pound-force lbf is the force unit. One pound-force is the force required to accelerate a mass of 32.174 lbm by 1 ft/s2. In other words, the weight of a 1-lbm mass at sea level is 1 lbf. 1-6C In this unit, the word light refers to the speed of light. The light-year unit is then the product of a velocity and time. Hence, this product forms a distance dimension and unit. 1-7C There is no acceleration, thus the net force is zero in both cases.

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1-8E The weight of a man on earth is given. His weight on the moon is to be determined. Analysis Applying Newton's second law to the weight force gives

W = mg ⎯ ⎯→ m =

W 180 lbf = g 32.10 ft/s 2

⎛ 32.174 lbm ⋅ ft/s 2 ⎜ ⎜ 1 lbf ⎝

⎞ ⎟ = 180.4 lbm ⎟ ⎠

Mass is invariant and the man will have the same mass on the moon. Then, his weight on the moon will be 1 lbf ⎞ ⎛ W = mg = (180.4 lbm)(5.47 ft/s 2 )⎜ ⎟ = 30.7 lbf 2 ⎝ 32.174 lbm ⋅ ft/s ⎠

1-9 The interior dimensions of a room are given. The mass and weight of the air in the room are to be determined. Assumptions The density of air is constant throughout the room. Properties The density of air is given to be ρ = 1.16 kg/m3. Analysis The mass of the air in the room is

m = ρV = (1.16 kg/m 3 )(6 × 6 × 8 m 3 ) = 334.1 kg

ROOM AIR 6X6X8 m3

Thus,

⎛ 1N W = mg = (334.1 kg)(9.81 m/s 2 )⎜ ⎜ 1 kg ⋅ m/s 2 ⎝

⎞ ⎟ = 3277 N ⎟ ⎠

1-10 The variation of gravitational acceleration above the sea level is given as a function of altitude. The height at which the weight of a body will decrease by 1% is to be determined. z Analysis The weight of a body at the elevation z can be expressed as W = mg = m(9.807 − 3.32 × 10−6 z )

In our case, W = 0.99Ws = 0.99mgs = 0.99(m)(9.807)

Substituting, 0.99(9.81) = (9.81 − 3.32 × 10 −6 z) ⎯ ⎯→ z = 29,539 m

0 Sea level

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1-11E The mass of an object is given. Its weight is to be determined. Analysis Applying Newton's second law, the weight is determined to be 1 lbf ⎛ ⎞ W = mg = (10 lbm)(32.0 ft/s 2 )⎜ ⎟ = 9.95 lbf 2 ⎝ 32.174 lbm ⋅ ft/s ⎠

1-12 The acceleration of an aircraft is given in g’s. The net upward force acting on a man in the aircraft is to be determined. Analysis From the Newton's second law, the force applied is ⎛ 1N F = ma = m(6 g) = (90 kg)(6 × 9.81 m/s 2 )⎜ ⎜ 1 kg ⋅ m/s 2 ⎝

⎞ ⎟ = 5297 N ⎟ ⎠

1-13 A rock is thrown upward with a specified force. The acceleration of the rock is to be determined. Analysis The weight of the rock is

⎛ 1N W = mg = (5 kg)(9.79 m/s 2 )⎜ ⎜ 1 kg ⋅ m/s 2 ⎝

⎞ ⎟ = 48.95 N ⎟ ⎠

Then the net force that acts on the rock is

Fnet = Fup − Fdown = 150 − 48.95 = 101.05 N

Stone

From the Newton's second law, the acceleration of the rock becomes a=

F 101.05 N ⎛⎜ 1 kg ⋅ m/s 2 = m 5 kg ⎜⎝ 1 N

⎞ ⎟ = 20.2 m/s 2 ⎟ ⎠

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1-14 EES Problem 1-13 is reconsidered. The entire EES solution is to be printed out, including the numerical results with proper units. Analysis The problem is solved using EES, and the solution is given below. W=m*g "[N]" m=5 [kg] g=9.79 [m/s^2] "The force balance on the rock yields the net force acting on the rock as" F_net = F_up - F_down"[N]" F_up=150 [N] F_down=W"[N]" "The acceleration of the rock is determined from Newton's second law." F_net=a*m "To Run the program, press F2 or click on the calculator icon from the Calculate menu" SOLUTION a=20.21 [m/s^2] F_down=48.95 [N] F_net=101.1 [N] F_up=150 [N] g=9.79 [m/s^2] m=5 [kg] W=48.95 [N]

1-15 Gravitational acceleration g and thus the weight of bodies decreases with increasing elevation. The percent reduction in the weight of an airplane cruising at 13,000 m is to be determined. Properties The gravitational acceleration g is given to be 9.807 m/s2 at sea level and 9.767 m/s2 at an altitude of 13,000 m. Analysis Weight is proportional to the gravitational acceleration g, and thus the percent reduction in weight is equivalent to the percent reduction in the gravitational acceleration, which is determined from Δg 9.807 − 9.767 × 100 = × 100 = 0.41% g 9.807 Therefore, the airplane and the people in it will weight 0.41% less at 13,000 m altitude. %Reduction in weight = %Reduction in g =

Discussion Note that the weight loss at cruising altitudes is negligible.

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Systems, Properties, State, and Processes 1-16C This system is a region of space or open system in that mass such as air and food can cross its control boundary. The system can also interact with the surroundings by exchanging heat and work across its control boundary. By tracking these interactions, we can determine the energy conversion characteristics of this system. 1-17C The system is taken as the air contained in the piston-cylinder device. This system is a closed or fixed mass system since no mass enters or leaves it. 1-18C Carbon dioxide is generated by the combustion of fuel in the engine. Any system selected for this analysis must include the fuel and air while it is undergoing combustion. The volume that contains this airfuel mixture within piston-cylinder device can be used for this purpose. One can also place the entire engine in a control boundary and trace the system-surroundings interactions to determine the rate at which the engine generates carbon dioxide. 1-19C When analyzing the control volume selected, we must account for all forms of water entering and leaving the control volume. This includes all streams entering or leaving the lake, any rain falling on the lake, any water evaporated to the air above the lake, any seepage to the underground earth, and any springs that may be feeding water to the lake. 1-20C Intensive properties do not depend on the size (extent) of the system but extensive properties do. 1-21C The original specific weight is

γ1 =

W

V

If we were to divide the system into two halves, each half weighs W/2 and occupies a volume of V/2. The specific weight of one of these halves is

γ =

W /2 = γ1 V /2

which is the same as the original specific weight. Hence, specific weight is an intensive property. 1-22C The number of moles of a substance in a system is directly proportional to the number of atomic particles contained in the system. If we divide a system into smaller portions, each portion will contain fewer atomic particles than the original system. The number of moles is therefore an extensive property. 1-23C For a system to be in thermodynamic equilibrium, the temperature has to be the same throughout but the pressure does not. However, there should be no unbalanced pressure forces present. The increasing pressure with depth in a fluid, for example, should be balanced by increasing weight.

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1-24C A process during which a system remains almost in equilibrium at all times is called a quasiequilibrium process. Many engineering processes can be approximated as being quasi-equilibrium. The work output of a device is maximum and the work input to a device is minimum when quasi-equilibrium processes are used instead of nonquasi-equilibrium processes. 1-25C A process during which the temperature remains constant is called isothermal; a process during which the pressure remains constant is called isobaric; and a process during which the volume remains constant is called isochoric. 1-26C The state of a simple compressible system is completely specified by two independent, intensive properties. 1-27C In order to describe the state of the air, we need to know the value of all its properties. Pressure, temperature, and water content (i.e., relative humidity or dew point temperature) are commonly cited by weather forecasters. But, other properties like wind speed and chemical composition (i.e., pollen count and smog index, for example} are also important under certain circumstances.

Assuming that the air composition and velocity do not change and that no pressure front motion occurs during the day, the warming process is one of constant pressure (i.e., isobaric). 1-28C A process is said to be steady-flow if it involves no changes with time anywhere within the system or at the system boundaries.

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1-29 EES The variation of density of atmospheric air with elevation is given in tabular form. A relation for the variation of density with elevation is to be obtained, the density at 7 km elevation is to be calculated, and the mass of the atmosphere using the correlation is to be estimated. Assumptions 1 Atmospheric air behaves as an ideal gas. 2 The earth is perfectly sphere with a radius of 6377 km, and the thickness of the atmosphere is 25 km. Properties The density data are given in tabular form as 1.4 1.2 1 3

ρ, kg/m3 1.225 1.112 1.007 0.9093 0.8194 0.7364 0.6601 0.5258 0.4135 0.1948 0.08891 0.04008

ρ, kg/m

z, km 0 1 2 3 4 5 6 8 10 15 20 25

r, km 6377 6378 6379 6380 6381 6382 6383 6385 6387 6392 6397 6402

0.8 0.6 0.4 0.2 0 0

5

10

15

20

25

z, km

Analysis Using EES, (1) Define a trivial function rho= a+z in equation window, (2) select new parametric table from Tables, and type the data in a two-column table, (3) select Plot and plot the data, and (4) select plot and click on “curve fit” to get curve fit window. Then specify 2nd order polynomial and enter/edit equation. The results are: ρ(z) = a + bz + cz2 = 1.20252 – 0.101674z + 0.0022375z2

for the unit of kg/m3,

(or, ρ(z) = (1.20252 – 0.101674z + 0.0022375z2)×109 for the unit of kg/km3) where z is the vertical distance from the earth surface at sea level. At z = 7 km, the equation would give ρ = 0.60 kg/m3. (b) The mass of atmosphere can be evaluated by integration to be m=

∫

V

ρdV =

∫

h

z =0

(a + bz + cz 2 )4π (r0 + z ) 2 dz = 4π

[

∫

h

z =0

(a + bz + cz 2 )(r02 + 2r0 z + z 2 )dz

= 4π ar02 h + r0 (2a + br0 )h 2 / 2 + (a + 2br0 + cr02 )h 3 / 3 + (b + 2cr0 )h 4 / 4 + ch 5 / 5

]

where r0 = 6377 km is the radius of the earth, h = 25 km is the thickness of the atmosphere, and a = 1.20252, b = -0.101674, and c = 0.0022375 are the constants in the density function. Substituting and multiplying by the factor 109 for the density unity kg/km3, the mass of the atmosphere is determined to be m = 5.092×1018 kg Discussion Performing the analysis with excel would yield exactly the same results.

EES Solution for final result: a=1.2025166; b=-0.10167 c=0.0022375; r=6377; h=25 m=4*pi*(a*r^2*h+r*(2*a+b*r)*h^2/2+(a+2*b*r+c*r^2)*h^3/3+(b+2*c*r)*h^4/4+c*h^5/5)*1E+9

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Temperature 1-30C The zeroth law of thermodynamics states that two bodies are in thermal equilibrium if both have the same temperature reading, even if they are not in contact. 1-31C They are Celsius (°C) and kelvin (K) in the SI, and fahrenheit (°F) and rankine (R) in the English system. 1-32C Probably, but not necessarily. The operation of these two thermometers is based on the thermal expansion of a fluid. If the thermal expansion coefficients of both fluids vary linearly with temperature, then both fluids will expand at the same rate with temperature, and both thermometers will always give identical readings. Otherwise, the two readings may deviate.

1-33 A temperature is given in °C. It is to be expressed in K. Analysis The Kelvin scale is related to Celsius scale by

T(K] = T(°C) + 273 Thus,

T(K] = 37°C + 273 = 310 K

1-34E A temperature is given in °C. It is to be expressed in °F, K, and R. Analysis Using the conversion relations between the various temperature scales,

T(K] = T(°C) + 273 = 18°C + 273 = 291 K T(°F] = 1.8T(°C) + 32 = (1.8)(18) + 32 = 64.4°F T(R] = T(°F) + 460 = 64.4 + 460 = 524.4 R

1-35 A temperature change is given in °C. It is to be expressed in K. Analysis This problem deals with temperature changes, which are identical in Kelvin and Celsius scales. Thus, ΔT(K] = ΔT(°C) = 15 K

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1-36E The temperature of steam given in K unit is to be converted to °F unit. Analysis Using the conversion relations between the various temperature scales,

T (°C) = T (K ) − 273 = 300 − 273 = 27°C T (°F) = 1.8T (°C) + 32 = (1.8)(27) + 32 = 80.6°F

1-37E The temperature of oil given in °F unit is to be converted to °C unit. Analysis Using the conversion relation between the temperature scales, T (°C) =

T (°F) − 32 150 − 32 = = 65.6°C 1. 8 1. 8

1-38E The temperature of air given in °C unit is to be converted to °F unit. Analysis Using the conversion relation between the temperature scales, T (°F) = 1.8T (°C) + 32 = (1.8)(150) + 32 = 302°F

1-39E A temperature range given in °F unit is to be converted to °C unit and the temperature difference in °F is to be expressed in K, °C, and R. Analysis The lower and upper limits of comfort range in °C are T (°C) =

T (°F) − 32 65 − 32 = = 18.3°C 1. 8 1. 8

T (°C) =

T (°F) − 32 75 − 32 = = 23.9 °C 1. 8 1. 8

A temperature change of 10°F in various units are ΔT (R ) = ΔT (°F) = 10 R ΔT (°F) 10 ΔT (°C) = = = 5.6°C 1.8 1.8 ΔT (K ) = ΔT (°C) = 5.6 K

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Pressure, Manometer, and Barometer 1-40C The pressure relative to the atmospheric pressure is called the gage pressure, and the pressure relative to an absolute vacuum is called absolute pressure. 1-41C The blood vessels are more restricted when the arm is parallel to the body than when the arm is perpendicular to the body. For a constant volume of blood to be discharged by the heart, the blood pressure must increase to overcome the increased resistance to flow.

1-42C No, the absolute pressure in a liquid of constant density does not double when the depth is doubled. It is the gage pressure that doubles when the depth is doubled. 1-43C If the lengths of the sides of the tiny cube suspended in water by a string are very small, the magnitudes of the pressures on all sides of the cube will be the same. 1-44C Pascal’s principle states that the pressure applied to a confined fluid increases the pressure throughout by the same amount. This is a consequence of the pressure in a fluid remaining constant in the horizontal direction. An example of Pascal’s principle is the operation of the hydraulic car jack.

1-45E The maximum pressure of a tire is given in English units. It is to be converted to SI units. Assumptions The listed pressure is gage pressure. Analysis Noting that 1 atm = 101.3 kPa = 14.7 psi, the listed maximum pressure can be expressed in SI units as ⎛ 101.3 kPa ⎞ ⎟⎟ = 241 kPa Pmax = 35 psi = (35 psi )⎜⎜ ⎝ 14.7 psi ⎠

Discussion We could also solve this problem by using the conversion factor 1 psi = 6.895 kPa.

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1-46 The pressure in a tank is given. The tank's pressure in various units are to be determined. Analysis Using appropriate conversion factors, we obtain

(a)

⎛ 1 kN/m 2 P = (1500 kPa )⎜ ⎜ 1 kPa ⎝

⎞ ⎟ = 1500 kN/m 2 ⎟ ⎠

(b)

⎛ 1 kN/m 2 P = (1500 kPa )⎜ ⎜ 1 kPa ⎝

⎞⎛ 1000 kg ⋅ m/s 2 ⎟⎜ ⎟⎜ 1 kN ⎠⎝

⎞ ⎟ = 1,500,000 kg/m ⋅ s 2 ⎟ ⎠

(c)

⎛ 1 kN/m 2 P = (1500 kPa )⎜ ⎜ 1 kPa ⎝

⎞⎛ 1000 kg ⋅ m/s 2 ⎟⎜ ⎟⎜ 1 kN ⎠⎝

⎞⎛ 1000 m ⎞ ⎟⎜ = 1,500,000, 000 kg/km ⋅ s 2 ⎟⎝ 1 km ⎟⎠ ⎠

1-47E The pressure given in kPa unit is to be converted to psia. Analysis Using the kPa to psia units conversion factor, ⎛ 1 psia ⎞ P = (200 kPa )⎜ ⎟ = 29.0 psia ⎝ 6.895 kPa ⎠

1-48E A manometer measures a pressure difference as inches of water. This is to be expressed in psia unit. Properties The density of water is taken to be 62.4 lbm/ft3 (Table A-3E). Analysis Applying the hydrostatic equation, ΔP = ρgh 1 lbf ⎛ = (62.4 lbm/ft 3 )(32.174 ft/s 2 )(40/12 ft)⎜ ⎝ 32.174 lbm ⋅ ft/s 2

⎞⎛⎜ 1 ft 2 ⎟⎜ ⎠⎝ 144 in 2

⎞ ⎟ ⎟ ⎠

= 1.44 lbf/in 2 = 1.44 psia

1-49 The pressure given in mm Hg unit is to be converted to kPa. Analysis Using the mm Hg to kPa units conversion factor, ⎛ 0.1333 kPa ⎞ ⎟⎟ = 133.3 kPa P = (1000 mm Hg)⎜⎜ ⎝ 1 mm Hg ⎠

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1-50 The pressure in a pressurized water tank is measured by a multi-fluid manometer. The gage pressure of air in the tank is to be determined. Assumptions The air pressure in the tank is uniform (i.e., its variation with elevation is negligible due to its low density), and thus we can determine the pressure at the air-water interface. Properties The densities of mercury, water, and oil are given to be 13,600, 1000, and 850 kg/m3, respectively. Analysis Starting with the pressure at point 1 at the air-water interface, and moving along the tube by adding (as we go down) or subtracting (as we go up) th e ρgh terms until we reach point 2, and setting the result equal to Patm since the tube is open to the atmosphere gives

P1 + ρ water gh1 + ρ oil gh2 − ρ mercury gh3 = Patm Solving for P1,

P1 = Patm − ρ water gh1 − ρ oil gh2 + ρ mercury gh3 or,

P1 − Patm = g ( ρ mercury h3 − ρ water h1 − ρ oil h2 ) Noting that P1,gage = P1 - Patm and substituting, P1,gage = (9.81 m/s 2 )[(13,600 kg/m 3 )(0.46 m) − (1000 kg/m 3 )(0.2 m) ⎛ 1N − (850 kg/m 3 )(0.3 m)]⎜ ⎜ 1 kg ⋅ m/s 2 ⎝ = 56.9 kPa

⎞⎛ 1 kPa ⎞ ⎟⎜ ⎟⎝ 1000 N/m 2 ⎟⎠ ⎠

Discussion Note that jumping horizontally from one tube to the next and realizing that pressure remains the same in the same fluid simplifies the analysis greatly.

1-51 The barometric reading at a location is given in height of mercury column. The atmospheric pressure is to be determined. Properties The density of mercury is given to be 13,600 kg/m3. Analysis The atmospheric pressure is determined directly from Patm = ρgh ⎛ 1N = (13,600 kg/m 3 )(9.81 m/s 2 )(0.750 m)⎜ ⎜ 1 kg ⋅ m/s 2 ⎝ = 100.1 kPa

⎞⎛ 1 kPa ⎞ ⎟⎜ ⎟⎝ 1000 N/m 2 ⎟⎠ ⎠

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1-52 The gage pressure in a liquid at a certain depth is given. The gage pressure in the same liquid at a different depth is to be determined. Assumptions The variation of the density of the liquid with depth is negligible. Analysis The gage pressure at two different depths of a liquid can be expressed as P1 = ρgh1

and

P2 = ρgh2

Taking their ratio,

h1

P2 ρgh2 h2 = = P1 ρgh1 h1

1

h2

Solving for P2 and substituting gives

2

h 9m P2 = 2 P1 = (28 kPa) = 84 kPa h1 3m

Discussion Note that the gage pressure in a given fluid is proportional to depth.

1-53 The absolute pressure in water at a specified depth is given. The local atmospheric pressure and the absolute pressure at the same depth in a different liquid are to be determined. Assumptions The liquid and water are incompressible. Properties The specific gravity of the fluid is given to be SG = 0.85. We take the density of water to be 1000 kg/m3. Then density of the liquid is obtained by multiplying its specific gravity by the density of water,

ρ = SG × ρ H 2O = (0.85)(100 0 kg/m 3 ) = 850 kg/m 3 Analysis (a) Knowing the absolute pressure, the atmospheric pressure can be determined from

Patm

Patm = P − ρgh

⎛ 1 kPa = (145 kPa) − (1000 kg/m 3 )(9.81 m/s 2 )(5 m)⎜ ⎜ 1000 N/m 2 ⎝ = 96.0 kPa

⎞ ⎟ ⎟ ⎠

h P

(b) The absolute pressure at a depth of 5 m in the other liquid is P = Patm + ρgh

⎛ 1 kPa = (96.0 kPa) + (850 kg/m 3 )(9.81 m/s 2 )(5 m)⎜ ⎜ 1000 N/m 2 ⎝ = 137.7 kPa

⎞ ⎟ ⎟ ⎠

Discussion Note that at a given depth, the pressure in the lighter fluid is lower, as expected.

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1-14

1-54E It is to be shown that 1 kgf/cm2 = 14.223 psi . Analysis Noting that 1 kgf = 9.80665 N, 1 N = 0.22481 lbf, and 1 in = 2.54 cm, we have ⎛ 0.22481 lbf 1 kgf = 9.80665 N = (9.80665 N )⎜⎜ 1N ⎝

⎞ ⎟⎟ = 2.20463 lbf ⎠

and 2

⎛ 2.54 cm ⎞ ⎟⎟ = 14.223 lbf/in 2 = 14.223 psi 1 kgf/cm = 2.20463 lbf/cm = (2.20463 lbf/cm )⎜⎜ ⎝ 1 in ⎠ 2

2

2

1-55E The pressure in chamber 3 of the two-piston cylinder shown in the figure is to be determined. Analysis The area upon which pressure 1 acts is

A1 = π

D12 (3 in) 2 =π = 7.069 in 2 4 4

F2

and the area upon which pressure 2 acts is

A2 = π

D22 4

=π

F3

2

(2 in) = 3.142 in 2 4

The area upon which pressure 3 acts is given by

A3 = A1 − A2 = 7.069 − 3.142 = 3.927 in 2 The force produced by pressure 1 on the piston is then

⎛ 1 lbf/in 2 F1 = P1 A1 = (150 psia )⎜ ⎜ 1 psia ⎝

F1

⎞ ⎟(7.069 in 2 ) = 1060 lbf ⎟ ⎠

while that produced by pressure 2 is

F1 = P2 A2 = (200 psia )(3.142 in 2 ) = 628 lbf According to the vertical force balance on the piston free body diagram F3 = F1 − F2 = 1060 − 628 = 432 lbf

Pressure 3 is then P3 =

F3 432 lbf = = 110 psia A3 3.927 in 2

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1-56 The pressure in chamber 2 of the two-piston cylinder shown in the figure is to be determined. Analysis Summing the forces acting on the piston in the vertical direction gives F2 + F3 = F1 P2 A2 + P3 ( A1 − A2 ) = P1 A1

F2

which when solved for P2 gives P2 = P1

⎛A ⎞ A1 − P3 ⎜⎜ 1 − 1⎟⎟ A2 ⎝ A2 ⎠

F3

since the areas of the piston faces are given by A = πD 2 / 4 the above equation becomes ⎛D P2 = P1 ⎜⎜ 1 ⎝ D2

2 ⎡⎛ D ⎞ ⎟⎟ − P3 ⎢⎜⎜ 1 ⎢⎝ D 2 ⎠ ⎣

2 ⎤ ⎞ ⎟⎟ − 1⎥ ⎥ ⎠ ⎦

F1

2 ⎡⎛ 10 ⎞ 2 ⎤ ⎛ 10 ⎞ = (1000 kPa)⎜ ⎟ − (500 kPa) ⎢⎜ ⎟ − 1⎥ ⎝ 4⎠ ⎢⎣⎝ 4 ⎠ ⎥⎦ = 3625 kPa

1-57 The pressure in chamber 1 of the two-piston cylinder shown in the figure is to be determined. Analysis Summing the forces acting on the piston in the vertical direction gives F2 + F3 = F1 P2 A2 + P3 ( A1 − A2 ) = P1 A1

F2

which when solved for P1 gives P1 = P2

⎛ A2 A ⎞ + P3 ⎜⎜1 − 2 ⎟⎟ A1 A1 ⎠ ⎝

F3

since the areas of the piston faces are given by A = πD 2 / 4 the above equation becomes ⎛D P1 = P2 ⎜⎜ 2 ⎝ D1

2 ⎡ ⎛D ⎞ ⎟⎟ + P3 ⎢1 − ⎜⎜ 2 ⎢ ⎝ D1 ⎠ ⎣

⎞ ⎟⎟ ⎠

2⎤

⎥ ⎥ ⎦

F1

2 ⎡ ⎛ 4 ⎞2 ⎤ ⎛ 4⎞ = (2000 kPa)⎜ ⎟ + (700 kPa) ⎢1 − ⎜ ⎟ ⎥ ⎝ 10 ⎠ ⎢⎣ ⎝ 10 ⎠ ⎥⎦

= 908 kPa

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1-16

1-58 The mass of a woman is given. The minimum imprint area per shoe needed to enable her to walk on the snow without sinking is to be determined. Assumptions 1 The weight of the person is distributed uniformly on the imprint area of the shoes. 2 One foot carries the entire weight of a person during walking, and the shoe is sized for walking conditions (rather than standing). 3 The weight of the shoes is negligible. Analysis The mass of the woman is given to be 70 kg. For a pressure of 0.5 kPa on the snow, the imprint area of one shoe must be A= =

W mg = P P (70 kg)(9.81 m/s 2 ) ⎛⎜ 1N 2 ⎜ 0.5 kPa ⎝ 1 kg ⋅ m/s

⎞⎛ 1 kPa ⎞ ⎟⎜ = 1.37 m 2 ⎟⎝ 1000 N/m 2 ⎟⎠ ⎠

Discussion This is a very large area for a shoe, and such shoes would be impractical to use. Therefore, some sinking of the snow should be allowed to have shoes of reasonable size.

1-59 The vacuum pressure reading of a tank is given. The absolute pressure in the tank is to be determined. Properties The density of mercury is given to be ρ = 13,590 kg/m3. Analysis The atmospheric (or barometric) pressure can be expressed as Patm = ρ g h

⎛ 1N = (13,590 kg/m )(9.807 m/s )(0.750 m)⎜ ⎜ 1 kg ⋅ m/s 2 ⎝ = 100.0 kPa 3

2

⎞⎛ 1 kPa ⎟⎜ ⎟⎜ 1000 N/m 2 ⎠⎝

⎞ ⎟ ⎟ ⎠

Pabs

15 kPa

Patm = 750 mmHg

Then the absolute pressure in the tank becomes Pabs = Patm − Pvac = 100.0 − 15 = 85.0 kPa

1-60E A pressure gage connected to a tank reads 50 psi. The absolute pressure in the tank is to be determined. Properties The density of mercury is given to be ρ = 848.4 lbm/ft3. Analysis The atmospheric (or barometric) pressure can be expressed as Patm = ρ g h

⎛ 1 lbf = (848.4 lbm/ft 3 )(32.2 ft/s 2 )(29.1/12 ft)⎜ ⎜ 32.2 lbm ⋅ ft/s 2 ⎝ = 14.29 psia

⎞⎛ 1 ft 2 ⎟⎜ ⎟⎜ 144 in 2 ⎠⎝

⎞ ⎟ ⎟ ⎠

Pabs

50 psi

Then the absolute pressure in the tank is

Pabs = Pgage + Patm = 50 + 14.29 = 64.3 psia

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1-17

1-61 A pressure gage connected to a tank reads 500 kPa. The absolute pressure in the tank is to be determined. Analysis The absolute pressure in the tank is determined from

500 kPa

Pabs

Pabs = Pgage + Patm = 500 + 94 = 594 kPa

Patm = 94 kPa

1-62 A mountain hiker records the barometric reading before and after a hiking trip. The vertical distance climbed is to be determined.

780 mbar

Assumptions The variation of air density and the gravitational acceleration with altitude is negligible. Properties The density of air is given to be ρ = 1.20 kg/m3.

h=?

Analysis Taking an air column between the top and the bottom of the mountain and writing a force balance per unit base area, we obtain

930 mbar

Wair / A = Pbottom − Ptop ( ρgh) air = Pbottom − Ptop ⎛ 1N (1.20 kg/m 3 )(9.81 m/s 2 )(h)⎜ ⎜ 1 kg ⋅ m/s 2 ⎝

⎞⎛ 1 bar ⎟⎜ ⎟⎜ 100,000 N/m 2 ⎠⎝

⎞ ⎟ = (0.930 − 0.780) bar ⎟ ⎠

It yields h = 1274 m which is also the distance climbed.

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1-18

1-63 A barometer is used to measure the height of a building by recording reading at the bottom and at the top of the building. The height of the building is to be determined. Assumptions The variation of air density with altitude is negligible.

730 mmHg

Properties The density of air is given to be ρ = 1.18 kg/m3. The density of mercury is 13,600 kg/m3. Analysis Atmospheric pressures at the top and at the bottom of the building are

Ptop = ( ρ g h) top

⎛ 1N = (13,600 kg/m 3 )(9.807 m/s 2 )(0.730 m)⎜ ⎜ 1 kg ⋅ m/s 2 ⎝ = 97.36 kPa

Pbottom = ( ρ g h) bottom

⎛ 1N = (13,600 kg/m 3 )(9.807 m/s 2 )(0.755 m)⎜ ⎜ 1kg ⋅ m/s 2 ⎝ = 100.70 kPa

⎞⎛ 1 kPa ⎟⎜ ⎟⎜ 1000 N/m 2 ⎠⎝

h ⎞ ⎟ ⎟ ⎠ 755 mmHg

⎞⎛ 1 kPa ⎟⎜ ⎟⎜ 1000 N/m 2 ⎠⎝

⎞ ⎟ ⎟ ⎠

Taking an air column between the top and the bottom of the building and writing a force balance per unit base area, we obtain Wair / A = Pbottom − Ptop ( ρgh) air = Pbottom − Ptop ⎛ 1N (1.18 kg/m 3 )(9.807 m/s 2 )(h)⎜ ⎜ 1 kg ⋅ m/s 2 ⎝

It yields

⎞⎛ 1 kPa ⎟⎜ ⎟⎜ 1000 N/m 2 ⎠⎝

⎞ ⎟ = (100.70 − 97.36) kPa ⎟ ⎠

h = 288.6 m

which is also the height of the building.

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1-19

1-64 EES Problem 1-63 is reconsidered. The entire EES solution is to be printed out, including the numerical results with proper units. Analysis The problem is solved using EES, and the solution is given below. P_bottom=755 [mmHg] P_top=730 [mmHg] g=9.807 [m/s^2] "local acceleration of gravity at sea level" rho=1.18 [kg/m^3] DELTAP_abs=(P_bottom-P_top)*CONVERT('mmHg','kPa')"[kPa]" "Delta P reading from the barometers, converted from mmHg to kPa." DELTAP_h =rho*g*h/1000 "[kPa]" "Equ. 1-16. Delta P due to the air fluid column height, h, between the top and bottom of the building." "Instead of dividing by 1000 Pa/kPa we could have multiplied rho*g*h by the EES function, CONVERT('Pa','kPa')" DELTAP_abs=DELTAP_h SOLUTION Variables in Main DELTAP_abs=3.333 [kPa] DELTAP_h=3.333 [kPa] g=9.807 [m/s^2] h=288 [m] P_bottom=755 [mmHg] P_top=730 [mmHg] rho=1.18 [kg/m^3]

1-65 A diver is moving at a specified depth from the water surface. The pressure exerted on the surface of the diver by water is to be determined. Assumptions The variation of the density of water with depth is negligible. Properties The specific gravity of seawater is given to be SG = 1.03. We take the density of water to be 1000 kg/m3. Analysis The density of the seawater is obtained by multiplying its specific gravity by the density of water which is taken to be 1000 kg/m3: 3

ρ = SG × ρ H 2 O = (1.03)(100 0 kg/m ) = 1030 kg/m

⎛ 1 kPa = (101 kPa) + (1030 kg/m 3 )(9.807 m/s 2 )(30 m)⎜ ⎜ 1000 N/m 2 ⎝ = 404.0 kPa

Sea h

3

The pressure exerted on a diver at 30 m below the free surface of the sea is the absolute pressure at that location: P = Patm + ρgh

Patm

P

⎞ ⎟ ⎟ ⎠

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1-20

1-66 A gas contained in a vertical piston-cylinder device is pressurized by a spring and by the weight of the piston. The pressure of the gas is to be determined. Analysis Drawing the free body diagram of the piston and balancing the vertical forces yield

Fspring

PA = Patm A + W + Fspring

Patm

Thus,

P = Patm +

mg + Fspring

A (4 kg)(9.81 m/s 2 ) + 60 N ⎛⎜ 1 kPa = (95 kPa) + ⎜ 1000 N/m 2 35 × 10 − 4 m 2 ⎝ = 123.4 kPa

P

⎞ ⎟ ⎟ ⎠

W = mg

1-67 EES Problem 1-66 is reconsidered. The effect of the spring force in the range of 0 to 500 N on the pressure inside the cylinder is to be investigated. The pressure against the spring force is to be plotted, and results are to be discussed. Analysis The problem is solved using EES, and the solution is given below. g=9.807 [m/s^2] P_atm= 95 [kPa] m_piston=4 [kg] {F_spring=60 [N]} A=35*CONVERT('cm^2','m^2')"[m^2]" W_piston=m_piston*g"[N]" F_atm=P_atm*A*CONVERT('kPa','N/m^2')"[N]" "From the free body diagram of the piston, the balancing vertical forces yield:" F_gas= F_atm+F_spring+W_piston"[N]" P_gas=F_gas/A*CONVERT('N/m^2','kPa')"[kPa]" 260

Pgas [kPa] 106.2 122.1 138 153.8 169.7 185.6 201.4 217.3 233.2 249.1

240 220

P gas [kPa]

Fspring [N] 0 55.56 111.1 166.7 222.2 277.8 333.3 388.9 444.4 500

200 180 160 140 120 100 0

100

200

300

F

[N]

spring

400

500

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1-21

1-68 [Also solved by EES on enclosed CD] Both a gage and a manometer are attached to a gas to measure its pressure. For a specified reading of gage pressure, the difference between the fluid levels of the two arms of the manometer is to be determined for mercury and water. Properties The densities of water and mercury are given to be ρwater = 1000 kg/m3 and be ρHg = 13,600 kg/m3. Analysis The gage pressure is related to the vertical distance h between the two fluid levels by

⎯→ h = Pgage = ρ g h ⎯

Pgage

ρg

(a) For mercury, h= =

Pgage

ρ Hg g ⎛ 1 kN/m 2 ⎜ (13,600 kg/m 3 )(9.81 m/s 2 ) ⎜⎝ 1 kPa 80 kPa

⎞⎛ 1000 kg/m ⋅ s 2 ⎟⎜ ⎟⎜ 1 kN ⎠⎝

⎞ ⎟ = 0.60 m ⎟ ⎠

(b) For water, h=

Pgage

ρ H 2O g

=

⎛ 1 kN/m 2 ⎜ (1000 kg/m 3 )(9.81 m/s 2 ) ⎜⎝ 1 kPa 80 kPa

⎞⎛ 1000 kg/m ⋅ s 2 ⎟⎜ ⎟⎜ 1 kN ⎠⎝

⎞ ⎟ = 8.16 m ⎟ ⎠

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1-22

1-69 EES Problem 1-68 is reconsidered. The effect of the manometer fluid density in the range of 800 to 13,000 kg/m3 on the differential fluid height of the manometer is to be investigated. Differential fluid height against the density is to be plotted, and the results are to be discussed. Analysis The problem is solved using EES, and the solution is given below. Function fluid_density(Fluid$) If fluid$='Mercury' then fluid_density=13600 else fluid_density=1000 end {Input from the diagram window. If the diagram window is hidden, then all of the input must come from the equations window. Also note that brackets can also denote comments - but these comments do not appear in the formatted equations window.} {Fluid$='Mercury' P_atm = 101.325 "kpa" DELTAP=80 "kPa Note how DELTAP is displayed on the Formatted Equations Window."} g=9.807 "m/s2, local acceleration of gravity at sea level" rho=Fluid_density(Fluid$) "Get the fluid density, either Hg or H2O, from the function" "To plot fluid height against density place {} around the above equation. Then set up the parametric table and solve." DELTAP = RHO*g*h/1000 "Instead of dividing by 1000 Pa/kPa we could have multiplied by the EES function, CONVERT('Pa','kPa')" h_mm=h*convert('m','mm') "The fluid height in mm is found using the built-in CONVERT function." P_abs= P_atm + DELTAP

ρ [kg/m3]

10197

800

3784

2156

2323

3511

1676

4867

1311

6222

1076

7578

913.1

8933

792.8

10289

700.5

11644

627.5

13000

Manometer Fluid Height vs Manometer Fluid Density 11000 8800

hmm [mm]

hmm [mm]

6600 4400 2200 0 0

2000

4000

6000

8000

10000 12000 14000

ρ [kg/m^3]

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1-23

1-70 The air pressure in a tank is measured by an oil manometer. For a given oil-level difference between the two columns, the absolute pressure in the tank is to be determined. Properties The density of oil is given to be ρ = 850 kg/m3. Analysis The absolute pressure in the tank is determined from P = Patm + ρgh

⎛ 1kPa = (98 kPa) + (850 kg/m 3 )(9.81m/s 2 )(0.60 m)⎜ ⎜ 1000 N/m 2 ⎝ = 103 kPa

⎞ ⎟ ⎟ ⎠

0.60 m

AIR

Patm = 98 kPa

1-71 The air pressure in a duct is measured by a mercury manometer. For a given mercury-level difference between the two columns, the absolute pressure in the duct is to be determined. Properties The density of mercury is given to be ρ = 13,600 kg/m3. Analysis (a) The pressure in the duct is above atmospheric pressure since the fluid column on the duct side is at a lower level.

(b) The absolute pressure in the duct is determined from P = Patm + ρgh

⎛ 1N = (100 kPa) + (13,600 kg/m 3 )(9.81 m/s 2 )(0.015 m)⎜ ⎜ 1 kg ⋅ m/s 2 ⎝ = 102 kPa

⎞⎛ 1 kPa ⎟⎜ ⎟⎜ 1000 N/m 2 ⎠⎝

⎞ ⎟ ⎟ ⎠

1-72 The air pressure in a duct is measured by a mercury manometer. For a given mercury-level difference between the two columns, the absolute pressure in the duct is to be determined. Properties The density of mercury is given to be ρ = 13,600 kg/m3. Analysis (a) The pressure in the duct is above atmospheric pressure since the fluid column on the duct side is at a lower level.

AIR

(b) The absolute pressure in the duct is determined from P = Patm + ρgh

⎛ 1N = (100 kPa) + (13,600 kg/m 3 )(9.81 m/s 2 )(0.045 m)⎜ ⎜ 1 kg ⋅ m/s 2 ⎝ = 106 kPa

45 mm

P ⎞⎛ 1 kPa ⎟⎜ ⎟⎜ 1000 N/m 2 ⎠⎝

⎞ ⎟ ⎟ ⎠

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1-24

1-73E The systolic and diastolic pressures of a healthy person are given in mmHg. These pressures are to be expressed in kPa, psi, and meter water column. Assumptions Both mercury and water are incompressible substances. Properties We take the densities of water and mercury to be 1000 kg/m3 and 13,600 kg/m3, respectively. Analysis Using the relation P = ρgh for gage pressure, the high and low pressures are expressed as ⎞⎛ 1 kPa ⎞ ⎛ 1N ⎟ = 16.0 kPa ⎟⎜ Phigh = ρghhigh = (13,600 kg/m 3 )(9.81 m/s 2 )(0.12 m)⎜⎜ 2 ⎟⎜ 2⎟ ⎝ 1 kg ⋅ m/s ⎠⎝ 1000 N/m ⎠ ⎞⎛ 1 kPa ⎞ ⎛ 1N ⎟ = 10.7 kPa ⎟⎜ Plow = ρghlow = (13,600 kg/m 3 )(9.81 m/s 2 )(0.08 m)⎜⎜ 2 ⎟⎜ 2⎟ ⎝ 1 kg ⋅ m/s ⎠⎝ 1000 N/m ⎠

Noting that 1 psi = 6.895 kPa, ⎛ 1 psi ⎞ ⎟⎟ = 2.32 psi Phigh = (16.0 Pa)⎜⎜ ⎝ 6.895 kPa ⎠

and

⎛ 1 psi ⎞ ⎟⎟ = 1.55 psi Plow = (10.7 Pa)⎜⎜ ⎝ 6.895 kPa ⎠

For a given pressure, the relation P = ρgh can be expressed for mercury and water as P = ρ water gh water and P = ρ mercury ghmercury . Setting these two relations equal to each other and solving for water height gives

P = ρ water ghwater = ρ mercury ghmercury → h water =

ρ mercury ρ water

hmercury

h

Therefore,

hwater, high = hwater, low =

ρ mercury ρ water ρ mercury ρ water

hmercury, high = hmercury, low =

13,600 kg/m 3 1000 kg/m 3

13,600 kg/m 3 1000 kg/m 3

(0.12 m) = 1.63 m

(0.08 m) = 1.09 m

Discussion Note that measuring blood pressure with a “water” monometer would involve differential fluid heights higher than the person, and thus it is impractical. This problem shows why mercury is a suitable fluid for blood pressure measurement devices.

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1-25

1-74 A vertical tube open to the atmosphere is connected to the vein in the arm of a person. The height that the blood will rise in the tube is to be determined. Assumptions 1 The density of blood is constant. 2 The gage pressure of blood is 120 mmHg. Properties The density of blood is given to be ρ = 1050 kg/m3. Analysis For a given gage pressure, the relation P = ρgh can be expressed

Blood

for mercury and blood as P = ρ blood ghblood and P = ρ mercury ghmercury .

h

Setting these two relations equal to each other we get

P = ρ blood ghblood = ρ mercury ghmercury Solving for blood height and substituting gives hblood =

ρ mercury ρ blood

hmercury =

13,600 kg/m 3 1050 kg/m 3

(0.12 m) = 1.55 m

Discussion Note that the blood can rise about one and a half meters in a tube connected to the vein. This explains why IV tubes must be placed high to force a fluid into the vein of a patient.

1-75 A man is standing in water vertically while being completely submerged. The difference between the pressures acting on the head and on the toes is to be determined. Assumptions Water is an incompressible substance, and thus the density does not change with depth.

hhead

Properties We take the density of water to be ρ =1000 kg/m3. Analysis The pressures at the head and toes of the person can be expressed as Phead = Patm + ρghhead

and

Ptoe = Patm + ρghtoe

where h is the vertical distance of the location in water from the free surface. The pressure difference between the toes and the head is determined by subtracting the first relation above from the second,

htoe

Ptoe − Phead = ρgh toe − ρghhead = ρg (h toe − hhead )

Substituting, ⎛ 1N Ptoe − Phead = (1000 kg/m 3 )(9.81 m/s 2 )(1.80 m - 0)⎜ ⎜ 1kg ⋅ m/s 2 ⎝

⎞⎛ 1kPa ⎟⎜ ⎟⎜ 1000 N/m 2 ⎠⎝

⎞ ⎟ = 17.7 kPa ⎟ ⎠

Discussion This problem can also be solved by noting that the atmospheric pressure (1 atm = 101.325 kPa) is equivalent to 10.3-m of water height, and finding the pressure that corresponds to a water height of 1.8 m.

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1-26

1-76 Water is poured into the U-tube from one arm and oil from the other arm. The water column height in one arm and the ratio of the heights of the two fluids in the other arm are given. The height of each fluid in that arm is to be determined. Assumptions Both water and oil are incompressible substances.

Water

Properties The density of oil is given to be ρ = 790 kg/m . We take the density of water to be ρ =1000 kg/m3. 3

ha

Analysis The height of water column in the left arm of the monometer is given to be hw1 = 0.70 m. We let the height of water and oil in the right arm to be hw2 and ha, respectively. Then, ha = 4hw2. Noting that both arms are open to the atmosphere, the pressure at the bottom of the U-tube can be expressed as Pbottom = Patm + ρ w gh w1

oil

hw1

hw2

Pbottom = Patm + ρ w gh w2 + ρ a gha

and

Setting them equal to each other and simplifying,

ρ w gh w1 = ρ w gh w2 + ρ a gha

→

ρ w h w1 = ρ w h w2 + ρ a ha

→

h w1 = h w2 + ( ρ a / ρ w )ha

Noting that ha = 4hw2, the water and oil column heights in the second arm are determined to be 0.7 m = h w2 + (790/1000) 4 hw 2 →

h w2 = 0.168 m

0.7 m = 0.168 m + (790/1000) ha →

ha = 0.673 m

Discussion Note that the fluid height in the arm that contains oil is higher. This is expected since oil is lighter than water.

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1-27

1-77 Fresh and seawater flowing in parallel horizontal pipelines are connected to each other by a double Utube manometer. The pressure difference between the two pipelines is to be determined. Assumptions 1 All the liquids are incompressible. 2 The effect of air column on pressure is negligible.

Air

Properties The densities of seawater and mercury are given to be ρsea = 1035 kg/m3 and ρHg = 13,600 kg/m3. We take the density of water to be ρ w =1000 kg/m3. Analysis Starting with the pressure in the fresh water pipe (point 1) and moving along the tube by adding (as we go down) or subtracting (as we go up) the ρgh terms until we reach the sea water pipe (point 2), and setting the result equal to P2 gives

hsea

Fresh Water

Sea Water

hair hw

P1 + ρ w ghw − ρ Hg ghHg − ρ air ghair + ρ sea ghsea = P2 Rearranging and neglecting the effect of air column on pressure,

hHg Mercury

P1 − P2 = − ρ w ghw + ρ Hg ghHg − ρ sea ghsea = g ( ρ Hg hHg − ρ w hw − ρ sea hsea ) Substituting, P1 − P2 = (9.81 m/s 2 )[(13600 kg/m 3 )(0.1 m) ⎛ 1 kN − (1000 kg/m 3 )(0.6 m) − (1035 kg/m 3 )(0.4 m)]⎜ ⎜ 1000 kg ⋅ m/s 2 ⎝

⎞ ⎟ ⎟ ⎠

= 3.39 kN/m 2 = 3.39 kPa

Therefore, the pressure in the fresh water pipe is 3.39 kPa higher than the pressure in the sea water pipe. Discussion A 0.70-m high air column with a density of 1.2 kg/m3 corresponds to a pressure difference of 0.008 kPa. Therefore, its effect on the pressure difference between the two pipes is negligible.

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1-28

1-78 Fresh and seawater flowing in parallel horizontal pipelines are connected to each other by a double Utube manometer. The pressure difference between the two pipelines is to be determined. Assumptions All the liquids are incompressible.

Oil

Properties The densities of seawater and mercury are given to be ρsea = 1035 kg/m3 and ρHg = 13,600 kg/m3. We take the density of water to be ρ w =1000 kg/m3. The specific gravity of oil is given to be 0.72, and thus its density is 720 kg/m3. Analysis Starting with the pressure in the fresh water pipe (point 1) and moving along the tube by adding (as we go down) or subtracting (as we go up) the ρgh terms until we reach the sea water pipe (point 2), and setting the result equal to P2 gives

hsea

Fresh Water

Sea Water

hoil hw hHg

P1 + ρ w ghw − ρ Hg ghHg − ρ oil ghoil + ρ sea ghsea = P2

Mercury

Rearranging, P1 − P2 = − ρ w gh w + ρ Hg ghHg + ρ oil ghoil − ρ sea ghsea = g ( ρ Hg hHg + ρ oil hoil − ρ w h w − ρ sea hsea )

Substituting, P1 − P2 = (9.81 m/s 2 )[(13600 kg/m 3 )(0.1 m) + (720 kg/m 3 )(0.7 m) − (1000 kg/m 3 )(0.6 m) ⎛ 1 kN − (1035 kg/m 3 )(0.4 m)]⎜ ⎜ 1000 kg ⋅ m/s 2 ⎝

⎞ ⎟ ⎟ ⎠

= 8.34 kN/m 2 = 8.34 kPa

Therefore, the pressure in the fresh water pipe is 8.34 kPa higher than the pressure in the sea water pipe.

1-79 The pressure indicated by a manometer is to be determined. Properties The specific weights of fluid A and fluid B are given to be 10 kN/m3 and 8 kN/m3, respectively. Analysis The absolute pressure P1 is determined from P1 = Patm + ( ρgh) A + ( ρgh) B = Patm + γ A h A + γ B h B ⎛ 0.1333 kPa ⎞ ⎟⎟ = (758 mm Hg)⎜⎜ ⎝ 1 mm Hg ⎠

= hB hA =

+ (10 kN/m 3 )(0.05 m) + (8 kN/m 3 )(0.15 m) = 102.7 kPa

Note that 1 kPa = 1 kN/m2.

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1-29

1-80 The pressure indicated by a manometer is to be determined. Properties The specific weights of fluid A and fluid B are given to be 100 kN/m3 and 8 kN/m3, respectively. Analysis The absolute pressure P1 is determined from P1 = Patm + ( ρgh) A + ( ρgh) B = Patm + γ A h A + γ B h B = 90 kPa + (100 kN/m 3 )(0.05 m) + (8 kN/m 3 )(0.15 m) = 96.2 kPa

Note that 1 kPa = 1 kN/m2.

= hB hA =

100 kN/m3

1-81 The pressure indicated by a manometer is to be determined. Properties The specific weights of fluid A and fluid B are given to be 10 kN/m3 and 20 kN/m3, respectively. Analysis The absolute pressure P1 is determined from P1 = Patm + ( ρgh) A + ( ρgh) B = Patm + γ A h A + γ B h B

= hB

⎛ 0.1333 kPa ⎞ ⎟⎟ = (745 mm Hg)⎜⎜ ⎝ 1 mm Hg ⎠ 3

hA = 3

+ (10 kN/m )(0.05 m) + (20 kN/m )(0.15 m) = 102.8 kPa

20 kN/m3

Note that 1 kPa = 1 kN/m2.

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1-30

1-82 The gage pressure of air in a pressurized water tank is measured simultaneously by both a pressure gage and a manometer. The differential height h of the mercury column is to be determined. Assumptions The air pressure in the tank is uniform (i.e., its variation with elevation is negligible due to its low density), and thus the pressure at the air-water interface is the same as the indicated gage pressure. Properties We take the density of water to be ρw =1000 kg/m3. The specific gravities of oil and mercury are given to be 0.72 and 13.6, respectively. Analysis Starting with the pressure of air in the tank (point 1), and moving along the tube by adding (as we go down) or subtracting (as we go u p) the ρgh terms until we reach the free surface of oil where the oil tube is exposed to the atmosphere, and setting the result equal to Patm gives

P1 + ρ w ghw − ρ Hg ghHg − ρ oil ghoil = Patm Rearranging,

P1 − Patm = ρ oil ghoil + ρ Hg ghHg − ρ w ghw or,

P1,gage

ρw g

= SG oil hoil + SG Hg hHg − hw

Substituting, ⎛ ⎞⎛ 1000 kg ⋅ m/s 2 80 kPa ⎜ ⎟⎜ ⎜ (1000 kg/m 3 )(9.81 m/s 2 ) ⎟⎜ 1 kPa. ⋅ m 2 ⎝ ⎠⎝

⎞ ⎟ = 0.72 × (0.75 m) + 13.6 × hHg − 0.3 m ⎟ ⎠

Solving for hHg gives hHg = 0.582 m. Therefore, the differential height of the mercury column must be 58.2 cm. Discussion Double instrumentation like this allows one to verify the measurement of one of the instruments by the measurement of another instrument.

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1-31

1-83 The gage pressure of air in a pressurized water tank is measured simultaneously by both a pressure gage and a manometer. The differential height h of the mercury column is to be determined. Assumptions The air pressure in the tank is uniform (i.e., its variation with elevation is negligible due to its low density), and thus the pressure at the air-water interface is the same as the indicated gage pressure. Properties We take the density of water to be ρ w =1000 kg/m3. The specific gravities of oil and mercury are given to be 0.72 and 13.6, respectively. Analysis Starting with the pressure of air in the tank (point 1), and moving along the tube by adding (as we go down) or subtracting (as we go up) the ρgh terms until we reach the free surface of oil where the oil tube is exposed to the atmosphere, and setting the result equal to Patm gives

P1 + ρ w ghw − ρ Hg ghHg − ρ oil ghoil = Patm 40 kPa

Rearranging,

P1 − Patm = ρ oil ghoil + ρ Hg ghHg − ρ w ghw P1,gage

or,

ρw g

AIR hoil

= SG oil hoil + SG

Hg hHg − h w

Water hHg

hw

Substituting, ⎤⎛ 1000 kg ⋅ m/s 2 ⎡ 40 kPa ⎜ ⎢ 3 2 ⎥ 2 ⎢⎣ (1000 kg/m )(9.81 m/s ) ⎥⎦⎜⎝ 1 kPa. ⋅ m

⎞ ⎟ = 0.72 × (0.75 m) + 13.6 × hHg − 0.3 m ⎟ ⎠

Solving for hHg gives hHg = 0.282 m. Therefore, the differential height of the mercury column must be 28.2 cm. Discussion Double instrumentation like this allows one to verify the measurement of one of the instruments by the measurement of another instrument.

1-84 The top part of a water tank is divided into two compartments, and a fluid with an unknown density is poured into one side. The levels of the water and the liquid are measured. The density of the fluid is to be determined. Assumptions 1 Both water and the added liquid are incompressible substances. 2 The added liquid does not mix with water. Properties We take the density of water to be ρ =1000 kg/m3. Analysis Both fluids are open to the atmosphere. Noting that the pressure of both water and the added fluid is the same at the contact surface, the pressure at this surface can be expressed as

Fluid Water hf

hw

Pcontact = Patm + ρ f ghf = Patm + ρ w ghw

Simplifying and solving for ρf gives

ρ f ghf = ρ w ghw →

ρf =

hw 45 cm ρw = (1000 kg/m 3 ) = 562.5 kg/m 3 80 cm hf

Discussion Note that the added fluid is lighter than water as expected (a heavier fluid would sink in water).

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1-32

1-85 The fluid levels in a multi-fluid U-tube manometer change as a result of a pressure drop in the trapped air space. For a given pressure drop and brine level change, the area ratio is to be determined. Assumptions 1 All the liquids are incompressible. 2 Pressure in the brine pipe remains constant. 3 The variation of pressure in the trapped air space is negligible. Properties The specific gravities are given to be 13.56 for mercury and 1.1 for brine. We take the standard density of water to be ρw =1000 kg/m3.

A Air

Area, A1

B Brine pipe

Water

SG=1.1

Analysis It is clear from the problem statement and the figure that the brine pressure is much higher than the air pressure, and when the air pressure drops by 0.7 kPa, the pressure difference between the brine and the air space increases also by the same amount.

Mercury SG=13.56

Δhb = 5 mm

Area, A2

Starting with the air pressure (point A) and moving along the tube by adding (as we go down) or subtracting (as we go up) the ρgh terms until we reach the brine pipe (point B), and setting the result equal to PB before and after the pressure change of air give Before:

PA1 + ρ w ghw + ρ Hg ghHg, 1 − ρ br ghbr,1 = PB

After:

PA2 + ρ w ghw + ρ Hg ghHg, 2 − ρ br ghbr,2 = PB

Subtracting,

PA2 − PA1 + ρ Hg gΔhHg − ρ br gΔhbr = 0 →

PA1 − PA2 = SG Hg ΔhHg − SG br Δh br = 0 ρwg

(1)

where ΔhHg and Δhbr are the changes in the differential mercury and brine column heights, respectively, due to the drop in air pressure. Both of these are positive quantities since as the mercury-brine interface drops, the differential fluid heights for both mercury and brine increase. Noting also that the volume of mercury is constant, we have A1ΔhHg,left = A2 ΔhHg,right and

PA2 − PA1 = −0.7 kPa = −700 N/m 2 = −700 kg/m ⋅ s 2 Δh br = 0.005 m

ΔhHg = ΔhHg,right + ΔhHg,left = Δhbr + Δhbr A2 /A 1 = Δhbr (1 + A2 /A 1 ) Substituting, 700 kg/m ⋅ s 2 (1000 kg/m 3 )(9.81 m/s 2 )

= [13.56 × 0.005(1 + A2 /A1 ) − (1.1× 0.005)] m

It gives A2/A1 = 0.134

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1-33

1-86 A multi-fluid container is connected to a U-tube. For the given specific gravities and fluid column heights, the gage pressure at A and the height of a mercury column that would create the same pressure at A are to be determined. Assumptions 1 All the liquids are incompressible. 2 The multi-fluid container is open to the atmosphere. Properties The specific gravities are given to be 1.26 for glycerin and 0.90 for oil. We take the standard density of water to be ρw =1000 kg/m3, and the specific gravity of mercury to be 13.6. Analysis Starting with the atmospheric pressure on the top surface of the container and moving along the tube by adding (as we go down) or subtracting (as we go up) the ρgh terms until we reach point A, and setting the result equal to PA give

70 cm

Oil SG=0.90

30 cm

Water

20 cm

Glycerin SG=1.26

A

90 cm

15 cm

Patm + ρ oil ghoil + ρ w ghw − ρ gly ghgly = PA Rearranging and using the definition of specific gravity,

PA − Patm = SG oil ρ w ghoil + SG w ρ w ghw − SG gly ρ w ghgly or

PA,gage = gρ w (SG oil hoil + SG w hw − SG gly hgly ) Substituting, ⎛ 1 kN PA,gage = (9.81 m/s 2 )(1000 kg/m 3 )[0.90(0.70 m) + 1(0.3 m) − 1.26(0.70 m)]⎜ ⎜ 1000 kg ⋅ m/s 2 ⎝

⎞ ⎟ ⎟ ⎠

= 0.471 kN/m 2 = 0.471 kPa

The equivalent mercury column height is

hHg =

PA,gage

ρ Hg g

=

⎛ 1000 kg ⋅ m/s 2 ⎜ 1 kN (13,600 kg/m 3 )(1000 kg/m 3 )(9.81 m/s 2 ) ⎜⎝ 0.471 kN/m 2

⎞ ⎟ = 0.00353 m = 0.353 cm ⎟ ⎠

Discussion Note that the high density of mercury makes it a very suitable fluid for measuring high pressures in manometers.

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1-34

Solving Engineering Problems and EES

1-87C Despite the convenience and capability the engineering software packages offer, they are still just tools, and they will not replace the traditional engineering courses. They will simply cause a shift in emphasis in the course material from mathematics to physics. They are of great value in engineering practice, however, as engineers today rely on software packages for solving large and complex problems in a short time, and perform optimization studies efficiently.

1-88 EES Determine a positive real root of the following equation using EES: 2x3 – 10x0.5 – 3x = -3 Solution by EES Software (Copy the following line and paste on a blank EES screen to verify solution): 2*x^3-10*x^0.5-3*x = -3

Answer: x = 2.063 (using an initial guess of x=2)

1-89 EES Solve the following system of 2 equations with 2 unknowns using EES: x3 – y2 = 7.75 3xy + y = 3.5 Solution by EES Software (Copy the following lines and paste on a blank EES screen to verify solution): x^3-y^2=7.75 3*x*y+y=3.5

Answer x=2 y=0.5

1-90 EES Solve the following system of 3 equations with 3 unknowns using EES: 2x – y + z = 5 3x2 + 2y = z + 2 xy + 2z = 8

Solution by EES Software (Copy the following lines and paste on a blank EES screen to verify solution): 2*x-y+z=5 3*x^2+2*y=z+2 x*y+2*z=8

Answer x=1.141, y=0.8159, z=3.535

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1-35

1-91 EES Solve the following system of 3 equations with 3 unknowns using EES: x2y – z = 1 x – 3y0.5 + xz = - 2 x+y–z=2

Solution by EES Software (Copy the following lines and paste on a blank EES screen to verify solution): x^2*y-z=1 x-3*y^0.5+x*z=-2 x+y-z=2

Answer x=1, y=1, z=0

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1-36

1-92E EES Specific heat of water is to be expressed at various units using unit conversion capability of EES. Analysis The problem is solved using EES, and the solution is given below.

EQUATION WINDOW "GIVEN" C_p=4.18 [kJ/kg-C] "ANALYSIS" C_p_1=C_p*Convert(kJ/kg-C, kJ/kg-K) C_p_2=C_p*Convert(kJ/kg-C, Btu/lbm-F) C_p_3=C_p*Convert(kJ/kg-C, Btu/lbm-R) C_p_4=C_p*Convert(kJ/kg-C, kCal/kg-C)

FORMATTED EQUATIONS WINDOW

GIVEN C p = 4.18

[kJ/kg-C]

ANALYSIS kJ/kg–K

C p,1 =

Cp ·

1 ·

C p,2 =

Cp ·

0.238846 ·

C p,3 =

Cp ·

0.238846 ·

C p,4 =

Cp ·

0.238846 ·

kJ/kg–C Btu/lbm–F kJ/kg–C Btu/lbm–R kJ/kg–C kCal/kg–C kJ/kg–C

SOLUTION WINDOW C_p=4.18 [kJ/kg-C] C_p_1=4.18 [kJ/kg-K] C_p_2=0.9984 [Btu/lbm-F] C_p_3=0.9984 [Btu/lbm-R] C_p_4=0.9984 [kCal/kg-C]

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1-37

Review Problems

1-93 The weight of a lunar exploration module on the moon is to be determined. Analysis Applying Newton's second law, the weight of the module on the moon can be determined from W moon = mg moon =

Wearth 4000 N g moon = (1.64 m/s 2 ) = 669 N g earth 9.8 m/s 2

1-94 The deflection of the spring of the two-piston cylinder with a spring shown in the figure is to be determined. Analysis Summing the forces acting on the piston in the vertical direction gives

F2

Fs + F2 + F3 = F1 kx + P2 A2 + P3 ( A1 − A2 ) = P1 A1

which when solved for the deflection of the spring and substituting A = πD 2 / 4 gives

[P D − P D − P ( D − D )] π [5000 × 0.08 − 10,000 × 0.03 = 4 × 800

x=

π

4k

1

2 1

2

2 2

3

2

= 0.0172 m

2 1

F3

Fs

2 2

2

− 1000(0.08 2 − 0.03 2 )

] F1

= 1.72 cm

We expressed the spring constant k in kN/m, the pressures in kPa (i.e., kN/m2) and the diameters in m units.

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1-38

1-95 The pressure in chamber 1 of the two-piston cylinder with a spring shown in the figure is to be determined. Analysis Summing the forces acting on the piston in the vertical direction gives Fs + F1 = F2 + F3

F2

kx + P1 A1 = P2 A2 + P3 ( A1 − A2 )

which when solved for the P3 and substituting A = πD 2 / 4 gives P1 = P2

F3

Fs

⎛ A ⎞ kx A2 + P3 ⎜⎜1 − 2 ⎟⎟ − A1 A1 ⎠ A1 ⎝

⎛D = P2 ⎜⎜ 2 ⎝ D1

2 ⎡ ⎛D ⎞ ⎟⎟ + P3 ⎢1 − ⎜⎜ 2 ⎢ ⎝ D1 ⎠ ⎣

⎞ ⎟⎟ ⎠

2⎤

4kx ⎥− ⎥ πD12 ⎦

F1 2⎤

⎡ ⎛3⎞ 4(1200 kN/m)(0.05 m) ⎛3⎞ = (8000 kPa)⎜ ⎟ + (300 kPa) ⎢1 − ⎜ ⎟ ⎥ − π (0.07 m) 2 ⎝7⎠ ⎢⎣ ⎝ 7 ⎠ ⎥⎦ = 13,880 kPa = 13.9 MPa 2

1-96E The pressure in chamber 2 of the two-piston cylinder with a spring shown in the figure is to be determined. Analysis The areas upon which pressures act are

A1 = π

D12 (5 in) 2 =π = 19.63 in 2 4 4

A2 = π

D22 (2 in) 2 =π = 3.142 in 2 4 4

A3 = A1 − A2 = 19.63 − 3.142 = 16.49 in

F2

F3

Fs 2

The forces generated by pressure 1 and 3 are

⎛ 1 lbf/in 2 F1 = P1 A1 = (100 psia )⎜ ⎜ 1 psia ⎝

⎞ ⎟(19.63 in 2 ) = 1963 lbf ⎟ ⎠

F1

2

F3 = P2 A2 = (20 psia )(16.49 in ) = 330 lbf The force exerted by the spring is Fs = kx = (200 lbf/in )(2 in ) = 400 lbf

Summing the vertical forces acting on the piston gives F2 = F1 − F3 − Fs = 1963 − 330 − 400 = 1233 lbf

The pressure at 2 is then P2 =

F2 1233 lbf = = 392 psia A2 3.142 in 2

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1-39

1-97 An airplane is flying over a city. The local atmospheric pressure in that city is to be determined. Assumptions The gravitational acceleration does not change with altitude. Properties The densities of air and mercury are given to be 1.15 kg/m3 and 13,600 kg/m3. Analysis The local atmospheric pressure is determined from Patm = Pplane + ρgh ⎛ 1 kN = 58 kPa + (1.15 kg/m 3 )(9.81 m/s 2 )(3000 m)⎜ ⎜ 1000 kg ⋅ m/s 2 ⎝

⎞ ⎟ = 91.84 kN/m 2 = 91.8 kPa ⎟ ⎠

The atmospheric pressure may be expressed in mmHg as

hHg =

Patm 91.8 kPa ⎛ 1000 Pa ⎞⎛ 1000 mm ⎞ = ⎜ ⎟⎜ ⎟ = 688 mmHg ρg (13,600 kg/m 3 )(9.81 m/s 2 ) ⎝ 1 kPa ⎠⎝ 1 m ⎠

1-98 The gravitational acceleration changes with altitude. Accounting for this variation, the weights of a body at different locations are to be determined. Analysis The weight of an 80-kg man at various locations is obtained by substituting the altitude z (values in m) into the relation

⎛ 1N W = mg = (80kg)(9.807 − 3.32 × 10 −6 z m/s 2 )⎜ ⎜ 1kg ⋅ m/s 2 ⎝

⎞ ⎟ ⎟ ⎠

Sea level:

(z = 0 m): W = 80×(9.807-3.32x10-6×0) = 80×9.807 = 784.6 N

Denver:

(z = 1610 m): W = 80×(9.807-3.32x10-6×1610) = 80×9.802 = 784.2 N

Mt. Ev.:

(z = 8848 m): W = 80×(9.807-3.32x10-6×8848) = 80×9.778 = 782.2 N

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1-40

1-99 A man is considering buying a 12-oz steak for $3.15, or a 320-g steak for $2.80. The steak that is a better buy is to be determined. Assumptions The steaks are of identical quality. Analysis To make a comparison possible, we need to express the cost of each steak on a common basis. Let us choose 1 kg as the basis for comparison. Using proper conversion factors, the unit cost of each steak is determined to be

12 ounce steak: ⎛ $3.15 ⎞ ⎛ 16 oz ⎞ ⎛ 1 lbm ⎞ ⎟⎟ = $9.26/kg Unit Cost = ⎜ ⎟⎜ ⎟ ⎜⎜ ⎝ 12 oz ⎠ ⎝ 1 lbm ⎠ ⎝ 0.45359 kg ⎠

320 gram steak: ⎛ $2.80 ⎞ ⎛ 1000 g ⎞ ⎟⎟ ⎜⎜ ⎟⎟ = $8.75/kg Unit Cost = ⎜⎜ ⎝ 320 g ⎠ ⎝ 1 kg ⎠

Therefore, the steak at the international market is a better buy.

1-100E The thrust developed by the jet engine of a Boeing 777 is given to be 85,000 pounds. This thrust is to be expressed in N and kgf. Analysis Noting that 1 lbf = 4.448 N and 1 kgf = 9.81 N, the thrust developed can be expressed in two other units as

Thrust in N:

⎛ 4.448 N ⎞ 5 Thrust = (85,000 lbf )⎜ ⎟ = 3.78 × 10 N 1 lbf ⎝ ⎠

Thrust in kgf:

⎛ 1 kgf ⎞ 4 Thrust = (37.8 × 10 5 N )⎜ ⎟ = 3.85 × 10 kgf ⎝ 9.81 N ⎠

1-101E The efficiency of a refrigerator increases by 3% per °C rise in the minimum temperature. This increase is to be expressed per °F, K, and R rise in the minimum temperature. Analysis The magnitudes of 1 K and 1°C are identical, so are the magnitudes of 1 R and 1°F. Also, a change of 1 K or 1°C in temperature corresponds to a change of 1.8 R or 1.8°F. Therefore, the increase in efficiency is

(a) 3% for each K rise in temperature, and (b), (c) 3/1.8 = 1.67% for each R or °F rise in temperature.

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1-41

1-102E The boiling temperature of water decreases by 3°C for each 1000 m rise in altitude. This decrease in temperature is to be expressed in °F, K, and R. Analysis The magnitudes of 1 K and 1°C are identical, so are the magnitudes of 1 R and 1°F. Also, a change of 1 K or 1°C in temperature corresponds to a change of 1.8 R or 1.8°F. Therefore, the decrease in the boiling temperature is

(a) 3 K for each 1000 m rise in altitude, and (b), (c) 3×1.8 = 5.4°F = 5.4 R for each 1000 m rise in altitude.

1-103E Hyperthermia of 5°C is considered fatal. This fatal level temperature change of body temperature is to be expressed in °F, K, and R. Analysis The magnitudes of 1 K and 1°C are identical, so are the magnitudes of 1 R and 1°F. Also, a change of 1 K or 1°C in temperature corresponds to a change of 1.8 R or 1.8°F. Therefore, the fatal level of hypothermia is

(a) 5 K (b) 5×1.8 = 9°F (c) 5×1.8 = 9 R

1-104E A house is losing heat at a rate of 4500 kJ/h per °C temperature difference between the indoor and the outdoor temperatures. The rate of heat loss is to be expressed per °F, K, and R of temperature difference between the indoor and the outdoor temperatures. Analysis The magnitudes of 1 K and 1°C are identical, so are the magnitudes of 1 R and 1°F. Also, a change of 1 K or 1°C in temperature corresponds to a change of 1.8 R or 1.8°F. Therefore, the rate of heat loss from the house is

(a) 4500 kJ/h per K difference in temperature, and (b), (c) 4500/1.8 = 2500 kJ/h per R or °F rise in temperature.

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1-42

1-105 The average temperature of the atmosphere is expressed as Tatm = 288.15 – 6.5z where z is altitude in km. The temperature outside an airplane cruising at 12,000 m is to be determined. Analysis Using the relation given, the average temperature of the atmosphere at an altitude of 12,000 m is determined to be

Tatm = 288.15 - 6.5z = 288.15 - 6.5×12 = 210.15 K = - 63°C Discussion This is the “average” temperature. The actual temperature at different times can be different.

1-106 A new “Smith” absolute temperature scale is proposed, and a value of 1000 S is assigned to the boiling point of water. The ice point on this scale, and its relation to the Kelvin scale are to be determined. Analysis All linear absolute temperature scales read zero at absolute zero pressure, and are constant multiples of each other. For example, T(R) = 1.8 T(K). That is, multiplying a temperature value in K by 1.8 will give the same temperature in R.

The proposed temperature scale is an acceptable absolute temperature scale since it differs from the other absolute temperature scales by a constant only. The boiling temperature of water in the Kelvin and the Smith scales are 315.15 K and 1000 K, respectively. Therefore, these two temperature scales are related to each other by

T (S ) =

S

K

1000

373.15

1000 T ( K ) = 2.6799 T(K ) 373.15

The ice point of water on the Smith scale is

0

T(S)ice = 2.6799 T(K)ice = 2.6799×273.15 = 732.0 S

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1-43

1-107E An expression for the equivalent wind chill temperature is given in English units. It is to be converted to SI units. Analysis The required conversion relations are 1 mph = 1.609 km/h and T(°F) = 1.8T(°C) + 32. The first thought that comes to mind is to replace T(°F) in the equation by its equivalent 1.8T(°C) + 32, and V in mph by 1.609 km/h, which is the “regular” way of converting units. However, the equation we have is not a regular dimensionally homogeneous equation, and thus the regular rules do not apply. The V in the equation is a constant whose value is equal to the numerical value of the velocity in mph. Therefore, if V is given in km/h, we should divide it by 1.609 to convert it to the desired unit of mph. That is, Tequiv (° F) = 91.4 − [ 91.4 − Tambient (° F)][ 0.475 − 0.0203(V / 1.609) + 0.304 V / 1.609 ]

or Tequiv (° F) = 91.4 − [ 91.4 − Tambient (° F)][ 0.475 − 0.0126V + 0.240 V ]

where V is in km/h. Now the problem reduces to converting a temperature in °F to a temperature in °C, using the proper convection relation: 18 . Tequiv (° C ) + 32 = 914 . − [914 . − (18 . Tambient (° C ) + 32 )][0.475 − 0.0126V + 0.240 V ]

which simplifies to Tequiv (° C) = 33.0 − ( 33.0 − Tambient )(0.475 − 0.0126V + 0.240 V )

where the ambient air temperature is in °C.

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1-44

1-108E EES Problem 1-107E is reconsidered. The equivalent wind-chill temperatures in °F as a function of wind velocity in the range of 4 mph to 100 mph for the ambient temperatures of 20, 40, and 60°F are to be plotted, and the results are to be discussed. Analysis The problem is solved using EES, and the solution is given below. "Obtain V and T_ambient from the Diagram Window" {T_ambient=10 V=20} V_use=max(V,4) T_equiv=91.4-(91.4-T_ambient)*(0.475 - 0.0203*V_use + 0.304*sqrt(V_use)) "The parametric table was used to generate the plot, Fill in values for T_ambient and V (use Alter Values under Tables menu) then use F3 to solve table. Plot the first 10 rows and then overlay the second ten, and so on. Place the text on the plot using Add Text under the Plot menu." V [mph]

20

10 10 10 10 10 10 10 10 10 10 20 20 20 20 20 20 20 20 20 20 30 30 30 30 30 30 30 30 30 30 40 40 40 40 40 40 40 40 40 40

10

W ind Chill Temperature

0 -10

W ind speed =10 mph

-20

T W indChill

Tambient [F] -25 -20 -15 -10 -5 0 5 10 15 20 -25 -20 -15 -10 -5 0 5 10 15 20 -25 -20 -15 -10 -5 0 5 10 15 20 -25 -20 -15 -10 -5 0 5 10 15 20

-30

20 mph

-40 -50

30 mph -60 -70

40 mph

-80 -30

-20

-10

0

10

20

T ambient 60 50

Tamb = 60F

40 30

Tequiv [F]

Tequiv [F] -52 -46 -40 -34 -27 -21 -15 -9 -3 3 -75 -68 -61 -53 -46 -39 -32 -25 -18 -11 -87 -79 -72 -64 -56 -49 -41 -33 -26 -18 -93 -85 -77 -69 -61 -54 -46 -38 -30 -22

20

Tamb = 40F 10 0 -10

Tamb = 20F -20 0

20

40

60

80

V [mph]

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100

1-45

1-109 One section of the duct of an air-conditioning system is laid underwater. The upward force the water will exert on the duct is to be determined. Assumptions 1 The diameter given is the outer diameter of the duct (or, the thickness of the duct material is negligible). 2 The weight of the duct and the air in is negligible. Properties The density of air is given to be ρ = 1.30 kg/m3. We take the density of water to be 1000 kg/m3. Analysis Noting that the weight of the duct and the air in it is negligible, the net upward force acting on the duct is the buoyancy force exerted by water. The volume of the underground section of the duct is

D =15 cm L = 20 m FB

V = AL = (πD 2 / 4) L = [π (0.15 m) 2 /4](20 m) = 0.353 m 3 Then the buoyancy force becomes ⎛ 1 kN FB = ρgV = (1000 kg/m 3 )(9.81 m/s 2 )(0.353 m 3 )⎜ ⎜ 1000 kg ⋅ m/s 2 ⎝

⎞ ⎟ = 3.46 kN ⎟ ⎠

Discussion The upward force exerted by water on the duct is 3.46 kN, which is equivalent to the weight of a mass of 353 kg. Therefore, this force must be treated seriously.

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1-46

1-110 A helium balloon tied to the ground carries 2 people. The acceleration of the balloon when it is first released is to be determined. Assumptions The weight of the cage and the ropes of the balloon is negligible. Properties The density of air is given to be ρ = 1.16 kg/m3. The density of helium gas is 1/7th of this. Analysis The buoyancy force acting on the balloon is

V balloon = 4π r 3 /3 = 4 π(5 m) 3 /3 = 523.6 m 3 FB = ρ air gV balloon

⎛ 1N = (1.16 kg/m 3 )(9.81m/s 2 )(523.6 m 3 )⎜ ⎜ 1 kg ⋅ m/s 2 ⎝

⎞ ⎟ = 5958 N ⎟ ⎠

The total mass is

⎛ 1.16 ⎞ kg/m 3 ⎟(523.6 m 3 ) = 86.8 kg m He = ρ HeV = ⎜ 7 ⎝ ⎠ m total = m He + m people = 86.8 + 2 × 70 = 226.8 kg The total weight is ⎛ 1N W = m total g = (226.8 kg)(9.81 m/s 2 )⎜ ⎜ 1 kg ⋅ m/s 2 ⎝

⎞ ⎟ = 2225 N ⎟ ⎠

Thus the net force acting on the balloon is Fnet = FB − W = 5958 − 2225 = 3733 N

Then the acceleration becomes a=

Fnet 3733 N ⎛⎜ 1kg ⋅ m/s 2 = m total 226.8 kg ⎜⎝ 1 N

⎞ ⎟ = 16.5 m/s 2 ⎟ ⎠

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1-47

1-111 EES Problem 1-110 is reconsidered. The effect of the number of people carried in the balloon on acceleration is to be investigated. Acceleration is to be plotted against the number of people, and the results are to be discussed. Analysis The problem is solved using EES, and the solution is given below. "Given Data:" rho_air=1.16"[kg/m^3]" "density of air" g=9.807"[m/s^2]" d_balloon=10"[m]" m_1person=70"[kg]" {NoPeople = 2} "Data suppied in Parametric Table" "Calculated values:" rho_He=rho_air/7"[kg/m^3]" "density of helium" r_balloon=d_balloon/2"[m]" V_balloon=4*pi*r_balloon^3/3"[m^3]" m_people=NoPeople*m_1person"[kg]" m_He=rho_He*V_balloon"[kg]" m_total=m_He+m_people"[kg]" "The total weight of balloon and people is:" W_total=m_total*g"[N]" "The buoyancy force acting on the balloon, F_b, is equal to the weight of the air displaced by the balloon." F_b=rho_air*V_balloon*g"[N]" "From the free body diagram of the balloon, the balancing vertical forces must equal the product of the total mass and the vertical acceleration:" F_b- W_total=m_total*a_up 30

NoPeople 1 2 3 4 5 6 7 8 9 10

25 20

a up [m /s^2]

Aup [m/s2] 28.19 16.46 10.26 6.434 3.831 1.947 0.5204 -0.5973 -1.497 -2.236

15 10 5 0 -5 1

2

3

4

5

6

7

8

9

10

NoPeople

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1-48

1-112 A balloon is filled with helium gas. The maximum amount of load the balloon can carry is to be determined. Assumptions The weight of the cage and the ropes of the balloon is negligible. Properties The density of air is given to be ρ = 1.16 kg/m3. The density of helium gas is 1/7th of this. Analysis In the limiting case, the net force acting on the balloon will be zero. That is, the buoyancy force and the weight will balance each other: W = mg = FB m total

Helium balloon

F 5958 N = B = = 607.3 kg g 9.81 m/s 2

Thus,

m

m people = m total − m He = 607.3 − 86.8 = 520.5 kg

1-113E The pressure in a steam boiler is given in kgf/cm2. It is to be expressed in psi, kPa, atm, and bars. Analysis We note that 1 atm = 1.03323 kgf/cm2, 1 atm = 14.696 psi, 1 atm = 101.325 kPa, and 1 atm = 1.01325 bar (inner cover page of text). Then the desired conversions become:

In atm:

⎛ 1 atm P = (92 kgf/cm 2 )⎜ ⎜ 1.03323 kgf/cm 2 ⎝

⎞ ⎟ = 89.04 atm ⎟ ⎠

In psi:

⎛ 1 atm P = (92 kgf/cm 2 )⎜ ⎜ 1.03323 kgf/cm 2 ⎝

⎞⎛ 14.696 psi ⎞ ⎟⎜ ⎟ = 1309 psi ⎟⎜ 1 atm ⎟ ⎠ ⎠⎝

In kPa:

⎛ 1 atm P = (92 kgf/cm 2 )⎜ ⎜ 1.03323 kgf/cm 2 ⎝

⎞⎛ 101.325 kPa ⎞ ⎟⎜ ⎟ = 9022 kPa ⎟⎜ 1 atm ⎟ ⎠ ⎠⎝

In bars:

⎛ 1 atm P = (92 kgf/cm 2 )⎜ ⎜ 1.03323 kgf/cm 2 ⎝

⎞⎛ 1.01325 bar ⎞ ⎟⎜ ⎟ = 90.22 bar ⎟⎜ 1 atm ⎟ ⎠ ⎠⎝

Discussion Note that the units atm, kgf/cm2, and bar are almost identical to each other.

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1-49

1-114 A 10-m high cylindrical container is filled with equal volumes of water and oil. The pressure difference between the top and the bottom of the container is to be determined. Properties The density of water is given to be ρ = 1000 kg/m3. The specific gravity of oil is given to be 0.85. Analysis The density of the oil is obtained by multiplying its specific gravity by the density of water,

Oil SG = 0.85

ρ = SG × ρ H 2O = (0.85)(100 0 kg/m 3 ) = 850 kg/m 3

h = 10 m Water

The pressure difference between the top and the bottom of the cylinder is the sum of the pressure differences across the two fluids, ΔPtotal = ΔPoil + ΔPwater = ( ρgh) oil + ( ρgh) water

[

]

⎛ 1 kPa = (850 kg/m 3 )(9.81 m/s 2 )(5 m) + (1000 kg/m 3 )(9.81 m/s 2 )(5 m) ⎜ ⎜ 1000 N/m 2 ⎝ = 90.7 kPa

⎞ ⎟ ⎟ ⎠

1-115 The pressure of a gas contained in a vertical piston-cylinder device is measured to be 250 kPa. The mass of the piston is to be determined. Assumptions There is no friction between the piston and the cylinder.

Patm

Analysis Drawing the free body diagram of the piston and balancing the vertical forces yield W = PA − Patm A mg = ( P − Patm ) A

⎛ 1000 kg/m ⋅ s 2 ( m)(9.81 m/s 2 ) = (250 − 100 kPa)(30 × 10 − 4 m 2 )⎜ ⎜ 1kPa ⎝

It yields

⎞ ⎟ ⎟ ⎠

P

W = mg

m = 45.9 kg

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1-50

1-116 The gage pressure in a pressure cooker is maintained constant at 100 kPa by a petcock. The mass of the petcock is to be determined. Assumptions There is no blockage of the pressure release valve. Analysis Atmospheric pressure is acting on all surfaces of the petcock, which balances itself out. Therefore, it can be disregarded in calculations if we use the gage pressure as the cooker pressure. A force balance on the petcock (ΣFy = 0) yields

Patm

W = Pgage A m=

Pgage A g

=

(100 kPa)(4 × 10 − 6 m 2 ) ⎛⎜ 1000 kg/m ⋅ s 2 ⎜ 1 kPa 9.81 m/s 2 ⎝

P

W = mg

⎞ ⎟ ⎟ ⎠

= 0.0408 kg

1-117 A glass tube open to the atmosphere is attached to a water pipe, and the pressure at the bottom of the tube is measured. It is to be determined how high the water will rise in the tube. Properties The density of water is given to be ρ = 1000 kg/m3. Analysis The pressure at the bottom of the tube can be expressed as P = Patm + ( ρ g h) tube

Solving for h, h=

Patm= 92 kPa

P − Patm ρg

⎛ 1 kg ⋅ m/s 2 ⎜ = 1N (1000 kg/m 3 )(9.81 m/s 2 ) ⎜⎝ = 2.34 m (115 − 92) kPa

⎞⎛ 1000 N/m 2 ⎟⎜ ⎟⎜ 1 kPa ⎠⎝

⎞ ⎟ ⎟ ⎠

h

Water

1-118 The average atmospheric pressure is given as Patm = 101.325(1 − 0.02256z )5.256 where z is the altitude in km. The atmospheric pressures at various locations are to be determined. Analysis The atmospheric pressures at various locations are obtained by substituting the altitude z values in km into the relation Patm = 101325 . (1 − 0.02256z )5.256

Atlanta:

(z = 0.306 km): Patm = 101.325(1 - 0.02256×0.306)5.256 = 97.7 kPa

Denver:

(z = 1.610 km): Patm = 101.325(1 - 0.02256×1.610)5.256 = 83.4 kPa

M. City:

(z = 2.309 km): Patm = 101.325(1 - 0.02256×2.309)5.256 = 76.5 kPa

Mt. Ev.:

(z = 8.848 km): Patm = 101.325(1 - 0.02256×8.848)5.256 = 31.4 kPa

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1-51

1-119 The air pressure in a duct is measured by an inclined manometer. For a given vertical level difference, the gage pressure in the duct and the length of the differential fluid column are to be determined. Assumptions The manometer fluid is an incompressible substance. Properties The density of the liquid is given to be ρ = 0.81 kg/L = 810 kg/m3. Analysis The gage pressure in the duct is determined from

Pgage = Pabs − Patm = ρgh

⎛ 1N = (810 kg/m 3 )(9.81 m/s 2 )(0.08 m)⎜ ⎜ 1 kg ⋅ m/s 2 ⎝ = 636 Pa

⎞⎛ 1 Pa ⎟⎜ ⎟⎜ 1 N/m 2 ⎠⎝

⎞ ⎟ ⎟ ⎠

The length of the differential fluid column is L = h / sin θ = (8 cm ) / sin 35° = 13.9 cm

Discussion Note that the length of the differential fluid column is extended considerably by inclining the manometer arm for better readability.

1-120E Equal volumes of water and oil are poured into a U-tube from different arms, and the oil side is pressurized until the contact surface of the two fluids moves to the bottom and the liquid levels in both arms become the same. The excess pressure applied on the oil side is to be determined. Assumptions 1 Both water and oil are incompressible substances. 2 Oil does not mix with water. 3 The cross-sectional area of the U-tube is constant. Properties The density of oil is given to be ρoil = 49.3 lbm/ft3. We take the density of water to be ρw = 62.4 lbm/ft3. Analysis Noting that the pressure of both the water and the oil is the same at the contact surface, the pressure at this surface can be expressed as Pcontact = Pblow + ρ a gha = Patm + ρ w gh w

Noting that ha = hw and rearranging, Pgage,blow = Pblow − Patm = ( ρ w − ρ oil ) gh

⎛ 1 lbf = (62.4 - 49.3 lbm/ft 3 )(32.2 ft/s 2 )(30/12 ft)⎜ ⎜ 32.2 lbm ⋅ ft/s 2 ⎝ = 0.227 psi

⎞⎛ 1 ft 2 ⎟⎜ ⎟⎜ 144 in 2 ⎠⎝

⎞ ⎟ ⎟ ⎠

Discussion When the person stops blowing, the oil will rise and some water will flow into the right arm. It can be shown that when the curvature effects of the tube are disregarded, the differential height of water will be 23.7 in to balance 30-in of oil.

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1-52

1-121 It is given that an IV fluid and the blood pressures balance each other when the bottle is at a certain height, and a certain gage pressure at the arm level is needed for sufficient flow rate. The gage pressure of the blood and elevation of the bottle required to maintain flow at the desired rate are to be determined. Assumptions 1 The IV fluid is incompressible. 2 The IV bottle is open to the atmosphere. Properties The density of the IV fluid is given to be ρ = 1020 kg/m3. Analysis (a) Noting that the IV fluid and the blood pressures balance each other when the bottle is 1.2 m above the arm level, the gage pressure of the blood in the arm is simply equal to the gage pressure of the IV fluid at a depth of 1.2 m,

Pgage, arm = Pabs − Patm = ρgharm- bottle

⎛ 1 kN = (1020 kg/m 3 )(9.81 m/s 2 )(1.20 m)⎜ ⎜ 1000 kg ⋅ m/s 2 ⎝ = 12.0 k Pa

⎞⎛ 1 kPa ⎟⎜ ⎟⎜ 1 kN/m 2 ⎠⎝

⎞ ⎟ ⎟ ⎠

(b) To provide a gage pressure of 20 kPa at the arm level, the height of the bottle from the arm level is again determined from Pgage, arm = ρgharm-bottle to be harm - bottle = =

Pgage, arm

ρg

⎛ 1000 kg ⋅ m/s 2 ⎜ 1 kN (1020 kg/m 3 )(9.81 m/s 2 ) ⎜⎝ 20 kPa

⎞⎛ 1 kN/m 2 ⎟⎜ ⎟⎜ 1 kPa ⎠⎝

⎞ ⎟ = 2.0 m ⎟ ⎠

Discussion Note that the height of the reservoir can be used to control flow rates in gravity driven flows. When there is flow, the pressure drop in the tube due to friction should also be considered. This will result in raising the bottle a little higher to overcome pressure drop.

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1-53

1-122E A water pipe is connected to a double-U manometer whose free arm is open to the atmosphere. The absolute pressure at the center of the pipe is to be determined. Assumptions 1 All the liquids are incompressible. 2 The solubility of the liquids in each other is negligible. Properties The specific gravities of mercury and oil are given to be 13.6 and 0.80, respectively. We take the density of water to be ρw = 62.4 lbm/ft3. Analysis Starting with the pressure at the center of the water pipe, and moving along the tube by adding (as we go down) or subtracting (as we go up) the ρgh terms until we reach the free surface of oil where the oil tube is exposed to the atmosphere, and setting the result equal to Patm gives

Pwater pipe − ρ water ghwater + ρ oil ghoil − ρ Hg ghHg − ρ oil ghoil = Patm Solving for Pwater

pipe,

Pwater pipe = Patm + ρ water g (hwater − SGoil hoil + SG Hg hHg + SGoil hoil ) Substituting, Pwater pipe = 14.2 psia + (62.4 lbm/ft 3 )(32.2 ft/s 2 )[(35/12 ft) − 0.8(60/12 ft) + 13.6(15/12 ft) ⎛ 1 lbf + 0.8(40/12 ft)] × ⎜ ⎜ 32.2 lbm ⋅ ft/s 2 ⎝ = 22.3 psia

⎞⎛ 1 ft 2 ⎟⎜ ⎟⎜ 144 in 2 ⎠⎝

⎞ ⎟ ⎟ ⎠

Therefore, the absolute pressure in the water pipe is 22.3 psia. Discussion Note that jumping horizontally from one tube to the next and realizing that pressure remains the same in the same fluid simplifies the analysis greatly.

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1-54

1-123 The temperature of the atmosphere varies with altitude z as T = T0 − βz , while the gravitational acceleration varies by g ( z ) = g 0 /(1 + z / 6,370,320) 2 . Relations for the variation of pressure in atmosphere are to be obtained (a) by ignoring and (b) by considering the variation of g with altitude. Assumptions The air in the troposphere behaves as an ideal gas. Analysis (a) Pressure change across a differential fluid layer of thickness dz in the vertical z direction is dP = − ρgdz

From the ideal gas relation, the air density can be expressed as ρ = dP = −

P P = . Then, RT R (T0 − βz )

P gdz R(T0 − β z )

Separating variables and integrating from z = 0 where P = P0 to z = z where P = P,

∫

P

P0

dP =− P

∫

z

0

gdz R(T0 − βz )

Performing the integrations. T − βz g P ln ln 0 = P0 Rβ T0 Rearranging, the desired relation for atmospheric pressure for the case of constant g becomes g

⎛ β z ⎞ βR P = P0 ⎜⎜1 − ⎟⎟ ⎝ T0 ⎠ (b) When the variation of g with altitude is considered, the procedure remains the same but the expressions become more complicated, g0 P dP = − dz R(T0 − βz ) (1 + z / 6,370,320) 2

Separating variables and integrating from z = 0 where P = P0 to z = z where P = P,

∫

P

P0

dP =− P

z

g 0 dz

0

R (T0 − βz )(1 + z / 6,370,320) 2

∫

Performing the integrations, P

ln P

P0

=

g0 1 1 1 + kz ln − Rβ (1 + kT0 / β )(1 + kz ) (1 + kT0 / β ) 2 T0 − βz

z

0

where R = 287 J/kg⋅K = 287 m2/s2⋅K is the gas constant of air. After some manipulations, we obtain ⎡ ⎛ 1 g0 1 1 + kz ⎜ P = P0 exp ⎢− ⎜ 1 + 1 / kz + 1 + kT / β ln 1 − βz / T R ( β kT ) + 0 ⎝ 0 0 ⎣⎢

⎞⎤ ⎟⎥ ⎟ ⎠⎦⎥

where T0 = 288.15 K, β = 0.0065 K/m, g0 = 9.807 m/s2, k = 1/6,370,320 m-1, and z is the elevation in m.. Discussion When performing the integration in part (b), the following expression from integral tables is used, together with a transformation of variable x = T0 − βz ,

dx

∫ x(a + bx)

2

=

1 1 a + bx − 2 ln a(a + bx ) a x

Also, for z = 11,000 m, for example, the relations in (a) and (b) give 22.62 and 22.69 kPa, respectively.

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1-55

1-124 The variation of pressure with density in a thick gas layer is given. A relation is to be obtained for pressure as a function of elevation z. Assumptions The property relation P = Cρ n is valid over the entire region considered. Analysis The pressure change across a differential fluid layer of thickness dz in the vertical z direction is given as, dP = − ρgdz

Also, the relation P = Cρ n can be expressed as C = P / ρ n = P0 / ρ 0n , and thus

ρ = ρ 0 ( P / P0 ) 1 / n Substituting,

dP = − gρ 0 ( P / P0 ) 1 / n dz Separating variables and integrating from z = 0 where P = P0 = Cρ 0n to z = z where P = P,

∫

P

P0

∫

z

( P / P0 ) −1 / n dP = − ρ 0 g dz 0

Performing the integrations. P0

( P / P0 ) −1 / n +1 − 1/ n + 1

P

= − ρ 0 gz P0

→

⎛ P ⎜ ⎜P ⎝ 0

⎞ ⎟ ⎟ ⎠

( n −1) / n

−1 = −

n − 1 ρ 0 gz n P0

Solving for P,

⎛ n − 1 ρ 0 gz ⎞ ⎟ P = P0 ⎜⎜1 − n P0 ⎟⎠ ⎝

n /( n −1)

which is the desired relation. Discussion The final result could be expressed in various forms. The form given is very convenient for calculations as it facilitates unit cancellations and reduces the chance of error.

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1-56

1-125 A pressure transducers is used to measure pressure by generating analogue signals, and it is to be calibrated by measuring both the pressure and the electric current simultaneously for various settings, and the results are tabulated. A calibration curve in the form of P = aI + b is to be obtained, and the pressure corresponding to a signal of 10 mA is to be calculated. Assumptions Mercury is an incompressible liquid. Properties The specific gravity of mercury is given to be 13.56, and thus its density is 13,560 kg/m3. Analysis For a given differential height, the pressure can be calculated from

P = ρgΔh For Δh = 28.0 mm = 0.0280 m, for example, ⎛ 1 kN P = 13.56(1000 kg/m 3 )(9.81 m/s 2 )(0.0280 m)⎜⎜ 1000 kg ⋅ m/s 2 ⎝

⎞⎛ 1 kPa ⎞ ⎟⎜ ⎟⎝ 1 kN/m 2 ⎟⎠ = 3.75 kPa ⎠

Repeating the calculations and tabulating, we have Δh(mm)

28.0

181.5

297.8

413.1

765.9

1027

1149

1362

1458

1536

P(kPa)

3.73

24.14

39.61

54.95

101.9

136.6

152.8

181.2

193.9

204.3

I (mA)

4.21

5.78

6.97

8.15

11.76

14.43

15.68

17.86

18.84

19.64

A plot of P versus I is given below. It is clear that the pressure varies linearly with the current, and using EES, the best curve fit is obtained to be P = 13.00I - 51.00

(kPa)

for 4.21 ≤ I ≤ 19.64 .

For I = 10 mA, for example, we would get P = 79.0 kPa

225

180

P, kPa

135

90

45

0 4

6

8

10

12

14

16

18

20

I, mA Discussion Note that the calibration relation is valid in the specified range of currents or pressures.

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1-57

Fundamentals of Engineering (FE) Exam Problems 1-126 Consider a fish swimming 5 m below the free surface of water. The increase in the pressure exerted on the fish when it dives to a depth of 45 m below the free surface is

(a) 392 Pa

(b) 9800 Pa

(c) 50,000 Pa

(d) 392,000 Pa

(e) 441,000 Pa

Answer (d) 392,000 Pa Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). rho=1000 "kg/m3" g=9.81 "m/s2" z1=5 "m" z2=45 "m" DELTAP=rho*g*(z2-z1) "Pa" "Some Wrong Solutions with Common Mistakes:" W1_P=rho*g*(z2-z1)/1000 "dividing by 1000" W2_P=rho*g*(z1+z2) "adding depts instead of subtracting" W3_P=rho*(z1+z2) "not using g" W4_P=rho*g*(0+z2) "ignoring z1"

1-127 The atmospheric pressures at the top and the bottom of a building are read by a barometer to be 96.0 and 98.0 kPa. If the density of air is 1.0 kg/m3, the height of the building is

(a) 17 m

(b) 20 m

(c) 170 m

(d) 204 m

(e) 252 m

Answer (d) 204 m Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). rho=1.0 "kg/m3" g=9.81 "m/s2" P1=96 "kPa" P2=98 "kPa" DELTAP=P2-P1 "kPa" DELTAP=rho*g*h/1000 "kPa" "Some Wrong Solutions with Common Mistakes:" DELTAP=rho*W1_h/1000 "not using g" DELTAP=g*W2_h/1000 "not using rho" P2=rho*g*W3_h/1000 "ignoring P1" P1=rho*g*W4_h/1000 "ignoring P2"

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1-58 1-128 An apple loses 4.5 kJ of heat as it cools per °C drop in its temperature. The amount of heat loss from the apple per °F drop in its temperature is (a) 1.25 kJ (b) 2.50 kJ (c) 5.0 kJ (d) 8.1 kJ (e) 4.1 kJ

Answer (b) 2.50 kJ Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). Q_perC=4.5 "kJ" Q_perF=Q_perC/1.8 "kJ" "Some Wrong Solutions with Common Mistakes:" W1_Q=Q_perC*1.8 "multiplying instead of dividing" W2_Q=Q_perC "setting them equal to each other"

1-129 Consider a 2-m deep swimming pool. The pressure difference between the top and bottom of the pool is

(a) 12.0 kPa

(b) 19.6 kPa

(c) 38.1 kPa

(d) 50.8 kPa

(e) 200 kPa

Answer (b) 19.6 kPa Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). rho=1000 "kg/m^3" g=9.81 "m/s2" z1=0 "m" z2=2 "m" DELTAP=rho*g*(z2-z1)/1000 "kPa" "Some Wrong Solutions with Common Mistakes:" W1_P=rho*(z1+z2)/1000 "not using g" W2_P=rho*g*(z2-z1)/2000 "taking half of z" W3_P=rho*g*(z2-z1) "not dividing by 1000"

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1-59

1-130 At sea level, the weight of 1 kg mass in SI units is 9.81 N. The weight of 1 lbm mass in English units is

(a) 1 lbf

(b) 9.81 lbf

(c) 32.2 lbf

(d) 0.1 lbf

(e) 0.031 lbf

Answer (a) 1 lbf Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m=1 "lbm" g=32.2 "ft/s2" W=m*g/32.2 "lbf" "Some Wrong Solutions with Common Mistakes:" gSI=9.81 "m/s2" W1_W= m*gSI "Using wrong conversion" W2_W= m*g "Using wrong conversion" W3_W= m/gSI "Using wrong conversion" W4_W= m/g "Using wrong conversion"

1-131 During a heating process, the temperature of an object rises by 20°C. This temperature rise is equivalent to a temperature rise of

(a) 20°F

(b) 52°F

(c) 36 K

(d) 36 R

(e) 293 K

Answer (d) 36 R Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T_inC=20 "C" T_inR=T_inC*1.8 "R" "Some Wrong Solutions with Common Mistakes:" W1_TinF=T_inC "F, setting C and F equal to each other" W2_TinF=T_inC*1.8+32 "F, converting to F " W3_TinK=1.8*T_inC "K, wrong conversion from C to K" W4_TinK=T_inC+273 "K, converting to K"

1-132 … 1-134 Design, Essay, and Experiment Problems

KJ

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2-1

Chapter 2 ENERGY, ENERGY TRANSFER, AND GENERAL ENERGY ANALYSIS Forms of Energy

2-1C Initially, the rock possesses potential energy relative to the bottom of the sea. As the rock falls, this potential energy is converted into kinetic energy. Part of this kinetic energy is converted to thermal energy as a result of frictional heating due to air resistance, which is transferred to the air and the rock. Same thing happens in water. Assuming the impact velocity of the rock at the sea bottom is negligible, the entire potential energy of the rock is converted to thermal energy in water and air.

2-2C Hydrogen is also a fuel, since it can be burned, but it is not an energy source since there are no hydrogen reserves in the world. Hydrogen can be obtained from water by using another energy source, such as solar or nuclear energy, and then the hydrogen obtained can used as a fuel to power cars or generators. Therefore, it is more proper to view hydrogen is an energy carrier than an energy source.

2-3C The macroscopic forms of energy are those a system possesses as a whole with respect to some outside reference frame. The microscopic forms of energy, on the other hand, are those related to the molecular structure of a system and the degree of the molecular activity, and are independent of outside reference frames.

2-4C The sum of all forms of the energy a system possesses is called total energy. In the absence of magnetic, electrical and surface tension effects, the total energy of a system consists of the kinetic, potential, and internal energies.

2-5C Thermal energy is the sensible and latent forms of internal energy, and it is referred to as heat in daily life.

2-6C The mechanical energy is the form of energy that can be converted to mechanical work completely and directly by a mechanical device such as a propeller. It differs from thermal energy in that thermal energy cannot be converted to work directly and completely. The forms of mechanical energy of a fluid stream are kinetic, potential, and flow energies.

2-7E The specific kinetic energy of a mass whose velocity is given is to be determined. Analysis According to the definition of the specific kinetic energy, ke =

V 2 (100 ft/s) 2 ⎛ 1 Btu/lbm ⎞ ⎜⎜ ⎟ = 0.200 Btu/lbm = 2 2⎟ 2 2 ⎝ 25,037 ft /s ⎠

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2-2

2-8 The specific kinetic energy of a mass whose velocity is given is to be determined. Analysis Substitution of the given data into the expression for the specific kinetic energy gives

ke =

V 2 (30 m/s) 2 ⎛ 1 kJ/kg ⎞ = ⎜ ⎟ = 0.45 kJ/kg 2 2 ⎝ 1000 m 2 /s 2 ⎠

2-9E The total potential energy of an object that is below a reference level is to be determined. Analysis Substituting the given data into the potential energy expression gives ⎛ 1 Btu/lbm PE = mgz = (100 lbm)(31.7 ft/s 2 )(−20 ft)⎜⎜ 2 2 ⎝ 25,037 ft /s

⎞ ⎟ = −2.53 Btu ⎟ ⎠

2-10 The specific potential energy of an object is to be determined. Analysis The specific potential energy is given by ⎛ 1 kJ/kg ⎞ pe = gz = (9.8 m/s 2 )(50 m)⎜ ⎟ = 0.49 kJ/kg ⎝ 1000 m 2 /s 2 ⎠

2-11 The total potential energy of an object is to be determined. Analysis Substituting the given data into the potential energy expression gives ⎛ 1 kJ/kg ⎞ PE = mgz = (100 kg)(9.8 m/s 2 )(20 m)⎜ ⎟ = 19.6 kJ ⎝ 1000 m 2 /s 2 ⎠

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2-3

2-12 A river is flowing at a specified velocity, flow rate, and elevation. The total mechanical energy of the river water per unit mass, and the power generation potential of the entire river are to be determined. Assumptions 1 The elevation given is the elevation of the free surface of the river. 2 The velocity given is the average velocity. 3 The mechanical energy of water at the turbine exit is negligible. Properties We take the density of water to be ρ = 1000 kg/m3.

River

3 m/s

Analysis Noting that the sum of the flow energy and the potential energy is constant for a given fluid body, we can take the elevation of the entire river water to be the elevation of the free surface, and ignore the flow energy. Then the total mechanical energy of the river water per unit mass becomes

emech = pe + ke = gh +

90 m

V 2 ⎛⎜ (3 m/s) 2 ⎞⎟⎛ 1 kJ/kg ⎞ = (9.81 m/s 2 )(90 m) + = 0.887 kJ/kg ⎜ ⎟⎜⎝ 1000 m 2 /s 2 ⎟⎠ 2 2 ⎝ ⎠

The power generation potential of the river water is obtained by multiplying the total mechanical energy by the mass flow rate,

m& = ρV& = (1000 kg/m 3 )(500 m 3 /s) = 500,000 kg/s

W&max = E& mech = m& emech = (500,000 kg/s)(0.887 kJ/kg) = 444,000 kW = 444 MW Therefore, 444 MW of power can be generated from this river as it discharges into the lake if its power potential can be recovered completely. Discussion Note that the kinetic energy of water is negligible compared to the potential energy, and it can be ignored in the analysis. Also, the power output of an actual turbine will be less than 444 MW because of losses and inefficiencies.

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2-4

2-13 A hydraulic turbine-generator is to generate electricity from the water of a large reservoir. The power generation potential is to be determined. Assumptions 1 The elevation of the reservoir remains constant. 2 The mechanical energy of water at the turbine exit is negligible. Analysis The total mechanical energy water in a reservoir possesses is equivalent to the potential energy of water at the free surface, and it can be converted to work entirely. Therefore, the power potential of water is its potential energy, which is gz per unit mass, and m& gz for a given mass flow rate.

120 m

Turbine

Generator

⎛ 1 kJ/kg ⎞ emech = pe = gz = (9.81 m/s 2 )(120 m)⎜ ⎟ = 1.177 kJ/kg 2 2 ⎝ 1000 m /s ⎠

Then the power generation potential becomes

⎛ 1 kW ⎞ W& max = E& mech = m& e mech = (1500 kg/s)(1.177 kJ/kg)⎜ ⎟ = 1766 kW ⎝ 1 kJ/s ⎠ Therefore, the reservoir has the potential to generate 1766 kW of power. Discussion This problem can also be solved by considering a point at the turbine inlet, and using flow energy instead of potential energy. It would give the same result since the flow energy at the turbine inlet is equal to the potential energy at the free surface of the reservoir.

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2-5

2-14 Wind is blowing steadily at a certain velocity. The mechanical energy of air per unit mass and the power generation potential are to be determined. Assumptions The wind is blowing steadily at a constant uniform velocity. Properties The density of air is given to be ρ = 1.25 kg/m3.

10 m/s

Analysis Kinetic energy is the only form of mechanical energy the wind possesses, and it can be converted to work entirely. Therefore, the power potential of the wind is its kinetic energy, which is V2/2 per unit mass, and m& V 2 / 2 for a given mass flow rate: e mech = ke =

Wind turbine

Wind

60 m

V 2 (10 m/s ) 2 ⎛ 1 kJ/kg ⎞ = ⎜ ⎟ = 0.050 kJ/kg 2 2 ⎝ 1000 m 2 /s 2 ⎠

m& = ρVA = ρV

πD 2 4

= (1.25 kg/m3 )(10 m/s)

π (60 m)2 4

= 35,340 kg/s

W& max = E& mech = m& e mech = (35,340 kg/s)(0.050 kJ/kg) = 1770 kW Therefore, 1770 kW of actual power can be generated by this wind turbine at the stated conditions. Discussion The power generation of a wind turbine is proportional to the cube of the wind velocity, and thus the power generation will change strongly with the wind conditions.

2-15 A water jet strikes the buckets located on the perimeter of a wheel at a specified velocity and flow rate. The power generation potential of this system is to be determined. Assumptions Water jet flows steadily at the specified speed and flow rate. Analysis Kinetic energy is the only form of harvestable mechanical energy the water jet possesses, and it can be converted to work entirely. Therefore, the power potential of the water jet is its kinetic energy, which is V2/2 per unit mass, and m& V 2 / 2 for a given mass flow rate:

emech = ke =

V (60 m/s) ⎛ 1 kJ/kg ⎞ = = 1.8 kJ/kg ⎜ 2 2⎟ 2 2 ⎝ 1000 m /s ⎠ 2

Shaft

2

Nozzle

W&max = E& mech = m& emech ⎛ 1 kW ⎞ = (120 kg/s)(1.8 kJ/kg)⎜ ⎟ = 216 kW ⎝ 1 kJ/s ⎠

Vj

Therefore, 216 kW of power can be generated by this water jet at the stated conditions. Discussion An actual hydroelectric turbine (such as the Pelton wheel) can convert over 90% of this potential to actual electric power.

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2-6

2-16 Two sites with specified wind data are being considered for wind power generation. The site better suited for wind power generation is to be determined. Assumptions 1The wind is blowing steadily at specified velocity during specified times. 2 The wind power generation is negligible during other times. Properties We take the density of air to be ρ = 1.25 kg/m3 (it does not affect the final answer). Analysis Kinetic energy is the only form of mechanical energy the wind possesses, and it can be converted to work entirely. Therefore, the power potential of the wind is its kinetic energy, which is V2/2 per unit mass, and m& V 2 / 2 for a given mass flow rate. Considering a unit flow area (A = 1 m2), the maximum wind power and power generation becomes

Wind

Wind turbine

V, m/s

e mech, 1 = ke1 =

V12 (7 m/s ) 2 ⎛ 1 kJ/kg ⎞ = ⎜ ⎟ = 0.0245 kJ/kg 2 2 ⎝ 1000 m 2 /s 2 ⎠

e mech, 2 = ke 2 =

V 22 (10 m/s ) 2 ⎛ 1 kJ/kg ⎞ = ⎜ ⎟ = 0.050 kJ/kg 2 2 ⎝ 1000 m 2 /s 2 ⎠

W& max, 1 = E& mech, 1 = m& 1e mech, 1 = ρV1 Ake1 = (1.25 kg/m 3 )(7 m/s)(1 m 2 )(0.0245 kJ/kg) = 0.2144 kW W& max, 2 = E& mech, 2 = m& 2 e mech, 2 = ρV 2 Ake 2 = (1.25 kg/m 3 )(10 m/s)(1 m 2 )(0.050 kJ/kg) = 0.625 kW

since 1 kW = 1 kJ/s. Then the maximum electric power generations per year become E max, 1 = W& max, 1 Δt1 = (0.2144 kW)(3000 h/yr) = 643 kWh/yr (per m 2 flow area) E max, 2 = W& max, 2 Δt 2 = (0.625 kW)(2000 h/yr) = 1250 kWh/yr (per m 2 flow area)

Therefore, second site is a better one for wind generation. Discussion Note the power generation of a wind turbine is proportional to the cube of the wind velocity, and thus the average wind velocity is the primary consideration in wind power generation decisions.

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2-7

2-17 A river flowing steadily at a specified flow rate is considered for hydroelectric power generation by collecting the water in a dam. For a specified water height, the power generation potential is to be determined. Assumptions 1 The elevation given is the elevation of the free surface of the river. 2 The mechanical energy of water at the turbine exit is negligible. Properties We take the density of water to be ρ = 1000 kg/m3. Analysis The total mechanical energy the water in a dam possesses is equivalent to the potential energy of water at the free surface of the dam (relative to free surface of discharge water), and it can be converted to work entirely. Therefore, the power potential of water is its potential energy, which is gz per unit mass, and m& gz for a given mass flow rate.

River

50 m

⎛ 1 kJ/kg ⎞ e mech = pe = gz = (9.81 m/s 2 )(50 m)⎜ ⎟ = 0.4905 kJ/kg ⎝ 1000 m 2 /s 2 ⎠ The mass flow rate is

m& = ρV& = (1000 kg/m 3 )(240 m 3 /s) = 240,000 kg/s Then the power generation potential becomes

⎛ 1 MW ⎞ W& max = E& mech = m& e mech = (240,000 kg/s)(0.4905 kJ/kg)⎜ ⎟ = 118 MW ⎝ 1000 kJ/s ⎠ Therefore, 118 MW of power can be generated from this river if its power potential can be recovered completely. Discussion Note that the power output of an actual turbine will be less than 118 MW because of losses and inefficiencies.

2-18 A person with his suitcase goes up to the 10th floor in an elevator. The part of the energy of the elevator stored in the suitcase is to be determined. Assumptions 1 The vibrational effects in the elevator are negligible. Analysis The energy stored in the suitcase is stored in the form of potential energy, which is mgz. Therefore,

⎛ 1 kJ/kg ⎞ ΔE suitcase = ΔPE = mgΔz = (30 kg )(9.81 m/s 2 )(35 m)⎜ ⎟ = 10.3 kJ ⎝ 1000 m 2 /s 2 ⎠ Therefore, the suitcase on 10th floor has 10.3 kJ more energy compared to an identical suitcase on the lobby level. Discussion Noting that 1 kWh = 3600 kJ, the energy transferred to the suitcase is 10.3/3600 = 0.0029 kWh, which is very small.

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2-8

Energy Transfer by Heat and Work

2-19C Energy can cross the boundaries of a closed system in two forms: heat and work.

2-20C The form of energy that crosses the boundary of a closed system because of a temperature difference is heat; all other forms are work.

2-21C An adiabatic process is a process during which there is no heat transfer. A system that does not exchange any heat with its surroundings is an adiabatic system.

2-22C Point functions depend on the state only whereas the path functions depend on the path followed during a process. Properties of substances are point functions, heat and work are path functions.

2-23C The caloric theory is based on the assumption that heat is a fluid-like substance called the "caloric" which is a massless, colorless, odorless substance. It was abandoned in the middle of the nineteenth century after it was shown that there is no such thing as the caloric.

2-24C (a) The car's radiator transfers heat from the hot engine cooling fluid to the cooler air. No work interaction occurs in the radiator.

(b) The hot engine transfers heat to cooling fluid and ambient air while delivering work to the transmission. (c) The warm tires transfer heat to the cooler air and to some degree to the cooler road while no work is produced. No work is produced since there is no motion of the forces acting at the interface between the tire and road. (d) There is minor amount of heat transfer between the tires and road. Presuming that the tires are hotter than the road, the heat transfer is from the tires to the road. There is no work exchange associated with the road since it cannot move. (e) Heat is being added to the atmospheric air by the hotter components of the car. Work is being done on the air as it passes over and through the car.

2-25C When the length of the spring is changed by applying a force to it, the interaction is a work interaction since it involves a force acting through a displacement. A heat interaction is required to change the temperature (and, hence, length) of the spring.

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2-9

2-26C (a) From the perspective of the contents, heat must be removed in order to reduce and maintain the content's temperature. Heat is also being added to the contents from the room air since the room air is hotter than the contents.

(b) Considering the system formed by the refrigerator box when the doors are closed, there are three interactions, electrical work and two heat transfers. There is a transfer of heat from the room air to the refrigerator through its walls. There is also a transfer of heat from the hot portions of the refrigerator (i.e., back of the compressor where condenser is placed) system to the room air. Finally, electrical work is being added to the refrigerator through the refrigeration system. (c) Heat is transferred through the walls of the room from the warm room air to the cold winter air. Electrical work is being done on the room through the electrical wiring leading into the room.

2-27C (a) As one types on the keyboard, electrical signals are produced and transmitted to the processing unit. Simultaneously, the temperature of the electrical parts is increased slightly. The work done on the keys when they are depressed is work done on the system (i.e., keyboard). The flow of electrical current (with its voltage drop) does work on the keyboard. Since the temperature of the electrical parts of the keyboard is somewhat higher than that of the surrounding air, there is a transfer of heat from the keyboard to the surrounding air.

(b) The monitor is powered by the electrical current supplied to it. This current (and voltage drop) is work done on the system (i.e., monitor). The temperatures of the electrical parts of the monitor are higher than that of the surrounding air. Hence there is a heat transfer to the surroundings. (c) The processing unit is like the monitor in that electrical work is done on it while it transfers heat to the surroundings. (d) The entire unit then has electrical work done on it, and mechanical work done on it to depress the keys. It also transfers heat from all its electrical parts to the surroundings.

2-28 The power produced by an electrical motor is to be expressed in different units. Analysis Using appropriate conversion factors, we obtain

(a)

⎛ 1 J/s ⎞⎛ 1 N ⋅ m ⎞ W& = (10 W )⎜ ⎟⎜ ⎟ = 10 N ⋅ m/s ⎝ 1 W ⎠⎝ 1 J ⎠

(b)

⎛ 1 J/s ⎞⎛ 1 N ⋅ m ⎞⎛⎜ 1 kg ⋅ m/s W& = (10 W)⎜ ⎟⎜ ⎟ ⎝ 1 W ⎠⎝ 1 J ⎠⎜⎝ 1 N

2

⎞ ⎟ = 10 kg ⋅ m 2 /s 3 ⎟ ⎠

2-29E The power produced by a model aircraft engine is to be expressed in different units. Analysis Using appropriate conversion factors, we obtain

(a)

⎛ 1 Btu/s ⎞⎛ 778.169 lbf ⋅ ft/s ⎞ W& = (10 W )⎜ ⎟⎜ ⎟ = 7.38 lbf ⋅ ft/s 1 Btu/s ⎝ 1055.056 W ⎠⎝ ⎠

(b)

⎛ 1 hp ⎞ W& = (10 W )⎜ ⎟ = 0.0134 hp ⎝ 745.7 W ⎠

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2-10

Mechanical Forms of Work

2-30C The work done is the same, but the power is different.

2-31 A car is accelerated from rest to 100 km/h. The work needed to achieve this is to be determined. Analysis The work needed to accelerate a body the change in kinetic energy of the body, ⎛ ⎛ 100,000 m ⎞ 2 ⎞⎛ ⎞ 1 1 1 kJ 2 2 ⎟ = 309 kJ ⎟⎟ − 0 ⎟⎜ Wa = m(V2 − V1 ) = (800 kg)⎜ ⎜⎜ 2 2⎟ ⎜ ⎜ ⎟ 2 2 3600 s ⎠ 1000 kg ⋅ m /s ⎠ ⎝⎝ ⎠⎝

2-32E A construction crane lifting a concrete beam is considered. The amount of work is to be determined considering (a) the beam and (b) the crane as the system. Analysis (a) The work is done on the beam and it is determined from 1 lbf ⎛ ⎞ W = mgΔz = (2 × 2000 lbm)(32.174 ft/s 2 )⎜ ⎟(18 ft ) 2 ⎝ 32.174 lbm ⋅ ft/s ⎠ = 72,000 lbf ⋅ ft 1 Btu ⎛ ⎞ = (72,000 lbf ⋅ ft)⎜ ⎟ = 92.5 Btu ⎝ 778.169 lbf ⋅ ft ⎠

(b) Since the crane must produce the same amount of work as is required to lift the beam, the work done by the crane is

18 ft

4000 lbf

W = 72,000 lbf ⋅ ft = 92.5 Btu

2-33 A man is pushing a cart with its contents up a ramp that is inclined at an angle of 20° from the horizontal. The work needed to move along this ramp is to be determined considering (a) the man and (b) the cart and its contents as the system. Analysis (a) Considering the man as the system, letting l be the displacement along the ramp, and letting θ be the inclination angle of the ramp, ⎛ 1 kJ/kg ⎞ W = Fl sin θ = mgl sin θ = (100 + 100 kg )(9.8 m/s 2 )(100 m)sin(20)⎜ ⎟ = 67.0 kJ ⎝ 1000 m 2 /s 2 ⎠

This is work that the man must do to raise the weight of the cart and contents, plus his own weight, a distance of lsinθ. (b) Applying the same logic to the cart and its contents gives ⎛ 1 kJ/kg ⎞ W = Fl sin θ = mgl sin θ = (100 kg )(9.8 m/s 2 )(100 m)sin(20)⎜ ⎟ = 33.5 kJ ⎝ 1000 m 2 /s 2 ⎠

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2-11

2-34E The work required to compress a spring is to be determined. Analysis Since there is no preload, F = kx. Substituting this into the work expression gives 2

2

∫

2

∫

∫

W = Fds = kxdx = k xdx = 1

1

1

[

F

k 2 ( x 2 − x12 ) 2

]

x

200 lbf/in ⎛ 1 ft ⎞ = (1 in ) 2 − 0 2 ⎜ ⎟ = 8.33 lbf ⋅ ft 2 ⎝ 12 in ⎠ 1 Btu ⎛ ⎞ = (8.33 lbf ⋅ ft)⎜ ⎟ = 0.0107 Btu ⋅ 778 . 169 lbf ft ⎝ ⎠

2-35 As a spherical ammonia vapor bubble rises in liquid ammonia, its diameter increases. The amount of work produced by this bubble is to be determined. Assumptions 1 The bubble is treated as a spherical bubble. 2 The surface tension coefficient is taken constant. Analysis Executing the work integral for a constant surface tension coefficient gives 2

∫

W = σ dA = σ ( A2 − A1 ) = σ 4π (r22 − r12 ) 1

[

= 4π (0.02 N/m) (0.015 m) 2 − (0.005 m) 2 = 5.03 × 10

−5

]

N⋅m

⎛ 1 kJ ⎞ −8 = (5.03 × 10 −5 N ⋅ m)⎜ ⎟ = 5.03 × 10 kJ ⎝ 1000 N ⋅ m ⎠

2-36 The work required to stretch a steel rod in a specified length is to be determined. Assumptions The Young’s modulus does not change as the rod is stretched. Analysis The original volume of the rod is

V0 =

πD 2 4

L=

π (0.005 m) 2 4

(10 m) = 1.963 × 10 − 4 m 3

The work required to stretch the rod 3 cm is

V0 E

(ε 22 − ε 12 ) 2 (1.963 × 10 − 4 m 3 )(21× 10 4 kN/m 2 ) = (0.03 m) 2 − 0 2 2 = 0.01855 kN ⋅ m = 0.01855 kJ

W=

[

]

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2-12

2-37E The work required to compress a spring is to be determined. Analysis The force at any point during the deflection of the spring is given by F = F0 + kx, where F0 is the initial force and x is the deflection as measured from the point where the initial force occurred. From the perspective of the spring, this force acts in the direction opposite to that in which the spring is deflected. Then, 2

∫

2

∫

F

W = Fds = ( F0 + kx)dx 1

1

x

k 2 ( x 2 − x12 ) 2 200 lbf/in 2 (1 − 0 2 )in 2 = (100 lbf)[(1 − 0)in ] + 2 = 200 lbf ⋅ in = F0 ( x 2 − x1 ) +

1 Btu ⎛ ⎞⎛ 1 ft ⎞ = (200 lbf ⋅ in)⎜ ⎟⎜ ⎟ = 0.0214 Btu ⎝ 778.169 lbf ⋅ ft ⎠⎝ 12 in ⎠

2-38 The work required to compress a spring is to be determined. Analysis Since there is no preload, F = kx. Substituting this into the work expression gives 2

2

2

∫

∫

∫

1

1

1

W = Fds = kxdx = k xdx =

[

300 kN/m (0.03 m) 2 − 0 2 2 = 0.135 kN ⋅ m =

k 2 ( x 2 − x12 ) 2

]

F x

⎛ 1 kJ ⎞ = (0.135 kN ⋅ m)⎜ ⎟ = 0.135 kJ ⎝ 1 kN ⋅ m ⎠

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2-13

2-39 A ski lift is operating steadily at 10 km/h. The power required to operate and also to accelerate this ski lift from rest to the operating speed are to be determined. Assumptions 1 Air drag and friction are negligible. 2 The average mass of each loaded chair is 250 kg. 3 The mass of chairs is small relative to the mass of people, and thus the contribution of returning empty chairs to the motion is disregarded (this provides a safety factor). Analysis The lift is 1000 m long and the chairs are spaced 20 m apart. Thus at any given time there are 1000/20 = 50 chairs being lifted. Considering that the mass of each chair is 250 kg, the load of the lift at any given time is

Load = (50 chairs)(250 kg/chair) = 12,500 kg Neglecting the work done on the system by the returning empty chairs, the work needed to raise this mass by 200 m is

⎛ 1 kJ W g = mg (z 2 − z1 ) = (12,500 kg)(9.81 m/s 2 )(200 m)⎜ ⎜ 1000 kg ⋅ m 2 /s 2 ⎝

⎞ ⎟ = 24,525 kJ ⎟ ⎠

At 10 km/h, it will take Δt =

distance 1 km = = 0.1 h = 360 s velocity 10 km / h

to do this work. Thus the power needed is W& g =

Wg Δt

=

24,525 kJ 360 s

= 68.1 kW

The velocity of the lift during steady operation, and the acceleration during start up are ⎛ 1 m/s ⎞ V = (10 km/h)⎜ ⎟ = 2.778 m/s ⎝ 3.6 km/h ⎠ a=

ΔV 2.778 m/s - 0 = = 0.556 m/s 2 Δt 5s

During acceleration, the power needed is

⎛ 1 kJ/kg 1 1 W& a = m(V22 − V12 ) / Δt = (12,500 kg) (2.778 m/s) 2 − 0 ⎜ ⎜ 1000 m 2 /s 2 2 2 ⎝

(

)

⎞ ⎟/(5 s) = 9.6 kW ⎟ ⎠

Assuming the power applied is constant, the acceleration will also be constant and the vertical distance traveled during acceleration will be h=

1 2 1 200 m 1 at sin α = at 2 = (0.556 m/s 2 )(5 s) 2 (0.2) = 1.39 m 2 2 1000 m 2

and

⎛ 1 kJ/kg W& g = mg (z 2 − z1 ) / Δt = (12,500 kg)(9.81 m/s 2 )(1.39 m)⎜ ⎜ 1000 kg ⋅ m 2 /s 2 ⎝

⎞ ⎟ /(5 s) = 34.1 kW ⎟ ⎠

Thus, W& total = W& a + W& g = 9.6 + 34.1 = 43.7 kW

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2-14

2-40 A car is to climb a hill in 10 s. The power needed is to be determined for three different cases. Assumptions Air drag, friction, and rolling resistance are negligible. Analysis The total power required for each case is the sum of the rates of changes in potential and kinetic energies. That is, W& total = W& a + W& g

(a) W& a = 0 since the velocity is constant. Also, the vertical rise is h = (100 m)(sin 30°) = 50 m. Thus,

⎛ 1 kJ W& g = mg ( z 2 − z1 ) / Δt = (2000 kg)(9.81 m/s 2 )(50 m)⎜ ⎜ 1000 kg ⋅ m 2 /s 2 ⎝ and

⎞ ⎟/(10 s) = 98.1 kW ⎟ ⎠

W& total = W& a + W& g = 0 + 98.1 = 98.1 kW

(b) The power needed to accelerate is

[

1 1 W&a = m(V22 − V12 ) / Δt = (2000 kg) (30 m/s )2 − 0 2 2

and

]⎛⎜⎜ 1000 kg1 kJ⋅ m /s 2

⎝

2

⎞ ⎟ /(10 s) = 90 kW ⎟ ⎠

W& total = W& a + W& g = 90 + 98.1 = 188.1 kW

(c) The power needed to decelerate is

[

1 1 W&a = m(V22 − V12 ) / Δt = (2000 kg) (5 m/s )2 − (35 m/s )2 2 2

and

]⎛⎜⎜ 1000 kg1 kJ⋅ m /s ⎝

2

2

⎞ ⎟ /(10 s) = −120 kW ⎟ ⎠

W& total = W& a + W& g = −120 + 98.1 = −21.9 kW (breaking power)

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2-15

2-41 A damaged car is being towed by a truck. The extra power needed is to be determined for three different cases. Assumptions Air drag, friction, and rolling resistance are negligible. Analysis The total power required for each case is the sum of the rates of changes in potential and kinetic energies. That is, W& total = W& a + W& g

(a) Zero. (b) W& a = 0 . Thus, Δz W& total = W& g = mg ( z 2 − z1 ) / Δt = mg = mgV z = mgV sin 30 o Δt ⎛ 50,000 m ⎞⎛ 1 kJ/kg ⎞ ⎟(0.5) = 81.7 kW ⎟⎜ = (1200 kg)(9.81m/s 2 )⎜⎜ 2 2 ⎟ ⎟⎜ ⎝ 3600 s ⎠⎝ 1000 m /s ⎠

(c) W& g = 0 . Thus, ⎛ ⎛ 90,000 m ⎞ 2 ⎞⎛ 1 kJ/kg ⎞ 1 1 ⎟ /(12 s) = 31.3 kW ⎟⎟ − 0 ⎟⎜ W& total = W&a = m(V22 − V12 ) / Δt = (1200 kg)⎜ ⎜⎜ ⎜ 3600 s ⎟⎜ 1000 m 2 /s 2 ⎟ 2 2 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

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2-16

The First Law of Thermodynamics

2-42C No. This is the case for adiabatic systems only.

2-43C Warmer. Because energy is added to the room air in the form of electrical work.

2-44C Energy can be transferred to or from a control volume as heat, various forms of work, and by mass transport.

2-45 The high rolling resistance tires of a car are replaced by low rolling resistance ones. For a specified unit fuel cost, the money saved by switching to low resistance tires is to be determined. Assumptions 1The low rolling resistance tires deliver 2 mpg over all velocities. 2 The car is driven 15,000 miles per year. Analysis The annual amount of fuel consumed by this car on high- and low-rolling resistance tires are

Annual Fuel Consumption High =

Miles driven per year 15,000 miles/year = = 600 gal/year Miles per gallon 25 miles/gal

Annual Fuel Consumption Low =

Miles driven per year 15,000 miles/year = = 555.5 gal/year Miles per gallon 27 miles/gal

Then the fuel and money saved per year become

Fuel Savings = Annual Fuel Consumption High − Annual Fuel Consumption Low = 600 gal/year - 555.5 gal/year = 44.5 gal/year Cost savings = (Fuel savings)( Unit cost of fuel) = (44.5 gal/year)($2.20/gal) = $97.9/year

Discussion A typical tire lasts about 3 years, and thus the low rolling resistance tires have the potential to save about $300 to the car owner over the life of the tires, which is comparable to the installation cost of the tires.

2-46 The specific energy change of a system which is accelerated is to be determined. Analysis Since the only property that changes for this system is the velocity, only the kinetic energy will change. The change in the specific energy is

Δke =

V 22 − V12 (30 m/s) 2 − (0 m/s) 2 = 2 2

⎛ 1 kJ/kg ⎞ ⎜ ⎟ = 0.45 kJ/kg ⎝ 1000 m 2 /s 2 ⎠

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2-17

2-47 The specific energy change of a system which is raised is to be determined. Analysis Since the only property that changes for this system is the elevation, only the potential energy will change. The change in the specific energy is then ⎛ 1 kJ/kg ⎞ Δpe = g ( z 2 − z1 ) = (9.8 m/s 2 )(100 − 0) m⎜ ⎟ = 0.98 kJ/kg ⎝ 1000 m 2 /s 2 ⎠

2-48E A water pump increases water pressure. The power input is to be determined. Analysis The power input is determined from

50 psia

W& = V& ( P2 − P1 ) ⎛ 1 Btu = (1.2 ft 3 /s)(50 − 10)psia ⎜ ⎜ 5.404 psia ⋅ ft 3 ⎝ = 12.6 hp

⎞⎛ 1 hp ⎞ ⎟⎜ ⎟⎝ 0.7068 Btu/s ⎟⎠ ⎠

Water 10 psia

The water temperature at the inlet does not have any significant effect on the required power.

2-49 An automobile moving at a given velocity is considered. The power required to move the car and the area of the effective flow channel behind the car are to be determined. Analysis The absolute pressure of the air is ⎛ 0.1333 kPa ⎞ ⎟⎟ = 99.98 kPa P = (750 mm Hg)⎜⎜ ⎝ 1 mm Hg ⎠

and the specific volume of the air is

v=

RT (0.287 kPa ⋅ m 3 /kg ⋅ K)(303 K) = = 0.8698 m 3 /kg P 99.98 kPa

The mass flow rate through the control volume is m& =

A1V1

v

=

(3 m 2 )(90/3.6 m/s) 0.8698 m 3 /kg

= 86.23 kg/s

The power requirement is

V 2 − V 22 (90 / 3.6 m/s) 2 − (82 / 3.6 m/s) 2 ⎛ 1 kJ/kg ⎞ W& = m& 1 = (86.23 kg/s) ⎜ ⎟ = 4.578 kW 2 2 ⎝ 1000 m 2 /s 2 ⎠ The outlet area is

m& =

A2V 2

v

⎯ ⎯→ A2 =

m& v (86.23 kg/s)(0.8698 m 3 /kg) = = 3.29 m 2 (82/3.6) m/s V2

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2-18

2-50 A classroom is to be air-conditioned using window air-conditioning units. The cooling load is due to people, lights, and heat transfer through the walls and the windows. The number of 5-kW window air conditioning units required is to be determined. Assumptions There are no heat dissipating equipment (such as computers, TVs, or ranges) in the room. Analysis The total cooling load of the room is determined from Q& cooling = Q& lights + Q& people + Q& heat gain

where Q& lights = 10 × 100 W = 1 kW Q& people = 40 × 360 kJ / h = 4 kW Q& heat gain = 15,000 kJ / h = 4.17 kW

Room 15,000 kJ/h

40 people 10 bulbs

·

Qcool

Substituting, Q& cooling = 1 + 4 + 4.17 = 9.17 kW

Thus the number of air-conditioning units required is 9.17 kW 5 kW/unit

= 1.83 ⎯ ⎯→ 2 units

2-51 The lighting energy consumption of a storage room is to be reduced by installing motion sensors. The amount of energy and money that will be saved as well as the simple payback period are to be determined. Assumptions The electrical energy consumed by the ballasts is negligible. Analysis The plant operates 12 hours a day, and thus currently the lights are on for the entire 12 hour period. The motion sensors installed will keep the lights on for 3 hours, and off for the remaining 9 hours every day. This corresponds to a total of 9×365 = 3285 off hours per year. Disregarding the ballast factor, the annual energy and cost savings become

Energy Savings = (Number of lamps)(Lamp wattage)(Reduction of annual operating hours) = (24 lamps)(60 W/lamp )(3285 hours/year) = 4730 kWh/year Cost Savings = (Energy Savings)(Unit cost of energy) = (4730 kWh/year)($0.08/kWh) = $378/year The implementation cost of this measure is the sum of the purchase price of the sensor plus the labor, Implementation Cost = Material + Labor = $32 + $40 = $72 This gives a simple payback period of Simple payback period =

Implementation cost $72 = = 0.19 year (2.3 months) Annual cost savings $378 / year

Therefore, the motion sensor will pay for itself in about 2 months.

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2-19

2-52 The classrooms and faculty offices of a university campus are not occupied an average of 4 hours a day, but the lights are kept on. The amounts of electricity and money the campus will save per year if the lights are turned off during unoccupied periods are to be determined. Analysis The total electric power consumed by the lights in the classrooms and faculty offices is E& lighting, classroom = (Power consumed per lamp) × (No. of lamps) = (200 × 12 × 110 W) = 264,000 = 264 kW E& lighting, offices = (Power consumed per lamp) × (No. of lamps) = (400 × 6 × 110 W) = 264,000 = 264 kW E& lighting, total = E& lighting, classroom + E& lighting, offices = 264 + 264 = 528 kW

Noting that the campus is open 240 days a year, the total number of unoccupied work hours per year is Unoccupied hours = (4 hours/day)(240 days/year) = 960 h/yr Then the amount of electrical energy consumed per year during unoccupied work period and its cost are Energy savings = ( E& lighting, total )( Unoccupied hours) = (528 kW)(960 h/yr) = 506,880 kWh Cost savings = (Energy savings)(Unit cost of energy) = (506,880 kWh/yr)($0.082/kWh) = $41,564/yr

Discussion Note that simple conservation measures can result in significant energy and cost savings.

2-53 A room contains a light bulb, a TV set, a refrigerator, and an iron. The rate of increase of the energy content of the room when all of these electric devices are on is to be determined. Assumptions 1 The room is well sealed, and heat loss from the room is negligible. 2 All the appliances are kept on. Analysis Taking the room as the system, the rate form of the energy balance can be written as

E& − E& 1in424out 3

Rate of net energy transfer by heat, work, and mass

=

dE system / dt 14243

→

dE room / dt = E& in

Rate of change in internal, kinetic, potential, etc. energies

since no energy is leaving the room in any form, and thus E& out = 0 . Also, E& in = E& lights + E& TV + E& refrig + E& iron = 100 + 110 + 200 + 1000 W = 1410 W

Electricity

ROOM

- Lights - TV - Refrig - Iron

Substituting, the rate of increase in the energy content of the room becomes

dE room / dt = E& in = 1410 W Discussion Note that some appliances such as refrigerators and irons operate intermittently, switching on and off as controlled by a thermostat. Therefore, the rate of energy transfer to the room, in general, will be less.

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2-20

2-54 A fan is to accelerate quiescent air to a specified velocity at a specified flow rate. The minimum power that must be supplied to the fan is to be determined. Assumptions The fan operates steadily. Properties The density of air is given to be ρ = 1.18 kg/m3. Analysis A fan transmits the mechanical energy of the shaft (shaft power) to mechanical energy of air (kinetic energy). For a control volume that encloses the fan, the energy balance can be written as

E& − E& 1in424out 3

Rate of net energy transfer by heat, work, and mass

= dE system / dt ©0 (steady) = 0 144424443

→

E& in = E& out

Rate of change in internal, kinetic, potential, etc. energies

V2 W& sh, in = m& air ke out = m& air out 2 where

m& air = ρV& = (1.18 kg/m 3 )(4 m 3 /s) = 4.72 kg/s Substituting, the minimum power input required is determined to be V2 (10 m/s) 2 W& sh, in = m& air out = (4.72 kg/s) 2 2

⎛ 1 J/kg ⎞ ⎜ ⎟ = 236 J/s = 236 W ⎝ 1 m 2 /s 2 ⎠

Discussion The conservation of energy principle requires the energy to be conserved as it is converted from one form to another, and it does not allow any energy to be created or destroyed during a process. In reality, the power required will be considerably higher because of the losses associated with the conversion of mechanical shaft energy to kinetic energy of air.

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2-21

2-55E A fan accelerates air to a specified velocity in a square duct. The minimum electric power that must be supplied to the fan motor is to be determined. Assumptions 1 The fan operates steadily. 2 There are no conversion losses. Properties The density of air is given to be ρ = 0.075 lbm/ft3. Analysis A fan motor converts electrical energy to mechanical shaft energy, and the fan transmits the mechanical energy of the shaft (shaft power) to mechanical energy of air (kinetic energy). For a control volume that encloses the fan-motor unit, the energy balance can be written as

E& − E& 1in424out 3

Rate of net energy transfer by heat, work, and mass

= dE system / dt ©0 (steady) = 0 144424443

→

E& in = E& out

Rate of change in internal, kinetic, potential, etc. energies

V2 W& elect, in = m& air ke out = m& air out 2 where

m& air = ρVA = (0.075 lbm/ft3 )(3 × 3 ft 2 )(22 ft/s) = 14.85 lbm/s Substituting, the minimum power input required is determined to be V2 (22 ft/s) 2 W& in = m& air out = (14.85 lbm/s) 2 2

⎛ 1 Btu/lbm ⎜⎜ 2 2 ⎝ 25,037 ft /s

⎞ ⎟⎟ = 0.1435 Btu/s = 151 W ⎠

since 1 Btu = 1.055 kJ and 1 kJ/s = 1000 W. Discussion The conservation of energy principle requires the energy to be conserved as it is converted from one form to another, and it does not allow any energy to be created or destroyed during a process. In reality, the power required will be considerably higher because of the losses associated with the conversion of electrical-to-mechanical shaft and mechanical shaft-to-kinetic energy of air.

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2-22

2-56 A gasoline pump raises the pressure to a specified value while consuming electric power at a specified rate. The maximum volume flow rate of gasoline is to be determined. Assumptions 1 The gasoline pump operates steadily. 2 The changes in kinetic and potential energies across the pump are negligible. Analysis For a control volume that encloses the pump-motor unit, the energy balance can be written as

E& − E& out 1in 424 3

Rate of net energy transfer by heat, work, and mass

= dEsystem / dt ©0 (steady) 144424443

=0

→

E&in = E& out 5.2 kW

Rate of change in internal, kinetic, potential, etc. energies

W&in + m& ( Pv )1 = m& ( Pv ) 2 → W&in = m& ( P2 − P1 )v = V& ΔP since m& = V&/v and the changes in kinetic and potential energies of gasoline are negligible, Solving for volume flow rate and substituting, the maximum flow rate is determined to be

V&max =

W& in 5.2 kJ/s ⎛ 1 kPa ⋅ m ⎜ = ΔP 5 kPa ⎜⎝ 1 kJ

3

⎞ ⎟ = 1.04 m 3 /s ⎟ ⎠

PUMP

Motor

Pump inlet

Discussion The conservation of energy principle requires the energy to be conserved as it is converted from one form to another, and it does not allow any energy to be created or destroyed during a process. In reality, the volume flow rate will be less because of the losses associated with the conversion of electricalto-mechanical shaft and mechanical shaft-to-flow energy.

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2-23

2-57 An inclined escalator is to move a certain number of people upstairs at a constant velocity. The minimum power required to drive this escalator is to be determined. Assumptions 1 Air drag and friction are negligible. 2 The average mass of each person is 75 kg. 3 The escalator operates steadily, with no acceleration or breaking. 4 The mass of escalator itself is negligible. Analysis At design conditions, the total mass moved by the escalator at any given time is

Mass = (30 persons)(75 kg/person) = 2250 kg The vertical component of escalator velocity is Vvert = V sin 45° = (0.8 m/s)sin45°

Under stated assumptions, the power supplied is used to increase the potential energy of people. Taking the people on elevator as the closed system, the energy balance in the rate form can be written as E& − E& out 1in 424 3

Rate of net energy transfer by heat, work, and mass

=

dEsystem / dt 14243

=0

→

ΔEsys E&in = dEsys / dt ≅ Δt

Rate of change in internal, kinetic, potential, etc. energies

ΔPE mgΔz W&in = = = mgVvert Δt Δt

That is, under stated assumptions, the power input to the escalator must be equal to the rate of increase of the potential energy of people. Substituting, the required power input becomes ⎛ 1 kJ/kg ⎞ ⎟ = 12.5 kJ/s = 12.5 kW W&in = mgVvert = (2250 kg)(9.81 m/s 2 )(0.8 m/s)sin45°⎜⎜ 2 2⎟ ⎝ 1000 m /s ⎠

When the escalator velocity is doubled to V = 1.6 m/s, the power needed to drive the escalator becomes ⎛ 1 kJ/kg ⎞ ⎟ = 25.0 kJ/s = 25.0 kW W&in = mgVvert = (2250 kg)(9.81 m/s 2 )(1.6 m/s)sin45°⎜⎜ 2 2⎟ ⎝ 1000 m /s ⎠

Discussion Note that the power needed to drive an escalator is proportional to the escalator velocity.

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2-24

Energy Conversion Efficiencies

2-58C Mechanical efficiency is defined as the ratio of the mechanical energy output to the mechanical energy input. A mechanical efficiency of 100% for a hydraulic turbine means that the entire mechanical energy of the fluid is converted to mechanical (shaft) work.

2-59C The combined pump-motor efficiency of a pump/motor system is defined as the ratio of the increase in the mechanical energy of the fluid to the electrical power consumption of the motor,

η pump-motor = η pumpη motor =

W& pump E& mech,out − E& mech,in ΔE& mech,fluid = = W& elect,in W& elect,in W& elect,in

The combined pump-motor efficiency cannot be greater than either of the pump or motor efficiency since both pump and motor efficiencies are less than 1, and the product of two numbers that are less than one is less than either of the numbers.

2-60C The turbine efficiency, generator efficiency, and combined turbine-generator efficiency are defined as follows:

η turbine =

W& shaft,out Mechanical energy output = Mechanical energy extracted from the fluid | ΔE& mech,fluid |

η generator =

Electrical power output W& elect,out = Mechanical power input W& shaft,in

η turbine -gen = η turbineηgenerator = & E

W&elect,out − E&

mech,in

mech,out

=

W&elect,out | ΔE& mech,fluid |

2-61C No, the combined pump-motor efficiency cannot be greater that either of the pump efficiency of the motor efficiency. This is because η pump - motor = η pumpη motor , and both η pump and η motor are less than one,

and a number gets smaller when multiplied by a number smaller than one.

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2-25

2-62 A hooded electric open burner and a gas burner are considered. The amount of the electrical energy used directly for cooking and the cost of energy per “utilized” kWh are to be determined. Analysis The efficiency of the electric heater is given to be 73 percent. Therefore, a burner that consumes 3-kW of electrical energy will supply

η gas = 38% η electric = 73% Q& utilized = (Energy input) × (Efficiency) = (3 kW)(0.73) = 2.19 kW of useful energy. The unit cost of utilized energy is inversely proportional to the efficiency, and is determined from Cost of utilized energy =

Cost of energy input $0.07 / kWh = = $0.096/kWh Efficiency 0.73

Noting that the efficiency of a gas burner is 38 percent, the energy input to a gas burner that supplies utilized energy at the same rate (2.19 kW) is Q& input, gas =

2.19 kW Q& utilized = = 5.76 kW (= 19,660 Btu/h) Efficiency 0.38

since 1 kW = 3412 Btu/h. Therefore, a gas burner should have a rating of at least 19,660 Btu/h to perform as well as the electric unit. Noting that 1 therm = 29.3 kWh, the unit cost of utilized energy in the case of gas burner is determined the same way to be Cost of utilized energy =

Cost of energy input $1.20 /( 29.3 kWh) = = $0.108/kWh Efficiency 0.38

2-63 A worn out standard motor is replaced by a high efficiency one. The reduction in the internal heat gain due to the higher efficiency under full load conditions is to be determined. Assumptions 1 The motor and the equipment driven by the motor are in the same room. 2 The motor operates at full load so that fload = 1. Analysis The heat generated by a motor is due to its inefficiency, and the difference between the heat generated by two motors that deliver the same shaft power is simply the difference between the electric power drawn by the motors,

W& in, electric, standard = W& shaft / η motor = (75 × 746 W)/0.91 = 61,484 W W& in, electric, efficient = W& shaft / η motor = (75 × 746 W)/0.954 = 58,648 W Then the reduction in heat generation becomes Q& reduction = W& in, electric, standard − W& in, electric, efficient = 61,484 − 58,648 = 2836 W

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2-26

2-64 An electric car is powered by an electric motor mounted in the engine compartment. The rate of heat supply by the motor to the engine compartment at full load conditions is to be determined. Assumptions The motor operates at full load so that the load factor is 1. Analysis The heat generated by a motor is due to its inefficiency, and is equal to the difference between the electrical energy it consumes and the shaft power it delivers, W& in, electric = W& shaft / η motor = (90 hp)/0.91 = 98.90 hp Q& generation = W& in, electric − W& shaft out = 98.90 − 90 = 8.90 hp = 6.64 kW

since 1 hp = 0.746 kW. Discussion Note that the electrical energy not converted to mechanical power is converted to heat.

2-65 A worn out standard motor is to be replaced by a high efficiency one. The amount of electrical energy and money savings as a result of installing the high efficiency motor instead of the standard one as well as the simple payback period are to be determined. Assumptions The load factor of the motor remains constant at 0.75. Analysis The electric power drawn by each motor and their difference can be expressed as W& electric in, standard = W& shaft / η standard = (Power rating)(Load factor) / η standard W& electric in, efficient = W& shaft / η efficient = (Power rating)(Load factor) / η efficient Power savings = W& electric in, standard − W& electric in, efficient = (Power rating)(Load factor)[1 / η standard − 1 / η efficient ]

where ηstandard is the efficiency of the standard motor, and ηefficient is the efficiency of the comparable high efficiency motor. Then the annual energy and cost savings associated with the installation of the high efficiency motor are determined to be Energy Savings = (Power savings)(Operating Hours) = (Power Rating)(Operating Hours)(Load Factor)(1/ηstandard- 1/ηefficient) = (75 hp)(0.746 kW/hp)(4,368 hours/year)(0.75)(1/0.91 - 1/0.954) = 9,290 kWh/year

η old = 91.0% η new = 95.4%

Cost Savings = (Energy savings)(Unit cost of energy) = (9,290 kWh/year)($0.08/kWh) = $743/year The implementation cost of this measure consists of the excess cost the high efficiency motor over the standard one. That is, Implementation Cost = Cost differential = $5,520 - $5,449 = $71 This gives a simple payback period of Simple payback period =

Implementation cost $71 = = 0.096 year (or 1.1 months) Annual cost savings $743 / year

Therefore, the high-efficiency motor will pay for its cost differential in about one month.

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2-27

2-66E The combustion efficiency of a furnace is raised from 0.7 to 0.8 by tuning it up. The annual energy and cost savings as a result of tuning up the boiler are to be determined. Assumptions The boiler operates at full load while operating. Analysis The heat output of boiler is related to the fuel energy input to the boiler by

Boiler output = (Boiler input)(Combustion efficiency)

or

Q& out = Q& inη furnace

The current rate of heat input to the boiler is given to be Q& in, current = 3.6 × 10 6 Btu/h . Then the rate of useful heat output of the boiler becomes

Q& out = (Q& inη furnace ) current = (3.6 ×10 6 Btu/h)(0.7) = 2.52 ×10 6 Btu/h The boiler must supply useful heat at the same rate after the tune up. Therefore, the rate of heat input to the boiler after the tune up and the rate of energy savings become

Boiler 70% 3.6×106

Q& in, new = Q& out / η furnace, new = ( 2.52 × 10 6 Btu/h)/0.8 = 3.15 × 10 6 Btu/h Q& in, saved = Q& in, current − Q& in, new = 3.6 × 10 6 − 3.15 × 10 6 = 0.45 × 10 6 Btu/h

Then the annual energy and cost savings associated with tuning up the boiler become Energy Savings = Q& in, saved (Operation hours) = (0.45×106 Btu/h)(1500 h/year) = 675×106 Btu/yr Cost Savings = (Energy Savings)(Unit cost of energy) = (675×106 Btu/yr)($4.35 per 106 Btu) = $2936/year Discussion Notice that tuning up the boiler will save $2936 a year, which is a significant amount. The implementation cost of this measure is negligible if the adjustment can be made by in-house personnel. Otherwise it is worthwhile to have an authorized representative of the boiler manufacturer to service the boiler twice a year.

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2-28

2-67E EES Problem 2-66E is reconsidered. The effects of the unit cost of energy and combustion efficiency on the annual energy used and the cost savings as the efficiency varies from 0.6 to 0.9 and the unit cost varies from $4 to $6 per million Btu are the investigated. The annual energy saved and the cost savings are to be plotted against the efficiency for unit costs of $4, $5, and $6 per million Btu. Analysis The problem is solved using EES, and the solution is given below. "Knowns:" eta_boiler_current = 0.7 eta_boiler_new = 0.8 Q_dot_in_current = 3.6E+6 "[Btu/h]" DELTAt = 1500 "[h/year]" UnitCost_energy = 5E-6 "[dollars/Btu]" "Analysis: The heat output of boiler is related to the fuel energy input to the boiler by Boiler output = (Boiler input)(Combustion efficiency) Then the rate of useful heat output of the boiler becomes" Q_dot_out=Q_dot_in_current*eta_boiler_current "[Btu/h]" "The boiler must supply useful heat at the same rate after the tune up. Therefore, the rate of heat input to the boiler after the tune up and the rate of energy savings become " Q_dot_in_new=Q_dot_out/eta_boiler_new "[Btu/h]" Q_dot_in_saved=Q_dot_in_current - Q_dot_in_new "[Btu/h]" "Then the annual energy and cost savings associated with tuning up the boiler become" EnergySavings =Q_dot_in_saved*DELTAt "[Btu/year]" CostSavings = EnergySavings*UnitCost_energy "[dollars/year]" "Discussion Notice that tuning up the boiler will save $2936 a year, which is a significant amount. The implementation cost of this measure is negligible if the adjustment can be made by in-house personnel. Otherwise it is worthwhile to have an authorized representative of the boiler manufacturer to service the boiler twice a year. " CostSavings [dollars/year] -4500 0 3375 6000

ηboiler,new

EnergySavings [Btu/year] -9.000E+08 0 6.750E+08 1.200E+09

0.6 0.7 0.8 0.9

6000 4000 2000

1,250x109

Unit Cost of Energy 4E-6 $/Btu 5E-6 $/Btu

EnergySavings [Btu/year]

CostSavings [dollars/year]

8000

6E-6 $/Btu

0 -2000 -4000 -6000 0.6

0.65

0.7

0.75

η

boiler,new

0.8

0.85

0.9

8,000x108

Unit Cost = $5/106 Btu

3,500x108 -1,000x108 -5,500x108 -1,000x109 0,6

0,65

0,7

0,75

0,8

0,85

ηboiler;new

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0,9

2-29

2-68 Several people are working out in an exercise room. The rate of heat gain from people and the equipment is to be determined. Assumptions The average rate of heat dissipated by people in an exercise room is 525 W. Analysis The 8 weight lifting machines do not have any motors, and thus they do not contribute to the internal heat gain directly. The usage factors of the motors of the treadmills are taken to be unity since they are used constantly during peak periods. Noting that 1 hp = 746 W, the total heat generated by the motors is Q& motors = ( No. of motors) × W& motor × f load × f usage / η motor = 4 × (2.5 × 746 W) × 0.70 × 1.0/0.77 = 6782 W

The heat gain from 14 people is Q& people = ( No. of people) × Q& person = 14 × (525 W) = 7350 W

Then the total rate of heat gain of the exercise room during peak period becomes Q& total = Q& motors + Q& people = 6782 + 7350 = 14,132 W

2-69 A classroom has a specified number of students, instructors, and fluorescent light bulbs. The rate of internal heat generation in this classroom is to be determined. Assumptions 1 There is a mix of men, women, and children in the classroom. 2 The amount of light (and thus energy) leaving the room through the windows is negligible. Properties The average rate of heat generation from people seated in a room/office is given to be 100 W. Analysis The amount of heat dissipated by the lamps is equal to the amount of electrical energy consumed by the lamps, including the 10% additional electricity consumed by the ballasts. Therefore, Q& lighting = (Energy consumed per lamp) × (No. of lamps) = (40 W)(1.1)(18) = 792 W & Qpeople = ( No. of people) × Q& person = 56 × (100 W) = 5600 W

Then the total rate of heat gain (or the internal heat load) of the classroom from the lights and people become Q& total = Q& lighting + Q& people = 792 + 5600 = 6392 W

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2-30

2-70 A room is cooled by circulating chilled water through a heat exchanger, and the air is circulated through the heat exchanger by a fan. The contribution of the fan-motor assembly to the cooling load of the room is to be determined. Assumptions The fan motor operates at full load so that fload = 1. Analysis The entire electrical energy consumed by the motor, including the shaft power delivered to the fan, is eventually dissipated as heat. Therefore, the contribution of the fan-motor assembly to the cooling load of the room is equal to the electrical energy it consumes, Q& internal generation = W& in, electric = W& shaft / η motor = (0.25 hp)/0.54 = 0.463 hp = 345 W

since 1 hp = 746 W.

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2-31

2-71 A hydraulic turbine-generator is generating electricity from the water of a large reservoir. The combined turbine-generator efficiency and the turbine efficiency are to be determined. Assumptions 1 The elevation of the reservoir remains constant. 2 The mechanical energy of water at the turbine exit is negligible. Analysis We take the free surface of the reservoir to be point 1 and the turbine exit to be point 2. We also take the turbine exit as the reference level (z2 = 0), and thus the potential energy at points 1 and 2 are pe1 = gz1 and pe2 = 0. The flow energy P/ρ at both points is zero since both 1 and 2 are open to the atmosphere (P1 = P2 = Patm). Further, the kinetic energy at both points is zero (ke1 = ke2 = 0) since the water at point 1 is essentially motionless, and the kinetic energy of water at turbine exit is assumed to be negligible. The potential energy of water at point 1 is ⎛ 1 kJ/kg ⎞ pe1 = gz1 = (9.81 m/s 2 )(70 m)⎜ ⎟ = 0.687 kJ/kg ⎝ 1000 m 2 /s 2 ⎠

Then the rate at which the mechanical energy of the fluid is supplied to the turbine become

1

ΔE& mech,fluid = m& (emech,in − emech,out ) = m& ( pe1 − 0) = m& pe1 = (1500 kg/s)(0.687 kJ/kg) = 1031 kW

The combined turbine-generator and the turbine efficiency are determined from their definitions,

η turbine-gen = η turbine =

750 kW

70 m

W& elect,out 750 kW = = 0.727 or 72.7% & 1031 kW | ΔE mech,fluid |

Turbine

Generator 2

W& shaft,out 800 kW = = 0.776 or 77.6% & | ΔE mech,fluid | 1031 kW

Therefore, the reservoir supplies 1031 kW of mechanical energy to the turbine, which converts 800 kW of it to shaft work that drives the generator, which generates 750 kW of electric power. Discussion This problem can also be solved by taking point 1 to be at the turbine inlet, and using flow energy instead of potential energy. It would give the same result since the flow energy at the turbine inlet is equal to the potential energy at the free surface of the reservoir.

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2-32

2-72 Wind is blowing steadily at a certain velocity. The mechanical energy of air per unit mass, the power generation potential, and the actual electric power generation are to be determined. Assumptions 1 The wind is blowing steadily at a constant uniform velocity. 2 The efficiency of the wind turbine is independent of the wind speed. Properties The density of air is given to be ρ = 1.25 kg/m3. Analysis Kinetic energy is the only form of mechanical energy the wind possesses, and it can be converted to work entirely. Therefore, the power potential of the wind is its kinetic energy, which is V2/2 per unit mass, and m& V 2 / 2 for a given mass flow rate:

emech = ke =

V 2 (12 m/s)2 = 2 2

m& = ρVA = ρV

πD 2 4

Wind 12 m/s

Wind turbine 50 m

⎛ 1 kJ/kg ⎞ = 0.072 kJ/kg ⎜ 2 2⎟ ⎝ 1000 m /s ⎠

= (1.25 kg/m3 )(12 m/s)

π (50 m)2 4

= 29,450 kg/s

W&max = E& mech = m& emech = (29,450 kg/s)(0.072 kJ/kg) = 2121 kW The actual electric power generation is determined by multiplying the power generation potential by the efficiency,

W&elect = ηwind turbineW&max = (0.30)(2121 kW) = 636 kW Therefore, 636 kW of actual power can be generated by this wind turbine at the stated conditions. Discussion The power generation of a wind turbine is proportional to the cube of the wind velocity, and thus the power generation will change strongly with the wind conditions.

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2-33

2-73 EES Problem 2-72 is reconsidered. The effect of wind velocity and the blade span diameter on wind power generation as the velocity varies from 5 m/s to 20 m/s in increments of 5 m/s, and the diameter varies from 20 m to 80 m in increments of 20 m is to be investigated. Analysis The problem is solved using EES, and the solution is given below. D1=20 [m] D2=40 [m] D3=60 [m] D4=80 [m] Eta=0.30 rho=1.25 [kg/m^3] m1_dot=rho*V*(pi*D1^2/4); W1_Elect=Eta*m1_dot*(V^2/2)/1000 "kW" m2_dot=rho*V*(pi*D2^2/4); W2_Elect=Eta*m2_dot*(V^2/2)/1000 "kW" m3_dot=rho*V*(pi*D3^2/4); W3_Elect=Eta*m3_dot*(V^2/2)/1000 "kW" m4_dot=rho*V*(pi*D4^2/4); W4_Elect=Eta*m4_dot*(V^2/2)/1000 "kW"

D, m 20

40

60

80

V, m/s 5 10 15 20 5 10 15 20 5 10 15 20 5 10 15 20

m, kg/s 1,963 3,927 5,890 7,854 7,854 15,708 23,562 31,416 17,671 35,343 53,014 70,686 31,416 62,832 94,248 125,664

Welect, kW 7 59 199 471 29 236 795 1885 66 530 1789 4241 118 942 3181 7540

8000 D = 80 m

7000 6000

WElect

5000 D = 60 m

4000 3000

D = 40 m

2000 1000 0 4

D = 20 m 6

8

10

12

14

16

18

20

V, m/s

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2-34

2-74 A wind turbine produces 180 kW of power. The average velocity of the air and the conversion efficiency of the turbine are to be determined. Assumptions The wind turbine operates steadily. Properties The density of air is given to be 1.31 kg/m3. Analysis (a) The blade diameter and the blade span area are Vtip D= = πn&

A=

πD 2 4

=

⎛ 1 m/s ⎞ (250 km/h)⎜ ⎟ ⎝ 3.6 km/h ⎠ = 88.42 m ⎛ 1 min ⎞ π (15 L/min)⎜ ⎟ ⎝ 60 s ⎠

π (88.42 m) 2 4

= 6140 m 2

Then the average velocity of air through the wind turbine becomes V=

42,000 kg/s m& = = 5.23 m/s ρA (1.31 kg/m 3 )(6140 m 2 )

(b) The kinetic energy of the air flowing through the turbine is

KE& =

1 1 m& V 2 = (42,000 kg/s)(5.23 m/s) 2 = 574.3 kW 2 2

Then the conversion efficiency of the turbine becomes

η=

W& 180 kW = = 0.313 = 31.3% KE& 574.3 kW

Discussion Note that about one-third of the kinetic energy of the wind is converted to power by the wind turbine, which is typical of actual turbines.

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2-35

2-75 Water is pumped from a lake to a storage tank at a specified rate. The overall efficiency of the pumpmotor unit and the pressure difference between the inlet and the exit of the pump are to be determined. Assumptions 1 The elevations of the tank and the lake remain constant. 2 Frictional losses in the pipes are negligible. 3 The changes in kinetic energy are negligible. 4 The elevation difference across the pump is negligible. Properties We take the density of water to be ρ = 1000 kg/m3. Analysis (a) We take the free surface of the lake to be point 1 and the free surfaces of the storage tank to be point 2. We also take the lake surface as the reference level (z1 = 0), and thus the potential energy at points 1 and 2 are pe1 = 0 and pe2 = gz2. The flow energy at both points is zero since both 1 and 2 are open to the atmosphere (P1 = P2 = Patm). Further, the kinetic energy at both points is zero (ke1 = ke2 = 0) since the water at both locations is essentially stationary. The mass flow rate of water and its potential energy at point 2 are

2 Storage tank

20 m

Pump

1

m& = ρV& = (1000 kg/m 3 )(0.070 m 3/s) = 70 kg/s

⎛ 1 kJ/kg ⎞ pe 2 = gz 2 = (9.81 m/s 2 )(20 m)⎜ ⎟ = 0.196 kJ/kg ⎝ 1000 m 2 /s 2 ⎠ Then the rate of increase of the mechanical energy of water becomes ΔE& mech,fluid = m& (e mech,out − e mech,in ) = m& ( pe 2 − 0) = m& pe 2 = (70 kg/s)(0.196 kJ/kg) = 13.7 kW

The overall efficiency of the combined pump-motor unit is determined from its definition,

η pump-motor =

ΔE& mech,fluid 13.7 kW = = 0.672 or 67.2% 20.4 kW W& elect,in

(b) Now we consider the pump. The change in the mechanical energy of water as it flows through the pump consists of the change in the flow energy only since the elevation difference across the pump and the change in the kinetic energy are negligible. Also, this change must be equal to the useful mechanical energy supplied by the pump, which is 13.7 kW: ΔE& mech,fluid = m& (e mech,out − e mech,in ) = m&

P2 − P1

ρ

= V&ΔP

Solving for ΔP and substituting, ΔP =

ΔE& mech,fluid 13.7 kJ/s ⎛ 1 kPa ⋅ m 3 ⎜ = 0.070 m 3 /s ⎜⎝ 1 kJ V&

⎞ ⎟ = 196 kPa ⎟ ⎠

Therefore, the pump must boost the pressure of water by 196 kPa in order to raise its elevation by 20 m. Discussion Note that only two-thirds of the electric energy consumed by the pump-motor is converted to the mechanical energy of water; the remaining one-third is wasted because of the inefficiencies of the pump and the motor.

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2-36

2-76 A large wind turbine is installed at a location where the wind is blowing steadily at a certain velocity. The electric power generation, the daily electricity production, and the monetary value of this electricity are to be determined. Assumptions 1 The wind is blowing steadily at a constant uniform velocity. 2 The efficiency of the wind turbine is independent of the wind speed.

Wind turbine

Wind

Properties The density of air is given to be ρ = 1.25 kg/m3.

8 m/s

100 m

Analysis Kinetic energy is the only form of mechanical energy the wind possesses, and it can be converted to work entirely. Therefore, the power potential of the wind is its kinetic energy, which is V2/2 per unit mass, and m& V 2 / 2 for a given mass flow rate: e mech = ke =

V 2 (8 m/s ) 2 ⎛ 1 kJ/kg ⎞ = ⎟ = 0.032 kJ/kg ⎜ 2 2 ⎝ 1000 m 2 /s 2 ⎠

m& = ρVA = ρV

πD 2 4

= (1.25 kg/m 3 )(8 m/s)

π (100 m) 2 4

= 78,540 kg/s

W& max = E& mech = m& e mech = (78,540 kg/s)(0.032 kJ/kg) = 2513 kW The actual electric power generation is determined from

W& elect = η wind turbineW& max = (0.32)(2513 kW) = 804.2 kW Then the amount of electricity generated per day and its monetary value become Amount of electricity = (Wind power)(Operating hours)=(804.2 kW)(24 h) =19,300 kWh Revenues = (Amount of electricity)(Unit price) = (19,300 kWh)($0.06/kWh) = $1158 (per day) Discussion Note that a single wind turbine can generate several thousand dollars worth of electricity every day at a reasonable cost, which explains the overwhelming popularity of wind turbines in recent years.

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2-37

2-77E A water pump raises the pressure of water by a specified amount at a specified flow rate while consuming a known amount of electric power. The mechanical efficiency of the pump is to be determined.

ΔP = 1.2 psi

Assumptions 1 The pump operates steadily. 2 The changes in velocity and elevation across the pump are negligible. 3 Water is incompressible. Analysis To determine the mechanical efficiency of the pump, we need to know the increase in the mechanical energy of the fluid as it flows through the pump, which is ΔE& mech,fluid = m& (emech,out − emech,in ) = m& [( Pv ) 2 − ( Pv )1 ] = m& ( P2 − P1 )v

3 hp PUMP

Pump inlet

⎛ ⎞ 1 Btu ⎟ = 1.776 Btu/s = 2.51 hp = V& ( P2 − P1 ) = (8 ft 3 /s)(1.2 psi)⎜⎜ 3⎟ ⎝ 5.404 psi ⋅ ft ⎠

since 1 hp = 0.7068 Btu/s, m& = ρV& = V& / v , and there is no change in kinetic and potential energies of the fluid. Then the mechanical efficiency of the pump becomes

η pump =

ΔE& mech,fluid 2.51 hp = = 0.838 or 83.8% 3 hp W& pump, shaft

Discussion The overall efficiency of this pump will be lower than 83.8% because of the inefficiency of the electric motor that drives the pump.

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2-38

2-78 Water is pumped from a lower reservoir to a higher reservoir at a specified rate. For a specified shaft power input, the power that is converted to thermal energy is to be determined. Assumptions 1 The pump operates steadily. 2 The elevations of the reservoirs remain constant. 3 The changes in kinetic energy are negligible.

2

Properties We take the density of water to be ρ = 1000 kg/m3.

Reservoir

Analysis The elevation of water and thus its potential energy changes during pumping, but it experiences no changes in its velocity and pressure. Therefore, the change in the total mechanical energy of water is equal to the change in its potential energy, which is gz per unit mass, and m& gz for a given mass flow rate. That is,

45 m

Pump

1 Reservoir

ΔE& mech = m& Δe mech = m& Δpe = m& gΔz = ρV&gΔz ⎛ 1N = (1000 kg/m 3 )(0.03 m 3 /s)(9.81 m/s 2 )(45 m)⎜⎜ 2 ⎝ 1 kg ⋅ m/s

⎞⎛ 1 kW ⎞ ⎟⎜ ⎟⎝ 1000 N ⋅ m/s ⎟⎠ = 13.2 kW ⎠

Then the mechanical power lost because of frictional effects becomes W& frict = W& pump, in − ΔE& mech = 20 − 13.2 kW = 6.8 kW

Discussion The 6.8 kW of power is used to overcome the friction in the piping system. The effect of frictional losses in a pump is always to convert mechanical energy to an equivalent amount of thermal energy, which results in a slight rise in fluid temperature. Note that this pumping process could be accomplished by a 13.2 kW pump (rather than 20 kW) if there were no frictional losses in the system. In this ideal case, the pump would function as a turbine when the water is allowed to flow from the upper reservoir to the lower reservoir and extract 13.2 kW of power from the water.

2-79 The mass flow rate of water through the hydraulic turbines of a dam is to be determined. Analysis The mass flow rate is determined from W& = m& g ( z 2 − z1 ) ⎯ ⎯→ m& =

W& = g ( z 2 − z1 )

100,000 kJ/s ⎛ 1 kJ/kg ⎞ (9.8 m/s 2 )(206 − 0) m⎜ ⎟ ⎝ 1000 m 2 /s 2 ⎠

= 49,500 kg/s

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2-39

2-80 A pump is pumping oil at a specified rate. The pressure rise of oil in the pump is measured, and the motor efficiency is specified. The mechanical efficiency of the pump is to be determined. Assumptions 1 The flow is steady and incompressible. 2 The elevation difference across the pump is negligible. Properties The density of oil is given to be ρ = 860 kg/m3. Analysis Then the total mechanical energy of a fluid is the sum of the potential, flow, and kinetic energies, and is expressed per unit mass as emech = gh + Pv + V 2 / 2 . To determine the mechanical efficiency of the pump, we need to know the increase in the mechanical energy of the fluid as it flows through the pump, which is

⎛ V2 V2 ΔE& mech,fluid = m& (e mech,out − e mech,in ) = m& ⎜ ( Pv) 2 + 2 − ( Pv ) 1 − 1 ⎜ 2 2 ⎝

⎞ &⎛ V 2 − V12 ⎟ = V ⎜ ( P2 − P1 ) + ρ 2 ⎟ ⎜ 2 ⎠ ⎝

since m& = ρV& = V& / v , and there is no change in the potential energy of the fluid. Also,

V1 = V2 =

V& A1

V& A2

V&

=

πD12 / 4

=

V& πD22 / 4

=

0.1 m 3 /s

π (0.08 m) 2 / 4

=

3

0.1 m /s

π (0.12 m) 2 / 4

2

⎞ ⎟ ⎟ ⎠

35 kW

= 19.9 m/s PUMP

Motor

= 8.84 m/s Pump inlet

Substituting, the useful pumping power is determined to be 1

W& pump,u = ΔE& mech,fluid ⎛ (8.84 m/s) 2 − (19.9 m/s) 2 = (0.1 m 3 /s)⎜ 400 kN/m 2 + (860 kg/m 3 ) ⎜ 2 ⎝ = 26.3 kW

⎛ 1 kN ⎜ ⎜ 1000 kg ⋅ m/s 2 ⎝

⎞ ⎞⎛ 1 kW ⎞ ⎟ ⎟⎜ ⎟ ⎟⎝ 1 kN ⋅ m/s ⎟⎠ ⎠⎠

Then the shaft power and the mechanical efficiency of the pump become W& pump,shaft = η motor W& electric = (0.90)(35 kW) = 31.5 kW

η pump =

W& pump, u W&

pump, shaft

=

26.3 kW = 0.836 = 83.6% 31.5 kW

Discussion The overall efficiency of this pump/motor unit is the product of the mechanical and motor efficiencies, which is 0.9×0.836 = 0.75.

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2-40

2-81E Water is pumped from a lake to a nearby pool by a pump with specified power and efficiency. The mechanical power used to overcome frictional effects is to be determined. Assumptions 1 The flow is steady and incompressible. 2 The elevation difference between the lake and the free surface of the pool is constant. 3 The average flow velocity is constant since pipe diameter is constant. Properties We take the density of water to be ρ = 62.4 lbm/ft3. Analysis The useful mechanical pumping power delivered to water is W& pump, u = η pump W& pump = (0.73)(12 hp) = 8.76 hp

The elevation of water and thus its potential energy changes during pumping, but it experiences no changes in its velocity and pressure. Therefore, the change in the total mechanical energy of water is equal to the change in its potential energy, which is gz per unit mass, and m& gz for a given mass flow rate. That is,

Pool

2

Pump

35 ft Lake

1

ΔE& mech = m& Δe mech = m& Δpe = m& gΔz = ρV&gΔz Substituting, the rate of change of mechanical energy of water becomes 1 lbf ⎛ ΔE& mech = (62.4 lbm/ft 3 )(1.2 ft 3 /s)(32.2 ft/s 2 )(35 ft )⎜ ⎝ 32.2 lbm ⋅ ft/s 2

1 hp ⎞⎛ ⎞ ⎟⎜ ⎟ = 4.76 hp ⋅ 550 lbf ft/s ⎝ ⎠ ⎠

Then the mechanical power lost in piping because of frictional effects becomes W& frict = W& pump, u − ΔE& mech = 8.76 − 4.76 hp = 4.0 hp

Discussion Note that the pump must supply to the water an additional useful mechanical power of 4.0 hp to overcome the frictional losses in pipes.

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2-41

Energy and Environment

2-82C Energy conversion pollutes the soil, the water, and the air, and the environmental pollution is a serious threat to vegetation, wild life, and human health. The emissions emitted during the combustion of fossil fuels are responsible for smog, acid rain, and global warming and climate change. The primary chemicals that pollute the air are hydrocarbons (HC, also referred to as volatile organic compounds, VOC), nitrogen oxides (NOx), and carbon monoxide (CO). The primary source of these pollutants is the motor vehicles.

2-83C Smog is the brown haze that builds up in a large stagnant air mass, and hangs over populated areas on calm hot summer days. Smog is made up mostly of ground-level ozone (O3), but it also contains numerous other chemicals, including carbon monoxide (CO), particulate matter such as soot and dust, volatile organic compounds (VOC) such as benzene, butane, and other hydrocarbons. Ground-level ozone is formed when hydrocarbons and nitrogen oxides react in the presence of sunlight in hot calm days. Ozone irritates eyes and damage the air sacs in the lungs where oxygen and carbon dioxide are exchanged, causing eventual hardening of this soft and spongy tissue. It also causes shortness of breath, wheezing, fatigue, headaches, nausea, and aggravate respiratory problems such as asthma.

2-84C Fossil fuels include small amounts of sulfur. The sulfur in the fuel reacts with oxygen to form sulfur dioxide (SO2), which is an air pollutant. The sulfur oxides and nitric oxides react with water vapor and other chemicals high in the atmosphere in the presence of sunlight to form sulfuric and nitric acids. The acids formed usually dissolve in the suspended water droplets in clouds or fog. These acid-laden droplets are washed from the air on to the soil by rain or snow. This is known as acid rain. It is called “rain” since it comes down with rain droplets.

As a result of acid rain, many lakes and rivers in industrial areas have become too acidic for fish to grow. Forests in those areas also experience a slow death due to absorbing the acids through their leaves, needles, and roots. Even marble structures deteriorate due to acid rain.

2-85C Carbon dioxide (CO2), water vapor, and trace amounts of some other gases such as methane and nitrogen oxides act like a blanket and keep the earth warm at night by blocking the heat radiated from the earth. This is known as the greenhouse effect. The greenhouse effect makes life on earth possible by keeping the earth warm. But excessive amounts of these gases disturb the delicate balance by trapping too much energy, which causes the average temperature of the earth to rise and the climate at some localities to change. These undesirable consequences of the greenhouse effect are referred to as global warming or global climate change. The greenhouse effect can be reduced by reducing the net production of CO2 by consuming less energy (for example, by buying energy efficient cars and appliances) and planting trees.

2-86C Carbon monoxide, which is a colorless, odorless, poisonous gas that deprives the body's organs from getting enough oxygen by binding with the red blood cells that would otherwise carry oxygen. At low levels, carbon monoxide decreases the amount of oxygen supplied to the brain and other organs and muscles, slows body reactions and reflexes, and impairs judgment. It poses a serious threat to people with heart disease because of the fragile condition of the circulatory system and to fetuses because of the oxygen needs of the developing brain. At high levels, it can be fatal, as evidenced by numerous deaths caused by cars that are warmed up in closed garages or by exhaust gases leaking into the cars.

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2-87E A person trades in his Ford Taurus for a Ford Explorer. The extra amount of CO2 emitted by the Explorer within 5 years is to be determined. Assumptions The Explorer is assumed to use 940 gallons of gasoline a year compared to 715 gallons for Taurus. Analysis The extra amount of gasoline the Explorer will use within 5 years is

Extra Gasoline

= (Extra per year)(No. of years) = (940 – 715 gal/yr)(5 yr) = 1125 gal

Extra CO2 produced

= (Extra gallons of gasoline used)(CO2 emission per gallon) = (1125 gal)(19.7 lbm/gal) = 22,163 lbm CO2

Discussion Note that the car we choose to drive has a significant effect on the amount of greenhouse gases produced.

2-88 A power plant that burns natural gas produces 0.59 kg of carbon dioxide (CO2) per kWh. The amount of CO2 production that is due to the refrigerators in a city is to be determined. Assumptions The city uses electricity produced by a natural gas power plant. Properties 0.59 kg of CO2 is produced per kWh of electricity generated (given). Analysis Noting that there are 200,000 households in the city and each household consumes 700 kWh of electricity for refrigeration, the total amount of CO2 produced is Amount of CO 2 produced = (Amount of electricity consumed)(Amount of CO 2 per kWh) = (200,000 household)(700 kWh/year household)(0.59 kg/kWh) = 8.26 × 107 CO 2 kg/year = 82,600 CO2 ton/year

Therefore, the refrigerators in this city are responsible for the production of 82,600 tons of CO2.

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2-89 A power plant that burns coal, produces 1.1 kg of carbon dioxide (CO2) per kWh. The amount of CO2 production that is due to the refrigerators in a city is to be determined. Assumptions The city uses electricity produced by a coal power plant. Properties 1.1 kg of CO2 is produced per kWh of electricity generated (given). Analysis Noting that there are 200,000 households in the city and each household consumes 700 kWh of electricity for refrigeration, the total amount of CO2 produced is Amount of CO 2 produced = ( Amount of electricity consumed)(Amount of CO 2 per kWh) = (200,000 household)(700 kWh/household)(1.1 kg/kWh) = 15.4 × 10 7 CO 2 kg/year = 154,000 CO 2 ton/year

Therefore, the refrigerators in this city are responsible for the production of 154,000 tons of CO2.

2-90E A household uses fuel oil for heating, and electricity for other energy needs. Now the household reduces its energy use by 20%. The reduction in the CO2 production this household is responsible for is to be determined. Properties The amount of CO2 produced is 1.54 lbm per kWh and 26.4 lbm per gallon of fuel oil (given). Analysis Noting that this household consumes 11,000 kWh of electricity and 1500 gallons of fuel oil per year, the amount of CO2 production this household is responsible for is Amount of CO 2 produced = (Amount of electricity consumed)(Amount of CO 2 per kWh) + (Amount of fuel oil consumed)(Amount of CO 2 per gallon) = (11,000 kWh/yr)(1.54 lbm/kWh) + (1500 gal/yr)(26.4 lbm/gal) = 56,540 CO 2 lbm/year

Then reducing the electricity and fuel oil usage by 15% will reduce the annual amount of CO2 production by this household by Reduction in CO 2 produced = (0.15)(Current amount of CO 2 production) = (0.15)(56,540 CO 2 kg/year) = 8481 CO 2 lbm/year

Therefore, any measure that saves energy also reduces the amount of pollution emitted to the environment.

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2-91 A household has 2 cars, a natural gas furnace for heating, and uses electricity for other energy needs. The annual amount of NOx emission to the atmosphere this household is responsible for is to be determined. Properties The amount of NOx produced is 7.1 g per kWh, 4.3 g per therm of natural gas, and 11 kg per car (given). Analysis Noting that this household has 2 cars, consumes 1200 therms of natural gas, and 9,000 kWh of electricity per year, the amount of NOx production this household is responsible for is

Amount of NO x produced = ( No. of cars)(Amount of NO x produced per car) + ( Amount of electricity consumed)(Amount of NO x per kWh) + ( Amount of gas consumed)(Amount of NO x per gallon) = (2 cars)(11 kg/car) + (9000 kWh/yr)(0.0071 kg/kWh) + (1200 therms/yr)(0.0043 kg/therm) = 91.06 NOx kg/year Discussion Any measure that saves energy will also reduce the amount of pollution emitted to the atmosphere.

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2-45

Special Topic: Mechanisms of Heat Transfer

2-92C The three mechanisms of heat transfer are conduction, convection, and radiation.

2-93C No. It is purely by radiation.

2-94C Diamond has a higher thermal conductivity than silver, and thus diamond is a better conductor of heat.

2-95C In forced convection, the fluid is forced to move by external means such as a fan, pump, or the wind. The fluid motion in natural convection is due to buoyancy effects only.

2-96C Emissivity is the ratio of the radiation emitted by a surface to the radiation emitted by a blackbody at the same temperature. Absorptivity is the fraction of radiation incident on a surface that is absorbed by the surface. The Kirchhoff's law of radiation states that the emissivity and the absorptivity of a surface are equal at the same temperature and wavelength.

2-97C A blackbody is an idealized body that emits the maximum amount of radiation at a given temperature, and that absorbs all the radiation incident on it. Real bodies emit and absorb less radiation than a blackbody at the same temperature.

2-98 The inner and outer surfaces of a brick wall are maintained at specified temperatures. The rate of heat transfer through the wall is to be determined. Assumptions 1 Steady operating conditions exist since the surface temperatures of the wall remain constant at the specified values. 2 Thermal properties of the wall are constant. Properties The thermal conductivity of the wall is given to be k = 0.69 W/m⋅°C. Analysis Under steady conditions, the rate of heat transfer through the wall is ΔT (20 − 5)°C Q& cond = kA = (0.69 W/m ⋅ °C)(5 × 6 m 2 ) = 1035 W L 0.3 m

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2-46

2-99 The inner and outer surfaces of a window glass are maintained at specified temperatures. The amount of heat transferred through the glass in 5 h is to be determined. Assumptions 1 Steady operating conditions exist since the surface temperatures of the glass remain constant at the specified values. 2 Thermal properties of the glass are constant. Properties The thermal conductivity of the glass is given to be k = 0.78 W/m⋅°C.

Glass

Analysis Under steady conditions, the rate of heat transfer through the glass by conduction is ΔT (10 − 3)°C Q& cond = kA = (0.78 W/m ⋅ °C)(2 × 2 m 2 ) = 4368 W L 0.005 m

Then the amount of heat transferred over a period of 5 h becomes

Q = Q& condΔt = (4.368 kJ/s)(5 × 3600s) = 78,600 kJ If the thickness of the glass is doubled to 1 cm, then the amount of heat transferred will go down by half to 39,300 kJ.

10°C

3°C

0.5 cm

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2-47

2-100 EES Reconsider Prob. 2-99. Using EES (or other) software, investigate the effect of glass thickness on heat loss for the specified glass surface temperatures. Let the glass thickness vary from 0.2 cm to 2 cm. Plot the heat loss versus the glass thickness, and discuss the results. Analysis The problem is solved using EES, and the solution is given below. FUNCTION klookup(material$) If material$='Glass' then klookup:=0.78 If material$='Brick' then klookup:=0.72 If material$='Fiber Glass' then klookup:=0.043 If material$='Air' then klookup:=0.026 If material$='Wood(oak)' then klookup:=0.17 END L=2 [m] W=2 [m] {material$='Glass' T_in=10 [C] T_out=3 [C] k=0.78 [W/m-C] t=5 [hr] thickness=0.5 [cm]} k=klookup(material$) "[W/m-K]" A=L*W"[m^2]" Q_dot_loss=A*k*(T_in-T_out)/(thickness*convert(cm,m))"[W]" Q_loss_total=Q_dot_loss*t*convert(hr,s)*convert(J,kJ)"[kJ]"

Thickness [cm] 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

200000

Heat loss through glass "w all" in 5 hours

160000

Q loss,total [kJ]

Qloss,total [kJ] 196560 98280 65520 49140 39312 32760 28080 24570 21840 19656

120000

80000

40000

0 0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

thickness [cm ]

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2

2-48

2-101 Heat is transferred steadily to boiling water in the pan through its bottom. The inner surface temperature of the bottom of the pan is given. The temperature of the outer surface is to be determined. Assumptions 1 Steady operating conditions exist since the surface temperatures of the pan remain constant at the specified values. 2 Thermal properties of the aluminum pan are constant. Properties The thermal conductivity of the aluminum is given to be k = 237 W/m⋅°C. Analysis The heat transfer surface area is

A = π r² = π(0.1 m)² = 0.0314 m² Under steady conditions, the rate of heat transfer through the bottom of the pan by conduction is

105°C

ΔT T −T Q& = kA = kA 2 1 L L Substituting,

T − 105o C 500 W = (237 W / m⋅o C)(0.0314 m2 ) 2 0.004 m

which gives

T2 = 105.3°C

500 W 0.4 cm

2-102 A person is standing in a room at a specified temperature. The rate of heat transfer between a person and the surrounding air by convection is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer by radiation is not considered. 3 The environment is at a uniform temperature.

Q& Ts =34°C

Analysis The heat transfer surface area of the person is

A = πDL = π(0.3 m)(1.70 m) = 1.60 m² Under steady conditions, the rate of heat transfer by convection is

Q& conv = hAΔT = (15 W/m2 ⋅ °C)(1.60 m2 )(34 − 20)°C = 336 W

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2-103 A spherical ball whose surface is maintained at a temperature of 70°C is suspended in the middle of a room at 20°C. The total rate of heat transfer from the ball is to be determined. Assumptions 1 Steady operating conditions exist since the ball surface and the surrounding air and surfaces remain at constant temperatures. 2 The thermal properties of the ball and the convection heat transfer coefficient are constant and uniform.

Air 20°C 70°C

Properties The emissivity of the ball surface is given to be ε = 0.8. Analysis The heat transfer surface area is

A = πD² = 3.14x(0.05 m)² = 0.007854 m²

D = 5 cm

. Q

Under steady conditions, the rates of convection and radiation heat transfer are

Q& conv = hAΔT = (15 W/m2 ⋅o C)(0.007854 m 2 )(70 − 20)o C = 5.89 W Q& rad = εσA(Ts4 − To4 ) = 0.8(0.007854 m 2 )(5.67 × 10−8 W/m2 ⋅ K 4 )[(343 K)4 − (293 K)4 ] = 2.31 W Therefore,

Q& total = Q& conv + Q& rad = 5.89 + 2.31 = 8.20 W

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2-50

2-104 EES Reconsider Prob. 2-103. Using EES (or other) software, investigate the effect of the convection heat transfer coefficient and surface emissivity on the heat transfer rate from the ball. Let the heat transfer coefficient vary from 5 W/m2.°C to 30 W/m2.°C. Plot the rate of heat transfer against the convection heat transfer coefficient for the surface emissivities of 0.1, 0.5, 0.8, and 1, and discuss the results. Analysis The problem is solved using EES, and the solution is given below. sigma=5.67e-8 [W/m^2-K^4] {T_sphere=70 [C] T_room=20 [C] D_sphere=5 [cm] epsilon=0.1 h_c=15 [W/m^2-K]} A=4*pi*(D_sphere/2)^2*convert(cm^2,m^2)"[m^2]" Q_dot_conv=A*h_c*(T_sphere-T_room)"[W]" Q_dot_rad=A*epsilon*sigma*((T_sphere+273)^4-(T_room+273)^4)"[W]" Q_dot_total=Q_dot_conv+Q_dot_rad"[W]"

hc [W/m2-K] 5 10 15 20 25 30

Qtotal [W] 2.252 4.215 6.179 8.142 10.11 12.07

15

Qtotal [W]

13 11

ε = 1.0

9 7

ε = 0.1

5 3 1 5

10

15

20

25

30

hc [W/m^2-K]

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2-51

2-105 Hot air is blown over a flat surface at a specified temperature. The rate of heat transfer from the air to the plate is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer by radiation is not considered. 3 The convection heat transfer coefficient is constant and uniform over the surface.

80°C Air

Analysis Under steady conditions, the rate of heat transfer by convection is Q& conv = hAΔT

30°C

= (55 W/m ⋅ °C)(2 × 4 m )(80 − 30) C 2

2

o

= 22,000 W = 22 kW

2-106 A 1000-W iron is left on the iron board with its base exposed to the air at 20°C. The temperature of the base of the iron is to be determined in steady operation. Assumptions 1 Steady operating conditions exist. 2 The thermal properties of the iron base and the convection heat transfer coefficient are constant and uniform. 3 The temperature of the surrounding surfaces is the same as the temperature of the surrounding air.

Iron 1000 W Air 20°C

Properties The emissivity of the base surface is given to be ε = 0.6. Analysis At steady conditions, the 1000 W of energy supplied to the iron will be dissipated to the surroundings by convection and radiation heat transfer. Therefore, Q& total = Q& conv + Q& rad = 1000 W

where

Q& conv = hAΔT = (35 W/m 2 ⋅ K)(0.02 m 2 )(Ts − 293 K) = 0.7(Ts − 293 K) W

and

Q& rad = εσA(Ts4 − To4 ) = 0.6(0.02 m 2 )(5.67 × 10 −8 W / m 2 ⋅ K 4 )[Ts4 − (293 K) 4 ] = 0.06804 × 10 −8 [Ts4 − (293 K) 4 ] W Substituting,

1000 W = 0.7(Ts − 293 K ) + 0.06804 ×10 −8 [Ts4 − (293 K) 4 ]

Solving by trial and error gives

Ts = 947 K = 674°C

Discussion We note that the iron will dissipate all the energy it receives by convection and radiation when its surface temperature reaches 947 K.

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2-52

2-107 The backside of the thin metal plate is insulated and the front side is exposed to solar radiation. The surface temperature of the plate is to be determined when it stabilizes. Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the insulated side of the plate is negligible. 3 The heat transfer coefficient is constant and uniform over the plate. 4 Heat loss by radiation is negligible. Properties The solar absorptivity of the plate is given to be α = 0.6. Analysis When the heat loss from the plate by convection equals the solar radiation absorbed, the surface temperature of the plate can be determined from

Q& solarabsorbed = Q& conv

αQ& solar = hA(Ts − To ) 0.6 × A × 700W/m 2 = (50W/m 2 ⋅ o C) A(Ts − 25)

700 W/m2 α = 0.6 25°C

Canceling the surface area A and solving for Ts gives

Ts = 33.4 o C

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2-53

2-108 EES Reconsider Prob. 2-107. Using EES (or other) software, investigate the effect of the convection heat transfer coefficient on the surface temperature of the plate. Let the heat transfer coefficient vary from 10 W/m2.°C to 90 W/m2.°C. Plot the surface temperature against the convection heat transfer coefficient, and discuss the results. Analysis The problem is solved using EES, and the solution is given below. sigma=5.67e-8 [W/m^2-K^4] "The following variables are obtained from the Diagram Window." {T_air=25 [C] S=700 [W/m^2] alpha_solar=0.6 h_c=50 [W/m^2-C]} "An energy balance on the plate gives:" Q_dot_solar=Q_dot_conv"[W]" "The absorbed solar per unit area of plate" Q_dot_solar =S*alpha_solar"[W]" "The leaving energy by convection per unit area of plate" Q_dot_conv=h_c*(T_plate-T_air)"[W]"

Tplate [C] 67 46 39 35.5 33.4 32 31 30.25 29.67

70 65 60 55

T plate [C]

hc [W/m2-K] 10 20 30 40 50 60 70 80 90

50 45 40 35 30 25 10

20

30

40

h

50

c

60

70

80

90

[W /m^2]

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2-54

2-109 A hot water pipe at 80°C is losing heat to the surrounding air at 5°C by natural convection with a heat transfer coefficient of 25 W/ m2.°C. The rate of heat loss from the pipe by convection is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer by radiation is not considered. 3 The convection heat transfer coefficient is constant and uniform over the surface. Analysis The heat transfer surface area is

A = (πD)L = 3.14x(0.05 m)(10 m) = 1.571 m²

80°C

D = 5 cm L = 10 m Q Air, 5°C

Under steady conditions, the rate of heat transfer by convection is

Q& conv = hAΔT = (25 W/m2 ⋅ °C)(1.571 m 2 )(80 − 5)°C = 2945 W = 2.95 kW

2-110 A spacecraft in space absorbs solar radiation while losing heat to deep space by thermal radiation. The surface temperature of the spacecraft is to be determined when steady conditions are reached.. Assumptions 1 Steady operating conditions exist since the surface temperatures of the wall remain constant at the specified values. 2 Thermal properties of the spacecraft are constant. Properties The outer surface of a spacecraft has an emissivity of 0.8 and an absorptivity of 0.3. Analysis When the heat loss from the outer surface of the spacecraft by radiation equals the solar radiation absorbed, the surface temperature can be determined from Q& solar absorbed = Q& rad 4 ) αQ& solar = εσA(Ts4 − Tspace

1000 W/m2 α = 0.3 ε = 0.8

0.3 × A × (1000 W/m 2 ) = 0.8 × A × (5.67 × 10 −8 W/m 2 ⋅ K 4 )[Ts4 − (0 K) 4 ]

Canceling the surface area A and solving for Ts gives T s = 285 K

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2-55

2-111 EES Reconsider Prob. 2-110. Using EES (or other) software, investigate the effect of the surface emissivity and absorptivity of the spacecraft on the equilibrium surface temperature. Plot the surface temperature against emissivity for solar absorptivities of 0.1, 0.5, 0.8, and 1, and discuss the results. Analysis The problem is solved using EES, and the solution is given below. "Knowns" sigma=5.67e-8 [W/m^2-K^4] "The following variables are obtained from the Diagram Window." {T_space=10 [C] S=1000[W/m^2] alpha_solar=0.3 epsilon=0.8} "Solution" "An energy balance on the spacecraft gives:" Q_dot_solar=Q_dot_out "The absorbed solar" Q_dot_solar =S*alpha_solar "The net leaving radiation leaving the spacecraft:" Q_dot_out=epsilon*sigma*((T_spacecraft+273)^4-(T_space+273)^4) 140

ε

Tspacecraft [C] 218.7 150 117.2 97.2 83.41 73.25 65.4 59.13 54 49.71

T spacecraft [C]

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Surface em issivity = 0.8 120 100 80 60 40 20 0.1

0.2

0.3

0.4

0.5

α

0.6

0.7

0.8

0.9

1

solar

225

solar absorptivity = 0.3

T spacecraft [C ]

185

145

105

65

25 0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

ε

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2-56

2-112 A hollow spherical iron container is filled with iced water at 0°C. The rate of heat loss from the sphere and the rate at which ice melts in the container are to be determined. Assumptions 1 Steady operating conditions exist since the surface temperatures of the wall remain constant at the specified values. 2 Heat transfer through the shell is one-dimensional. 3 Thermal properties of the iron shell are constant. 4 The inner surface of the shell is at the same temperature as the iced water, 0°C. Properties The thermal conductivity of iron is k = 80.2 W/m⋅°C (Table 2-3). The heat of fusion of water is at 1 atm is 333.7 kJ/kg. 5°C Analysis This spherical shell can be approximated as a plate of

thickness 0.4 cm and surface area A = πD² = 3.14×(0.2 m)² = 0.126 m² Then the rate of heat transfer through the shell by conduction is ΔT (5 − 0)°C Q& cond = kA = (80.2 W/m⋅o C)(0.126 m 2 ) = 12,632 W L 0.004 m

Iced water 0°C

0.4 cm

Considering that it takes 333.7 kJ of energy to melt 1 kg of ice at 0°C, the rate at which ice melts in the container can be determined from

m& ice =

Q& 12.632 kJ/s = = 0.038 kg/s hif 333.7 kJ/kg

Discussion We should point out that this result is slightly in error for approximating a curved wall as a plain wall. The error in this case is very small because of the large diameter to thickness ratio. For better accuracy, we could use the inner surface area (D = 19.2 cm) or the mean surface area (D = 19.6 cm) in the calculations.

2-113 The inner and outer glasses of a double pane window with a 1-cm air space are at specified temperatures. The rate of heat transfer through the window is to be determined. Assumptions 1 Steady operating conditions exist since the surface temperatures of the glass remain constant at the specified values. 2 Heat transfer through the window is one-dimensional. 3 Thermal properties of the air are constant. 4 The air trapped between the two glasses is still, and thus heat transfer is by conduction only. Properties The thermal conductivity of air at room temperature is k = 0.026 W/m.°C (Table 2-3).

18°C

Analysis Under steady conditions, the rate of heat transfer through the window by conduction is

Air 6°C · Q

o

ΔT (18 − 6) C = (0.026 W/m⋅o C)(2 × 2 m 2 ) Q& cond = kA 0.01 m L = 125 W = 0.125 kW

1cm

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2-57

2-114 Two surfaces of a flat plate are maintained at specified temperatures, and the rate of heat transfer through the plate is measured. The thermal conductivity of the plate material is to be determined. Assumptions 1 Steady operating conditions exist since the surface temperatures of the plate remain constant at the specified values. 2 Heat transfer through the plate is one-dimensional. 3 Thermal properties of the plate are constant. Analysis The thermal conductivity is determined directly from the steady one-dimensional heat conduction relation to be

T −T Q& = kA 1 2 L & (Q / A) L (500 W/m 2 )(0.02 m) = = 0.1 W/m.°C k= (100 - 0)°C T1 − T2

Plate 2 cm

100°C

0°C 500 W/m2

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2-58

Review Problems

2-115 A decision is to be made between a cheaper but inefficient natural gas heater and an expensive but efficient natural gas heater for a house. Assumptions The two heaters are comparable in all aspects other than the initial cost and efficiency. Analysis Other things being equal, the logical choice is the heater that will cost less during its lifetime. The total cost of a system during its lifetime (the initial, operation, maintenance, etc.) can be determined by performing a life cycle cost analysis. A simpler alternative is to determine the simple payback period.

The annual heating cost is given to be $1200. Noting that the existing heater is 55% efficient, only 55% of that energy (and thus money) is delivered to the house, and the rest is wasted due to the inefficiency of the heater. Therefore, the monetary value of the heating load of the house is

Gas Heater η1 = 82% η2 = 95%

Cost of useful heat = (55%)(Current annual heating cost) = 0.55×($1200/yr)=$660/yr This is how much it would cost to heat this house with a heater that is 100% efficient. For heaters that are less efficient, the annual heating cost is determined by dividing $660 by the efficiency:

82% heater:

Annual cost of heating = (Cost of useful heat)/Efficiency = ($660/yr)/0.82 = $805/yr

95% heater:

Annual cost of heating = (Cost of useful heat)/Efficiency = ($660/yr)/0.95 = $695/yr

Annual cost savings with the efficient heater = 805 - 695 = $110 Excess initial cost of the efficient heater = 2700 - 1600 = $1100 The simple payback period becomes

Simple payback period =

Excess initial cost $1100 = = 10 years Annaul cost savings $110 / yr

Therefore, the more efficient heater will pay for the $1100 cost differential in this case in 10 years, which is more than the 8-year limit. Therefore, the purchase of the cheaper and less efficient heater is a better buy in this case.

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2-116 A wind turbine is rotating at 20 rpm under steady winds of 30 km/h. The power produced, the tip speed of the blade, and the revenue generated by the wind turbine per year are to be determined. Assumptions 1 Steady operating conditions exist. 2 The wind turbine operates continuously during the entire year at the specified conditions. Properties The density of air is given to be ρ = 1.20 kg/m3. Analysis (a) The blade span area and the mass flow rate of air through the turbine are

A = πD 2 / 4 = π (80 m) 2 / 4 = 5027 m 2 ⎛ 1000 m ⎞⎛ 1 h ⎞ V = (30 km/h)⎜ ⎟⎜ ⎟ = 8.333 m/s ⎝ 1 km ⎠⎝ 3600 s ⎠ m& = ρAV = (1.2 kg/m 3 )(5027 m 2 )(8.333 m/s) = 50,270 kg/s Noting that the kinetic energy of a unit mass is V2/2 and the wind turbine captures 35% of this energy, the power generated by this wind turbine becomes 1 ⎛ 1 kJ/kg ⎞ ⎛1 ⎞ W& = η ⎜ m& V 2 ⎟ = (0.35) (50,270 kg/s)(8.333 m/s ) 2 ⎜ ⎟ = 610.9 kW 2 ⎝2 ⎠ ⎝ 1000 m 2 /s 2 ⎠ (b) Noting that the tip of blade travels a distance of πD per revolution, the tip velocity of the turbine blade for an rpm of n& becomes

V tip = πDn& = π (80 m)(20 / min) = 5027 m/min = 83.8 m/s = 302 km/h (c) The amount of electricity produced and the revenue generated per year are Electricity produced = W& Δt = (610.9 kW)(365 × 24 h/year) = 5.351× 10 6 kWh/year Revenue generated = (Electricity produced)(Unit price) = (5.351× 10 6 kWh/year)($0.06/kWh) = $321,100/y ear

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2-117 A wind turbine is rotating at 20 rpm under steady winds of 25 km/h. The power produced, the tip speed of the blade, and the revenue generated by the wind turbine per year are to be determined. Assumptions 1 Steady operating conditions exist. 2 The wind turbine operates continuously during the entire year at the specified conditions. Properties The density of air is given to be ρ = 1.20 kg/m3. Analysis (a) The blade span area and the mass flow rate of air through the turbine are

A = πD 2 / 4 = π (80 m)2 / 4 = 5027 m 2 ⎛ 1000 m ⎞⎛ 1 h ⎞ V = (25 km/h)⎜ ⎟⎜ ⎟ = 6.944 m/s ⎝ 1 km ⎠⎝ 3600 s ⎠ m& = ρAV = (1.2 kg/m3 )(5027 m 2 )(6.944 m/s) = 41,891 kg/s Noting that the kinetic energy of a unit mass is V2/2 and the wind turbine captures 35% of this energy, the power generated by this wind turbine becomes 1 ⎛1 ⎞ ⎛ 1 kJ/kg ⎞ W& = η ⎜ m& V 2 ⎟ = (0.35) (41,891 kg/s)(6.944 m/s) 2 ⎜ ⎟ = 353.5 kW 2 2 2 ⎝2 ⎠ ⎝ 1000 m /s ⎠

(b) Noting that the tip of blade travels a distance of πD per revolution, the tip velocity of the turbine blade for an rpm of n& becomes

Vtip = πDn& = π (80 m)(20 / min) = 5027 m/min = 83.8 m/s = 302 km/h (c) The amount of electricity produced and the revenue generated per year are Electricity produced = W& Δt = (353.5 kW)(365 × 24 h/year) = 3,096,660 kWh/year Revenue generated = (Electricity produced)(Unit price) = (3,096,660 kWh/year)($0.06/kWh) = $185,800/y ear

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2-118E The energy contents, unit costs, and typical conversion efficiencies of various energy sources for use in water heaters are given. The lowest cost energy source is to be determined. Assumptions The differences in installation costs of different water heaters are not considered. Properties The energy contents, unit costs, and typical conversion efficiencies of different systems are given in the problem statement. Analysis The unit cost of each Btu of useful energy supplied to the water heater by each system can be determined from

Unit cost of useful energy =

Unit cost of energy supplied Conversion efficiency

Substituting, ⎛ 1 ft 3 ⎞ ⎟ = $21.3 × 10 −6 / Btu ⎜ ⎜ 1025 Btu ⎟ ⎠ ⎝

Natural gas heater:

Unit cost of useful energy =

$0.012/ft 3 0.55

Heating by oil heater:

Unit cost of useful energy =

$1.15/gal ⎛ 1 gal ⎞ ⎜⎜ ⎟ = $15.1× 10 − 6 / Btu 0.55 ⎝ 138,700 Btu ⎟⎠

Electric heater:

Unit cost of useful energy =

$0.084/kWh) ⎛ 1 kWh ⎞ −6 ⎜ ⎟ = $27.4 × 10 / Btu 0.90 ⎝ 3412 Btu ⎠

Therefore, the lowest cost energy source for hot water heaters in this case is oil.

2-119 A home owner is considering three different heating systems for heating his house. The system with the lowest energy cost is to be determined. Assumptions The differences in installation costs of different heating systems are not considered. Properties The energy contents, unit costs, and typical conversion efficiencies of different systems are given in the problem statement. Analysis The unit cost of each Btu of useful energy supplied to the house by each system can be determined from

Unit cost of useful energy =

Unit cost of energy supplied Conversion efficiency

Substituting, Natural gas heater:

Unit cost of useful energy =

$1.24/therm ⎛ 1 therm ⎞ ⎜⎜ ⎟⎟ = $13.5 × 10 −6 / kJ 0.87 ⎝ 105,500 kJ ⎠

Heating oil heater:

Unit cost of useful energy =

$1.25/gal ⎛ 1 gal ⎞ ⎜ ⎟ = $10.4 × 10 − 6 / kJ 0.87 ⎜⎝ 138,500 kJ ⎟⎠

Electric heater:

Unit cost of useful energy =

$0.09/kWh) ⎛ 1 kWh ⎞ −6 ⎜ ⎟ = $25.0 × 10 / kJ 1.0 ⎝ 3600 kJ ⎠

Therefore, the system with the lowest energy cost for heating the house is the heating oil heater.

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2-120 The heating and cooling costs of a poorly insulated house can be reduced by up to 30 percent by adding adequate insulation. The time it will take for the added insulation to pay for itself from the energy it saves is to be determined. Assumptions It is given that the annual energy usage of a house is $1200 a year, and 46% of it is used for heating and cooling. The cost of added insulation is given to be $200.

Heat loss

Analysis The amount of money that would be saved per year is determined directly from Money saved = ($1200 / year)(0.46)(0.30) = $166 / yr

Then the simple payback period becomes Payback period =

House

Cost $200 = = 1.2 yr Money saved $166/yr

Therefore, the proposed measure will pay for itself in less than one and a half year.

2-121 Caulking and weather-stripping doors and windows to reduce air leaks can reduce the energy use of a house by up to 10 percent. The time it will take for the caulking and weather-stripping to pay for itself from the energy it saves is to be determined. Assumptions It is given that the annual energy usage of a house is $1100 a year, and the cost of caulking and weather-stripping a house is $50. Analysis The amount of money that would be saved per year is determined directly from Money saved = ($1100 / year)(0.10) = $110 / yr

Then the simple payback period becomes Payback period =

Cost $50 = = 0.45 yr Money saved $110/yr

Therefore, the proposed measure will pay for itself in less than half a year.

2-122 The work required to compress a spring is to be determined. Analysis (a) With the preload, F = F0 + kx. Substituting this into the work integral gives 2

∫

2

∫

W = Fds = (kx + F0 )dx 1

F

1

k 2 ( x 2 − x12 ) + F0 ( x 2 − x1 ) 2 300 N/cm = (1 cm) 2 − 0 2 + (100 N)[(1 cm) − 0] 2 = 250 N ⋅ cm = 2.50 N ⋅ m = 2.50 J = 0.0025 kJ

=

[

x

]

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2-123 The work required to compress a gas in a gas spring is to be determined. Assumptions All forces except that generated by the gas spring will be neglected. Analysis When the expression given in the problem statement is substituted into the work integral relation, and advantage is taken of the fact that the force and displacement vectors are collinear, the result is 2

2

∫

∫

1

1

W = Fds =

Constant xk

dx F

Constant 1− k = ( x 2 − x11− k ) 1− k 1000 N ⋅ m 1.3 = (0.3 m) −0.3 − (0.1 m) −0.3 1 − 1.3 = 1867 N ⋅ m = 1867 J = 1.87 kJ

[

x

]

2-124E A man pushes a block along a horizontal plane. The work required to move the block is to be determined considering (a) the man and (b) the block as the system. Analysis The work applied to the block to overcome the friction is found by using the work integral, 2

∫

W = Fds = 1

2

∫ fW ( x

2

− x1 )

x

1

= (0.2)(100 lbf )(100 ft) = 2000 lbf ⋅ ft

fW

1 Btu ⎛ ⎞ = (2000 lbf ⋅ ft)⎜ ⎟ = 2.57 Btu ⎝ 778.169 lbf ⋅ ft ⎠

The man must then produce the amount of work W = 2.57 Btu

fW W

2-125E The power required to pump a specified rate of water to a specified elevation is to be determined. Properties The density of water is taken to be 62.4 lbm/ft3 (Table A-3E). Analysis The required power is determined from W& = m& g ( z 2 − z1 ) = ρV&g ( z 2 − z1 ) ⎛ 35.315 ft 3 /s ⎞ 1 lbf ⎞ ⎟(32.174 ft/s 2 )(300 ft)⎛⎜ = (62.4 lbm/ft 3 )(200 gal/min)⎜⎜ ⎟ 2 ⎟ ⎝ 32.174 lbm ⋅ ft/s ⎠ ⎝ 15,850 gal/min ⎠ 1 kW ⎛ ⎞ = 8342 lbf ⋅ ft/s = (8342 lbf ⋅ ft/s)⎜ ⎟ = 11.3 kW ⎝ 737.56 lbf ⋅ ft/s ⎠

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2-126 The power that could be produced by a water wheel is to be determined. Properties The density of water is taken to be 1000 m3/kg (Table A-3). Analysis The power production is determined from

W& = m& g ( z 2 − z1 ) = ρV&g ( z 2 − z1 ) ⎛ 1 kJ/kg ⎞ = (1000 kg/m 3 )(0.400/60 m 3 /s)(9.81 m/s 2 )(10 m)⎜ ⎟ = 0.654 kW ⎝ 1000 m 2 /s 2 ⎠

2-127 The flow of air through a flow channel is considered. The diameter of the wind channel downstream from the rotor and the power produced by the windmill are to be determined. Analysis The specific volume of the air is

v=

RT (0.287 kPa ⋅ m 3 /kg ⋅ K)(293 K) = = 0.8409 m 3 /kg P 100 kPa

The diameter of the wind channel downstream from the rotor is A1V1 = A2V 2 ⎯ ⎯→(πD12 / 4)V1 = (πD 22 / 4)V 2 ⎯ ⎯→ D 2 = D1

V1 10 m/s = (7 m) = 7.38 m 9 m/s V2

The mass flow rate through the wind mill is m& =

A1V1

v

=

π (7 m) 2 (10 m/s) 4(0.8409 m 3 /kg)

= 457.7 kg/s

The power produced is then

V 2 − V 22 (10 m/s) 2 − (9 m/s) 2 ⎛ 1 kJ/kg ⎞ = (457.7 kg/s) W& = m& 1 ⎜ ⎟ = 4.35 kW 2 2 ⎝ 1000 m 2 /s 2 ⎠

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2-128 The available head, flow rate, and efficiency of a hydroelectric turbine are given. The electric power output is to be determined. Assumptions 1 The flow is steady. 2 Water levels at the reservoir and the discharge site remain constant. 3 Frictional losses in piping are negligible. Properties We take the density of water to be ρ = 1000 kg/m3 = 1 kg/L. Analysis The total mechanical energy the water in a dam possesses is equivalent to the potential energy of water at the free surface of the dam (relative to free surface of discharge water), and it can be converted to work entirely. Therefore, the power potential of water is its potential energy, which is gz per unit mass, and m& gz for a given mass flow rate.

1

120 m

⎛ 1 kJ/kg ⎞ emech = pe = gz = (9.81 m/s 2 )(120 m)⎜ ⎟ = 1.177 kJ/kg ⎝ 1000 m 2 /s 2 ⎠

ηoverall = 80%

Generator

Turbin 2

The mass flow rate is

m& = ρV& = (1000 kg/m3 )(100 m3/s) = 100,000 kg/s Then the maximum and actual electric power generation become ⎛ 1 MW ⎞ W&max = E& mech = m& emech = (100,000 kg/s)(1.177 kJ/kg)⎜ ⎟ = 117.7 MW ⎝ 1000 kJ/s ⎠ W& =η W& = 0.80(117.7 MW) = 94.2 MW electric

overall

max

Discussion Note that the power generation would increase by more than 1 MW for each percentage point improvement in the efficiency of the turbine–generator unit.

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2-129 An entrepreneur is to build a large reservoir above the lake level, and pump water from the lake to the reservoir at night using cheap power, and let the water flow from the reservoir back to the lake during the day, producing power. The potential revenue this system can generate per year is to be determined. Assumptions 1 The flow in each direction is steady and incompressible. 2 The elevation difference between the lake and the reservoir can be taken to be constant, and the elevation change of reservoir during charging and discharging is disregarded. 3 Frictional losses in piping are negligible. 4 The system operates every day of the year for 10 hours in each mode. Properties We take the density of water to be ρ = 1000 kg/m3. Analysis The total mechanical energy of water in an upper reservoir relative to water in a lower reservoir is equivalent to the potential energy of water at the free surface of this reservoir relative to free surface of the lower reservoir. Therefore, the power potential of water is its potential energy, which is gz per unit mass, and m& gz for a given mass flow rate. This also represents the minimum power required to pump water from the lower reservoir to the higher reservoir.

2 Reservoir

Pumpturbine

40 m Lake

1

W&max, turbine = W&min, pump = W&ideal = ΔE& mech = m& Δemech = m& Δpe = m& gΔz = ρV&gΔz ⎛ 1N ⎞⎛ 1 kW ⎞ ⎟ = (1000 kg/m3 )(2 m3/s)(9.81 m/s2 )(40 m)⎜⎜ ⎟ 2 ⎟⎜ ⋅ 1000 N m/s ⋅ ⎠ ⎝ 1 kg m/s ⎠⎝ = 784.8 kW The actual pump and turbine electric powers are W& pump, elect =

W& ideal

η pump -motor

=

784.8 kW = 1046 kW 0.75

W& turbine = η turbine -gen W& ideal = 0.75(784.8 kW) = 588.6 kW

Then the power consumption cost of the pump, the revenue generated by the turbine, and the net income (revenue minus cost) per year become Cost = W& pump, elect Δt × Unit price = (1046 kW)(365 × 10 h/year)($ 0.03/kWh) = $114,500/y ear

Reveue = W& turbine Δt × Unit price = (588.6 kW)(365× 10 h/year)($0.08/kWh) = $171,900/year Net income = Revenue – Cost = 171,900 –114,500 = $57,400/year Discussion It appears that this pump-turbine system has a potential to generate net revenues of about $57,000 per year. A decision on such a system will depend on the initial cost of the system, its life, the operating and maintenance costs, the interest rate, and the length of the contract period, among other things.

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2-130 A diesel engine burning light diesel fuel that contains sulfur is considered. The rate of sulfur that ends up in the exhaust and the rate of sulfurous acid given off to the environment are to be determined. Assumptions 1 All of the sulfur in the fuel ends up in the exhaust. 2 For one kmol of sulfur in the exhaust, one kmol of sulfurous acid is added to the environment. Properties The molar mass of sulfur is 32 kg/kmol. Analysis The mass flow rates of fuel and the sulfur in the exhaust are m& fuel =

m& air (336 kg air/h) = = 18.67 kg fuel/h AF (18 kg air/kg fuel)

m& Sulfur = (750 × 10 -6 )m& fuel = (750 × 10 -6 )(18.67 kg/h) = 0.014 kg/h The rate of sulfurous acid given off to the environment is m& H2SO3 =

M H2SO3 2 × 1 + 32 + 3 × 16 m& Sulfur = (0.014 kg/h) = 0.036 kg/h M Sulfur 32

Discussion This problem shows why the sulfur percentage in diesel fuel must be below certain value to satisfy regulations.

2-131 Lead is a very toxic engine emission. Leaded gasoline contains lead that ends up in the exhaust. The amount of lead put out to the atmosphere per year for a given city is to be determined. Assumptions 35% of lead is exhausted to the environment. Analysis The gasoline consumption and the lead emission are

Gasoline Consumption = (10,000 cars)(15,000 km/car - year)(10 L/100 km) = 1.5 × 107 L/year Lead Emission = (GaolineConsumption)mlead f lead = (1.5 × 107 L/year)(0.15 × 10-3 kg/L)(0.35) = 788 kg/year Discussion Note that a huge amount of lead emission is avoided by the use of unleaded gasoline.

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Fundamentals of Engineering (FE) Exam Problems

2-132 A 2-kW electric resistance heater in a room is turned on and kept on for 30 min. The amount of energy transferred to the room by the heater is (a) 1 kJ

(b) 60 kJ

(c) 1800 kJ

(d) 3600 kJ

(e) 7200 kJ

Answer (d) 3600 kJ

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). We= 2 "kJ/s" time=30*60 "s" We_total=We*time "kJ" "Some Wrong Solutions with Common Mistakes:" W1_Etotal=We*time/60 "using minutes instead of s" W2_Etotal=We "ignoring time"

2-133 In a hot summer day, the air in a well-sealed room is circulated by a 0.50-hp (shaft) fan driven by a 65% efficient motor. (Note that the motor delivers 0.50 hp of net shaft power to the fan). The rate of energy supply from the fan-motor assembly to the room is (a) 0.769 kJ/s

(b) 0.325 kJ/s

(c) 0.574 kJ/s

(d) 0.373 kJ/s

(e) 0.242 kJ/s

Answer (c) 0.574 kJ/s

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). Eff=0.65 W_fan=0.50*0.7457 "kW" E=W_fan/Eff "kJ/s" "Some Wrong Solutions with Common Mistakes:" W1_E=W_fan*Eff "Multiplying by efficiency" W2_E=W_fan "Ignoring efficiency" W3_E=W_fan/Eff/0.7457 "Using hp instead of kW"

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2-134 A fan is to accelerate quiescent air to a velocity to 12 m/s at a rate of 3 m3/min. If the density of air is 1.15 kg/m3, the minimum power that must be supplied to the fan is (a) 248 W

(b) 72 W

(c) 497 W

(d) 216 W

(e) 162 W

Answer (a) 248 W

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). rho=1.15 V=12 Vdot=3 "m3/s" mdot=rho*Vdot "kg/s" We=mdot*V^2/2 "Some Wrong Solutions with Common Mistakes:" W1_We=Vdot*V^2/2 "Using volume flow rate" W2_We=mdot*V^2 "forgetting the 2" W3_We=V^2/2 "not using mass flow rate"

2-135 A 900-kg car cruising at a constant speed of 60 km/h is to accelerate to 100 km/h in 6 s. The additional power needed to achieve this acceleration is (a) 41 kW

(b) 222 kW

(c) 1.7 kW

(d) 26 kW

(e) 37 kW

Answer (e) 37 kW

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m=900 "kg" V1=60 "km/h" V2=100 "km/h" Dt=6 "s" Wa=m*((V2/3.6)^2-(V1/3.6)^2)/2000/Dt "kW" "Some Wrong Solutions with Common Mistakes:" W1_Wa=((V2/3.6)^2-(V1/3.6)^2)/2/Dt "Not using mass" W2_Wa=m*((V2)^2-(V1)^2)/2000/Dt "Not using conversion factor" W3_Wa=m*((V2/3.6)^2-(V1/3.6)^2)/2000 "Not using time interval" W4_Wa=m*((V2/3.6)-(V1/3.6))/1000/Dt "Using velocities"

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2-136 The elevator of a large building is to raise a net mass of 400 kg at a constant speed of 12 m/s using an electric motor. Minimum power rating of the motor should be (a) 0 kW

(b) 4.8 kW

(c) 47 kW

(d) 12 kW

(e) 36 kW

Answer (c) 47 kW

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m=400 "kg" V=12 "m/s" g=9.81 "m/s2" Wg=m*g*V/1000 "kW" "Some Wrong Solutions with Common Mistakes:" W1_Wg=m*V "Not using g" W2_Wg=m*g*V^2/2000 "Using kinetic energy" W3_Wg=m*g/V "Using wrong relation"

2-137 Electric power is to be generated in a hydroelectric power plant that receives water at a rate of 70 m3/s from an elevation of 65 m using a turbine–generator with an efficiency of 85 percent. When frictional losses in piping are disregarded, the electric power output of this plant is (a) 3.9 MW

(b) 38 MW

(c) 45 MW

(d) 53 MW

(e) 65 MW

Answer (b) 38 MW

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). Vdot=70 "m3/s" z=65 "m" g=9.81 "m/s2" Eff=0.85 rho=1000 "kg/m3" We=rho*Vdot*g*z*Eff/10^6 "MW" "Some Wrong Solutions with Common Mistakes:" W1_We=rho*Vdot*z*Eff/10^6 "Not using g" W2_We=rho*Vdot*g*z/Eff/10^6 "Dividing by efficiency" W3_We=rho*Vdot*g*z/10^6 "Not using efficiency"

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2-138 A 75 hp (shaft) compressor in a facility that operates at full load for 2500 hours a year is powered by an electric motor that has an efficiency of 88 percent. If the unit cost of electricity is $0.06/kWh, the annual electricity cost of this compressor is (a) $7382

(b) $9900

(c) $12,780

(d) $9533

(e) $8389

Answer (d) $9533

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). Wcomp=75 "hp" Hours=2500 “h/year” Eff=0.88 price=0.06 “$/kWh” We=Wcomp*0.7457*Hours/Eff Cost=We*price "Some Wrong Solutions with Common Mistakes:" W1_cost= Wcomp*0.7457*Hours*price*Eff “multiplying by efficiency” W2_cost= Wcomp*Hours*price/Eff “not using conversion” W3_cost= Wcomp*Hours*price*Eff “multiplying by efficiency and not using conversion” W4_cost= Wcomp*0.7457*Hours*price “Not using efficiency”

2-139 Consider a refrigerator that consumes 320 W of electric power when it is running. If the refrigerator runs only one quarter of the time and the unit cost of electricity is $0.09/kWh, the electricity cost of this refrigerator per month (30 days) is (a) $3.56

(b) $5.18

(c) $8.54

(d) $9.28

(e) $20.74

Answer (b) $5.18

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). We=0.320 "kW" Hours=0.25*(24*30) "h/year" price=0.09 "$/kWh" Cost=We*hours*price "Some Wrong Solutions with Common Mistakes:" W1_cost= We*24*30*price "running continuously"

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2-140 A 2-kW pump is used to pump kerosene (ρ = 0.820 kg/L) from a tank on the ground to a tank at a higher elevation. Both tanks are open to the atmosphere, and the elevation difference between the free surfaces of the tanks is 30 m. The maximum volume flow rate of kerosene is (a) 8.3 L/s

(b) 7.2 L/s

(c) 6.8 L/s

(d) 12.1 L/s

(e) 17.8 L/s

Answer (a) 8.3 L/s

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). W=2 "kW" rho=0.820 "kg/L" z=30 "m" g=9.81 "m/s2" W=rho*Vdot*g*z/1000 "Some Wrong Solutions with Common Mistakes:" W=W1_Vdot*g*z/1000 "Not using density"

2-141 A glycerin pump is powered by a 5-kW electric motor. The pressure differential between the outlet and the inlet of the pump at full load is measured to be 211 kPa. If the flow rate through the pump is 18 L/s and the changes in elevation and the flow velocity across the pump are negligible, the overall efficiency of the pump is (a) 69%

(b) 72%

(c) 76%

(d) 79%

(e) 82%

Answer (c) 76%

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). We=5 "kW" Vdot= 0.018 "m3/s" DP=211 "kPa" Emech=Vdot*DP Emech=Eff*We

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2-73

The following problems are based on the optional special topic of heat transfer

2-142 A 10-cm high and 20-cm wide circuit board houses on its surface 100 closely spaced chips, each generating heat at a rate of 0.08 W and transferring it by convection to the surrounding air at 40°C. Heat transfer from the back surface of the board is negligible. If the convection heat transfer coefficient on the surface of the board is 10 W/m2.°C and radiation heat transfer is negligible, the average surface temperature of the chips is (a) 80°C

(b) 54°C

(c) 41°C

(d) 72°C

(e) 60°C

Answer (a) 80°C

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). A=0.10*0.20 "m^2" Q= 100*0.08 "W" Tair=40 "C" h=10 "W/m^2.C" Q= h*A*(Ts-Tair) "W" "Some Wrong Solutions with Common Mistakes:" Q= h*(W1_Ts-Tair) "Not using area" Q= h*2*A*(W2_Ts-Tair) "Using both sides of surfaces" Q= h*A*(W3_Ts+Tair) "Adding temperatures instead of subtracting" Q/100= h*A*(W4_Ts-Tair) "Considering 1 chip only"

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2-74

2-143 A 50-cm-long, 0.2-cm-diameter electric resistance wire submerged in water is used to determine the boiling heat transfer coefficient in water at 1 atm experimentally. The surface temperature of the wire is measured to be 130°C when a wattmeter indicates the electric power consumption to be 4.1 kW. Then the heat transfer coefficient is (a) 43,500 W/m2.°C

(b) 137 W/m2.°C

(c) 68,330 W/m2.°C

(d) 10,038 W/m2.°C

(e) 37,540 W/m2.°C Answer (a) 43,500 W/m2.°C

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). L=0.5 "m" D=0.002 "m" A=pi*D*L "m^2" We=4.1 "kW" Ts=130 "C" Tf=100 "C (Boiling temperature of water at 1 atm)" We= h*A*(Ts-Tf) "W" "Some Wrong Solutions with Common Mistakes:" We= W1_h*(Ts-Tf) "Not using area" We= W2_h*(L*pi*D^2/4)*(Ts-Tf) "Using volume instead of area" We= W3_h*A*Ts "Using Ts instead of temp difference"

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2-75

2-144 A 3-m2 hot black surface at 80°C is losing heat to the surrounding air at 25°C by convection with a convection heat transfer coefficient of 12 W/m2.°C, and by radiation to the surrounding surfaces at 15°C. The total rate of heat loss from the surface is (a) 1987 W

(b) 2239 W

(c) 2348 W

(d) 3451 W

(e) 3811 W

Answer (d) 3451 W

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). sigma=5.67E-8 "W/m^2.K^4" eps=1 A=3 "m^2" h_conv=12 "W/m^2.C" Ts=80 "C" Tf=25 "C" Tsurr=15 "C" Q_conv=h_conv*A*(Ts-Tf) "W" Q_rad=eps*sigma*A*((Ts+273)^4-(Tsurr+273)^4) "W" Q_total=Q_conv+Q_rad "W" "Some Wrong Solutions with Common Mistakes:" W1_Ql=Q_conv "Ignoring radiation" W2_Q=Q_rad "ignoring convection" W3_Q=Q_conv+eps*sigma*A*(Ts^4-Tsurr^4) "Using C in radiation calculations" W4_Q=Q_total/A "not using area"

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2-76

2-145 Heat is transferred steadily through a 0.2-m thick 8 m by 4 m wall at a rate of 1.6 kW. The inner and outer surface temperatures of the wall are measured to be 15°C to 5°C. The average thermal conductivity of the wall is (a) 0.001 W/m.°C

(b) 0.5 W/m.°C

(c) 1.0 W/m.°C

(d) 2.0 W/m.°C

(e) 5.0 W/m.°C

Answer (c) 1.0 W/m.°C

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). A=8*4 "m^2" L=0.2 "m" T1=15 "C" T2=5 "C" Q=1600 "W" Q=k*A*(T1-T2)/L "W" "Some Wrong Solutions with Common Mistakes:" Q=W1_k*(T1-T2)/L "Not using area" Q=W2_k*2*A*(T1-T2)/L "Using areas of both surfaces" Q=W3_k*A*(T1+T2)/L "Adding temperatures instead of subtracting" Q=W4_k*A*L*(T1-T2) "Multiplying by thickness instead of dividing by it"

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2-77

2-146 The roof of an electrically heated house is 7 m long, 10 m wide, and 0.25 m thick. It is made of a flat layer of concrete whose thermal conductivity is 0.92 W/m.°C. During a certain winter night, the temperatures of the inner and outer surfaces of the roof are measured to be 15°C and 4°C, respectively. The average rate of heat loss through the roof that night was (a) 41 W

(b) 177 W

(c) 4894 W

(d) 5567 W

(e) 2834 W

Answer (e) 2834 W

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). A=7*10 "m^2" L=0.25 "m" k=0.92 "W/m.C" T1=15 "C" T2=4 "C" Q_cond=k*A*(T1-T2)/L "W" "Some Wrong Solutions with Common Mistakes:" W1_Q=k*(T1-T2)/L "Not using area" W2_Q=k*2*A*(T1-T2)/L "Using areas of both surfaces" W3_Q=k*A*(T1+T2)/L "Adding temperatures instead of subtracting" W4_Q=k*A*L*(T1-T2) "Multiplying by thickness instead of dividing by it"

2-147 … 2-153 Design and Essay Problems

KJ

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3-1

Chapter 3 PROPERTIES OF PURE SUBSTANCES Pure Substances, Phase Change Processes, Property Diagrams 3-1C Yes, since the chemical composition throughout the tank remain the same. 3-2C A liquid that is about to vaporize is saturated liquid; otherwise it is compressed liquid. 3-3C A vapor that is about to condense is saturated vapor; otherwise it is superheated vapor. 3-4C No. 3-5C The temperature will also increase since the boiling or saturation temperature of a pure substance depends on pressure. 3-6C Because one cannot be varied while holding the other constant. In other words, when one changes, so does the other one. 3-7C At critical point the saturated liquid and the saturated vapor states are identical. At triple point the three phases of a pure substance coexist in equilibrium. 3-8C Yes. 3-9C Case (c) when the pan is covered with a heavy lid. Because the heavier the lid, the greater the pressure in the pan, and thus the greater the cooking temperature. 3-10C At supercritical pressures, there is no distinct phase change process. The liquid uniformly and gradually expands into a vapor. At subcritical pressures, there is always a distinct surface between the phases.

Property Tables 3-11C A perfectly fitting pot and its lid often stick after cooking as a result of the vacuum created inside as the temperature and thus the corresponding saturation pressure inside the pan drops. An easy way of removing the lid is to reheat the food. When the temperature rises to boiling level, the pressure rises to atmospheric value and thus the lid will come right off.

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3-12C The molar mass of gasoline (C8H18) is 114 kg/kmol, which is much larger than the molar mass of air that is 29 kg/kmol. Therefore, the gasoline vapor will settle down instead of rising even if it is at a much higher temperature than the surrounding air. As a result, the warm mixture of air and gasoline on top of an open gasoline will most likely settle down instead of rising in a cooler environment 3-13C Ice can be made by evacuating the air in a water tank. During evacuation, vapor is also thrown out, and thus the vapor pressure in the tank drops, causing a difference between the vapor pressures at the water surface and in the tank. This pressure difference is the driving force of vaporization, and forces the liquid to evaporate. But the liquid must absorb the heat of vaporization before it can vaporize, and it absorbs it from the liquid and the air in the neighborhood, causing the temperature in the tank to drop. The process continues until water starts freezing. The process can be made more efficient by insulating the tank well so that the entire heat of vaporization comes essentially from the water. 3-14C Yes. Otherwise we can create energy by alternately vaporizing and condensing a substance. 3-15C No. Because in the thermodynamic analysis we deal with the changes in properties; and the changes are independent of the selected reference state. 3-16C The term hfg represents the amount of energy needed to vaporize a unit mass of saturated liquid at a specified temperature or pressure. It can be determined from hfg = hg - hf . 3-17C Yes; the higher the temperature the lower the hfg value. 3-18C Quality is the fraction of vapor in a saturated liquid-vapor mixture. It has no meaning in the superheated vapor region. 3-19C Completely vaporizing 1 kg of saturated liquid at 1 atm pressure since the higher the pressure, the lower the hfg . 3-20C Yes. It decreases with increasing pressure and becomes zero at the critical pressure. 3-21C No. Quality is a mass ratio, and it is not identical to the volume ratio. 3-22C The compressed liquid can be approximated as a saturated liquid at the given temperature. Thus v T ,P ≅ v f @ T .

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3-3

3-23 [Also solved by EES on enclosed CD] Complete the following table for H2 O: T, °C

P, kPa

v, m3 / kg

Phase description

50

12.352

4.16

Saturated mixture

120.21

200

0.8858

Saturated vapor

250

400

0.5952

Superheated vapor

110

600

0.001051

Compressed liquid

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3-4

3-24 EES Problem 3-23 is reconsidered. The missing properties of water are to be determined using EES, and the solution is to be repeated for refrigerant-134a, refrigerant-22, and ammonia. Analysis The problem is solved using EES, and the solution is given below. $Warning off {$Arrays off} Procedure Find(Fluid$,Prop1$,Prop2$,Value1,Value2:T,p,h,s,v,u,x,State$) "Due to the very general nature of this problem, a large number of 'if-then-else' statements are necessary." If Prop1$='Temperature, C' Then T=Value1 If Prop2$='Temperature, C' then Call Error('Both properties cannot be Temperature, T=xxxF2',T) if Prop2$='Pressure, kPa' then p=value2 h=enthalpy(Fluid$,T=T,P=p) s=entropy(Fluid$,T=T,P=p) v=volume(Fluid$,T=T,P=p) u=intenergy(Fluid$,T=T,P=p) x=quality(Fluid$,T=T,P=p) endif if Prop2$='Enthalpy, kJ/kg' then h=value2 p=Pressure(Fluid$,T=T,h=h) s=entropy(Fluid$,T=T,h=h) v=volume(Fluid$,T=T,h=h) u=intenergy(Fluid$,T=T,h=h) x=quality(Fluid$,T=T,h=h) endif if Prop2$='Entropy, kJ/kg-K' then s=value2 p=Pressure(Fluid$,T=T,s=s) h=enthalpy(Fluid$,T=T,s=s) v=volume(Fluid$,T=T,s=s) u=intenergy(Fluid$,T=T,s=s) x=quality(Fluid$,T=T,s=s) endif if Prop2$='Volume, m^3/kg' then v=value2 p=Pressure(Fluid$,T=T,v=v) h=enthalpy(Fluid$,T=T,v=v) s=entropy(Fluid$,T=T,v=v) u=intenergy(Fluid$,T=T,v=v) x=quality(Fluid$,T=T,v=v) endif if Prop2$='Internal Energy, kJ/kg' then u=value2 p=Pressure(Fluid$,T=T,u=u) h=enthalpy(Fluid$,T=T,u=u) s=entropy(Fluid$,T=T,u=u) v=volume(Fluid$,T=T,s=s) x=quality(Fluid$,T=T,u=u) endif if Prop2$='Quality' then

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3-5

x=value2 p=Pressure(Fluid$,T=T,x=x) h=enthalpy(Fluid$,T=T,x=x) s=entropy(Fluid$,T=T,x=x) v=volume(Fluid$,T=T,x=x) u=IntEnergy(Fluid$,T=T,x=x) endif Endif If Prop1$='Pressure, kPa' Then p=Value1 If Prop2$='Pressure, kPa' then Call Error('Both properties cannot be Pressure, p=xxxF2',p) if Prop2$='Temperature, C' then T=value2 h=enthalpy(Fluid$,T=T,P=p) s=entropy(Fluid$,T=T,P=p) v=volume(Fluid$,T=T,P=p) u=intenergy(Fluid$,T=T,P=p) x=quality(Fluid$,T=T,P=p) endif if Prop2$='Enthalpy, kJ/kg' then h=value2 T=Temperature(Fluid$,p=p,h=h) s=entropy(Fluid$,p=p,h=h) v=volume(Fluid$,p=p,h=h) u=intenergy(Fluid$,p=p,h=h) x=quality(Fluid$,p=p,h=h) endif if Prop2$='Entropy, kJ/kg-K' then s=value2 T=Temperature(Fluid$,p=p,s=s) h=enthalpy(Fluid$,p=p,s=s) v=volume(Fluid$,p=p,s=s) u=intenergy(Fluid$,p=p,s=s) x=quality(Fluid$,p=p,s=s) endif if Prop2$='Volume, m^3/kg' then v=value2 T=Temperature(Fluid$,p=p,v=v) h=enthalpy(Fluid$,p=p,v=v) s=entropy(Fluid$,p=p,v=v) u=intenergy(Fluid$,p=p,v=v) x=quality(Fluid$,p=p,v=v) endif if Prop2$='Internal Energy, kJ/kg' then u=value2 T=Temperature(Fluid$,p=p,u=u) h=enthalpy(Fluid$,p=p,u=u) s=entropy(Fluid$,p=p,u=u) v=volume(Fluid$,p=p,s=s) x=quality(Fluid$,p=p,u=u) endif if Prop2$='Quality' then x=value2 T=Temperature(Fluid$,p=p,x=x) h=enthalpy(Fluid$,p=p,x=x) s=entropy(Fluid$,p=p,x=x) PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-6

v=volume(Fluid$,p=p,x=x) u=IntEnergy(Fluid$,p=p,x=x) endif Endif If Prop1$='Enthalpy, kJ/kg' Then h=Value1 If Prop2$='Enthalpy, kJ/kg' then Call Error('Both properties cannot be Enthalpy, h=xxxF2',h) if Prop2$='Pressure, kPa' then p=value2 T=Temperature(Fluid$,h=h,P=p) s=entropy(Fluid$,h=h,P=p) v=volume(Fluid$,h=h,P=p) u=intenergy(Fluid$,h=h,P=p) x=quality(Fluid$,h=h,P=p) endif if Prop2$='Temperature, C' then T=value2 p=Pressure(Fluid$,T=T,h=h) s=entropy(Fluid$,T=T,h=h) v=volume(Fluid$,T=T,h=h) u=intenergy(Fluid$,T=T,h=h) x=quality(Fluid$,T=T,h=h) endif if Prop2$='Entropy, kJ/kg-K' then s=value2 p=Pressure(Fluid$,h=h,s=s) T=Temperature(Fluid$,h=h,s=s) v=volume(Fluid$,h=h,s=s) u=intenergy(Fluid$,h=h,s=s) x=quality(Fluid$,h=h,s=s) endif if Prop2$='Volume, m^3/kg' then v=value2 p=Pressure(Fluid$,h=h,v=v) T=Temperature(Fluid$,h=h,v=v) s=entropy(Fluid$,h=h,v=v) u=intenergy(Fluid$,h=h,v=v) x=quality(Fluid$,h=h,v=v) endif if Prop2$='Internal Energy, kJ/kg' then u=value2 p=Pressure(Fluid$,h=h,u=u) T=Temperature(Fluid$,h=h,u=u) s=entropy(Fluid$,h=h,u=u) v=volume(Fluid$,h=h,s=s) x=quality(Fluid$,h=h,u=u) endif if Prop2$='Quality' then x=value2 p=Pressure(Fluid$,h=h,x=x) T=Temperature(Fluid$,h=h,x=x) s=entropy(Fluid$,h=h,x=x) v=volume(Fluid$,h=h,x=x) u=IntEnergy(Fluid$,h=h,x=x) endif endif PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-7

If Prop1$='Entropy, kJ/kg-K' Then s=Value1 If Prop2$='Entropy, kJ/kg-K' then Call Error('Both properties cannot be Entrolpy, h=xxxF2',s) if Prop2$='Pressure, kPa' then p=value2 T=Temperature(Fluid$,s=s,P=p) h=enthalpy(Fluid$,s=s,P=p) v=volume(Fluid$,s=s,P=p) u=intenergy(Fluid$,s=s,P=p) x=quality(Fluid$,s=s,P=p) endif if Prop2$='Temperature, C' then T=value2 p=Pressure(Fluid$,T=T,s=s) h=enthalpy(Fluid$,T=T,s=s) v=volume(Fluid$,T=T,s=s) u=intenergy(Fluid$,T=T,s=s) x=quality(Fluid$,T=T,s=s) endif if Prop2$='Enthalpy, kJ/kg' then h=value2 p=Pressure(Fluid$,h=h,s=s) T=Temperature(Fluid$,h=h,s=s) v=volume(Fluid$,h=h,s=s) u=intenergy(Fluid$,h=h,s=s) x=quality(Fluid$,h=h,s=s) endif if Prop2$='Volume, m^3/kg' then v=value2 p=Pressure(Fluid$,s=s,v=v) T=Temperature(Fluid$,s=s,v=v) h=enthalpy(Fluid$,s=s,v=v) u=intenergy(Fluid$,s=s,v=v) x=quality(Fluid$,s=s,v=v) endif if Prop2$='Internal Energy, kJ/kg' then u=value2 p=Pressure(Fluid$,s=s,u=u) T=Temperature(Fluid$,s=s,u=u) h=enthalpy(Fluid$,s=s,u=u) v=volume(Fluid$,s=s,s=s) x=quality(Fluid$,s=s,u=u) endif if Prop2$='Quality' then x=value2 p=Pressure(Fluid$,s=s,x=x) T=Temperature(Fluid$,s=s,x=x) h=enthalpy(Fluid$,s=s,x=x) v=volume(Fluid$,s=s,x=x) u=IntEnergy(Fluid$,s=s,x=x) endif Endif if x1 then State$='in the superheated region.' If (x0) then State$='in the two-phase region.' If (x=1) then State$='a saturated vapor.' PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-8

if (x=0) then State$='a saturated liquid.' end "Input from the diagram window" {Fluid$='Steam' Prop1$='Temperature' Prop2$='Pressure' Value1=50 value2=101.3} Call Find(Fluid$,Prop1$,Prop2$,Value1,Value2:T,p,h,s,v,u,x,State$) T[1]=T ; p[1]=p ; h[1]=h ; s[1]=s ; v[1]=v ; u[1]=u ; x[1]=x "Array variables were used so the states can be plotted on property plots." ARRAYS TABLE x h P s T u v KJ/kg kPa kJ/kgK C KJ/kg m3/kg 2964.5 400 7.3804 250 2726.4 0.5952 100

Steam

700 600

T [C]

500 400 300

8600 kPa 2600 kPa

200

500 kPa

100 0 0,0

45 kPa

1,0

2,0

3,0

4,0

5,0

6,0

7,0

8,0

9,0

10,0

s [kJ/kg-K]

Steam

700 600

T [C]

500 400 300

8600 kPa 2600 kPa

200

500 kPa

100 0 10-4

45 kPa

10-3

10-2

10-1

100

101

102

103

v [m3/kg] PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-9

Steam

105

104 250 C

P [kPa]

103

170 C

110 C

102

75 C

101

100 10-3

10-2

10-1

100

101

102

v [m3/kg]

Steam

105

104 250 C

P [kPa]

103

170 C

110 C

102

75 C 1

10

100 0

500

1000

1500

2000

2500

3000

h [kJ/kg]

Steam

4000

h [kJ/kg]

8600 kPa

3500

2600 kPa

3000

500 kPa

2500

45 kPa

2000 1500 1000 500 0 0,0

1,0

2,0

3,0

4,0

5,0

6,0

7,0

8,0

9,0

10,0

s [kJ/kg-K]

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3-10

3-25E Complete the following table for H2 O: T, °F

P, psia

u, Btu / lbm

Phase description

300

67.03

782

Saturated mixture

267.22

40

236.02

Saturated liquid

500

120

1174.4

Superheated vapor

400

400

373.84

Compressed liquid

3-26E EES Problem 3-25E is reconsidered. The missing properties of water are to be determined using EES, and the solution is to be repeated for refrigerant-134a, refrigerant-22, and ammonia. Analysis The problem is solved using EES, and the solution is given below. "Given" T[1]=300 [F] u[1]=782 [Btu/lbm] P[2]=40 [psia] x[2]=0 T[3]=500 [F] P[3]=120 [psia] T[4]=400 [F] P[4]=420 [psia] "Analysis" Fluid$='steam_iapws' P[1]=pressure(Fluid$, T=T[1], u=u[1]) x[1]=quality(Fluid$, T=T[1], u=u[1]) T[2]=temperature(Fluid$, P=P[2], x=x[2]) u[2]=intenergy(Fluid$, P=P[2], x=x[2]) u[3]=intenergy(Fluid$, P=P[3], T=T[3]) x[3]=quality(Fluid$, P=P[3], T=T[3]) u[4]=intenergy(Fluid$, P=P[4], T=T[4]) x[4]=quality(Fluid$, P=P[4], T=T[4]) "x = 100 for superheated vapor and x = -100 for compressed liquid" Solution for steam T, ºF P, psia 300 67.028 267.2 40 500 120 400 400

x 0.6173 0 100 -100

u, Btu/lbm 782 236 1174 373.8

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3-11

3-27 Complete the following table for H2 O: T, °C

P, kPa

h, kJ / kg

x

Phase description

120.21

200

2045.8

0.7

Saturated mixture

140

361.53

1800

0.565

Saturated mixture

177.66

950

752.74

0.0

Saturated liquid

80

500

335.37

---

Compressed liquid

350.0

800

3162.2

---

Superheated vapor

3-28 Complete the following table for Refrigerant-134a:

T, °C

P, kPa

v, m3 / kg

Phase description

-8

320

0.0007569

Compressed liquid

30

770.64

0.015

Saturated mixture

-12.73

180

0.11041

Saturated vapor

80

600

0.044710

Superheated vapor

3-29 Complete the following table for Refrigerant-134a:

T, °C

P, kPa

u, kJ / kg

Phase description

20

572.07

95

Saturated mixture

-12

185.37

35.78

Saturated liquid

86.24

400

300

Superheated vapor

8

600

62.26

Compressed liquid

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3-12

3-30E Complete the following table for Refrigerant-134a:

T, °F

P, psia

h, Btu / lbm

x

Phase description

65.89

80

78

0.566

Saturated mixture

15

29.759

69.92

0.6

Saturated mixture

10

70

15.35

---

Compressed liquid

160

180

129.46

---

Superheated vapor

110

161.16

117.23

1.0

Saturated vapor

3-31 A piston-cylinder device contains R-134a at a specified state. Heat is transferred to R-134a. The final pressure, the volume change of the cylinder, and the enthalpy change are to be determined. Analysis (a) The final pressure is equal to the initial pressure, which is determined from

P2 = P1 = Patm +

mpg

πD 2 /4

= 88 kPa +

(12 kg)(9.81 m/s 2 ) ⎛⎜ 1 kN 2 ⎜ π (0.25 m) /4 ⎝ 1000 kg.m/s 2

⎞ ⎟ = 90.4 kPa ⎟ ⎠

(b) The specific volume and enthalpy of R-134a at the initial state of 90.4 kPa and -10°C and at the final state of 90.4 kPa and 15°C are (from EES)

v1 = 0.2302 m3/kg 3

v 2 = 0.2544 m /kg

h1 = 247.76 kJ/kg h2 = 268.16 kJ/kg

The initial and the final volumes and the volume change are

V1 = mv 1 = (0.85 kg)(0.2302 m 3 /kg) = 0.1957 m 3 V 2 = mv 2 = (0.85 kg)(0.2544 m 3 /kg) = 0.2162 m 3

R-134a 0.85 kg -10°C

Q

ΔV = 0.2162 − 0.1957 = 0.0205 m 3 (c) The total enthalpy change is determined from ΔH = m(h2 − h1 ) = (0.85 kg)(268.16 − 247.76) kJ/kg = 17.4 kJ/kg

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3-13

3-32E A rigid container that is filled with water is cooled. The initial temperature and final pressure are to be determined. Analysis The initial state is superheated vapor. The temperature is determined to be P1 = 250 psia

⎫ ⎬ T = 550°F 3 v 1 = 2.29 ft /lbm ⎭ 1

H2O 250 psia 1 lbm 2.29 ft3

(Table A - 6E)

This is a constant volume cooling process (v = V /m = constant). The final state is saturated mixture and thus the pressure is the saturation pressure at the final temperature:

P

1

T2 = 100°F

⎫ ⎬ P = Psat @ 100° F = 0.9505 psia (Table A - 4E) 3 v 2 = v 1 = 2.29 ft /lbm ⎭ 2

2

v

3-33 A piston-cylinder device that is filled with R-134a is heated. The final volume is to be determined. Analysis The initial specific volume is

v1 =

V1 m

=

0.14 m 3 = 0.14 m 3 /kg 1 kg

R-134a −26.4°C 1 kg 0.14 m3

This is a constant-pressure process. The initial state is determined to be a mixture, and thus the pressure is the saturation pressure at the given temperature

P1 = P2 = Psat @ -26.4°C = 100 kPa (Table A - 12)

P

The final state is superheated vapor and the specific volume is P2 = 100 kPa ⎫ 3 ⎬ v 2 = 0.30138 m /kg (Table A - 13) T2 = 100°C ⎭

The final volume is then

1

2

v

V 2 = mv 2 = (1 kg)(0.30138 m 3 /kg) = 0.30138 m 3

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3-14

3-34 Left chamber of a partitioned system contains water at a specified state while the right chamber is evacuated. The partition is now ruptured and heat is transferred from the water. The pressure at the final state is to be determined. Analysis The initial specific volume is

v1 =

V1 1.1989 m 3 = = 1.1989 m 3 /kg m 1 kg

At the final state, the water occupies three times the initial volume. Then,

Water 200 kPa 1 kg 1.1989 m3

Evacuated

v 2 = 3v 1 = 3(1.1989 m 3 /kg) = 3.5967 m 3 /kg Based on this specific volume and the final temperature, the final state is a saturated mixture and the pressure is

P2 = Psat @ 3°C = 0.768 kPa (Table A - 4)

3-35E A piston-cylinder device that is filled with water is cooled. The final pressure and volume of the water are to be determined. Analysis The initial specific volume is

v1 =

V1 m

=

2.3615 ft 3 = 2.3615 ft 3 /lbm 1 lbm

H2O 400°F 1 lbm 2.3615 ft3

This is a constant-pressure process. The initial state is determined to be superheated vapor and thus the pressure is determined to be T1 = 400°F ⎫ ⎬ P = P2 = 200 psia (Table A - 6E) 3 v 1 = 2.3615 ft /lbm ⎭ 1

The saturation temperature at 200 psia is 381.8°F. Since the final temperature is less than this temperature, the final state is compressed liquid. Using the incompressible liquid approximation,

v 2 = v f @ 100° F = 0.01613 ft 3 /lbm (Table A - 4E)

P

2

1

v

The final volume is then

V 2 = mv 2 = (1 lbm)(0.01613 ft 3 /lbm) = 0.01613 ft 3

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3-15

3-36 A piston-cylinder device that is filled with R-134a is heated. The final volume is to be determined. Analysis This is a constant pressure process. The initial specific volume is

v1 =

V 1.595 m 3 = = 0.1595 m 3 /kg m 10 kg

R-134a -26.4°C 10 kg 1.595 m3

The initial state is determined to be a mixture, and thus the pressure is the saturation pressure at the given temperature

P1 = Psat @ -26.4°C = 100 kPa (Table A - 12) The final state is superheated vapor and the specific volume is P2 = 100 kPa ⎫ 3 ⎬ v 2 = 0.30138 m /kg (Table A - 13) T2 = 100°C ⎭

P

1

2

The final volume is then

V 2 = mv 2 = (10 kg)(0.30138 m 3 /kg) = 3.0138 m 3

v

3-37 The internal energy of water at a specified state is to be determined. Analysis The state of water is superheated vapor. From the steam tables, P = 50 kPa ⎫ ⎬ u = 2660.0 kJ/kg (Table A - 6) T = 200°C ⎭

3-38 The specific volume of water at a specified state is to be determined using the incompressible liquid approximation and it is to be compared to the more accurate value. Analysis The state of water is compressed liquid. From the steam tables, P = 5 MPa ⎫ 3 ⎬ v = 0.001041 m /kg (Table A - 7) T = 100°C ⎭

Based upon the incompressible liquid approximation, P = 2 MPa ⎫ ⎬ v ≅v T = 100°C ⎭

f @ 100°C

= 0.001043 m 3 /kg (Table A - 4)

The error involved is Percent Error =

0.001043 − 0.001041 × 100 = 0.19% 0.001041

which is quite acceptable in most engineering calculations.

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3-16

3-39E The total internal energy and enthalpy of water in a container are to be determined. Analysis The specific volume is

v=

V m

=

2 ft 3 = 2 ft 3 /lbm 1 lbm

Water 100 psia 2 ft3

At this specific volume and the given pressure, the state is a saturated mixture. The quality, internal energy, and enthalpy at this state are (Table A-5E) x=

v −v f

=

(2 − 0.01774) ft 3 /lbm

= 0.4490 (4.4327 − 0.01774) ft 3 /lbm u = u f + xu fg = 298.19 + (0.4490)(807.29) = 660.7 Btu/lbm h = h f + xh fg = 298.51 + (0.4490)(888.99) = 697.7 Btu/lbm

v fg

The total internal energy and enthalpy are then

U = mu = (1 lbm)(660.7 Btu/lbm) = 660.7 Btu H = mh = (1 lbm)(697.7 Btu/lbm) = 697.7 Btu

3-40 The volume of a container that contains water at a specified state is to be determined. Analysis The specific volume is determined from steam tables by interpolation to be P = 100 kPa ⎫ 3 ⎬ v = 2.9172 m /kg (Table A - 6) T = 360°C ⎭

The volume of the container is then

V = mv = (3 kg)(2.9172 m 3 /kg) = 8.752 m 3

Water 3 kg 100 kPa 360°C

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3-17

3-41 A rigid container that is filled with R-134a is heated. The temperature and total enthalpy are to be determined at the initial and final states. Analysis This is a constant volume process. The specific volume is

v1 = v 2 =

V m

=

R-134a 300 kPa 10 kg 14 L

3

0.014 m = 0.0014 m 3 /kg 10 kg

The initial state is determined to be a mixture, and thus the temperature is the saturation temperature at the given pressure. From Table A-12 by interpolation

P

T1 = Tsat @ 300 kPa = 0.61°C

2

and x1 =

v1 −v f v fg

=

(0.0014 − 0.0007736) m 3 /kg 3

(0.067978 − 0.0007736) m /kg

= 0.009321

1

h1 = h f + x1 h fg = 52.67 + (0.009321)(198.13) = 54.52 kJ/kg

The total enthalpy is then H 1 = mh1 = (10 kg )(54.52 kJ/kg ) = 545.2 kJ

The final state is also saturated mixture. Repeating the calculations at this state,

T2 = Tsat @ 600 kPa = 21.55°C x2 =

v 2 −v f v fg

=

(0.0014 − 0.0008199) m 3 /kg (0.034295 − 0.0008199) m 3 /kg

= 0.01733

h2 = h f + x 2 h fg = 81.51 + (0.01733)(180.90) = 84.64 kJ/kg

H 2 = mh2 = (10 kg )(84.64 kJ/kg ) = 846.4 kJ

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v

3-18

3-42 A piston-cylinder device that is filled with R-134a is cooled at constant pressure. The final temperature and the change of total internal energy are to be determined. Analysis The initial specific volume is

v1 =

V m

=

12.322 m 3 = 0.12322 m 3 /kg 100 kg

R-134a 200 kPa 100 kg 12.322 m3

The initial state is superheated and the internal energy at this state is P1 = 200 kPa

⎫ ⎬ u = 263.08 kJ/kg (Table A - 13) v 1 = 0.12322 m /kg ⎭ 1 3

P

The final specific volume is

v2 =

v1 2

=

0.12322 m 3 / kg = 0.06161 m 3 /kg 2

2

1

This is a constant pressure process. The final state is determined to be saturated mixture whose temperature is

T2 = Tsat @ 200 kPa = −10.09°C (Table A - 12) The internal energy at the final state is (Table A-12) v 2 −v f (0.06161 − 0.0007533) m 3 /kg x2 = = = 0.6140 v fg (0.099867 − 0.0007533) m 3 /kg u 2 = u f + x 2 u fg = 38.28 + (0.6140)(186.21) = 152.61 kJ/kg

Hence, the change in the internal energy is Δu = u 2 − u1 = 152.61 − 263.08 = −110.47 kJ/kg

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v

3-19

3-43 A spring-loaded piston-cylinder device is filled with water. The water now undergoes a process until its volume is one-half of the original volume. The final temperature and the entropy are to be determined. Analysis From the steam tables, P1 = 4 MPa ⎫ 3 ⎬ v 1 = 0.07343 m /kg (Table A - 6) T1 = 400°C ⎭

The process experienced by this system is a linear P-v process. The equation for this line is P − P1 = c(v − v 1 )

where P1 is the system pressure when its specific volume is v1. The spring equation may be written as P − P1 =

Fs − Fs ,1 A

=k

x − x1 kA k km = 2 ( x − x1 ) = 2 (V −V1 ) = 2 (v − v 1 ) A A A A

Constant c is hence c=

km A

=

2

4 2 km 2

π D

4

=

(16)(90 kN/m)(0.5 kg)

π 2 (0.2 m) 4

= 45,595 kN ⋅ kg/m 5

The final pressure is then c ⎛v ⎞ P2 = P1 + c(v 2 − v 1 ) = P1 + c⎜⎜ 1 − v 1 ⎟⎟ = P1 − v 1 2 ⎝ 2 ⎠ 45,595 kN ⋅ kg/m 5 = 4000 kPa − (0.07343 m 3 /kg ) = 2326 kPa 2

and

v2 =

v1 2

=

0.07343 m 3 / kg = 0.03672 m 3 /kg 2

P

The final state is a mixture and the temperature is

1

T2 = Tsat @ 2326 kPa ≅ 220°C (Table A - 5)

2

The quality and the entropy at the final state are x2 =

v 2 −v f v fg

=

3

(0.03672 − 0.001190) m /kg (0.086094 − 0.001190) m 3 /kg

v

= 0.4185

h2 = h f + x 2 h fg = 943.55 + (0.4185)(1857.4) = 1720.9 kJ/kg

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3-20

3-44E The local atmospheric pressure, and thus the boiling temperature, changes with the weather conditions. The change in the boiling temperature corresponding to a change of 0.3 in of mercury in atmospheric pressure is to be determined. Properties The saturation pressures of water at 200 and 212°F are 11.538 and 14.709 psia, respectively (Table A-4E). One in. of mercury is equivalent to 1 inHg = 3.387 kPa = 0.491 psia (inner cover page). Analysis A change of 0.3 in of mercury in atmospheric pressure corresponds to ⎛ 0.491 psia ⎞ ⎟⎟ = 0.147 psia ΔP = (0.3 inHg)⎜⎜ ⎝ 1 inHg ⎠

P ± 0.3 inHg

At about boiling temperature, the change in boiling temperature per 1 psia change in pressure is determined using data at 200 and 212°F to be (212 − 200)°F ΔT = = 3.783 °F/psia ΔP (14.709 − 11.538) psia

Then the change in saturation (boiling) temperature corresponding to a change of 0.147 psia becomes

ΔTboiling = (3.783 °F/psia)ΔP = (3.783 °F/psia)(0.147 psia) = 0.56°F which is very small. Therefore, the effect of variation of atmospheric pressure on the boiling temperature is negligible.

3-45 A person cooks a meal in a pot that is covered with a well-fitting lid, and leaves the food to cool to the room temperature. It is to be determined if the lid will open or the pan will move up together with the lid when the person attempts to open the pan by lifting the lid up. Assumptions 1 The local atmospheric pressure is 1 atm = 101.325 kPa. 2 The weight of the lid is small and thus its effect on the boiling pressure and temperature is negligible. 3 No air has leaked into the pan during cooling. Properties The saturation pressure of water at 20°C is 2.3392 kPa (Table A-4). Analysis Noting that the weight of the lid is negligible, the reaction force F on the lid after cooling at the pan-lid interface can be determined from a force balance on the lid in the vertical direction to be

PA +F = PatmA or, F = A( Patm − P) = (πD 2 / 4)( Patm − P ) =

π (0.3 m) 2

P

2.3392 kPa

(101,325 − 2339.2) Pa

4 = 6997 m 2 Pa = 6997 N (since 1 Pa = 1 N/m 2 )

Patm = 1 atm

The weight of the pan and its contents is

W = mg = (8 kg)(9.81 m/s2 ) = 78.5 N which is much less than the reaction force of 6997 N at the pan-lid interface. Therefore, the pan will move up together with the lid when the person attempts to open the pan by lifting the lid up. In fact, it looks like the lid will not open even if the mass of the pan and its contents is several hundred kg.

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3-21

3-46 Water is boiled at 1 atm pressure in a pan placed on an electric burner. The water level drops by 10 cm in 45 min during boiling. The rate of heat transfer to the water is to be determined. Properties The properties of water at 1 atm and thus at a saturation temperature of Tsat = 100°C are hfg = 2256.5 kJ/kg and vf = 0.001043 m3/kg (Table A-4). Analysis The rate of evaporation of water is mevap = m& evap =

Vevap (πD 2 / 4) L [π (0.25 m)2 / 4](0.10 m) = = = 4.704 kg vf vf 0.001043 mevap Δt

=

H2O 1 atm

4.704 kg = 0.001742 kg/s 45 × 60 s

Then the rate of heat transfer to water becomes Q& = m& evap h fg = (0.001742 kg/s)(2256 .5 kJ/kg) = 3.93 kW

3-47 Water is boiled at a location where the atmospheric pressure is 79.5 kPa in a pan placed on an electric burner. The water level drops by 10 cm in 45 min during boiling. The rate of heat transfer to the water is to be determined. Properties The properties of water at 79.5 kPa are Tsat = 93.3°C, hfg = 2273.9 kJ/kg and vf = 0.001038 m3/kg (Table A-5). Analysis The rate of evaporation of water is m evap = m& evap =

V evap vf m evap Δt

= =

2

(πD / 4) L

vf

=

2

[π (0.25 m) / 4](0.10 m) = 4.727 kg 0.001038

H2O 79.5 kPa

4.727 kg = 0.001751 kg/s 45 × 60 s

Then the rate of heat transfer to water becomes Q& = m& evap h fg = (0.001751 kg/s)(2273 .9 kJ/kg) = 3.98 kW

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3-22

3-48 Saturated steam at Tsat = 30°C condenses on the outer surface of a cooling tube at a rate of 45 kg/h. The rate of heat transfer from the steam to the cooling water is to be determined. Assumptions 1 Steady operating conditions exist. 2 The condensate leaves the condenser as a saturated liquid at 30°C. Properties The properties of water at the saturation temperature of 30°C are hfg = 2429.8 kJ/kg (Table A4). Analysis Noting that 2429.8 kJ of heat is released as 1 kg of saturated vapor at 30°C condenses, the rate of heat transfer from the steam to the cooling water in the tube is determined directly from

30°C L = 35 m

Q& = m& evap h fg

D = 3 cm

= (45 kg/h)(2429.8 kJ/kg) = 109,341 kJ/h = 30.4 kW

3-49 The boiling temperature of water in a 5-cm deep pan is given. The boiling temperature in a 40-cm deep pan is to be determined. Assumptions Both pans are full of water. Properties The density of liquid water is approximately ρ = 1000 kg/m3. Analysis The pressure at the bottom of the 5-cm pan is the saturation pressure corresponding to the boiling temperature of 98°C: P = Psat@98o C = 94.39 kPa

40 cm 5 cm

(Table A-4)

The pressure difference between the bottoms of two pans is ⎛ 1 kPa ⎜ 1000 kg/m ⋅ s 2 ⎝

ΔP = ρ g h = (1000 kg/m 3 )(9.807 m/s 2 )(0.35 m)⎜

⎞ ⎟ = 3.43 kPa ⎟ ⎠

Then the pressure at the bottom of the 40-cm deep pan is P = 94.39 + 3.43 = 97.82 kPa Then the boiling temperature becomes

Tboiling = [email protected]

kPa

= 99.0°C

(Table A-5)

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3-23

3-50 A vertical piston-cylinder device is filled with water and covered with a 20-kg piston that serves as the lid. The boiling temperature of water is to be determined. Analysis The pressure in the cylinder is determined from a force balance on the piston,

PA = PatmA + W Patm

or, P = Patm +

mg A

= (100 kPa) + = 119.61 kPa

⎞ (20 kg)(9.81 m/s 2 ) ⎛ 1 kPa ⎜ ⎟ 2 2⎟ ⎜ 0.01 m ⎝ 1000 kg/m ⋅ s ⎠

P

W = mg

The boiling temperature is the saturation temperature corresponding to this pressure,

T = Tsat @119.61 kPa = 104.7°C

(Table A-5)

3-51 A rigid tank that is filled with saturated liquid-vapor mixture is heated. The temperature at which the liquid in the tank is completely vaporized is to be determined, and the T-v diagram is to be drawn. Analysis This is a constant volume process (v = V /m = constant),

H2O 75°C

and the specific volume is determined to be

v=

V m

=

2.5 m 3 = 0.1667 m 3 /kg 15 kg

When the liquid is completely vaporized the tank will contain saturated vapor only. Thus,

2

v 2 = v g = 0.1667 m 3 /kg The temperature at this point is the temperature that corresponds to this vg value, T = Tsat @v

g = 0.1667

m 3 /kg

= 187.0°C

T

(Table A-4)

1

v

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3-24

3-52 A rigid vessel is filled with refrigerant-134a. The total volume and the total internal energy are to be determined. Properties The properties of R-134a at the given state are (Table A-13).

P = 800 kPa ⎫ u = 327.87 kJ/kg T = 120 o C ⎬⎭ v = 0.037625 m 3 /kg Analysis The total volume and internal energy are determined from

V = mv = (2 kg)(0.037625 m 3 /kg) = 0.0753 m 3

R-134a 2 kg 800 kPa 120°C

U = mu = (2 kg)(327.87 kJ/kg) = 655.7 kJ

3-53 A rigid vessel contains R-134a at specified temperature. The pressure, total internal energy, and the volume of the liquid phase are to be determined. Analysis (a) The specific volume of the refrigerant is

v=

V m

=

0.5 m 3 = 0.05 m 3 /kg 10 kg

At -20°C, vf = 0.0007362 m3/kg and vg = 0.14729 m3/kg (Table A-11). Thus the tank contains saturated liquid-vapor mixture since vf < v < vg , and the pressure must be the saturation pressure at the specified temperature,

R-134a 10 kg -20°C

P = Psat @ − 20o C = 132.82 kPa

(b) The quality of the refrigerant-134a and its total internal energy are determined from x=

v −v f v fg

=

0.05 − 0.0007362 = 0.3361 0.14729 − 0.0007362

u = u f + xu fg = 25.39 + 0.3361× 193.45 = 90.42 kJ/kg U = mu = (10 kg)(90.42 kJ/kg) = 904.2 kJ (c) The mass of the liquid phase and its volume are determined from

m f = (1 − x)mt = (1 − 0.3361) × 10 = 6.639 kg

V f = m f v f = (6.639 kg)(0.0007362 m3/kg) = 0.00489 m 3

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3-25

3-54 [Also solved by EES on enclosed CD] A piston-cylinder device contains a saturated liquid-vapor mixture of water at 800 kPa pressure. The mixture is heated at constant pressure until the temperature rises to 350°C. The initial temperature, the total mass of water, the final volume are to be determined, and the Pv diagram is to be drawn. Analysis (a) Initially two phases coexist in equilibrium, thus we have a saturated liquid-vapor mixture. Then the temperature in the tank must be the saturation temperature at the specified pressure,

T = Tsat @800 kPa = 170.41°C (b) The total mass in this case can easily be determined by adding the mass of each phase, mf = mg =

Vf vf Vg vg

= =

0.1 m 3 0.001115 m 3 /kg 0.9 m

3

0.24035 m 3 /kg

= 89.704 kg P

= 3.745 kg 1

m t = m f + m g = 89.704 + 3.745 = 93.45 kg

(c) At the final state water is superheated vapor, and its specific volume is P2 = 800 kPa ⎫ 3 ⎬ v = 0.35442 m /kg T2 = 350 o C ⎭ 2

2

v

(Table A-6)

Then,

V 2 = mt v 2 = (93.45 kg)(0.35442 m 3 /kg) = 33.12 m 3

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-26

3-55 EES Problem 3-54 is reconsidered. The effect of pressure on the total mass of water in the tank as the pressure varies from 0.1 MPa to 1 MPa is to be investigated. The total mass of water is to be plotted against pressure, and results are to be discussed. Analysis The problem is solved using EES, and the solution is given below. P[1]=800 [kPa] P[2]=P[1] T[2]=350 [C] V_f1 = 0.1 [m^3] V_g1=0.9 [m^3] spvsat_f1=volume(Steam_iapws, P=P[1],x=0) "sat. liq. specific volume, m^3/kg" spvsat_g1=volume(Steam_iapws,P=P[1],x=1) "sat. vap. specific volume, m^3/kg" m_f1=V_f1/spvsat_f1 "sat. liq. mass, kg" m_g1=V_g1/spvsat_g1 "sat. vap. mass, kg" m_tot=m_f1+m_g1 V[1]=V_f1+V_g1 spvol[1]=V[1]/m_tot "specific volume1, m^3" T[1]=temperature(Steam_iapws, P=P[1],v=spvol[1])"C" "The final volume is calculated from the specific volume at the final T and P" spvol[2]=volume(Steam_iapws, P=P[2], T=T[2]) "specific volume2, m^3/kg" V[2]=m_tot*spvol[2] mtot [kg] 96.39 95.31 94.67 94.24 93.93 93.71 93.56 93.45 93.38 93.34

Steam

10 5

P1 [kPa] 100 200 300 400 500 600 700 800 900 1000

10 4

350 C

P [kPa]

10 3

1

2 P=800 kPa

10 2

10 1

10 0 10 -3

10 -2

10 -1

10 0

10 1

3

v [m /kg]

96.5 96

m tot [kg]

95.5 95 94.5 94 93.5 93 100

200

300

400

500

600

700

800

900

1000

P[1] [kPa]

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10 2

3-27

3-56E Superheated water vapor cools at constant volume until the temperature drops to 250°F. At the final state, the pressure, the quality, and the enthalpy are to be determined. Analysis This is a constant volume process (v = V/m = constant), and the initial specific volume is determined to be P1 = 180 psia ⎫ 3 ⎬ v = 3.0433 ft /lbm T1 = 500 o F ⎭ 1 3

(Table A-6E) H2O 180 psia 500°F

3

At 250°F, vf = 0.01700 ft /lbm and vg = 13.816 ft /lbm. Thus at the final state, the tank will contain saturated liquid-vapor mixture since vf < v < vg , and the final pressure must be the saturation pressure at the final temperature, P = Psat @ 250o F = 29.84 psia

T

1

(b) The quality at the final state is determined from x2 =

v 2 −v f v fg

=

3.0433 − 0.01700 = 0.219 13.816 − 0.01700

(c) The enthalpy at the final state is determined from

2 v

h = h f + xh fg = 218.63 + 0.219 × 945.41 = 426.0 Btu/lbm

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3-28

3-57E EES Problem 3-56E is reconsidered. The effect of initial pressure on the quality of water at the final state as the pressure varies from 100 psi to 300 psi is to be investigated. The quality is to be plotted against initial pressure, and the results are to be discussed. Analysis The problem is solved using EES, and the solution is given below. T[1]=500 [F] P[1]=180 [psia] T[2]=250 [F] v[ 1]=volume(steam_iapws,T=T[1],P=P[1]) v[2]=v[1] P[2]=pressure(steam_iapws,T=T[2],v=v[2]) h[2]=enthalpy(steam_iapws,T=T[2],v=v[2]) x[2]=quality(steam_iapws,T=T[2],v=v[2])

P1 [psia] 100 122.2 144.4 166.7 188.9 211.1 233.3 255.6 277.8 300

Steam

1400

x2 0.4037 0.3283 0.2761 0.2378 0.2084 0.1853 0.1665 0.1510 0.1379 0.1268

1200

1.21.31.4 1.5 Btu/lbm-R

T [°F]

1000 800 600 400

1600 psia 780 psia

1

180 psia

2

29.82 psia

200

0.050.1 0.2 0.5

0 10 -2

10 -1

10 0

10 1

10 2

10 3

10 4

3

v [ft /lb ] m

0.45 0.4

x[2]

0.35 0.3 0.25 0.2 0.15 0.1 100

140

180

220

260

300

P[1] [psia]

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3-29

3-58 A rigid vessel that contains a saturated liquid-vapor mixture is heated until it reaches the critical state. The mass of the liquid water and the volume occupied by the liquid at the initial state are to be determined. Analysis This is a constant volume process (v = V /m = constant) to the critical state, and thus the initial specific volume will be equal to the final specific volume, which is equal to the critical specific volume of water,

v 1 = v 2 = v cr = 0.003106 m 3 /kg

(last row of Table A-4)

The total mass is

T

0.3 m 3 V m= = = 96.60 kg v 0.003106 m 3 /kg

H2O 150°C

3

At 150°C, vf = 0.001091 m /kg and vg = 0.39248 m3/kg (Table A-4). Then the quality of water at the initial state is x1 =

v1 −v f v fg

CP

0.003106 − 0.001091 = = 0.005149 0.39248 − 0.001091

vcr

v

Then the mass of the liquid phase and its volume at the initial state are determined from

m f = (1 − x1 )mt = (1 − 0.005149)(96.60) = 96.10 kg

V f = m f v f = (96.10 kg)(0.001091 m3/kg) = 0.105 m 3

3-59 The properties of compressed liquid water at a specified state are to be determined using the compressed liquid tables, and also by using the saturated liquid approximation, and the results are to be compared. Analysis Compressed liquid can be approximated as saturated liquid at the given temperature. Then from Table A-4,

T = 100°C ⇒

v ≅ v f @ 100°C = 0.001043 m 3 /kg (0.72% error) u ≅ u f @ 100°C = 419.06 kJ/kg h ≅ h f @ 100°C = 419.17 kJ/kg

(1.02% error) (2.61% error)

From compressed liquid table (Table A-7),

v = 0.001036 m 3 /kg P = 15 MPa ⎫ u = 414.85 kJ/kg T = 100°C ⎬⎭ h = 430.39 kJ/kg The percent errors involved in the saturated liquid approximation are listed above in parentheses.

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3-30

3-60 EES Problem 3-59 is reconsidered. Using EES, the indicated properties of compressed liquid are to be determined, and they are to be compared to those obtained using the saturated liquid approximation. Analysis The problem is solved using EES, and the solution is given below. Fluid$='Steam_IAPWS' T = 100 [C] P = 15000 [kPa] v = VOLUME(Fluid$,T=T,P=P) u = INTENERGY(Fluid$,T=T,P=P) h = ENTHALPY(Fluid$,T=T,P=P) v_app = VOLUME(Fluid$,T=T,x=0) u_app = INTENERGY(Fluid$,T=T,x=0) h_app_1 = ENTHALPY(Fluid$,T=T,x=0) h_app_2 = ENTHALPY(Fluid$,T=T,x=0)+v_app*(P-pressure(Fluid$,T=T,x=0)) SOLUTION Fluid$='Steam_IAPWS' h=430.4 [kJ/kg] h_app_1=419.2 [kJ/kg] h_app_2=434.7 [kJ/kg] P=15000 [kPa] T=100 [C] u=414.9 [kJ/kg] u_app=419.1 [kJ/kg] v=0.001036 [m^3/kg] v_app=0.001043 [m^3/kg]

3-61 Superheated steam in a piston-cylinder device is cooled at constant pressure until half of the mass condenses. The final temperature and the volume change are to be determined, and the process should be shown on a T-v diagram. Analysis (b) At the final state the cylinder contains saturated liquid-vapor mixture, and thus the final temperature must be the saturation temperature at the final pressure,

T = Tsat@1 MPa = 179.88°C

H2O 300°C 1 MPa

(Table A-5)

(c) The quality at the final state is specified to be x2 = 0.5. The specific volumes at the initial and the final states are P1 = 1.0 MPa T1 = 300 o C

⎫ 3 ⎬ v 1 = 0.25799 m /kg ⎭

P2 = 1.0 MPa x2 = 0.5

⎫ ⎬ v 2 = v f + x2v fg ⎭ = 0.001127 + 0.5 × (0.19436 − 0.001127 ) = 0.09775 m3/kg

(Table A-6)

Thus,

T 1 2

v

ΔV = m(v 2 − v 1 ) = (0.8 kg)(0.09775 − 0.25799)m 3 /kg = −0.1282 m 3

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3-31

3-62 The water in a rigid tank is cooled until the vapor starts condensing. The initial pressure in the tank is to be determined. Analysis This is a constant volume process (v = V /m = constant), and the initial specific volume is equal to the final specific volume that is

T °C

v 1 = v 2 = v g @150°C = 0.39248 m 3 /kg (Table A-4) since the vapor starts condensing at 150°C. Then from Table A-6, T1 = 250°C ⎫ ⎬ P = 0.60 MPa v1 = 0.39248 m3/kg ⎭ 1

H2O T1= 250°C P1 = ?

1

25 15

2

v

3-63 Heat is supplied to a piston-cylinder device that contains water at a specified state. The volume of the tank, the final temperature and pressure, and the internal energy change of water are to be determined. Properties The saturated liquid properties of water at 200°C are: vf = 0.001157 m3/kg and uf = 850.46 kJ/kg (Table A-4). Analysis (a) The cylinder initially contains saturated liquid water. The volume of the cylinder at the initial state is

V1 = mv 1 = (1.4 kg)(0.001157 m 3 /kg) = 0.001619 m 3 The volume at the final state is

V = 4(0.001619) = 0.006476 m 3

Water 1.4 kg, 200°C sat. liq.

(b) The final state properties are

v2 =

V m

=

0.006476 m3 = 0.004626 m3 / kg 1.4 kg

v 2 = 0.004626 m3 / kg ⎫⎪ x2 = 1

Q

T2 = 371.3°C

⎬ P2 = 21,367 kPa ⎪⎭ u2 = 2201.5 kJ/kg

(Table A-4 or A-5 or EES)

(c) The total internal energy change is determined from ΔU = m(u 2 − u1 ) = (1.4 kg)(2201.5 - 850.46) kJ/kg = 1892 kJ

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3-32

3-64 Heat is lost from a piston-cylinder device that contains steam at a specified state. The initial temperature, the enthalpy change, and the final pressure and quality are to be determined. Analysis (a) The saturation temperature of steam at 3.5 MPa is

[email protected] MPa = 242.6°C (Table A-5) Then, the initial temperature becomes T1 = 242.6+5 = 247.6°C Also,

P1 = 3.5 MPa ⎫ ⎬h1 = 2821.1 kJ/kg T1 = 247.6°C ⎭

(Table A-6)

Steam 3.5 MPa

Q

(b) The properties of steam when the piston first hits the stops are P2 = P1 = 3.5 MPa ⎫ h2 = 1049.7 kJ/kg ⎬ 3 x2 = 0 ⎭ v 2 = 0.001235 m /kg

(Table A-5)

Then, the enthalpy change of steam becomes Δh = h2 − h1 = 1049.7 − 2821.1 = -1771 kJ/kg

(c) At the final state

v 3 = v 2 = 0.001235 m3/kg ⎫⎪ P3 = 1555 kPa T3 = 200°C

⎬ ⎪⎭ x3 = 0.0006

(Table A-4 or EES)

The cylinder contains saturated liquid-vapor mixture with a small mass of vapor at the final state.

3-65E The error involved in using the enthalpy of water by the incompressible liquid approximation is to be determined. Analysis The state of water is compressed liquid. From the steam tables, P = 1500 psia ⎫ ⎬ h = 376.51 Btu/lbm (Table A - 7E) T = 400°F ⎭

Based upon the incompressible liquid approximation, P = 1500 psia ⎫ ⎬ h ≅ h f @ 400° F = 375.04 Btu/lbm (Table A - 4E) T = 400°F ⎭

The error involved is Percent Error =

376.51 − 375.04 × 100 = 0.39% 376.51

which is quite acceptable in most engineering calculations.

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3-33

3-66 The errors involved in using the specific volume and enthalpy of water by the incompressible liquid approximation are to be determined. Analysis The state of water is compressed liquid. From the steam tables, P = 10 MPa ⎫ v = 0.0010385 m 3 /kg ⎬ T = 100°C ⎭ h = 426.62 kJ/kg

(Table A - 7)

Based upon the incompressible liquid approximation, P = 10 MPa ⎫ v ≅ v f @ 100°C = 0.001043 m 3 /kg (Table A - 4) ⎬ T = 100°C ⎭ h ≅ h f @ 100°C = 419.17 kJ/kg

The errors involved are Percent Error (specific volume) = Percent Error (enthalpy) =

0.001043 − 0.0010385 × 100 = 0.43% 0.0010385 426.62 − 419.17 × 100 = 1.75% 426.62

which are quite acceptable in most engineering calculations.

3-67 The specific volume and internal energy of R-134a at a specified state are to be determined. Analysis The state of R-134a is compressed liquid. Based upon the incompressible liquid approximation, P = 700 kPa ⎫ v ≅ v f @ 20°C = 0.0008161 m 3 /kg (Table A - 11) ⎬ T = 20°C ⎭ u ≅ u f @ 20°C = 78.86 kJ/kg

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3-34

3-68 A piston-cylinder device that is filled with R-134a is heated. The volume change is to be determined. Analysis The initial specific volume is P1 = 60 kPa ⎫ 3 ⎬ v 1 = 0.33608 m /kg (Table A - 13) T1 = −20°C ⎭

R-134a 60 kPa -20°C 100 g

and the initial volume is

V1 = mv 1 = (0.100 kg)(0.33608 m 3 /kg) = 0.033608 m 3 At the final state, we have

P

P2 = 60 kPa ⎫ 3 ⎬ v 2 = 0.50410 m /kg (Table A - 13) T2 = 100°C ⎭

1

V 2 = mv 2 = (0.100 kg)(0.50410 m 3 /kg) = 0.050410 m 3

2

The volume change is then

v

ΔV = V 2 −V1 = 0.050410 − 0.033608 = 0.0168 m 3

3-69 EES The Pessure-Enthalpy diagram of R-134a showing some constant-temperature and constantentropy lines are obtained using Property Plot feature of EES.

P [kPa]

1.2 k

1

0.8

0.5

0.3

0.2

104

J/k g-K

R134a

105

70°C

3

10

40°C 10°C -10°C

2

10

101 -100

-30°C

0

100

200

300

400

500

h [kJ/kg]

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3-35

Ideal Gas 3-70C Propane (molar mass = 44.1 kg/kmol) poses a greater fire danger than methane (molar mass = 16 kg/kmol) since propane is heavier than air (molar mass = 29 kg/kmol), and it will settle near the floor. Methane, on the other hand, is lighter than air and thus it will rise and leak out. 3-71C A gas can be treated as an ideal gas when it is at a high temperature or low pressure relative to its critical temperature and pressure. 3-72C Ru is the universal gas constant that is the same for all gases whereas R is the specific gas constant that is different for different gases. These two are related to each other by R = Ru / M, where M is the molar mass of the gas. 3-73C Mass m is simply the amount of matter; molar mass M is the mass of one mole in grams or the mass of one kmol in kilograms. These two are related to each other by m = NM, where N is the number of moles.

3-74E The specific volume of oxygen at a specified state is to be determined. Assumptions At specified conditions, oxygen behaves as an ideal gas. Properties The gas constant of oxygen is R = 0.3353 psia⋅ft3/lbm⋅R (Table A-1E). Analysis According to the ideal gas equation of state,

v=

RT (0.3353 psia ⋅ ft 3 /lbm ⋅ R)(80 + 460 R) = = 7.242 ft 3 /lbm P 25 psia

3-75 The pressure in a container that is filled with air is to be determined. Assumptions At specified conditions, air behaves as an ideal gas. Properties The gas constant of air is R = 0.287 kJ/kg⋅K (Table A-1). Analysis The definition of the specific volume gives

v=

V m

=

0.100 m 3 = 0.100 m 3 /kg 1 kg

Using the ideal gas equation of state, the pressure is P=

RT

v

=

(0.287 kPa ⋅ m 3 /kg ⋅ K)(27 + 273 K) 0.100 m 3 /kg

= 861 kPa

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3-36

3-76E The volume of a tank that is filled with argon at a specified state is to be determined. Assumptions At specified conditions, argon behaves as an ideal gas. Properties The gas constant of argon is R = 0.2686 psia⋅ft3/lbm⋅R (Table A-1E) Analysis According to the ideal gas equation of state,

V =

mRT (1 lbm)(0.2686 psia ⋅ ft 3 /lbm ⋅ R)(100 + 460 R) = = 0.7521 ft 3 P 200 psia

3-77 A balloon is filled with helium gas. The mole number and the mass of helium in the balloon are to be determined. Assumptions At specified conditions, helium behaves as an ideal gas. Properties The universal gas constant is Ru = 8.314 kPa.m3/kmol.K. The molar mass of helium is 4.0 kg/kmol (Table A-1). Analysis The volume of the sphere is

V =

4 3 4 π r = π (3 m) 3 = 113.1 m 3 3 3

Assuming ideal gas behavior, the mole numbers of He is determined from

N=

PV (200 kPa)(113.1 m3 ) = = 9.28 kmol RuT (8.314 kPa ⋅ m3/kmol ⋅ K)(293 K)

He D=6m 20°C 200 kPa

Then the mass of He can be determined from m = NM = (9.28 kmol)(4.0 kg/kmol) = 37.15 kg

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3-37

3-78 EES Problem 3-77 is to be reconsidered. The effect of the balloon diameter on the mass of helium contained in the balloon is to be determined for the pressures of (a) 100 kPa and (b) 200 kPa as the diameter varies from 5 m to 15 m. The mass of helium is to be plotted against the diameter for both cases. Analysis The problem is solved using EES, and the solution is given below. "Given Data" {D=6 [m]} {P=200 [kPa]} T=20 [C] P=200 [kPa] R_u=8.314 [kJ/kmol-K] "Solution" P*V=N*R_u*(T+273) V=4*pi*(D/2)^3/3 m=N*MOLARMASS(Helium)

D [m] 5 6.111 7.222 8.333 9.444 10.56 11.67 12.78 13.89 15

m [kg] 21.51 39.27 64.82 99.57 145 202.4 273.2 359 461 580.7

600

500

m [kg]

400

300

P=200 kPa

200

P=100 kPa

100

0 5

7

9

11

13

15

D [m]

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3-38

3-79 An automobile tire is inflated with air. The pressure rise of air in the tire when the tire is heated and the amount of air that must be bled off to reduce the temperature to the original value are to be determined. Assumptions 1 At specified conditions, air behaves as an ideal gas. 2 The volume of the tire remains constant. Properties The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A-1). Analysis Initially, the absolute pressure in the tire is

P1 = Pg + Patm = 210 + 100 = 310kPa

Tire 25°C

Treating air as an ideal gas and assuming the volume of the tire to remain constant, the final pressure in the tire can be determined from P1V1 P2V 2 T 323 K = ⎯ ⎯→ P2 = 2 P1 = (310 kPa) = 336 kPa T1 298 K T1 T2

Thus the pressure rise is ΔP = P2 − P1 = 336 − 310 = 26 kPa

The amount of air that needs to be bled off to restore pressure to its original value is m1 =

P1V (310 kPa)(0.025 m3 ) = = 0.0906 kg RT1 (0.287 kPa ⋅ m3/kg ⋅ K)(298 K)

m2 =

P1V (310 kPa)(0.025 m3 ) = = 0.0836 kg RT2 (0.287 kPa ⋅ m3/kg ⋅ K)(323 K) Δm = m1 − m2 = 0.0906 − 0.0836 = 0.0070 kg

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3-39

3-80 Two rigid tanks connected by a valve to each other contain air at specified conditions. The volume of the second tank and the final equilibrium pressure when the valve is opened are to be determined. Assumptions At specified conditions, air behaves as an ideal gas. Properties The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A-1). Analysis Let's call the first and the second tanks A and B. Treating air as an ideal gas, the volume of the second tank and the mass of air in the first tank are determined to be ⎛ m1RT1 ⎞ (5 kg)(0.287 kPa ⋅ m 3/kg ⋅ K)(308 K) ⎟ = = 2.21 m 3 ⎟ 200 kPa ⎝ P1 ⎠ B

V B = ⎜⎜

⎛ PV ⎞ (500 kPa)(1.0 m3 ) m A = ⎜⎜ 1 ⎟⎟ = = 5.846 kg 3 ⎝ RT1 ⎠ A (0.287 kPa ⋅ m /kg ⋅ K)(298 K)

A

Thus,

B

Air

V = V A + V B = 1.0 + 2.21 = 3.21 m

3

m = m A + mB = 5.846 + 5.0 = 10.846 kg

V = 1 m3

×

T = 25°C P = 500 kPa

Air m = 5 kg T = 35°C P = 200 kPa

Then the final equilibrium pressure becomes

P2 =

mRT2

V

=

(10.846 kg)(0.287 kPa ⋅ m3 /kg ⋅ K)(293 K) 3.21 m3

= 284.1 kPa

3-81E The validity of a statement that tires lose roughly 1 psi of pressure for every 10°F drop in outside temperature is to be investigated. Assumptions 1The air in the tire is an ideal gas. 2 The volume of air in the tire is constant. 3 The tire is in thermal equilibrium with the outside air. 4 The atmospheric conditions are 70°F and 1 atm = 14.7 psia. Analysis The pressure in a tire should be checked at least once a month when a vehicle has sat for at least one hour to ensure that the tires are cool. The recommended gage pressure in cool tires is typically above 30 psi. Taking the initial gage pressure to be 32 psi, the gage pressure after the outside temperature drops by 10°F is determined from the ideal gas relation to be

P1V P2V = T1 T2

→

P2 =

T2 (60 + 460) R P1 = (32 + 14.7 psia ) = 45.8 psia = 31.1 psig (gage) T1 (70 + 460) R

Then the drop in pressure corresponding to a drop of 10°F in temperature becomes ΔP = P1 − P2 = 32.0 − 31.1 = 0.9 psi

which is sufficiently close to 1 psi. Therefore, the statement is valid. Discussion Note that we used absolute temperatures and pressures in ideal gas calculations. Using gage pressures would result in pressure drop of 0.6 psi, which is considerably lower than 1 psi. Therefore, it is important to use absolute temperatures and pressures in the ideal gas relation.

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3-40

3-82 A piston-cylinder device containing oxygen is cooled. The change of the volume is to be determined. Assumptions At specified conditions, oxygen behaves as an ideal gas. Properties The gas constant of oxygen is R = 0.2598 kJ/kg⋅K (Table A-1). Analysis According to the ideal gas equation of state, the initial volume of the oxygen is

V1 =

mRT1 (0.010 kg)(0.2598 kPa ⋅ m 3 /kg ⋅ K)(100 + 273 K) = = 0.04845 m 3 20 kPa P1

Similarly, the final volume is

mRT2 (0.010 kg)(0.2598 kPa ⋅ m 3 /kg ⋅ K)(0 + 273 K) V2 = = = 0.03546 m 3 20 kPa P2 The change of volume is then

Oxygen 10 g 20 kPa 100°C

ΔV = V 2 −V1 = 0.03546 − 0.04845 = −0.013 m 3

3-83 A rigid vessel containing helium is heated. The temperature chang is to be determined. Assumptions At specified conditions, helium behaves as an ideal gas. Properties The gas constant of helium is R = 2.0769 kJ/kg⋅K (Table A-1). Analysis According to the ideal gas equation of state, the initial temperature is T1 =

3

P1V (350 kPa)(0.2 m ) = = 337 K mR (0.1 kg)(2.0769 kPa ⋅ m 3 /kg ⋅ K)

Helium 0.1 kg 0.2 m3 350 kPa

Since the specific volume remains constant, the ideal gas equation gives

v1 =

RT1 RT2 P 700 kPa =v2 = ⎯ ⎯→ T2 = T1 2 = (337 K ) = 674 K P1 P2 P1 350 kPa

The temperature change is then ΔT = T2 − T1 = 674 − 337 = 337 K

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3-41

3-84 A piston-cylinder device containing argon undergoes an isothermal process. The final pressure is to be determined. Assumptions At specified conditions, argon behaves as an ideal gas. Properties The gas constant of argon is R = 0.2081 kJ/kg⋅K (Table A-1). Analysis Since the temperature remains constant, the ideal gas equation gives PV PV m= 1 1 = 2 2 ⎯ ⎯→ P1V1 = P2V 2 RT RT

which when solved for final pressure becomes P2 = P1

Argon 0.2 kg 0.05 m3 400 kPa

V1 V = P1 1 = 0.5P1 = 0.5(400 kPa ) = 200 kPa V2 2V1

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3-42

Compressibility Factor

3-85C It represent the deviation from ideal gas behavior. The further away it is from 1, the more the gas deviates from ideal gas behavior. 3-86C All gases have the same compressibility factor Z at the same reduced temperature and pressure. 3-87C Reduced pressure is the pressure normalized with respect to the critical pressure; and reduced temperature is the temperature normalized with respect to the critical temperature.

3-88 The specific volume of steam is to be determined using the ideal gas relation, the compressibility chart, and the steam tables. The errors involved in the first two approaches are also to be determined. Properties The gas constant, the critical pressure, and the critical temperature of water are, from Table A-1,

R = 0.4615 kPa·m3/kg·K,

Tcr = 647.1 K,

Pcr = 22.06 MPa

Analysis (a) From the ideal gas equation of state,

v=

RT (0.4615 kPa ⋅ m 3 /kg ⋅ K)(673 K) = = 0.03106 m 3 /kg (17.6% error) P (10,000 kPa)

(b) From the compressibility chart (Fig. A-15), 10 MPa P ⎫ = = 0.453 ⎪ Pcr 22.06 MPa ⎪ ⎬ Z = 0.84 673 K T ⎪ TR = = = 1.04 ⎪⎭ Tcr 647.1 K PR =

H2O 10 MPa 400°C

Thus,

v = Zv ideal = (0.84)(0.03106 m 3 /kg) = 0.02609 m 3 /kg (1.2% error) (c) From the superheated steam table (Table A-6),

P = 10 MPa T = 400°C

} v = 0.02644 m /kg 3

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3-43

3-89 EES Problem 3-88 is reconsidered. The problem is to be solved using the general compressibility factor feature of EES (or other) software. The specific volume of water for the three cases at 10 MPa over the temperature range of 325°C to 600°C in 25°C intervals is to be compared, and the %error involved in the ideal gas approximation is to be plotted against temperature. Analysis The problem is solved using EES, and the solution is given below. P=10 [MPa]*Convert(MPa,kPa) {T_Celsius= 400 [C]} T=T_Celsius+273 "[K]" T_critical=T_CRIT(Steam_iapws) P_critical=P_CRIT(Steam_iapws) {v=Vol/m} P_table=P; P_comp=P;P_idealgas=P T_table=T; T_comp=T;T_idealgas=T v_table=volume(Steam_iapws,P=P_table,T=T_table) "EES data for steam as a real gas" {P_table=pressure(Steam_iapws, T=T_table,v=v)} {T_sat=temperature(Steam_iapws,P=P_table,v=v)} MM=MOLARMASS(water) R_u=8.314 [kJ/kmol-K] "Universal gas constant" R=R_u/MM "[kJ/kg-K], Particular gas constant" P_idealgas*v_idealgas=R*T_idealgas "Ideal gas equation" z = COMPRESS(T_comp/T_critical,P_comp/P_critical) P_comp*v_comp=z*R*T_comp "generalized Compressibility factor" Error_idealgas=Abs(v_table-v_idealgas)/v_table*Convert(, %) Error_comp=Abs(v_table-v_comp)/v_table*Convert(, %) Errorcomp [%] 6.088 2.422 0.7425 0.129 0.6015 0.8559 0.9832 1.034 1.037 1.01 0.9652 0.9093

Errorideal gas [%] 38.96 28.2 21.83 17.53 14.42 12.07 10.23 8.755 7.55 6.55 5.712 5

TCelcius [C] 325 350 375 400 425 450 475 500 525 550 575 600

40

Percent Error [%]

Specific Volum e

35

Steam at 10 MPa

30

Ideal Gas

25 Compressibility Factor 20 15 10 5 0 300

350

400

T

450

Celsius

500

550

600

[C]

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3-44

3-90 The specific volume of R-134a is to be determined using the ideal gas relation, the compressibility chart, and the R-134a tables. The errors involved in the first two approaches are also to be determined. Properties The gas constant, the critical pressure, and the critical temperature of refrigerant-134a are, from Table A-1,

R = 0.08149 kPa·m3/kg·K,

Tcr = 374.2 K,

Pcr = 4.059 MPa

Analysis (a) From the ideal gas equation of state,

v=

RT (0.08149 kPa ⋅ m3/kg ⋅ K)(343 K) = = 0.03105 m3 /kg P 900 kPa

(13.3% error )

(b) From the compressibility chart (Fig. A-15), 0.9 MPa P ⎫ = = 0.222 ⎪ Pcr 4.059 MPa ⎪ ⎬ Z = 0.894 343 K T TR = = = 0.917 ⎪ ⎪⎭ Tcr 374.2 K

R-134a 0.9 MPa

PR =

70°C

Thus,

v = Zv ideal = (0.894)(0.03105 m 3 /kg) = 0.02776 m 3 /kg

(1.3%error)

(c) From the superheated refrigerant table (Table A-13),

}

P = 0.9 MPa v = 0.027413 m3 /kg T = 70°C

3-91 The specific volume of nitrogen gas is to be determined using the ideal gas relation and the compressibility chart. The errors involved in these two approaches are also to be determined. Properties The gas constant, the critical pressure, and the critical temperature of nitrogen are, from Table A-1,

R = 0.2968 kPa·m3/kg·K,

Tcr = 126.2 K,

Pcr = 3.39 MPa

Analysis (a) From the ideal gas equation of state,

v=

RT (0.2968 kPa ⋅ m 3 /kg ⋅ K)(150 K) = = 0.004452 m 3 /kg P 10,000 kPa

(86.4% error)

(b) From the compressibility chart (Fig. A-15), 10 MPa P ⎫ = = 2.95 ⎪ Pcr 3.39 MPa ⎪ ⎬ Z = 0.54 150 K T TR = = = 1.19 ⎪ ⎪⎭ Tcr 126.2 K PR =

N2 10 MPa 150 K

Thus,

v = Zv ideal = (0.54)(0.004452 m 3 /kg) = 0.002404 m 3 /kg

(0.7% error)

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3-45

3-92 The specific volume of steam is to be determined using the ideal gas relation, the compressibility chart, and the steam tables. The errors involved in the first two approaches are also to be determined. Properties The gas constant, the critical pressure, and the critical temperature of water are, from Table A-1,

R = 0.4615 kPa·m3/kg·K,

Tcr = 647.1 K,

Pcr = 22.06 MPa

Analysis (a) From the ideal gas equation of state,

v=

RT (0.4615 kPa ⋅ m3/kg ⋅ K)(723 K) = = 0.09533 m3 /kg 3500 kPa P

(3.7% error)

(b) From the compressibility chart (Fig. A-15), P 3.5 MPa ⎫ = = 0.159 ⎪ Pcr 22.06 MPa ⎪ ⎬ Z = 0.961 T 723 K ⎪ TR = = = 1.12 ⎪⎭ Tcr 647.1 K

H2O 3.5 MPa

PR =

450°C

Thus,

v = Zv ideal = (0.961)(0.09533 m 3 /kg) = 0.09161 m 3 /kg

(0.4% error)

(c) From the superheated steam table (Table A-6),

}

P = 3.5 MPa v = 0.09196 m3 /kg T = 450°C

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3-46

3-93E Ethane in a rigid vessel is heated. The final pressure is to be determined using the compressibility chart. Properties The gas constant, the critical pressure, and the critical temperature of ethane are, from Table A1E,

R = 0.3574 psia·ft3/lbm·R,

Tcr = 549.8 R,

Pcr = 708 psia

Analysis From the compressibility chart at the initial state (Fig. A-15), ⎫ ⎪ ⎪ ⎬ Z1 = 0.977 P1 50 psia = = = 0.0706 ⎪ ⎪⎭ Pcr 708 psia

T R1 = PR1

T1 560 R = = 1.019 Tcr 549.8 R

Ethane 50 psia

Q

100°F

The specific volume does not change during the process. Then,

v1 = v 2 =

Z 1 RT1 (0.977)(0.3574 psia ⋅ ft 3 /lbm ⋅ R)(560 R) = = 3.911 ft 3 /lbm P1 50 psia

At the final state,

⎫ ⎪ ⎪ ⎬ Z 2 = 1.0 3 v 2,actual 3.911ft /lbm = = = 14.09 ⎪ ⎪ RTcr /Pcr (0.3574 psia ⋅ ft 3 /lbm ⋅ R)(549.8 R)/(708 psia) ⎭

TR 2 =

v R2

T2 1060 R = = 1.928 Tcr 549.8 R

Thus,

P2 =

Z 2 RT2

v2

=

(1.0)(0.3574 psia ⋅ ft 3 /lbm ⋅ R)(1060 R) 3.911 ft 3 /lbm

= 96.9 psia

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3-47

3-94 Ethylene is heated at constant pressure. The specific volume change of ethylene is to be determined using the compressibility chart. Properties The gas constant, the critical pressure, and the critical temperature of ethane are, from Table A1,

R = 0.2964 kPa·m3/kg·K,

Tcr = 282.4 K,

Pcr = 5.12 MPa

Analysis From the compressibility chart at the initial and final states (Fig. A-15), ⎫ ⎪ ⎪ ⎬ Z1 = 0.56 P1 5 MPa = = = 0.977 ⎪ ⎪⎭ Pcr 5.12 MPa

T R1 = PR1

T1 293 K = = 1.038 Tcr 282.4 K

T2 473 K ⎫ = = 1.675 ⎪ Tcr 282.4 KR ⎬ Z1 = 0.961 ⎪ = PR1 = 0.977 ⎭

TR 2 = PR 2

Ethylene 5 MPa 20°C

Q

The specific volume change is R ( Z 2 T2 − Z 1T1 ) P 0.2964 kPa ⋅ m 3 /kg ⋅ K [(0.961)(473 K) − (0.56)(293 K)] = 5000 kPa = 0.0172 m 3 /kg

Δv =

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3-48

3-95 Water vapor is heated at constant pressure. The final temperature is to be determined using ideal gas equation, the compressibility charts, and the steam tables. Properties The gas constant, the critical pressure, and the critical temperature of water are, from Table A-1,

R = 0.4615 kPa·m3/kg·K,

Tcr = 647.1 K,

Pcr = 22.06 MPa

Analysis (a) From the ideal gas equation, T2 = T1

v2 = (350 + 273 K )(2) = 1246 K v1

(b) The pressure of the steam is Water 350°C sat. vapor

P1 = P2 = Psat@350°C = 16,529 kPa From the compressibility chart at the initial state (Fig. A-15), ⎫ ⎪ ⎪ ⎬ Z1 = 0.593, v R1 = 0.75 P 16.529 MPa = 1 = = 0.749 ⎪ ⎪⎭ Pcr 22.06 MPa

T R1 = PR1

Q

T1 623 K = = 0.963 Tcr 647.1 KR

At the final state, PR 2 = PR1 = 0.749

⎫ ⎬ Z 2 = 0.88

v R 2 = 2v R1 = 2(0.75) = 1.50 ⎭ Thus, T2 =

P2v 2 P v T 16,529 kPa (1.50)(647.1 K) = 2 R 2 cr = = 826 K Z 2 R Z 2 Pcr 0.88 22,060 kPa

(c) From the superheated steam table, T1 = 350°C ⎫ 3 ⎬ v 1 = 0.008806 m /kg x1 = 1 ⎭ P2 = 16,529 kPa

(Table A-4)

⎫ ⎬ T = 477°C = 750 K v 2 = 2v 1 = 0.01761 m /kg ⎭ 2 3

(from Table A-6 or EES)

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3-49

3-96E Water vapor is heated at constant pressure. The final temperature is to be determined using ideal gas equation, the compressibility charts, and the steam tables. Properties The critical pressure and the critical temperature of water are, from Table A-1E,

R = 0.5956 psia·ft3/lbm·R,

Tcr = 1164.8 R,

Pcr = 3200 psia

Analysis (a) From the ideal gas equation, T2 = T1

v2 = (400 + 460 R )(2) = 1720 R v1

(b) The properties of steam are (Table A-4E)

Water 400°F sat. vapor

P1 = P2 = Psat@400° F = 247.26 psia

v 1 = v g@400°F = 1.8639 ft 3 /lbm

Q

v 2 = 2v 1 = 3.7278 ft 3 /lbm At the final state, from the compressibility chart (Fig. A-15), ⎫ ⎪ ⎪ ⎬ Z 2 = 0.985 3 v 2,actual 3.7278 ft /lbm = = = 17.19 ⎪ ⎪ RTcr /Pcr (0.5956 psia ⋅ ft 3 /lbm ⋅ R)(1164.8 R)/(3200 psia) ⎭

PR 2 =

v R2

P2 247.26 psia = = 0.0773 Pcr 3200 psia

Thus, T2 =

P2v 2 (247.26 psia)(3.7278 ft 3 /lbm) = = 1571 R Z 2 R (0.985)(0.5956 psia ⋅ ft 3 /lbm ⋅ R)

(c) From the superheated steam table, P2 = 247.26 psia

⎫ ⎬ T = 1100°F = 1560 R v 2 = 3.7278 ft /lbm ⎭ 2 3

(from Table A-6E or EES)

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3-50

3-97 Methane is heated at constant pressure. The final temperature is to be determined using ideal gas equation and the compressibility charts. Properties The gas constant, the critical pressure, and the critical temperature of methane are, from Table A-1,

R = 0.5182 kPa·m3/kg·K,

Tcr = 191.1 K,

Pcr = 4.64 MPa

Analysis From the ideal gas equation, T2 = T1

v2 = (300 K )(1.5) = 450 K v1

From the compressibility chart at the initial state (Fig. A-15), T R1 PR1

Methane 8 MPa 300 K

⎫ ⎪ ⎪ ⎬ Z1 = 0.88, v R1 = 0.80 P1 8 MPa = = = 1.724 ⎪ ⎪⎭ Pcr 4.64 MPa T 300 K = 1 = = 1.570 Tcr 191.1 K

Q

At the final state, PR 2 = PR1 = 1.724

v R 2 = 1.5v R1

⎫ ⎬ Z 2 = 0.975 = 1.5(0.80) = 1.2 ⎭

Thus, T2 =

P2v 2 P v T 8000 kPa (1.2)(191.1 K) = 2 R 2 cr = = 406 K Z 2 R Z 2 Pcr 0.975 4640 kPa

Of these two results, the accuracy of the second result is limited by the accuracy with which the charts may be read. Accepting the error associated with reading charts, the second temperature is the more accurate.

3-98 The percent error involved in treating CO2 at a specified state as an ideal gas is to be determined. Properties The critical pressure, and the critical temperature of CO2 are, from Table A-1, Tcr = 304.2K and Pcr = 7.39MPa

Analysis From the compressibility chart (Fig. A-15), P 3 MPa ⎫ = = 0.406 ⎪ Pcr 7.39 MPa ⎪ ⎬ Z = 0.80 T 283 K ⎪ TR = = = 0.93 ⎪⎭ Tcr 304.2 K PR =

CO2 3 MPa 10°C

Then the error involved in treating CO2 as an ideal gas is Error =

v − v ideal 1 1 = 1− = 1− = −0.25 or 25.0% v Z 0.80

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3-51

3-99 CO2 gas flows through a pipe. The volume flow rate and the density at the inlet and the volume flow rate at the exit of the pipe are to be determined.

3 MPa 500 K 2 kg/s

CO2

450 K

Properties The gas constant, the critical pressure, and the critical temperature of CO2 are (Table A-1)

R = 0.1889 kPa·m3/kg·K,

Tcr = 304.2 K,

Pcr = 7.39 MPa

Analysis (a) From the ideal gas equation of state,

V&1 =

m& RT1 (2 kg/s)(0.1889 kPa ⋅ m 3 /kg ⋅ K)(500 K) = = 0.06297 m 3 /kg (2.1% error) (3000 kPa) P1

ρ1 =

P1 (3000 kPa) = = 31.76 kg/m 3 (2.1% error) RT1 (0.1889 kPa ⋅ m 3 /kg ⋅ K)(500 K)

V&2 =

m& RT2 (2 kg/s)(0.1889 kPa ⋅ m 3 /kg ⋅ K)(450 K) = = 0.05667 m 3 /kg (3.6% error) (3000 kPa) P2

(b) From the compressibility chart (EES function for compressibility factor is used) P1 3 MPa ⎫ = = 0.407 ⎪ Pcr 7.39 MPa ⎪ ⎬ Z1 = 0.9791 T1 500 K = = = 1.64 ⎪ ⎪⎭ Tcr 304.2 K

PR = TR ,1

P2 3 MPa ⎫ = = 0.407 ⎪ Pcr 7.39 MPa ⎪ ⎬ Z 2 = 0.9656 T2 450 K = = = 1.48 ⎪ ⎪⎭ Tcr 304.2 K

PR = TR , 2

Thus,

V&1 =

Z 1 m& RT1 (0.9791)(2 kg/s)(0.1889 kPa ⋅ m 3 /kg ⋅ K)(500 K) = = 0.06165 m 3 /kg (3000 kPa) P1

ρ1 =

P1 (3000 kPa) = = 32.44 kg/m 3 Z 1 RT1 (0.9791)(0.1889 kPa ⋅ m 3 /kg ⋅ K)(500 K)

V&2 =

Z 2 m& RT2 (0.9656)(2 kg/s)(0.1889 kPa ⋅ m 3 /kg ⋅ K)(450 K) = = 0.05472 m 3 /kg (3000 kPa) P2

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3-52

Other Equations of State

3-100C The constant a represents the increase in pressure as a result of intermolecular forces; the constant b represents the volume occupied by the molecules. They are determined from the requirement that the critical isotherm has an inflection point at the critical point.

3-101E Carbon dioxide is heated in a constant pressure apparatus. The final volume of the carbon dioxide is to be determined using the ideal gas equation and the Benedict-Webb-Rubin equation of state. Properties The gas constant and molar mass of CO2 are (Table A-1E)

R = 0.2438 psia·ft3/lbm·R, M = 44.01 lbm/lbmol Analysis (a) From the ideal gas equation of state,

V2 =

mRT2 (1 lbm)(0.2438 psia ⋅ ft 3 /lbm ⋅ R)(1260 R) = = 0.3072 ft 3 1000 psia P

CO2 1000 psia 200°F

Q

(b) Using the coefficients of Table 3-4 for carbon dioxide and the given data in SI units, the Benedict-Webb-Rubin equation of state for state 2 is ⎛ C ⎞ 1 bR T − a aα c ⎛ γ ⎞ + ⎜⎜ B0 RuT2 − A0 − 02 ⎟⎟ 2 + u 23 + 6 + 3 2 ⎜1 + 2 ⎟ exp(−γ / v 2 ) v2 ⎝ T2 ⎠ v v v v T2 ⎝ v ⎠ (8.314)(700) ⎛⎜ 1.404 ×107 ⎞⎟ 1 0.007210× 8.314 × 700 − 13.86 6895 = + 0.04991× 8.314 × 700 − 277.30 − + ⎜ v2 7002 ⎟⎠ v 2 v3 ⎝ P2 =

+

RuT2

13.86 × 8.470 ×10−5

v6

+

1.511×106 ⎛ 0.00539 ⎞ 2 ⎜1 + ⎟ exp(−0.00539 / v ) v 3 (700)2 ⎝ v2 ⎠

The solution of this equation by an equation solver such as EES gives

v 2 = 0.8477 m 3 /kmol Then,

v2 =

v2 M

=

0.8477 m 3 /kmol = 0.01926 m 3 /kg 44.01 kg/kmol

V 2 = mv 2 = (1 / 2.2046 kg)(0.01927 m 3 /kg) = 0.008741 m 3 ⎛ 35.315 ft 3 = (0.008741 m 3 )⎜ ⎜ 1m3 ⎝

⎞ ⎟ = 0.3087 ft 3 ⎟ ⎠

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3-53

3-102 Methane is heated in a rigid container. The final pressure of the methane is to be determined using the ideal gas equation and the Benedict-Webb-Rubin equation of state. Analysis (a) From the ideal gas equation of state, P2 = P1

T2 673 K = (100 kPa) = 229.7 kPa T1 293 K

The specific molar volume of the methane is

v1 = v 2 =

Methane 100 kPa

Q

20°C

Ru T1 (8.314 kPa ⋅ m 3 /kmol ⋅ K)(293 K) = = 24.36 m 3 /kmol P1 100 kPa

(b) The specific molar volume of the methane is

v1 = v 2 =

Ru T1 (8.314 kPa ⋅ m 3 /kmol ⋅ K)(293 K) = = 24.36 m 3 /kmol P1 100 kPa

Using the coefficients of Table 3-4 for methane and the given data, the Benedict-Webb-Rubin equation of state for state 2 gives ⎛ γ ⎞ C ⎞ 1 bR T − a aα c ⎛ + 6 + 3 2 ⎜1 + 2 ⎟ exp(−γ / v 2 ) + ⎜⎜ B0 RuT2 − A0 − 02 ⎟⎟ 2 + u 23 v2 ⎝ v v v T2 ⎝ v ⎠ T2 ⎠ v (8.314)(673) ⎛⎜ 2.286 ×106 ⎞⎟ 1 0.003380× 8.314 × 673 − 5.00 + = + 0.04260× 8.314 × 673 − 187.91 − 2 2 ⎟ ⎜ 24.36 673 24.363 ⎠ 24.36 ⎝

P2 =

RuT2

5.00 ×1.244 ×10− 4 2.578 ×105 ⎛ 0.0060 ⎞ 2 + ⎜1 + ⎟ exp(−0.0060 / 24.36 ) 6 24.36 24.363 (673)2 ⎝ 24.362 ⎠ = 229.8 kPa +

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3-54

3-103E Carbon monoxide is heated in a rigid container. The final pressure of the CO is to be determined using the ideal gas equation and the Benedict-Webb-Rubin equation of state. Properties The gas constant and molar mass of CO are (Table A-1)

R = 0.2968 kPa·m3/kg·K, M = 28.011 kg/kmol Analysis (a) From the ideal gas equation of state, T 1260 R = 34.95 psia P2 = P1 2 = (14.7 psia) 530 R T1

CO 14.7 psia

Q

70°F

The specific molar volume of the CO in SI units is

v1 = v 2 =

Ru T1 (8.314 kPa ⋅ m 3 /kmol ⋅ K)(294 K) = = 24.20 m 3 /kmol P1 101 kPa

(b) The specific molar volume of the CO in SI units is

v1 = v 2 =

Ru T1 (8.314 kPa ⋅ m 3 /kmol ⋅ K)(294 K) = = 24.20 m 3 /kmol P1 101 kPa

Using the coefficients of Table 3-4 for CO and the given data, the Benedict-Webb-Rubin equation of state for state 2 gives P2 = =

RuT2

v2

⎛ C ⎞ 1 bR T − a aα c ⎛ γ ⎞ + ⎜⎜ B0 RuT2 − A0 − 02 ⎟⎟ 2 + u 23 + 6 + 3 2 ⎜1 + 2 ⎟ exp(−γ / v 2 ) T T v v v v v ⎝ ⎠ 2 ⎠ 2 ⎝

(8.314)(700) ⎛⎜ 8.673 ×105 ⎞⎟ 1 0.002632× 8.314 × 700 − 3.71 + + 0.05454× 8.314 × 700 − 135.87 − 2 2 ⎟ ⎜ 24.20 700 24.203 ⎠ 24.20 ⎝

3.71×1.350 ×10−4 1.054 ×105 ⎛ 0.0060 ⎞ 2 + ⎜1 + ⎟ exp(−0.0060 / 24.20 ) 24.206 24.203 (700)2 ⎝ 24.202 ⎠ = 240.8 kPa +

The pressure in English unit is ⎛ 1 psia ⎞ P2 = (240.8 kPa)⎜ ⎟ = 34.92 psia ⎝ 6.8948 kPa ⎠

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3-55

3-104 Carbon dioxide is compressed in a piston-cylinder device in a polytropic process. The final temperature is to be determined using the ideal gas and van der Waals equations. Properties The gas constant, molar mass, critical pressure, and critical temperature of carbon dioxide are (Table A-1)

R = 0.1889 kPa·m3/kg·K,

M = 44.01 kg/kmol,

Tcr = 304.2 K,

Pcr = 7.39 MPa

Analysis (a) The specific volume at the initial state is

v1 =

RT1 (0.1889 kPa ⋅ m 3 /kg ⋅ K)(473 K) = = 0.08935 m 3 /kg P1 1000 kPa

According to process specification, ⎛ P1 ⎝ P2

v 2 = v 1 ⎜⎜

⎞ ⎟⎟ ⎠

1/ n

⎛ 1000 kPa ⎞ = (0.08935 m 3 /kg)⎜ ⎟ ⎝ 3000 kPa ⎠

1 / 1.2

= 0.03577 m 3 /kg

CO2 1 MPa 200°C

The final temperature is then T2 =

P2v 2 (3000 kPa)(0.03577 m 3 /kg) = = 568 K R 0.1889 kPa ⋅ m 3 /kg ⋅ K

(b) The van der Waals constants for carbon dioxide are determined from a=

27 R 2 Tcr2 (27)(0.1889 kPa ⋅ m 3 /kg ⋅ K) 2 (304.2 K) 2 = = 0.1885 m 6 ⋅ kPa/kg 2 64 Pcr (64)(7390 kPa)

b=

RTcr (0.1889 kPa ⋅ m 3 /kg ⋅ K)(304.2 K) = = 0.0009720 m 3 /kg 8 Pcr 8 × 7390 kPa

Applying the van der Waals equation to the initial state, a ⎞ ⎛ ⎜ P + 2 ⎟(v − b) = RT v ⎠ ⎝ 0.1885 ⎞ ⎛ ⎜1000 + ⎟(v − 0.0009720) = (0.1889)(473) v2 ⎠ ⎝

Solving this equation by trial-error or by EES gives

v 1 = 0.08821 m 3 /kg According to process specification,

⎛ P1 ⎝ P2

v 2 = v 1 ⎜⎜

⎞ ⎟⎟ ⎠

1/ n

⎛ 1000 kPa ⎞ = (0.08821 m 3 /kg)⎜ ⎟ ⎝ 3000 kPa ⎠

1 / 1.2

= 0.03531 m 3 /kg

Applying the van der Waals equation to the final state, a ⎞ ⎛ ⎜ P + 2 ⎟(v − b) = RT v ⎠ ⎝ 0.1885 ⎞ ⎛ ⎜ 3000 + ⎟(0.03531 − 0.0009720) = (0.1889)T 0.035312 ⎠ ⎝

Solving for the final temperature gives T2 = 573 K

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3-56

3-105E The temperature of R-134a in a tank at a specified state is to be determined using the ideal gas relation, the van der Waals equation, and the refrigerant tables. Properties The gas constant, critical pressure, and critical temperature of R-134a are (Table A-1E)

R = 0.1052 psia·ft3/lbm·R,

Tcr = 673.6 R,

Pcr = 588. 7 psia

Analysis (a) From the ideal gas equation of state,

T=

Pv (100 psia)(0.54022 ft 3/lbm) = = 513.5 R R 0.1052 psia ⋅ ft 3/lbm ⋅ R

(b) The van der Waals constants for the refrigerant are determined from

Then,

a=

27 R 2Tcr2 (27)(0.1052 psia ⋅ ft 3 /lbm ⋅ R) 2 (673.6 R) 2 = = 3.591 ft 6 ⋅ psia/lbm 2 64 Pcr (64)(588.7 psia)

b=

RTcr (0.1052 psia ⋅ ft 3 /lbm ⋅ R)(673.6 R) = = 0.0150 ft 3 /lbm 8 Pcr 8 × 588.7 psia

T=

a ⎞ 1⎛ 1 ⎛ 3.591 ⎞ ⎟(0.54022 − 0.0150) = 560.7 R ⎜100 + ⎜ P + 2 ⎟(v − b ) = ⎜ R⎝ 0.1052 ⎝ v ⎠ (0.54022)2 ⎟⎠

(c) From the superheated refrigerant table (Table A-13E), P = 100 psia

⎫

v = 0.54022 ft 3/lbm ⎬⎭

T = 120°F (580R)

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3-57

3-106 [Also solved by EES on enclosed CD] The pressure of nitrogen in a tank at a specified state is to be determined using the ideal gas relation and the Beattie-Bridgeman equation. The error involved in each case is to be determined. Properties The gas constant and molar mass of nitrogen are (Table A-1)

R = 0.2968 kPa·m3/kg·K and M = 28.013 kg/kmol Analysis (a) From the ideal gas equation of state,

P=

RT

v

=

(0.2968 kPa ⋅ m3 /kg ⋅ K)(150 K) 0.041884 m3/kg

= 1063 kPa (6.3% error)

N2 0.041884 m3/kg 150 K

(b) The constants in the Beattie-Bridgeman equation are ⎛ a⎞ ⎛ 0.02617 ⎞ A = Ao ⎜1 − ⎟ = 136.2315⎜1 − ⎟ = 133.193 1.1733 ⎠ v ⎝ ⎠ ⎝ ⎛ b⎞ ⎛ − 0.00691 ⎞ B = Bo ⎜1 − ⎟ = 0.05046⎜1 − ⎟ = 0.05076 1.1733 ⎠ ⎝ v ⎠ ⎝ c = 4.2 × 10 4 m 3 ⋅ K 3 /kmol

since

v = Mv = (28.013 kg/kmol)(0.041884 m 3 /kg) = 1.1733 m 3 /kmol .

Substituting, RuT ⎛ c ⎞ A 8.314 × 150 ⎛⎜ 4.2 × 10 4 ⎞⎟ ( ) (1.1733 + 0.05076) − 133.1932 + − = − − v B 1 1 ⎜ ⎟ 2 3 2 2 ⎜ 3⎟ v ⎝ vT ⎠ v (1.1733) ⎝ 1.1733 × 150 ⎠ (1.1733) = 1000.4 kPa (negligible error)

P=

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3-58

3-107 EES Problem 3-106 is reconsidered. Using EES (or other) software, the pressure results of the ideal gas and Beattie-Bridgeman equations with nitrogen data supplied by EES are to be compared. The temperature is to be plotted versus specific volume for a pressure of 1000 kPa with respect to the saturated liquid and saturated vapor lines of nitrogen over the range of 110 K < T < 150 K. Analysis The problem is solved using EES, and the solution is given below. Function BeattBridg(T,v,M,R_u) v_bar=v*M "Conversion from m^3/kg to m^3/kmol" "The constants for the Beattie-Bridgeman equation of state are found in text" Ao=136.2315; aa=0.02617; Bo=0.05046; bb=-0.00691; cc=4.20*1E4 B=Bo*(1-bb/v_bar) A=Ao*(1-aa/v_bar) "The Beattie-Bridgeman equation of state is" BeattBridg:=R_u*T/(v_bar**2)*(1-cc/(v_bar*T**3))*(v_bar+B)-A/v_bar**2 End T=150 [K] v=0.041884 [m^3/kg] P_exper=1000 [kPa] T_table=T; T_BB=T;T_idealgas=T P_table=PRESSURE(Nitrogen,T=T_table,v=v) "EES data for nitrogen as a real gas" {T_table=temperature(Nitrogen, P=P_table,v=v)} M=MOLARMASS(Nitrogen) R_u=8.314 [kJ/kmol-K] "Universal gas constant" R=R_u/M "Particular gas constant" P_idealgas=R*T_idealgas/v "Ideal gas equation" P_BB=BeattBridg(T_BB,v,M,R_u) "Beattie-Bridgeman equation of state Function" PBB [kPa] 1000 1000 1000 1000 1000 1000 1000

Ptable [kPa] 1000 1000 1000 1000 1000 1000 1000 160

Pidealgas [kPa] 1000 1000 1000 1000 1000 1000 1000

v [m3/kg] 0.01 0.02 0.025 0.03 0.035 0.04 0.05

TBB [K] 91.23 95.52 105 116.8 130.1 144.4 174.6

Tideal gas [K] 33.69 67.39 84.23 101.1 117.9 134.8 168.5

Ttable [K] 103.8 103.8 106.1 117.2 130.1 144.3 174.5

Nitrogen, T vs v for P=1000 kPa Ideal Gas

150 140

Beattie-Bridgem an EES Table Value

T [K]

130 120 110 1000 kPa

100 90 80 70 10 -3

10 -2

10 -1

3

v [m /kg]

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3-59

Special Topic: Vapor Pressure and Phase Equilibrium 3-108 A glass of water is left in a room. The vapor pressures at the free surface of the water and in the room far from the glass are to be determined. Assumptions The water in the glass is at a uniform temperature. Properties The saturation pressure of water is 2.339 kPa at 20°C, and 1.706 kPa at 15°C (Table A-4). Analysis The vapor pressure at the water surface is the saturation pressure of water at the water temperature,

Pv, water surface = Psat @ Twater = Psat@15°C = 1.706 kPa Noting that the air in the room is not saturated, the vapor pressure in the room far from the glass is

H2O 15°C

Pv, air = φPsat @ Tair = φPsat@20°C = (0.6)(2.339 kPa) = 1.404 kPa

3-109 The vapor pressure in the air at the beach when the air temperature is 30°C is claimed to be 5.2 kPa. The validity of this claim is to be evaluated. Properties The saturation pressure of water at 30°C is 4.247 kPa (Table A-4). Analysis The maximum vapor pressure in the air is the saturation pressure of water at the given temperature, which is

30°C WATER

Pv, max = Psat @ Tair = Psat@30°C = 4.247 kPa which is less than the claimed value of 5.2 kPa. Therefore, the claim is false.

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3-60

3-110 The temperature and relative humidity of air over a swimming pool are given. The water temperature of the swimming pool when phase equilibrium conditions are established is to be determined. Assumptions The temperature and relative humidity of air over the pool remain constant. Properties The saturation pressure of water at 20°C is 2.339 kPa (Table A-4). Analysis The vapor pressure of air over the swimming pool is

Pv, air = φPsat @ Tair = φPsat@20°C = (0.4)(2.339 kPa) = 0.9357 kPa Phase equilibrium will be established when the vapor pressure at the water surface equals the vapor pressure of air far from the surface. Therefore,

Patm, 20°C

POOL

Pv , water surface = Pv , air = 0.9357 kPa

and

Twater = Tsat @ Pv = Tsat @ 0.9357 kPa = 6.0°C

Discussion Note that the water temperature drops to 6.0°C in an environment at 20°C when phase equilibrium is established.

3-111 Two rooms are identical except that they are maintained at different temperatures and relative humidities. The room that contains more moisture is to be determined. Properties The saturation pressure of water is 2.339 kPa at 20°C, and 4.247 kPa at 30°C (Table A-4). Analysis The vapor pressures in the two rooms are

Room 1:

Pv1 = φ1 Psat @ T1 = φ1 Psat@30°C = (0.4)(4.247 kPa) = 1.699 kPa

Room 2:

Pv 2 = φ 2 Psat @ T2 = φ 2 Psat@20°C = (0.7)(2.339 kPa) = 1.637 kPa

Therefore, room 1 at 30°C and 40% relative humidity contains more moisture.

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3-61

3-112E A thermos bottle half-filled with water is left open to air in a room at a specified temperature and pressure. The temperature of water when phase equilibrium is established is to be determined. Assumptions The temperature and relative humidity of air over the bottle remain constant. Properties The saturation pressure of water at 70°F is 0.3633 psia (Table A-4E). Analysis The vapor pressure of air in the room is

Pv, air = φPsat @ Tair = φPsat@70°F = (0.35)(0.3633 psia) = 0.1272 psia Phase equilibrium will be established when the vapor pressure at the water surface equals the vapor pressure of air far from the surface. Therefore, Pv , water surface = Pv , air = 0.1272 psia

Thermos bottle 70°F 35%

and

Twater = Tsat @ Pv = Tsat @ 0.1272 psia = 41.1°F Discussion Note that the water temperature drops to 41°F in an environment at 70°F when phase equilibrium is established.

3-113 A person buys a supposedly cold drink in a hot and humid summer day, yet no condensation occurs on the drink. The claim that the temperature of the drink is below 10°C is to be evaluated. Properties The saturation pressure of water at 35°C is 5.629 kPa (Table A-4). Analysis The vapor pressure of air is

Pv, air = φPsat @ Tair = φPsat@35°C = (0.7)(5.629 kPa) = 3.940 kPa

35°C 70%

The saturation temperature corresponding to this pressure (called the dew-point temperature) is

Tsat = Tsat @ Pv = [email protected] kPa = 28.7°C That is, the vapor in the air will condense at temperatures below 28.7°C. Noting that no condensation is observed on the can, the claim that the drink is at 10°C is false.

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3-62

Review Problems 3-114 The cylinder conditions before the heat addition process is specified. The pressure after the heat addition process is to be determined. Assumptions 1 The contents of cylinder are approximated by the air properties. 2 Air is an ideal gas. Analysis The final pressure may be determined from the ideal gas relation P2 =

Combustion chamber 1.8 MPa 450°C

T2 ⎛ 1300 + 273 K ⎞ P1 = ⎜ ⎟(1800 kPa) = 3916 kPa T1 ⎝ 450 + 273 K ⎠

3-115 A rigid tank contains an ideal gas at a specified state. The final temperature is to be determined for two different processes. Analysis (a) The first case is a constant volume process. When half of the gas is withdrawn from the tank, the final temperature may be determined from the ideal gas relation as T2 =

m1 P2 ⎛ 100 kPa ⎞ T1 = (2 )⎜ ⎟(600 K) = 400 K m 2 P1 ⎝ 300 kPa ⎠

(b) The second case is a constant volume and constant mass process. The ideal gas relation for this case yields P2 =

Ideal gas 300 kPa 600 K

T2 ⎛ 400 K ⎞ P1 = ⎜ ⎟(300 kPa) = 200 kPa T1 ⎝ 600 K ⎠

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3-63

3-116 Carbon dioxide flows through a pipe at a given state. The volume and mass flow rates and the density of CO2 at the given state and the volume flow rate at the exit of the pipe are to be determined. Analysis (a) The volume and mass flow rates may be determined from ideal gas relation as

3 MPa 500 K 0.4 kmol/s

CO2

450 K

V&1 =

N& Ru T1 (0.4 kmol/s)(8.314 kPa.m 3 /kmol.K)(500 K) = = 0.5543 m 3 /s 3000 kPa P

m& 1 =

(3000 kPa)(0.5543 m3 / s) P1V&1 = = 17.60 kg/s RT1 (0.1889 kPa.m3 /kg.K)(500 K)

The density is

ρ1 =

m& 1 (17.60 kg/s) = = 31.76 kg/m 3 3 & V1 (0.5543 m /s)

(b) The volume flow rate at the exit is

V&2 =

N& Ru T2 (0.4 kmol/s)(8.314 kPa.m 3 /kmol.K)(450 K) = = 0.4988 m 3 /s 3000 kPa P

3-117 The cylinder conditions before the heat addition process is specified. The temperature after the heat addition process is to be determined. Assumptions 1 The contents of cylinder is approximated by the air properties. 2 Air is an ideal gas. Analysis The ratio of the initial to the final mass is m1 AF 22 22 = = = m2 AF + 1 22 + 1 23

The final temperature may be determined from ideal gas relation T2 =

3 m1 V 2 ⎛ 22 ⎞⎛ 150 cm T1 = ⎜ ⎟⎜ m 2 V1 ⎝ 23 ⎠⎜⎝ 75 cm 3

Combustion chamber 950 K 75 cm3

⎞ ⎟(950 K) = 1817 K ⎟ ⎠

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3-64

3-118 A rigid container that is filled with R-13a is heated. The initial pressure and the final temperature are to be determined. Analysis The initial specific volume is 0.1450 m3/kg. Using this with the initial temperature reveals that the initial state is a mixture. The initial pressure is then the saturation pressure, T1 = −40°C ⎫ ⎬ P = Psat @ - 40°C = 51.25 kPa (Table A - 11) 3 v 1 = 0.1450 m /kg ⎭ 1

R-134a -40°C 1 kg 0.1450 m3 P

This is a constant volume cooling process (v = V /m = constant). The final state is superheated vapor and the final temperature is then

2

P2 = 200 kPa ⎫ ⎬ T = 90°C (Table A - 13) 3 v 2 = v 1 = 0.1450 m /kg ⎭ 2

1

v

3-119E A piston-cylinder device that is filled with water is cooled. The final pressure and volume of the water are to be determined. Analysis The initial specific volume is

v1 =

V1 m

=

2.649 ft 3 = 2.649 ft 3 /lbm 1 lbm

H2O 400°F 1 lbm 2.649 ft3

This is a constant-pressure process. The initial state is determined to be superheated vapor and thus the pressure is determined to be T1 = 400°F ⎫ ⎬ P = P2 = 180 psia (Table A - 6E) 3 v 1 = 2.649 ft /lbm ⎭ 1

The saturation temperature at 180 psia is 373.1°F. Since the final temperature is less than this temperature, the final state is compressed liquid. Using the incompressible liquid approximation,

P

2

1

v 2 = v f @ 100° F = 0.01613 ft 3 /lbm (Table A - 4E) The final volume is then

V 2 = mv 2 = (1 lbm)(0.01613 ft 3 /lbm) = 0.01613 ft 3

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v

3-65

3-120 The volume of chamber 1 of the two-piston cylinder shown in the figure is to be determined. Assumptions At specified conditions, helium behaves as an ideal gas. Properties The gas constant of helium is R = 2.0769 kJ/kg⋅K (Table A-1). Analysis Since the water vapor in chamber 2 is condensing, the pressure in this chamber is the saturation pressure,

P2 = Psat @ 200°C = 1555 kPa

P2A2

(Table A-4)

Summing the forces acting on the piston in the vertical direction gives ⎛D A P1 = P2 2 = P2 ⎜⎜ 2 A1 ⎝ D1

2

2

⎞ ⎛ 4⎞ ⎟⎟ = (1555 kPa)⎜ ⎟ = 248.8 kPa ⎝ 10 ⎠ ⎠

According to the ideal gas equation of state,

P1A1

mRT (1 kg)(2.0769 kPa ⋅ m 3 /kg ⋅ K)(200 + 273 K) V1 = = = 3.95 m 3 248.8 kPa P1

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3-66

3-121E The volume of chamber 1 of the two-piston cylinder shown in the figure is to be determined. Assumptions At specified conditions, air behaves as an ideal gas. Properties The gas constant of air is R = 0.3704 psia⋅ft3/lbm⋅R (Table A-1E). Analysis Since R-134a in chamber 2 is condensing, the pressure in this chamber is the saturation pressure,

P2 = Psat @ 120°F = 186.0 psia (Table A-11E) Summing the forces acting on the piston in the vertical direction gives F2 + F3 = F1 P2 A2 + P3 ( A1 − A2 ) = P1 A1

F2

which when solved for P1 gives P1 = P2

⎛ A ⎞ A2 + P3 ⎜⎜1 − 2 ⎟⎟ A1 A1 ⎠ ⎝

F3

since the areas of the piston faces are given by A = πD 2 / 4 the above equation becomes

⎛D P1 = P2 ⎜⎜ 2 ⎝ D1

2 ⎡ ⎛D ⎞ ⎟⎟ + P3 ⎢1 − ⎜⎜ 2 ⎢ ⎝ D1 ⎠ ⎣

⎞ ⎟⎟ ⎠

2⎤

⎥ ⎥ ⎦

F1

2 ⎡ ⎛ 2 ⎞2 ⎤ ⎛2⎞ = (186.0 psia)⎜ ⎟ + (30 psia) ⎢1 − ⎜ ⎟ ⎥ 3 ⎝3⎠ ⎣⎢ ⎝ ⎠ ⎦⎥

= 99.33 psia According to the ideal gas equation of state,

V1 =

mRT (0.5 lbm)(0.3704 psia ⋅ ft 3 /lbm ⋅ R)(120 + 460 R) = = 1.08 ft 3 99.33 psia P1

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3-67

3-122E The difference in the volume of chamber 1 for two cases of pressure in chamber 3 is to be determined. Assumptions At specified conditions, air behaves as an ideal gas. Properties The gas constant of air is R = 0.3704 psia⋅ft3/lbm⋅R (Table A-1). Analysis Since R-134a in chamber 2 is condensing, the pressure in this chamber is the saturation pressure,

P2 = Psat @ 120°F = 186.0 psia (Table A-11E) Summing the forces acting on the piston in the vertical direction gives F2 + F3 = F1 P2 A2 + P3 ( A1 − A2 ) = P1 A1

F2

which when solved for P1 gives P1 = P2

⎛ A ⎞ A2 + P3 ⎜⎜1 − 2 ⎟⎟ A1 A1 ⎠ ⎝

F3

since the areas of the piston faces are given by A = πD 2 / 4 the above equation becomes ⎛D P1 = P2 ⎜⎜ 2 ⎝ D1

2 ⎡ ⎛D ⎞ ⎟⎟ + P3 ⎢1 − ⎜⎜ 2 ⎢ ⎝ D1 ⎠ ⎣

⎞ ⎟⎟ ⎠

2⎤

⎥ ⎥ ⎦

F1

2 ⎡ ⎛ 2 ⎞2 ⎤ ⎛2⎞ = (186.0 psia)⎜ ⎟ + (100 kPa) ⎢1 − ⎜ ⎟ ⎥ 3 ⎝3⎠ ⎣⎢ ⎝ ⎠ ⎦⎥

= 138.2 psia According to the ideal gas equation of state,

V1 =

mRT (0.5 lbm)(0.3704 psia ⋅ ft 3 /lbm ⋅ R)(120 + 460 R) = = 0.777 ft 3 P1 138.2 psia

For a chamber 3 pressure of 30 psia, the volume of chamber 1 was determined to be 1.08 ft3. Then the change in the volume of chamber 1 is

ΔV = V 2 −V1 = 1.08 − 0.777 = 0.303 ft 3

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3-68

3-123 Ethane is heated at constant pressure. The final temperature is to be determined using ideal gas equation and the compressibility charts. Properties The gas constant, the critical pressure, and the critical temperature of ethane are, from Table A1,

R = 0.2765 kPa·m3/kg·K,

Tcr = 305.5 K,

Pcr = 4.48 MPa

Analysis From the ideal gas equation, T2 = T1

v2 = (373 K )(1.6) = 596.8 K v1

From the compressibility chart at the initial state (Fig. A-15), T R1 PR1

⎫ ⎪ ⎪ ⎬ Z1 = 0.61, v R1 = 0.35 P1 10 MPa = = = 2.232 ⎪ ⎪⎭ Pcr 4.48 MPa

T 373 K = 1 = = 1.221 Tcr 305.5 K

Ethane 10 MPa 100°C

Q

At the final state, PR 2 = PR1 = 2.232

v R 2 = 1.6v R1

⎫ ⎬ Z 2 = 0.83 = 1.6(0.35) = 0.56 ⎭

Thus, T2 =

P2v 2 P2 v R 2 Tcr 10,000 kPa (0.56)(305.5 K) = = = 460 K 0.83 4480 kPa Z 2 R Z 2 Pcr

Of these two results, the accuracy of the second result is limited by the accuracy with which the charts may be read. Accepting the error associated with reading charts, the second temperature is the more accurate.

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3-69

3-124 (a) On the P-v diagram, the constant temperature process through the state P= 300 kPa, v = 0.525 m3/kg as pressure changes from P1 = 200 kPa to P2 = 400 kPa is to be sketched. The value of the temperature on the process curve on the P-v diagram is to be placed.

SteamIAPWS

106

105

P [kPa]

104

103

102

400 300 200

2 133.5°C

1

101

0.525 100 10-4

10-3

10-2

10-1

100

101

102

3

v [m /kg]

(b) On the T-v diagram the constant specific vol-ume process through the state T = 120°C, v = 0.7163 m3/kg from P1= 100 kPa to P2 = 300 kPa is to be sketched.. For this data set, the temperature values at states 1 and 2 on its axis is to be placed. The value of the specific volume on its axis is also to be placed.

SteamIAPWS

700

600

300 kPa

T [°C]

500

198.7 kPa

400

100 kPa 300

200

2

100

1 0.7163

0 10-3

10-2

10-1

100

101

102

3

v [m /kg]

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3-70

3-125 The pressure in an automobile tire increases during a trip while its volume remains constant. The percent increase in the absolute temperature of the air in the tire is to be determined. Assumptions 1 The volume of the tire remains constant. 2 Air is an ideal gas. Properties The local atmospheric pressure is 90 kPa.

TIRE 200 kPa 0.035 m3

Analysis The absolute pressures in the tire before and after the trip are P1 = Pgage,1 + Patm = 200 + 90 = 290 kPa P2 = Pgage,2 + Patm = 220 + 90 = 310 kPa

Noting that air is an ideal gas and the volume is constant, the ratio of absolute temperatures after and before the trip are P1V1 P2V 2 T P 310 kPa = 1.069 = → 2 = 2 = T1 T2 T1 P1 290 kPa

Therefore, the absolute temperature of air in the tire will increase by 6.9% during this trip.

3-126 The rigid tank contains saturated liquid-vapor mixture of water. The mixture is heated until it exists in a single phase. For a given tank volume, it is to be determined if the final phase is a liquid or a vapor. Analysis This is a constant volume process (v = V /m = constant), and thus the final specific volume will be equal to the initial specific volume,

v 2 = v1 The critical specific volume of water is 0.003106 m3/kg. Thus if the final specific volume is smaller than this value, the water will exist as a liquid, otherwise as a vapor.

V = 4 L ⎯⎯→v =

V m

V = 400 L ⎯⎯→v =

=

V m

0.004 m3 = 0.002 m3/kg < v cr Thus, liquid. 2 kg =

H2O

V=4L m = 2 kg T = 50°C

0.4 m3 = 0.2 m3/kg > v cr . Thus, vapor. 2 kg

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3-71

3-127 Two rigid tanks that contain hydrogen at two different states are connected to each other. Now a valve is opened, and the two gases are allowed to mix while achieving thermal equilibrium with the surroundings. The final pressure in the tanks is to be determined. Properties The gas constant for hydrogen is 4.124 kPa·m3/kg·K (Table A-1). Analysis Let's call the first and the second tanks A and B. Treating H2 as an ideal gas, the total volume and the total mass of H2 are

A

V = V A + V B = 0.5 + 0.5 = 1.0 m 3 ⎛ PV ⎞ (600 kPa)(0.5 m 3 ) = 0.248 kg m A = ⎜⎜ 1 ⎟⎟ = 3 ⎝ RT1 ⎠ A (4.124 kPa ⋅ m /kg ⋅ K)(293 K) ⎛ PV ⎞ (150 kPa)(0.5 m 3 ) m B = ⎜⎜ 1 ⎟⎟ = = 0.060 kg 3 ⎝ RT1 ⎠ B (4.124 kPa ⋅ m /kg ⋅ K)(303 K) m = m A + m B = 0.248 + 0.060 = 0.308kg

B

H2

V = 0.5 m3 T=20°C P=600 kPa

×

H2

V = 0.5 m3 T=30°C P=150 kPa

Then the final pressure can be determined from P=

mRT2

V

=

(0.308 kg)(4.124 kPa ⋅ m 3 /kg ⋅ K)(288 K) 1.0 m 3

= 365.8 kPa

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3-72

3-128 EES Problem 3-127 is reconsidered. The effect of the surroundings temperature on the final equilibrium pressure in the tanks is to be investigated. The final pressure in the tanks is to be plotted versus the surroundings temperature, and the results are to be discussed. Analysis The problem is solved using EES, and the solution is given below. "Given Data" V_A=0.5 [m^3] T_A=20 [C] P_A=600 [kPa] V_B=0.5 [m^3] T_B=30 [C] P_B=150 [kPa] {T_2=15 [C]} "Solution" R=R_u/MOLARMASS(H2) R_u=8.314 [kJ/kmol-K] V_total=V_A+V_B m_total=m_A+m_B P_A*V_A=m_A*R*(T_A+273) P_B*V_B=m_B*R*(T_B+273) P_2*V_total=m_total*R*(T_2+273)

T2 [C] -10 -5 0 5 10 15 20 25 30

390 380 370

P 2 [kPa]

P2 [kPa] 334.4 340.7 347.1 353.5 359.8 366.2 372.5 378.9 385.2

360 350 340 330 -10

-5

0

5

10

T

2

15

20

25

30

[C]

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-73

3-129 A large tank contains nitrogen at a specified temperature and pressure. Now some nitrogen is allowed to escape, and the temperature and pressure of nitrogen drop to new values. The amount of nitrogen that has escaped is to be determined. Properties The gas constant for nitrogen is 0.2968 kPa·m3/kg·K (Table A-1). Analysis Treating N2 as an ideal gas, the initial and the final masses in the tank are determined to be m1 =

P1V (600 kPa)(20 m 3 ) = = 136.6 kg RT1 (0.2968kPa ⋅ m 3 /kg ⋅ K)(296 K)

m2 =

P2V (400 kPa)(20 m 3 ) = = 92.0 kg RT2 (0.2968 kPa ⋅ m 3 /kg ⋅ K)(293 K)

Thus the amount of N2 that escaped is

Δm = m1 − m 2 = 136.6 − 92.0 = 44.6 kg

N2 600 kPa 23°C 20 m3

3-130 The temperature of steam in a tank at a specified state is to be determined using the ideal gas relation, the generalized chart, and the steam tables. Properties The gas constant, the critical pressure, and the critical temperature of water are, from Table A-1,

R = 0.4615 kPa ⋅ m3/kg ⋅ K,

Tcr = 647.1 K,

Pcr = 22.06 MPa

Analysis (a) From the ideal gas equation of state,

P=

RT

v

=

(0.4615 kPa ⋅ m3/kg ⋅ K)(673 K) = 15,529 kPa 0.02 m3/kg

H2O 0.02 m3/kg 400°C

(b) From the compressibility chart (Fig. A-15a), ⎫ ⎪ ⎪ ⎬ PR = 0.57 (0.02 m3 /kg)(22,060 kPa) v actual = = 1.48 ⎪ vR = ⎪ RTcr / Pcr (0.4615 kPa ⋅ m3/kg ⋅ K)(647.1 K) ⎭ TR =

Thus,

673 K T = = 1.040 Tcr 647.1 K

P = PR Pcr = 0.57 × 22,060 = 12,574 kPa

(c) From the superheated steam table,

T = 400°C

⎫ P = 12,576 kPa

v = 0.02 m 3 /kg ⎬⎭

(from EES)

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3-74

3-131 One section of a tank is filled with saturated liquid R-134a while the other side is evacuated. The partition is removed, and the temperature and pressure in the tank are measured. The volume of the tank is to be determined. Analysis The mass of the refrigerant contained in the tank is m=

V1 0.01 m 3 = = 11.82 kg v 1 0.0008458 m 3 /kg

since

v 1 = v f @ 0.8 MPa = 0.0008458 m 3 /kg

R-134a P=0.8 MPa V =0.01 m3

Evacuated

At the final state (Table A-13), P2 = 400 kPa T2 = 20°C

Thus,

}v

2

= 0.05421 m3/kg

V tank = V 2 = mv 2 = (11.82 kg)(0.05421 m 3 /kg) = 0.641 m 3

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3-75

3-132 EES Problem 3-131 is reconsidered. The effect of the initial pressure of refrigerant-134 on the volume of the tank is to be investigated as the initial pressure varies from 0.5 MPa to 1.5 MPa. The volume of the tank is to be plotted versus the initial pressure, and the results are to be discussed. Analysis The problem is solved using EES, and the solution is given below. "Given Data" x_1=0.0 Vol_1=0.01[m^3] P_1=800 [kPa] T_2=20 [C] P_2=400 [kPa] "Solution" v_1=volume(R134a,P=P_1,x=x_1) Vol_1=m*v_1 v_2=volume(R134a,P=P_2,T=T_2) Vol_2=m*v_2 P1 [kPa] 500 600 700 800 900 1000 1100 1200 1300 1400 1500

Vol2 [m3] 0.6727 0.6612 0.6507 0.641 0.6318 0.6231 0.6148 0.6068 0.599 0.5914 0.584

m [kg] 12.41 12.2 12 11.82 11.65 11.49 11.34 11.19 11.05 10.91 10.77

0.68

3

Vol2 [m ]

0.66

0.64

0.62

0.6

0.58 500

700

900

1100

1300

1500

P1 [kPa]

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3-76

3-133 A propane tank contains 5 L of liquid propane at the ambient temperature. Now a leak develops at the top of the tank and propane starts to leak out. The temperature of propane when the pressure drops to 1 atm and the amount of heat transferred to the tank by the time the entire propane in the tank is vaporized are to be determined. Properties The properties of propane at 1 atm are Tsat = -42.1°C, ρ = 581 kg / m 3 , and hfg = 427.8 kJ/kg (Table A-3). Analysis The temperature of propane when the pressure drops to 1 atm is simply the saturation pressure at that temperature,

T = Tsat @1 atm = −42.1° C

Propane 5L

The initial mass of liquid propane is

m = ρV = (581 kg/m3 )(0.005 m3 ) = 2.905 kg The amount of heat absorbed is simply the total heat of vaporization,

20°C Leak

Qabsorbed = mh fg = (2.905 kg)(427.8 kJ / kg) = 1243 kJ

3-134 An isobutane tank contains 5 L of liquid isobutane at the ambient temperature. Now a leak develops at the top of the tank and isobutane starts to leak out. The temperature of isobutane when the pressure drops to 1 atm and the amount of heat transferred to the tank by the time the entire isobutane in the tank is vaporized are to be determined. Properties The properties of isobutane at 1 atm are Tsat = -11.7°C, ρ = 593.8 kg / m 3 , and hfg = 367.1 kJ/kg (Table A-3). Analysis The temperature of isobutane when the pressure drops to 1 atm is simply the saturation pressure at that temperature,

T = Tsat @1 atm = −11.7° C Isobutane 5L

The initial mass of liquid isobutane is

m = ρV = (593.8 kg/m 3 )(0.005 m 3 ) = 2.969kg

20°C

The amount of heat absorbed is simply the total heat of vaporization, Qabsorbed = mh fg = (2.969 kg)(367.1 kJ / kg) = 1090 kJ

Leak

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3-77

3-135 A tank contains helium at a specified state. Heat is transferred to helium until it reaches a specified temperature. The final gage pressure of the helium is to be determined. Assumptions 1 Helium is an ideal gas. Properties The local atmospheric pressure is given to be 100 kPa. Analysis Noting that the specific volume of helium in the tank remains constant, from ideal gas relation, we have P2 = P1

Helium 100ºC 10 kPa gage

T2 (300 + 273)K = (10 + 100 kPa) = 169.0 kPa T1 (100 + 273)K

Then the gage pressure becomes

Q

Pgage,2 = P2 − Patm = 169.0 − 100 = 69.0 kPa

3-136 The first eight virial coefficients of a Benedict-Webb-Rubin gas are to be obtained. Analysis The Benedict-Webb-Rubin equation of state is given by P=

γ ⎞ C ⎞ 1 bR T − a aα c ⎛ ⎛ + 6 + 3 2 ⎜1 + 2 ⎟ exp(−γ / v 2 ) + ⎜ B0 RuT − A0 − 02 ⎟ 2 + u 3 v v v v T ⎝ v ⎠ T ⎠v ⎝

RuT

Expanding the last term in a series gives exp(−γ / v 2 ) = 1 −

1 γ2 1 γ3 γ + − + .... v 2 2! v 4 3! v 6

Substituting this into the Benedict-Webb-Rubin equation of state and rearranging the first terms gives P=

Ru T

v

+

R u TB 0 − A0 − C 0 / T 2

v

2

+

bR u T − a

v

3

+

c (1 + γ )

v 5T 2

+

aα

v

6

−

c γ (1 + γ )

v 7T 2

+

1 c γ 2 (1 + γ ) 2! v 9 T 2

The virial equation of state is P=

Ru T

v

+

a (T )

v

2

+

b (T )

v

3

+

c (T )

v

4

+

d (T )

v

5

+

e (T )

v

6

+

f (T )

v

7

+

g (T )

v

8

+

h (T )

v

9

...

Comparing the Benedict-Webb-Rubin equation of state to the virial equation of state, the virial coefficients are a (T ) = R u TB 0 − A0 − C 0 / T 2 b (T ) = bR u T − a c (T ) = 0 d (T ) = c (1 + γ ) / T 2 e (T ) = a α f (T ) = cγ (1 + γ ) / T 2 g (T ) = 0 h (T ) =

1 cγ 2 (1 + γ ) 2! T2

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3-78

3-137 The specific volume of oxygen at a given state is to be determined using the ideal gas relation, the Beattie-Bridgeman equation, and the compressibility factor. Properties The properties of oxygen are (Table A-1)

R = 0.2598 kPa·m3/kg·K,

M = 31.999 kg/kmol,

Tcr = 154.8 K,

Pcr = 5.08 MPa

Analysis (a) From the ideal gas equation of state,

v=

RT (0.2598 kPa ⋅ m 3 /kg ⋅ K)(293 K) = = 0.01903 m 3 /kg 4000 kPa P

(b) The constants in the Beattie-Bridgeman equation are expressed as ⎛ a⎞ ⎛ 0.02562 ⎞ A = Ao ⎜1 − ⎟ = 151.0857⎜1 − ⎟ v ⎝ v ⎠ ⎝ ⎠ ⎛ b⎞ ⎛ 0.004208 ⎞ B = Bo ⎜1 − ⎟ = 0.04624⎜1 − ⎟ v v ⎝ ⎠ ⎝ ⎠

Oxygen 4 MPa, 20°C

c = 4.80 × 10 4 m 3 ⋅ K 3 /kmol

Substituting these coefficients into the Beattie-Bridgeman equation P=

Ru T ⎛ c ⎞ A ⎜1 − ⎟(v + B ) − 2 2 3 v ⎝ vT ⎠ v

and solving using an equation solver such as EES gives

v = 0.5931 m 3 /kmol and

v=

v M

=

0.5931 m 3 /kmol = 0.01853 m 3 /kg 31.999 kg/kmol

(c) From the compressibility chart (Fig. A-15), ⎫ ⎪ ⎪ ⎬ Z = 0.975 P 4 MPa = = 0.787 ⎪ PR = ⎪⎭ Pcr 5.08 MPa

TR =

293 K T = = 1.893 Tcr 154.8 K

Thus,

v = Zv ideal = (0.975)(0.01903 m 3 /kg) = 0.01855 m 3 /kg

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3-79

3-138E The specific volume of nitrogen at a given state is to be determined using the ideal gas relation, the Benedict-Webb-Rubin equation, and the compressibility factor. Properties The properties of nitrogen are (Table A-1E)

R = 0.3830 psia·ft3/lbm·R, M = 28.013 lbm/lbmol,

Tcr = 227.1 R,

Pcr = 492 psia

Analysis (a) From the ideal gas equation of state,

v=

RT (0.3830 psia ⋅ ft 3 /lbm ⋅ R)(360 R) = = 0.3447 ft 3 /lbm 400 psia P

Nitrogen 400 psia, -100°F

(b) Using the coefficients of Table 3-4 for nitrogen and the given data in SI units, the Benedict-Webb-Rubin equation of state is ⎛ C ⎞ 1 bR T − a aα c ⎛ γ ⎞ + ⎜ B0 RuT − A0 − 02 ⎟ + u 3 + 6 + 3 2 ⎜1 + 2 ⎟ exp(−γ / v 2 ) ⎟v ⎜ v T T v v v v ⎝ ⎠ ⎠ ⎝ (8.314)(200) ⎛⎜ 8.164 ×105 ⎞⎟ 1 0.002328× 8.314 × 200 − 2.54 + 0.04074× 8.314 × 200 − 106.73 − 2758 = + ⎜ v2 2002 ⎟⎠ v 2 v3 ⎝ P=

+

RuT

2.54 × 1.272 ×10−4

v6

+

7.379 ×104 ⎛ 0.0053 ⎞ 1+ exp(−0.0053 / v 2 ) 3 2 ⎜ 2 ⎟ v (200) ⎝ v ⎠

The solution of this equation by an equation solver such as EES gives

v 2 = 0.5666 m 3 /kmol Then,

v2 =

v2 M

=

0.5666 m 3 /kmol ⎛⎜ 16.02 ft 3 /lbm ⎞⎟ = 0.3240 ft 3 /lbm 28.013 kg/kmol ⎜⎝ 1 m 3 /kg ⎟⎠

(c) From the compressibility chart (Fig. A-15), 360 R T ⎫ = = 1.585 ⎪ Tcr 227.1 R ⎪ ⎬ Z = 0.94 400 psia P = = 0.813 ⎪ PR = ⎪⎭ Pcr 492 psia

TR =

Thus,

v = Zv ideal = (0.94)(0.3447 ft 3 /lbm) = 0.3240 ft 3 /lbm

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3-80

3-139 Complete the following table for H2 O:

P, kPa

T, °C

v, m3 / kg

u, kJ/kg

Phase description

200

30

0.001004

125.71

Compressed liquid

270.3

130

-

-

Insufficient information

200

400

1.5493

2967.2

Superheated steam

300

133.52

0.500

2196.4

Saturated mixture, x=0.825

500

473.1

0.6858

3084

Superheated steam

3-140 Complete the following table for R-134a:

P, kPa

T, °C

v, m3 / kg

u, kJ/kg

Phase description

320

-12

0.0007497

35.72

Compressed liquid

1000

39.37

-

-

Insufficient information

140

40

0.17794

263.79

Superheated vapor

180

-12.73

0.0700

153.66

Saturated mixture, x=0.6315

200

22.13

0.1152

249

Superheated vapor

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3-81

3-141 (a) On the P-v diagram the constant temperature process through the state P = 280 kPa, v = 0.06 m3/kg as pressure changes from P1 = 400 kPa to P2 = 200 kPa is to be sketched. The value of the temperature on the process curve on the P-v diagram is to be placed.

R134a

105

P [kPa]

104

103

1 400 280 200

-1.25°C

2

102

0.06 101 10-4

10-3

10-2

10-1

100

101

3

v [m /kg]

(b) On the T-v diagram the constant specific volume process through the state T = 20°C, v = 0.02 m3/kg from P1 = 1200 kPa to P2 = 300 kPa is to be sketched. For this data set the temperature values at states 1 and 2 on its axis is to be placed. The value of the specific volume on its axis is also to be placed.

R134a

250

200

T [°C]

150

100

1

75 50

1200 kPa

20 0 0.6

572 kPa 300 kPa

2

-50

0.02 -100 10-4

10-3

10-2

10-1

100

3

v [m /kg]

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3-82

Fundamentals of Engineering (FE) Exam Problems 3-142 A rigid tank contains 6 kg of an ideal gas at 3 atm and 40°C. Now a valve is opened, and half of mass of the gas is allowed to escape. If the final pressure in the tank is 2.2 atm, the final temperature in the tank is (a) 186°C

(b) 59°C

(c) -43°C

(d) 20°C

(e) 230°C

Answer (a) 186°C

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). "When R=constant and V= constant, P1/P2=m1*T1/m2*T2" m1=6 "kg" P1=3 "atm" P2=2.2 "atm" T1=40+273 "K" m2=0.5*m1 "kg" P1/P2=m1*T1/(m2*T2) T2_C=T2-273 "C" "Some Wrong Solutions with Common Mistakes:" P1/P2=m1*(T1-273)/(m2*W1_T2) "Using C instead of K" P1/P2=m1*T1/(m1*(W2_T2+273)) "Disregarding the decrease in mass" P1/P2=m1*T1/(m1*W3_T2) "Disregarding the decrease in mass, and not converting to deg. C" W4_T2=(T1-273)/2 "Taking T2 to be half of T1 since half of the mass is discharged"

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3-83

3-143 The pressure of an automobile tire is measured to be 190 kPa (gage) before a trip and 215 kPa (gage) after the trip at a location where the atmospheric pressure is 95 kPa. If the temperature of air in the tire before the trip is 25°C, the air temperature after the trip is (a) 51.1°C

(b) 64.2°C

(c) 27.2°C

(d) 28.3°C

(e) 25.0°C

Answer (a) 51.1°C

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). "When R, V, and m are constant, P1/P2=T1/T2" Patm=95 P1=190+Patm "kPa" P2=215+Patm "kPa" T1=25+273 "K" P1/P2=T1/T2 T2_C=T2-273 "C" "Some Wrong Solutions with Common Mistakes:" P1/P2=(T1-273)/W1_T2 "Using C instead of K" (P1-Patm)/(P2-Patm)=T1/(W2_T2+273) "Using gage pressure instead of absolute pressure" (P1-Patm)/(P2-Patm)=(T1-273)/W3_T2 "Making both of the mistakes above" W4_T2=T1-273 "Assuming the temperature to remain constant"

3-144 A 300-m3 rigid tank is filled with saturated liquid-vapor mixture of water at 200 kPa. If 25% of the mass is liquid and the 75% of the mass is vapor, the total mass in the tank is (a) 451 kg

(b) 556 kg

(c) 300 kg

(d) 331 kg

(e) 195 kg

Answer (a) 451 kg

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). V_tank=300 "m3" P1=200 "kPa" x=0.75 v_f=VOLUME(Steam_IAPWS, x=0,P=P1) v_g=VOLUME(Steam_IAPWS, x=1,P=P1) v=v_f+x*(v_g-v_f) m=V_tank/v "kg" "Some Wrong Solutions with Common Mistakes:" R=0.4615 "kJ/kg.K" T=TEMPERATURE(Steam_IAPWS,x=0,P=P1) P1*V_tank=W1_m*R*(T+273) "Treating steam as ideal gas" P1*V_tank=W2_m*R*T "Treating steam as ideal gas and using deg.C" W3_m=V_tank "Taking the density to be 1 kg/m^3"

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3-84

3-145 Water is boiled at 1 atm pressure in a coffee maker equipped with an immersion-type electric heating element. The coffee maker initially contains 1 kg of water. Once boiling started, it is observed that half of the water in the coffee maker evaporated in 18 minutes. If the heat loss from the coffee maker is negligible, the power rating of the heating element is (a) 0.90 kW

(b) 1.52 kW

(c) 2.09 kW

(d) 1.05 kW

(e) 1.24 kW

Answer (d) 1.05 kW

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m_1=1 "kg" P=101.325 "kPa" time=18*60 "s" m_evap=0.5*m_1 Power*time=m_evap*h_fg "kJ" h_f=ENTHALPY(Steam_IAPWS, x=0,P=P) h_g=ENTHALPY(Steam_IAPWS, x=1,P=P) h_fg=h_g-h_f "Some Wrong Solutions with Common Mistakes:" W1_Power*time=m_evap*h_g "Using h_g" W2_Power*time/60=m_evap*h_g "Using minutes instead of seconds for time" W3_Power=2*Power "Assuming all the water evaporates"

3-146 A 1-m3 rigid tank contains 10 kg of water (in any phase or phases) at 160°C. The pressure in the tank is (a) 738 kPa

(b) 618 kPa

(c) 370 kPa

(d) 2000 kPa

(e) 1618 kPa

Answer (b) 618 kPa

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). V_tank=1 "m^3" m=10 "kg" v=V_tank/m T=160 "C" P=PRESSURE(Steam_IAPWS,v=v,T=T) "Some Wrong Solutions with Common Mistakes:" R=0.4615 "kJ/kg.K" W1_P*V_tank=m*R*(T+273) "Treating steam as ideal gas" W2_P*V_tank=m*R*T "Treating steam as ideal gas and using deg.C"

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3-85

3-147 Water is boiling at 1 atm pressure in a stainless steel pan on an electric range. It is observed that 2 kg of liquid water evaporates in 30 minutes. The rate of heat transfer to the water is (a) 2.51 kW

(b) 2.32 kW

(c) 2.97 kW

(d) 0.47 kW

(e) 3.12 kW

Answer (a) 2.51 kW

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m_evap=2 "kg" P=101.325 "kPa" time=30*60 "s" Q*time=m_evap*h_fg "kJ" h_f=ENTHALPY(Steam_IAPWS, x=0,P=P) h_g=ENTHALPY(Steam_IAPWS, x=1,P=P) h_fg=h_g-h_f "Some Wrong Solutions with Common Mistakes:" W1_Q*time=m_evap*h_g "Using h_g" W2_Q*time/60=m_evap*h_g "Using minutes instead of seconds for time" W3_Q*time=m_evap*h_f "Using h_f"

3-148 Water is boiled in a pan on a stove at sea level. During 10 min of boiling, its is observed that 200 g of water has evaporated. Then the rate of heat transfer to the water is (a) 0.84 kJ/min

(b) 45.1 kJ/min

(c) 41.8 kJ/min

(d) 53.5 kJ/min

(e) 225.7 kJ/min

Answer (b) 45.1 kJ/min

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m_evap=0.2 "kg" P=101.325 "kPa" time=10 "min" Q*time=m_evap*h_fg "kJ" h_f=ENTHALPY(Steam_IAPWS, x=0,P=P) h_g=ENTHALPY(Steam_IAPWS, x=1,P=P) h_fg=h_g-h_f "Some Wrong Solutions with Common Mistakes:" W1_Q*time=m_evap*h_g "Using h_g" W2_Q*time*60=m_evap*h_g "Using seconds instead of minutes for time" W3_Q*time=m_evap*h_f "Using h_f"

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3-86

3-149 A rigid 3-m3 rigid vessel contains steam at 10 MPa and 500°C. The mass of the steam is (a) 3.0 kg

(b) 19 kg

(c) 84 kg

(d) 91 kg

(e) 130 kg

Answer (d) 91 kg

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). V=3 "m^3" m=V/v1 "m^3/kg" P1=10000 "kPa" T1=500 "C" v1=VOLUME(Steam_IAPWS,T=T1,P=P1) "Some Wrong Solutions with Common Mistakes:" R=0.4615 "kJ/kg.K" P1*V=W1_m*R*(T1+273) "Treating steam as ideal gas" P1*V=W2_m*R*T1 "Treating steam as ideal gas and using deg.C"

3-150 Consider a sealed can that is filled with refrigerant-134a. The contents of the can are at the room temperature of 25°C. Now a leak developes, and the pressure in the can drops to the local atmospheric pressure of 90 kPa. The temperature of the refrigerant in the can is expected to drop to (rounded to the nearest integer) (a) 0°C

(b) -29°C

(c) -16°C

(d) 5°C

(e) 25°C

Answer (b) -29°C

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T1=25 "C" P2=90 "kPa" T2=TEMPERATURE(R134a,x=0,P=P2) "Some Wrong Solutions with Common Mistakes:" W1_T2=T1 "Assuming temperature remains constant"

3-151 … 3-153 Design, Essay and Experiment Problems

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4-1

Chapter 4 ENERGY ANALYSIS OF CLOSED SYSTEMS Moving Boundary Work 4-1C It represents the boundary work for quasi-equilibrium processes. 4-2C Yes. 4-3C The area under the process curve, and thus the boundary work done, is greater in the constant pressure case.

4-4C 1 kPa ⋅ m 3 = 1 k(N / m 2 ) ⋅ m 3 = 1 kN ⋅ m = 1 kJ

4-5 Helium is compressed in a piston-cylinder device. The initial and final temperatures of helium and the work required to compress it are to be determined. Assumptions The process is quasi-equilibrium. Properties The gas constant of helium is R = 2.0769 kJ/kg⋅K (Table A-1). Analysis The initial specific volume is

v1 =

V1 m

=

5 m3 = 5 m 3 /kg 1 kg

Using the ideal gas equation, T1 =

P1v 1 (200 kPa)(5 m 3 /kg ) = = 481.5 K 2.0769 kJ/kg ⋅ K R

P (kPa) 200

Since the pressure stays constant, T2 =

2

1

3

5

V (m3)

V2 3 m3 T1 = (481.5 K ) = 288.9 K V1 5 m3

and the work integral expression gives Wb,out =

That is,

∫

2

1

⎛ 1 kJ P dV = P(V 2 −V 1 ) = (200 kPa)(3 − 5) m 3 ⎜ ⎜ 1 kPa ⋅ m 3 ⎝

⎞ ⎟ = −400 kJ ⎟ ⎠

Wb,in = 400 kJ

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

4-2

4-6 The boundary work done during the process shown in the figure is to be determined. Assumptions The process is quasi-equilibrium.

P (kPa)

Analysis No work is done during the process 2-3 since the area under process line is zero. Then the work done is equal to the area under the process line 1-2: P1 + P2 m(v 2 − v 1 ) 2 ⎛ 1 kJ (100 + 500)kPa (2 kg)(1.0 − 0.5)m 3 /kg⎜ = ⎜ 1 kPa ⋅ m 3 2 ⎝ = 300 kJ

1

500 400

3

Wb,out = Area =

⎞ ⎟ ⎟ ⎠

100

2 0.5

v (m3/kg)

1

4-7E The boundary work done during the process shown in the figure is to be determined. Assumptions The process is quasi-equilibrium.

P (psia)

Analysis The work done is equal to the area under the process line 1-2: P1 + P2 (V 2 −V 1 ) 2 ⎛ (100 + 500)psia 1 Btu = (4.0 − 2.0)ft 3 ⎜ 3 ⎜ 2 ⎝ 5.404 psia ⋅ ft = 111 Btu

2

500

Wb,out = Area =

⎞ ⎟ ⎟ ⎠

100

1 2

4

V (ft3)

4-8 A piston-cylinder device contains nitrogen gas at a specified state. The boundary work is to be determined for the polytropic expansion of nitrogen. Properties The gas constant for nitrogen is 0.2968 kJ/kg.K (Table A-2). Analysis The mass and volume of nitrogen at the initial state are m=

V2 =

P1V1 (130 kPa)(0.07 m 3 ) = = 0.07802 kg RT1 (0.2968 kJ/kg.K)(120 + 273 K) mRT2 (0.07802 kg)(0.2968 kPa.m 3 /kg.K)(100 + 273 K) = = 0.08637 m 3 P2 100 kPa

N2 130 kPa 120°C

The polytropic index is determined from P1V1n = P2V 2n ⎯ ⎯→(130 kPa)(0.07 m 3 ) n = (100 kPa)(0.08637 m 3 ) n ⎯ ⎯→ n = 1.249

The boundary work is determined from Wb =

P2V 2 − P1V 1 (100 kPa)(0.08637 m 3 ) − (130 kPa)(0.07 m 3 ) = = 1.86 kJ 1− n 1 − 1.249

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

4-3

4-9 A piston-cylinder device with a set of stops contains steam at a specified state. Now, the steam is cooled. The compression work for two cases and the final temperature are to be determined. Analysis (a) The specific volumes for the initial and final states are (Table A-6) P1 = 1 MPa ⎫ 3 ⎬v1 = 0.30661 m /kg T1 = 400°C⎭

P2 = 1 MPa ⎫ 3 ⎬v 2 = 0.23275 m /kg T2 = 250°C⎭

Noting that pressure is constant during the process, the boundary work is determined from Wb = mP (v1 − v 2 ) = (0.3 kg)(1000 kPa)(0.30661 − 0.23275)m 3/kg = 22.16 kJ

(b) The volume of the cylinder at the final state is 60% of initial volume. Then, the boundary work becomes

Steam 0.3 kg 1 MPa 400°C

Q

Wb = mP (v1 − 0.60v1 ) = (0.3 kg)(1000 kPa)(0.30661 − 0.60 × 0.30661)m3/kg = 36.79 kJ

The temperature at the final state is P2 = 0.5 MPa

⎫⎪ ⎬T2 = 151.8°C (Table A-5) v 2 = (0.60 × 0.30661) m3/kg ⎪⎭

4-10 A piston-cylinder device contains nitrogen gas at a specified state. The final temperature and the boundary work are to be determined for the isentropic expansion of nitrogen. Properties The properties of nitrogen are R = 0.2968 kJ/kg.K , k = 1.4 (Table A-2a) Analysis The mass and the final volume of nitrogen are m=

P1V1 (130 kPa)(0.07 m 3 ) = = 0.07802 kg RT1 (0.2968 kJ/kg.K)(120 + 273 K)

N2 130 kPa 120°C

P1V1k = P2V 2k ⎯ ⎯→(130 kPa)(0.07 m 3 )1.4 = (100 kPa)V 21.4 ⎯ ⎯→V 2 = 0.08443 m 3

The final temperature and the boundary work are determined as T2 =

P2V 2 (100 kPa)(0.08443 m 3 ) = = 364.6 K mR (0.07802 kg)(0.2968 kPa.m 3 /kg.K)

Wb =

P2V 2 − P1V 1 (100 kPa)(0.08443 m 3 ) − (130 kPa)(0.07 m 3 ) = = 1.64 kJ 1− k 1 − 1.4

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

4-4

4-11 Saturated water vapor in a cylinder is heated at constant pressure until its temperature rises to a specified value. The boundary work done during this process is to be determined. Assumptions The process is quasi-equilibrium. Properties Noting that the pressure remains constant during this process, the specific volumes at the initial and the final states are (Table A-4 through A-6) P1 = 300 kPa ⎫ 3 ⎬ v 1 = v g @ 300 kPa = 0.60582 m /kg Sat. vapor ⎭ P2 = 300 kPa ⎫ 3 ⎬ v 2 = 0.71643 m /kg T2 = 200°C ⎭

P (kPa 1

300

2

Analysis The boundary work is determined from its definition to be Wb,out =

∫

2

1

V

P dV = P (V 2 − V1 ) = mP (v 2 − v1 )

⎛ 1 kJ ⎞ ⎟ = (5 kg)(300 kPa)(0.71643 − 0.60582) m3/kg⎜⎜ 3⎟ ⎝ 1 kPa ⋅ m ⎠ = 165.9 kJ

Discussion The positive sign indicates that work is done by the system (work output).

4-12 Refrigerant-134a in a cylinder is heated at constant pressure until its temperature rises to a specified value. The boundary work done during this process is to be determined. Assumptions The process is quasi-equilibrium. Properties Noting that the pressure remains constant during this process, the specific volumes at the initial and the final states are (Table A-11 through A-13) P1 = 900 kPa ⎫ 3 ⎬ v 1 = v f @ 900 kPa = 0.0008580 m /kg Sat. liquid ⎭ P2 = 900 kPa ⎫ 3 ⎬ v 2 = 0.027413 m /kg T2 = 70°C ⎭

P (kPa) 900

1

2

Analysis The boundary work is determined from its definition to be m=

V1 0.2 m 3 = = 233.1 kg v 1 0.0008580 m 3 /kg

and Wb,out =

∫

2

1

P dV = P (V 2 − V1 ) = mP(v 2 − v1 )

⎛ 1 kJ ⎞ ⎟ = (233.1 kg)(900 kPa)(0.027413 − 0.0008580)m3/kg⎜⎜ 3⎟ ⎝ 1 kPa ⋅ m ⎠ = 5571 kJ

Discussion The positive sign indicates that work is done by the system (work output).

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v

4-5

4-13 EES Problem 4-12 is reconsidered. The effect of pressure on the work done as the pressure varies from 400 kPa to 1200 kPa is to be investigated. The work done is to be plotted versus the pressure. Analysis The problem is solved using EES, and the solution is given below. "Knowns" Vol_1L=200 [L] x_1=0 "saturated liquid state" P=900 [kPa] T_2=70 [C] "Solution" Vol_1=Vol_1L*convert(L,m^3) "The work is the boundary work done by the R-134a during the constant pressure process." W_boundary=P*(Vol_2-Vol_1) R134a

150

"The mass is:"

125

P [kPa] 400 500 600 700 800 900 1000 1100 1200

Wboundary [kJ] 6643 6405 6183 5972 5769 5571 5377 5187 4999

T [°C]

"Plot information:" v[1]=v_1 v[2]=v_2 P[1]=P P[2]=P T[1]=temperature(R134a,P=P,x=x_1) T[2]=T_2

100 75

2

50 25

1

900 kPa

0 -25 -50 10-4

10-3

10-2

10-1

3

v [m /kg] R134a

105

104

P [kPa]

Vol_1=m*v_1 v_1=volume(R134a,P=P,x=x_1) Vol_2=m*v_2 v_2=volume(R134a,P=P,T=T_2)

2 103

1

102

101 10-4

10-3

10-2

10-1

3

v [m /kg]

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4-6

7250

P = 800 kPa Wboundary [kJ]

6800 6350 5900 5450 5000 50

60

70

80

90

100

110

120

130

T[2] [C] 7500

T2 = 100 C Wboundary [kJ]

7150 6800 6450 6100 5750 400

500

600

700

800

900

1000 1100 1200

P [kPa] 7000

Wboundary [kJ]

6500

T2 = 70 C

6000

5500

5000

4500 400

500

600

700

800

900

1000 1100 1200

P [kPa]

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4-7

4-14E Superheated water vapor in a cylinder is cooled at constant pressure until 70% of it condenses. The boundary work done during this process is to be determined. Assumptions The process is quasi-equilibrium. Properties Noting that the pressure remains constant during this process, the specific volumes at the initial and the final states are (Table A-4E through A-6E) P1 = 40 psia ⎫ 3 ⎬v 1 = 15.686 ft /lbm T1 = 600°F ⎭ P2 = 40 psia ⎫ ⎬ v 2 = v f + x 2v fg x 2 = 0.3 ⎭ = 0.01715 + 0.3(10.501 − 0.01715)

P (psia) 2

40

1

= 3.1623 ft 3 /lbm

v

Analysis The boundary work is determined from its definition to be Wb,out =

∫

2

1

P dV = P(V 2 − V1 ) = mP(v 2 − v1 )

⎞ ⎛ 1 Btu ⎟ = (16 lbm)(40 psia)(3.1623 − 15.686)ft 3/lbm⎜⎜ 3⎟ ⎝ 5.4039 psia ⋅ ft ⎠ = −1483 Btu

Discussion The negative sign indicates that work is done on the system (work input).

4-15 Air in a cylinder is compressed at constant temperature until its pressure rises to a specified value. The boundary work done during this process is to be determined. Assumptions 1 The process is quasi-equilibrium. 2 Air is an ideal gas. Properties The gas constant of air is R = 0.287 kJ/kg.K (Table A-1). Analysis The boundary work is determined from its definition to be

Wb,out =

∫

2

1

P dV = P1V1 ln

P 2

V2 P = mRT ln 1 V1 P2

T = 12°C 1

150 kPa = (2.4 kg)(0.287 kJ/kg ⋅ K)(285 K)ln 600 kPa = −272 kJ

Discussion The negative sign indicates that work is done on the system (work input).

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V

4-8

4-16E A gas in a cylinder is heated and is allowed to expand to a specified pressure in a process during which the pressure changes linearly with volume. The boundary work done during this process is to be determined. Assumptions The process is quasi-equilibrium. Analysis (a) The pressure of the gas changes linearly with volume, and thus the process curve on a P-V diagram will be a straight line. The boundary work during this process is simply the area under the process curve, which is a trapezoidal. Thus, At state 1:

P (psia)

P1 = aV1 + b 15 psia = (5 psia/ft 3 )(7 ft 3 ) + b b = −20 psia

P = aV + b 2

100

At state 2:

1

15 P2 = aV 2 + b

100 psia = (5 psia/ft 3 )V 2 + (−20 psia)

V 2 = 24 ft

V

7

3

(ft3)

and, Wb,out = Area =

⎛ P1 + P2 1 Btu (100 + 15)psia (24 − 7)ft 3 ⎜ (V 2 −V1 ) = 3 ⎜ 2 2 ⎝ 5.4039 psia ⋅ ft

⎞ ⎟ ⎟ ⎠

= 181 Btu

Discussion The positive sign indicates that work is done by the system (work output).

4-17 [Also solved by EES on enclosed CD] A gas in a cylinder expands polytropically to a specified volume. The boundary work done during this process is to be determined. Assumptions The process is quasi-equilibrium. Analysis The boundary work for this polytropic process can be determined directly from ⎛V P2 = P1 ⎜⎜ 1 ⎝V 2

n

⎛ 0.03 m 3 ⎞ ⎟⎟ = (150 kPa)⎜ ⎜ 0.2 m 3 ⎠ ⎝

⎞ ⎟ ⎟ ⎠

1.3

= 12.74 kPa

P (kPa) 15

and, Wb,out = =

∫

2

1

P dV =

P2V 2 − P1V1 1− n

(12.74 × 0.2 − 150 × 0.03) kPa ⋅ m 3 1 − 1.3

= 6.51 kJ

1

PV ⎛ 1 kJ ⎜ ⎜ 1 kPa ⋅ m 3 ⎝

⎞ ⎟ ⎟ ⎠

2 0.0

0.2

V

(m3)

Discussion The positive sign indicates that work is done by the system (work output).

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4-9

4-18 EES Problem 4-17 is reconsidered. The process described in the problem is to be plotted on a P-V diagram, and the effect of the polytropic exponent n on the boundary work as the polytropic exponent varies from 1.1 to 1.6 is to be plotted. Analysis The problem is solved using EES, and the solution is given below. Function BoundWork(P[1],V[1],P[2],V[2],n) "This function returns the Boundary Work for the polytropic process. This function is required since the expression for boundary work depens on whether n=1 or n1" If n1 then BoundWork:=(P[2]*V[2]-P[1]*V[1])/(1-n)"Use Equation 3-22 when n=1" else BoundWork:= P[1]*V[1]*ln(V[2]/V[1]) "Use Equation 3-20 when n=1" endif end "Inputs from the diagram window" {n=1.3 P[1] = 150 [kPa] V[1] = 0.03 [m^3] V[2] = 0.2 [m^3] Gas$='AIR'} "System: The gas enclosed in the piston-cylinder device." "Process: Polytropic expansion or compression, P*V^n = C" P[2]*V[2]^n=P[1]*V[1]^n "n = 1.3" "Polytropic exponent" "Input Data" W_b = BoundWork(P[1],V[1],P[2],V[2],n)"[kJ]" "If we modify this problem and specify the mass, then we can calculate the final temperature of the fluid for compression or expansion" m[1] = m[2] "Conservation of mass for the closed system" "Let's solve the problem for m[1] = 0.05 kg" m[1] = 0.05 [kg] "Find the temperatures from the pressure and specific volume." T[1]=temperature(gas$,P=P[1],v=V[1]/m[1]) T[2]=temperature(gas$,P=P[2],v=V[2]/m[2])

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

4-10

160 140 120

P [kPa]

100 80 60 40 20 0 0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

0.2

3

V [m ]

Wb [kJ] 7.776 7.393 7.035 6.7 6.387 6.094 5.82 5.564 5.323 5.097

8 7.5 7

Wb [kJ]

n 1.1 1.156 1.211 1.267 1.322 1.378 1.433 1.489 1.544 1.6

6.5 6 5.5 5 1.1

1.2

1.3

1.4

1.5

1.6

n

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4-11

4-19 Nitrogen gas in a cylinder is compressed polytropically until the temperature rises to a specified value. The boundary work done during this process is to be determined. Assumptions 1 The process is quasi-equilibrium. 2 Nitrogen is an ideal gas. Properties The gas constant for nitrogen is R = 0.2968 kJ/kg.K (Table A-2a) Analysis The boundary work for this polytropic process can be determined from P V − PV mR(T2 − T1 ) = P dV = 2 2 1 1 = 1 1− n 1− n (2 kg)(0.2968 kJ/kg ⋅ K)(360 − 300)K = 1 − 1.4 = −89.0 kJ

∫

Wb,out

P

2

2

PV n =C 1

Discussion The negative sign indicates that work is done on the system (work input).

V

4-20 [Also solved by EES on enclosed CD] A gas whose equation of state is v ( P + 10 / v 2 ) = Ru T expands in a cylinder isothermally to a specified volume. The unit of the quantity 10 and the boundary work done during this process are to be determined. Assumptions The process is quasi-equilibrium. Analysis (a) The term 10 / v since it is added to P.

2

must have pressure units

P

Thus the quantity 10 must have the unit kPa·m6/kmol2.

T = 300 K

(b) The boundary work for this process can be determined from P=

Ru T

v

−

10

v

2

=

Ru T NRu T 10 N 2 10 − = − V / N (V / N ) 2 V V2

2

4

V

and 2⎞ ⎛ ⎞ ⎜ NRuT − 10 N ⎟dV = NRuT ln V 2 + 10 N 2 ⎜ 1 − 1 ⎟ 2 ⎜ ⎟ ⎜ ⎟ 1 1 V1 V ⎠ ⎝ V 2 V1 ⎠ ⎝ V 4 m3 = (0.5 kmol)(8.314 kJ/kmol ⋅ K)(300 K)ln 2 m3 ⎛ 1 1 ⎞⎛ 1 kJ ⎞ ⎟⎜ ⎟ + (10 kPa ⋅ m 6 /kmol2 )(0.5kmol)2 ⎜⎜ − 3 4m 2 m3 ⎟⎠⎜⎝ 1 kPa ⋅ m3 ⎟⎠ ⎝ = 864 kJ

Wb,out =

∫

2

P dV =

∫

2⎛

Discussion The positive sign indicates that work is done by the system (work output).

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

4-12

4-21 EES Problem 4-20 is reconsidered. Using the integration feature, the work done is to be calculated and compared, and the process is to be plotted on a P-V diagram. Analysis The problem is solved using EES, and the solution is given below. "Input Data" N=0.5 [kmol] v1_bar=2/N "[m^3/kmol]" v2_bar=4/N "[m^3/kmol]" T=300 [K] R_u=8.314 [kJ/kmol-K] "The quation of state is:" v_bar*(P+10/v_bar^2)=R_u*T "P is in kPa" "using the EES integral function, the boundary work, W_bEES, is" W_b_EES=N*integral(P,v_bar, v1_bar, v2_bar,0.01) "We can show that W_bhand= integeral of Pdv_bar is (one should solve for P=F(v_bar) and do the integral 'by hand' for practice)." W_b_hand = N*(R_u*T*ln(v2_bar/v1_bar) +10*(1/v2_bar-1/v1_bar)) "To plot P vs v_bar, define P_plot =f(v_bar_plot, T) as" {v_bar_plot*(P_plot+10/v_bar_plot^2)=R_u*T} " P=P_plot and v_bar=v_bar_plot just to generate the parametric table for plotting purposes. To plot P vs v_bar for a new temperature or v_bar_plot range, remove the '{' and '}' from the above equation, and reset the v_bar_plot values in the Parametric Table. Then press F3 or select Solve Table from the Calculate menu. Next select New Plot Window under the Plot menu to plot the new data."

vplot 4 4.444 4.889 5.333 5.778 6.222 6.667 7.111 7.556 8

P vs v bar

650 600

1

550 500

T = 300 K

450

P plot [kPa]

Pplot 622.9 560.7 509.8 467.3 431.4 400.6 373.9 350.5 329.9 311.6

400 350

2

300 250

Area = W boundary

200 150 100 50 0 3.5

4.0

4.5

5.0

v

5.5

plot

6.0

6.5

7.0

7.5

8.0

[m ^3/km ol]

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

8.5

4-13

4-22 CO2 gas in a cylinder is compressed until the volume drops to a specified value. The pressure changes during the process with volume as P = aV −2 . The boundary work done during this process is to be determined. Assumptions The process is quasi-equilibrium.

P

Analysis The boundary work done during this process is determined from Wb,out =

2

∫ PdV = ∫ 1

2

⎛ 1 a ⎞ 1 ⎞ − ⎟⎟ ⎜ 2 ⎟dV = − a⎜⎜ ⎝V ⎠ ⎝ V 2 V1 ⎠

2⎛

1

⎛ 1 1 = −(8 kPa ⋅ m 6 )⎜ − ⎜ 0.1 m 3 0.3 m 3 ⎝ = −53.3 kJ

⎞⎛ 1 kJ ⎟⎜ ⎟⎜ 1 kPa ⋅ m 3 ⎠⎝

P = aV--2 ⎞ ⎟ ⎟ ⎠

1 0.1

0.3

V

(m3)

Discussion The negative sign indicates that work is done on the system (work input).

4-23 Several sets of pressure and volume data are taken as a gas expands. The boundary work done during this process is to be determined using the experimental data. Assumptions The process is quasi-equilibrium. Analysis Plotting the given data on a P-V diagram on a graph paper and evaluating the area under the process curve, the work done is determined to be 0.25 kJ.

4-24 A piston-cylinder device contains nitrogen gas at a specified state. The boundary work is to be determined for the isothermal expansion of nitrogen. Properties The properties of nitrogen are R = 0.2968 kJ/kg.K , k = 1.4 (Table A-2a). Analysis We first determine initial and final volumes from ideal gas relation, and find the boundary work using the relation for isothermal expansion of an ideal gas

V1 =

mRT (0.25 kg)(0.2968 kJ/kg.K)(120 + 273 K) = = 0.2243 m 3 P1 (130 kPa)

V2 =

mRT (0.25 kg)(0.2968 kJ/kg.K)(120 + 273 K) = = 0.2916 m 3 P2 (100 kPa)

⎛V Wb = P1V1 ln⎜⎜ 2 ⎝ V1

⎛ 0.2916 m 3 ⎞ ⎟⎟ = (130 kPa)(0.2243 m 3 ) ln⎜ ⎜ 0.2243 m 3 ⎠ ⎝

⎞ ⎟ = 7.65 kJ ⎟ ⎠

N2 130 kPa 120°C

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

4-14

4-25 A piston-cylinder device contains air gas at a specified state. The air undergoes a cycle with three processes. The boundary work for each process and the net work of the cycle are to be determined. Properties The properties of air are R = 0.287 kJ/kg.K , k = 1.4 (Table A-2a). Analysis For the isothermal expansion process:

V1 = V2 =

mRT (0.15 kg)(0.287 kJ/kg.K)(350 + 273 K) = = 0.01341 m 3 P1 (2000 kPa) mRT (0.15 kg)(0.287 kJ/kg.K)(350 + 273 K) = = 0.05364 m 3 P2 (500 kPa)

Air 2 MPa 350°C

⎛ 0.05364 m3 ⎞ ⎛V ⎞ ⎟ = 37.18 kJ Wb,1− 2 = P1V1 ln⎜⎜ 2 ⎟⎟ = (2000 kPa)(0.01341 m3 ) ln⎜ ⎜ 0.01341 m3 ⎟ ⎝ V1 ⎠ ⎝ ⎠

For the polytropic compression process: P2V 2n = P3V 3n ⎯ ⎯→(500 kPa)(0.05364 m 3 )1.2 = (2000 kPa)V 31.2 ⎯ ⎯→V 3 = 0.01690 m 3 Wb , 2 − 3 =

P3V 3 − P2V 2 (2000 kPa)(0.01690 m 3 ) − (500 kPa)(0.05364 m 3 ) = = -34.86 kJ 1− n 1 − 1.2

For the constant pressure compression process: Wb,3−1 = P3 (V 1 −V 3 ) = (2000 kPa)(0.01341 − 0.01690)m 3 = -6.97 kJ

The net work for the cycle is the sum of the works for each process Wnet = Wb,1− 2 + Wb,2−3 + Wb,3−1 = 37.18 + (−34.86) + (−6.97) = -4.65 kJ

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

4-15

4-26 A saturated water mixture contained in a spring-loaded piston-cylinder device is heated until the pressure and temperature rises to specified values. The work done during this process is to be determined. Assumptions The process is quasi-equilibrium. Analysis The initial state is saturated mixture at 90°C. The pressure and the specific volume at this state are (Table A-4), P1 = 70.183 kPa

P 2

800 kPa

v 1 = v f + xv fg = 0.001036 + (0.10)(2.3593 − 0.001036)

1

3

= 0.23686 m /kg

v

The final specific volume at 800 kPa and 250°C is (Table A-6)

v 2 = 0.29321 m 3 /kg Since this is a linear process, the work done is equal to the area under the process line 1-2: P1 + P2 m(v 2 − v 1 ) 2 (70.183 + 800)kPa ⎛ 1 kJ ⎞ (1 kg)(0.29321 − 0.23686)m 3 ⎜ = ⎟ 2 ⎝ 1 kPa ⋅ m 3 ⎠ = 24.52 kJ

Wb,out = Area =

4-27 A saturated water mixture contained in a spring-loaded piston-cylinder device is cooled until it is saturated liquid at a specified temperature. The work done during this process is to be determined. Assumptions The process is quasi-equilibrium. Analysis The initial state is saturated mixture at 1 MPa. The specific volume at this state is (Table A-5),

P

v 1 = v f + xv fg = 0.001127 + (0.10)(0.19436 − 0.001127)

1 MPa

1

3

= 0.020450 m /kg

2

The final state is saturated liquid at 100°C (Table A-4) P2 = 101.42 kPa

v

v 2 = v f = 0.001043 m 3 /kg Since this is a linear process, the work done is equal to the area under the process line 1-2: P1 + P2 m(v 2 − v 1 ) 2 (1000 + 101.42)kPa ⎛ 1 kJ ⎞ = (0.5 kg)(0.001043 − 0.020450)m 3 ⎜ ⎟ 2 ⎝ 1 kPa ⋅ m 3 ⎠ = −5.34 kJ

Wb,out = Area =

The negative sign shows that the work is done on the system in the amount of 5.34 kJ.

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

4-16

4-28 Argon is compressed in a polytropic process. The final temperature is to be determined. Assumptions The process is quasi-equilibrium. Analysis For a polytropic expansion or compression process, Pv n = Constant

For an ideal gas, Pv = RT

Combining these equations produces ⎛P T2 = T1 ⎜⎜ 2 ⎝ P1

⎞ ⎟⎟ ⎠

( n-1 )/n

⎛ 1200 kPa ⎞ = (303 K)⎜ ⎟ ⎝ 120 kPa ⎠

0.2 / 1.2

= 444.7 K

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

4-17

Closed System Energy Analysis 4-29 Saturated water vapor is isothermally condensed to a saturated liquid in a piston-cylinder device. The heat transfer and the work done are to be determined. Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2 There are no work interactions involved other than the boundary work. 3 The thermal energy stored in the cylinder itself is negligible. 4 The compression or expansion process is quasi-equilibrium. Analysis We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as E −E 1in424out 3

Net energy transfer by heat, work, and mass

=

ΔE system 1 424 3

Change in internal, kinetic, potential, etc. energies

Wb,in − Qout = ΔU = m(u 2 − u1 )

Water 200°C sat. vapor

(since KE = PE = 0)

Qout = Wb,in − m(u 2 − u1 )

Heat

The properties at the initial and final states are (Table A-4) T1 = 200°C ⎫ v 1 = v g = 0.12721 m 3 / kg ⎬ x1 = 1 ⎭ u1 = u g = 2594.2 kJ/kg P1 = P2 = 1554.9 kPa T2 = 200°C ⎫ v 2 = v f = 0.001157 m 3 / kg ⎬ x2 = 0 ⎭ u 2 = u f = 850.46 kJ/kg

T

2

1

v

The work done during this process is wb,out =

∫

2

1

⎛ 1 kJ P dV = P (v 2 − v 1 ) = (1554.9 kPa)(0.001157 − 0.12721) m 3 /kg⎜ ⎜ 1 kPa ⋅ m 3 ⎝

⎞ ⎟ = −196.0 kJ/kg ⎟ ⎠

That is,

wb,in = 196.0 kJ/kg Substituting the energy balance equation, we get q out = wb,in − (u 2 − u1 ) = wb,in + u fg = 196.0 + 1743.7 = 1940 kJ/kg

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

4-18

4-30E The heat transfer during a process that a closed system undergoes without any internal energy change is to be determined. Assumptions 1 The system is stationary and thus the kinetic and potential energy changes are zero. 2 The compression or expansion process is quasi-equilibrium. Analysis The energy balance for this stationary closed system can be expressed as E −E 1in424out 3

=

Net energy transfer by heat, work, and mass

ΔE system 1 424 3

Change in internal, kinetic, potential, etc. energies

Qin − Wout = ΔU = 0

(since KE = PE = 0)

Qin = Wout

Then, 1 Btu ⎛ ⎞ Qin = 1.6 × 10 6 lbf ⋅ ft ⎜ ⎟ = 2056 Btu ⎝ 778.17 lbf ⋅ ft ⎠

4-31 The table is to be completed using conservation of energy principle for a closed system. Analysis The energy balance for a closed system can be expressed as E −E 1in424out 3

Net energy transfer by heat, work, and mass

=

ΔE system 1 424 3

Change in internal, kinetic, potential, etc. energies

Qin − Wout = E 2 − E1 = m(e 2 − e1 )

Application of this equation gives the following completed table:

Qin

Wout

E1

E2

m

(kJ)

(kJ)

(kJ)

(kJ)

(kg)

e2 − e1 (kJ/kg)

280

440

1020

860

3

-53.3

-350

130

550

70

5

-96

-40

260

300

0

2

-150

300

550

750

500

1

-250

-400

-200

500

300

2

-100

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

4-19

4-32 A substance is contained in a well-insulated, rigid container that is equipped with a stirring device. The change in the internal energy of this substance for a given work input is to be determined. Assumptions 1 The tank is stationary and thus the kinetic and potential energy changes are zero. 2 The tank is insulated and thus heat transfer is negligible. Analysis This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as E −E 1in424out 3

=

Net energy transfer by heat, work, and mass

ΔE system 1 424 3

Change in internal, kinetic, potential, etc. energies

Wsh,in = ΔU

(since KE = PE = 0)

Then, ΔU = 15 kJ

4-33 Motor oil is contained in a rigid container that is equipped with a stirring device. The rate of specific energy increase is to be determined. Analysis This is a closed system since no mass enters or leaves. The energy balance for closed system can be expressed as E −E 1in424out 3

Net energy transfer by heat, work, and mass

=

ΔE system 1 424 3

Change in internal, kinetic, potential, etc. energies

Q& in + W& sh,in = ΔE&

Then, ΔE& = Q& in + W& sh,in = 1 + 1.5 = 2.5 = 2.5 W

Dividing this by the mass in the system gives

Δe& =

ΔE& 2.5 J/s = = 1.67 J/kg ⋅ s m 1.5 kg

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

4-20

4-34E R-134a contained in a rigid vessel is heated. The heat transfer is to be determined. Assumptions 1 The system is stationary and thus the kinetic and potential energy changes are zero. 2 There are no work interactions involved 3 The thermal energy stored in the vessel itself is negligible. Analysis We take R-134a as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as E −E 1in424out 3

Net energy transfer by heat, work, and mass

=

R-134a 1 ft3 −20°F x = 0.277

ΔE system 1 424 3

Change in internal, kinetic, potential, etc. energies

Qin = ΔU = m(u 2 − u1 )

(since KE = PE = 0)

Q

The properties at the initial and final states are (Tables A-11E, A-13E) T1 = −20°F ⎫ v 1 = v f + xv fg = 0.01156 + (0.277)(3.4426 − 0.01156) = 0.96196 ft 3 / lbm ⎬ x1 = 0.277 ⎭ u1 = u f + xu fg = 6.019 + (0.277)(85.874) = 29.81 Btu/lbm T2 = 100°F

⎫⎪ ⎬ u 2 = 111.30 Btu/lbm v 2 = v 1 = 0.96196 ft / lbm ⎪⎭ 3

T

Note that the final state is superheated vapor and the internal energy at this state should be obtained by interpolation using 50 psia and 60 psia mini tables (100°F line) in Table A-13E. The mass in the system is m=

V1 1 ft 3 = = 1.0395 lbm v 1 0.96196 ft 3 /lbm

2

1

v

Substituting, Qin = m(u 2 − u1 ) = (1.0395 lbm)(111.30 − 29.81) Btu/lbm = 84.7 Btu

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4-21

4-35 An insulated rigid tank is initially filled with a saturated liquid-vapor mixture of water. An electric heater in the tank is turned on, and the entire liquid in the tank is vaporized. The length of time the heater was kept on is to be determined, and the process is to be shown on a P-v diagram. Assumptions 1 The tank is stationary and thus the kinetic and potential energy changes are zero. 2 The device is well-insulated and thus heat transfer is negligible. 3 The energy stored in the resistance wires, and the heat transferred to the tank itself is negligible. Analysis We take the contents of the tank as the system. This is a closed system since no mass enters or leaves. Noting that the volume of the system is constant and thus there is no boundary work, the energy balance for this stationary closed system can be expressed as E − Eout 1in424 3

Net energy transfer by heat, work, and mass

=

ΔEsystem 1 424 3

H2O

Change in internal, kinetic, potential, etc. energies

We,in = ΔU = m(u2 − u1 )

V = const.

(since Q = KE = PE = 0)

VIΔt = m(u2 − u1 )

We

The properties of water are (Tables A-4 through A-6) P1 = 100kPa ⎫ v f = 0.001043, v g = 1.6941 m3 /kg ⎬ x1 = 0.25 ⎭ u f = 417.40, u fg = 2088.2 kJ/kg

T

v1 = v f + x1v fg = 0.001043 + [0.25 × (1.6941 − 0.001043)] = 0.42431 m3/kg

2

v 2 = v1 = 0.42431 m3/kg ⎫⎪

1

u1 = u f + x1u fg = 417.40 + (0.25 × 2088.2 ) = 939.4 kJ/kg

sat.vapor

⎬ u2 = u g @ 0.42431m 3 /kg = 2556.2 kJ/kg ⎪⎭

Substituting, ⎛ 1000 VA ⎞ ⎟⎟ (110 V)(8 A)Δt = (5 kg)(2556.2 − 939.4)kJ/kg⎜⎜ ⎝ 1 kJ/s ⎠ Δt = 9186 s ≅ 153.1 min

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v

4-22

4-36 EES Problem 4-35 is reconsidered. The effect of the initial mass of water on the length of time required to completely vaporize the liquid as the initial mass varies from 1 kg to 10 kg is to be investigated. The vaporization time is to be plotted against the initial mass. Analysis The problem is solved using EES, and the solution is given below. PROCEDURE P2X2(v[1]:P[2],x[2]) Fluid$='Steam_IAPWS' If v[1] > V_CRIT(Fluid$) then P[2]=pressure(Fluid$,v=v[1],x=1) x[2]=1 else P[2]=pressure(Fluid$,v=v[1],x=0) x[2]=0 EndIf End "Knowns" {m=5 [kg]} P[1]=100 [kPa] y=0.75 "moisture" Volts=110 [V] I=8 [amp] "Solution" "Conservation of Energy for the closed tank:" E_dot_in-E_dot_out=DELTAE_dot E_dot_in=W_dot_ele "[kW]" W_dot_ele=Volts*I*CONVERT(J/s,kW) "[kW]" E_dot_out=0 "[kW]" DELTAE_dot=m*(u[2]-u[1])/DELTAt_s "[kW]" DELTAt_min=DELTAt_s*convert(s,min) "[min]" "The quality at state 1 is:" Fluid$='Steam_IAPWS' x[1]=1-y u[1]=INTENERGY(Fluid$,P=P[1], x=x[1]) "[kJ/kg]" v[1]=volume(Fluid$,P=P[1], x=x[1]) "[m^3/kg]" T[1]=temperature(Fluid$,P=P[1], x=x[1]) "[C]" "Check to see if state 2 is on the saturated liquid line or saturated vapor line:" Call P2X2(v[1]:P[2],x[2]) u[2]=INTENERGY(Fluid$,P=P[2], x=x[2]) "[kJ/kg]" v[2]=volume(Fluid$,P=P[2], x=x[2]) "[m^3/kg]" T[2]=temperature(Fluid$,P=P[2], x=x[2]) "[C]"

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4-23

S te a m

700 600

T [°C]

500 400 300 200

2 43 7.9 kP a

100

10 0 kP a

1

0 1 0 -3

0.0 5

1 0 -2

1 0 -1

0.1

100

0 .2

0 .5

101

3

102

103

v [m /k g ]

Δtmin [min] 30.63 61.26 91.89 122.5 153.2 183.8 214.4 245 275.7 306.3

m [kg] 1 2 3 4 5 6 7 8 9 10

350 300

Δ t m in [m in]

250 200 150 100 50 0 1

2

3

4

5

6

7

8

9

10

m [kg]

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4-24

4-37 A cylinder is initially filled with R-134a at a specified state. The refrigerant is cooled at constant pressure. The amount of heat loss is to be determined, and the process is to be shown on a T-v diagram. Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2 There are no work interactions involved other than the boundary work. 3 The thermal energy stored in the cylinder itself is negligible. 4 The compression or expansion process is quasi-equilibrium. Analysis We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as E − Eout 1in 424 3

Net energy transfer by heat, work, and mass

=

ΔEsystem 1 424 3

Change in internal, kinetic, potential, etc. energies

− Qout − Wb,out = ΔU = m(u2 − u1 )

(since KE = PE = 0)

− Qout = m(h2 − h1 )

Q

R-134a 800 kPa

since ΔU + Wb = ΔH during a constant pressure quasiequilibrium process. The properties of R-134a are (Tables A-11 through A-13) P1 = 800 kPa ⎫ ⎬ h1 = 306.88 kJ/kg T1 = 70°C ⎭ P2 = 800 kPa ⎫ ⎬ h2 = h f @15°C = 72.34 kJ/kg T2 = 15°C ⎭

Substituting,

T 1 2

v

Qout = - (5 kg)(72.34 - 306.88) kJ/kg = 1173 kJ

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4-25

4-38E A cylinder contains water initially at a specified state. The water is heated at constant pressure. The final temperature of the water is to be determined, and the process is to be shown on a T-v diagram. Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2 The thermal energy stored in the cylinder itself is negligible. 3 The compression or expansion process is quasiequilibrium. Analysis We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as E −E 1in424out 3

Net energy transfer by heat, work, and mass

=

ΔE system 1 424 3

Change in internal, kinetic, potential, etc. energies

Qin − Wb,out = ΔU = m(u 2 − u1 )

(since KE = PE = 0)

Qin = m(h2 − h1 )

H2O 120 psia

since ΔU + Wb = ΔH during a constant pressure quasi-equilibrium process. The properties of water are (Tables A-6E)

v1 =

V1 m

=

2 ft 3 = 4 ft 3 /lbm 0.5 lbm

P1 = 120 psia ⎫⎪ ⎬ h1 = 1217.0 Btu/lbm v 1 = 4 ft 3 /lbm ⎪⎭

Q

T 2 1

Substituting, 200 Btu = (0.5 lbm)(h2 − 1217.0)Btu/lbm

v

h2 = 1617.0 Btu/lbm

Then, P2 = 120 psia

⎫ ⎬ T2 = 1161.4°F h2 = 1617.0 Btu/lbm ⎭

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

4-26

4-39 A cylinder is initially filled with saturated liquid water at a specified pressure. The water is heated electrically as it is stirred by a paddle-wheel at constant pressure. The voltage of the current source is to be determined, and the process is to be shown on a P-v diagram. Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2 The cylinder is well-insulated and thus heat transfer is negligible. 3 The thermal energy stored in the cylinder itself is negligible. 4 The compression or expansion process is quasi-equilibrium. Analysis We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as E −E 1in424out 3

=

Net energy transfer by heat, work, and mass

ΔE system 1 424 3

Change in internal, kinetic, potential, etc. energies

We,in + W pw,in − W b,out = ΔU

(since Q = KE = PE = 0)

H2O P = const.

We,in + W pw,in = m(h2 − h1 ) ( VIΔt ) + W pw,in = m(h2 − h1 )

since ΔU + Wb = ΔH during a constant pressure quasi-equilibrium process. The properties of water are (Tables A-4 through A-6)

Wpw

We

P1 = 175 kPa ⎫ h1 = h f @175 kPa = 487.01 kJ/kg ⎬ 3 sat.liquid ⎭ v1 = v f @175 kPa = 0.001057 m /kg P2 = 175 kPa ⎫ ⎬ h2 = h f + x2 h fg = 487.01 + (0.5 × 2213.1) = 1593.6 kJ/kg x2 = 0.5 ⎭ m=

0.005 m3 V1 = = 4.731 kg v1 0.001057 m3/kg

P

Substituting, VIΔt + (400kJ) = (4.731 kg)(1593.6 − 487.01)kJ/kg VIΔt = 4835 kJ V=

⎛ 1000 VA ⎞ 4835 kJ ⎜ ⎟ = 223.9 V (8 A)(45 × 60 s) ⎜⎝ 1 kJ/s ⎟⎠

1

2

v

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4-27

4-40 [Also solved by EES on enclosed CD] A cylinder equipped with an external spring is initially filled with steam at a specified state. Heat is transferred to the steam, and both the temperature and pressure rise. The final temperature, the boundary work done by the steam, and the amount of heat transfer are to be determined, and the process is to be shown on a P-v diagram. Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2 The thermal energy stored in the cylinder itself is negligible. 3 The compression or expansion process is quasiequilibrium. 4 The spring is a linear spring. Analysis We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. Noting that the spring is not part of the system (it is external), the energy balance for this stationary closed system can be expressed as E −E 1in424out 3

=

Net energy transfer by heat, work, and mass

ΔE system 1 424 3

Change in internal, kinetic, potential, etc. energies

Qin − Wb,out = ΔU = m(u 2 − u1 )

(since KE = PE = 0)

H2O 200 kPa 200°C

Qin = m(u 2 − u1 ) + Wb,out

The properties of steam are (Tables A-4 through A-6) P1 = 200 kPa ⎫ v 1 = 1.08049 m 3 /kg ⎬ T1 = 200°C ⎭ u1 = 2654.6 kJ/kg m=

P 2

3

V1 0.5 m = = 0.4628 kg v 1 1.08049 m 3 /kg

v2 =

Q

V2 m

=

1

0.6 m 3 = 1.2966 m 3 /kg 0.4628 kg

P2 = 500 kPa

⎫⎪ T2 = 1132°C ⎬ v 2 = 1.2966 m /kg ⎪⎭ u 2 = 4325.2 kJ/kg

v

3

(b) The pressure of the gas changes linearly with volume, and thus the process curve on a P-V diagram will be a straight line. The boundary work during this process is simply the area under the process curve, which is a trapezoidal. Thus, Wb = Area =

⎛ P1 + P2 (V 2 −V1 ) = (200 + 500)kPa (0.6 − 0.5)m 3 ⎜⎜ 1 kJ 3 2 2 ⎝ 1 kPa ⋅ m

⎞ ⎟ = 35 kJ ⎟ ⎠

(c) From the energy balance we have Qin = (0.4628 kg)(4325.2 - 2654.6)kJ/kg + 35 kJ = 808 kJ

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

4-28

4-41 EES Problem 4-40 is reconsidered. The effect of the initial temperature of steam on the final temperature, the work done, and the total heat transfer as the initial temperature varies from 150°C to 250°C is to be investigated. The final results are to be plotted against the initial temperature. Analysis The problem is solved using EES, and the solution is given below. "The process is given by:" "P[2]=P[1]+k*x*A/A, and as the spring moves 'x' amount, the volume changes by V[2]-V[1]." P[2]=P[1]+(Spring_const)*(V[2] - V[1]) "P[2] is a linear function of V[2]" "where Spring_const = k/A, the actual spring constant divided by the piston face area" "Conservation of mass for the closed system is:" m[2]=m[1] "The conservation of energy for the closed system is" "E_in - E_out = DeltaE, neglect DeltaKE and DeltaPE for the system" Q_in - W_out = m[1]*(u[2]-u[1]) DELTAU=m[1]*(u[2]-u[1]) "Input Data" P[1]=200 [kPa] V[1]=0.5 [m^3] "T[1]=200 [C]" P[2]=500 [kPa] V[2]=0.6 [m^3] Fluid$='Steam_IAPWS' m[1]=V[1]/spvol[1] spvol[1]=volume(Fluid$,T=T[1], P=P[1]) u[1]=intenergy(Fluid$, T=T[1], P=P[1]) spvol[2]=V[2]/m[2] "The final temperature is:" T[2]=temperature(Fluid$,P=P[2],v=spvol[2]) u[2]=intenergy(Fluid$, P=P[2], T=T[2]) Wnet_other = 0 W_out=Wnet_other + W_b "W_b = integral of P[2]*dV[2] for 0.5 T4 s

where T4s is the temperature that would occur if the expansion were reversible and adiabatic (n=k). This can only occur when n≤k

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9-28

9-47 An ideal Otto cycle is considered. The heat rejection, the net work production, the thermal efficiency, and the mean effective pressure are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are R = 0.287 kJ/kg.K, cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-2a). Analysis The mass in this system is m=

P1V1 (90 kPa)(0.004 m 3 ) = = 0.004181 kg RT1 (0.287 kPa ⋅ m 3 /kg ⋅ K )(300 K)

P

3

⎛v T2 = T1 ⎜⎜ 1 ⎝v 2 ⎛v T4 = T3 ⎜⎜ 3 ⎝v 4

⎞ ⎟⎟ ⎠

k −1

⎞ ⎟⎟ ⎠

k −1

2

⎛1⎞ = T3 ⎜ ⎟ ⎝r⎠

⎛1⎞ = (1400 K)⎜ ⎟ ⎝7⎠

qout 1

= T1 r k −1 = (300 K)(7 )1.4−1 = 653.4 K k −1

4

qin

The two unknown temperatures are

v

1.4 −1

= 642.8 K

Application of the first law to four cycle processes gives W1− 2 = mcv (T2 − T1 ) = (0.004181 kg)(0.718 kJ/kg ⋅ K )(653.4 − 300)K = 1.061 kJ Q 2−3 = mcv (T3 − T2 ) = (0.004181 kg)(0.718 kJ/kg ⋅ K )(1400 − 653.4)K = 2.241 kJ W3− 4 = mcv (T3 − T4 ) = (0.004181 kg)(0.718 kJ/kg ⋅ K )(1400 − 642.8)K = 2.273 kJ Q 4−1 = mcv (T4 − T1 ) = (0.004181 kg)(0.718 kJ/kg ⋅ K )(642.8 − 300)K = 1.029 kJ

The net work is W net = W3− 4 − W1− 2 = 2.273 − 1.061 = 1.212 kJ

The thermal efficiency is then

η th =

W net 1.212 kJ = = 0.541 Qin 2.241 kJ

The minimum volume of the cycle occurs at the end of the compression

V2 =

V1 r

=

0.004 m 3 = 0.0005714 m 3 7

The engine’s mean effective pressure is then MEP =

W net 1.212 kJ = V1 −V 2 (0.004 − 0.0005714) m 3

⎛ 1 kPa ⋅ m 3 ⎜ ⎜ 1 kJ ⎝

⎞ ⎟ = 354 kPa ⎟ ⎠

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9-29

9-48 The power produced by an ideal Otto cycle is given. The rate of heat addition is to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are R = 0.287 kPa·m3/kg.K, cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-2a). Analysis The compression ratio is r=

v1 v1 = = 6.667 v 2 0.15v 1

P

qin

and the thermal efficiency is

η th = 1 −

1

r k −1

= 1−

3

2 1 6.667 1.4 −1

= 0.5318

qout

4 1

v

The rate at which heat must be added to this engine is then W& 90 hp ⎛ 0.7457 kW ⎞ ⎜ ⎟⎟ = 126.2 kW Q& in = net = η th 0.5318 ⎜⎝ 1 hp ⎠

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9-30

Diesel Cycle

9-49C A diesel engine differs from the gasoline engine in the way combustion is initiated. In diesel engines combustion is initiated by compressing the air above the self-ignition temperature of the fuel whereas it is initiated by a spark plug in a gasoline engine.

9-50C The Diesel cycle differs from the Otto cycle in the heat addition process only; it takes place at constant volume in the Otto cycle, but at constant pressure in the Diesel cycle.

9-51C The gasoline engine.

9-52C Diesel engines operate at high compression ratios because the diesel engines do not have the engine knock problem.

9-53C Cutoff ratio is the ratio of the cylinder volumes after and before the combustion process. As the cutoff ratio decreases, the efficiency of the diesel cycle increases.

9-54 An expression for cutoff ratio of an ideal diesel cycle is to be developed. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Analysis Employing the isentropic process equations, T2 = T1 r k −1

P

2

qin

3

while the ideal gas law gives T3 = T2 rc = rc r k −1T1

When the first law and the closed system work integral is applied to the constant pressure heat addition, the result is q in = c p (T3 − T2 ) = c p (rc r k −1T1 − r k −1T1 )

4 qout 1

v

When this is solved for cutoff ratio, the result is rc = 1 +

q in c p rc r k −1T1

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9-31

9-55 An ideal diesel cycle has a compression ratio of 18 and a cutoff ratio of 1.5. The maximum temperature of the air and the rate of heat addition are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are R = 0.287 kPa·m3/kg⋅K, cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-2a). Analysis We begin by using the process types to fix the temperatures of the states.

P

k −1

⎛v T2 = T1 ⎜⎜ 1 ⎝v 2

⎞ ⎟⎟ ⎠

⎛v T3 = T2 ⎜⎜ 3 ⎝v 2

⎞ ⎟⎟ = T2 rc = (921.5 K)(1.5) = 1382 K ⎠

= T1 r k −1 = (290 K)(18)1.4−1 = 921.5 K

1 r k −1

qin

3 4

Combining the first law as applied to the various processes with the process equations gives

η th = 1 −

2

qout 1

v

rck − 1 1 1.51.4 − 1 = 1 − 1.4 −1 = 0.6565 1.4(1.5 − 1) k (rc − 1) 18

According to the definition of the thermal efficiency, W& 200 hp ⎛ 0.7457 kW ⎞ ⎜ ⎟⎟ = 227.2 kW Q& in = net = 0.6565 ⎜⎝ 1 hp η th ⎠

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9-32

9-56 A Diesel cycle with non-isentropic compression and expansion processes is considered. The maximum temperature of the air and the rate of heat addition are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are R = 0.287 kPa·m3/kg⋅K, cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-2a). Analysis We begin by determining the temperatures of the cycle states qin using the process equations and component efficiencies. The ideal P 2 3 temperature at the end of the compression is then k −1

⎛v ⎞ T2 s = T1 ⎜⎜ 1 ⎟⎟ = T1 r k −1 = (290 K)(18)1.4−1 = 921.5 K ⎝v 2 ⎠ With the isentropic compression efficiency, the actual temperature at the end of the compression is T −T T −T (921.5 − 290) K = 991.7 K η = 2 s 1 ⎯⎯→ T2 = T1 + 2 s 1 = (290 K) + T2 − T1 0.90 η

4 qout 1

v

The maximum temperature is ⎛v ⎞ T3 = T2 ⎜⎜ 3 ⎟⎟ = T2 rc = (991.7 K)(1.5) = 1488 K ⎝v 2 ⎠ For the isentropic expansion process, ⎛v T4 s = T3 ⎜⎜ 3 ⎝v 4

⎞ ⎟⎟ ⎠

k −1

⎛r = T3 ⎜⎜ c ⎝ r

⎞ ⎟⎟ ⎠

k −1

⎛ 1.5 ⎞ = (1488 K)⎜ ⎟ ⎝ 18 ⎠

1.4 −1

= 550.7 K

since ⎫ ⎪ ⎪ rc v 3 / v 2 v 3 = = ⎬ ⎪ r v 4 /v 2 v 4 ⎪⎭ The actual temperature at the end of expansion process is then T −T η = 3 4 ⎯⎯→ T4 = T3 − η (T3 − T4 s ) = (1488 K) − (0.95)(1488 − 550.7) K = 597.6 K T3 − T4 s

v3 v2 v r= 4 v2 rc =

The net work production is the difference between the work produced by the expansion and that used by the compression, wnet = cv (T3 − T4 ) − cv (T2 − T1 ) = (0.718 kJ/kg ⋅ K )(1488 − 597.6)K − (0.718 kJ/kg ⋅ K )(991.7 − 290)K = 135.5 kJ/kg

The heat addition occurs during process 2-3, q in = c p (T3 − T2 ) = (1.005 kJ/kg ⋅ K )(1488 − 991.7)K = 498.8 kJ/kg The thermal efficiency of this cycle is then w 135.5 kJ/kg η th = net = = 0.2717 q in 498.8 kJ/kg According to the definition of the thermal efficiency, W& 200 hp ⎛ 0.7457 kW ⎞ ⎟⎟ = 548.9 kW ⎜ Q& in = net = η th 0.2717 ⎜⎝ 1 hp ⎠

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9-33

9-57 An ideal diesel cycle has a a cutoff ratio of 1.2. The power produced is to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4 (Table A-2a). Analysis The specific volume of the air at the start of the compression is

v1 =

P

2

qin

RT1 (0.287 kPa ⋅ m 3 /kg ⋅ K )(288 K) = = 0.8701 m 3 /kg P1 95 kPa

The total air mass taken by all 8 cylinders when they are charged is m = N cyl

ΔV

v1

= N cyl

π (0.10 m) 2 (0.12 m)/4 πB 2 S / 4 = (8) = 0.008665 kg v1 0.8701 m 3 /kg

3 4 qout 1

v

The rate at which air is processed by the engine is determined from m& =

(0.008665 kg/cycle)(1600/60 rev/s) mn& = = 0.1155 kg/s N rev 2 rev/cycle

since there are two revolutions per cycle in a four-stroke engine. The compression ratio is r=

1 = 20 0.05

At the end of the compression, the air temperature is T2 = T1 r k −1 = (288 K)(20 )1.4 −1 = 954.6 K

Application of the first law and work integral to the constant pressure heat addition gives q in = c p (T3 − T2 ) = (1.005 kJ/kg ⋅ K )(2273 − 954.6)K = 1325 kJ/kg

while the thermal efficiency is

η th = 1 −

1 r k −1

rck − 1 1 1.21.4 − 1 = 1 − 1.4 −1 = 0.6867 1.4(1.2 − 1) k ( rc − 1) 20

The power produced by this engine is then W& net = m& wnet = m& η th q in = (0.1155 kg/s)(0.6867)(1325 kJ/kg) = 105.1 kW

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-34

9-58E An ideal dual cycle has a compression ratio of 20 and cutoff ratio of 1.3. The thermal efficiency, amount of heat added, and the maximum gas pressure and temperature are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are R = 0.3704 psia·ft3/lbm.R (Table A-1E), cp = 0.240 Btu/lbm·R, cv = 0.171 Btu/lbm·R, and k = 1.4 (Table A-2Ea). Analysis Working around the cycle, the germane properties at the various states are ⎛v T2 = T1 ⎜⎜ 1 ⎝v 2 ⎛v P2 = P1 ⎜⎜ 1 ⎝v 2

⎞ ⎟⎟ ⎠

k −1

P

= T1 r k −1 = (530 R)(20 )1.4−1 = 1757 R

x 2

3 qin

4 qout

k

⎞ ⎟⎟ = P1 r k = (14 psia)(20)1.4 = 928 psia ⎠

1

v

Px = P3 = r p P2 = (1.2)(928 psia) = 1114 psia ⎛P T x = T2 ⎜⎜ x ⎝ P2

⎞ ⎛ 1114 psia ⎞ ⎟⎟ = (1757 R)⎜⎜ ⎟⎟ = 2109 R ⎝ 928 psia ⎠ ⎠

⎛v T3 = T x ⎜⎜ 3 ⎝v x

⎞ ⎟ = T x rc = (2109 R)(1.3) = 2742 R ⎟ ⎠

⎛v T4 = T3 ⎜⎜ 3 ⎝v4

⎞ ⎟⎟ ⎠

k −1

⎛r = T3 ⎜⎜ c ⎝ r

⎞ ⎟⎟ ⎠

k −1

⎛ 1.3 ⎞ = (2742 R)⎜ ⎟ ⎝ 20 ⎠

1.4 −1

= 918.8 R

Applying the first law and work expression to the heat addition processes gives q in = cv (T x − T2 ) + c p (T3 − T x ) = (0.171 Btu/lbm ⋅ R )(2109 − 1757)R + (0.240 Btu/lbm ⋅ R )(2742 − 2109)R = 212.1 Btu/lbm

The heat rejected is q out = cv (T4 − T1 ) = (0.171 Btu/lbm ⋅ R )(918.8 − 530)R = 66.48 Btu/lbm

Then,

η th = 1 −

q out 66.48 Btu/lbm = 1− = 0.687 212.1 Btu/lbm q in

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-35

9-59E An ideal dual cycle has a compression ratio of 12 and cutoff ratio of 1.3. The thermal efficiency, amount of heat added, and the maximum gas pressure and temperature are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are R = 0.3704 psia·ft3/lbm.R (Table A-1E), cp = 0.240 Btu/lbm·R, cv = 0.171 Btu/lbm·R, and k = 1.4 (Table A-2Ea). Analysis Working around the cycle, the germane properties at the various states are ⎛v T2 = T1 ⎜⎜ 1 ⎝v 2 ⎛v P2 = P1 ⎜⎜ 1 ⎝v 2

⎞ ⎟⎟ ⎠

k −1

P

= T1 r k −1 = (530 R)(12 )1.4 −1 = 1432 R

x 2

3 qin

4 qout

k

⎞ ⎟⎟ = P1 r k = (14 psia)(12)1.4 = 453.9 psia ⎠

1

v

Px = P3 = r p P2 = (1.2)(453.9 psia) = 544.7 psia ⎛P T x = T2 ⎜⎜ x ⎝ P2

⎞ ⎛ 544.7 psia ⎞ ⎟⎟ = (1432 R)⎜⎜ ⎟⎟ = 1718 R ⎝ 453.9 psia ⎠ ⎠

⎛v T3 = T x ⎜⎜ 3 ⎝v x

⎞ ⎟ = T x rc = (1718 R)(1.3) = 2233 R ⎟ ⎠

⎛v T4 = T3 ⎜⎜ 3 ⎝v 4

⎞ ⎟⎟ ⎠

k −1

⎛r = T3 ⎜⎜ c ⎝ r

⎞ ⎟⎟ ⎠

k −1

⎛ 1.3 ⎞ = (2233 R)⎜ ⎟ ⎝ 12 ⎠

1.4 −1

= 917.9 R

Applying the first law and work expression to the heat addition processes gives q in = cv (T x − T2 ) + c p (T3 − T x ) = (0.171 Btu/lbm ⋅ R )(1718 − 1432)R + (0.240 Btu/lbm ⋅ R )(2233 − 1718)R = 172.5 Btu/lbm

The heat rejected is q out = cv (T4 − T1 ) = (0.171 Btu/lbm ⋅ R )(917.9 − 530)R = 66.33 Btu/lbm

Then,

η th = 1 −

q out 66.33 Btu/lbm = 1− = 0.615 172.5 Btu/lbm q in

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-36

9-60E An air-standard Diesel cycle with a compression ratio of 18.2 is considered. The cutoff ratio, the heat rejection per unit mass, and the thermal efficiency are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The properties of air are given in Table A-17E. Analysis (a) Process 1-2: isentropic compression. ⎯→ T1 = 540 R ⎯

v r2 =

u1 = 92.04 Btu/lbm

v r1 = 144.32

T = 1623.6 R v2 1 1 (144.32) = 7.93 ⎯⎯→ 2 v r1 = v r1 = h2 = 402.05 Btu/lbm v1 r 18.2

Process 2-3: P = constant heat addition. P3v 3 P2v 2 T v 3000 R = ⎯ ⎯→ 3 = 3 = = 1.848 T3 T2 v 2 T2 1623.6 R

(b)

T3 = 3000 R ⎯ ⎯→

h3 = 790.68 Btu/lbm

v r3 = 1.180

q in = h3 − h2 = 790.68 − 402.05 = 388.63 Btu/lbm

P 2

qin

3 3000 R

4 qout 1

v

Process 3-4: isentropic expansion.

v r4 =

v4 v4 r 18.2 vr = vr = (1.180) = 11.621 ⎯⎯→ u 4 = 250.91 Btu/lbm vr = v 3 3 1.848v 2 3 1.848 3 1.848

Process 4-1: v = constant heat rejection. q out = u 4 − u1 = 250.91 − 92.04 = 158.87 Btu/lbm

(c)

η th = 1 −

q out 158.87 Btu/lbm = 1− = 59.1% q in 388.63 Btu/lbm

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-37

9-61E An air-standard Diesel cycle with a compression ratio of 18.2 is considered. The cutoff ratio, the heat rejection per unit mass, and the thermal efficiency are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 0.240 Btu/lbm.R, cv = 0.171 Btu/lbm.R, and k = 1.4 (Table A-2E). Analysis (a) Process 1-2: isentropic compression. ⎛v T2 = T1 ⎜⎜ 1 ⎝v 2

⎞ ⎟⎟ ⎠

k −1

P 2

= (540R )(18.2 )

0.4

qin

3

= 1724R

4

Process 2-3: P = constant heat addition. 1

P3v 3 P2v 2 T v 3000 R = ⎯ ⎯→ 3 = 3 = = 1.741 T3 T2 v 2 T2 1724 R

(b)

v

q in = h3 − h2 = c p (T3 − T2 ) = (0.240 Btu/lbm.R )(3000 − 1724 )R = 306 Btu/lbm

Process 3-4: isentropic expansion. ⎛v T4 = T3 ⎜⎜ 3 ⎝v4

⎞ ⎟⎟ ⎠

k −1

⎛ 1.741v 2 = T3 ⎜⎜ ⎝ v4

⎞ ⎟⎟ ⎠

k −1

⎛ 1.741 ⎞ = (3000 R )⎜ ⎟ ⎝ 18.2 ⎠

0.4

= 1173 R

Process 4-1: v = constant heat rejection. q out = u 4 − u1 = cv (T4 − T1 )

= (0.171 Btu/lbm.R )(1173 − 540 )R = 108 Btu/lbm

(c)

η th = 1 −

q out 108 Btu/lbm = 1− = 64.6% q in 306 Btu/lbm

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-38

9-62 An ideal diesel engine with air as the working fluid has a compression ratio of 20. The thermal efficiency and the mean effective pressure are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4 (Table A-2). Analysis (a) Process 1-2: isentropic compression. ⎛V T2 = T1 ⎜⎜ 1 ⎝V 2

⎞ ⎟⎟ ⎠

k −1

P 2

= (293 K )(20 )0.4 = 971.1 K

qin

3

4

Process 2-3: P = constant heat addition.

qout

P3V 3 P2V 2 T V 2200K = ⎯ ⎯→ 3 = 3 = = 2.265 T3 T2 V 2 T2 971.1K

1

v

Process 3-4: isentropic expansion. ⎞ ⎟⎟ ⎠

⎛V T4 = T3 ⎜⎜ 3 ⎝V 4

k −1

⎛ 2.265V 2 = T3 ⎜⎜ ⎝ V4

⎞ ⎟⎟ ⎠

k −1

⎛ 2.265 ⎞ = T3 ⎜ ⎟ ⎝ r ⎠

k −1

⎛ 2.265 ⎞ = (2200 K )⎜ ⎟ ⎝ 20 ⎠

0.4

= 920.6 K

q in = h3 − h2 = c p (T3 − T2 ) = (1.005 kJ/kg ⋅ K )(2200 − 971.1)K = 1235 kJ/kg q out = u 4 − u1 = cv (T4 − T1 ) = (0.718 kJ/kg ⋅ K )(920.6 − 293)K = 450.6 kJ/kg wnet,out = q in − q out = 1235 − 450.6 = 784.4 kJ/kg

η th =

(b)

v1 =

wnet,out q in

=

784.4 kJ/kg = 63.5% 1235 kJ/kg

(

)

RT1 0.287 kPa ⋅ m 3 /kg ⋅ K (293 K ) = = 0.885 m 3 /kg = v max P1 95 kPa

v min = v 2 = MEP =

v max

r wnet,out

v1 −v 2

=

wnet,out

v 1 (1 − 1 / r )

=

⎛ kPa ⋅ m 3 ⎜ 0.885 m 3 /kg (1 − 1/20 ) ⎜⎝ kJ

(

784.4 kJ/kg

)

⎞ ⎟ = 933 kPa ⎟ ⎠

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-39

9-63 A diesel engine with air as the working fluid has a compression ratio of 20. The thermal efficiency and the mean effective pressure are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = P 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-2). qin 2 3 Analysis (a) Process 1-2: isentropic compression. Polytropic k −1 ⎛ V1 ⎞ 0.4 ⎟ ⎜ T2 = T1 ⎜ ⎟ = (293 K )(20 ) = 971.1 K ⎝V 2 ⎠ 4 qout Process 2-3: P = constant heat addition. 1 P3V 3 P2V 2 T V 2200 K = ⎯ ⎯→ 3 = 3 = = 2.265 T3 T2 V 2 T2 971.1 K v

Process 3-4: polytropic expansion. ⎛V T4 = T3 ⎜⎜ 3 ⎝V 4

⎞ ⎟⎟ ⎠

n −1

⎞ ⎟⎟ ⎠

⎛ 2.265V 2 = T3 ⎜⎜ ⎝ V4

n −1

⎛ 2.265 ⎞ = T3 ⎜ ⎟ ⎝ r ⎠

n −1

⎛ 2.265 ⎞ = (2200 K )⎜ ⎟ ⎝ 20 ⎠

0.35

= 1026 K

q in = h3 − h2 = c p (T3 − T2 ) = (1.005 kJ/kg ⋅ K )(2200 − 971.1) K = 1235 kJ/kg q out = u 4 − u1 = cv (T4 − T1 ) = (0.718 kJ/kg ⋅ K )(1026 − 293) K = 526.3 kJ/kg

Note that qout in this case does not represent the entire heat rejected since some heat is also rejected during the polytropic process, which is determined from an energy balance on process 3-4: R(T4 − T3 ) (0.287 kJ/kg ⋅ K )(1026 − 2200 ) K w34,out = = = 963 kJ/kg 1− n 1 − 1.35 E in − E out = ΔE system ⎯→ q 34,in = w34,out + cv (T4 − T3 ) q 34,in − w34,out = u 4 − u 3 ⎯

= 963 kJ/kg + (0.718 kJ/kg ⋅ K )(1026 − 2200 ) K

= 120.1 kJ/kg which means that 120.1 kJ/kg of heat is transferred to the combustion gases during the expansion process. This is unrealistic since the gas is at a much higher temperature than the surroundings, and a hot gas loses heat during polytropic expansion. The cause of this unrealistic result is the constant specific heat assumption. If we were to use u data from the air table, we would obtain q 34,in = w34,out + (u 4 − u 3 ) = 963 + (781.3 − 1872.4) = −128.1 kJ/kg

which is a heat loss as expected. Then qout becomes q out = q 34,out + q 41,out = 128.1 + 526.3 = 654.4 kJ/kg and wnet,out = q in − q out = 1235 − 654.4 = 580.6 kJ/kg

η th = (b)

v1 =

wnet,out q in

=

(

580.6 kJ/kg = 47.0% 1235 kJ/kg

)

RT1 0.287 kPa ⋅ m 3 /kg ⋅ K (293 K ) = = 0.885 m 3 /kg = v max P1 95 kPa

v min = v 2 = MEP =

v max r

wnet,out

v1 −v 2

=

wnet,out

v 1 (1 − 1 / r )

=

⎛ 1 kPa ⋅ m 3 ⎜ kJ 0.885 m 3 /kg (1 − 1/20 ) ⎜⎝

(

580.6 kJ/kg

)

⎞ ⎟ = 691 kPa ⎟ ⎠

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-40

9-64 EES Problem 9-63 is reconsidered. The effect of the compression ratio on the net work output, mean effective pressure, and thermal efficiency is to be investigated. Also, T-s and P-v diagrams for the cycle are to be plotted. Analysis Using EES, the problem is solved as follows: Procedure QTotal(q_12,q_23,q_34,q_41: q_in_total,q_out_total) q_in_total = 0 q_out_total = 0 IF (q_12 > 0) THEN q_in_total = q_12 ELSE q_out_total = - q_12 If q_23 > 0 then q_in_total = q_in_total + q_23 else q_out_total = q_out_total - q_23 If q_34 > 0 then q_in_total = q_in_total + q_34 else q_out_total = q_out_total - q_34 If q_41 > 0 then q_in_total = q_in_total + q_41 else q_out_total = q_out_total - q_41 END "Input Data" T[1]=293 [K] P[1]=95 [kPa] T[3] = 2200 [K] n=1.35 {r_comp = 20} "Process 1-2 is isentropic compression" s[1]=entropy(air,T=T[1],P=P[1]) s[2]=s[1] T[2]=temperature(air, s=s[2], P=P[2]) P[2]*v[2]/T[2]=P[1]*v[1]/T[1] P[1]*v[1]=R*T[1] R=0.287 [kJ/kg-K] V[2] = V[1]/ r_comp "Conservation of energy for process 1 to 2" q_12 - w_12 = DELTAu_12 q_12 =0"isentropic process" DELTAu_12=intenergy(air,T=T[2])-intenergy(air,T=T[1]) "Process 2-3 is constant pressure heat addition" P[3]=P[2] s[3]=entropy(air, T=T[3], P=P[3]) P[3]*v[3]=R*T[3] "Conservation of energy for process 2 to 3" q_23 - w_23 = DELTAu_23 w_23 =P[2]*(V[3] - V[2])"constant pressure process" DELTAu_23=intenergy(air,T=T[3])-intenergy(air,T=T[2]) "Process 3-4 is polytropic expansion" P[3]/P[4] =(V[4]/V[3])^n s[4]=entropy(air,T=T[4],P=P[4]) P[4]*v[4]=R*T[4] "Conservation of energy for process 3 to 4" q_34 - w_34 = DELTAu_34 "q_34 is not 0 for the ploytropic process" DELTAu_34=intenergy(air,T=T[4])-intenergy(air,T=T[3]) P[3]*V[3]^n = Const w_34=(P[4]*V[4]-P[3]*V[3])/(1-n) "Process 4-1 is constant volume heat rejection" V[4] = V[1] "Conservation of energy for process 4 to 1" q_41 - w_41 = DELTAu_41 w_41 =0 "constant volume process"

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-41 DELTAu_41=intenergy(air,T=T[1])-intenergy(air,T=T[4]) Call QTotal(q_12,q_23,q_34,q_41: q_in_total,q_out_total) w_net = w_12+w_23+w_34+w_41 Eta_th=w_net/q_in_total*100 "Thermal efficiency, in percent" "The mean effective pressure is:" MEP = w_net/(V[1]-V[2])

ηth 47.69 50.14 52.16 53.85 55.29 56.54

wnet [kJ/kg] 797.9 817.4 829.8 837.0 840.6 841.5

Air

0.0 44

0.8 8m

3/k g

59 20 kP a

3

2

4

1 4.5

5.0

95

kP a

0.1

2400 2200 2000 1800 1600 1400 1200 1000 800 600 400 200 4.0

MEP [kPa] 970.8 985 992.6 995.4 994.9 992

34 0.1 kP a

T [K]

rcomp 14 16 18 20 22 24

5.5

6.0

6.5

7.0

7.5

s [kJ/kg-K] Air

104

104

P [kPa]

103

103 293 K

2200 K

102

102

1049 K

6.7 g J/k 4k

9 5.6

-K

101 10-2

10-1

100

101

101 102

3

v [m /kg]

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9-42

850

wnet [kJ/kg]

840 830 820 810 800 790 14

16

18

20

22

24

rcomp 57

ηth

55 53 51 49 47 14

16

18

20

22

24

rcomp 1000

MEP [kPa]

995 990 985 980 975 970 14

16

18

20

22

24

rcomp

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-43

9-65 A four-cylinder ideal diesel engine with air as the working fluid has a compression ratio of 17 and a cutoff ratio of 2.2. The power the engine will deliver at 1500 rpm is to be determined. Assumptions 1 The cold air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4 (Table A-2). Analysis Process 1-2: isentropic compression. ⎛V T2 = T1 ⎜⎜ 1 ⎝V 2

⎞ ⎟⎟ ⎠

k −1

P 2

= (328 K )(17 )0.4 = 1019 K

Qin

3

Process 2-3: P = constant heat addition.

4 Qout

P3v 3 P2v 2 v = ⎯ ⎯→ T3 = 3 T2 = 2.2T2 = (2.2)(1019 K ) = 2241 K v2 T3 T2

1

v

Process 3-4: isentropic expansion. ⎛V T4 = T3 ⎜⎜ 3 ⎝V 4 m=

⎞ ⎟⎟ ⎠

k −1

⎛ 2.2V 2 = T3 ⎜⎜ ⎝ V4

⎞ ⎟⎟ ⎠

k −1

⎛ 2.2 ⎞ = T3 ⎜ ⎟ ⎝ r ⎠

k −1

⎛ 2.2 ⎞ = (2241 K )⎜ ⎟ ⎝ 17 ⎠

0.4

= 989.2 K

P1V1 (97 kPa )(0.0024 m 3 ) = = 2.473 × 10 −3 kg RT1 (0.287 kPa ⋅ m 3 /kg ⋅ K )(328 K )

Qin = m(h3 − h2 ) = mc p (T3 − T2 ) = (2.473 × 10 −3 kg)(1.005 kJ/kg ⋅ K )(2241 − 1019)K = 3.038 kJ Qout = m(u 4 − u1 ) = mc v (T4 − T1 ) = 2.473 × 10 −3 kg (0.718 kJ/kg ⋅ K )(989.2 − 328)K = 1.174 kJ

(

)

W net,out = Qin − Qout = 3.038 − 1.174 = 1.864 kJ/rev W& net,out = n& W net,out = (1500/60 rev/s)(1.864 kJ/rev ) = 46.6 kW

Discussion Note that for 2-stroke engines, 1 thermodynamic cycle is equivalent to 1 mechanical cycle (and thus revolutions).

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9-44

9-66 A four-cylinder ideal diesel engine with nitrogen as the working fluid has a compression ratio of 17 and a cutoff ratio of 2.2. The power the engine will deliver at 1500 rpm is to be determined. Assumptions 1 The air-standard assumptions are applicable with nitrogen as the working fluid. 2 Kinetic and potential energy changes are negligible. 3 Nitrogen is an ideal gas with constant specific heats. Properties The properties of nitrogen at room temperature are cp = 1.039 kJ/kg·K, cv = 0.743 kJ/kg·K, R = 0.2968 kJ/kg·K, and k = 1.4 (Table A-2). Analysis Process 1-2: isentropic compression. ⎛V T2 = T1 ⎜⎜ 1 ⎝V 2

⎞ ⎟⎟ ⎠

k −1

P 2

= (328 K )(17 )0.4 = 1019 K

Qin

3

4

Process 2-3: P = constant heat addition.

Qout

P3v 3 P2v 2 v = ⎯ ⎯→ T3 = 3 T2 = 2.2T2 = (2.2)(1019 K ) = 2241 K v2 T3 T2

1

v

Process 3-4: isentropic expansion. ⎛V T4 = T3 ⎜⎜ 3 ⎝V 4 m=

⎞ ⎟⎟ ⎠

k −1

⎛ 2.2V 2 = T3 ⎜⎜ ⎝ V4

(

⎞ ⎟⎟ ⎠

k −1

⎛ 2.2 ⎞ = T3 ⎜ ⎟ ⎝ r ⎠

k −1

)

⎛ 2.2 ⎞ = (2241 K )⎜ ⎟ ⎝ 17 ⎠

0.4

= 989.2 K

P1V 1 (97 kPa ) 0.0024 m 3 = = 2.391×10 −3 kg 3 RT1 0.2968 kPa ⋅ m /kg ⋅ K (328 K )

(

Qin = m(h3 − h2 ) = mc p (T3 − T2 )

(

)

(

)

)

= 2.391× 10 −3 kg (1.039 kJ/kg ⋅ K )(2241 − 1019 )K = 3.037 kJ

Qout = m(u 4 − u1 ) = mcv (T4 − T1 ) = 2.391× 10 −3 kg (0.743 kJ/kg ⋅ K )(989.2 − 328)K = 1.175 kJ W net,out = Qin − Qout = 3.037 − 1.175 = 1.863 kJ/rev W& net,out = n& W net,out = (1500/60 rev/s)(1.863 kJ/rev ) = 46.6 kW

Discussion Note that for 2-stroke engines, 1 thermodynamic cycle is equivalent to 1 mechanical cycle (and thus revolutions).

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9-45

9-67 An ideal dual cycle has a compression ratio of 18 and cutoff ratio of 1.1. The power produced by the cycle is to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4 (Table A-2a). Analysis We begin by fixing the temperatures at all states. ⎛v T2 = T1 ⎜⎜ 1 ⎝v 2 ⎛v P2 = P1 ⎜⎜ 1 ⎝v 2

⎞ ⎟⎟ ⎠

k −1

= T1 r k −1 = (291 K)(18)1.4−1 = 924.7 K

P

x 2

3 qin

qout

⎞ ⎟⎟ = P1 r k = (90 kPa)(18)1.4 = 5148 kPa ⎠

1

v

Px = P3 = r p P2 = (1.1)(5148 kPa) = 5663 kPa ⎛P T x = T2 ⎜⎜ x ⎝ P2

4

k

⎞ ⎛ 5663 kPa ⎞ ⎟⎟ = (924.7 K)⎜ ⎟ = 1017 K ⎝ 5148 kPa ⎠ ⎠

T3 = rc T x = (1.1)(1017 K) = 1119 K ⎛v T4 = T3 ⎜⎜ 3 ⎝v 4

⎞ ⎟⎟ ⎠

k −1

⎛r = T3 ⎜⎜ c ⎝ r

⎞ ⎟⎟ ⎠

k −1

⎛ 1.1 ⎞ = (1119 K)⎜ ⎟ ⎝ 18 ⎠

1.4 −1

= 365.8 K

Applying the first law to each of the processes gives w1− 2 = cv (T2 − T1 ) = (0.718 kJ/kg ⋅ K )(924.7 − 291)K = 455.0 kJ/kg q x −3 = c p (T3 − T x ) = (1.005 kJ/kg ⋅ K )(1119 − 1017)K = 102.5 kJ/kg

w x −3 = q x −3 − cv (T3 − T x ) = 102.5 − (0.718 kJ/kg ⋅ K )(1119 − 1017)K = 29.26 kJ/kg w3− 4 = cv (T3 − T4 ) = (0.718 kJ/kg ⋅ K )(1119 − 365.8)K = 540.8 kJ/kg

The net work of the cycle is wnet = w3− 4 + w x −3 − w1− 2 = 540.8 + 29.26 − 455.0 = 115.1 kJ/kg

The mass in the device is given by m=

P1V1 (90 kPa)(0.003 m 3 ) = = 0.003233 kg RT1 (0.287 kPa ⋅ m 3 /kg ⋅ K )(291 K)

The net power produced by this engine is then W& net = mwnet n& = (0.003233 kg/cycle)(115.1 kJ/kg)(4000/60 cycle/s) = 24.8 kW

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9-46

9-68 A dual cycle with non-isentropic compression and expansion processes is considered. The power produced by the cycle is to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4 (Table A-2a). Analysis We begin by fixing the temperatures at all states. ⎛v T2 s = T1 ⎜⎜ 1 ⎝v 2

η=

⎞ ⎟⎟ ⎠

k −1

= T1 r k −1 = (291 K)(18)1.4−1 = 924.7 K

T2 s − T1 T − T1 (924.7 − 291) K ⎯ ⎯→ T2 = T1 + 2 s = (291 K) + = 1037 K η 0.85 T2 − T1

⎛v P2 = P1 ⎜⎜ 1 ⎝v 2

k

⎞ ⎟⎟ = P1 r k = (90 kPa)(18)1.4 = 5148 kPa ⎠

Px = P3 = r p P2 = (1.1)(5148 kPa) = 5663 kPa ⎛P T x = T2 ⎜⎜ x ⎝ P2

3

x

P

qin 2

⎞ ⎛ 5663 kPa ⎞ ⎟⎟ = (1037 K)⎜ ⎟ = 1141 K ⎝ 5148 kPa ⎠ ⎠

qout 1

v

T3 = rc T x = (1.1)(1141 K) = 1255 K ⎛v T4 s = T3 ⎜⎜ 3 ⎝v 4

η=

⎞ ⎟⎟ ⎠

k −1

⎛r = T3 ⎜⎜ c ⎝ r

⎞ ⎟⎟ ⎠

k −1

⎛ 1.1 ⎞ = (1255 K)⎜ ⎟ ⎝ 18 ⎠

4

1.4 −1

= 410.3 K

T3 − T4 ⎯ ⎯→ T4 = T3 − η (T3 − T4 s ) = (1255 K) − (0.90)(1255 − 410.3) K = 494.8 K T3 − T4 s

Applying the first law to each of the processes gives w1− 2 = cv (T2 − T1 ) = (0.718 kJ/kg ⋅ K )(1037 − 291)K = 535.6 kJ/kg q x −3 = c p (T3 − T x ) = (1.005 kJ/kg ⋅ K )(1255 − 1141)K = 114.6 kJ/kg

w x −3 = q x −3 − cv (T3 − T x ) = 114.6 − (0.718 kJ/kg ⋅ K )(1255 − 1141)K = 32.75 kJ/kg w3− 4 = cv (T3 − T4 ) = (0.718 kJ/kg ⋅ K )(1255 − 494.8)K = 545.8 kJ/kg

The net work of the cycle is wnet = w3− 4 + w x −3 − w1− 2 = 545.8 + 32.75 − 535.6 = 42.95 kJ/kg

The mass in the device is given by m=

P1V1 (90 kPa)(0.003 m 3 ) = = 0.003233 kg RT1 (0.287 kPa ⋅ m 3 /kg ⋅ K )(291 K)

The net power produced by this engine is then W& net = mwnet n& = (0.003233 kg/cycle)(42.95 kJ/kg)(4000/60 cycle/s) = 9.26 kW

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9-47

9-69E An ideal dual cycle has a compression ratio of 15 and cutoff ratio of 1.4. The net work, heat addition, and the thermal efficiency are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are R = 0.3704 psia·ft3/lbm.R (Table A-1E), cp = 0.240 Btu/lbm·R, cv = 0.171 Btu/lbm·R, and k = 1.4 (Table A-2Ea). Analysis Working around the cycle, the germane properties at the various states are k −1

⎛v T2 = T1 ⎜⎜ 1 ⎝v 2

⎞ ⎟⎟ ⎠

⎛v P2 = P1 ⎜⎜ 1 ⎝v 2

⎞ ⎟⎟ = P1 r k = (14.2 psia)(15)1.4 = 629.2 psia ⎠

= T1 r k −1 = (535 R)(15)1.4 −1 = 1580 R

x

P k

2

3 qin

4 qout

Px = P3 = r p P2 = (1.1)(629.2 psia) = 692.1 psia

1

⎛P T x = T2 ⎜⎜ x ⎝ P2

⎞ ⎛ 692.1 psia ⎞ ⎟⎟ = (1580 R)⎜⎜ ⎟⎟ = 1738 R ⎝ 629.2 psia ⎠ ⎠

v

⎛v T3 = T x ⎜⎜ 3 ⎝v x

⎞ ⎟ = T x rc = (1738 R)(1.4) = 2433 R ⎟ ⎠

⎛v T4 = T3 ⎜⎜ 3 ⎝v 4

⎞ ⎟⎟ ⎠

k −1

⎛r = T3 ⎜⎜ c ⎝ r

⎞ ⎟⎟ ⎠

k −1

⎛ 1.4 ⎞ = (2433 R)⎜ ⎟ ⎝ 15 ⎠

1.4 −1

= 942.2 R

Applying the first law to each of the processes gives w1− 2 = cv (T2 − T1 ) = (0.171 Btu/lbm ⋅ R )(1580 − 535)R = 178.7 Btu/lbm q 2− x = cv (T x − T2 ) = (0.171 Btu/lbm ⋅ R )(1738 − 1580)R = 27.02 Btu/lbm q x −3 = c p (T3 − T x ) = (0.240 Btu/lbm ⋅ R )(2433 − 1738)R = 166.8 Btu/lbm

w x −3 = q x −3 − cv (T3 − T x ) = 166.8 Btu/lbm − (0.171 Btu/lbm ⋅ R )(2433 − 1738)R = 47.96 Btu/lbm w3− 4 = cv (T3 − T4 ) = (0.171 Btu/lbm ⋅ R )(2433 − 942.2)R = 254.9 Btu/lbm

The net work of the cycle is wnet = w3− 4 + w x −3 − w1− 2 = 254.9 + 47.96 − 178.7 = 124.2 Btu/lbm

and the net heat addition is q in = q 2− x + q x −3 = 27.02 + 166.8 = 193.8 Btu/lbm

Hence, the thermal efficiency is

η th =

wnet 124.2 Btu/lbm = = 0.641 q in 193.8 Btu/lbm

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-48

9-70 An expression for the thermal efficiency of a dual cycle is to be developed and the thermal efficiency for a given case is to be calculated. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4 (Table A-2) Analysis The thermal efficiency of a dual cycle may be expressed as

η th = 1 −

q out cv (T4 − T1 ) = 1− q in cv (T x − T2 ) + c p (T3 − T x )

By applying the isentropic process relations for ideal gases with constant specific heats to the processes 1-2 and 3-4, as well as the ideal gas equation of state, the temperatures may be eliminated from the thermal efficiency expression. This yields the result

η th = 1 −

⎤ r p rck − 1 1 ⎡ ⎥ ⎢ r k −1 ⎢⎣ kr p (rc − 1) + r p − 1 ⎥⎦

P

x

where rp =

Px v and rc = 3 P2 vx

When rc = rp, we obtain

η th = 1 −

2

3 qin

4 qout 1

v

⎞ −1 1 ⎛⎜ ⎟ 2 ⎜ ⎟ r k ( r r ) r 1 − + − p p p ⎝ ⎠ r pk +1

k −1

For the case r = 20 and rp = 2,

η th = 1 −

1 201.4 −1

⎞ ⎛ 21.4+1 − 1 ⎜ ⎟ = 0.660 ⎜ 1.4(2 2 − 2) + 2 − 1 ⎟ ⎝ ⎠

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-49

9-71 An expression regarding the thermal efficiency of a dual cycle for a special case is to be obtained. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Analysis The thermal efficiency of a dual cycle may be expressed as

η th = 1 −

q out cv (T4 − T1 ) = 1− q in cv (T x − T2 ) + c p (T3 − T x )

By applying the isentropic process relations for ideal gases with constant specific heats to the processes 1-2 and 3-4, as well as the ideal gas equation of state, the temperatures may be eliminated from the thermal efficiency expression. This yields the result

η th

⎤ r p rck − 1 1 ⎡ ⎥ = 1 − k −1 ⎢ ⎢⎣ kr p (rc − 1) + r p − 1 ⎥⎦ r

P

x

where P v r p = x and rc = 3 P2 vx

When rc = rp, we obtain

η th

⎞ r pk +1 − 1 1 ⎛ ⎟ = 1 − k −1 ⎜ ⎜ k (r 2 − r ) + r − 1 ⎟ r p p p ⎝ ⎠

2

3 qin

4 qout 1

v

Rearrangement of this result gives r pk +1 − 1 k (r p2 − r p ) + r p − 1

= (1 − η th )r k −1

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-50

9-72 A six-cylinder compression ignition engine operates on the ideal Diesel cycle. The maximum temperature in the cycle, the cutoff ratio, the net work output per cycle, the thermal efficiency, the mean effective pressure, the net power output, and the specific fuel consumption are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at 850 K are cp = 1.110 kJ/kg·K, cv = 0.823 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.349 (Table A-2b). Analysis (a) Process 1-2: Isentropic compression k −1

⎛v T2 = T1 ⎜⎜ 1 ⎝v 2

⎞ ⎟⎟ ⎠

⎛v P2 = P1 ⎜⎜ 1 ⎝v 2

⎞ ⎟⎟ = (95 kPa )(17 )1.349 = 4341 kPa ⎠

= (328 K )(17 )1.349-1 = 881.7 K

k

Qin 2

3

The clearance volume and the total volume of the engine at the beginning of compression process (state 1) are

V c +V d V + 0.0045 m 3 ⎯ ⎯→ 17 = c Vc Vc

r=

V c = 0.0002813 m

4 Qout 1

3

V1 = V c +V d = 0.0002813 + 0.0045 = 0.004781 m 3 The total mass contained in the cylinder is m=

P1V1 (95 kPa)(0.004781 m 3 ) = = 0.004825 kg RT1 0.287 kPa ⋅ m 3 /kg ⋅ K (328 K )

(

)

The mass of fuel burned during one cycle is AF =

(0.004825 kg) − m f ma m − m f = ⎯ ⎯→ 24 = ⎯ ⎯→ m f = 0.000193 kg mf mf mf

Process 2-3: constant pressure heat addition Qin = m f q HVη c = (0.000193 kg)(42,500 kJ/kg)(0.98) = 8.039 kJ Qin = mcv (T3 − T2 ) ⎯ ⎯→ 8.039 kJ = (0.004825 kg)(0.823 kJ/kg.K)(T3 − 881.7)K ⎯ ⎯→ T3 = 2383 K

The cutoff ratio is

β=

(b)

T3 2383 K = = 2.7 T2 881.7 K

V2 =

V1 r

=

0.004781 m 3 = 0.0002813 m 3 17

V 3 = βV 2 = (2.70)(0.0002813 m 3 ) = 0.00076 m 3 V 4 = V1 P3 = P2

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-51

Process 3-4: isentropic expansion. k −1

1.349-1

⎛V T4 = T3 ⎜⎜ 3 ⎝V 4

⎞ ⎟⎟ ⎠

⎛ 0.00076 m 3 = (2383 K )⎜ ⎜ 0.004781 m 3 ⎝

⎞ ⎟ ⎟ ⎠

⎛V P4 = P3 ⎜⎜ 3 ⎝V 4

⎛ 0.00076 m 3 ⎞ ⎟⎟ = (4341 kPa )⎜ ⎜ 0.004781 m 3 ⎠ ⎝

⎞ ⎟ ⎟ ⎠

k

= 1254 K 1.349

= 363.2 kPa

Process 4-1: constant voume heat rejection. Qout = mcv (T4 − T1 ) = (0.004825 kg)(0.823 kJ/kg ⋅ K )(1254 − 328)K = 3.677 kJ

The net work output and the thermal efficiency are Wnet,out = Qin − Qout = 8.039 − 3.677 = 4.361 kJ

η th =

Wnet,out Qin

=

4.361 kJ = 0.543 8.039 kJ

(c) The mean effective pressure is determined to be MEP =

Wnet,out

V 1 −V 2

=

⎛ kPa ⋅ m 3 ⎜ (0.004781 − 0.0002813)m 3 ⎜⎝ kJ 4.361 kJ

⎞ ⎟ = 969.2 kPa ⎟ ⎠

(d) The power for engine speed of 2000 rpm is 2000 (rev/min) ⎛ 1 min ⎞ n& W& net = Wnet = (4.361 kJ/cycle) ⎜ ⎟ = 72.7 kW 2 (2 rev/cycle) ⎝ 60 s ⎠

Note that there are two revolutions in one cycle in four-stroke engines. (e) Finally, the specific fuel consumption is sfc =

mf Wnet

=

0.000193 kg ⎛ 1000 g ⎞⎛ 3600 kJ ⎞ ⎜ ⎟⎜ ⎟ = 159.3 g/kWh 4.361 kJ/kg ⎜⎝ 1 kg ⎟⎠⎝ 1 kWh ⎠

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9-52

Stirling and Ericsson Cycles

9-73C The efficiencies of the Carnot and the Stirling cycles would be the same, the efficiency of the Otto cycle would be less.

9-74C The efficiencies of the Carnot and the Ericsson cycles would be the same, the efficiency of the Diesel cycle would be less.

9-75C The Stirling cycle.

9-76C The two isentropic processes of the Carnot cycle are replaced by two constant pressure regeneration processes in the Ericsson cycle.

9-77 An ideal steady-flow Ericsson engine with air as the working fluid is considered. The maximum pressure in the cycle, the net work output, and the thermal efficiency of the cycle are to be determined. Assumptions Air is an ideal gas. Properties The gas constant of air is R = 0.287 kJ/kg.K (Table A-1). Analysis (a) The entropy change during process 3-4 is s 4 − s3 = −

q 34,out T0

=−

150 kJ/kg = −0.5 kJ/kg ⋅ K 300 K

T

qin 2

1

1200 K

and s 4 − s 3 = c p ln

T4 T3

©0

− Rln

P4 P3

300 K

P4 = −(0.287 kJ/kg ⋅ K )ln = −0.5 kJ/kg ⋅ K 120 kPa

It yields

4

3 qout

P4 = 685.2 kPa

(b) For reversible cycles, q out T L T 1200 K = ⎯ ⎯→ q in = H q out = (150 kJ/kg ) = 600 kJ/kg q in TH TL 300 K

Thus, wnet,out = q in − q out = 600 − 150 = 450 kJ/kg

(c) The thermal efficiency of this totally reversible cycle is determined from

η th = 1 −

TL 300K = 1− = 75.0% TH 1200K

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s

9-53

9-78 An ideal Stirling engine with air as the working fluid operates between the specified temperature and pressure limits. The net work produced per cycle and the thermal efficiency of the cycle are to be determined. Assumptions Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4 (Table A-2a). Analysis Since the specific volume is constant during process 2-3,

P2 = P3

T

T2 ⎛ 800 K ⎞ = (100 kPa)⎜ ⎟ = 266.7 kPa T3 ⎝ 300 K ⎠

1

800 K

qin

2

Heat is only added to the system during reversible process 1-2. Then, s 2 − s1 = cv ln

T2 T1

©0

− R ln

P2 P1

300 K

⎛ 266.7 kPa ⎞ = 0 − (0.287 kJ/kg ⋅ K )ln⎜ ⎟ ⎝ 2000 kPa ⎠ = 0.5782 kJ/kg ⋅ K

4

3 qout s

q in = T1 ( s 2 − s1 ) = (800 K )(0.5782 kJ/kg ⋅ K ) = 462.6 kJ/kg

The thermal efficiency of this totally reversible cycle is determined from

η th = 1 −

TL 300 K = 1− = 0.625 TH 800 K

Then, W net = η th mq in = (0.625)(1 kg)(462.6 kJ/kg) = 289.1 kJ

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9-54

9-79 An ideal Stirling engine with air as the working fluid operates between the specified temperature and pressure limits. The power produced and the rate of heat input are to be determined. Assumptions Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4 (Table A-2a). Analysis Since the specific volume is constant during process 2-3,

P2 = P3

T2 ⎛ 800 K ⎞ = (100 kPa)⎜ ⎟ = 266.7 kPa T3 ⎝ 300 K ⎠

T 1

800 K

qin

2

Heat is only added to the system during reversible process 1-2. Then, s 2 − s1 = cv ln

T2 T1

©0

− R ln

P2 P1

300 K

⎛ 266.7 kPa ⎞ = 0 − (0.287 kJ/kg ⋅ K )ln⎜ ⎟ ⎝ 2000 kPa ⎠ = 0.5782 kJ/kg ⋅ K

4

3 qout s

q in = T1 ( s 2 − s1 ) = (800 K )(0.5782 kJ/kg ⋅ K ) = 462.6 kJ/kg

The thermal efficiency of this totally reversible cycle is determined from

η th = 1 −

TL 300 K = 1− = 0.625 TH 800 K

Then, W net = η th mq in = (0.625)(1 kg)(462.6 kJ/kg) = 289.1 kJ

The rate at which heat is added to this engine is Q& in = mq in n& = (1 kg/cycle)(462.6 kJ/kg)(500/60 cycle/s) = 3855 kW

while the power produced by the engine is W& net = W net n& =)(289.1 kJ/cycle)(500/60 cycle/s) = 2409 kW

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9-55

9-80E An ideal Stirling engine with hydrogen as the working fluid operates between the specified temperature limits. The amount of external heat addition, external heat rejection, and heat transfer between the working fluid and regenerator per cycle are to be determined. Assumptions Hydrogen is an ideal gas with constant specific heats. Properties The properties of hydrogen at room temperature are R = 5.3224 psia·ft3/lbm.R = 0.9851 Btu/lbm·R, cp = 3.43 Btu/lbm·R, cv = 2.44 Btu/lbm·R, and k = 1.404 (Table A-2Ea). Analysis The mass of the air contained in this engine is m=

P1V1 (400 psia)(0.1 ft 3 ) = = 0.006832 lbm RT1 (5.3224 psia ⋅ ft 3 /lbm ⋅ R )(1100 R)

At the end of the compression, the pressure will be ⎛ 0.1 ft V P2 = P1 1 = (400 psia)⎜ ⎜ 1 ft 3 V2 ⎝

3

⎞ ⎟ = 40 psia ⎟ ⎠

T 1

1100R

qin

2

The entropy change is T s 2 − s1 = s 3 − s 4 = cv ln 2 T1

©0

500 R

4

P − R ln 2 P1

3 qout s

⎛ 40 psia ⎞ ⎟⎟ = 2.268 Btu/lbm ⋅ R = 0 − (0.9851 Btu/lbm ⋅ R )ln⎜⎜ ⎝ 400 psia ⎠

Since the processes are reversible, Qin = mT1 ( s 2 − s1 ) = (0.006832 lbm)(1100 R )(2.268 Btu/lbm ⋅ R ) = 17.0 Btu Qout = mT3 ( s 4 − s 3 ) = (0.006832 lbm)(500 R )(2.268 Btu/lbm ⋅ R ) = 7.75 Btu

Applying the first law to the process where the gas passes through the regenerator gives Q regen = mcv (T1 − T4 ) = (0.006832 lbm)(2.44 Btu/lbm ⋅ R )(1100 − 500)R = 10.0 Btu

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9-56

9-81E An ideal Stirling engine with air as the working fluid operates between specified pressure limits. The heat added to and rejected by this cycle, and the net work produced by the cycle are to be determined. Assumptions Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are R = 0.3704 psia·ft3/lbm.R = 0.06855 Btu/lbm·R, cp = 0.240 Btu/lbm·R, cv = 0.171 Btu/lbm·R, and k = 1.4 (Table A-2Ea). Analysis Applying the ideal gas equation to the isothermal process 3-4 gives P4 = P3

v3 = (10 psia)(10) = 100 psia v4

T

qin

1

2

Since process 4-1 is a constant volume process,, ⎛P T1 = T4 ⎜⎜ 1 ⎝ P4

⎞ ⎛ 600 psia ⎞ ⎟⎟ = (560 R)⎜⎜ ⎟⎟ = 3360 R ⎝ 100 psia ⎠ ⎠

560 R

4

3 qout

According to first law and work integral, q in = w1− 2

s

v = RT1 ln 2 = (0.06855 Btu/lbm ⋅ R )(3360 R)ln(10) = 530.3 Btu/lbm v1

and q out = w3− 4 = RT3 ln

v4 ⎛1⎞ = (0.06855 Btu/lbm ⋅ R )(560 R)ln⎜ ⎟ = 88.4 Btu/lbm v3 ⎝ 10 ⎠

The net work is then wnet = q in − q out = 530.3 − 88.4 = 441.9 Btu/lbm

9-82E An ideal Stirling engine with air as the working fluid operates between specified pressure limits. The heat transfer in the regenerator is to be determined. Assumptions Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are R = 0.3704 psia·ft3/lbm.R, cp = 0.240 Btu/lbm·R, cv = 0.171 Btu/lbm·R, and k = 1.4 (Table A-2Ea). Analysis Applying the ideal gas equation to the isothermal process 1-2 gives P2 = P1

v1 ⎛1⎞ = (600 psia)⎜ ⎟ = 60 psia v2 ⎝ 10 ⎠

T 1

qin

2

Since process 2-3 is a constant-volume process, ⎛P T2 = T3 ⎜⎜ 2 ⎝ P3

⎞ ⎛ 60 psia ⎞ ⎟ = (560 R)⎜⎜ ⎟⎟ = 3360 R ⎟ ⎝ 10 psia ⎠ ⎠

560 R

4

Application of the first law to process 2-3 gives

3 qout s

q regen = cv (T2 − T3 ) = (0.171 Btu/lbm ⋅ R )(3360 − 560)R = 478.8 Btu/lbm

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9-57

9-83 An ideal Ericsson cycle operates between the specified temperature limits. The rate of heat addition is to be determined. Analysis The thermal efficiency of this totally reversible cycle is determined from

η th = 1 −

TL 280 K = 1− = 0.6889 TH 900 K

According to the general definition of the thermal efficiency, the rate of heat addition is

T 1

900 K

280 K

qin

4

2

3 qout

W& 500 kW = 726 kW Q& in = net = η th 0.6889

s

9-84 An ideal Ericsson cycle operates between the specified temperature limits. The power produced by the cycle is to be determined. Analysis The power output is 500 kW when the cycle is repeated 2000 times per minute. Then the work per cycle is W net

T 1

900 K

qin

2

W& 500 kJ/s = net = = 15 kJ/cycle n& (2000/60) cycle/s

When the cycle is repeated 3000 times per minute, the power output will be W& net = n& W net = (3000/60 cycle/s)(15 kJ/cycle) = 750 kW

280 K

4

3 qout s

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9-58

Ideal and Actual Gas-Turbine (Brayton) Cycles

9-85C In gas turbine engines a gas is compressed, and thus the compression work requirements are very large since the steady-flow work is proportional to the specific volume.

9-86C They are (1) isentropic compression (in a compressor), (2) P = constant heat addition, (3) isentropic expansion (in a turbine), and (4) P = constant heat rejection.

9-87C For fixed maximum and minimum temperatures, (a) the thermal efficiency increases with pressure ratio, (b) the net work first increases with pressure ratio, reaches a maximum, and then decreases.

9-88C Back work ratio is the ratio of the compressor (or pump) work input to the turbine work output. It is usually between 0.40 and 0.6 for gas turbine engines.

9-89C As a result of turbine and compressor inefficiencies, (a) the back work ratio increases, and (b) the thermal efficiency decreases.

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9-59

9-90E A simple ideal Brayton cycle with air as the working fluid has a pressure ratio of 10. The air temperature at the compressor exit, the back work ratio, and the thermal efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with variable specific heats. Properties The properties of air are given in Table A-17E. Analysis (a) Noting that process 1-2 is isentropic, T1 = 520 R

Pr 2 =

T

h1 = 124.27 Btu / lbm

⎯ ⎯→

Pr 1 = 1.2147

qin

T2 = 996.5 R P2 Pr1 = (10 )(1.2147 ) = 12.147 ⎯ ⎯→ h2 = 240.11 Btu/lbm P1

(b) Process 3-4 is isentropic, and thus T3 = 2000 R ⎯ ⎯→

3

2000 2

4 520

h3 = 504.71 Btu/lbm

1

qout

s

Pr3 = 174.0

P4 ⎛1⎞ Pr3 = ⎜ ⎟(174.0) = 17.4 ⎯ ⎯→ h4 = 265.83 Btu/lbm P3 ⎝ 10 ⎠ = h2 − h1 = 240.11 − 124.27 = 115.84 Btu/lbm

Pr 4 = wC,in

wT,out = h3 − h4 = 504.71 − 265.83 = 238.88 Btu/lbm

Then the back-work ratio becomes rbw =

(c)

wC,in wT,out

=

115.84 Btu/lbm = 48.5% 238.88 Btu/lbm

q in = h3 − h2 = 504.71 − 240.11 = 264.60 Btu/lbm wnet,out = wT,out − wC,in = 238.88 − 115.84 = 123.04 Btu/lbm

η th =

wnet,out q in

=

123.04 Btu/lbm = 46.5% 264.60 Btu/lbm

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9-60

9-91 [Also solved by EES on enclosed CD] A simple Brayton cycle with air as the working fluid has a pressure ratio of 8. The air temperature at the turbine exit, the net work output, and the thermal efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with variable specific heats.

T 3

1160 K qin

Properties The properties of air are given in Table A-17.

2 2s

Analysis (a) Noting that process 1-2s is isentropic, T1 = 310 K

⎯ ⎯→

h1 = 310.24 kJ / kg

310 K

Pr 1 = 1.5546

1

qout 4s

Pr 2 =

P2 ⎯→ h2 s = 562.58 kJ/kg and T2 s = 557.25 K Pr = (8)(1.5546 ) = 12.44 ⎯ P1 1

ηC =

h2 s − h1 h − h1 ⎯ ⎯→ h2 = h1 + 2 s h2 − h1 ηC

4 s

562.58 − 310.24 = 646.7 kJ/kg 0.75 h3 = 1230.92 kJ/kg = 310.24 +

⎯→ T3 = 1160 K ⎯

Pr3 = 207.2

P4 ⎛1⎞ ⎯→ h4 s = 692.19 kJ/kg and T4 s = 680.3 K Pr3 = ⎜ ⎟(207.2) = 25.90 ⎯ P3 ⎝8⎠ h −h η T = 3 4 ⎯⎯→ h4 = h3 − η T (h3 − h4 s ) h3 − h 4 s = 1230.92 − (0.82 )(1230.92 − 692.19 )

Pr 4 =

= 789.16 kJ/kg

Thus, T4 = 770.1 K (b)

q in = h3 − h2 = 1230.92 − 646.7 = 584.2 kJ/kg q out = h4 − h1 = 789.16 − 310.24 = 478.92 kJ/kg wnet,out = q in − q out = 584.2 − 478.92 = 105.3 kJ/kg

(c)

η th =

wnet,out q in

=

105.3 kJ/kg = 18.0% 584.2 kJ/kg

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9-61

9-92 EES Problem 9-91 is reconsidered. The mass flow rate, pressure ratio, turbine inlet temperature, and the isentropic efficiencies of the turbine and compressor are to be varied and a general solution for the problem by taking advantage of the diagram window method for supplying data to EES is to be developed. Analysis Using EES, the problem is solved as follows: "Input data - from diagram window" {P_ratio = 8} {T[1] = 310 [K] P[1]= 100 [kPa] T[3] = 1160 [K] m_dot = 20 [kg/s] Eta_c = 75/100 Eta_t = 82/100} "Inlet conditions" h[1]=ENTHALPY(Air,T=T[1]) s[1]=ENTROPY(Air,T=T[1],P=P[1]) "Compressor anaysis" s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor" P_ratio=P[2]/P[1]"Definition of pressure ratio - to find P[2]" T_s[2]=TEMPERATURE(Air,s=s_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exit" h_s[2]=ENTHALPY(Air,T=T_s[2]) Eta_c =(h_s[2]-h[1])/(h[2]-h[1]) "Compressor adiabatic efficiency; Eta_c = W_dot_c_ideal/W_dot_c_actual. " m_dot*h[1] +W_dot_c=m_dot*h[2] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" "External heat exchanger analysis" P[3]=P[2]"process 2-3 is SSSF constant pressure" h[3]=ENTHALPY(Air,T=T[3]) m_dot*h[2] + Q_dot_in= m_dot*h[3]"SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0" "Turbine analysis" s[3]=ENTROPY(Air,T=T[3],P=P[3]) s_s[4]=s[3] "For the ideal case the entropies are constant across the turbine" P_ratio= P[3] /P[4] T_s[4]=TEMPERATURE(Air,s=s_s[4],P=P[4]) "Ts[4] is the isentropic value of T[4] at turbine exit" h_s[4]=ENTHALPY(Air,T=T_s[4]) "Eta_t = W_dot_t /Wts_dot turbine adiabatic efficiency, Wts_dot > W_dot_t" Eta_t=(h[3]-h[4])/(h[3]-h_s[4]) m_dot*h[3] = W_dot_t + m_dot*h[4] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" "Cycle analysis" W_dot_net=W_dot_t-W_dot_c"Definition of the net cycle work, kW" Eta=W_dot_net/Q_dot_in"Cycle thermal efficiency" Bwr=W_dot_c/W_dot_t "Back work ratio" "The following state points are determined only to produce a T-s plot" T[2]=temperature('air',h=h[2]) T[4]=temperature('air',h=h[4]) s[2]=entropy('air',T=T[2],P=P[2]) s[4]=entropy('air',T=T[4],P=P[4]) PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-62

η 0.1 0.1644 0.1814 0.1806 0.1702 0.1533 0.131 0.1041 0.07272 0.03675

Bwr 0.5229 0.6305 0.7038 0.7611 0.8088 0.85 0.8864 0.9192 0.9491 0.9767

Pratio 2 4 6 8 10 12 14 16 18 20

Wc [kW] 1818 4033 5543 6723 7705 8553 9304 9980 10596 11165

Wnet [kW] 1659 2364 2333 2110 1822 1510 1192 877.2 567.9 266.1

Wt [kW] 3477 6396 7876 8833 9527 10063 10496 10857 11164 11431

Qin [kW] 16587 14373 12862 11682 10700 9852 9102 8426 7809 7241

1500 Air Standard Brayton Cycle Pressure ratio = 8 and Tmax = 1160K 3

T [K]

1000

4

2 2s

4s

500 800 kPa 100 kPa

0 5.0

1

5.5

6.0

6.5

7.0

7.5

s [kJ/kg-K] 0.25

2500

η

η Cycle efficiency,

0.15

0.00 2

1500

η = 0.82 t η = 0.75 c Tmax=1160 K

0.10

0.05

2000

Wnet

1000

500

Note Pratio for maximum work and η

4

6

8

10 12 Pratio

14

Wnet [kW]

0.20

16

18

0 20

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-63

9-93 A simple Brayton cycle with air as the working fluid has a pressure ratio of 8. The air temperature at the turbine exit, the net work output, and the thermal efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2). Analysis (a) Using the compressor and turbine efficiency relations, ⎛P T2 s = T1 ⎜⎜ 2 ⎝ P1

⎞ ⎟⎟ ⎠

(k −1) / k

(k −1) / k

⎛P ⎞ ⎛1⎞ T4 s = T3 ⎜⎜ 4 ⎟⎟ = 640.4 K = (1160 K )⎜ ⎟ P ⎝8⎠ ⎝ 3⎠ c p (T2 s − T1 ) T − T1 h −h η C = 2s 1 = ⎯ ⎯→ T2 = T1 + 2 s ηC h2 − h1 c p (T2 − T1 ) 0.4/1.4

= 310 +

ηT =

(b)

T

= (310 K )(8)0.4/1.4 = 561.5 K

3

1160 K qin 2s 310 K

561.5 − 310 = 645.3 K 0.75

2

1

qout

4s

4

c p (T3 − T4 ) h3 − h4 ⎯ ⎯→ T4 = T3 − η T (T3 − T4 s ) = h3 − h4 s c p (T3 − T4 s ) = 1160 − (0.82)(1160 − 640.4 ) = 733.9 K

q in = h3 − h2 = c p (T3 − T2 ) = (1.005 kJ/kg ⋅ K )(1160 − 645.3)K = 517.3 kJ/kg q out = h4 − h1 = c p (T4 − T1 ) = (1.005 kJ/kg ⋅ K )(733.9 − 310)K = 426.0 kJ/kg wnet,out = q in − q out = 517.3 − 426.0 = 91.3 kJ/kg

(c)

η th =

wnet,out q in

=

91.3 kJ/kg = 17.6% 517.3 kJ/kg

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

s

9-64

9-94 A simple ideal Brayton cycle with air as the working fluid operates between the specified temperature and pressure limits. The net work and the thermal efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2a). Analysis Using the isentropic relations for an ideal gas, ⎛P T2 = T1 ⎜⎜ 2 ⎝ P1

⎞ ⎟⎟ ⎠

( k −1) / k

⎞ ⎟ ⎟ ⎠

( k −1) / k

⎛ 2000 kPa ⎞ = (300 K)⎜ ⎟ ⎝ 100 kPa ⎠

0.4/1.4

= 706.1 K

T 3

1000 K

Similarly,

qin 2

⎛P T4 = T3 ⎜⎜ 4 ⎝ P3

⎛ 100 kPa ⎞ = (1000 K)⎜ ⎟ ⎝ 2000 kPa ⎠

0.4/1.4

= 424.9 K

300 K

4 1

Applying the first law to the constant-pressure heat addition process 2-3 produces

qout s

q in = h3 − h2 = c p (T3 − T2 ) = (1.005 kJ/kg ⋅ K )(1000 − 706.1)K = 295.4 kJ/kg

Similarly, q out = h4 − h1 = c p (T4 − T1 ) = (1.005 kJ/kg ⋅ K )(424.9 − 300)K = 125.5 kJ/kg

The net work production is then wnet = q in − q out = 295.4 − 125.5 = 169.9 kJ/kg

and the thermal efficiency of this cycle is

η th =

wnet 169.9 kJ/kg = = 0.575 q in 295.4 kJ/kg

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9-65

9-95 A simple Brayton cycle with air as the working fluid operates between the specified temperature and pressure limits. The net work and the thermal efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2a). Analysis Using the isentropic relations for an ideal gas, ⎛P T2 = T1 ⎜⎜ 2 ⎝ P1

⎞ ⎟⎟ ⎠

( k −1) / k

⎛ 2000 kPa ⎞ = (300 K)⎜ ⎟ ⎝ 100 kPa ⎠

0.4/1.4

T

= 706.1 K

3

1000 K

qin

For the expansion process, T4 s

⎛P = T3 ⎜⎜ 4 ⎝ P3

ηT =

⎞ ⎟ ⎟ ⎠

( k −1) / k

2 ⎛ 100 kPa ⎞ = (1000 K)⎜ ⎟ ⎝ 2000 kPa ⎠

0.4/1.4

= 424.9 K

300 K

1

4

qout 4s

c p (T3 − T4 ) h3 − h4 = ⎯ ⎯→ T4 = T3 − η T (T3 − T4 s ) h3 − h4 s c p (T3 − T4 s ) = 1000 − (0.90)(1000 − 424.9) = 482.4 K

s

Applying the first law to the constant-pressure heat addition process 2-3 produces Qin = m(h3 − h2 ) = mc p (T3 − T2 ) = (1 kg )(1.005 kJ/kg ⋅ K )(1000 − 706.1)K = 295.4 kJ

Similarly, Qout = m(h4 − h1 ) = mc p (T4 − T1 ) = (1 kg )(1.005 kJ/kg ⋅ K )(482.4 − 300)K = 183.3 kJ

The net work production is then W net = Qin − Qout = 295.4 − 183.3 = 112.1 kJ

and the thermal efficiency of this cycle is

η th =

W net 112.1 kJ = = 0.379 Qin 295.4 kJ

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-66

9-96 A simple Brayton cycle with air as the working fluid operates between the specified temperature and pressure limits. The net work and the thermal efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2a).

T

Analysis For the compression process, ⎛P T2 s = T1 ⎜⎜ 2 ⎝ P1

⎞ ⎟⎟ ⎠

( k −1) / k

1000 K

⎛ 2000 kPa ⎞ = (300 K)⎜ ⎟ ⎝ 100 kPa ⎠

0.4/1.4

2s 2

= 706.1 K

c p (T2 s − T1 ) h −h T − T1 ⎯ ⎯→ T2 = T1 + 2 s η C = 2s 1 = h2 − h1 c p (T2 − T1 ) ηC = 300 +

3

qin

300 K

1

4

qout 4s s

706.1 − 300 = 807.6 K 0.80

For the expansion process, ⎛P T4 s = T3 ⎜⎜ 4 ⎝ P3

ηT =

⎞ ⎟ ⎟ ⎠

( k −1) / k

⎛ 100 kPa ⎞ = (1000 K)⎜ ⎟ ⎝ 2000 kPa ⎠

0.4/1.4

= 424.9 K

c p (T3 − T4 ) h3 − h 4 = ⎯ ⎯→ T4 = T3 − η T (T3 − T4 s ) h3 − h4 s c p (T3 − T4 s ) = 1000 − (0.90)(1000 − 424.9) = 482.4 K

Applying the first law to the constant-pressure heat addition process 2-3 produces Qin = m(h3 − h2 ) = mc p (T3 − T2 ) = (1 kg )(1.005 kJ/kg ⋅ K )(1000 − 807.6)K = 193.4 kJ

Similarly, Qout = m(h4 − h1 ) = mc p (T4 − T1 ) = (1 kg )(1.005 kJ/kg ⋅ K )(482.4 − 300)K = 183.3 kJ

The net work production is then W net = Qin − Qout = 193.4 − 183.3 = 10.1 kJ

and the thermal efficiency of this cycle is

η th =

W net 10.1 kJ = = 0.0522 193.4 kJ Qin

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-67

9-97 A simple Brayton cycle with air as the working fluid operates between the specified temperature and pressure limits. The net work and the thermal efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2a).

T

Analysis For the compression process, ⎛P T2 s = T1 ⎜⎜ 2 ⎝ P1

ηC =

⎞ ⎟⎟ ⎠

( k −1) / k

1000 K

⎛ 2000 kPa ⎞ = (300 K)⎜ ⎟ ⎝ 100 kPa ⎠

0.4/1.4

2s 2

= 706.1 K

h2 s − h1 c p (T2 s − T1 ) T − T1 = ⎯ ⎯→ T2 = T1 + 2 s h2 − h1 c p (T2 − T1 ) ηC = 300 +

3

qin

300 K

1

4

qout 4s

706.1 − 300 = 807.6 K 0.80

s

For the expansion process, ⎛P T4 s = T3 ⎜⎜ 4 ⎝ P3

ηT =

⎞ ⎟ ⎟ ⎠

( k −1) / k

⎛ 100 kPa ⎞ = (1000 K)⎜ ⎟ ⎝ 1950 kPa ⎠

0.4/1.4

= 428.0 K

c p (T3 − T4 ) h3 − h 4 = ⎯ ⎯→ T4 = T3 − η T (T3 − T4 s ) h3 − h4 s c p (T3 − T4 s ) = 1000 − (0.90)(1000 − 428.0) = 485.2 K

Applying the first law to the constant-pressure heat addition process 2-3 produces Qin = m(h3 − h2 ) = mc p (T3 − T2 ) = (1 kg )(1.005 kJ/kg ⋅ K )(1000 − 807.6)K = 193.4 kJ

Similarly, Qout = m(h4 − h1 ) = mc p (T4 − T1 ) = (1 kg )(1.005 kJ/kg ⋅ K )(485.2 − 300)K = 186.1 kJ

The net work production is then W net = Qin − Qout = 193.4 − 186.1 = 7.3 kJ

and the thermal efficiency of this cycle is

η th =

W net 7.3 kJ = = 0.0377 193.4 kJ Qin

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-68

9-98 A gas turbine power plant that operates on the simple Brayton cycle with air as the working fluid has a specified pressure ratio. The required mass flow rate of air is to be determined for two cases. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2). Analysis (a) Using the isentropic relations, ⎛P T2 s = T1 ⎜⎜ 2 ⎝ P1

⎞ ⎟⎟ ⎠

⎛P = T3 ⎜⎜ 4 ⎝ P3

⎞ ⎟ ⎟ ⎠

T4 s

(k −1) / k

(k −1) / k

T

= (300 K )(12 )0.4/1.4 = 610.2 K ⎛ 1⎞ = (1000 K )⎜ ⎟ ⎝ 12 ⎠

3

1000 K

0.4/1.4

= 491.7 K

2s 300 K

2

1

4s

4

ws,C,in = h2 s − h1 = c p (T2 s − T1 ) = (1.005 kJ/kg ⋅ K )(610.2 − 300 )K = 311.75 kJ/kg ws,T, out = h3 − h4 s = c p (T3 − T4 s ) = (1.005 kJ/kg ⋅ K )(1000 − 491.7 )K = 510.84 kJ/kg ws, net,out = ws,T,out − ws,C,in = 510.84 − 311.75 = 199.1 kJ/kg m& s =

W& net,out ws, net,out

=

70,000 kJ/s = 352 kg/s 199.1 kJ/kg

(b) The net work output is determined to be wa,net,out = wa,T,out − wa,C,in = η T ws,T, out − ws,C,in / η C = (0.85)(510.84 ) − 311.75 0.85 = 67.5 kJ/kg

m& a =

W& net,out wa,net,out

=

70,000 kJ/s = 1037 kg/s 67.5 kJ/kg

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

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9-69

9-99 A stationary gas-turbine power plant operates on a simple ideal Brayton cycle with air as the working fluid. The power delivered by this plant is to be determined assuming constant and variable specific heats. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas. T Analysis (a) Assuming constant specific heats, T2 s

⎛P = T1 ⎜⎜ 2 ⎝ P1

⎛P T4 s = T3 ⎜⎜ 4 ⎝ P3

η th

⎞ ⎟⎟ ⎠ ⎞ ⎟ ⎟ ⎠

(k −1) / k

(k −1) / k

= (290 K )(8)

3

1100 K 0.4/1.4

qin

= 525.3 K

2 ⎛1⎞ = (1100 K )⎜ ⎟ ⎝8⎠

0.4/1.4

= 607.2 K

290 K

4 1

qout

c p (T4 − T1 ) q T −T 607.2 − 290 = 1 − out = 1 − = 1− 4 1 = 1− = 0.448 q in c p (T3 − T2 ) T3 − T2 1100 − 525.3

W& net,out = η th Q& in = (0.448)(35,000 kW ) = 15,680 kW

(b) Assuming variable specific heats (Table A-17), T1 = 290 K ⎯ ⎯→ Pr 2 =

h1 = 290.16 kJ/kg Pr1 = 1.2311

P2 Pr = (8)(1.2311) = 9.8488 ⎯ ⎯→ h2 = 526.12 kJ/kg P1 1

T3 = 1100 K ⎯ ⎯→

h3 = 1161.07 kJ/kg Pr3 = 167.1

P4 ⎛1⎞ Pr = ⎜ ⎟(167.1) = 20.89 ⎯ ⎯→ h4 = 651.37 kJ/kg P3 3 ⎝ 8 ⎠ q h −h 651.37 − 290.16 = 1 − out == 1 − 4 1 = 1 − = 0.431 q in h3 − h2 1161.07 − 526.11

Pr 4 =

η th

W& net,out = η T Q& in = (0.431)(35,000 kW ) = 15,085 kW

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

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9-70

9-100 An actual gas-turbine power plant operates at specified conditions. The fraction of the turbine work output used to drive the compressor and the thermal efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with variable specific heats. Properties The properties of air are given in Table A-17.

T

Analysis (a) Using the isentropic relations, T1 = 300 K

⎯ ⎯→

h1 = 300.19 kJ / kg

T2 = 580 K

⎯ ⎯→

h2 = 586.04 kJ / kg

rp =

P2 700 = =7 P1 100

950 kJ/kg 580 K 300 K

2s 1

3

2 4s

4

⎯→ h3 = 950 + 586.04 = 1536.04kJ/kg q in = h3 − h2 ⎯ → Pr3 = 474.11 Pr 4 =

P4 ⎛1⎞ Pr = ⎜ ⎟(474.11) = 67.73 ⎯ ⎯→ h4 s = 905.83 kJ/kg P3 3 ⎝ 7 ⎠

wC,in = h2 − h1 = 586.04 − 300.19 = 285.85 kJ/kg wT,out = η T (h3 − h4 s ) = (0.86 )(1536.04 − 905.83) = 542.0 kJ/kg

Thus, rbw =

(b)

wC,in wT,out

=

285.85 kJ/kg = 52.7% 542.0 kJ/kg

wnet.out = wT,out − wC,in = 542.0 − 285.85 = 256.15 kJ/kg

η th =

wnet,out q in

=

256.15 kJ/kg = 27.0% 950 kJ/kg

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9-71

9-101 A gas-turbine power plant operates at specified conditions. The fraction of the turbine work output used to drive the compressor and the thermal efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 The airstandard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2).

T 950 kJ/kg 580 K

2s

300 K

1

Analysis (a) Using constant specific heats, rp =

P2 700 = =7 P1 100

3

2 4s

4 s

q in = h3 − h2 = c p (T3 − T2 )⎯ ⎯→ T3 = T2 + q in /c p

= 580 K + (950 kJ/kg )/ (1.005 kJ/kg ⋅ K ) = 1525.3 K

⎛P T4s = T3 ⎜⎜ 4 ⎝ P3

⎞ ⎟ ⎟ ⎠

(k −1)/k

⎛1⎞ = (1525.3 K )⎜ ⎟ ⎝7⎠

0.4/1.4

= 874.8 K

wC,in = h2 − h1 = c p (T2 − T1 ) = (1.005kJ/kg ⋅ K )(580 − 300)K = 281.4 kJ/kg wT,out = η T (h3 − h4 s ) = η T c p (T3 − T4 s ) = (0.86 )(1.005 kJ/kg ⋅ K )(1525.3 − 874.8)K = 562.2 kJ/kg

Thus, rbw =

(b)

wC,in wT,out

=

281.4 kJ/kg = 50.1% 562.2 kJ/kg

wnet,out = wT,out − wC,in = 562.2 − 281.4 = 280.8 kJ/kg

η th =

wnet,out q in

=

280.8 kJ/kg = 29.6% 950 kJ/kg

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-72

9-102E A simple ideal Brayton cycle with argon as the working fluid operates between the specified temperature and pressure limits. The rate of heat addition, the power produced, and the thermal efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 Argon is an ideal gas with constant specific heats. Properties The properties of argon at room temperature are R = 0.2686 psia⋅ft3/lbm·R (Table A-1E), cp = 0.1253 Btu/lbm·R and k = 1.667 (Table A-2Ea). Analysis At the compressor inlet,

v1 = m& =

RT1 (0.2686 psia ⋅ ft 3 /lbm ⋅ R )(540 R) = = 9.670 ft 3 /lbm P1 15 psia A1V1

v1

(3 ft 2 )(200 ft/s)

=

9.670 ft 3 /lbm

⎞ ⎟⎟ ⎠

( k −1) / k

⎛P T4 = T3 ⎜⎜ 4 ⎝ P3

⎞ ⎟ ⎟ ⎠

( k −1) / k

3

1660 R

qin 2

= 62.05 lbm/s

4

According to the isentropic process expressions for an ideal gas, ⎛P T2 = T1 ⎜⎜ 2 ⎝ P1

T

⎛ 150 psia ⎞ ⎟⎟ = (540 R)⎜⎜ ⎝ 15 psia ⎠

540 R

1

qout

0.667/1.667

⎛ 15 psia ⎞ ⎟⎟ = (1660 R)⎜⎜ ⎝ 150 psia ⎠

= 1357 R 0.667/1.667

= 660.7 R

Applying the first law to the constant-pressure heat addition process 2-3 gives

Q& in = m& c p (T3 − T2 ) = (62.05 lbm/s)(0.1253 Btu/lbm ⋅ R )(1660 − 1357)R = 2356 Btu/s The net power output is W& net = m& c p (T3 − T4 + T1 − T2 ) = (62.05 lbm/s)(0.1253 Btu/lbm ⋅ R )(1660 − 660.7 + 540 − 1357)R = 1417 Btu/s

The thermal efficiency of this cycle is then

η th =

W& net 1417 Btu/s = = 0.601 2356 Btu/s Q& in

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

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9-73

9-103 An aircraft engine operates as a simple ideal Brayton cycle with air as the working fluid. The pressure ratio and the rate of heat input are given. The net power and the thermal efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2a). Analysis For the isentropic compression process,

T2 = T1 r p( k −1) / k = (273 K)(10) 0.4/1.4 = 527.1 K

T 3

The heat addition is q in =

qin

Q& in 500 kW = = 500 kJ/kg m& 1 kg/s

2

Applying the first law to the heat addition process,

273 K

q in = c p (T3 − T2 )

4 1

qout s

q 500 kJ/kg = 1025 K T3 = T2 + in = 527.1 K + cp 1.005 kJ/kg ⋅ K

The temperature at the exit of the turbine is ⎛ 1 T4 = T3 ⎜ ⎜ rp ⎝

⎞ ⎟ ⎟ ⎠

( k −1) / k

⎛1⎞ = (1025 K)⎜ ⎟ ⎝ 10 ⎠

0.4/1.4

= 530.9 K

Applying the first law to the adiabatic turbine and the compressor produce wT = c p (T3 − T4 ) = (1.005 kJ/kg ⋅ K )(1025 − 530.9)K = 496.6 kJ/kg wC = c p (T2 − T1 ) = (1.005 kJ/kg ⋅ K )(527.1 − 273)K = 255.4 kJ/kg

The net power produced by the engine is then W& net = m& ( wT − wC ) = (1 kg/s)(496.6 − 255.4)kJ/kg = 241.2 kW

Finally the thermal efficiency is

η th =

W& net 241.2 kW = = 0.482 500 kW Q& in

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-74

9-104 An aircraft engine operates as a simple ideal Brayton cycle with air as the working fluid. The pressure ratio and the rate of heat input are given. The net power and the thermal efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2a). Analysis For the isentropic compression process,

T2 = T1 r p( k −1) / k = (273 K)(15) 0.4/1.4 = 591.8 K

T

The heat addition is q in =

3 qin

Q& in 500 kW = = 500 kJ/kg m& 1 kg/s

2

Applying the first law to the heat addition process,

273 K

q in = c p (T3 − T2 ) q 500 kJ/kg = 1089 K T3 = T2 + in = 591.8 K + cp 1.005 kJ/kg ⋅ K

4 1

qout s

The temperature at the exit of the turbine is ⎛ 1 T4 = T3 ⎜ ⎜ rp ⎝

⎞ ⎟ ⎟ ⎠

( k −1) / k

⎛1⎞ = (1089 K)⎜ ⎟ ⎝ 15 ⎠

0.4/1.4

= 502.3 K

Applying the first law to the adiabatic turbine and the compressor produce wT = c p (T3 − T4 ) = (1.005 kJ/kg ⋅ K )(1089 − 502.3)K = 589.6 kJ/kg wC = c p (T2 − T1 ) = (1.005 kJ/kg ⋅ K )(591.8 − 273)K = 320.4 kJ/kg

The net power produced by the engine is then W& net = m& ( wT − wC ) = (1 kg/s)(589.6 − 320.4)kJ/kg = 269.2 kW

Finally the thermal efficiency is

η th =

W& net 269.2 kW = = 0.538 500 kW Q& in

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-75

9-105 A gas-turbine plant operates on the simple Brayton cycle. The net power output, the back work ratio, and the thermal efficiency are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The gas constant of air is R = 0.287 kJ/kg·K (Table A-1). Analysis (a) For this problem, we use the properties from EES software. Remember that for an ideal gas, Combustion enthalpy is a function of temperature only whereas chamber entropy is functions of both temperature and pressure. 3 Process 1-2: Compression 2 1.2 MPa ⎯→ h = 303.60 kJ/kg T = 30°C ⎯ 1

1

T1 = 30°C

⎫ ⎬s1 = 5.7159 kJ/kg ⋅ K P1 = 100 kPa ⎭

Compress.

P2 = 1200 kPa

⎫ ⎬h2 s = 617.37 kJ/kg s 2 = s1 = 5.7159 kJ/kg.K ⎭

ηC =

1

100 kPa 30°C

Turbine 500°C

4

h2 s − h1 617.37 − 303.60 ⎯ ⎯→ 0.82 = ⎯ ⎯→ h2 = 686.24 kJ/kg h2 − h1 h2 − 303.60

Process 3-4: Expansion T4 = 500°C ⎯ ⎯→ h4 = 792.62 kJ/kg

ηT =

h3 − h 4 h − 792.62 ⎯ ⎯→ 0.88 = 3 h3 − h4 s h3 − h4 s

We cannot find the enthalpy at state 3 directly. However, using the following lines in EES together with the isentropic efficiency relation, we find h3 = 1404.7 kJ/kg, T3 = 1034ºC, s3 = 6.5699 kJ/kg.K. The solution by hand would require a trial-error approach. h_3=enthalpy(Air, T=T_3) s_3=entropy(Air, T=T_3, P=P_2) h_4s=enthalpy(Air, P=P_1, s=s_3)

The mass flow rate is determined from P V& (100 kPa)(150/60 m 3 / s) = 2.875 kg/s m& = 1 1 = RT1 0.287 kPa ⋅ m 3 /kg ⋅ K (30 + 273 K )

(

)

The net power output is W& C,in = m& (h2 − h1 ) = (2.875 kg/s)(686.24 − 303.60)kJ/kg = 1100 kW W& T,out = m& (h3 − h4 ) = (2.875 kg/s)(1404.7 − 792.62)kJ/kg = 1759 kW W& net = W& T,out − W& C,in = 1759 − 1100 = 659 kW

(b) The back work ratio is W& C,in 1100 kW = = 0.625 rbw = W& T, out 1759 kW (c) The rate of heat input and the thermal efficiency are Q& = m& (h − h ) = (2.875 kg/s)(1404.7 − 686.24)kJ/kg = 2065 kW in

3

2

W& 659 kW η th = net = = 0.319 2065 kW Q& in

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-76

Brayton Cycle with Regeneration 9-106C Regeneration increases the thermal efficiency of a Brayton cycle by capturing some of the waste heat from the exhaust gases and preheating the air before it enters the combustion chamber.

9-107C Yes. At very high compression ratios, the gas temperature at the turbine exit may be lower than the temperature at the compressor exit. Therefore, if these two streams are brought into thermal contact in a regenerator, heat will flow to the exhaust gases instead of from the exhaust gases. As a result, the thermal efficiency will decrease. 9-108C The extent to which a regenerator approaches an ideal regenerator is called the effectiveness ε, and is defined as ε = qregen, act /qregen, max.

9-109C (b) turbine exit.

9-110C The steam injected increases the mass flow rate through the turbine and thus the power output. This, in turn, increases the thermal efficiency since η = W / Qin and W increases while Qin remains constant. Steam can be obtained by utilizing the hot exhaust gases.

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-77

9-111 A Brayton cycle with regeneration produces 150 kW power. The rates of heat addition and rejection are to be determined. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K and k = 1.4 (Table A-2a). Analysis According to the isentropic process expressions for an ideal gas,

T2 = T1 r p( k −1) / k = (293 K)(8) 0.4/1.4 = 530.8 K ⎛ 1 T5 = T4 ⎜ ⎜ rp ⎝

⎞ ⎟ ⎟ ⎠

( k −1) / k

⎛1⎞ = (1073 K)⎜ ⎟ ⎝8⎠

T

0.4/1.4

= 592.3 K

3

When the first law is applied to the heat exchanger, the result is

5

2

T3 − T2 = T5 − T6

while the regenerator temperature specification gives

4

qin

1073 K

293 K

1

T3 = T5 − 10 = 592.3 − 10 = 582.3 K

6 qout s

The simultaneous solution of these two results gives T6 = T5 − (T3 − T2 ) = 592.3 − (582.3 − 530.8) = 540.8 K

Application of the first law to the turbine and compressor gives wnet = c p (T4 − T5 ) − c p (T2 − T1 ) = (1.005 kJ/kg ⋅ K )(1073 − 592.3) K − (1.005 kJ/kg ⋅ K )(530.8 − 293) K = 244.1 kJ/kg

Then, m& =

W& net 150 kW = = 0.6145 kg/s wnet 244.1 kJ/kg

Applying the first law to the combustion chamber produces Q& in = m& c p (T4 − T3 ) = (0.6145 kg/s)(1.005 kJ/kg ⋅ K )(1073 − 582.3)K = 303.0 kW Similarly, Q& out = m& c p (T6 − T1 ) = (0.6145 kg/s)(1.005 kJ/kg ⋅ K )(540.8 − 293)K = 153.0 kW

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-78

9-112 A Brayton cycle with regeneration produces 150 kW power. The rates of heat addition and rejection are to be determined. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K and k = 1.4 (Table A-2a). Analysis For the compression and expansion processes we have

T2 s = T1 r p( k −1) / k = (293 K)(8) 0.4/1.4 = 530.8 K

ηC =

c p (T2 s − T1 ) c p (T2 − T1 )

⎯ ⎯→ T2 = T1 +

T2 s − T1

ηC

= 293 +

T5 s

⎛ 1 = T4 ⎜ ⎜ rp ⎝

ηT =

⎞ ⎟ ⎟ ⎠

( k −1) / k

c p (T4 − T5 ) c p (T4 − T5 s )

T

530.8 − 293 = 566.3 K 0.87

2s 293 K

⎛1⎞ = (1073 K)⎜ ⎟ ⎝8⎠

0.4/1.4

4

qin

1073 K 2

5

3

1

6

5s

qout

= 592.3 K

s

⎯ ⎯→ T5 = T4 − η T (T4 − T5 s ) = 1073 − (0.93)(1073 − 592.3) = 625.9 K

When the first law is applied to the heat exchanger, the result is T3 − T2 = T5 − T6

while the regenerator temperature specification gives T3 = T5 − 10 = 625.9 − 10 = 615.9 K

The simultaneous solution of these two results gives T6 = T5 − (T3 − T2 ) = 625.9 − (615.9 − 566.3) = 576.3 K

Application of the first law to the turbine and compressor gives w net = c p (T4 − T5 ) − c p (T2 − T1 ) = (1.005 kJ/kg ⋅ K )(1073 − 625.9) K − (1.005 kJ/kg ⋅ K )(566.3 − 293) K = 174.7 kJ/kg

Then, m& =

W& net 150 kW = = 0.8586 kg/s wnet 174.7 kJ/kg

Applying the first law to the combustion chamber produces Q& in = m& c p (T4 − T3 ) = (0.8586 kg/s)(1.005 kJ/kg ⋅ K )(1073 − 615.9)K = 394.4 kW

Similarly, Q& out = m& c p (T6 − T1 ) = (0.8586 kg/s)(1.005 kJ/kg ⋅ K )(576.3 − 293)K = 244.5 kW

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-79

9-113 A Brayton cycle with regeneration is considered. The thermal efficiencies of the cycle for parallelflow and counter-flow arrangements of the regenerator are to be compared. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are cp = 1.005 kJ/kg⋅K and k = 1.4 (Table A-2a). Analysis According to the isentropic process expressions for an ideal gas,

T2 = T1 r p( k −1) / k = (293 K)(7) 0.4/1.4 = 510.9 K ⎛ 1 T5 = T4 ⎜ ⎜ rp ⎝

⎞ ⎟ ⎟ ⎠

( k −1) / k

⎛1⎞ = (1000 K)⎜ ⎟ ⎝7⎠

0.4/1.4

T

= 573.5 K

When the first law is applied to the heat exchanger as originally arranged, the result is T3 − T2 = T5 − T6

4

qin

1000 K 3

5

2 293 K

1

while the regenerator temperature specification gives

6 qout s

T3 = T5 − 6 = 573.5 − 6 = 567.5 K

The simultaneous solution of these two results gives T6 = T5 − T3 + T2 = 573.5 − 567.5 + 510.9 = 516.9 K

The thermal efficiency of the cycle is then

η th = 1 −

q out T −T 516.9 − 293 = 1− 6 1 = 1− = 0.482 1000 − 567.5 q in T4 − T3

For the rearranged version of this cycle, T3 = T6 − 6

T 4

qin

1000 K

An energy balance on the heat exchanger gives T3 − T2 = T5 − T6

The solution of these two equations is T3 = 539.2 K

2 293 K

1

3

6

5

qout s

T6 = 545.2 K The thermal efficiency of the cycle is then

η th = 1 −

q out T −T 545.2 − 293 = 1− 6 1 = 1− = 0.453 1000 − 539.2 q in T4 − T3

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9-80

9-114E An ideal Brayton cycle with regeneration has a pressure ratio of 8. The thermal efficiency of the cycle is to be determined with and without regenerator cases. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are cp = 0.24 Btu/lbm⋅R and k = 1.4 (Table A-2Ea). Analysis According to the isentropic process expressions for an ideal gas,

T2 = T1 r p( k −1) / k = (510 R)(8) 0.4/1.4 = 923.8 R T ⎛ 1 T5 = T4 ⎜ ⎜ rp ⎝

⎞ ⎟ ⎟ ⎠

( k −1) / k

⎛1⎞ = (1960 R)⎜ ⎟ ⎝8⎠

0.4/1.4

= 1082 R

qin

1960 R

4

3

T3 = T5 = 1082 R

5

2

The regenerator is ideal (i.e., the effectiveness is 100%) and thus, 510 R

6 qout

1

T6 = T2 = 923.8 R

s

The thermal efficiency of the cycle is then

η th = 1 −

q out T −T 923.8 − 510 = 1− 6 1 = 1− = 0.529 1960 − 1082 q in T4 − T3

T

The solution without a regenerator is as follows: T2 = T1 r p( k −1) / k = (510 R)(8) 0.4/1.4 = 923.8 R ⎛ 1 T4 = T3 ⎜ ⎜ rp ⎝

η th = 1 −

⎞ ⎟ ⎟ ⎠

( k −1) / k

⎛1⎞ = (1960 R)⎜ ⎟ ⎝8⎠

3

1960 R

qin 2

0.4/1.4

= 1082 R

q out T −T 1082 − 510 = 1− 4 1 = 1− = 0.448 1960 − 923.8 q in T3 − T2

510 R

4 1

qout s

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9-81

9-115 An expression for the thermal efficiency of an ideal Brayton cycle with an ideal regenerator is to be developed. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. Analysis The expressions for the isentropic compression and expansion processes are T2 = T1 r p( k −1) / k ⎛ 1 T4 = T3 ⎜ ⎜ rp ⎝

⎞ ⎟ ⎟ ⎠

( k −1) / k

T qin

For an ideal regenerator, T5 = T4

1

The thermal efficiency of the cycle is

η th

5 4

2

T6 = T 2

3

6 qout s

q T −T T (T / T ) − 1 = 1 − out = 1 − 6 1 = 1 − 1 6 1 q in T3 − T5 T3 1 − (T5 / T3 ) = 1−

T1 (T2 / T1 ) − 1 T3 1 − (T4 / T3 )

= 1−

( k −1) / k −1 T1 r p ( 1 k − − T3 1 − r p ) / k

= 1−

T1 ( k −1) / k rp T3

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9-82

9-116E A car is powered by a gas turbine with a pressure ratio of 4. The thermal efficiency of the car and the mass flow rate of air for a net power output of 95 hp are to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with variable specific heats. 3 The ambient air is 540 R and 14.5 psia. 4 The effectiveness of the regenerator is 0.9, and the isentropic efficiencies for both the compressor and the turbine are 80%. 5 The combustion gases can be treated as air. Properties The properties of air at the compressor and turbine inlet temperatures can be obtained from Table A-17E. Analysis The gas turbine cycle with regeneration can be analyzed as follows: T1 = 540 R ⎯ ⎯→ Pr 2 =

h1 = 129.06 Btu/lbm Pr1 = 1.386

P2 Pr = (4 )(1.386 ) = 5.544 ⎯ ⎯→ h2 s = 192.0 Btu/lbm P1 1

T3 = 2160 R ⎯ ⎯→

T qin

2160 R

h3 = 549.35 Btu/lbm

5

Pr3 = 230.12

P ⎛1⎞ Pr 4 = 4 Pr3 = ⎜ ⎟(230.12 ) = 57.53 ⎯ ⎯→ h4 s = 372.2 Btu/lbm P3 ⎝4⎠

3

2

4 4s

2s 540 R

1 s

and

η comp =

h2 s − h1 192.0 − 129.06 → 0.80 = → h2 = 207.74 Btu/lbm h2 − h1 h2 − 129.06

η turb =

h3 − h4 549.35 − h4 → 0.80 = → h4 = 407.63 Btu/lbm h3 − h4 s 549.35 − 372.2

Then the thermal efficiency of the gas turbine cycle becomes q regen = ε (h4 − h2 ) = 0.9(407.63 − 207.74) = 179.9 Btu/lbm q in = (h3 − h2 ) − q regen = (549.35 − 207.74) − 179.9 = 161.7 Btu/lbm w net,out = wT,out − wC,in = (h3 − h4 ) − (h2 − h1 ) = (549.35 − 407.63) − (207.74 − 129.06) = 63.0 Btu/lbm

η th =

wnet,out q in

=

63.0 Btu/lbm = 0.39 = 39% 161.7 Btu/lbm

Finally, the mass flow rate of air through the turbine becomes m& air =

W& net ⎛ 0.7068 Btu/s ⎞ 95 hp ⎟⎟ = 1.07 lbm/s ⎜ = wnet 63.0 Btu/lbm ⎜⎝ 1 hp ⎠

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9-83

9-117 [Also solved by EES on enclosed CD] The thermal efficiency and power output of an actual gas turbine are given. The isentropic efficiency of the turbine and of the compressor, and the thermal efficiency of the gas turbine modified with a regenerator are to be determined. Assumptions 1 Air is an ideal gas with variable specific heats. 2 Kinetic and potential energy changes are negligible. 3 The mass flow rates of air and of the combustion gases are the same, and the properties of combustion gases are the same as those of air. Properties The properties of air are given in Table A-17. Analysis The properties at various states are ⎯→ T1 = 20°C = 293 K ⎯ Pr 2 =

Pr1 = 1.2765

qin

1561 K 2 2s

h3 = 1710.0 kJ/kg Pr3 = 712.5

3

5

P2 Pr = (14.7 )(1.2765) = 18.765 ⎯ ⎯→ h2 s = 643.3 kJ/kg P1 1

T3 = 1288°C = 1561 K ⎯ ⎯→ Pr 4 =

T

h1 = 293.2 kJ/kg

293 K

4 4s

1

P4 ⎛ 1 ⎞ Pr = ⎜ ⎯→ h4 s = 825.23 kJ/kg ⎟(712.5) = 48.47 ⎯ P3 3 ⎝ 14.7 ⎠

The net work output and the heat input per unit mass are W& net 159,000 kW ⎛ 3600 s ⎞ = ⎜ ⎟ = 372.66 kJ/kg m& 1,536,000 kg/h ⎝ 1 h ⎠ w 372.66 kJ/kg = net = = 1038.0 kJ/kg η th 0.359

wnet = q in

q in = h3 − h2 → h2 = h3 − q in = 1710 − 1038 = 672.0 kJ/kg q out = q in − wnet = 1038.0 − 372.66 = 665.34 kJ/kg q out = h4 − h1 → h4 = q out + h1 = 665.34 + 293.2 = 958.54 kJ/kg → T4 = 650°C

Then the compressor and turbine efficiencies become

ηT =

h3 − h4 1710 − 958.54 = 0.849 = h3 − h4 s 1710 − 825.23

ηC =

h2 s − h1 643.3 − 293.2 = = 0.924 672 − 293.2 h2 − h1

When a regenerator is added, the new heat input and the thermal efficiency become q regen = ε (h4 − h2 ) = (0.80)(958.54 - 672.0) = 286.54 kJ/kg q in, new = q in − q regen = 1038 − 286.54 = 751.46 kJ/kg

η th,new =

wnet 372.66 kJ/kg = = 0.496 q in, new 751.46 kJ/kg

Discussion Note an 80% efficient regenerator would increase the thermal efficiency of this gas turbine from 35.9% to 49.6%.

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9-84

9-118 EES Problem 9-117 is reconsidered. A solution that allows different isentropic efficiencies for the compressor and turbine is to be developed and the effect of the isentropic efficiencies on net work done and the heat supplied to the cycle is to be studied. Also, the T-s diagram for the cycle is to be plotted. Analysis Using EES, the problem is solved as follows: "Input data" T[3] = 1288 [C] Pratio = 14.7 T[1] = 20 [C] P[1]= 100 [kPa] {T[4]=589 [C]} {W_dot_net=159 [MW] }"We omit the information about the cycle net work" m_dot = 1536000 [kg/h]*Convert(kg/h,kg/s) {Eta_th_noreg=0.359} "We omit the information about the cycle efficiency." Eta_reg = 0.80 Eta_c = 0.892 "Compressor isentorpic efficiency" Eta_t = 0.926 "Turbien isentropic efficiency" "Isentropic Compressor anaysis" s[1]=ENTROPY(Air,T=T[1],P=P[1]) s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor" P[2] = Pratio*P[1] s_s[2]=ENTROPY(Air,T=T_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exit" Eta_c = W_dot_compisen/W_dot_comp "compressor adiabatic efficiency, W_dot_comp > W_dot_compisen" "Conservation of energy for the compressor for the isentropic case: E_dot_in - E_dot_out = DELTAE_dot=0 for steady-flow" m_dot*h[1] + W_dot_compisen = m_dot*h_s[2] h[1]=ENTHALPY(Air,T=T[1]) h_s[2]=ENTHALPY(Air,T=T_s[2]) "Actual compressor analysis:" m_dot*h[1] + W_dot_comp = m_dot*h[2] h[2]=ENTHALPY(Air,T=T[2]) s[2]=ENTROPY(Air,T=T[2], P=P[2]) "External heat exchanger analysis" "SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0 E_dot_in - E_dot_out =DELTAE_dot_cv =0 for steady flow" m_dot*h[2] + Q_dot_in_noreg = m_dot*h[3] q_in_noreg=Q_dot_in_noreg/m_dot h[3]=ENTHALPY(Air,T=T[3]) P[3]=P[2]"process 2-3 is SSSF constant pressure" "Turbine analysis" s[3]=ENTROPY(Air,T=T[3],P=P[3]) s_s[4]=s[3] "For the ideal case the entropies are constant across the turbine" P[4] = P[3] /Pratio s_s[4]=ENTROPY(Air,T=T_s[4],P=P[4])"T_s[4] is the isentropic value of T[4] at turbine exit" Eta_t = W_dot_turb /W_dot_turbisen "turbine adiabatic efficiency, W_dot_turbisen > W_dot_turb" "SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0 E_dot_in -E_dot_out = DELTAE_dot_cv = 0 for steady-flow" PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-85 m_dot*h[3] = W_dot_turbisen + m_dot*h_s[4] h_s[4]=ENTHALPY(Air,T=T_s[4]) "Actual Turbine analysis:" m_dot*h[3] = W_dot_turb + m_dot*h[4] h[4]=ENTHALPY(Air,T=T[4]) s[4]=ENTROPY(Air,T=T[4], P=P[4]) "Cycle analysis" "Using the definition of the net cycle work and 1 MW = 1000 kW:" W_dot_net*1000=W_dot_turb-W_dot_comp "kJ/s" Eta_th_noreg=W_dot_net*1000/Q_dot_in_noreg"Cycle thermal efficiency" Bwr=W_dot_comp/W_dot_turb"Back work ratio" "With the regenerator the heat added in the external heat exchanger is" m_dot*h[5] + Q_dot_in_withreg = m_dot*h[3] q_in_withreg=Q_dot_in_withreg/m_dot h[5]=ENTHALPY(Air, T=T[5]) s[5]=ENTROPY(Air,T=T[5], P=P[5]) P[5]=P[2] "The regenerator effectiveness gives h[5] and thus T[5] as:" Eta_reg = (h[5]-h[2])/(h[4]-h[2]) "Energy balance on regenerator gives h[6] and thus T[6] as:" m_dot*h[2] + m_dot*h[4]=m_dot*h[5] + m_dot*h[6] h[6]=ENTHALPY(Air, T=T[6]) s[6]=ENTROPY(Air,T=T[6], P=P[6]) P[6]=P[4] "Cycle thermal efficiency with regenerator" Eta_th_withreg=W_dot_net*1000/Q_dot_in_withreg "The following data is used to complete the Array Table for plotting purposes." s_s[1]=s[1] T_s[1]=T[1] s_s[3]=s[3] T_s[3]=T[3] s_s[5]=ENTROPY(Air,T=T[5],P=P[5]) T_s[5]=T[5] s_s[6]=s[6] T_s[6]=T[6]

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-86

ηt

ηc

ηth,noreg

ηth,withreg

0.7 0.75 0.8 0.85 0.9 0.95 1

0.892 0.892 0.892 0.892 0.892 0.892 0.892

0.2309 0.2736 0.3163 0.359 0.4016 0.4443 0.487

0.3405 0.3841 0.4237 0.4599 0.493 0.5234 0.5515

Qinnoreg [kW] 442063 442063 442063 442063 442063 442063 442063

Qinwithreg [kW] 299766 314863 329960 345056 360153 375250 390346

Wnet [kW] 102.1 120.9 139.8 158.7 177.6 196.4 215.3

T-s Diagram for Gas Turbine with Regeneration 1700 1500

1470 kPa

3

1300

T [C]

1100 900

100 kPa

700

5 2

500

2s

6

300 100 -100 4.5

4 4s

1 5.0

5.5

6.0

6.5

7.0

7.5

8.0

8.5

0.95

1

s [kJ/kg-K]

220

Wnet [kW]

200 180 160 140 120 100 0.7

0.75

0.8

0.85

0.9

ηt

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-87

450000

Q dot,in

415000

no regeneration

380000

with regeneration 345000 310000 275000 0.7

0.75

0.8

0.85

0.9

0.95

1

ηt

0.6 0.55 0.5

with regeneration

Eta th

0.45 0.4 0.35 0.3

no regeneration

0.25 0.2 0.7

0.75

0.8

0.85

0.9

0.95

1

ηt

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-88

9-119 A Brayton cycle with regeneration using air as the working fluid is considered. The air temperature at the turbine exit, the net work output, and the thermal efficiency are to be determined. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of air are given in Table A-17.

T qin

1150 K 5

Analysis (a) The properties of air at various states are T1 = 310 K ⎯ ⎯→

h1 = 310.24 kJ/kg Pr1 = 1.5546

P Pr2 = 2 Pr1 = (7 )(1.5546 ) = 10.88 ⎯ ⎯→ h2 s = 541.26 kJ/kg P1

ηC =

3

2s 310 K

4 4s

2 6

1

h2 s − h1 ⎯ ⎯→ h2 = h1 + (h2 s − h1 ) / η C = 310.24 + (541.26 − 310.24 )/ (0.75) = 618.26 kJ/kg h2 − h1

T3 = 1150 K ⎯ ⎯→

h3 = 1219.25 kJ/kg Pr3 = 200.15

Pr4 =

P4 ⎛1⎞ ⎯→ h4 s = 711.80 kJ/kg Pr = ⎜ ⎟(200.15) = 28.59 ⎯ P3 3 ⎝ 7 ⎠

ηT =

h3 − h4 ⎯ ⎯→ h4 = h3 − η T (h3 − h4 s ) = 1219.25 − (0.82 )(1219.25 − 711.80) = 803.14 kJ/kg h3 − h4 s

Thus, T4 = 782.8 K (b)

w net = wT,out − wC,in = (h3 − h4 ) − (h2 − h1 )

= (1219.25 − 803.14 ) − (618.26 − 310.24 ) = 108.09 kJ/kg

(c)

ε=

h5 − h 2 ⎯ ⎯→ h5 = h2 + ε (h4 − h2 ) h4 − h2 = 618.26 + (0.65)(803.14 − 618.26) = 738.43 kJ/kg

Then, q in = h3 − h5 = 1219.25 − 738.43 = 480.82 kJ/kg

η th =

wnet 108.09 kJ/kg = = 22.5% q in 480.82 kJ/kg

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

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9-89

9-120 A stationary gas-turbine power plant operating on an ideal regenerative Brayton cycle with air as the working fluid is considered. The power delivered by this plant is to be determined for two cases. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas. 3 Kinetic and potential energy changes are negligible. Properties When assuming constant specific heats, the properties of air at room temperature are cp = 1.005 kJ/kg.K and k = 1.4 (Table A-2a). When assuming variable specific heats, the properties of air are obtained from Table A-17. T Analysis (a) Assuming constant specific heats, ⎛P T2 = T1 ⎜⎜ 2 ⎝ P1

⎞ ⎟⎟ ⎠

⎛P T4 = T3 ⎜⎜ 4 ⎝ P3

⎞ ⎟ ⎟ ⎠

(k −1) / k

(k −1) / k

= (290 K )(8)0.4/1.4 = 525.3 K ⎛1⎞ = (1100 K )⎜ ⎟ ⎝8⎠

1100 K

0.4/1.4

2

= 607.2 K

290 K

1

75,000 kW

3

5

4 6

qout

ε = 100% ⎯⎯→ T5 = T4 = 607.2 K and T6 = T2 = 525.3 K η th = 1 −

c p (T6 − T1 ) q out T −T 525.3 − 290 = 1− = 1− 6 1 = 1− = 0.5225 T3 − T5 q in c p (T3 − T5 ) 1100 − 607.2

W& net = η T Q& in = (0.5225)(75,000 kW ) = 39,188 kW

(b) Assuming variable specific heats, T1 = 290K ⎯ ⎯→ Pr 2 =

h1 = 290.16 kJ/kg Pr1 = 1.2311

P2 Pr = (8)(1.2311) = 9.8488 ⎯ ⎯→ h2 = 526.12 kJ/kg P1 1

⎯→ T3 = 1100K ⎯ Pr 4 =

h3 = 1161.07 kJ/kg Pr3 = 167.1

P4 ⎛1⎞ Pr3 = ⎜ ⎟(167.1) = 20.89 ⎯ ⎯→ h4 = 651.37 kJ/kg P3 ⎝8⎠

ε = 100% ⎯⎯→ h5 = h4 = 651.37 kJ/kg and h6 = h2 = 526.12 kJ/kg q out h −h 526.12 − 290.16 = 1− 6 1 = 1− = 0.5371 1161.07 − 651.37 q in h3 − h5 = η T Q& in = (0.5371)(75,000 kW ) = 40,283 kW

η th = 1 − W& net

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9-90

9-121 A regenerative gas-turbine engine using air as the working fluid is considered. The amount of heat transfer in the regenerator and the thermal efficiency are to be determined. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of air are given in Table A-17.

T

Analysis (a) The properties at various states are r p = P2 / P1 = 800 / 100 = 8

qin

1200 K

3

5

T1 = 300 K ⎯ ⎯→ h1 = 300.19 kJ/kg

4

2

T2 = 580 K ⎯ ⎯→ h2 = 586.04 kJ/kg

580 K

2s

T3 = 1200 K ⎯ ⎯→ h3 = 1277.79 kJ/kg Pr3 = 238.0

300 K

1

4s 6 s

P4 ⎛1⎞ Pr = ⎜ ⎟(238.0 ) = 29.75 ⎯ ⎯→ h4 s = 719.75 kJ/kg P3 3 ⎝ 8 ⎠ h −h η T = 3 4 ⎯⎯→ h4 = h3 − η T (h3 − h4 s ) h3 − h 4 s = 1277.79 − (0.86 )(1277.79 − 719.75) = 797.88 kJ/kg q regen = ε (h4 − h2 ) = (0.72 )(797.88 − 586.04) = 152.5 kJ/kg Pr 4 =

(b)

w net = wT,out − wC,in = (h3 − h4 ) − (h2 − h1 ) = (1277.79 − 797.88) − (586.04 − 300.19 ) = 194.06 kJ/kg q in = (h3 − h2 ) − q regen = (1277.79 − 586.04 ) − 152.52 = 539.23 kJ/kg

η th =

wnet 194.06 kJ/kg = = 36.0% 539.23 kJ/kg q in

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-91

9-122 A regenerative gas-turbine engine using air as the working fluid is considered. The amount of heat transfer in the regenerator and the thermal efficiency are to be determined. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K and k = 1.4 (Table A-2a). Analysis (a) Using the isentropic relations and turbine efficiency, r p = P2 / P1 = 800 / 100 = 8

T qin

1200 K 5 580 K

2

2s 6 (k −1) / k 0.4 / 1.4 ⎛ P4 ⎞ 1 ⎛ ⎞ 300 K = (1200 K )⎜ ⎟ = 662.5 K T4 s = T3 ⎜⎜ ⎟⎟ 1 ⎝8⎠ ⎝ P3 ⎠ c p (T3 − T4 ) h −h ⎯ ⎯→ T4 = T3 − η T (T3 − T4 s ) ηT = 3 4 = h3 − h4 s c p (T3 − T4 s ) = 1200 − (0.86)(1200 − 662.5) = 737.8 K q regen = ε (h4 − h2 ) = ε c p (T4 − T2 ) = (0.72 )(1.005 kJ/kg ⋅ K )(737.8 − 580)K = 114.2 kJ/kg (b)

3 4 4s

s

wnet = wT,out − wC,in = c p (T3 − T4 ) − c p (T2 − T1 )

= (1.005 kJ/kg ⋅ K )[(1200 − 737.8) − (580 − 300 )]K = 183.1 kJ/kg

q in = (h3 − h2 ) − q regen = c p (T3 − T2 ) − q regen = (1.005 kJ/kg ⋅ K )(1200 − 580 )K − 114.2 = 508.9 kJ/kg

η th =

wnet 183.1 kJ/kg = = 36.0% q in 508.9 kJ/kg

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-92

9-123 A regenerative gas-turbine engine using air as the working fluid is considered. The amount of heat transfer in the regenerator and the thermal efficiency are to be determined. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of air are given in Table A-17.

T

Analysis (a) The properties of air at various states are r p = P2 / P1 = 800 / 100 = 8 ⎯→ h1 = 300.19kJ/kg T1 = 300K ⎯ ⎯→ h2 = 586.04kJ/kg T2 = 580K ⎯ ⎯→ h3 = 1277.79kJ/kg T3 = 1200K ⎯ Pr3 = 238.0

qin

1200 K

3

5

4

2

580 K

2s

300 K

1

4s 6 s

P4 ⎛1⎞ ⎯→ h4 s = 719.75 kJ/kg Pr = ⎜ ⎟(238.0) = 29.75 ⎯ P3 3 ⎝ 8 ⎠ h −h η T = 3 4 ⎯⎯→ h4 = h3 − η T (h3 − h4 s ) = 1277.79 − (0.86)(1277.79 − 719.75) = 797.88 kJ/kg h3 − h 4 s Pr 4 =

q regen = ε (h3 − h2 ) = (0.70)(797.88 − 586.04) = 148.3 kJ/kg

(b)

w net = wT,out − wC,in = (h3 − h4 ) − (h2 − h1 ) = (1277.79 − 797.88) − (586.04 − 300.19) = 194.06 kJ/kg q in = (h3 − h2 ) − q regen = (1277.79 − 586.04 ) − 148.3 = 543.5 kJ/kg

η th =

wnet 194.06 kJ/kg = = 35.7% 543.5 kJ/kg q in

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-93

Brayton Cycle with Intercooling, Reheating, and Regeneration

9-124C As the number of compression and expansion stages are increased and regeneration is employed, the ideal Brayton cycle will approach the Ericsson cycle.

9-125C (a) decrease, (b) decrease, and (c) decrease.

9-126C (a) increase, (b) decrease, and (c) decrease.

9-127C (a) increase, (b) decrease, (c) decrease, and (d) increase.

9-128C (a) increase, (b) decrease, (c) increase, and (d) decrease.

9-129C Because the steady-flow work is proportional to the specific volume of the gas. Intercooling decreases the average specific volume of the gas during compression, and thus the compressor work. Reheating increases the average specific volume of the gas, and thus the turbine work output.

9-130C (c) The Carnot (or Ericsson) cycle efficiency.

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9-94

9-131 An ideal gas-turbine cycle with two stages of compression and two stages of expansion is considered. The back work ratio and the thermal efficiency of the cycle are to be determined for the cases of with and without a regenerator. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible.

T 1200 K

qin

Properties The properties of air are given in Table A-17.

9

Analysis (a) The work inputs to each stage of compressor are identical, so are the work outputs of each stage of the turbine since this is an ideal cycle. Then, ⎯→ T1 = 300 K ⎯ Pr 2 =

h1 = 300.19 kJ/kg

300 K

Pr1 = 1.386

4

2

3

1

5

7

6

8

10

P2 ⎯→ h2 = h4 = 411.26 kJ/kg Pr = (3)(1.386) = 4.158 ⎯ P1 1

⎯→ T5 = 1200 K ⎯

h5 = h7 = 1277.79 kJ/kg Pr5 = 238

P6 ⎛1⎞ ⎯→ h6 = h8 = 946.36 kJ/kg Pr = ⎜ ⎟(238) = 79.33 ⎯ P5 5 ⎝ 3 ⎠ = 2(h2 − h1 ) = 2(411.26 − 300.19 ) = 222.14 kJ/kg

Pr6 = wC,in

wT,out = 2(h5 − h6 ) = 2(1277.79 − 946.36) = 662.86 kJ/kg

Thus, rbw =

wC,in wT,out

=

222.14 kJ/kg = 33.5% 662.86 kJ/kg

q in = (h5 − h4 ) + (h7 − h6 ) = (1277.79 − 411.26 ) + (1277.79 − 946.36 ) = 1197.96 kJ/kg w net = wT,out − wC,in = 662.86 − 222.14 = 440.72 kJ/kg

η th =

wnet 440.72 kJ/kg = = 36.8% q in 1197.96 kJ/kg

(b) When a regenerator is used, rbw remains the same. The thermal efficiency in this case becomes q regen = ε (h8 − h4 ) = (0.75)(946.36 − 411.26) = 401.33 kJ/kg q in = q in,old − q regen = 1197.96 − 401.33 = 796.63 kJ/kg

η th =

wnet 440.72 kJ/kg = = 55.3% q in 796.63 kJ/kg

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

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9-95

9-132 A gas-turbine cycle with two stages of compression and two stages of expansion is considered. The back work ratio and the thermal efficiency of the cycle are to be determined for the cases of with and without a regenerator. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of air are given in Table A-17. Analysis (a) The work inputs to each stage of compressor are identical, so are the work outputs of each stage of the turbine. Then, T1 = 300 K ⎯ ⎯→ h1 = 300.19 kJ/kg Pr1 = 1.386 Pr 2 =

ηC

T

P2 Pr = (3)(1.386) = 4.158 ⎯ ⎯→ h2 s = h4 s = 411.26 kJ/kg P1 1

h − h1 = 2s ⎯ ⎯→ h2 = h4 = h1 + (h2 s − h1 ) / η C h2 − h1 = 300.19 + (411.26 − 300.19) / (0.80) = 439.03 kJ/kg

⎯→ h5 = h7 = 1277.79 kJ/kg T5 = 1200 K ⎯

5

qin

6 8 6s 8

9

4 3

4

2 2s

7

1

1

Pr5 = 238 P6 ⎛1⎞ ⎯→ h6 = h8 = 946.36 kJ/kg Pr5 = ⎜ ⎟(238) = 79.33 ⎯ P5 ⎝3⎠ h −h η T = 5 6 ⎯⎯→ h6 = h8 = h5 − η T (h5 − h6 s ) h5 − h6 s = 1277.79 − (0.85)(1277.79 − 946.36) = 996.07 kJ/kg Pr6 =

wC,in = 2(h2 − h1 ) = 2(439.03 − 300.19 ) = 277.68 kJ/kg

wT,out = 2(h5 − h6 ) = 2(1277.79 − 996.07 ) = 563.44 kJ/kg

Thus, rbw =

wC,in wT,out

=

277.68 kJ/kg = 49.3% 563.44 kJ/kg

q in = (h5 − h4 ) + (h7 − h6 ) = (1277.79 − 439.03) + (1277.79 − 996.07 ) = 1120.48 kJ/kg wnet = wT,out − wC,in = 563.44 − 277.68 = 285.76 kJ/kg

η th =

wnet 285.76 kJ/kg = = 25.5% 1120.48 kJ/kg q in

(b) When a regenerator is used, rbw remains the same. The thermal efficiency in this case becomes q regen = ε (h8 − h4 ) = (0.75)(996.07 − 439.03) = 417.78 kJ/kg q in = q in,old − q regen = 1120.48 − 417.78 = 702.70 kJ/kg

η th =

wnet 285.76 kJ/kg = = 40.7% 702.70 kJ/kg q in

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9-96

9-133E A regenerative gas-turbine cycle with two stages of compression and two stages of expansion is considered. The mass flow rate of air and the rates of heat addition and rejection for a specified net power output are to be determined. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are cp = 0.24 Btu/lbm⋅R and k = 1.4 (Table A-2Ea). Analysis According to the isentropic process expressions for an ideal gas,

T2 = T4 = T1 r p( k −1) / k = (520 R)(3) 0.4/1.4 = 711.7 R ⎛ 1 T7 = T9 = T6 ⎜ ⎜ rp ⎝

⎞ ⎟ ⎟ ⎠

( k −1) / k

⎛1⎞ = (1400 R)⎜ ⎟ ⎝3⎠

0.4/1.4

= 1023 R

T 1400 R 5

6

8

7

9

The regenerator is ideal (i.e., the effectiveness is 100%) and thus, T5 = T7 = 1023 R T10 = T2 = 711.7 R

520 R

4

2

3

1

The net work output is determined as follows

1 s

wC,in = 2c p (T2 − T1 ) = 2(0.24 Btu/lbm ⋅ R)(711.7 − 520) R = 92.02 Btu/lbm wT,out = 2c p (T6 − T7 ) = 2(0.24 Btu/lbm ⋅ R)(1400 − 1023) R = 180.96 Btu/lbm wnet = wT, out − wC,in = 180.96 − 92.02 = 88.94 Btu/lbm

The mass flow rate is then m& =

W& net ⎛ 0.7068 Btu/s ⎞ 1000 hp ⎟⎟ = 7.947 lbm/s ⎜ = wnet 88.94 Btu/lbm ⎜⎝ 1 hp ⎠

Applying the first law to the heat addition processes gives Q& in = m& c p (T6 − T5 ) + m& c p (T8 − T7 ) = (7.947 lbm/s)(0.24 Btu/lbm ⋅ R)(1400 − 1023 + 1400 − 1023) R = 1438 Btu/s

Similarly, Q& out = m& c p (T10 − T1 ) + m& c p (T2 − T3 ) = (7.947 lbm/s)(0.24 Btu/lbm ⋅ R)(711.7 − 520 + 711.7 − 520) R = 731 Btu/s

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-97

9-134E A regenerative gas-turbine cycle with two stages of compression and two stages of expansion is considered. The mass flow rate of air and the rates of heat addition and rejection for a specified net power output are to be determined. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are cp = 0.24 Btu/lbm⋅R and k = 1.4 (Table A-2Ea). Analysis For the compression and expansion processes, we have

T2 s = T4 s = T1 r p( k −1) / k = (520 R)(3) 0.4/1.4 = 711.7 R

ηC =

c p (T2 s − T1 ) c p (T2 − T1 )

⎯ ⎯→ T2 = T4 = T1 + = 520 +

⎛ 1 T7 s = T9 s = T6 ⎜ ⎜ rp ⎝

ηT =

c p (T6 − T7 ) c p (T6 − T7 s )

⎞ ⎟ ⎟ ⎠

( k −1) / k

T2 s − T1

T 5

ηC

7s

711.7 − 520 = 737.8 R 0.88

⎛1⎞ = (1400 R)⎜ ⎟ ⎝ 3⎠

6

1400 R

0.4/1.4

= 1023 R

4 2 4s 2s 520 R

3

7

8 9s

9

1

1 s

⎯ ⎯→ T7 = T9 = T6 − η T (T6 − T7 s ) = 1400 − (0.93)(1400 − 1023) = 1049 R

The regenerator is ideal (i.e., the effectiveness is 100%) and thus, T5 = T7 = 1049 R T10 = T2 = 737.8 R

The net work output is determined as follows wC,in = 2c p (T2 − T1 ) = 2(0.24 Btu/lbm ⋅ R)(737.8 − 520) R = 104.54 Btu/lbm wT,out = 2c p (T6 − T7 ) = 2(0.24 Btu/lbm ⋅ R)(1400 − 1049) R = 168.48 Btu/lbm wnet = wT, out − wC,in = 168.48 − 104.54 = 63.94 Btu/lbm

The mass flow rate is then m& =

W& net ⎛ 0.7068 Btu/s ⎞ 1000 hp ⎟⎟ = 11.05 lbm/s ⎜ = 1 hp wnet 63.94 Btu/lbm ⎜⎝ ⎠

The rate of heat addition is then Q& in = m& c p (T6 − T5 ) + m& c p (T8 − T7 ) = (11.05 lbm/s)(0.24 Btu/lbm ⋅ R)(1400 − 1049 + 1400 − 1049) R = 1862 Btu/s

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-98

9-135 A regenerative gas-turbine cycle with two stages of compression and two stages of expansion is considered. The thermal efficiency of the cycle is to be determined. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are cp = 1.005 kJ/kg⋅K and k = 1.4 (Table A-2a). Analysis The temperatures at various states are obtained as follows

T2 = T4 = T1 r p( k −1) / k = (290 K)(4) 0.4/1.4 = 430.9 K T5 = T4 + 20 = 430.9 + 20 = 450.9 K

T

8 6

q in = c p (T6 − T5 ) T6 = T5 + ⎛ 1 T 7 = T6 ⎜ ⎜ rp ⎝ T8 = T7 +

q in cp ⎞ ⎟ ⎟ ⎠

q in cp

⎛ 1 T9 = T8 ⎜ ⎜ rp ⎝

⎞ ⎟ ⎟ ⎠

= 450.9 K +

( k −1) / k

⎛1⎞ = (749.4 K)⎜ ⎟ ⎝4⎠

= 504.3 K + ( k −1) / k

300 kJ/kg = 749.4 K 1.005 kJ/kg ⋅ K

7 5 4

2

3

1

9

1

0.4/1.4

= 504.3 K

290 K

s

300 kJ/kg = 802.8 K 1.005 kJ/kg ⋅ K

⎛1⎞ = (802.8 K)⎜ ⎟ ⎝4⎠

0.4/1.4

= 540.2 K

T10 = T9 − 20 = 540.2 − 20 = 520.2 K

The heat input is q in = 300 + 300 = 600 kJ/kg

The heat rejected is q out = c p (T10 − T1 ) + c p (T2 − T3 ) = (1.005 kJ/kg ⋅ K)(520.2 − 290 + 430.9 − 290) R = 373.0 kJ/kg

The thermal efficiency of the cycle is then

η th = 1 −

q out 373.0 = 1− = 0.378 q in 600

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-99

9-136 A regenerative gas-turbine cycle with three stages of compression and three stages of expansion is considered. The thermal efficiency of the cycle is to be determined. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are cp = 1.005 kJ/kg⋅K and k = 1.4 (Table A-2a). Analysis The temperatures at various states are obtained as follows

T2 = T4 = T6 = T1 r p( k −1) / k = (290 K)(4) 0.4/1.4 = 430.9 K

T

T7 = T6 + 20 = 430.9 + 20 = 450.9 K q in = c p (T8 − T7 ) q in

T8 = T7 +

cp

⎛ 1 T9 = T8 ⎜ ⎜ rp ⎝ T10 = T9 +

⎞ ⎟ ⎟ ⎠

cp

⎛ 1 T11 = T10 ⎜ ⎜ rp ⎝ T12 = T11 +

T13

⎞ ⎟ ⎟ ⎠

q in cp

⎛ 1 = T12 ⎜ ⎜ rp ⎝

= 450.9 K +

( k −1) / k

q in

⎞ ⎟ ⎟ ⎠

9 7

300 kJ/kg = 749.4 K 1.005 kJ/kg ⋅ K

6 290 K

⎛1⎞ = (749.4 K)⎜ ⎟ ⎝4⎠

= 504.3 K + ( k −1) / k

= 504.3 K

5

2

3

1

13

11 14

s

300 kJ/kg = 802.8 K 1.005 kJ/kg ⋅ K

⎛1⎞ = (802.8 K)⎜ ⎟ ⎝4⎠

= 540.2 K + ( k −1) / k

0.4/1.4

4

12

1

8

0.4/1.4

= 540.2 K

300 kJ/kg = 838.7 K 1.005 kJ/kg ⋅ K

⎛1⎞ = (838.7 K)⎜ ⎟ ⎝4⎠

0.4/1.4

= 564.4 K

T14 = T13 − 20 = 564.4 − 20 = 544.4 K

The heat input is q in = 300 + 300 + 300 = 900 kJ/kg

The heat rejected is q out = c p (T14 − T1 ) + c p (T2 − T3 ) + c p (T4 − T5 ) = (1.005 kJ/kg ⋅ K)(544.4 − 290 + 430.9 − 290 + 430.9 − 290) R = 538.9 kJ/kg

The thermal efficiency of the cycle is then

η th = 1 −

q out 538.9 = 1− = 0.401 q in 900

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-100

9-137 A regenerative gas-turbine cycle with three stages of compression and three stages of expansion is considered. The thermal efficiency of the cycle is to be determined. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are cp = 1.005 kJ/kg⋅K and k = 1.4 (Table A-2a). Analysis Since all compressors share the same compression ratio and begin at the same temperature,

T2 = T4 = T6 = T1 r p( k −1) / k = (290 K)(4) 0.4/1.4 = 430.9 K From the problem statement, T7 = T13 − 40

The relations for heat input and expansion processes are ⎯→ T8 = T7 + q in = c p (T8 − T7 ) ⎯ ⎛ 1 T9 = T8 ⎜ ⎜ rp ⎝ T10 = T9 +

T12 = T11 +

⎞ ⎟ ⎟ ⎠

T

q in cp

9

( k −1) / k

q in cp q in cp

7 6 ( k −1) / k

,

⎛ 1 T11 = T10 ⎜ ⎜ rp ⎝

⎞ ⎟ ⎟ ⎠

,

⎛ 1 = T12 ⎜ ⎜ rp ⎝

⎞ ⎟ ⎟ ⎠

T13

290 K

5

4

2

3

1

12

1

8

13

11 14

s

( k −1) / k

The simultaneous solution of above equations using EES software gives the following results T7 = 556.7 K,

T8 = 855.2 K,

T10 = 874.0 K,

T11 = 588.2 K,

T9 = 575.5 K T12 = 886.7 K,

T13 = 596.7 K

From am energy balance on the regenerator, T7 − T6 = T13 − T14 (T13 − 40) − T6 = T13 − T14 ⎯ ⎯→ T14 = T6 + 40 = 430.9 + 40 = 470.9 K

The heat input is q in = 300 + 300 + 300 = 900 kJ/kg

The heat rejected is q out = c p (T14 − T1 ) + c p (T2 − T3 ) + c p (T4 − T5 ) = (1.005 kJ/kg ⋅ K)(470.9 − 290 + 430.9 − 290 + 430.9 − 290) R = 465.0 kJ/kg

The thermal efficiency of the cycle is then

η th = 1 −

q out 465.0 = 1− = 0.483 q in 900

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-101

Jet-Propulsion Cycles

9-138C The power developed from the thrust of the engine is called the propulsive power. It is equal to thrust times the aircraft velocity.

9-139C The ratio of the propulsive power developed and the rate of heat input is called the propulsive efficiency. It is determined by calculating these two quantities separately, and taking their ratio.

9-140C It reduces the exit velocity, and thus the thrust.

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-102

9-141E A turboprop engine operating on an ideal cycle is considered. The thrust force generated is to be determined. Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with constant specific heats at room temperature. 4 The turbine work output is equal to the compressor work input. Properties The properties of air at room temperature are R = 0.3704 psia⋅ft3/lbm⋅R (Table A-1E), cp = 0.24 Btu/lbm⋅R and k = 1.4 (Table A-2Ea). Analysis Working across the two isentropic processes of the cycle yields

T2 = T1 r p( k −1) / k = (450 R)(10) 0.4/1.4 = 868.8 R ⎛ 1 T5 = T3 ⎜ ⎜ rp ⎝

⎞ ⎟ ⎟ ⎠

( k −1) / k

⎛1⎞ = (1400 R)⎜ ⎟ ⎝ 10 ⎠

T

3

qin

0.4/1.4

= 725.1 R

4 2 5

Since the work produced by expansion 3-4 equals that used by compression 1-2, an energy balance gives

qout

1

T4 = T3 − (T2 − T1 ) = 1400 − (868.8 − 450) = 981.2 R

s

The excess enthalpy generated by expansion 4-5 is used to increase the kinetic energy of the flow through the propeller, m& e c p (T4 − T5 ) = m& p

2 2 Vexit − Vinlet 2

which when solved for the velocity at which the air leaves the propeller gives Vexit

⎡ m& ⎤ 2 = ⎢2 e c p (T4 − T5 ) + Vinlet ⎥ & ⎣⎢ m p ⎦⎥

1/ 2

⎡ 1 ⎛ 25,037 ft 2 /s 2 = ⎢2 (0.24 Btu/lbm ⋅ R )(981.2 − 725.1)R ⎜ ⎜ 1 Btu/lbm ⎢⎣ 20 ⎝ = 716.9 ft/s

⎤ ⎞ ⎟ + (600 ft/s) 2 ⎥ ⎟ ⎥⎦ ⎠

1/ 2

The mass flow rate through the propeller is

v1 = m& p =

RT (0.3704 psia ⋅ ft 3 )(450 R) = = 20.84 ft 3 /lbm P 8 psia AV1

v1

=

600 ft/s πD 2 V1 π (10 ft) 2 = = 2261 lbm/s 4 v1 4 20.84 ft 3 /lbm

The thrust force generated by this propeller is then ⎛ 1 lbf F = m& p (Vexit − Vinlet ) = (2261 lbm/s)(716.9 − 600)ft/s⎜ ⎜ 32.174 lbm ⋅ft/s 2 ⎝

⎞ ⎟ = 8215 lbf ⎟ ⎠

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-103

9-142E A turboprop engine operating on an ideal cycle is considered. The thrust force generated is to be determined. Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with constant specific heats at room temperature. 4 The turbine work output is equal to the compressor work input. Properties The properties of air at room temperature are R = 0.3704 psia⋅ft3/lbm⋅R (Table A-1E), cp = 0.24 Btu/lbm⋅R and k = 1.4 (Table A-2Ea). Analysis Working across the two isentropic processes of the cycle yields

T2 = T1 r p( k −1) / k = (450 R)(10) 0.4/1.4 = 868.8 R ⎛ 1 T5 = T3 ⎜ ⎜ rp ⎝

⎞ ⎟ ⎟ ⎠

( k −1) / k

⎛1⎞ = (1400 R)⎜ ⎟ ⎝ 10 ⎠

0.4/1.4

= 725.1 R

Since the work produced by expansion 3-4 equals that used by compression 1-2, an energy balance gives

T

3

qin

4 2

T4 = T3 − (T2 − T1 ) = 1400 − (868.8 − 450) = 981.2 R

5 1

qout s

The mass flow rate through the propeller is

v1 = m& p =

RT (0.3704 psia ⋅ ft 3 )(450 R) = = 20.84 ft 3 /lbm P 8 psia AV1

v1

=

600 ft/s πD 2 V1 π (8 ft) 2 = = 1447 lbm/s 4 v1 4 20.84 ft 3 /lbm

According to the previous problem, m& e =

m& p 20

=

2261 lbm/s = 113.1 lbm/s 20

The excess enthalpy generated by expansion 4-5 is used to increase the kinetic energy of the flow through the propeller, m& e c p (T4 − T5 ) = m& p

2 2 Vexit − Vinlet 2

which when solved for the velocity at which the air leaves the propeller gives ⎡ m& ⎤ 2 Vexit = ⎢2 e c p (T4 − T5 ) + Vinlet ⎥ & ⎣⎢ m p ⎦⎥

1/ 2

⎡ 113.1 lbm/s ⎛ 25,037 ft 2 /s 2 = ⎢2 (0.24 Btu/lbm ⋅ R )(981.2 − 725.1)R ⎜ ⎜ 1 Btu/lbm ⎝ ⎣⎢ 1447 lbm/s

⎤ ⎞ ⎟ + (600 ft/s) 2 ⎥ ⎟ ⎠ ⎦⎥

1/ 2

= 775.0 ft/s

The thrust force generated by this propeller is then ⎛ 1 lbf F = m& p (Vexit − Vinlet ) = (1447 lbm/s)(775 − 600)ft/s⎜ ⎜ 32.174 lbm ⋅ft/s 2 ⎝

⎞ ⎟ = 7870 lbf ⎟ ⎠

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-104

9-143 A turbofan engine operating on an ideal cycle produces 50,000 N of thrust. The air temperature at the fan outlet needed to produce this thrust is to be determined. Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with constant specific heats at room temperature. 4 The turbine work output is equal to the compressor work input. Properties The properties of air at room temperature are R = 0.287 kPa⋅m3/kg⋅K, cp = 1.005 kJ/kg⋅K and k = 1.4 (Table A-2a). Analysis The total mass flow rate is RT (0.287 kPa ⋅ m 3 )(253 K) = = 1.452 m 3 /kg P 50 kPa AV1 πD 2 V1 π (2.5 m) 2 200 m/s = = = 676.1 kg/s m& = v1 4 v1 4 1.452 m 3 /kg

v1 =

T

3

qin

4 2 5

Now, m& e =

1

m& 676.1 kg/s = = 84.51 kg/s 8 8

qout s

The mass flow rate through the fan is m& f = m& − m& e = 676.1 − 84.51 = 591.6 kg/s

In order to produce the specified thrust force, the velocity at the fan exit will be F = m& f (Vexit − Vinlet ) Vexit = Vinlet +

F 50,000 N ⎛⎜ 1 kg ⋅m/s 2 = (200 m/s) + m& f 591.6 kg/s ⎜⎝ 1 N

⎞ ⎟ = 284.5 m/s ⎟ ⎠

An energy balance on the stream passing through the fan gives 2 2 − Vinlet Vexit 2 2 V 2 − Vinlet T5 = T4 − exit 2c p

c p (T4 − T5 ) =

= 253 K −

(284.5 m/s) 2 − (200 m/s) 2 ⎛ 1 kJ/kg ⎞ ⎜ ⎟ 2(1.005 kJ/kg ⋅ K ) ⎝ 1000 m 2 /s 2 ⎠

= 232.6 K

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-105

9-144 A pure jet engine operating on an ideal cycle is considered. The velocity at the nozzle exit and the thrust produced are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with constant specific heats at room temperature. 4 The turbine work output is equal to the compressor work input. Properties The properties of air at room temperature are R = 0.287 kPa⋅m3/kg⋅K, cp = 1.005 kJ/kg⋅K and k = 1.4 (Table A-2a). Analysis Working across the two isentropic processes of the cycle yields

T2 = T1 r p( k −1) / k = (273 K)(10) 0.4/1.4 = 527.1 K ⎛ 1 T5 = T3 ⎜ ⎜ rp ⎝

⎞ ⎟ ⎟ ⎠

( k −1) / k

⎛1⎞ = (723 K)⎜ ⎟ ⎝ 10 ⎠

T

0.4/1.4

= 374.5 K

3

qin

4 2

Since the work produced by expansion 3-4 equals that used by compression 1-2, an energy balance gives

5

T4 = T3 − (T2 − T1 ) = 723 − (527.1 − 273) = 468.9 K

qout

1

s

The excess enthalpy generated by expansion 4-5 is used to increase the kinetic energy of the flow through the propeller, c p (T4 − T5 ) =

2 2 Vexit − Vinlet 2

which when solved for the velocity at which the air leaves the propeller gives

[

2 Vexit = 2c p (T4 − T5 ) + Vinlet

]

1/ 2

⎡ ⎛ 1000 m 2 /s 2 = ⎢2(1.005 kJ/kg ⋅ K )(468.9 − 374.5)K⎜ ⎜ 1 kJ/kg ⎢⎣ ⎝ = 528.9 m/s

⎤ ⎞ ⎟ + (300 m/s) 2 ⎥ ⎟ ⎥⎦ ⎠

1/ 2

The mass flow rate through the engine is RT (0.287 kPa ⋅ m 3 )(273 K) = = 1.306 m 3 /kg 60 kPa P AV1 πD 2 V1 π (2 m) 2 300 m/s = = = 721.7 kg/s m& = v1 4 v1 4 1.306 m 3 /kg

v1 =

The thrust force generated is then ⎛ 1N F = m& (V exit − Vinlet ) = (721.7 kg/s)(528.9 − 300)m/s⎜ ⎜ 1 kg ⋅ m/s 2 ⎝

⎞ ⎟ = 165,200 N ⎟ ⎠

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-106

9-145 A turbojet aircraft flying at an altitude of 9150 m is operating on the ideal jet propulsion cycle. The velocity of exhaust gases, the propulsive power developed, and the rate of fuel consumption are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with constant specific heats at room temperature. 4 Kinetic and potential energies are negligible, except at the diffuser inlet and the nozzle exit. 5 The turbine work output is equal to the compressor work input. Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K and k = 1.4 (Table A-2a).

T · Qi

Analysis (a) We assume the aircraft is stationary and the air is moving towards the aircraft at a velocity of V 1 = 320 m/s. Ideally, the air will leave the diffuser with a negligible velocity (V 2 ≅ 0).

5 3 2

Diffuser:

− V12 2

V12 (320 m/s)2 ⎛⎜ 1 kJ/kg = 241 K + (2)(1.005 kJ/kg ⋅ K ) ⎜⎝ 1000 m 2 /s 2 2c p

⎛T P2 = P1 ⎜⎜ 2 ⎝ T1

Compressor:

s

©0 V 22

0 = c p (T2 − T1 ) − V12 / 2 T2 = T1 +

⎞ ⎟⎟ ⎠

k / (k −1)

6

1

⎯→ E& in = E& out E& in − E& out = ΔE& system ©0 (steady) ⎯ ⎯→ 0 = h2 − h1 + h1 + V12 / 2 = h2 + V 22 / 2 ⎯

4

⎛ 291.9 K ⎞ ⎟⎟ = (32 kPa )⎜⎜ ⎝ 241 K ⎠

⎞ ⎟ = 291.9 K ⎟ ⎠

1.4/0.4

= 62.6 kPa

( )

P3 = P4 = r p (P2 ) = (12 )(62.6 kPa ) = 751.2 kPa ⎛P T3 = T2 ⎜⎜ 3 ⎝ P2

⎞ ⎟⎟ ⎠

(k −1) / k

= (291.9 K )(12)0.4/1.4 = 593.7 K

Turbine: wcomp,in = w turb,out ⎯ ⎯→ h3 − h2 = h4 − h5 ⎯ ⎯→ c p (T3 − T2 ) = c p (T4 − T5 )

or T5 = T4 − T3 + T2 = 1400 − 593.7 + 291.9 = 1098.2K

Nozzle: (k −1) / k

⎛P ⎞ ⎛ 32 kPa ⎞ ⎟⎟ = (1400 K )⎜⎜ T6 = T4 ⎜⎜ 6 ⎟⎟ ⎝ 751.2 kPa ⎠ ⎝ P4 ⎠ E& in − E& out = ΔE& system ©0 (steady) ⎯ ⎯→ E& in = E& out

0.4/1.4

= 568.2 K

h5 + V52 / 2 = h6 + V 62 / 2 V 2 − V52 0 = h6 − h5 + 6 2

©0

⎯ ⎯→ 0 = c p (T6 − T5 ) + V 62 / 2

or,

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9-107

V6 =

(b) (c)

⎛

/s 2 1 kJ/kg

(2)(1.005 kJ/kg ⋅ K )(1098.2 − 568.2)K⎜⎜ 1000 m ⎝

2

⎞ ⎟ = 1032 m/s ⎟ ⎠

⎛ 1 kJ/kg W& p = m& (V exit − Vinlet )V aircraft = (60 kg/s )(1032 − 320 )m/s(320 m/s )⎜ ⎜ 1000 m 2 /s 2 ⎝

⎞ ⎟ = 13,670 kW ⎟ ⎠

Q& in = m& (h4 − h3 ) = m& c p (T4 − T3 ) = (60 kg/s )(1.005 kJ/kg ⋅ K )(1400 − 593.7 )K = 48,620 kJ/s m& fuel =

Q& in 48,620 kJ/s = = 1.14 kg/s HV 42,700 kJ/kg

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-108

9-146 A turbojet aircraft is flying at an altitude of 9150 m. The velocity of exhaust gases, the propulsive power developed, and the rate of fuel consumption are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with constant specific heats at room temperature. 4 Kinetic and potential energies are negligible, except at the diffuser inlet and the nozzle exit. Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K and k = 1.4 (Table A-2a). Analysis (a) For convenience, we assume the aircraft is stationary and the air is moving towards the aircraft at a velocity of V 1 = 320 m/s. Ideally, the air will leave the diffuser with a negligible velocity (V 2 ≅ 0).

Diffuser: T

E& in − E& out = ΔE& system ©0 (steady)

5 3

h1 + V12 / 2 = h2 + V 22 / 2 ©0

− V12 2 0 = c p (T2 − T1 ) − V12 / 2 0 = h2 − h1 +

T2 = T1 +

V 22

2

⎞ ⎟⎟ ⎠

k / (k −1)

5s 6

1 s

V12 (320 m/s)2 ⎛⎜ 1 kJ/kg = 241 K + (2)(1.005 kJ/kg ⋅ K ) ⎜⎝ 1000 m 2 /s 2 2c p

⎛T P2 = P1 ⎜⎜ 2 ⎝ T1

Compressor:

4

· Qin

E& in = E& out

⎛ 291.9 K ⎞ ⎟⎟ = (32 kPa )⎜⎜ ⎝ 241 K ⎠

⎞ ⎟ = 291.9 K ⎟ ⎠

1.4/0.4

= 62.6 kPa

( )

P3 = P4 = r p (P2 ) = (12 )(62.6 kPa ) = 751.2 kPa T3s

⎛P = T2 ⎜⎜ 3 ⎝ P2

⎞ ⎟⎟ ⎠

(k −1) / k

= (291.9 K )(12 )0.4/1.4 = 593.7 K

h3s − h2 c p (T3s − T2 ) = h3 − h 2 c p (T3 − T2 )

ηC =

T3 = T2 + (T3s − T2 ) / η C = 291.9 + (593.7 − 291.9 )/ (0.80 ) = 669.2 K

Turbine: wcomp,in = w turb,out ⎯ ⎯→ h3 − h2 = h4 − h5 ⎯ ⎯→ c p (T3 − T2 ) = c p (T4 − T5 )

or, T5 = T4 − T3 + T2 = 1400 − 669.2 + 291.9 = 1022.7 K

ηT =

c p (T4 − T5 ) h4 − h5 = h4 − h5 s c p (T4 − T5 s )

T5 s = T4 − (T4 − T5 ) / η T = 1400 − (1400 − 1022.7 ) / 0.85 = 956.1 K ⎛T P5 = P4 ⎜⎜ 5 s ⎝ T4

⎞ ⎟⎟ ⎠

k / (k −1)

⎛ 956.1 K ⎞ ⎟⎟ = (751.2 kPa )⎜⎜ ⎝ 1400 K ⎠

1.4/0.4

= 197.7 kPa

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-109

Nozzle: (k −1) / k

⎛P ⎞ ⎛ 32 kPa ⎞ ⎟⎟ = (1022.7 K )⎜⎜ T6 = T5 ⎜⎜ 6 ⎟⎟ ⎝ 197.7 kPa ⎠ ⎝ P5 ⎠ E& in − E& out = ΔE& system ©0 (steady)

0.4/1.4

= 607.8 K

E& in = E& out h5 + V52 / 2 = h6 + V 62 / 2 ©0

V 2 − V52 0 = h6 − h5 + 6 2 0 = c p (T6 − T5 ) + V62 / 2

or, V6 =

⎛

/s 2 1 kJ/kg

(2)(1.005 kJ/kg ⋅ K )(1022.7 − 607.8)K⎜⎜ 1000 m ⎝

2

⎞ ⎟ = 913.2 m/s ⎟ ⎠

(b)

W& p = m& (Vexit − Vinlet )Vaircraft

(c)

Q& in = m& (h4 − h3 ) = m& c p (T4 − T3 ) = (60 kg/s )(1.005 kJ/kg ⋅ K )(1400 − 669.2 )K = 44,067 kJ/s

⎛ 1 kJ/kg = (60 kg/s )(913.2 − 320 )m/s(320 m/s )⎜ ⎜ 1000 m 2 /s 2 ⎝ = 11,390 kW

m& fuel =

⎞ ⎟ ⎟ ⎠

Q& in 44,067 kJ/s = = 1.03 kg/s HV 42,700 kJ/kg

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-110

9-147 A turbojet aircraft that has a pressure rate of 12 is stationary on the ground. The force that must be applied on the brakes to hold the plane stationary is to be determined. Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with variable specific heats. 4 Kinetic and potential energies are negligible, except at the nozzle exit. Properties The properties of air are given in Table A17.

T

Analysis (a) Using variable specific heats for air,

qin

Compressor:

4

T1 = 300 K ⎯ ⎯→ h1 = 300.19 kJ/kg

2

Pr1 = 1.386 Pr 2 =

3

5

P2 ⎯→ h2 = 610.65 kJ/kg Pr = (12 )(1.386 ) = 16.63 ⎯ P1 1

1

s

Q& in = m& fuel × HV = (0.2 kg/s )(42,700 kJ/kg ) = 8540 kJ/s q in =

Q& in 8540 kJ/s = = 854 kJ/kg 10 kg/s m&

q in = h3 − h2 ⎯ ⎯→ h3 = h2 + q in = 610.65 + 854 = 1464.65 kJ/kg ⎯ ⎯→ Pr3 = 396.27

Turbine: wcomp,in = w turb,out ⎯ ⎯→ h2 − h1 = h3 − h4

or, h4 = h3 − h2 + h1 = 1464.65 − 610.65 + 300.19 = 741.17 kJ/kg

Nozzle: ⎛P ⎞ ⎛1⎞ Pr5 = Pr3 ⎜⎜ 5 ⎟⎟ = (396.27 )⎜ ⎟ = 33.02 ⎯ ⎯→ h5 = 741.79 kJ/kg ⎝ 12 ⎠ ⎝ P3 ⎠ E& in − E& out = ΔE& system ©0 (steady) E& in = E& out h4 + V 42 / 2 = h5 + V52 / 2 0 = h5 − h4 +

V52 − V 42 2

©0

or, V5 = 2(h4 − h5 ) =

⎛

/s 2 1 kJ/kg

(2)(1154.19 − 741.17 )kJ/kg⎜⎜ 1000 m ⎝

2

⎞ ⎟ = 908.9 m/s ⎟ ⎠

⎛ 1N Brake force = Thrust = m& (Vexit − Vinlet ) = (10 kg/s )(908.9 − 0 )m/s⎜ ⎜ 1 kg ⋅ m/s 2 ⎝

⎞ ⎟ = 9089 N ⎟ ⎠

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9-111

9-148 EES Problem 9-147 is reconsidered. The effect of compressor inlet temperature on the force that must be applied to the brakes to hold the plane stationary is to be investigated. Analysis Using EES, the problem is solved as follows: P_ratio = 12 T_1 = 27 [C] T[1] = T_1+273 "[K]" P[1]= 95 [kPa] P[5]=P[1] Vel[1]=0 [m/s] V_dot[1] = 9.063 [m^3/s] HV_fuel = 42700 [kJ/kg] m_dot_fuel = 0.2 [kg/s] Eta_c = 1.0 Eta_t = 1.0 Eta_N = 1.0 "Inlet conditions" h[1]=ENTHALPY(Air,T=T[1]) s[1]=ENTROPY(Air,T=T[1],P=P[1]) v[1]=volume(Air,T=T[1],P=P[1]) m_dot = V_dot[1]/v[1] "Compressor anaysis" s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor" P_ratio=P[2]/P[1]"Definition of pressure ratio - to find P[2]" T_s[2]=TEMPERATURE(Air,s=s_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exit" h_s[2]=ENTHALPY(Air,T=T_s[2]) Eta_c =(h_s[2]-h[1])/(h[2]-h[1]) "Compressor adiabatic efficiency; Eta_c = W_dot_c_ideal/W_dot_c_actual. " m_dot*h[1] +W_dot_c=m_dot*h[2] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" "External heat exchanger analysis" P[3]=P[2]"process 2-3 is SSSF constant pressure" h[3]=ENTHALPY(Air,T=T[3]) Q_dot_in = m_dot_fuel*HV_fuel m_dot*h[2] + Q_dot_in= m_dot*h[3]"SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0" "Turbine analysis" s[3]=ENTROPY(Air,T=T[3],P=P[3]) s_s[4]=s[3] "For the ideal case the entropies are constant across the turbine" {P_ratio= P[3] /P[4]} T_s[4]=TEMPERATURE(Air,h=h_s[4]) "Ts[4] is the isentropic value of T[4] at turbine exit" {h_s[4]=ENTHALPY(Air,T=T_s[4])} "Eta_t = W_dot_t /Wts_dot turbine adiabatic efficiency, Wts_dot > W_dot_t" Eta_t=(h[3]-h[4])/(h[3]-h_s[4]) m_dot*h[3] = W_dot_t + m_dot*h[4] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" T[4]=TEMPERATURE(Air,h=h[4]) P[4]=pressure(Air,s=s_s[4],h=h_s[4]) "Cycle analysis" W_dot_net=W_dot_t-W_dot_c"Definition of the net cycle work, kW" W_dot_net = 0 [kW] PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-112

"Exit nozzle analysis:" s[4]=entropy('air',T=T[4],P=P[4]) s_s[5]=s[4] "For the ideal case the entropies are constant across the nozzle" T_s[5]=TEMPERATURE(Air,s=s_s[5], P=P[5]) "T_s[5] is the isentropic value of T[5] at nozzle exit" h_s[5]=ENTHALPY(Air,T=T_s[5]) Eta_N=(h[4]-h[5])/(h[4]-h_s[5]) m_dot*h[4] = m_dot*(h_s[5] + Vel_s[5]^2/2*convert(m^2/s^2,kJ/kg)) m_dot*h[4] = m_dot*(h[5] + Vel[5]^2/2*convert(m^2/s^2,kJ/kg)) T[5]=TEMPERATURE(Air,h=h[5]) s[5]=entropy('air',T=T[5],P=P[5]) "Brake Force to hold the aircraft:" Thrust = m_dot*(Vel[5] - Vel[1]) "[N]" BrakeForce = Thrust "[N]" "The following state points are determined only to produce a T-s plot" T[2]=temperature('air',h=h[2]) s[2]=entropy('air',T=T[2],P=P[2]) Air 1600

m [kg/s]

T3 [K]

T1 [C]

11.86 11.41 10.99 10.6 10.24 9.9

1164 1206 1247 1289 1330 1371

-20 -10 0 10 20 30

1400

3

4s 95

kP

11

1000

a

40

kP

a

1200

T [K]

Brake Force [N] 9971 9764 9568 9383 9207 9040

800 5s

2s

600 400 1

200 4.5

5.0

5.5

6.0

6.5

7.0

7.5

s [kJ/kg-K] 10000

BrakeForce [N]

9800

9600

9400

9200

9000 -20

-10

0

10

20

30

T 1 [C]

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9-113

9-149 Air enters a turbojet engine. The thrust produced by this turbojet engine is to be determined. Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with variable specific heats. 4 Kinetic and potential energies are negligible, except at the diffuser inlet and the nozzle exit. Properties The properties of air are given in Table A-17. Analysis We assume the aircraft is stationary and the air is moving towards the aircraft at a velocity of V 1 = 300 m/s. Taking the entire engine as our control volume and writing the steady-flow energy balance yield T1 = 280 K

⎯ ⎯→

h1 = 28013 . kJ / kg

T2 = 700 K

⎯ ⎯→

h2 = 713.27 kJ / kg

E& in − E& out = ΔE& system ©0 (steady)

7°C 300 m/s 16 kg/s

E& in = E& out Q& in + m& (h1 + V12

/ 2) =

m& (h2 + V 22

/ 2)

⎛ Q& in = m& ⎜ h2 − h1 + ⎜ ⎝

V 22

− V12 2

15,000 kJ/s

427°C 1

2

⎞ ⎟ ⎟ ⎠

⎡ V 2 − (300 m/s )2 15,000 kJ/s = (16 kg/s )⎢713.27 − 280.13 + 2 2 ⎢⎣

⎛ 1 kJ/kg ⎜ ⎜ 1000 m 2 /s 2 ⎝

⎞⎤ ⎟⎥ ⎟⎥ ⎠⎦

It gives V 2 = 1048 m/s Thus, Fp = m& (V2 − V1 ) = (16 kg/s )(1048 − 300 )m/s = 11,968 N

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-114

Second-Law Analysis of Gas Power Cycles

9-150E The exergy destruction associated with the heat rejection process of the Diesel cycle described in Prob. 9-60 and the exergy at the end of the expansion stroke are to be determined. Analysis From Prob. 9-60E, qout = 158.9 Btu/lbm, T1 = 540 R, T4 = 1420.6 R, and v 4 = v 1. At Tavg = (T4 + T1)/2 = (1420.6 + 540)/2 = 980.3 R, we have cv,avg = 0.180 Btu/lbm·R. The entropy change during process 4-1 is s1 − s4 = cv ln

540 R v ©0 T1 + R ln 1 = (0.180 Btu/lbm ⋅ R )ln = −0.1741 Btu/lbm ⋅ R v4 1420.6 R T4

Thus, qR, 41 ⎞ ⎛ ⎛ 158.9 Btu/lbm ⎞ ⎟ = (540R )⎜ − 0.1741 Btu/lbm ⋅ R + ⎟⎟ = 64.9 Btu/lbm xdestroyed, 41 = T0 ⎜⎜ s1 − s4 + ⎜ ⎟ T 540 R R ⎠ ⎝ ⎠ ⎝ Noting that state 4 is identical to the state of the surroundings, the exergy at the end of the power stroke (state 4) is determined from

φ 4 = (u 4 − u 0 ) − T0 (s 4 − s 0 ) + P0 (v 4 − v 0 ) where u 4 − u 0 = u 4 − u1 = q out = 158.9 Btu/lbm ⋅ R

v 4 −v 0 = v 4 −v1 = 0 s 4 − s 0 = s 4 − s1 = 0.1741 Btu/lbm ⋅ R

Thus,

φ 4 = (158.9 Btu/lbm ) − (540R )(0.1741 Btu/lbm ⋅ R ) + 0 = 64.9 Btu/lbm Discussion Note that the exergy at state 4 is identical to the exergy destruction for the process 4-1 since state 1 is identical to the dead state, and the entire exergy at state 4 is wasted during process 4-1.

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9-115

9-151 The exergy destruction associated with each of the processes of the Brayton cycle described in Prob. 9-91 is to be determined. Analysis From Prob. 9-91, qin = 584.62 kJ/kg, qout = 478.92 kJ/kg, and ⎯ ⎯→ s1o = 1.73498kJ/kg ⋅ K

T1 = 310K

h2 = 646.3kJ/kg ⎯ ⎯→ s 2o = 2.47256kJ/kg ⋅ K T3 = 1160K

⎯ ⎯→ s 3o = 3.13916kJ/kg ⋅ K

h4 = 789.16kJ/kg ⎯ ⎯→ s 4o = 2.67602kJ/kg ⋅ K

Thus, ⎛ P ⎞ x destroyed,12 = T0 s gen,12 = T0 (s 2 − s1 ) = T0 ⎜⎜ s 2o − s1o − Rln 2 ⎟⎟ = P1 ⎠ ⎝ = (310 K )(2.47256 − 1.73498 − (0.287 kJ/kg ⋅ K )ln(8)) = 43.6 kJ/kg q R ,23 ⎛ x destroyed, 23 = T0 s gen,23 = T0 ⎜⎜ s 3 − s 2 + TR ⎝

⎛ ⎞ P ⎟ = T0 ⎜ s 3o − s 2o − Rln 3 ⎟ ⎜ P2 ⎠ ⎝

©0

+

− q in TH

⎞ ⎟ ⎟ ⎠

⎛ 584.62 kJ/kg ⎞ ⎟ = 93.4 kJ/kg = (310 K )⎜⎜ 3.13916 − 2.47256 − 1600 K ⎟⎠ ⎝ ⎛ P ⎞ x destroyed, 34 = T0 s gen,34 = T0 (s 4 − s 3 ) = T0 ⎜⎜ s 4o − s 3o − Rln 4 ⎟⎟ = P3 ⎠ ⎝ = (310 K )(2.67602 − 3.13916 − (0.287 kJ/kg ⋅ K )ln(1/8)) = 41.4 kJ/kg ⎛ q R,41 ⎞ ⎛ P ⎟ = T0 ⎜ s1o − s 4o − Rln 1 x destroyed, 41 = T0 s gen,41 = T0 ⎜ s1 − s 4 + ⎟ ⎜ ⎜ TR ⎠ P4 ⎝ ⎝

©0

+

q out TL

⎞ ⎟ ⎟ ⎠

⎛ 478.92 kJ/kg ⎞ ⎟⎟ = 220 kJ/kg = (310 K )⎜⎜1.73498 − 2.67602 + 310 K ⎝ ⎠

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9-116

9-152E The exergy loss of an ideal dual cycle described in Prob. 9-58E is to be determined. Analysis From Prob. 9-58E, qout = 66.48 Btu/lbm, T1 = 530 R, T2 = 1757 R, Tx = 2109 R, T3 = 2742 R, and T4 = 918.8 R. Also, q in , 2− x = cv (T x − T2 ) = (0.171 Btu/lbm ⋅ R )(2109 − 1757)R = 60.19 Btu/lbm q in , x −3 = cv (T3 − T x ) = (0.240 Btu/lbm ⋅ R )(2742 − 2109)R = 151.9 Btu/lbm q out = cv (T4 − T1 ) = (0.718 kJ/kg ⋅ K )(494.8 − 291)K = 146.3 kJ/kg

The exergy destruction during a process of the cycle is ⎛ q q x dest = T0 s gen = T0 ⎜⎜ Δs − in + out T T source sink ⎝

⎞ ⎟ ⎟ ⎠

Application of this equation for each process of the cycle gives

3

x

P

qin 2

x dest,1- 2 = 0 (isentropic process) s x − s 2 = cv ln

Tx v + Rln x v2 T2

s 3 − s x = c p ln

qout 1

v

= (0.171 Btu/lbm ⋅ R ) ln q in, 2- x ⎛ x dest, 2- x = T0 ⎜⎜ s x − s 2 − Tsource ⎝

4

2109 R + 0 = 0.03123 Btu/lbm ⋅ R 1757 R

⎞ 60.19 Btu/lbm ⎞ ⎟ = (530 R)⎛⎜ 0.03123 Btu/lbm ⋅ R − ⎟ = 4.917 Btu/lbm ⎟ 2742 R ⎝ ⎠ ⎠

T3 P 2742 R − Rln 3 = (0.240 Btu/lbm ⋅ R ) ln − 0 = 0.06299 Btu/lbm ⋅ R Tx Px 2109 R

q in, x -3 ⎛ x dest, x -3 = T0 ⎜⎜ s 3 − s x − Tsource ⎝

⎞ 151.9 Btu/lbm ⎞ ⎟ = (530 R)⎛⎜ 0.06299 Btu/lbm ⋅ R − ⎟ = 4.024 Btu/lbm ⎟ 2742 R ⎝ ⎠ ⎠

x dest,3-4 = 0 (isentropic process) s1 − s 4 = cv ln

T1 v 530 R + Rln 1 = (0.171 Btu/lbm ⋅ R ) ln + 0 = −0.09408 Btu/lbm ⋅ R v4 T4 918.8 R

⎛ q x dest,4-1 = T0 ⎜⎜ s1 − s 4 + out Tsink ⎝

⎞ 66.48 Btu/lbm ⎞ ⎟ = (530 R)⎛⎜ − 0.09408 Btu/lbm ⋅ R + ⎟ = 16.62 Btu/lbm ⎟ 530 R ⎝ ⎠ ⎠

The largest exergy destruction in the cycle occurs during the heat-rejection process s. The total exergy destruction in the cycle is x dest, total = 4.917 + 4.024 + 16.62 = 25.6 Btu/lbm

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9-117

9-153E The entropy generated by the Brayton cycle of Prob. 9-102E is to be determined. Analysis From Prob. 9-102E, Q& in = 2356 Btu/s Q& out = Q& in − W& net = 2356 − 1417 = 939 Btu/s T H = 1660 R T L = 540 R

No entropy is generated by the working fluid since it always returns to its original state. Then, Q& Q& 939 Btu/s 2356 Btu/s S& gen = out − in = − = 0.320 Btu/s ⋅ R TL TH 540 R 1660 R

9-154 The exergy loss of each process for a regenerative Brayton cycle described in Prob. 9-112 is to be determined. Analysis From Prob. 9-112, T1 = 293 K, T2 = 566.3 K, T3 = 1073 K, T4 = 625.9 K, T5 = 615.9 K, T6 = 576.3 K, and rp = 8. Also,

T

q in = c p (T3 − T5 ) = (1.005 kJ/kg ⋅ K )(1073 − 615.9)K = 659.4 kJ/kg q out = c p (T6 − T1 ) = (1.005 kJ/kg ⋅ K )(576.3 − 293)K = 284.7 kJ/kg

The exergy destruction during a process of a stream from an inlet state to exit state is given by ⎛ q q x dest = T0 s gen = T0 ⎜⎜ s e − s i − in + out Tsource Tsink ⎝

2s 293 K

3

qin

1073 K

1

2

4

5 6

4s

qout s

⎞ ⎟ ⎟ ⎠

Application of this equation for each process of the cycle gives ⎛ T P ⎞ 566.3 ⎡ ⎤ x dest, 1- 2 = T0 ⎜⎜ c p ln 2 − R ln 2 ⎟⎟ = (293) ⎢(1.005)ln − (0.287) ln(8)⎥ = 19.2 kJ/kg 293 T1 P1 ⎠ ⎣ ⎦ ⎝ ⎛ T P q x dest, 5-3 = T0 ⎜⎜ c p ln 3 − Rln 3 − in T5 P5 Tsource ⎝

⎞ 1073 459.4 ⎤ ⎟ = (293) ⎡⎢(1.005)ln −0− = 38.0 kJ/kg ⎟ 615.9 1073 ⎥⎦ ⎣ ⎠

⎛ T P ⎞ ⎡ 625.9 ⎛ 1 ⎞⎤ x dest, 3-4 = T0 ⎜⎜ c p ln 4 − Rln 4 ⎟⎟ = (293 K) ⎢(1.005)ln − (0.287) ln⎜ ⎟⎥ = 16.1 kJ/kg 1073 T P ⎝ 8 ⎠⎦ ⎣ 3 3 ⎠ ⎝ ⎛ q T P x dest, 6-1 = T0 ⎜⎜ c p ln 1 − Rln 1 + out T6 P6 Tsin k ⎝

⎞ 293 284.7 ⎤ ⎟ = (293) ⎡⎢(1.005)ln −0+ = 85.5 kJ/kg ⎟ 576.3 293 ⎥⎦ ⎣ ⎠

⎛ T T ⎞ x dest, regen = T0 (Δs 2−5 + Δs 4 −6 = T0 ⎜⎜ c p ln 5 + c p ln 6 ⎟⎟ T2 T4 ⎠ ⎝ 615.9 576.3 ⎤ ⎡ = (293) ⎢(1.005)ln + (1.005)ln = 0.41 kJ/kg 566.3 625.9 ⎥⎦ ⎣

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9-118

9-155 The total exergy destruction associated with the Brayton cycle described in Prob. 9-119 and the exergy at the exhaust gases at the turbine exit are to be determined. Properties The gas constant of air is R = 0.287 kJ/kg·K (Table A-1). Analysis From Prob. 9-119, qin = 480.82, qout = 372.73 kJ/kg, and T1 = 310 K

T

⎯ ⎯→ s1o = 1.73498 kJ/kg ⋅ K

h2 = 618.26 kJ/kg ⎯ ⎯→ s 2o = 2.42763 kJ/kg ⋅ K T3 = 1150 K

qin

1150 K

3

5

⎯ ⎯→ s 3o = 3.12900 kJ/kg ⋅ K

h4 = 803.14 kJ/kg ⎯ ⎯→

s 4o

= 2.69407 kJ/kg ⋅ K

h5 = 738.43 kJ/kg ⎯ ⎯→

s 5o

= 2.60815 kJ/kg ⋅ K

2s 310 K

4 4s

2 6

1

s

and, from an energy balance on the heat exchanger, h5 − h2 = h4 − h6 ⎯ ⎯→ h6 = 803.14 − (738.43 − 618.26) = 682.97 kJ/kg ⎯ ⎯→ s 6o = 2.52861 kJ/kg ⋅ K

Thus, ⎛ P ⎞ x destroyed,12 = T0 s gen,12 = T0 (s 2 − s1 ) = T0 ⎜⎜ s 2o − s1o − Rln 2 ⎟⎟ P1 ⎠ ⎝ = (310 K )(2.42763 − 1.73498 − (0.287 kJ/kg ⋅ K )ln (7 )) = 41.59 kJ/kg ⎛ P ⎞ x destroyed, 34 = T0 s gen,34 = T0 (s 4 − s 3 ) = T0 ⎜⎜ s 4o − s 3o − Rln 4 ⎟⎟ P3 ⎠ ⎝ = (310 K )(2.69407 − 3.12900 − (0.287 kJ/kg ⋅ K )ln (1/7 )) = 38.30 kJ/kg

[(

) (

x destroyed, regen = T0 s gen,regen = T0 [(s 5 − s 2 ) + (s 6 − s 4 )] = T0 s 5o − s 2o + s 6o − s 4o

)]

= (310 K )(2.60815 − 2.42763 + 2.52861 − 2.69407 ) = 4.67 kJ/kg

q R ,53 ⎛ x destroyed, 53 = T0 s gen,53 = T0 ⎜ s 3 − s 5 − ⎜ TR ⎝

⎛ ⎞ P ©0 q in ⎞⎟ ⎟ = T0 ⎜ s 3o − s 5o − Rln 3 − ⎟ ⎜ P5 T H ⎟⎠ ⎠ ⎝

⎛ 480.82 kJ/kg ⎞ ⎟ = 78.66 kJ/kg = (310 K )⎜⎜ 3.12900 − 2.60815 − 1800 K ⎟⎠ ⎝ ⎛ q R ,61 ⎞ ⎛ P ⎟ = T0 ⎜ s1o − s 6o − Rln 1 x destroyed, 61 = T0 s gen,61 = T0 ⎜ s1 − s 6 + ⎟ ⎜ ⎜ TR ⎠ P6 ⎝ ⎝

©0

+

q out TL

⎞ ⎟ ⎟ ⎠

⎛ 372.73 kJ/kg ⎞ ⎟⎟ = 126.7 kJ/kg = (310 K )⎜⎜1.73498 − 2.52861 + 310 K ⎝ ⎠

Noting that h0 = h@ 310 K = 310.24 kJ/kg, the stream exergy at the exit of the regenerator (state 6) is determined from

φ6 = (h6 − h0 ) − T0 (s6 − s0 ) +

V62 2

©0

+ gz6©0 P6 P1

©0

where

s 6 − s 0 = s 6 − s1 = s 6o − s1o − R ln

Thus,

φ 6 = 682.97 − 310.24 − (310 K )(0.79363 kJ/kg ⋅ K ) = 126.7 kJ/kg

= 2.52861 − 1.73498 = 0.79363 kJ/kg ⋅ K

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9-119

9-156 EES Prob. 9-155 is reconsidered. The effect of the cycle pressure on the total irreversibility for the cycle and the exergy of the exhaust gas leaving the regenerator is to be investigated. Analysis Using EES, the problem is solved as follows: "Given" T[1]=310 [K] P[1]=100 [kPa] Ratio_P=7 P[2]=Ratio_P*P[1] T[3]=1150 [K] eta_C=0.75 eta_T=0.82 epsilon=0.65 T_H=1800 [K] T0=310 [K] P0=100 [kPa] "Analysis for Problem 9-156" q_in=h[3]-h[5] q_out=h[6]-h[1] h[5]-h[2]=h[4]-h[6] s[2]=entropy(Fluid$, P=P[2], h=h[2]) s[4]=entropy(Fluid$, h=h[4], P=P[4]) s[5]=entropy(Fluid$, h=h[5], P=P[5]) P[5]=P[2] s[6]=entropy(Fluid$, h=h[6], P=P[6]) P[6]=P[1] h[0]=enthalpy(Fluid$, T=T0) s[0]=entropy(Fluid$, T=T0, P=P0) x_destroyed_12=T0*(s[2]-s[1]) x_destroyed_34=T0*(s[4]-s[3]) x_destroyed_regen=T0*(s[5]-s[2]+s[6]-s[4]) x_destroyed_53=T0*(s[3]-s[5]-q_in/T_H) x_destroyed_61=T0*(s[1]-s[6]+q_out/T0) x_total=x_destroyed_12+x_destroyed_34+x_destroyed_regen+x_destroyed_53+x_destroyed_61 x6=h[6]-h[0]-T0*(s[6]-s[0]) "since state 0 and state 1 are identical" "Analysis for Problem 9-119" Fluid$='air' "(a)" h[1]=enthalpy(Fluid$, T=T[1]) s[1]=entropy(Fluid$, T=T[1], P=P[1]) s_s[2]=s[1] "isentropic compression" h_s[2]=enthalpy(Fluid$, P=P[2], s=s_s[2]) eta_C=(h_s[2]-h[1])/(h[2]-h[1]) h[3]=enthalpy(Fluid$, T=T[3]) s[3]=entropy(Fluid$, T=T[3], P=P[3]) P[3]=P[2] s_s[4]=s[3] "isentropic expansion" h_s[4]=enthalpy(Fluid$, P=P[4], s=s_s[4]) P[4]=P[1] eta_T=(h[3]-h[4])/(h[3]-h_s[4]) q_regen=epsilon*(h[4]-h[2])

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9-120

"(b)" w_C_in=(h[2]-h[1]) w_T_out=h[3]-h[4] w_net_out=w_T_out-w_C_in q_in=(h[3]-h[2])-q_regen eta_th=w_net_out/q_in xtotal [kJ/kg] 279.8 289.9 299.8 309.5 318.8 327.9 336.7 345.2 353.4

Ratio_P 6 7 8 9 10 11 12 13 14

x6 [kJ/kg] 120.7 126.7 132.5 138 143.4 148.6 153.7 158.6 163.4

360

180 175

350

170 165

330

160 155

320

150 310

145

300

x6 [kJ/kg]

xtotal [kJ/kg]

340

140 135

290

130 280

125

270 6

7

8

9

10

11

12

13

120 14

RatioP

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-121

9-157E The exergy loss of each process for a reheat-regenerative Brayton cycle with intercooling described in Prob. 9-134E is to be determined. Analysis From Prob. 9-134E,

T

T1 = T3 = 520 R,

6

1400 R

T2 = T4 = T10 = 737.8 R,

5 7s

T5 = T7 = T9 = 1049 K, T6 = T8 = 1400 R, and rp = 3.

4 2 4s 2s

Also, 520 R

q in,5-6 = q in,7 -8 = c p (T6 − T5 )

3

7

8 9 9s

10

1

= (0.240 Btu/lbm ⋅ R )(1400 − 1049)R = 84.24 Btu/lbm

s

q out,10-1 = q out,2-3 = c p (T10 − T1 ) = (0.240 Btu/lbm ⋅ R )(737.8 − 520)R = 52.27 Btu/lbm

The exergy destruction during a process of a stream from an inlet state to exit state is given by ⎛ q q x dest = T0 s gen = T0 ⎜⎜ s e − s i − in + out Tsource Tsink ⎝

⎞ ⎟ ⎟ ⎠

Application of this equation for each process of the cycle gives ⎛ T P ⎞ 737.8 ⎡ ⎤ x dest, 1- 2 = x dest, 3- 4 = T0 ⎜⎜ c p ln 2 − Rln 2 ⎟⎟ = (520) ⎢(0.24)ln − (0.06855) ln(3)⎥ = 4.50 Btu/lbm T1 P1 ⎠ 520 ⎣ ⎦ ⎝ q in,5-6 ⎛ T P x dest, 5-6 = x dest, 7 -8 = T0 ⎜⎜ c p ln 6 − Rln 6 − T5 P5 Tsource ⎝

⎞ 1400 84.24 ⎤ ⎟ = (520) ⎡⎢(0.24)ln −0− = 4.73 Btu/lbm ⎟ 1049 1400 ⎥⎦ ⎣ ⎠

⎛ T P ⎞ ⎡ 1049 ⎛ 1 ⎞⎤ x dest, 6-7 = x dest, 8-9 = T0 ⎜⎜ c p ln 7 − R ln 7 ⎟⎟ = (520) ⎢(0.24)ln − (0.06855) ln⎜ ⎟⎥ = 3.14 Btu/lbm T P 1400 ⎝ 3 ⎠⎦ ⎣ 6 6 ⎠ ⎝ ⎛ q T P x dest, 10-1 = x dest, 2-3 = T0 ⎜⎜ c p ln 1 − Rln 1 + out T10 P10 Tsink ⎝

⎞ 520 52.27 ⎤ ⎟ = (520) ⎡⎢(0.24)ln −0+ = 8.61 Btu/lbm ⎟ 737.8 520 ⎥⎦ ⎣ ⎠

⎛ T T ⎞ x dest,regen = T0 (Δs 4 −5 + Δs 9−10 = T0 ⎜⎜ c p ln 5 + c p ln 10 ⎟⎟ T4 T9 ⎠ ⎝ 1049 737.8 ⎤ ⎡ = (520) ⎢(0.24)ln + (0.24)ln = 0 Btu/lbm 737.8 1049 ⎥⎦ ⎣

The greatest exergy destruction occurs during the heat rejection processes.

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-122

9-158 The exergy loss of each process for a regenerative Brayton cycle with three stages of reheating and intercooling described in Prob. 9-137 is to be determined. Analysis From Prob. 9-137, rp = 4, qin,7-8 = qin,9-10 = qin,11-12 = 300 kJ/kg, T 12 10 qout,14-1 = 181.8 kJ/kg, qout,2-3 = qout,4-5 = 141.6 kJ/kg, 8 T1 = T3 = T5 = 290 K , T2 = T4 = T6 = 430.9 K T7 = 556.7 K,

T8 = 855.2 K,

T10 = 874.0 K,

T11 = 588.2 K,

T13 = 596.7 K,

T14 = 470.9 K

T9 = 575.5 K

9

T12 = 886.7 K,

7 6

The exergy destruction during a process of a stream from an inlet state to exit state is given by x dest = T0 s gen

⎛ q q = T0 ⎜⎜ s e − s i − in + out Tsource Tsink ⎝

11

5

4

2

3

1

13

14

⎞ ⎟ ⎟ ⎠

s

Application of this equation for each process of the cycle gives ⎛ T P ⎞ x dest, 1- 2 = x dest, 3- 4 = x dest, 5-6 = T0 ⎜⎜ c p ln 2 − Rln 2 ⎟⎟ T P1 ⎠ 1 ⎝ 430.9 ⎡ ⎤ = (290) ⎢(1.005)ln − (0.287) ln(4)⎥ = 0.03 kJ/kg ≈ 0 290 ⎣ ⎦ q in,7 -8 ⎛ T P x dest, 7 -8 = T0 ⎜⎜ c p ln 8 − Rln 8 − T7 P7 Tsource ⎝

⎞ 855.2 300 ⎤ ⎟ = (290) ⎡⎢(1.005)ln −0− = 27.0 kJ/kg ⎟ 556.7 886.7 ⎥⎦ ⎣ ⎠

⎛ T P q in,9-10 x dest, 9-10 = T0 ⎜⎜ c p ln 10 − Rln 8 − T9 P7 Tsource ⎝

⎞ 874.0 300 ⎤ ⎟ = (290) ⎡⎢(1.005)ln −0− = 23.7 kJ/kg ⎟ 575.5 886.7 ⎥⎦ ⎣ ⎠

q in,11-12 ⎛ T P x dest, 11-12 = T0 ⎜⎜ c p ln 12 − Rln 12 − T11 P11 Tsource ⎝

⎞ 886.7 300 ⎤ ⎟ = (290) ⎡⎢(1.005)ln −0− = 21.5 kJ/kg ⎟ 588.2 886.7 ⎥⎦ ⎣ ⎠

⎛ T P ⎞ ⎡ 575.5 ⎛ 1 ⎞⎤ x dest, 8-9 = T0 ⎜⎜ c p ln 9 − R ln 9 ⎟⎟ = (290) ⎢(1.005)ln − (0.287) ln⎜ ⎟⎥ = −0.06 kJ/kg ≈ 0 855.2 T8 P8 ⎠ ⎝ 4 ⎠⎦ ⎣ ⎝ ⎛ T P ⎞ ⎡ 588.2 ⎛ 1 ⎞⎤ x dest, 10-11 = T0 ⎜⎜ c p ln 11 − Rln 11 ⎟⎟ = (290) ⎢(1.005)ln − (0.287) ln⎜ ⎟⎥ = 0.42 kJ/kg ≈ 0 874.0 T10 P10 ⎠ ⎝ 4 ⎠⎦ ⎣ ⎝ ⎛ T P ⎞ ⎡ 596.7 ⎛ 1 ⎞⎤ x dest, 12-13 = T0 ⎜⎜ c p ln 13 − Rln 13 ⎟⎟ = (290) ⎢(1.005)ln − (0.287) ln⎜ ⎟⎥ = −0.05 kJ/kg ≈ 0 886.7 T12 P12 ⎠ ⎝ 4 ⎠⎦ ⎣ ⎝ q out,14-1 ⎞ ⎛ T P 290 181.8 ⎤ ⎟ = (290) ⎡⎢(1.005)ln x dest, 14-1 = T0 ⎜⎜ c p ln 1 − Rln 1 + −0+ = 40.5 kJ/kg ⎟ 470.9 290 ⎥⎦ T14 P14 Tsink ⎠ ⎣ ⎝ q out,2-3 ⎛ T P x dest, 2-3 = x dest, 4-5 = T0 ⎜⎜ c p ln 3 − Rln 3 + T2 P2 Tsink ⎝

⎞ 290 141.6 ⎤ ⎟ = (290) ⎡⎢(1.005)ln −0+ = 26.2 kJ/kg ⎟ 430.9 290 ⎥⎦ ⎣ ⎠

⎛ T T ⎞ x dest,regen = T0 (Δs 6 − 7 + Δs13−14 ) = T0 ⎜⎜ c p ln 7 + c p ln 14 ⎟⎟ T6 T13 ⎠ ⎝ 556.7 470.9 ⎤ ⎡ = (290) ⎢(1.005)ln + (1.005)ln = 5.65 kJ/kg 430.9 596.7 ⎥⎦ ⎣

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9-123

9-159 A gas-turbine plant uses diesel fuel and operates on simple Brayton cycle. The isentropic efficiency of the compressor, the net power output, the back work ratio, the thermal efficiency, and the second-law efficiency are to be determined. Assumptions 1 The air-standard assumptions Diesel fuel Combustion are applicable. 2 Kinetic and potential energy chamber changes are negligible. 3 Air is an ideal gas 3 with constant specific heats. 700 kPa 2 Properties The properties of air at 500ºC = 773 260°C K are cp = 1.093 kJ/kg·K, cv = 0.806 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.357 (Table ATurbine Compress. 2b). Analysis (a) The isentropic efficiency of the compressor may be determined if we first 100 kPa 4 1 calculate the exit temperature for the isentropic 30°C case ( k −1) / k

⎛P ⎞ ⎛ 700 kPa ⎞ T2 s = T1 ⎜⎜ 2 ⎟⎟ = (303 K )⎜ = 505.6 K ⎟ ⎝ 100 kPa ⎠ ⎝ P1 ⎠ T −T (505.6 − 303)K η C = 2s 1 = = 0.881 T2 − T1 (533 − 303)K (b) The total mass flowing through the turbine and the rate of heat input are m& 12.6 kg/s = 12.6 kg/s + 0.21 kg/s = 12.81 kg/s m& t = m& a + m& f = m& a + a = 12.6 kg/s + AF 60 Q& in = m& f q HVη c = (0.21 kg/s)(42,000 kJ/kg)(0.97) = 8555 kW (1.357-1)/1.357

The temperature at the exit of combustion chamber is Q& in = m& c p (T3 − T2 ) ⎯ ⎯→ 8555 kJ/s = (12.81 kg/s)(1.093 kJ/kg.K)(T3 − 533)K ⎯ ⎯→ T3 = 1144 K The temperature at the turbine exit is determined using isentropic efficiency relation ( k −1) / k

(1.357-1)/1.357 ⎛P ⎞ ⎛ 100 kPa ⎞ T4 s = T3 ⎜⎜ 4 ⎟⎟ = (1144 K )⎜ = 685.7 K ⎟ ⎝ 700 kPa ⎠ ⎝ P3 ⎠ T − T4 (1144 − T4 )K ηT = 3 ⎯ ⎯→ 0.85 = ⎯ ⎯→ T4 = 754.4 K T3 − T4 s (1144 − 685.7)K The net power and the back work ratio are W&C, in = m& a c p (T2 − T1 ) = (12.6 kg/s)(1.093 kJ/kg.K)(533 − 303)K = 3168 kW

W&T, out = m& c p (T3 − T4 ) = (12.81 kg/s)(1.093 kJ/kg.K)(1144 − 754.4)K = 5455 kW W& net = W&T, out − W&C, in = 5455 − 3168 = 2287 kW W& 3168 kW rbw = & C,in = = 0.581 WT,out 5455 kW

(c) The thermal efficiency is

η th =

W& net 2287 kW = = 0.267 8555 kW Q& in

The second-law efficieny of the cycle is defined as the ratio of actual thermal efficiency to the maximum possible thermal efficiency (Carnot efficiency). The maximum temperature for the cycle can be taken to be the turbine inlet temperature. That is, T 303 K η max = 1 − 1 = 1 − = 0.735 T3 1144 K and

η II =

η th 0.267 = = 0.364 η max 0.735

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9-124

9-160 A modern compression ignition engine operates on the ideal dual cycle. The maximum temperature in the cycle, the net work output, the thermal efficiency, the mean effective pressure, the net power output, the second-law efficiency of the cycle, and the rate of exergy of the exhaust gases are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at 850 K are cp = 1.110 kJ/kg·K, cv = 0.823 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.349 (Table A-2b). Analysis (a) The clearance volume and the total volume of the engine at the beginning of compression process (state 1) are r=

Vc + V d V + 0.0028 m3 ⎯ ⎯→14 = c ⎯ ⎯→V c = 0.0002154 m3 = V 2 = V x Vc Vc

V 1 = V c +V d = 0.0002154 + 0.0028 = 0.003015 m 3 = V 4

P

Process 1-2: Isentropic compression k −1

3

x

⎛v T2 = T1 ⎜⎜ 1 ⎝v 2

⎞ ⎟⎟ ⎠

⎛v P2 = P1 ⎜⎜ 1 ⎝v 2

⎞ ⎟⎟ = (95 kPa )(14 )1.349 = 3341 kPa ⎠

= (328 K )(14 )1.349-1 = 823.9 K

2

Qin

k

4

Process 2-x and x-3: Constant-volume and constant pressure heat addition processes: T x = T2

Qout 1

Px 9000 kPa = (823.9 K) = 2220 K P2 3341 kPa

V

q 2- x = cv (T x − T2 ) = (0.823 kJ/kg.K)(2220 − 823.9)K = 1149 kJ/kg q 2− x = q x -3 = c p (T3 − T x ) ⎯ ⎯→ 1149 kJ/kg = (0.823 kJ/kg.K)(T3 − 2220)K ⎯ ⎯→ T3 = 3254 K

(b)

q in = q 2− x + q x -3 = 1149 + 1149 = 2298 kJ/kg

V3 =V x

T3 3254 K = (0.0002154 m 3 ) = 0.0003158 m 3 Tx 2220 K

Process 3-4: isentropic expansion. k −1

1.349 -1

⎛V T4 = T3 ⎜⎜ 3 ⎝V 4

⎞ ⎟⎟ ⎠

⎛ 0.0003158 m 3 = (3254 K )⎜ ⎜ 0.003015 m 3 ⎝

⎞ ⎟ ⎟ ⎠

⎛V P4 = P3 ⎜⎜ 3 ⎝V 4

⎛ 0.0003158 m 3 ⎞ ⎟⎟ = (9000 kPa )⎜ ⎜ 0.003015 m 3 ⎠ ⎝

⎞ ⎟ ⎟ ⎠

k

= 1481 K 1.349

= 428.9 kPa

Process 4-1: constant voume heat rejection. q out = cv (T4 − T1 ) = (0.823 kJ/kg ⋅ K )(1481 − 328)K = 948.7 kJ/kg

The net work output and the thermal efficiency are wnet,out = q in − q out = 2298 − 948.7 = 1349 kJ/kg

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9-125

η th =

wnet,out q in

=

1349 kJ/kg = 0.587 2298 kJ/kg

(c) The mean effective pressure is determined to be m=

P1V1 (95 kPa)(0.003015 m 3 ) = = 0.003043 kg RT1 0.287 kPa ⋅ m 3 /kg ⋅ K (328 K )

(

MEP =

mwnet,out

V 1 −V 2

)

=

(0.003043 kg)(1349 kJ/kg) ⎛ kPa ⋅ m 3 ⎜ (0.003015 − 0.0002154)m 3 ⎜⎝ kJ

⎞ ⎟ = 1466 kPa ⎟ ⎠

(d) The power for engine speed of 3500 rpm is 3500 (rev/min) ⎛ 1 min ⎞ n& W& net = mwnet = (0.003043 kg)(1349 kJ/kg) ⎜ ⎟ = 120 kW (2 rev/cycle) ⎝ 60 s ⎠ 2

Note that there are two revolutions in one cycle in four-stroke engines. (e) The second-law efficieny of the cycle is defined as the ratio of actual thermal efficiency to the maximum possible thermal efficiency (Carnot efficiency). We take the dead state temperature and pressure to be 25ºC and 100 kPa.

η max = 1 −

T0 (25 + 273) K =1− = 0.908 T3 3254 K

and

η II =

η th 0.587 = = 0.646 η max 0.908

The rate of exergy of the exhaust gases is determined as follows ⎡ T P ⎤ x 4 = u 4 − u 0 − T0 ( s 4 − s 0 ) = cv (T4 − T0 ) − T0 ⎢c p ln 4 − R ln 4 ⎥ T0 P0 ⎦ ⎣ 428.9 ⎤ 1481 ⎡ = (0.823)(1481 − 298)− (298) ⎢(1.110 kJ/kg.Kln − 0.287 ln = 567.6 kJ/kg 100 ⎥⎦ 298 ⎣

3500 (rev/min) ⎛ 1 min ⎞ n& X& 4 = mx 4 = (0.003043 kg)(567.6 kJ/kg) ⎜ ⎟ = 50.4 kW 2 (2 rev/cycle) ⎝ 60 s ⎠

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9-126

Review Problems

9-161 A turbocharged four-stroke V-16 diesel engine produces 3500 hp at 1200 rpm. The amount of power produced per cylinder per mechanical and per thermodynamic cycle is to be determined. Analysis Noting that there are 16 cylinders and each thermodynamic cycle corresponds to 2 mechanical cycles (revolutions), we have

(a) Total power produced (No. of cylinders)(No. of mechanical cycles)

wmechanical =

⎛ 42.41 Btu/min ⎞ 3500 hp ⎜ ⎟⎟ (16 cylinders)(1200 rev/min) ⎜⎝ 1 hp ⎠

=

= 7.73 Btu/cyl ⋅ mech cycle (= 8.16 kJ/cyl ⋅ mech cycle)

(b) wthermodynamic =

Total power produced (No. of cylinders)(No. of thermodynamic cycles)

=

⎛ 42.41 Btu/min ⎞ 3500 hp ⎜ ⎟⎟ (16 cylinders)(1200/2 rev/min) ⎜⎝ 1 hp ⎠

= 15.46 Btu/cyl ⋅ therm cycle (= 16.31 kJ/cyl ⋅ therm cycle)

9-162 A simple ideal Brayton cycle operating between the specified temperature limits is considered. The pressure ratio for which the compressor and the turbine exit temperature of air are equal is to be determined. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats. Properties The specific heat ratio of air is k =1.4 (Table A-2). Analysis We treat air as an ideal gas with constant specific heats. Using the isentropic relations, the temperatures at the compressor and turbine exit can be expressed as ⎛P ⎞ T2 = T1⎜⎜ 2 ⎟⎟ ⎝ P1 ⎠

(k −1) / k

⎛P ⎞ T4 = T3 ⎜⎜ 4 ⎟⎟ ⎝ P3 ⎠

( )

= T1 rp (k −1) / k

(k −1) / k

⎛1⎞ = T3 ⎜ ⎟ ⎜ rp ⎟ ⎝ ⎠

T

(k −1) / k

3

T3

qin 2

Setting T2 = T4 and solving for rp gives ⎛T ⎞ r p = ⎜⎜ 3 ⎟⎟ ⎝ T1 ⎠

k / 2 (k −1)

⎛ 1500 K ⎞ ⎟⎟ = ⎜⎜ ⎝ 300 K ⎠

4 T1

1.4/0.8

= 16.7

1

qout

s

Therefore, the compressor and turbine exit temperatures will be equal when the compression ratio is 16.7.

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9-127

9-163 The three processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-s diagrams, and the maximum temperature in the cycle and the thermal efficiency are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The properties of air are given in Table A-17.

P

Analysis (b) We treat air as an ideal gas with variable specific heats,

2

T1 = 300 K ⎯ ⎯→ u1 = 214.07 kJ/kg

qin

3

Pr1 = 1.386 Pr2 =

⎛ 700 kPa ⎞ P2 ⎟⎟(1.386 ) = 9.702 ⎯ ⎯→ h2 = 523.90 kJ/kg Pr1 = ⎜⎜ P1 ⎝ 100 kPa ⎠

⎛ 700 kPa ⎞ P3v 3 P1v 1 P ⎟⎟(300 K ) = 2100 K = ⎯ ⎯→ Tmax = T3 = 3 T1 = ⎜⎜ T3 T1 P1 ⎝ 100 kPa ⎠ ⎯→ u 3 = 1775.3 kJ/kg T3 = 2100 K ⎯ h3 = 2377.7 kJ/kg

(c)

qout 1

v

T qin

3

2

q in = h3 − h2 = 2377.7 − 523.9 = 1853.8 kJ/kg q out = u 3 − u1 = 1775.3 − 214.07 = 1561.23kJ/kg

η th = 1 −

1

qout

q out 1561.23 kJ/kg = 1− = 15.8% 1853.8 kJ/kg q in

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s

9-128

9-164 All three processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-s diagrams, and the maximum temperature in the cycle and the thermal efficiency are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats.

P

Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-2).

2

qin

3 qout

Analysis (b) We treat air as an ideal gas with constant specific heats.

Process 1-2 is isentropic: ⎛P ⎞ T2 = T1⎜⎜ 2 ⎟⎟ ⎝ P1 ⎠

(k −1) / k

⎛ 700 kPa ⎞ ⎟⎟ = (300 K )⎜⎜ ⎝ 100 kPa ⎠

1 0.4/1.4

= 523.1 K

⎛ 700 kPa ⎞ P3v 3 P1v1 P ⎟⎟(300 K ) = 2100 K = ⎯⎯→ Tmax = T3 = 3 T1 = ⎜⎜ T3 T1 P1 ⎝ 100 kPa ⎠

(c)

v

T qin

3

2

q in = h3 − h2 = c p (T3 − T2 )

= (1.005 kJ/kg ⋅ K )(2100 − 523.1)K = 1584.8 kJ/kg

q out = u 3 − u1 = c v (T3 − T1 )

1

qout

= (0.718 kJ/kg ⋅ K )(2100 − 300)K = 1292.4 kJ/kg

η th = 1 −

q out 1292.4 kJ/kg = 1− = 18.5% q in 1584.8 kJ/kg

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s

9-129

9-165 [Also solved by EES on enclosed CD] A four-cylinder spark-ignition engine with a compression ratio of 8 is considered. The amount of heat supplied per cylinder, the thermal efficiency, and the rpm for a net power output of 60 kW are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The gas constant of air is R = 0.287 kJ/kg·K (Table A-1). The properties of air are given in Table A-17. Analysis (a) Process 1-2: isentropic compression.

P

⎯→ u1 = 206.91 kJ/kg T1 = 290 K ⎯

v r1 = 676.1 v r2 =

1800 K

v2 1 1 v r1 = v r1 = (676.1) = 84.51 v1 8 r

3 Qin

⎯ ⎯→ u 2 = 475.11 kJ/kg

4 Qout

2

Process 2-3: v = constant heat addition.

1

T3 = 1800 K ⎯ ⎯→ u 3 = 1487.2 kJ/kg

v

v r3 = 3.994 m=

(

)

P1V1 (98 kPa ) 0.0006 m 3 = = 7.065 × 10 − 4 kg RT1 0.287 kPa ⋅ m 3 /kg ⋅ K (290 K )

(

(

)

)

Qin = m(u 3 − u 2 ) = 7.065 × 10 − 4 kg (1487.2 − 475.11)kJ/kg = 0.715 kJ

(b) Process 3-4: isentropic expansion.

v r4 =

v4 v r = rv r3 = (8)(3.994 ) = 31.95 ⎯⎯→ u 4 = 693.23 kJ/kg v3 3

Process 4-1: v = constant heat rejection.

(

)

Qout = m(u 4 − u1 ) = 7.065 × 10 -4 kg (693.23 − 206.91)kJ/kg = 0.344 kJ W net = Qin − Qout = 0.715 − 0.344 = 0.371 kJ

η th =

(c)

n& = 2

W net 0.371 kJ = = 51.9% 0.715 kJ Qin

⎛ 60 s ⎞ W& net 60 kJ/s ⎜ ⎟ = 4852 rpm = (2 rev/cycle) 4 × (0.371 kJ/cycle) ⎜⎝ 1 min ⎟⎠ n cyl W net,cyl

Note that for four-stroke cycles, there are two revolutions per cycle.

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9-130

9-166 EES Problem 9-165 is reconsidered. The effect of the compression ratio net work done and the efficiency of the cycle is to be investigated. Also, the T-s and P-v diagrams for the cycle are to be plotted. Analysis Using EES, the problem is solved as follows: "Input Data" T[1]=(17+273) [K] P[1]=98 [kPa] T[3]=1800 [K] V_cyl=0.6 [L]*Convert(L, m^3) r_v=8 "Compression ratio" W_dot_net = 60 [kW] N_cyl=4 "number of cyclinders" v[1]/v[2]=r_v "The first part of the solution is done per unit mass." "Process 1-2 is isentropic compression" s[1]=entropy(air,T=T[1],P=P[1]) s[2]=s[1] s[2]=entropy(air, T=T[2], v=v[2]) P[2]*v[2]/T[2]=P[1]*v[1]/T[1] P[1]*v[1]=R*T[1] R=0.287 [kJ/kg-K] "Conservation of energy for process 1 to 2: no heat transfer (s=const.) with work input" w_in = DELTAu_12 DELTAu_12=intenergy(air,T=T[2])-intenergy(air,T=T[1]) "Process 2-3 is constant volume heat addition" s[3]=entropy(air, T=T[3], P=P[3]) {P[3]*v[3]/T[3]=P[2]*v[2]/T[2]} P[3]*v[3]=R*T[3] v[3]=v[2] "Conservation of energy for process 2 to 3: the work is zero for v=const, heat is added" q_in = DELTAu_23 DELTAu_23=intenergy(air,T=T[3])-intenergy(air,T=T[2]) "Process 3-4 is isentropic expansion" s[4]=entropy(air,T=T[4],P=P[4]) s[4]=s[3] P[4]*v[4]/T[4]=P[3]*v[3]/T[3] {P[4]*v[4]=R*T[4]} "Conservation of energy for process 3 to 4: no heat transfer (s=const) with work output" - w_out = DELTAu_34 DELTAu_34=intenergy(air,T=T[4])-intenergy(air,T=T[3]) "Process 4-1 is constant volume heat rejection" v[4]=v[1] "Conservation of energy for process 2 to 3: the work is zero for v=const; heat is rejected" - q_out = DELTAu_41 DELTAu_41=intenergy(air,T=T[1])-intenergy(air,T=T[4]) w_net = w_out - w_in Eta_th=w_net/q_in*Convert(, %) "Thermal efficiency, in percent" "The mass contained in each cylinder is found from the volume of the cylinder:" V_cyl=m*v[1] "The net work done per cycle is:" W_dot_net=m*w_net"kJ/cyl"*N_cyl*N_dot"mechanical cycles/min"*1"min"/60"s"*1"thermal cycle"/2"mechanical cycles"

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9-131

ηth [%] 42.81 46.39 49.26 51.63 53.63 55.35 56.85

wnet [kJ/kg] 467.1 492.5 509.8 521.7 529.8 535.2 538.5

rv 5 6 7 8 9 10 11

Air Otto Cycle P-v Diagram 8000 3 s = const

P [kPa]

2 1000

4

100

1

50 10-2

10-1

1800 K

290 K

100

101

102

v [m3/kg]

Air Otto Cycle T-s Diagram 2500

4866 kPa

2000

3 T [K]

1500 98 kPa 1000

4 2

500

v = const

1 0 4.0

4.5

5.0

5.5

6.0

6.5

7.0

7.5

8.0

8.5

s [kJ/kg-K]

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9-132

540 530

w net [kJ/kg]

520 510 500 490 480 470 460 5

6

7

8

r

9

10

11

v

58 56 54

η th [%]

52 50 48 46 44 42 5

6

7

8

r

9

10

11

v

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9-133

9-167 An ideal gas Carnot cycle with helium as the working fluid is considered. The pressure ratio, compression ratio, and minimum temperature of the energy source are to be determined. Assumptions 1 Kinetic and potential energy changes are negligible. 2 Helium is an ideal gas with constant specific heats. Properties The specific heat ratio of helium is k = 1.667 (Table A-2a). Analysis From the definition of the thermal efficiency of a Carnot heat engine,

η th,Carnot = 1 −

TL TL (15 + 273) K ⎯ ⎯→ T H = = = 576 K TH 1 − η th,Carnot 1 − 0.50

An isentropic process for an ideal gas is one in which Pvk remains constant. Then, the pressure ratio is P2 ⎛ T2 =⎜ P1 ⎜⎝ T1

⎞ ⎟⎟ ⎠

k /( k −1)

⎛ 576 K ⎞ =⎜ ⎟ ⎝ 288 K ⎠

qin 2

1

TH

1.667 /(1.667 −1)

= 5.65

Based on the process equation, the compression ratio is

v 1 ⎛ P2 ⎞ =⎜ ⎟ v 2 ⎜⎝ P1 ⎟⎠

T

288 K

4

qout

1/ k

3 s

= (5.65)1 / 1.667 = 2.83

9-168E An ideal gas Carnot cycle with helium as the working fluid is considered. The pressure ratio, compression ratio, and minimum temperature of the energy-source reservoir are to be determined. Assumptions 1 Kinetic and potential energy changes are negligible. 2 Helium is an ideal gas with constant specific heats. Properties The specific heat ratio of helium is k = 1.667 (Table A-2Ea). Analysis From the definition of the thermal efficiency of a Carnot heat engine,

η th,Carnot = 1 −

TL TL (60 + 460) R ⎯ ⎯→ T H = = = 1300 R TH 1 − η th,Carnot 1 − 0.60

An isentropic process for an ideal gas is one in which Pvk remains constant. Then, the pressure ratio is P2 ⎛ T2 ⎞ =⎜ ⎟ P1 ⎜⎝ T1 ⎟⎠

k /( k −1)

⎛ 1300 R ⎞ =⎜ ⎟ ⎝ 520 R ⎠

1.667 /(1.667 −1)

1/ k

= (9.88)1 / 1.667 = 3.95

qin

TH

1

520 R

4

2

= 9.88

Based on the process equation, the compression ratio is

v 1 ⎛ P2 ⎞ =⎜ ⎟ v 2 ⎜⎝ P1 ⎟⎠

T

qout

3 s

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-134

9-169 The compression ratio required for an ideal Otto cycle to produce certain amount of work when consuming a given amount of fuel is to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. 4 The combustion efficiency is 100 percent. Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-2). Analysis The heat input to the cycle for 0.043 grams of fuel consumption is

P

3

Qin = m fuel q HV = (0.043 × 10 −3 kg)(42,000 kJ/kg) = 1.806 kJ

qin

The thermal efficiency is then

η th

4 qout

2

1

W 1 kJ = net = = 0.5537 Qin 1.806 kJ

v

From the definition of thermal efficiency, we obtain the required compression ratio to be

η th = 1 −

1 r

k −1

⎯ ⎯→ r =

1 (1 − η th )

1 /( k −1)

=

1 (1 − 0.5537)1 /(1.4 −1)

= 7.52

9-170 An equation is to be developed for q in /(cv T1 r k −1 ) in terms of k, rc and rp. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Analysis The temperatures at various points of the dual cycle are given by T2 = T1 r k −1 ⎛P T x = T2 ⎜⎜ x ⎝ P2

⎞ ⎟⎟ = T2 r p = r p T1 r k −1 ⎠

⎛v T3 = T x ⎜⎜ 3 ⎝v x

⎞ ⎟ = T x rc = r p rc T1 r k −1 ⎟ ⎠

Application of the first law to the two heat addition processes gives

3

x

P

qin 2

4 qout 1

v

q in = cv (T x − T2 ) + c p (T3 − T x ) = cv (r p T1 r k −1 − T1 r k −1 ) + c p (r p rc T1 r k −1 − r p T1 r k −1 )

or upon rearrangement q in cv T1 r k −1

= (r p − 1) + kr p (rc − 1)

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-135

9-171 An ideal Otto cycle with air as the working fluid with a compression ratio of 9.2 is considered. The amount of heat transferred to the air, the net work output, the thermal efficiency, and the mean effective pressure are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The gas constant of air is R = 0.287 kJ/kg·K (Table A-1). The properties of air are given in Table A-17.

P

Analysis (a) Process 1-2: isentropic compression.

3

T1 = 300 K ⎯ ⎯→ u1 = 214.07 kJ/kg

qin

v r1 = 621.2

4 qout

2

v r2

v 1 1 (621.2) = 67.52 ⎯⎯→ T2 = 708.3 K = 2 v r1 = v r1 = v1 r 9.2

1

v

u 2 = 518.9 kJ/kg

⎛ 708.3 K ⎞ v T P2v 2 P1v 1 ⎟⎟(98 kPa ) = 2129 kPa = ⎯ ⎯→ P2 = 1 2 P1 = (9.2 )⎜⎜ T2 T1 v 2 T1 ⎝ 300 K ⎠

Process 2-3: v = constant heat addition. P3v 3 P2v 2 P = ⎯ ⎯→ T3 = 3 T2 = 2T2 = (2 )(708.3) = 1416.6 K ⎯ ⎯→ u 3 = 1128.7 kJ/kg T3 T2 P2 v r3 = 8.593 q in = u 3 − u 2 = 1128.7 − 518.9 = 609.8 kJ/kg

(b) Process 3-4: isentropic expansion.

v r4 =

v4 v r = rv r3 = (9.2)(8.593) = 79.06 ⎯⎯→ u 4 = 487.75 kJ/kg v3 3

Process 4-1: v = constant heat rejection. q out = u 4 − u1 = 487.75 − 214.07 = 273.7 kJ/kg wnet = q in − q out = 609.8 − 273.7 = 336.1 kJ/kg wnet 336.1 kJ/kg = = 55.1% 609.8 kJ/kg q in

(c)

η th =

(d)

v max = v 1 = v min = v 2 = MEP =

(

)

RT1 0.287 kPa ⋅ m 3 /kg ⋅ K (300 K ) = = 0.879 m 3 /kg 98 kPa P1

v max r

⎛ 1 kPa ⋅ m 3 wnet wnet 336.1 kJ/kg ⎜ = = v 1 − v 2 v 1 (1 − 1 / r ) 0.879 m 3 /kg (1 − 1/9.2 ) ⎜⎝ 1 kJ

(

)

⎞ ⎟ = 429 kPa ⎟ ⎠

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-136

9-172 An ideal Otto cycle with air as the working fluid with a compression ratio of 9.2 is considered. The amount of heat transferred to the air, the net work output, the thermal efficiency, and the mean effective pressure are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats.

P

Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-2).

qin

Analysis (a) Process 1-2 is isentropic compression: ⎛v T2 = T1 ⎜⎜ 1 ⎝v 2

⎞ ⎟⎟ ⎠

k −1

= (300 K )(9.2 )

3 4 qout

2 0.4

1

= 728.8 K

v

⎛ 728.8 K ⎞ v T P2v 2 P1v 1 ⎟⎟(98 kPa ) = 2190 kPa = ⎯ ⎯→ P2 = 1 2 P1 = (9.2 )⎜⎜ v 2 T1 T2 T1 ⎝ 300 K ⎠

Process 2-3: v = constant heat addition. P3v 3 P2v 2 P = ⎯ ⎯→ T3 = 3 T2 = 2T2 = (2 )(728.8) = 1457.6 K T3 T2 P2 q in = u 3 − u 2 = cv (T3 − T2 ) = (0.718 kJ/kg ⋅ K )(1457.6 − 728.8)K = 523.3 kJ/kg

(b) Process 3-4: isentropic expansion. ⎛v T4 = T3 ⎜⎜ 3 ⎝v4

⎞ ⎟⎟ ⎠

k −1

⎛ 1 ⎞ = (1457.6 K )⎜ ⎟ ⎝ 9.2 ⎠

0.4

= 600.0 K

Process 4-1: v = constant heat rejection. q out = u 4 − u1 = cv (T4 − T1 ) = (0.718 kJ/kg ⋅ K )(600 − 300 )K = 215.4 kJ/kg wnet = q in − q out = 523.3 − 215.4 = 307.9 kJ/kg wnet 307.9 kJ/kg = = 58.8% 523.3 kJ/kg q in

(c)

η th =

(d)

v max = v 1 = v min = v 2 = MEP =

(

)

RT1 0.287 kPa ⋅ m 3 /kg ⋅ K (300 K ) = = 0.879 m 3 /kg P1 98 kPa

v max r

⎛ 1 kPa ⋅ m 3 wnet wnet 307.9 kJ/kg ⎜ = = v 1 − v 2 v 1 (1 − 1 / r ) 0.879 m 3 /kg (1 − 1/9.2 ) ⎜⎝ 1 kJ

(

)

⎞ ⎟ = 393 kPa ⎟ ⎠

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-137

9-173 An engine operating on the ideal diesel cycle with air as the working fluid is considered. The pressure at the beginning of the heat-rejection process, the net work per cycle, and the mean effective pressure are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The gas constant of air is R = 0.287 kJ/kg·K (Table A-1). The properties of air are given in Table A-17. Analysis (a) The compression and the cutoff ratios are

V 1200 cm r= 1 = = 16 V2 75 cm3

P 2

V 150 cm rc = 3 = =2 V 2 75 cm3

3

3

qin

4

Process 1-2: isentropic compression.

qout

T1 = 290 K ⎯ ⎯→ u1 = 206.91 kJ/kg

1

v r1 = 676.1 v r2 =

3

v

v2 1 1 v r = v r = (676.1) = 42.256 ⎯⎯→ T2 = 837.3 K v 1 1 r 1 16

h2 = 863.03 kJ/kg

Process 2-3: P = constant heat addition. P3v 3 P2v 2 v = ⎯ ⎯→ T3 = 3 T2 = 2T2 = (2 )(837.3) = 1674.6 K T3 T2 v2 ⎯ ⎯→ h3 = 1848.9 kJ/kg

v r3 = 5.002 Process 3-4: isentropic expansion.

v r4 =

r v4 v ⎛ 16 ⎞ v r 3 = 4 v r 3 = v r 3 = ⎜ ⎟(5.002) = 40.016 ⎯⎯→ T4 = 853.4 K v3 2v 2 2 ⎝ 2⎠

u4 = 636.00 kJ/kg

Process 4-1: v = constant heat rejection. ⎛ 853.4 K ⎞ P4v 4 P1v1 T ⎟⎟(100 kPa ) = 294.3 kPa = ⎯ ⎯→ P4 = 4 P1 = ⎜⎜ T4 T1 T1 ⎝ 290 K ⎠

(b)

m=

(

)

P1V 1 (100 kPa ) 0.0012 m 3 = = 1.442 × 10 −3 kg 3 RT1 0.287 kPa ⋅ m /kg ⋅ K (290 K )

(

( − u ) = (1.442 × 10

)

) kg )(636.00 − 206.91)kJ/kg = 0.619 kJ

Qin = m(h3 − h2 ) = 1.442 × 10 -3 kg (1848.9 − 863.08) = 1.422 kJ Qout = m(u 4

1

-3

W net = Qin − Qout = 1.422 − 0.619 = 0.803 kJ

(c)

MEP =

⎛ 1 kPa ⋅ m 3 W net W net 0.803 kJ ⎜ = = V1 −V 2 V1 (1 − 1 / r ) 0.0012m 3 (1 − 1/16 ) ⎜⎝ 1 kJ

(

)

⎞ ⎟ = 714 kPa ⎟ ⎠

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-138

9-174 An engine operating on the ideal diesel cycle with argon as the working fluid is considered. The pressure at the beginning of the heat-rejection process, the net work per cycle, and the mean effective pressure are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Argon is an ideal gas with constant specific heats. Properties The properties of argon at room temperature are cp = 0.5203 kJ/kg.K, cv = 0.3122 kJ/kg·K, R = 0.2081 kJ/kg·K and k = 1.667 (Table A-2). Analysis (a) The compression and the cutoff ratios are

V 1200 cm3 r= 1 = = 16 V2 75 cm3

P

V 150 cm3 rc = 3 = =2 V 2 75 cm3

2

Qin

3

Process 1-2: isentropic compression. ⎛V ⎞ T2 = T1⎜⎜ 2 ⎟⎟ ⎝ V1 ⎠

k −1

4 Qout

= (290 K )(16)0.667 = 1843 K

1

v

Process 2-3: P = constant heat addition. P3v 3 P2v 2 v = ⎯⎯→ T3 = 3 T2 = 2T2 = (2)(1843) = 3686 K T3 T2 v2

Process 3-4: isentropic expansion. ⎛V ⎞ T4 = T3 ⎜⎜ 3 ⎟⎟ ⎝ V4 ⎠

k −1

⎛ 2V ⎞ = T3 ⎜⎜ 2 ⎟⎟ ⎝ V4 ⎠

k −1

⎛2⎞ = T3 ⎜ ⎟ ⎝r⎠

k −1

⎛ 2⎞ = (3686 K )⎜ ⎟ ⎝ 16 ⎠

0.667

= 920.9 K

Process 4-1: v = constant heat rejection. ⎛ 920.9 K ⎞ P4v 4 P1v1 T ⎟⎟(100 kPa ) = 317.6 kPa = ⎯ ⎯→ P4 = 4 P1 = ⎜⎜ T4 T1 T1 ⎝ 290 K ⎠

(b)

m=

(

)

P1V1 (100 kPa ) 0.0012 m3 = = 1.988 × 10− 3 kg RT1 0.2081 kPa ⋅ m3 /kg ⋅ K (290 K )

(

)

( − T ) = (1.988 × 10

) kg )(0.3122 kJ/kg ⋅ K )(920.9 − 290 )K = 0.392 kJ

Qin = m(h3 − h2 ) = mc p (T3 − T2 ) = 1.988 × 10 -3 kg (0.5203 kJ/kg ⋅ K )(3686 − 1843)K = 1.906 kJ Qout = m(u 4 − u1 ) = mcv (T4

-3

1

W net = Qin − Qout = 1.906 − 0.392 = 1.514 kJ

(c)

MEP =

⎛ 1 kPa ⋅ m 3 W net W net 1.514 kJ ⎜ = = V1 −V 2 V1 (1 − 1 / r ) 0.0012 m 3 (1 − 1/16 ) ⎜⎝ 1 kJ

(

)

⎞ ⎟ = 1346 kPa ⎟ ⎠

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-139

9-175E An ideal dual cycle with air as the working fluid with a compression ratio of 12 is considered. The thermal efficiency of the cycle is to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 0.240 Btu/lbm.R, cv = 0.171 Btu/lbm.R, and k = 1.4 (Table A-2E). Analysis The mass of air is m=

(

)

P1V1 (14.7 psia ) 75/1728 ft 3 = = 3.132 × 10− 3 lbm RT1 0.3704 psia ⋅ ft 3 /lbm ⋅ R (550 R )

(

)

P

Process 1-2: isentropic compression. ⎛V ⎞ T2 = T1 ⎜⎜ 1 ⎟⎟ ⎝ V2 ⎠

k −1

x

1.1 Btu

0.3 Btu

2

= (550 R )(12)

0.4

3

4 Qout

= 1486 R

1

v

Process 2-x: v = constant heat addition, Q 2 − x ,in = m(u x − u 2 ) = mcv (T x − T2 )

(

)

0.3 Btu = 3.132 × 10 −3 lbm (0.171 Btu/lbm ⋅ R )(Tx − 1486 )R ⎯ ⎯→ T x = 2046 R

Process x-3: P = constant heat addition. Q x −3,in = m(h3 − h x ) = mc p (T3 − T x )

(

)

1.1 Btu = 3.132 × 10 −3 lbm (0.240 Btu/lbm ⋅ R )(T3 − 2046 )R ⎯ ⎯→ T3 = 3509 R P3V 3 PxV x T V 3509 R = ⎯ ⎯→ rc = 3 = 3 = = 1.715 T3 Tx V x T x 2046 R

Process 3-4: isentropic expansion. ⎛V T4 = T3 ⎜⎜ 3 ⎝V 4

⎞ ⎟⎟ ⎠

k −1

⎛ 1.715V 1 ⎞ ⎟⎟ = T3 ⎜⎜ ⎝ V4 ⎠

k −1

⎛ 1.715 ⎞ = T3 ⎜ ⎟ ⎝ r ⎠

k −1

⎛ 1.715 ⎞ = (3509 R )⎜ ⎟ ⎝ 12 ⎠

0.4

= 1611 R

Process 4-1: v = constant heat rejection. Qout = m(u 4 − u1 ) = mcv (T4 − T1 )

(

)

= 3.132 × 10 −3 lbm (0.171 Btu/lbm ⋅ R )(1611 − 550 )R = 0.568 Btu

η th = 1 −

Qout 0.568 Btu = 1− = 59.4% Qin 1.4 Btu

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-140

9-176 An ideal Stirling cycle with air as the working fluid is considered. The maximum pressure in the cycle and the net work output are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats.

qin = 900 kJ/kg

T

2

1

1800 K

Properties The properties of air at room temperature are R = 0.287 kJ/kg.K, cp = 1.005 kJ/kg.K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-2). Analysis (a) The entropy change during process 1-2 is s 2 − s1 =

350 K

4

q12 900 kJ/kg = = 0.5 kJ/kg ⋅ K 1800 K TH

3 qout s

and s 2 − s1 = cv ln

T2 T1

©0

+ Rln

v2 v v ⎯ ⎯→ 0.5 kJ/kg ⋅ K = (0.287 kJ/kg ⋅ K ) ln 2 ⎯ ⎯→ 2 = 5.710 v1 v1 v1

P3v 3 P1v 1 ⎛ 1800 K ⎞ v T v T ⎟⎟ = 5873 kPa = ⎯ ⎯→ P1 = P3 3 1 = P3 2 1 = (200 kPa )(5.710 )⎜⎜ T3 T1 v 1 T3 v 1 T3 ⎝ 350 K ⎠

(b)

⎛ T w net = η th q in = ⎜⎜1 − L ⎝ TH

⎞ ⎛ 350 K ⎞ ⎟⎟q in = ⎜⎜1 − ⎟⎟(900 kJ/kg ) = 725 kJ/kg ⎝ 1800 K ⎠ ⎠

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-141

9-177 A simple ideal Brayton cycle with air as the working fluid is considered. The changes in the net work output per unit mass and the thermal efficiency are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The properties of air are given in Table A-17.

T

Analysis The properties at various states are T1 = 300 K

⎯ ⎯→

3′

h1 = 300.19 kJ / kg

2′

Pr 1 = 1.386 T3 = 1300 K ⎯ ⎯→

h3 = 1395.97 kJ / kg Pr 3 = 330.9

3

qin

2 4 1

For rp = 6,

qout s

P ⎯→ h2 = 501.40 kJ/kg Pr2 = 2 Pr1 = (6 )(1.386) = 8.316 ⎯ P1 Pr4 =

P4 ⎛1⎞ ⎯→ h4 = 855.3 kJ/kg Pr = ⎜ ⎟(330.9) = 55.15 ⎯ P3 3 ⎝ 6 ⎠

q in = h3 − h2 = 1395.97 − 501.40 = 894.57 kJ/kg q out = h4 − h1 = 855.3 − 300.19 = 555.11 kJ/kg w net = q in − q out = 894.57 − 555.11 = 339.46 kJ/kg

η th =

wnet 339.46 kJ/kg = = 37.9% 894.57 kJ/kg q in

For rp = 12, Pr2 =

P2 ⎯→ h2 = 610.6 kJ/kg Pr = (12 )(1.386) = 16.63 ⎯ P1 1

Pr4 =

P4 ⎛ 1⎞ ⎯→ h4 = 704.6 kJ/kg Pr3 = ⎜ ⎟(330.9 ) = 27.58 ⎯ P3 ⎝ 12 ⎠

q in = h3 − h2 = 1395.97 − 610.60 = 785.37 kJ/kg q out = h4 − h1 = 704.6 − 300.19 = 404.41 kJ/kg w net = q in − q out = 785.37 − 404.41 = 380.96 kJ/kg

η th =

w net 380.96 kJ/kg = = 48.5% 785.37 kJ/kg q in

Thus, (a) (b)

Δwnet = 380.96 − 339.46 = 41.5 kJ/kg (increase) Δη th = 48.5% − 37.9% = 10.6%

(increase)

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-142

9-178 A simple ideal Brayton cycle with air as the working fluid is considered. The changes in the net work output per unit mass and the thermal efficiency are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are R = 0.287 kJ/kg.K, cp = 1.005 kJ/kg.K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-2). Analysis Processes 1-2 and 3-4 are isentropic. Therefore, For rp = 6, ⎛P T2 = T1 ⎜⎜ 2 ⎝ P1 ⎛P T4 = T3 ⎜⎜ 4 ⎝ P3

⎞ ⎟⎟ ⎠

(k −1) / k

= (300 K )(6)0.4/1.4 = 500.6 K

T 3′

⎞ ⎟ ⎟ ⎠

(k −1) / k

⎛1⎞ = (1300 K )⎜ ⎟ ⎝6⎠

0.4/1.4

= 779.1 K

2

q in = h3 − h2 = c p (T3 − T2 )

2

q out = h4 − h1 = c p (T4 − T1 )

1

= (1.005 kJ/kg ⋅ K )(1300 − 500.6 )K = 803.4 kJ/kg

3

qin

4 qout

= (1.005 kJ/kg ⋅ K )(779.1 − 300)K = 481.5 kJ/kg

wnet = q in − q out = 803.4 − 481.5 = 321.9 kJ/kg

η th =

wnet 321.9 kJ/kg = = 40.1% 803.4 kJ/kg q in

For rp = 12, (k −1) / k

⎛P T2 = T1 ⎜⎜ 2 ⎝ P1

⎞ ⎟⎟ ⎠

⎛P T4 = T3 ⎜⎜ 4 ⎝ P3

⎞ ⎟ ⎟ ⎠

(k −1) / k

= (300 K )(12)0.4/1.4 = 610.2 K ⎛1⎞ = (1300 K )⎜ ⎟ ⎝ 12 ⎠

0.4/1.4

= 639.2 K

q in = h3 − h2 = c p (T3 − T2 )

= (1.005 kJ/kg ⋅ K )(1300 − 610.2)K = 693.2 kJ/kg

q out = h4 − h1 = c p (T4 − T1 )

= (1.005 kJ/kg ⋅ K )(639.2 − 300)K = 340.9 kJ/kg

wnet = q in − q out = 693.2 − 340.9 = 352.3 kJ/kg

η th =

wnet 352.3 kJ/kg = = 50.8% 693.2 kJ/kg q in

Thus, (a)

Δwnet = 352.3 − 321.9 = 30.4 kJ/kg (increase)

(b)

Δη th = 50.8% − 40.1% = 10.7%

(increase)

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

s

9-143

9-179 A regenerative gas-turbine engine operating with two stages of compression and two stages of expansion is considered. The back work ratio and the thermal efficiency are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats.

T 6

Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-2).

5

Analysis The work inputs to each stage of compressor are identical, so are the work outputs of each stage of the turbine. ⎛P T4 s = T2 s = T1 ⎜⎜ 2 ⎝ P1

ηC

⎞ ⎟⎟ ⎠

(k −1) / k

4 = (300 K )(3.5)0.4/1.4 = 429.1 K

4

3

2s

T9 s = T7 s

ηT =

ε=

⎞ ⎟ ⎟ ⎠

(k −1) / k

⎛ 1 ⎞ = (1200 K )⎜ ⎟ ⎝ 3.5 ⎠

7 9 7s 9

2 1

h − h1 c p (T2 s − T1 ) = ⎯ ⎯→ T4 = T2 = T1 + (T2 s − T1 ) / η C = 2s h2 − h1 c p (T2 − T1 ) = 300 + (429.1 − 300) / (0.78) = 465.5 K ⎛P = T6 ⎜⎜ 7 ⎝ P6

8

s

0.4/1.4

= 838.9 K

c p (T6 − T7 ) h6 − h7 = ⎯ ⎯→ T9 = T7 = T6 − η T (T6 − T7 s ) h6 − h7 s c p (T6 − T7 s ) = 1200 − (0.86)(1200 − 838.9) = 889.5 K h5 − h4 c p (T5 − T4 ) = ⎯ ⎯→ T5 = T4 + ε (T9 − T4 ) h9 − h4 c p (T9 − T4 ) = 465.5 + (0.72)(889.5 − 465.5) = 770.8 K

wC,in = 2(h2 − h1 ) = 2c p (T2 − T1 ) = 2(1.005 kJ/kg ⋅ K )(465.5 − 300 )K = 332.7 kJ/kg wT,out = 2(h6 − h7 ) = 2c p (T6 − T7 ) = 2(1.005 kJ/kg ⋅ K )(1200 − 889.5)K = 624.1 kJ/kg

Thus, rbw =

wC,in wT,out

=

332.7 kJ/kg = 53.3% 624.1 kJ/kg

q in = (h6 − h5 ) + (h8 − h7 ) = c p [(T6 − T5 ) + (T8 − T7 )]

= (1.005 kJ/kg ⋅ K )[(1200 − 770.8) + (1200 − 889.5)]K = 743.4 kJ/kg

wnet = wT,out − wC,in = 624.1 − 332.7 = 291.4 kJ/kg

η th =

wnet 291.4 kJ/kg = = 39.2% q in 743.4 kJ/kg

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-144

9-180 EES Problem 9-179 is reconsidered. The effect of the isentropic efficiencies for the compressor and turbine and regenerator effectiveness on net work done and the heat supplied to the cycle is to be investigated. Also, the T-s diagram for the cycle is to be plotted. Analysis Using EES, the problem is solved as follows: "Input data" T[6] = 1200 [K] T[8] = T[6] Pratio = 3.5 T[1] = 300 [K] P[1]= 100 [kPa] T[3] = T[1] Eta_reg = 0.72 "Regenerator effectiveness" Eta_c =0.78 "Compressor isentorpic efficiency" Eta_t =0.86 "Turbien isentropic efficiency" "LP Compressor:" "Isentropic Compressor anaysis" s[1]=ENTROPY(Air,T=T[1],P=P[1]) s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor" P[2] = Pratio*P[1] s_s[2]=ENTROPY(Air,T=T_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exit" Eta_c = w_compisen_LP/w_comp_LP "compressor adiabatic efficiency, W_comp > W_compisen" "Conservation of energy for the LP compressor for the isentropic case: e_in - e_out = DELTAe=0 for steady-flow" h[1] + w_compisen_LP = h_s[2] h[1]=ENTHALPY(Air,T=T[1]) h_s[2]=ENTHALPY(Air,T=T_s[2]) "Actual compressor analysis:" h[1] + w_comp_LP = h[2] h[2]=ENTHALPY(Air,T=T[2]) s[2]=ENTROPY(Air,T=T[2], P=P[2]) "HP Compressor:" s[3]=ENTROPY(Air,T=T[3],P=P[3]) s_s[4]=s[3] "For the ideal case the entropies are constant across the HP compressor" P[4] = Pratio*P[3] P[3] = P[2] s_s[4]=ENTROPY(Air,T=T_s[4],P=P[4]) "T_s[4] is the isentropic value of T[4] at compressor exit" Eta_c = w_compisen_HP/w_comp_HP "compressor adiabatic efficiency, W_comp > W_compisen" "Conservation of energy for the compressor for the isentropic case: e_in - e_out = DELTAe=0 for steady-flow" h[3] + w_compisen_HP = h_s[4] h[3]=ENTHALPY(Air,T=T[3]) h_s[4]=ENTHALPY(Air,T=T_s[4]) "Actual compressor analysis:" h[3] + w_comp_HP = h[4] PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-145 h[4]=ENTHALPY(Air,T=T[4]) s[4]=ENTROPY(Air,T=T[4], P=P[4]) "Intercooling heat loss:" h[2] = q_out_intercool + h[3] "External heat exchanger analysis" "SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0 e_in - e_out =DELTAe_cv =0 for steady flow" h[4] + q_in_noreg = h[6] h[6]=ENTHALPY(Air,T=T[6]) P[6]=P[4]"process 4-6 is SSSF constant pressure" "HP Turbine analysis" s[6]=ENTROPY(Air,T=T[6],P=P[6]) s_s[7]=s[6] "For the ideal case the entropies are constant across the turbine" P[7] = P[6] /Pratio s_s[7]=ENTROPY(Air,T=T_s[7],P=P[7])"T_s[7] is the isentropic value of T[7] at HP turbine exit" Eta_t = w_turb_HP /w_turbisen_HP "turbine adiabatic efficiency, w_turbisen > w_turb" "SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0 e_in -e_out = DELTAe_cv = 0 for steady-flow" h[6] = w_turbisen_HP + h_s[7] h_s[7]=ENTHALPY(Air,T=T_s[7]) "Actual Turbine analysis:" h[6] = w_turb_HP + h[7] h[7]=ENTHALPY(Air,T=T[7]) s[7]=ENTROPY(Air,T=T[7], P=P[7]) "Reheat Q_in:" h[7] + q_in_reheat = h[8] h[8]=ENTHALPY(Air,T=T[8]) "HL Turbine analysis" P[8]=P[7] s[8]=ENTROPY(Air,T=T[8],P=P[8]) s_s[9]=s[8] "For the ideal case the entropies are constant across the turbine" P[9] = P[8] /Pratio s_s[9]=ENTROPY(Air,T=T_s[9],P=P[9])"T_s[9] is the isentropic value of T[9] at LP turbine exit" Eta_t = w_turb_LP /w_turbisen_LP "turbine adiabatic efficiency, w_turbisen > w_turb" "SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0 e_in -e_out = DELTAe_cv = 0 for steady-flow" h[8] = w_turbisen_LP + h_s[9] h_s[9]=ENTHALPY(Air,T=T_s[9]) "Actual Turbine analysis:" h[8] = w_turb_LP + h[9] h[9]=ENTHALPY(Air,T=T[9]) s[9]=ENTROPY(Air,T=T[9], P=P[9]) "Cycle analysis" w_net=w_turb_HP+w_turb_LP - w_comp_HP - w_comp_LP q_in_total_noreg=q_in_noreg+q_in_reheat Eta_th_noreg=w_net/(q_in_total_noreg)*Convert(, %) "[%]" "Cycle thermal efficiency" PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-146 Bwr=(w_comp_HP + w_comp_LP)/(w_turb_HP+w_turb_LP)"Back work ratio" "With the regenerator, the heat added in the external heat exchanger is" h[5] + q_in_withreg = h[6] h[5]=ENTHALPY(Air, T=T[5]) s[5]=ENTROPY(Air,T=T[5], P=P[5]) P[5]=P[4] "The regenerator effectiveness gives h[5] and thus T[5] as:" Eta_reg = (h[5]-h[4])/(h[9]-h[4]) "Energy balance on regenerator gives h[10] and thus T[10] as:" h[4] + h[9]=h[5] + h[10] h[10]=ENTHALPY(Air, T=T[10]) s[10]=ENTROPY(Air,T=T[10], P=P[10]) P[10]=P[9] "Cycle thermal efficiency with regenerator" q_in_total_withreg=q_in_withreg+q_in_reheat Eta_th_withreg=w_net/(q_in_total_withreg)*Convert(, %) "[%]" "The following data is used to complete the Array Table for plotting purposes." s_s[1]=s[1] T_s[1]=T[1] s_s[3]=s[3] T_s[3]=T[3] s_s[5]=ENTROPY(Air,T=T[5],P=P[5]) T_s[5]=T[5] s_s[6]=s[6] T_s[6]=T[6] s_s[8]=s[8] T_s[8]=T[8] s_s[10]=s[10] T_s[10]=T[10]

ηC

ηreg

ηt

0.78 0.78 0.78 0.78 0.78 0.78 0.78 0.78 0.78

0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1

0.86 0.86 0.86 0.86 0.86 0.86 0.86 0.86 0.86

ηth,noreg [%] 27.03 27.03 27.03 27.03 27.03 27.03 27.03 27.03 27.03

ηth,withreg [%] 36.59 37.7 38.88 40.14 41.48 42.92 44.45 46.11 47.88

qin,total,noreg [kJ/kg] 1130 1130 1130 1130 1130 1130 1130 1130 1130

qin,total,withreg [kJ/kg] 834.6 810 785.4 760.8 736.2 711.6 687 662.4 637.8

wnet [kJ/kg] 305.4 305.4 305.4 305.4 305.4 305.4 305.4 305.4 305.4

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-147

Air kP 10 0

8

kP a

35 0

6

1200

T [K]

12 25

1400

a

kP a

1600

1000 5

600

4

9

7

800 2

10

400 1

3 200 4.5

5.0

5.5

6.0

6.5

7.0

7.5

s [kJ/kg-K] 50

η

45

η

40

reg c

= 0.72

= 0.78

W ith regeneration 35

η th

30 25 20

No regeneration

15 10 0.6

0.65

0.7

0.75

0.8

η

0.85

0.9

0.95

1

t

1200 1100

q in,total

1000 900

η

reg

η

No regeneration

c

= 0.72 = 0.78

W ith regeneration

800 700 600 0.6

0.65

0.7

0.75

0.8

η

0.85

0.9

0.95

1

t

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9-148

450 400

η

reg

η

w net [kJ/kg]

350

c

= 0.72 = 0.78

300 250 200 150 100 0.6

0.65

0.7

0.75

0.8

η

0.85

0.9

0.95

1

t

50

η 45

= 0.78

c

η

t

= 0.86

η th

40

W ith regenertion

35

No regeneration 30

25 0.6

0.65

0.7

0.75

0.8

η

0.85

0.9

0.95

1

reg

50 45 40

η η

reg t

= 0.72

= 0.86 W ith regeneration

η th

35 30 25

No regeneration

20 15 0.6

0.65

0.7

0.75

0.8

η

0.85

0.9

0.95

1

c

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-149

9-181 A regenerative gas-turbine engine operating with two stages of compression and two stages of expansion is considered. The back work ratio and the thermal efficiency are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Helium is an ideal gas with constant specific heats.

T

Properties The properties of helium at room temperature are cp = 5.1926 kJ/kg.K and k = 1.667 (Table A-2).

qin 5

Analysis The work inputs to each stage of compressor are identical, so are the work outputs of each stage of the turbine. ⎛P T4 s = T2 s = T1 ⎜⎜ 2 ⎝ P1

ηC

⎞ ⎟⎟ ⎠

(k −1) / k

4 = (300 K )(3.5)0.667/1.667 = 495.2 K

4

3

2s

T9 s = T7 s

ηT =

ε=

⎞ ⎟ ⎟ ⎠

(k −1) / k

⎛ 1 ⎞ = (1200 K )⎜ ⎟ ⎝ 3.5 ⎠

8

7 9 7s 9

2 1

h − h1 c p (T2 s − T1 ) = ⎯ ⎯→ T4 = T2 = T1 + (T2 s − T1 ) / η C = 2s h2 − h1 c p (T2 − T1 ) = 300 + (495.2 − 300) / (0.78) = 550.3 K ⎛P = T6 ⎜⎜ 7 ⎝ P6

6

s

0.667/1.667

= 726.9 K

c p (T6 − T7 ) h6 − h7 = ⎯ ⎯→ T9 = T7 = T6 − η T (T6 − T7 s ) h6 − h7 s c p (T6 − T7 s ) = 1200 − (0.86)(1200 − 726.9) = 793.1 K h5 − h4 c p (T5 − T4 ) = ⎯ ⎯→ T5 = T4 + ε (T9 − T4 ) h9 − h4 c p (T9 − T4 ) = 550.3 + (0.72)(793.1 − 550.3) = 725.1 K

wC,in = 2(h2 − h1 ) = 2c p (T2 − T1 ) = 2(5.1926 kJ/kg ⋅ K )(550.3 − 300 )K = 2599.4 kJ/kg wT,out = 2(h6 − h7 ) = 2c p (T6 − T7 ) = 2(5.1926 kJ/kg ⋅ K )(1200 − 793.1)K = 4225.7 kJ/kg

Thus, rbw =

wC,in wT,out

=

2599.4 kJ/kg = 61.5% 4225.7 kJ/kg

q in = (h6 − h5 ) + (h8 − h7 ) = c p [(T6 − T5 ) + (T8 − T7 )]

= (5.1926 kJ/kg ⋅ K )[(1200 − 725.1) + (1200 − 793.1)]K = 4578.8 kJ/kg

wnet = wT,out − wC,in = 4225.7 − 2599.4 = 1626.3 kJ/kg

η th =

wnet 1626.3 kJ/kg = = 35.5% q in 4578.8 kJ/kg

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-150

9-182 An ideal gas-turbine cycle with one stage of compression and two stages of expansion and regeneration is considered. The thermal efficiency of the cycle as a function of the compressor pressure ratio and the high-pressure turbine to compressor inlet temperature ratio is to be determined, and to be compared with the efficiency of the standard regenerative cycle. Analysis The T-s diagram of the cycle is as shown in the figure. If the overall pressure ratio of the cycle is rp, which is the pressure ratio across the compressor, then the pressure ratio across each turbine stage in the ideal case becomes √ rp. Using the isentropic relations, the temperatures at the compressor and turbine exit can be expressed as ⎛P ⎞ T5 = T2 = T1⎜⎜ 2 ⎟⎟ ⎝ P1 ⎠

(k −1) / k

⎛P ⎞ T7 = T4 = T3 ⎜⎜ 4 ⎟⎟ ⎝ P3 ⎠ ⎛P ⎞ T6 = T5 ⎜⎜ 6 ⎟⎟ ⎝ P5 ⎠

(k −1) / k

Then,

3

7 4

( )

⎛ 1 ⎞ ⎟ = T3 ⎜ ⎜ r ⎟ ⎝ p⎠

⎛ 1 ⎞ ⎟ = T5 ⎜ ⎜ r ⎟ ⎝ p⎠

qin

(k −1) / k

= T1 rp (k −1) / k

T

2

5

1

6 qout

(k −1) / k

= T3rp

(1− k ) / 2 k

s

(k −1) / k

= T2 rp (1− k ) / 2 k = T1rp (k −1) / k rp (1− k ) / 2 k = T1rp (k −1) / 2 k

( − T ) = c T (r (

) − 1)

q in = h3 − h7 = c p (T3 − T7 ) = c p T3 1 − r p (1− k ) / 2 k q out = h6 − h1 = c p (T6

and thus

η th = 1 −

1

( (

p 1

p

k −1) / 2 k

) )

c p T1 r p (k −1) / 2 k − 1 q out = 1− q in c p T3 1 − r p (1− k ) / 2 k

which simplifies to

η th = 1 −

T1 (k −1) / 2 k rp T3

The thermal efficiency of the single stage ideal regenerative cycle is given as

η th = 1 −

T1 (k −1) / k rp T3

Therefore, the regenerative cycle with two stages of expansion has a higher thermal efficiency than the standard regenerative cycle with a single stage of expansion for any given value of the pressure ratio rp.

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-151

9-183 A gas-turbine plant operates on the regenerative Brayton cycle with reheating and intercooling. The back work ratio, the net work output, the thermal efficiency, the second-law efficiency, and the exergies at the exits of the combustion chamber and the regenerator are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The gas constant of air is R = 0.287 kJ/kg·K. Analysis (a) For this problem, we use the properties from EES software. Remember that for an ideal gas, enthalpy is a function of temperature only whereas entropy is functions of both temperature and pressure.

Optimum intercooling and reheating pressure is P2 = P1 P4 = (100)(1200) = 346.4 kPa

T

Process 1-2, 3-4: Compression ⎯→ h1 = 300.43 kJ/kg T1 = 300 K ⎯ T1 = 300 K ⎫ ⎬s1 = 5.7054 kJ/kg ⋅ K P1 = 100 kPa ⎭ P2 = 346.4 kPa ⎫ ⎬h2 s = 428.79 kJ/kg s 2 = s1 = 5.7054 kJ/kg.K ⎭

ηC =

5

qin

6 8 6s 8s

9 4 2 4s 2s 3

7

10

1 s

h2 s − h1 428.79 − 300.43 ⎯ ⎯→ h2 = 460.88 kJ/kg ⎯ ⎯→ 0.80 = h2 − h1 h2 − 300.43

⎯→ h3 = 350.78 kJ/kg T3 = 350 K ⎯ T3 = 350 K ⎫ ⎬s 3 = 5.5040 kJ/kg ⋅ K P3 = 346.4 kPa ⎭ P4 = 1200 kPa ⎫ ⎬h4 s = 500.42 kJ/kg s 4 = s 3 = 5.5040 kJ/kg.K ⎭

ηC =

h 4 s − h3 500.42 − 350.78 ⎯ ⎯→ h4 = 537.83 kJ/kg ⎯ ⎯→ 0.80 = h4 − h3 h4 − 350.78

Process 6-7, 8-9: Expansion ⎯→ h6 = 1514.9 kJ/kg T6 = 1400 K ⎯ T6 = 1400 K ⎫ ⎬s 6 = 6.6514 kJ/kg ⋅ K P6 = 1200 kPa ⎭ P7 = 346.4 kPa ⎫ ⎬h7 s = 1083.9 kJ/kg s 7 = s 6 = 6.6514 kJ/kg.K ⎭

ηT =

1514.9 − h7 h6 − h7 ⎯ ⎯→ h7 = 1170.1 kJ/kg ⎯ ⎯→ 0.80 = 1514.9 − 1083.9 h6 − h7 s

⎯→ h8 = 1395.6 kJ/kg T8 = 1300 K ⎯ T8 = 1300 K ⎫ ⎬s 8 = 6.9196 kJ/kg ⋅ K P8 = 346.4 kPa ⎭ PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-152

P9 = 100 kPa ⎫ ⎬h9 s = 996.00 kJ/kg s 9 = s8 = 6.9196 kJ/kg.K ⎭

ηT =

1395.6 − h9 h8 − h9 ⎯ ⎯→ h9 = 1075.9 kJ/kg ⎯ ⎯→ 0.80 = 1395.6 − 996.00 h8 − h9 s

Cycle analysis: wC,in = h2 − h1 + h4 − h3 = 460.88 − 300.43 + 537.83 − 350.78 = 347.50 kJ/kg

wT,out = h6 − h7 + h8 − h9 = 1514.9 − 1170.1 + 1395.6 − 1075.9 = 664.50 kJ/kg rbw =

wC,in wT, out

=

347.50 = 0.523 664.50

wnet = wT,out − wC,in = 664.50 − 347.50 = 317.0 kJ/kg Regenerator analysis:

ε regen =

1075.9 − h10 h9 − h10 ⎯ ⎯→ h10 = 672.36 kJ/kg ⎯ ⎯→ 0.75 = 1075.9 − 537.83 h9 − h4

h10 = 672.36 K ⎫ ⎬s10 = 6.5157 kJ/kg ⋅ K P10 = 100 kPa ⎭ q regen = h9 − h10 = h5 − h4 ⎯ ⎯→ 1075.9 − 672.36 = h5 − 537.83 ⎯ ⎯→ h5 = 941.40 kJ/kg

(b)

q in = h6 − h5 = 1514.9 − 941.40 = 573.54 kJ/kg

η th =

wnet 317.0 = = 0.553 q in 573.54

(c) The second-law efficieny of the cycle is defined as the ratio of actual thermal efficiency to the maximum possible thermal efficiency (Carnot efficiency). The maximum temperature for the cycle can be taken to be the turbine inlet temperature. That is,

η max = 1 −

T1 300 K =1− = 0.786 T6 1400 K

and

η II =

η th 0.553 = = 0.704 η max 0.786

(d) The exergies at the combustion chamber exit and the regenerator exit are x 6 = h 6 − h 0 − T0 ( s 6 − s 0 ) = (1514.9 − 300.43)kJ/kg − (300 K )(6.6514 − 5.7054)kJ/kg.K = 930.7 kJ/kg x10 = h10 − h0 − T0 ( s10 − s 0 ) = (672.36 − 300.43)kJ/kg − (300 K )(6.5157 − 5.7054)kJ/kg.K = 128.8 kJ/kg

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9-153

9-184 The thermal efficiency of a two-stage gas turbine with regeneration, reheating and intercooling to that of a three-stage gas turbine is to be compared. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are cp = 1.005 kJ/kg⋅K and k = 1.4 (Table A-2a). Analysis

Two Stages:

T

The pressure ratio across each stage is 873 K

r p = 16 = 4

6

8

7

9

5

The temperatures at the end of compression and expansion are Tc = Tmin r p( k −1) / k = (283 K)(4) 0.4/1.4 = 420.5 K ⎛ 1 Te = Tmax ⎜ ⎜ rp ⎝

⎞ ⎟ ⎟ ⎠

( k −1) / k

283 K ⎛1⎞ = (873 K)⎜ ⎟ ⎝4⎠

0.4/1.4

4

2

3

1

10

= 587.5 K

s

The heat input and heat output are q in = 2c p (Tmax − Te ) = 2(1.005 kJ/kg ⋅ K)(873 − 587.5) K = 573.9 kJ/kg

q out = 2c p (Tc − Tmin ) = 2(1.005 kJ/kg ⋅ K)(420.5 − 283) K = 276.4 kJ/kg

The thermal efficiency of the cycle is then

η th = 1 −

q out 276.4 = 1− = 0.518 q in 573.9

Three Stages:

T

The pressure ratio across each stage is r p = 16

1/ 3

8

= 2.520

7 9 11

The temperatures at the end of compression and expansion are Tc = Tmin r p( k −1) / k = (283 K)(2.520) 0.4/1.4 = 368.5 K ⎛ 1 Te = Tmax ⎜ ⎜ rp ⎝

⎞ ⎟ ⎟ ⎠

( k −1) / k

10

⎛ 1 ⎞ = (873 K)⎜ ⎟ ⎝ 2.520 ⎠

0.4/1.4

= 670.4 K

6 5

4

2

3

1

12 13

14 s

The heat input and heat output are q in = 3c p (Tmax − Te ) = 3(1.005 kJ/kg ⋅ K)(873 − 670.4) K = 610.8 kJ/kg q out = 3c p (Tc − Tmin ) = 3(1.005 kJ/kg ⋅ K)(368.5 − 283) K = 257.8 kJ/kg

The thermal efficiency of the cycle is then

η th = 1 −

q out 257.8 = 1− = 0.578 q in 610.8

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-154

9-185E A pure jet engine operating on an ideal cycle is considered. The thrust force produced per unit mass flow rate is to be determined. Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with constant specific heats at room temperature. 4 The turbine work output is equal to the compressor work input. Properties The properties of air at room temperature are R = 0.3704 psia⋅ft3/lbm⋅R (Table A-1E), cp = 0.24 Btu/lbm⋅R and k = 1.4 (Table A-2Ea). Analysis Working across the two isentropic processes of the cycle yields

T2 = T1 r p( k −1) / k = (490 R)(9) 0.4/1.4 = 918.0 R ⎛ 1 T5 = T3 ⎜ ⎜ rp ⎝

⎞ ⎟ ⎟ ⎠

( k −1) / k

⎛1⎞ = (1160 R)⎜ ⎟ ⎝9⎠

T

0.4/1.4

3

qin

= 619.2 R

4 2

The work input to the compressor is

5

wC = c p (T2 − T1 ) = (0.24 Btu/lbm ⋅ R)(918.0 - 490)R = 102.7 Btu/lbm

An energy balance gives the excess enthalpy to be

1

qout s

Δh = c p (T3 − T5 ) − wC = (0.24 Btu/lbm ⋅ R)(1160 − 619.2)R − 102.7 Btu/lbm = 27.09 Btu/lbm

The velocity of the air at the engine exit is determined from Δh =

2 2 Vexit − Vinlet 2

Rearranging,

(

2 Vexit = 2Δh + Vinlet

)

1/ 2

⎡ ⎛ 25,037 ft 2 /s 2 = ⎢2(27.09 Btu/lbm)⎜ ⎜ 1 Btu/lbm ⎢⎣ ⎝ = 1672 ft/s

⎤ ⎞ ⎟ + (1200 ft/s) 2 ⎥ ⎟ ⎥⎦ ⎠

1/ 2

The specific impulse is then F = V exit − Vinlet = 1672 − 1200 = 472 m/s m&

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9-155

9-186 A simple ideal Brayton cycle with air as the working fluid operates between the specified temperature limit. The net work is to be determined using constant and variable specific heats. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2a). Analysis (a) Constant specific heats: T2 = T1 r p( k −1) / k = (273 K)(15) 0.4/1.4 = 591.8 K ⎛ 1 T4 = T3 ⎜ ⎜ rp ⎝

⎞ ⎟ ⎟ ⎠

T

( k −1) / k

⎛1⎞ = (873 K)⎜ ⎟ ⎝ 15 ⎠

0.4/1.4

= 402.7 K

3

873 K 2

wnet = w turb − wcomp = c p (T3 − T4 ) − c p (T2 − T1 ) = c p (T3 − T4 + T1 − T2 ) = (1.005 kJ/kg ⋅ K )(873 − 402.7 + 273 − 591.8)K = 152.3 kJ/kg

qin

273

4 1

qout s

(b) Variable specific heats: (using air properties from Table A-17) T1 = 273 K ⎯ ⎯→ Pr 2 =

h1 = 273.12 kJ/kg Pr1 = 0.9980

P2 Pr1 = (15)(0.9980) = 14.97 ⎯ ⎯→ h2 = 592.69 kJ/kg P1

⎯→ T3 = 873 K ⎯ Pr 4 =

h3 = 902.76 kJ/kg Pr 3 = 66.92

P4 ⎛1⎞ ⎯→ h4 = 419.58 kJ/kg Pr 3 = ⎜ ⎟(66.92) = 4.461 ⎯ P3 ⎝ 15 ⎠

wnet = w turb − wcomp = (h3 − h4 ) − (h2 − h1 ) = (902.76 − 419.58) − (592.69 − 273.12) = 163.6 kJ/kg

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9-156

9-187 An Otto cycle with a compression ratio of 8 is considered. The thermal efficiency is to be determined using constant and variable specific heats. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are R = 0.287 kPa·m3/kg·K, cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-2a). Analysis (a) Constant specific heats:

η th = 1 −

1 r

k −1

1

= 1− 8

1.4 −1

P = 0.5647

(b) Variable specific heats: (using air properties from Table A-17) Process 1-2: isentropic compression. ⎯→ T1 = 283 K ⎯

v r2 =

u1 = 201.9 kJ/kg

v r1 = 718.9

3 4 2 1

v

v2 1 1 v r 2 = v r 2 = (718.9) = 89.86 ⎯⎯→ u 2 = 463.76 kJ/kg r 8 v1

Process 2-3: v = constant heat addition. T3 = 1173 K ⎯ ⎯→

u 3 = 909.39 kJ/kg

v r 3 = 15.529

q in = u 3 − u 2 = 909.39 − 463.76 = 445.63 kJ/kg

Process 3-4: isentropic expansion.

v r4 =

v4 v r 3 = rv r 3 = (8)(15.529) = 124.2 ⎯⎯→ u 4 = 408.06 kJ/kg v3

Process 4-1: v = constant heat rejection. q out = u 4 − u1 = 408.06 − 201.9 = 206.16 kJ/kg

η th = 1 −

q out 206.16 kJ/kg = 1− = 0.5374 445.63 kJ/kg q in

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9-157

9-188 An ideal diesel engine with air as the working fluid has a compression ratio of 22. The thermal efficiency is to be determined using constant and variable specific heats. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4 (Table A-2a). Analysis (a) Constant specific heats: P qin Process 1-2: isentropic compression. 2 3 k −1

⎛V ⎞ T2 = T1 ⎜⎜ 1 ⎟⎟ = (288 K)(22) 0.4 = 991.7 K V ⎝ 2⎠ Process 2-3: P = constant heat addition. V P3V 3 P2V 2 T 1473 K = ⎯ ⎯→ 3 = 3 = = 1.485 T3 T2 V 2 T2 991.7 K

4 qout 1

v

Process 3-4: isentropic expansion. ⎛V T4 = T3 ⎜⎜ 3 ⎝V 4

⎞ ⎟⎟ ⎠

k −1

⎛ 1.485V 2 = T3 ⎜⎜ ⎝ V4

⎞ ⎟⎟ ⎠

k −1

⎛ 1.485 ⎞ = T3 ⎜ ⎟ ⎝ r ⎠

k −1

⎛ 1.485 ⎞ = (1473 K)⎜ ⎟ ⎝ 22 ⎠

0.4

= 501.1 K

q in = h3 − h2 = c p (T3 − T2 ) = (1.005 kJ/kg ⋅ K )(1473 − 991.7 )K = 483.7 kJ/kg q out = u 4 − u1 = cv (T4 − T1 ) = (0.718 kJ/kg ⋅ K )(501.1 − 288)K = 153.0 kJ/kg w net,out = q in − q out = 483.7 − 153.0 = 330.7 kJ/kg

η th =

w net,out q in

=

330.7 kJ/kg = 0.698 473.7 kJ/kg

(b) Variable specific heats: (using air properties from Table A-17) Process 1-2: isentropic compression. u1 = 205.48 kJ/kg ⎯→ T1 = 288 K ⎯ v r1 = 688.1

v r2 =

T = 929.2 K v2 1 1 v r1 = v r1 = (688.1) = 31.28 ⎯ ⎯→ 2 h2 = 965.73 kJ/kg r 22 v1

Process 2-3: P = constant heat addition. v P3v 3 P2v 2 T 1473 K = ⎯ ⎯→ 3 = 3 = = 1.585 T3 T2 v 2 T2 929.2 K T3 = 1473 K ⎯ ⎯→

h3 = 1603.33 kJ/kg

v r 3 = 7.585

q in = h3 − h2 = 1603.33 − 965.73 = 637.6 kJ/kg

Process 3-4: isentropic expansion.

v r4 =

v4 v4 r 22 v r3 = v r3 = v r3 = (7.585) = 105.3 ⎯ ⎯→ u 4 = 435.61 kJ/kg 1.585v 2 1.585 1.585 v3

Process 4-1: v = constant heat rejection. q out = u 4 − u1 = 435.61 − 205.48 = 230.13 kJ/kg Then

η th = 1 −

230.13 kJ/kg qout = 1− = 0.639 637.6 kJ/kg qin

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9-158

9-189 The electricity and the process heat requirements of a manufacturing facility are to be met by a cogeneration plant consisting of a gas-turbine and a heat exchanger for steam production. The mass flow rate of the air in the cycle, the back work ratio, the thermal efficiency, the rate at which steam is produced in the heat exchanger, and the utilization efficiency of the cogeneration plant are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Analysis (a) For this problem, we use the properties of air from EES software. Remember that for an ideal gas, enthalpy is a function of temperature only whereas entropy is functions of both temperature and pressure.

350°C

25°C

5 Combustion chamber

Heat exchanger

3 2

1.2 MPa

Compress.

1

Process 1-2: Compression

4 Turbine

500°C Sat. vap. 200°C

100 kPa 30°C

⎯→ h1 = 303.60 kJ/kg T1 = 30°C ⎯ T1 = 30°C

⎫ ⎬s1 = 5.7159 kJ/kg ⋅ K P1 = 100 kPa ⎭

P2 = 1200 kPa

⎫ ⎬h2 s = 617.37 kJ/kg s 2 = s1 = 5.7159 kJ/kg.K ⎭

ηC =

h2 s − h1 617.37 − 303.60 ⎯ ⎯→ 0.82 = ⎯ ⎯→ h2 = 686.24 kJ/kg h2 − h1 h2 − 303.60

Process 3-4: Expansion T4 = 500°C ⎯ ⎯→ h4 = 792.62 kJ/kg

ηT =

h3 − h4 h − 792.62 ⎯ ⎯→ 0.82 = 3 h3 − h4 s h3 − h4 s

We cannot find the enthalpy at state 3 directly. However, using the following lines in EES together with the isentropic efficiency relation, we find h3 = 1404.7 kJ/kg, T3 = 1034ºC, s3 = 6.5699 kJ/kg.K. The solution by hand would require a trial-error approach. h_3=enthalpy(Air, T=T_3) s_3=entropy(Air, T=T_3, P=P_2) h_4s=enthalpy(Air, P=P_1, s=s_3)

Also, T5 = 350°C ⎯ ⎯→ h5 = 631.44 kJ/kg

The inlet water is compressed liquid at 25ºC and at the saturation pressure of steam at 200ºC (1555 kPa). This is not available in the tables but we can obtain it in EES. The alternative is to use saturated liquid enthalpy at the given temperature.

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9-159

Tw1 = 25°C

⎫ ⎬hw1 = 106.27 kJ/kg P1 = 1555 kPa ⎭

Tw2 = 200°C⎫ ⎬hw2 = 2792.0 kJ/kg x2 = 1 ⎭

The net work output is wC,in = h2 − h1 = 686.24 − 303.60 = 382.64 kJ/kg wT,out = h3 − h4 = 1404.7 − 792.62 = 612.03 kJ/kg wnet = wT,out − wC,in = 612.03 − 382.64 = 229.39 kJ/kg

The mass flow rate of air is W& net 800 kJ/s = = 3.487 kg/s wnet 229.39 kJ/kg

m& a =

(b) The back work ratio is rbw =

wC,in wT, out

=

382.64 = 0.625 612.03

The rate of heat input and the thermal efficiency are Q& in = m& a (h3 − h2 ) = (3.487 kg/s)(1404.7 − 686.24)kJ/kg = 2505 kW

η th =

W& net 800 kW = = 0.319 & 2505 kW Qin

(c) An energy balance on the heat exchanger gives m& a (h4 − h5 ) = m& w (hw2 − hw1 ) (3.487 kg/s)(792.62 − 631.44)kJ/kg = m& w (2792.0 − 106.27)kJ/kg ⎯ ⎯→ m& w = 0.2093 kg/s

(d) The heat supplied to the water in the heat exchanger (process heat) and the utilization efficiency are Q& p = m& w (hw 2 − hw1 ) = (0.2093 kg/s)(2792.0 − 106.27)kJ/kg = 562.1 kW

εu =

W& net + Q& p 800 + 562.1 = = 0.544 2505 kW Q& in

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9-160

9-190 A turbojet aircraft flying is considered. The pressure of the gases at the turbine exit, the mass flow rate of the air through the compressor, the velocity of the gases at the nozzle exit, the propulsive power, and the propulsive efficiency of the cycle are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The gas constant of air is R = 0.287 kJ/kg·K (Table A-1). Analysis (a) For this problem, we use the properties from EES software. Remember that for an ideal gas, enthalpy is a function of temperature only whereas entropy is functions of both temperature and pressure.

T

Diffuser, Process 1-2: T1 = −35°C ⎯ ⎯→ h1 = 238.23 kJ/kg

4

qin

5 3 2 1

6 qout

s V12 V22 h1 + = h2 + 2 2 (900/3.6 m/s) 2 ⎛ 1 kJ/kg ⎞ (15 m/s) 2 ⎛ 1 kJ/kg ⎞ ⎯→ h2 = 269.37 kJ/kg (238.23 kJ/kg) + ⎜ ⎟ = h2 + ⎜ ⎟⎯ 2 2 2 2 2 2 ⎝ 1000 m /s ⎠ ⎝ 1000 m /s ⎠ h2 = 269.37 kJ/kg ⎫ ⎬s 2 = 5.7951 kJ/kg ⋅ K P2 = 50 kPa ⎭

Compressor, Process 2-3: P3 = 450 kPa ⎫ ⎬h3s = 505.19 kJ/kg s 3 = s 2 = 5.7951 kJ/kg.K ⎭

ηC =

h3s − h2 505.19 − 269.37 ⎯ ⎯→ 0.83 = ⎯ ⎯→ h3 = 553.50 kJ/kg h3 − h2 h3 − 269.37

Turbine, Process 3-4: T4 = 950°C ⎯⎯→ h4 = 1304.8 kJ/kg h3 − h2 = h4 − h5 ⎯ ⎯→ 553.50 − 269.37 = 1304.8 − h5 ⎯ ⎯→ h5 = 1020.6 kJ/kg

where the mass flow rates through the compressor and the turbine are assumed equal.

ηT =

h4 − h5 1304.8 − 1020.6 ⎯ ⎯→ 0.83 = ⎯ ⎯→ h5 s = 962.45 kJ/kg h4 − h5 s 1304.8 − h5 s

T4 = 950°C ⎫ ⎬s 4 = 6.7725 kJ/kg ⋅ K P4 = 450 kPa ⎭ h5 s = 962.45 kJ/kg ⎫ ⎬ P5 = 147.4 kPa s 5 = s 4 = 6.7725 kJ/kg ⋅ K ⎭

(b) The mass flow rate of the air through the compressor is m& =

W& C 500 kJ/s = = 1.760 kg/s h3 − h2 (553.50 − 269.37) kJ/kg

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9-161

(c) Nozzle, Process 5-6: h5 = 1020.6 kJ/kg ⎫ ⎬s 5 = 6.8336 kJ/kg ⋅ K P5 = 147.4 kPa ⎭ P6 = 40 kPa ⎫ ⎬h6 s = 709.66 kJ/kg s 6 = s 5 = 6.8336 kJ/kg.K ⎭

ηN =

h5 − h6 1020.6 − h6 ⎯ ⎯→ 0.83 = ⎯ ⎯→ h6 = 762.52 kJ/kg h5 − h6 s 1020.6 − 709.66 h5 +

V52 V2 = h6 + 6 2 2

(1020.6 kJ/kg) + 0 = 762.52 kJ/kg +

V62 ⎛ 1 kJ/kg ⎞ ⎯→V6 = 718.5 m/s ⎜ ⎟⎯ 2 ⎝ 1000 m 2 /s 2 ⎠

where the velocity at nozzle inlet is assumed zero. (d) The propulsive power and the propulsive efficiency are ⎛ 1 kJ/kg ⎞ W& p = m& (V6 − V1 )V1 = (1.76 kg/s)(718.5 m/s − 250 m/s)(250 m/s)⎜ ⎟ = 206.1 kW 2 2 ⎝ 1000 m /s ⎠ Q& in = m& (h4 − h3 ) = (1.76 kg/s)(1304.8 − 553.50)kJ/kg = 1322 kW

ηp =

W& p 206.1 kW = = 0.156 Q& in 1322 kW

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9-162

9-191 EES The effects of compression ratio on the net work output and the thermal efficiency of the Otto cycle for given operating conditions is to be investigated. Analysis Using EES, the problem is solved as follows: "Input Data" T[1]=300 [K] P[1]=100 [kPa] T[3] = 2000 [K] r_comp = 12 "Process 1-2 is isentropic compression" s[1]=entropy(air,T=T[1],P=P[1]) s[2]=s[1] T[2]=temperature(air, s=s[2], P=P[2]) P[2]*v[2]/T[2]=P[1]*v[1]/T[1] P[1]*v[1]=R*T[1] R=0.287 [kJ/kg-K] V[2] = V[1]/ r_comp "Conservation of energy for process 1 to 2" q_12 - w_12 = DELTAu_12 q_12 =0"isentropic process" DELTAu_12=intenergy(air,T=T[2])-intenergy(air,T=T[1]) "Process 2-3 is constant volume heat addition" v[3]=v[2] s[3]=entropy(air, T=T[3], P=P[3]) P[3]*v[3]=R*T[3] "Conservation of energy for process 2 to 3" q_23 - w_23 = DELTAu_23 w_23 =0"constant volume process" DELTAu_23=intenergy(air,T=T[3])-intenergy(air,T=T[2]) "Process 3-4 is isentropic expansion" s[4]=s[3] s[4]=entropy(air,T=T[4],P=P[4]) P[4]*v[4]=R*T[4] "Conservation of energy for process 3 to 4" q_34 -w_34 = DELTAu_34 q_34 =0"isentropic process" DELTAu_34=intenergy(air,T=T[4])-intenergy(air,T=T[3]) "Process 4-1 is constant volume heat rejection" V[4] = V[1] "Conservation of energy for process 4 to 1" q_41 - w_41 = DELTAu_41 w_41 =0 "constant volume process" DELTAu_41=intenergy(air,T=T[1])-intenergy(air,T=T[4]) q_in_total=q_23 q_out_total = -q_41 w_net = w_12+w_23+w_34+w_41 Eta_th=w_net/q_in_total*Convert(, %) "Thermal efficiency, in percent"

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9-163

ηth [%] 45.83 48.67 51.03 53.02 54.74 56.24 57.57 58.75 59.83 60.8

wnet [kJ/kg] 567.4 589.3 604.9 616.2 624.3 630 633.8 636.3 637.5 637.9

rcomp 6 7 8 9 10 11 12 13 14 15

640 630

w net [kJ/kg]

620 610 600 590 580 570 560 6

7

8

9

10

r

11

12

13

14

15

com p

62.5

59

η th [%]

55.5

52

48.5

45 6

7

8

9

10

11

12

13

14

15

r comp

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9-164

9-192 The effects of pressure ratio on the net work output and the thermal efficiency of a simple Brayton cycle is to be investigated. The pressure ratios at which the net work output and the thermal efficiency are maximum are to be determined. Analysis Using EES, the problem is solved as follows: P_ratio = 8 T[1] = 300 [K] P[1]= 100 [kPa] T[3] = 1800 [K] m_dot = 1 [kg/s] Eta_c = 100/100 Eta_t = 100/100 "Inlet conditions" h[1]=ENTHALPY(Air,T=T[1]) s[1]=ENTROPY(Air,T=T[1],P=P[1]) "Compressor anaysis" s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor" P_ratio=P[2]/P[1]"Definition of pressure ratio - to find P[2]" T_s[2]=TEMPERATURE(Air,s=s_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exit" h_s[2]=ENTHALPY(Air,T=T_s[2]) Eta_c =(h_s[2]-h[1])/(h[2]-h[1]) "Compressor adiabatic efficiency; Eta_c = W_dot_c_ideal/W_dot_c_actual. " m_dot*h[1] +W_dot_c=m_dot*h[2] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" "External heat exchanger analysis" P[3]=P[2]"process 2-3 is SSSF constant pressure" h[3]=ENTHALPY(Air,T=T[3]) m_dot*h[2] + Q_dot_in= m_dot*h[3]"SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0" "Turbine analysis" s[3]=ENTROPY(Air,T=T[3],P=P[3]) s_s[4]=s[3] "For the ideal case the entropies are constant across the turbine" P_ratio= P[3] /P[4] T_s[4]=TEMPERATURE(Air,s=s_s[4],P=P[4]) "Ts[4] is the isentropic value of T[4] at turbine exit" h_s[4]=ENTHALPY(Air,T=T_s[4]) "Eta_t = W_dot_t /Wts_dot turbine adiabatic efficiency, Wts_dot > W_dot_t" Eta_t=(h[3]-h[4])/(h[3]-h_s[4]) m_dot*h[3] = W_dot_t + m_dot*h[4] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" "Cycle analysis" W_dot_net=W_dot_t-W_dot_c"Definition of the net cycle work, kW" Eta=W_dot_net/Q_dot_in"Cycle thermal efficiency" Bwr=W_dot_c/W_dot_t "Back work ratio" "The following state points are determined only to produce a T-s plot" T[2]=temperature('air',h=h[2]) T[4]=temperature('air',h=h[4]) s[2]=entropy('air',T=T[2],P=P[2]) s[4]=entropy('air',T=T[4],P=P[4])

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9-165

Bwr

η

Pratio

0.254 0.2665 0.2776 0.2876 0.2968 0.3052 0.313 0.3203 0.3272 0.3337 0.3398 0.3457 0.3513 0.3567 0.3618 0.3668 0.3716 0.3762 0.3806 0.385 0.3892 0.3932 0.3972 0.401 0.4048 0.4084 0.412 0.4155 0.4189 0.4222

0.3383 0.3689 0.3938 0.4146 0.4324 0.4478 0.4615 0.4736 0.4846 0.4945 0.5036 0.512 0.5197 0.5269 0.5336 0.5399 0.5458 0.5513 0.5566 0.5615 0.5663 0.5707 0.575 0.5791 0.583 0.5867 0.5903 0.5937 0.597 0.6002

5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34

Wc [kW] 175.8 201.2 223.7 244.1 262.6 279.7 295.7 310.6 324.6 337.8 350.4 362.4 373.9 384.8 395.4 405.5 415.3 424.7 433.8 442.7 451.2 459.6 467.7 475.5 483.2 490.7 498 505.1 512.1 518.9

725

Wnet [kW] 516.3 553.7 582.2 604.5 622.4 637 649 659.1 667.5 674.7 680.8 685.9 690.3 694.1 697.3 700 702.3 704.3 705.9 707.2 708.3 709.2 709.8 710.3 710.6 710.7 710.8 710.7 710.4 710.1

Wt [kW] 692.1 754.9 805.9 848.5 885 916.7 944.7 969.6 992.1 1013 1031 1048 1064 1079 1093 1106 1118 1129 1140 1150 1160 1169 1177 1186 1194 1201 1209 1216 1223 1229

Qin [kW] 1526 1501 1478 1458 1439 1422 1406 1392 1378 1364 1352 1340 1328 1317 1307 1297 1287 1277 1268 1259 1251 1243 1234 1227 1219 1211 1204 1197 1190 1183

0.65

700

0.6

675

Wnet [kW]

625

0.5

600

0.45

575

η th

0.55

650

0.4

550 0.35

525 500 5

10

15

20

25

30

0.3 35

Pratio

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-166

9-193 EES The effects of pressure ratio on the net work output and the thermal efficiency of a simple Brayton cycle is to be investigated assuming adiabatic efficiencies of 85 percent for both the turbine and the compressor. The pressure ratios at which the net work output and the thermal efficiency are maximum are to be determined. Analysis Using EES, the problem is solved as follows: P_ratio = 8 T[1] = 300 [K] P[1]= 100 [kPa] T[3] = 1800 [K] m_dot = 1 [kg/s] Eta_c = 85/100 Eta_t = 85/100 "Inlet conditions" h[1]=ENTHALPY(Air,T=T[1]) s[1]=ENTROPY(Air,T=T[1],P=P[1]) "Compressor anaysis" s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor" P_ratio=P[2]/P[1]"Definition of pressure ratio - to find P[2]" T_s[2]=TEMPERATURE(Air,s=s_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exit" h_s[2]=ENTHALPY(Air,T=T_s[2]) Eta_c =(h_s[2]-h[1])/(h[2]-h[1]) "Compressor adiabatic efficiency; Eta_c = W_dot_c_ideal/W_dot_c_actual. " m_dot*h[1] +W_dot_c=m_dot*h[2] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" "External heat exchanger analysis" P[3]=P[2]"process 2-3 is SSSF constant pressure" h[3]=ENTHALPY(Air,T=T[3]) m_dot*h[2] + Q_dot_in= m_dot*h[3]"SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0" "Turbine analysis" s[3]=ENTROPY(Air,T=T[3],P=P[3]) s_s[4]=s[3] "For the ideal case the entropies are constant across the turbine" P_ratio= P[3] /P[4] T_s[4]=TEMPERATURE(Air,s=s_s[4],P=P[4]) "Ts[4] is the isentropic value of T[4] at turbine exit" h_s[4]=ENTHALPY(Air,T=T_s[4]) "Eta_t = W_dot_t /Wts_dot turbine adiabatic efficiency, Wts_dot > W_dot_t" Eta_t=(h[3]-h[4])/(h[3]-h_s[4]) m_dot*h[3] = W_dot_t + m_dot*h[4] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" "Cycle analysis" W_dot_net=W_dot_t-W_dot_c"Definition of the net cycle work, kW" Eta=W_dot_net/Q_dot_in"Cycle thermal efficiency" Bwr=W_dot_c/W_dot_t "Back work ratio" "The following state points are determined only to produce a T-s plot" T[2]=temperature('air',h=h[2]) T[4]=temperature('air',h=h[4]) s[2]=entropy('air',T=T[2],P=P[2]) s[4]=entropy('air',T=T[4],P=P[4])

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Bwr

η

Pratio

0.3515 0.3689 0.3843 0.3981 0.4107 0.4224 0.4332 0.4433 0.4528 0.4618 0.4704 0.4785 0.4862 0.4937 0.5008 0.5077 0.5143 0.5207 0.5268 0.5328

0.2551 0.2764 0.2931 0.3068 0.3182 0.3278 0.3361 0.3432 0.3495 0.355 0.3599 0.3643 0.3682 0.3717 0.3748 0.3777 0.3802 0.3825 0.3846 0.3865

5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

Wc [kW] 206.8 236.7 263.2 287.1 309 329.1 347.8 365.4 381.9 397.5 412.3 426.4 439.8 452.7 465.1 477.1 488.6 499.7 510.4 520.8

Wnet [kW] 381.5 405 421.8 434.1 443.3 450.1 455.1 458.8 461.4 463.2 464.2 464.7 464.7 464.4 463.6 462.6 461.4 460 458.4 456.6

470

0.38

W net

η th

440

0.36 0.34

430 0.32 0.3

410 400 390 380 5

9

13

17

P

P ratio for

0.28

W

0.26

net,m ax 21

th

420

η

W net [kW ]

Qin [kW] 1495 1465 1439 1415 1393 1373 1354 1337 1320 1305 1290 1276 1262 1249 1237 1225 1214 1202 1192 1181

0.4

460 450

Wt [kW] 588.3 641.7 685 721.3 752.2 779.2 803 824.2 843.3 860.6 876.5 891.1 904.6 917.1 928.8 939.7 950 959.6 968.8 977.4

0.24 25

ratio

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9-168

9-194 EES The effects of pressure ratio, maximum cycle temperature, and compressor and turbine inefficiencies on the net work output per unit mass and the thermal efficiency of a simple Brayton cycle with air as the working fluid is to be investigated. Constant specific heats at room temperature are to be used. Analysis Using EES, the problem is solved as follows: Procedure ConstPropResult(T[1],P[1],r_comp,T[3]:Eta_th_ConstProp,Eta_th_easy) "For Air:" C_V = 0.718 [kJ/kg-K] k = 1.4 T2 = T[1]*r_comp^(k-1) P2 = P[1]*r_comp^k q_in_23 = C_V*(T[3]-T2) T4 = T[3]*(1/r_comp)^(k-1) q_out_41 = C_V*(T4-T[1]) Eta_th_ConstProp = (1-q_out_41/q_in_23)*Convert(, %) "[%]" "The Easy Way to calculate the constant property Otto cycle efficiency is:" Eta_th_easy = (1 - 1/r_comp^(k-1))*Convert(, %) "[%]" END "Input Data" T[1]=300 [K] P[1]=100 [kPa] {T[3] = 1000 [K]} r_comp = 12 "Process 1-2 is isentropic compression" s[1]=entropy(air,T=T[1],P=P[1]) s[2]=s[1] T[2]=temperature(air, s=s[2], P=P[2]) P[2]*v[2]/T[2]=P[1]*v[1]/T[1] P[1]*v[1]=R*T[1] R=0.287 [kJ/kg-K] V[2] = V[1]/ r_comp "Conservation of energy for process 1 to 2" q_12 - w_12 = DELTAu_12 q_12 =0"isentropic process" DELTAu_12=intenergy(air,T=T[2])-intenergy(air,T=T[1]) "Process 2-3 is constant volume heat addition" v[3]=v[2] s[3]=entropy(air, T=T[3], P=P[3]) P[3]*v[3]=R*T[3] "Conservation of energy for process 2 to 3" q_23 - w_23 = DELTAu_23 w_23 =0"constant volume process" DELTAu_23=intenergy(air,T=T[3])-intenergy(air,T=T[2]) "Process 3-4 is isentropic expansion" s[4]=s[3] s[4]=entropy(air,T=T[4],P=P[4]) P[4]*v[4]=R*T[4] "Conservation of energy for process 3 to 4" q_34 -w_34 = DELTAu_34 q_34 =0"isentropic process" DELTAu_34=intenergy(air,T=T[4])-intenergy(air,T=T[3])

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9-169

"Process 4-1 is constant volume heat rejection" V[4] = V[1] "Conservation of energy for process 4 to 1" q_41 - w_41 = DELTAu_41 w_41 =0 "constant volume process" DELTAu_41=intenergy(air,T=T[1])-intenergy(air,T=T[4]) q_in_total=q_23 q_out_total = -q_41 w_net = w_12+w_23+w_34+w_41 Eta_th=w_net/q_in_total*Convert(, %) "Thermal efficiency, in percent" Call ConstPropResult(T[1],P[1],r_comp,T[3]:Eta_th_ConstProp,Eta_th_easy) PerCentError = ABS(Eta_th - Eta_th_ConstProp)/Eta_th*Convert(, %) "[%]" PerCentError [%] 3.604 6.681 9.421 11.64

ηth [%] 60.8 59.04 57.57 56.42

rcomp 12 12 12 12

P e rc e n t E rro r = | η

th

ηth,ConstProp [%] 62.99 62.99 62.99 62.99

ηth,easy [%] 62.99 62.99 62.99 62.99

T3 [K] 1000 1500 2000 2500

- η |/η th ,C o n s tP ro p th

4 .3

PerCentError [%]

4 .2

T

m ax

= 1000 K

4 .1 4 3 .9 3 .8 3 .7 3 .6 6

7

8

9

r

10

11

12

com p

15

PerCentError [%]

r

comp

=6

12.8

=12 10.6

8.4

6.2

4 1000

1200

1400

1600

1800

2000

2200

2400

2600

T[3] [K]

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9-170

9-195 EES The effects of pressure ratio, maximum cycle temperature, and compressor and turbine efficiencies on the net work output per unit mass and the thermal efficiency of a simple Brayton cycle with air as the working fluid is to be investigated. Variable specific heats are to be used. Analysis Using EES, the problem is solved as follows: "Input data - from diagram window" {P_ratio = 8} {T[1] = 300 [K] P[1]= 100 [kPa] T[3] = 800 [K] m_dot = 1 [kg/s] Eta_c = 75/100 Eta_t = 82/100} "Inlet conditions" h[1]=ENTHALPY(Air,T=T[1]) s[1]=ENTROPY(Air,T=T[1],P=P[1]) "Compressor anaysis" s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor" P_ratio=P[2]/P[1]"Definition of pressure ratio - to find P[2]" T_s[2]=TEMPERATURE(Air,s=s_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exit" h_s[2]=ENTHALPY(Air,T=T_s[2]) Eta_c =(h_s[2]-h[1])/(h[2]-h[1]) "Compressor adiabatic efficiency; Eta_c = W_dot_c_ideal/W_dot_c_actual. " m_dot*h[1] +W_dot_c=m_dot*h[2] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" "External heat exchanger analysis" P[3]=P[2]"process 2-3 is SSSF constant pressure" h[3]=ENTHALPY(Air,T=T[3]) m_dot*h[2] + Q_dot_in= m_dot*h[3]"SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0" "Turbine analysis" s[3]=ENTROPY(Air,T=T[3],P=P[3]) s_s[4]=s[3] "For the ideal case the entropies are constant across the turbine" P_ratio= P[3] /P[4] T_s[4]=TEMPERATURE(Air,s=s_s[4],P=P[4]) "Ts[4] is the isentropic value of T[4] at turbine exit" h_s[4]=ENTHALPY(Air,T=T_s[4]) "Eta_t = W_dot_t /Wts_dot turbine adiabatic efficiency, Wts_dot > W_dot_t" Eta_t=(h[3]-h[4])/(h[3]-h_s[4]) m_dot*h[3] = W_dot_t + m_dot*h[4] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" "Cycle analysis" W_dot_net=W_dot_t-W_dot_c"Definition of the net cycle work, kW" Eta=W_dot_net/Q_dot_in"Cycle thermal efficiency" Bwr=W_dot_c/W_dot_t "Back work ratio" "The following state points are determined only to produce a T-s plot" T[2]=temperature('air',h=h[2]) T[4]=temperature('air',h=h[4]) s[2]=entropy('air',T=T[2],P=P[2]) s[4]=entropy('air',T=T[4],P=P[4])

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9-171

Bwr

η

Pratio

0.5229 0.6305 0.7038 0.7611 0.8088 0.85 0.8864 0.9192 0.9491 0.9767

0.1 0.1644 0.1814 0.1806 0.1702 0.1533 0.131 0.1041 0.07272 0.03675

2 4 6 8 10 12 14 16 18 20

Wc [kW] 1818 4033 5543 6723 7705 8553 9304 9980 10596 11165

Wnet [kW] 1659 2364 2333 2110 1822 1510 1192 877.2 567.9 266.1

Wt [kW] 3477 6396 7876 8833 9527 10063 10496 10857 11164 11431

Qin [kW] 16587 14373 12862 11682 10700 9852 9102 8426 7809 7241

1500 Air Standard Brayton Cycle Pressure ratio = 8 and T max = 1160K 3

T [K]

1000

4

2 2s

4s

500 800 kPa 100 kPa

0 5.0

1

5.5

6.0

6.5

7.0

7.5

s [kJ/kg-K]

0.25

2500

η

Cycle efficiency,

η

W

0.15

T

0.00 2

t c

0.10

0.05

1500

η η

Note P

4

ratio

6

2000 net

m ax

= 0.82 = 0.75

1000

=1160 K 500

for m axim um w ork and η

8

10

12 P

14

W net [kW ]

0.20

16

18

0 20

ratio

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9-172

9-196 EES The effects of pressure ratio, maximum cycle temperature, and compressor and turbine efficiencies on the net work output per unit mass and the thermal efficiency of a simple Brayton cycle with helium as the working fluid is to be investigated. Analysis Using EES, the problem is solved as follows: Function hFunc(WorkFluid$,T,P) "The EES functions teat helium as a real gas; thus, T and P are needed for helium's enthalpy." IF WorkFluid$ = 'Air' then hFunc:=enthalpy(Air,T=T) ELSE hFunc: = enthalpy(Helium,T=T,P=P) endif END Procedure EtaCheck(Eta_th:EtaError$) If Eta_th < 0 then EtaError$ = 'Why are the net work done and efficiency < 0?' Else EtaError$ = '' END "Input data - from diagram window" {P_ratio = 8} {T[1] = 300 [K] P[1]= 100 [kPa] T[3] = 800 [K] m_dot = 1 [kg/s] Eta_c = 0.8 Eta_t = 0.8 WorkFluid$ = 'Helium'} "Inlet conditions" h[1]=hFunc(WorkFluid$,T[1],P[1]) s[1]=ENTROPY(WorkFluid$,T=T[1],P=P[1]) "Compressor anaysis" s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor" P_ratio=P[2]/P[1]"Definition of pressure ratio - to find P[2]" T_s[2]=TEMPERATURE(WorkFluid$,s=s_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exit" h_s[2]=hFunc(WorkFluid$,T_s[2],P[2]) Eta_c =(h_s[2]-h[1])/(h[2]-h[1]) "Compressor adiabatic efficiency; Eta_c = W_dot_c_ideal/W_dot_c_actual. " m_dot*h[1] +W_dot_c=m_dot*h[2] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" "External heat exchanger analysis" P[3]=P[2]"process 2-3 is SSSF constant pressure" h[3]=hFunc(WorkFluid$,T[3],P[3]) m_dot*h[2] + Q_dot_in= m_dot*h[3]"SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0" "Turbine analysis" s[3]=ENTROPY(WorkFluid$,T=T[3],P=P[3]) s_s[4]=s[3] "For the ideal case the entropies are constant across the turbine" P_ratio= P[3] /P[4] T_s[4]=TEMPERATURE(WorkFluid$,s=s_s[4],P=P[4]) "Ts[4] is the isentropic value of T[4] at turbine exit" h_s[4]=hFunc(WorkFluid$,T_s[4],P[4]) "Eta_t = W_dot_t /Wts_dot turbine adiabatic efficiency, Wts_dot > W_dot_t" Eta_t=(h[3]-h[4])/(h[3]-h_s[4]) m_dot*h[3] = W_dot_t + m_dot*h[4] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" "Cycle analysis" W_dot_net=W_dot_t-W_dot_c"Definition of the net cycle work, kW"

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9-173

Eta_th=W_dot_net/Q_dot_in"Cycle thermal efficiency" Call EtaCheck(Eta_th:EtaError$) Bwr=W_dot_c/W_dot_t "Back work ratio" "The following state points are determined only to produce a T-s plot" T[2]=temperature('air',h=h[2]) T[4]=temperature('air',h=h[4]) s[2]=entropy('air',T=T[2],P=P[2]) s[4]=entropy('air',T=T[4],P=P[4]) Bwr

η

Pratio

0.5229 0.6305 0.7038 0.7611 0.8088 0.85 0.8864 0.9192 0.9491 0.9767

0.1 0.1644 0.1814 0.1806 0.1702 0.1533 0.131 0.1041 0.07272 0.03675

2 4 6 8 10 12 14 16 18 20

Wc [kW] 1818 4033 5543 6723 7705 8553 9304 9980 10596 11165

Wnet [kW] 1659 2364 2333 2110 1822 1510 1192 877.2 567.9 266.1

Wt [kW] 3477 6396 7876 8833 9527 10063 10496 10857 11164 11431

Qin [kW] 16587 14373 12862 11682 10700 9852 9102 8426 7809 7241

1500 B ra yto n C yc le P r e s s u r e r a tio = 8 a n d T

m ax

= 1160K 3

T [K]

1000

4

2 2

4

s

s

500 800 kP a 100 kP a

0 5 .0

1

5 .5

6 .0

6 .5

7 .0

7 .5

s [k J /k g -K ]

0.25

Brayton Cycle using Air m air = 20 kg/s

0.20

2000

η

η

Wnet 0.15

0.10

0.05

0.00 2

1500

η = 0.82 t η = 0.75 c

1000

Tmax=1160 K

500

Note Pratio for maximum work and η

4

6

8

10 12 Pratio

14

Wnet [kW]

Cycle efficiency,

2500

16

18

0 20

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-174

9-197 EES The effect of the number of compression and expansion stages on the thermal efficiency of an ideal regenerative Brayton cycle with multistage compression and expansion and air as the working fluid is to be investigated. Analysis Using EES, the problem is solved as follows: "Input data for air" C_P = 1.005 [kJ/kg-K] k = 1.4 "Nstages is the number of compression and expansion stages" Nstages = 1 T_6 = 1200 [K] Pratio = 12 T_1 = 300 [K] P_1= 100 [kPa] Eta_reg = 1.0 "regenerator effectiveness" Eta_c =1.0 "Compressor isentorpic efficiency" Eta_t =1.0 "Turbine isentropic efficiency" R_p = Pratio^(1/Nstages) "Isentropic Compressor anaysis" T_2s = T_1*R_p^((k-1)/k) P_2 = R_p*P_1 "T_2s is the isentropic value of T_2 at compressor exit" Eta_c = w_compisen/w_comp "compressor adiabatic efficiency, W_comp > W_compisen" "Conservation of energy for the compressor for the isentropic case: e_in - e_out = DELTAe=0 for steady-flow" w_compisen = C_P*(T_2s-T_1) "Actual compressor analysis:" w_comp = C_P*(T_2 - T_1) "Since intercooling is assumed to occur such that T_3 = T_1 and the compressors have the same pressure ratio, the work input to each compressor is the same. The total compressor work is:" w_comp_total = Nstages*w_comp "External heat exchanger analysis" "SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0 e_in - e_out =DELTAe_cv =0 for steady flow" "The heat added in the external heat exchanger + the reheat between turbines is" q_in_total = C_P*(T_6 - T_5) +(Nstages - 1)*C_P*(T_8 - T_7) "Reheat is assumed to occur until:" T_8 = T_6 "Turbine analysis" P_7 = P_6 /R_p "T_7s is the isentropic value of T_7 at turbine exit" T_7s = T_6*(1/R_p)^((k-1)/k) "Turbine adiabatic efficiency, w_turbisen > w_turb" Eta_t = w_turb /w_turbisen "SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0 e_in -e_out = DELTAe_cv = 0 for steady-flow" w_turbisen = C_P*(T_6 - T_7s) "Actual Turbine analysis:" w_turb = C_P*(T_6 - T_7) w_turb_total = Nstages*w_turb

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9-175

"Cycle analysis" w_net=w_turb_total-w_comp_total "[kJ/kg]" Bwr=w_comp/w_turb "Back work ratio" P_4=P_2 P_5=P_4 P_6=P_5 T_4 = T_2 "The regenerator effectiveness gives T_5 as:" Eta_reg = (T_5 - T_4)/(T_9 - T_4) T_9 = T_7 "Energy balance on regenerator gives T_10 as:" T_4 + T_9=T_5 + T_10 "Cycle thermal efficiency with regenerator" Eta_th_regenerative=w_net/q_in_total*Convert(, %) "[%]" "The efficiency of the Ericsson cycle is the same as the Carnot cycle operating between the same max and min temperatures, T_6 and T_1 for this problem." Eta_th_Ericsson = (1 - T_1/T_6)*Convert(, %) "[%]"

ηth,Ericksson [%] 75 75 75 75 75 75 75 75

ηth,Regenerative [%] 49.15 64.35 68.32 70.14 72.33 73.79 74.05 74.18

Nstages 1 2 3 4 7 15 19 22

80

70

Ericsson η th [%]

60

Ideal Regenerative Brayton

50

40 0

2

4

6

8

10

12

14

16

18

20

22

24

Nstages

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9-176

9-198 EES The effect of the number of compression and expansion stages on the thermal efficiency of an ideal regenerative Brayton cycle with multistage compression and expansion and helium as the working fluid is to be investigated. Analysis Using EES, the problem is solved as follows: "Input data for Helium" C_P = 5.1926 [kJ/kg-K] k = 1.667 "Nstages is the number of compression and expansion stages" {Nstages = 1} T_6 = 1200 [K] Pratio = 12 T_1 = 300 [K] P_1= 100 [kPa] Eta_reg = 1.0 "regenerator effectiveness" Eta_c =1.0 "Compressor isentorpic efficiency" Eta_t =1.0 "Turbine isentropic efficiency" R_p = Pratio^(1/Nstages) "Isentropic Compressor anaysis" T_2s = T_1*R_p^((k-1)/k) P_2 = R_p*P_1 "T_2s is the isentropic value of T_2 at compressor exit" Eta_c = w_compisen/w_comp "compressor adiabatic efficiency, W_comp > W_compisen" "Conservation of energy for the compressor for the isentropic case: e_in - e_out = DELTAe=0 for steady-flow" w_compisen = C_P*(T_2s-T_1) "Actual compressor analysis:" w_comp = C_P*(T_2 - T_1) "Since intercooling is assumed to occur such that T_3 = T_1 and the compressors have the same pressure ratio, the work input to each compressor is the same. The total compressor work is:" w_comp_total = Nstages*w_comp "External heat exchanger analysis" "SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0 e_in - e_out =DELTAe_cv =0 for steady flow" "The heat added in the external heat exchanger + the reheat between turbines is" q_in_total = C_P*(T_6 - T_5) +(Nstages - 1)*C_P*(T_8 - T_7) "Reheat is assumed to occur until:" T_8 = T_6 "Turbine analysis" P_7 = P_6 /R_p "T_7s is the isentropic value of T_7 at turbine exit" T_7s = T_6*(1/R_p)^((k-1)/k) "Turbine adiabatic efficiency, w_turbisen > w_turb" Eta_t = w_turb /w_turbisen "SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0 e_in -e_out = DELTAe_cv = 0 for steady-flow" w_turbisen = C_P*(T_6 - T_7s) "Actual Turbine analysis:" w_turb = C_P*(T_6 - T_7) w_turb_total = Nstages*w_turb

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9-177

"Cycle analysis" w_net=w_turb_total-w_comp_total Bwr=w_comp/w_turb "Back work ratio" P_4=P_2 P_5=P_4 P_6=P_5 T_4 = T_2 "The regenerator effectiveness gives T_5 as:" Eta_reg = (T_5 - T_4)/(T_9 - T_4) T_9 = T_7 "Energy balance on regenerator gives T_10 as:" T_4 + T_9=T_5 + T_10 "Cycle thermal efficiency with regenerator" Eta_th_regenerative=w_net/q_in_total*Convert(, %) "[%]" "The efficiency of the Ericsson cycle is the same as the Carnot cycle operating between the same max and min temperatures, T_6 and T_1 for this problem." Eta_th_Ericsson = (1 - T_1/T_6)*Convert(, %) "[%]"

ηth,Ericksson [%] 75 75 75 75 75 75 75 75

ηth,Regenerative [%] 32.43 58.9 65.18 67.95 71.18 73.29 73.66 73.84

Nstages 1 2 3 4 7 15 19 22

80

70

Ericsson

η th [%]

60

Ideal Regenerative Brayton

50

40

30 0

2

4

6

8

10

12

14

16

18

20

22

24

Nstages

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9-178

Fundamentals of Engineering (FE) Exam Problems

9-199 An Otto cycle with air as the working fluid has a compression ratio of 8.2. Under cold air standard conditions, the thermal efficiency of this cycle is

(a) 24%

(b) 43%

(c) 52%

(d) 57%

(e) 75%

Answer (d) 57% Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). r=8.2 k=1.4 Eta_Otto=1-1/r^(k-1) "Some Wrong Solutions with Common Mistakes:" W1_Eta = 1/r "Taking efficiency to be 1/r" W2_Eta = 1/r^(k-1) "Using incorrect relation" W3_Eta = 1-1/r^(k1-1); k1=1.667 "Using wrong k value"

9-200 For specified limits for the maximum and minimum temperatures, the ideal cycle with the lowest thermal efficiency is

(a) Carnot

(b) Stirling

(c) Ericsson

(d) Otto

(e) All are the same

Answer (d) Otto

9-201 A Carnot cycle operates between the temperatures limits of 300 K and 2000 K, and produces 600 kW of net power. The rate of entropy change of the working fluid during the heat addition process is

(a) 0

(b) 0.300 kW/K

(c) 0.353 kW/K

(d) 0.261 kW/K

(e) 2.0 kW/K

Answer (c) 0.353 kW/K Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). TL=300 "K" TH=2000 "K" Wnet=600 "kJ/s" Wnet= (TH-TL)*DS "Some Wrong Solutions with Common Mistakes:" W1_DS = Wnet/TH "Using TH instead of TH-TL" W2_DS = Wnet/TL "Using TL instead of TH-TL" W3_DS = Wnet/(TH+TL) "Using TH+TL instead of TH-TL"

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9-179

9-202 Air in an ideal Diesel cycle is compressed from 3 L to 0.15 L, and then it expands during the constant pressure heat addition process to 0.30 L. Under cold air standard conditions, the thermal efficiency of this cycle is

(a) 35%

(b) 44%

(c) 65%

(d) 70%

(e) 82%

Answer (c) 65% Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). V1=3 "L" V2= 0.15 "L" V3= 0.30 "L" r=V1/V2 rc=V3/V2 k=1.4 Eta_Diesel=1-(1/r^(k-1))*(rc^k-1)/k/(rc-1) "Some Wrong Solutions with Common Mistakes:" W1_Eta = 1-(1/r1^(k-1))*(rc^k-1)/k/(rc-1); r1=V1/V3 "Wrong r value" W2_Eta = 1-Eta_Diesel "Using incorrect relation" W3_Eta = 1-(1/r^(k1-1))*(rc^k1-1)/k1/(rc-1); k1=1.667 "Using wrong k value" W4_Eta = 1-1/r^(k-1) "Using Otto cycle efficiency"

9-203 Helium gas in an ideal Otto cycle is compressed from 20°C and 2.5 L to 0.25 L, and its temperature increases by an additional 700°C during the heat addition process. The temperature of helium before the expansion process is

(a) 1790°C

(b) 2060°C

(c) 1240°C

(d) 620°C

(e) 820°C

Answer (a) 1790°C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). k=1.667 V1=2.5 V2=0.25 r=V1/V2 T1=20+273 "K" T2=T1*r^(k-1) T3=T2+700-273 "C" "Some Wrong Solutions with Common Mistakes:" W1_T3 =T22+700-273; T22=T1*r^(k1-1); k1=1.4 "Using wrong k value" W2_T3 = T3+273 "Using K instead of C" W3_T3 = T1+700-273 "Disregarding temp rise during compression" W4_T3 = T222+700-273; T222=(T1-273)*r^(k-1) "Using C for T1 instead of K"

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9-180

9-204 In an ideal Otto cycle, air is compressed from 1.20 kg/m3 and 2.2 L to 0.26 L, and the net work output of the cycle is 440 kJ/kg. The mean effective pressure (MEP) for this cycle is

(a) 612 kPa

(b) 599 kPa

(c) 528 kPa

(d) 416 kPa

(e) 367 kPa

Answer (b) 599 kPa Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). rho1=1.20 "kg/m^3" k=1.4 V1=2.2 V2=0.26 m=rho1*V1/1000 "kg" w_net=440 "kJ/kg" Wtotal=m*w_net MEP=Wtotal/((V1-V2)/1000) "Some Wrong Solutions with Common Mistakes:" W1_MEP = w_net/((V1-V2)/1000) "Disregarding mass" W2_MEP = Wtotal/(V1/1000) "Using V1 instead of V1-V2" W3_MEP = (rho1*V2/1000)*w_net/((V1-V2)/1000); "Finding mass using V2 instead of V1" W4_MEP = Wtotal/((V1+V2)/1000) "Adding V1 and V2 instead of subtracting"

9-205 In an ideal Brayton cycle, air is compressed from 95 kPa and 25°C to 800 kPa. Under cold air standard conditions, the thermal efficiency of this cycle is

(a) 46%

(b) 54%

(c) 57%

(d) 39%

(e) 61%

Answer (a) 46% Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=95 "kPa" P2=800 "kPa" T1=25+273 "K" rp=P2/P1 k=1.4 Eta_Brayton=1-1/rp^((k-1)/k) "Some Wrong Solutions with Common Mistakes:" W1_Eta = 1/rp "Taking efficiency to be 1/rp" W2_Eta = 1/rp^((k-1)/k) "Using incorrect relation" W3_Eta = 1-1/rp^((k1-1)/k1); k1=1.667 "Using wrong k value"

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9-181

9-206 Consider an ideal Brayton cycle executed between the pressure limits of 1200 kPa and 100 kPa and temperature limits of 20°C and 1000°C with argon as the working fluid. The net work output of the cycle is

(a) 68 kJ/kg

(b) 93 kJ/kg

(c) 158 kJ/kg

(d) 186 kJ/kg

(e) 310 kJ/kg

Answer (c) 158 kJ/kg Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=100 "kPa" P2=1200 "kPa" T1=20+273 "K" T3=1000+273 "K" rp=P2/P1 k=1.667 Cp=0.5203 "kJ/kg.K" Cv=0.3122 "kJ/kg.K" T2=T1*rp^((k-1)/k) q_in=Cp*(T3-T2) Eta_Brayton=1-1/rp^((k-1)/k) w_net=Eta_Brayton*q_in "Some Wrong Solutions with Common Mistakes:" W1_wnet = (1-1/rp^((k-1)/k))*qin1; qin1=Cv*(T3-T2) "Using Cv instead of Cp" W2_wnet = (1-1/rp^((k-1)/k))*qin2; qin2=1.005*(T3-T2) "Using Cp of air instead of argon" W3_wnet = (1-1/rp^((k1-1)/k1))*Cp*(T3-T22); T22=T1*rp^((k1-1)/k1); k1=1.4 "Using k of air instead of argon" W4_wnet = (1-1/rp^((k-1)/k))*Cp*(T3-T222); T222=(T1-273)*rp^((k-1)/k) "Using C for T1 instead of K"

9-207 An ideal Brayton cycle has a net work output of 150 kJ/kg and a backwork ratio of 0.4. If both the turbine and the compressor had an isentropic efficiency of 85%, the net work output of the cycle would be

(a) 74 kJ/kg

(b) 95 kJ/kg

(c) 109 kJ/kg

(d) 128 kJ/kg

(e) 177 kJ/kg

Answer (b) 95 kJ/kg Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). wcomp/wturb=0.4 wturb-wcomp=150 "kJ/kg" Eff=0.85 w_net=Eff*wturb-wcomp/Eff "Some Wrong Solutions with Common Mistakes:" W1_wnet = Eff*wturb-wcomp*Eff "Making a mistake in Wnet relation" W2_wnet = (wturb-wcomp)/Eff "Using a wrong relation" W3_wnet = wturb/eff-wcomp*Eff "Using a wrong relation"

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9-182

9-208 In an ideal Brayton cycle, air is compressed from 100 kPa and 25°C to 1 MPa, and then heated to 1200°C before entering the turbine. Under cold air standard conditions, the air temperature at the turbine exit is

(a) 490°C

(b) 515°C

(c) 622°C

(d) 763°C

(e) 895°C

Answer (a) 490°C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=100 "kPa" P2=1000 "kPa" T1=25+273 "K" T3=1200+273 "K" rp=P2/P1 k=1.4 T4=T3*(1/rp)^((k-1)/k)-273 "Some Wrong Solutions with Common Mistakes:" W1_T4 = T3/rp "Using wrong relation" W2_T4 = (T3-273)/rp "Using wrong relation" W3_T4 = T4+273 "Using K instead of C" W4_T4 = T1+800-273 "Disregarding temp rise during compression"

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9-183

9-209 In an ideal Brayton cycle with regeneration, argon gas is compressed from 100 kPa and 25°C to 400 kPa, and then heated to 1200°C before entering the turbine. The highest temperature that argon can be heated in the regenerator is

(a) 246°C

(b) 846°C

(c) 689°C

(d) 368°C

(e) 573°C

Answer (e) 573°C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). k=1.667 Cp=0.5203 "kJ/kg.K" P1=100 "kPa" P2=400 "kPa" T1=25+273 "K" T3=1200+273 "K" "The highest temperature that argon can be heated in the regenerator is the turbine exit temperature," rp=P2/P1 T2=T1*rp^((k-1)/k) T4=T3/rp^((k-1)/k)-273 "Some Wrong Solutions with Common Mistakes:" W1_T4 = T3/rp "Using wrong relation" W2_T4 = (T3-273)/rp^((k-1)/k) "Using C instead of K for T3" W3_T4 = T4+273 "Using K instead of C" W4_T4 = T2-273 "Taking compressor exit temp as the answer"

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9-184

9-210 In an ideal Brayton cycle with regeneration, air is compressed from 80 kPa and 10°C to 400 kPa and 175°C, is heated to 450°C in the regenerator, and then further heated to 1000°C before entering the turbine. Under cold air standard conditions, the effectiveness of the regenerator is

(a) 33%

(b) 44%

(c) 62%

(d) 77%

(e) 89%

Answer (d) 77% Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). k=1.4 Cp=1.005 "kJ/kg.K" P1=80 "kPa" P2=400 "kPa" T1=10+273 "K" T2=175+273 "K" T3=1000+273 "K" T5=450+273 "K" "The highest temperature that the gas can be heated in the regenerator is the turbine exit temperature," rp=P2/P1 T2check=T1*rp^((k-1)/k) "Checking the given value of T2. It checks." T4=T3/rp^((k-1)/k) Effective=(T5-T2)/(T4-T2) "Some Wrong Solutions with Common Mistakes:" W1_eff = (T5-T2)/(T3-T2) "Using wrong relation" W2_eff = (T5-T2)/(T44-T2); T44=(T3-273)/rp^((k-1)/k) "Using C instead of K for T3" W3_eff = (T5-T2)/(T444-T2); T444=T3/rp "Using wrong relation for T4"

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9-185

9-211 Consider a gas turbine that has a pressure ratio of 6 and operates on the Brayton cycle with regeneration between the temperature limits of 20°C and 900°C. If the specific heat ratio of the working fluid is 1.3, the highest thermal efficiency this gas turbine can have is

(a) 38%

(b) 46%

(c) 62%

(d) 58%

(e) 97%

Answer (c) 62% Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). k=1.3 rp=6 T1=20+273 "K" T3=900+273 "K" Eta_regen=1-(T1/T3)*rp^((k-1)/k) "Some Wrong Solutions with Common Mistakes:" W1_Eta = 1-((T1-273)/(T3-273))*rp^((k-1)/k) "Using C for temperatures instead of K" W2_Eta = (T1/T3)*rp^((k-1)/k) "Using incorrect relation" W3_Eta = 1-(T1/T3)*rp^((k1-1)/k1); k1=1.4 "Using wrong k value (the one for air)"

9-212 An ideal gas turbine cycle with many stages of compression and expansion and a regenerator of 100 percent effectiveness has an overall pressure ratio of 10. Air enters every stage of compressor at 290 K, and every stage of turbine at 1200 K. The thermal efficiency of this gas-turbine cycle is

(a) 36%

(b) 40%

(c) 52%

(d) 64%

(e) 76%

Answer (e) 76% Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). k=1.4 rp=10 T1=290 "K" T3=1200 "K" Eff=1-T1/T3 "Some Wrong Solutions with Common Mistakes:" W1_Eta = 100 W2_Eta = 1-1/rp^((k-1)/k) "Using incorrect relation" W3_Eta = 1-(T1/T3)*rp^((k-1)/k) "Using wrong relation" W4_Eta = T1/T3 "Using wrong relation"

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9-186

9-213 Air enters a turbojet engine at 260 m/s at a rate of 30 kg/s, and exits at 800 m/s relative to the aircraft. The thrust developed by the engine is

(a) 8 kN

(b) 16 kN

(c) 24 kN

(d) 20 kN

(e) 32 kN

Answer (b) 16 kN Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). Vel1=260 "m/s" Vel2=800 "m/s" Thrust=m*(Vel2-Vel1)/1000 "kN" m= 30 "kg/s" "Some Wrong Solutions with Common Mistakes:" W1_thrust = (Vel2-Vel1)/1000 "Disregarding mass flow rate" W2_thrust = m*Vel2/1000 "Using incorrect relation"

9-214 ··· 9-220 Design and Essay Problems.

KJ

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10-1

Chapter 10 VAPOR AND COMBINED POWER CYCLES Carnot Vapor Cycle 10-1C Because excessive moisture in steam causes erosion on the turbine blades. The highest moisture content allowed is about 10%. 10-2C The Carnot cycle is not a realistic model for steam power plants because (1) limiting the heat transfer processes to two-phase systems to maintain isothermal conditions severely limits the maximum temperature that can be used in the cycle, (2) the turbine will have to handle steam with a high moisture content which causes erosion, and (3) it is not practical to design a compressor that will handle two phases.

10-3E A steady-flow Carnot engine with water as the working fluid operates at specified conditions. The thermal efficiency, the quality at the end of the heat rejection process, and the net work output are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) We note that TH = Tsat @180 psia = 373.1°F = 833.1 R

T

TL = Tsat @14.7 psia = 212.0°F = 672.0 R

and ηth,C

672.0 R T = 1− L = 1− = 19.3% 833.1 R TH

1 180 psia 2 qin 14.7 psia

(b) Noting that s4 = s1 = sf @ 180 psia = 0.53274 Btu/lbm·R, x4 =

s4 − s f s fg

=

0.53274 − 0.31215 = 0.153 1.44441

4

3 s

(c) The enthalpies before and after the heat addition process are

h1 = h f @ 180 psia = 346.14 Btu/lbm

h2 = h f + x 2 h fg = 346.14 + (0.90)(851.16) = 1112.2 Btu/lbm Thus, q in = h2 − h1 = 1112.2 − 346.14 = 766.0 Btu/lbm

and,

wnet = η th q in = (0.1934)(766.0 Btu/lbm) = 148.1 Btu/lbm

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10-2

10-4 A steady-flow Carnot engine with water as the working fluid operates at specified conditions. The thermal efficiency, the amount of heat rejected, and the net work output are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) Noting that TH = 250°C = 523 K and TL = Tsat @ 20 kPa = 60.06°C = 333.1 K, the thermal efficiency becomes

η th,C = 1 −

TL 333.1 K = 1− = 0.3632 = 36.3% 523 K TH

(b) The heat supplied during this cycle is simply the enthalpy of vaporization, q in = h fg @ 250oC = 1715.3 kJ/kg

T

250°C

2

1 qin

Thus, q out = q L =

20 kPa

⎛ 333.1 K ⎞ TL ⎟⎟(1715.3 kJ/kg ) = 1092.3 kJ/kg q in = ⎜⎜ TH ⎝ 523 K ⎠

4

qout

3 s

(c) The net work output of this cycle is wnet = η th q in = (0.3632 )(1715.3 kJ/kg ) = 623.0 kJ/kg

10-5 A steady-flow Carnot engine with water as the working fluid operates at specified conditions. The thermal efficiency, the amount of heat rejected, and the net work output are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) Noting that TH = 250°C = 523 K and TL = Tsat @ 10 kPa = 45.81°C = 318.8 K, the thermal efficiency becomes

η th, C = 1 −

TL 318.8 K =1− = 39.04% 523 K TH

(b) The heat supplied during this cycle is simply the enthalpy of vaporization, q in = h fg @ 250°C = 1715.3 kJ/kg

T

250°C

2

1 qin

Thus, q out = q L =

⎛ 318.8 K ⎞ TL ⎟⎟(1715.3 kJ/kg ) = 1045.6 kJ/kg q in = ⎜⎜ TH ⎝ 523 K ⎠

10 kPa 4

qout

3

(c) The net work output of this cycle is wnet = η th q in = (0.3904)(1715.3 kJ/kg ) = 669.7 kJ/kg

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s

10-3

10-6 A steady-flow Carnot engine with water as the working fluid operates at specified conditions. The thermal efficiency, the pressure at the turbine inlet, and the net work output are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) The thermal efficiency is determined from

η th, C = 1 −

TL 60 + 273 K = 1− = 46.5% 350 + 273 K TH

T

(b) Note that s2 = s3 = sf + x3sfg

350°C

1

2

4

3

= 0.8313 + 0.891 × 7.0769 = 7.1368 kJ/kg·K Thus,

60°C T2 = 350°C

⎫ ⎬ P2 ≅ 1.40 MPa (Table A-6) s 2 = 7.1368 kJ/kg ⋅ K ⎭

s

(c) The net work can be determined by calculating the enclosed area on the T-s diagram, s 4 = s f + x 4 s fg = 0.8313 + (0.1)(7.0769) = 1.5390 kJ/kg ⋅ K

Thus,

wnet = Area = (TH − TL )(s 3 − s 4 ) = (350 − 60)(7.1368 − 1.5390) = 1623 kJ/kg

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10-4

The Simple Rankine Cycle 10-7C The four processes that make up the simple ideal cycle are (1) Isentropic compression in a pump, (2) P = constant heat addition in a boiler, (3) Isentropic expansion in a turbine, and (4) P = constant heat rejection in a condenser. 10-8C Heat rejected decreases; everything else increases. 10-9C Heat rejected decreases; everything else increases. 10-10C The pump work remains the same, the moisture content decreases, everything else increases. 10-11C The actual vapor power cycles differ from the idealized ones in that the actual cycles involve friction and pressure drops in various components and the piping, and heat loss to the surrounding medium from these components and piping. 10-12C The boiler exit pressure will be (a) lower than the boiler inlet pressure in actual cycles, and (b) the same as the boiler inlet pressure in ideal cycles. 10-13C We would reject this proposal because wturb = h1 - h2 - qout, and any heat loss from the steam will adversely affect the turbine work output. 10-14C Yes, because the saturation temperature of steam at 10 kPa is 45.81°C, which is much higher than the temperature of the cooling water.

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10-5

10-15E A simple ideal Rankine cycle with water as the working fluid operates between the specified pressure limits. The rates of heat addition and rejection, and the thermal efficiency of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From the steam tables (Tables A-4E, A-5E, and A-6E), h1 = h f @ 6 psia = 138.02 Btu/lbm

v 1 = v f @ 6 psia = 0.01645 ft 3 /lbm

T

wp,in = v 1 ( P2 − P1 )

⎛ 1 Btu = (0.01645 ft 3 /lbm)(500 − 6)psia ⎜ ⎜ 5.404 psia ⋅ ft 3 ⎝ = 1.50 Btu/lbm

h2 = h1 + wp,in = 138.02 + 1.50 = 139.52 Btu/lbm

⎞ ⎟ ⎟ ⎠

500 psia 2

3

qin 6 psia

1 qout P3 = 500 psia ⎫ h3 = 1630.0 Btu/lbm ⎬ T3 = 1200°F ⎭ s 3 = 1.8075 Btu/lbm ⋅ R s 4 − s f 1.8075 − 0.24739 P4 = 6 psia ⎫ x 4 = = = 0.9864 s fg 1.58155 ⎬ s 4 = s3 ⎭ h = h + x h = 138.02 + (0.9864)(995.88) = 1120.4 Btu/lbm 4 f 4 fg

4 s

Knowing the power output from the turbine the mass flow rate of steam in the cycle is determined from W& T,out = m& (h3 − h4 ) ⎯ ⎯→ m& =

W& T,out h3 − h4

=

500 kJ/s ⎛ 0.94782 Btu ⎞ ⎜ ⎟ = 0.9300 lbm/s (1630.0 − 1120.4)Btu/lbm ⎝ 1 kJ ⎠

The rates of heat addition and rejection are Q& in = m& (h3 − h2 ) = (0.9300 lbm/s)(1630.0 − 139.52)Btu/lbm = 1386 Btu/s Q& out = m& (h4 − h1 ) = (0.9300 lbm/s)(1120.4 − 138.02)Btu/lbm = 913.6 Btu/s

and the thermal efficiency of the cycle is

η th = 1 −

Q& out 913.6 = 1− = 0.341 & 1386 Q in

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10-6

10-16 A simple ideal Rankine cycle with water as the working fluid operates between the specified pressure limits. The maximum thermal efficiency of the cycle for a given quality at the turbine exit is to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis For maximum thermal efficiency, the quality at state 4 would be at its minimum of 85% (most closely approaches the Carnot cycle), and the properties at state 4 would be (Table A-5) P4 = 30 kPa ⎫ h4 = h f + x 4 h fg = 289.27 + (0.85)(2335.3) = 2274.3 kJ/kg ⎬ x 4 = 0.85 ⎭ s 4 = s f + x 4 s fg = 0.9441 + (0.85)(6.8234) = 6.7440 kJ/kg ⋅ K

Since the expansion in the turbine is isentropic, P3 = 3000 kPa ⎫ ⎬ h3 = 3115.5 kJ/kg s 3 = s 4 = 6.7440 kJ/kg ⋅ K ⎭

T 3

Other properties are obtained as follows (Tables A-4, A-5, and A-6), h1 = h f @ 30 kPa = 289.27 kJ/kg

v 1 = v f @ 30 kPa = 0.001022 m 3 /kg wp,in = v 1 ( P2 − P1 )

⎛ 1 kJ ⎞ = (0.001022 m 3 /kg )(3000 − 30)kPa ⎜ ⎟ 1 kPa ⋅ m 3 ⎠ ⎝ = 3.04 kJ/kg h2 = h1 + wp,in = 289.27 + 3.04 = 292.31 kJ/kg

3 MPa 2

qin 30 kPa 1

qout

4

Thus, q in = h3 − h2 = 3115.5 − 292.31 = 2823.2 kJ/kg q out = h4 − h1 = 2274.3 − 289.27 = 1985.0 kJ/kg and the thermal efficiency of the cycle is

η th = 1 −

q out 1985.0 = 1− = 0.297 2823.2 q in

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10-7

10-17 A simple ideal Rankine cycle with water as the working fluid operates between the specified pressure limits. The power produced by the turbine and consumed by the pump are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 20 kPa = 251.42 kJ/kg

v 1 = v f @ 20 kPa = 0.001017 m 3 /kg

T

wp,in = v 1 ( P2 − P1 )

⎛ 1 kJ ⎞ = (0.001017 m 3 /kg)(4000 − 20)kPa ⎜ ⎟ 1 kPa ⋅ m 3 ⎠ ⎝ = 4.05 kJ/kg h2 = h1 + wp,in = 251.42 + 4.05 = 255.47 kJ/kg

4 MPa 2

3

qin 20 kPa 1

qout P3 = 4000 kPa ⎫ h3 = 3906.3 kJ/kg ⎬ T3 = 700°C ⎭ s 3 = 7.6214 kJ/kg ⋅ K s4 − s f 7.6214 − 0.8320 P4 = 20 kPa ⎫ x 4 = = = 0.9596 s fg 7.0752 ⎬ s 4 = s3 ⎭ h = h + x h = 251.42 + (0.9596)(2357.5) = 2513.7 kJ/kg 4 f 4 fg

4 s

The power produced by the turbine and consumed by the pump are W& T,out = m& (h3 − h4 ) = (50 kg/s)(3906.3 − 2513.7)kJ/kg = 69,630 kW W& P,in = m& wP,in = (50 kg/s)(4.05 kJ/kg) = 203 kW

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10-8

10-18E A simple ideal Rankine cycle with water as the working fluid operates between the specified pressure limits. The turbine inlet temperature and the thermal efficiency of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

T

Analysis From the steam tables (Tables A-4E, A-5E, and A-6E),

2500 psia

h1 = h f @ 5 psia = 130.18 Btu/lbm

v 1 = v f @ 5 psia = 0.01641 ft 3 /lbm wp,in = v 1 ( P2 − P1 )

⎛ 1 Btu = (0.01641 ft /lbm)(2500 − 5)psia ⎜ ⎜ 5.404 psia ⋅ ft 3 ⎝ = 7.58 Btu/lbm 3

2 ⎞ ⎟ ⎟ ⎠

3

qin 5 psia 1

qout

4

h2 = h1 + wp,in = 130.18 + 7.58 = 137.76 Btu/lbm

s

P4 = 5 psia ⎫ h4 = h f + x 4 h fg = 130.18 + (0.80)(1000.5) = 930.58 Btu/lbm ⎬ x 4 = 0.80 ⎭ s 4 = s f + x 4 s fg = 0.23488 + (0.80)(1.60894) = 1.52203 Btu/lbm ⋅ R P3 = 2500 psia ⎫ h3 = 1450.8 Btu/lbm ⎬ s 3 = s 4 = 1.52203 Btu/lbm ⋅ R ⎭ T3 = 989.2 °F

Thus, q in = h3 − h2 = 1450.8 − 137.76 = 1313.0 Btu/lbm q out = h4 − h1 = 930.58 − 130.18 = 800.4 Btu/lbm

The thermal efficiency of the cycle is

η th = 1 −

q out 800.4 = 1− = 0.390 1313.0 q in

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10-9

10-19 A simple ideal Rankine cycle with water as the working fluid operates between the specified pressure limits. The power produced by the turbine, the heat added in the boiler, and the thermal efficiency of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 100 kPa = 417.51 kJ/kg

v 1 = v f @ 100 kPa = 0.001043 m 3 /kg

T 15 MPa

w p,in = v 1 ( P2 − P1 )

⎛ 1 kJ ⎞ = (0.001043 m 3 /kg )(15,000 − 100)kPa ⎜ ⎟ 1 kPa ⋅ m 3 ⎠ ⎝ = 15.54 kJ/kg h2 = h1 + wp,in = 417.51 + 15.54 = 433.05 kJ/kg

2

qin

3

100 kPa

1 qout 4 P3 = 15,000 kPa ⎫ h3 = 2610.8 kJ/kg ⎬ x3 = 1 ⎭ s 3 = 5.3108 kJ/kg ⋅ K s4 − s f 5.3108 − 1.3028 P4 = 100 kPa ⎫ x 4 = = = 0.6618 6.0562 s fg ⎬ s 4 = s3 ⎭ h = h + x h = 417.51 + (0.6618)(2257.5) = 1911.5 kJ/kg 4 f 4 fg Thus, wT,out = h3 − h4 = 2610.8 − 1911.5 = 699.3 kJ/kg q in = h3 − h2 = 2610.8 − 433.05 = 2177.8 kJ/kg q out = h4 − h1 = 1911.5 − 417.51 = 1494.0 kJ/kg

The thermal efficiency of the cycle is

η th = 1 −

q out 1494.0 = 1− = 0.314 q in 2177.8

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10-10

10-20 A simple Rankine cycle with water as the working fluid operates between the specified pressure limits. The isentropic efficiency of the turbine, and the thermal efficiency of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 100 kPa = 417.51 kJ/kg

v 1 = v f @ 100 kPa = 0.001043 m 3 /kg

T 15 MPa

wp,in = v 1 ( P2 − P1 )

⎛ 1 kJ ⎞ = (0.001043 m /kg )(15,000 − 100)kPa ⎜ ⎟ 1 kPa ⋅ m 3 ⎠ ⎝ = 15.54 kJ/kg h2 = h1 + wp,in = 417.51 + 15.54 = 433.05 kJ/kg 3

2

qin

3

100 kPa

1 qout 4s 4 P3 = 15,000 kPa ⎫ h3 = 2610.8 kJ/kg ⎬ x3 = 1 ⎭ s 3 = 5.3108 kJ/kg ⋅ K s4 − s f 5.3108 − 1.3028 P4 = 100 kPa ⎫ x 4 s = = = 0.6618 s fg 6.0562 ⎬ s 4 = s3 ⎭ h = h + x h = 417.51 + (0.6618)(2257.5) = 1911.5 kJ/kg f 4s 4 s fg

s

P4 = 100 kPa ⎫ ⎬ h4 = h f + x 4 h fg = 417.51 + (0.70)(2257.5) = 1997.8 kJ/kg x 4 = 0.70 ⎭

The isentropic efficiency of the turbine is

ηT =

h3 − h4 2610.8 − 1997.8 = = 0.877 h3 − h4 s 2610.8 − 1911.5

Thus, q in = h3 − h2 = 2610.8 − 433.05 = 2177.8 kJ/kg q out = h4 − h1 = 1997.8 − 417.51 = 1580.3 kJ/kg The thermal efficiency of the cycle is

η th = 1 −

q out 1580.3 = 1− = 0.274 q in 2177.8

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10-11

10-21E A simple steam Rankine cycle operates between the specified pressure limits. The mass flow rate, the power produced by the turbine, the rate of heat addition, and the thermal efficiency of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From the steam tables (Tables A-4E, A-5E, and A-6E), h1 = h f @ 1 psia = 69.72 Btu/lbm

v 1 = v f @ 6 psia = 0.01614 ft 3 /lbm wp,in = v 1 ( P2 − P1 )

⎛ 1 Btu = (0.01614 ft 3 /lbm)(2500 − 1)psia ⎜ ⎜ 5.404 psia ⋅ ft 3 ⎝ = 7.46 Btu/lbm h2 = h1 + wp,in = 69.72 + 7.46 = 77.18 Btu/lbm

T ⎞ ⎟ ⎟ ⎠

2500 psia

2

qin 1 psia

1 qout P3 = 2500 psia ⎫ h3 = 1302.0 Btu/lbm ⎬ T3 = 800°F ⎭ s 3 = 1.4116 Btu/lbm ⋅ R s 4 − s f 1.4116 − 0.13262 P4 = 1 psia ⎫ x 4 s = = = 0.6932 s fg 1.84495 ⎬ s 4 = s3 ⎭ h = h + x h = 69.72 + (0.6932)(1035.7) = 787.70 Btu/lbm 4s f 4 s fg

ηT =

3

4s 4

h3 − h4 ⎯ ⎯→ h4 = h3 − η T (h3 − h4s ) = 1302.0 − (0.90)(1302.0 − 787.70) = 839.13 kJ/kg h3 − h4 s

Thus, q in = h3 − h2 = 1302.0 − 77.18 = 1224.8 Btu/lbm q out = h4 − h1 = 839.13 − 69.72 = 769.41 Btu/lbm wnet = q in − q out = 1224.8 − 769.41 = 455.39 Btu/lbm

The mass flow rate of steam in the cycle is determined from W& 1000 kJ/s ⎛ 0.94782 Btu ⎞ ⎯→ m& = net = W& net = m& wnet ⎯ ⎜ ⎟ = 2.081 lbm/s wnet 455.39 Btu/lbm ⎝ 1 kJ ⎠

The power output from the turbine and the rate of heat addition are 1 kJ ⎛ ⎞ W& T,out = m& (h3 − h4 ) = (2.081 lbm/s)(1302.0 − 839.13)Btu/lbm⎜ ⎟ = 1016 kW 0.94782 Btu ⎝ ⎠ Q& in = m& q in = (2.081 lbm/s)(1224.8 Btu/lbm) = 2549 Btu/s

and the thermal efficiency of the cycle is

η th =

W& net 1000 kJ/s ⎛ 0.94782 Btu ⎞ = ⎜ ⎟ = 0.3718 & 2549 Btu/s ⎝ 1 kJ Qin ⎠

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10-12

10-22E A simple steam Rankine cycle operates between the specified pressure limits. The mass flow rate, the power produced by the turbine, the rate of heat addition, and the thermal efficiency of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From the steam tables (Tables A-4E, A-5E, and A-6E), h1 = h f @ 1 psia = 69.72 Btu/lbm

T

v 1 = v f @ 6 psia = 0.01614 ft 3 /lbm wp,in = v 1 ( P2 − P1 )

⎛ 1 Btu = (0.01614 ft 3 /lbm)(2500 − 1)psia ⎜ ⎜ 5.404 psia ⋅ ft 3 ⎝ = 7.46 Btu/lbm

h2 = h1 + wp,in = 69.72 + 7.46 = 77.18 Btu/lbm

⎞ ⎟ ⎟ ⎠

2500 psia 2

qin 1 psia

1 qout P3 = 2500 psia ⎫ h3 = 1302.0 Btu/lbm ⎬ T3 = 800°F ⎭ s 3 = 1.4116 Btu/lbm ⋅ R s 4 − s f 1.4116 − 0.13262 P4 = 1 psia ⎫ x 4 s = = = 0.6932 s fg 1.84495 ⎬ s 4 = s3 ⎭ h = h + x h = 69.72 + (0.6932)(1035.7) = 787.70 Btu/lbm 4s f 4 s fg

ηT =

3

4s 4 s

h3 − h4 ⎯ ⎯→ h4 = h3 − η T (h3 − h4s ) = 1302.0 − (0.90)(1302.0 − 787.70) = 839.13 kJ/kg h3 − h4 s

The mass flow rate of steam in the cycle is determined from W& net 1000 kJ/s ⎛ 0.94782 Btu ⎞ W& net = m& (h3 − h4 ) ⎯ ⎯→ m& = = ⎜ ⎟ = 2.048 lbm/s h3 − h4 (1302.0 − 839.13) Btu/lbm ⎝ 1 kJ ⎠

The rate of heat addition is 1 kJ ⎛ ⎞ Q& in = m& (h3 − h2 ) = (2.048 lbm/s)(1302.0 − 77.18)Btu/lbm⎜ ⎟ = 2508 Btu/s 0.94782 Btu ⎝ ⎠ and the thermal efficiency of the cycle is

η th =

W& net 1000 kJ/s ⎛ 0.94782 Btu ⎞ = ⎜ ⎟ = 0.3779 2508 Btu/s ⎝ 1 kJ Q& in ⎠

The thermal efficiency in the previous problem was determined to be 0.3718. The error in the thermal efficiency caused by neglecting the pump work is then Error =

0.3779 − 0.3718 × 100 = 1.64% 0.3718

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10-13

10-23 A 300-MW coal-fired steam power plant operates on a simple ideal Rankine cycle between the specified pressure limits. The overall plant efficiency and the required rate of the coal supply are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 25 kPa = 271.96 kJ/kg

v 1 = v f @ 25 kPa = 0.001020 m 3 /kg w p ,in = v 1 (P2 − P1 )

(

)

⎛ 1 kJ = 0.00102 m /kg (5000 − 25 kPa )⎜ ⎜ 1 kPa ⋅ m 3 ⎝ = 5.07 kJ/kg h2 = h1 + w p ,in = 271.96 + 5.07 = 277.03 kJ/kg 3

T ⎞ ⎟ ⎟ ⎠

P3 = 5 MPa ⎫ h3 = 3317.2 kJ/kg ⎬ T3 = 450°C ⎭ s 3 = 6.8210 kJ/kg ⋅ K

3 2 1

5 MPa · Qin 25 kPa · Qout

4

s 4 − s f 6.8210 − 0.8932 P4 = 25 kPa ⎫ = = 0.8545 ⎬ x4 = s 4 = s3 s fg 6.9370 ⎭

h4 = h f + x 4 h fg = 271.96 + (0.8545)(2345.5) = 2276.2 kJ/kg

The thermal efficiency is determined from qin = h3 − h2 = 3317.2 − 277.03 = 3040.2 kJ/kg qout = h4 − h1 = 2276.2 − 271.96 = 2004.2 kJ/kg

and

η th = 1 − Thus,

q out 2004.2 = 1− = 0.3407 q in 3040.2

η overall = η th ×η comb ×η gen = (0.3407 )(0.75)(0.96 ) = 24.5%

(b) Then the required rate of coal supply becomes

and

W& net 300,000 kJ/s = = 1,222,992 kJ/s Q& in = 0.2453 η overall m& coal =

Q& in 1,222,992 kJ/s ⎛ 1 ton ⎞ ⎜ ⎟ = 0.04174 tons/s = 150.3 tons/h = 29,300 kJ/kg ⎜⎝ 1000 kg ⎟⎠ C coal

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10-14

10-24 A solar-pond power plant that operates on a simple ideal Rankine cycle with refrigerant-134a as the working fluid is considered. The thermal efficiency of the cycle and the power output of the plant are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the refrigerant tables (Tables A-11, A-12, and A-13), h1 = h f @ 0.7 MPa = 88.82 kJ/kg

v 1 = v f @ 0.7 MPa = 0.0008331 m 3 /kg

T

w p ,in = v 1 (P2 − P1 )

(

)

⎛ 1 kJ = 0.0008331 m 3 /kg (1400 − 700 kPa )⎜ ⎜ 1 kPa ⋅ m 3 ⎝ = 0.58 kJ/kg

⎞ ⎟ ⎟ ⎠

2

h2 = h1 + w p ,in = 88.82 + 0.58 = 89.40 kJ/kg P3 = 1.4 MPa ⎫ h3 = h g @ 1.4 MPa = 276.12 kJ/kg ⎬ sat.vapor ⎭ s 3 = s g @ 1.4 MPa = 0.9105 kJ/kg ⋅ K

1.4 MPa qin R-134a

3

0.7 MPa 1

qout

4 s

s 4 − s f 0.9105 − 0.33230 P4 = 0.7 MPa ⎫ = = 0.9839 ⎬ x4 = s 4 = s3 s fg 0.58763 ⎭ h4 = h f + x 4 h fg = 88.82 + (0.9839)(176.21) = 262.20 kJ/kg

Thus , q in = h3 − h2 = 276.12 − 89.40 = 186.72 kJ/kg q out = h4 − h1 = 262.20 − 88.82 = 173.38 kJ/kg wnet = q in − q out = 186.72 − 173.38 = 13.34 kJ/kg

and

η th = (b)

wnet 13.34 kJ/kg = = 7.1% q in 186.72 kJ/kg

W& net = m& wnet = (3 kg/s )(13.34 kJ/kg ) = 40.02 kW

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-15

10-25 A steam power plant operates on a simple ideal Rankine cycle between the specified pressure limits. The thermal efficiency of the cycle, the mass flow rate of the steam, and the temperature rise of the cooling water are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 10 kPa = 191.81 kJ/kg

v 1 = v f @ 10 kPa = 0.00101 m 3 /kg

T

w p ,in = v 1 (P2 − P1 )

(

)

⎛ 1 kJ = 0.00101 m 3 /kg (7,000 − 10 kPa )⎜ ⎜ 1 kPa ⋅ m 3 ⎝ = 7.06 kJ/kg

⎞ ⎟ ⎟ ⎠

h2 = h1 + w p ,in = 191.81 + 7.06 = 198.87 kJ/kg

3 7 MPa qin

2

10 kPa 1

qout

4

P3 = 7 MPa ⎫ h3 = 3411.4 kJ/kg ⎬ T3 = 500°C ⎭ s 3 = 6.8000 kJ/kg ⋅ K s 4 − s f 6.8000 − 0.6492 P4 = 10 kPa ⎫ = = 0.8201 ⎬ x4 = s 4 = s3 s fg 7.4996 ⎭

h4 = h f + x 4 h fg = 191.81 + (0.8201)(2392.1) = 2153.6 kJ/kg

Thus, q in = h3 − h2 = 3411.4 − 198.87 = 3212.5 kJ/kg q out = h4 − h1 = 2153.6 − 191.81 = 1961.8 kJ/kg wnet = q in − q out = 3212.5 − 1961.8 = 1250.7 kJ/kg

and

η th = (b)

m& =

wnet 1250.7 kJ/kg = = 38.9% q in 3212.5 kJ/kg

W&net 45,000 kJ/s = = 36.0 kg/s wnet 1250.7 kJ/kg

(c) The rate of heat rejection to the cooling water and its temperature rise are Q& out = m& q out = (35.98 kg/s )(1961.8 kJ/kg ) = 70,586 kJ/s Q& out 70,586 kJ/s ΔTcooling water = = = 8.4°C (m& c) cooling water (2000 kg/s )(4.18 kJ/kg ⋅ °C )

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10-16

10-26 A steam power plant operates on a simple nonideal Rankine cycle between the specified pressure limits. The thermal efficiency of the cycle, the mass flow rate of the steam, and the temperature rise of the cooling water are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 10 kPa = 191.81 kJ/kg

T

v 1 = v f @ 10 kPa = 0.00101 m 3 /kg w p ,in = v 1 (P2 − P1 ) / η p

(

)

⎛ 1 kJ = 0.00101 m /kg (7,000 − 10 kPa )⎜ ⎜ 1 kPa ⋅ m 3 ⎝ = 8.11 kJ/kg 3

⎞ ⎟ / (0.87 ) ⎟ ⎠

h2 = h1 + w p ,in = 191.81 + 8.11 = 199.92 kJ/kg P3 = 7 MPa ⎫ h3 = 3411.4 kJ/kg ⎬ T3 = 500°C ⎭ s 3 = 6.8000 kJ/kg ⋅ K

2

2

7 MPa qin

3

10 kPa 1

qout

4 4

s 4 − s f 6.8000 − 0.6492 P4 = 10 kPa ⎫ = = 0.8201 ⎬ x4 = s 4 = s3 s fg 7.4996 ⎭

h4 s = h f + x 4 h fg = 191.81 + (0.820)(2392.1) = 2153.6 kJ/kg

ηT =

h3 − h4 ⎯ ⎯→ h4 = h3 − ηT (h3 − h4 s ) h3 − h4 s = 3411.4 − (0.87 )(3411.4 − 2153.6) = 2317.1 kJ/kg

Thus, qin = h3 − h2 = 3411.4 − 199.92 = 3211.5 kJ/kg qout = h4 − h1 = 2317.1 − 191.81 = 2125.3 kJ/kg wnet = qin − qout = 3211.5 − 2125.3 = 1086.2 kJ/kg

and

η th = (b)

m& =

wnet 1086.2 kJ/kg = = 33.8% q in 3211.5 kJ/kg

45,000 kJ/s W&net = = 41.43 kg/s wnet 1086.2 kJ/kg

(c) The rate of heat rejection to the cooling water and its temperature rise are Q& out = m& q out = (41.43 kg/s )(2125.3 kJ/kg ) = 88,051 kJ/s ΔTcooling water =

Q& out (m& c) cooling water

=

88,051 kJ/s = 10.5°C (2000 kg/s )(4.18 kJ/kg ⋅ °C)

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10-17

10-27 The net work outputs and the thermal efficiencies for a Carnot cycle and a simple ideal Rankine cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) Rankine cycle analysis: From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 20 kPa = 251.42 kJ/kg

v1 = v f @ 20 kPa = 0.001017 m 3 /kg w p ,in = v1 (P2 − P1 )

(

Rankine cycle

T

)

⎛ 1 kJ ⎞ ⎟ = 0.001017 m3 /kg (10,000 − 20 ) kPa ⎜⎜ 1 kPa ⋅ m 3 ⎟⎠ ⎝ = 10.15 kJ/kg

h2 = h1 + w p ,in = 251.42 + 10.15 = 261.57 kJ/kg P3 = 10 MPa ⎫ h3 = 2725.5 kJ/kg ⎬ x3 = 1 ⎭ s 3 = 5.6159 kJ/kg ⋅ K

3 2 4

1

s

s4 − s f P4 = 20 kPa ⎫ 5.6159 − 0.8320 = = 0.6761 ⎬ x4 = s 4 = s3 7.0752 s fg ⎭ h4 = h f + x 4 h fg = 251.42 + (0.6761)(2357.5) = 1845.3 kJ/kg q in = h3 − h2 = 2725.5 − 261.57 = 2463.9 kJ/kg q out = h4 − h1 = 1845.3 − 251.42 = 1594.0 kJ/kg wnet = q in − q out = 2463.9 − 1594.0 = 869.9 kJ/kg

η th = 1 −

q out 1594.0 = 1− = 0.353 2463.9 q in

(b) Carnot Cycle analysis: P3 = 10 MPa ⎫ h3 = 2725.5 kJ/kg ⎬ x3 = 1 ⎭ T3 = 311.0 °C T2 = T3 = 311.0 °C ⎫ h2 = 1407.8 kJ/kg ⎬ x2 = 0 ⎭ s 2 = 3.3603 kJ/kg ⋅ K x1 =

s1 − s f

=

P1 = 20 kPa ⎫ s fg ⎬h = h +x h s1 = s 2 f 1 fg ⎭ 1

3.3603 − 0.8320 = 0.3574 7.0752

T

Carnot cycle 2

3

1

4 s

= 251.42 + (0.3574)(2357.5) = 1093.9 kJ/kg q in = h3 − h2 = 2725.5 − 1407.8 = 1317.7 kJ/kg q out = h4 − h1 = 1845.3 − 1093.9 = 751.4 kJ/kg wnet = q in − q out = 1317.7 − 752.3 = 565.4 kJ/kg

η th = 1 −

q out 751.4 = 1− = 0.430 1317.7 q in

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-18

10-28 A single-flash geothermal power plant uses hot geothermal water at 230ºC as the heat source. The mass flow rate of steam through the turbine, the isentropic efficiency of the turbine, the power output from the turbine, and the thermal efficiency of the plant are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) We use properties of water for geothermal water (Tables A-4 through A-6) T1 = 230°C⎫ ⎬ h1 = 990.14 kJ/kg x1 = 0 ⎭ h2 − h f P2 = 500 kPa ⎫ ⎬x2 = h2 = h1 = 990.14 kJ/kg ⎭ h fg 990.14 − 640.09 = 2108 2 = 0.1661

The mass flow rate of steam through the turbine is

separator

4

condenser Flash chamber

= (0.1661)(230 kg/s)

(b) Turbine:

steam turbine

6

m& 3 = x 2 m& 1 = 38.20 kg/s

3

production well

1

5

reinjection well

P3 = 500 kPa ⎫ h3 = 2748.1 kJ/kg ⎬ x3 = 1 ⎭ s 3 = 6.8207 kJ/kg ⋅ K P4 = 10 kPa ⎫ ⎬h4 s = 2160.3 kJ/kg s 4 = s3 ⎭ P4 = 10 kPa ⎫ ⎬h4 = h f + x 4 h fg = 191.81 + (0.90)(2392.1) = 2344.7 kJ/kg x 4 = 0.90 ⎭

ηT =

h3 − h4 2748.1 − 2344.7 = = 0.686 h3 − h4 s 2748.1 − 2160.3

(c) The power output from the turbine is W& T,out = m& 3 (h3 − h4 ) = (38.20 kJ/kg)(2748.1 − 2344.7)kJ/kg = 15,410 kW

(d) We use saturated liquid state at the standard temperature for dead state enthalpy T0 = 25°C⎫ ⎬ h0 = 104.83 kJ/kg x0 = 0 ⎭ E& in = m& 1 (h1 − h0 ) = (230 kJ/kg)(990.14 − 104.83)kJ/kg = 203,622 kW

η th =

W& T,out 15,410 = = 0.0757 = 7.6% & 203,622 E in

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-19

10-29 A double-flash geothermal power plant uses hot geothermal water at 230ºC as the heat source. The temperature of the steam at the exit of the second flash chamber, the power produced from the second turbine, and the thermal efficiency of the plant are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) We use properties of water for geothermal water (Tables A-4 through A-6) T1 = 230°C⎫ ⎬ h1 = 990.14 kJ/kg x1 = 0 ⎭ P2 = 500 kPa ⎫ ⎬ x 2 = 0.1661 h2 = h1 = 990.14 kJ/kg ⎭

3

m& 3 = x2 m& 1 = (0.1661)(230 kg/s) = 38.20 kg/s m& 6 = m& 1 − m& 3 = 230 − 0.1661 = 191.80 kg/s P3 = 500 kPa ⎫ ⎬ h3 = 2748.1 kJ/kg x3 = 1 ⎭ P4 = 10 kPa ⎫ ⎬h4 = 2344.7 kJ/kg x 4 = 0.90 ⎭

steam turbine

8

4

separator 2

P6 = 500 kPa ⎫ ⎬ h6 = 640.09 kJ/kg x6 = 0 ⎭ P7 = 150 kPa ⎫ T7 = 111.35 °C ⎬ h7 = h6 ⎭ x 7 = 0.0777 P8 = 150 kPa ⎫ ⎬h8 = 2693.1 kJ/kg x8 = 1 ⎭

6 Flash chamber

7 separator Flash chamber

1 production well

condenser 5

9

reinjection well

(b) The mass flow rate at the lower stage of the turbine is m& 8 = x7 m& 6 = (0.0777)(191.80 kg/s) = 14.90 kg/s

The power outputs from the high and low pressure stages of the turbine are W&T1, out = m& 3 (h3 − h4 ) = (38.20 kJ/kg)(2748.1 − 2344.7)kJ/kg = 15,410 kW W&T2, out = m& 8 (h8 − h4 ) = (14.90 kJ/kg)(2693.1 − 2344.7)kJ/kg = 5191 kW

(c) We use saturated liquid state at the standard temperature for the dead state enthalpy T0 = 25°C⎫ ⎬ h0 = 104.83 kJ/kg x0 = 0 ⎭ E& in = m& 1 (h1 − h0 ) = (230 kg/s)(990.14 − 104.83)kJ/kg = 203,621 kW

η th =

W& T, out 15,410 + 5193 = = 0.101 = 10.1% 203,621 E& in

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-20

10-30 A combined flash-binary geothermal power plant uses hot geothermal water at 230ºC as the heat source. The mass flow rate of isobutane in the binary cycle, the net power outputs from the steam turbine and the binary cycle, and the thermal efficiencies for the binary cycle and the combined plant are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) We use properties of water for geothermal water (Tables A-4 through A-6) T1 = 230°C⎫ ⎬ h1 = 990.14 kJ/kg x1 = 0 ⎭ P2 = 500 kPa ⎫ ⎬ x 2 = 0.1661 h2 = h1 = 990.14 kJ/kg ⎭

m& 3 = x2 m& 1 = (0.1661)(230 kg/s) = 38.20 kg/s m& 6 = m& 1 − m& 3 = 230 − 38.20 = 191.80 kg/s P3 = 500 kPa ⎫ ⎬ h3 = 2748.1 kJ/kg x3 = 1 ⎭

3 separator

steam turbine

P4 = 10 kPa ⎫ ⎬h4 = 2344.7 kJ/kg x 4 = 0.90 ⎭

condenser

4 6

P6 = 500 kPa ⎫ ⎬ h6 = 640.09 kJ/kg x6 = 0 ⎭

9

1

isobutane turbine

2

T7 = 90°C ⎫ ⎬ h7 = 377.04 kJ/kg x7 = 0 ⎭

BINARY CYCLE

8

The isobutane properties are obtained from EES:

pump

heat exchanger flash chamber

P8 = 3250 kPa ⎫ ⎬ h8 = 755.05 kJ/kg T8 = 145°C ⎭

1

1

7

production well

P9 = 400 kPa ⎫ ⎬ h9 = 691.01 kJ/kg T9 = 80°C ⎭

5

air-cooled condenser

reinjection well

P10 = 400 kPa ⎫ h10 = 270.83 kJ/kg ⎬ 3 x10 = 0 ⎭ v 10 = 0.001839 m /kg w p ,in = v10 (P11 − P10 ) / η p

(

)

⎛ 1 kJ ⎞ ⎟ / 0.90 = 0.001819 m 3/kg (3250 − 400 ) kPa ⎜⎜ 3⎟ 1 kPa m ⋅ ⎝ ⎠ = 5.82 kJ/kg.

h11 = h10 + w p ,in = 270.83 + 5.82 = 276.65 kJ/kg

An energy balance on the heat exchanger gives m& 6 (h6 − h7 ) = m& iso (h8 − h11 ) (191.81 kg/s)(640.09 - 377.04)kJ/kg = m& iso (755.05 - 276.65)kJ/kg ⎯ ⎯→ m& iso = 105.46 kg/s

(b) The power outputs from the steam turbine and the binary cycle are W&T,steam = m& 3 (h3 − h4 ) = (38.19 kJ/kg)(2748.1 − 2344.7)kJ/kg = 15,410 kW

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10-21

W& T,iso = m& iso (h8 − h9 ) = (105.46 kJ/kg)(755.05 − 691.01)kJ/kg = 6753 kW W& net,binary = W& T,iso − m& iso w p ,in = 6753 − (105.46 kg/s)(5.82 kJ/kg ) = 6139 kW

(c) The thermal efficiencies of the binary cycle and the combined plant are

Q& in,binary = m& iso (h8 − h11 ) = (105.46 kJ/kg)(755.05 − 276.65)kJ/kg = 50,454 kW

η th,binary =

W& net, binary 6139 = = 0.122 = 12.2% & 50,454 Qin, binary

T0 = 25°C⎫ ⎬ h0 = 104.83 kJ/kg x0 = 0 ⎭ E& in = m& 1 (h1 − h0 ) = (230 kJ/kg)(990.14 − 104.83)kJ/kg = 203,622 kW

η th, plant =

W& T,steam + W& net, binary 15,410 + 6139 = = 0.106 = 10.6% 203,622 E& in

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-22

The Reheat Rankine Cycle

10-31C The pump work remains the same, the moisture content decreases, everything else increases. 10-32C The T-s diagram shows two reheat cases for the reheat Rankine cycle similar to the one shown in Figure 10-11. In the first case there is expansion through the high-pressure turbine from 6000 kPa to 4000 kPa between states 1 and 2 with reheat at 4000 kPa to state 3 and finally expansion in the low-pressure turbine to state 4. In the second case there is expansion through the high-pressure turbine from 6000 kPa to 500 kPa between states 1 and 5 with reheat at 500 kPa to state 6 and finally expansion in the low-pressure turbine to state 7. Increasing the pressure for reheating increases the average temperature for heat addition makes the energy of the steam more available for doing work, see the reheat process 2 to 3 versus the reheat process 5 to 6. Increasing the reheat pressure will increase the cycle efficiency. However, as the reheating pressure increases, the amount of condensation increases during the expansion process in the lowpressure turbine, state 4 versus state 7. An optimal pressure for reheating generally allows for the moisture content of the steam at the low-pressure turbine exit to be in the range of 10 to 15% and this corresponds to quality in the range of 85 to 90%. SteamIAPWS

900

800

700

T [K]

1 600

3

6

2 6000 kPa 4000 kPa

500

5

500 kPa

400 0.2

300

0.4

20 kPa

0.6

0.8

4

7

200 0

20

40

60

80

100

120

140

160

180

s [kJ/kmol-K]

10-33C The thermal efficiency of the simple ideal Rankine cycle will probably be higher since the average temperature at which heat is added will be higher in this case.

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10-23

10-34 [Also solved by EES on enclosed CD] A steam power plant that operates on the ideal reheat Rankine cycle is considered. The turbine work output and the thermal efficiency of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 20 kPa = 251.42 kJ/kg

v1 = v f @ 20 kPa = 0.001017 m3 /kg w p ,in = v1 (P2 − P1 )

(

T 3

)

⎛ 1 kJ ⎞ ⎟ = 0.001017 m 3 /kg (8000 − 20 kPa )⎜⎜ 1 kPa ⋅ m3 ⎟⎠ ⎝ = 8.12 kJ/kg

h2 = h1 + w p ,in = 251.42 + 8.12 = 259.54 kJ/kg P3 = 8 MPa ⎫ h3 = 3399.5 kJ/kg ⎬ T3 = 500°C ⎭ s3 = 6.7266 kJ/kg ⋅ K

5

8 MPa 4 2 20 kPa 1

6 s

P4 = 3 MPa ⎫ ⎬ h4 = 3105.1 kJ/kg s4 = s3 ⎭ P5 = 3 MPa ⎫ h5 = 3457.2 kJ/kg ⎬ T5 = 500°C ⎭ s5 = 7.2359 kJ/kg ⋅ K s6 − s f 7.2359 − 0.8320 = = 0.9051 P6 = 20 kPa ⎫ x6 = s fg 7.0752 ⎬ s6 = s5 ⎭ h6 = h f + x6 h fg = 251.42 + (0.9051)(2357.5) = 2385.2 kJ/kg

The turbine work output and the thermal efficiency are determined from wT,out = (h3 − h4 ) + (h5 − h6 ) = 3399.5 − 3105.1 + 3457.2 − 2385.2 = 1366.4 kJ/kg

and

q in = (h3 − h2 ) + (h5 − h4 ) = 3399.5 − 259.54 + 3457.2 − 3105.1 = 3492.0 kJ/kg wnet = wT ,out − w p ,in = 1366.4 − 8.12 = 1358.3 kJ/kg

Thus,

η th =

wnet 1358.3 kJ/kg = = 38.9% 3492.5 kJ/kg q in

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-24

10-35 EES Problem 10-34 is reconsidered. The problem is to be solved by the diagram window data entry feature of EES by including the effects of the turbine and pump efficiencies and reheat on the steam quality at the low-pressure turbine exit Also, the T-s diagram is to be plotted. Analysis The problem is solved using EES, and the solution is given below. "Input Data - from diagram window" {P[6] = 20 [kPa] P[3] = 8000 [kPa] T[3] = 500 [C] P[4] = 3000 [kPa] T[5] = 500 [C] Eta_t = 100/100 "Turbine isentropic efficiency" Eta_p = 100/100 "Pump isentropic efficiency"} "Pump analysis" function x6$(x6) "this function returns a string to indicate the state of steam at point 6" x6$='' if (x6>1) then x6$='(superheated)' if (x6 w_comp_isen" h[8] + w_gas_comp_isen =hs9"SSSF conservation of energy for the isentropic compressor, assuming: adiabatic, ke=pe=0 per unit gas mass flow rate in kg/s" h[8]=ENTHALPY(Air,T=T[8]) hs9=ENTHALPY(Air,T=Ts9) h[8] + w_gas_comp = h[9]"SSSF conservation of energy for the actual compressor, assuming: adiabatic, ke=pe=0" T[9]=temperature(Air,h=h[9]) s[9]=ENTROPY(Air,T=T[9],P=P[9]) "Gas Cycle External heat exchanger analysis" h[9] + q_in = h[10]"SSSF conservation of energy for the external heat exchanger, assuming W=0, ke=pe=0" h[10]=ENTHALPY(Air,T=T[10]) P[10]=P[9] "Assume process 9-10 is SSSF constant pressure" Q_dot_in"MW"*1000"kW/MW"=m_dot_gas*q_in "Gas Turbine analysis" s[10]=ENTROPY(Air,T=T[10],P=P[10]) ss11=s[10] "For the ideal case the entropies are constant across the turbine" P[11] = P[10] /Pratio Ts11=temperature(Air,s=ss11,P=P[11])"Ts11 is the isentropic value of T[11] at gas turbine exit" Eta_gas_turb = w_gas_turb /w_gas_turb_isen "gas turbine adiabatic efficiency, w_gas_turb_isen > w_gas_turb"

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86

h[10] = w_gas_turb_isen + hs11"SSSF conservation of energy for the isentropic gas turbine, assuming: adiabatic, ke=pe=0" hs11=ENTHALPY(Air,T=Ts11) h[10] = w_gas_turb + h[11]"SSSF conservation of energy for the actual gas turbine, assuming: adiabatic, ke=pe=0" T[11]=temperature(Air,h=h[11]) s[11]=ENTROPY(Air,T=T[11],P=P[11]) "Gas-to-Steam Heat Exchanger" "SSSF conservation of energy for the gas-to-steam heat exchanger, assuming: adiabatic, W=0, ke=pe=0" m_dot_gas*h[11] + m_dot_steam*h[4] = m_dot_gas*h[12] + m_dot_steam*h[5] h[12]=ENTHALPY(Air, T=T[12]) s[12]=ENTROPY(Air,T=T[12],P=P[12]) "STEAM CYCLE ANALYSIS" "Steam Condenser exit pump or Pump 1 analysis" Fluid$='Steam_IAPWS' P[1] = P[7] P[2]=P[6] h[1]=enthalpy(Fluid$,P=P[1],x=0) {Saturated liquid} v1=volume(Fluid$,P=P[1],x=0) s[1]=entropy(Fluid$,P=P[1],x=0) T[1]=temperature(Fluid$,P=P[1],x=0) w_pump1_s=v1*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume" w_pump1=w_pump1_s/Eta_pump "Definition of pump efficiency" h[1]+w_pump1= h[2] "Steady-flow conservation of energy" s[2]=entropy(Fluid$,P=P[2],h=h[2]) T[2]=temperature(Fluid$,P=P[2],h=h[2]) "Open Feedwater Heater analysis" y*h[6] + (1-y)*h[2] = 1*h[3] "Steady-flow conservation of energy" P[3]=P[6] h[3]=enthalpy(Fluid$,P=P[3],x=0) "Condensate leaves heater as sat. liquid at P[3]" T[3]=temperature(Fluid$,P=P[3],x=0) s[3]=entropy(Fluid$,P=P[3],x=0) "Boiler condensate pump or Pump 2 analysis" P[4] = P[5] v3=volume(Fluid$,P=P[3],x=0) w_pump2_s=v3*(P[4]-P[3])"SSSF isentropic pump work assuming constant specific volume" w_pump2=w_pump2_s/Eta_pump "Definition of pump efficiency" h[3]+w_pump2= h[4] "Steady-flow conservation of energy" s[4]=entropy(Fluid$,P=P[4],h=h[4]) T[4]=temperature(Fluid$,P=P[4],h=h[4]) w_steam_pumps = (1-y)*w_pump1+ w_pump2 "Total steam pump work input/ mass steam" "Steam Turbine analysis" h[5]=enthalpy(Fluid$,T=T[5],P=P[5]) s[5]=entropy(Fluid$,P=P[5],T=T[5]) ss6=s[5] hs6=enthalpy(Fluid$,s=ss6,P=P[6]) Ts6=temperature(Fluid$,s=ss6,P=P[6]) h[6]=h[5]-Eta_steam_turb*(h[5]-hs6)"Definition of steam turbine efficiency" T[6]=temperature(Fluid$,P=P[6],h=h[6]) s[6]=entropy(Fluid$,P=P[6],h=h[6]) ss7=s[5] hs7=enthalpy(Fluid$,s=ss7,P=P[7]) Ts7=temperature(Fluid$,s=ss7,P=P[7]) PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

87

h[7]=h[5]-Eta_steam_turb*(h[5]-hs7)"Definition of steam turbine efficiency" T[7]=temperature(Fluid$,P=P[7],h=h[7]) s[7]=entropy(Fluid$,P=P[7],h=h[7]) "SSSF conservation of energy for the steam turbine: adiabatic, neglect ke and pe" h[5] = w_steam_turb + y*h[6] +(1-y)*h[7] "Steam Condenser analysis" (1-y)*h[7]=q_out+(1-y)*h[1]"SSSF conservation of energy for the Condenser per unit mass" Q_dot_out*Convert(MW, kW)=m_dot_steam*q_out "Cycle Statistics" MassRatio_gastosteam =m_dot_gas/m_dot_steam W_dot_net*Convert(MW, kW)=m_dot_gas*(w_gas_turb-w_gas_comp)+ m_dot_steam*(w_steam_turb - w_steam_pumps)"definition of the net cycle work" Eta_th=W_dot_net/Q_dot_in*Convert(, %) "Cycle thermal efficiency, in percent" Bwr=(m_dot_gas*w_gas_comp + m_dot_steam*w_steam_pumps)/(m_dot_gas*w_gas_turb + m_dot_steam*w_steam_turb) "Back work ratio" W_dot_net_steam = m_dot_steam*(w_steam_turb - w_steam_pumps) W_dot_net_gas = m_dot_gas*(w_gas_turb - w_gas_comp) NetWorkRatio_gastosteam = W_dot_net_gas/W_dot_net_steam

Pratio

MassRatio

Wnetgas [kW] 342944 349014 354353 359110 363394 367285 370849 374135 377182 380024 382687

gastosteam

10 11 12 13 14 15 16 17 18 19 20

7.108 7.574 8.043 8.519 9.001 9.492 9.993 10.51 11.03 11.57 12.12

ηth [%] 59.92 60.65 61.29 61.86 62.37 62.83 63.24 63.62 63.97 64.28 64.57

Wnetsteam [kW] 107056 100986 95647 90890 86606 82715 79151 75865 72818 69976 67313

NetWorkRatio gastosteam

3.203 3.456 3.705 3.951 4.196 4.44 4.685 4.932 5.18 5.431 5.685

C o m b in ed G as an d S team P o w er C ycle 1600 1500

10

1400 1300

G as C ycle

1200 1100

T [K]

1000

S te am C yc le

900

11

800

9

700

5

600

8000 kP a

500 400

12

3,4

600 kP a

6

1,2 20 kP a

300 200 0.0

8 1.1

2.2

3.3

4.4

5.5

7 6.6

7.7

8.8

9.9

11.0

s [kJ/kg -K ]

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

88

66

6 3 .8

η th [%]

6 1 .6

5 9 .4

5 7 .2

55 5

9

13

17

21

25

P ra tio W 6.5

dot,gas

/W

dot,steam

vs Gas Pressure Ratio

NetW orkRatio gastosteam

6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 5

9

14

18

23

Pratio

Ratio of Gas Flow Rate to Steam Flow Rate vs Gas Pressure Ratio 14.0 13.0

M assRatio gastosteam

12.0 11.0 10.0 9.0 8.0 7.0 6.0 5.0 5

9

14

18

23

Pratio

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89

10-86 A combined gas-steam power cycle uses a simple gas turbine for the topping cycle and simple Rankine cycle for the bottoming cycle. The mass flow rate of air for a specified power output is to be determined. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable fo Brayton cycle. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2a). Analysis Working around the topping cycle gives the following results: T6 s

⎛P = T5 ⎜⎜ 6 ⎝ P5

ηC =

⎞ ⎟ ⎟ ⎠

( k −1) / k

= (293 K)(8)

0.4/1.4

= 530.8 K

T 7

1373 K · Qin

c p (T6 s − T5 )

h6 s − h5 = h 6 − h5 c p (T6 − T5 )

⎯ ⎯→ T6 = T5 +

6

T6 s − T5

6s

ηC

T8 s

⎞ ⎟ ⎟ ⎠

( k −1) / k

⎛1⎞ = (1373 K)⎜ ⎟ ⎝8⎠

8s 6 MPa

530.8 − 293 = 293 + = 572.8 K 0.85 ⎛P = T7 ⎜⎜ 8 ⎝ P7

GAS CYCLE

8 3 320°C

9

0.4/1.4

= 758.0 K

293 K

5

2 1

STEAM CYCLE 20 kPa · 4s Qout

4

c p (T7 − T8 ) h −h ηT = 7 8 = ⎯ ⎯→ T8 = T7 − η T (T7 − T8 s ) h7 − h8 s c p (T7 − T8 s ) = 1373 − (0.90)(1373 − 758.0) = 819.5 K T9 = Tsat @ 6000 kPa = 275.6°C = 548.6 K

Fixing the states around the bottom steam cycle yields (Tables A-4, A-5, A-6): h1 = h f @ 20 kPa = 251.42 kJ/kg

v 1 = v f @ 20 kPa = 0.001017 m 3 /kg wp,in = v 1 ( P2 − P1 )

⎛ 1 kJ ⎞ = (0.001017 m 3 /kg )(6000 − 20)kPa ⎜ ⎟ ⎝ 1 kPa ⋅ m 3 ⎠ = 6.08 kJ/kg h2 = h1 + wp,in = 251.42 + 6.08 = 257.5 kJ/kg

P3 = 6000 kPa ⎫ h3 = 2953.6 kJ/kg ⎬ ⎭ s 3 = 6.1871 kJ/kg ⋅ K P4 = 20 kPa ⎫ ⎬ h4 s = 2035.8 kJ/kg s 4 = s3 ⎭

T3 = 320°C

ηT =

h3 − h4 ⎯ ⎯→h4 = h3 − η T (h3 − h4 s ) h3 − h4 s = 2953.6 − (0.90)(2953.6 − 2035.8) = 2127.6 kJ/kg

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

s

90

The net work outputs from each cycle are wnet, gas cycle = wT, out − wC,in = c p (T7 − T8 ) − c p (T6 − T5 ) = (1.005 kJ/kg ⋅ K )(1373 − 819.5 − 572.7 + 293)K = 275.2 kJ/kg w net, steam cycle = wT,out − wP,in = (h3 − h4 ) − w P,in = (2953.6 − 2127.6) − 6.08 = 819.9 kJ/kg

An energy balance on the heat exchanger gives m& a c p (T8 − T9 ) = m& w (h3 -h2 ) ⎯ ⎯→ m& w =

c p (T8 − T9 ) h3 -h2

m& a =

(1.005)(819.5 − 548.6) = 0.1010m& a 2953.6 − 257.5

That is, 1 kg of exhaust gases can heat only 0.1010 kg of water. Then, the mass flow rate of air is m& a =

W& net 100,000 kJ/s = = 279.3 kg/s wnet (1× 275.2 + 0.1010 × 819.9) kJ/kg air

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91

10-87 A combined gas-steam power cycle uses a simple gas turbine for the topping cycle and simple Rankine cycle for the bottoming cycle. The mass flow rate of air for a specified power output is to be determined. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable fo Brayton cycle. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2a). Analysis With an ideal regenerator, the temperature of the air at the compressor exit will be heated to the to the temperature at the turbine exit. Representing this state by “6a”

T 7

1373 K · Qin 6a

T6 a = T8 = 819.5 K

6

The rate of heat addition in the cycle is

8s

6s

6 MPa

Q& in = m& a c p (T7 − T6 a ) = 155,370 kW

η th

W& 100,000 kW = net = = 0.6436 & 155,370 kW Qin

8 3 320°C

9

= (279.3 kg/s)(1.005 kJ/kg ⋅ °C)(1373 − 819.5) K

The thermal efficiency of the cycle is then

GAS CYCLE

293 K

5

2 1

STEAM CYCLE 20 kPa · 4s Qout

4

Without the regenerator, the rate of heat addition and the thermal efficiency are Q& in = m& a c p (T7 − T6 ) = (279.3 kg/s)(1.005 kJ/kg ⋅ °C)(1373 − 572.7) K = 224,640 kW

η th =

W& net 100,000 kW = = 0.4452 224,640 kW Q& in

The change in the thermal efficiency due to using the ideal regenerator is Δη th = 0.6436 − 0.4452 = 0.1984

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s

92

10-88 The component of the combined cycle with the largest exergy destruction of the component of the combined cycle in Prob. 10-86 is to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

T 7

1373 K

Analysis From Problem 10-86,

· Qin

Tsource, gas cycle = 1373 K Tsource, steam cycle = T8 = 819.5 K Tsink = 293 K

GAS CYCLE

6 6s

s1 = s 2 = s f @ 20 kPa = 0.8320 kJ/kg ⋅ K

8s 6 MPa

s 3 = 6.1871 kJ/kg ⋅ K

8 3 320°C

9

s 4 = 6.4627 kJ/kg ⋅ K q in,67 = c p (T7 − T6 ) = 804.3 kJ/kg q in,23 = h3 − h2 = 2696.1 kJ/kg

293 K

q out = h4 − h1 = 1876.2 kJ/kg m& w = 0.1010m& a = 0.1010(279.3) = 28.21 kg/s

5

2 1

STEAM CYCLE 20 kPa · 4s Qout

4 s

X& destroyed,12 = 0 (isentropic process) X& destroyed, 34 = m& wT0 (s 4 − s 3 ) = (28.21 kg/s)(293 K )(6.4627 − 6.1871) = 2278 kW ⎛ q X& destroyed, 41 = m& wT0 ⎜⎜ s1 − s 4 + out Tsink ⎝

⎞ ⎟ ⎟ ⎠

1876.2 kJ/kg ⎞ ⎛ = (28.21 kg/s)(293 K )⎜ 0.8320 − 6.1871 + ⎟ = 8665 kW 293 K ⎝ ⎠ ⎛ T ⎞ X& destroyed,heat exchanger = m& a T0 Δs 89 + m& wT0 Δs 23 = m& a T0 ⎜⎜ c p ln 9 ⎟⎟ + m& a T0 ( s 3 − s 2 ) T8 ⎠ ⎝ 548.6 ⎤ ⎡ + (28.21)(293)(6.1871 − 0.8320) = (279.3)(293) ⎢(1.005) ln 819.5 ⎥⎦ ⎣ = 11260 kW ⎛ T P ⎞ 572.7 ⎡ ⎤ X& destroyed, 56 = m& a T0 ⎜⎜ c p ln 6 − Rln 6 ⎟⎟ = (279.3)(293) ⎢(1.005)ln − (0.287) ln(8)⎥ = 6280 kW 293 T P ⎣ ⎦ 5 5 ⎠ ⎝ ⎛ T q X& destroyed, 67 = m& a T0 ⎜⎜ c p ln 7 − in T6 Tsource ⎝

⎞ 1373 804.3 ⎤ ⎟ = (279.3)(293) ⎡⎢(1.005)ln = 23,970 kW − ⎟ 572.7 1373 ⎥⎦ ⎣ ⎠

⎛ T P ⎞ ⎡ 819.5 ⎛ 1 ⎞⎤ X& destroyed, 78 = m& a T0 ⎜⎜ c p ln 8 − Rln 8 ⎟⎟ = (279.3)(293) ⎢(1.005)ln − (0.287) ln⎜ ⎟⎥ = 6396 kW T7 P7 ⎠ 1373 ⎝ 8 ⎠⎦ ⎣ ⎝

The largest exergy destruction occurs during the heat addition process in the combustor of the gas cycle.

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93

10-89 A 450-MW combined gas-steam power plant is considered. The topping cycle is a gas-turbine cycle and the bottoming cycle is a nonideal Rankine cycle with an open feedwater heater. The mass flow rate of air to steam, the required rate of heat input in the combustion chamber, and the thermal efficiency of the combined cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats.

T 10

Analysis (a) Using the properties of air from Table A-17, the analysis of gas cycle yields

· Qin

T8 = 300 K ⎯ ⎯→ h8 = 300.19 kJ/kg Pr8 = 1.386 P9 Pr9 = Pr = (14 )(1.386 ) = 19.40 ⎯ ⎯→ h9 s = 635.5 kJ/kg P8 8 ηC =

9s

h9 s − h8 ⎯ ⎯→ h9 = h8 + (h9 s − h8 ) / ηC h9 − h8 = 300.19 + (635.5 − 300.19 ) / (0.82 )

9

2 3 8

T10 = 1400 K ⎯ ⎯→ h10 = 1515.42 kJ/kg Pr10 = 450.5

1

Pr11 =

P11 ⎛ 1⎞ Pr10 = ⎜ ⎟ (450.5) = 32.18 ⎯ ⎯→ h11s = 735.8 kJ/kg P10 ⎝ 14 ⎠

ηT =

h10 − h11 ⎯ ⎯→ h11 = h10 − ηT (h10 − h11s ) h10 − h11s = 1515.42 − (0.86 )(1515.42 − 735.8)

11 5

11s

4 12

= 709.1 kJ/kg

GAS CYCLE

STEAM CYCLE 6s · Qout

6

7s 7

= 844.95 kJ/kg

T12 = 460 K ⎯ ⎯→ h12 = 462.02 kJ/kg

From the steam tables (Tables A-4, A-5, and A-6), h1 = h f

@ 20 kPa

v1 = v f

@ 20 kPa

= 251.42 kJ/kg = 0.001017 m 3 /kg

wpI,in = v 1 (P2 − P1 )

(

)

⎛ 1 kJ = 0.001017 m 3 /kg (600 − 20 kPa )⎜ ⎜ 1 kPa ⋅ m 3 ⎝ = 0.59 kJ/kg

⎞ ⎟ ⎟ ⎠

h2 = h1 + wpI,in = 251.42 + 0.59 = 252.01 kJ/kg h3 = h f @ 0.6 MPa = 670.38 kJ/kg v 3 = v f @ 0.6 MPa = 0.001101 m 3 /kg w pII,in = v 3 (P4 − P3 )

(

)

⎛ 1 kJ = 0.001101 m 3 /kg (8,000 − 600 kPa )⎜ ⎜ 1 kPa ⋅ m 3 ⎝ = 8.15 kJ/kg

⎞ ⎟ ⎟ ⎠

h4 = h3 + wpI,in = 670.38 + 8.15 = 678.52 kJ/kg P5 = 8 MPa ⎫ h5 = 3139.4 kJ/kg T5 = 400°C ⎬⎭ s 5 = 6.3658 kJ/kg ⋅ K

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s

94 s 6s − s f 6.3658 − 1.9308 P6 = 0.6 MPa ⎫ x 6 s = = = 0.9184 s fg 4.8285 ⎬ s 6s = s5 ⎭ h = h + x h = 670.38 + (0.9184 )(2085.8) = 2585.9 kJ/kg 6s f 6 s fg

ηT =

h5 − h 6 ⎯ ⎯→ h6 = h5 − η T (h5 − h6 s ) = 3139.4 − (0.86 )(3139.4 − 2585.9 ) = 2663.3 kJ/kg h5 − h 6 s

s7 − s f 6.3658 − 0.8320 P7 = 20 kPa ⎫ x 7 s = = = 0.7820 s fg 7.0752 ⎬ s7 = s5 ⎭ h = h + x h = 251.42 + (0.7820 )(2357.5) = 2095.1 kJ/kg 7s f 7 fg

ηT =

h5 − h7 ⎯ ⎯→ h7 = h5 − η T (h5 − h7 s ) = 3139.4 − (0.86 )(3139.4 − 2095.1) = 2241.3 kJ/kg h5 − h7 s

Noting that Q& ≅ W& ≅ Δke ≅ Δpe ≅ 0 for the heat exchanger, the steady-flow energy balance equation yields E& in − E& out = ΔE& system ©0 (steady) = 0 E& in = E& out

∑ m& h = ∑ m& h i i

e e

⎯ ⎯→ m& s (h5 − h4 ) = m& air (h11 − h12 )

m& air h − h4 3139.4 − 678.52 = 5 = = 6.425 kg air / kg steam & ms h11 − h12 844.95 − 462.02

(b) Noting that Q& ≅ W& ≅ Δke ≅ Δpe ≅ 0 for the open FWH, the steady-flow energy balance equation yields E& in − E& out = ΔE& system©0 (steady) = 0 → E& in = E& out

∑ m& h = ∑ m& h i i

e e

⎯ ⎯→ m& 2 h2 + m& 6 h6 = m& 3 h3 ⎯ ⎯→ yh6 + (1 − y )h2 = (1)h3

Thus, h3 − h2 670.38 − 252.01 = = 0.1735 h6 − h2 2663.3 − 252.01

y=

(the

fraction of steam extracted )

wT = η T [h5 − h6 + (1 − y )(h6 − h7 )] = (0.86 )[3139.4 − 2663.3 + (1 − 0.1735)(2663.3 − 2241.3)] = 824.5 kJ/kg w net,steam = wT − w p,in = wT − (1 − y )wp, I − wp, II = 824.5 − (1 − 0.1735)(0.59 ) − 8.15 = 815.9 kJ/kg w net,gas = wT − wC ,in = (h10 − h11 ) − (h9 − h8 ) = 1515.42 − 844.95 − (709.1 − 300.19 ) = 261.56 kJ/kg

The net work output per unit mass of gas is wnet = wnet,gas +

m& air = and (c)

1 1 (815.9) = 388.55 kJ/kg wnet,steam = 261.56 + 6.425 6.425

W& net 450,000 kJ/s = = 1158.2 kg/s wnet 388.55 kJ/kg

Q& in = m& air (h10 − h9 ) = (1158.2 kg/s )(1515.42 − 709.1) kJ/kg = 933,850 kW

η th =

W& net 450,000 kW = = 48.2% 933,850 kW Q& in

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95

10-90 EES Problem 10-89 is reconsidered. The effect of the gas cycle pressure ratio on the ratio of gas flow rate to steam flow rate and cycle thermal efficiency is to be investigated. Analysis The problem is solved using EES, and the solution is given below. "Input data" T[8] = 300 [K] P[8] = 14.7 [kPa] "Pratio = 14" T[10] = 1400 [K] T[12] = 460 [K] P[12] = P[8] W_dot_net=450 [MW] Eta_comp = 0.82 Eta_gas_turb = 0.86 Eta_pump = 1.0 Eta_steam_turb = 0.86 P[5] = 8000 [kPa] T[5] =(400+273) "K" P[6] = 600 [kPa] P[7] = 20 [kPa]

"Gas compressor inlet" "Assumed air inlet pressure" "Pressure ratio for gas compressor" "Gas turbine inlet" "Gas exit temperature from Gas-to-steam heat exchanger " "Assumed air exit pressure"

"Steam turbine inlet" "Steam turbine inlet" "Extraction pressure for steam open feedwater heater" "Steam condenser pressure"

"GAS POWER CYCLE ANALYSIS" "Gas Compressor anaysis" s[8]=ENTROPY(Air,T=T[8],P=P[8]) ss9=s[8] "For the ideal case the entropies are constant across the compressor" P[9] = Pratio*P[8] Ts9=temperature(Air,s=ss9,P=P[9])"Ts9 is the isentropic value of T[9] at compressor exit" Eta_comp = w_gas_comp_isen/w_gas_comp "compressor adiabatic efficiency, w_comp > w_comp_isen" h[8] + w_gas_comp_isen =hs9"SSSF conservation of energy for the isentropic compressor, assuming: adiabatic, ke=pe=0 per unit gas mass flow rate in kg/s" h[8]=ENTHALPY(Air,T=T[8]) hs9=ENTHALPY(Air,T=Ts9) h[8] + w_gas_comp = h[9]"SSSF conservation of energy for the actual compressor, assuming: adiabatic, ke=pe=0" T[9]=temperature(Air,h=h[9]) s[9]=ENTROPY(Air,T=T[9],P=P[9]) "Gas Cycle External heat exchanger analysis" h[9] + q_in = h[10]"SSSF conservation of energy for the external heat exchanger, assuming W=0, ke=pe=0" h[10]=ENTHALPY(Air,T=T[10]) P[10]=P[9] "Assume process 9-10 is SSSF constant pressure" Q_dot_in"MW"*1000"kW/MW"=m_dot_gas*q_in "Gas Turbine analysis" s[10]=ENTROPY(Air,T=T[10],P=P[10]) ss11=s[10] "For the ideal case the entropies are constant across the turbine" P[11] = P[10] /Pratio Ts11=temperature(Air,s=ss11,P=P[11])"Ts11 is the isentropic value of T[11] at gas turbine exit" Eta_gas_turb = w_gas_turb /w_gas_turb_isen "gas turbine adiabatic efficiency, w_gas_turb_isen > w_gas_turb"

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96

h[10] = w_gas_turb_isen + hs11"SSSF conservation of energy for the isentropic gas turbine, assuming: adiabatic, ke=pe=0" hs11=ENTHALPY(Air,T=Ts11) h[10] = w_gas_turb + h[11]"SSSF conservation of energy for the actual gas turbine, assuming: adiabatic, ke=pe=0" T[11]=temperature(Air,h=h[11]) s[11]=ENTROPY(Air,T=T[11],P=P[11]) "Gas-to-Steam Heat Exchanger" "SSSF conservation of energy for the gas-to-steam heat exchanger, assuming: adiabatic, W=0, ke=pe=0" m_dot_gas*h[11] + m_dot_steam*h[4] = m_dot_gas*h[12] + m_dot_steam*h[5] h[12]=ENTHALPY(Air, T=T[12]) s[12]=ENTROPY(Air,T=T[12],P=P[12]) "STEAM CYCLE ANALYSIS" "Steam Condenser exit pump or Pump 1 analysis" Fluid$='Steam_IAPWS' P[1] = P[7] P[2]=P[6] h[1]=enthalpy(Fluid$,P=P[1],x=0) {Saturated liquid} v1=volume(Fluid$,P=P[1],x=0) s[1]=entropy(Fluid$,P=P[1],x=0) T[1]=temperature(Fluid$,P=P[1],x=0) w_pump1_s=v1*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume" w_pump1=w_pump1_s/Eta_pump "Definition of pump efficiency" h[1]+w_pump1= h[2] "Steady-flow conservation of energy" s[2]=entropy(Fluid$,P=P[2],h=h[2]) T[2]=temperature(Fluid$,P=P[2],h=h[2]) "Open Feedwater Heater analysis" y*h[6] + (1-y)*h[2] = 1*h[3] "Steady-flow conservation of energy" P[3]=P[6] h[3]=enthalpy(Fluid$,P=P[3],x=0) "Condensate leaves heater as sat. liquid at P[3]" T[3]=temperature(Fluid$,P=P[3],x=0) s[3]=entropy(Fluid$,P=P[3],x=0) "Boiler condensate pump or Pump 2 analysis" P[4] = P[5] v3=volume(Fluid$,P=P[3],x=0) w_pump2_s=v3*(P[4]-P[3])"SSSF isentropic pump work assuming constant specific volume" w_pump2=w_pump2_s/Eta_pump "Definition of pump efficiency" h[3]+w_pump2= h[4] "Steady-flow conservation of energy" s[4]=entropy(Fluid$,P=P[4],h=h[4]) T[4]=temperature(Fluid$,P=P[4],h=h[4]) w_steam_pumps = (1-y)*w_pump1+ w_pump2 "Total steam pump work input/ mass steam" "Steam Turbine analysis" h[5]=enthalpy(Fluid$,T=T[5],P=P[5]) s[5]=entropy(Fluid$,P=P[5],T=T[5]) ss6=s[5] hs6=enthalpy(Fluid$,s=ss6,P=P[6]) Ts6=temperature(Fluid$,s=ss6,P=P[6]) h[6]=h[5]-Eta_steam_turb*(h[5]-hs6)"Definition of steam turbine efficiency" T[6]=temperature(Fluid$,P=P[6],h=h[6]) s[6]=entropy(Fluid$,P=P[6],h=h[6]) ss7=s[5] hs7=enthalpy(Fluid$,s=ss7,P=P[7]) Ts7=temperature(Fluid$,s=ss7,P=P[7]) PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

97

h[7]=h[5]-Eta_steam_turb*(h[5]-hs7)"Definition of steam turbine efficiency" T[7]=temperature(Fluid$,P=P[7],h=h[7]) s[7]=entropy(Fluid$,P=P[7],h=h[7]) "SSSF conservation of energy for the steam turbine: adiabatic, neglect ke and pe" h[5] = w_steam_turb + y*h[6] +(1-y)*h[7] "Steam Condenser analysis" (1-y)*h[7]=q_out+(1-y)*h[1]"SSSF conservation of energy for the Condenser per unit mass" Q_dot_out*Convert(MW, kW)=m_dot_steam*q_out "Cycle Statistics" MassRatio_gastosteam =m_dot_gas/m_dot_steam W_dot_net*Convert(MW, kW)=m_dot_gas*(w_gas_turb-w_gas_comp)+ m_dot_steam*(w_steam_turb - w_steam_pumps)"definition of the net cycle work" Eta_th=W_dot_net/Q_dot_in*Convert(, %) "Cycle thermal efficiency, in percent" Bwr=(m_dot_gas*w_gas_comp + m_dot_steam*w_steam_pumps)/(m_dot_gas*w_gas_turb + m_dot_steam*w_steam_turb) "Back work ratio" W_dot_net_steam = m_dot_steam*(w_steam_turb - w_steam_pumps) W_dot_net_gas = m_dot_gas*(w_gas_turb - w_gas_comp) NetWorkRatio_gastosteam = W_dot_net_gas/W_dot_net_steam

Pratio

MassRatio gastosteam

6 8 10 12 14 15 16 18 20 22

4.463 5.024 5.528 5.994 6.433 6.644 6.851 7.253 7.642 8.021

Wnetgas [kW] 262595 279178 289639 296760 301809 303780 305457 308093 309960 311216

ηth [%] 45.29 46.66 47.42 47.82 47.99 48.01 47.99 47.87 47.64 47.34

Wnetsteam [kW] 187405 170822 160361 153240 148191 146220 144543 141907 140040 138784

NetWorkRatio gastosteam

1.401 1.634 1.806 1.937 2.037 2.078 2.113 2.171 2.213 2.242

Com bined Gas and Steam Pow er Cycle 1600 1500

10

1400 1300

Gas Cycle

1200 1100

T [K]

1000

Steam Cycle

900

11

800

9

700

5

600

8000 kPa

500 400

12

3,4

600 kPa

6

1,2 20 kPa

300 200 0.0

8 1.1

2.2

3.3

4.4

5.5

7 6.6

7.7

8.8

9.9

11.0

s [kJ/kg-K]

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98

Cycle Therm al Efficiency vs Gas Cycle Pressure Ratio 48.5 48.0 47.5

η th [% ]

47.0 46.5 46.0 45.5 45.0 5

9

12

16

19

23

Pratio 2.3

W dot,gas / W dot,steam vs Gas Pressure Ratio

NetW orkRatio gastosteam

2.2 2.1 2.0 1.9 1.8 1.7 1.6 1.5 1.4 5

9

12

16

19

23

Pratio

Ratio of Gas Flow Rate to Steam Flow Rate vs Gas Pressure Ratio 8.5 8.0

M assRatio gastosteam

7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 5

9

12

16

19

23

Pratio

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99

10-91 A combined gas-steam power plant is considered. The topping cycle is a gas-turbine cycle and the bottoming cycle is a nonideal reheat Rankine cycle. The moisture percentage at the exit of the lowpressure turbine, the steam temperature at the inlet of the high-pressure turbine, and the thermal efficiency of the combined cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Analysis (a) We obtain the air properties from EES. The analysis of gas cycle is as follows T7 = 15°C ⎯ ⎯→ h7 = 288.50 kJ/kg

Combustion chamber

8

9

T7 = 15°C

⎫ ⎬s 7 = 5.6648 kJ/kg P7 = 100 kPa ⎭ P8 = 700 kPa ⎫ ⎬h8 s = 503.47 kJ/kg s8 = s 7 ⎭

Compressor

Gas turbine

7 11

h − h7 ⎯ ⎯→ h8 = h7 + (h8 s − h7 ) / η C η C = 8s h8 − h7 = 290.16 + (503.47 − 290.16 ) / (0.80 ) = 557.21 kJ/kg

10

Heat exchanger 3

Steam turbine 4

T9 = 950°C ⎯ ⎯→ h9 = 1304.8 kJ/kg

6

T9 = 950°C ⎫ ⎬s 9 = 6.6456 kJ/kg P9 = 700 kPa ⎭

5

P10 = 100 kPa ⎫ ⎬h10 s = 763.79 kJ/kg s10 = s 9 ⎭ h −h η T = 9 10 ⎯⎯→ h10 = h9 − η T (h9 − h10 s ) h9 − h10 s = 1304.8 − (0.80 )(1304.8 − 763.79 )

Condenser

pump

2

1

= 871.98 kJ/kg

T

T11 = 200 °C ⎯ ⎯→ h11 = 475.62 kJ/kg

From the steam tables (Tables A-4, A-5, and A-6 or from EES), h1 = h f v1 = v f

9

950°C · Qin

= 191.81 kJ/kg 3 @ 10 kPa = 0.00101 m /kg

@ 10 kPa

wpI,in = v1 (P2 − P1 ) / η p

(

)

⎛ 1 kJ ⎞ ⎟ / 0.80 = 0.00101 m3 /kg (6000 − 10 kPa )⎜⎜ 1 kPa ⋅ m 3 ⎟⎠ ⎝ = 7.56 kJ/kg

8s

10 10s 3

8 6 MPa

h2 = h1 + wpI,in = 191.81 + 7.65 = 199.37 kJ/kg P5 = 1 MPa ⎫ h5 = 3264.5 kJ/kg T5 = 400°C ⎬⎭ s 5 = 7.4670 kJ/kg ⋅ K P6 = 10 kPa ⎫ x 6 s ⎬ s 6s = s5 ⎭h 6s

15°C

7

2

1 7.4670 − 0.6492 = = = 0.9091 s fg 7.4996 = h f + x 6 s h fg = 191.81 + (0.9091)(2392.1) = 2366.4 kJ/kg s6s − s f

GAS CYCLE

1 MPa 5

11 STEAM 4 CYCLE 4s 10 kPa · 6s 6 Qout

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s

100

ηT =

h5 − h6 ⎯ ⎯→ h6 = h5 − η T (h5 − h6 s ) h5 − h6 s = 3264.5 − (0.80 )(3264.5 − 2366.4) = 2546.0 kJ/kg

P6 = 10 kPa ⎫ x = 0.9842 h6 = 2546.5 kJ/kg ⎬⎭ 6 Moisture Percentage = 1 − x 6 = 1 − 0.9842 = 0.0158 = 1.6%

(b) Noting that Q& ≅ W& ≅ Δke ≅ Δpe ≅ 0 for the heat exchanger, the steady-flow energy balance equation yields E& in = E& out

∑ m& h = ∑ m& h i i

e e

m& s (h3 − h2 ) + m& s (h5 − h4 ) = m& air (h10 − h11 )

(1.15)[(3346.5 − 199.37) + (3264.5 − h4 )] = (10)(871.98 − 475.62) ⎯ ⎯→ h4 = 2965.0 kJ/kg

Also, P3 = 6 MPa ⎫ h3 = ⎬ T3 = ? ⎭ s3 =

ηT =

P4 = 1 MPa ⎫ ⎬ h4 s = s 4s = s3 ⎭

h3 − h4 ⎯ ⎯→ h4 = h3 − η T (h3 − h4 s ) h3 − h4 s

The temperature at the inlet of the high-pressure turbine may be obtained by a trial-error approach or using EES from the above relations. The answer is T3 = 468.0ºC. Then, the enthalpy at state 3 becomes: h3 = 3346.5 kJ/kg (c)

W& T,gas = m& air (h9 − h10 ) = (10 kg/s )(1304.8 − 871.98) kJ/kg = 4328 kW W& C,gas = m& air (h8 − h7 ) = (10 kg/s )(557.21 − 288.50 ) kJ/kg = 2687 kW W& net,gas = W& T,gas − W& C,gas = 4328 − 2687 = 1641 kW

W& T,steam = m& s (h3 − h4 + h5 − h6 ) = (1.15 kg/s )(3346.5 − 2965.0 + 3264.5 − 2546.0) kJ/kg = 1265 kW

W& P,steam = m& s w pump = (1.15 kg/s )(7.564) kJ/kg = 8.7 kW W& net,steam = W& T,steam − W& P,steam = 1265 − 8.7 = 1256 kW W& net,plant = W& net,gas + W& net,steam = 1641 + 1256 = 2897 kW

(d)

Q& in = m& air (h9 − h8 ) = (10 kg/s )(1304.8 − 557.21) kJ/kg = 7476 kW

η th =

W& net, plant 2897 kW = = 0.388 = 38.8% 7476 kW Q& in

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101

Special Topic: Binary Vapor Cycles 10-92C Binary power cycle is a cycle which is actually a combination of two cycles; one in the high temperature region, and the other in the low temperature region. Its purpose is to increase thermal efficiency. 10-93C Consider the heat exchanger of a binary power cycle. The working fluid of the topping cycle (cycle A) enters the heat exchanger at state 1 and leaves at state 2. The working fluid of the bottoming cycle (cycle B) enters at state 3 and leaves at state 4. Neglecting any changes in kinetic and potential energies, and assuming the heat exchanger is well-insulated, the steady-flow energy balance relation yields E& in − E& out = ΔE& system ©0 (steady) = 0 E& in = E& out

∑ m& h = ∑ m& h e e

i i

m& A h2 + m& B h4 = m& A h1 + m& B h3 or m& A (h2 − h1 ) = m& B (h3 − h4 )

Thus, m& A h3 − h4 = m& B h2 − h1

10-94C Steam is not an ideal fluid for vapor power cycles because its critical temperature is low, its saturation dome resembles an inverted V, and its condenser pressure is too low. 10-95C Because mercury has a high critical temperature, relatively low critical pressure, but a very low condenser pressure. It is also toxic, expensive, and has a low enthalpy of vaporization. 10-96C In binary vapor power cycles, both cycles are vapor cycles. In the combined gas-steam power cycle, one of the cycles is a gas cycle.

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102

Review Problems 10-97 It is to be demonstrated that the thermal efficiency of a combined gas-steam power plant ηcc can be expressed as η cc = η g + ηs − η gηs where η g = Wg / Qin and η s = Ws / Qg,out are the thermal efficiencies of

the gas and steam cycles, respectively, and the efficiency of a combined cycle is to be obtained. Analysis The thermal efficiencies of gas, steam, and combined cycles can be expressed as

η cc = ηg = ηs =

Wtotal Q = 1 − out Qin Qin Wg Qin

= 1−

Qg,out Qin

Ws Q = 1 − out Qg,out Qg,out

where Qin is the heat supplied to the gas cycle, where Qout is the heat rejected by the steam cycle, and where Qg,out is the heat rejected from the gas cycle and supplied to the steam cycle. Using the relations above, the expression η g + η s − η gη s can be expressed as ⎛

Q

⎞ ⎛

Q

⎞ ⎛ Qg,out ⎞⎛ Q ⎟ − ⎜1 − ⎟⎜1 − out ⎟ ⎜ ⎟ ⎜ Qin ⎠⎝ Qg,out ⎠ ⎝ Qg,out Q Q −1+ + out − out Qin Qg,out Qin

g,out ⎟ + ⎜1 − out η g + η s − η gη s = ⎜⎜1 − Qin ⎟⎠ ⎜⎝ Qg,out ⎝

= 1− = 1−

Qg,out Qin

+1−

Qout Qg,out

⎞ ⎟ ⎟ ⎠

Qout Qin

= η cc

Therefore, the proof is complete. Using the relation above, the thermal efficiency of the given combined cycle is determined to be

η cc = η g + η s − η gη s = 0.4 + 0.30 − 0.40 × 0.30 = 0.58

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103

10-98 The thermal efficiency of a combined gas-steam power plant ηcc can be expressed in terms of the thermal efficiencies of the gas and the steam turbine cycles as ηcc = η g + η s − ηgη s . It is to be shown that

the value of η cc is greater than either of η g or η s . Analysis By factoring out terms, the relation ηcc = η g + η s − ηgη s can be expressed as

η cc = η g + η s − η gη s = η g + η s (1 − η g ) > η g 14243 Positive since η g η s 14243 Positive since η s 1) then x4$='(superheated)' if (x41) then x4$='(superheated)' if (x41) then x6$='(superheated)' if (x6imax) END "NoRHStages = 2" P[6] = 10"kPa" P[3] = 15000"kPa" P_extract = P[6] "Select a lower limit on the reheat pressure" T[3] = 500"C" T[5] = 500"C" Eta_t = 1.0 "Turbine isentropic efficiency" Eta_p = 1.0 "Pump isentropic efficiency" PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

141

Pratio = P[3]/P_extract P[4] = P[3]*(1/Pratio)^(1/(NoRHStages+1))"kPa" Fluid$='Steam_IAPWS' "Pump analysis" P[1] = P[6] P[2]=P[3] x[1]=0 "Sat'd liquid" h[1]=enthalpy(Fluid$,P=P[1],x=x[1]) v[1]=volume(Fluid$,P=P[1],x=x[1]) s[1]=entropy(Fluid$,P=P[1],x=x[1]) T[1]=temperature(Fluid$,P=P[1],x=x[1]) W_p_s=v[1]*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume" W_p=W_p_s/Eta_p h[2]=h[1]+W_p "SSSF First Law for the pump" v[2]=volume(Fluid$,P=P[2],h=h[2]) s[2]=entropy(Fluid$,P=P[2],h=h[2]) T[2]=temperature(Fluid$,P=P[2],h=h[2]) "High Pressure Turbine analysis" h[3]=enthalpy(Fluid$,T=T[3],P=P[3]) s[3]=entropy(Fluid$,T=T[3],P=P[3]) v[3]=volume(Fluid$,T=T[3],P=P[3]) s_s[4]=s[3] hs[4]=enthalpy(Fluid$,s=s_s[4],P=P[4]) Ts[4]=temperature(Fluid$,s=s_s[4],P=P[4]) Eta_t=(h[3]-h[4])/(h[3]-hs[4])"Definition of turbine efficiency" T[4]=temperature(Fluid$,P=P[4],h=h[4]) s[4]=entropy(Fluid$,h=h[4],P=P[4]) v[4]=volume(Fluid$,s=s[4],P=P[4]) h[3] =W_t_hp+h[4]"SSSF First Law for the high pressure turbine" "Low Pressure Turbine analysis" Call Reheat(P[3],T[3],T[5],h[4],NoRHStages,Pratio,Eta_t:Q_in_reheat,W_t_lp,h6) h[6]=h6 {P[5]=P[4] s[5]=entropy(Fluid$,T=T[5],P=P[5]) h[5]=enthalpy(Fluid$,T=T[5],P=P[5]) s_s[6]=s[5] hs[6]=enthalpy(Fluid$,s=s_s[6],P=P[6]) Ts[6]=temperature(Fluid$,s=s_s[6],P=P[6]) vs[6]=volume(Fluid$,s=s_s[6],P=P[6]) Eta_t=(h[5]-h[6])/(h[5]-hs[6])"Definition of turbine efficiency" h[5]=W_t_lp+h[6]"SSSF First Law for the low pressure turbine" x[6]=QUALITY(Fluid$,h=h[6],P=P[6]) W_t_lp_total = NoRHStages*W_t_lp Q_in_reheat = NoRHStages*(h[5] - h[4])} "Boiler analysis" Q_in_boiler + h[2]=h[3]"SSSF First Law for the Boiler" Q_in = Q_in_boiler+Q_in_reheat "Condenser analysis" h[6]=Q_out+h[1]"SSSF First Law for the Condenser" T[6]=temperature(Fluid$,h=h[6],P=P[6]) s[6]=entropy(Fluid$,h=h[6],P=P[6]) PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

142

x[6]=QUALITY(Fluid$,h=h[6],P=P[6]) x6s$=x6$(x[6]) "Cycle Statistics" W_net=W_t_hp+W_t_lp - W_p Eta_th=W_net/Q_in

0.4097 0.4122 0.4085 0.4018 0.3941 0.386 0.3779 0.3699 0.3621 0.3546

NoRH Stages 1 2 3 4 5 6 7 8 9 10

Qin [kJ/kg] 4085 4628 5020 5333 5600 5838 6058 6264 6461 6651

2400

Wnet [kJ/kg] 1674 1908 2051 2143 2207 2253 2289 2317 2340 2358

2300 2200

W net [kJ/kg]

ηth

2100 2000 1900 1800 1700 1600 1

2

3

4

5

6

7

8

9

NoRHStages 0.42 0.41

ηth

0.4 0.39 0.38 0.37 0.36 0.35 1

2

3

4

5

6

7

8

9

10

8

9

10

NoRHStages 7000 6500

Q in [kJ/kg]

6000 5500 5000 4500 4000 1

2

3

4

5

6

7

NoRHStages

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10

143

10-122 EES The effect of number of regeneration stages on the performance an ideal regenerative Rankine cycle with one open feedwater heater is to be investigated. Analysis The problem is solved using EES, and the solution is given below. Procedure Reheat(NoFwh,T[5],P[5],P_cond,Eta_turb,Eta_pump:q_in,w_net) Fluid$='Steam_IAPWS' Tcond = temperature(Fluid$,P=P_cond,x=0) Tboiler = temperature(Fluid$,P=P[5],x=0) P[7] = P_cond s[5]=entropy(Fluid$, T=T[5], P=P[5]) h[5]=enthalpy(Fluid$, T=T[5], P=P[5]) h[1]=enthalpy(Fluid$, P=P[7],x=0) P4[1] = P[5] "NOTICE THIS IS P4[i] WITH i = 1" DELTAT_cond_boiler = Tboiler - Tcond If NoFWH = 0 Then "the following are h7, h2, w_net, and q_in for zero feedwater heaters, NoFWH = 0" h7=enthalpy(Fluid$, s=s[5],P=P[7]) h2=h[1]+volume(Fluid$, P=P[7],x=0)*(P[5] - P[7])/Eta_pump w_net = Eta_turb*(h[5]-h7)-(h2-h[1]) q_in = h[5] - h2 else i=0 REPEAT i=i+1 "The following maintains the same temperature difference between any two regeneration stages." T_FWH[i] = (NoFWH +1 - i)*DELTAT_cond_boiler/(NoFWH + 1)+Tcond"[C]" P_extract[i] = pressure(Fluid$,T=T_FWH[i],x=0)"[kPa]" P3[i]=P_extract[i] P6[i]=P_extract[i] If i > 1 then P4[i] = P6[i - 1] UNTIL i=NoFWH P4[NoFWH+1]=P6[NoFWH] h4[NoFWH+1]=h[1]+volume(Fluid$, P=P[7],x=0)*(P4[NoFWH+1] - P[7])/Eta_pump i=0 REPEAT i=i+1 "Boiler condensate pump or the Pumps 2 between feedwater heaters analysis" h3[i]=enthalpy(Fluid$,P=P3[i],x=0) v3[i]=volume(Fluid$,P=P3[i],x=0) w_pump2_s=v3[i]*(P4[i]-P3[i])"SSSF isentropic pump work assuming constant specific volume" w_pump2[i]=w_pump2_s/Eta_pump "Definition of pump efficiency" h4[i]= w_pump2[i] +h3[i] "Steady-flow conservation of energy" s4[i]=entropy(Fluid$,P=P4[i],h=h4[i]) T4[i]=temperature(Fluid$,P=P4[i],h=h4[i]) PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

144

Until i = NoFWH i=0 REPEAT i=i+1 "Open Feedwater Heater analysis:" {h2[i] = h6[i]} s5[i] = s[5] ss6[i]=s5[i] hs6[i]=enthalpy(Fluid$,s=ss6[i],P=P6[i]) Ts6[i]=temperature(Fluid$,s=ss6[i],P=P6[i]) h6[i]=h[5]-Eta_turb*(h[5]-hs6[i])"Definition of turbine efficiency for high pressure stages" If i=1 then y[1]=(h3[1] - h4[2])/(h6[1] - h4[2]) "Steady-flow conservation of energy for the FWH" If i > 1 then js = i -1 j=0 sumyj = 0 REPEAT j = j+1 sumyj = sumyj + y[ j ] UNTIL j = js y[i] =(1- sumyj)*(h3[i] - h4[i+1])/(h6[i] - h4[i+1]) ENDIF T3[i]=temperature(Fluid$,P=P3[i],x=0) "Condensate leaves heater as sat. liquid at P[3]" s3[i]=entropy(Fluid$,P=P3[i],x=0) "Turbine analysis" T6[i]=temperature(Fluid$,P=P6[i],h=h6[i]) s6[i]=entropy(Fluid$,P=P6[i],h=h6[i]) yh6[i] = y[i]*h6[i] UNTIL i=NoFWH ss[7]=s6[i] hs[7]=enthalpy(Fluid$,s=ss[7],P=P[7]) Ts[7]=temperature(Fluid$,s=ss[7],P=P[7]) h[7]=h6[i]-Eta_turb*(h6[i]-hs[7])"Definition of turbine efficiency for low pressure stages" T[7]=temperature(Fluid$,P=P[7],h=h[7]) s[7]=entropy(Fluid$,P=P[7],h=h[7]) sumyi = 0 sumyh6i = 0 wp2i = W_pump2[1] i=0 REPEAT i=i+1 sumyi = sumyi + y[i] sumyh6i = sumyh6i + yh6[i] If NoFWH > 1 then wp2i = wp2i + (1- sumyi)*W_pump2[i] UNTIL i = NoFWH "Condenser Pump---Pump_1 Analysis:" P[2] = P6 [ NoFWH] P[1] = P_cond h[1]=enthalpy(Fluid$,P=P[1],x=0) {Sat'd liquid} v1=volume(Fluid$,P=P[1],x=0) s[1]=entropy(Fluid$,P=P[1],x=0) T[1]=temperature(Fluid$,P=P[1],x=0) w_pump1_s=v1*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume" PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

145

w_pump1=w_pump1_s/Eta_pump "Definition of pump efficiency" h[2]=w_pump1+ h[1] "Steady-flow conservation of energy" s[2]=entropy(Fluid$,P=P[2],h=h[2]) T[2]=temperature(Fluid$,P=P[2],h=h[2]) "Boiler analysis" q_in = h[5] - h4[1]"SSSF conservation of energy for the Boiler" w_turb = h[5] - sumyh6i - (1- sumyi)*h[7] "SSSF conservation of energy for turbine" "Condenser analysis" q_out=(1- sumyi)*(h[7] - h[1])"SSSF First Law for the Condenser" "Cycle Statistics" w_net=w_turb - ((1- sumyi)*w_pump1+ wp2i) endif END "Input Data" NoFWH = 2 P[5] = 15000 [kPa] T[5] = 600 [C] P_cond=5 [kPa] Eta_turb= 1.0 "Turbine isentropic efficiency" Eta_pump = 1.0 "Pump isentropic efficiency" P[1] = P_cond P[4] = P[5] "Condenser exit pump or Pump 1 analysis" Call Reheat(NoFwh,T[5],P[5],P_cond,Eta_turb,Eta_pump:q_in,w_net) Eta_th=w_net/q_in

ηth 0.4466 0.4806 0.4902 0.4983 0.5036 0.5073 0.5101 0.5123 0.5141 0.5155 0.5167

wnet [kJ/kg] 1532 1332 1243 1202 1175 1157 1143 1132 1124 1117 1111

qin [kJ/kg] 3430 2771 2536 2411 2333 2280 2240 2210 2186 2167 2151

Steam 700 600 500

5

T [°C]

No FWH 0 1 2 3 4 5 6 7 8 9 10

400 300 6000 kPa

4 200

2

3

6 400 kPa

100

1 0 0

7

10 kPa

2

4

6

8

10

12

s [kJ/kg-K]

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146

0.52 0.51 0.5 0.49

η th

0.48 0.47 0.46 0.45 0.44 0

2

4

6

8

10

8

10

NoFw h

1550 1500

w net [kJ/kg]

1450 1400 1350 1300 1250 1200 1150 1100 0

2

4

6

NoFw h

3600 3400

q in [kJ/kg]

3200 3000 2800 2600 2400 2200 2000 0

2

4

6

8

10

NoFw h

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147

Fundamentals of Engineering (FE) Exam Problems 10-123 Consider a steady-flow Carnot cycle with water as the working fluid executed under the saturation dome between the pressure limits of 8 MPa and 20 kPa. Water changes from saturated liquid to saturated vapor during the heat addition process. The net work output of this cycle is (a) 494 kJ/kg

(b) 975 kJ/kg

(c) 596 kJ/kg

(d) 845 kJ/kg

(e) 1148 kJ/kg

Answer (c) 596 kJ/kg Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=8000 "kPa" P2=20 "kPa" h_fg=ENTHALPY(Steam_IAPWS,x=1,P=P1)-ENTHALPY(Steam_IAPWS,x=0,P=P1) T1=TEMPERATURE(Steam_IAPWS,x=0,P=P1)+273 T2=TEMPERATURE(Steam_IAPWS,x=0,P=P2)+273 q_in=h_fg Eta_Carnot=1-T2/T1 w_net=Eta_Carnot*q_in "Some Wrong Solutions with Common Mistakes:" W1_work = Eta1*q_in; Eta1=T2/T1 "Taking Carnot efficiency to be T2/T1" W2_work = Eta2*q_in; Eta2=1-(T2-273)/(T1-273) "Using C instead of K" W3_work = Eta_Carnot*ENTHALPY(Steam_IAPWS,x=1,P=P1) "Using h_g instead of h_fg" W4_work = Eta_Carnot*q2; q2=ENTHALPY(Steam_IAPWS,x=1,P=P2)ENTHALPY(Steam_IAPWS,x=0,P=P2) "Using h_fg at P2"

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148

10-124 A simple ideal Rankine cycle operates between the pressure limits of 10 kPa and 3 MPa, with a turbine inlet temperature of 600°C. Disregarding the pump work, the cycle efficiency is (a) 24%

(b) 37%

(c) 52%

(d) 63%

(e) 71%

Answer (b) 37% Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=10 "kPa" P2=3000 "kPa" P3=P2 P4=P1 T3=600 "C" s4=s3 h1=ENTHALPY(Steam_IAPWS,x=0,P=P1) v1=VOLUME(Steam_IAPWS,x=0,P=P1) w_pump=v1*(P2-P1) "kJ/kg" h2=h1+w_pump h3=ENTHALPY(Steam_IAPWS,T=T3,P=P3) s3=ENTROPY(Steam_IAPWS,T=T3,P=P3) h4=ENTHALPY(Steam_IAPWS,s=s4,P=P4) q_in=h3-h2 q_out=h4-h1 Eta_th=1-q_out/q_in "Some Wrong Solutions with Common Mistakes:" W1_Eff = q_out/q_in "Using wrong relation" W2_Eff = 1-(h44-h1)/(h3-h2); h44 = ENTHALPY(Steam_IAPWS,x=1,P=P4) "Using h_g for h4" W3_Eff = 1-(T1+273)/(T3+273); T1=TEMPERATURE(Steam_IAPWS,x=0,P=P1) "Using Carnot efficiency" W4_Eff = (h3-h4)/q_in "Disregarding pump work"

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149

10-125 A simple ideal Rankine cycle operates between the pressure limits of 10 kPa and 5 MPa, with a turbine inlet temperature of 600°C. The mass fraction of steam that condenses at the turbine exit is (a) 6%

(b) 9%

(c) 12%

(d) 15%

(e) 18%

Answer (c) 12% Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=10 "kPa" P2=5000 "kPa" P3=P2 P4=P1 T3=600 "C" s4=s3 h3=ENTHALPY(Steam_IAPWS,T=T3,P=P3) s3=ENTROPY(Steam_IAPWS,T=T3,P=P3) h4=ENTHALPY(Steam_IAPWS,s=s4,P=P4) x4=QUALITY(Steam_IAPWS,s=s4,P=P4) moisture=1-x4 "Some Wrong Solutions with Common Mistakes:" W1_moisture = x4 "Taking quality as moisture" W2_moisture = 0 "Assuming superheated vapor"

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150

10-126 A steam power plant operates on the simple ideal Rankine cycle between the pressure limits of 10 kPa and 10 MPa, with a turbine inlet temperature of 600°C. The rate of heat transfer in the boiler is 800 kJ/s. Disregarding the pump work, the power output of this plant is (a) 243 kW

(b) 284 kW

(c) 508 kW

(d) 335 kW

(e) 800 kW

Answer (d) 335 kW Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=10 "kPa" P2=10000 "kPa" P3=P2 P4=P1 T3=600 "C" s4=s3 Q_rate=800 "kJ/s" m=Q_rate/q_in h1=ENTHALPY(Steam_IAPWS,x=0,P=P1) h2=h1 "pump work is neglected" "v1=VOLUME(Steam_IAPWS,x=0,P=P1) w_pump=v1*(P2-P1) h2=h1+w_pump" h3=ENTHALPY(Steam_IAPWS,T=T3,P=P3) s3=ENTROPY(Steam_IAPWS,T=T3,P=P3) h4=ENTHALPY(Steam_IAPWS,s=s4,P=P4) q_in=h3-h2 W_turb=m*(h3-h4) "Some Wrong Solutions with Common Mistakes:" W1_power = Q_rate "Assuming all heat is converted to power" W3_power = Q_rate*Carnot; Carnot = 1-(T1+273)/(T3+273); T1=TEMPERATURE(Steam_IAPWS,x=0,P=P1) "Using Carnot efficiency" W4_power = m*(h3-h44); h44 = ENTHALPY(Steam_IAPWS,x=1,P=P4) "Taking h4=h_g"

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151

10-127 Consider a combined gas-steam power plant. Water for the steam cycle is heated in a well-insulated heat exchanger by the exhaust gases that enter at 800 K at a rate of 60 kg/s and leave at 400 K. Water enters the heat exchanger at 200°C and 8 MPa and leaves at 350°C and 8 MPa. If the exhaust gases are treated as air with constant specific heats at room temperature, the mass flow rate of water through the heat exchanger becomes (a) 11 kg/s

(b) 24 kg/s

(c) 46 kg/s

(d) 53 kg/s

(e) 60 kg/s

Answer (a) 11 kg/s Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m_gas=60 "kg/s" Cp=1.005 "kJ/kg.K" T3=800 "K" T4=400 "K" Q_gas=m_gas*Cp*(T3-T4) P1=8000 "kPa" T1=200 "C" P2=8000 "kPa" T2=350 "C" h1=ENTHALPY(Steam_IAPWS,T=T1,P=P1) h2=ENTHALPY(Steam_IAPWS,T=T2,P=P2) Q_steam=m_steam*(h2-h1) Q_gas=Q_steam "Some Wrong Solutions with Common Mistakes:" m_gas*Cp*(T3 -T4)=W1_msteam*4.18*(T2-T1) "Assuming no evaporation of liquid water" m_gas*Cv*(T3 -T4)=W2_msteam*(h2-h1); Cv=0.718 "Using Cv for air instead of Cp" W3_msteam = m_gas "Taking the mass flow rates of two fluids to be equal" m_gas*Cp*(T3 -T4)=W4_msteam*(h2-h11); h11=ENTHALPY(Steam_IAPWS,x=0,P=P1) "Taking h1=hf@P1"

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152

10-128 An ideal reheat Rankine cycle operates between the pressure limits of 10 kPa and 8 MPa, with reheat occurring at 4 MPa. The temperature of steam at the inlets of both turbines is 500°C, and the enthalpy of steam is 3185 kJ/kg at the exit of the high-pressure turbine, and 2247 kJ/kg at the exit of the low-pressure turbine. Disregarding the pump work, the cycle efficiency is (a) 29%

(b) 32%

(c) 36%

(d) 41%

(e) 49%

Answer (d) 41% Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=10 "kPa" P2=8000 "kPa" P3=P2 P4=4000 "kPa" P5=P4 P6=P1 T3=500 "C" T5=500 "C" s4=s3 s6=s5 h1=ENTHALPY(Steam_IAPWS,x=0,P=P1) h2=h1 h44=3185 "kJ/kg - for checking given data" h66=2247 "kJ/kg - for checking given data" h3=ENTHALPY(Steam_IAPWS,T=T3,P=P3) s3=ENTROPY(Steam_IAPWS,T=T3,P=P3) h4=ENTHALPY(Steam_IAPWS,s=s4,P=P4) h5=ENTHALPY(Steam_IAPWS,T=T5,P=P5) s5=ENTROPY(Steam_IAPWS,T=T5,P=P5) h6=ENTHALPY(Steam_IAPWS,s=s6,P=P6) q_in=(h3-h2)+(h5-h4) q_out=h6-h1 Eta_th=1-q_out/q_in "Some Wrong Solutions with Common Mistakes:" W1_Eff = q_out/q_in "Using wrong relation" W2_Eff = 1-q_out/(h3-h2) "Disregarding heat input during reheat" W3_Eff = 1-(T1+273)/(T3+273); T1=TEMPERATURE(Steam_IAPWS,x=0,P=P1) "Using Carnot efficiency" W4_Eff = 1-q_out/(h5-h2) "Using wrong relation for q_in"

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153

10-129 Pressurized feedwater in a steam power plant is to be heated in an ideal open feedwater heater that operates at a pressure of 0.5 MPa with steam extracted from the turbine. If the enthalpy of feedwater is 252 kJ/kg and the enthalpy of extracted steam is 2665 kJ/kg, the mass fraction of steam extracted from the turbine is (a) 4%

(b) 10%

(c) 16%

(d) 27%

(e) 12%

Answer (c) 16% Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). h_feed=252 "kJ/kg" h_extracted=2665 "kJ/kg" P3=500 "kPa" h3=ENTHALPY(Steam_IAPWS,x=0,P=P3) "Energy balance on the FWH" h3=x_ext*h_extracted+(1-x_ext)*h_feed "Some Wrong Solutions with Common Mistakes:" W1_ext = h_feed/h_extracted "Using wrong relation" W2_ext = h3/(h_extracted-h_feed) "Using wrong relation" W3_ext = h_feed/(h_extracted-h_feed) "Using wrong relation"

10-130 Consider a steam power plant that operates on the regenerative Rankine cycle with one open feedwater heater. The enthalpy of the steam is 3374 kJ/kg at the turbine inlet, 2797 kJ/kg at the location of bleeding, and 2346 kJ/kg at the turbine exit. The net power output of the plant is 120 MW, and the fraction of steam bled off the turbine for regeneration is 0.172. If the pump work is negligible, the mass flow rate of steam at the turbine inlet is (a) 117 kg/s

(b) 126 kg/s

(c) 219 kg/s

(d) 288 kg/s

(e) 679 kg/s

Answer (b) 126 kg/s Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). h_in=3374 "kJ/kg" h_out=2346 "kJ/kg" h_extracted=2797 "kJ/kg" Wnet_out=120000 "kW" x_bleed=0.172 w_turb=(h_in-h_extracted)+(1-x_bleed)*(h_extracted-h_out) m=Wnet_out/w_turb "Some Wrong Solutions with Common Mistakes:" W1_mass = Wnet_out/(h_in-h_out) "Disregarding extraction of steam" W2_mass = Wnet_out/(x_bleed*(h_in-h_out)) "Assuming steam is extracted at trubine inlet" W3_mass = Wnet_out/(h_in-h_out-x_bleed*h_extracted) "Using wrong relation"

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154

10-131 Consider a simple ideal Rankine cycle. If the condenser pressure is lowered while keeping turbine inlet state the same, (select the correct statement) (a) the turbine work output will decrease. (b) the amount of heat rejected will decrease. (c) the cycle efficiency will decrease. (d) the moisture content at turbine exit will decrease. (e) the pump work input will decrease.

Answer (b) the amount of heat rejected will decrease.

10-132 Consider a simple ideal Rankine cycle with fixed boiler and condenser pressures. If the steam is superheated to a higher temperature, (select the correct statement) (a) the turbine work output will decrease. (b) the amount of heat rejected will decrease. (c) the cycle efficiency will decrease. (d) the moisture content at turbine exit will decrease. (e) the amount of heat input will decrease.

Answer (d) the moisture content at turbine exit will decrease.

10-133 Consider a simple ideal Rankine cycle with fixed boiler and condenser pressures . If the cycle is modified with reheating, (select the correct statement) (a) the turbine work output will decrease. (b) the amount of heat rejected will decrease. (c) the pump work input will decrease. (d) the moisture content at turbine exit will decrease. (e) the amount of heat input will decrease.

Answer (d) the moisture content at turbine exit will decrease.

10-134 Consider a simple ideal Rankine cycle with fixed boiler and condenser pressures . If the cycle is modified with regeneration that involves one open feed water heater, (select the correct statement per unit mass of steam flowing through the boiler) (a) the turbine work output will decrease. (b) the amount of heat rejected will increase. (c) the cycle thermal efficiency will decrease. (d) the quality of steam at turbine exit will decrease. (e) the amount of heat input will increase.

Answer (a) the turbine work output will decrease.

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155

10-135 Consider a cogeneration power plant modified with regeneration. Steam enters the turbine at 6 MPa and 450°C at a rate of 20 kg/s and expands to a pressure of 0.4 MPa. At this pressure, 60% of the steam is extracted from the turbine, and the remainder expands to a pressure of 10 kPa. Part of the extracted steam is used to heat feedwater in an open feedwater heater. The rest of the extracted steam is used for process heating and leaves the process heater as a saturated liquid at 0.4 MPa. It is subsequently mixed with the feedwater leaving the feedwater heater, and the mixture is pumped to the boiler pressure. The steam in the condenser is cooled and condensed by the cooling water from a nearby river, which enters the adiabatic condenser at a rate of 463 kg/s.

6 Turbine

Boiler

8 10 Process heater

5 9 4

7 11 Condenser

3

P II

PI

fwh

h1 = 191.81 h2 = 192.20 h3 = h4 = h9 = 604.66 h5 = 610.73 h6 = 3302.9 h7 = h8 = h10 = 2665.6 h11 = 2128.8

1

2

1. The total power output of the turbine is (a) 17.0 MW

(b) 8.4 MW

(c) 12.2 MW

(d) 20.0 MW

(e) 3.4 MW

Answer (a) 17.0 MW 2. The temperature rise of the cooling water from the river in the condenser is (a) 8.0°C

(b) 5.2°C

(c) 9.6°C

(d) 12.9°C

(e) 16.2°C

(d) 7.6 kg/s

(e) 10.4 kg/s

Answer (a) 8.0°C 3. The mass flow rate of steam through the process heater is (a) 1.6 kg/s

(b) 3.8 kg/s

(c) 5.2 kg/s

Answer (e) 10.4 kg/s 4. The rate of heat supply from the process heater per unit mass of steam passing through it is (a) 246 kJ/kg

(b) 893 kJ/kg

(c) 1344 kJ/kg

(d) 1891 kJ/kg

(e) 2060 kJ/kg

Answer (e) 2060 kJ/kg

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156

5. The rate of heat transfer to the steam in the boiler is (a) 26.0 MJ/s

(b) 53.8 MJ/s

(c) 39.5 MJ/s

(d) 62.8 MJ/s

(e) 125.4 MJ/s

Answer (b) 53.8 MJ/s Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). Note: The solution given below also evaluates all enthalpies given on the figure. P1=10 "kPa" P11=P1 P2=400 "kPa" P3=P2; P4=P2; P7=P2; P8=P2; P9=P2; P10=P2 P5=6000 "kPa" P6=P5 T6=450 "C" m_total=20 "kg/s" m7=0.6*m_total m_cond=0.4*m_total C=4.18 "kJ/kg.K" m_cooling=463 "kg/s" s7=s6 s11=s6 h1=ENTHALPY(Steam_IAPWS,x=0,P=P1) v1=VOLUME(Steam_IAPWS,x=0,P=P1) w_pump=v1*(P2-P1) h2=h1+w_pump h3=ENTHALPY(Steam_IAPWS,x=0,P=P3) h4=h3; h9=h3 v4=VOLUME(Steam_IAPWS,x=0,P=P4) w_pump2=v4*(P5-P4) h5=h4+w_pump2 h6=ENTHALPY(Steam_IAPWS,T=T6,P=P6) s6=ENTROPY(Steam_IAPWS,T=T6,P=P6) h7=ENTHALPY(Steam_IAPWS,s=s7,P=P7) h8=h7; h10=h7 h11=ENTHALPY(Steam_IAPWS,s=s11,P=P11) W_turb=m_total*(h6-h7)+m_cond*(h7-h11) m_cooling*C*T_rise=m_cond*(h11-h1) m_cond*h2+m_feed*h10=(m_cond+m_feed)*h3 m_process=m7-m_feed q_process=h8-h9 Q_in=m_total*(h6-h5)

10-136 ··· 10-143 Design and Essay Problems

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11-1

Chapter 11 REFRIGERATION CYCLES The Reversed Carnot Cycle

11-1C Because the compression process involves the compression of a liquid-vapor mixture which requires a compressor that will handle two phases, and the expansion process involves the expansion of high-moisture content refrigerant.

11-2 A steady-flow Carnot refrigeration cycle with refrigerant-134a as the working fluid is considered. The coefficient of performance, the amount of heat absorbed from the refrigerated space, and the net work input are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) Noting that TH = 30°C = 303 K and TL = Tsat @ 160 kPa = -15.60°C = 257.4 K, the COP of this Carnot refrigerator is determined from COPR,C =

1 1 = = 5.64 TH / TL − 1 (303 K ) / (257.4 K ) − 1

T

(b) From the refrigerant tables (Table A-11), h3 = h g @30°C = 266.66 kJ/kg h4 = h f @30°C = 93.58 kJ/kg

Thus,

4

QH

3 30°C

160 kPa q H = h3 − h4 = 266.66 − 93.58 = 173.08 kJ/kg

1 QL

2

and ⎛ 257.4 K ⎞ q H TH T ⎟⎟(173.08 kJ/kg ) = 147.03 kJ/kg = ⎯ ⎯→ q L = L q H = ⎜⎜ q L TL TH ⎝ 303 K ⎠

(c) The net work input is determined from wnet = q H − q L = 173.08 − 147.03 = 26.05 kJ/kg

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s

11-2

11-3E A steady-flow Carnot refrigeration cycle with refrigerant-134a as the working fluid is considered. The coefficient of performance, the quality at the beginning of the heat-absorption process, and the net work input are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) Noting that TH = Tsat @ 90 psia = 72.78°F = 532.8 R and TL = Tsat @ 30 psia = 15.37°F = 475.4 R. COPR,C =

1 1 = = 8.28 TH / TL − 1 (532.8 R )/ (475.4 R ) − 1

T

(b) Process 4-1 is isentropic, and thus

(

s1 = s 4 = s f + x 4 s fg

) @ 90 psia = 0.07481 + (0.05)(0.14525)

= 0.08207 Btu/lbm ⋅ R ⎛ s1 − s f x1 = ⎜ ⎜ s fg ⎝

⎞ 0.08207 − 0.03793 ⎟ = = 0.2374 ⎟ 0.18589 ⎠ @ 30 psia

(c) Remembering that on a T-s diagram the area enclosed represents the net work, and s3 = sg @ 90 psia = 0.22006 Btu/lbm·R,

4

1

QH

QL

3

2

s

w net,in = (T H − T L )(s 3 − s 4 ) = (72.78 − 15.37)(0.22006 − 0.08207 ) Btu/lbm ⋅ R = 7.92 Btu/lbm

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11-3

Ideal and Actual Vapor-Compression Cycles

11-4C Yes; the throttling process is an internally irreversible process.

11-5C To make the ideal vapor-compression refrigeration cycle more closely approximate the actual cycle.

11-6C No. Assuming the water is maintained at 10°C in the evaporator, the evaporator pressure will be the saturation pressure corresponding to this pressure, which is 1.2 kPa. It is not practical to design refrigeration or air-conditioning devices that involve such extremely low pressures.

11-7C Allowing a temperature difference of 10°C for effective heat transfer, the condensation temperature of the refrigerant should be 25°C. The saturation pressure corresponding to 25°C is 0.67 MPa. Therefore, the recommended pressure would be 0.7 MPa.

11-8C The area enclosed by the cyclic curve on a T-s diagram represents the net work input for the reversed Carnot cycle, but not so for the ideal vapor-compression refrigeration cycle. This is because the latter cycle involves an irreversible process for which the process path is not known.

11-9C The cycle that involves saturated liquid at 30°C will have a higher COP because, judging from the T-s diagram, it will require a smaller work input for the same refrigeration capacity.

11-10C The minimum temperature that the refrigerant can be cooled to before throttling is the temperature of the sink (the cooling medium) since heat is transferred from the refrigerant to the cooling medium.

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11-4

11-11 A commercial refrigerator with refrigerant-134a as the working fluid is considered. The quality of the refrigerant at the evaporator inlet, the refrigeration load, the COP of the refrigerator, and the theoretical maximum refrigeration load for the same power input to the compressor are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From refrigerant-134a tables (Tables A-11 through A-13) P1 = 60 kPa ⎫ ⎬ h1 = 230.03 kJ/kg T1 = −34°C ⎭ P2 = 1200 kPa ⎫ ⎬ h2 = 295.16 kJ/kg T2 = 65°C ⎭ P3 = 1200 kPa ⎫ ⎬ h3 = 111.23 kJ/kg T3 = 42°C ⎭ h4 = h3 = 111.23 kJ/kg

Water 18°C

26°C QH

1.2 MPa 65°C

42°C

P4 = 60 kPa

⎫ ⎬ x 4 = 0.4795 h4 = 111.23 kJ/kg ⎭

Using saturated liquid enthalpy at the given temperature, for water we have (Table A-4)

Condenser 3

2

Expansion valve

Win Compressor

4

60 kPa -34°C

1 Evaporator

hw1 = h f @ 18°C = 75.47 kJ/kg

QL

hw 2 = h f @ 26°C = 108.94 kJ/kg

(b) The mass flow rate of the refrigerant may be determined from an energy balance on the compressor m& R (h2 − h3 ) = m& w (hw 2 − hw1 ) m& R (295.16 − 111.23)kJ/kg = (0.25 kg/s)(108.94 − 75.47)kJ/kg ⎯ ⎯→ m& R = 0.0455 kg/s

The waste heat transferred from the refrigerant, the compressor power input, and the refrigeration load are Q& H = m& R (h2 − h3 ) = (0.0455 kg/s)(295.16 − 111.23)kJ/kg = 8.367 kW W& in = m& R (h2 − h1 ) − Q& in = (0.0455 kg/s)(295.16 − 230.03)kJ/kg − 0.45 kW = 2.513 kW Q& L = Q& H − W& in = 8.367 − 2.513 = 5.85 kW

(c) The COP of the refrigerator is determined from its definition

T

Q& 5.85 COP = L = = 2.33 & Win 2.513

2 · QH

2 · Win

3

(d) The reversible COP of the refrigerator for the same temperature limits is COPmax =

1 1 = = 5.063 T H / T L − 1 (18 + 273) /(−30 + 273) − 1

4

· QL

1 s

Then, the maximum refrigeration load becomes Q& L,max = COPmax W& in = (5.063)(2.513 kW) = 12.72 kW

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11-5

11-12 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The COP and the power requirement are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-11, A-12, and A-13), T1 = 4°C ⎫ h1 = h g @ 4°C = 252.77 kJ/kg ⎬ sat. vapor ⎭ s1 = s g @ 4°C = 0.92927 kJ/kg ⋅ K P2 = 1 MPa ⎫ ⎬ h2 = 275.29 kJ/kg s 2 = s1 ⎭ P3 = 1 MPa ⎫ ⎬ h = hf sat. liquid ⎭ 3

@ 1 MPa

T · QH

2

3 1 MPa

· Win

= 107.32 kJ/kg

h4 ≅ h3 = 107.32 kJ/kg ( throttling)

4°C 4s

4

· QL

1

The mass flow rate of the refrigerant is Q& L = m& (h1 − h4 ) ⎯ ⎯→ m& =

Q& L 400 kJ/s = = 2.750 kg/s h1 − h4 (252.77 − 107.32) kJ/kg

s

The power requirement is W& in = m& (h2 − h1 ) = (2.750 kg/s)(275.29 − 252.77) kJ/kg = 61.93 kW

The COP of the refrigerator is determined from its definition, COPR =

Q& L 400 kW = = 6.46 & Win 61.93 kW

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11-6

11-13 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The mass flow rate of the refrigerant and the power requirement are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-11, A-12, and A-13), P1 = 400 kPa ⎫ h1 = h g @ 400 kPa = 255.55 kJ/kg ⎬ s =s sat. vapor g @ 400 kPa = 0.92691 kJ/kg ⋅ K ⎭ 1 P2 = 800 kPa ⎫ ⎬ h2 = 269.90 kJ/kg s 2 = s1 ⎭ P3 = 800 kPa ⎫ ⎬ h3 = h f sat. liquid ⎭

@ 800 kPa

T · QH

2

3 0.8 MPa

· Win

= 95.47 kJ/kg

0.4 MPa

h4 ≅ h3 = 95.47 kJ/kg ( throttling )

4s

4

· QL

1

The mass flow rate of the refrigerant is determined from Q& L = m& (h1 − h4 ) ⎯ ⎯→ m& =

Q& L 10 kJ/s = = 0.06247 kg/s h1 − h4 (255.55 − 95.47) kJ/kg

s

The power requirement is W& in = m& (h2 − h1 ) = (0.06247 kg/s)(269.90 − 255.55) kJ/kg = 0.8964 kW

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11-7

11-14E An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The mass flow rate of the refrigerant and the power requirement are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-11E, A-12E, and A13E), T1 = −10°F ⎫ h1 = h g @ −10° F = 101.61 Btu/lbm ⎬ sat. vapor ⎭ s1 = s g @ −10°F = 0.22660 Btu/lbm ⋅ R P2 = 100 psia ⎫ ⎬ h2 = 117.57 Btu/lbm s 2 = s1 ⎭ P3 = 100 psia ⎫ ⎬ h3 = h f sat. liquid ⎭

@ 100 psia

T · QH

= 37.869 Btu/lbm

h4 ≅ h3 = 37.869 Btu/lbm ( throttling )

2

3 100 psia

· Win

-10°F 4s

4

· QL

1

The mass flow rate of the refrigerant is

s

Q& L 24,000 Btu/h Q& L = m& (h1 − h4 ) ⎯ ⎯→ m& = = = 376.5 lbm/h h1 − h4 (101.61 − 37.869) Btu/lbm

The power requirement is

1 kW ⎛ ⎞ W& in = m& (h2 − h1 ) = (376.5 lbm/h)(117.57 − 101.61) Btu/lbm⎜ ⎟ = 1.761 kW 3412.14 Btu/h ⎝ ⎠

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11-8

11-15E Problem 11-14E is to be repeated if ammonia is used as the refrigerant. Analysis The problem is solved using EES, and the solution is given below. "Given" x[1]=1 T[1]=-10 [F] x[3]=0 P[3]=100 [psia] Q_dot_L=24000 [Btu/h] "Analysis" Fluid$='ammonia' "compressor" h[1]=enthalpy(Fluid$, T=T[1], x=x[1]) s[1]=entropy(Fluid$, T=T[1], x=x[1]) s[2]=s[1] P[2]=P[3] h[2]=enthalpy(Fluid$, P=P[2], s=s[2]) "expansion valve" h[3]=enthalpy(Fluid$, P=P[3], x=x[3]) h[4]=h[3] "cycle" m_dot_R=Q_dot_L/(h[1]-h[4]) W_dot_in=m_dot_R*(h[2]-h[1])*Convert(Btu/h, kW) Solution for ammonia Fluid$='ammonia' h[2]=701.99 [Btu/lb_m] m_dot_R=47.69 [lbm/h] Q_dot_L=24000 [Btu/h] T[1]=-10 [F] x[3]=0

COP_R=5.847 h[3]=112.67 [Btu/lb_m] P[2]=100 [psia] s[1]=1.42220 [Btu/lb_m-R] W_dot_in=1.203 [kW]

h[1]=615.92 [Btu/lb_m] h[4]=112.67 [Btu/lb_m] P[3]=100 [psia] s[2]=1.42220 [Btu/lb_m-R] x[1]=1

Solution for R-134a Fluid$='R134a' h[2]=117.58 [Btu/lb_m] m_dot_R=376.5 [lbm/h] Q_dot_L=24000 [Btu/h] T[1]=-10 [F] x[3]=0

COP_R=3.993 h[3]=37.87 [Btu/lb_m] P[2]=100 [psia] s[1]=0.22662 [Btu/lb_m-R] W_dot_in=1.761 [kW]

h[1]=101.62 [Btu/lb_m] h[4]=37.87 [Btu/lb_m] P[3]=100 [psia] s[2]=0.22662 [Btu/lb_m-R] x[1]=1

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11-9

11-16 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The rate of heat removal from the refrigerated space, the power input to the compressor, the rate of heat rejection to the environment, and the COP are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-12 and A-13), P1 = 120 kPa sat. vapor

⎫ h1 = h g @ 120 kPa = 236.97 kJ/kg ⎬s = s g @ 120 kPa = 0.94779 kJ/kg ⋅ K ⎭ 1

P2 = 0.7 MPa s 2 = s1

⎫ ⎬ h2 = 273.50 kJ/kg (T2 = 34.95°C ) ⎭

P3 = 0.7 MPa sat. liquid

⎫ ⎬ h3 = h f ⎭

@ 0.7 MPa

T · QH

2

3 0.7 MPa

· Win

= 88.82 kJ/kg

h4 ≅ h3 = 88.82 kJ/kg (throttling )

Then the rate of heat removal from the refrigerated space and the power input to the compressor are determined from

0.12 4s

4

· QL

1

Q& L = m& (h1 − h4 ) = (0.05 kg/s )(236.97 − 88.82) kJ/kg = 7.41 kW

and W& in = m& (h2 − h1 ) = (0.05 kg/s )(273.50 − 236.97 ) kJ/kg = 1.83 kW

(b) The rate of heat rejection to the environment is determined from Q& H = Q& L + W& in = 7.41 + 1.83 = 9.23 kW

(c) The COP of the refrigerator is determined from its definition, COPR =

Q& L 7.41 kW = = 4.06 & 1.83 kW Win

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s

11-10

11-17 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The rate of heat removal from the refrigerated space, the power input to the compressor, the rate of heat rejection to the environment, and the COP are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-12 and A-13), P1 = 120 kPa sat. vapor

⎫ h1 = h g @ 120 kPa = 236.97 kJ/kg ⎬s = s g @ 120 kPa = 0.94779 kJ/kg ⋅ K ⎭ 1

P2 = 0.9 MPa s 2 = s1

⎫ ⎬ h2 = 278.93 kJ/kg (T2 = 44.45°C ) ⎭

P3 = 0.9 MPa sat. liquid

⎫ ⎬ h3 = h f ⎭

@ 0.9 MPa

T · QH

2

3 0.9 MPa

· Win

= 101.61 kJ/kg

h4 ≅ h3 = 101.61 kJ/kg (throttling )

Then the rate of heat removal from the refrigerated space and the power input to the compressor are determined from

0.12 MPa 4s

4

· QL

1 s

Q& L = m& (h1 − h4 ) = (0.05 kg/s )(236.97 − 101.61) kJ/kg = 6.77 kW

and W& in = m& (h2 − h1 ) = (0.05 kg/s )(278.93 − 236.97 ) kJ/kg = 2.10 kW

(b) The rate of heat rejection to the environment is determined from Q& H = Q& L + W& in = 6.77 + 2.10 = 8.87 kW

(c) The COP of the refrigerator is determined from its definition, COPR =

Q& L 6.77 kW = = 3.23 & Win 2.10 kW

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11-11

11-18 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The throttling valve in the cycle is replaced by an isentropic turbine. The percentage increase in the COP and in the rate of heat removal from the refrigerated space due to this replacement are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

T

Analysis If the throttling valve in the previous problem is replaced by an isentropic turbine, we would have s4s = s3 = sf @ 0.7 MPa = 0.33230 kJ/kg·K, and the enthalpy at the turbine exit would be ⎛ s3 − s f x4s = ⎜ ⎜ s fg ⎝

(

⎞ 0.33230 − 0.09275 ⎟ = = 0.2802 ⎟ 0.85503 ⎠ @ 120 kPa

h4 s = h f + x 4 s h fg

· QH

2

3 0.7 MPa

· Win

0.12 MPa 4s

4

· QL

1 s

)@ 120 kPa = 22.49 + (0.2802)(214.48) = 82.58 kJ/kg

Then, Q& L = m& (h1 − h4 s ) = (0.05 kg/s )(236.97 − 82.58) kJ/kg = 7.72 kW

and COPR =

Q& L 7.72 kW = = 4.23 & Win 1.83 kW

Then the percentage increase in Q& and COP becomes ΔQ& L 7.72 − 7.41 Increase in Q& L = = = 4.2% 7.41 Q& L Increase in COPR =

ΔCOPR 4.23 − 4.06 = = 4.2% COPR 4.06

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11-12

11-19 A refrigerator with refrigerant-134a as the working fluid is considered. The rate of heat removal from the refrigerated space, the power input to the compressor, the isentropic efficiency of the compressor, and the COP of the refrigerator are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the refrigerant tables (Tables A-12 and A-13), P1 = 0.14 MPa ⎫ h1 = 246.36 kJ/kg ⎬ s = 0.97236 kJ/kg ⋅ K T1 = −10°C ⎭ 1

T

P2 = 0.7 MPa ⎫ ⎬ h2 = 288.53 kJ/kg T2 = 50°C ⎭

0.65 MPa

P2 s = 0.7 MPa ⎫ ⎬ h2 s = 281.16 kJ/kg s 2 s = s1 ⎭ P3 = 0.65 MPa ⎫ ⎬ h3 = h f T3 = 24°C ⎭

@ 24°C

· QH

2 0.7 MPa 2 50°C · Win

3

= 84.98 kJ/kg

h4 ≅ h3 = 84.98 kJ/kg (throttling )

0.15 MP 4

· QL

1

0.14 MPa s

Then the rate of heat removal from the refrigerated space and the power input to the compressor are determined from Q& L = m& (h1 − h4 ) = (0.12 kg/s )(246.36 − 84.98) kJ/kg = 19.4 kW

and W& in = m& (h2 − h1 ) = (0.12 kg/s )(288.53 − 246.36) kJ/kg = 5.06 kW

(b) The adiabatic efficiency of the compressor is determined from

ηC =

h2 s − h1 281.16 − 246.36 = = 82.5% 288.53 − 246.36 h2 − h1

(c) The COP of the refrigerator is determined from its definition, COPR =

Q& L 19.4 kW = = 3.83 W& in 5.06 kW

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11-13

11-20 An air conditioner operating on the ideal vapor-compression refrigeration cycle with refrigerant134a as the working fluid is considered. The COP of the system is to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. The evaporating temperature will be 22-2=20°C. From the refrigerant tables (Tables A-11, A-12, and A-13), T1 = 20°C ⎫ h1 = h g @ 20°C = 261.59 kJ/kg ⎬ sat. vapor ⎭ s1 = s g @ 20°C = 0.92234 kJ/kg ⋅ K P2 = 1 MPa ⎫ ⎬ h2 = 273.11 kJ/kg s 2 = s1 ⎭ P3 = 1 MPa ⎫ ⎬ h = hf sat. liquid ⎭ 3

@ 1 MPa

· QH

2

3 1 MPa

· Win

= 107.32 kJ/kg

h4 ≅ h3 = 107.32 kJ/kg ( throttling)

The COP of the air conditioner is determined from its definition, COPAC =

T

qL h -h 261.59 − 107.32 = 1 4 = = 13.39 win h2 -h1 273.11 − 261.59

20°C 4s

4

· QL

1

s

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11-14

11-21E A refrigerator operating on the ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The increase in the COP if the throttling process were replaced by an isentropic expansion is to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-11E, A-12E, and A13E), T1 = 20°F ⎫ h1 = h g @ 20° F = 105.98 Btu/lbm ⎬ sat. vapor ⎭ s1 = s g @ 20°F = 0.22341 Btu/lbm ⋅ R P2 = 300 psia ⎫ ⎬ h2 = 125.68 Btu/lbm s 2 = s1 ⎭

T · QH

P3 = 300 psia ⎫ h3 = h f @ 300 psia = 66.339 Btu/lbm ⎬ s =s sat. liquid f @ 300 psia = 0.12715 Btu/lbm ⋅ R ⎭ 3 h4 ≅ h3 = 66.339 Btu/lbm ( throttling) T4 = 15°F ⎫ ⎬ h = 59.80 Btu/lbm (isentropic expansion) s 4 = s3 ⎭ 4s

2

3 300 psia

· Win

20°F 4s

4

· QL

1

s

The COP of the refrigerator for the throttling case is COPR =

qL h -h 105.98 − 66.339 = 1 4 = = 2.012 win h2 -h1 125.68 − 105.98

The COP of the refrigerator for the isentropic expansion case is COPR =

h -h qL 105.98 − 59.80 = 1 4s = = 2.344 win h2 -h1 125.68 − 105.98

The increase in the COP by isentropic expansion is 16.5%.

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11-15

11-22 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The COP of the system and the cooling load are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-11, A-12, and A-13), T1 = −10°C ⎫ h1 = h g @ −10°C = 244.51 kJ/kg ⎬ sat. vapor ⎭ s1 = s g @ −10°C = 0.93766 kJ/kg ⋅ K

T

P2 = 600 kPa ⎫ ⎬ h2 = 267.12 kJ/kg s 2 = s1 ⎭ P3 = 600 kPa ⎫ ⎬ h3 = h f sat. liquid ⎭

@ 600 kPa

· QH 3 600 kPa

= 81.51 kJ/kg

h4 ≅ h3 = 81.51 kJ/kg ( throttling)

The COP of the air conditioner is determined from its definition, COPAC =

qL h -h 244.51 − 81.51 = 1 4 = = 7.21 win h2 -h1 267.12 − 244.51

2 · Win

-10°C 4s

4

· QL

1

s

The cooling load is Q& L = m& (h1 − h4 ) = (7 kg/s)(244.51 − 81.51) kJ/kg = 1141 kW

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11-16

11-23 A refrigerator with refrigerant-134a as the working fluid is considered. The power input to the compressor, the rate of heat removal from the refrigerated space, and the pressure drop and the rate of heat gain in the line between the evaporator and the compressor are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the refrigerant tables (Tables A-12 and A-13), h1 = 246.36 kJ/kg P1 = 140 kPa ⎫ s = 0.97236 kJ/kg ⋅ K T1 = −10°C ⎬⎭ 1 v 1 = 0.14605 m 3 /kg

T

P2 = 1.0 MPa ⎫ ⎬ h2 s = 289.20 kJ/kg s 2 s = s1 ⎭ P3 = 0.95 MPa ⎫ ⎬ h3 ≅ h f T3 = 30°C ⎭

@ 30 °C

2 0.95 MPa

2s

· QH

1 MPa · Win

3

= 93.58 kJ/kg

h4 ≅ h3 = 93.58 kJ/kg (throttling ) T5 = −18.5°C ⎫ P5 = 0.14165 MPa ⎬ h = 239.33 kJ/kg sat. vapor ⎭ 5

0.15 MPa 4

· QL

1

0.14 MPa -10°C -18.5°C s

Then the mass flow rate of the refrigerant and the power input becomes m& =

0.3/60 m3/s V&1 = = 0.03423 kg/s v1 0.14605 m3/kg

W&in = m& (h2 s − h1 ) /ηC = (0.03423 kg/s )[(289.20 − 246.36) kJ/kg ]/ (0.78) = 1.88 kW

(b) The rate of heat removal from the refrigerated space is Q& L = m& (h5 − h4 ) = (0.03423 kg/s )(239.33 − 93.58) kJ/kg = 4.99 kW

(c) The pressure drop and the heat gain in the line between the evaporator and the compressor are ΔP = P5 − P1 = 141.65 − 140 = 1.65

and Q& gain = m& (h1 − h5 ) = (0.03423 kg/s )(246.36 − 239.33) kJ/kg = 0.241 kW

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11-17

11-24 EES Problem 11-23 is reconsidered. The effects of the compressor isentropic efficiency and the compressor inlet volume flow rate on the power input and the rate of refrigeration are to be investigated. Analysis The problem is solved using EES, and the solution is given below. "Input Data" "T[5]=-18.5 [C] P[1]=140 [kPa] T[1] = -10 [C]} V_dot[1]=0.1 [m^3/min] P[2] = 1000 [kPa] P[3]=950 [kPa] T[3] = 30 [C] Eta_c=0.78 Fluid$='R134a'" "Compressor" h[1]=enthalpy(Fluid$,P=P[1],T=T[1]) "properties for state 1" s[1]=entropy(Fluid$,P=P[1],T=T[1]) v[1]=volume(Fluid$,P=P[1],T=T[1])"[m^3/kg]" m_dot=V_dot[1]/v[1]*convert(m^3/min,m^3/s)"[kg/s]" h2s=enthalpy(Fluid$,P=P[2],s=s[1]) "Identifies state 2s as isentropic" h[1]+Wcs=h2s "energy balance on isentropic compressor" Wc=Wcs/Eta_c"definition of compressor isentropic efficiency" h[1]+Wc=h[2] "energy balance on real compressor-assumed adiabatic" s[2]=entropy(Fluid$,h=h[2],P=P[2]) "properties for state 2" T[2]=temperature(Fluid$,h=h[2],P=P[2]) W_dot_c=m_dot*Wc "Condenser" h[3]=enthalpy(Fluid$,P=P[3],T=T[3]) "properties for state 3" s[3]=entropy(Fluid$,P=P[3],T=T[3]) h[2]=q_out+h[3] "energy balance on condenser" Q_dot_out=m_dot*q_out "Throttle Valve" h[4]=h[3] "energy balance on throttle - isenthalpic" x[4]=quality(Fluid$,h=h[4],P=P[4]) "properties for state 4" s[4]=entropy(Fluid$,h=h[4],P=P[4]) T[4]=temperature(Fluid$,h=h[4],P=P[4]) "Evaporator" P[4]=pressure(Fluid$,T=T[5],x=0)"pressure=Psat at evaporator exit temp." P[5] = P[4] h[5]=enthalpy(Fluid$,T=T[5],x=1) "properties for state 5" q_in + h[4]=h[5] "energy balance on evaporator" Q_dot_in=m_dot*q_in COP=Q_dot_in/W_dot_c "definition of COP" COP_plot = COP W_dot_in = W_dot_c Q_dot_line5to1=m_dot*(h[1]-h[5])"[kW]"

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11-18

Win[kW] 0.8149 0.6985 0.6112 0.5433 0.4889

COPplot 2.041 2.381 2.721 3.062 3.402

ηc 0.6 0.7 0.8 0.9 1

Qin [kW] 1.663 1.663 1.663 1.663 1.663

9

3

V 1 m /m in 1.0 0.5

8 7

0.1

6

W in

5 4 3 2 1 0 0.6

0.65

0.7

0.75

0.8

0.85

0.9

0.95

1

ηc 4 3.5 3

COP plot

2.5

3

2

V 1 m /m in 1.0

1.5

0.5

0.1

1 0.5 0 0.6

0.65

0.7

0.75

0.8

0.85

0.9

0.95

1

ηc 18

3

V 1 m /m in

Q in [kW ]

14.4

1.0

10.8

0.5

0.1

7.2

3.6

0 0.6

0.65

0.7

0.75

0.8

0.85

0.9

0.95

1

ηc

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11-19

11-25 A refrigerator uses refrigerant-134a as the working fluid and operates on the ideal vaporcompression refrigeration cycle. The mass flow rate of the refrigerant, the condenser pressure, and the COP of the refrigerator are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) (b) From the refrigerant-134a tables (Tables A-11 through A-13) . P4 = 120 kPa ⎫ QH ⎬ h4 = 86.83 kJ/kg x 4 = 0.30 ⎭ h3 = h4 Condenser h3 = 86.83 kJ/kg ⎫ ⎬ P3 = 671.8 kPa x 3 = 0 (sat. liq.) ⎭ P2 = P3 P2 = 671.8 kPa ⎫ ⎬ h2 = 298.87 kJ/kg T2 = 60°C ⎭ P1 = P4 = 120 kPa ⎫ ⎬ h1 = 236.97 kJ/kg x1 = 1 (sat. vap.) ⎭

3

2

Expansion valve

. Win

Compressor 1

4 Evaporator 120 kPa x=0.3

The mass flow rate of the refrigerant is determined from m& =

60°C

W& in 0.45 kW = = 0.00727 kg/s h2 − h1 (298.87 − 236.97)kJ/kg

. QL T · QH 3

2 · Win

(c) The refrigeration load and the COP are Q& L = m& (h1 − h4 ) = (0.0727 kg/s)(236.97 − 86.83)kJ/kg = 1.091 kW COP =

Q& L 1.091 kW = = 2.43 0.45 kW W& in

120 kPa 4s

4

· QL

1 s

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

11-20

Selecting the Right Refrigerant

11-26C The desirable characteristics of a refrigerant are to have an evaporator pressure which is above the atmospheric pressure, and a condenser pressure which corresponds to a saturation temperature above the temperature of the cooling medium. Other desirable characteristics of a refrigerant include being nontoxic, noncorrosive, nonflammable, chemically stable, having a high enthalpy of vaporization (minimizes the mass flow rate) and, of course, being available at low cost.

11-27C The minimum pressure that the refrigerant needs to be compressed to is the saturation pressure of the refrigerant at 30°C, which is 0.771 MPa. At lower pressures, the refrigerant will have to condense at temperatures lower than the temperature of the surroundings, which cannot happen.

11-28C Allowing a temperature difference of 10°C for effective heat transfer, the evaporation temperature of the refrigerant should be -20°C. The saturation pressure corresponding to -20°C is 0.133 MPa. Therefore, the recommended pressure would be 0.12 MPa.

11-29 A refrigerator that operates on the ideal vapor-compression cycle with refrigerant-134a is considered. Reasonable pressures for the evaporator and the condenser are to be selected. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis Allowing a temperature difference of 10°C for effective heat transfer, the evaporation and condensation temperatures of the refrigerant should be -20°C and 35°C, respectively. The saturation pressures corresponding to these temperatures are 0.133 MPa and 0.888 MPa. Therefore, the recommended evaporator and condenser pressures are 0.133 MPa and 0.888 MPa, respectively.

11-30 A heat pump that operates on the ideal vapor-compression cycle with refrigerant-134a is considered. Reasonable pressures for the evaporator and the condenser are to be selected. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis Allowing a temperature difference of 10°C for effective heat transfer, the evaporation and condensation temperatures of the refrigerant should be 0°C and 32°C, respectively. The saturation pressures corresponding to these temperatures are 0.293 MPa and 0.816 MPa. Therefore, the recommended evaporator and condenser pressures are 0.293 MPa and 0.816 MPa, respectively.

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11-21

Heat Pump Systems

11-31C A heat pump system is more cost effective in Miami because of the low heating loads and high cooling loads at that location.

11-32C A water-source heat pump extracts heat from water instead of air. Water-source heat pumps have higher COPs than the air-source systems because the temperature of water is higher than the temperature of air in winter.

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11-22

11-33 An actual heat pump cycle with R-134a as the refrigerant is considered. The isentropic efficiency of the compressor, the rate of heat supplied to the heated room, the COP of the heat pump, and the COP and the rate of heat supplied to the heated room if this heat pump operated on the ideal vapor-compression cycle between the same pressure limits are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) The properties of refrigerant-134a are (Tables A-11 through A-13) P2 = 800 kPa ⎫ ⎬ h2 = 291.76 kJ/kg T2 = 55°C ⎭ T3 = Tsat@750 kPa = 29.06°C

. QH

800 kPa 55°C

750 kPa

P3 = 750 kPa

⎫ ⎬ h3 = 87.91 kJ/kg T3 = (29.06 − 3)°C⎭ h4 = h3 = 87.91 kJ/kg Tsat@200 kPa = −10.09°C P1 = 200 kPa

⎫h1 = 247.87 kJ/kg ⎬ T1 = (−10.09 + 4)°C⎭ s1 = 0.9506 kJ/kg

Condenser 3

2

Expansion valve

. Win

Compressor 1

4 Evaporator . QL

P2 = 800 kPa ⎫ ⎬h2 s = 277.26 s 2 = s1 ⎭

The isentropic efficiency of the compressor is

ηC =

h2 s − h1 277.26 − 247.87 = 0.670 = h2 − h1 291.76 − 247.87

T

2 2

· Q

· Win

(b) The rate of heat supplied to the room is 3

Q& H = m& (h2 − h3 ) = (0.018 kg/s)(291.76 − 87.91)kJ/kg = 3.67 kW

(c) The power input and the COP are

COP =

· QL

4

W& in = m& (h2 − h1 ) = (0.018 kg/s)(291.76 − 247.87)kJ/kg = 0.790 kW

1 s

Q& H 3.67 = = 4.64 & 0 .790 Win

(d) The ideal vapor-compression cycle analysis of the cycle is as follows:

T

h1 = h g @ 200 kPa = 244.46 kJ/kg

· QH

s1 = s g @ 200 kPa = 0.9377 kJ/kg.K P2 = 800 MPa ⎫ ⎬ h2 = 273.25 kJ/kg s 2 = s1 ⎭ h3 = h f @ 800 kPa = 95.47 kJ/kg h4 = h3 COP =

2

3 0.8 MPa

· Win

0.2 MPa 4s

4

· QL

1

h2 − h3 273.25 − 95.47 = 6.18 = h2 − h1 273.25 − 244.46

Q& H = m& (h2 − h3 ) = (0.018 kg/s)(273.25 − 95.47)kJ/kg = 3.20 kW

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s

11-23

11-34 A heat pump operating on the ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The COP and the rate of heat supplied to the evaporator are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-11, A-12, and A-13), P1 = 280 kPa ⎫ h1 = h g @ 280 kPa = 249.72 kJ/kg ⎬ s =s sat. vapor g @ 280 kPa = 0.93210 kJ/kg ⋅ K ⎭ 1 P2 = 1200 kPa ⎫ ⎬ h2 = 280.00 kJ/kg s 2 = s1 ⎭ P3 = 1200 kPa ⎫ ⎬ h3 = h f sat. liquid ⎭

@ 1200 kPa

T · QH 3 1.2 MPa

2 · Win

= 117.77 kJ/kg

h4 ≅ h3 = 117.77 kJ/kg ( throttling)

280 kPa 4s

4

· QL

1

The mass flow rate of the refrigerant is determined from W& in 20 kJ/s ⎯→ m& = = = 0.6605 kg/s W& in = m& (h2 − h1 ) ⎯ h2 − h1 (280.00 − 249.72) kJ/kg

s

Then the rate of heat supplied to the evaporator is Q& L = m& (h1 − h4 ) = (0.6605 kg/s)(249.72 − 117.77) kJ/kg = 87.15 kW

The COP of the heat pump is determined from its definition, COPHP =

h -h qH 280.00 − 117.77 = 5.36 = 2 3 = win h2 -h1 280.00 − 249.72

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11-24

11-35 A heat pump operating on a vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The effect of compressor irreversibilities on the COP of the cycle is to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis In this cycle, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. The compression process is not isentropic. The saturation pressure of refrigerant at −1.25°C is 280 kPa. From the refrigerant tables (Tables A-11, A-12, and A-13), P1 = 280 kPa ⎫ h1 = h g @ 280 kPa = 249.72 kJ/kg ⎬ s =s sat. vapor g @ 280 kPa = 0.93210 kJ/kg ⋅ K ⎭ 1

T

P2 = 800 kPa ⎫ ⎬ h2 s = 271.50 kJ/kg s 2 = s1 ⎭ P3 = 800 kPa ⎫ ⎬ h3 = h f sat. liquid ⎭

@ 800 kPa

· QH 3 0.8 MPa

· Win

= 95.47 kJ/kg

h4 ≅ h3 = 95.47 kJ/kg ( throttling)

-1.25°C 4s

4

· QL

The actual enthalpy at the compressor exit is determined by using the compressor efficiency:

ηC =

2

2s

1

s

h − h1 h2 s − h1 271.50 − 249.72 ⎯ ⎯→ h2 = h1 + 2 s = 249.72 + = 275.34 kJ/kg ηC 0.85 h2 − h1

The COPs of the heat pump for isentropic and irreversible compression cases are COPHP, ideal =

h − h3 qH 271.50 − 95.47 = 8.082 = = 2s win h2 s − h1 271.50 − 249.72

COPHP, actual =

h − h3 qH 275.34 − 95.47 = 7.021 = = 2 win h2 − h1 275.34 − 249.72

The irreversible compressor decreases the COP by 13.1%.

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11-25

11-36 A heat pump operating on a vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The effect of superheating at the compressor inlet on the COP of the cycle is to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis In this cycle, the compression process is isentropic and leaves the condenser as saturated liquid at the condenser pressure. The refrigerant entering the compressor is superheated by 2°C. The saturation pressure of refrigerant at −1.25°C is 280 kPa. From the refrigerant tables (Tables A-11, A-12, and A-13), P1 = 280 kPa ⎫ h1 = 251.96 kJ/kg ⎬ T1 = −1.25 + 2 = 1.25°C ⎭ s1 = 0.9403 kJ/kg ⋅ K P2 = 800 kPa ⎫ ⎬ h2 = 274.04 kJ/kg s 2 = s1 ⎭ P3 = 800 kPa ⎫ ⎬ h3 = h f sat. liquid ⎭

@ 800 kPa

T

= 95.47 kJ/kg

· QH 3 0.8 MPa

h4 ≅ h3 = 95.47 kJ/kg ( throttling)

The states at the inlet and exit of the compressor when the refrigerant enters the compressor as a saturated vapor are P1 = 280 kPa ⎫ h1 = h g @ 280 kPa = 249.72 kJ/kg ⎬ s =s sat. vapor g @ 280 kPa = 0.93210 kJ/kg ⋅ K ⎭ 1

2 · Win

-1.25°C 4s

4

· QL

1

s

P2 = 800 kPa ⎫ ⎬ h2 s = 271.50 kJ/kg s 2 = s1 ⎭

The COPs of the heat pump for the two cases are COPHP, ideal =

h − h3 qH 271.50 − 95.47 = 8.082 = = 2s win h2 s − h1 271.50 − 249.72

COPHP, actual =

h − h3 qH 274.04 − 95.47 = 8.087 = = 2 win h2 − h1 274.04 − 251.96

The effect of superheating on the COP is negligible.

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11-26

11-37E A heat pump operating on the vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The effect of subcooling at the exit of the condenser on the power requirement is to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis In this cycle, the compression process is isentropic, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as subcooled liquid. From the refrigerant tables (Tables A-11E, A-12E, and A-13E), P1 = 50 psia ⎫ h1 = h g @ 50 psia = 108.81 Btu/lbm ⎬ sat. vapor ⎭ s1 = s g @ 50 psia = 0.22188 kJ/kg ⋅ K P2 = 160 psia ⎫ ⎬ h2 = 119.19 Btu/lbm s 2 = s1 ⎭ T3 = Tsat @ 160 psia − 9.5 = 109.5 − 9.5 = 100°F P3 = 160 psia ⎫ ⎬ h ≅ hf T3 = 100°F ⎭ 3

@ 100° F

T · QH

2

3 160 psia

· Win

= 45.124 Btu/lbm

50 psia

h4 ≅ h3 = 45.124 Btu/lbm ( throttling)

4s

4

· QL

1

The states at the inlet and exit of the expansion valve when the refrigerant is saturated liquid at the condenser exit are P3 = 160 psia ⎫ ⎬ h3 = h f sat. liquid ⎭

@ 160 psia

s

= 48.519 Btu/lbm

h4 ≅ h3 = 48.519 Btu/lbm ( throttling)

The mass flow rate of the refrigerant in the ideal case is Q& L = m& (h1 − h4 ) ⎯ ⎯→ m& =

Q& H 100,000 Btu/h = = 1415.0 lbm/h h2 − h3 (119.19 − 48.519) Btu/lbm

The power requirement is 1 kW ⎛ ⎞ W& in = m& (h2 − h1 ) = (1415.0 lbm/h)(119.19 − 108.81) Btu/lbm⎜ ⎟ = 4.305 kW 3412.14 Btu/h ⎝ ⎠ With subcooling, the mass flow rate and the power input are Q& L = m& (h1 − h4 ) ⎯ ⎯→ m& =

Q& H 100,000 Btu/h = = 1350.1 lbm/h h2 − h3 (119.19 − 45.124) Btu/lbm

1 kW ⎛ ⎞ W& in = m& (h2 − h1 ) = (1350.1 lbm/h)(119.19 − 108.81) Btu/lbm⎜ ⎟ = 4.107 kW ⎝ 3412.14 Btu/h ⎠ Subcooling decreases the power requirement by 4.6%.

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11-27

11-38E A heat pump operating on the vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The effect of superheating at the compressor inlet on the power requirement is to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis In this cycle, the compression process is isentropic and leaves the condenser as saturated liquid at the condenser pressure. The refrigerant entering the compressor is superheated by 10°F. The saturation temperature of the refrigerant at 50 psia is 40.23°F. From the refrigerant tables (Tables A-11E, A-12E, and A-13E), P1 = 50 psia ⎫ h1 = 110.99 Btu/lbm ⎬ T1 = 40.23 + 10 = 50.2°F ⎭ s1 = 0.2262 kJ/kg ⋅ K

T

P2 = 160 psia ⎫ ⎬ h2 = 121.71 Btu/lbm s 2 = s1 ⎭ P3 = 160 psia ⎫ ⎬ h3 = h f sat. liquid ⎭

· QH

2

3 160 psia @ 160 psia

· Win

= 48.519 Btu/lbm

h4 ≅ h3 = 48.519 Btu/lbm ( throttling)

The states at the inlet and exit of the compressor when the refrigerant enters the compressor as a saturated vapor are

50 psia 4s

4

· QL

1

P1 = 50 psia ⎫ h1 = h g @ 50 psia = 108.81 Btu/lbm ⎬ sat. vapor ⎭ s1 = s g @ 50 psia = 0.22188 kJ/kg ⋅ K

s

P2 = 160 psia ⎫ ⎬ h2 = 119.19 Btu/lbm s 2 = s1 ⎭

The mass flow rate of the refrigerant in the ideal case is Q& L = m& (h1 − h4 ) ⎯ ⎯→ m& =

Q& H 100,000 Btu/h = = 1415.0 lbm/h h2 − h3 (119.19 − 48.519) Btu/lbm

The power requirement is 1 kW ⎛ ⎞ W& in = m& (h2 − h1 ) = (1415.0 lbm/h)(119.19 − 108.81) Btu/lbm⎜ ⎟ = 4.305 kW ⎝ 3412.14 Btu/h ⎠ With superheating, the mass flow rate and the power input are Q& L = m& (h1 − h4 ) ⎯ ⎯→ m& =

Q& H 100,000 Btu/h = = 1366.3 lbm/h h2 − h3 (121.71 − 48.519) Btu/lbm

1 kW ⎛ ⎞ W& in = m& (h2 − h1 ) = (1366.3 lbm/h)(121.71 − 110.99) Btu/lbm⎜ ⎟ = 4.293 kW ⎝ 3412.14 Btu/h ⎠ Superheating decreases the power requirement slightly.

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11-28

11-39 A geothermal heat pump is considered. The degrees of subcooling done on the refrigerant in the condenser, the mass flow rate of the refrigerant, the heating load, the COP of the heat pump, the minimum power input are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the refrigerant-134a tables . (Tables A-11 through A-13) QH T4 = 20°C⎫ P4 = 572.1 kPa Condenser ⎬ 1.4 MPa x 4 = 0.23 ⎭ h4 = 121.24 kJ/kg 3 s2 = s1 2 . h3 = h4 Expansion Win P1 = 572.1 kPa ⎫ h1 = 261.59 kJ/kg valve Compressor ⎬ = 0 . 9223 kJ/kg x1 = 1 (sat. vap.)⎭ s1 P2 = 1400 kPa ⎫ ⎬ h2 = 280.00 kJ/kg s 2 = s1 ⎭

From the steam tables (Table A-4) hw1 = h f @ 50°C = 209.34 kJ/kg

4

1

Evaporator

sat. vap.

20°C x=0.23

40°C Water 50°C

hw 2 = h f @ 40°C = 167.53 kJ/kg

The saturation temperature at the condenser pressure of 1400 kPa and the actual temperature at the condenser outlet are Tsat @ 1400 kPa = 52.40°C

T

P3 = 1400 kPa ⎫ ⎬T3 = 48.59°C (from EES) h3 = 121.24 kJ ⎭

Then, the degrees of subcooling is ΔTsubcool = Tsat − T3 = 52.40 − 48.59 = 3.81°C

· QH

2

1.4 MPa

· Win

3

4s

4

· QL

(b) The rate of heat absorbed from the geothermal water in the evaporator is Q& = m& (h − h ) = (0.065 kg/s)(209.34 − 167.53)kJ/kg = 2.718 kW L

w

w1

1 s

w2

This heat is absorbed by the refrigerant in the evaporator Q& L 2.718 kW = = 0.01936 kg/s m& R = h1 − h4 (261.59 − 121.24)kJ/kg (c) The power input to the compressor, the heating load and the COP are W& = m& (h − h ) + Q& = (0.01936 kg/s)(280.00 − 261.59)kJ/kg = 0.6564 kW in

R

2

1

out

Q& H = m& R (h2 − h3 ) = (0.01936 kg/s)(280.00 − 121.24)kJ/kg = 3.074 kW COP =

Q& H 3.074 kW = = 4.68 W& in 0.6564 kW

(d) The reversible COP of the cycle is 1 1 COPrev = = = 12.92 1 − T L / T H 1 − (25 + 273) /(50 + 273) The corresponding minimum power input is Q& H 3.074 kW W& in, min = = = 0.238 kW COPrev 12.92

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11-29

Innovative Refrigeration Systems

11-40C Performing the refrigeration in stages is called cascade refrigeration. In cascade refrigeration, two or more refrigeration cycles operate in series. Cascade refrigerators are more complex and expensive, but they have higher COP's, they can incorporate two or more different refrigerants, and they can achieve much lower temperatures.

11-41C Cascade refrigeration systems have higher COPs than the ordinary refrigeration systems operating between the same pressure limits.

11-42C The saturation pressure of refrigerant-134a at -32°C is 77 kPa, which is below the atmospheric pressure. In reality a pressure below this value should be used. Therefore, a cascade refrigeration system with a different refrigerant at the bottoming cycle is recommended in this case.

11-43C We would favor the two-stage compression refrigeration system with a flash chamber since it is simpler, cheaper, and has better heat transfer characteristics.

11-44C Yes, by expanding the refrigerant in stages in several throttling devices.

11-45C To take advantage of the cooling effect by throttling from high pressures to low pressures.

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11-30

11-46 [Also solved by EES on enclosed CD] A two-stage compression refrigeration system with refrigerant-134a as the working fluid is considered. The fraction of the refrigerant that evaporates as it is throttled to the flash chamber, the rate of heat removed from the refrigerated space, and the COP are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 The flash chamber is adiabatic. Analysis (a) The enthalpies of the refrigerant at several states are determined from the refrigerant tables (Tables A-11, A-12, and A-13) to be h1 = 239.16 kJ/kg,

h2 = 265.31 kJ/kg

T

1 MPa

h3 = 259.30 kJ/kg, h5 = 107.32 kJ/kg,

h6 = 107.32 kJ/kg

h7 = 73.33 kJ/kg,

h8 = 73.33 kJ/kg

The fraction of the refrigerant that evaporates as it is throttled to the flash chamber is simply the quality at state 6, x6 =

h6 − h f h fg

4 0.5 MPa

5

2

A 7

107.32 − 73.33 = = 0.1828 185.98

6

B

8

0.14 MPa

3 9 · QL

1 s

(b) The enthalpy at state 9 is determined from an energy balance on the mixing chamber: E& in − E& out = ΔE& system ©0 (steady) = 0 E& in = E& out

∑ m& h = ∑ m& h e e

(1)h9

i i

= x6 h3 + (1 − x6 )h2 h9 = (0.1828)(259.30 ) + (1 − 0.1828)(265.31) = 264.21 kJ/kg

also, P4 = 1 MPa

⎫ ⎬ h4 = 278.97 kJ/kg s 4 = s 9 = 0.94083 kJ/kg ⋅ K ⎭

Then the rate of heat removed from the refrigerated space and the compressor work input per unit mass of refrigerant flowing through the condenser are m& B = (1 − x 6 )m& A = (1 − 0.1828)(0.25 kg/s ) = 0.2043 kg/s Q& L = m& B (h1 − h8 ) = (0.2043 kg/s )(239.16 − 73.33) kJ/kg = 33.88 kW W& in = W& compI,in + W& compII,in = m& A (h4 − h9 ) + m& B (h2 − h1 ) = (0.25 kg/s )(278.97 − 264.21) kJ/kg + (0.2043 kg/s )(265.31 − 239.16) kJ/kg = 9.03 kW

(c) The coefficient of performance is determined from COPR =

Q& L 33.88 kW = = 3.75 & 9.03 kW Wnet,in

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

11-31

11-47 EES Problem 11-46 is reconsidered. The effects of the various refrigerants in EES data bank for compressor efficiencies of 80, 90, and 100 percent is to be investigated. Analysis The problem is solved using EES, and the results are tabulated and plotted below. "Input Data" "P[1]=140 [kPa] P[4] = 1000 [kPa] P[6]=500 [kPa] Eta_compB =1.0 Eta_compA =1.0" m_dot_A=0.25 [kg/s] "High Pressure Compressor A" P[9]=P[6] h4s=enthalpy(R134a,P=P[4],s=s[9]) "State 4s is the isentropic value of state 4" h[9]+w_compAs=h4s "energy balance on isentropic compressor" w_compA=w_compAs/Eta_compA"definition of compressor isentropic efficiency" h[9]+w_compA=h[4] "energy balance on real compressor-assumed adiabatic" s[4]=entropy(R134a,h=h[4],P=P[4]) "properties for state 4" T[4]=temperature(R134a,h=h[4],P=P[4]) W_dot_compA=m_dot_A*w_compA "Condenser" P[5]=P[4] "neglect pressure drops across condenser" T[5]=temperature(R134a,P=P[5],x=0) "properties for state 5, assumes sat. liq. at cond. exit" h[5]=enthalpy(R134a,T=T[5],x=0) "properties for state 5" s[5]=entropy(R134a,T=T[5],x=0) h[4]=q_out+h[5] "energy balance on condenser" Q_dot_out = m_dot_A*q_out "Throttle Valve A" h[6]=h[5] "energy balance on throttle - isenthalpic" x6=quality(R134a,h=h[6],P=P[6]) "properties for state 6" s[6]=entropy(R134a,h=h[6],P=P[6]) T[6]=temperature(R134a,h=h[6],P=P[6]) "Flash Chamber" m_dot_B = (1-x6) * m_dot_A P[7] = P[6] h[7]=enthalpy(R134a, P=P[7], x=0) s[7]=entropy(R134a,h=h[7],P=P[7]) T[7]=temperature(R134a,h=h[7],P=P[7]) "Mixing Chamber" x6*m_dot_A*h[3] + m_dot_B*h[2] =(x6* m_dot_A + m_dot_B)*h[9] P[3] = P[6] h[3]=enthalpy(R134a, P=P[3], x=1) "properties for state 3" s[3]=entropy(R134a,P=P[3],x=1) T[3]=temperature(R134a,P=P[3],x=x1) s[9]=entropy(R134a,h=h[9],P=P[9]) "properties for state 9" T[9]=temperature(R134a,h=h[9],P=P[9]) "Low Pressure Compressor B" x1=1 "assume flow to compressor inlet to be saturated vapor"

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11-32

h[1]=enthalpy(R134a,P=P[1],x=x1) "properties for state 1" T[1]=temperature(R134a,P=P[1], x=x1) s[1]=entropy(R134a,P=P[1],x=x1) P[2]=P[6] h2s=enthalpy(R134a,P=P[2],s=s[1]) " state 2s is isentropic state at comp. exit" h[1]+w_compBs=h2s "energy balance on isentropic compressor" w_compB=w_compBs/Eta_compB"definition of compressor isentropic efficiency" h[1]+w_compB=h[2] "energy balance on real compressor-assumed adiabatic" s[2]=entropy(R134a,h=h[2],P=P[2]) "properties for state 2" T[2]=temperature(R134a,h=h[2],P=P[2]) W_dot_compB=m_dot_B*w_compB "Throttle Valve B" h[8]=h[7] "energy balance on throttle - isenthalpic" x8=quality(R134a,h=h[8],P=P[8]) "properties for state 8" s[8]=entropy(R134a,h=h[8],P=P[8]) T[8]=temperature(R134a,h=h[8],P=P[8]) "Evaporator" P[8]=P[1] "neglect pressure drop across evaporator" q_in + h[8]=h[1] "energy balance on evaporator" Q_dot_in=m_dot_B*q_in "Cycle Statistics" W_dot_in_total = W_dot_compA + W_dot_compB COP=Q_dot_in/W_dot_in_total "definition of COP" ηcompA 0,8 0,8333 0,8667 0,9 0,9333 0,9667 1

ηcompB 0,8 0,8333 0,8667 0,9 0,9333 0,9667 1

Qout 45.32 44.83 44.39 43.97 43.59 43.24 42.91

COP 2.963 3.094 3.225 3.357 3.488 3.619 3.751

R134a

125 100

T [C]

75 50

4

5 1000 kPa

25

7

-50 0,0

3

6

0 -25

500 kPa

140 kPa

8 0,2

0,4

0,6

2 9

1

0,8

1,0

1,2

s [kJ/kg-K]

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11-33

45,5

3,8 3,7

45 3,6 3,5 3,4 44 3,3 43,5

COP

Qout [kW]

44,5

3,2 3,1

43 3 42,5 0,8

0,84

0,88

0,92

2,9 1

0,96

ηcomp

COP vs Flash Cham ber Pressure, P 6 3.80 3.75 3.70

COP

3.65 3.60 3.55 3.50 3.45 200

300

400

500

600

700

800

900

1000

1100

P[6] [kPa]

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

11-34

11-48 [Also solved by EES on enclosed CD] A two-stage compression refrigeration system with refrigerant-134a as the working fluid is considered. The fraction of the refrigerant that evaporates as it is throttled to the flash chamber, the rate of heat removed from the refrigerated space, and the COP are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 The flash chamber is adiabatic. Analysis (a) The enthalpies of the refrigerant at several states are determined from the refrigerant tables (Tables A-11, A-12, and A-13) to be h1 = 239.16 kJ/kg,

h2 = 255.90 kJ/kg

h3 = 251.88 kJ/kg, h5 = 107.32 kJ/kg,

h6 = 107.32 kJ/kg

h7 = 55.16 kJ/kg,

h8 = 55.16 kJ/kg

T

The fraction of the refrigerant that evaporates as it is throttled to the flash chamber is simply the quality at state 6, x6 =

h6 − h f h fg

=

1 MPa 4 0.32 MPa

5

2

A 7

107.32 − 55.16 = 0.2651 196.71

6

B

8

(b) The enthalpy at state 9 is determined from an energy balance on the mixing chamber: E& in − E& out = ΔE& system

©0 (steady)

0.14 MPa

3 9 · QL

1 s

=0

E& in = E& out

∑ m& h = ∑ m& h e e

(1)h9

i i

= x 6 h3 + (1 − x 6 )h2 h9 = (0.2651)(251.88) + (1 − 0.2651)(255.90 ) = 254.84 kJ/kg

and P9 = 0.32 MPa

⎫ ⎬ s 9 = 0.94074 kJ/kg ⋅ K h9 = 254.84 kJ/kg ⎭

also,

P4 = 1 MPa

⎫ ⎬ h4 = 278.94 kJ/kg s 4 = s 9 = 0.94074 kJ/kg ⋅ K ⎭

Then the rate of heat removed from the refrigerated space and the compressor work input per unit mass of refrigerant flowing through the condenser are m& B = (1 − x6 )m& A = (1 − 0.2651)(0.25 kg/s ) = 0.1837 kg/s Q& L = m& B (h1 − h8 ) = (0.1837 kg/s )(239.16 − 55.16 ) kJ/kg = 33.80 kW W& in = W& compI,in + W& compII,in = m& A (h4 − h9 ) + m& B (h2 − h1 ) = (0.25 kg/s )(278.94 − 254.84) kJ/kg + (0.1837 kg/s )(255.90 − 239.16) kJ/kg = 9.10 kW

(c) The coefficient of performance is determined from Q& L 33.80 kW COPR = = = 3.71 & 9.10 kW Wnet,in

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

11-35

11-49 A two-stage cascade refrigeration cycle is considered. The mass flow rate of the refrigerant through the upper cycle, the rate of heat removal from the refrigerated space, and the COP of the refrigerator are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) The properties are to be obtained from the refrigerant tables (Tables A-11 through A-13): h1 = h g @ 200 kPa = 244.46 kJ/kg

. QH

s1 = s g @ 200 kPa = 0.9377 kJ/kg.K P2 = 500 kPa ⎫ ⎬ h2 s = 263.30 kJ/kg s 2 = s1 ⎭

h −h η C = 2s 1 h2 − h1

Condenser 7 Expansion valve

263.30 − 244.46 0.80 = ⎯ ⎯→ h2 = 268.01 kJ/kg h2 − 244.46

6 Compressor

8

5

h3 = h f @ 500 kPa = 73.33 kJ/kg

Evaporator

h4 = h3 = 73.33 kJ/kg

Condenser

h5 = h g @ 400 kPa = 255.55 kJ/kg s 5 = s g @ 400 kPa = 0.9269 kJ/kg.K

3 Expansion valve

P6 = 1200 kPa ⎫ ⎬ h6 s = 278.33 kJ/kg s6 = s5 ⎭

ηC = 0.80 =

. Win

2

. Win

Compressor

4

1 Evaporator

h 6 s − h5 h6 − h5 278.33 − 255.55 ⎯ ⎯→ h6 = 284.02 kJ/kg h6 − 255.55

. QL

h7 = h f @ 1200 kPa = 117.77 kJ/kg h8 = h7 = 117.77 kJ/kg

The mass flow rate of the refrigerant through the upper cycle is determined from an energy balance on the heat exchanger m& A (h5 − h8 ) = m& B (h2 − h3 ) m& A (255.55 − 117.77)kJ/kg = (0.15 kg/s)(268.01 − 73.33)kJ/kg ⎯ ⎯→ m& A = 0.212 kg/s

(b) The rate of heat removal from the refrigerated space is Q& L = m& B (h1 − h4 ) = (0.15 kg/s)(244.46 − 73.33)kJ/kg = 25.67 kW

(c) The power input and the COP are W& in = m& A (h6 − h5 ) + m& B (h2 − h1 ) = (0.15 kg/s)(284.02 − 255.55)kJ/kg + (0.212 kg/s)(268.01 − 244.46)kJ/kg = 9.566 kW COP =

Q& L 25.67 = = 2.68 W& in 9.566

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

11-36

11-50 A two-evaporator compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The cooling rate of the high-temperature evaporator, the power required by the compressor, and the COP of the system are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

Condenser

2

T 3

· QH

Compressor 1

Expansion valve

Expansion valve Evaporator 1 4

5 7

2

3 800 kPa 0°C

4

· Win 5

-26.4°C Expansion valve

6

Evaporator 2

· 7 1 QL s

6 Analysis From the refrigerant tables (Tables A-11, A-12, and A-13), P3 = 800 kPa ⎫ ⎬ h3 = h f sat. liquid ⎭

@ 800 kPa

= 95.47 kJ/kg

h4 = h6 ≅ h3 = 95.47 kJ/kg ( throttling) T5 = 0°C ⎫ ⎬ h = h g @ 0°C = 250.45 kJ/kg sat. vapor ⎭ 5 T7 = −26.4°C ⎫ ⎬ h7 = h g @ − 26.4°C = 234.44 kJ/kg sat. vapor ⎭

The mass flow rate through the low-temperature evaporator is found by Q& L = m& 2 (h7 − h6 ) ⎯ ⎯→ m& 2 =

Q& L 8 kJ/s = = 0.05757 kg/s h7 − h6 (234.44 − 95.47) kJ/kg

The mass flow rate through the warmer evaporator is then m& 1 = m& − m& 2 = 0.1 − 0.05757 = 0.04243 kg/s

Applying an energy balance to the point in the system where the two evaporator streams are recombined gives m& 1 h5 + m& 2 h7 = m& h1 ⎯ ⎯→ h1 =

m& 1 h5 + m& 2 h7 (0.04243)(250.45) + (0.05757)(234.44) = = 241.23 kJ/kg m& 0.1

Then,

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11-37

P1 = Psat @ − 26.4°C ≅ 100 kPa ⎫ ⎬ s1 = 0.9789 kJ/kg ⋅ K h1 = 241.23 kJ/kg ⎭ P2 = 800 kPa ⎫ ⎬ h2 = 286.26 kJ/kg s 2 = s1 ⎭

The cooling rate of the high-temperature evaporator is Q& L = m& 1 (h5 − h4 ) = (0.04243 kg/s )(250.45 − 95.47) kJ/kg = 6.58 kW

The power input to the compressor is W& in = m& (h2 − h1 ) = (0.1 kg/s)(286.26 − 241.23) kJ/kg = 4.50 kW

The COP of this refrigeration system is determined from its definition, COPR =

Q& L (8 + 6.58) kW = = 3.24 4.50 kW W& in

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

11-38

11-51E A two-evaporator compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The power required by the compressor and the COP of the system are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

Condenser

2

T 3

· QH

Compressor 1

Expansion valve

Expansion valve

3 140 psia

Evaporator 1

7

30 psia 4

4

5

2

Expansion valve

· Win 5

15 psia 6

· 7 1 QL

Evaporator 2 6

s

Analysis From the refrigerant tables (Tables A-11E, A-12E, and A-13E), P3 = 140 psia ⎫ ⎬ h3 = h f @ 140 psia = 45.304 Btu/lbm sat. liquid ⎭ h4 = h6 ≅ h3 = 45.304 Btu/lbm ( throttling) P5 = 30 psia ⎫ ⎬ h = h g @ 30 psia = 105.32 Btu/lbm sat. vapor ⎭ 5 P7 = 15 psia ⎫ ⎬ h = h g @ 15 psia = 100.99 Btu/lbm sat. vapor ⎭ 7 The mass flow rates through the high-temperature and low-temperature evaporators are found by Q& L,1 3000 Btu/h Q& L,1 = m& 1 (h5 − h4 ) ⎯ ⎯→ m& 1 = = = 49.99 lbm/h h5 − h4 (105.32 − 45.304) Btu/lbm Q& L, 2 10,000 Btu/h Q& L, 2 = m& 2 (h7 − h6 ) ⎯ ⎯→ m& 2 = = = 179.58 lbm/h h7 − h6 (100.99 − 45.304) Btu/lbm Applying an energy balance to the point in the system where the two evaporator streams are recombined gives m& h + m& 2h7 (49.99)(105.32) + (179.58)(100.99) m& 1h5 + m& 2h7 = (m& 1 + m& 2 )h1 ⎯ ⎯→ h1 = 1 5 = = 101.93 Btu/lbm m& 1 + m& 2 49.99 + 179.58 Then, P1 = 15 psia ⎫ ⎬ s = 0.2293 Btu/lbm ⋅ R h1 = 101.93 Btu/lbm ⎭ 1 P2 = 140 psia ⎫ ⎬ h2 = 122.24 Btu/lbm s 2 = s1 ⎭

The power input to the compressor is 1 kW ⎛ ⎞ W&in = (m& 1 + m& 2 )(h2 − h1) = (49.99 + 179.58) lbm/h(122.24 − 101.93) Btu/lbm⎜ ⎟ = 1.366 kW ⎝ 3412.14 Btu/h ⎠ The COP of this refrigeration system is determined from its definition, Q& (10,000 + 3000) Btu/h ⎛ 1 kW ⎞ COPR = L = ⎜ ⎟ = 2.79 & 1.366 kW Win ⎝ 3412.14 Btu/h ⎠

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

11-39

11-52E A two-evaporator compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The power required by the compressor and the COP of the system are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

Condenser

2

T 3

· QH

Compressor 1

Expansion valve

Expansion valve

3 140 psia 60 psia

Evaporator II

7

4

4

5

2

Expansion valve Evaporator I 6

· Win 5

15 psia 6

· 7 1 QL s

Analysis From the refrigerant tables (Tables A-11E, A-12E, and A-13E), P3 = 140 psia ⎫ ⎬ h3 = h f @ 140 psia = 45.304 Btu/lbm sat. liquid ⎭ h4 = h6 ≅ h3 = 45.304 Btu/lbm ( throttling ) P5 = 60 psia ⎫ ⎬ h = h g @ 60 psia = 110.11 Btu/lbm sat. vapor ⎭ 5 P7 = 15 psia ⎫ ⎬ h = h g @ 15 psia = 100.99 Btu/lbm sat. vapor ⎭ 7 The mass flow rates through the high-temperature and low-temperature evaporators are found by Q& L,1 48,000 Btu/h Q& L,1 = m& 1 (h5 − h4 ) ⎯ ⎯→ m& 1 = = = 740.7 lbm/h h5 − h4 (110.11 − 45.304) Btu/lbm Q& L, 2 10,000 Btu/h Q& L, 2 = m& 2 (h7 − h6 ) ⎯ ⎯→ m& 2 = = = 179.6 lbm/h h7 − h6 (100.99 − 45.304) Btu/lbm Applying an energy balance to the point in the system where the two evaporator streams are recombined gives m& h + m& 2h7 (740.7)(110.11) + (179.58)(100.99) m& 1h5 + m& 2h7 = (m& 1 + m& 2 )h1 ⎯ ⎯→ h1 = 1 5 = = 108.33 Btu/lbm m& 1 + m& 2 740.7 + 179.58 Then, P1 = 15 psia ⎫ ⎬ s = 0.2430 Btu/lbm ⋅ R h1 = 108.33 Btu/lbm ⎭ 1 P2 = 140 psia ⎫ ⎬ h2 = 130.45 Btu/lbm s 2 = s1 ⎭

The power input to the compressor is 1 kW ⎛ ⎞ W&in = (m& 1 + m& 2 )(h2 − h1) = (740.7 + 179.6) lbm/h(130.45 − 108.33) Btu/lbm⎜ ⎟ = 5.966 kW ⎝ 3412.14 Btu/h ⎠ The COP of this refrigeration system is determined from its definition, Q& (48,000 + 10,000) Btu/h ⎛ 1 kW ⎞ COPR = L = ⎜ ⎟ = 2.85 & 5.966 kW Win ⎝ 3412.14 Btu/h ⎠

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11-40

11-53 A two-stage compression refrigeration system with a separation unit is considered. The mass flow rate through the two compressors, the power used by the compressors, and the system’s COP are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

Condenser

2

T1 = −10.1°C ⎫ h1 = h g @ −10.1°C = 244.46 kJ/kg ⎬ s =s sat. vapor g @ −10.1°C = 0.93773 kJ/kg ⋅ K ⎭ 1 P2 = 800 kPa ⎫ ⎬ h2 = 273.24 kJ/kg s 2 = s1 ⎭ P3 = 800 kPa ⎫ ⎬ h3 = h f sat. liquid ⎭

@800 kPa

= 95.47 kJ/kg

3

Compressor

Analysis From the refrigerant tables (Tables A-11, A-12, and A-13),

1 8

4 5 Expansion valve

Compressor 7

Expansion valve

Separator

Evaporator

6

h4 ≅ h3 = 95.47 kJ/kg ( throttling) T5 = −10.1°C ⎫ ⎬ h5 = h f sat. liquid ⎭

@ −10.1° C

= 38.43 kJ/kg

T 2

h6 ≅ h5 = 38.43 kJ/kg ( throttling) T7 = −40°C ⎫ h7 = h g @ − 40°C = 225.86 kJ/kg ⎬ sat. vapor ⎭ s 7 = s g @ − 40°C = 0.96866 kJ/kg ⋅ K P8 = Psat @ −10.1°C = 200 kPa ⎫ ⎬ h8 = 252.74 kJ/kg s8 = s 7 ⎭

3

0.8 MPa 8 -10.1°C

5

1

4 -40°C 6

The mass flow rate through the evaporator is determined from

· QL

7 s

Q& L 30 kJ/s Q& L = m& 6 (h7 − h6 ) ⎯ ⎯→ m& 6 = = = 0.1601 kg/s h7 − h6 (225.86 − 38.43) kJ/kg

An energy balance on the separator gives m& 6 (h8 − h5 ) = m& 2 (h1 − h4 ) ⎯ ⎯→ m& 2 = m& 6

h8 − h5 252.74 − 38.43 = (0.1601) = 0.2303 kg/s h1 − h4 244.46 − 95.47

The total power input to the compressors is W& in = m& 6 (h8 − h7 ) + m& 2 (h2 − h1 ) = (0.1601 kg/s)(252.74 − 225.86) kJ/kg + (0.2303 kg/s)(273.24 − 244.46) kJ/kg = 10.93 kW

The COP of this refrigeration system is determined from its definition, COPR =

Q& L 30 kW = = 2.74 W& in 10.93 kW

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

11-41

11-54E A two-stage compression refrigeration system with a separation unit is considered. The cooling load and the system’s COP are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

Condenser

2

3

Compressor Analysis From the refrigerant tables (Tables A-11E, A-12E, and A-13E),

1

P1 = 120 psia ⎫ h1 = h g @ 120 psia = 115.16 Btu/lbm ⎬ s =s sat. vapor g @ 120 psia = 0.21924 Btu/lbm ⋅ R ⎭ 1

8

P2 = 300 psia ⎫ ⎬ h2 = 123.06 Btu/lbm s 2 = s1 ⎭ P3 = 300 psia ⎫ ⎬ h3 = h f sat. liquid ⎭

@ 300 psia

Separator

4 5 Expansion valve

Compressor

= 66.339 Btu/lbm

7

Expansion valve

Evaporator

6

h 4 ≅ h3 = 66.339 Btu/lbm ( throttling) P5 = 120 psia ⎫ ⎬ h5 = h f sat. liquid ⎭

T @ 120 psia

= 41.787 Btu/lbm

2 3

h6 ≅ h5 = 41.787 kJ/kg ( throttling) P7 = 60 psia ⎫ h7 = h g @ 60 psia = 110.11 Btu/lbm ⎬ sat. vapor ⎭ s 7 = s g @ 60 psia = 0.22127 Btu/lbm ⋅ R P8 = 120 psia ⎫ ⎬ h8 = 116.28 Btu/lbm s8 = s 7 ⎭

5

300 psia

8 120 psia 1

4

60 psia 6

·

QL

7 s

An energy balance on the separator gives m& 6 (h8 − h5 ) = m& 2 (h1 − h4 ) ⎯ ⎯→ m& 2 = m& 6

h8 − h5 116.28 − 41.787 = m& 6 = 1.5258m& 6 115.16 − 66.339 h1 − h4

The total power input to the compressors is given by W& in = m& 6 (h8 − h7 ) + m& 2 (h2 − h1 ) = m& 6 (h8 − h7 ) + 1.5258m& 6 (h2 − h1 )

Solving for m& 6 , m& 6 =

W& in (25 × 3412.14) Btu/h = = 4681 lbm/h (h8 − h7 ) + 1.5258(h2 − h1 ) (116.28 − 110.11) + 1.5258(123.06 − 115.16) Btu/lbm

The cooling effect produced by this system is then Q& L = m& 6 ( h7 − h6 ) = ( 4681 lbm/h)(110.11 − 41.787) Btu/lbm = 319,800 Btu/h

The COP of this refrigeration system is determined from its definition, COPR =

Q& L (319,800/3412.14) kW = = 3.75 25 kW W& in

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11-42

11-55 A two-stage cascade refrigeration system is considered. Each stage operates on the ideal vaporcompression cycle with upper cycle using water and lower cycle using refrigerant-134a as the working fluids. The mass flow rate of R-134a and water in their respective cycles and the overall COP of this system are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 The heat exchanger is adiabatic. Analysis From the water and refrigerant tables (Tables A-4, A-5, A-6, A-11, A-12, and A-13), T1 = 5°C ⎫ h1 = h g @ 5°C = 2510.1 kJ/kg ⎬ sat. vapor ⎭ s1 = s g @ 5°C = 9.0249 kJ/kg ⋅ K P2 = 1.6 MPa ⎫ ⎬ h2 = 5083.4 kJ/kg s 2 = s1 ⎭ P3 = 1.6 MPa ⎫ ⎬ h3 = h f sat. liquid ⎭

@ 1.6 MPa

T 2 3

= 858.44 kJ/kg

h4 ≅ h3 = 858.44 kJ/kg ( throttling)

7

1.6 MPa 6 5°C 4 400 kPa 1

T5 = −40°C ⎫ h5 = h g @ − 40°C = 225.86 kJ/kg ⎬ sat. vapor ⎭ s 5 = s g @ − 40°C = 0.96866 kJ/kg ⋅ K

8 -40°C · QL

P6 = 400 kPa ⎫ ⎬ h6 = 267.59 kJ/kg s6 = s5 ⎭ P7 = 400 kPa ⎫ ⎬ h7 = h f sat. liquid ⎭

@ 400 kPa

5 s

= 63.94 kJ/kg

h8 ≅ h7 = 63.94 kJ/kg ( throttling)

The mass flow rate of R-134a is determined from ⎯→ m& R = Q& L = m& R (h5 − h8 ) ⎯

Q& L 20 kJ/s = = 0.1235 kg/s h5 − h8 (225.86 − 63.94) kJ/kg

An energy balance on the heat exchanger gives the mass flow rate of water m& R (h6 − h7 ) = m& w (h1 − h4 ) ⎯ ⎯→ m& w = m& R

h6 − h7 267.59 − 63.94 = (0.1235 kg/s) = 0.01523 kg/s 2510.1 − 858.44 h1 − h4

The total power input to the compressors is W& in = m& R (h6 − h5 ) + m& w (h2 − h1 ) = (0.1235 kg/s)(267.59 − 225.86) kJ/kg + (0.01523 kg/s)(5083.4 − 2510.1) kJ/kg = 44.35 kJ/s

The COP of this refrigeration system is determined from its definition, COPR =

Q& L 20 kJ/s = = 0.451 & 44.35 kJ/s Win

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

11-43

11-56 A two-stage vapor-compression refrigeration system with refrigerant-134a as the working fluid is considered. The process with the greatest exergy destruction is to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From Prob. 11-55 and the water and refrigerant tables (Tables A-4, A-5, A-6, A-11, A-12, and A-13), s1 = s 2 = 9.0249 kJ/kg ⋅ K s 3 = 2.3435 kJ/kg ⋅ K s 4 = 3.0869 kJ/kg ⋅ K s 5 = s 6 = 0.96866 kJ/kg ⋅ K s 7 = 0.24757 kJ/kg ⋅ K s 8 = 0.27423 kJ/kg ⋅ K m& R = 0.1235 kg/s m& w = 0.01751 kg/s q L = h5 − h8 = 161.92 kJ/kg q H = h2 − h3 = 4225.0 kJ/kg T L = −30°C = 243 K T H = 30°C = 303 K T0 = 30°C = 303 K

T 2 3 7

1.6 MPa 6 5°C 4 400 kPa 1 8 -40°C · QL

5 s

The exergy destruction during a process of a stream from an inlet state to exit state is given by ⎛ q q x dest = T0 s gen = T0 ⎜⎜ s e − s i − in + out Tsource Tsink ⎝

⎞ ⎟ ⎟ ⎠

Application of this equation for each process of the cycle gives ⎛ q ⎞ X& destroyed, 23 = m& w T0 ⎜⎜ s 3 − s 2 + H ⎟⎟ TH ⎠ ⎝ 4225.0 ⎞ ⎛ = (0.01751)(303 K )⎜ 2.3435 − 9.0249 + ⎟ = 38.53 kJ/s 303 ⎠ ⎝ X& destroyed, 34 = m& w T0 ( s 4 − s 3 ) = (0.01751)(303)(3.0869 − 2.3435) = 3.94 kJ/s X& destroyed, 78 = m& R T0 ( s8 − s 7 ) = (0.1235)(303)(0.27423 − 0.24757) = 1.00 kJ/s ⎛ q ⎞ X& destroyed, 85 = m& R T0 ⎜⎜ s 5 − s8 − L ⎟⎟ TL ⎠ ⎝ 161.92 ⎞ ⎛ = (0.1235)(303)⎜ 0.96866 − 0.27423 − ⎟ = 1.05 kJ/s 243 ⎠ ⎝ X& destroyed, heat exch = T0 [m& w ( s1 − s 4 ) + m& R ( s 7 − s 6 )]

= (303)[(0.01751)(9.0249 − 3.0869) + (0.1235)(0.24757 − 0.96866)] = 4.52 kJ/s

For isentropic processes, the exergy destruction is zero: X& destroyed,12 = 0 X& destroyed, 56 = 0

Note that heat is absorbed from a reservoir at -30°C (243 K) and rejected to a reservoir at 30°C (303 K), which is also taken as the dead state temperature. Alternatively, one may use the standard 25°C (298 K) as the dead state temperature, and perform the calculations accordingly. The greatest exergy destruction occurs in the condenser.

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

11-44

Gas Refrigeration Cycles

11-57C The ideal gas refrigeration cycle is identical to the Brayton cycle, except it operates in the reversed direction.

11-58C The reversed Stirling cycle is identical to the Stirling cycle, except it operates in the reversed direction. Remembering that the Stirling cycle is a totally reversible cycle, the reversed Stirling cycle is also totally reversible, and thus its COP is COPR, Stirling =

1 TH / TL − 1

11-59C In the ideal gas refrigeration cycle, the heat absorption and the heat rejection processes occur at constant pressure instead of at constant temperature.

11-60C In aircraft cooling, the atmospheric air is compressed by a compressor, cooled by the surrounding air, and expanded in a turbine. The cool air leaving the turbine is then directly routed to the cabin.

11-61C No; because h = h(T) for ideal gases, and the temperature of air will not drop during a throttling (h1 = h2) process.

11-62C By regeneration.

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11-45

11-63 [Also solved by EES on enclosed CD] An ideal-gas refrigeration cycle with air as the working fluid is considered. The rate of refrigeration, the net power input, and the COP are to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible. Analysis (a) We assume both the turbine and the compressor to be isentropic, the turbine inlet temperature to be the temperature of the surroundings, and the compressor inlet temperature to be the temperature of the refrigerated space. From the air table (Table A-17), T1 = 285 K T1 = 320 K

⎯ ⎯→ ⎯ ⎯→

h1 = 28514 . kJ / kg Pr 1 = 11584 . h3 = 320.29 kJ / kg Pr 3 = 1.7375

Thus,

T

2 · QH 47°C 12°C

Pr2 =

P2 ⎛ 250 ⎞ Pr = ⎜ ⎯→ T2 = 450.4 K ⎟(1.1584 ) = 5.792 ⎯ P1 1 ⎝ 50 ⎠ h2 = 452.17 kJ/kg

Pr4 =

P4 ⎛ 50 ⎞ Pr = ⎜ ⎯→ T4 = 201.8 K ⎟(1.7375) = 0.3475 ⎯ P3 3 ⎝ 250 ⎠ h4 = 201.76 kJ/kg

3 1 · Q Refrig 4

Then the rate of refrigeration is Q& refrig = m& (qL ) = m& (h1 − h4 ) = (0.08 kg/s)(285.14 − 201.76) kJ/kg = 6.67 kW

(b) The net power input is determined from

W& net, in = W& comp, in − W& turb, out where W&comp,in = m& (h2 − h1 ) = (0.08 kg/s)(452.17 − 285.14 ) kJ/kg = 13.36 kW W& turb,out = m& (h3 − h4 ) = (0.08 kg/s)(320.29 − 201.76 ) kJ/kg = 9.48 kW

Thus,

W& net, in = 13.36 − 9.48 = 3.88 kW (c) The COP of this ideal gas refrigeration cycle is determined from Q& 6.67 kW COPR = & L = = 1.72 Wnet, in 3.88 kW

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s

11-46

11-64 EES Problem 11-63 is reconsidered. The effects of compressor and turbine isentropic efficiencies on the rate of refrigeration, the net power input, and the COP are to be investigated. Analysis The problem is solved using EES, and the solution is given below. "Input data" T[1] = 12 [C] P[1]= 50 [kPa] T[3] = 47 [C] P[3]=250 [kPa] m_dot=0.08 [kg/s] Eta_comp = 1.00 Eta_turb = 1.0 "Compressor anaysis" s[1]=ENTROPY(Air,T=T[1],P=P[1]) s2s=s[1] "For the ideal case the entropies are constant across the compressor" P[2] = P[3] s2s=ENTROPY(Air,T=Ts2,P=P[2])"Ts2 is the isentropic value of T[2] at compressor exit" Eta_comp = W_dot_comp_isen/W_dot_comp "compressor adiabatic efficiency, W_dot_comp > W_dot_comp_isen" m_dot*h[1] + W_dot_comp_isen = m_dot*hs2"SSSF First Law for the isentropic compressor, assuming: adiabatic, ke=pe=0, m_dot is the mass flow rate in kg/s" h[1]=ENTHALPY(Air,T=T[1]) hs2=ENTHALPY(Air,T=Ts2) m_dot*h[1] + W_dot_comp = m_dot*h[2]"SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" h[2]=ENTHALPY(Air,T=T[2]) s[2]=ENTROPY(Air,h=h[2],P=P[2]) "Heat Rejection Process 2-3, assumed SSSF constant pressure process" m_dot*h[2] + Q_dot_out = m_dot*h[3]"SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0" h[3]=ENTHALPY(Air,T=T[3]) "Turbine analysis" s[3]=ENTROPY(Air,T=T[3],P=P[3]) s4s=s[3] "For the ideal case the entropies are constant across the turbine" P[4] = P[1] s4s=ENTROPY(Air,T=Ts4,P=P[4])"Ts4 is the isentropic value of T[4] at turbine exit" Eta_turb = W_dot_turb /W_dot_turb_isen "turbine adiabatic efficiency, W_dot_turb_isen > W_dot_turb" m_dot*h[3] = W_dot_turb_isen + m_dot*hs4"SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0" hs4=ENTHALPY(Air,T=Ts4) m_dot*h[3] = W_dot_turb + m_dot*h[4]"SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" h[4]=ENTHALPY(Air,T=T[4]) s[4]=ENTROPY(Air,h=h[4],P=P[4]) "Refrigeration effect:" m_dot*h[4] + Q_dot_Refrig = m_dot*h[1] "Cycle analysis" W_dot_in_net=W_dot_comp-W_dot_turb"External work supplied to compressor" COP= Q_dot_Refrig/W_dot_in_net "The following is for plotting data only:" Ts[1]=Ts2 ss[1]=s2s Ts[2]=Ts4 ss[2]=s4s

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11-47

COP

ηcomp

ηturb

0.6937 0.9229 1.242 1.717

0.7 0.8 0.9 1

1 1 1 1

QRefrig [kW] 6.667 6.667 6.667 6.667

Winnet [kW] 9.612 7.224 5.368 3.882

7 6

QRefrig [kW]

5 4 3

η turb

2

0.7

1

0.85 1.0

0 0,7

0,75

0,8

0,85

0,9

0,95

1

ηcomp

12

Win;net [kW]

10 8 6 4 2 0 0,7

η turb 0.7 0.85 1.0 0,75

0,8

0,85

0,9

0,95

1

0,9

0,95

1

ηcomp

2

η turb

COP

1,5

1

0.7 0.85 1.0

0,5

0 0,7

0,75

0,8

0,85

ηcomp

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11-48

11-65 [Also solved by EES on enclosed CD] An ideal-gas refrigeration cycle with air as the working fluid is considered. The rate of refrigeration, the net power input, and the COP are to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible. Analysis (a) We assume the turbine inlet temperature to be the temperature of the surroundings, and the compressor inlet temperature to be the temperature of the refrigerated space. From the air table (Table A17), T1 = 285 K T1 = 320 K

⎯ ⎯→ ⎯ ⎯→

h1 = 28514 . kJ / kg Pr 1 = 11584 . h3 = 320.29 kJ / kg Pr 3 = 1.7375

Thus,

2 T

· QH 47°C 12°C

Pr2 =

P2 ⎛ 250 ⎞ Pr = ⎜ ⎯→ T2 s = 450.4 K ⎟(1.1584 ) = 5.792 ⎯ P1 1 ⎝ 50 ⎠ h2 s = 452.17 kJ/kg

2

3 1 · 4 QRefrig 4s

P ⎛ 50 ⎞ Pr4 = 4 Pr3 = ⎜ ⎯→ T4 s = 201.8 K ⎟(1.7375) = 0.3475 ⎯ P3 ⎝ 250 ⎠ h4 s = 201.76 kJ/kg

Also, ηT =

h3 − h 4 ⎯ ⎯→ h4 = h3 − ηT (h3 − h4 s ) h3 − h4 s = 320.29 − (0.85)(320.29 − 201.76) = 219.54 kJ/kg

Then the rate of refrigeration is Q& refrig = m& (qL ) = m& (h1 − h4 ) = (0.08 kg/s)(285.14 − 219.54 ) kJ/kg = 5.25 kW

(b) The net power input is determined from W& net, in = W& comp, in − W& turb, out

where W&comp,in = m& (h2 − h1 ) = m& (h2 s − h1 ) / ηC = (0.08 kg/s)[(452.17 − 285.14 ) kJ/kg]/ (0.80) = 16.70 kW W& turb,out = m& (h3 − h4 ) = (0.08 kg/s)(320.29 − 219.54 ) kJ/kg = 8.06 kW

Thus, W& net, in = 16.70 − 8.06 = 8.64 kW

(c) The COP of this ideal gas refrigeration cycle is determined from COPR =

Q& L 5.25 kW = = 0.61 & Wnet, in 8.64 kW

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s

11-49

11-66 A gas refrigeration cycle with helium as the working fluid is considered. The minimum temperature in the cycle, the COP, and the mass flow rate of the helium are to be determined. Assumptions 1 Steady operating conditions exist. 2 Helium is an ideal gas with constant specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of helium are cp = 5.1926 kJ/kg·K and k = 1.667 (Table A-2).

2 T

· QH

Analysis (a) From the isentropic relations, (k −1) / k

⎛P T2 s = T1 ⎜⎜ 2 ⎝ P1

⎞ ⎟⎟ ⎠

⎛P = T3 ⎜⎜ 4 ⎝ P3

⎞ ⎟ ⎟ ⎠

T4 s

(k −1) / k

= (263K )(3)0.667 / 1.667 = 408.2K ⎛1⎞ = (323K )⎜ ⎟ ⎝3⎠

50°C -10°C

0.667 / 1.667

= 208.1K

2

3 1 · 4 QRefrig 4s s

and

ηT =

h3 − h4 T − T4 = 3 ⎯ ⎯→ T4 = T3 − ηT (T3 − T4 s ) = 323 − (0.80 )(323 − 208.1) h3 − h4 s T3 − T4 s = 231.1 K = T min

h − h1 T2 s − T1 ηC = 2s = ⎯ ⎯→ T2 = T1 + (T2 s − T1 ) / η C = 263 + (408.2 − 263) / (0.80 ) h2 − h1 T2 − T1 = 444.5 K

(b) The COP of this gas refrigeration cycle is determined from COPR =

qL qL = wnet,in wcomp,in − wturb,out

=

h1 − h4 (h2 − h1 ) − (h3 − h4 )

=

T1 − T4 (T2 − T1 ) − (T3 − T4 )

=

263 − 231.1

(444.5 − 263) − (323 − 231.1)

= 0.356

(c) The mass flow rate of helium is determined from m& =

Q& refrig qL

=

Q& refrig h1 − h4

=

Q& refrig

c p (T1 − T4 )

=

18 kJ/s

(5.1926 kJ/kg ⋅ K )(263 − 231.1) K

= 0.109 kg/s

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

11-50

11-67 A regenerative gas refrigeration cycle using air as the working fluid is considered. The effectiveness of the regenerator, the rate of heat removal from the refrigerated space, the COP of the cycle, and the refrigeration load and the COP if this system operated on the simple gas refrigeration cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2). Analysis (a) From the isentropic relations, ⎛P T2 s = T1 ⎜⎜ 2 ⎝ P1

ηC = 0.80 =

⎞ ⎟⎟ ⎠

(k −1) / k

= (273.2 K )(5)0.4 / 1.4 = 432.4 K

h2 s − h1 T2 s − T1 = h2 − h1 T2 − T1

T5 s

Heat Exch.

432.4 − 273.2 ⎯ ⎯→ T2 = 472.5 K T2 − 273.2

The temperature at state 4 can be determined by solving the following two equations simultaneously: ⎛P = T4 ⎜⎜ 5 ⎝ P4

⎞ ⎟⎟ ⎠

(k −1) / k

⎛1⎞ = T4 ⎜ ⎟ ⎝5⎠

. QL

6 Regenerator

Heat Exch.

3 5

1

. QH

4

0.4 / 1.4

2

Turbine

h −h T − 193.2 η T = 4 5 → 0.85 = 4 h4 − h5 s T4 − T5 s

Compressor

Using EES, we obtain T4 = 281.3 K. An energy balance on the regenerator may be written as

2

T

· QH

m& c p (T3 − T4 ) = m& c p (T1 − T6 ) ⎯ ⎯→ T3 − T4 = T1 − T6

3

35°C

or, T6 = T1 − T3 + T4 = 273.2 − 308.2 + 281.3 = 246.3 K

0°C

The effectiveness of the regenerator is

ε regen =

2s

Qrege

1

4

h3 − h4 T3 − T4 308.2 − 281.3 = = = 0.434 h3 − h6 T3 − T6 308.2 − 246.3

-80°C 5

·6 5 QRefrig

s

(b) The refrigeration load is Q& L = m& c p (T6 − T5 ) = (0.4 kg/s)(1.005 kJ/kg.K)(246.3 − 193.2)K = 21.36 kW

(c) The turbine and compressor powers and the COP of the cycle are W& C,in = m& c p (T2 − T1 ) = (0.4 kg/s)(1.005 kJ/kg.K)(472.5 − 273.2)K = 80.13 kW W& T,out = m& c p (T4 − T5 ) = (0.4 kg/s)(1.005 kJ/kg.K)(281.3 − 193.2)kJ/kg = 35.43 kW COP =

Q& L W&

net,in

=

Q& L W& C,in − W& T,out

=

21.36 = 0.478 80.13 − 35.43

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11-51

(d) The simple gas refrigeration cycle analysis is as follows: ⎛1⎞ T4 s = T3 ⎜ ⎟ ⎝r⎠

ηT =

(k −1) / k

⎛1⎞ = (308.2 K )⎜ ⎟ ⎝5⎠

2

0.4 / 1.4

= 194.6 K

T3 − T4 308.2 − T4 ⎯ ⎯→ 0.85 = ⎯ ⎯→ T4 = 211.6 K 308.2 − 194.6 T3 − T4 s

Q& L = m& c p (T1 − T4 )

T

35°C 0°C

= (0.4 kg/s)(1.005 kJ/kg.K)(273.2 − 211.6)kJ/kg = 24.74 kW

· QH

2

3 1 · 4 QRefrig 4s

W& net,in = m& c p (T2 − T1 ) − m& c p (T3 − T4 )

= (0.4 kg/s)(1.005 kJ/kg.K)[(472.5 − 273.2) − (308.2 − 211.6)kJ/kg ] = 41.32 kW

COP =

Q& L W&

net,in

=

24.74 = 0.599 41.32

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11-52

11-68E An ideal gas refrigeration cycle with air as the working fluid has a compression ratio of 4. The COP of the cycle is to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are cp = 0.240 Btu/lbm·R and k = 1.4 (Table A-2Ea). Analysis From the isentropic relations, ⎛P T2 = T1 ⎜⎜ 2 ⎝ P1

⎞ ⎟⎟ ⎠

( k −1) / k

⎛P T4 = T3 ⎜⎜ 4 ⎝ P3

⎞ ⎟ ⎟ ⎠

( k −1) / k

= (450 R)(4) 0.4 / 1.4 = 668.7 R ⎛1⎞ = (560 R)⎜ ⎟ ⎝4⎠

T

0.4 / 1.4

= 376.8 R

The COP of this ideal gas refrigeration cycle is determined from qL qL = COPR = wnet,in wcomp,in − w turb,out h1 − h4 = (h2 − h1 ) − (h3 − h4 ) =

T1 − T4 (T2 − T1 ) − (T3 − T4 )

=

450 − 376.8 = 2.06 (668.7 − 450) − (560 − 376.8)

100°F -10°F

· QH

2

3 1 · 4 QRefrig s

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11-53

11-69E An gas refrigeration cycle with air as the working fluid has a compression ratio of 4. The COP of the cycle is to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are cp = 0.240 Btu/lbm·R and k = 1.4 (Table A-2Ea).

T

Analysis From the isentropic relations, ⎛P T2 s = T1 ⎜⎜ 2 ⎝ P1

⎞ ⎟⎟ ⎠

( k −1) / k

⎛P = T3 ⎜⎜ 4 ⎝ P3

⎞ ⎟ ⎟ ⎠

( k −1) / k

T4 s

2

= (450 R)(4) 0.4 / 1.4 = 668.7 R

100°F -10°F

⎛ 6 psia ⎞ ⎟⎟ = (560 R)⎜⎜ ⎝ 19 psia ⎠

0.4 / 1.4

= 402.9 R

· QH

2s

3 1 · 4 4s QRefrig s

and h3 − h4 T − T4 ⎯ ⎯→ T4 = T3 − η T (T3 − T4 s ) = 560 − (0.94)(560 − 402.9) = 3 h3 − h4 s T3 − T4 s = 412.3 R h −h T −T η C = 2 s 1 = 2 s 1 ⎯⎯→ T2 = T1 + (T2 s − T1 ) / η C = 450 + (668.7 − 450) /(0.87) h2 − h1 T2 − T1 = 701.4 R

ηT =

The COP of this gas refrigeration cycle is determined from COPR =

qL qL = wnet,in wcomp,in − w turb,out

=

h1 − h4 (h2 − h1 ) − (h3 − h4 )

=

T1 − T4 (T2 − T1 ) − (T3 − T4 )

=

450 − 412.3 = 0.364 (701.4 − 450) − (560 − 412.3)

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11-54

11-70 An ideal gas refrigeration cycle with air as the working fluid provides 15 kW of cooling. The mass flow rate of air and the rates of heat addition and rejection are to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2a).

T

Analysis From the isentropic relations, ⎛P T2 = T1 ⎜⎜ 2 ⎝ P1

⎞ ⎟⎟ ⎠

( k −1) / k

⎛P T4 = T3 ⎜⎜ 4 ⎝ P3

⎞ ⎟ ⎟ ⎠

( k −1) / k

⎛ 500 kPa ⎞ = (293 K)⎜ ⎟ ⎝ 100 kPa ⎠

0.4 / 1.4

⎛ 100 kPa ⎞ = (303 K)⎜ ⎟ ⎝ 500 kPa ⎠

0.4 / 1.4

= 464.1 K

30°C

· QH 3

20°C

= 191.3 K

Q& Refrig c p (T1 − T4 )

1 · 4 QRefrig s

The mass flow rate of the air is determined from ⎯→ m& = Q& Refrig = m& c p (T1 − T4 ) ⎯

2

=

15 kJ/s = 0.1468 kg/s (1.005 kJ/kg ⋅ K)(293 − 191.3) K

The rate of heat addition to the cycle is the same as the rate of cooling, Q& in = Q& Refrig = 15 kW

The rate of heat rejection from the cycle is Q& H = m& c p (T2 − T3 ) = (0.1468 kg/s)(1.005 kJ/kg ⋅ K)(464.1 − 303)K = 23.8 kW

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11-55

11-71 An ideal gas refrigeration cycle with air as the working fluid is considered. The minimum pressure ratio for this system to operate properly is to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2a).

T

Analysis An energy balance on process 4-1 gives q Refrig = c p (T1 − T4 ) T4 = T1 −

q Refrig cp

20°C = 268 K −

20 kJ/kg = 248.1 K 1.005 kJ/kg ⋅ K

-5°C

The minimum temperature at the turbine inlet would be the same as that to which the heat is rejected. That is,

· QH

2

3 1 · 4 QRefrig s

T3 = 293 K

Then the minimum pressure ratio is determined from the isentropic relation to be P3 ⎛ T3 ⎞ =⎜ ⎟ P4 ⎜⎝ T4 ⎟⎠

k /( k −1)

⎛ 293 K ⎞ =⎜ ⎟ ⎝ 248.1 K ⎠

1.4 / 0.4

= 1.79

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11-56

11-72 An ideal gas refrigeration cycle with with two stages of compression with intercooling using air as the working fluid is considered. The COP of this system and the mass flow rate of air are to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2a).

. Q

. Q 4 T

Heat exch.

Heat exch. 3

5 4

2

10°C

2 5 3

-18°C

1

Turbine 6 Heat exch. 1

6

. Q

Analysis From the isentropic relations, ⎛P T2 = T1 ⎜⎜ 2 ⎝ P1

⎞ ⎟⎟ ⎠

( k −1) / k

⎛P T4 = T3 ⎜⎜ 4 ⎝ P3

⎞ ⎟ ⎟ ⎠

( k −1) / k

⎛P T6 = T5 ⎜⎜ 6 ⎝ P5

⎞ ⎟ ⎟ ⎠

( k −1) / k

= (255 K)(4) 0.4 / 1.4 = 378.9 K = (283 K)(4) 0.4 / 1.4 = 420.5 K ⎛1⎞ = (283 K)⎜ ⎟ ⎝ 16 ⎠

0.4 / 1.4

= 128.2 K

The COP of this ideal gas refrigeration cycle is determined from COPR =

qL qL = w net,in wcomp,in − w turb,out

=

h1 − h6 ( h2 − h1 ) + ( h4 − h3 ) − ( h5 − h6 )

=

T1 − T6 (T2 − T1 ) + (T4 − T3 ) − (T5 − T6 )

=

255 − 128.2 = 1.19 (378.9 − 255) + (420.5 − 283) − (283 − 128.2)

The mass flow rate of the air is determined from ⎯→ m& = Q& Refrig = m& c p (T1 − T6 ) ⎯

Q& Refrig c p (T1 − T6 )

=

(75,000 / 3600) kJ/s = 0.163 kg/s (1.005 kJ/kg ⋅ K)(255 − 128.2) K

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s

11-57

11-73 A gas refrigeration cycle with with two stages of compression with intercooling using air as the working fluid is considered. The COP of this system and the mass flow rate of air are to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2a). . . Q Q 4 4s 2s 2 T Heat exch. Heat exch. 3 5 5 10°C 4 2 3 -18°C Turbine 1

6s Heat exch. 1 Analysis From the isentropic relations, ⎛P T2 s = T1 ⎜⎜ 2 ⎝ P1

⎞ ⎟⎟ ⎠

( k −1) / k

⎛P T4 s = T3 ⎜⎜ 4 ⎝ P3

⎞ ⎟ ⎟ ⎠

( k −1) / k

⎛P T6 s = T5 ⎜⎜ 6 ⎝ P5

⎞ ⎟ ⎟ ⎠

( k −1) / k

6

6 s

. Q

= (255 K)(4) 0.4 / 1.4 = 378.9 K = (283 K)(4) 0.4 / 1.4 = 420.5 K ⎛1⎞ = (283 K)⎜ ⎟ ⎝ 16 ⎠

0.4 / 1.4

= 128.2 K

and

ηC =

h2 s − h1 T2 s − T1 = ⎯ ⎯→ T2 = T1 + (T2 s − T1 ) / η C = 255 + (378.9 − 255) / 0.85 = 400.8 K h2 − h1 T2 − T1

ηC =

h4 s − h3 T4 s − T3 = ⎯ ⎯→ T4 = T3 + (T4 s − T3 ) / η C = 283 + (420.5 − 283) / 0.85 = 444.8 K h 4 − h3 T4 − T3

h5 − h 6 T − T6 = 5 ⎯ ⎯→ T6 = T5 − η T (T5 − T6 s ) = 283 − (0.95)(283 − 128.2) = 135.9 K h5 − h6 s T5 − T6 s The COP of this ideal gas refrigeration cycle is determined from qL qL COPR = = wnet,in wcomp,in − w turb,out

ηT =

=

h1 − h6 ( h2 − h1 ) + ( h4 − h3 ) − ( h5 − h6 )

=

T1 − T6 (T2 − T1 ) + (T4 − T3 ) − (T5 − T6 )

255 − 135.9 = 0.742 (400.8 − 255) + (444.8 − 283) − (283 − 135.9) The mass flow rate of the air is determined from Q& Refrig (75,000 / 3600) kJ/s Q& Refrig = m& c p (T1 − T6 ) ⎯ ⎯→ m& = = = 0.174 kg/s c p (T1 − T6 ) (1.005 kJ/kg ⋅ K)(255 − 135.9) K =

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11-58

Absorption Refrigeration Systems

11-74C Absorption refrigeration is the kind of refrigeration that involves the absorption of the refrigerant during part of the cycle. In absorption refrigeration cycles, the refrigerant is compressed in the liquid phase instead of in the vapor form.

11-75C The main advantage of absorption refrigeration is its being economical in the presence of an inexpensive heat source. Its disadvantages include being expensive, complex, and requiring an external heat source.

11-76C In absorption refrigeration, water can be used as the refrigerant in air conditioning applications since the temperature of water never needs to fall below the freezing point.

11-77C The fluid in the absorber is cooled to maximize the refrigerant content of the liquid; the fluid in the generator is heated to maximize the refrigerant content of the vapor.

11-78C The coefficient of performance of absorption refrigeration systems is defined as

COPR =

QL Q desiredoutput = ≅ L requiredinput Qgen + Wpump,in Qgen

11-79C The rectifier separates the water from NH3 and returns it to the generator. The regenerator transfers some heat from the water-rich solution leaving the generator to the NH3-rich solution leaving the pump.

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11-59

11-80 The COP of an absorption refrigeration system that operates at specified conditions is given. It is to be determined whether the given COP value is possible. Analysis The maximum COP that this refrigeration system can have is ⎛ T ⎞⎛ TL ⎞ ⎛ 300 K ⎞⎛ 268 ⎞ ⎟⎟ = ⎜1 − COPR,max = ⎜⎜1 − 0 ⎟⎟⎜⎜ ⎟⎜ ⎟ = 2.14 ⎝ Ts ⎠⎝ T0 − TL ⎠ ⎝ 403 K ⎠⎝ 300 − 268 ⎠

which is slightly greater than 2. Thus the claim is possible, but not probable.

11-81 The conditions at which an absorption refrigeration system operates are specified. The maximum COP this absorption refrigeration system can have is to be determined. Analysis The maximum COP that this refrigeration system can have is ⎛ T COPR,max = ⎜⎜1 − 0 ⎝ Ts

⎞⎛ TL ⎟⎜ ⎟⎜ T − T L ⎠⎝ 0

⎞ ⎛ 298 K ⎞⎛ 273 ⎞ ⎟ = ⎜1 − ⎟ ⎝ 393 K ⎟⎠⎜⎝ 298 − 273 ⎟⎠ = 2.64 ⎠

11-82 The conditions at which an absorption refrigeration system operates are specified. The maximum rate at which this system can remove heat from the refrigerated space is to be determined. Analysis The maximum COP that this refrigeration system can have is ⎛ T ⎞⎛ TL ⎞ ⎛ 298 K ⎞⎛ 243 ⎞ ⎟⎟ = ⎜⎜1 − ⎟⎟⎜ COPR,max = ⎜⎜1 − 0 ⎟⎟⎜⎜ ⎟ = 1.15 ⎝ Ts ⎠⎝ T0 − TL ⎠ ⎝ 403 K ⎠⎝ 298 − 243 ⎠

Thus,

(

)

Q& L,max = COPR, maxQ& gen = (1.15) 5 × 105 kJ/h = 5.75 × 105 kJ/h

11-83E The conditions at which an absorption refrigeration system operates are specified. The COP is also given. The maximum rate at which this system can remove heat from the refrigerated space is to be determined. Analysis For a COP = 0.55, the rate at which this system can remove heat from the refrigerated space is

(

)

Q& L = COPR Q& gen = (0.55) 10 5 Btu/h = 0.55 × 10 5 Btu/h

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11-60

11-84 A reversible absorption refrigerator consists of a reversible heat engine and a reversible refrigerator. The rate at which the steam condenses, the power input to the reversible refrigerator, and the second law efficiency of an actual chiller are to be determined. Properties The enthalpy of vaporization of water at 200°C is hfg = 1939.8 kJ/kg (Table A-4). Analysis (a) The thermal efficiency of the reversible heat engine is

η th,rev = 1 −

T0 (25 + 273.15) K = 1− = 0.370 Ts (200 + 273.15) K

Ts

T0

Rev. HE

Rev. Ref.

T0

TL

The COP of the reversible refrigerator is COPR,rev =

TL (−10 + 273.15) K = = 7.52 T0 − T L (25 + 273.15) − (−10 + 273.15) K

The COP of the reversible absorption refrigerator is COPabs, rev = η th,rev COPR, rev = (0.370)(7.52) = 2.78

The heat input to the reversible heat engine is Q& in =

Q& L 22 kW = = 7.911 kW COPabs,rev 2.78

Then, the rate at which the steam condenses becomes m& s =

Q& in 7.911 kJ/s = = 0.00408 kg/s h fg 1939.8 kJ/kg

(b) The power input to the refrigerator is equal to the power output from the heat engine W& in, R = W& out, HE = η th, rev Q& in = (0.370)(7.911 kW ) = 2.93 kW

(c) The second-law efficiency of an actual absorption chiller with a COP of 0.7 is

η II =

COPactual 0.7 = = 0.252 COPabs,rev 2.78

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11-61

Special Topic: Thermoelectric Power Generation and Refrigeration Systems

11-85C The circuit that incorporates both thermal and electrical effects is called a thermoelectric circuit.

11-86C When two wires made from different metals joined at both ends (junctions) forming a closed circuit and one of the joints is heated, a current flows continuously in the circuit. This is called the Seebeck effect. When a small current is passed through the junction of two dissimilar wires, the junction is cooled. This is called the Peltier effect.

11-87C No.

11-88C No.

11-89C Yes.

11-90C When a thermoelectric circuit is broken, the current will cease to flow, and we can measure the voltage generated in the circuit by a voltmeter. The voltage generated is a function of the temperature difference, and the temperature can be measured by simply measuring voltages.

11-91C The performance of thermoelectric refrigerators improves considerably when semiconductors are used instead of metals.

11-92C The efficiency of a thermoelectric generator is limited by the Carnot efficiency because a thermoelectric generator fits into the definition of a heat engine with electrons serving as the working fluid.

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11-62

11-93E A thermoelectric generator that operates at specified conditions is considered. The maximum thermal efficiency this thermoelectric generator can have is to be determined. Analysis The maximum thermal efficiency of this thermoelectric generator is the Carnot efficiency,

η th,max = η th,Carnot = 1 −

TL 550R = 1− = 31.3% 800R TH

11-94 A thermoelectric refrigerator that operates at specified conditions is considered. The maximum COP this thermoelectric refrigerator can have and the minimum required power input are to be determined. Analysis The maximum COP of this thermoelectric refrigerator is the COP of a Carnot refrigerator operating between the same temperature limits, COPmax = COPR,Carnot =

(TH

1 1 = = 10.72 / T L ) − 1 (293 K ) / (268 K ) − 1

Thus, W& in, min =

Q& L 130 W = = 12.1 W COPmax 10.72

11-95 A thermoelectric cooler that operates at specified conditions with a given COP is considered. The required power input to the thermoelectric cooler is to be determined. Analysis The required power input is determined from the definition of COPR, Q& COPR = & L Win

⎯ ⎯→

Q& L 180 W W&in = = = 1200 W COPR 0.15

11-96E A thermoelectric cooler that operates at specified conditions with a given COP is considered. The required power input to the thermoelectric cooler is to be determined. Analysis The required power input is determined from the definition of COPR, COPR =

Q& L Q& L 20 Btu/min ⎯ ⎯→ W& in = = = 133.3 Btu/min = 3.14 hp & COPR 0.15 Win

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11-63

11-97 A thermoelectric refrigerator powered by a car battery cools 9 canned drinks in 12 h. The average COP of this refrigerator is to be determined. Assumptions Heat transfer through the walls of the refrigerator is negligible. Properties The properties of canned drinks are the same as those of water at room temperature, ρ = 1 kg/L and cp = 4.18 kJ/kg·°C (Table A-3). Analysis The cooling rate of the refrigerator is simply the rate of decrease of the energy of the canned drinks, m = ρV = 9 × (1 kg/L)(0.350 L) = 3.15 kg Qcooling = mcΔT = (3.15 kg)(4.18 kJ/kg ⋅ °C)(25 - 3)°C = 290 kJ Q& cooling =

Qcooling Δt

=

290 kJ = 0.00671 kW = 6.71 W 12 × 3600 s

The electric power consumed by the refrigerator is W& in = VI = (12 V)(3 A) = 36 W

Then the COP of the refrigerator becomes COP =

Q& cooling 6.71 W = = 0.186 ≈ 0.20 36 W W& in

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11-64

11-98E A thermoelectric cooler is said to cool a 12-oz drink or to heat a cup of coffee in about 15 min. The average rate of heat removal from the drink, the average rate of heat supply to the coffee, and the electric power drawn from the battery of the car are to be determined. Assumptions Heat transfer through the walls of the refrigerator is negligible. Properties The properties of canned drinks are the same as those of water at room temperature, cp = 1.0 Btu/lbm.°F (Table A-3E). Analysis (a) The average cooling rate of the refrigerator is simply the rate of decrease of the energy content of the canned drinks, Qcooling = mc p ΔT = (0.771 lbm)(1.0 Btu/lbm ⋅ °F)(78 - 38)°F = 30.84 Btu Q& cooling =

Qcooling Δt

=

30.84 Btu ⎛ 1055 J ⎞ ⎜ ⎟ = 36.2 W 15 × 60 s ⎝ 1 Btu ⎠

(b) The average heating rate of the refrigerator is simply the rate of increase of the energy content of the canned drinks, Q heating = mc p ΔT = (0.771 lbm)(1.0 Btu/lbm ⋅ °F)(130 - 75)°F = 42.4 Btu Q& heating =

Q heating Δt

=

42.4 Btu ⎛ 1055 J ⎞ ⎜ ⎟ = 49.7 W 15 × 60 s ⎝ 1 Btu ⎠

(c) The electric power drawn from the car battery during cooling and heating is W& in,cooling =

Q& cooling COPcooling

=

36.2 W = 181 W 0.2

COPheating = COPcooling + 1 = 0.2 + 1 = 1.2 W& in, heating =

Q& heating COPheating

=

49.7 W = 41.4 W 1.2

11-99 The maximum power a thermoelectric generator can produce is to be determined. Analysis The maximum thermal efficiency this thermoelectric generator can have is ηth,max = 1 −

TL 303 K = 1− = 0.142 TH 353 K

Thus, W&out, max = η th, maxQ&in = (0.142)(106 kJ/h) = 142,000 kJ/h = 39.4 kW

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11-65

Review Problems

11-100 A steady-flow Carnot refrigeration cycle with refrigerant-134a as the working fluid is considered. The COP, the condenser and evaporator pressures, and the net work input are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) The COP of this refrigeration cycle is determined from COPR,C

T

1 1 = = = 5.06 (TH / TL ) − 1 (303 K )/ (253 K ) − 1

(b) The condenser and evaporative pressures are (Table A-11) Pevap = Psat @ − 20°C = 132.82 kPa Pcond = Psat @30°C = 770.64 kPa (c) The net work input is determined from

30°C

-20°C

4

1

3

qL

2

s

( )@−20°C = 25.49 + (0.15)(212.91) = 57.43 kJ/kg h2 = (h f + x 2 h fg )@ − 20°C = 25.49 + (0.80 )(212.91) = 195.82 kJ/kg h1 = h f + x1 h fg

q L = h2 − h1 = 195.82 − 57.43 = 138.4kJ/kg qL 138.4 kJ/kg = = 27.35 kJ/kg w net,in = COPR 5.06

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11-66

11-101 A heat pump that operates on the ideal vapor-compression cycle with refrigerant-134a as the working fluid is used to heat a house. The rate of heat supply to the house, the volume flow rate of the refrigerant at the compressor inlet, and the COP of this heat pump are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-12 and A-13), h1 = h g @ 200 kPa = 244.46 kJ/kg P1 = 200 kPa ⎫ ⎬ s1 = s g @ 200 kPa = 0.93773 kJ/kg ⋅ K sat. vapor ⎭ v 1 = v g @ 200 kPa = 0.099867 m 3 /kg P2 = 0.9 MPa s 2 = s1

⎫ ⎬ h2 = 275.75 kJ/kg ⎭

P3 = 0.9 MPa sat. liquid

⎫ ⎬ h3 = h f ⎭

@ 0.9 MPa

T House · QH

2

3 0.9 MPa

· Win

= 101.61 kJ/kg

200 kPa

h4 ≅ h3 = 101.61 kJ/kg (throttling )

4

The rate of heat supply to the house is determined from

· QL

1 s

Q& H = m& (h2 − h3 ) = (0.32 kg/s )(275.75 − 101.61) kJ/kg = 55.73 kW

(b) The volume flow rate of the refrigerant at the compressor inlet is

(

)

V&1 = m& v 1 = (0.32 kg/s ) 0.099867 m 3 /kg = 0.0320 m 3 /s (c) The COP of t his heat pump is determined from COPR =

q L h2 − h3 275.75 − 101.61 = = = 5.57 win h2 − h1 275.75 − 244.46

11-102 A relation for the COP of the two-stage refrigeration system with a flash chamber shown in Fig. 11-12 is to be derived. Analysis The coefficient of performance is determined from COPR =

qL win

where qL = (1 − x6 )(h1 − h8 ) with x6 =

h6 − h f h fg

win = wcompI,in + wcompII,in = (1 − x6 )(h2 − h1 ) + (1)(h4 − h9 )

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11-67

11-103 A two-stage compression refrigeration system using refrigerant-134a as the working fluid is considered. The fraction of the refrigerant that evaporates as it is throttled to the flash chamber, the amount of heat removed from the refrigerated space, the compressor work, and the COP are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 The flashing chamber is adiabatic. Analysis (a) The enthalpies of the refrigerant at several states are determined from the refrigerant tables to be (Tables A-11, A-12, and A-13) h1 = 239.16 kJ/kg,

h2 = 260.58 kJ/kg

h3 = 255.55 kJ/kg,

T

h5 =95.47 kJ/kg,

h6 = 95.47 kJ/kg

h7 = 63.94 kJ/kg,

h8 = 63.94 kJ/kg

The fraction of the refrigerant that evaporates as it is throttled to the flash chamber is simply the quality at state 6, x6 =

h6 − h f h fg

95.47 − 63.94 = = 0.1646 191.62

0.8 MPa 4 0.4 MPa

5

2

A 7

6

B

8

0.14 MPa

3 9 1 qL

(b) The enthalpy at state 9 is determined from an energy balance on the mixing chamber:

s

E& in − E& out = ΔE& system ©0 (steady) = 0 → E& in = E& out

∑ m& h = ∑ m& h e e

i i

(1)h9 = x6 h3 + (1 − x6 )h2 h9 = (0.1646 )(255.55) + (1 − 0.1646 )(260.58) = 259.75 kJ/kg

P9 = 0.4 MPa

⎫ ⎬ s 9 = 0.94168 kJ/kg ⋅ K h9 = 259.75 kJ/kg ⎭

Also, P4 = 0.8 MPa

⎫ ⎬ h4 = 274.47 kJ/kg s 4 = s 9 = 0.94168 kJ/kg ⋅ K ⎭

Then the amount of heat removed from the refrigerated space and the compressor work input per unit mass of refrigerant flowing through the condenser are q L = (1 − x 6 )(h1 − h8 ) = (1 − 0.1646)(239.16 − 63.94) kJ/kg = 146.4 kJ/kg win = wcompI,in + wcompII,in = (1 − x 6 )(h2 − h1 ) + (1)(h4 − h9 ) = (1 − 0.1646)(260.58 − 239.16 ) kJ/kg + (274.47 − 259.75) kJ/kg = 32.6 kJ/kg

(c) The coefficient of performance is determined from COPR =

q L 146.4 kJ/kg = = 4.49 win 32.6 kJ/kg

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11-68

11-104 A refrigerator operating on a vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The process with the greatest exergy loss is to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis In this cycle, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. The compression process is not isentropic. From the refrigerant tables (Tables A-11, A-12, and A-13), T1 = −10°C ⎫ h1 = h g @ −10°C = 244.51 kJ/kg ⎬ sat. vapor ⎭ s1 = s g @ −10°C = 0.93766 kJ/kg ⋅ K

T

P2 = 700 kPa ⎫ ⎬ h2 s = 270.38 kJ/kg s 2 = s1 ⎭

· QH

P3 = 700 kPa ⎫ h3 = h f @ 700 kPa = 88.82 kJ/kg ⎬ s =s sat. liquid f @ 700 kPa = 0.33230 kJ/kg ⋅ K ⎭ 3

2

2s

3 0.7 MPa

h4 ≅ h3 = 88.82 kJ/kg ( throttling)

· Win

-10°C

T4 = −10°C ⎫ ⎬ s = 0.34605 kJ/kg ⋅ K h4 = 88.82 kJ/kg ⎭ 4

4s

4

· QL

1

The actual enthalpy at the compressor exit is determined by using the compressor efficiency:

ηC = and

s

h − h1 h2 s − h1 270.38 − 244.51 ⎯ ⎯→ h2 = h1 + 2 s = 244.51 + = 274.95 kJ/kg ηC 0.85 h2 − h1

P2 = 700 kPa ⎫ ⎬ s = 0.95252 kJ/kg h2 = 274.95 kJ/kg ⎭ 2

The heat added in the evaporator and that rejected in the condenser are q L = h1 − h4 = (244.51 − 88.82) kJ/kg = 155.69 kJ/kg q H = h2 − h3 = (274.95 − 88.82) kJ/kg = 186.13 kJ/kg The exergy destruction during a process of a stream from an inlet state to exit state is given by ⎛ q q x dest = T0 s gen = T0 ⎜⎜ s e − s i − in + out Tsource Tsink ⎝

⎞ ⎟ ⎟ ⎠

Application of this equation for each process of the cycle gives x destroyed,12 = T0 ( s 2 − s1 ) = (295 K )(0.95252 − 0.93766) kJ/kg ⋅ K = 4.38 kJ/kg ⎛ q ⎞ 186.13 kJ/kg ⎞ ⎛ x destroyed, 23 = T0 ⎜⎜ s 3 − s 2 + H ⎟⎟ = (295 K )⎜ 0.33230 − 0.95252 + ⎟ = 3.17 kJ/kg 295 K TH ⎠ ⎠ ⎝ ⎝ x destroyed, 34 = T0 ( s 4 − s 3 ) = (295 K )(0.34605 − 0.33230) kJ/kg ⋅ K = 4.06 kJ/kg ⎛ q ⎞ 155.69 kJ/kg ⎞ ⎛ x destroyed, 41 = T0 ⎜⎜ s1 − s 4 − L ⎟⎟ = (295 K )⎜ 0.93766 − 0.34605 − ⎟ = 6.29 kJ/kg T 273 K ⎠ ⎝ L ⎠ ⎝

The greatest exergy destruction occurs in the evaporator. Note that heat is absorbed from freezing water at 0°C (273 K) and rejected to the ambient air at 22°C (295 K), which is also taken as the dead state temperature. Alternatively, one may use the standard 25°C (298 K) as the dead state temperature, and perform the calculations accordingly.

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11-69

11-105 A refrigerator operating on a vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The process with the greatest exergy loss is to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From the refrigerant tables (Tables A-11, A-12, and A-13), T1 = −10°C ⎫ h1 = h g @ −10°C = 244.51 kJ/kg ⎬ sat. vapor ⎭ s1 = s g @ −10°C = 0.93766 kJ/kg ⋅ K

T

P2 = 700 kPa ⎫ ⎬ h2 s = 270.38 kJ/kg s 2 = s1 ⎭

· QH

⎫ ⎪ h3 ≅ h f @ 24°C = 84.98 kJ/kg ⎬ s ≅s 3 f @ 24°C = 0.31958 kJ/kg ⋅ K = 26.7 − 2.7 = 24°C ⎪⎭

P3 = 700 kPa T3 = Tsat @ 700 kPa − 2.7

h4 ≅ h3 = 84.98 kJ/kg ( throttling)

2

2s

0.7 MPa

· Win

3 -10°C 4

T4 = −10°C ⎫ ⎬ s = 0.33146 kJ/kg ⋅ K h4 = 84.98 kJ/kg ⎭ 4

· QL

1

s

The actual enthalpy at the compressor exit is determined by using the compressor efficiency:

ηC = and

h − h1 h2 s − h1 270.38 − 244.51 ⎯ ⎯→ h2 = h1 + 2 s = 244.51 + = 274.95 kJ/kg ηC 0.85 h2 − h1

P2 = 700 kPa ⎫ ⎬ s = 0.95252 kJ/kg h2 = 274.95 kJ/kg ⎭ 2

The heat added in the evaporator and that rejected in the condenser are q L = h1 − h4 = (244.51 − 84.98) kJ/kg = 159.53 kJ/kg q H = h2 − h3 = (274.95 − 84.98) kJ/kg = 189.97 kJ/kg

The exergy destruction during a process of a stream from an inlet state to exit state is given by ⎛ q q x dest = T0 s gen = T0 ⎜⎜ s e − s i − in + out Tsource Tsink ⎝

⎞ ⎟ ⎟ ⎠

Application of this equation for each process of the cycle gives x destroyed,12 = T0 ( s 2 − s1 ) = (295 K )(0.95252 − 0.93766) kJ/kg ⋅ K = 4.38 kJ/kg ⎛ q ⎞ 189.97 kJ/kg ⎞ ⎛ x destroyed, 23 = T0 ⎜⎜ s 3 − s 2 + H ⎟⎟ = (295 K )⎜ 0.31958 − 0.95252 + ⎟ = 3.25 kJ/kg 295 K TH ⎠ ⎠ ⎝ ⎝ x destroyed, 34 = T0 ( s 4 − s 3 ) = (295 K )(0.33146 − 0.31958) kJ/kg ⋅ K = 3.50 kJ/kg ⎛ q ⎞ 159.53 kJ/kg ⎞ ⎛ x destroyed, 41 = T0 ⎜⎜ s1 − s 4 − L ⎟⎟ = (295 K )⎜ 0.93766 − 0.33146 − ⎟ = 6.44 kJ/kg T 273 K ⎠ ⎝ L ⎠ ⎝

The greatest exergy destruction occurs in the evaporator. Note that heat is absorbed from freezing water at 0°C (273 K) and rejected to the ambient air at 22°C (295 K), which is also taken as the dead state temperature. Alternatively, one may use the standard 25°C (298 K) as the dead state temperature, and perform the calculations accordingly.

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11-70

11-106 A refrigerator operating on a vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The process with the greatest exergy loss is to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis In this cycle, the refrigerant leaves the condenser as saturated liquid at the condenser pressure. The compression process is not isentropic. From the refrigerant tables (Tables A-11, A-12, and A-13), P1 = Psat @ −37°C = 60 kPa ⎫ h1 = 233.09 kJ/kg ⎬ s = 0.9865 kJ/kg ⋅ K T1 = −37 + 7 = −30°C ⎭ 1

T

P2 = 1.2 MPa ⎫ ⎬ h2 s = 298.11 kJ/kg s 2 = s1 ⎭ P3 = 1.2 MPa ⎫ h3 = h f @ 1.2 MPa = 117.77 kJ/kg ⎬ s =s sat. liquid f @ 1.2 MPa = 0.42441 kJ/kg ⋅ K ⎭ 3

2

2s

· QH 3 1.2 MPa

h4 ≅ h3 = 117.77 kJ/kg ( throttling)

· Win

-37°C 4s

T4 = −37°C ⎫ x 4 = 0.5089 ⎬ h4 = 117.77 kJ/kg ⎭ s 4 = 0.4988 kJ/kg ⋅ K

4

· QL

1

The actual enthalpy at the compressor exit is determined by using the compressor efficiency:

ηC = and

s

h − h1 h2 s − h1 298.11 − 233.09 ⎯ ⎯→ h2 = h1 + 2 s = 233.09 + = 305.33 kJ/kg ηC 0.90 h2 − h1

P2 = 1.2 MPa ⎫ ⎬ s = 1.0075 kJ/kg ⋅ K h2 = 305.33 Btu/lbm⎭ 2

The heat added in the evaporator and that rejected in the condenser are q L = h1 − h4 = (233.09 − 117.77) kJ/kg = 115.32 kJ/kg q H = h2 − h3 = (305.33 − 117.77) kJ/kg = 187.56 kJ/kg

The exergy destruction during a process of a stream from an inlet state to exit state is given by ⎛ q q x dest = T0 s gen = T0 ⎜⎜ s e − s i − in + out Tsource Tsink ⎝

⎞ ⎟ ⎟ ⎠

Application of this equation for each process of the cycle gives x destroyed,12 = T0 ( s 2 − s1 ) = (303 K )(1.0075 − 0.9865) kJ/kg ⋅ K = 6.36 kJ/kg ⎛ q ⎞ 187.56 kJ/kg ⎞ ⎛ x destroyed, 23 = T0 ⎜⎜ s 3 − s 2 + H ⎟⎟ = (303 K )⎜ 0.42441 − 1.0075 + ⎟ = 10.88 kJ/kg T 303 K ⎠ ⎝ H ⎠ ⎝ x destroyed, 34 = T0 ( s 4 − s 3 ) = (303 K )(0.4988 − 0.42441) kJ/kg ⋅ K = 22.54 kJ/kg ⎛ q x destroyed, 41 = T0 ⎜⎜ s1 − s 4 − L TL ⎝

⎞ ⎛ 115.32 kJ/kg ⎞ ⎟⎟ = (303 K )⎜⎜ 0.9865 − 0.4988 − ⎟ = 1.57 kJ/kg (−34 + 273) K ⎟⎠ ⎝ ⎠

The greatest exergy destruction occurs in the expansion valve. Note that heat is absorbed from fruits at 34°C (239 K) and rejected to the ambient air at 30°C (303 K), which is also taken as the dead state temperature. Alternatively, one may use the standard 25°C (298 K) as the dead state temperature, and perform the calculations accordingly.

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11-71

11-107 A refrigerator operating on a vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The process with the greatest exergy loss is to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From the refrigerant tables (Tables A-11, A-12, and A-13), P1 = Psat @ −37°C = 60 kPa ⎫ h1 = 233.09 kJ/kg ⎬ s = 0.9865 kJ/kg ⋅ K T1 = −37 + 7 = −30°C ⎭ 1

T

P2 = 1.2 MPa ⎫ ⎬ h2 s = 298.11 kJ/kg s 2 = s1 ⎭ ⎫ P3 = 1.2 MPa ⎪ h3 ≅ h f @ 40°C = 108.26 kJ/kg T3 = Tsat @ 1.2 MPa − 6.3 ⎬ s 3 ≅ s f @ 40°C = 0.39486 kJ/kg ⋅ K = 46.3 − 6.3 = 40°C ⎪⎭ h4 ≅ h3 = 108.26 kJ/kg ( throttling)

2

2s

· QH 1.2 MPa

· Win

3 -37°C 4

T4 = −37°C ⎫ x 4 = 0.4665 ⎬ h4 = 108.26 kJ/kg ⎭ s 4 = 0.4585 kJ/kg ⋅ K

· QL

1

s

The actual enthalpy at the compressor exit is determined by using the compressor efficiency:

ηC = and

h − h1 h2 s − h1 298.11 − 233.09 ⎯ ⎯→ h2 = h1 + 2 s = 233.09 + = 305.33 kJ/kg ηC 0.90 h2 − h1

P2 = 1.2 MPa ⎫ ⎬ s = 1.0075 kJ/kg ⋅ K h2 = 305.33 Btu/lbm⎭ 2

The heat added in the evaporator and that rejected in the condenser are q L = h1 − h4 = (233.09 − 108.26) kJ/kg = 124.83 kJ/kg q H = h2 − h3 = (305.33 − 108.26) kJ/kg = 197.07 kJ/kg The exergy destruction during a process of a stream from an inlet state to exit state is given by ⎛ q q x dest = T0 s gen = T0 ⎜⎜ s e − s i − in + out Tsource Tsink ⎝

⎞ ⎟ ⎟ ⎠

Application of this equation for each process of the cycle gives x destroyed,12 = T0 ( s 2 − s1 ) = (303 K )(1.0075 − 0.9865) kJ/kg ⋅ K = 6.36 kJ/kg ⎛ q ⎞ 197.07 kJ/kg ⎞ ⎛ x destroyed, 23 = T0 ⎜⎜ s 3 − s 2 + H ⎟⎟ = (303 K )⎜ 0.39486 − 1.0075 + ⎟ = 11.44 kJ/kg T 303 K ⎠ ⎝ H ⎠ ⎝ x destroyed, 34 = T0 ( s 4 − s 3 ) = (303 K )(0.4585 − 0.39486) kJ/kg ⋅ K = 19.28 kJ/kg ⎛ q x destroyed, 41 = T0 ⎜⎜ s1 − s 4 − L TL ⎝

⎞ ⎛ 124.83 kJ/kg ⎞ ⎟⎟ = (303 K )⎜⎜ 0.9865 − 0.4585 − ⎟ = 1.73 kJ/kg (−34 + 273) K ⎟⎠ ⎝ ⎠

The greatest exergy destruction occurs in the expansion valve. Note that heat is absorbed from fruits at 34°C (239 K) and rejected to the ambient air at 30°C (303 K), which is also taken as the dead state temperature. Alternatively, one may use the standard 25°C (298 K) as the dead state temperature, and perform the calculations accordingly.

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11-72

11-108E A two-evaporator compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The cooling load of both evaporators per unit of flow through the compressor and the COP of the system are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

Condenser

2

3

T · QH

Compressor 1

Expansion valve

Expansion valve

3 160 psia

Evaporator 1

30°F 4

4

5

Expansion valve

7

2 · Win 5

-29.5°F 6

Evaporator 2 6

· 7 1 QL s

Analysis From the refrigerant tables (Tables A-11E, A-12E, and A-13E), P3 = 160 psia ⎫ ⎬ h3 = h f sat. liquid ⎭

@ 160 psia

= 48.519 Btu/lbm

h4 = h6 ≅ h3 = 48.519 Btu/lbm ( throttling) T5 = 30°F ⎫ ⎬ h = h g @ 30°F = 107.40 Btu/lbm sat. vapor ⎭ 5 T7 = −29.5°F ⎫ ⎬ h7 = h g @ − 29.5°F = 98.68 Btu/lbm sat. vapor ⎭

For a unit mass flowing through the compressor, the fraction of mass flowing through Evaporator II is denoted by x and that through Evaporator I is y (y = 1-x). From the cooling loads specification, Q& L,e vap 1 = 2Q& L ,e vap 2 x(h5 − h4 ) = 2 y (h7 − h6 )

where x = 1− y

Combining these results and solving for y gives y=

h5 − h 4 107.40 − 48.519 = = 0.3698 2(h7 − h6 ) + (h5 − h4 ) 2(98.68 − 48.519) + (107.40 − 48.519)

Then, x = 1 − y = 1 − 0.3698 = 0.6302

Applying an energy balance to the point in the system where the two evaporator streams are recombined gives

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11-73

xh5 + yh7 = h1 ⎯ ⎯→ h1 =

xh5 + yh7 (0.6302)(107.40) + (0.3698)(98.68) = = 104.18 Btu/lbm 1 1

Then, P1 = Psat @ − 29.5° F ≅ 10 psia ⎫ ⎬ s1 = 0.2418 Btu/lbm ⋅ R h1 = 104.18 Btu/lbm ⎭ P2 = 160 psia ⎫ ⎬ h2 = 131.14 Btu/lbm s 2 = s1 ⎭

The cooling load of both evaporators per unit mass through the compressor is q L = x ( h5 − h 4 ) + y ( h 7 − h 6 ) = (0.6302)(107.40 − 48.519) Btu/lbm + (0.3698)(98.68 − 48.519) Btu/lbm = 55.66 Btu/lbm

The work input to the compressor is win = h2 − h1 = (131.14 − 104.18) Btu/lbm = 26.96 Btu/lbm

The COP of this refrigeration system is determined from its definition, COPR =

qL 55.66 Btu/lbm = = 2.06 win 26.96 Btu/lbm

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

11-74

11-109E A two-evaporator compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The process with the greatest exergy destruction is to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From Prob. 11-107E and the refrigerant tables (Tables A-11E, A-12E, and A-13E), s1 = s 2 = 0.2418 Btu/lbm ⋅ R s 3 = 0.09774 Btu/lbm ⋅ R s 4 = 0.10238 Btu/lbm ⋅ R s 5 = 0.22260 Btu/lbm ⋅ R s 6 = 0.11286 Btu/lbm ⋅ R s 7 = 0.22948 Btu/lbm ⋅ R x = 0.6302 y = 1 − x = 0.3698 q L, 45 = h5 − h4 = 58.881 Btu/lbm q L,67 = h7 − h6 = 50.161 Btu/lbm q H = 82.621 Btu/lbm

T · QH

2

3 160 psia

· Win 5

30°F 4

-29.5°F 6

· 7 1 QL

The exergy destruction during a process of a stream from an inlet state to exit state is given by ⎛ q q x dest = T0 s gen = T0 ⎜⎜ s e − s i − in + out Tsource Tsink ⎝

s

⎞ ⎟ ⎟ ⎠

Application of this equation for each process of the cycle gives the exergy destructions per unit mass flowing through the compressor: ⎛ q ⎞ 82.621 Btu/lbm ⎞ ⎛ x destroyed, 23 = T0 ⎜⎜ s 3 − s 2 + H ⎟⎟ = (555 R )⎜ 0.09774 − 0.2418 + ⎟ = 2.67 Btu/lbm T 555 R ⎝ ⎠ H ⎠ ⎝ x destroyed, 346 = T0 ( xs 4 + ys 6 − s 3 ) = (555 R )(0.6302 × 0.10238 + 0.3698 × 0.11286 − 0.09774) Btu/lbm ⋅ R = 4.73 Btu/lbm q L , 45 ⎞ ⎛ ⎟ x destroyed, 45 = xT0 ⎜⎜ s 5 − s 4 − T L ⎟⎠ ⎝ 58.881 Btu/lbm ⎞ ⎛ = (0.6302)(555 R )⎜ 0.22260 − 0.10238 − ⎟ = 0.44 Btu/lbm 495 R ⎝ ⎠ q L ,67 ⎞ ⎛ ⎟ x destroyed, 67 = yT0 ⎜⎜ s 7 − s 6 − T L ⎟⎠ ⎝ 50.161 Btu/lbm ⎞ ⎛ = (0.3698)(555 R )⎜ 0.22948 − 0.11286 − ⎟ = 0.54 Btu/lbm 440 R ⎝ ⎠ X& destroyed, mixing = T0 ( s1 − xs 5 − ys 6 )

= (555 R )[0.2418 − (0.6302)(0.22260) − (0.3698)(0.11286)] = 3.18 Btu/lbm

For isentropic processes, the exergy destruction is zero: X& destroyed,12 = 0

The greatest exergy destruction occurs during the mixing process. Note that heat is absorbed in evaporator 2 from a reservoir at -20°F (440 R), in evaporator 1 from a reservoir at 35°F (495 R), and rejected to a reservoir at 95°F (555 R), which is also taken as the dead state temperature.

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11-75

11-110 A two-stage compression refrigeration system with a separation unit is considered. The rate of cooling and the power requirement are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

Condenser

2

3

Compressor

T 2 3

1

Separator

8

Evaporator

5

8 8.9°C 1

4 -32°C

4 5 Expansion valve

Compressor 7

Expansion valve

1.4 MPa

6

· QL

7 s

6

Analysis From the refrigerant tables (Tables A-11, A-12, and A-13), T1 = 8.9°C ⎫ h1 = h g @ 8.9°C = 255.55 kJ/kg ⎬ sat. vapor ⎭ s1 = s g @ 8.9°C = 0.92691 kJ/kg ⋅ K P2 = 1400 kPa ⎫ ⎬ h2 = 281.49 kJ/kg s 2 = s1 ⎭ P3 = 1400 kPa ⎫ ⎬ h3 = h f sat. liquid ⎭

@ 1400 kPa

= 127.22 kJ/kg

h4 ≅ h3 = 127.22 kJ/kg ( throttling) T5 = 8.9°C ⎫ ⎬ h = hf sat. liquid ⎭ 5

@ 8.9°C

= 63.94 kJ/kg

h6 ≅ h5 = 63.94 kJ/kg ( throttling) T7 = −32°C ⎫ h7 = h g @ −32°C = 230.91 kJ/kg ⎬ sat. vapor ⎭ s 7 = s g @ −32°C = 0.95813 kJ/kg ⋅ K P8 = Psat @ 8.9°C = 400 kPa ⎫ ⎬ h8 = 264.51 kJ/kg s8 = s 7 ⎭

An energy balance on the separator gives m& 6 (h8 − h5 ) = m& 2 (h1 − h4 ) ⎯ ⎯→ m& 6 = m& 2

h1 − h4 255.55 − 127.22 = (2 kg/s) = 1.280 kg/s h8 − h5 264.51 − 63.94

The rate of cooling produced by this system is then Q& = m& (h − h ) = (1.280 kg/s)(230.91 − 63.94) kJ/kg = 213.7 kJ/s L

6

7

6

The total power input to the compressors is W& = m& (h − h ) + m& (h − h ) in

6

8

7

2

2

1

= (1.280 kg/s)(264.51 − 230.91) kJ/kg + (2 kg/s)(281.49 − 255.55) kJ/kg = 94.89 kW

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11-76

11-111 A two-stage vapor-compression refrigeration system with refrigerant-134a as the working fluid is considered. The process with the greatest exergy destruction is to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From Prob. 11-109 and the refrigerant tables (Tables A-11, A-12, and A-13), s1 = s 2 = 0.92691 kJ/kg ⋅ K s 3 = 0.45315 kJ/kg ⋅ K s 4 = 0.4720 kJ/kg ⋅ K s 5 = 0.24761 kJ/kg ⋅ K s 6 = 0.2658 kJ/kg ⋅ K s 7 = s 8 = 0.95813 kJ/kg ⋅ K m& upper = 2 kg/s m& lower = 1.280 kg/s q L = h7 − h6 = 166.97 kJ/kg q H = h2 − h3 = 154.27 kJ/kg T L = −18 + 273 = 255 K T H = T0 = 25 + 273 = 298 K

T 2 3 5

1.4 MPa 8 8.9°C 4 -32°C

1

6

· QL

7 s

The exergy destruction during a process of a stream from an inlet state to exit state is given by ⎛ q q x dest = T0 s gen = T0 ⎜⎜ s e − s i − in + out Tsource Tsink ⎝

⎞ ⎟ ⎟ ⎠

Application of this equation for each process of the cycle gives ⎛ q ⎞ X& destroyed, 23 = m& upper T0 ⎜⎜ s 3 − s 2 + H ⎟⎟ TH ⎠ ⎝ 154.27 kJ/kg ⎞ ⎛ = (2 kg/s)(298 K )⎜ 0.45315 − 0.92691 + ⎟ = 26.18 kW 298 K ⎝ ⎠ & X destroyed, 34 = m& upper T0 ( s 4 − s 3 ) = (2 kg/s)(298 K )(0.4720 − 0.45315) kJ/kg ⋅ K = 11.23 kW X& destroyed, 56 = m& lower T0 ( s 6 − s 5 ) = (1.280 kg/s)(298 K )(0.2658 − 0.24761) kJ/kg ⋅ K = 6.94 kW ⎛ q ⎞ X& destroyed, 67 = m& lower T0 ⎜⎜ s 7 − s 6 − L ⎟⎟ TL ⎠ ⎝ 166.97 kJ/kg ⎞ ⎛ = (1.280 kg/s)(298 K )⎜ 0.95813 − 0.2658 − ⎟ = 14.32 kW 255 K ⎝ ⎠ & X destroyed, separator = T0 m& lower ( s 5 − s 8 ) − m& upper ( s1 − s 4 )

[

]

= (298 K )[(1.280 kg/s)(0.24761 − 0.95813) + (2 kg/s)(0.92691 − 0.4720)] = 0.11 kW

For isentropic processes, the exergy destruction is zero: X& destroyed,12 = 0 X& destroyed, 78 = 0

Note that heat is absorbed from a reservoir at 0°F (460 R) and rejected to the standard ambient air at 77°F (537 R), which is also taken as the dead state temperature. The greatest exergy destruction occurs during the condensation process.

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11-77

11-112 A regenerative gas refrigeration cycle with helium as the working fluid is considered. The temperature of the helium at the turbine inlet, the COP of the cycle, and the net power input required are to be determined. Assumptions 1 Steady operating conditions exist. 2 Helium is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. Properties The properties of helium are cp = 5.1926 kJ/kg·K and k = 1.667 (Table A-2).

T

Analysis (a) The temperature of the helium at the turbine inlet is determined from an energy balance on the regenerator,

· QH 3

20°C -10°C

E& in − E& out = ΔE& system©0 (steady) = 0 E& in = E& out

1 4

∑ m& h = ∑ m& h e e

i i

⎯ ⎯→ m& (h3 − h4 ) = m& (h1 − h6 )

-25°C

or,

2

5

· Qregen ·6 QRefrig

m& c p (T3 − T4 ) = m& c p (T1 − T6 ) ⎯ ⎯→ T3 − T4 = T1 − T6

Thus,

T4 = T3 − T1 + T6 = 20°C − (− 10°C ) + (− 25°C ) = 5°C = 278 K

(b) From the isentropic relations, ⎛P ⎞ T2 = T1 ⎜⎜ 2 ⎟⎟ ⎝ P1 ⎠

(k −1) / k

⎛P ⎞ T5 = T4 ⎜⎜ 5 ⎟⎟ ⎝ P4 ⎠

(k −1) / k

= (263 K )(3)0.667 / 1.667 = 408.2 K = 135.2°C ⎛1⎞ = (278 K )⎜ ⎟ ⎝ 3⎠

0.667 / 1.667

= 179.1 K = −93.9°C

Then the COP of this ideal gas refrigeration cycle is determined from COPR = =

h6 − h5 qL qL = = wnet,in wcomp,in − w turb,out (h2 − h1 ) − (h4 − h5 )

T6 − T5 − 25°C − (− 93.9°C ) = = 1.49 (T2 − T1 ) − (T4 − T5 ) [135.2 − (− 10)]°C − [5 − (− 93.9)]°C

(c) The net power input is determined from W& net,in = W& comp,in − W& turb,out = m& [(h2 − h1 ) − (h4 − h5 )] = m& c p [(T2 − T1 ) − (T4 − T5 )]

= (0.45 kg/s )(5.1926 kJ/kg ⋅ °C )([135.2 − (− 10)] − [5 − (− 93.9 )]) = 108.2 kW

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s

11-78

11-113 An absorption refrigeration system operating at specified conditions is considered. The minimum rate of heat supply required is to be determined. Analysis The maximum COP that this refrigeration system can have is ⎛ T COPR, max = ⎜⎜1 − 0 ⎝ Ts

⎞⎛ T L ⎟⎜ ⎟⎜ T − T L ⎠⎝ 0

⎞ ⎛ 298K ⎞⎛ 263 ⎞ ⎟ = ⎜⎜1 − ⎟⎟⎜ ⎟ = 1.259 ⎟ ⎠ ⎝ 358K ⎠⎝ 298 − 263 ⎠

Thus, Q& gen,min =

Q& L 12 kW = = 9.53 kW COPR, max 1.259

11-114 EES Problem 11-113 is reconsidered. The effect of the source temperature on the minimum rate of heat supply is to be investigated. Analysis The problem is solved using EES, and the solution is given below. "Input Data:" T_L = -10 [C] T_0 = 25 [C] T_s = 85 [C] Q_dot_L = 8 [kW] "The maximum COP that this refrigeration system can have is:" COP_R_max = (1-(T_0+273)/(T_s+273))*((T_L+273)/(T_0 - T_L)) "The minimum rate of heat supply is:" Q_dot_gen_min = Q_dot_L/COP_R_max

Ts [C] 50 65 80 100 125 150 200 250

14 12

Qgen;min [kW]

Qgenmin [kW] 13.76 8.996 6.833 5.295 4.237 3.603 2.878 2.475

10 8 6 4 2 50

90

130

170

210

250

Ts [C]

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11-79

11-115 A room is cooled adequately by a 5000 Btu/h window air-conditioning unit. The rate of heat gain of the room when the air-conditioner is running continuously is to be determined. Assumptions 1 The heat gain includes heat transfer through the walls and the roof, infiltration heat gain, solar heat gain, internal heat gain, etc. 2 Steady operating conditions exist. Analysis The rate of heat gain of the room in steady operation is simply equal to the cooling rate of the airconditioning system, Q& heat gain = Q& cooling = 5,000 Btu / h

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

11-80

11-116 A regenerative gas refrigeration cycle using air as the working fluid is considered. The effectiveness of the regenerator, the rate of heat removal from the refrigerated space, the COP of the cycle, and the refrigeration load and the COP if this system operated on the simple gas refrigeration cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Analysis (a) For this problem, we use the properties of air from EES: T1 = 0°C ⎯ ⎯→ h1 = 273.40 kJ/kg

. QL

P1 = 100 kPa ⎫ ⎬ s1 = 5.6110 kJ/kg.K T1 = 0°C ⎭

Heat Exch.

P2 = 500 kPa ⎫ ⎬ h2 s = 433.50 kJ/kg ⎭

6

s 2 = s1

ηC =

h2 s − h1 h2 − h1

Regenerator 3 5

1

. QH

4

433.50 − 273.40 0.80 = h2 − 273.40

Heat Exch. 2

h2 = 473.52 kJ/kg T3 = 35°C ⎯ ⎯→ h3 = 308.63 kJ/kg

Turbine Compressor

For the turbine inlet and exit we have T5 = −80°C ⎯ ⎯→ h5 = 193.45 kJ/kg

ηT =

2

T

⎯→ h4 = T4 = ? ⎯ h4 − h5 h4 − h5 s

· QH 3

35°C

P1 = 100 kPa ⎫ ⎬ s1 = 5.6110 kJ/kg.K ⎭ P4 = 500 kPa ⎫ ⎬ s4 = T4 = ? ⎭

0°C

T1 = 0°C

P5 = 500 kPa ⎫ ⎬ h5 s = s5 = s 4 ⎭

2s

Qregen

1

4 -80°C 5s

·6 5 QRefrig s

We can determine the temperature at the turbine inlet from EES using the above relations. A hand solution would require a trial-error approach. T4 = 281.8 K, h4 = 282.08 kJ/kg An energy balance on the regenerator gives h6 = h1 − h3 + h4 = 273.40 − 308.63 + 282.08 = 246.85 kJ/kg

The effectiveness of the regenerator is determined from

ε regen =

h3 − h4 308.63 − 282.08 = = 0.430 h3 − h6 308.63 − 246.85

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11-81

(b) The refrigeration load is Q& L = m& (h6 − h5 ) = (0.4 kg/s)(246.85 − 193.45)kJ/kg = 21.36 kW

(c) The turbine and compressor powers and the COP of the cycle are W& C,in = m& (h2 − h1 ) = (0.4 kg/s)(473.52 − 273.40)kJ/kg = 80.05 kW W& T,out = m& (h4 − h5 ) = (0.4 kg/s)(282.08 − 193.45)kJ/kg = 35.45 kW COP =

Q& L W&

=

net,in

Q& L W& C,in − W& T, out

=

21.36 = 0.479 80.05 − 35.45

(d) The simple gas refrigeration cycle analysis is as follows: h1 = 273.40 kJ/kg h2 = 473.52 kJ/kg

2 T

h3 = 308.63 kJ/kg P3 = 500 kPa ⎫ ⎬ s 3 = 5.2704 kJ/kg T3 = 35°C ⎭

35°C 0°C

P1 = 100 kPa ⎫ ⎬ h4 s = 194.52 kJ/kg.K s 4 = s3 ⎭

ηT =

· QH

2

3 1 · 4 QRefrig 4s s

h3 − h4 308.63 − h4 ⎯ ⎯→ 0.85 = ⎯ ⎯→ h4 = 211.64 kJ/kg h3 − h4 s 308.63 − 194.52

Q& L = m& (h1 − h4 ) = (0.4 kg/s)(273.40 − 211.64)kJ/kg = 24.70 kW W& net,in = m& (h2 − h1 ) − m& (h3 − h4 ) = (0.4 kg/s)[(473.52 − 273.40) − (308.63 − 211.64)kJ/kg ] = 41.25 kW COP =

Q& L W&

net,in

=

24.70 = 0.599 41.25

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

11-82

11-117 A heat pump water heater has a COP of 2.2 and consumes 2 kW when running. It is to be determined if this heat pump can be used to meet the cooling needs of a room by absorbing heat from it. Assumptions The COP of the heat pump remains constant whether heat is absorbed from the outdoor air or room air. Analysis The COP of the heat pump is given to be 2.2. Then the COP of the air-conditioning system becomes COPair-cond = COPheat pump − 1 = 2.2 − 1 = 1.2

Then the rate of cooling (heat absorption from the air) becomes Q& cooling = COPair-cond W& in = (1.2)(2 kW) = 2.4 kW = 8640 kJ / h

since 1 kW = 3600 kJ/h. We conclude that this heat pump can meet the cooling needs of the room since its cooling rate is greater than the rate of heat gain of the room.

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

11-83

11-118 An innovative vapor-compression refrigeration system with a heat exchanger is considered. The system’s COP is to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

2 Condenser

T

Compressor

3

· QH

1

2

3 800 kPa Heat exchanger

· Win

4 4

-10.1°C

6

Throttle valve 5

Evaporator

5

· 6 QL

1

s

Analysis From the refrigerant tables (Tables A-11, A-12, and A-13), P3 = 800 kPa ⎫ ⎬ h3 = h f sat. liquid ⎭

@ 800 kPa

= 95.47 kJ/kg

T4 = Tsat @ 800 kPa − 11.3 ⎫ ⎪ = 31.3 − 11.3 = 20°C ⎬ h4 ≅ h f @ 20°C = 79.32 kJ/kg ⎪ P4 = 800 kPa ⎭ h5 ≅ h4 = 79.32 kJ/kg ( throttling) T6 = −10.1°C ⎫ h6 = h g @ −10.1°C = 244.46 kJ/kg ⎬ sat. vapor ⎭ P6 = Psat @ −10.1°C = 200 kPa

An energy balance on the heat exchanger gives m& (h1 − h6 ) = m& (h3 − h4 ) ⎯ ⎯→ h1 = h3 − h4 + h6 = 95.47 − 79.32 + 244.46 = 260.61 kJ/kg

Then, P1 = 200 kPa ⎫ ⎬ s = 0.9970 kJ/kg ⋅ K h1 = 260.61 kJ/kg ⎭ 1 P2 = 800 kPa ⎫ ⎬ h2 = 292.17 kJ/kg s 2 = s1 ⎭

The COP of this refrigeration system is determined from its definition, COPR =

h − h5 qL 244.46 − 79.32 = 6 = = 5.23 win h2 − h1 292.17 − 260.61

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

11-84

11-119 An innovative vapor-compression refrigeration system with a heat exchanger is considered. The system’s COP is to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

2 Condenser

T

Compressor

· QH

3

1

2

3 800 kPa

· Win

4

Heat exchanger 4

200 kPa

6

Throttle valve 5

Evaporator

5

· 6 QL

1

s

Analysis From the refrigerant tables (Tables A-11, A-12, and A-13), P3 = 800 kPa ⎫ ⎬ h3 = h f sat. liquid ⎭

@ 800 kPa

= 95.47 kJ/kg

T4 = Tsat @ 800 kPa − 21.3 ⎫ ⎪ = 31.3 − 21.3 = 10°C ⎬ h4 ≅ h f @ 10°C = 65.43 kJ/kg ⎪ P4 = 800 kPa ⎭ h5 ≅ h4 = 65.43 kJ/kg ( throttling) T6 = −10.1°C ⎫ h6 = h g @ −10.1°C = 244.46 kJ/kg ⎬ sat. vapor ⎭ P6 = Psat @ −10.1°C = 200 kPa

An energy balance on the heat exchanger gives m& (h1 − h6 ) = m& (h3 − h4 ) ⎯ ⎯→ h1 = h3 − h4 + h6 = 95.47 − 65.43 + 244.46 = 274.50 kJ/kg

Then, P1 = 200 kPa ⎫ ⎬ s = 1.0449 kJ/kg ⋅ K h1 = 274.50 kJ/kg ⎭ 1 P2 = 800 kPa ⎫ ⎬ h2 = 308.28 kJ/kg s 2 = s1 ⎭

The COP of this refrigeration system is determined from its definition, COPR =

h − h5 qL 244.46 − 65.43 = 6 = = 5.30 win h2 − h1 308.28 − 274.50

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11-85

11-120 An ideal gas refrigeration cycle with with three stages of compression with intercooling using air as the working fluid is considered. The COP of this system is to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2a). Analysis From the isentropic relations, ⎛P T2 = T1 ⎜⎜ 2 ⎝ P1

⎞ ⎟⎟ ⎠

( k −1) / k

= (253 K)(5) 0.4 / 1.4 = 400.7 K

⎛P T4 = T6 = T3 ⎜⎜ 4 ⎝ P3 ⎛P T8 = T7 ⎜⎜ 8 ⎝ P7

⎞ ⎟ ⎟ ⎠

⎞ ⎟ ⎟ ⎠

( k −1) / k

( k −1) / k

= (288 K)(5) 0.4 / 1.4 = 456.1 K

15°C

⎛ 1 ⎞ = (288 K)⎜ ⎟ ⎝ 5× 5× 5 ⎠

4 2

7 5

-20°C

3 1

0.4 / 1.4

= 72.5 K

The COP of this ideal gas refrigeration cycle is determined from COPR =

6

T

8 s

qL qL = w net,in wcomp,in − w turb,out

=

h1 − h8 (h2 − h1 ) + ( h4 − h3 ) + ( h6 − h5 ) − ( h7 − h8 )

=

T1 − T8 (T2 − T1 ) + 2(T4 − T3 ) − (T7 − T8 )

=

253 − 72.5 = 0.673 (400.7 − 253) + 2(456.1 − 288) − (288 − 72.5)

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11-86

11-121 An ideal vapor-compression refrigeration cycle with refrigerant-22 as the working fluid is considered. The evaporator is located inside the air handler of building. The hardware and the T-s diagram for this heat pump application are to be sketched. The COP of the unit and the ratio of volume flow rate of air entering the air handler to mass flow rate of R-22 through the air handler are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. From the refrigerant-22 data from the problem statement,

Air T

45°C

· QH

Condenser 3 2 Expansion valve Compressor 4

Evaporator

3

· Win

Win -5°C

1

4s

sat. vap.

-5°C

2

45°C

QL

4

· QL

1

s

T1 = −5°C ⎫ h1 = h g @ −5°C = 248.1 kJ/kg ⎬ sat. vapor ⎭ s1 = s g @ −5°C = 0.9344 kJ/kg ⋅ K P2 = 1728 kPa ⎫ ⎬ h2 = 283.7 kJ/kg s 2 = s1 ⎭ P3 = 1728 kPa ⎫ ⎬ h3 = h f sat. liquid ⎭

@ 1728 kPa

= 101 kJ/kg

h4 ≅ h3 = 101 kJ/kg ( throttling)

(b) The COP of the heat pump is determined from its definition, COPHP =

h − h3 qH 283.7 − 101 = 2 = = 5.13 win h2 − h1 283.7 − 248.1

(c) An energy balance on the condenser gives

V& Q& H = m& R (h2 − h3 ) = m& a c p ΔT = a c p ΔT va

Rearranging, we obtain the ratio of volume flow rate of air entering the air handler to mass flow rate of R22 through the air handler

V&a m& R

=

h 2 − h3 (283.7 − 101) kJ/kg = (1 / v )c p ΔT (1 / 0.847 m 3 /kg )(1.005 kJ/kg ⋅ K))(20 K )

= 7.699 (m 3 air/s)/(kg R22/s) = 462 (m 3 air/min)/(kg R22/s)

Note that the specific volume of air is obtained from ideal gas equation taking the pressure of air to be 101 kPa and using the room temperature of air (25°C = 298 K) to be 0.847 m3/kg.

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11-87

11-122 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. Cooling water flows through the water jacket surrounding the condenser. To produce ice, potable water is supplied to the chiller section of the refrigeration cycle. The hardware and the T-s diagram for this refrigerant-ice making system are to be sketched. The mass flow rates of the refrigerant and the potable water are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. From the refrigerant-134a data from the problem statement,

Water 200 kg/s T 1.2 MPa

· QH

Condenser 3 2 Expansion valve Compressor 4

3

140 kPa 4s

4

sat. vap.

140 kPa

· Win

Win

1

Evaporator

2

1.2 MPa

· QL

1

s

Potable water T1 = 140 kPa ⎫ h1 = h g @ 140 kPa = 239.16 kJ/kg ⎬ s =s sat. vapor g @ 140 kPa = 0.94456 kJ/kg ⋅ K ⎭ 1 P2 = 1200 kPa ⎫ ⎬ h2 = 284.07 kJ/kg s 2 = s1 ⎭ P3 = 1200 kPa ⎫ ⎬ h3 = h f sat. liquid ⎭

@ 1200 kPa

= 117.77 kJ/kg

h4 ≅ h3 = 117.77 kJ/kg ( throttling)

(b) An energy balance on the condenser gives Q& H = m& R (h2 − h3 ) = m& w c p ΔT

Solving for the mass flow rate of the refrigerant m& R =

m& w c p ΔT h2 − h3

=

(200 kg/s)(4.18 kJ/kg ⋅ K))(10 K ) = 50.3 kg/s (284.07 − 117.77)kJ/kg

(c) An energy balance on the evaporator gives Q& L = m& R (h1 − h4 ) = m& w hif Solving for the mass flow rate of the potable water m& w =

m& R (h1 − h4 ) (50.3 kg/s)(239.16 − 117.77)kJ/kg = = 18.3 kg/s hif 333 kJ/kg

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11-88

11-123 A vortex tube receives compressed air at 500 kPa and 300 K, and supplies 25 percent of it as cold air and the rest as hot air. The COP of the vortex tube is to be compared to that of a reversed Brayton cycle for the same pressure ratio; the exit temperature of the hot fluid stream and the COP are to be determined; and it is to be shown if this process violates the second law. Assumptions 1 The vortex tube is adiabatic. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Steady operating conditions exist. Properties The gas constant of air is 0.287 kJ/kg.K (Table A-1). The specific heat of air at room temperature is cp = 1.005 kJ/kg.K (Table A-2). The enthalpy of air at absolute temperature T can be expressed in terms of specific heats as h = cpT. Analysis (a) The COP of the vortex tube is much lower than the COP of a reversed Brayton cycle of the same pressure ratio since the vortex tube involves vortices, which are highly irreversible. Owing to this irreversibility, the minimum temperature that can be obtained by the vortex tube is not as low as the one that can be obtained by the revered Brayton cycle.

(b) We take the vortex tube as the system. This is a steady flow system with one inlet and two exits, and it involves no heat or work interactions. Then the steady-flow energy balance equation for this system E& in = E& out for a unit mass flow rate at the inlet (m& 1 = 1 kg / s) can be expressed as m& 1 h1 = m& 2 h2 + m& 3 h3 m& 1 c p T1 = m& 2 c p T2 + m& 3 c p T3

Compressed air

1c p T1 = 0.25c p T2 + 0.75c p T3

Canceling cp and solving for T3 gives

1

Cold air

T − 0.25T2 T3 = 1 0.75 300 − 0.25 × 278 = = 307.3 K 0.75

2

Warm air

3

Therefore, the hot air stream will leave the vortex tube at an average temperature of 307.3 K. (c) The entropy balance for this steady flow system S&in − S&out + S&gen = 0 can be expressed as with one inlet and two exits, and it involves no heat or work interactions. Then the steady-flow entropy balance equation for this system for a unit mass flow rate at the inlet (m& 1 = 1 kg / s) can be expressed S& gen = S& out − S& in = m& 2 s 2 + m& 3 s 3 − m& 1 s1 = m& 2 s 2 + m& 3 s 3 − (m& 2 + m& 3 ) s1 = m& 2 ( s 2 − s1 ) + m& 3 ( s 3 − s1 ) = 0.25( s 2 − s1 ) + 0.75( s 3 − s1 ) ⎛ ⎛ T P ⎞ T P ⎞ = 0.25⎜⎜ c p ln 2 − R ln 2 ⎟⎟ + 0.75⎜⎜ c p ln 3 − R ln 3 ⎟⎟ T1 P1 ⎠ T1 P1 ⎠ ⎝ ⎝

Substituting the known quantities, the rate of entropy generation is determined to be 278 K 100 kPa ⎞ ⎛ S& gen = 0.25⎜ (1.005 kJ/kg.K) ln − (0.287 kJ/kg.K) ln ⎟ 300 K 500 kPa ⎠ ⎝ 307.3 K 100 kPa ⎞ ⎛ + 0.75⎜ (1.005 kJ/kg.K) ln − (0.287 kJ/kg.K) ln ⎟ 300 K 500 kPa ⎠ ⎝ = 0.461 kW/K > 0

which is a positive quantity. Therefore, this process satisfies the 2nd law of thermodynamics.

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

11-89 (d) For a unit mass flow rate at the inlet (m& 1 = 1 kg / s) , the cooling rate and the power input to the compressor are determined to Q& cooling = m& c (h1 − hc ) = m& c c p (T1 − Tc ) = (0.25 kg/s)(1.005 kJ/kg.K)(300 - 278)K = 5.53 kW W& comp,in = =

m& 0 RT0 (k − 1)η comp

⎡⎛ P ⎢⎜ 1 ⎢⎜⎝ P0 ⎣

⎞ ⎟ ⎟ ⎠

( k −1) / k

⎤ − 1⎥ ⎥ ⎦

(1 kg/s)(0.287 kJ/kg.K)(300 K) ⎡⎛ 500 kPa ⎞ ⎢⎜ ⎟ (1.4 − 1)0.80 ⎢⎣⎝ 100 kPa ⎠

(1.4 −1) / 1.4

⎤ − 1⎥ = 157.1 kW ⎥⎦

Then the COP of the vortex refrigerator becomes COP =

Q& cooling 5.53 kW = = 0.035 & Wcomp, in 157.1 kW

The COP of a Carnot refrigerator operating between the same temperature limits of 300 K and 278 K is COPCarnot =

TL 278 K = = 12.6 TH − TL (300 − 278) K

Discussion Note that the COP of the vortex refrigerator is a small fraction of the COP of a Carnot refrigerator operating between the same temperature limits.

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

11-90

11-124 A vortex tube receives compressed air at 600 kPa and 300 K, and supplies 25 percent of it as cold air and the rest as hot air. The COP of the vortex tube is to be compared to that of a reversed Brayton cycle for the same pressure ratio; the exit temperature of the hot fluid stream and the COP are to be determined; and it is to be shown if this process violates the second law. Assumptions 1 The vortex tube is adiabatic. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Steady operating conditions exist. Properties The gas constant of air is 0.287 kJ/kg.K (Table A-1). The specific heat of air at room temperature is cp = 1.005 kJ/kg.K (Table A-2). The enthalpy of air at absolute temperature T can be expressed in terms of specific heats as h = cp T. Analysis (a) The COP of the vortex tube is much lower than the COP of a reversed Brayton cycle of the same pressure ratio since the vortex tube involves vortices, which are highly irreversible. Owing to this irreversibility, the minimum temperature that can be obtained by the vortex tube is not as low as the one that can be obtained by the revered Brayton cycle.

(b) We take the vortex tube as the system. This is a steady flow system with one inlet and two exits, and it involves no heat or work interactions. Then the steady-flow entropy balance equation for this system E& in = E& out for a unit mass flow rate at the inlet (m& 1 = 1 kg / s) can be expressed as m& 1 h1 = m& 2 h2 + m& 3 h3 m& 1 c p T1 = m& 2 c p T2 + m& 3 c p T3

Compressed air

1c p T1 = 0.25c p T2 + 0.75c p T3

Canceling cp and solving for T3 gives T1 − 0.25T2 0.75 300 − 0.25 × 278 = = 307.3 K 0.75

T3 =

1

Cold air 2

Warm air 3

Therefore, the hot air stream will leave the vortex tube at an average temperature of 307.3 K. (c) The entropy balance for this steady flow system S&in − S&out + S&gen = 0 can be expressed as with one inlet and two exits, and it involves no heat or work interactions. Then the steady-flow energy balance equation for this system for a unit mass flow rate at the inlet (m& 1 = 1 kg / s) can be expressed S& gen = S& out − S& in = m& 2 s 2 + m& 3 s 3 − m& 1 s1 = m& 2 s 2 + m& 3 s 3 − ( m& 2 + m& 3 ) s1 = m& 2 ( s 2 − s1 ) + m& 3 ( s 3 − s1 ) = 0.25( s 2 − s1 ) + 0.75( s 3 − s1 ) ⎛ ⎛ P ⎞ T P ⎞ T = 0.25⎜⎜ c p ln 2 − R ln 2 ⎟⎟ + 0.75⎜⎜ c p ln 3 − R ln 3 ⎟⎟ P1 ⎠ T P T 1 1 ⎠ 1 ⎝ ⎝

Substituting the known quantities, the rate of entropy generation is determined to be 100 kPa ⎞ 278 K ⎛ S&gen = 0.25⎜ (1.005 kJ/kg.K)ln − (0.287 kJ/kg.K)ln ⎟ 600 kPa ⎠ 300 K ⎝ 100 kPa ⎞ 307.3 K ⎛ + 0.75⎜ (1.005 kJ/kg.K)ln − (0.287 kJ/kg.K)ln ⎟ 600 kPa ⎠ 300 K ⎝ = 0.513 kW/K > 0

which is a positive quantity. Therefore, this process satisfies the 2nd law of thermodynamics.

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

11-91 (d) For a unit mass flow rate at the inlet (m& 1 = 1 kg / s) , the cooling rate and the power input to the compressor are determined to Q& cooling = m& c (h1 − hc ) = m& c c p (T1 − Tc ) = (0.25 kg/s)(1.005 kJ/kg.K)(300 - 278)K = 5.53 kW W& comp,in = =

m& 0 RT0 (k − 1)η comp

⎡⎛ P ⎢⎜ 1 ⎢⎜⎝ P0 ⎣

⎞ ⎟ ⎟ ⎠

( k −1) / k

⎤ − 1⎥ ⎥ ⎦

(1 kg/s)(0.287 kJ/kg.K)(300 K) ⎡⎛ 600 kPa ⎞ ⎢⎜ ⎟ (1.4 − 1)0.80 ⎢⎣⎝ 100 kPa ⎠

(1.4 −1) / 1.4

⎤ − 1⎥ = 179.9 kW ⎥⎦

Then the COP of the vortex refrigerator becomes COP =

Q& cooling 5.53 kW = = 0.031 & Wcomp, in 179.9 kW

The COP of a Carnot refrigerator operating between the same temperature limits of 300 K and 278 K is COPCarnot =

TL 278 K = = 12.6 TH − TL (300 − 278) K

Discussion Note that the COP of the vortex refrigerator is a small fraction of the COP of a Carnot refrigerator operating between the same temperature limits.

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11-92

11-125 EES The effect of the evaporator pressure on the COP of an ideal vapor-compression refrigeration cycle with R-134a as the working fluid is to be investigated. Analysis The problem is solved using EES, and the solution is given below.

COP 1.851 2.863 4.014 5.462 7.424

ηc 0.7 0.7 0.7 0.7 0.7

P1 [kPa] 100 200 300 400 500

COP

"Input Data" P[1]=100 [kPa] P[2] = 1000 [kPa] Fluid$='R134a' Eta_c=0.7 "Compressor isentropic efficiency" "Compressor" h[1]=enthalpy(Fluid$,P=P[1],x=1) "properties for state 1" s[1]=entropy(Fluid$,P=P[1],x=1) T[1]=temperature(Fluid$,h=h[1],P=P[1]) h2s=enthalpy(Fluid$,P=P[2],s=s[1]) "Identifies state 2s as isentropic" h[1]+Wcs=h2s "energy balance on isentropic compressor" W_c=Wcs/Eta_c"definition of compressor isentropic efficiency" h[1]+W_c=h[2] "energy balance on real compressor-assumed adiabatic" s[2]=entropy(Fluid$,h=h[2],P=P[2]) "properties for state 2" T[2]=temperature(Fluid$,h=h[2],P=P[2]) "Condenser" P[3] = P[2] h[3]=enthalpy(Fluid$,P=P[3],x=0) "properties for state 3" s[3]=entropy(Fluid$,P=P[3],x=0) h[2]=Qout+h[3] "energy balance on condenser" "Throttle Valve" h[4]=h[3] "energy balance on throttle - isenthalpic" x[4]=quality(Fluid$,h=h[4],P=P[4]) "properties for state 4" s[4]=entropy(Fluid$,h=h[4],P=P[4]) T[4]=temperature(Fluid$,h=h[4],P=P[4]) "Evaporator" P[4]= P[1] Q_in + h[4]=h[1] "energy balance on evaporator" "Coefficient of Performance:" COP=Q_in/W_c "definition of COP"

10

η comp

8

1.0 0.7

6 4 2

0 100 150 200 250 300 350 400 450 500

P[1] [kPa]

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

11-93

11-126 EES The effect of the condenser pressure on the COP of an ideal vapor-compression refrigeration cycle with R-134a as the working fluid is to be investigated. Analysis The problem is solved using EES, and the solution is given below. "Input Data" P[1]=120 [kPa] P[2] = 400 [kPa] Fluid$='R134a' Eta_c=0.7 "Compressor isentropic efficiency" "Compressor" h[1]=enthalpy(Fluid$,P=P[1],x=1) "properties for state 1" s[1]=entropy(Fluid$,P=P[1],x=1) T[1]=temperature(Fluid$,h=h[1],P=P[1]) h2s=enthalpy(Fluid$,P=P[2],s=s[1]) "Identifies state 2s as isentropic" h[1]+Wcs=h2s "energy balance on isentropic compressor" W_c=Wcs/Eta_c"definition of compressor isentropic efficiency" h[1]+W_c=h[2] "energy balance on real compressor-assumed adiabatic" s[2]=entropy(Fluid$,h=h[2],P=P[2]) "properties for state 2" T[2]=temperature(Fluid$,h=h[2],P=P[2]) "Condenser" P[3] = P[2] h[3]=enthalpy(Fluid$,P=P[3],x=0) "properties for state 3" s[3]=entropy(Fluid$,P=P[3],x=0) h[2]=Qout+h[3] "energy balance on condenser" "Throttle Valve" h[4]=h[3] "energy balance on throttle - isenthalpic" x[4]=quality(Fluid$,h=h[4],P=P[4]) "properties for state 4" s[4]=entropy(Fluid$,h=h[4],P=P[4]) T[4]=temperature(Fluid$,h=h[4],P=P[4]) "Evaporator" P[4]= P[1] Q_in + h[4]=h[1] "energy balance on evaporator" "Coefficient of Performance:" COP=Q_in/W_c "definition of COP"

8 ηc 0.7 0.7 0.7 0.7 0.7

P2 [kPa] 400 650 900 1150 1400

7

η comp

6

1.0 0.7

5

COP

COP 4.935 3.04 2.258 1.803 1.492

4 3 2 1 0 400

600

800

1000

1200

1400

P[2] [kPa]

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11-94

Fundamentals of Engineering (FE) Exam Problems

11-127 Consider a heat pump that operates on the reversed Carnot cycle with R-134a as the working fluid executed under the saturation dome between the pressure limits of 140 kPa and 800 kPa. R-134a changes from saturated vapor to saturated liquid during the heat rejection process. The net work input for this cycle is (a) 28 kJ/kg

(b) 34 kJ/kg

(c) 49 kJ/kg

(d) 144 kJ/kg

(e) 275 kJ/kg

Answer (a) 28 kJ/kg

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=800 "kPa" P2=140 "kPa" h_fg=ENTHALPY(R134a,x=1,P=P1)-ENTHALPY(R134a,x=0,P=P1) TH=TEMPERATURE(R134a,x=0,P=P1)+273 TL=TEMPERATURE(R134a,x=0,P=P2)+273 q_H=h_fg COP=TH/(TH-TL) w_net=q_H/COP "Some Wrong Solutions with Common Mistakes:" W1_work = q_H/COP1; COP1=TL/(TH-TL) "Using COP of regrigerator" W2_work = q_H/COP2; COP2=(TH-273)/(TH-TL) "Using C instead of K" W3_work = h_fg3/COP; h_fg3= ENTHALPY(R134a,x=1,P=P2)-ENTHALPY(R134a,x=0,P=P2) "Using h_fg at P2" W4_work = q_H*TL/TH "Using the wrong relation"

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

11-95

11-128 A refrigerator removes heat from a refrigerated space at –5°C at a rate of 0.35 kJ/s and rejects it to an environment at 20°C. The minimum required power input is (a) 30 W

(b) 33 W

(c) 56 W

(d) 124 W

(e) 350 W

Answer (b) 33 W

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). TH=20+273 TL=-5+273 Q_L=0.35 "kJ/s" COP_max=TL/(TH-TL) w_min=Q_L/COP_max "Some Wrong Solutions with Common Mistakes:" W1_work = Q_L/COP1; COP1=TH/(TH-TL) "Using COP of heat pump" W2_work = Q_L/COP2; COP2=(TH-273)/(TH-TL) "Using C instead of K" W3_work = Q_L*TL/TH "Using the wrong relation" W4_work = Q_L "Taking the rate of refrigeration as power input"

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

11-96

11-129 A refrigerator operates on the ideal vapor compression refrigeration cycle with R-134a as the working fluid between the pressure limits of 120 kPa and 800 kPa. If the rate of heat removal from the refrigerated space is 32 kJ/s, the mass flow rate of the refrigerant is (a) 0.19 kg/s

(b) 0.15 kg/s

(c) 0.23 kg/s

(d) 0.28 kg/s

(e) 0.81 kg/s

Answer (c) 0.23 kg/s

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=120 "kPa" P2=800 "kPa" P3=P2 P4=P1 s2=s1 Q_refrig=32 "kJ/s" m=Q_refrig/(h1-h4) h1=ENTHALPY(R134a,x=1,P=P1) s1=ENTROPY(R134a,x=1,P=P1) h2=ENTHALPY(R134a,s=s2,P=P2) h3=ENTHALPY(R134a,x=0,P=P3) h4=h3 "Some Wrong Solutions with Common Mistakes:" W1_mass = Q_refrig/(h2-h1) "Using wrong enthalpies, for W_in" W2_mass = Q_refrig/(h2-h3) "Using wrong enthalpies, for Q_H" W3_mass = Q_refrig/(h1-h44); h44=ENTHALPY(R134a,x=0,P=P4) "Using wrong enthalpy h4 (at P4)" W4_mass = Q_refrig/h_fg; h_fg=ENTHALPY(R134a,x=1,P=P2) - ENTHALPY(R134a,x=0,P=P2) "Using h_fg at P2"

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

11-97

11-130 A heat pump operates on the ideal vapor compression refrigeration cycle with R-134a as the working fluid between the pressure limits of 0.32 MPa and 1.2 MPa. If the mass flow rate of the refrigerant is 0.193 kg/s, the rate of heat supply by the heat pump to the heated space is (a) 3.3 kW

(b) 23 kW

(c) 26 kW

(d) 31 kW

(e) 45 kW

Answer (d) 31 kW

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=320 "kPa" P2=1200 "kPa" P3=P2 P4=P1 s2=s1 m=0.193 "kg/s" Q_supply=m*(h2-h3) "kJ/s" h1=ENTHALPY(R134a,x=1,P=P1) s1=ENTROPY(R134a,x=1,P=P1) h2=ENTHALPY(R134a,s=s2,P=P2) h3=ENTHALPY(R134a,x=0,P=P3) h4=h3 "Some Wrong Solutions with Common Mistakes:" W1_Qh = m*(h2-h1) "Using wrong enthalpies, for W_in" W2_Qh = m*(h1-h4) "Using wrong enthalpies, for Q_L" W3_Qh = m*(h22-h4); h22=ENTHALPY(R134a,x=1,P=P2) "Using wrong enthalpy h2 (hg at P2)" W4_Qh = m*h_fg; h_fg=ENTHALPY(R134a,x=1,P=P1) - ENTHALPY(R134a,x=0,P=P1) "Using h_fg at P1"

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

11-98

11-131 An ideal vapor compression refrigeration cycle with R-134a as the working fluid operates between the pressure limits of 120 kPa and 1000 kPa. The mass fraction of the refrigerant that is in the liquid phase at the inlet of the evaporator is (a) 0.65

(b) 0.60

(c) 0.40

(d) 0.55

(e) 0.35

Answer (b) 0.60

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=120 "kPa" P2=1000 "kPa" P3=P2 P4=P1 h1=ENTHALPY(R134a,x=1,P=P1) h3=ENTHALPY(R134a,x=0,P=P3) h4=h3 x4=QUALITY(R134a,h=h4,P=P4) liquid=1-x4 "Some Wrong Solutions with Common Mistakes:" W1_liquid = x4 "Taking quality as liquid content" W2_liquid = 0 "Assuming superheated vapor" W3_liquid = 1-x4s; x4s=QUALITY(R134a,s=s3,P=P4) "Assuming isentropic expansion" s3=ENTROPY(R134a,x=0,P=P3)

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

11-99

11-132 Consider a heat pump that operates on the ideal vapor compression refrigeration cycle with R-134a as the working fluid between the pressure limits of 0.32 MPa and 1.2 MPa. The coefficient of performance of this heat pump is (a) 0.17

(b) 1.2

(c) 3.1

(d) 4.9

(e) 5.9

Answer (e) 5.9

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=320 "kPa" P2=1200 "kPa" P3=P2 P4=P1 s2=s1 h1=ENTHALPY(R134a,x=1,P=P1) s1=ENTROPY(R134a,x=1,P=P1) h2=ENTHALPY(R134a,s=s2,P=P2) h3=ENTHALPY(R134a,x=0,P=P3) h4=h3 COP_HP=qH/Win Win=h2-h1 qH=h2-h3 "Some Wrong Solutions with Common Mistakes:" W1_COP = (h1-h4)/(h2-h1) "COP of refrigerator" W2_COP = (h1-h4)/(h2-h3) "Using wrong enthalpies, QL/QH" W3_COP = (h22-h3)/(h22-h1); h22=ENTHALPY(R134a,x=1,P=P2) "Using wrong enthalpy h2 (hg at P2)"

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

11-100

11-133 An ideal gas refrigeration cycle using air as the working fluid operates between the pressure limits of 80 kPa and 280 kPa. Air is cooled to 35°C before entering the turbine. The lowest temperature of this cycle is (a) –58°C

(b) -26°C

(c) 0°C

(d) 11°C

(e) 24°C

Answer (a) –58°C

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). k=1.4 P1= 80 "kPa" P2=280 "kPa" T3=35+273 "K" "Mimimum temperature is the turbine exit temperature" T4=T3*(P1/P2)^((k-1)/k) - 273 "Some Wrong Solutions with Common Mistakes:" W1_Tmin = (T3-273)*(P1/P2)^((k-1)/k) "Using C instead of K" W2_Tmin = T3*(P1/P2)^((k-1)) - 273 "Using wrong exponent" W3_Tmin = T3*(P1/P2)^k - 273 "Using wrong exponent"

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11-101

11-134 Consider an ideal gas refrigeration cycle using helium as the working fluid. Helium enters the compressor at 100 kPa and –10°C and is compressed to 250 kPa. Helium is then cooled to 20°C before it enters the turbine. For a mass flow rate of 0.2 kg/s, the net power input required is (a) 9.3 kW

(b) 27.6 kW

(c) 48.8 kW

(d) 93.5 kW

(e) 119 kW

Answer (b) 27.6 kW

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). k=1.667 Cp=5.1926 "kJ/kg.K" P1= 100 "kPa" T1=-10+273 "K" P2=250 "kPa" T3=20+273 "K" m=0.2 "kg/s" "Mimimum temperature is the turbine exit temperature" T2=T1*(P2/P1)^((k-1)/k) T4=T3*(P1/P2)^((k-1)/k) W_netin=m*Cp*((T2-T1)-(T3-T4)) "Some Wrong Solutions with Common Mistakes:" W1_Win = m*Cp*((T22-T1)-(T3-T44)); T22=T1*P2/P1; T44=T3*P1/P2 "Using wrong relations for temps" W2_Win = m*Cp*(T2-T1) "Ignoring turbine work" W3_Win=m*1.005*((T2B-T1)-(T3-T4B)); T2B=T1*(P2/P1)^((kB-1)/kB); T4B=T3*(P1/P2)^((kB1)/kB); kB=1.4 "Using air properties" W4_Win=m*Cp*((T2A-(T1-273))-(T3-273-T4A)); T2A=(T1-273)*(P2/P1)^((k-1)/k); T4A=(T3273)*(P1/P2)^((k-1)/k) "Using C instead of K"

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

11-102

11-135 An absorption air-conditioning system is to remove heat from the conditioned space at 20°C at a rate of 150 kJ/s while operating in an environment at 35°C. Heat is to be supplied from a geothermal source at 140°C. The minimum rate of heat supply required is (a) 86 kJ/s

(b) 21 kJ/s

(c) 30 kJ/s

(d) 61 kJ/s

(e) 150 kJ/s

Answer (c) 30 kJ/s

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). TL=20+273 "K" Q_refrig=150 "kJ/s" To=35+273 "K" Ts=140+273 "K" COP_max=(1-To/Ts)*(TL/(To-TL)) Q_in=Q_refrig/COP_max "Some Wrong Solutions with Common Mistakes:" W1_Qin = Q_refrig "Taking COP = 1" W2_Qin = Q_refrig/COP2; COP2=TL/(Ts-TL) "Wrong COP expression" W3_Qin = Q_refrig/COP3; COP3=(1-To/Ts)*(Ts/(To-TL)) "Wrong COP expression, COP_HP" W4_Qin = Q_refrig*COP_max "Multiplying by COP instead of dividing"

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

11-103

11-136 Consider a refrigerator that operates on the vapor compression refrigeration cycle with R-134a as the working fluid. The refrigerant enters the compressor as saturated vapor at 160 kPa, and exits at 800 kPa and 50°C, and leaves the condenser as saturated liquid at 800 kPa. The coefficient of performance of this refrigerator is (a) 2.6

(b) 1.0

(c) 4.2

(d) 3.2

(e) 4.4

Answer (d) 3.2

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=160 "kPa" P2=800 "kPa" T2=50 "C" P3=P2 P4=P1 h1=ENTHALPY(R134a,x=1,P=P1) s1=ENTROPY(R134a,x=1,P=P1) h2=ENTHALPY(R134a,T=T2,P=P2) h3=ENTHALPY(R134a,x=0,P=P3) h4=h3 COP_R=qL/Win Win=h2-h1 qL=h1-h4 "Some Wrong Solutions with Common Mistakes:" W1_COP = (h2-h3)/(h2-h1) "COP of heat pump" W2_COP = (h1-h4)/(h2-h3) "Using wrong enthalpies, QL/QH" W3_COP = (h1-h4)/(h2s-h1); h2s=ENTHALPY(R134a,s=s1,P=P2) "Assuming isentropic compression"

11-137 ··· 11-145 Design and Essay Problems

KJ

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

12-1

Chapter 12 THERMODYNAMIC PROPERTY RELATIONS Partial Derivatives and Associated Relations

12-1C z

∂x ≡ dx ∂y ≡ dy dz = (∂z ) x + (∂z ) y

dz (∂z)y (∂z)x

y + dy

y

y

dy

x dx x+dx

x 12-2C For functions that depend on one variable, they are identical. For functions that depend on two or more variable, the partial differential represents the change in the function with one of the variables as the other variables are held constant. The ordinary differential for such functions represents the total change as a result of differential changes in all variables. 12-3C (a) (∂x)y = dx ; (b) (∂z) y ≤ dz; and (c) dz = (∂z)x + (∂z) y 12-4C Yes. 12-5C Yes.

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12-2

12-6 Air at a specified temperature and specific volume is considered. The changes in pressure corresponding to a certain increase of different properties are to be determined. Assumptions Air is an ideal gas. Properties The gas constant of air is R = 0.287 kPa·m3/kg·K (Table A-1). Analysis An ideal gas equation can be expressed as P = RT/v. Noting that R is a constant and P = P(T, v), R dT RT dv ⎛ ∂P ⎞ ⎛ ∂P ⎞ dP = ⎜ − ⎟ dT + ⎜ ⎟ dv = ∂ T ∂ v v v2 ⎝ ⎠v ⎝ ⎠T

(a) The change in T can be expressed as dT ≅ ΔT = 400 × 0.01 = 4.0 K. At v = constant,

(dP )v

=

R dT

v

=

(0.287 kPa ⋅ m 3 /kg ⋅ K)(4.0 K) 0.90 m 3 /kg

= 1.276 kPa

(b) The change in v can be expressed as dv ≅ Δv = 0.90 × 0.01 = 0.009 m3/kg. At T = constant,

(dP )T

=−

RT dv

v2

=−

(0.287 kPa ⋅ m 3 /kg ⋅ K)(400K)(0.009 m 3 /kg) (0.90 m 3 /kg) 2

= −1.276 kPa

(c) When both v and T increases by 1%, the change in P becomes dP = (dP)v + (dP )T = 1.276 + (−1.276) = 0

Thus the changes in T and v balance each other.

12-7 Helium at a specified temperature and specific volume is considered. The changes in pressure corresponding to a certain increase of different properties are to be determined. Assumptions Helium is an ideal gas Properties The gas constant of helium is R = 2.0769 kPa·m3/kg·K (Table A-1). Analysis An ideal gas equation can be expressed as P = RT/v. Noting that R is a constant and P = P(T, v ), R dT RT dv ⎛ ∂P ⎞ ⎛ ∂P ⎞ dP = ⎜ − ⎟ dT + ⎜ ⎟ dv = v ⎝ ∂T ⎠v v2 ⎝ ∂v ⎠ T

(a) The change in T can be expressed as dT ≅ ΔT = 400 × 0.01 = 4.0 K. At v = constant,

(dP )v

=

R dT

v

=

(2.0769 kPa ⋅ m 3 /kg ⋅ K)(4.0 K) 0.90 m 3 /kg

= 9.231 kPa

(b) The change in v can be expressed as d v ≅ Δ v = 0.90 × 0.01 = 0.009 m3/kg. At T = constant,

(dP )T

=−

RT dv

v2

=

(2.0769 kPa ⋅ m 3 /kg ⋅ K)(400 K)(0.009 m 3 ) (0.90 m 3 /kg) 2

= −9.231 kPa

(c) When both v and T increases by 1%, the change in P becomes dP = (dP) v + (dP ) T = 9.231 + (−9.231) = 0

Thus the changes in T and v balance each other.

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

12-3

12-8 It is to be proven for an ideal gas that the P = constant lines on a T- v diagram are straight lines and that the high pressure lines are steeper than the low-pressure lines. Analysis (a) For an ideal gas Pv = RT or T = Pv/R. Taking the partial derivative of T with respect to v holding P constant yields P ⎛ ∂T ⎞ ⎟ = ⎜ v ∂ ⎠P R ⎝

T

which remains constant at P = constant. Thus the derivative (∂T/∂v)P, which represents the slope of the P = const. lines on a T-v diagram, remains constant. That is, the P = const. lines are straight lines on a T-v diagram. (b) The slope of the P = const. lines on a T-v diagram is equal to P/R, which is proportional to P. Therefore, the high pressure lines are steeper than low pressure lines on the T-v diagram.

P = const

v

12-9 A relation is to be derived for the slope of the v = constant lines on a T-P diagram for a gas that obeys the van der Waals equation of state. Analysis The van der Waals equation of state can be expressed as

T=

1⎛ a ⎜P + 2 R⎝ v

⎞ ⎟(v − b ) ⎠

Taking the derivative of T with respect to P holding v constant, 1 v −b ⎛ ∂T ⎞ ⎟ = (1 + 0)(v − b ) = ⎜ R ⎝ ∂P ⎠v R

which is the slope of the v = constant lines on a T-P diagram.

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12-4

12-10 Nitrogen gas at a specified state is considered. The cp and cv of the nitrogen are to be determined using Table A-18, and to be compared to the values listed in Table A-2b. Analysis The cp and cv of ideal gases depends on temperature only, and are expressed as cp(T) = dh(T)/dT and cv(T) = du(T)/dT. Approximating the differentials as differences about 400 K, the cp and cv values are determined to be ⎛ dh(T ) ⎞ ⎛ Δh(T ) ⎞ ≅⎜ c p (400 K ) = ⎜ ⎟ ⎟ dT ⎝ ⎠ T = 400 K ⎝ ΔT ⎠ T ≅ 400 K =

h(410 K ) − h(390 K ) (410 − 390)K

=

(11,932 − 11,347)/28.0 kJ/kg (410 − 390)K

h

cp

= 1.045 kJ/kg ⋅ K

(Compare: Table A-2b at 400 K → cp = 1.044 kJ/kg·K)

T

⎛ du (T ) ⎞ ⎛ Δu (T ) ⎞ ≅⎜ cv (400K ) = ⎜ ⎟ ⎟ ⎝ dT ⎠ T = 400 K ⎝ ΔT ⎠ T ≅ 400 K =

u (410 K ) − u (390 K ) (410 − 390)K

=

(8,523 − 8,104)/28.0 kJ/kg = 0.748 kJ/kg ⋅ K (410 − 390)K

(Compare: Table A-2b at 400 K → cv = 0.747 kJ/kg·K)

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12-5

12-11E Nitrogen gas at a specified state is considered. The cp and cv of the nitrogen are to be determined using Table A-18E, and to be compared to the values listed in Table A-2Eb. Analysis The cp and cv of ideal gases depends on temperature only, and are expressed as cp(T) = dh(T)/dT and cv(T) = du(T)/dT. Approximating the differentials as differences about 600 R, the cp and cv values are determined to be ⎛ dh(T ) ⎞ ⎛ Δh(T ) ⎞ ≅⎜ c p (600 R ) = ⎜ ⎟ ⎟ dT ⎝ ⎠T = 600 R ⎝ ΔT ⎠T ≅ 600 R =

h(620 R ) − h(580 R ) (620 − 580)R

=

(4,307.1 − 4,028.7)/28.0 Btu/lbm = 0.249 Btu/lbm ⋅ R (620 − 580)R

(Compare: Table A-2Eb at 600 R → cp = 0.248 Btu/lbm·R ) ⎛ du (T ) ⎞ ⎛ Δu (T ) ⎞ ≅⎜ cv (600 R ) = ⎜ ⎟ ⎟ ⎝ dT ⎠T = 600 R ⎝ ΔT ⎠T ≅ 600 R =

u (620 R ) − u (580 R ) (620 − 580)R

=

(3,075.9 − 2,876.9)/28.0 Btu/lbm = 0.178 Btu/lbm ⋅ R (620 − 580) R

(Compare: Table A-2Eb at 600 R → cv = 0.178 Btu/lbm·R)

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12-6

12-12 The state of an ideal gas is altered slightly. The change in the specific volume of the gas is to be determined using differential relations and the ideal-gas relation at each state. Assumptions The gas is air and air is an ideal gas. Properties The gas constant of air is R = 0.287 kPa·m3/kg·K (Table A-1). Analysis (a) The changes in T and P can be expressed as dT ≅ ΔT = (404 − 400)K = 4 K dP ≅ ΔP = (96 − 100)kPa = −4 kPa

The ideal gas relation Pv = RT can be expressed as v = RT/P. Note that R is a constant and v = v (T, P). Applying the total differential relation and using average values for T and P, R dT RT dP ⎛ ∂v ⎞ ⎛ ∂v ⎞ − dv = ⎜ ⎟ dP = ⎟ dT + ⎜ P T P ∂ ∂ ⎠T P2 ⎝ ⎝ ⎠P ⎛ 4K (402 K)(−4 kPa) ⎞⎟ = (0.287 kPa ⋅ m 3 /kg ⋅ K)⎜ − ⎜ 98 kPa ⎟ (98 kPa) 2 ⎝ ⎠ = (0.0117 m 3 /kg) + (0.04805 m 3 /kg) = 0.0598 m 3 /kg

(b) Using the ideal gas relation at each state,

v1 =

RT1 (0.287 kPa ⋅ m 3 /kg ⋅ K)(400 K) = = 1.1480 m 3 /kg P1 100 kPa

v2 =

RT2 (0.287 kPa ⋅ m 3 /kg ⋅ K)(404 K) = = 1.2078 m 3 /kg P2 96 kPa

Thus, Δv = v 2 − v1 = 1.2078 − 1.1480 = 0.0598 m3 /kg

The two results are identical.

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12-7

12-13 Using the equation of state P(v-a) = RT, the cyclic relation, and the reciprocity relation at constant v are to be verified. Analysis (a) This equation of state involves three variables P, v, and T. Any two of these can be taken as the independent variables, with the remaining one being the dependent variable. Replacing x, y, and z by P, v, and T, the cyclic relation can be expressed as ⎛ ∂P ⎞ ⎛ ∂v ⎞ ⎛ ∂T ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = −1 ⎝ ∂v ⎠T ⎝ ∂T ⎠ P ⎝ ∂P ⎠v

where − RT RT P ⎛ ∂P ⎞ ⎯ ⎯→ ⎜ =− ⎟ = v −a v −a ⎝ ∂v ⎠ T (v − a )2 RT R ⎛ ∂v ⎞ +a ⎯ ⎯→ ⎜ v= ⎟ = ∂ P T ⎝ ⎠P P P (v − a) v −a ⎛ ∂T ⎞ ⎯ ⎯→ ⎜ T= ⎟ = ∂ R P R ⎝ ⎠v P=

Substituting, P ⎞⎛ R ⎞⎛ v − a ⎞ ⎛ ∂P ⎞ ⎛ ∂v ⎞ ⎛ ∂T ⎞ ⎛ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = ⎜− ⎟⎜ ⎟⎜ ⎟ = −1 ⎝ ∂v ⎠T ⎝ ∂T ⎠ P ⎝ ∂P ⎠v ⎝ v − a ⎠⎝ P ⎠⎝ R ⎠

which is the desired result. (b) The reciprocity rule for this gas at v = constant can be expressed as 1 ⎛ ∂P ⎞ ⎜ ⎟ = ⎝ ∂T ⎠v (∂T / ∂P)v P (v − a) v −a ⎛ ∂T ⎞ ⎯⎯→ ⎜ T= ⎟ = R R ⎝ ∂P ⎠v ∂ RT P R ⎛ ⎞ ⎯⎯→ ⎜ P= ⎟ = v −a ⎝ ∂T ⎠v v − a

We observe that the first differential is the inverse of the second one. Thus the proof is complete.

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12-8

The Maxwell Relations

12-14 The validity of the last Maxwell relation for refrigerant-134a at a specified state is to be verified. Analysis We do not have exact analytical property relations for refrigerant-134a, and thus we need to replace the differential quantities in the last Maxwell relation with the corresponding finite quantities. Using property values from the tables about the specified state, ⎛ ∂s ⎞ ? ⎛ ∂v ⎞ ⎜ ⎟ =− ⎜ ⎟ ⎝ ∂P ⎠ T ⎝ ∂T ⎠ P ? ⎛ Δs ⎞ ⎛ Δv ⎞ ≅−⎜ ⎜ ⎟ ⎟ P Δ ⎝ ⎠ T =80°C ⎝ ΔT ⎠ P =1200 kPa

⎛ s1400 kPa − s1000 kPa ⎜ ⎜ (1400 − 1000 )kPa ⎝

? ⎛ v 100°C − v 60° C ⎞ ⎟ ≅ −⎜⎜ ⎟ ⎠ T =80°C ⎝ (100 − 60 )°C

⎞ ⎟ ⎟ ⎠ P =1200kPa

(1.0056 − 1.0458)kJ/kg ⋅ K ? (0.022442 − 0.018404)m 3 /kg ≅− (1400 − 1000)kPa (100 − 60)°C − 1.005 × 10 − 4 m 3 /kg ⋅ K ≅ −1.0095 × 10 − 4 m 3 /kg ⋅ K

since kJ ≡ kPa·m³, and K ≡ °C for temperature differences. Thus the last Maxwell relation is satisfied.

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12-9

12-15 EES Problem 12-14 is reconsidered. The validity of the last Maxwell relation for refrigerant 134a at the specified state is to be verified. Analysis The problem is solved using EES, and the solution is given below. "Input Data:" T=80 [C] P=1200 [kPa] P_increment = 200 [kPa] T_increment = 20 [C] P[2]=P+P_increment P[1]=P-P_increment T[2]=T+T_increment T[1]=T-T_increment DELTAP = P[2]-P[1] DELTAT = T[2]-T[1] v[1]=volume(R134a,T=T[1],P=P) v[2]=volume(R134a,T=T[2],P=P) s[1]=entropy(R134a,T=T,P=P[1]) s[2]=entropy(R134a,T=T,P=P[2]) DELTAs=s[2] - s[1] DELTAv=v[2] - v[1] "The partial derivatives in the last Maxwell relation (Eq. 11-19) is associated with the Gibbs function and are approximated by the ratio of ordinary differentials:" LeftSide =DELTAs/DELTAP*Convert(kJ,m^3-kPa) "[m^3/kg-K]" "at T = Const." RightSide=-DELTAv/DELTAT "[m^3/kg-K]" "at P = Const." SOLUTION DELTAP=400 [kPa] DELTAs=-0.04026 [kJ/kg-K] DELTAT=40 [C] DELTAv=0.004038 [m^3/kg] LeftSide=-0.0001007 [m^3/kg-K] P=1200 [kPa] P[1]=1000 [kPa] P[2]=1400 [kPa] P_increment=200 [kPa]

RightSide=-0.000101 [m^3/kg-K] s[1]=1.046 [kJ/kg-K] s[2]=1.006 [kJ/kg-K] T=80 [C] T[1]=60 [C] T[2]=100 [C] T_increment=20 [C] v[1]=0.0184 [m^3/kg] v[2]=0.02244 [m^3/kg]

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12-10

12-16E The validity of the last Maxwell relation for steam at a specified state is to be verified. Analysis We do not have exact analytical property relations for steam, and thus we need to replace the differential quantities in the last Maxwell relation with the corresponding finite quantities. Using property values from the tables about the specified state, ⎛ ∂s ⎞ ? ⎛ ∂v ⎞ ⎟ ⎜ ⎟ =− ⎜ ⎝ ∂P ⎠ T ⎝ ∂T ⎠ P ? ⎛ Δs ⎞ ⎛ Δv ⎞ ≅− ⎜ ⎜ ⎟ ⎟ ⎝ ΔP ⎠ T =800°F ⎝ ΔT ⎠ P = 400psia

⎛ s 450 psia − s 350 psia ⎜ ⎜ (450 − 350)psia ⎝

? ⎛ v 900° F − v 700° F ⎞ ⎟ ≅ −⎜⎜ ⎟ ⎠ T =800°F ⎝ (900 − 700 )°F

⎞ ⎟ ⎟ ⎠ P = 400psia

(1.6706 − 1.7009)Btu/lbm ⋅ R ? (1.9777 − 1.6507)ft 3 /lbm ≅− (450 − 350)psia (900 − 700)°F − 1.639 × 10 −3 ft 3 /lbm ⋅ R ≅ −1.635 × 10 −3 ft 3 /lbm ⋅ R

since 1 Btu ≡ 5.4039 psia·ft3, and R ≡ °F for temperature differences. Thus the fourth Maxwell relation is satisfied.

12-17 Using the Maxwell relations, a relation for (∂s/∂P)T for a gas whose equation of state is P(v-b) = RT is to be obtained. Analysis This equation of state can be expressed as v =

RT + b . Then, P

R ⎛ ∂v ⎞ ⎟ = ⎜ T ∂ ⎝ ⎠P P

From the fourth Maxwell relation, R ⎛ ∂v ⎞ ⎛ ∂s ⎞ ⎜ ⎟ = −⎜ ⎟ =− P ⎝ ∂P ⎠ T ⎝ ∂T ⎠ P

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12-11

12-18 Using the Maxwell relations, a relation for (∂s/∂v)T for a gas whose equation of state is (P-a/v2)(v-b) = RT is to be obtained.

Analysis This equation of state can be expressed as P =

RT a + 2 . Then, v −b v

R ⎛ ∂P ⎞ ⎜ ⎟ = ⎝ ∂T ⎠ v v − b

From the third Maxwell relation, R ⎛ ∂s ⎞ ⎛ ∂P ⎞ ⎜ ⎟ =⎜ ⎟ = ⎝ ∂v ⎠ T ⎝ ∂T ⎠ v v − b

12-19 Using the Maxwell relations and the ideal-gas equation of state, a relation for (∂s/∂v)T for an ideal gas is to be obtained.

Analysis The ideal gas equation of state can be expressed as P =

RT

v

. Then,

R ⎛ ∂P ⎞ ⎜ ⎟ = ⎝ ∂T ⎠ v v

From the third Maxwell relation, R ⎛ ∂s ⎞ ⎛ ∂P ⎞ ⎜ ⎟ =⎜ ⎟ = ⎝ ∂v ⎠ T ⎝ ∂T ⎠ v v

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12-12

k ⎛ ∂P ⎞ ⎛ ∂P ⎞ 12-20 It is to be proven that ⎜ ⎟ = ⎜ ⎟ ⎝ ∂T ⎠ s k − 1 ⎝ ∂T ⎠ v

Analysis Using the definition of cv , ⎛ ∂s ⎞ ⎛ ∂s ⎞ ⎛ ∂P ⎞ cv = T ⎜ ⎟ = T⎜ ⎟ ⎜ ⎟ ∂ T ⎝ ⎠v ⎝ ∂P ⎠ v ⎝ ∂T ⎠ v ⎛ ∂s ⎞ ⎛ ∂v ⎞ Substituting the first Maxwell relation ⎜ ⎟ = −⎜ ⎟ , ⎝ ∂P ⎠ v ⎝ ∂T ⎠ s

⎛ ∂v ⎞ ⎛ ∂P ⎞ cv = −T ⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠ s ⎝ ∂T ⎠ v Using the definition of cp, ⎛ ∂s ⎞ ⎛ ∂s ⎞ ⎛ ∂v ⎞ c p = T⎜ ⎟ = T⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠ P ⎝ ∂v ⎠ P ⎝ ∂T ⎠ P

⎛ ∂P ⎞ ⎛ ∂s ⎞ Substituting the second Maxwell relation ⎜ ⎟ =⎜ ⎟ , ⎝ ∂v ⎠ P ⎝ ∂T ⎠ s ⎛ ∂P ⎞ ⎛ ∂v ⎞ cp = T⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠ s ⎝ ∂T ⎠ P

From Eq. 12-46, 2

⎛ ∂v ⎞ ⎛ ∂P ⎞ c p − cv = −T ⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠ P ⎝ ∂v ⎠ T

Also, cp k = k −1 c p − cv

Then, ⎛ ∂P ⎞ ⎛ ∂v ⎞ ⎜ ⎟ ⎜ ⎟ k ⎝ ∂T ⎠ s ⎝ ∂T ⎠ P ⎛ ∂P ⎞ ⎛ ∂T ⎞ ⎛ ∂v ⎞ =− = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ 2 k −1 ⎝ ∂T ⎠ s ⎝ ∂v ⎠ P ⎝ ∂P ⎠ T ⎛ ∂v ⎞ ⎛ ∂P ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠ P ⎝ ∂v ⎠ T

Substituting this into the original equation in the problem statement produces ⎛ ∂P ⎞ ⎛ ∂P ⎞ ⎛ ∂T ⎞ ⎛ ∂v ⎞ ⎛ ∂P ⎞ ⎜ ⎟ = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠ s ⎝ ∂T ⎠ s ⎝ ∂v ⎠ P ⎝ ∂P ⎠ T ⎝ ∂T ⎠ v

But, according to the cyclic relation, the last three terms are equal to −1. Then, ⎛ ∂P ⎞ ⎛ ∂P ⎞ ⎜ ⎟ =⎜ ⎟ ⎝ ∂T ⎠ s ⎝ ∂T ⎠ s

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12-13

12-21 It is to be shown how T, v, u, a, and g could be evaluated from the thermodynamic function h = h(s, P).

Analysis Forming the differential of the given expression for h produces ⎛ ∂h ⎞ ⎛ ∂h ⎞ dh = ⎜ ⎟ ds + ⎜ ⎟ dP ⎝ ∂s ⎠ P ⎝ ∂P ⎠ s

Solving the dh Gibbs equation gives

dh = Tds + vdP Comparing the coefficient of these two expressions ⎛ ∂h ⎞ T =⎜ ⎟ ⎝ ∂s ⎠ P ⎛ ∂h ⎞ ⎟ ⎝ ∂P ⎠ s

v =⎜

both of which can be evaluated for a given P and s. From the definition of the enthalpy, ⎛ ∂h ⎞ u = h − Pv = h-P⎜ ⎟ ⎝ ∂P ⎠ s

Similarly, the definition of the Helmholtz function,

⎛ ∂h ⎞ ⎛ ∂h ⎞ a = u − Ts = h − P⎜ ⎟ − s⎜ ⎟ ⎝ ∂P ⎠ s ⎝ ∂s ⎠ P while the definition of the Gibbs function gives ⎛ ∂h ⎞ q = h − Ts = h − s⎜ ⎟ ⎝ ∂s ⎠ P

All of these can be evaluated for a given P and s and the fundamental h(s,P) equation.

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12-14

The Clapeyron Equation 12-22C It enables us to determine the enthalpy of vaporization from hfg at a given temperature from the P, v, T data alone. 12-23C It is assumed that vfg ≅ vg ≅ RT/P, and hfg ≅ constant for small temperature intervals.

12-24 Using the Clapeyron equation, the enthalpy of vaporization of steam at a specified pressure is to be estimated and to be compared to the tabulated data.

Analysis From the Clapeyron equation, ⎛ dP ⎞ h fg = Tv fg ⎜ ⎟ ⎝ dT ⎠ sat ⎛ ΔP ⎞ ≅ T (v g − v f ) @300 kPa ⎜ ⎟ ⎝ ΔT ⎠ sat, 300 kPa ⎛ ⎞ (325 − 275)kPa ⎟ = Tsat @300 kPa (v g − v f ) @300 kPa ⎜ ⎜ Tsat @325 kPa − Tsat @275 kPa ⎟ ⎝ ⎠ ⎛ ⎞ 50 kPa 3 ⎟⎟ = (133.52 + 273.15 K)(0.60582 − 0.001073 m /kg)⎜⎜ ⎝ (136.27 − 130.58)°C ⎠ = 2159.9 kJ/kg

The tabulated value of hfg at 300 kPa is 2163.5 kJ/kg.

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12-15

12-25 The hfg and sfg of steam at a specified temperature are to be calculated using the Clapeyron equation and to be compared to the tabulated data.

Analysis From the Clapeyron equation, ⎛ dP ⎞ h fg = Tv fg ⎜ ⎟ ⎝ dT ⎠ sat ⎛ ΔP ⎞ ≅ T (v g − v f ) @120°C ⎜ ⎟ ⎝ ΔT ⎠ sat,120°C ⎛ Psat @125°C − Psat @115°C = T (v g − v f ) @120°C ⎜⎜ 125°C − 115°C ⎝

⎞ ⎟ ⎟ ⎠

⎛ (232.23 − 169.18)kPa ⎞ ⎟⎟ = (120 + 273.15 K)(0.89133 − 0.001060 m 3 /kg)⎜⎜ 10 K ⎝ ⎠ = 2206.8 kJ/kg

Also, s fg =

h fg T

=

2206.8 kJ/kg = 5.6131 kJ/kg ⋅ K (120 + 273.15) K

The tabulated values at 120°C are hfg = 2202.1 kJ/kg and sfg = 5.6013 kJ/kg·K.

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12-16

12-26E [Also solved by EES on enclosed CD] The hfg of refrigerant-134a at a specified temperature is to be calculated using the Clapeyron equation and Clapeyron-Clausius equation and to be compared to the tabulated data.

Analysis (a) From the Clapeyron equation, ⎛ dP ⎞ h fg = Tv fg ⎜ ⎟ ⎝ dT ⎠ sat ⎛ ΔP ⎞ ≅ T (v g − v f ) @ 50°F ⎜ ⎟ ⎝ ΔT ⎠ sat, 50° F ⎛ Psat @ 60 ° F − Psat @ 40 °F = T (v g − v f ) @ 50°F ⎜⎜ 60°F − 40°F ⎝

⎞ ⎟ ⎟ ⎠

⎛ (72.152 − 49.776) psia ⎞ ⎟⎟ = (50 + 459.67 R)(0.79136 − 0.01270 ft 3 /lbm)⎜⎜ 20 R ⎠ ⎝ = 444.0 psia ⋅ ft 3 /lbm = 82.16 Btu/lbm

(0.2% error)

since 1 Btu = 5.4039 psia·ft3. (b) From the Clapeyron-Clausius equation, ⎛P ln⎜⎜ 2 ⎝ P1

h fg ⎛ 1 ⎞ 1 ⎞ ⎟⎟ ≅ ⎜⎜ − ⎟⎟ R ⎝ T1 T2 ⎠ sat ⎠ sat

h fg ⎛ 72.152 psia ⎞ 1 1 ⎛ ⎞ ⎟⎟ ≅ ln⎜⎜ − ⎜ ⎟ ⎝ 49.776 psia ⎠ 0.01946 Btu/lbm ⋅ R ⎝ 40 + 459.67 R 60 + 459.67 R ⎠ h fg = 93.80 Btu/lbm (14.4% error)

The tabulated value of hfg at 50°F is 82.00 Btu/lbm.

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12-17

12-27 EES The enthalpy of vaporization of steam as a function of temperature using Clapeyron equation and steam data in EES is to be plotted.

Analysis The enthalpy of vaporization is determined using Clapeyron equation from h fg ,Clapeyron = Tv fg

ΔP ΔT

At 100ºC, for an increment of 5ºC, we obtain T1 = T − Tincrement = 100 − 5 = 95°C T2 = T + Tincrement = 100 + 5 = 105°C P1 = Psat @ 95°C = 84.61 kPa P2 = Psat @ 105°C = 120.90 kPa ΔT = T2 − T1 = 105 − 95 = 10°C ΔP = P2 − P1 = 120.90 − 84.61 = 36.29 kPa

v f @ 100°C = 0.001043 m 3 /kg v g @ 100°C = 1.6720 m 3 /kg v fg = v g − v f = 1.6720 − 0.001043 = 1.6710 m 3 /kg Substituting, h fg ,Clapeyron = Tv fg

36.29 kPa ΔP = (100 + 273.15 K)(1.6710 m 3 /kg) = 2262.8 kJ/kg 10 K ΔT

The enthalpy of vaporization from steam table is

h fg @ 100°C = 2256.4 m 3 /kg The percent error in using Clapeyron equation is PercentError =

2262.8 − 2256.4 × 100 = 0.28% 2256.4

We repeat the analysis over the temperature range 10 to 200ºC using EES. Below, the copy of EES solution is provided: "Input Data:" "T=100" "[C]" T_increment = 5"[C]" T[2]=T+T_increment"[C]" T[1]=T-T_increment"[C]" P[1] = pressure(Steam_iapws,T=T[1],x=0)"[kPa]" P[2] = pressure(Steam_iapws,T=T[2],x=0)"[kPa]" DELTAP = P[2]-P[1]"[kPa]" DELTAT = T[2]-T[1]"[C]" v_f=volume(Steam_iapws,T=T,x=0)"[m^3/kg]" v_g=volume(Steam_iapws,T=T,x=1)"[m^3/kg]" h_f=enthalpy(Steam_iapws,T=T,x=0)"[kJ/kg]" h_g=enthalpy(Steam_iapws,T=T,x=1)"[kJ/kg]" h_fg=h_g - h_f"[kJ/kg-K]" v_fg=v_g - v_f"[m^3/kg]"

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12-18 "The Clapeyron equation (Eq. 11-22) provides a means to calculate the enthalpy of vaporization, h_fg at a given temperature by determining the slope of the saturation curve on a P-T diagram and the specific volume of the saturated liquid and satruated vapor at the temperature." h_fg_Clapeyron=(T+273.15)*v_fg*DELTAP/DELTAT*Convert(m^3-kPa,kJ)"[kJ/kg]" PercentError=ABS(h_fg_Clapeyron-h_fg)/h_fg*100"[%]" hfg [kJ/kg] 2477.20 2429.82 2381.95 2333.04 2282.51 2229.68 2173.73 2113.77 2014.17 1899.67 1765.50

hfg,Clapeyron [kJ/kg] 2508.09 2451.09 2396.69 2343.47 2290.07 2235.25 2177.86 2116.84 2016.15 1900.98 1766.38

PercentError [%] 1.247 0.8756 0.6188 0.4469 0.3311 0.25 0.1903 0.1454 0.09829 0.06915 0.05015

T [C] 10 30 50 70 90 110 130 150 180 210 240

2600 2500

hfg calculated by Clapeyron equation

hfg [kJ/kg]

2400 2300 2200

hfg calculated by EES

2100 2000 1900 1800 1700 0

50

100

150

200

250

T [C]

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12-19

12-28 A substance is heated in a piston-cylinder device until it turns from saturated liquid to saturated vapor at a constant pressure and temperature. The boiling temperature of this substance at a different pressure is to be estimated.

Analysis From the Clapeyron equation, h fg (5 kPa ⋅ m 3 )/(0.002 kg) ⎛ dP ⎞ = = 14.16 kPa/K ⎜ ⎟ = ⎝ dT ⎠ sat Tv fg (353 K)(1× 10 −3 m 3 )/(0.002 kg)

Weight

Using the finite difference approximation, ⎛ P − P1 ⎞ ⎛ dP ⎞ ⎟⎟ ⎜ ⎟ ≈ ⎜⎜ 2 ⎝ dT ⎠ sat ⎝ T2 − T1 ⎠ sat

200 kPa 80°C 2 grams Sat. liquid

Solving for T2,

T2 = T1 +

Q

P2 − P1 (180 − 200)kPa = 353 K + = 351.6 K dP / dT 14.16 kPa/K

12-29 A substance is heated in a piston-cylinder device until it turns from saturated liquid to saturated vapor at a constant pressure and temperature. The saturation pressure of this substance at a different temperature is to be estimated.

Analysis From the Clapeyron equation,

Weight 3

h fg (5 kPa ⋅ m )/(0.002 kg) ⎛ dP ⎞ = = 14.16 kPa/K ⎟ = ⎜ dT T v (353 K)(1× 10 −3 m 3 )/(0.002 kg) ⎠ sat ⎝ fg

Using the finite difference approximation, ⎛ P − P1 ⎞ ⎛ dP ⎞ ⎟⎟ ⎜ ⎟ ≈ ⎜⎜ 2 dT ⎠ sat ⎝ T2 − T1 ⎠ sat ⎝

200 kPa 80°C 2 grams Sat. liquid

Solving for P2, P2 = P1 +

dP (T2 − T1 ) = 200 kPa + (14.16 kPa/K )(373 − 353)K = 483.2 kPa dT

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Q

12-20

12-30 A substance is heated in a piston-cylinder device until it turns from saturated liquid to saturated vapor at a constant pressure and temperature. The sfg of this substance at a different temperature is to be estimated.

Analysis From the Clapeyron equation,

Weight

h fg s fg ⎛ dP ⎞ = ⎟ = ⎜ ⎝ dT ⎠ sat Tv fg v fg

Solving for sfg, s fg =

h fg T

=

(5 kJ)/(0.002 kg) = 7.082 kJ/kg ⋅ K 353 K

200 kPa 80°C 2 grams Sat. liquid

Q

Alternatively, 1× 10 -3 m 3 ⎛ dP ⎞ = 7.08 kPa ⋅ m 3 /kg ⋅ K = 7.08 kJ/kg ⋅ K s fg = ⎜ ⎟ v fg = (14.16 kPa/K ) 0.002 kg dT ⎠ sat ⎝

⎛ ∂ (h fg / T ) ⎞ ⎟ + v fg ⎛⎜ ∂P ⎞⎟ . 12-31 It is to be shown that c p , g − c p , f = T ⎜⎜ ⎟ ∂ T ⎝ ∂T ⎠ sat ⎝ ⎠P

Analysis The definition of specific heat and Clapeyron equation are ⎛ ∂h ⎞ cp = ⎜ ⎟ ⎝ ∂T ⎠ P h fg ⎛ dP ⎞ ⎟ = ⎜ ⎝ dT ⎠ sat Tv fg

According to the definition of the enthalpy of vaporization, h fg T

=

hg T

−

hf T

Differentiating this expression gives ⎛ ∂h fg / T ⎜ ⎜ ∂T ⎝

⎞ ⎛ ∂h / T ⎟ =⎜ g ⎟ ⎜ ⎠ P ⎝ ∂T

⎞ ⎟ ⎟ ⎠P

h ⎞ ⎛ ∂h ⎟ − g −1⎜ f 2 ⎟ T ⎜⎝ ∂T ⎠P T c p, g c p, f h g − h f = − − T T T2 =

1 T

⎛ ∂h g ⎜ ⎜ ∂T ⎝

⎞ ⎛ ∂h / T ⎟ −⎜ f ⎟ ⎜ ⎠ P ⎝ ∂T

h ⎞ ⎟ + f 2 ⎟ ⎠P T

Using Clasius-Clapeyron to replace the last term of this expression and solving for the specific heat difference gives ⎛ ∂ (h fg / T ) ⎞ ⎟ + v fg ⎛⎜ ∂P ⎞⎟ c p , g − c p , f = T ⎜⎜ ⎟ ∂T ⎝ ∂T ⎠ sat ⎠P ⎝

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12-21

12-32E A table of properties for methyl chloride is given. The saturation pressure is to be estimated at two different temperatures.

Analysis The Clapeyron equation is h fg ⎛ dP ⎞ ⎟ = ⎜ ⎝ dT ⎠ sat Tv fg

Using the finite difference approximation, h fg ⎛ P2 − P1 ⎞ ⎛ dP ⎞ ⎟⎟ = ⎟ ≈ ⎜⎜ ⎜ ⎝ dT ⎠ sat ⎝ T2 − T1 ⎠ sat Tv fg

Solving this for the second pressure gives for T2 = 110°F P2 = P1 +

h fg Tv fg

(T2 − T1 )

= 116.7 psia +

⎛ 5.404 psia ⋅ ft 3 ⎜ 1 Btu (560 R)(0.86332 ft 3 /lbm) ⎜⎝ 154.85 Btu/lbm

⎞ ⎟(110 − 100)R ⎟ ⎠

= 134.0 psia

When T2 = 90°F P2 = P1 +

h fg Tv fg

(T2 − T1 )

= 116.7 psia +

⎛ 5.404 psia ⋅ ft 3 ⎜ 1 Btu (560 R)(0.86332 ft 3 /lbm) ⎜⎝ 154.85 Btu/lbm

⎞ ⎟(90 − 100)R ⎟ ⎠

= 99.4 psia

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12-22

12-33 Saturation properties for R-134a at a specified temperature are given. The saturation pressure is to be estimated at two different temperatures.

Analysis From the Clapeyron equation, h fg 225.86 kPa ⋅ m 3 /kg ⎛ dP ⎞ = = 2.692 kPa/K ⎜ ⎟ = ⎝ dT ⎠ sat Tv fg (233 K)(0.36010 m 3 /kg)

Using the finite difference approximation, ⎛ P − P1 ⎞ ⎛ dP ⎞ ⎟⎟ ⎜ ⎟ ≈ ⎜⎜ 2 ⎝ dT ⎠ sat ⎝ T2 − T1 ⎠ sat

Solving for P2 at −50°C P2 = P1 +

dP (T2 − T1 ) = 51.25 kPa + (2.692 kPa/K )(223 − 233)K = 24.33 kPa dT

Solving for P2 at −30°C P2 = P1 +

dP (T2 − T1 ) = 51.25 kPa + (2.692 kPa/K )(243 − 233)K = 78.17 kPa dT

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12-23

General Relations for du, dh, ds, cv, and cp 12-34C Yes, through the relation

⎛ ∂c p ⎜ ⎜ ∂P ⎝

⎛ 2 ⎞ ⎟ = −T ⎜ ∂ v ⎟ ⎜ ∂T 2 ⎠T ⎝

⎞ ⎟ ⎟ ⎠P

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12-24

12-35 General expressions for Δu, Δh, and Δs for a gas whose equation of state is P(v-a) = RT for an isothermal process are to be derived.

Analysis (a) A relation for Δu is obtained from the general relation Δu = u 2 − u1 =

∫

T2

T1

⎛ ⎛ ∂P ⎞ ⎞ ⎜T ⎜ ⎟ ⎜ ⎝ ∂T ⎟⎠ − P ⎟dv v ⎝ ⎠

v2

∫v

cv dT +

1

The equation of state for the specified gas can be expressed as RT R ⎛ ∂P ⎞ ⎯⎯→ ⎜ ⎟ = v −a ⎝ ∂T ⎠v v − a

P=

Thus, RT ⎛ ∂P ⎞ T⎜ −P = P−P = 0 ⎟ −P = ∂ T −a v ⎝ ⎠v

∫

Δu =

Substituting,

T2

T1

cv dT

(b) A relation for Δh is obtained from the general relation

Δh = h2 − h1 =

∫

T2

T1

c P dT +

∫

P2

P1

⎛ ∂v ⎞ ⎞⎟ ⎜v − T ⎛⎜ ⎟ dP ⎜ ⎝ ∂T ⎠ P ⎟⎠ ⎝

The equation of state for the specified gas can be expressed as

v=

RT R ⎛ ∂v ⎞ +a ⎯ ⎯→ ⎜ ⎟ = P ⎝ ∂T ⎠ P P

Thus, R ⎛ ∂v ⎞ ⎟ = v − T = v − (v − a ) = a ∂ T P ⎝ ⎠P

v −T⎜ Substituting,

Δh =

∫

T2

T1

c p dT +

∫

P2

a dP =

P1

∫

T2

T1

c p dT + a(P2 − P1 )

(c) A relation for Δs is obtained from the general relation Δs = s 2 − s1 =

T2

cp

T1

T

∫

dT −

∫

P2

P1

⎛ ∂v ⎞ ⎜ ⎟ dP ⎝ ∂T ⎠ P

Substituting (∂v/∂T)P = R/T, Δs =

T2

cp

T1

T

∫

dT −

∫

P2

P1

⎛R⎞ ⎜ ⎟ dP = ⎝ P ⎠P

T2

cp

T1

T

∫

dT − R ln

P2 P1

For an isothermal process dT = 0 and these relations reduce to Δu = 0,

Δh = a(P2 − P1 ),

and

Δs = − Rln

P2 P1

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12-25

12-36 General expressions for (∂u/∂P)T and (∂h/∂v)T in terms of P, v, and T only are to be derived.

Analysis The general relation for du is ⎛ ⎛ ∂P ⎞ ⎞ du = cv dT + ⎜⎜ T ⎜ ⎟ − P ⎟⎟dv ⎝ ⎝ ∂T ⎠ v ⎠

Differentiating each term in this equation with respect to P at T = constant yields ⎛ ⎛ ∂P ⎞ ⎞⎛ ∂v ⎞ ⎛ ∂u ⎞ ⎛ ∂v ⎞ ⎛ ∂P ⎞ ⎛ ∂v ⎞ ⎟ − P ⎟⎟⎜ ⎜ ⎟ = 0 + ⎜⎜ T ⎜ ⎟ = T⎜ ⎟ ⎜ ⎟ − P⎜ ⎟ ∂ ∂ P T ∂ P ∂ T ∂ P ⎝ ⎠T ⎠v ⎠T ⎝ ⎠v ⎝ ⎠T ⎝ ∂P ⎠ T ⎝ ⎝ ⎠⎝

Using the properties P, T, v, the cyclic relation can be expressed as

⎛ ∂P ⎞ ⎛ ∂T ⎞ ⎛ ∂v ⎞ ⎛ ∂P ⎞ ⎛ ∂v ⎞ ⎛ ∂v ⎞ ⎯→ ⎜ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = −1 ⎯ ⎟ ⎜ ⎟ = −⎜ ⎟ ⎝ ∂T ⎠v ⎝ ∂v ⎠ P ⎝ ∂P ⎠ T ⎝ ∂T ⎠ v ⎝ ∂P ⎠ T ⎝ ∂T ⎠ P Substituting, we get ⎛ ∂u ⎞ ⎛ ∂v ⎞ ⎛ ∂v ⎞ ⎜ ⎟ = −T ⎜ ⎟ − P⎜ ⎟ ⎝ ∂P ⎠ T ⎝ ∂T ⎠ P ⎝ ∂P ⎠ T

The general relation for dh is

⎛ ⎛ ∂v ⎞ ⎞⎟ dh = c p dT + ⎜⎜v − T ⎜ ⎟ dP ⎝ ∂T ⎠ P ⎟⎠ ⎝ Differentiating each term in this equation with respect to v at T = constant yields

⎛ ⎛ ∂v ⎞ ⎛ ∂P ⎞ ⎛ ∂h ⎞ ⎛ ∂v ⎞ ⎞⎟⎛ ∂P ⎞ ⎛ ∂P ⎞ ⎜ ⎟ = 0 + ⎜⎜v − T ⎜ ⎟ ⎟⎜ ⎟ =v⎜ ⎟ −T⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠ P ⎝ ∂v ⎠ T ⎝ ∂v ⎠ T ⎝ ∂T ⎠ P ⎠⎝ ∂v ⎠ T ⎝ ∂v ⎠ T ⎝ Using the properties v, T, P, the cyclic relation can be expressed as ⎛ ∂v ⎞ ⎛ ∂T ⎞ ⎛ ∂P ⎞ ⎛ ∂v ⎞ ⎛ ∂P ⎞ ⎛ ∂T ⎞ ⎯→ ⎜ ⎟ ⎜ ⎟ = −1 ⎯ ⎟ ⎜ ⎟ = −⎜ ⎟ ⎜ ⎟ ⎜ ⎝ ∂T ⎠ P ⎝ ∂P ⎠ v ⎝ ∂v ⎠ T ⎝ ∂T ⎠ P ⎝ ∂v ⎠ T ⎝ ∂P ⎠ v

Substituting, we get ⎛ ∂h ⎞ ⎛ ∂P ⎞ ⎛ ∂T ⎞ ⎜ ⎟ =v⎜ ⎟ +T⎜ ⎟ ⎝ ∂v ⎠ T ⎝ ∂P ⎠ v ⎝ ∂v ⎠ T

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12-26

12-37E The specific heat difference cp-cv for liquid water at 1000 psia and 150°F is to be estimated.

Analysis The specific heat difference cp - cv is given as 2

⎛ ∂v ⎞ ⎛ ∂P ⎞ c p − cv = −T ⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠ P ⎝ ∂v ⎠ T Approximating differentials by differences about the specified state, 2

⎛ Δv ⎞ ⎛ ΔP ⎞ c p − cv ≅ −T ⎜ ⎟ ⎜ ⎟ ⎝ ΔT ⎠ P =1000 psia ⎝ Δv ⎠ T =150°F ⎛ v 175° F − v 125° F = −(150 + 459.67 R )⎜⎜ ⎝ (175 − 125)°F

2

⎛ (1500 − 500)psia ⎞ ⎜ ⎟ ⎟ ⎜ ⎠ P =1000psia ⎝ v 1500 psia − v 500 psia

⎛ (0.016427 − 0.016177)ft 3 /lbm ⎞ ⎟ = −(609.67 R)⎜ ⎜ ⎟ 50 R ⎝ ⎠

2

⎞ ⎟ ⎟ ⎠ T =150° F

⎛ ⎞ 1000 psia ⎜ ⎟ ⎜ (0.016267 − 0.016317)ft 3 /lbm ⎟ ⎝ ⎠

= 0.3081 psia ⋅ ft 3 /lbm ⋅ R = 0.0570 Btu/lbm ⋅ R (1 Btu = 5.4039 psia ⋅ ft 3 )

12-38 The volume expansivity β and the isothermal compressibility α of refrigerant-134a at 200 kPa and 30°C are to be estimated.

Analysis The volume expansivity and isothermal compressibility are expressed as

β=

1 ⎛ ∂v ⎞ 1 ⎛ ∂v ⎞ ⎜ ⎟ and α = − ⎜ ⎟ v ⎝ ∂T ⎠ P v ⎝ ∂P ⎠ T

Approximating differentials by differences about the specified state,

β≅ =

1 ⎛ Δv ⎞ 1 ⎛ v 40°C − v 20°C = ⎜⎜ ⎟ ⎜ v ⎝ ΔT ⎠ P = 200kPa v ⎝ (40 − 20)°C

⎞ ⎟ ⎟ ⎠ P = 200 kPa

⎛ (0.12322 − 0.11418) m 3 /kg ⎞ ⎜ ⎟ ⎟ 20 K 0.11874 m 3 /kg ⎜⎝ ⎠ 1

= 0.00381 K −1

and

α ≅− =−

1 ⎛ Δv ⎞ 1 ⎛ v 240 kPa − v 180 kPa = − ⎜⎜ ⎜ ⎟ v ⎝ ΔP ⎠ T =30°C v ⎝ (240 − 180)kPa

⎞ ⎟ ⎟ ⎠ T =30°C

⎛ (0.09812 − 0.13248)m 3 /kg ⎞ ⎜ ⎟ ⎟ 60 kPa 0.11874 m /kg ⎜⎝ ⎠ 1

3

= 0.00482 kPa −1

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12-27

⎛ ∂P ⎞ ⎛ ∂v ⎞ 12-39 It is to be shown that c p − cv = T ⎜ ⎟ ⎜ ⎟ . ⎝ ∂T ⎠ v ⎝ ∂T ⎠ P

Analysis We begin by taking the entropy to be a function of specific volume and temperature. The differential of the entropy is then ⎛ ∂s ⎞ ⎛ ∂s ⎞ ds = ⎜ ⎟ dT + ⎜ ⎟ dv T ∂ ⎝ ⎠v ⎝ ∂v ⎠ T c ⎛ ∂s ⎞ Substituting ⎜ ⎟ = v from Eq. 12-28 and the third Maxwell equation changes this to T ⎝ ∂T ⎠ v

ds =

cv ⎛ ∂P ⎞ dT + ⎜ ⎟ dv T ⎝ ∂T ⎠v

Taking the entropy to be a function of pressure and temperature, ⎛ ∂s ⎞ ⎛ ∂s ⎞ ds = ⎜ ⎟ dT + ⎜ ⎟ dP ⎝ ∂T ⎠ P ⎝ ∂P ⎠ T cp ⎛ ∂s ⎞ from Eq. 12-34 and the fourth Maxwell equation produces Combining this result with ⎜ ⎟ = T ⎝ ∂T ⎠ P ds =

⎛ ∂v ⎞ dT − ⎜ ⎟ dP T ⎝ ∂T ⎠ P

cp

Equating the two previous ds expressions and solving the result for the specific heat difference, ⎛ ∂v ⎞ ⎛ ∂P ⎞ (c p − cv )dT = T ⎜ ⎟ dP + ⎜ ⎟ dv T ∂ ⎝ ⎠P ⎝ ∂T ⎠ v Taking the pressure to be a function of temperature and volume, ⎛ ∂P ⎞ ⎛ ∂P ⎞ dP = ⎜ ⎟ dT + ⎜ ⎟ dv ⎝ ∂T ⎠ v ⎝ ∂v ⎠ T

When this is substituted into the previous expression, the result is ⎡⎛ ∂v ⎞ ⎛ ∂P ⎞ ⎛ ∂P ⎞ ⎤ ⎛ ∂v ⎞ ⎛ ∂P ⎞ (c p − cv )dT = T ⎜ ⎟ ⎜ ⎟ dT + T ⎢⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎥ dv ⎝ ∂T ⎠ P ⎝ ∂T ⎠ v ⎣⎝ ∂T ⎠ P ⎝ ∂v ⎠ T ⎝ ∂T ⎠ v ⎦

According to the cyclic relation, the term in the bracket is zero. Then, canceling the common dT term, ⎛ ∂P ⎞ ⎛ ∂v ⎞ c p − cv = T ⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠ v ⎝ ∂T ⎠ P

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12-28

12-40 It is to be proven that the definition for temperature T = (∂u / ∂s ) v reduces the net entropy change of two constant-volume systems filled with simple compressible substances to zero as the two systems approach thermal equilibrium.

Analysis The two constant-volume systems form an isolated system shown here

For the isolated system dS tot = dS A + dS B ≥ 0

VA =const.

Assume S = S (u , v )

TA

Then, ⎛ ∂s ⎞ ⎛ ∂s ⎞ ds = ⎜ ⎟ du + ⎜ ⎟ dv ⎝ ∂u ⎠ v ⎝ ∂v ⎠ u

Isolated system boundary

VB =const. TB

Since v = const. and dv = 0 , ⎛ ∂s ⎞ ds = ⎜ ⎟ du ⎝ ∂u ⎠ v

and from the definition of temperature from the problem statement, du du = (∂u / ∂s ) v T

Then, dS tot = m A

du A du B + mB TA TB

The first law applied to the isolated system yields E in − E out = dU 0 = dU ⎯ ⎯→ m A du A + m B du B = 0 ⎯ ⎯→ m B du B = −m A du A

Now, the entropy change may be expressed as ⎛ 1 1 dS tot = m A du A ⎜⎜ − ⎝ T A TB

⎞ ⎛ T − TA ⎟⎟ = m A du A ⎜⎜ B ⎠ ⎝ T AT B

⎞ ⎟⎟ ⎠

As the two systems approach thermal equilibrium, lim dS tot = 0 T A → TB

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12-29

12-41 The internal energy change of air between two specified states is to be compared for two equations of states.

Assumptions Constant specific heats for air can be used. Properties For air at the average temperature (20+300)/2=160°C=433 K, cv = 0.731 kJ/kg⋅K (Table A-2b). Analysis Solving the equation of state for P gives

P=

RT v −a

Then, R ⎛ ∂P ⎞ ⎜ ⎟ = ⎝ ∂T ⎠v v − a

Using equation 12-29, ⎡ ⎛ ∂P ⎞ ⎤ du = cv dT + ⎢T ⎜ ⎟ − P ⎥ dv T ∂ ⎠v ⎣ ⎝ ⎦

Substituting, RT ⎞ ⎛ RT du = cv dT + ⎜ − ⎟dv − −a⎠ a v v ⎝ = cv dT

Integrating this result between the two states with constant specific heats gives u 2 − u1 = cv (T2 − T1 ) = (0.731 kJ/kg ⋅ K)(300 − 20)K = 205 kJ/kg

The ideal gas model for the air gives du = cv dT

which gives the same answer.

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12-30

12-42 The enthalpy change of air between two specified states is to be compared for two equations of states.

Assumptions Constant specific heats for air can be used. Properties For air at the average temperature (20+300)/2=160°C=433 K, cp = 1.018 kJ/kg⋅K (Table A-2b). Analysis Solving the equation of state for v gives

v=

RT +a P

Then, R ⎛ ∂v ⎞ ⎜ ⎟ = T ∂ ⎝ ⎠P P Using equation 12-35, ⎡ ⎛ ∂v ⎞ ⎤ dh = c p dT + ⎢v − T ⎜ ⎟ ⎥ dP ⎝ ∂T ⎠ P ⎦ ⎣

Substituting, RT ⎞ ⎛ RT dh = c p dT + ⎜ +a− ⎟dP P ⎠ ⎝ P = c p dT + adP

Integrating this result between the two states with constant specific heats gives h2 − h1 = c p (T2 − T1 ) + a ( P2 − P1 ) = (1.018 kJ/kg ⋅ K)(300 − 20)K + (0.10 m 3 /kg)(600 − 100)kPa = 335.0 kJ/kg

For an ideal gas, dh = c p dT

which when integrated gives h2 − h1 = c p (T2 − T1 ) = (1.018 kJ/kg ⋅ K)(300 − 20)K = 285.0 kJ/kg

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12-31

12-43 The entropy change of air between two specified states is to be compared for two equations of states.

Assumptions Constant specific heats for air can be used. Properties For air at the average temperature (20+300)/2=160°C=433 K, cp = 1.018 kJ/kg⋅K (Table A-2b) and R = 0.287 kJ/kg⋅K (Table A-1). Analysis Solving the equation of state for v gives

v=

RT +a P

Then, R ⎛ ∂v ⎞ ⎜ ⎟ = T ∂ ⎝ ⎠P P The entropy differential is dT ⎛ ∂v ⎞ −⎜ ⎟ dP T ⎝ ∂T ⎠ P dT dP = cp −R T P

ds = c p

which is the same as that of an ideal gas. Integrating this result between the two states with constant specific heats gives s 2 − s1 = c p ln

T2 P − R ln 2 T1 P1

= (1.018 kJ/kg ⋅ K)ln

573 K 600 kPa − (0.287 kJ/kg ⋅ K)ln 293 K 100 kPa

= 0.1686 kJ/kg ⋅ K

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12-32

12-44 The internal energy change of helium between two specified states is to be compared for two equations of states.

Properties For helium, cv = 3.1156 kJ/kg⋅K (Table A-2a). Analysis Solving the equation of state for P gives P=

RT v −a

Then, R ⎛ ∂P ⎞ ⎜ ⎟ = ⎝ ∂T ⎠v v − a Using equation 12-29, ⎡ ⎛ ∂P ⎞ ⎤ du = cv dT + ⎢T ⎜ ⎟ − P ⎥ dv ⎣ ⎝ ∂T ⎠v ⎦

Substituting, RT ⎞ ⎛ RT du = cv dT + ⎜ − ⎟dv ⎝v − a v − a ⎠ = cv dT

Integrating this result between the two states gives u 2 − u1 = cv (T2 − T1 ) = (3.1156 kJ/kg ⋅ K)(300 − 20)K = 872.4 kJ/kg

The ideal gas model for the helium gives du = cv dT

which gives the same answer.

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12-33

12-45 The enthalpy change of helium between two specified states is to be compared for two equations of states.

Properties For helium, cp = 5.1926 kJ/kg⋅K (Table A-2a). Analysis Solving the equation of state for v gives

v=

RT +a P

Then, R ⎛ ∂v ⎞ ⎜ ⎟ = ⎝ ∂T ⎠ P P

Using equation 12-35, ⎡ ⎛ ∂v ⎞ ⎤ dh = c p dT + ⎢v − T ⎜ ⎟ ⎥ dP ⎝ ∂T ⎠ P ⎦ ⎣

Substituting, RT ⎞ ⎛ RT dh = c p dT + ⎜ +a− ⎟dP P ⎠ ⎝ P = c p dT + adP

Integrating this result between the two states gives h2 − h1 = c p (T2 − T1 ) + a ( P2 − P1 ) = (5.1926 kJ/kg ⋅ K)(300 − 20)K + (0.10 m 3 /kg)(600 − 100)kPa = 1504 kJ/kg

For an ideal gas, dh = c p dT

which when integrated gives h2 − h1 = c p (T2 − T1 ) = (5.1926 kJ/kg ⋅ K)(300 − 20)K = 1454 kJ/kg

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12-34

12-46 The entropy change of helium between two specified states is to be compared for two equations of states.

Properties For helium, cp = 5.1926 kJ/kg⋅K and R = 2.0769 kJ/kg⋅K (Table A-2a). Analysis Solving the equation of state for v gives

v=

RT +a P

Then, R ⎛ ∂v ⎞ ⎟ = ⎜ ⎝ ∂T ⎠ P P

The entropy differential is dT ⎛ ∂v ⎞ −⎜ ⎟ dP T ⎝ ∂T ⎠ P dT dP = cp −R T P

ds = c p

which is the same as that of an ideal gas. Integrating this result between the two states gives s 2 − s1 = c p ln

T2 P − R ln 2 T1 P1

= (5.1926 kJ/kg ⋅ K)ln

573 K 600 kPa − (2.0769 kJ/kg ⋅ K)ln 293 K 100 kPa

= −0.2386 kJ/kg ⋅ K

12-47 An expression for the volume expansivity of a substance whose equation of state is P (v − a) = RT is to be derived.

Analysis Solving the equation of state for v gives

v=

RT +a P

The specific volume derivative is then R ⎛ ∂v ⎞ ⎜ ⎟ = T ∂ ⎝ ⎠P P The definition for volume expansivity is β=

1 ⎛ ∂v ⎞ ⎜ ⎟ v ⎝ ∂T ⎠ P

Combining these two equations gives β=

R RT + aP

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12-35

12-48 The Helmholtz function of a substance has the form a = − RT ln

⎛ v T T T ⎞ − cT0 ⎜⎜1 − + ln ⎟⎟ . It is to v0 ⎝ T0 T0 T0 ⎠

be shown how to obtain P, h, s, cv, and cp from this expression. Analysis Taking the Helmholtz function to be a function of temperature and specific volume yields

⎛ ∂a ⎞ ⎛ ∂a ⎞ da = ⎜ ⎟ dT + ⎜ ⎟ dv ⎝ ∂T ⎠ v ⎝ ∂v ⎠ T while the applicable Helmholtz equation is da = − Pdv − sdT

Equating the coefficients of the two results produces ⎛ ∂a ⎞ P = −⎜ ⎟ ⎝ ∂v ⎠ T ⎛ ∂a ⎞ s = −⎜ ⎟ ⎝ ∂T ⎠ v

Taking the indicated partial derivatives of the Helmholtz function given in the problem statement reduces these expressions to P=

RT

v

s = R ln

v T + c ln T0 v0

The definition of the enthalpy (h = u + Pv) and Helmholtz function (a = u−Ts) may be combined to give h = u + Pv = a + Ts + Pv ⎛ ∂a ⎞ ⎛ ∂a ⎞ = a −T⎜ ⎟ ⎟ −v ⎜ ∂ T ⎝ ⎠v ⎝ ∂v ⎠ T = − RT ln

⎛ v v T T T ⎞ T − cT0 ⎜⎜1 − + − cT ln + RT ln ⎟⎟ + RT ln T T T T v0 v 0 0 0 ⎠ 0 0 ⎝

= cT0 + cT + RT

c ⎛ ∂s ⎞ According to ⎜ ⎟ = v given in the text (Eq. 12-28), T ⎝ ∂T ⎠ v c ⎛ ∂s ⎞ cv = T ⎜ ⎟ =T =c T ⎝ ∂T ⎠ v

The preceding expression for the temperature indicates that the equation of state for the substance is the same as that of an ideal gas. Then, c p = R + cv = R + c

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12-36

12-49 An expression for the volume expansivity of a substance whose equation of state RT a is P = is to be derived. − v − b v (v + b)T 1 / 2

Analysis The definition for volume expansivity is

β=

1 ⎛ ∂v ⎞ ⎜ ⎟ v ⎝ ∂T ⎠ P

According to the cyclic relation, ⎛ ∂v ⎞ ⎛ ∂P ⎞ ⎛ ∂T ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = −1 ⎝ ∂T ⎠ P ⎝ ∂v ⎠ T ⎝ ∂P ⎠ v

which on rearrangement becomes ⎛ ∂P ⎞ ⎜ ⎟ ⎝ ∂T ⎠ v ⎛ ∂v ⎞ ⎜ ⎟ =− ⎛ ∂P ⎞ ⎝ ∂T ⎠ P ⎜ ⎟ ⎝ ∂v ⎠ T

Proceeding to perform the differentiations gives R a ⎛ ∂P ⎞ + ⎜ ⎟ = ⎝ ∂T ⎠ v v − b 2v (v + b)T 3 / 2

and ⎤ RT a ⎡ 1 1 ⎛ ∂P ⎞ + 1/ 2 ⎢ 2 − ⎜ ⎟ =− ⎥ 2 2 ⎝ ∂v ⎠ T (v − b) bT (v + b) ⎦⎥ ⎣⎢ v =−

RT (v − b) 2

+

a

2v + b

T 1 / 2 v 2 (v + b) 2

Substituting these results into the definition of the volume expansivity produces R a + v − b 2v (v + b)T 3 / 2 1 β=− a 2v + b v − RT + 1/ 2 2 2 (v − b) T v (v + b) 2

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12-37

12-50 An expression for the specific heat difference of a substance whose equation of state RT a is P = is to be derived. − v − b v (v + b)T 1 / 2

Analysis The specific heat difference is expressed by 2

⎛ ∂v ⎞ ⎛ ∂P ⎞ c p − cv = −T ⎜ ⎟ ⎟ ⎜ ⎝ ∂T ⎠ P ⎝ ∂v ⎠ T

According to the cyclic relation, ⎛ ∂v ⎞ ⎛ ∂P ⎞ ⎛ ∂T ⎞ ⎟ ⎜ ⎟ = −1 ⎜ ⎟ ⎜ ⎝ ∂T ⎠ P ⎝ ∂v ⎠ T ⎝ ∂P ⎠ v which on rearrangement becomes −1

⎛ ∂P ⎞ ⎛ ∂P ⎞ ⎛ ∂v ⎞ ⎟ ⎟ ⎜ ⎜ ⎟ = −⎜ ⎝ ∂T ⎠ v ⎝ ∂v ⎠ T ⎝ ∂T ⎠ P

Substituting this result into the expression for the specific heat difference gives −1

2

⎛ ∂P ⎞ ⎛ ∂P ⎞ c p − cv = −T ⎜ ⎟ ⎟ ⎜ ⎝ ∂T ⎠ v ⎝ ∂v ⎠ T

The appropriate partial derivatives of the equation of state are R a/2 ⎛ ∂P ⎞ + ⎜ ⎟ = ⎝ ∂T ⎠v v − b v (v + b)T 3 / 2

⎤ 1 RT a ⎡ 1 ⎛ ∂P ⎞ + 1/ 2 ⎢ 2 − ⎜ ⎟ =− ⎥ 2 2 (v + b) ⎦⎥ (v − b) bT ⎝ ∂v ⎠ T ⎣⎢ v =−

RT (v − b)

2

a

+ T

1/ 2

2v + b 2

v (v + b) 2

The difference in the specific heats is then 2

⎡ R ⎤ a/2 −T ⎢ + 3/ 2 ⎥ ⎣⎢ v − b v (v + b)T ⎦⎥ c p − cv = 2v + b RT a − + (v − b) 2 T 1 / 2 v 2 (v + b) 2

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12-38

12-51 An expression for the volume expansivity of a substance whose equation of state is P =

RT

−

is to be derived. Analysis The definition for volume expansivity is β=

a

v − b v 2T

1 ⎛ ∂v ⎞ ⎜ ⎟ v ⎝ ∂T ⎠ P

According to the cyclic relation, ⎛ ∂v ⎞ ⎛ ∂P ⎞ ⎛ ∂T ⎞ ⎟ ⎜ ⎟ = −1 ⎜ ⎟ ⎜ ⎝ ∂T ⎠ P ⎝ ∂v ⎠ T ⎝ ∂P ⎠ v which on rearrangement becomes ⎛ ∂P ⎞ ⎜ ⎟ v ∂ ⎝ ∂T ⎠ v ⎛ ⎞ ⎜ ⎟ =− ⎛ ∂P ⎞ ⎝ ∂T ⎠ P ⎜ ⎟ ⎝ ∂v ⎠ T

Proceeding to perform the differentiations gives R a ⎛ ∂P ⎞ + 2 2 ⎜ ⎟ = ∂ − v T b ⎝ ⎠v v T

and 2a RT ⎛ ∂P ⎞ + 3 ⎜ ⎟ =− 2 (v − b) v T ⎝ ∂v ⎠ T

Substituting these results into the definition of the volume expansivity produces R a + 1 v − b v 2T 2 β=− 2a v − RT + 3 2 (v − b) v T

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12-39

12-52 An expression for the isothermal compressibility of a substance whose equation of state RT a is P = is to be derived. − v − b v 2T

Analysis The definition for the isothermal compressibility is

α=−

1 ⎛ ∂v ⎞ ⎜ ⎟ v ⎝ ∂P ⎠ T

Now, 2a − RT ⎛ ∂P ⎞ + 3 ⎜ ⎟ = 2 ⎝ ∂v ⎠ T (v − b) v T

Using the partial derivative properties, α=−

1 ⎛ ∂P ⎞ v⎜ ⎟ ⎝ ∂v ⎠ T

=−

1 − RTv (v − b)

2

+

2a

v 2T

⎛ ∂P ⎞ 12-53 It is to be shown that β = α ⎜ ⎟ . ⎝ ∂T ⎠ v

Analysis The definition for the volume expansivity is β=

1 ⎛ ∂v ⎞ ⎜ ⎟ v ⎝ ∂T ⎠ P

The definition for the isothermal compressibility is α=−

1 ⎛ ∂v ⎞ ⎜ ⎟ v ⎝ ∂P ⎠ T

According to the cyclic relation, ⎛ ∂v ⎞ ⎛ ∂P ⎞ ⎛ ∂T ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = −1 ⎝ ∂T ⎠ P ⎝ ∂v ⎠ T ⎝ ∂P ⎠ v which on rearrangement becomes ⎛ ∂v ⎞ ⎛ ∂v ⎞ ⎛ ∂P ⎞ ⎜ ⎟ = −⎜ ⎟ ⎜ ⎟ ∂ T ⎝ ⎠P ⎝ ∂P ⎠ T ⎝ ∂T ⎠ v

When this is substituted into the definition of the volume expansivity, the result is β=−

1 ⎛ ∂v ⎞ ⎛ ∂P ⎞ ⎟ ⎜ ⎟ ⎜ v ⎝ ∂P ⎠ T ⎝ ∂T ⎠v

⎛ ∂P ⎞ = − α⎜ ⎟ ⎝ ∂T ⎠ v

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12-40

12-54 It is to be demonstrated that k =

cp cv

=−

vα

(∂v / ∂P )s

.

Analysis The relations for entropy differential are ds = cv

dT ⎛ ∂P ⎞ +⎜ ⎟ dv T ⎝ ∂T ⎠ v

ds = c p

dT ⎛ ∂v ⎞ −⎜ ⎟ dP T ⎝ ∂T ⎠ P

For fixed s, these basic equations reduce to cv

dT ⎛ ∂P ⎞ = −⎜ ⎟ dv T ⎝ ∂T ⎠ v

cp

dT ⎛ ∂v ⎞ =⎜ ⎟ dP T ⎝ ∂T ⎠ P

Also, when s is fixed, ∂v ⎛ ∂v ⎞ =⎜ ⎟ ∂P ⎝ ∂P ⎠ s Forming the specific heat ratio from these expressions gives ⎛ ∂v ⎞ ⎛ ∂T ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠ P ⎝ ∂P ⎠ v k=− ⎛ ∂v ⎞ ⎟ ⎜ ⎝ ∂P ⎠ s

The cyclic relation is ⎛ ∂v ⎞ ⎛ ∂P ⎞ ⎛ ∂T ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = −1 ⎝ ∂T ⎠ P ⎝ ∂v ⎠ T ⎝ ∂P ⎠ v

Solving this for the numerator of the specific heat ratio expression and substituting the result into this numerator produces ⎛ ∂v ⎞ ⎜ ⎟ vα ⎝ ∂P ⎠ T k= =− ⎛ ∂v ⎞ ⎛ ∂v ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ∂P ⎠ s ⎝ ∂P ⎠ s

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12-41

12-55 An expression for the specific heat difference of a substance whose equation of state is P =

RT

−

is to be derived. Analysis The specific heat difference is expressed by 2

⎛ ∂v ⎞ ⎛ ∂P ⎞ c p − cv = −T ⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠ P ⎝ ∂v ⎠ T

According to the cyclic relation, ⎛ ∂v ⎞ ⎛ ∂P ⎞ ⎛ ∂T ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = −1 ⎝ ∂T ⎠ P ⎝ ∂v ⎠ T ⎝ ∂P ⎠ v

which on rearrangement becomes −1

⎛ ∂v ⎞ ⎛ ∂P ⎞ ⎛ ∂P ⎞ ⎜ ⎟ = −⎜ ⎟ ⎜ ⎟ T ∂ ⎝ ⎠P ⎝ ∂T ⎠ v ⎝ ∂v ⎠ T

Substituting this result into the expression for the specific heat difference gives 2

a

v − b v 2T

−1

⎛ ∂P ⎞ ⎛ ∂P ⎞ c p − cv = −T ⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠ v ⎝ ∂v ⎠ T

The appropriate partial derivatives of the equation of state are R a ⎛ ∂P ⎞ + 2 2 ⎜ ⎟ = ⎝ ∂T ⎠ v v − b v T 2a RT ⎛ ∂P ⎞ + 3 ⎜ ⎟ =− 2 ∂ v ⎝ ⎠T (v − b) v T

The difference in the specific heats is then 2

a ⎤ ⎡ R ⎢v − b + 2 2 ⎥ v T ⎦ ⎣ c p − cv = −T 2a RT − + 3 2 (v − b) v T

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12-42

The Joule-Thomson Coefficient 12-56C It represents the variation of temperature with pressure during a throttling process. 12-57C The line that passes through the peak points of the constant enthalpy lines on a T-P diagram is called the inversion line. The maximum inversion temperature is the highest temperature a fluid can be cooled by throttling. 12-58C No. The temperature may even increase as a result of throttling. 12-59C Yes. 12-60C No. Helium is an ideal gas and h = h(T) for ideal gases. Therefore, the temperature of an ideal gas remains constant during a throttling (h = constant) process.

12-61E [Also solved by EES on enclosed CD] The Joule-Thompson coefficient of nitrogen at two states is to be estimated.

Analysis (a) The enthalpy of nitrogen at 200 psia and 500 R is, from EES, h = -10.564 Btu/lbm. Note that in EES, by default, the reference state for specific enthalpy and entropy is 0 at 25ºC (77ºF) and 1 atm. Approximating differentials by differences about the specified state, the Joule-Thomson coefficient is expressed as ⎛ ∂T ⎞ ⎛ ΔT ⎞ μ=⎜ ⎟ ≅⎜ ⎟ P ∂ ⎝ ⎠ h ⎝ ΔP ⎠ h = −10.564 Btu/lbm

Considering a throttling process from 210 psia to 190 psia at h = -10.564 Btu/lbm, the Joule-Thomson coefficient is determined to be ⎛ T190 psia − T210 psia μ = ⎜⎜ ⎝ (190 − 210) psia

⎞ (499.703 − 500.296) R ⎟ = = 0.0297 R/psia ⎟ (190 − 210) psia ⎠ h = −10.564 Btu/lbm

(b) The enthalpy of nitrogen at 2000 psia and 400 R is, from EES, h = -55.321 Btu/lbm. Approximating differentials by differences about the specified state, the Joule-Thomson coefficient is expressed as ⎛ ∂T ⎞ ⎛ ΔT ⎞ μ=⎜ ⎟ ≅⎜ ⎟ ⎝ ∂P ⎠ h ⎝ ΔP ⎠ h = −55.321 Btu/lbm

Considering a throttling process from 2010 psia to 1990 psia at h = -55.321 Btu/lbm, the Joule-Thomson coefficient is determined to be ⎛ T1999 psia − T2001 psia μ = ⎜⎜ ⎝ (1990 − 2010) psia

⎞ (399.786 - 400.213) R ⎟ = = 0.0213 R/psia ⎟ (1990 − 2010) psia ⎠ h = −55.321 Btu/lbm

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12-43

12-62E EES Problem 12-61E is reconsidered. The Joule-Thompson coefficient for nitrogen over the pressure range 100 to 1500 psia at the enthalpy values 100, 175, and 225 Btu/lbm is to be plotted. Analysis The problem is solved using EES, and the results are tabulated and plotted below. Gas$ = 'Nitrogen' {P_ref=200 [psia] T_ref=500 [R] P= P_ref} h=50 [Btu/lbm] {h=enthalpy(Gas$, T=T_ref, P=P_ref)} dP = 10 [psia] T = temperature(Gas$, P=P, h=h) P[1] = P + dP P[2] = P - dP T[1] = temperature(Gas$, P=P[1], h=h) T[2] = temperature(Gas$, P=P[2], h=h) Mu = DELTAT/DELTAP "Approximate the differential by differences about the state at h=const." DELTAT=T[2]-T[1] DELTAP=P[2]-P[1]

h = 100 Btu/lbm P [psia] µ [R/psia] 100 0.003675 275 0.003277 450 0.002899 625 0.00254 800 0.002198 975 0.001871 1150 0.001558 1325 0.001258 1500 0.0009699

0.004 0.003 0.002 h = 100 Btu/lbm

μ [R/psia]

0.001 0 -0.001 -0.002

h = 175 Btu/lbm

-0.003 -0.004

h = 225 Btu/lbm

-0.005 -0.006 0

200

400

600

800

1000

1200

1400

1600

P [psia]

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12-44

12-63 The Joule-Thompson coefficient of refrigerant-134a at a specified state is to be estimated. Analysis The enthalpy of refrigerant-134a at 0.7 MPa and T = 50°C is h = 288.53 kJ/kg. Approximating differentials by differences about the specified state, the Joule-Thomson coefficient is expressed as ⎛ ∂T ⎞ ⎛ ΔT ⎞ μ=⎜ ⎟ ≅⎜ ⎟ ⎝ ∂P ⎠ h ⎝ ΔP ⎠ h = 288.53 kJ/kg

Considering a throttling process from 0.8 MPa to 0.6 MPa at h = 288.53 kJ/kg, the Joule-Thomson coefficient is determined to be ⎛ T0.8 MPa − T0.6 MPa μ = ⎜⎜ ⎝ (0.8 − 0.6)MPa

⎞ (51.81 − 48.19)°C ⎟ = = 18.1 °C/MPa ⎟ (0.8 − 0.6)MPa ⎠ h = 288.53 kJ/kg

12-64 Steam is throttled slightly from 1 MPa and 300°C. It is to be determined if the temperature of the steam will increase, decrease, or remain the same during this process. Analysis The enthalpy of steam at 1 MPa and T = 300°C is h = 3051.6 kJ/kg. Now consider a throttling process from this state to 0.8 MPa, which is the next lowest pressure listed in the tables. The temperature of the steam at the end of this throttling process will be

P = 0.8 MPa ⎫ ⎬ T2 = 297.52°C h = 3051.6 kJ/kg ⎭ Therefore, the temperature will decrease.

12-65 It is to be demonstrated that the Joule-Thomson coefficient is given by μ =

T 2 ⎛ ∂ (v / T ) ⎞ ⎜ ⎟ . c p ⎝ ∂T ⎠ P

Analysis From Eq. 12-52 of the text, cp =

⎤ 1 ⎡ ⎛ ∂v ⎞ ⎟ −v ⎥ ⎢T ⎜ μ ⎣ ⎝ ∂T ⎠ P ⎦

Expanding the partial derivative of v/T produces 1 ⎛ ∂v ⎞ v ⎛ ∂v / T ⎞ ⎜ ⎟ = ⎜ ⎟ − 2 ⎝ ∂T ⎠ P T ⎝ ∂T ⎠ P T When this is multiplied by T2, the right-hand side becomes the same as the bracketed quantity above. Then,

μ=

T 2 ⎛ ∂ (v / T ) ⎞ ⎜ ⎟ c p ⎝ ∂T ⎠ P

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12-45

12-66 The most general equation of state for which the Joule-Thomson coefficient is always zero is to be determined. Analysis From Eq. 12-52 of the text, cp =

⎤ 1 ⎡ ⎛ ∂v ⎞ ⎟ −v ⎥ ⎢T ⎜ μ ⎣ ⎝ ∂T ⎠ P ⎦

When μ = 0, this equation becomes

v ⎛ ∂v ⎞ ⎜ ⎟ = ⎝ ∂T ⎠ P T This can only be satisfied by an equation of state of the form

v T

= f ( P)

where f(P) is an arbitrary function of the pressure.

12-67E The Joule-Thomson coefficient of refrigerant-134a at a given state is to be estimated. Analysis The Joule-Thomson coefficient is defined as ⎛ ∂T ⎞ ⎟ ⎝ ∂P ⎠ h

μ =⎜

We use a finite difference approximation as

μ≅

T2 − T1 (at constant enthalpy) P2 − P1

At the given state (we call it state 1), the enthalpy of R-134a is P1 = 30 psia ⎫ ⎬ h1 = 106.27 Btu/lbm (Table A - 13E) T1 = 20°F ⎭

The second state will be selected for a pressure of 20 psia. At this pressure and the same enthalpy, we have P2 = 20 psia h2 = h1 = 106.27 Btu/lbm

⎫ ⎬ T2 = 15.65°F (Table A - 13E) ⎭

Substituting,

μ≅

T2 − T1 (15.65 − 20)R = = 0.435 R/psia P2 − P1 (20 − 30)psia

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12-46

12-68 The Joule-Thomson coefficient of refrigerant-134a at a given state is to be estimated. Analysis The Joule-Thomson coefficient is defined as ⎛ ∂T ⎞ ⎟ ⎝ ∂P ⎠ h

μ =⎜

We use a finite difference approximation as

μ≅

T2 − T1 (at constant enthalpy) P2 − P1

At the given state (we call it state 1), the enthalpy of R-134a is P1 = 200 kPa ⎫ ⎬ h1 = 270.18 kJ/kg (Table A - 13) T1 = 20°C ⎭

The second state will be selected for a pressure of 180 kPa. At this pressure and the same enthalpy, we have P2 = 180 kPa h2 = h1 = 270.18 kJ/kg

⎫ ⎬ T2 = 19.51°C (Table A - 13) ⎭

Substituting,

μ≅

T2 − T1 (19.51 − 20)K = = 0.0245 K/kPa P2 − P1 (180 − 200)kPa

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12-47

RT a − + b . An equation for the Joule-Thomson P T coefficient inversion line using this equation is to be derived.

12-69 The equation of state of a gas is given by v =

Analysis From Eq. 12-52 of the text, cp =

⎤ 1 ⎡ ⎛ ∂v ⎞ ⎟ −v ⎥ ⎢T ⎜ μ ⎣ ⎝ ∂T ⎠ P ⎦

When μ = 0 as it does on the inversion line, this equation becomes ⎛ ∂v ⎞ T⎜ ⎟ =v ⎝ ∂T ⎠ P Using the equation of state to evaluate the partial derivative, R a ⎛ ∂v ⎞ ⎜ ⎟ = + 2 T ∂ ⎝ ⎠P P T

Substituting this result into the previous expression produces ⎛ R a ⎞ RT a T⎜ + 2 ⎟ = − +b ⎝P T ⎠ P T Solving this for the temperature gives T=

2a b

as the condition along the inversion line.

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12-48

The Δh, Δu, and Δs of Real Gases

12-70C It is the variation of enthalpy with pressure at a fixed temperature. 12-71C As PR approaches zero, the gas approaches ideal gas behavior. As a result, the deviation from ideal gas behavior diminishes. 12-72C So that a single chart can be used for all gases instead of a single particular gas.

12-73 The errors involved in the enthalpy and internal energy of CO2 at 350 K and 10 MPa if it is assumed to be an ideal gas are to be determined. Analysis (a) The enthalpy departure of CO2 at the specified state is determined from the generalized chart to be (Fig. A-29) 350 T ⎫ = = 1.151 ⎪ (hideal − h ) T , P Tcr 304.2 ⎪ ⎯→ Z h = = 1.5 ⎬⎯ 10 P Ru Tcr PR = = = 1.353 ⎪ ⎪⎭ Pcr 7.39

TR =

and

CO2 350 K 10 MPa

Thus, h = hideal − Z h Ru Tcr = 11,351 − [(1.5)(8.314)(304.2)] = 7,557kJ/kmol

and, Error =

(hideal − h ) T , P h

=

11,351 − 7,557 = 50.2% 7,557

(b) At the calculated TR and PR the compressibility factor is determined from the compressibility chart to be Z = 0.65. Then using the definition of enthalpy, the internal energy is determined to be u = h − Pv = h − ZRu T = 7557 − [(0.65)(8.314)(350)] = 5,666kJ/kmol

and, Error =

u ideal − u 8,439 − 5,666 = = 48.9% u 5,666

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12-49

12-74 The enthalpy and entropy changes of nitrogen during a process are to be determined assuming ideal gas behavior and using generalized charts. Analysis (a) Using data from the ideal gas property table of nitrogen (Table A-18), (h2 − h1 ) ideal = h2,ideal − h1,ideal = 9306 − 6,537 = 2769 kJ/kmol

and ( s 2 − s1 ) ideal = s 2o − s1o − Ru ln

P2 12 = 193.562 − 183.289 − 8.314 × ln = 4.510 kJ/kmol ⋅ K P1 6

(b) The enthalpy and entropy departures of nitrogen at the specified states are determined from the generalized charts to be (Figs. A-29, A-30) T1 225 ⎫ = = 1.783 ⎪ Tcr 126.2 ⎪ ⎯→ Z h1 = 0.6 and Z s1 = 0.25 ⎬⎯ P1 6 ⎪ = = = 1.770 ⎪⎭ Pcr 3.39

T R1 = PR1

and T2 320 ⎫ = = 2.536 ⎪ Tcr 126.2 ⎪ ⎯→ Z h 2 = 0.4 and Z s 2 = 0.15 ⎬⎯ P2 12 ⎪ = = = 2.540 ⎪⎭ Pcr 3.39

TR 2 = PR 2

Substituting, h2 − h1 = Ru Tcr ( Z h1 − Z h 2 ) + ( h2 − h1 ) ideal = (8.314 )(126.2 )(0.6 − 0.4) + 2769 = 2979 kJ/kmol

s 2 − s1 = Ru ( Z s1 − Z s 2 ) + ( s 2 − s1 ) ideal = (8.314)(0.25 − 0.15) + 4.510 = 5.341 kJ/kmol ⋅ K

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12-50

12-75 Methane is compressed adiabatically by a steady-flow compressor. The required power input to the compressor is to be determined using the generalized charts. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis The steady-flow energy balance equation for this compressor can be expressed as E& in − E& out = ΔE& system ©0 (steady) = 0 E& in = E& out

10 MPa 110°C

W& C,in + m& h1 = m& h2 W& C,in = m& (h2 − h1 )

The enthalpy departures of CH4 at the specified states are determined from the generalized charts to be (Fig. A-29) T1 263 ⎫ = = 1.376 ⎪ Tcr 191.1 ⎪ ⎯→ Z h1 = 0.21 ⎬⎯ P1 2 ⎪ = = = 0.431 ⎪⎭ Pcr 4.64

CH4 · = 0.55 kg/s m

T R1 = PR1

2 MPa -10 °C

and T2 383 ⎫ = = 2.00 ⎪ Tcr 191.1 ⎪ ⎯→ Z h 2 = 0.50 ⎬⎯ P2 10 ⎪ = = = 2.155 ⎪⎭ Pcr 4.64

TR 2 = PR 2

Thus, h2 − h1 = RTcr ( Z h1 − Z h 2 ) + (h2 − h1 ) ideal

= (0.5182)(191.1)(0.21 − 0.50 ) + 2.2537(110 − (− 10 )) = 241.7 kJ/kg

Substituting, W& C,in = (0.55 kg/s )(241.7 kJ/kg ) = 133 kW

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12-51

12-76E The enthalpy and entropy changes of water vapor during a change of state are to be determined using the departure charts and the property tables. Properties The properties of water are (Table A-1E) M = 18.015 lbm/lbmol, Tcr = 1164.8 R, Pcr = 3200 psia

Analysis (a) Using data from the ideal gas property table of water vapor (Table A-23E), (h2 − h1 ) ideal = h2,ideal − h1,ideal = 12,178.8 − 17,032.5 = −4853.7 Btu/lbmol

and ( s 2 − s1 ) ideal = s 2o − s1o − Ru ln

P2 1000 = 53.556 − 56.411 − 1.9858 × ln = −0.6734 Btu/lbmol ⋅ R P1 3000

The enthalpy and entropy departures of water vapor at the specified states are determined from the generalized charts to be (Figs. A-29, A-30 or from EES) T1 1960 ⎫ = = 1.683 ⎪ Tcr 1164.8 ⎪ ⎯→ Z h1 = 0.387 and Z s1 = 0.188 ⎬⎯ P1 3000 = = = 0.9375 ⎪ ⎪⎭ Pcr 3200

T R1 = PR1

and T2 1460 ⎫ = = 1.253 ⎪ Tcr 1164.8 ⎪ ⎯→ Z h 2 = 0.233 and Z s 2 = 0.134 ⎬⎯ P 1000 = 2 = = 0.3125 ⎪ ⎪⎭ Pcr 3200

TR 2 = PR 2

The enthalpy and entropy changes per mole basis are h2 − h1 = (h2 − h1 ) ideal − Ru Tcr ( Z h 2 − Z h1 ) = −4853.7 − (1.9858)(1164.8)(0.233 − 0.387) = −4497.5 Btu/lbmol s 2 − s1 = ( s 2 − s1 ) ideal − Ru ( Z s 2 − Z s1 ) = −0.6734 − (1.9858)(0.134 − 0.188) = −0.5662 Btu/lbmol ⋅ R

The enthalpy and entropy changes per mass basis are h2 − h1 − 4497.5 Btu/lbmol = = −249.7 Btu/lbm M 18.015 lbm/lbmol s −s − 0.5662 Btu/lbmol ⋅ R = −0.0314 Btu/lbm ⋅ R s 2 − s1 = 2 1 = 18.015 lbm/lbmol M

h2 − h1 =

(b) Using water tables (Table A-6E) P1 = 3000 psia ⎫ h1 = 1764.6 Btu/lbm ⎬ T1 = 1500°F ⎭ s1 = 1.6883 Btu/lbm ⋅ R P2 = 1000 psia ⎫ h2 = 1506.2 Btu/lbm ⎬ T2 = 1000°F ⎭ s 2 = 1.6535 Btu/lbm ⋅ R

The enthalpy and entropy changes are h2 − h1 = 1506.2 − 1764.6 = −258.4 Btu/lbm s 2 − s1 = 1.6535 − 1.6883 = −0.0348 Btu/lbm ⋅ R

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12-52

12-77 The enthalpy and entropy changes of water vapor during a change of state are to be determined using the departure charts and the property tables. Properties The properties of water are (Table A-1) M = 18.015 kg/kmol, Tcr = 647.1 K, Pcr = 22.06 MPa

Analysis Using data from the ideal gas property table of water vapor (Table A-23), (h2 − h1 ) ideal = h2,ideal − h1,ideal = 23,082 − 30,754 = −7672 kJ/kmol

and ( s 2 − s1 ) ideal = s 2o − s1o − Ru ln

P2 500 = 217.141 − 227.109 − 8.314 × ln = −4.2052 kJ/kmol ⋅ K P1 1000

The enthalpy and entropy departures of water vapor at the specified states are determined from the generalized charts to be (Figs. A-29, A-30 or from EES) T1 873 ⎫ = = 1.349 ⎪ Tcr 647.1 ⎪ ⎯→ Z h1 = 0.0288 and Z s1 = 0.0157 ⎬⎯ P1 1 = = = 0.0453⎪ ⎪⎭ Pcr 22.06

T R1 = PR1

and T2 673 ⎫ = = 1.040 ⎪ Tcr 647.1 ⎪ ⎯→ Z h 2 = 0.0223 and Z s 2 = 0.0146 ⎬⎯ P 0.5 = 2 = = 0.0227 ⎪ ⎪⎭ Pcr 22.06

TR 2 = PR 2

The enthalpy and entropy changes per mole basis are h2 − h1 = (h2 − h1 ) ideal − Ru Tcr ( Z h 2 − Z h1 ) = −7672 − (8.314)(647.1)(0.0223 − 0.0288) = −7637 kJ/kmol s 2 − s1 = ( s 2 − s1 ) ideal − Ru ( Z s 2 − Z s1 ) = −4.2052 − (8.314)(0.0146 − 0.0157) = −4.1961 kJ/kmol ⋅ K

The enthalpy and entropy changes per mass basis are h2 − h1 − 7637 kJ/kmol = = −423.9 kJ/kg 18.015 kg/kmol M s −s − 4.1961 kJ/kmol ⋅ K = −0.2329 kJ/kg ⋅ K s 2 − s1 = 2 1 = 18.015 kg/kmol M

h2 − h1 =

The inlet and exit state properties of water vapor from Table A-6 are P1 = 1000 kPa ⎫ h1 = 3698.6 kJ/kg ⎬ T1 = 600°C ⎭ s1 = 8.0311 kJ/kg ⋅ K P2 = 500 kPa ⎫ h2 = 3272.4 kJ/kg ⎬ T2 = 400°C ⎭ s 2 = 7.7956 kJ/kg ⋅ K

The enthalpy and entropy changes are h2 − h1 = 3272.4 − 3698.6 = −426.2 kJ/kg s 2 − s1 = 7.7956 − 8.0311 = −0.2355 kJ/kg ⋅ K

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12-53

12-78 The enthalpy and entropy changes of water vapor during a change of state are to be determined using the departure charts and the property tables. Properties The properties of water are (Table A-1) M = 18.015 kg/kmol, Tcr = 647.1 K, Pcr = 22.06 MPa Analysis (a) The pressure of water vapor during this process is P1 = P2 = Psat @ 300°C = 8588 kPa

Using data from the ideal gas property table of water vapor (Table A-23), (h2 − h1 ) ideal = h2,ideal − h1,ideal = 34,775 − 19,426 = 15,349 kJ/kmol

and ( s 2 − s1 ) ideal = s 2o − s1o − Ru ln

P2 = 231.473 − 211.263 − 0 = 20.210 kJ/kmol ⋅ K P1

The enthalpy and entropy departures of water vapor at the specified states are determined from the generalized charts to be (Figs. A-29, A-30 or from EES) T 573 ⎫ T R1 = 1 = = 0.885 ⎪ Tcr 647.1 ⎪ ⎯→ Z h1 = 0.609 and Z s1 = 0.481 ⎬⎯ P 8.588 PR1 = 1 = = 0.389 ⎪ ⎪⎭ Pcr 22.06 and T 973 ⎫ TR 2 = 2 = = 1.504 ⎪ Tcr 647.1 ⎪ ⎯→ Z h 2 = 0.204 and Z s 2 = 0.105 ⎬⎯ P 8.588 PR 2 = 2 = = 0.389⎪ ⎪⎭ Pcr 22.06 The enthalpy and entropy changes per mole basis are h2 − h1 = (h2 − h1 ) ideal − Ru Tcr ( Z h 2 − Z h1 ) = 15,349 − (8.314)(647.1)(0.204 − 0.609) = 17,528 kJ/kmol s 2 − s1 = ( s 2 − s1 ) ideal − Ru ( Z s 2 − Z s1 ) = 20.210 − (8.314)(0.105 − 0.481) = 23.336 kJ/kmol ⋅ K

The enthalpy and entropy changes per mass basis are h2 − h1 17,528 kJ/kmol = = 973.0 kJ/kg 18.015 kg/kmol M s −s 23.336 kJ/kmol ⋅ K = 1.2954 kJ/kg ⋅ K s 2 − s1 = 2 1 = 18.015 kg/kmol M

h2 − h1 =

(b) The inlet and exit state properties of water vapor from Table A-6 are T1 = 300°C⎫ h1 = 2749.6 kJ/kg ⎬ x1 = 1 ⎭ s1 = 5.7059 kJ/kg ⋅ K P2 = 8588 kPa ⎫ h2 = 3878.6 kJ/kg (from EES) ⎬ T2 = 700°C ⎭ s 2 = 7.2465 kJ/kg ⋅ K

The enthalpy and entropy changes are h2 − h1 = 3878.6 − 2749.6 = 1129 kJ/kg s 2 − s1 = 7.2465 − 5.7059 = 1.5406 kJ/kg ⋅ K

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12-54

12-79 Propane is to be adiabatically and reversibly compressed in a steady-flow device. The specific work required for this compression is to be determined using the departure charts and treating the propane as an ideal gas with temperature variable specific heats. Properties The properties of propane are (Table A-1) M = 44.097 kg/kmol, R = 0.1885 kJ/kg ⋅ K, Tcr = 370 K, Pcr = 4.26 MPa

Analysis The temperature at the exit state may be determined by the fact that the process is isentropic and the molar entropy change between the inlet and exit is zero. When the entropy change equation is integrated with variable specific heats, it becomes 2

( s 2 − s1 ) ideal =

∫ 1

cp

P dT − Ru ln 2 P1 T

When the expression of Table A-2c is substituted for cp and the integration performed, we obtain 2

( s 2 − s1 ) ideal =

cp

∫T

Propane

500 kPa 100°C

2

dT − Ru ln

1

= a ln

4 MPa

P P2 ⎞ ⎛a = ⎜ + b + cT + dT 2 ⎟dT − Ru ln 2 P1 P1 ⎠ ⎝T 1

∫

P T2 c d + b(T2 − T1 ) + (T22 − T12 ) + (T23 − T13 ) − Ru ln 2 P1 T1 2 3

⎡⎛ T ⎞ 3 ⎤ ⎡⎛ T ⎞ 2 ⎤ T2 ⎞ ⎛T + 30.48⎜⎜ 2 − 3.73 ⎟⎟ − 0.786⎢⎜⎜ 2 ⎟⎟ − 3.73 2 ⎥ + 0.01058⎢⎜⎜ 2 ⎟⎟ − 3.73 3 ⎥ 373 ⎢⎣⎝ 100 ⎠ ⎥⎦ ⎢⎣⎝ 100 ⎠ ⎥⎦ ⎠ ⎝ 100 4000 − (8.314) ln 500

0 = −4.04 ln

Solving this equation by EES or an iterative solution by hand gives T2 = 446 K

When en energy balance is applied to the compressor, it becomes 2

∫

2

∫

win = (h2 − h1 ) ideal = c p dT = (a + bT + cT 2 + dT 3 )dT 1

1

b c d = a (T2 − T1 ) + (T22 − T12 ) + (T23 − T13 ) + (T24 − T14 ) 2 3 4 2 2 = −4.04(446 − 373) + 0.1524(446 − 373 ) − 52.4(4.46 3 − 3.73 3 ) + 0.7935(4.46 4 − 3.73 4 ) = 7048 kJ/kmol

The work input per unit mass basis is win =

win 7048 kJ/kmol = = 159.8 kJ/kg M 44.097 kg/kmol

The enthalpy departures of propane at the specified states are determined from the generalized charts to be (Fig. A-29 or from EES)

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12-55

T1 373 ⎫ = = 1.008 ⎪ Tcr 370 ⎪ ⎯→ Z h1 = 0.124 ⎬⎯ P 0.5 = 1 = = 0.117⎪ ⎪⎭ Pcr 4.26

T R1 = PR1

and T2 446 ⎫ = = 1.205 ⎪ Tcr 370 ⎪ ⎯→ Z h 2 = 0.826 ⎬⎯ P2 4 ⎪ = = = 0.939 ⎪⎭ Pcr 4.26

TR 2 = PR 2

The work input (i.e., enthalpy change) is determined to be win = h2 − h1 = (h2 − h1 ) ideal − RTcr ( Z h 2 − Z h1 ) = 159.8 − (0.1885)(370)(0.826 − 0.124) = 110.8 kJ/kg

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

12-56

12-80E Oxygen is to be adiabatically and reversibly expanded in a nozzle. The exit velocity is to be determined using the departure charts and treating the oxygen as an ideal gas with temperature variable specific heats. Properties The properties of oxygen are (Table A-1) M = 31.999 lbm/lbmol, R = 0.06206 Btu/lbm ⋅ R, Tcr = 278.6 R, Pcr = 736 psia Analysis The temperature at the exit state may be determined by the fact that the process is isentropic and the molar entropy change between the inlet and exit is zero. From the entropy change equation for an ideal gas with variable specific heats: ( s 2 − s1 ) ideal = 0 200 psia O2 600°F P 70 70 psia s 2o − s1o = Ru ln 2 = (1.9858) ln = −2.085 Btu/lbmol ⋅ R ≈ 0 ft/s P 200 1

Then from Table A-19E, T1 = 1060 R ⎯ ⎯→ h1,ideal = 7543.6 Btu/lbmol, s1o = 53.921 Btu/lbmol ⋅ R s 2o = s1o − 2.085 = 53.921 − 2.085 = 51.836 Btu/lbmol ⋅ R s 2o = 51.836 Btu/lbmol ⋅ R ⎯ ⎯→ T2 = 802 R, h2,ideal = 5614.1 Btu/lbmol

The enthalpy change per mole basis is (h2 − h1 ) ideal = h2,ideal − h1,ideal = 5614.1 − 7543.6 = −1929.5 Btu/lbmol The enthalpy change per mass basis is (h2 − h1 ) ideal − 1929.5 Btu/lbmol = = −60.30 Btu/lbm M 31.999 lbm/lbmol An energy balance on the nozzle gives E& = E& (h2 − h1 ) ideal =

in

out

m& (h1 + V12 / 2) = m& (h2 + V 22 /2) h1 + V12 / 2 = h2 + V 22 /2

Solving for the exit velocity, 0.5

⎡ ⎛ 25,037 ft 2 /s 2 ⎞⎤ ⎟⎥ = 1738 ft/s + 2(h1 − h2 ) = ⎢(0 ft/s) 2 + 2(60.30 Btu/lbm)⎜ V2 = ⎜ 1 Btu/lbm ⎟⎥ ⎢⎣ ⎝ ⎠⎦ The enthalpy departures of oxygen at the specified states are determined from the generalized charts to be (Fig. A-29 or from EES) T T 1060 802 ⎫ ⎫ = 3.805 ⎪ = 2.879 ⎪ T R1 = 1 = TR 2 = 2 = Tcr 278.6 T 278 . 6 ⎪ ⎪ cr ⎬ Z h1 = 0.000759 ⎬ Z h 2 = 0.00894 P1 P 200 70 2 ⎪ = = 0.272 = = 0.0951 ⎪ PR1 = PR 2 = ⎪⎭ ⎪⎭ Pcr 736 Pcr 736 The enthalpy change is h2 − h1 = ( h2 − h1 ) ideal − RTcr ( Z h 2 − Z h1 )

[

]

0.5

V12

= −60.30 Btu/lbm − (0.06206 Btu/lbm ⋅ R )(278.6 R )(0.00894 − 0.000759) = −60.44 Btu/lbm The exit velocity is

V2 =

[

V12

+ 2(h1 − h2 )

]

0.5

⎡ ⎛ 25,037 ft 2 /s 2 = ⎢(0 ft/s) 2 + 2(60.44 Btu/lbm)⎜ ⎜ 1 Btu/lbm ⎢⎣ ⎝

⎞⎤ ⎟⎥ ⎟⎥ ⎠⎦

0.5

= 1740 ft/s

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

12-57

12-81 [Also solved by EES on enclosed CD] Propane is compressed isothermally by a piston-cylinder device. The work done and the heat transfer are to be determined using the generalized charts. Assumptions 1 The compression process is quasi-equilibrium. 2 Kinetic and potential energy changes are negligible. Analysis (a) The enthalpy departure and the compressibility factors of propane at the initial and the final states are determined from the generalized charts to be (Figs. A-29, A-15) T1 373 ⎫ = = 1.008 ⎪ Tcr 370 ⎪ ⎯→ Z h1 = 0.28 and Z 1 = 0.92 ⎬⎯ P 1 = 1 = = 0.235⎪ ⎪⎭ Pcr 4.26

T R1 = PR1

and T2 373 ⎫ = = 1.008 ⎪ Tcr 370 ⎪ ⎯→ Z h 2 = 1.8 and Z 2 = 0.50 ⎬⎯ P 4 = 2 = = 0.939⎪ ⎪⎭ Pcr 4.26

TR 2 = PR 2

Propane 1 MPa 100 °C

Treating propane as a real gas with Zavg = (Z1+Z2)/2 = (0.92 + 0.50)/2 = 0.71, Pv = ZRT ≅ Z avg RT = C = constant

Then the boundary work becomes

∫

2

wb,in = − P dv = − 1

∫

2

1

C

v

dv = −C ln

v2 Z P Z RT / P2 = Z avg RT ln 2 = − Z ave RT ln 2 1 v1 Z 1 RT / P1 Z 1 P2

= −(0.71)(0.1885 kJ/kg ⋅ K )(373 K )ln

(0.50)(1) = 99.6 kJ/kg (0.92)(4)

Also, h2 − h1 = RTcr ( Z h1 − Z h 2 ) + (h2 − h1 ) ideal = (0.1885)(370)(0.28 − 1.8) + 0 = −106 kJ/kg u 2 − u1 = (h2 − h1 ) − R( Z 2 T2 − Z 1T1 ) = −106 − (0.1885)[(0.5)(373) − (0.92 )(373)] = −76.5 kJ/kg

Then the heat transfer for this process is determined from the closed system energy balance to be E in − E out = ΔE system q in + wb,in = Δu = u 2 − u1

q in = (u 2 − u1 ) − wb,in = −76.5 − 99.6 = −176.1 kJ/kg → q out = 176.1 kJ/kg

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Q

12-58

12-82 EES Problem 12-81 is reconsidered. This problem is to be extended to compare the solutions based on the ideal gas assumption, generalized chart data and real fluid (EES) data. Also, the solution is to be extended to carbon dioxide, nitrogen and methane. Analysis The problem is solved using EES, and the solution is given below. Procedure INFO(Name$, T[1] : Fluid$, T_critical, p_critical) If Name$='Propane' then T_critical=370 ; p_critical=4620 ; Fluid$='C3H8'; goto 10 endif If Name$='Methane' then T_critical=191.1 ; p_critical=4640 ; Fluid$='CH4'; goto 10 endif If Name$='Nitrogen' then T_critical=126.2 ; p_critical=3390 ; Fluid$='N2'; goto 10 endif If Name$='Oxygen' then T_critical=154.8 ; p_critical=5080 ; Fluid$='O2'; goto 10 endif If Name$='CarbonDioxide' then T_critical=304.2 ; p_critical=7390 ; Fluid$='CO2' ; goto 10 endif If Name$='n-Butane' then T_critical=425.2 ; p_critical=3800 ; Fluid$='C4H10' ; goto 10 endif 10: If T[1] 0.

Using the Tds relation:

⎯→ dh = T ds + v dP ⎯

(1) P = constant:

⎛∂ h⎞ ⎟⎟ = T ⎜⎜ ⎝ ∂ s ⎠P

(2) T = constant:

⎛ ∂P ⎞ ⎛ ∂h ⎞ ⎟ ⎜ ⎟ = T +v ⎜ ⎝ ∂s ⎠ T ⎝ ∂s ⎠ T

But the 4th Maxwell relation: Substituting:

dh dP = T +v ds ds

⎛ ∂T ⎞ ⎛ ∂P ⎞ ⎟ ⎜ ⎟ = −⎜ ⎝ ∂v ⎠ P ⎝ ∂s ⎠ T

1 ⎛ ∂T ⎞ ⎛ ∂h ⎞ ⎟ =T − ⎜ ⎟ = T −v⎜ β ⎝ ∂v ⎠ P ⎝ ∂s ⎠ T

Therefore, the slope of P = constant lines is greater than the slope of T = constant lines. (3) v = constant:

⎛ ∂P ⎞ ⎛ ∂h ⎞ ⎜ ⎟ = T + v ⎜ ⎟ (a) ⎝ ∂s ⎠ v ⎝ ∂s ⎠ v

From the ds relation:

Divide by dP holding v constant:

ds =

cv ⎛ ∂P ⎞ dT + ⎜ ⎟ dv T ⎝ ∂T ⎠ v

c ⎛ ∂s ⎞ ⎜ ⎟ = v T ⎝ ∂P ⎠v

⎛ ∂T ⎞ ⎜ ⎟ ⎝ ∂P ⎠v

or

T ⎛ ∂P ⎞ ⎛ ∂P ⎞ ⎜ ⎟ = ⎜ ⎟ (b) ⎝ ∂s ⎠ v cv ⎝ ∂T ⎠ v

Using the properties P, T, v, the cyclic relation can be expressed as ⎛ 1 ⎞ β ⎛ ∂v ⎞ ⎛ ∂P ⎞ ⎛ ∂P ⎞ ⎛ ∂P ⎞ ⎛ ∂T ⎞ ⎛ ∂v ⎞ ⎯→ ⎜ ⎟= ⎟ = (− β v )⎜ ⎟ ⎜ ⎟ = −⎜ ⎟ = −1 ⎯ ⎟ ⎜ ⎟ ⎜ ⎜ ⎝ − αv ⎠ α ⎝ ∂T ⎠ P ⎝ ∂v ⎠ T ⎝ ∂T ⎠ v ⎝ ∂T ⎠ v ⎝ ∂v ⎠ P ⎝ ∂P ⎠ T

(c )

where we used the definitions of α and β. Substituting (b) and (c) into (a),

Tβ v ⎛ ∂P ⎞ ⎛ ∂h ⎞ >T ⎜ ⎟ = T +v ⎜ ⎟ = T + cv α ⎝ ∂s ⎠ v ⎝ ∂s ⎠ v Here α is positive for all phases of all substances. T is the absolute temperature that is also positive, so is cv. Therefore, the second term on the right is always a positive quantity since β is given to be positive. Then we conclude that the slope of P = constant lines is less than the slope of v = constant lines.

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12-66

12-88 Using the cyclic relation and the first Maxwell relation, the other three Maxwell relations are to be obtained. Analysis (1) Using the properties P, s, v, the cyclic relation can be expressed as ⎛ ∂P ⎞ ⎛ ∂s ⎞ ⎛ ∂v ⎞ ⎟ = −1 ⎟ ⎜ ⎜ ⎟ ⎜ ⎝ ∂s ⎠ v ⎝ ∂v ⎠ P ⎝ ∂P ⎠ s

Substituting the first Maxwell relation,

⎛ ∂T ⎞ ⎛ ∂P ⎞ ⎜ ⎟ = −⎜ ⎟ , ⎝ ∂v ⎠ s ⎝ ∂s ⎠ v

⎛ ∂v ⎞ ⎛ ∂T ⎞ ⎛ ∂T ⎞ ⎛ ∂s ⎞ ⎛ ∂T ⎞ ⎛ ∂s ⎞ ⎛ ∂v ⎞ ⎯→ ⎜ ⎯→ ⎜ −⎜ ⎟ ⎟ =⎜ ⎟ =1 ⎯ ⎟ ⎜ ⎟ = −1 ⎯ ⎟ ⎜ ⎟ ⎜ ⎝ ∂P ⎠ s ⎝ ∂s ⎠ P ⎝ ∂P ⎠ s ⎝ ∂v ⎠ P ⎝ ∂v ⎠ s ⎝ ∂v ⎠ P ⎝ ∂P ⎠ s

(2) Using the properties T, v, s, the cyclic relation can be expressed as ⎛ ∂T ⎞ ⎛ ∂v ⎞ ⎛ ∂s ⎞ ⎟ = −1 ⎟ ⎜ ⎟ ⎜ ⎜ ⎝ ∂v ⎠ s ⎝ ∂s ⎠ T ⎝ ∂T ⎠ v

Substituting the first Maxwell relation,

⎛ ∂P ⎞ ⎛ ∂T ⎞ ⎜ ⎟ = −⎜ ⎟ , ⎝ ∂v ⎠ s ⎝ ∂s ⎠ v

⎛ ∂P ⎞ ⎛ ∂v ⎞ ⎛ ∂P ⎞ ⎛ ∂v ⎞ ⎛ ∂s ⎞ ⎯→ ⎯→ ⎜ −⎜ ⎟ =1 ⎯ ⎟ ⎜ ⎟ = −1 ⎯ ⎟ ⎜ ⎟ ⎜ ∂ ∂ ∂ s s T ⎝ ∂T ⎠ v ⎝ ∂s ⎠ T ⎠v ⎠T ⎝ ⎠v ⎝ ⎝

⎛ ∂P ⎞ ⎛ ∂s ⎞ ⎟ ⎟ =⎜ ⎜ v ∂ ⎠ T ⎝ ∂T ⎠ v ⎝

(3) Using the properties P, T, v, the cyclic relation can be expressed as

⎛ ∂P ⎞ ⎛ ∂T ⎞ ⎛ ∂v ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = −1 ⎝ ∂T ⎠v ⎝ ∂v ⎠ P ⎝ ∂P ⎠ T Substituting the third Maxwell relation,

⎛ ∂P ⎞ ⎛ ∂s ⎞ ⎟ , ⎜ ⎟ =⎜ ∂ v ⎝ ⎠ T ⎝ ∂T ⎠ v

⎛ ∂s ⎞ ⎛ ∂T ⎞ ⎛ ∂s ⎞ ⎛ ∂T ⎞ ⎛ ∂v ⎞ ⎯→ ⎯→ ⎜ ⎟ = −1 ⎯ ⎟ ⎜ ⎟ = −1 ⎯ ⎟ ⎜ ⎟ ⎜ ⎜ P v v ∂ ∂ ∂ ⎝ ∂P ⎠ T ⎝ ∂v ⎠ P ⎠T ⎠P ⎝ ⎠T ⎝ ⎝

⎛ ∂v ⎞ ⎛ ∂s ⎞ ⎟ ⎟ = −⎜ ⎜ P ∂ ⎝ ∂T ⎠ P ⎠T ⎝

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12-67

12-89 It is to be shown that the slope of a constant-pressure line on an h-s diagram is constant in the saturation region and increases with temperature in the superheated region. Analysis For P = constant, dP = 0 and the given relation reduces to dh = Tds, which can also be expressed as

⎛∂ h⎞ ⎜⎜ ⎟⎟ = T ⎝ ∂ s ⎠P

h P = const.

Thus the slope of the P = constant lines on an h-s diagram is equal to the temperature. (a) In the saturation region, T = constant for P = constant lines, and the slope remains constant. (b) In the superheat region, the slope increases with increasing temperature since the slope is equal temperature.

s

12-90 It is to be shown that ⎛ ∂v ⎞ ⎛ ∂P ⎞ ⎛ ∂P ⎞ ⎛ ∂v ⎞ cv = −T ⎜ ⎟ and c p = T ⎜ ⎟ ⎟ ⎜ ⎟ ⎜ ⎝ ∂T ⎠ s ⎝ ∂T ⎠ v ⎝ ∂T ⎠ s ⎝ ∂T ⎠ P

Analysis Using the definition of cv , ⎛ ∂s ⎞ ⎛ ∂P ⎞ ⎛ ∂s ⎞ cv = T ⎜ ⎟ ⎟ = T⎜ ⎟ ⎜ ∂ T ⎝ ∂P ⎠ v ⎝ ∂T ⎠ v ⎠v ⎝ ⎛ ∂v ⎞ ⎛ ∂s ⎞ Substituting the first Maxwell relation ⎜ ⎟ = −⎜ ⎟ , P ∂ ⎝ ∂T ⎠ s ⎝ ⎠v

⎛ ∂v ⎞ ⎛ ∂P ⎞ cv = −T ⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠ s ⎝ ∂T ⎠ v Using the definition of cp, ⎛ ∂s ⎞ ⎛ ∂v ⎞ ⎛ ∂s ⎞ c p = T⎜ ⎟ ⎟ ⎜ ⎟ = T⎜ T ∂ ⎝ ∂v ⎠ P ⎝ ∂T ⎠ P ⎠P ⎝

⎛ ∂s ⎞ ⎛ ∂P ⎞ Substituting the second Maxwell relation ⎜ ⎟ =⎜ ⎟ , ⎝ ∂v ⎠ P ⎝ ∂T ⎠ s ⎛ ∂P ⎞ ⎛ ∂v ⎞ cp = T⎜ ⎟ ⎟ ⎜ ⎝ ∂T ⎠ s ⎝ ∂T ⎠ P

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12-68

P ⎛ ∂s ⎞ 12-91 It is to be proven that for a simple compressible substance ⎜ ⎟ = . ⎝ ∂v ⎠ u T

Analysis The proof is simply obtained as ⎛ ∂u ⎞ −⎜ ⎟ −P P ⎝ ∂v ⎠ s ⎛ ∂s ⎞ =− = ⎟ = ⎜ T T v ∂ u ∂ ⎞ ⎛ ⎠u ⎝ ⎜ ⎟ ⎝ ∂s ⎠ v

⎛ ∂u ⎞ ⎛ ∂u ⎞ 12-92 It is to be proven by using the definitions of pressure and temperature, T = ⎜ ⎟ and P = −⎜ ⎟ ∂ s ⎝ ⎠v ⎝ ∂v ⎠ s

⎛ ∂h ⎞ that for ideal gases, the development of the constant-pressure specific heat yields ⎜ ⎟ = 0 ⎝ ∂P ⎠ T Analysis The definition for enthalpy is h = u + Pv

Then,

⎛ ∂u ⎞ ⎛ ∂v ⎞ ⎛ ∂P ⎞ ⎛ ∂h ⎞ ⎜ ⎟ = ⎜ ⎟ + P⎜ ⎟ +v ⎜ ⎟ ⎝ ∂P ⎠ T ⎝ ∂P ⎠ T ⎝ ∂P ⎠ T ⎝ ∂P ⎠ T Assume u = u (s, v) Then, ⎛ ∂u ⎞ ⎛ ∂u ⎞ du = ⎜ ⎟ ds + ⎜ ⎟ dv ⎝ ∂v ⎠ s ⎝ ∂s ⎠ v

⎛ ∂u ⎞ ⎛ ∂v ⎞ ⎛ ∂u ⎞ ⎛ ∂u ⎞ ⎛ ∂s ⎞ ⎜ ⎟ = ⎜ ⎟ ⎜ ⎟ +⎜ ⎟ ⎜ ⎟ P ∂ s ∂ P ∂ ⎝ ⎠ T ⎝ ⎠ v ⎝ ⎠ T ⎝ ∂v ⎠ s ⎝ ∂P ⎠ T ⎡ ⎛ ∂v ⎞ ⎤ ⎛ ∂v ⎞ ⎛ ∂v ⎞ ⎛ ∂u ⎞ ⎟ ⎟ = −(T + P )⎜ ⎟ ⎥ − P⎜ ⎜ ⎟ = T ⎢− ⎜ ⎝ ∂P ⎠ T ⎝ ∂P ⎠ T ⎝ ∂P ⎠ T ⎣ ⎝ ∂T ⎠ P ⎦ ⎛ ∂v ⎞ ⎛ ∂v ⎞ ⎛ ∂v ⎞ ⎛ ∂h ⎞ ⎟ +v ⎟ + v = −T ⎜ ⎟ + P⎜ ⎜ ⎟ = −(T + P)⎜ P P P ∂ ∂ ∂ ⎝ ∂P ⎠ T ⎠T ⎝ ⎠T ⎝ ⎝ ⎠T

For ideal gases

v=

RT R ⎛ ∂v ⎞ and ⎜ ⎟ = P ⎝ ∂P ⎠ T P

Then,

TR ⎛ ∂h ⎞ + v = −v + v = 0 ⎜ ⎟ =− P ⎝ ∂P ⎠ T

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12-69

⎛ ∂u ⎞ ⎛ ∂u ⎞ 12-93 It is to be proven by using the definitions of pressure and temperature, T = ⎜ ⎟ and P = −⎜ ⎟ ⎝ ∂s ⎠ v ⎝ ∂v ⎠ s

⎛ ∂u ⎞ that for ideal gases, the development of the constant-volume specific heat yields ⎜ ⎟ =0. ⎝ ∂v ⎠ T Analysis Assume u = u (s, v)

Then, ⎛ ∂u ⎞ ⎛ ∂u ⎞ du = ⎜ ⎟ ds + ⎜ ⎟ dv s ∂ ⎝ ∂v ⎠ s ⎝ ⎠v ⎛ ∂u ⎞ ⎛ ∂v ⎞ ⎛ ∂u ⎞ ⎛ ∂s ⎞ ⎛ ∂u ⎞ ⎟ ⎟ ⎜ ⎟ +⎜ ⎟ =⎜ ⎟ ⎜ ⎜ s v v ∂ ∂ ∂ ⎠ T ⎝ ∂v ⎠ s ⎝ ∂v ⎠ T ⎠ T ⎝ ⎠v ⎝ ⎝ ⎛ ∂s ⎞ = T⎜ ⎟ +P ⎝ ∂v ⎠ T

From Maxwell equation,

⎛ ∂P ⎞ ⎛ ∂s ⎞ T⎜ ⎟ + P = T⎜ ⎟ −P ⎝ ∂v ⎠ T ⎝ ∂T ⎠ v For ideal gases P=

RT

v

R ⎛ ∂P ⎞ and ⎜ ⎟ = ⎝ ∂T ⎠ v v

Then,

R ⎛ ∂u ⎞ ⎜ ⎟ =T −P = P−P =0 v v ∂ ⎝ ⎠T

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12-70

12-94 Expressions for h, u, so, Pr, and vr for an ideal gas whose cpo is given by ⎛ β /T ⎞ c op = ai T i−n + a0 e β / T ⎜ β / T ⎟ are to be developed. −1 ⎠ ⎝e

∑

Analysis The enthalpy of this substance relative to a reference state is given by T

h=

∫c

p dT

=

Tref

∑ i − n + 1 (T a1

i − n +1

)

⎡ ⎛ β ⎛ β⎞ β /T i − n +1 − Tref + a 0 ⎢e β / T − e ref − eE1 ⎜1 − ⎟ − eE1 ⎜⎜1 − T T ⎝ ⎠ ref ⎝ ⎣⎢

⎞⎤ ⎟⎥ ⎟ ⎠⎦⎥

where E1(x) is the exponential integral function of order 1. Similarly, s0 is given by T

so =

cp

∫T

dT =

Tref

∑ i − n (T a1

i−n

)

⎡ ⎛ β ⎛ β⎞ i−n − Tref −a 0 ⎢e β / T − e β / Tref − eE1 ⎜1 − ⎟ − eE1 ⎜⎜1 − ⎝ T⎠ ⎢⎣ ⎝ Tref

⎞⎤ ⎟⎥ ⎟ ⎠⎥⎦

With these two results, u = h − Pv Pr = e s

o

/R

According to the du form of Gibbs equations, du dv = −R T v

Noting that for ideal gases, cv = c p − R and du = cv dT , this expression reduces to (c p − R )

dT dv = −R T v

When this is integrated between the reference and actual states, the result is

∫c

p

dT T v − R ln = − R ln T Tref v ref

Solving this for the specific volume ratio gives

v v ref

⎛ so T ⎞ T ⎜e − eR ⎟ eR o ⎜ ⎟ so T Tref es T T e ref ⎠ ⎝ =− = − + = − + = − exp( s o − R ) + R R R R T T e e e e ref ref

The ratio of the specific volumes at two states which have the same entropy is then T v 2 v r,2 = = − exp( s 2o − s1o − R ) + 2 T1 v 1 v r,1

Inspection of this result gives

v r = − exp( s o − R ) + T

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12-71

12-95 The cp of nitrogen at 300 kPa and 400 K is to be estimated using the relation given and its definition, and the results are to be compared to the value listed in Table A-2b. Analysis (a) We treat nitrogen as an ideal gas with R = 0.297 kJ/kg·K and k = 1.397. Note that PT-k/(k-1) = C = constant for the isentropic processes of ideal gases. The cp relation is given as ⎛ ∂P ⎞ ⎛ ∂v ⎞ cp = T⎜ ⎟ ⎟ ⎜ ⎝ ∂T ⎠ s ⎝ ∂T ⎠ P

v=

RT R ⎛ ∂v ⎞ ⎯ ⎯→ ⎜ ⎟ = P ⎝ ∂T ⎠ P P

(

)

kP k k ⎛ ∂P ⎞ PT − k /( k −1) T k /( k −1) −1 = ⎯→ ⎜ CT k /( k −1) −1 = P = CT k /( k −1) ⎯ ⎟ = T (k − 1) k −1 ⎝ ∂T ⎠ s k − 1

Substituting, ⎛ kP ⎞⎛ R ⎞ kR 1.397(0.297 kJ/kg ⋅ K) ⎟⎟⎜ ⎟ = = 1.045 kJ/kg ⋅ K = c p = T ⎜⎜ 1.397 − 1 ⎝ T (k − 1) ⎠⎝ P ⎠ k − 1

⎛ ∂h ⎞ (b) The cp is defined as cp = ⎜ ⎟ . Replacing the differentials by differences, ⎝ ∂T ⎠ P h(410 K ) − h(390 K ) (11,932 − 11,347 )/28.0 kJ/kg ⎛ Δh ⎞ = 1.045 kJ/kg ⋅ K cp ≅ ⎜ = = ⎟ T (410 − 390)K (410 − 390)K Δ ⎠ P =300 kPa ⎝

(Compare: Table A-2b at 400 K → cp = 1.044 kJ/kg·K)

12-96 The temperature change of steam and the average Joule-Thompson coefficient during a throttling process are to be estimated. Analysis The enthalpy of steam at 4.5 MPa and T = 300°C is h = 2944.2 kJ/kg. Now consider a throttling process from this state to 2.5 MPa. The temperature of the steam at the end of this throttling process will be P = 2.5 MPa ⎫ ⎬T2 = 273.72°C h = 2944.2 kJ/kg ⎭

Thus the temperature drop during this throttling process is

ΔT = T2 − T1 = 273.72 − 300 = −26.28°C The average Joule-Thomson coefficient for this process is determined from

(273.72 − 300)°C = 13.14°C/MPa ⎛ ΔT ⎞ ⎛ ∂T ⎞ = μ=⎜ ⎟ ⎟ ≅⎜ (2.5 − 4.5)MPa P P Δ ∂ ⎠ h =3204.7 kJ/kg ⎠h ⎝ ⎝

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12-72

12-97 Argon enters a turbine at a specified state and leaves at another specified state. Power output of the turbine and exergy destruction during this process are to be determined using the generalized charts. Properties The gas constant and critical properties of Argon are R = 0.2081 kJ/kg.K, Tcr = 151 K, and Pcr = 4.86 MPa (Table A-1). Analysis (a) The enthalpy and entropy departures of argon at the specified states are determined from the generalized charts to be T 600 ⎫ P1 = 7 MPa = 3.97 ⎪ T R1 = 1 = T1 = 600 K Tcr 151 ⎪ ⎬ Z h1 ≅ 0 and Z s1 ≅ 0 V1 = 100 m/s P 7 60 kW = 1.44⎪ PR1 = 1 = ⎪ Pcr 4.86 ⎭ Thus argon behaves as an ideal gas at turbine inlet. Also, T 280 ⎫ = 1.85 ⎪ T R2 = 2 = Ar · Tcr 151 ⎪ m = 5 kg/s Z = 0 . 04 and Z = 0 . 02 ⎬ h2 s2 · P2 1 ⎪ W = = 0.206 PR2 = ⎪⎭ Pcr 4.86 T0 = 25°C h2 − h1 = RTcr Z h1 − Z h2 + (h2 − h1 )ideal Thus, = (0.2081)(151)(0 − 0.04) + 0.5203(280 − 600) = −167.8 kJ/kg P2 = 1 MPa The power output of the turbine is to be determined from the T2 = 280 K energy balance equation, V2 = 150 m/s E& − E& = ΔE& = 0 (steady) → E& = E&

(

in

)

out

m& (h1 + V12

system

/ 2) =

m& ( h2 + V 22

in

out

/ 2) + Q& out + W& out

⎡ V 2 − V12 ⎤ & W& out = − m& ⎢(h2 − h1 ) + 2 ⎥ − Qout 2 ⎢⎣ ⎥⎦

Substituting, ⎛ (150 m/s) 2 − (100 m/s) 2 ⎛⎜ 1 kJ/kg ⎞⎟ ⎞⎟ − 60 kJ/s = 747.8 kW W& out = −(5 kg/s)⎜ − 167.8 + ⎜ 1000 m 2 /s 2 ⎟ ⎟ ⎜ 2 ⎝ ⎠⎠ ⎝ (b) Under steady conditions, the rate form of the entropy balance for the turbine simplifies to Ê0 S& − S& + S& = ΔS& =0 in

out

gen

Q& & 1 − ms & 2 − out + S&gen = 0 ms Tb,out

system

→

Q& & ( s2 − s2 ) + out S&gen = m T0

The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 S gen ,

where

⎛ Q& X& destroyed = T0 S& gen = T0 ⎜⎜ m& ( s 2 − s 2 ) + out T0 ⎝ s 2 − s1 = R Z s1 − Z s2 + (s 2 − s1 )ideal

and

(s 2 − s1 )ideal = c p ln

Thus,

s 2 − s1 = R Z s1 − Z s2 + (s 2 − s1 )ideal = (0.2081)[0 − (0.02)] + 0.0084 = 0.0042 kJ/kg ⋅ K

(

Substituting,

(

)

⎞ ⎟ ⎟ ⎠

P T2 1 280 − R ln 2 = 0.5203 ln − 0.2081 ln = 0.0084 kJ/kg ⋅ K P1 7 T1 600

)

⎛ 60 kW ⎞ ⎟ = 66.3 kW X& destroyed = (298 K )⎜⎜ (5 kg/s )(0.0042 kJ/kg ⋅ K ) + 298 K ⎟⎠ ⎝

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12-73

12-98 EES Problem 12-97 is reconsidered. The problem is to be solved assuming steam is the working fluid by using the generalized chart method and EES data for steam. The power output and the exergy destruction rate for these two calculation methods against the turbine exit pressure are to be plotted. Analysis The problem is solved using EES, and the results are tabulated and plotted below. " Input Data " T[1]=600 [K] P[1]=7000 [kPa] Vel[1]=100 [m/s] T[2]=455 [K] P[2]=1000 [kPa] Vel[2]=150 [m/s] Q_dot_out=60 [kW] T_o=25+273 "[K]" m_dot=5 [kg/s] Name$='Steam_iapws' T_critical=647.3 [K] P_critical=22090 [kPa] Fluid$='H2O' R_u=8.314 M=molarmass(Fluid$) R=R_u/M "****** IDEAL GAS SOLUTION ******" "State 1" h_ideal[1]=enthalpy(Fluid$,T=T[1]) "Enthalpy of ideal gas" s_ideal[1]=entropy(Fluid$, T=T[1], P=P[1]) "Entropy of ideal gas" "State 2" h_ideal[2]=enthalpy(Fluid$,T=T[2]) "Enthalpy of ideal gas" s_ideal[2]=entropy(Fluid$, T=T[2], P=P[2]) "Entropy of ideal gas" "Conservation of Energy, Steady-flow: " "E_dot_in=E_dot_out" m_dot*(h_ideal[1]+Vel[1]^2/2*convert(m^2/s^2,kJ/kg))=m_dot*(h_ideal[2]+Vel[2]^2/2*convert(m^2 /s^2,kJ/kg))+Q_dot_out+W_dot_out_ideal "Second Law analysis:" "S_dot_in-S_dot_out+S_dot_gen = 0" m_dot*s_ideal[1] - m_dot*s_ideal[2] - Q_dot_out/T_o + S_dot_gen_ideal = 0 "Exergy Destroyed:" X_dot_destroyed_ideal = T_o*S_dot_gen_ideal "***** COMPRESSABILITY CHART SOLUTION ******" "State 1" Tr[1]=T[1]/T_critical Pr[1]=P[1]/P_critical Z[1]=COMPRESS(Tr[1], Pr[1]) DELTAh[1]=ENTHDEP(Tr[1], Pr[1])*R*T_critical "Enthalpy departure" h_chart[1]=h_ideal[1]-DELTAh[1] "Enthalpy of real gas using charts" DELTAs[1]=ENTRDEP(Tr[1], Pr[1])*R "Entropy departure" s_chart[1]=s_ideal[1]-DELTAs[1] "Entropy of real gas using charts" "State 2"

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12-74 Tr[2]=T[2]/T_critical Pr[2]=P[2]/P_critical Z[2]=COMPRESS(Tr[2], Pr[2]) DELTAh[2]=ENTHDEP(Tr[2], Pr[2])*R*T_critical "Enthalpy departure" DELTAs[2]=ENTRDEP(Tr[2], Pr[2])*R "Entropy departure" h_chart[2]=h_ideal[2]-DELTAh[2] "Enthalpy of real gas using charts" s_chart[2]=s_ideal[2]-DELTAs[2] "Entropy of real gas using charts" "Conservation of Energy, Steady-flow: " "E_dot_in=E_dot_out" m_dot*(h_chart[1]+Vel[1]^2/2*convert(m^2/s^2,kJ/kg))=m_dot*(h_chart[2]+Vel[2]^2/2*convert(m^ 2/s^2,kJ/kg))+Q_dot_out+W_dot_out_chart "Second Law analysis:" "S_dot_in-S_dot_out+S_dot_gen = 0" m_dot*s_chart[1] - m_dot*s_chart[2] - Q_dot_out/T_o + S_dot_gen_chart = 0 "Exergy Destroyed:" X_dot_destroyed_chart = T_o*S_dot_gen_chart"[kW]" "***** SOLUTION USING EES BUILT-IN PROPERTY DATA *****" "At state 1" h_ees[1]=enthalpy(Name$,T=T[1],P=P[1]) s_ees[1]=entropy(Name$,T=T[1],P=P[1]) "At state 2" h_ees[2]=enthalpy(Name$,T=T[2],P=P[2]) s_ees[2]=entropy(Name$,T=T[2],P=P[2]) "Conservation of Energy, Steady-flow: " "E_dot_in=E_dot_out" m_dot*(h_ees[1]+Vel[1]^2/2*convert(m^2/s^2,kJ/kg))=m_dot*(h_ees[2]+Vel[2]^2/2*convert(m^2/s ^2,kJ/kg))+Q_dot_out+W_dot_out_ees "Second Law analysis:" "S_dot_in-S_dot_out+S_dot_gen = 0" m_dot*s_ees[1] - m_dot*s_ees[2] - Q_dot_out/T_o + S_dot_gen_ees= 0 "Exergy Destroyed:" X_dot_destroyed_ees = T_o*S_dot_gen_ees

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12-75

P2 [kPa] 100 200 300 400 500 600 700 800 900 1000

T2 [K] 455 455 455 455 455 455 455 455 455 455

Woutchart [kW] 713.3 725.2 737.3 749.5 761.7 774.1 786.5 799.1 811.8 824.5

Woutees [kW] 420.6 448.1 476.5 505.8 536.1 567.5 600 633.9 669.3 706.6

Woutideal [kW] 1336 1336 1336 1336 1336 1336 1336 1336 1336 1336

Xdestroyedchart [kW] 2383 1901 1617 1415 1256 1126 1014 917.3 831 753.1

Xdestroyedees [kW] 2519 2029 1736 1523 1354 1212 1090 980.1 880.6 788.4

Xdestroyedideal [kW] 2171 1694 1416 1218 1064 939 833 741.2 660.2 587.7

Wout;ees [kW]

1600

Solution Method EES Chart Ideal gas

1200

800

400 100

200

300

400

500

600

700

800

900 1000

P[2] [kPa] 2800

Xdestroyed [kW]

2400

Solution Method EES Charts Ideal Gas

2000

1600

1200

800

400 100

200

300

400

500

600

700

800

900

1000

P[2] [kPa]

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12-76

12-99 An adiabatic storage tank that is initially evacuated is connected to a supply line that carries nitrogen. A valve is opened, and nitrogen flows into the tank. The final temperature in the tank is to be determined by treating nitrogen as an ideal gas and using the generalized charts, and the results are to be compared to the given actual value. Assumptions 1 Uniform flow conditions exist. 2 Kinetic and potential energies are negligible. Analysis We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: min − mout = Δmsystem Energy balance:

→

Ein − Eout = ΔEsystem

Combining the two balances:

mi = m2

(since mout = minitial = 0)

→ 0 + mi hi = m2 u2

u2 = hi

(a) From the ideal gas property table of nitrogen, at 225 K we read

N2

10 MPa

u2 = hi = h@ 225 K = 6,537 kJ / kmol

V1 = 0.2 m3

The temperature that corresponds to this u2 value is T2 = 314.8 K

(7.4% error)

(b) Using the generalized enthalpy departure chart, hi is determined to be Ti ⎫ 225 = = 1.78 ⎪ hi ,ideal − hi Tcr 126.2 ⎪ = 0.9 ⎬ Z h ,i = Pi Ru Tcr 10 ⎪ = = = 2.95 ⎪⎭ Pcr 3.39

Initially evacuated

T R ,i = PR ,i

(Fig. A-29)

Thus, hi = hi ,ideal − 0.9 Ru Tcr = 6,537 − (0.9 )(8.314 )(126.2) = 5,593 kJ/kmol

and u 2 = hi = 5,593 kJ/kmol

Try T2 = 280 K. Then at PR2 = 2.95 and TR2 = 2.22 we read Z2 = 0.98 and (h2,ideal − h2 ) / Ru Tcr = 0.55 Thus, h2 = h2,ideal − 0.55Ru Tcr = 8,141 − (0.55)(8.314 )(126.2) = 7,564 kJ/kmol u 2 = h2 − ZRu T2 = 7,564 − (0.98)(8.314)(280) = 5,283 kJ/kmol

Try T2 = 300 K. Then at PR2 = 2.95 and TR2 = 2.38 we read Z2 = 1.0 and (h2,ideal − h2 ) / Ru Tcr = 0.50 Thus, h2 = h2,ideal − 0.50 Ru Tcr = 8,723 − (0.50 )(8.314)(126.2 ) = 8,198 kJ/kmol u 2 = h2 − ZRu T2 = 8,198 − (1.0)(8.314 )(300) = 5,704 kJ/kmol

By linear interpolation, T2 = 294.7 K

(0.6% error)

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12-77

12-100 Propane is compressed in a steady-flow device. The entropy change and the specific work required for this compression are to be determined using the departure charts and treating the propane as an ideal gas with temperature variable specific heats. Properties The properties of propane are (Table A-1, A-2a) M = 44.097 kg/kmol, R = 0.1885 kJ/kg ⋅ K, Tcr = 370 K, Pcr = 4.26 MPa, c p = 1.6794 kJ/kg ⋅ K

Analysis (a) Using empirical correlation for the cp of propane as given in Table A-2c gives 2

∫

2

∫

win = (h2 − h1 ) ideal = c p dT = (a + bT + cT 2 + dT 3 )dT 1

1

b c d = a(T2 − T1 ) + (T22 − T12 ) + (T23 − T13 ) + (T24 − T14 ) 2 3 4 0.3048 15.72 ×10 −5 31.74 ×10 −9 = −4.04(773 − 373) + (7732 − 3732 ) − (7733 − 3733 ) + (7734 − 3734 ) 2 3 4 = 49,440 kJ/kmol 4 MPa The work input per unit mass is 500°C win 49,440 kJ/kmol = = 1121 kJ/kg win = 44.097 kg/kmol M Propane

Similarly, the entropy change is given by 2

( s 2 − s1 ) ideal =

∫ 1

2

P P ⎛a ⎞ dT − Ru ln 2 = ⎜ + b + cT + dT 2 ⎟dT − Ru ln 2 P1 P1 T ⎝T ⎠ 1

cp

∫

P T c d = a ln 2 + b(T2 − T1 ) + (T22 − T12 ) + (T23 − T13 ) − Ru ln 2 2 3 P1 T1

500 kPa 100°C

773 15.72 ×10 −5 31.74 × 10 −9 (773 2 − 373 2 ) + (773 3 − 373 3 ) + 0.3048(773 − 373) − 373 2 3 4000 − (8.314) ln 500 = 69.995 kJ/kmol ⋅ K = −4.04 ln

The entropy change per unit mass is ( s 2 − s1 ) ideal =

( s 2 − s1 ) ideal 69.995 kJ/kmol ⋅ K = = 1.587 kJ/kg ⋅ K 44.097 kg/kmol M

(b) The enthalpy and entropy departures of propane at the specified states are determined from the generalized charts to be (Fig. A-29, A-30 or from EES) T1 373 ⎫ = = 1.008 ⎪ Tcr 370 ⎪ Z h1 = 0.124 ⎬ Z = 0.0837 P1 0.5 = = = 0.117 ⎪ s1 ⎪⎭ Pcr 4.26

T2 773 ⎫ = = 2.089 ⎪ Tcr 370 ⎪ Z h 2 = 0.233 ⎬ Z = 0.105 P2 4 = = = 0.939⎪ s 2 ⎪⎭ Pcr 4.26

T R1 =

TR 2 =

PR1

PR 2

The work input is determined from win = h2 − h1 = (h2 − h1 ) ideal − RTcr ( Z h 2 − Z h1 )

= 1121 − (0.1885)(370)(0.233 − 0.124) = 1113 kJ/kg

and the entropy change is determined to be PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

12-78

s 2 − s1 = ( s 2 − s1 ) ideal − R( Z s 2 − Z s1 )

= 1.587 − (0.1885)(0.105 − 0.0837) = 1.583 kJ/kg ⋅ K

Discussion Let us see what happens when constant specific heats for propane at the room temperature are used when calculating enthalpy and entropy changes under ideal gas assumption. The entropy and enthalpy changes are determined from (h2 − h1 ) ideal = c p (T2 − T1 ) = (1.6794 kJ/kg ⋅ K )(500 − 100)K = 671.8 kJ/kg ( s 2 − s1 ) ideal = c p ln

T2 P − R ln 2 T1 P1

= (1.6794) ln

773 4000 − (0.1885) ln = 0.8318 kJ/kg ⋅ K 373 500

With departure factors, win = h2 − h1 = (h2 − h1 ) ideal − RTcr ( Z h 2 − Z h1 ) = 671.8 − (0.1885)(370)(0.233 − 0.124) = 664.2 kJ/kg s 2 − s1 = ( s 2 − s1 ) ideal − R ( Z s 2 − Z s1 ) = 0.8318 − (0.1885)(0.105 − 0.0837) = 0.8278 kJ/kg ⋅ K

These are not any close to the results obtained using variable specific heats. This shows that using constant specific heats may yield unacceptable results.

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12-79

12-101 Propane is compressed in a steady-flow device. The second-law efficiency of the compression process is to be determined. Properties The properties of propane are (Table A-1, A-2a) M = 44.097 kg/kmol, R = 0.1885 kJ/kg ⋅ K, Tcr = 370 K, Pcr = 4.26 MPa, c p = 1.6794 kJ/kg ⋅ K

Analysis Using the variable specific heat results of the previous problem, the actual and reversible works are win = h2 − h1 = 1113 kJ/kg w rev,in = h2 − h1 − T0 ( s 2 − s1 ) = 1113 kJ/kg − (298 K )(1.583 kJ/kg ⋅ K ) = 641.3 kJ/kg

The second-law efficiency is then

η II =

wrev,in wrev,in

=

641.3 = 0.576 1113

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12-80

12-102E Methane is to be adiabatically and reversibly compressed in a steady-flow device. The specific work required for this compression is to be determined using the departure charts and treating the methane as an ideal gas with temperature variable specific heats. Properties The properties of methane are (Table A-1E) M = 16.043 lbm/lbmol, R = 0.1238 Btu/lbm ⋅ R, Tcr = 343.9 R, Pcr = 673 psia

Analysis The temperature at the exit state may be determined by the fact that the process is isentropic and the molar entropy change between the inlet and exit is zero. When the expression of Table A-2Ec is substituted for cp and the integration performed, we obtain 2

( s 2 − s1 ) ideal =

cp

∫T

Methane

2

dT − Ru ln

1

= a ln

500 psia

P2 P ⎛a ⎞ = ⎜ + b + cT + dT 2 ⎟dT − Ru ln 2 P1 P1 ⎝T ⎠ 1

∫

T2 P c d + b(T2 − T1 ) + (T22 − T12 ) + (T23 − T13 ) − Ru ln 2 T1 2 3 P1

50 psia 100°F

Substituting, T2 0.09352 × 10 −5 2 (T2 − 560 2 ) + 0.006666(T2 − 560) + 560 2 0.4510 × 10 −9 3 500 (T2 − 560 3 ) − (1.9858) ln − 3 50

0 = 4.75 ln

Solving this equation by EES or an iterative solution gives T2 = 892 R

When en energy balance is applied to the compressor, it becomes 2

∫

2

∫

win = (h2 − h1 ) ideal = c p dT = (a + bT + cT 2 + dT 3 )dT 1

1

b c d = a (T2 − T1 ) + (T22 − T12 ) + (T23 − T13 ) + (T24 − T14 ) 2 3 4 0.006666 0.09352 × 10 −5 = 4.75(892 − 560) + (892 2 − 560 2 ) + (892 3 − 560 3 ) 2 3 0.4510 × 10 −9 − (892 4 − 560 4 ) 4 = 3290 Btu/lbmol

The work input per unit mass basis is win =

win 3290 Btu/lbmol = = 205.1 Btu/lbm M 16.043 lbm/lbmol

The enthalpy departures of propane at the specified states are determined from the generalized charts to be (Fig. A-29 or from EES) T1 560 ⎫ = = 1.628 ⎪ Tcr 343.9 ⎪ ⎯→ Z h1 = 0.0332 ⎬⎯ P1 50 ⎪ = = = 0.0743 ⎪⎭ Pcr 673

T R1 = PR1

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12-81

and T2 892 ⎫ = = 2.594 ⎪ Tcr 343.9 ⎪ ⎯→ Z h 2 = 0.0990 ⎬⎯ P2 500 ⎪ = = = 0.743 ⎪⎭ Pcr 673

TR 2 = PR 2

The work input is determined to be win = h2 − h1 = (h2 − h1 ) ideal − RTcr ( Z h 2 − Z h1 ) = 205.1 Btu/lbm − (0.1238 Btu/lbm ⋅ R )(343.9 R )(0.0990 − 0.0332) = 202.3 Btu/lbm

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12-82

12-103 The volume expansivity of water is given. The change in volume of water when it is heated at constant pressure is to be determined. Properties The volume expansivity of water is given to be 0.207×10-6 K-1 at 20°C. Analysis We take v = v (P, T). Its total differential is ⎛ ∂v ⎞ ⎛ ∂v ⎞ dv = ⎜ ⎟ dT + ⎜ ⎟ dP ∂ T ⎝ ∂P ⎠ T ⎠P ⎝

which, for a constant pressure process, reduces to ⎛ ∂v ⎞ dv = ⎜ ⎟ dT ⎝ ∂T ⎠ P Dividing by v and using the definition of β, dv

v

=

1 ⎛ ∂v ⎞ ⎟ dT = β dT ⎜ v ⎝ ∂T ⎠ P

Taking β to be a constant, integration from 1 to 2 yields ln

v2 = β (T2 − T1 ) v1

or

v2 = exp[β (T2 − T1 )] v1 Substituting the given values and noting that for a fixed mass V2/V1 = v2/v1,

(

) [(

)

V 2 = V 1 exp[β (T2 − T1 )] = 1 m 3 exp 0.207 × 10 −6 K −1 (30 − 10 )°C = 1.00000414 m

]

3

Therefore, ΔV = V 2 −V1 = 1.00000414 − 1 = 0.00000414 m 3 = 4.14 cm 3

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12-83

12-104 It is to be shown that the position of the Joule-Thompson coefficient inversion curve on the T-P plane is given by (∂Z/∂T)P = 0. Analysis The inversion curve is the locus of the points at which the Joule-Thompson coefficient μ is zero, μ=

1 cp

⎛ ⎛ ∂v ⎞ ⎞ ⎜T ⎜ ⎟ ⎜ ⎝ ∂T ⎟⎠ − v ⎟ = 0 P ⎝ ⎠

which can also be written as ZRT ⎛ ∂v ⎞ =0 T⎜ ⎟ − P ⎝ ∂T ⎠ P

(a)

since it is given that

v=

ZRT P

(b)

Taking the derivative of (b) with respect to T holding P constant gives ⎞ R ⎛ ⎛ ∂Z ⎞ ⎛ ∂ (ZRT / P ) ⎞ ⎛ ∂v ⎞ ⎟ = ⎜⎜ T ⎜ ⎟ + Z ⎟⎟ ⎟ =⎜ ⎜ ∂T ⎠ P P ⎝ ⎝ ∂T ⎠ P ⎝ ∂T ⎠ P ⎝ ⎠

Substituting in (a), ⎞ ZRT TR ⎛ ⎛ ∂Z ⎞ ⎜T ⎜ =0 ⎟ + Z ⎟⎟ − ⎜ P P ⎝ ⎝ ∂T ⎠ P ⎠ ⎛ ∂Z ⎞ T⎜ ⎟ +Z −Z =0 ⎝ ∂T ⎠ P ⎛ ∂Z ⎞ ⎟ =0 ⎜ ⎝ ∂T ⎠ P

which is the desired relation.

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12-84

12-105 It is to be shown that for an isentropic expansion or compression process Pv k = constant. It is also to be shown that the isentropic expansion exponent k reduces to the specific heat ratio cp/cv for an ideal gas. Analysis We note that ds = 0 for an isentropic process. Taking s = s(P, v), the total differential ds can be expressed as ⎛ ∂s ⎞ ⎛ ∂s ⎞ ds = ⎜ ⎟ dv = 0 ⎟ dP + ⎜ P ∂ ⎝ ∂v ⎠ P ⎠v ⎝

(a)

We now substitute the Maxwell relations below into (a) ⎛ ∂s ⎞ ⎛ ∂v ⎞ ⎛ ∂s ⎞ ⎛ ∂P ⎞ ⎜ ⎟ = −⎜ ⎟ and ⎜ ⎟ = ⎜ ⎟ ⎝ ∂P ⎠ v ⎝ ∂T ⎠ s ⎝ ∂v ⎠ P ⎝ ∂T ⎠ s

to get ⎛ ∂v ⎞ ⎛ ∂P ⎞ −⎜ ⎟ dP + ⎜ ⎟ dv = 0 ⎝ ∂T ⎠ s ⎝ ∂T ⎠ s

Rearranging, ⎛ ∂T ⎞ ⎛ ∂P ⎞ ⎛ ∂P ⎞ ⎯→ dP − ⎜ dP − ⎜ ⎟ ⎜ ⎟ dv = 0 ⎯ ⎟ dv = 0 ∂ ∂ v T ⎝ ⎠s ⎝ ⎠s ⎝ ∂v ⎠ s

Dividing by P,

dP 1 ⎛ ∂P ⎞ − ⎜ ⎟ dv = 0 P P ⎝ ∂v ⎠ s

(b)

We now define isentropic expansion exponent k as k =−

v ⎛ ∂P ⎞ ⎜ ⎟ P ⎝ ∂v ⎠ s

Substituting in (b), dP dv +k =0 P v

Taking k to be a constant and integrating, ln P + k ln v = constant ⎯ ⎯→ ln Pv k = constant

Thus, Pv k = constant

To show that k = cp/cv for an ideal gas, we write the cyclic relations for the following two groups of variables:

(s, T , v ) ⎯⎯→ ⎛⎜ ∂s ⎞⎟

c ⎛ ∂v ⎞ ⎛ ∂T ⎞ ⎛ ∂v ⎞ ⎛ ∂T ⎞ ⎯→ v ⎜ ⎜ ⎟ ⎜ ⎟ = −1 ⎯ ⎟ ⎜ ⎟ = −1 (c) T ⎝ ∂s ⎠ T ⎝ ∂v ⎠ s ⎝ ∂T ⎠v ⎝ ∂s ⎠ T ⎝ ∂v ⎠ s c (s, T , P ) ⎯⎯→ ⎛⎜ ∂s ⎞⎟ ⎛⎜ ∂P ⎞⎟ ⎛⎜ ∂T ⎞⎟ = −1 ⎯⎯→ p ⎛⎜ ∂P ⎞⎟ ⎛⎜ ∂T ⎞⎟ = −1 (d ) T ⎝ ∂s ⎠ T ⎝ ∂P ⎠ s ⎝ ∂T ⎠ P ⎝ ∂s ⎠ T ⎝ ∂P ⎠ s

where we used the relations ⎛ ∂s ⎞ ⎛ ∂s ⎞ cv = T ⎜ ⎟ and c p = T ⎜ ⎟ ⎝ ∂T ⎠v ⎝ ∂T ⎠ P

Setting Eqs. (c) and (d) equal to each other,

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12-85

c p ⎛ ∂P ⎞ ⎛ ∂T ⎞ c ⎛ ∂v ⎞ ⎛ ∂T ⎞ ⎜ ⎟ ⎜ ⎟ = v ⎜ ⎟ ⎜ ⎟ T ⎝ ∂s ⎠ T ⎝ ∂P ⎠ s T ⎝ ∂s ⎠ T ⎝ ∂v ⎠ s

or, cp cv

⎛ ∂s ⎞ ⎛ ∂P ⎞ ⎛ ∂v ⎞ ⎛ ∂T ⎞ ⎛ ∂s ∂v ⎞ ⎛ ∂P ∂T ⎞ ⎛ ∂v ⎞ ⎛ ∂P ⎞ =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ =⎜ ⎟ ⎜ ⎟ =⎜ ⎟ ⎜ ⎟ ⎝ ∂P ⎠ T ⎝ ∂T ⎠ s ⎝ ∂s ⎠ T ⎝ ∂v ⎠ s ⎝ ∂P ∂s ⎠ T ⎝ ∂T ∂v ⎠ s ⎝ ∂P ⎠ T ⎝ ∂v ⎠ s

but

v ⎛ ∂ (RT / P ) ⎞ ⎛ ∂v ⎞ ⎟ =− ⎜ ⎟ =⎜ ∂P P ⎝ ∂P ⎠ T ⎝ ⎠T Substituting, cp cv

=−

v ⎛ ∂P ⎞ ⎜ ⎟ =k P ⎝ ∂v ⎠ s

which is the desired relation.

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12-86

12-106 EES The work done by the refrigerant 134a as it undergoes an isothermal process in a closed system is to be determined using the tabular (EES) data and the generalized charts. Analysis The solution using EES built-in property data is as follows: T1 = 60°C ⎫ u1 = 135.65 kJ/kg ⎬ P1 = 3 MPa ⎭ s1 = 0.4828 kJ/kg.K T2 = 60°C

⎫ u 2 = 280.35 kJ/kg ⎬ P2 = 0.1 MPa ⎭ s 2 = 1.2035 kJ/kg.K

Δs EES = s 2 − s1 = 1.2035 − 0.4828 = 0.7207 kJ/kg.K q EES = T1 Δs EES = (60 + 273.15 K )(0.7207 kJ/kg.K ) = 240.11 kJ/kg wEES = q EES − (u 2 − u1 ) = 240.1 − (280.35 − 135.65) = 95.40 kJ/kg

For the generalized chart solution we first determine the following factors using EES as T1 333.15 ⎫ = = 0.8903 ⎪ Tcr 374.2 ⎪ ⎯→ Z1 = 0.1292, Z h1 = 4.475 and Z s1 = 4.383 ⎬ ⎯ 3 P PR1 = 1 = = 0.7391 ⎪ ⎪⎭ Pcr 4.059

TR1 =

T2 333.15 ⎫ = = 0.8903 ⎪ 374.2 Tcr ⎪ ⎯→ Z 2 = 0.988, Z h 2 = 0.03091 and Z s 2 = 0.02281 ⎬⎯ P 0.1 = 2 = = 0.02464 ⎪ ⎪⎭ Pcr 4.059

TR 2 = PR 2

Then, Δh1 = Z h1 RTcr = (4.475)(0.08148 kJ/kg.K)(374.2 K) = 136.43 kJ/kg Δs1 = Z s1 R = (4.383)(0.08148 kJ/kg.K) = 0.3572 kJ/kg.K Δh2 = Z h 2 RTcr = (0.03091)(0.08148 kJ/kg.K)(374.2 K) = 0.94 kJ/kg Δs 2 = Z s 2 R = (0.02281)(0.08148 kJ/kg.K) = 0.001858 kJ/kg.K

Δs ideal = R ln

P2 ⎛ 0.1 ⎞ = (0.08148 kJ/kg.K)ln⎜ ⎟ = 0.2771 kJ/kg ⋅ K P1 ⎝ 3 ⎠

Δs chart = Δs ideal − (Δs 2 − Δs1 ) = 0.2771 − (0.001858 − 0.3572) = 0.6324 kJ/kg ⋅ K q chart = T1 Δs chart = (60 + 273.15 K )(0.6324 kJ/kg.K ) = 210.70 kJ/kg Δu chart = Δhideal − (Δh2 − Δh1 ) − ( Z 2 RT2 − Z 1 RT1 )

= 0 − (0.94 − 136.43) − [(0.988)(0.08148)(333) − (0.1292)(0.08148)(333)] = 112.17 kJ/kg

wchart = q chart − Δu chart = 210.70 − 112.17 = 98.53 kJ/kg

The copy of the EES solution of this problem is given next. "Input data" T_critical=T_CRIT(R134a) "[K]" P_critical=P_CRIT(R134a) "[kpa]" T[1]=60+273.15"[K]" PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

12-87

T[2]=T[1]"[K]" P[1]=3000"[kPa]" P[2]=100"[kPa]" R_u=8.314"[kJ/kmol-K]" M=molarmass(R134a) R=R_u/M"[kJ/kg-K]" "***** SOLUTION USING EES BUILT-IN PROPERTY DATA *****" "For the isothermal process, the heat transfer is T*(s[2] - s[1]):" DELTAs_EES=(entropy(R134a,T=T[2],P=P[2])-entropy(R134a,T=T[1],P=P[1])) q_EES=T[1]*DELTAs_EES s_2=entropy(R134a,T=T[2],P=P[2]) s_1=entropy(R134a,T=T[1],P=P[1]) "Conservation of energy for the closed system:" DELTAu_EES=intEnergy(R134a,T=T[2],p=P[2])-intEnergy(R134a,T=T[1],P=P[1]) q_EES-w_EES=DELTAu_EES u_1=intEnergy(R134a,T=T[1],P=P[1]) u_2=intEnergy(R134a,T=T[2],p=P[2]) "***** COMPRESSABILITY CHART SOLUTION ******" "State 1" Tr[1]=T[1]/T_critical pr[1]=p[1]/p_critical Z[1]=COMPRESS(Tr[1], Pr[1]) DELTAh[1]=ENTHDEP(Tr[1], Pr[1])*R*T_critical"Enthalpy departure" Z_h1=ENTHDEP(Tr[1], Pr[1]) DELTAs[1]=ENTRDEP(Tr[1], Pr[1])*R "Entropy departure" Z_s1=ENTRDEP(Tr[1], Pr[1]) "State 2" Tr[2]=T[2]/T_critical Pr[2]=P[2]/P_critical Z[2]=COMPRESS(Tr[2], Pr[2]) DELTAh[2]=ENTHDEP(Tr[2], Pr[2])*R*T_critical"Enthalpy departure" Z_h2=ENTHDEP(Tr[2], Pr[2]) DELTAs[2]=ENTRDEP(Tr[2], Pr[2])*R "Entropy departure" Z_s2=ENTRDEP(Tr[2], Pr[2]) "Entropy Change" DELTAs_ideal= -R*ln(P[2]/P[1]) DELTAs_chart=DELTAs_ideal-(DELTAs[2]-DELTAs[1]) "For the isothermal process, the heat transfer is T*(s[2] - s[1]):" q_chart=T[1]*DELTAs_chart "Conservation of energy for the closed system:" DELTAh_ideal=0 DELTAu_chart=DELTAh_ideal-(DELTAh[2]-DELTAh[1])-(Z[2]*R*T[2]-Z[1]*R*T[1]) q_chart-w_chart=DELTAu_chart

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12-88

SOLUTION DELTAh[1]=136.43 DELTAh[2]=0.94 DELTAh_ideal=0 DELTAs[1]=0.3572 DELTAs[2]=0.001858 DELTAs_chart=0.6324 [kJ/kg-K] DELTAs_EES=0.7207 [kJ/kg-K] DELTAs_ideal=0.2771 [kJ/kg-K] DELTAu_chart=112.17 DELTAu_EES=144.7 M=102 [kg/kmol] P[1]=3000 [kPa] P[2]=100 [kPa] pr[1]=0.7391 Pr[2]=0.02464 P_critical=4059 [kpa] q_chart=210.70 [kJ/kg] q_EES=240.11 [kJ/kg] R=0.08148 [kJ/kg-K

R_u=8.314 [kJ/kmol-K] s_1=0.4828 [kJ/kg-K] s_2=1.2035 [kJ/kg-K] T[1]=333.2 [K] T[2]=333.2 [K] Tr[1]=0.8903 Tr[2]=0.8903 T_critical=374.2 [K] u_1=135.65 [kJ/kg] u_2=280.35 [kJ/kg] w_chart=98.53 [kJ/kg] w_EES=95.42 [kJ/kg] Z[1]=0.1292 Z[2]=0.988 Z_h1=4.475 Z_h2=0.03091 Z_s1=4.383 Z_s2=0.02281

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12-89

12-107 The heat transfer, work, and entropy changes of methane during a process in a piston-cylinder device are to be determined assuming ideal gas behavior, using generalized charts, and real fluid (EES) data. Analysis The ideal gas solution: (Properties are obtained from EES) State 1: T1 = 100°C ⎯ ⎯→ h1 = −4492 kJ/kg T1 = 100°C, P1 = 4 MPa ⎯ ⎯→ s1 = 10.22 kJ/kg.K u1 = h1 − RT1 = (−4492) − (0.5182)(100 + 273.15) = −4685 kJ/kg

v1 = R

T1 ⎛ 100 + 273.15 K ⎞ 3 = (0.5182 kJ/kg.K )⎜ ⎟ = 0.04834 m /kg P1 4000 kPa ⎝ ⎠

State 2: T2 = 350°C ⎯ ⎯→ h2 = −3770 kJ/kg T2 = 350°C, P2 = 4 MPa ⎯ ⎯→ s 2 = 11.68 kJ/kg.K u 2 = h2 − RT2 = (−3770) − (0.5182)(350 + 273.15) = −4093 kJ/kg

v2 = R

T2 ⎛ 350 + 273.15 K ⎞ 3 = (0.5182 kJ/kg.K )⎜ ⎟ = 0.08073 m /kg P2 ⎝ 4000 kPa ⎠

wideal = P (v 2 − v 1 ) = (4000 kPa)(0.08073 - 0.04834)m 3 /kg = 129.56 kJ/kg

q ideal = wideal + (u 2 − u1 ) = 129.56 + [(−4093) − (−4685)] = 721.70 kJ/kg Δs ideal = s 2 − s1 = 11.68 − 10.22 = 1.46 kJ/kg

For the generalized chart solution we first determine the following factors using EES as T1 373 ⎫ = = 1.227 ⎪ Tcr 304.2 ⎪ ⎯→ Z 1 = 0.9023, Z h1 = 0.4318 and Z s1 = 0.2555 ⎬⎯ P1 4 ⎪ = = = 0.5413 ⎪⎭ Pcr 7.39

T R1 = PR1

T2 623 ⎫ = = 2.048 ⎪ Tcr 304.2 ⎪ ⎯→ Z 2 = 0.995, Z h 2 = 0.1435 and Z s 2 = 0.06446 ⎬⎯ P2 4 ⎪ = = = 0.5413 ⎪⎭ Pcr 7.39

TR 2 = PR 2

State 1: Δh1 = Z h1 RTcr = (0.4318)(0.5182 kJ/kg.K)(304.2 K) = 68.07 kJ/kg h1 = h1,ideal − Δh1 = (−4492) − 68.07 = −4560 kJ/kg u1 = h1 − Z 1 RT1 = (−4560) − (0.9023)(0.5182)(373.15) = −4734 kJ/kg

v 1 = Z1 R

T1 373.15 = (0.9023)(0.5182) = 0.04362 m 3 /kg P1 4000

Δs1 = Z s1 R = (0.2555)(0.5182 kJ/kg.K) = 0.1324 kJ/kg.K s1 = s1,ideal − Δs1 = 10.22 − 0.1324 = 10.09 kJ/kg.K State 2: PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

12-90 Δh2 = Z h 2 RTcr = (0.1435)(0.5182 kJ/kg.K)(304.2 K) = 22.62 kJ/kg h2 = h2,ideal − Δh2 = (−3770) − 22.62 = −3793 kJ/kg u 2 = h2 − Z 2 RT2 = (−3793) − (0.995)(0.5182)(623.15) = −4114 kJ/kg

v 2 = Z2R

T2 623.15 = (0.995)(0.5182) = 0.08033 m 3 /kg P2 4000

Δs 2 = Z s 2 R = (0.06446)(0.5182 kJ/kg.K) = 0.03341 kJ/kg.K s 2 = s 2,ideal − Δs 2 = 11.68 − 0.03341 = 11.65 kJ/kg.K Then, wchart = P (v 2 − v 1 ) = (4000 kPa)(0.08033 - 0.04362)m 3 /kg = 146.84 kJ/kg

q chart = wchart + (u 2 − u1 ) = 146.84 + [(−4114) − (−4734)] = 766.84 kJ/kg Δs chart = s 2 − s1 = 11.65 − 10.09 = 1.56 kJ/kg

The solution using EES built-in property data is as follows:

v = 0.04717 m 3 /kg T1 = 100°C ⎫ 1 ⎬ u1 = −39.82 kJ/kg P1 = 4 MPa ⎭ s1 = −1.439 kJ/kg.K v = 0.08141 m 3 /kg T2 = 350°C ⎫ 2 ⎬ u 2 = 564.52 kJ/kg P2 = 4 MPa ⎭ s 2 = 0.06329 kJ/kg.K w EES = P (v 2 − v 1 ) = (4000 kPa)(0.08141 - 0.04717)m 3 /kg = 136.96 kJ/kg

q EES = wEES + (u 2 − u1 ) = 136.97 + [564.52 − (−39.82)] = 741.31 kJ/kg Δs EES = s 2 − s1 = 0.06329 − (−1.439) = 1.50 kJ/kg

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12-91

Fundamentals of Engineering (FE) Exam Problems 12-108 A substance whose Joule-Thomson coefficient is negative is throttled to a lower pressure. During this process, (select the correct statement) (a) the temperature of the substance will increase. (b) the temperature of the substance will decrease. (c) the entropy of the substance will remain constant. (d) the entropy of the substance will decrease. (e) the enthalpy of the substance will decrease. Answer (a) the temperature of the substance will increase.

12-109 Consider the liquid-vapor saturation curve of a pure substance on the P-T diagram. The magnitude of the slope of the tangent line to this curve at a temperature T (in Kelvin) is (a) proportional to the enthalpy of vaporization hfg at that temperature, (b) proportional to the temperature T, (c) proportional to the square of the temperature T, (d) proportional to the volume change vfg at that temperature, (e) inversely proportional to the entropy change sfg at that temperature, Answer (a) proportional to the enthalpy of vaporization hfg at that temperature,

12-110 Based on the generalized charts, the error involved in the enthalpy of CO2 at 350 K and 8 MPa if it is assumed to be an ideal gas is (a) 0

(b) 20%

(c) 33%

(d) 26%

(e) 65%

Answer (c) 33%

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T=350 "K" P=8000 "kPa" Pcr=P_CRIT(CarbonDioxide) Tcr=T_CRIT(CarbonDioxide) Tr=T/Tcr Pr=P/Pcr hR=ENTHDEP(Tr, Pr) h_ideal=11351/Molarmass(CO2) "Table A-20 of the text" h_chart=h_ideal-R*Tcr*hR R=0.1889 Error=(h_chart-h_ideal)/h_chart*Convert(, %)

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12-92

12-111 Based on data from the refrigerant-134a tables, the Joule-Thompson coefficient of refrigerant-134a at 0.8 MPa and 100°C is approximately (a) 0

(b) -5°C/MPa

(c) 11°C/MPa

(d) 8°C/MPa

(e) 26°C/MPa

Answer (c) 11°C/MPa

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T1=100 "C" P1=800 "kPa" h1=ENTHALPY(R134a,T=T1,P=P1) Tlow=TEMPERATURE(R134a,h=h1,P=P1+100) Thigh=TEMPERATURE(R134a,h=h1,P=P1-100) JT=(Tlow-Thigh)/200

12-112 For a gas whose equation of state is P(v - b) = RT, the specific heat difference cp – cv is equal to (a) R

(b) R – b

(c) R + b

(d) 0

(e) R(1 + v/b)

Answer (a) R

Solution The general relation for the specific heat difference cp - cv is 2

⎛ ∂v ⎞ ⎛ ∂P ⎞ c p − cv = −T ⎜ ⎟ ⎟ ⎜ ⎝ ∂T ⎠ P ⎝ ∂v ⎠ T For the given gas, P(v - b) = RT. Then, RT R ⎛ ∂v ⎞ v= +b ⎯ ⎯→ ⎜ ⎟ = P T ∂ ⎠P P ⎝ P=

RT

v −b

RT P ⎛ ∂P ⎞ =− ⎯ ⎯→ ⎜ ⎟ =− v −b (v − b) 2 ⎝ ∂v ⎠ T

Substituting, 2

2

P ⎞ TR ⎛R⎞ ⎛ =R c p − cv = −T ⎜ ⎟ ⎜ − ⎟= ⎝ P ⎠ ⎝ v − b ⎠ P (v − b)

12-113 ··· 12-115 Design and Essay Problems

KJ

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13-1

Chapter 13 GAS MIXTURES Composition of Gas Mixtures

13-1C It is the average or the equivalent gas constant of the gas mixture. No.

13-2C No. We can do this only when each gas has the same mole fraction.

13-3C It is the average or the equivalent molar mass of the gas mixture. No.

13-4C The mass fractions will be identical, but the mole fractions will not.

13-5C Yes.

13-6C The ratio of the mass of a component to the mass of the mixture is called the mass fraction (mf), and the ratio of the mole number of a component to the mole number of the mixture is called the mole fraction (y).

13-7C From the definition of mass fraction, mf i =

⎛M mi N M = i i = y i ⎜⎜ i mm N m M m ⎝ Mm

⎞ ⎟ ⎟ ⎠

13-8C Yes, because both CO2 and N2O has the same molar mass, M = 44 kg/kmol.

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13-2

13-9 A mixture consists of two gases. Relations for mole fractions when mass fractions are known are to be obtained . Analysis The mass fractions of A and B are expressed as mf A =

mA N M MA = A A = yA mm N m M m Mm

and

mf B = y B

MB Mm

Where m is mass, M is the molar mass, N is the number of moles, and y is the mole fraction. The apparent molar mass of the mixture is

Mm =

mm N A M A + N B M B = = y AM A + yB M B Nm Nm

Combining the two equation above and noting that y A + y B = 1 gives the following convenient relations for converting mass fractions to mole fractions, yA =

MB M A (1 / mf A − 1) + M B

and

yB = 1 − y A

which are the desired relations.

13-10 The definitions for the mass fraction, weight, and the weight fractions are (mf) i =

mi m total

W = mg Wi ( wf) i = W total

Since the total system consists of one mass unit, the mass of the ith component in this mixture is xi. The weight of this one component is then Wi = g (mf) i

Hence, the weight fraction for this one component is

( wf) i =

g (mf) i

∑ g (mf)

= (mf) i i

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13-3

13-11 The moles of components of a gas mixture are given. The mole fractions and the apparent molecular weight are to be determined. Properties The molar masses of He, O2, N2, and H2O are 4.0, 32.0, 28.0 and 18.0 lbm/lbmol, respectively (Table A-1). Analysis The total mole number of the mixture is N m = N He + N O2 + N H2O + N N2 = 1 + 2 + 0.1 + 1.5 = 4.6 lbmol

and the mole fractions are y He =

N He 1 lbmol = = 0.217 4.6 lbmol Nm

y O2 =

N O2 2 lbmol = = 0.435 4.6 lbmol Nm

y H2O = y N2 =

N H2O 0.1 lbmol = = 0.0217 4.6 lbmol Nm

1 lbmol He 2 lbmol O2 0.1 lbmol H2O 1.5 lbmol N2

N N2 1.5 lbmol = = 0.326 4.6 lbmol Nm

The total mass of the mixture is m m = m He + m O2 + m H2O + + m N2 = N He M He + N O2 M O2 + N H2O M H2O + N N2 M N2 = (1 lbm)(4 lbm/lbmol) + (2 lbm)(32 lbm/lbmol) + (0.1 lbm)(18 lbm/lbmol) + (1.5 lbm)(28 lbm/lbmol) = 111.8 kg

Then the apparent molecular weight of the mixture becomes Mm =

m m 111.8 lbm = = 24.3 lbm/lbmol N m 4.6 lbmol

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13-4

13-12 The masses of the constituents of a gas mixture are given. The mass fractions, the mole fractions, the average molar mass, and gas constant are to be determined. Properties The molar masses of O2, N2, and CO2 are 32.0, 28.0 and 44.0 kg/kmol, respectively (Table A-1) Analysis (a) The total mass of the mixture is m m = m O 2 + m N 2 + m CO 2 = 5 kg + 8 kg + 10 kg = 23 kg

Then the mass fraction of each component becomes mf O 2 = mf N 2 = mf CO 2 =

mO2

5 kg = = 0.217 23 kg

mm m N2

8 kg = 0.348 23 kg

=

mm m CO 2

=

mm

5 kg O2 8 kg N2 10 kg CO2

10 kg = 0.435 23 kg

(b) To find the mole fractions, we need to determine the mole numbers of each component first, N O2 = N N2 = N CO 2 =

mO2 M O2 m N2 M N2

=

5 kg = 0.156 kmol 32 kg/kmol

=

8 kg = 0.286 kmol 28 kg/kmol

m CO 2

=

M CO 2

10 kg = 0.227 kmol 44 kg/kmol

Thus, N m = N O 2 + N N 2 + N CO 2 = 0.156 kmol + 0.286 kmol + 0.227 kmol = 0.669 kmol

and y O2 = y N2 = y CO 2 =

N O2 Nm N N2 Nm N CO 2 Nm

=

0.156 kmol = 0.233 0.699 kmol

=

0.286 kmol = 0.428 0.669 kmol

=

0.227 kmol = 0.339 0.669 kmol

(c) The average molar mass and gas constant of the mixture are determined from their definitions: Mm =

mm 23 kg = = 34.4 kg/kmol N m 0.669 kmol

and Rm =

Ru 8.314 kJ/kmol ⋅ K = = 0.242 kJ/kg ⋅ K Mm 34.4 kg/kmol

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13-5

13-13 The mass fractions of the constituents of a gas mixture are given. The mole fractions of the gas and gas constant are to be determined. Properties The molar masses of CH4, and CO2 are 16.0 and 44.0 kg/kmol, respectively (Table A-1) Analysis For convenience, consider 100 kg of the mixture. Then the number of moles of each component and the total number of moles are ⎯→ N CH 4 = m CH 4 = 75 kg ⎯ ⎯→ N CO 2 = m CO 2 = 25 kg ⎯

m CH 4 M CH 4 m CO 2 M CO 2

=

75 kg = 4.688 kmol 16 kg/kmol

mass

=

25 kg = 0.568 kmol 44 kg/kmol

75% CH4 25% CO2

N m = N CH 4 + N CO 2 = 4.688 kmol + 0.568 kmol = 5.256 kmol

Then the mole fraction of each component becomes y CH 4 = y CO 2 =

N CH 4 Nm N CO 2 Nm

=

4.688 kmol = 0.892 or 89.2% 5.256 kmol

=

0.568 kmol = 0.108 or 10.8% 5.256 kmol

The molar mass and the gas constant of the mixture are determined from their definitions, Mm =

mm 100 kg = = 19.03 kg/kmol N m 5.256 kmol

and Rm =

Ru 8.314 kJ/kmol ⋅ K = = 0.437 kJ/kg ⋅ K 19.03 kg/kmol Mm

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13-6

13-14 The mole numbers of the constituents of a gas mixture are given. The mass of each gas and the apparent gas constant are to be determined. Properties The molar masses of H2, and N2 are 2.0 and 28.0 kg/kmol, respectively (Table A-1) Analysis The mass of each component is determined from N H 2 = 8 kmol ⎯ ⎯→ m H 2 = N H 2 M H 2 = (8 kmol)(2.0 kg/kmol) = 16 kg N N 2 = 2 kmol ⎯ ⎯→ m N 2 = N N 2 M N 2 = (2 kmol)(28 kg/kmol) = 56 kg

The total mass and the total number of moles are

8 kmol H2 2 kmol N2

m m = m H 2 + m N 2 = 16 kg + 56 kg = 72 kg N m = N H 2 + N N 2 = 8 kmol + 2 kmol = 10 kmol

The molar mass and the gas constant of the mixture are determined from their definitions, Mm =

mm 72 kg = = 7.2 kg/kmol N m 10 kmol

and Rm =

Ru 8.314 kJ/kmol ⋅ K = = 1.155 kJ/kg ⋅ K Mm 7.2 kg/kmol

13-15E The mole numbers of the constituents of a gas mixture are given. The mass of each gas and the apparent gas constant are to be determined. Properties The molar masses of H2, and N2 are 2.0 and 28.0 lbm/lbmol, respectively (Table A-1E). Analysis The mass of each component is determined from N H 2 = 5 lbmol ⎯ ⎯→ m H 2 = N H 2 M H 2 = (5 lbmol)(2.0 lbm/lbmol) = 10 lbm N N 2 = 4 lbmol ⎯ ⎯→ m N 2 = N N 2 M N 2 = (4 lbmol)(28 lbm/lbmol) = 112 lbm

5 lbmol H2 4 lbmol N2

The total mass and the total number of moles are m m = m H 2 + m N 2 = 10 lbm + 112 lbm = 122 lbm N m = N H 2 + N N 2 = 5 lbmol + 4 lbmol = 9 lbmol

The molar mass and the gas constant of the mixture are determined from their definitions, Mm =

m m 122 lbm = = 13.56 lbm/lbmol N m 9 lbmol

and Rm =

Ru 1.986 Btu/lbmol ⋅ R = = 0.1465 Btu/lbm ⋅ R Mm 13.56 lbm/lbmol

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13-7

13-16 The mass fractions of the constituents of a gas mixture are given. The volumetric analysis of the mixture and the apparent gas constant are to be determined. Properties The molar masses of O2, N2 and CO2 are 32.0, 28, and 44.0 kg/kmol, respectively (Table A-1) Analysis For convenience, consider 100 kg of the mixture. Then the number of moles of each component and the total number of moles are ⎯→ N O 2 = m O 2 = 20 kg ⎯ ⎯→ N N 2 = m N 2 = 20 kg ⎯

mO2 M O2 mN2 M N2

⎯→ N CO 2 = m CO 2 = 50 kg ⎯

=

20 kg = 0.625 kmol 32 kg/kmol

30 kg = = 1.071 kmol 28 kg/kmol

m CO 2 M CO 2

=

50 kg = 1.136 kmol 44 kg/kmol

mass 20% O2 30% N2 50% CO2

N m = N O 2 + N N 2 + N CO 2 = 0.625 + 1.071 + 1.136 = 2.832 kmol

Noting that the volume fractions are same as the mole fractions, the volume fraction of each component becomes y O2 = y N2 =

N O2 Nm

=

0.625 kmol = 0.221 or 22.1% 2.832 kmol

=

1.071 kmol = 0.378 or 37.8% 2.832 kmol

N N2 Nm

y CO 2 =

N CO 2 Nm

=

1.136 kmol = 0.401 or 40.1% 2.832 kmol

The molar mass and the gas constant of the mixture are determined from their definitions, Mm =

mm 100 kg = = 35.31 kg/kmol N m 2.832 kmol

Rm =

Ru 8.314 kJ/kmol ⋅ K = = 0.235 kJ/kg ⋅ K Mm 35.31 kg/kmol

and

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13-8

P-v-T Behavior of Gas Mixtures

13-17C Normally yes. Air, for example, behaves as an ideal gas in the range of temperatures and pressures at which oxygen and nitrogen behave as ideal gases.

13-18C The pressure of a gas mixture is equal to the sum of the pressures each gas would exert if existed alone at the mixture temperature and volume. This law holds exactly for ideal gas mixtures, but only approximately for real gas mixtures.

13-19C The volume of a gas mixture is equal to the sum of the volumes each gas would occupy if existed alone at the mixture temperature and pressure. This law holds exactly for ideal gas mixtures, but only approximately for real gas mixtures. 13-20C The P-v-T behavior of a component in an ideal gas mixture is expressed by the ideal gas equation of state using the properties of the individual component instead of the mixture, Pivi = RiTi. The P-v-T behavior of a component in a real gas mixture is expressed by more complex equations of state, or by Pivi = ZiRiTi, where Zi is the compressibility factor.

13-21C Component pressure is the pressure a component would exert if existed alone at the mixture temperature and volume. Partial pressure is the quantity yiPm, where yi is the mole fraction of component i. These two are identical for ideal gases.

13-22C Component volume is the volume a component would occupy if existed alone at the mixture temperature and pressure. Partial volume is the quantity yiVm, where yi is the mole fraction of component i. These two are identical for ideal gases.

13-23C The one with the highest mole number.

13-24C The partial pressures will decrease but the pressure fractions will remain the same.

13-25C The partial pressures will increase but the pressure fractions will remain the same.

13-26C No. The correct expression is “the volume of a gas mixture is equal to the sum of the volumes each gas would occupy if existed alone at the mixture temperature and pressure.”

13-27C No. The correct expression is “the temperature of a gas mixture is equal to the temperature of the individual gas components.”

13-28C Yes, it is correct.

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13-9

13-29C With Kay's rule, a real-gas mixture is treated as a pure substance whose critical pressure and temperature are defined in terms of the critical pressures and temperatures of the mixture components as Pcr′ , m =

∑y P

i cr ,i

and Tcr′ , m =

∑yT

i cr ,i

The compressibility factor of the mixture (Zm) is then easily determined using these pseudo-critical point values.

13-30 A tank contains a mixture of two gases of known masses at a specified pressure and temperature. The mixture is now heated to a specified temperature. The volume of the tank and the final pressure of the mixture are to be determined. Assumptions Under specified conditions both Ar and N2 can be treated as ideal gases, and the mixture as an ideal gas mixture. Analysis The total number of moles is N m = N Ar + N N 2 = 0.5 kmol + 2 kmol = 2.5 kmol

and

Vm =

0.5 kmol Ar 2 kmol N2

N m Ru Tm (2.5 kmol)(8.314 kPa ⋅ m /kmol ⋅ K)(280 K) = = 23.3 m 3 250 kPa Pm 3

Also,

Q

280 K 250 kPa

P2V 2 P1V1 T 400 K = ⎯ ⎯→ P2 = 2 P1 = (250 kPa ) = 357.1 kPa T2 T1 T1 280 K

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13-10

13-31 The masses of the constituents of a gas mixture at a specified pressure and temperature are given. The partial pressure of each gas and the apparent molar mass of the gas mixture are to be determined. Assumptions Under specified conditions both CO2 and CH4 can be treated as ideal gases, and the mixture as an ideal gas mixture. Properties The molar masses of CO2 and CH4 are 44.0 and 16.0 kg/kmol, respectively (Table A-1) Analysis The mole numbers of the constituents are mCO 2 = 1 kg

⎯ ⎯→

N CO 2 =

mCH 4 = 3 kg

⎯ ⎯→

N CH 4 =

mCO 2 MCO 2 mCH 4 MCH 4

=

1 kg = 0.0227 kmol 44 kg / kmol

=

3 kg = 0.1875 kmol 16 kg / kmol

N m = N CO 2 + N CH 4 = 0.0227 kmol + 0.1875 kmol = 0.2102 kmol

yCO 2 = yCH 4 =

N CO 2 Nm N CH 4 Nm

=

0.0227 kmol = 0108 . 0.2102 kmol

=

0.1875 kmol = 0.892 0.2102 kmol

1 kg CO2 3 kg CH4 300 K 200 kPa

Then the partial pressures become PCO 2 = y CO 2 Pm = (0.108)(200 kPa ) = 21.6 kPa

PCH 4 = y CH 4 Pm = (0.892)(200 kPa ) = 178.4 kPa

The apparent molar mass of the mixture is Mm =

mm 4 kg = = 19.03 kg / kmol N m 0.2102 kmol

13-32 The masses of the constituents of a gas mixture at a specified temperature are given. The partial pressure of each gas and the total pressure of the mixture are to be determined. Assumptions Under specified conditions both N2 and O2 can be treated as ideal gases, and the mixture as an ideal gas mixture. Analysis The partial pressures of constituent gases are PN 2

(0.6 kg)(0.2968 kPa ⋅ m 3 /kg ⋅ K)(300 K) ⎛ mRT ⎞ = 178.1 kPa =⎜ ⎟ = 0.3 m 3 ⎝ V ⎠ N2

(0.4 kg)(0.2598 kPa ⋅ m 3 /kg ⋅ K)(300 K) ⎛ mRT ⎞ PO 2 = ⎜ = 103.9 kPa ⎟ = 0.3 m 3 ⎝ V ⎠ O2

0.3 m3 0.6 kg N2 0.4 kg O2 300 K

and Pm = PN 2 + PO 2 = 178.1 kPa + 103.9 kPa = 282.0 kPa

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13-11

13-33 The masses, temperatures, and pressures of two gases contained in two tanks connected to each other are given. The valve connecting the tanks is opened and the final temperature is measured. The volume of each tank and the final pressure are to be determined. Assumptions Under specified conditions both N2 and O2 can be treated as ideal gases, and the mixture as an ideal gas mixture Properties The molar masses of N2 and O2 are 28.0 and 32.0 kg/kmol, respectively (Table A-1) Analysis The volumes of the tanks are (1 kg)(0.2968 kPa ⋅ m 3 /kg ⋅ K)(298 K) ⎛ mRT ⎞ = 0.295 m 3 ⎟ = 300 kPa ⎝ P ⎠ N2

V N2 = ⎜

(3 kg)(0.2598 kPa ⋅ m 3 /kg ⋅ K)(298 K) ⎛ mRT ⎞ = 0.465 m 3 ⎟ = 500 kPa P ⎝ ⎠ O2

V O2 = ⎜

1 kg N2

3 kg O2

25°C 300 kPa

25°C 500 kPa

V total = V N 2 +V O 2 = 0.295 m 3 + 0.465 m 3 = 0.76 m 3 Also, m N 2 = 1 kg ⎯ ⎯→ N N 2 = ⎯→ N O 2 = m O 2 = 3 kg ⎯

mN2 M N2 mO2 M O2

=

1 kg = 0.03571 kmol 28 kg/kmol

=

3 kg = 0.09375 kmol 32 kg/kmol

N m = N N 2 + N O 2 = 0.03571 kmol + 0.09375 kmol = 0.1295 kmol

Thus, ⎛ NRu T Pm = ⎜⎜ ⎝ V

(0.1295 kmol)(8.314 kPa ⋅ m 3 /kmol ⋅ K)(298 K) ⎞ ⎟⎟ = = 422.2 kPa 0.76 m 3 ⎠m

13-34 A container contains a mixture of two fluids whose volumes are given. The density of the mixture is to be determined. Assumptions The volume of the mixture is the sum of the volumes of the two constituents. Properties The specific volumes of the two fluids are given to be 0.0003 m3/kg and 0.00023 m3/kg. Analysis The mass of the two fluids are mA =

VA 0.001 m 3 = = 3.333 kg v A 0.0003 m 3 /kg

V 0.002 m 3 mB = B = = 8.696 kg v B 0.00023 m 3 /kg

1 L fluid A 2 L fluid B

The density of the mixture is then

ρ=

m A + m B (3.333 + 8.696) lbf = = 4010 kg/m 3 3 V A +V B (0.001 + 0.002) ft

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13-12

13-35E A mixture is obtained by mixing two gases at constant pressure and temperature. The volume and specific volume of the mixture are to be determined. Properties The densities of two gases are given in the problem statement. Analysis The volume of constituent gas A is

VA =

mA

ρA

=

1 lbm 0.001 lbm/ft 3

= 1000 ft 3

and the volume of constituent gas B is

VB =

mB

ρB

=

2 lbm 0.002 lbm/ft

3

= 1000 ft 3

1 lbm gas A 2 lbm gas B

Hence, the volume of the mixture is

V = V A + V B = 1000 + 1000 = 2000 ft 3 The specific volume of the mixture will then be

v=

V m

=

2000 ft 3 = 666.7 ft 3 /lbm (1 + 2) lbm

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13-13

13-36 The masses of components of a gas mixture are given. The apparent molecular weight of this mixture, the volume it occupies, the partial volume of the oxygen, and the partial pressure of the helium are to be determined. Properties The molar masses of O2, CO2, and He are 32.0, 44.0, and 4.0 kg/kmol, respectively (Table A-1). Analysis The total mass of the mixture is m m = m O2 + m CO2 + m He = 0.1 + 1 + 0.5 = 1.6 kg

The mole numbers of each component are N O2 =

m O2 0.1 kg = = 0.003125 kmol M O2 32 kg/kmol

N CO2 =

m CO2 1 kg = = 0.02273 kmol M CO2 44 kg/kmol

N He =

0.1 kg O2 1 kg CO2 0.5 kg He

m He 0.5 kg = = 0.125 kmol M He 4 kg/kmol

The mole number of the mixture is N m = N O2 + N CO2 + N He = 0.003125 + 0.02273 + 0.125 = 0.15086 kmol

Then the apparent molecular weight of the mixture becomes Mm =

mm 1.6 kg = = 10.61 kg/kmol N m 0.15086 kmol

The volume of this ideal gas mixture is

Vm =

N m Ru T (0.1509 kmol)(8.314 kPa ⋅ m 3 /kmol ⋅ K)(300 K) = = 3.764 m 3 P 100 kPa

The partial volume of oxygen in the mixture is

V O2 = y O2V m =

N O2 0.003125 kmol Vm = (3.764 m 3 ) = 0.07795 m 3 Nm 0.1509 kmol

The partial pressure of helium in the mixture is PHe = y He Pm =

N He 0.125 kmol Pm = (100 kPa) = 82.84 kPa Nm 0.1509 kmol

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13-14

13-37 The mass fractions of components of a gas mixture are given. The mole fractions of each constituent, the mixture’s apparent molecular weight, the partial pressure of each constituent, and the apparent specific heats of the mixture are to be determined. Properties The molar masses of N2, He, CH4, and C2H6 are 28.0, 4.0, 16.0, and 30.0 kg/kmol, respectively (Table A-1). The constant-pressure specific heats of these gases at room temperature are 1.039, 5.1926, 2.2537, and 1.7662 kJ/kg⋅K, respectively (Table A-2a). Analysis We consider 100 kg of this mixture. The mole numbers of each component are m 15 kg N N2 = N2 = = 0.5357 kmol M N2 28 kg/kmol N He = N CH4 = N C2H6 =

m He 5 kg = = 1.25 kmol M He 4 kg/kmol m CH4 60 kg = = 3.75 kmol M CH4 16 kg/kmol m C2H6 20 kg = = 0.6667 kmol M C2H6 30 kg/kmol

15% N2 5% He 60% CH4 20% C2H6 (by mass)

The mole number of the mixture is N m = N N2 + N He + N CH4 + N C2H6 = 0.5357 + 1.25 + 3.75 + 0.6667 = 6.2024 kmol and the mole fractions are N 0.5357 kmol = 0.08637 y N2 = N2 = 6.2024 kmol Nm y He =

N He 1.25 kmol = = 0.2015 6.2024 kmol Nm

y CH4 =

N CH4 3.75 kmol = = 0.6046 6.2024 kmol Nm

y C2H6 =

N C2H6 0.6667 kmol = = 0.1075 6.2024 kmol Nm

The apparent molecular weight of the mixture is m 100 kg Mm = m = = 16.12 kg/kmol N m 6.2024 kmol The partial pressure of each constituent for a mixture pressure of 1200 kPa are PN2 = y N2 Pm = (0.08637)(1200 kPa) = 103.6 kPa PHe = y He Pm = (0.2015)(1200 kPa) = 241.8 kPa PCH4 = y CH4 Pm = (0.6046)(1200 kPa) = 725.5 kPa PC2H6 = y C2H6 Pm = (0.1075)(1200 kPa) = 129.0 kPa

The constant-pressure specific heat of the mixture is determined from c p = mf N2 c p , N2 + mf He c p ,He + mf CH4 c p ,CH4 + mf C2H6 c p ,C2H6 = 0.15 ×1.039 + 0.05 × 5.1926 + 0.60 × 2.2537 + 0.20 × 1.7662 = 2.121 kJ/kg ⋅ K

The apparent gas constant of the mixture is R 8.314 kJ/kmol ⋅ K = 0.5158 kJ/kg ⋅ K R= u = 16.12 kg/kmol Mm Then the constant-volume specific heat is cv = c p − R = 2.121 − 0.5158 = 1.605 kJ/kg ⋅ K

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13-15

13-38 The volume fractions of components of a gas mixture are given. The mixture’s apparent molecular weight and the apparent specific heats of the mixture are to be determined. Properties The molar masses of O2, N2, CO2, and CH4 are 32.0, 28.0, 44.0, and 16.0 kg/kmol, respectively (Table A-1). The constant-pressure specific heats of these gases at room temperature are 0.918, 1.039, 0.846, and 2.2537 kJ/kg⋅K, respectively (Table A-2). Analysis We consider 100 kmol of this mixture. Noting that volume fractions are equal to the mole fractions, mass of each component are m O2 = N O2 M O2 = (30 kmol)(32 kg/kmol) = 960 kg m N2 = N N2 M N2 = (40 kmol)(28 kg/kmol) = 1120 kg m CO2 = N CO2 M CO2 = (10 kmol)(44 kg/kmol) = 440 kg m CH4 = N CH4 M CH4 = (20 kmol)(16 kg/kmol) = 320 kg

The total mass is m m = m O2 + m N2 + m CO2 + m CH4 = 960 + 1120 + 440 + 320 = 2840 kg

30% O2 40% N2 10% CO2 20% CH4 (by volume)

Then the mass fractions are mf O2 =

m O2 960 kg = = 0.3380 mm 2840 kg

mf N2 =

m N2 1120 kg = = 0.3944 2840 kg mm

mf CO2 =

m CO2 440 kg = = 0.1549 2840 kg mm

mf CH4 =

m CH4 320 kg = = 0.1127 2840 kg mm

The apparent molecular weight of the mixture is Mm =

mm 2840 kg = = 28.40 kg/kmol N m 100 kmol

The constant-pressure specific heat of the mixture is determined from c p = mf O2 c p ,O2 + mf N2 c p , N2 + mf CO2 c p ,CO2 + mf CH4 c p ,CH4 = 0.3380 × 0.918 + 0.3944 × 1.039 + 0.1549 × 0.846 + 0.1127 × 2.2537 = 1.1051 kJ/kg ⋅ K

The apparent gas constant of the mixture is R=

Ru 8.314 kJ/kmol ⋅ K = = 0.2927 kJ/kg ⋅ K 28.40 kg/kmol Mm

Then the constant-volume specific heat is cv = c p − R = 1.1051 − 0.2927 = 0.8124 kJ/kg ⋅ K

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13-16

13-39 The mass fractions of components of a gas mixture are given. The volume occupied by 100 kg of this mixture is to be determined. Properties The molar masses of CH4, C3H8, and C4H10 are 16.0, 44.0, and 58.0 kg/kmol, respectively (Table A-1). Analysis The mole numbers of each component are N CH4 =

m CH4 60 kg = = 3.75 kmol M CH4 16 kg/kmol

N C3H8 =

m C3H8 25 kg = = 0.5682 kmol M C3H8 44 kg/kmol

N C4H10 =

m C4H10 15 kg = = 0.2586 kmol M C4H10 58 kg/kmol

60% CH4 25% C3H8 15% C4H10 (by mass)

The mole number of the mixture is N m = N CH4 + N C3H8 + N C4H10 = 3.75 + 0.5682 + 0.2586 = 4.5768 kmol

The apparent molecular weight of the mixture is Mm =

mm 100 kg = = 21.85 kg/kmol N m 4.5768 kmol

Then the volume of this ideal gas mixture is

Vm =

N m Ru T (4.5768 kmol)(8.314 kPa ⋅ m 3 /kmol ⋅ K)(310 K) = = 3.93 m 3 3000 kPa P

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13-17

13-40E The mass fractions of components of a gas mixture are given. The mass of 5 ft3 of this mixture and the partial volumes of the components are to be determined. Properties The molar masses of N2, O2, and He are 28.0, 32.0, and 4.0 lbm/lbmol, respectively (Table A1E). Analysis We consider 100 lbm of this mixture for calculating the molar mass of the mixture. The mole numbers of each component are N N2 = N O2

m N2 60 lbm = = 2.1429 lbmol M N2 28 lbm/lbmol

m 30 lbm = O2 = = 0.9375 lbmol M O2 32 lbm/lbmol

N He =

m He 10 lbm = = 2.5 lbmol M He 4 lbm/lbmol

5 ft3 60% N2 30% O2 10% He (by mass)

The mole number of the mixture is N m = N N2 + N O2 + N He = 2.1429 + 0.9375 + 2.5 = 5.5804 lbmol

The apparent molecular weight of the mixture is Mm =

mm 100 lbm = = 17.92 lbm/lbmol N m 5.5804 lbmol

Then the mass of this ideal gas mixture is m=

PVM m (300 psia)(5 ft 3 )(17.92 lbm/lbmol) = = 4.727 lbm Ru T (10.73 psia ⋅ ft 3 /lbmol ⋅ R)(530 R)

The mole fractions are y N2 =

N N2 2.1429 lbmol = = 0.3840 5.5804 lbmol Nm

y O2 =

N O2 0.9375 lbmol = = 0.1680 5.5804 lbmol Nm

y He =

N He 2.5 lbmol = = 0.4480 5.5804 lbmol Nm

Noting that volume fractions are equal to mole fractions, the partial volumes are determined from

V N2 = y N2V m = (0.3840)(5 ft 3 ) = 1.92 ft 3 V O2 = y O2V m = (0.1680)(5 ft 3 ) = 0.84 ft 3 V He = y HeV m = (0.4480)(5 ft 3 ) = 2.24 ft 3

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13-18

13-41 The mass fractions of components of a gas mixture are given. The partial pressure of ethane is to be determined. Properties The molar masses of CH4 and C2H6 are 16.0 and 30.0 kg/kmol, respectively (Table A-1). Analysis We consider 100 kg of this mixture. The mole numbers of each component are N CH4 = N C2H6 =

m CH4 70 kg = = 4.375 kmol M CH4 16 kg/kmol m C2H6 30 kg = = 1.0 kmol M C2H6 30 kg/kmol

The mole number of the mixture is N m = N CH4 + N C2H6 = 4.375 + 1.0 = 5.375 kmol

70% CH4 30% C2H6 (by mass) 100 m3 130 kPa, 25°C

The mole fractions are y CH4 =

N CH4 4.375 kmol = = 0.8139 5.375 kmol Nm

y C2H6 =

N C2H6 1.0 kmol = = 0.1861 5.375 kmol Nm

The final pressure of ethane in the final mixture is PC2H6 = y C2H6 Pm = (0.1861)(130 kPa) = 24.19 kPa

13-42E The Orsat analysis (molar fractions) of components of a gas mixture are given. The mass flow rate of the mixture is to be determined. Properties The molar masses of CO2, O2, N2, and CO are 44.0, 32.0, 28.0, and 28.0 lbm/lbmol, respectively (Table A-1E). Analysis The molar fraction of N2 is y N2 = 1 − y CO2 − y O2 − y CO = 1 − 0.15 − 0.15 − 0.01 = 0.69

The molar mass of the mixture is determined from M m = y CO2 M CO2 + y O2 M O2 + y CO M CO + y N2 M N2 = 0.15 × 44 + 0.15 × 32 + 0.01× 28 + 0.69 × 28 = 31.00 lbm/lbmol

15% CO2 15% O2 1% CO 69% N2 (by mole)

The specific volume of the mixture is

v=

Ru T (10.73 psia ⋅ ft 3 /lbmol ⋅ R)(660 R) = = 15.54 ft 3 /lbm MmP (31.00 lbm/lbmol)(14.7 psia)

The mass flow rate of these gases is then m& =

AV

v

=

(10 ft 2 )(20 ft/s) 15.54 ft 3 /lbm

Mixture 20 ft/s, 1 atm 200°F

= 12.87 lbm/s

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13-19

13-43 The volumetric fractions of components of a gas mixture before and after a separation unit are given. The changes in partial pressures of the components in the mixture before and after the separation unit are to be determined. Analysis The partial pressures before the separation unit are PCH4 = y CH4 Pm = (0.60)(100 kPa) = 60 kPa PC2H6 = y C2H6 Pm = (0.20)(100 kPa) = 20 kPa PC3H8 = y C3H8 Pm = (0.10)(100 kPa) = 10 kPa

The mole fraction of propane is 0.10 after the separation unit. The corresponding mole fractions of methane and ethane are determined as follows: x = 0.01 ⎯ ⎯→ x = 0.00808 0.6 + 0.2 + x 0.60 y CH4 = = 0.7425 0.6 + 0.2 + 0.00808 0.20 y C2H6 = = 0.2475 0.6 + 0.2 + 0.00808 0.00808 y C3H8 = = 0.01 0.6 + 0.2 + 0.00808

60% CH4 20% C2H6 10% C3H8 (by volume)

The partial pressures after the separation unit are PCH4 = y CH4 Pm = (0.7425)(100 kPa) = 74.25 kPa PC2H6 = y C2H6 Pm = (0.2475)(100 kPa) = 24.75 kPa PC3H8 = y C3H8 Pm = (0.01)(100 kPa) = 1kPa

The changes in partial pressures are then ΔPCH4 = 74.25 − 60 = 14.25 kPa ΔPC2H6 = 24.75 − 20 = 4.75 kPa ΔPC3H8 = 1 − 10 = −9 kPa

13-44 The partial pressure of R-134a in atmospheric air to form a 100-ppm contaminant is to be determined. Analysis Noting that volume fractions and mole fractions are equal, the molar fraction of R-134a in air is y R134a =

100 10 6

= 0.0001

The partial pressure of R-134a in air is then PR134a = y R134a Pm = (0.0001)(100 kPa) = 0.01 kPa

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13-20

13-45E The volumetric analysis of a mixture of gases is given. The volumetric and mass flow rates are to be determined using three methods. Properties The molar masses of O2, N2, CO2, and CH4 are 32.0, 28.0, 44.0, and 16.0 lbm/lbmol, respectively (Table A-1E). Analysis (a) We consider 100 lbmol of this mixture. Noting that volume fractions are equal to the mole fractions, mass of each component are m O2 = N O2 M O2 = (30 lbmol)(32 lbm/lbmol) = 960 lbm m N2 = N N2 M N2 = (40 lbmol)(28 lbm/lbmol) = 1120 lbm

30% O2 40% N2 10% CO2 20% CH4 (by volume)

m CO2 = N CO2 M CO2 = (10 lbmol)(44 lbm/lbmol) = 440 lbm m CH4 = N CH4 M CH4 = (20 lbmol)(16 lbm/lbmol) = 320 lbm

The total mass is m m = m O2 + m N2 + m CO2 + m CH4 = 960 + 1120 + 440 + 320 = 2840 lbm

The apparent molecular weight of the mixture is Mm =

mm 2840 lbm = = 28.40 lbm/lbmol N m 100 lbmol

Mixture 1500 psia 70°F

The apparent gas constant of the mixture is R=

Ru 10.73 psia ⋅ ft 3 /lbmol ⋅ R = = 0.3778 psia ⋅ ft 3 /lbm ⋅ R 28.40 lbm/lbmol Mm

The specific volume of the mixture is

v=

RT (0.3778 psia ⋅ ft 3 /lbm ⋅ R)(530 R) = = 0.1335 ft 3 /lbm P 1500 psia

The volume flow rate is

V& = AV =

πD 2 4

V=

π (1/12 ft) 2 4

(10 ft/s) = 0.05454 ft 3 /s

and the mass flow rate is m& =

V& 0.05454 ft 3 /s = = 0.4085 lbm/s v 0.1335 ft 3 /lbm

(b) To use the Amagat’s law for this real gas mixture, we first need the mole fractions and the Z of each component at the mixture temperature and pressure. The compressibility factors are obtained using Fig. A15 to be ⎫ ⎪ ⎪ ⎬ Z O2 = 0.94 1500 psia = = 2.038 ⎪ ⎪ 736 psia ⎭

T R ,O2 =

Tm 530 R = = 1.902 Tcr,O2 278.6 R

T R , N2 =

PR ,O2 =

Pm Pcr,O2

PR ,CN

530 R = 2.334 227.1 R 1500 psia = = 3.049 492 psia

⎫ ⎪⎪ ⎬ Z N2 = 0.99 ⎪ ⎭⎪

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13-21

530 R = 0.968 547.5 R 1500 psia = = 1.401 1071 psia

T R ,CO2 = PR ,CO2

⎫ ⎪⎪ ⎬ Z CO2 = 0.21 ⎪ ⎭⎪

530 R = 1.541 343.9 R 1500 psia = = 2.229 673 psia

T R ,CH4 = PR ,CH4

⎫ ⎪⎪ ⎬ Z CO2 = 0.85 ⎪ ⎭⎪

and Zm =

∑y Z i

i

= y O2 Z O2 + y O2 Z O2 + y CO2 Z CO2 + y CH4 Z CH4

= (0.30)(0.94) + (0.40)(0.99) + (0.10)(0.21) + (0.20)(0.85) = 0.869

Then,

v=

Z m RT (0.869)(0.3778 psia ⋅ ft 3 /lbm ⋅ R)(530 R) = = 0.1160 ft 3 /lbm P 1500 psia

V& = 0.05454 ft 3 /s m& =

0.05454 ft 3 /s V& = = 0.4702 lbm/s v 0.1160 ft 3 /lbm

(c) To use Kay's rule, we need to determine the pseudo-critical temperature and pseudo-critical pressure of the mixture using the critical point properties of mixture gases. Tcr′ , m =

∑yT

i cr ,i

= y O2 Tcr ,O2 + y N2 Tcr , N2 + y CO2 Tcr ,CO2 + y CH4 Tcr ,CH4

= (0.30)(278.6 R) + (0.40)(227.1 R) + (0.10)(547.5 R) + (0.20)(343.9 R) = 298.0 R Pcr′ , m =

∑y P

i cr ,i

= y O2 Pcr ,O2 + + y N2 Pcr , N2 + y CO2 Pcr ,CO2 + y CH4 Pcr ,CH4

= (0.30)(736 psia) + (0.40)(492 psia) + (0.10)(1071 psia) + (0.20)(673 psia) = 659.3 psia

and TR = PR =

Tm ' Tcr, m

Pm ' Pcr, m

⎫ ⎪ ⎪ ⎬ Z m = 0.915 1500 psia = = 2.275 ⎪ ⎪ 659.3 psia ⎭

=

530 R = 1.779 298.0 R

(Fig. A-15)

Then,

v=

Z m RT (0.915)(0.3778 psia ⋅ ft 3 /lbm ⋅ R)(530 R) = = 0.1221 ft 3 /lbm P 1500 psia

V& = 0.05454 ft 3 /s m& =

0.05454 ft 3 /s V& = = 0.4467 lbm/s v 0.1221 ft 3 /lbm

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13-22

13-46 The volumes, temperatures, and pressures of two gases forming a mixture are given. The volume of the mixture is to be determined using three methods. Analysis (a) Under specified conditions both O2 and N2 will considerably deviate from the ideal gas behavior. Treating the mixture as an ideal gas, ⎛ PV N O 2 = ⎜⎜ ⎝ RuT

⎞ (8000 kPa)(0.3 m3 ) ⎟ = = 1.443 kmol 3 ⎟ ⎠O 2 (8.314 kPa ⋅ m /kmol ⋅ K)(200 K)

⎛ PV N N 2 = ⎜⎜ ⎝ RuT

⎞ (8000 kPa)(0.5 m 3 ) ⎟ = = 2.406 kmol 3 ⎟ ⎠ N 2 (8.314 kPa ⋅ m /kmol ⋅ K)(200 K)

N m = N O 2 + N N 2 = 1.443 kmol + 2.406 kmol = 3.849 kmol

Vm =

0.3 m3 O2 200 K 8 MPa 0.5 m3 N2 200 K 8 MPa

N 2 + O2 200 K 8 MPa

N m RuTm (3.849 kmol)(8.314 kPa ⋅ m3 /kmol ⋅ K)(200 K) = = 0.8 m 3 Pm 8000 kPa

(b) To use Kay's rule, we need to determine the pseudo-critical temperature and pseudo-critical pressure of the mixture using the critical point properties of O2 and N2 from Table A-1. But we first need to determine the Z and the mole numbers of each component at the mixture temperature and pressure (Fig. A-15), ⎫ ⎪ ⎪ ⎬ Z O 2 = 0.77 8 MPa = = 1.575 ⎪ ⎪ 5.08 MPa ⎭

T R ,O 2 =

Tm 200 K = = 1.292 Tcr,O 2 154.8 K

PR ,O 2 =

Pm Pcr,O 2

TR, N 2 =

Tm

O 2:

N 2: PR , N 2 =

⎫ ⎪ ⎪ ⎬ Z N 2 = 0.863 8 MPa = = 2.360 ⎪ ⎪ 3.39 MPa ⎭

=

Tcr, N 2 Pm Pcr, N 2

200K = 1.585 126.2K

⎛ PV N O 2 = ⎜⎜ ⎝ ZRu T

⎞ (8000 kPa)(0.3 m 3 ) ⎟ = = 1.874 kmol ⎟ 3 ⎠ O 2 (0.77)(8.314 kPa ⋅ m /kmol ⋅ K)(200 K)

⎛ PV N N 2 = ⎜⎜ ⎝ ZRu T

⎞ (8000 kPa)(0.5 m 3 ) ⎟ = = 2.787 kmol ⎟ 3 ⎠ N 2 (0.863)(8.314 kPa ⋅ m /kmol ⋅ K)(200 K)

N m = N O 2 + N N 2 = 1.874 kmol + 2.787 kmol = 4.661 kmol

The mole fractions are y O2 = y N2 = Tcr′ , m =

N O2 Nm N N2 Nm

=

1.874kmol = 0.402 4.661kmol

=

2.787kmol = 0.598 4.661kmol

∑yT

i cr ,i

= y O 2 Tcr ,O 2 + y N 2 Tcr , N 2

= (0.402)(154.8K) + (0.598)(126.2K) = 137.7K Pcr′ , m =

∑y P

i cr ,i

= y O 2 Pcr ,O 2 + y N 2 Pcr , N 2

= (0.402)(5.08MPa) + (0.598)(3.39MPa) = 4.07 MPa

Then, PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

13-23

TR = PR =

Tm

⎫ ⎪ ⎪ ⎬ Z m = 0.82 8 MPa = = 1.966 ⎪ ⎪ 4.07 MPa ⎭

=

' Tcr, O2

Pm ' Pcr, O2

200 K = 1.452 137.7 K

(Fig. A-15)

Thus,

Vm =

Z m N m Ru Tm (0.82)(4.661 kmol)(8.314 kPa ⋅ m 3 /kmol ⋅ K)(200 K) = = 0.79 m 3 Pm 8000 kPa

(c) To use the Amagat’s law for this real gas mixture, we first need the Z of each component at the mixture temperature and pressure, which are determined in part (b). Then, Zm =

∑y Z

Vm =

Z m N m Ru Tm (0.83)(4.661 kmol)(8.314 kPa ⋅ m 3 /kmol ⋅ K)(200 K) = = 0.80 m 3 Pm 8000 kPa

i

i

= y O 2 Z O 2 + y N 2 Z N 2 = (0.402 )(0.77 ) + (0.598)(0.863) = 0.83

Thus,

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13-24

13-47 [Also solved by EES on enclosed CD] The mole numbers, temperatures, and pressures of two gases forming a mixture are given. The final temperature is also given. The pressure of the mixture is to be determined using two methods. Analysis (a) Under specified conditions both Ar and N2 will considerably deviate from the ideal gas behavior. Treating the mixture as an ideal gas, Initial state : P1V1 = N1RuT1 ⎫ N 2T2 (4)(200 K) P1 = (5 MPa ) = 18.2 MPa ⎬ P2 = Final state : P2V 2 = N 2 RuT2 ⎭ N1T1 (1)(220 K)

(b) Initially, TR = PR =

T1 Tcr,Ar P1 Pcr,Ar

⎫ ⎪ ⎪ ⎬ Z Ar = 0.90 (Fig. A-15) 5 MPa = = 1.0278 ⎪ ⎪ 4.86 MPa ⎭

=

220 K = 1.457 151.0 K

1 kmol Ar 220 K 5 MPa

3 kmol N2 190 K 8 MPa

Then the volume of the tank is

V =

ZN Ar Ru T (0.90)(1 kmol)(8.314 kPa ⋅ m 3 /kmol ⋅ K)(220 K) = = 0.33 m 3 P 5000 kPa

After mixing, ⎫ ⎪ ⎪ ⎪ V m / N Ar v Ar ⎪ = = ⎬ PR = 0.90 Ru Tcr,Ar / Pcr,Ar Ru Tcr,Ar / Pcr,Ar ⎪ ⎪ (0.33 m 3 )/(1 kmol) ⎪ = = 1 . 278 ⎪ (8.314 kPa ⋅ m 3 /kmol ⋅ K)(151.0 K)/(4860 kPa) ⎭

(Fig. A-15)

⎫ ⎪ Tcr, N 2 ⎪ ⎪ v N2 V m / N N2 ⎪ = = ⎬ PR = 3.75 Ru Tcr, N 2 / Pcr, N 2 Ru Tcr, N 2 / Pcr, N 2 ⎪ ⎪ 3 (0.33 m )/(3 kmol) ⎪ 0 . 355 = = ⎪ (8.314 kPa ⋅ m 3 /kmol ⋅ K)(126.2 K)/(3390 kPa) ⎭

(Fig. A-15)

T R ,Ar =

Ar:

V R , Ar

TR, N 2 =

N 2:

V R, N 2

Tm 200K = = 1.325 Tcr,Ar 151.0K

Tm

=

200K = 1.585 126.2K

Thus, PAr = ( PR Pcr ) Ar = (0.90)(4.86 MPa) = 4.37 MPa PN 2 = ( PR Pcr ) N 2 = (3.75)(3.39 MPa) = 12.7 MPa

and Pm = PAr + PN 2 = 4.37 MPa + 12.7 MPa = 17.1 MPa

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13-25

13-48 EES Problem 13-47 is reconsidered. The effect of the moles of nitrogen supplied to the tank on the final pressure of the mixture is to be studied using the ideal-gas equation of state and the compressibility chart with Dalton's law. Analysis The problem is solved using EES, and the solution is given below. "Input Data" R_u = 8.314 [kJ/kmol-K] "universal Gas Constant" T_Ar = 220 [K] P_Ar = 5000 [kPa] "Pressure for only Argon in the tank initially." N_Ar = 1 [kmol] {N_N2 = 3 [kmol]} T_mix = 200 [K] T_cr_Ar=151.0 [K] "Critical Constants are found in Table A.1 of the text" P_cr_Ar=4860 [kPa] T_cr_N2=126.2 [K] P_cr_N2=3390 [kPa] "Ideal-gas Solution:" P_Ar*V_Tank_IG = N_Ar*R_u*T_Ar "Apply the ideal gas law the gas in the tank." P_mix_IG*V_Tank_IG = N_mix*R_u*T_mix "Ideal-gas mixture pressure" N_mix=N_Ar + N_N2 "Moles of mixture" "Real Gas Solution:" P_Ar*V_Tank_RG = Z_Ar_1*N_Ar*R_u*T_Ar "Real gas volume of tank" T_R=T_Ar/T_cr_Ar "Initial reduced Temp. of Ar" P_R=P_Ar/P_cr_Ar "Initial reduced Press. of Ar" Z_Ar_1=COMPRESS(T_R, P_R ) "Initial compressibility factor for Ar" P_Ar_mix*V_Tank_RG = Z_Ar_mix*N_Ar*R_u*T_mix "Real gas Ar Pressure in mixture" T_R_Ar_mix=T_mix/T_cr_Ar "Reduced Temp. of Ar in mixture" P_R_Ar_mix=P_Ar_mix/P_cr_Ar "Reduced Press. of Ar in mixture" Z_Ar_mix=COMPRESS(T_R_Ar_mix, P_R_Ar_mix ) "Compressibility factor for Ar in mixture" P_N2_mix*V_Tank_RG = Z_N2_mix*N_N2*R_u*T_mix "Real gas N2 Pressure in mixture" T_R_N2_mix=T_mix/T_cr_N2 "Reduced Temp. of N2 in mixture" P_R_N2_mix=P_N2_mix/P_cr_N2 "Reduced Press. of N2 in mixture" Z_N2_mix=COMPRESS(T_R_N2_mix, P_R_N2_mix ) "Compressibility factor for N2 in mixture" P_mix=P_R_Ar_mix*P_cr_Ar +P_R_N2_mix*P_cr_N2 "Mixture pressure by Dalton's law. 23800" 225000

Pmix [kPa] 9009 13276 17793 23254 30565 41067 56970 82372 126040 211047

Pmix,IG [kPa] 9091 13636 18182 22727 27273 31818 36364 40909 45455 50000

180000

S olu tio n M ethod P mix [kPa]

NN2 [kmol] 1 2 3 4 5 6 7 8 9 10

135000

C h art Ideal G as

90000

45000

0 1

2

3

4

5

6

7

8

9

N N 2 [km ol]

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10

13-26

Properties of Gas Mixtures

13-49C Yes. Yes (extensive property).

13-50C No (intensive property).

13-51C The answers are the same for entropy.

13-52C Yes. Yes (conservation of energy).

13-53C We have to use the partial pressure.

13-54C No, this is an approximate approach. It assumes a component behaves as if it existed alone at the mixture temperature and pressure (i.e., it disregards the influence of dissimilar molecules on each other.)

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13-27

13-55 Volumetric fractions of the constituents of a mixture are given. The mixture undergoes an adiabatic compression process. The makeup of the mixture on a mass basis and the internal energy change per unit mass of mixture are to be determined. Assumptions Under specified conditions all CO2, CO, O2, and N2 can be treated as ideal gases, and the mixture as an ideal gas mixture. Properties 1 The molar masses of CO2, CO, O2, and N2 are 44.0, 28.0, 32.0, and 28.0 kg/kmol, respectively (Table A-1). 2 The process is reversible. Analysis Noting that volume fractions are equal to mole fractions in ideal gas mixtures, the molar mass of the mixture is determined to be M m = y CO 2 M CO 2 + y CO M CO + y O 2 M O 2 + y N 2 M N 2 = (0.15)(44) + (0.05)(28) + (0.10)(32) + (0.70)(28) = 30.80 kg/kmol

The mass fractions are mf CO 2 = y CO 2 mf CO = y CO mf O 2 = y O 2 mf N 2 = y N 2

M CO 2 Mm

= (0.15)

44 kg/kmol = 0.2143 30.80 kg/kmol

M CO 28 kg/kmol = (0.05) = 0.0454 30.80 kg/kmol Mm M O2 Mm M N2 Mm

= (0.10)

32 kg/kmol = 0.1039 30.80 kg/kmol

= (0.70)

28 kg/kmol = 0.6364 30.80 kg/kmol

15% CO2 5% CO 10% O2 70% N2 300 K, 1 bar

The final pressure of mixture is expressed from ideal gas relation to be P2 = P1 r

T2 T = (100 kPa )(8) 2 = 2.667T2 300 K T1

(Eq. 1)

since the final temperature is not known. We assume that the process is reversible as well being adiabatic (i.e. isentropic). Using Dalton’s law to find partial pressures, the entropies at the initial state are determined from EES to be: ⎯→ s CO 2 ,1 = 5.2190 kJ/kg.K T = 300 K, P = (0.2143 × 100) = 21.43 kPa ⎯ ⎯→ s CO,1 = 79483 kJ/kg.K T = 300 K, P = (0.04545 × 100) = 4.55 kPa ⎯ ⎯→ s N 2 ,1 = 6.9485 kJ/kg.K T = 300 K, P = (0.1039 × 100) = 10.39 kPa ⎯ ⎯→ s O 2 ,1 = 7.0115 kJ/kg.K T = 300 K, P = (0.6364 × 100) = 63.64 kPa ⎯

The final state entropies cannot be determined at this point since the final pressure and temperature are not known. However, for an isentropic process, the entropy change is zero and the final temperature and the final pressure may be determined from Δs total = mf CO 2 Δs CO 2 + mf CO Δs CO + mf O 2 Δs O 2 + mf N 2 Δs N 2 = 0

and using Eq. (1). The solution may be obtained using EES to be T2 = 631.4 K, P2 = 1684 kPa The initial and final internal energies are (from EES)

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13-28

u CO 2 ,1 u CC,1 ⎯→ T1 = 300 K ⎯ u O 2 ,1 u N 2 ,1

= −8997 kJ/kg = −4033 kJ/kg = −76.24 kJ/kg = −87.11 kJ/kg,

u CO 2 , 2 u CO, 2 ⎯→ T2 = 631.4 K ⎯ u O 2 ,2 u N 2 ,2

= −8734 kJ/kg = −3780 kJ/kg = 156.8 kJ/kg = 163.9 kJ/kg

The internal energy change per unit mass of mixture is determined from Δu mixture = mf CO 2 (u CO 2 , 2 − u CO 2 ,1 ) + mf CO (u CO, 2 − u CO,1 ) + mf O 2 (u O 2 , 2 − u O 2 ,1 ) + mf N 2 (u N 2 , 2 − u N 2 ,1 ) = 0.2143[(−8734) − (−8997)] + 0.0454[(−3780) − (−4033)] + 0.1039[156.8 − (−76.24)]6 + 0.6364[163.9 − (−87.11)] = 251.8 kJ/kg

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13-29

13-56 Propane and air mixture is compressed isentropically in an internal combustion engine. The work input is to be determined. Assumptions Under specified conditions propane and air can be treated as ideal gases, and the mixture as an ideal gas mixture. Properties The molar masses of C3H8 and air are 44.0 and 28.97 kg/kmol, respectively (TableA-1). Analysis Given the air-fuel ratio, the mass fractions are determined to be mf air = mf C3H8

AF 16 = = 0.9412 AF + 1 17 1 1 = = = 0.05882 AF + 1 17

The molar mass of the mixture is determined to be Mm =

mf air M air

1 1 = = 29.56 kg/kmol mf C3H8 0.9412 0.05882 + + 28.97 kg/kmol 44.0 kg/kmol M C3H8

Propane Air 95 kPa 30ºC

The mole fractions are y air = mf air

Mm 29.56 kg/kmol = (0.9412) = 0.9606 M air 28.97 kg/kmol

y C3H8 = mf C3H8

Mm 29.56 kg/kmol = (0.05882) = 0.03944 M C3H8 44.0 kg/kmol

The final pressure is expressed from ideal gas relation to be P2 = P1 r

T2 T2 = (95 kPa )(9.5) = 2.977T2 T1 (30 + 273.15) K

(1)

since the final temperature is not known. Using Dalton’s law to find partial pressures, the entropies at the initial state are determined from EES to be: T = 30°C, P = (0.9606 × 95) = 91.26 kPa ⎯ ⎯→ s air ,1 = 5.7417 kJ/kg.K T = 30°C, P = (0.03944 × 95) = 3.75 kPa ⎯ ⎯→ s C3H8 ,1 = 6.7697 kJ/kg.K

The final state entropies cannot be determined at this point since the final pressure and temperature are not known. However, for an isentropic process, the entropy change is zero and the final temperature and the final pressure may be determined from Δs total = mf air Δs air + mf C3H8 Δs C3H 8 = 0

and using Eq. (1). The solution may be obtained using EES to be T2 = 654.9 K, P2 = 1951 kPa The initial and final internal energies are (from EES) ⎯→ T1 = 30°C ⎯

u air ,1 = 216.5 kJ/kg u C3H8 ,1 = −2404 kJ/kg

⎯→ T2 = 654.9 K ⎯

u air , 2 = 477.1 kJ/kg u C3H8 , 2 = −1607 kJ/kg

Noting that the heat transfer is zero, an energy balance on the system gives q in + win = Δu m ⎯ ⎯→ win = Δu m

where

Δu m = mf air (u air,2 − u air,1 ) + mf C3H8 (u C3H8 ,2 − u C3H 8 ,1 )

Substituting, win = Δu m = (0.9412)(477.1 − 216.5) + (0.05882)[(−1607) − (−2404)] = 292.2 kJ/kg

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13-30

13-57 The moles, temperatures, and pressures of two gases forming a mixture are given. The mixture temperature and pressure are to be determined. Assumptions 1 Under specified conditions both CO2 and H2 can be treated as ideal gases, and the mixture as an ideal gas mixture. 2 The tank is insulated and thus there is no heat transfer. 3 There are no other forms of work involved. Properties The molar masses and specific heats of CO2 and H2 are 44.0 kg/kmol, 2.0 kg/kmol, 0.657 kJ/kg.°C, and 10.183 kJ/kg.°C, respectively. (Tables A-1 and A-2b). Analysis (a) We take both gases as our system. No heat, work, or mass crosses the system boundary, therefore this is a closed system with Q = 0 and W = 0. Then the energy balance for this closed system reduces to E in − E out = ΔE system

0 = ΔU = ΔU CO 2 + ΔU H 2

0 =[mcv (Tm − T1 )]CO + [mcv (Tm − T1 )]H 2

2

Using cv values at room temperature and noting that m = NM, the final temperature of the mixture is determined to be

CO2 2.5 kmol 200 kPa 27°C

H2 7.5 kmol 400 kPa 40°C

(2.5 × 44 kg )(0.657 kJ/kg ⋅ °C)(Tm − 27°C) + (7.5 × 2 kg )(10.183 kJ/kg ⋅ °C)(Tm − 40°C) = 0 Tm = 35.8°C (308.8 K ) (b) The volume of each tank is determined from ⎛ NRu T1 ⎞ (2.5 kmol)(8.314 kPa ⋅ m 3 /kmol ⋅ K)(300 K) ⎟⎟ = = 31.18 m 3 P 200 kPa 1 ⎠ CO 2 ⎝

V CO 2 = ⎜⎜

⎛ NRu T1 ⎞ (7.5 kmol)(8.314 kPa ⋅ m 3 /kmol ⋅ K)(313 K) ⎟⎟ = = 48.79 m 3 P 400 kPa 1 ⎠H ⎝

V H 2 = ⎜⎜

2

Thus,

V m = V CO2 + V H 2 = 31.18 m 3 + 48.79 m 3 = 79.97 m 3 N m = N CO2 + N H 2 = 2.5 kmol + 7.5 kmol = 10.0 kmol

and Pm =

N m Ru Tm (10.0 kmol)(8.314 kPa ⋅ m 3 /kmol ⋅ K)(308.8 K) = = 321 kPa Vm 79.97 m 3

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13-31

13-58 [Also solved by EES on enclosed CD] The temperatures and pressures of two gases forming a mixture in a mixing chamber are given. The mixture temperature and the rate of entropy generation are to be determined. Assumptions 1 Under specified conditions both C2H6 and CH4 can be treated as ideal gases, and the mixture as an ideal gas mixture. 2 The mixing chamber is insulated and thus there is no heat transfer. 3 There are no other forms of work involved. 3 This is a steady-flow process. 4 The kinetic and potential energy changes are negligible. Properties The specific heats of C2H6 and CH4 are 1.7662 kJ/kg.°C and 2.2537 kJ/kg.°C, respectively. (Table A-2b).

20°C 9 kg/s C2H6

Analysis (a) The enthalpy of ideal gases is independent of pressure, and thus the two gases can be treated independently even after mixing. Noting that W& = Q& = 0 , the steady-flow energy balance equation reduces to

200 kPa 45°C CH 4 4.5 kg/s

E& in − E& out = ΔE& system ©0 (steady) = 0 E& in = E& out

∑ m& h = ∑ m& h 0 = ∑ m& h − ∑ m& h = m& 0 = [m& c (T − T )] + [m& c i i

e e e e

p

e

i i

i

C2H6

(he − hi )C H p (Te − Ti )]CH

C2H6

2

6

+ m& CH 4 (he − hi )CH

4

4

Using cp values at room temperature and substituting, the exit temperature of the mixture becomes 0 = (9 kg/s )(1.7662 kJ/kg ⋅ °C )(Tm − 20°C ) + (4.5 kg/s )(2.2537 kJ/kg ⋅ °C )(Tm − 45°C )

Tm = 29.7°C (302.7 K )

(b) The rate of entropy change associated with this process is determined from an entropy balance on the mixing chamber, S& in − S& out + S& gen = ΔS& system Ê0 = 0 [m& ( s1 − s 2 )] C 2 H 6 + [m& ( s1 − s 2 )] CH 4 + S& gen = 0 S& gen = [m& ( s 2 − s1 )] C 2 H 6 + [m& ( s 2 − s1 )] CH 4

The molar flow rate of the two gases in the mixture is 9 kg/s ⎛ m& ⎞ N& C 2 H 6 = ⎜ ⎟ = = 0.3 kmol/s M 30 kg/kmol ⎝ ⎠ C2H6 4.5 kg/s ⎛ m& ⎞ N& CH 4 = ⎜ ⎟ = = 0.2813 kmol/s M 16 kg/kmol ⎝ ⎠ CH 4

Then the mole fraction of each gas becomes 0.3 = 0.516 0.3 + 0.2813 0.2813 = = 0.484 0.3 + 0.2813

y C2H6 = y CH 4

Thus,

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13-32

y Pm,2 ⎛ T ( s 2 − s1 ) C 2 H 6 = ⎜⎜ c p ln 2 − R ln P1 T1 ⎝ = (1.7662 kJ/kg ⋅ K) ln

302.7 K − (0.2765 kJ/kg ⋅ K) ln(0.516) = 0.240 kJ/kg ⋅ K 293 K

y Pm,2 ⎛ T ( s 2 − s1 ) CH 4 = ⎜⎜ c p ln 2 − R ln P1 T1 ⎝ = (2.2537 kJ/kg ⋅ K) ln

⎞ ⎛ ⎞ T ⎟ = ⎜⎜ c p ln 2 − R ln y ⎟⎟ ⎟ T 1 ⎠ C2H6 ⎠ C2H6 ⎝

⎞ ⎛ ⎞ T ⎟ = ⎜⎜ c p ln 2 − R ln y ⎟⎟ ⎟ T1 ⎠ CH 4 ⎠ CH 4 ⎝

302.7 K − (0.5182 kJ/kg ⋅ K) ln(0.484) = 0.265 kJ/kg ⋅ K 318 K

Noting that Pm, 2 = Pi, 1 = 200 kPa and substituting, S&gen = (9 kg/s )(0.240 kJ/kg ⋅ K ) + (4.5 kg/s )(0.265 kJ/kg ⋅ K ) = 3.353 kW/K

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13-33

13-59 EES Problem 13-58 is reconsidered. The effect of the mass fraction of methane in the mixture on the mixture temperature and the rate of exergy destruction is to be investigated. Analysis The problem is solved using EES, and the solution is given below. "Input from the Diagram Window" {Fluid1$='C2H6' Fluid2$='CH4' m_dot_F1=9 [kg/s] m_dot_F2=m_dot_F1/2 T1=20 [C] T2=45 [C] P=200 [kPa]} {mf_F2=0.1} {m_dot_total =13.5 [kg/s] m_dot_F2 =mf_F2*m_dot_total} m_dot_total = m_dot_F1 + m_dot_F2 T_o = 25 [C] "Conservation of energy for this steady-state, steady-flow control volume is" E_dot_in=E_dot_out E_dot_in=m_dot_F1*enthalpy(Fluid1$,T=T1) +m_dot_F2 *enthalpy(Fluid2$,T=T2) E_dot_out=m_dot_F1*enthalpy(Fluid1$,T=T3) +m_dot_F2 *enthalpy(Fluid2$,T=T3) "For entropy calculations the partial pressures are used." Mwt_F1=MOLARMASS(Fluid1$) N_dot_F1=m_dot_F1/Mwt_F1 Mwt_F2=MOLARMASS(Fluid2$) N_dot_F2=m_dot_F2 /Mwt_F2 N_dot_tot=N_dot_F1+N_dot_F2 y_F1=IF(fluid1$,Fluid2$,N_dot_F1/N_dot_tot,1,N_dot_F1/N_dot_tot) y_F2=IF(fluid1$,Fluid2$,N_dot_F2/N_dot_tot,1,N_dot_F2/N_dot_tot) "If the two fluids are the same, the mole fractions are both 1 ." "The entropy change of each fluid is:" DELTAs_F1=entropy(Fluid1$, T=T3, P=y_F1*P)-entropy(Fluid1$, T=T1, P=P) DELTAs_F2=entropy(Fluid2$, T=T3, P=y_F2*P)-entropy(Fluid2$, T=T2, P=P) "And the entropy generation is:" S_dot_gen=m_dot_F1*DELTAs_F1+m_dot_F2*DELTAs_F2 "Then the exergy destroyed is:" X_dot_destroyed = (T_o+273)*S_dot_gen 38

Xdestroyed [kW] 20.48 24.08 27 29.2 30.92 32.3 33.43 34.38 35.18 35.87 36.41

2000

36 1600

34 32

1200

30 28

F1 = C2H6

26

F2 = CH4 mtotal = 13.5 kg/s

24

800

400

22 20 0

0,2

0,4

0,6

0,8

0 1

mfF2

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Xdestroyed [kW]

0.01 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 0.99

T3 [C] 95.93 502.5 761.4 948.5 1096 1219 1324 1415 1497 1570 1631

T3 [C]

mfF2

13-34

13-60 A mixture of hydrogen and oxygen is considered. The entropy change of this mixture between the two specified states is to be determined. Assumptions Hydrogen and oxygen are ideal gases. Properties The gas constants of hydrogen and oxygen are 4.124 and 0.2598 kJ/kg⋅K, respectively (Table A-1). Analysis The effective gas constant of this mixture is R = mf H2 R H2 + mf O2 RO2 = (0.33)(4.1240) + (0.67)(0.2598) = 1.5350 kJ/kg ⋅ K Since the temperature of the two states is the same, the entropy change is determined from s 2 − s1 = − R ln

P2 150 kPa = −(1.5350 kJ/kg ⋅ K ) ln = 2.470 kJ/kg ⋅ K 750 kPa P1

13-61 A mixture of nitrogen and carbon dioxide is heated at constant pressure in a closed system. The work produced is to be determined. Assumptions 1 Nitrogen and carbon dioxide are ideal gases. 2 The process is reversible. Properties The mole numbers of nitrogen and carbon dioxide are 28.0 and 44.0 kg/kmol, respectively (Table A-1). Analysis One kg of this mixture consists of 0.5 kg of nitrogen and 0.5 kg of carbon dioxide or 0.5 kg×28.0 kg/kmol=14.0 kmol of nitrogen and 0.5 kg×44.0 kg/kmol=22.0 kmol of carbon dioxide. The constituent mole fraction are then y N2 =

N N2 14 kmol = = 0.3889 N total 36 kmol

y CO2 =

N CO2 22 kmol = = 0.6111 N total 36 kmol

The effective molecular weight of this mixture is M = y N2 M N2 + y CO2 M CO2

50% N2 50% CO2 (by mass) 120 kPa, 30°C

Q

= (0.3889)(28) + (0.6111)(44) = 37.78 kg/kmol

The work done is determined from 2

∫

w = PdV = P2v 2 − P1v 1 = R(T2 − T1 ) 1

Ru 8.314 kJ/kmol ⋅ K (T2 − T1 ) = (200 − 30)K 37.78 kg/kmol M = 37.4 kJ/kg =

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13-35

13-62E The mass fractions of components of a gas mixture are given. This mixture is compressed in an isentropic process. The final mixture temperature and the work required per unit mass of the mixture are to be determined. Assumptions All gases will be modeled as ideal gases with constant specific heats. Properties The molar masses of N2, He, CH4, and C2H6 are 28.0, 4.0, 16.0, and 30.0 lbm/lbmol, respectively (Table A-1E). The constant-pressure specific heats of these gases at room temperature are 0.248, 1.25, 0.532, and 0.427 Btu/lbm⋅R, respectively (Table A-2Ea). Analysis We consider 100 lbm of this mixture. The mole numbers of each component are N N2 =

m N2 15 lbm = = 0.5357 lbmol M N2 28 lbm/lbmol

N He =

m He 5 lbm = = 1.25 lbmol M He 4 lbm/lbmol

N CH4 = N C2H6 =

15% N2 5% He 60% CH4 20% C2H6 (by mass)

m CH4 60 lbm = = 3.75 lbmol M CH4 16 lbm/lbmol m C2H6 20 lbm = = 0.6667 lbmol M C2H6 30 lbm/lbmol

20 psia, 100°F

The mole number of the mixture is N m = N N2 + N He + N CH4 + N C2H6 = 0.5357 + 1.25 + 3.75 + 0.6667 = 6.2024 lbmol

The apparent molecular weight of the mixture is Mm =

mm 100 lbm = = 16.12 lbm/lbmol N m 6.2024 lbmol

The constant-pressure specific heat of the mixture is determined from c p = mf N2 c p , N2 + mf He c p ,He + mf CH4 c p ,CH4 + mf C2H6 c p ,C2H6 = 0.15 × 0.248 + 0.05 × 1.25 + 0.60 × 0.532 + 0.20 × 0.427 = 0.5043 Btu/lbm ⋅ R

The apparent gas constant of the mixture is R=

Ru 1.9858 Btu/lbmol ⋅ R = = 0.1232 Btu/lbm ⋅ R 16.12 lbm/lbmol Mm

Then the constant-volume specific heat is cv = c p − R = 0.5043 − 0.1232 = 0.3811 Btu/lbm ⋅ R

The specific heat ratio is k=

cp cv

=

0.5043 = 1.323 0.3811

The temperature at the end of the compression is ⎛P T2 = T1 ⎜⎜ 2 ⎝ P1

⎞ ⎟⎟ ⎠

( k −1) / k

⎛ 200 psia ⎞ ⎟⎟ = (560 R )⎜⎜ ⎝ 20 psia ⎠

0.323/1.323

= 982 R

An energy balance on the adiabatic compression process gives win = c p (T2 − T1 ) = (0.5043 Btu/lbm ⋅ R )(982 − 560) R = 213 Btu/lbm

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13-36

13-63 The mass fractions of components of a gas mixture are given. This mixture is compressed in a reversible, isothermal, steady-flow compressor. The work and heat transfer for this compression per unit mass of the mixture are to be determined. Assumptions All gases will be modeled as ideal gases with constant specific heats. Properties The molar masses of CH4, C3H8, and C4H10 are 16.0, 44.0, and 58.0 lbm/lbmol, respectively (Table A-1E). Analysis The mole numbers of each component are N CH4 =

1 MPa

m CH4 60 lbm = = 3.75 lbmol M CH4 16 lbm/lbmol

N C3H8 =

m C3H8 25 lbm = = 0.5682 lbmol M C3H8 44 lbm/lbmol

N C4H10 =

m C4H10 15 lbm = = 0.2586 lbmol M C4H10 58 lbm/lbmol

qout

60% CH4 25% C3H8 15% C4H10 (by mass)

The mole number of the mixture is N m = N CH4 + N C3H8 + N C4H10

100 kPa 20°C

= 3.75 + 0.5682 + 0.2586 = 4.5768 lbmol

The apparent molecular weight of the mixture is Mm =

mm 100 lbm = = 21.85 lbm/lbmol N m 4.5768 lbmol

The apparent gas constant of the mixture is R=

Ru 8.314 kJ/kmol ⋅ K = = 0.3805 kJ/kg ⋅ K 21.85 kg/kmol Mm

For a reversible, isothermal process, the work input is ⎛P win = RT ln⎜⎜ 2 ⎝ P1

⎞ ⎛ 1000 kPa ⎞ ⎟⎟ = (0.3805 kJ/kg ⋅ K )(293 K)ln⎜ ⎟ = 257 kJ/kg ⎝ 100 kPa ⎠ ⎠

An energy balance on the control volume gives E& − E& 1in424out 3

=

Rate of net energy transfer by heat, work, and mass

ΔE& system Ê0 (steady) 1442444 3

=0

Rate of change in internal, kinetic, potential, etc. energies

E& in = E& out m& h1 + W& in = m& h2 + Q& out W& in − Q& out = m& (h2 − h1 ) win − q out = c p (T2 − T1 ) = 0 since T2 = T1 win = q out

That is, q out = win = 257 kJ/kg

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13-37

13-64 The masses of components of a gas mixture are given. This mixture is heated at constant pressure. The change in the volume of the mixture and the total heat transferred to the mixture are to be determined. Assumptions All gases will be modeled as ideal gases with constant specific heats. Properties The molar masses of O2, CO2, and He are 32.0, 44.0, and 4.0 kg/kmol, respectively (Table A-1). The constant-pressure specific heats of these gases at room temperature are 0.918, 0.846, and 5.1926 kJ/kg⋅K, respectively (Table A-2a). Analysis The total mass of the mixture is m m = m O2 + m CO2 + m He = 0.1 + 1 + 0.5 = 1.6 kg

The mole numbers of each component are N O2 =

m O2 0.1 kg = = 0.003125 kmol M O2 32 kg/kmol

N CO2 =

m CO2 1 kg = = 0.02273 kmol M CO2 44 kg/kmol

N He

0.1 kg O2 1 kg CO2 0.5 kg He 350 kPa, 10°C

m 0.5 kg = He = = 0.125 kmol M He 4 kg/kmol

The mole number of the mixture is N m = N O2 + N CO2 + N He = 0.003125 + 0.02273 + 0.125 = 0.15086 kmol

The apparent molecular weight of the mixture is Mm =

mm 1.6 kg = = 10.61 kg/kmol N m 0.15086 kmol

The apparent gas constant of the mixture is R=

Ru 8.314 kJ/kmol ⋅ K = = 0.7836 kJ/kg ⋅ K 10.61 kg/kmol Mm

The mass fractions are mf O2 =

m O2 0.1 kg = = 0.0625 m m 1.6 kg

mf CO2 =

m CO2 1 kg = = 0.625 1.6 kg mm

mf He =

m He 0.5 kg = = 0.3125 m m 1.6 kg

The constant-pressure specific heat of the mixture is determined from c p = mf O2 c p ,O2 + mf CO2 c p ,CO2 + mf He c p ,He = 0.0625 × 0.918 + 0.625 × 0.846 + 0.3125 × 5.1926 = 2.209 kJ/kg ⋅ K

The change in the volume of this ideal gas mixture is ΔV m =

m m RΔT (1.6 kg)(0.7836 kPa ⋅ m 3 /kg ⋅ K)(260 − 10) K = = 0.8955 m 3 350 kPa P

The heat transfer is determined to be q in = c p (T2 − T1 ) = (2.209 kJ/kg ⋅ K )(260 − 10) K = 552 kJ/kg

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13-38

13-65E The volume fractions of components of a gas mixture during the expansion process of the ideal Otto cycle are given. The thermal efficiency of this cycle is to be determined. Assumptions All gases will be modeled as ideal gases with constant specific heats. Properties The molar masses of N2, O2, H2O, and CO2 are 28.0, 32.0, 18.0, and 44.0 lbm/lbmol, respectively (Table A-1E). The constant-pressure specific heats of these gases at room temperature are 0.248, 0.219, 0.445, and 0.203 Btu/lbm⋅R, respectively. The air properties at room temperature are cp = 0.240 Btu/lbm⋅R, cv = 0.171 Btu/lbm⋅R, k = 1.4 (Table A-2Ea). Analysis We consider 100 lbmol of this mixture. Noting that volume fractions are equal to the mole fractions, mass of each component are m N2 = N N2 M N2 = (25 lbmol)(28 lbm/lbmol) = 700 lbm m O2 = N O2 M O2 = (7 lbmol)(32 lbm/lbmol) = 224 lbm

25% N2 7% O2 28% H2O 40% CO2 (by volume)

m H2O = N H2O M H2O = (28 lbmol)(18 lbm/lbmol) = 504 lbm m CO2 = N CO2 M CO2 = (40 lbmol)(44 lbm/lbmol) = 1760 lbm

The total mass is m m = m N2 + m O2 + m H2O + m CO2 = 700 + 224 + 504 + 1760 = 3188 lbm

Then the mass fractions are mf N2 =

m N2 700 lbm = = 0.2196 mm 3188 lbm

mf O2 =

m O2 224 lbm = = 0.07026 mm 3188 lbm

mf H2O

m 504 lbm = H2O = = 0.1581 mm 3188 lbm

mf CO2

m 1760 lbm = CO2 = = 0.5521 mm 3188 lbm

P

3 4 2

1

v

The constant-pressure specific heat of the mixture is determined from c p = mf N2 c p , N2 + mf O2 c p ,O2 + mf H2O c p ,H2O + mf CO2 c p ,CO2 = 0.2196 × 0.248 + 0.07026 × 0.219 + 0.1581× 0.445 + 0.5521× 0.203 = 0.2523 Btu/lbm ⋅ R

The apparent molecular weight of the mixture is Mm =

mm 3188 lbm = = 31.88 lbm/lbmol N m 100 lbmol

The apparent gas constant of the mixture is R=

Ru 1.9858 Btu/lbmol ⋅ R = = 0.06229 Btu/lbm ⋅ R Mm 31.88 lbm/lbmol

Then the constant-volume specific heat is cv = c p − R = 0.2523 − 0.06229 = 0.1900 Btu/lbm ⋅ R

The specific heat ratio is

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13-39

k=

cp cv

=

0.2523 = 1.328 0.1900

The average of the air properties at room temperature and combustion gas properties are c p ,avg = 0.5(0.2523 + 0.240) = 0.2462 Btu/lbm ⋅ R cv ,avg = 0.5(0.1900 + 0.171) = 0.1805 Btu/lbm ⋅ R k avg = 0.5(1.328 + 1.4) = 1.364

These average properties will be used for heat addition and rejection processes. For compression, the air properties at room temperature and during expansion, the mixture properties will be used. During the compression process, T2 = T1 r k −1 = (515 R )(7) 0.4 = 1122 R

During the heat addition process, q in = cv ,avg (T3 − T2 ) = (0.1805 Btu/lbm ⋅ R )(2060 − 1122) R = 169.3 Btu/lbm

During the expansion process, ⎛1⎞ T4 = T3 ⎜ ⎟ ⎝r⎠

k −1

⎛1⎞ = (2060 R )⎜ ⎟ ⎝7⎠

0.364

= 1014 R

During the heat rejection process, q out = cv ,avg (T4 − T1 ) = (0.1805 Btu/lbm ⋅ R )(1014 − 515) R = 90.1 Btu/lbm

The thermal efficiency of the cycle is then

η th = 1 −

q out 90.1 Btu/lbm = 1− = 0.468 169.3 Btu/lbm q in

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13-40

13-66E The thermal efficiency of the cycle in the previous problem is to be compared to that predicted by air standard analysis? Assumptions Air-standard assumptions are applicable. Properties The air properties at room temperature are cp = 0.240 Btu/lbm⋅R, cv = 0.171 Btu/lbm⋅R, k = 1.4 (Table A-2Ea). Analysis In the previous problem, the thermal efficiency of the cycle was determined to be 0.468 (46.8%). The thermal efficiency with airstandard model is determined from

η th = 1 −

1 r k −1

= 1−

1 7 0.4

= 0.541

which is significantly greater than that calculated with gas mixture analysis in the previous problem.

P

3 4 2

1

v

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13-41

13-67 The volume fractions of components of a gas mixture passing through the turbine of a simple ideal Brayton cycle are given. The thermal efficiency of this cycle is to be determined. Assumptions All gases will be modeled as ideal gases with constant specific heats. Properties The molar masses of N2, O2, H2O, and CO2 are 28.0, 32.0, 18.0, and 44.0 kg/kmol, respectively (Table A-1). The constant-pressure specific heats of these gases at room temperature are 1.039, 0.918, 1.8723, and 0.846 kJ/kg⋅K, respectively. The air properties at room temperature are cp = 1.005 kJ/kg⋅K, cv = 0.718 kJ/kg⋅K, k = 1.4 (Table A-2a). Analysis We consider 100 kmol of this mixture. Noting that volume fractions are equal to the mole fractions, mass of each component are 30% N2, 10% O2 m N2 = N N2 M N2 = (30 kmol)(28 kg/kmol) = 840 kg 20% H2O, 40% CO2 m O2 = N O2 M O2 = (10 kmol)(32 kg/kmol) = 320 kg (by volume) m H2O = N H2O M H2O = (20 kmol)(18 kg/kmol) = 360 kg m CO2 = N CO2 M CO2 = (40 kmol)(44 kg/kmol) = 1760 kg

The total mass is m m = m N2 + m O2 + m H2O + m CO2 = 840 + 320 + 360 + 1760 = 3280 kg

100 kPa

Then the mass fractions are mf N2 =

m N2 840 kg = = 0.2561 3280 kg mm

mf O2 =

m O2 320 kg = = 0.09756 3280 kg mm

mf H2O =

m H2O 360 kg = = 0.1098 3280 kg mm

mf CO2 =

m CO2 1760 kg = = 0.5366 3280 kg mm

T 3

1273 K

qin 2

293 K

4 1

qout s

The constant-pressure specific heat of the mixture is determined from c p = mf N2 c p , N2 + mf O2 c p ,O2 + mf H2O c p , H2O + mf CO2 c p ,CO2 = 0.2561× 1.039 + 0.09756 × 0.918 + 0.1098 × 1.8723 + 0.5366 × 0.846 = 1.015 kJ/kg ⋅ K

The apparent molecular weight of the mixture is Mm =

mm 3280 kg = = 32.80 kg/kmol N m 100 kmol

The apparent gas constant of the mixture is R=

Ru 8.314 kJ/kmol ⋅ K = = 0.2535 kJ/kg ⋅ K 32.80 kg/kmol Mm

Then the constant-volume specific heat is cv = c p − R = 1.015 − 0.2535 = 0.762 kJ/kg ⋅ K

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13-42

k=

cp cv

=

1.015 = 1.332 0.762

The average of the air properties at room temperature and combustion gas properties are c p ,avg = 0.5(1.015 + 1.005) = 1.010 kJ/kg ⋅ K cv ,avg = 0.5(0.762 + 0.718) = 0.740 kJ/kg ⋅ K k avg = 0.5(1.332 + 1.4) = 1.366

These average properties will be used for heat addition and rejection processes. For compression, the air properties at room temperature and during expansion, the mixture properties will be used. During the compression process, ⎛P T2 = T1 ⎜⎜ 2 ⎝ P1

⎞ ⎟⎟ ⎠

( k −1) / k

= (293 K )(8) 0.4/1.4 = 531 K

During the heat addition process, q in = c p ,avg (T3 − T2 ) = (1.010 kJ/kg ⋅ K )(1273 − 531) K = 749.4 kJ/kg

During the expansion process, ⎛P T4 = T3 ⎜⎜ 4 ⎝ P3

⎞ ⎟ ⎟ ⎠

( k −1) / k

⎛1⎞ = (1273 K )⎜ ⎟ ⎝8⎠

0.332/1.332

= 758 K

During the heat rejection process, q out = c p ,avg (T4 − T1 ) = (1.010 kJ/kg ⋅ K )(758 − 293) K = 469.7 kJ/kg

The thermal efficiency of the cycle is then η th = 1 −

q out 469.7 kJ/kg = 1− = 0.373 749.4 kJ/kg q in

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13-43

13-68 The thermal efficiency of the cycle in the previous problem is to be compared to that predicted by air standard analysis? Assumptions Air-standard assumptions are applicable. Properties The air properties at room temperature are cp = 1.005 kJ/kg⋅K, cv = 1.4 kJ/kg⋅K, k = 1.4 (Table A-2a). Analysis In the previous problem, the thermal efficiency of the cycle was determined to be 0.373 (37.3%). The thermal efficiency with air-standard model is determined from η th = 1 −

1 r p( k −1) / k

= 1−

1 8 0.4 / 1.4

T qin 2

= 0.448

which is significantly greater than that calculated with gas mixture analysis in the previous problem.

3

1273 K

4 293 K

1

qout s

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13-44

13-69E The mass fractions of a natural gas mixture at a specified pressure and temperature trapped in a geological location are given. This natural gas is pumped to the surface. The work required is to be determined using Kay's rule and the enthalpy-departure method. Properties The molar masses of CH4 and C2H6 are 16.0 and 30.0 lbm/lbmol, respectively. The critical properties are 343.9 R, 673 psia for CH4 and 549.8 R and 708 psia for C2H6 (Table A-1E). The constantpressure specific heats of these gases at room temperature are 0.532 and 0.427 Btu/lbm⋅R, respectively (Table A-2Ea). Analysis We consider 100 lbm of this mixture. Then the mole numbers of each component are N CH4 =

m CH4 75 lbm = = 4.6875 lbmol M CH4 16 lbm/lbmol

N C2H6 =

m C2H6 25 lbm = = 0.8333 lbmol M C2H6 30 lbm/lbmol

75% CH4 25% C2H6 (by mass) 2000 psia 300°F

The mole number of the mixture and the mole fractions are N m = 4.6875 + 0.8333 = 5.5208 lbmol y CH4 =

N CH4 4.6875 lbmol = = 0.8491 5.5208 lbmol Nm

y C2H6 =

N C2H6 0.8333 lbmol = = 0.1509 5.5208 lbmol Nm

Then the apparent molecular weight of the mixture becomes mm 100 lbm = = 18.11 lbm/lbmol N m 5.5208 lbmol

Mm =

The apparent gas constant of the mixture is R=

Ru 1.9858 Btu/lbmol ⋅ R = = 0.1097 Btu/lbm ⋅ R 18.11 lbm/lbmol Mm

The constant-pressure specific heat of the mixture is determined from c p = mf CH4 c p ,CH4 + mf C2H6 c p ,C2H6 = 0.75 × 0.532 + 0.25 × 0.427 = 0.506 Btu/lbm ⋅ R

To use Kay's rule, we need to determine the pseudo-critical temperature and pseudo-critical pressure of the mixture using the critical point properties of gases. Tcr′ ,m =

∑yT

i cr ,i

= y CH4 Tcr ,Ch4 + y C2H6 Tcr ,C2H6

= (0.8491)(343.9 R) + (0.1509)(549.8 R) = 375.0 R Pcr′ ,m =

∑y P

i cr ,i

= y Ch4 Pcr ,Ch4 + y C2H6 Pcr ,C2H6

= (0.8491)(673 psia) + (0.1509)(708 psia) = 678.3 psia

The compressibility factor of the gas mixture in the reservoir and the mass of this gas are TR = PR =

Tm ' Tcr, m

Pm ' Pcr, m

⎫ ⎪ ⎪ ⎬ Z m = 0.963 2000 psia = = 2.949 ⎪ ⎪ 678.3 psia ⎭

=

760 R = 2.027 375.0 R

(Fig. A-15)

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13-45

m=

(2000 psia)(1× 10 6 ft 3 ) PV = = 4.612 × 10 6 lbm Z m RT (0.963)(0.5925 psia ⋅ ft 3 /lbm ⋅ R)(760 R)

The enthalpy departure factors in the reservoir and the surface are (from EES or Fig. A-29) T R1 = PR1 =

TR 2 = PR 2 =

Tm ' Tcr, m

Pm ' Pcr, m

Tm ' Tcr, m

Pm ' Pcr, m

⎫ ⎪ ⎪ ⎬ Z h1 = 0.703 2000 psia = = 2.949 ⎪ ⎪ 678.3 psia ⎭

=

760 R = 2.027 375.0 R

⎫ ⎪ ⎪ ⎬ Z h 2 = 0.0112 20 psia = = 0.0295 ⎪ ⎪ 678.3 psia ⎭

=

660 R = 1.76 375.0 R

The enthalpy change for the ideal gas mixture is (h2 − h1 ) ideal = c p (T2 − T1 ) = (0.506 Btu/lbm ⋅ R )(760 − 660)R = 50.6 Btu/lbm

The enthalpy change with departure factors is h2 − h1 = (h2 − h1 ) ideal − RTcr′ ,m ( Z h 2 − Z h1 ) = 50.6 − (0.1096)(375)(0.0112 − 0.703) = 79.0 Btu/lbm

The work input is then Win = m(h2 − h1 ) = (4.612 × 10 6 lbm)(79.0 Btu/lbm) = 3.64 × 10 8 Btu

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13-46

13-70 In a liquid-oxygen plant, it is proposed that the pressure and temperature of air be adiabatically reduced. It is to be determined whether this process is possible and the work produced is to be determined using Kay's rule and the departure charts. Assumptions Air is a gas mixture with 21% O2 and 79% N2, by mole. Properties The molar masses of O2 and N2 are 32.0 and 28.0 kg/kmol, respectively. The critical properties are 154.8 K, 5.08 MPa for O2 and 126.2 K and 3.39 MPa for N2 (Table A-1). Analysis To use Kay's rule, we need to determine the pseudo-critical temperature and pseudo-critical pressure of the mixture using the critical point properties of gases. Tcr′ , m =

∑yT

i cr ,i

= y O2 Tcr ,O2 + y N2 Tcr , N2

= (0.21)(154.8 K) + (0.79)(126.2 K) = 132.2 K Pcr′ , m =

∑y P

i cr ,i

= y O2 Pcr ,O2 + y N2 Pcr , N2

= (0.21)(5.08 MPa) + (0.79)(3.39 MPa) = 3.745 MPa

The enthalpy and entropy departure factors at the initial and final states are (from EES) T R1 = PR1 =

TR 2 = PR 2 =

Tm1 ' Tcr, m

Pm1 ' Pcr, m

Tm 2 ' Tcr, m

Pm 2 ' Pcr, m

⎫ ⎪ ⎪ Z h1 = 0.513 ⎬ Z = 0.235 9 MPa = = 2.403 ⎪ s1 ⎪ 3.745 MPa ⎭

=

283 K = 2.141 132.2 K

⎫ ⎪ ⎪ Z h 2 = 0.0069 ⎬ Z = 0.0035 0.050 MPa = = 0.0134 ⎪ s 2 ⎪ 3.745 MPa ⎭

=

200 K = 1.513 132.2 K

21% O2 79% N2 (by mole) 9000 kPa 10°C

The enthalpy and entropy changes of the air under the ideal gas assumption is (Properties are from Table A-17) (h2 − h1 ) ideal = 199.97 − 283.14 = −83.2 kJ/kg ( s 2 − s1 ) ideal = s 2o − s1o − R ln

P2 50 = 1.29559 − 1.64345 − (0.287) ln = 1.1425 kJ/kg ⋅ K P1 9000

With departure factors, the enthalpy change (i.e., the work output) and the entropy change are wout = h1 − h2 = (h1 − h2 ) ideal − RTcr' ( Z h1 − Z h 2 ) = 83.2 − (0.287)(132.2)(0.513 − 0.0069) = 64.0 kJ/kg s 2 − s1 = ( s 2 − s1 ) ideal − R( Z s 2 − Z s1 ) = 1.1425 − (0.287)(0.0035 − 0.235) = 1.209 kJ/kg ⋅ K

The entropy change in this case is equal to the entropy generation during the process since the process is adiabatic. The positive value of entropy generation shows that this process is possible.

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13-47

13-71E [Also solved by EES on enclosed CD] A gas mixture with known mass fractions is accelerated through a nozzle from a specified state to a specified pressure. For a specified isentropic efficiency, the exit temperature and the exit velocity of the mixture are to be determined. Assumptions 1 Under specified conditions both N2 and CO2 can be treated as ideal gases, and the mixture as an ideal gas mixture. 2 The nozzle is adiabatic and thus heat transfer is negligible. 3 This is a steadyflow process. 4 Potential energy changes are negligible. Properties The specific heats of N2 and CO2 are cp,N2 = 0.248 Btu/lbm.R, cv,N2 = 0.177 Btu/lbm.R, cp,CO2 = 0.203 Btu/lbm.R, and cv,CO2 = 0.158 Btu/lbm.R. (Table A-2E). Analysis (a) Under specified conditions both N2 and CO2 can be treated as ideal gases, and the mixture as an ideal gas mixture. The cp, cv, and k values of this mixture are determined from c p,m =

∑ mf c

i p ,i

= mf N 2 c p , N 2 + mf CO 2 c p ,CO 2

= (0.8)(0.248) + (0.2)(0.203) = 0.239 Btu/lbm ⋅ R

cv ,m =

∑ mf cv i

,i

= mf N 2 cv , N 2 + mf CO 2 cv ,CO 2

= (0.8)(0.177 ) + (0.2)(0.158)

90 psia 1800 R

80% N2 20% CO2

12 psia

= 0.173 Btu/lbm ⋅ R km =

c p,m cv , m

=

0.239 Btu/lbm ⋅ R = 1.382 0.173 Btu/lbm ⋅ R

Therefore, the N2-CO2 mixture can be treated as a single ideal gas with above properties. Then the isentropic exit temperature can be determined from ⎛P T2 s = T1 ⎜⎜ 2 ⎝ P1

⎞ ⎟⎟ ⎠

(k −1) / k

⎛ 12 psia ⎞ ⎟⎟ = (1800 R )⎜⎜ ⎝ 90 psia ⎠

0.382/1.382

= 1031.3 R

From the definition of adiabatic efficiency,

ηN =

c p (T1 − T2 ) h1 − h2 1,800 − T2 = ⎯ ⎯→ 0.92 = ⎯ ⎯→ T2 = 1092.8 R h1 − h2 s c p (T1 − T2 s ) 1,800 − 1031.3

(b) Noting that, q = w = 0, from the steady-flow energy balance relation, E& in − E& out = ΔE& system ©0 (steady) = 0 E& in = E& out h1 + V12 / 2 = h2 + V 22 / 2 0 = c p (T2 − T1 ) +

V 22 − V12 2

©0

⎛ 25,037 ft 2 /s 2 V 2 = 2c p (T1 − T2 ) = 2(0.239 Btu/lbm ⋅ R )(1800 − 1092.8) R ⎜ ⎜ 1 Btu/lbm ⎝

⎞ ⎟ = 2,909 ft/s ⎟ ⎠

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13-48

13-72E EES Problem 13-71E is reconsidered. The problem is first to be solved and then, for all other conditions being the same, the problem is to be resolved to determine the composition of the nitrogen and carbon dioxide that is required to have an exit velocity of 2000 ft/s at the nozzle exit. Analysis The problem is solved using EES, and the solution is given below. "Input Data" mf_N2 = 0.8 "Mass fraction for the nitrogen, lbm_N2/lbm_mix" mf_CO2 = 0.2 "Mass fraction for the carbon dioxide, lbm_CO2/lbm_mix" T[1] = 1800 [R] P[1] = 90 [psia] Vel[1] = 0 [ft/s] P[2] = 12 [psia] Eta_N =0.92 "Nozzle adiabatic efficiency" "Enthalpy property data per unit mass of mixture:" " Note: EES calculates the enthalpy of ideal gases referenced to the enthalpy of formation as h = h_f + (h_T - h_537) where h_f is the enthalpy of formation such that the enthalpy of the elements or their stable compounds is zero at 77 F or 537 R, see Chapter 14. The enthalpy of formation is often negative; thus, the enthalpy of ideal gases can be negative at a given temperature. This is true for CO2 in this problem." h[1]= mf_N2* enthalpy(N2, T=T[1]) + mf_CO2* enthalpy(CO2, T=T[1]) h[2]= mf_N2* enthalpy(N2, T=T[2]) + mf_CO2* enthalpy(CO2, T=T[2]) "Conservation of Energy for a unit mass flow of mixture:" "E_in - E_out = DELTAE_cv Where DELTAE_cv = 0 for SSSF" h[1]+Vel[1]^2/2*convert(ft^2/s^2,Btu/lbm) - h[2] - Vel[2]^2/2*convert(ft^2/s^2,Btu/lbm) =0 "SSSF energy balance" "Nozzle Efficiency Calculation:" Eta_N=(h[1]-h[2])/(h[1]-h_s2) h_s2= mf_N2* enthalpy(N2, T=T_s2) + mf_CO2* enthalpy(CO2, T=T_s2) "The mixture isentropic exit temperature, T_s2, is calculated from setting the entropy change per unit mass of mixture equal to zero." DELTAs_mix=mf_N2 * DELTAs_N2 + mf_CO2 * DELTAs_CO2 DELTAs_N2 = entropy(N2, T=T_s2, P=P_2_N2) - entropy(N2, T=T[1], P=P_1_N2) DELTAs_CO2 = entropy(CO2, T=T_s2, P=P_2_CO2) - entropy(CO2, T=T[1], P=P_1_CO2) DELTAs_mix=0 "By Dalton's Law the partial pressures are:" P_1_N2 = y_N2 * P[1]; P_1_CO2 = y_CO2 * P[1] P_2_N2 = y_N2 * P[2]; P_2_CO2 = y_CO2 * P[2] "mass fractions, mf, and mole fractions, y, are related by:" M_N2 = molarmass(N2) M_CO2=molarmass(CO2) y_N2=mf_N2/M_N2/(mf_N2/M_N2 + mf_CO2/M_CO2) y_CO2=mf_CO2/M_CO2/(mf_N2/M_N2 + mf_CO2/M_CO2)

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13-49

SOLUTION of the stated problem DELTAs_CO2=-0.04486 [Btu/lbm-R] DELTAs_N2=0.01122 [Btu/lbm-R] h[1]=-439.7 [Btu/lbm] h_s2=-628.8 [Btu/lbm] mf_N2=0.8 [lbm_N2/lbm_mix] M_N2=28.01 [lbm/lbmol] P[2]=12 [psia] P_1_N2=77.64 [psia] P_2_N2=10.35 [psia] T[2]=1160 [R] Vel[1]=0 [ft/s] y_CO2=0.1373 [ft/s]

DELTAs_mix=0 [Btu/lbm-R] Eta_N=0.92 h[2]=-613.7 [Btu/lbm] mf_CO2=0.2 [lbm_CO2/lbm_mix] M_CO2=44.01 [lbm/lbmol] P[1]=90 [psia] P_1_CO2=12.36 [psia] P_2_CO2=1.647 [psia] T[1]=1800 [R] T_s2=1102 [R] Vel[2]=2952 [ft/s] y_N2=0.8627 [lbmol_N2/lbmol_mix]

SOLUTION of the problem with exit velocity of 2600 ft/s DELTAs_CO2=-0.005444 [Btu/lbm-R] DELTAs_N2=0.05015 [Btu/lbm-R] h[1]=-3142 [Btu/lbm] h_s2=-3288 [Btu/lbm] mf_N2=0.09793 [lbm_N2/lbm_mix] M_N2=28.01 [lbm/lbmol] P[2]=12 [psia] P_1_N2=13.11 [psia] P_2_N2=1.748 [psia] T[2]=1323 [R] Vel[1]=0 [ft/s] y_CO2=0.8543 [ft/s]

DELTAs_mix=0 [Btu/lbm-R] Eta_N=0.92 h[2]=-3277 [Btu/lbm] mf_CO2=0.9021 [lbm_CO2/lbm_mix] M_CO2=44.01 [lbm/lbmol] P[1]=90 [psia] P_1_CO2=76.89 [psia] P_2_CO2=10.25 [psia] T[1]=1800 [R] T_s2=1279 [R] Vel[2]=2600 [ft/s] y_N2=0.1457 [lbmol_N2/lbmol_mix]

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13-50

13-73 A piston-cylinder device contains a gas mixture at a given state. Heat is transferred to the mixture. The amount of heat transfer and the entropy change of the mixture are to be determined. Assumptions 1 Under specified conditions both H2 and N2 can be treated as ideal gases, and the mixture as an ideal gas mixture. 2 Kinetic and potential energy changes are negligible. Properties The constant pressure specific heats of H2 and N2 at 450 K are 14.501 kJ/kg.K and 1.049 kJ/kg.K, respectively. (Table A-2b). Analysis (a) Noting that P2 = P1 and V2 = 2V1, P2V 2 P1V1 2V = ⎯ ⎯→ T2 = 1 T1 = 2T1 = (2 )(300 K ) = 600 K T2 T1 V1

0.5 kg H2 1.6 kg N2 100 kPa 300 K

Also P = constant. Then from the closed system energy balance relation, Ein − Eout = ΔEsystem Qin − Wb,out = ΔU

→

Q

Qin = ΔH

since Wb and ΔU combine into ΔH for quasi-equilibrium constant pressure processes.

[

]

[

]

Qin = ΔH = ΔH H 2 + ΔH N 2 = mc p ,avg (T2 − T1 ) H + mc p ,avg (T2 − T1 ) N 2

2

= (0.5 kg )(14.501 kJ/kg ⋅ K )(600 − 300 )K + (1.6 kg )(1.049 kJ/kg ⋅ K )(600 − 300 )K

= 2679 kJ

(b) Noting that the total mixture pressure, and thus the partial pressure of each gas, remains constant, the entropy change of the mixture during this process is ⎛ ⎛ T P ©0 ⎞ T ⎞ ΔS H 2 = [m(s 2 − s1 )]H 2 = mH 2 ⎜ c p ln 2 − R ln 2 ⎟ = mH 2 ⎜⎜ c p ln 2 ⎟⎟ ⎜ ⎟ T P T1 ⎠ H 1 1 ⎝ ⎝ ⎠ H2 2 = (0.5 kg )(14.501 kJ/kg ⋅ K )ln = 5.026 kJ/K

600 K 300 K

⎛ ⎛ T P ©0 ⎞ T ΔS N 2 = [m(s 2 − s1 )]N 2 = m N 2 ⎜ c p ln 2 − R ln 2 ⎟ = m N 2 ⎜⎜ c p ln 2 ⎜ ⎟ T1 P1 ⎠ T1 ⎝ ⎝ N2 = (1.6 kg )(1.049 kJ/kg ⋅ K )ln = 1.163 kJ/K

⎞ ⎟⎟ ⎠ N2

600 K 300 K

ΔS total = ΔS H 2 + ΔS N 2 = 5.026 kJ/K + 1.163 kJ/K = 6.19 kJ/K

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13-51

13-74 Heat is transferred to a gas mixture contained in a piston cylinder device. The initial state and the final temperature are given. The heat transfer is to be determined for the ideal gas and non-ideal gas cases. Properties The molar masses of H2 and N2 are 2.0, and 28.0 kg/kmol. (Table A-1). Analysis From the energy balance relation, E in − E out = ΔE Qin − Wb,out = ΔU

(

Qin = ΔH = ΔH H 2 + ΔH N 2 = N H 2 h2 − h1

)H

2

(

+ N N 2 h2 − h1

)N

2

since Wb and ΔU combine into ΔH for quasi-equilibrium constant pressure processes mH 2 6 kg NH2 = = = 3 kmol M H 2 2 kg / kmol NN2 =

mN 2 MN 2

=

6 kg H2 21 kg N2 5 MPa 160 K Q

21 kg = 0.75 kmol 28 kg / kmol

(a) Assuming ideal gas behavior, the inlet and exit enthalpies of H2 and N2 are determined from the ideal gas tables to be

Thus,

H2 :

h1 = h@160 K = 4,535.4 kJ / kmol,

N2 :

h1 = h@160 K = 4,648 kJ / kmol,

h2 = h@ 200 K = 5,669.2 kJ / kmol h2 = h@ 200 K = 5,810 kJ / kmol

Qideal = 3 × (5,669.2 − 4,535.4 ) + 0.75 × (5,810 − 4,648) = 4273 kJ

(b) Using Amagat's law and the generalized enthalpy departure chart, the enthalpy change of each gas is determined to be ⎫ ⎪ Tcr,H 2 ⎪ ⎪ Z h1 ≅ 0 Pm 5 ⎪ PR1 ,H 2 = PR2 ,H 2 = H2: = = 3.846 ⎬ Pcr,H 2 1.30 ⎪Zh ≅ 0 ⎪ 2 Tm , 2 200 ⎪ T R2 , H 2 = = = 6.006 ⎪ Tcr,H 2 33.3 ⎭ Thus H2 can be treated as an ideal gas during this process. T R1 ,H 2 =

T R1 , N 2 =

N2:

Tm,1

=

Tm,1 Tcr, N 2

PR1 , N 2 = PR2 , N 2 T R2 , N 2 =

Tm, 2 Tcr, N 2

160 = 4.805 33.3

⎫ ⎪ ⎪ ⎪ Z h1 = 1.3 Pm 5 ⎪ = = = 1.47 ⎬ Pcr, N 2 3.39 ⎪ Z h = 0.7 ⎪ 2 200 ⎪ = = 1.58 ⎪ 126.2 ⎭

=

(Fig. A-29)

160 = 1.27 126.2

(Fig. A-29)

Therefore,

(h2 − h1 )H = (h2 − h1 )H ,ideal = 5,669.2 − 4,535.4 = 1,133.8kJ/kmol 2

(h2 − h1 )N

2

2

(

) (

= Ru Tcr Z h1 − Z h2 + h2 − h1

)ideal

= (8.314kPa ⋅ m 3 /kmol ⋅ K)(126.2K)(1.3 − 0.7) + (5,810 − 4,648)kJ/kmol = 1,791.5kJ/kmol

Substituting,

Qin = (3 kmol)(1,133.8 kJ/kmol) + (0.75 kmol)(1,791.5 kJ/kmol) = 4745 kJ

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

13-52

13-75 Heat is transferred to a gas mixture contained in a piston cylinder device discussed in previous problem. The total entropy change and the exergy destruction are to be determined for two cases. Analysis The entropy generated during this process is determined by applying the entropy balance on an extended system that includes the piston-cylinder device and its immediate surroundings so that the boundary temperature of the extended system is the environment temperature at all times. It gives S in − S out + S gen = ΔSsystem Qin + S gen = ΔS water Tboundary

→

S gen = m( s2 − s1 ) −

Qin Tsurr

Then the exergy destroyed during a process can be determined from its definition X destroyed = T0 Sgen . (a) Noting that the total mixture pressure, and thus the partial pressure of each gas, remains constant, the entropy change of a component in the mixture during this process is ⎛ T P ©0 ⎞ T ΔS i = mi ⎜ c p ln 2 − R ln 2 ⎟ = mi c p ,i ln 2 ⎟ ⎜ T P T1 1 1 ⎠i ⎝

Assuming ideal gas behavior and using cp values at the average temperature, the ΔS of H2 and N2 are determined from ΔS H 2 ,ideal = (6 kg )(13.60 kJ/kg ⋅ K ) ln

200 K = 18.21 kJ/K 160 K

ΔS N 2 ,ideal = (21 kg )(1.039 kJ/kg ⋅ K ) ln

200 K = 4.87 kJ/K 160 K

and 4273 kJ = 8.98 kJ/K 303 K = (303 K )(8.98 kJ/K ) = 2721 kJ

S gen = 18.21 kJ/K + 4.87 kJ/K − X destroyed = T0 S gen

(b) Using Amagat's law and the generalized entropy departure chart, the entropy change of each gas is determined to be T R1 ,H 2 =

H2:

Tm,1 Tcr,H 2

PR1 ,H 2 = PR2 , H 2 T R2 , H 2 =

Tm , 2 Tcr,H 2

⎫ ⎪ ⎪ ⎪ Z s1 ≅ 1 Pm 5 ⎪ = = = 3.846 ⎬ Pcr,H 2 1.30 ⎪Zs ≅1 ⎪ 2 200 ⎪ = = 6.006 ⎪ 33.3 ⎭

=

160 = 4.805 33.3

(Table A-30)

Thus H2 can be treated as an ideal gas during this process. T R1 , N 2 =

N2:

Tm,1 Tcr, N 2

PR1 , N 2 = PR2 , N 2 T R2 , N 2 =

Tm, 2 Tcr, N 2

⎫ ⎪ ⎪ ⎪ Z s1 = 0.8 Pm 5 ⎪ = = = 1.475 ⎬ Pcr, N 2 3.39 ⎪ Z s = 0.4 ⎪ 2 200 ⎪ = = 1.585 ⎪ 126.2 ⎭

=

160 = 1.268 126.2

(Table A-30)

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13-53

Therefore, ΔS H 2 = ΔS H 2 ,ideal = 18.21 kJ/K

(

)

ΔS N 2 = N N 2 Ru Z s1 − Z s2 + ΔS N 2 ,ideal = (0.75 kmol)(8.314 kPa ⋅ m 3 /kmol ⋅ K)(0.8 − 0.4) + (4.87 kJ/K ) = 7.37 kJ/K ΔS surr =

Qsurr − 4745 kJ = = −15.66 kJ/K T0 303 K

and 4745 kJ = 9.92 kJ/K 303 K = (303 K )(9.92 kJ/K ) = 3006 kJ

S gen = 18.21 kJ/K + 7.37 kJ/K − X destroyed = T0 S gen

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13-54

13-76 Air is compressed isothermally in a steady-flow device. The power input to the compressor and the rate of heat rejection are to be determined for ideal and non-ideal gas cases. Assumptions 1 This is a steady-flow process. 2 The kinetic and potential energy changes are negligible. Properties The molar mass of air is 29.0 kg/kmol. (Table A-1). Analysis The mass flow rate of air can be expressed in terms of the mole numbers as

200 K 8 MPa

2.90 kg / s m& = = 0.10 kmol / s N& = M 29.0 kg / kmol

(a) Assuming ideal gas behavior, the Δh and Δs of air during this process is Δh = 0 (isothermal process ) Δs = c p ln

T2 T1

©0

− Ru ln

79% N2 21% O2

P2 P = − Ru ln 2 P1 P1

= −(8.314 kJ/kg ⋅ K ) ln

8 MPa = −5.763 kJ/kmol ⋅ K 4 MPa

200 K 4 MPa

Disregarding any changes in kinetic and potential energies, the steady-flow energy balance equation for the isothermal process of an ideal gas reduces to E& in − E& out = ΔE& systemÊ0 (steady) = 0 E& in = E& out W& in + N& h1 = Q& out + N& h2 W& in − Q& out = N& Δh Ê0 = 0 ⎯ ⎯→ W& in = Q& out

Also for an isothermal, internally reversible process the heat transfer is related to the entropy change by

Q = TΔS = NTΔs , Q& = N& TΔs = (0.10 kmol/s )(200 K )(− 5.763 kJ/kmol ⋅ K ) = −115.3 kW → Q& out = 115.3 kW

Therefore, W& in = Q& out = 115.3 kW

(b) Using Amagat's law and the generalized charts, the enthalpy and entropy changes of each gas are determined from h2 − h1 = Ru Tcr ( Z h1 − Z h2 ) + (h2 − h1 ) ideal

Ê0

s2 − s1 = Ru ( Z s1 − Z s2 ) + (s2 − s1 ) ideal

where ⎫ ⎪ Pcr, N 2 ⎪ ⎪ Z h1 = 0.4, Z s1 = 0.2 T 220 ⎪ T R1 = T R2 = m = = 1.74 ⎬ Tcr, N 2 126.2 ⎪ Z h = 0.8, Z s = 0.35 2 ⎪ 2 Pm,2 8 ⎪ PR2 = = = 2.36 Pcr, N 2 3.39 ⎪⎭ PR1 =

N2:

Pm,1

=

4 = 1.18 3.39

(Tables A-29 and A-30)

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13-55

⎫ ⎪ Pcr,O 2 ⎪ ⎪ Z h1 = 0.4, Z s1 = 0.25 Tm 220 ⎪ = = 1.421 ⎬ T R1 = T R2 = Tcr,O 2 154.8 ⎪ Z h = 1.0, Z s = 0.5 2 ⎪ 2 Pm,2 8 ⎪ = = 1.575 PR2 = Pcr,O 2 5.08 ⎪⎭ PR1 =

O2:

Pm,1

=

4 = 0.787 5.08

(Tables A-29 and A-30)

Then, h2 − h1 = y i Δhi = y N 2 ( h2 − h1 ) N 2 + y O 2 ( h2 − h1 ) O 2 = (0.79)(8.314)(126.2)(0.4 − 0.8) + (0.21)(8.314)(154.8)(0.4 − 1.0) + 0 = −494kJ/kmol s 2 − s1 = y i Δs i = y N 2 ( s 2 − s1 ) N 2 + y O 2 ( s 2 − s1 ) O 2 = (0.79)(8.314)(0.2 − 0.35) + (0.21)(8.314)(0.25 − 0.5) + (−5.763) = −7.18kJ/kmol ⋅ K

Thus, Q& out = − N& TΔs = −(0.10 kmol/s)(200 K )(− 7.18 kJ/kmol ⋅ K ) = 143.6 kW E& in − E& out = ΔE& system ©0 (steady) = 0 E& in = E& out W& in + N& h1 = Q& out + N& h2 W& in = Q& out + N& ( h2 − h1 ) ⎯ ⎯→ W& in = 143.6 kW + (0.10kmol/s)(−494kJ/kmol) = 94.2 kW

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13-56

13-77 EES Problem 13-76 is reconsidered. The results obtained by assuming ideal behavior, real gas behavior with Amagat's law, and real gas behavior with EES data are to be compared. Analysis The problem is solved using EES, and the solution is given below. "Input Data:" y_N2 = 0.79 y_O2 = 0.21 T[1]=200 [K] "Inlet temperature" T[2]=200 [K] "Exit temmperature" P[1]=4000 [kPa] P[2]=8000 [kPa] m_dot = 2.9 [kg/s] R_u = 8.314 [kJ/kmol-K] DELTAe_bar_sys = 0 "Steady-flow analysis for all cases" m_dot = N_dot * (y_N2*molarmass(N2)+y_O2*molarmass(O2)) "Ideal gas:" e_bar_in_IG - e_bar_out_IG = DELTAe_bar_sys e_bar_in_IG =w_bar_in_IG + h_bar_IG[1] e_bar_out_IG = q_bar_out_IG +h_bar_IG[2] h_bar_IG[1] = y_N2*enthalpy(N2,T=T[1]) + y_O2*enthalpy(O2,T=T[1]) h_bar_IG[2] = y_N2*enthalpy(N2,T=T[2]) + y_O2*enthalpy(O2,T=T[2]) "The pocess is isothermal so h_bar_IG's are equal. q_bar_IG is found from the entropy change:" q_bar_out_IG = -T[1]*DELTAs_IG s_IG[2]= y_N2*entropy(N2,T=T[2],P=y_N2*P[2]) + y_O2*entropy(O2,T=T[2],P=y_O2*P[2]) s_IG[1] =y_N2*entropy(N2,T=T[1],P=y_N2*P[1]) + y_O2*entropy(O2,T=T[1],P=y_O2*P[1]) DELTAs_IG =s_IG[2]-s_IG[1] Q_dot_out_IG=N_dot*q_bar_out_IG W_dot_in_IG=N_dot*w_bar_in_IG "EES:" PN2[1]=y_N2*P[1] PO2[1]=y_O2*P[1] PN2[2]=y_N2*P[2] PO2[2]=y_O2*P[2] e_bar_in_EES - e_bar_out_EES = DELTAe_bar_sys e_bar_in_EES =w_bar_in_EES + h_bar_EES[1] e_bar_out_EES = q_bar_out_EES+h_bar_EES[2] h_bar_EES[1] = y_N2*enthalpy(Nitrogen,T=T[1], P=PN2[1]) + y_O2*enthalpy(Oxygen,T=T[1],P=PO2[1]) h_bar_EES[2] = y_N2*enthalpy(Nitrogen,T=T[2],P=PN2[2]) + y_O2*enthalpy(Oxygen,T=T[2],P=PO2[2]) q_bar_out_EES = -T[1]*DELTAs_EES DELTAs_EES =y_N2*entropy(Nitrogen,T=T[2],P=PN2[2]) + y_O2*entropy(Oxygen,T=T[2],P=PO2[2]) - y_N2*entropy(Nitrogen,T=T[1],P=PN2[1]) y_O2*entropy(Oxygen,T=T[1],P=PO2[1]) Q_dot_out_EES=N_dot*q_bar_out_EES W_dot_in_EES=N_dot*w_bar_in_EES "Amagat's Rule:" Tcr_N2=126.2 [K] "Table A.1" Tcr_O2=154.8 [K] Pcr_N2=3390 [kPa] "Table A.1" Pcr_O2=5080 [kPa] PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

13-57 e_bar_in_Zchart - e_bar_out_Zchart = DELTAe_bar_sys e_bar_in_Zchart=w_bar_in_Zchart + h_bar_Zchart[1] e_bar_out_Zchart =q_bar_out_Zchart + h_bar_Zchart[2] q_bar_out_Zchart = -T[1]*DELTAs_Zchart Q_dot_out_Zchart=N_dot*q_bar_out_Zchart W_dot_in_Zchart=N_dot*w_bar_in_Zchart "State 1by compressability chart" Tr_N2[1]=T[1]/Tcr_N2 Pr_N2[1]=y_N2*P[1]/Pcr_N2 Tr_O2[1]=T[1]/Tcr_O2 Pr_O2[1]=y_O2*P[1]/Pcr_O2 DELTAh_bar_1_N2=ENTHDEP(Tr_N2[1], Pr_N2[1])*R_u*Tcr_N2 "Enthalpy departure, N2" DELTAh_bar_1_O2=ENTHDEP(Tr_O2[1], Pr_O2[1])*R_u*Tcr_O2 "Enthalpy departure, O2" h_bar_Zchart[1]=h_bar_IG[1]-(y_N2*DELTAh_bar_1_N2+y_O2*DELTAh_bar_1_O2) "Enthalpy of real gas using charts" DELTAs_N2[1]=ENTRDEP(Tr_N2[1], Pr_N2[1])*R_u "Entropy departure, N2" DELTAs_O2[1]=ENTRDEP(Tr_O2[1], Pr_O2[1])*R_u "Entropy departure, O2" s[1]=s_IG[1]-(y_N2*DELTAs_N2[1]+y_O2*DELTAs_O2[1]) "Entropy of real gas using charts" "State 2 by compressability chart" Tr_N2[2]=T[2]/Tcr_N2 Pr_N2[2]=y_N2*P[2]/Pcr_N2 Tr_O2[2]=T[2]/Tcr_O2 Pr_O2[2]=y_O2*P[2]/Pcr_O2 DELTAh_bar_2_N2=ENTHDEP(Tr_N2[2], Pr_N2[2])*R_u*Tcr_N2 "Enthalpy departure, N2" DELTAh_bar_2_O2=ENTHDEP(Tr_O2[2], Pr_O2[2])*R_u*Tcr_O2 "Enthalpy departure, O2" h_bar_Zchart[2]=h_bar_IG[2]-(y_N2*DELTAh_bar_2_N2+y_O2*DELTAh_bar_2_O2) "Enthalpy of real gas using charts" DELTAs_N2[2]=ENTRDEP(Tr_N2[2], Pr_N2[2])*R_u "Entropy departure, N2" DELTAs_O2[2]=ENTRDEP(Tr_O2[2], Pr_O2[2])*R_u "Entropy departure, O2" s[2]=s_IG[2]-(y_N2*DELTAs_N2[2]+y_O2*DELTAs_O2[2]) "Entropy of real gas using charts" DELTAs_Zchart = s[2]-s[1] "[kJ/kmol-K]" SOLUTION DELTAe_bar_sys=0 [kJ/kmol] DELTAh_bar_1_O2=147.6 DELTAh_bar_2_O2=299.5 DELTAs_IG=-5.763 [kJ/kmol-K] DELTAs_N2[2]=3.644 DELTAs_O2[2]=1.094 e_bar_in_EES=-2173 [kJ/kmol] e_bar_in_Zchart=-2103 e_bar_out_IG=-1633 [kJ/kmol] h_bar_EES[1]=-3235 h_bar_IG[1]=-2785 h_bar_Zchart[1]=-3181 m_dot=2.9 [kg/s] Pcr_N2=3390 [kPa] P[1]=4000 [kPa] PN2[1]=3160 PO2[1]=840 Pr_N2[1]=0.9322 Pr_O2[1]=0.1654 q_bar_out_EES=1446 [kJ/kmol]

DELTAh_bar_1_N2=461.2 DELTAh_bar_2_N2=907.8 DELTAs_EES=-7.23 [kJ/kmol-K] DELTAs_N2[1]=1.831 DELTAs_O2[1]=0.5361 DELTAs_Zchart=-7.312 [kJ/kmol-K] e_bar_in_IG=-1633 [kJ/kmol] e_bar_out_EES=-2173 [kJ/kmol] e_bar_out_Zchart=-2103 h_bar_EES[2]=-3619 h_bar_IG[2]=-2785 h_bar_Zchart[2]=-3565 N_dot=0.1005 [kmol/s] Pcr_O2=5080 [kPa] P[2]=8000 [kPa] PN2[2]=6320 PO2[2]=1680 Pr_N2[2]=1.864 Pr_O2[2]=0.3307 q_bar_out_IG=1153 [kJ/kmol]

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13-58 q_bar_out_Zchart=1462 Q_dot_out_IG=115.9 [kW] R_u=8.314 [kJ/kmol-K] s[2]=147.8 s_IG[2]=150.9 Tcr_O2=154.8 [K] T[2]=200 [K] Tr_N2[2]=1.585 Tr_O2[2]=1.292 w_bar_in_IG=1153 [kJ/kmol] W_dot_in_EES=106.8 [kW] W_dot_in_Zchart=108.3 [kW] y_O2=0.21

Q_dot_out_EES=145.3 [kW] Q_dot_out_Zchart=147 [kW] s[1]=155.1 s_IG[1]=156.7 Tcr_N2=126.2 [K] T[1]=200 [K] Tr_N2[1]=1.585 Tr_O2[1]=1.292 w_bar_in_EES=1062 [kJ/kmol] w_bar_in_Zchart=1078 [kJ/kmmol] W_dot_in_IG=115.9 [kW] y_N2=0.79

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13-59

13-78 The volumetric fractions of the constituents of a mixture of products of combustion are given. The average molar mass of the mixture, the average specific heat, and the partial pressure of the water vapor in the mixture are to be determined. Assumptions Under specified conditions all N2, O2, H2O, and CO2 can be treated as ideal gases, and the mixture as an ideal gas mixture. Properties The molar masses of CO2, H2O, O2, and N2 are 44.0, 18.0, 32.0, and 28.0 kg/kmol, respectively (Table A-1). The specific heats of CO2, H2O, O2, and N2 at 600 K are 1.075, 2.015, 1.003, and 1.075 kJ/kg.K, respectively (Table A-2b). The specific heat of water vapor at 600 K is obtained from EES. Analysis For convenience, consider 100 kmol of mixture. Noting that volume fractions are equal to mole fractions in ideal gas mixtures, the average molar mass of the mixture is determined to be Mm =

N CO 2 M CO 2 + N H 2O M H 2O + N O 2 M O 2 + N N 2 M N 2 N CO 2 + N H 2O + N O 2 + N N 2

(4.89 kmol)(44 kg/kmol) + (6.50)(18) + (12.20)(32) + (76.41)(28) = (4.89 + 6.50 + 12.20 + 76.41) kmol = 28.62 kg/kmol

600 K 200 kPa

The average specific heat is determined from c p,m =

76.41% N2 12.20% O2 6.50% H2O 4.89% CO2

N CO2 c p,CO2 M CO2 + N H 2O c p,H 2O M H 2O + N O 2 c p,O2 M O2 + N N 2 c p, N 2 M N 2 N CO2 + N H 2O + N O2 + N N 2

(4.89 kmol)(1.075 kJ/kg.K)(44 kg/kmol) + (6.50)(2.015)(18) + (12.20)(1.003)(32) + (76.41)(1.075)(28) (4.89 + 6.50 + 12.20 + 76.41) kmol = 31.59 kJ/kmol.K =

The partial pressure of the water in the mixture is yv =

N H 2O N CO 2 + N H 2O + N O 2 + N N 2

=

6.50 kmol = 0.0650 (4.89 + 6.50 + 12.20 + 76.41) kmol

Pv = y v Pm = (0.0650)(200 kPa) = 13.0 kPa

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13-60

Special Topic: Chemical Potential and the Separation Work of Mixtures

13-79C No, a process that separates a mixture into its components without requiring any work (exergy) input is impossible since such a process would violate the 2nd law of thermodynamics.

13-80C Yes, the volume of the mixture can be more or less than the sum of the initial volumes of the mixing liquids because of the attractive or repulsive forces acting between dissimilar molecules.

13-81C The person who claims that the temperature of the mixture can be higher than the temperatures of the components is right since the total enthalpy of the mixture of two components at the same pressure and temperature, in general, is not equal to the sum of the total enthalpies of the individual components before mixing, the difference being the enthalpy (or heat) of mixing, which is the heat released or absorbed as two or more components are mixed isothermally.

13-82C Mixtures or solutions in which the effects of molecules of different components on each other are negligible are called ideal solutions (or ideal mixtures). The ideal-gas mixture is just one category of ideal solutions. For ideal solutions, the enthalpy change and the volume change due to mixing are zero, but the entropy change is not. The chemical potential of a component of an ideal mixture is independent of the identity of the other constituents of the mixture. The chemical potential of a component in an ideal mixture is equal to the Gibbs function of the pure component.

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13-61

13-83 Brackish water is used to produce fresh water. The minimum power input and the minimum height the brackish water must be raised by a pump for reverse osmosis are to be determined. Assumptions 1 The brackish water is an ideal solution since it is dilute. 2 The total dissolved solids in water can be treated as table salt (NaCl). 3 The environment temperature is also 12°C. Properties The molar masses of water and salt are Mw = 18.0 kg/kmol and Ms = 58.44 kg/kmol. The gas constant of pure water is Rw = 0.4615 kJ/kg⋅K (Table A-1). The density of fresh water is 1000 kg/m3. Analysis First we determine the mole fraction of pure water in brackish water using Eqs. 13-4 and 13-5. Noting that mfs = 0.00078 and mfw = 1- mfs = 0.99922, Mm =

1 1 1 = = = 18.01 kg/kmol mfi mf s mf w 0.00078 0.99922 + + 58.44 18.0 Mi Ms Mw

∑

yi = mfi

Mm Mi

→

yw = mf w

Mm 18.01 kg/kmol = (0.99922) = 0.99976 Mw 18.0 kg/kmol

The minimum work input required to produce 1 kg of freshwater from brackish water is

wmin, in = RwT0 ln(1 / yw ) = (0.4615 kJ/kg ⋅ K)(285.15 K) ln(1/0.99976) = 0.03159 kJ/kg fresh water Therefore, 0.03159 kJ of work is needed to produce 1 kg of fresh water is mixed with seawater reversibly. Therefore, the required power input to produce fresh water at the specified rate is ⎛ 1 kW ⎞ W& min, in = ρV&wmin, in = (1000 kg/m 3 )(0.280 m 3 /s)(0.03159 kJ/kg)⎜ ⎟ = 8.85 kW ⎝ 1 kJ/s ⎠

The minimum height to which the brackish water must be pumped is Δz min =

wmin,in g

⎛ 0.03159 kJ/kg ⎞⎛⎜ 1 kg.m/s 2 =⎜ ⎟ ⎝ 9.81 m/s 2 ⎠⎜⎝ 1 N

⎞⎛ 1000 N.m ⎞ ⎟⎜ = 3.22 m ⎟⎝ 1 kJ ⎟⎠ ⎠

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13-62

13-84 A river is discharging into the ocean at a specified rate. The amount of power that can be generated is to be determined. Assumptions 1 The seawater is an ideal solution since it is dilute. 2 The total dissolved solids in water can be treated as table salt (NaCl). 3 The environment temperature is also 15°C. Properties The molar masses of water and salt are Mw = 18.0 kg/kmol and Ms = 58.44 kg/kmol. The gas constant of pure water is Rw = 0.4615 kJ/kg⋅K (Table A-1). The density of river water is 1000 kg/m3. Analysis First we determine the mole fraction of pure water in ocean water using Eqs. 13-4 and 13-5. Noting that mfs = 0.035 and mfw = 1- mfs = 0.965, Mm =

1 1 1 = = = 18.45 kg/kmol mfi mf s mf w 0.035 0.965 + + 58.44 18.0 Mi Ms Mw

∑

yi = mfi

Mm Mi

→

yw = mf w

Mm 18.45 kg/kmol = (0.965) = 0.9891 Mw 18.0 kg/kmol

The maximum work output associated with mixing 1 kg of seawater (or the minimum work input required to produce 1 kg of freshwater from seawater) is wmax, out = RwT0 ln(1 / yw ) = (0.4615 kJ/kg ⋅ K)(288.15 K)ln(1/0.9891) = 1.46 kJ/kg fresh water Therefore, 1.46 kJ of work can be produced as 1 kg of fresh water is mixed with seawater reversibly. Therefore, the power that can be generated as a river with a flow rate of 400,000 m3/s mixes reversibly with seawater is ⎛ 1 kW ⎞ 6 W& max out = ρV&wmax out = (1000 kg/m 3 )(4 × 10 5 m 3 /s)(1.46 kJ/kg)⎜ ⎟ = 582 × 10 kW ⎝ 1 kJ/s ⎠

Discussion This is more power than produced by all nuclear power plants (112 of them) in the U.S., which shows the tremendous amount of power potential wasted as the rivers discharge into the seas.

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13-63

13-85 EES Problem 13-84 is reconsidered. The effect of the salinity of the ocean on the maximum power generated is to be investigated. Analysis The problem is solved using EES, and the solution is given below. "Properties:" M_w = 18.0 [kg/kmol] "Molar masses of water" M_s = 58.44 [kg/kmol] "Molar masses of salt" R_w = 0.4615 [kJ/kg-K] "Gas constant of pure water" roh_w = 1000 [kg/m^3] "density of river water" V_dot = 4E5 [m^3/s] T_0 = 15 [C] "Analysis: First we determine the mole fraction of pure water in ocean water using Eqs. 13-4 and 13-5. " mf_s = 0.035 "mass fraction of the salt in seawater = salinity" mf_w = 1- mf_s "mass fraction of the water in seawater" "Molar mass of the seawater is:" M_m=1/(mf_s/m_s+mf_w/M_w) "Mole fraction of the water is:" y_w=mf_w*M_m/M_w "The maximum work output associated with mixing 1 kg of seawater (or the minimum work input required to produce 1 kg of freshwater from seawater) is:" w_maxout =R_w*(T_0+273.15)*ln(1/y_w) "[kJ/kg fresh water]" "The power that can be generated as a river with a flow rate of 400,000 m^3/s mixes reversibly with seawater is" W_dot_max=roh_w*V_dot*w_maxout "Discussion This is more power than produced by all nuclear power plants (112 of them) in the US., which shows the tremendous amount of power potential wasted as the rivers discharge into the seas." 9.000 x 10 8

0 0.01 0.02 0.03 0.04 0.05

Wmax [kW] 0 1.652E+08 3.333E+08 5.043E+08 6.783E+08 8.554E+08

8.000 x 10 8 7.000 x 10 8

W m ax [kw ]

mfs

6.000 x 10 8 5.000 x 10 8 4.000 x 10 8 3.000 x 10 8 2.000 x 10 8 1.000 x 10 8 0 0

0.01

0.02

0.03

0.04

0.05

m fs

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13-64

13-86E Brackish water is used to produce fresh water. The mole fractions, the minimum work inputs required to separate 1 lbm of brackish water and to obtain 1 lbm of fresh water are to be determined. Assumptions 1 The brackish water is an ideal solution since it is dilute. 2 The total dissolved solids in water can be treated as table salt (NaCl). 3 The environment temperature is equal to the water temperature. Properties The molar masses of water and salt are Mw = 18.0 lbm/lbmol and Ms = 58.44 lbm/lbmol. The gas constant of pure water is Rw = 0.1102 Btu/lbm⋅R (Table A-1E). Analysis (a) First we determine the mole fraction of pure water in brackish water using Eqs. 13-4 and 135. Noting that mfs = 0.0012 and mfw = 1- mfs = 0.9988, Mm =

1 1 1 = = = 18.015 lbm/lbmol mfi mf s mf w 0.0012 0.9988 + + 58.44 18.0 Mi Ms Mw

∑

yi = mfi

Mm Mi

→

yw = mf w

Mm 18.015 lbm/lbmol = (0.9988) = 0.99963 Mw 18.0 lbm/lbmol

y s = 1 − y w = 1 − 0.99963 = 0.00037

(b) The minimum work input required to separate 1 lbmol of brackish water is wmin,in = − R wT0 ( y w ln y w + y s ln y s ) = −(0.1102 Btu/lbmol.R)(525 R)[0.99963 ln(0.99963) + 0.00037 ln(0.00037)] = −0.191 Btu/lbm brackish water

(c) The minimum work input required to produce 1 lbm of freshwater from brackish water is wmin, in = RwT0 ln(1 / yw ) = (0.1102 Btu/lbm ⋅ R)(525 R)ln(1/0.99963) = 0.0214 Btu/lbm fresh water Discussion Note that it takes about 9 times work to separate 1 lbm of brackish water into pure water and salt compared to producing 1 lbm of fresh water from a large body of brackish water.

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13-65

13-87 A desalination plant produces fresh water from seawater. The second law efficiency of the plant is to be determined. Assumptions 1 The seawater is an ideal solution since it is dilute. 2 The total dissolved solids in water can be treated as table salt (NaCl). 3 The environment temperature is equal to the seawater temperature. Properties The molar masses of water and salt are Mw = 18.0 kg/kmol and Ms = 58.44 kg/kmol. The gas constant of pure water is Rw = 0.4615 kJ/kg⋅K (Table A-1). The density of river water is 1000 kg/m3. Analysis First we determine the mole fraction of pure water in seawater using Eqs. 13-4 and 13-5. Noting that mfs = 0.032 and mfw = 1- mfs = 0.968, Mm =

1 1 1 = = = 18.41 kg/kmol mfi mf s mf w 0.032 0.968 + + Mi Ms Mw 58.44 18.0

∑

yi = mfi

Mm Mi

→

yw = mf w

18.41 kg/kmol Mm = (0.968) = 0.9900 18.0 kg/kmol Mw

The maximum work output associated with mixing 1 kg of seawater (or the minimum work input required to produce 1 kg of freshwater from seawater) is wmax, out = R w T0 ln(1 / y w ) = (0.4615 kJ/kg ⋅ K)(283.15 K)ln(1/0.990) = 1.313 kJ/kg fresh water

The power that can be generated as 1.4 m3/s fresh water mixes reversibly with seawater is ⎛ 1 kW ⎞ W& max out = ρV&wmax out = (1000 kg/m 3 )(1.4 m 3 /s)(1.313 kJ/kg)⎜ ⎟ = 1.84 kW ⎝ 1 kJ/s ⎠

Then the second law efficiency of the plant becomes

η II =

W& min,in 1.83 MW = = 0.216 = 21.6% 8.5 MW W& in

13-88 The power consumption and the second law efficiency of a desalination plant are given. The power that can be produced if the fresh water produced is mixed with the seawater reversibly is to be determined. Assumptions 1 This is a steady-flow process. 2 The kinetic and potential energy changes are negligible. Analysis From the definition of the second law efficiency

W&rev

ηII = & W

actual

→ 0.18 =

W&rev → W&rev = 0.594 MW 3.3 MW

which is the maximum power that can be generated.

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13-66

13-89E It is to be determined if it is it possible for an adiabatic liquid-vapor separator to separate wet steam at 100 psia and 90 percent quality, so that the pressure of the outlet streams is greater than 100 psia. Analysis Because the separator divides the inlet stream into the liquid and vapor portions, m& 2 = xm& 1 = 0.9m& 1 m& 3 = (1 − x)m& 1 = 0.1m& 1

(2) Vapor (1) Mixture

According to the water property tables at 100 psia (Table A-5E), s1 = s f + xs fg = 0.47427 + 0.9 × 1.12888 = 1.4903 Btu/lbm ⋅ R

(3) Liquid

When the increase in entropy principle is adapted to this system, it becomes m& 2 s 2 + m& 3 s 3 ≥ m& 1 s1 xm& 1 s 2 + (1 − x)m& 1 s 3 ≥ m& 1 s1 0.9 s 2 + 0.1s 3 ≥ s1 ≥ 1.4903 Btu/lbm ⋅ R

To test this hypothesis, let’s assume the outlet pressures are 110 psia. Then, s 2 = s g = 1.5954 Btu/lbm ⋅ R s 3 = s f = 0.48341 Btu/lbm ⋅ R

The left-hand side of the above equation is 0.9 s 2 + 0.1s 3 = 0.9 × 1.5954 + 0.1× 0.48341 = 1.4842 Btu/lbm ⋅ R

which is less than the minimum possible specific entropy. Hence, the outlet pressure cannot be 110 psia. Inspection of the water table in light of above equation proves that the pressure at the separator outlet cannot be greater than that at the inlet.

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13-67

Review Problems

13-90 The molar fractions of constituents of air are given. The gravimetric analysis of air and its molar mass are to be determined. Assumptions All the constituent gases and their mixture are ideal gases. Properties The molar masses of O2, N2, and Ar are 32.0, 28.0, and 40.0 kg/kmol. (Table A-1). Analysis For convenience, consider 100 kmol of air. Then the mass of each component and the total mass are ⎯→ m O 2 = N O 2 M O 2 = (21 kmol)(32 kg/kmol) = 672 kg N O 2 = 21 kmol ⎯

⎯→ m N 2 = N N 2 M N 2 = (78 kmol)(28 kg/kmol) = 2184 kg N N 2 = 78 kmol ⎯ ⎯→ m Ar = N Ar M Ar = (1 kmol)(40 kg/kmol) = 40 kg N Ar = 1 kmol ⎯ m m = m O 2 + m N 2 + m Ar = 672 kg + 2184 kg + 40 kg = 2896 kg

AIR 21% O2 78% N2 1% Ar

Then the mass fraction of each component (gravimetric analysis) becomes mf O 2 = mf N 2 = mf Ar =

mO 2 mm mN2 mm

=

672 kg = 0.232 or 23.2% 2896 kg

=

2184 kg = 0.754 or 75.4% 2896 kg

m Ar 40 kg = = 0.014 or 1.4% 2896 kg mm

The molar mass of the mixture is determined from its definitions, Mm =

mm 2,896 kg = = 28.96 kg / kmol N m 100 kmol

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13-68

k

13-91 Using Dalton’s law, it is to be shown that Z m =

∑y Z

i i

for a real-gas mixture.

i =1

Analysis Using the compressibility factor, the pressure of a component of a real-gas mixture and of the pressure of the gas mixture can be expressed as Pi =

Zi N i Ru Tm Vm

and

Dalton's law can be expressed as Pm = Z m N m Ru Tm = Vm

∑

Pm =

∑ P (T i

Z m N m Ru Tm Vm

m , Vm

) . Substituting,

Zi Ni Ru Tm Vm

Simplifying, Zm Nm =

∑Z N i

i

Dividing by Nm, Zm =

∑yZ

i i

where Zi is determined at the mixture temperature and volume.

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13-69

13-92 A mixture of carbon dioxide and nitrogen flows through a converging nozzle. The required make up of the mixture on a mass basis is to be determined. Assumptions Under specified conditions CO2 and N2 can be treated as ideal gases, and the mixture as an ideal gas mixture. Properties The molar masses of CO2 and N2 are 44.0 and 28.0 kg/kmol, respectively (Table A-1). The specific heat ratios of CO2 and N2 at 500 K are kCO2 = 1.229 and kN2 = 1.391 (Table A-2). Analysis The molar mass of the mixture is determined from M m = y CO 2 M CO 2 + y N 2 M N 2

The molar fractions are related to each other by y CO 2 + y N 2 = 1

The gas constant of the mixture is given by Rm =

CO2 N2

500 K 360 m/s

Ru Mm

The specific heat ratio of the mixture is expressed as k = mf CO 2 k CO 2 + mf N 2 k N 2

The mass fractions are mf CO 2 = y CO 2 mf N 2 = y N 2

M CO 2 Mm

M N2 Mm

The exit velocity equals the speed of sound at 500 K ⎛ 1000 m 2 /s 2 Vexit = kR m T ⎜⎜ ⎝ 1 kJ/kg

⎞ ⎟ ⎟ ⎠

Substituting the given values and known properties and solving the above equations simultaneously using EES, we find mf CO 2 = 0.838 mf N 2 = 0.162

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13-70

13-93 The mole numbers, pressure, and temperature of the constituents of a gas mixture are given. The volume of the tank containing this gas mixture is to be determined using three methods. Analysis (a) Under specified conditions both N2 and CH4 will considerably deviate from the ideal gas behavior. Treating the mixture as an ideal gas gives N m = N N 2 + N CH 4 = 2 kmol + 6 kmol = 8 kmol

and

Vm

N R T (8 kmol)(8.314 kPa ⋅ m 3 /kmol ⋅ K)(200 K) = m u m = = 1.11 m 3 12,000 kPa Pm

(b) To use Kay's rule, we first need to determine the pseudo-critical temperature and pseudo-critical pressure of the mixture using the critical point properties of N2 and CH4 from Table A-1, N N 2 2 kmol N CH 4 6 kmol = = 0.25 and y CH 4 = = = 0.75 y N2 = 8 kmol 8 kmol Nm Nm Tcr′ , m =

∑yT

= y N 2 Tcr , N 2 + y CH 4 Tcr ,CH 4

∑y P

= y N 2 Pcr , N 2 + y CH 4 Pcr ,CH 4

i cr ,i

2 kmol N2 6 kmol CH4 200 K 12 MPa

= (0.25)(126.2 K) + (0.75)(191.1K) = 174.9K Pcr′ , m =

i cr ,i

= (0.25)(3.39 MPa) + (0.75)(4.64 MPa) = 4.33 MPa

Then, TR = PR =

200 ⎫ = 1.144 ⎪ 174.9 ⎪ ⎬ Z m = 0.47 12 = = 2.77 ⎪ ⎪ 4.33 ⎭

Tm

=

Tcr' , m Pm Pcr' ,m

(Fig. A-15)

Thus,

Vm =

Z m N m Ru Tm = Z mV ideal = (0.47)(1.11 m 3 ) = 0.52 m 3 Pm

(c) To use the Amagat's law for this real gas mixture, we first need to determine the Z of each component at the mixture temperature and pressure, Tm

TR, N 2 =

N2: PR , N 2

Tcr, N 2 P = m Pcr, N 2

T R ,CH 4 =

CH4: PR ,CH 4

Mixture:

Tm

Tcr,CH 4 Pm = Pcr,CH 4

⎫ 200 = 1.585 ⎪ 126.2 ⎪ ⎬ Z N 2 = 0.85 12 = = 3.54 ⎪ ⎪⎭ 3.39

(Fig. A-15)

⎫ 200 = 1.047 ⎪ 191.1 ⎪ ⎬ Z CH 4 = 0.37 12 = = 2.586 ⎪ ⎪⎭ 4.64

(Fig. A-15)

=

=

Zm =

∑y Z

Vm =

Z m N m Ru Tm = Z mV ideal = (0.49)(1.11 m 3 ) = 0.544 m 3 Pm

i

i

= y N 2 Z N 2 + y CH 4 Z CH 4 = (0.25)(0.85) + (0.75)(0.37 ) = 0.49

Thus,

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13-71

13-94 A stream of gas mixture at a given pressure and temperature is to be separated into its constituents steadily. The minimum work required is to be determined. Assumptions 1 Both the N2 and CO2 gases and their mixture are ideal gases. 2 This is a steady-flow process. 3 The kinetic and potential energy changes are negligible. Properties The molar masses of N2 and CO2 are 28.0 and 44.0 kg/kmol. (Table A-1). Analysis The minimum work required to separate a gas mixture into its components is equal to the reversible work associated with the mixing process, which is equal to the exergy destruction (or irreversibility) associated with the mixing process since X destroyed = W rev,out − Wact ,u

Ê0

= W rev,out = T0 S gen

N2 18°C 50% N2 50% CO2 18°C

100 kPa

where Sgen is the entropy generation associated with the steady-flow mixing process. The entropy change associated with a constant pressure and temperature adiabatic mixing process is determined from s gen =

∑ Δs

i

= − Ru

∑y

i

CO2 18°C

ln y i = −(8.314 kJ/kmol ⋅ K )[0.5 ln(0.5) + 0.5 ln(0.5)]

= 5.763 kJ/kmol ⋅ K Mm = s gen =

∑y M i

s gen Mm

=

i

= (0.5)(28 kg/kmol) + (0.5)(44 kg/kmol) = 36 kg/kmol

5.763 kJ/kmol ⋅ K = 0.160 kJ/kg ⋅ K 36 kg/kmol

x destroyed = T0 s gen = (291 K )(0.160 kJ/kg ⋅ K ) = 46.6 kJ/kg

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13-72

13-95 A mixture of carbon dioxide, nitrogen, and oxygen is compressed isothermally. The required work is to be determined. Assumptions 1 Nitrogen, oxygen, and carbon dioxide are ideal gases. 2 The process is reversible. Properties The mole numbers of nitrogen, oxygen, and carbon dioxide are 28.0, 32.0, and 44.0 kg/kmol, respectively (Table A-1). Analysis The mole fractions are y CO2 =

N CO2 1 kmol = = 0.4348 N total 2.3 kmol

y N2 =

N N2 1 kmol = = 0.4348 N total 2.3 kmol

y O2 =

N O2 0.3 kmol = = 0.1304 N total 2.3 kmol

1 kmol CO2 1 kmol N2 0.3 kmol O2 10 kPa, 27°C

The gas constant for this mixture is then R= =

y CO2 M CO2

Ru + y N2 M N2 + y O2 M O2

8.314 kJ/kmol ⋅ K = 0.2343 kJ/kg ⋅ K (0.4348 × 44 + 0.4348 × 28 + 0.1304 × 32)kg/kmol

The mass of this mixture of gases is m = N CO2 M CO2 + N N2 M N2 + N O2 M O2 = 1× 44 + 1× 28 + 0.3 × 32 = 81.6 kg

Noting that Pv = RT for an ideal gas, the work done for this process is then 2

∫

Wout = m Pdv = mRT 1

2

dv

∫v 1

= mRT ln

v2 P = mRT ln 1 v1 P2

= (81.6 kg)(0.2343 kJ/kg ⋅ K )(300 K) ln

10 kPa 100 kPa

= −13,200 kJ

The negative sign shows that the work is done on the system.

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13-73

13-96E A mixture of nitrogen and oxygen is expanded isothermally. The work produced is to be determined. Assumptions 1 Nitrogen and oxygen are ideal gases. 2 The process is reversible. Properties The mole numbers of nitrogen and oxygen are 28.0 and 32.0 lbm/lbmol, respectively (Table A1E). Analysis The mole fractions are y N2 =

N N2 0.1 lbmol = = 0.3333 N total 0.3 lbmol

y O2 =

N O2 0.2 kmol = = 0.6667 N total 0.3 kmol

0.1 lbmol N2 0.2 lbmol O2 300 psia 5 ft3

The gas constant for this mixture is then R=

Ru y N2 M N2 + y O2 M O2

1.9858 Btu/lbmol ⋅ R (0.3333 × 28 + 0.6667 × 32)lbm/lbmol = 0.06475 Btu/lbm ⋅ R =

⎛ 5.404 psia ⋅ ft 3 = (0.06475 Btu/lbm ⋅ R )⎜ ⎜ 1 Btu ⎝

⎞ ⎟ ⎟ ⎠

= 0.3499 psia ⋅ ft 3 /lbm ⋅ R

The mass of this mixture of gases is m = N N2 M N2 + N O2 M O2 = 0.1× 28 + 0.2 × 32 = 9.2 lbm

The temperature of the mixture is T1 =

P1V1 (300 psia)(5 ft 3 ) = = 466.0 R mR (9.2 lbm)(0.3499 psia ⋅ ft 3 /lbm ⋅ R)

Noting that Pv = RT for an ideal gas, the work done for this process is then 2

∫

Wout = m Pdv = mRT 1

2

dv

∫v 1

= mRT ln

v2 v1

= (9.2 lbm)(0.06475 Btu/lbm ⋅ R )(466 R) ln

10 ft 3 5 ft 3

= 192.4 Btu

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13-74

13-97 The specific heat ratio and an apparent molecular weight of a mixture of ideal gases are given. The work required to compress this mixture isentropically in a closed system is to be determined. Analysis For an isentropic process of an ideal gas with constant specific heats, the work is expresses as 2

2

∫

∫

wout = Pdv = P1v 1k v − k dv = 1

=

P1v 1k 1− k

1

(v 121− k − v 11− k ) =

P1v 1k 1− k (v 21 − v 11− k ) 1− k

P1v 1k 1− k

⎡⎛ v ⎢⎜ 2 ⎢⎜⎝ v 1 ⎣

⎞ ⎟⎟ ⎠

1− k

⎤ − 1⎥ ⎥ ⎦

since P1v 1k = P1v k for an isentropic process. Also, P1v 1 = RT1

Gas mixture k=1.35 M=32 kg/kmol 100 kPa, 20°C

(v 2 / v 1 ) k = P1 / P2

Substituting, we obtain wout =

Ru T1 ⎡⎛ P2 ⎢⎜ M (1 − k ) ⎢⎜⎝ P1 ⎣

⎞ ⎟⎟ ⎠

( k −1) / k

⎤ − 1⎥ ⎥ ⎦

(1.35 −1) / 1.35 ⎤ (8.314)(293) ⎡⎛ 1000 ⎞ − 1⎥ ⎢⎜ ⎟ (32)(1 − 1.35) ⎢⎣⎝ 100 ⎠ ⎥⎦ = −177.6 kJ/kg

=

The negative sign shows that the work is done on the system.

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13-75

13-98 A mixture of gases is assembled by filling an evacuated tank with neon, oxygen, and nitrogen added one after another. The mass of each constituent in the resulting mixture, the apparent molecular weight of the mixture, and the fraction of the tank volume occupied by nitrogen are to be determined. Properties The molar masses of Ne, O2, and N2 are 20.18, 32.0, 28.0 kg/kmol, respectively and the gas constants are 0.4119, 0.2598, and 0.2968 kJ/kg⋅K, respectively (Table A-1). Analysis The mass of each constituent is calculated by m Ne =

P NeV m (35 kPa)(0.15 m 3 ) = = 0.03828 kg R NeT (0.4119 kPa ⋅ m 3 /kg ⋅ K)(333 K)

m O2 =

P O2V m (70 kPa)(0.15 m 3 ) = = 0.1214 kg R O2T (0.2598 kPa ⋅ m 3 /kg ⋅ K)(333 K)

m N2 =

P N2V m (35 kPa)(0.15 m 3 ) = = 0.05312 kg R N2T (0.2968 kPa ⋅ m 3 /kg ⋅ K)(333 K)

35 kPa Ne 70 kPa O2 35 kPa N2 0.15 m3 60°C

The mole number of each constituent is N Ne =

m Ne 0.03828 kg = = 0.001896 kmol M Ne 20.18 kg/kmol

N O2 =

m O2 0.1214 kg = = 0.003794 kmol M O2 32.0 kg/kmol

N N2 =

m N2 0.05312 kg = = 0.001897 kmol M N2 28.0 kg/kmol

The apparent molecular weight of the mixture is Mm =

mm (0.03828 + 0.1214 + 0.05312) kg 0.2128 kg = = 28.05 kg/kmol = N m (0.001896 + 0.003794 + 0.001897) kmol 0.007586 kmol

The mole fraction of nitrogen is y N2 =

PN2 35 kPa = = 0.25 Pm 140 kPa

The partial volume occupied by nitrogen is then

V N2 = y N2V m = (0.25)(0.15 m 3 ) = 0.0375 m 3

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13-76

13-99 A portion of the gas in the previous problem is placed in a spring-loaded piston-cylinder device. The device is now heated until the pressure rises to a specified value. The total work and heat transfer for this process are to be determined. Properties The molar masses of Ne, O2, and N2 are 20.18, 32.0, 28.0 kg/kmol, respectively and the gas constants are 0.4119, 0.2598, and 0.2968 kJ/kg⋅K, respectively (Table A-1). The constant-volume specific volumes are 0.6179, 0.658, and 0.743 kJ/kg⋅K, respectively (Table A-2a). Analysis Using the data from the previous problem, the mass fractions are mf Ne =

m Ne 0.03828 kg = = 0.1799 mm 0.2128 kg

mf O2 =

m O2 0.1214 kg = = 0.5705 mm 0.2128 kg

mf N2 =

m N2 0.05312 kg = = 0.2496 mm 0.2128 kg

25% Ne 50% O2 25% N2 (by pressure) 0.1 m3 10°C, 200 kPa

The constant-volume specific heat of the mixture is determined from cv = mf Ne cv , Ne + mf O2 cv ,O2 + mf N2 cv , N2 = 0.1799 × 0.6179 + 0.5705 × 0.658 + 0.2496 × 0.743 = 0.672 kJ/kg ⋅ K

Ru 8.314 kJ/kmol ⋅ K = = 0.2964 kJ/kg ⋅ K Mm 28.05 kg/kmol

200

The mass contained in the system is m=

2

500

The apparent gas constant of the mixture is R=

P (kPa)

1

0.1

V (m3)

P 1V1 (200 kPa)(0.1 m ) = = 0.2384 kg RT1 (0.2964 kPa ⋅ m 3 /kg ⋅ K)(283 K) 3

Noting that the pressure changes linearly with volume, the final volume is determined by linear interpolation to be 500 − 200 V 2 − 0.1 = ⎯ ⎯→V 2 = 0.4375 m 3 1000 − 200 1.0 − 0.1

The final temperature is T2 =

P 2V 2 (500 kPa)(0.4375 m 3 ) = = 3096 K mR (0.2384 kg)(0.2964 kPa ⋅ m 3 /kg ⋅ K)

The work done during this process is Wout =

P1 + P 2 (500 + 200) kPa (V 2 −V 1 ) = (0.4375 − 0.1) m 3 = 118 kJ 2 2

An energy balance on the system gives Qin = Wout + mcv (T2 − T1 ) = 118 + (0.2384 kg)(0.672 kJ/kg ⋅ K )(3096 − 283) K = 569 kJ

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Q

13-77

13-100 A spring-loaded piston-cylinder device is filled with a mixture of nitrogen and carbon dioxide whose mass fractions are given. The gas is heated until the volume has doubled. The total work and heat transfer for this process are to be determined. Properties The molar masses of N2 and CO2 are 28.0 and 44.0 kg/kmol, respectively (Table A-1). The constant-volume specific heats of these gases at room temperature are 0.743 and 0.657 kJ/kg⋅K, respectively (Table A-2a). Analysis We consider 100 kg of this mixture. The mole numbers of each component are m 70 kg = 2.5 kmol N N2 = N2 = M N2 28 kg/kmol N CO2 =

m CO2 30 kg = = 0.6818 kmol M CO2 44 kg/kmol

The mole number of the mixture is N m = N N2 + N CO2 = 2.5 + 0.6818 = 3.1818 kmol

70% N2 30% CO2 (by mass) 0.1 m3 30°C, 400 kPa

The apparent molecular weight of the mixture is m 100 kg = 31.43 kg/kmol Mm = m = N m 3.1818 kmol

Q

The constant-volume specific heat of the mixture is determined from cv = mf N2 cv , N2 + mf CO2 cv ,CO2 = 0.70 × 0.743 + 0.30 × 0.657 = 0.717 kJ/kg ⋅ K The apparent gas constant of the mixture is R 8.134 kJ/kmol ⋅ K = 0.2645 kJ/kg ⋅ K R= u = 31.43 kg/kmol Mm

P (kPa)

Noting that the pressure changes linearly with volume, the initial volume is determined by linear interpolation using the data of the previous problem to be 400 − 200 V1 − 0.1 = ⎯ ⎯→V1 = 0.325 m 3 1000 − 200 1.0 − 0.1 The final volume is

2

400

1

V (m3)

V 2 = 2V 1 = 2(0.325 m 3 ) = 0.650 m 3 The final pressure is similarly determined by linear interpolation using the data of the previous problem to be P2 − 200 0.650 − 0.1 = ⎯ ⎯→ P2 = 689 kPa 1000 − 200 1.0 − 0.1 The mass contained in the system is m=

P 1V1 (400 kPa)(0.325 m 3 ) = = 1.622 kg RT1 (0.2645 kPa ⋅ m 3 /kg ⋅ K)(303 K)

The final temperature is T2 =

P 2V 2 (689 kPa)(0.650 m 3 ) = = 1044 K mR (1.622 kg)(0.2645 kPa ⋅ m 3 /kg ⋅ K)

The work done during this process is P +P (400 + 689) kPa (0.650 − 0.325) m 3 = 177 kJ Wout = 1 2 (V 2 −V1 ) = 2 2 An energy balance on the system gives Qin = Wout + mcv (T2 − T1 ) = 177 + (1.622 kg)(0.717 kJ/kg ⋅ K )(1044 − 303) K = 1039 kJ

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13-78

13-101 A spring-loaded piston-cylinder device is filled with a mixture of nitrogen and carbon dioxide whose mass fractions are given. The gas is heated until the pressure has tripled. The total work and heat transfer for this process are to be determined. Properties The molar masses of N2 and CO2 are 28.0 and 44.0 kg/kmol, respectively (Table A-1). The constant-volume specific heats of these gases at room temperature are 0.743 and 0.657 kJ/kg⋅K, respectively (Table A-2a). Analysis We consider 100 kg of this mixture. The mole numbers of each component are m 70 kg N N2 = N2 = = 2.5 kmol M N2 28 kg/kmol N CO2 =

m CO2 30 kg = = 0.6818 kmol M CO2 44 kg/kmol

The mole number of the mixture is N m = N N2 + N CO2 = 2.5 + 0.6818 = 3.1818 kmol

70% N2 30% CO2 (by mass) 0.1 m3 30°C, 400 kPa

The apparent molecular weight of the mixture is m 100 kg = 31.43 kg/kmol Mm = m = N m 3.1818 kmol

Q

The constant-volume specific heat of the mixture is determined from cv = mf N2 cv , N2 + mf CO2 cv ,CO2 = 0.70 × 0.743 + 0.30 × 0.657 = 0.717 kJ/kg ⋅ K The apparent gas constant of the mixture is R 8.134 kJ/kmol ⋅ K = 0.2645 kJ/kg ⋅ K R= u = 31.43 kg/kmol Mm Noting that the pressure changes linearly with volume, the initial volume is determined by linear interpolation using the data of the earlier problem to be 400 − 200 V1 − 0.1 = ⎯ ⎯→V1 = 0.325 m 3 1000 − 200 1.0 − 0.1 The final pressure is P2 = 3P1 = 3(400 kPa ) = 1200 kPa

P (kPa) 2

400

1

V (m3)

The final volume is similarly determined by linear interpolation using the data of the earlier problem to be 1200 − 200 V 2 − 0.1 = ⎯ ⎯→V 2 = 1.225 m 3 1000 − 200 1.0 − 0.1 The mass contained in the system is m=

P 1V1 (400 kPa)(0.325 m 3 ) = = 1.622 kg RT1 (0.2645 kPa ⋅ m 3 /kg ⋅ K)(303 K)

The final temperature is T2 =

P 2V 2 (1200 kPa)(1.225 m 3 ) = = 3426 K mR (1.622 kg)(0.2645 kPa ⋅ m 3 /kg ⋅ K)

The work done during this process is P +P (400 + 1200) kPa (1.225 − 0.325) m 3 = 720 kJ Wout = 1 2 (V 2 −V1 ) = 2 2 An energy balance on the system gives Qin = Wout + mcv (T2 − T1 ) = 720 + (1.622 kg)(0.717 kJ/kg ⋅ K )(3426 − 303) K = 4352 kJ

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13-79

13-102 The masses of components of a gas mixture are given. This mixture is expanded in an adiabatic, steady-flow turbine of specified isentropic efficiency. The second law efficiency and the exergy destruction during this expansion process are to be determined. Assumptions All gases will be modeled as ideal gases with constant specific heats. Properties The molar masses of O2, CO2, and He are 32.0, 44.0, and 4.0 kg/kmol, respectively (Table A-1). The constant-pressure specific heats of these gases at room temperature are 0.918, 0.846, and 5.1926 kJ/kg⋅K, respectively (Table A-2a). Analysis The total mass of the mixture is m m = m O2 + m CO2 + m He = 0.1 + 1 + 0.5 = 1.6 kg

1000 kPa 327°C

The mole numbers of each component are N O2 =

m O2 0.1 kg = = 0.003125 kmol M O2 32 kg/kmol

N CO2 =

m CO2 1 kg = = 0.02273 kmol M CO2 44 kg/kmol

N He =

m He 0.5 kg = = 0.125 kmol M He 4 kg/kmol

O2, CO2, He mixture

100 kPa

The mole number of the mixture is N m = N O2 + N CO2 + N He = 0.003125 + 0.02273 + 0.125 = 0.15086 kmol

The apparent molecular weight of the mixture is Mm =

mm 1.6 kg = = 10.61 kg/kmol N m 0.15086 kmol

The apparent gas constant of the mixture is R=

Ru 8.314 kJ/kmol ⋅ K = = 0.7836 kJ/kg ⋅ K 10.61 kg/kmol Mm

The mass fractions are mf O2 =

m O2 0.1 kg = = 0.0625 m m 1.6 kg

mf CO2 =

m CO2 1 kg = = 0.625 1.6 kg mm

mf He =

m He 0.5 kg = = 0.3125 m m 1.6 kg

The constant-pressure specific heat of the mixture is determined from c p = mf O2 c p ,O2 + mf CO2 c p ,CO2 + mf He c p ,He = 0.0625 × 0.918 + 0.625 × 0.846 + 0.3125 × 5.1926 = 2.209 kJ/kg ⋅ K

Then the constant-volume specific heat is cv = c p − R = 2.209 − 0.7836 = 1.425 kJ/kg ⋅ K

The specific heat ratio is

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

13-80

k=

cp cv

=

2.209 = 1.550 1.425

The temperature at the end of the expansion for the isentropic process is T2 s

⎛P = T1 ⎜⎜ 2 ⎝ P1

⎞ ⎟⎟ ⎠

( k −1) / k

⎛ 100 kPa ⎞ = (600 K )⎜ ⎟ ⎝ 1000 kPa ⎠

0.55/1.55

= 265 K

Using the definition of turbine isentropic efficiency, the actual outlet temperature is T2 = T1 − η turb (T1 − T2 s ) = (600 K ) − (0.90)(600 − 265) = 299 K

The entropy change of the gas mixture is s 2 − s1 = c p ln

T2 P 299 100 − R ln 2 = (2.209) ln − (0.7836) ln = 0.2658 kJ/kg ⋅ K T1 P1 600 1000

The actual work produced is wout = h1 − h2 = c p (T1 − T2 ) = (2.209 kJ/kg ⋅ K )(600 − 299) K = 665 kJ/kg

The reversible work output is wrev,out = h1 − h2 − T0 ( s1 − s 2 ) = 665 kJ/kg − (298 K )(−0.2658 kJ/kg ⋅ K ) = 744 kJ/kg

The second-law efficiency and the exergy destruction are then η II =

wout 665 = = 0.894 wrev,out 744

x dest = wrev,out − wout = 744 − 665 = 79 kJ/kg

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13-81

13-103 The masses, pressures, and temperatures of the constituents of a gas mixture in a tank are given. Heat is transferred to the tank. The final pressure of the mixture and the heat transfer are to be determined. Assumptions He is an ideal gas and O2 is a nonideal gas. Properties The molar masses of He and O2 are 4.0 and 32.0 kg/kmol. (Table A-1) Analysis (a) The number of moles of each gas is m He 4 kg = = 1 kmol M He 4.0 kg/kmol

N He =

mO2

N O2 =

M O2

=

4 kg He 8 kg O2

8 kg = 0.25 kmol 32 kg/kmol

170 K 7 MPa

N m = N He + N O 2 = 1 kmol + 0.25 kmol = 1.25 kmol

Q

Then the partial volume of each gas and the volume of the tank are He:

V He =

N He Ru T1 (1 kmol)(8.314 kPa ⋅ m 3 /kmol ⋅ K)(170 K) = = 0.202 m 3 7000 kPa Pm,1 Pm,1

PR1 =

O2: T R1

Pcr ,O 2 T1 = Tcr ,O 2

V O2 =

⎫ 7 = 1.38 ⎪ 5.08 ⎪ ⎬ Z 1 = 0.53 170 = = 1.10 ⎪ ⎪ 154.8 ⎭ =

ZN O 2 Ru T1 Pm,1

=

(Fig. A-15)

(0.53)(0.25 kmol)(8.314 kPa ⋅ m 3 /kg ⋅ K)(170 K) = 0.027 m 3 7000 kPa

V tank = V He + V O 2 = 0.202 m 3 + 0.027 m 3 = 0.229 m 3 The partial pressure of each gas and the total final pressure is He:

PHe,2 =

v R ,O 2

V tank

=

(1 kmol)(8.314 kPa ⋅ m 3 /kmol ⋅ K)(220 K) 0.229 m 3

= 7987 kPa

⎫ ⎪ Tcr,O 2 ⎪ ⎪ v O2 V m / N O2 ⎪ = = ⎬ PR = 0.39 Ru Tcr,O 2 / Pcr,O 2 Ru Tcr,O 2 / Pcr,O 2 ⎪ 3 ⎪ (0.229 m )/(0.25 kmol) ⎪ 3 . 616 = = ⎪⎭ (8.314 kPa ⋅ m 3 /kmol ⋅ K)(154.8 K)/(5080 kPa)

T R2 =

O2:

N He Ru T2

T2

=

220 = 1.42 154.8

(Fig. A-15)

PO 2 = (PR Pcr )O = (0.39 )(5080 kPa ) = 1981 kPa = 1.981 MPa 2

Pm,2 = PHe + PO 2 = 7.987 MPa + 1.981 MPa = 9.97 MPa

(b) We take both gases as the system. No work or mass crosses the system boundary, therefore this is a closed system with no work interactions. Then the energy balance for this closed system reduces to Ein − Eout = ΔEsystem Qin = ΔU = ΔU He + ΔU O 2

He:

ΔU He = mcv (Tm − T1 ) = (4 kg )(3.1156 kJ/kg ⋅ K )(220 − 170)K = 623.1 kJ

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13-82

O2: T R1 = 1.10 ⎫ Z = 2.2 PR1 = 1.38 ⎬⎭ h1 T R2 = 1.42 ⎫ ⎪ 9.97 ⎬ Z h = 1.2 PR2 = = 1.963 ⎪ 2 5.08 ⎭

(Fig. A-29)

h2 − h1 = Ru Tcr ( Z h1 − Z h2 ) + (h2 − h1 ) ideal = (8.314 kJ/kmol ⋅ K)(154.8 K)(2.2 − 1.2) + (6404 − 4949)kJ/kmol = 2742 kJ/kmol

Also, PHe,1 =

N He Ru T1

V tank

=

PO 2 ,1 = Pm,1 − PHe,1

(1 kmol)(8.314 kPa ⋅ m 3 /kg ⋅ K)(170 K)

0.229 m 3 = 7000 kPa − 6172 kPa = 828 kPa

= 6,172 kPa

Thus, ΔU O 2 = N O 2 (h2 − h1 ) − ( P2V 2 − P1V 1 ) = N O 2 (h2 − h1 ) − ( PO 2 ,2 − PO 2 ,1 )V tank = (0.25 kmol)(2742 kJ/kmol) − (1981 − 828)(0.229)kPa ⋅ m 3 = 421.5 kJ

Substituting, Qin = 623.1 kJ + 421.5 kJ = 1045 kJ

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13-83

13-104 A mixture of carbon dioxide and methane expands through a turbine. The power produced by the mixture is to be determined using ideal gas approximation and Kay’s rule. Assumptions The expansion process is reversible and adiabatic (isentropic). Properties The molar masses of CO2 and CH4 are 44.0 and 16.0 kg/kmol and respectively. The critical properties are 304.2 K, 7390 kPa for CO2 and 191.1 K and 4640 kPa for CH4 (Table A-1). Analysis The molar mass of the mixture is determined to be M m = y CO 2 M CO 2 + y CH 4 M CH 4 = (0.60)(44) + (0.40)(16) = 32.80 kg/kmol

The gas constant is 1600 K 800 kPa 10 L/s

R 8.314 kJ/kmol.K = 0.2533 kJ/kg.K R= u = 32.8 kg/kmol Mm

The mass fractions are mf CO 2 = y CO 2 mf CH 4 = y CH 4

M CO 2 Mm M CH 4 Mm

= (0.60)

44 kg/kmol = 0.8049 32.8 kg/kmol

= (0.40)

16 kg/kmol = 0.1951 32.8 kg/kmol

60% CO2 40% CH4

100 kPa

Ideal gas solution:

Using Dalton’s law to find partial pressures, the entropies at the initial state are determined from EES to be: T = 1600 K, P = (0.60 × 800) = 480 kPa ⎯ ⎯→ s CO 2 ,1 = 6.424 kJ/kg.K T = 1600 K, P = (0.40 × 800) = 320 kPa ⎯ ⎯→ s CH 4 ,1 = 17.188 kJ/kg.K

The final state entropies cannot be determined at this point since the final temperature is not known. However, for an isentropic process, the entropy change is zero and the final temperature may be determined from Δs total = mf CO 2 Δs CO 2 + mf CH 4 Δs CH 4 0 = mf CO 2 ( s CO 2 , 2 − s CO 2 ,1 ) + mf CH 4 ( s CH 4 , 2 − s CH 4 ,1 )

The solution is obtained using EES to be

T2 = 1243 K The initial and final enthalpies and the changes in enthalpy are (from EES) T1 = 1600 K ⎯ ⎯→

hCO 2 ,1 = −7408 kJ/kg u CH 4 ,1 = 747.4 kJ/kg

T2 = 1243 K ⎯ ⎯→

hCO 2 , 2 = −7877 kJ/kg u CH 4 , 2 = −1136 kJ/kg

Noting that the heat transfer is zero, an energy balance on the system gives Q& in − W& out = m& Δhm ⎯ ⎯→ W& out = − m& Δhm

where Δhm = mf CO 2 (hCO 2 ,2 − hCO 2 ,1 ) + mf CH 4 (hCH 4 , 2 − hCH 4 ,1 )

= (0.8049)[(−7877) − (−7408)] + (0.1951)[(−1136) − (747.4)] = −745.9 kJ/kg

The mass flow rate is m& =

Substituting,

P1V&1 (800 kPa)(0.010 m 3 /s) = = 0.01974 kg/s RT1 (0.2533 kJ/kg.K)(1600 K) W& out = m& Δhm = −(0.01974)(−745.9 kJ/kg) = 14.72 kW

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13-84

Kay’s rule solution:

The critical temperature and pressure of the mixture is Tcr = y CO 2 Tcr,CO 2 + y CH 4 Tcr, CH = (0.60)(304.2 K) + (0.40)(191.1 K) = 259.0 K 4

Pcr = y CO 2 Pcr,CO 2 + y CH 4 Pcr, CH = (0.60)(7390 kPa) + (0.40)(4640 kPa) = 6290 kPa 4

State 1 properties: T1 1600 K ⎫ = = 6.178 ⎪ Z 1 = 1.002 Tcr 259.0 K ⎪ ⎬ Z h1 = −0.01025 (from EES) P 800 kPa = 1 = = 0.127 ⎪ Z s1 = 0.0001277 ⎪⎭ Pcr 6290 kPa

T R1 = PR1

Δh1 = Z h1 RTcr = (−0.01025)(0.2533 kJ/kg.K)(259.0 K) = −0.6714 kJ/kg h1 = mf CO 2 hCO 2 ,1 + mf CH 4 hCH 4 ,1 − Δh1

= (0.8049)(−7408) + (0.1951)(747.1) − (−0.6714) = −5813 kJ/kg Δs1 = Z s1 R = (0.0001277)(0.2533 kJ/kg.K) = 0.00003234 kJ/kg.K s1 = mf CO 2 s CO 2 ,1 + mf CH 4 s CH 4 ,1 − Δs1

= (0.8049)(6.424) + (0.1951)(17.188) − (0.00003234) = 8.529 kJ/kg.K

The final state entropies cannot be determined at this point since the final temperature is not known. However, for an isentropic process, the entropy change is zero and the final temperature may be determined from Δs total = mf CO 2 Δs CO 2 + mf CH 4 Δs CH 4 0 = mf CO 2 ( s CO 2 , 2 − s CO 2 ,1 ) + mf CH 4 ( s CH 4 , 2 − s CH 4 ,1 )

The solution is obtained using EES to be

T2 = 1243 K The initial and final enthalpies and the changes in enthalpy are ⎫ ⎪ Z = −0.00007368 ⎪ h2 (from EES) ⎬ Z = 0.0001171 P 100 kPa = 2 = = 0.016 ⎪ s 2 ⎪⎭ Pcr 6290 kPa

TR 2 = PR 2

T2 1243 K = = 4.80 Tcr 259.0 K

Δh2 = Z h 2 RTcr = (−0.000007368)(0.2533 kJ/kg.K)(259.0 K) = −0.04828 kJ/kg h2 = mf CO 2 hCO 2 ,2 + mf CH 4 hCH 4 , 2 − Δh2

= (0.8049)(−7877) + (0.1951)(−1136) − (−0.4828) = −6559 kJ/kg

Noting that the heat transfer is zero, an energy balance on the system gives Q& in − W& out = m& Δhm ⎯ ⎯→ W& out = −m& (h2 − h1 )

where the mass flow rate is m& =

P1V&1 (800 kPa)(0.010 m 3 /s) = = 0.01970 kg/s Z1 RT1 (1.002)(0.2533 kJ/kg.K)(1600 K)

Substituting, W& out = −(0.01970 kg/s)[(−6559) − (−5813) kJ/kg ] = 14.71 kW

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

13-85

13-105 EES A program is to be written to determine the mole fractions of the components of a mixture of three gases with known molar masses when the mass fractions are given, and to determine the mass fractions of the components when the mole fractions are given. Also, the program is to be run for a sample case. Analysis The problem is solved using EES, and the solution is given below. Procedure Fractions(Type$,A$,B$,C$,A,B,C:mf_A,mf_B,mf_C,y_A,y_B,y_C) {If Type$ ('mass fraction' OR 'mole fraction' ) then Call ERROR('Type$ must be set equal to "mass fraction" or "mole fraction".') GOTO 10 endif} Sum = A+B+C If ABS(Sum - 1) > 0 then goto 20 MM_A = molarmass(A$) MM_B = molarmass(B$) MM_C = molarmass(C$) If Type$ = 'mass fraction' then mf_A = A mf_B = B mf_C = C sumM_mix = mf_A/MM_A+ mf_B/MM_B+ mf_C/MM_C y_A = mf_A/MM_A/sumM_mix y_B = mf_B/MM_B/sumM_mix y_C = mf_C/MM_C/sumM_mix GOTO 10 endif if Type$ = 'mole fraction' then y_A = A y_B = B y_C = C MM_mix = y_A*MM_A+ y_B*MM_B+ y_C*MM_C mf_A = y_A*MM_A/MM_mix mf_B = y_B*MM_B/MM_mix mf_C = y_C*MM_C/MM_mix GOTO 10 Endif Call ERROR('Type$ must be either mass fraction or mole fraction.') GOTO 10 20: Call ERROR('The sum of the mass or mole fractions must be 1') 10: END "Either the mole fraction y_i or the mass fraction mf_i may be given by setting the parameter Type$='mole fraction' when the mole fractions are given or Type$='mass fraction' is given" {Input Data in the Diagram Window} {Type$='mole fraction' A$ = 'N2' B$ = 'O2' C$ = 'Argon' A = 0.71 "When Type$='mole fraction' A, B, C are the mole fractions" B = 0.28 "When Type$='mass fraction' A, B, C are the mass fractions" C = 0.01} Call Fractions(Type$,A$,B$,C$,A,B,C:mf_A,mf_B,mf_C,y_A,y_B,y_C) SOLUTION A=0.71 C=0.01 mf_C=0.014 y_B=0.280

A$='N2' B=0.28 C$='Argon' mf_A=0.680 Type$='mole fraction' y_C=0.010

B$='O2' mf_B=0.306 y_A=0.710

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13-86

13-106 EES A program is to be written to determine the entropy change of a mixture of 3 ideal gases when the mole fractions and other properties of the constituent gases are given. Also, the program is to be run for a sample case. Analysis The problem is solved using EES, and the solution is given below. T1=300 [K] T2=600 [K] P1=100 [kPa] P2=500 [kPa] A$ = 'N2' B$ = 'O2' C$ = 'Argon' y_A = 0.71 y_B = 0.28 y_C = 0.01 MM_A = molarmass(A$) MM_B = molarmass(B$) MM_C = molarmass(C$) MM_mix = y_A*MM_A+ y_B*MM_B+ y_C*MM_C mf_A = y_A*MM_A/MM_mix mf_B = y_B*MM_B/MM_mix mf_C = y_C*MM_C/MM_mix DELTAs_mix=mf_A*(entropy(A$,T=T2,P=y_B*P2)entropy(A$,T=T1,P=y_A*P1))+mf_B*(entropy(B$,T=T2,P=y_B*P2)entropy(B$,T=T1,P=y_B*P1))+mf_C*(entropy(C$,T=T2,P=y_C*P2)entropy(C$,T=T1,P=y_C*P1)) SOLUTION A$='N2' B$='O2' C$='Argon' DELTAs_mix=12.41 [kJ/kg-K] mf_A=0.68 mf_B=0.3063 mf_C=0.01366 MM_A=28.01 [kg/kmol] MM_B=32 [kg/kmol] MM_C=39.95 [kg/kmol] MM_mix=29.25 [kJ/kmol] P1=100 [kPa] P2=500 [kPa] T1=300 [K] T2=600 [K] y_A=0.71 y_B=0.28 y_C=0.01

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

13-87

Fundamentals of Engineering (FE) Exam Problems

13-107 An ideal gas mixture whose apparent molar mass is 36 kg/kmol consists of nitrogen N2 and three other gases. If the mole fraction of nitrogen is 0.30, its mass fraction is

(a) 0.15

(b) 0.23

(c) 0.30

(d) 0.39

(e) 0.70

Answer (b) 0.23 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). M_mix=36 "kg/kmol" M_N2=28 "kg/kmol" y_N2=0.3 mf_N2=(M_N2/M_mix)*y_N2 "Some Wrong Solutions with Common Mistakes:" W1_mf = y_N2 "Taking mass fraction to be equal to mole fraction" W2_mf= y_N2*(M_mix/M_N2) "Using the molar mass ratio backwords" W3_mf= 1-mf_N2 "Taking the complement of the mass fraction"

13-108 An ideal gas mixture consists of 2 kmol of N2 and 6 kmol of CO2. The mass fraction of CO2 in the mixture is

(a) 0.175

(b) 0.250

(c) 0.500

(d) 0.750

(e) 0.825

Answer (e) 0.825 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). N1=2 "kmol" N2=6 "kmol" N_mix=N1+N2 MM1=28 "kg/kmol" MM2=44 "kg/kmol" m_mix=N1*MM1+N2*MM2 mf2=N2*MM2/m_mix "Some Wrong Solutions with Common Mistakes:" W1_mf = N2/N_mix "Using mole fraction" W2_mf = 1-mf2 "The wrong mass fraction"

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13-88

13-109 An ideal gas mixture consists of 2 kmol of N2 and 4 kmol of CO2. The apparent gas constant of the mixture is

(a) 0.215 kJ/kg⋅K (e) 1.24 kJ/kg⋅K

(b) 0.225 kJ/kg⋅K

(c) 0.243 kJ/kg⋅K

(d) 0.875 kJ/kg⋅K

Answer (a) 0.215 kJ/kg⋅K Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). Ru=8.314 "kJ/kmol.K" N1=2 "kmol" N2=4 "kmol" MM1=28 "kg/kmol" MM2=44 "kg/kmol" R1=Ru/MM1 R2=Ru/MM2 N_mix=N1+N2 y1=N1/N_mix y2=N2/N_mix MM_mix=y1*MM1+y2*MM2 R_mix=Ru/MM_mix "Some Wrong Solutions with Common Mistakes:" W1_Rmix =(R1+R2)/2 "Taking the arithmetic average of gas constants" W2_Rmix= y1*R1+y2*R2 "Using wrong relation for Rmixture"

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13-89

13-110 A rigid tank is divided into two compartments by a partition. One compartment contains 3 kmol of N2 at 600 kPa pressure and the other compartment contains 7 kmol of CO2 at 200 kPa. Now the partition is removed, and the two gases form a homogeneous mixture at 300 kPa. The partial pressure of N2 in the mixture is

(a) 75 kPa

(b) 90 kPa

(c) 150 kPa

(d) 175 kPa

(e) 225 kPa

Answer (b) 90 kPa Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1 = 600 "kPa" P2 = 200 "kPa" P_mix=300 "kPa" N1=3 "kmol" N2=7 "kmol" MM1=28 "kg/kmol" MM2=44 "kg/kmol" N_mix=N1+N2 y1=N1/N_mix y2=N2/N_mix P_N2=y1*P_mix "Some Wrong Solutions with Common Mistakes:" W1_P1= P_mix/2 "Assuming equal partial pressures" W2_P1= mf1*P_mix; mf1=N1*MM1/(N1*MM1+N2*MM2) "Using mass fractions" W3_P1 = P_mix*N1*P1/(N1*P1+N2*P2) "Using some kind of weighed averaging"

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13-90

13-111 An 80-L rigid tank contains an ideal gas mixture of 5 g of N2 and 5 g of CO2 at a specified pressure and temperature. If N2 were separated from the mixture and stored at mixture temperature and pressure, its volume would be

(a) 32 L

(b) 36 L

(c) 40 L

(d) 49 L

(e) 80 L

Answer (d) 49 L Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). V_mix=80 "L" m1=5 "g" m2=5 "g" MM1=28 "kg/kmol" MM2=44 "kg/kmol" N1=m1/MM1 N2=m2/MM2 N_mix=N1+N2 y1=N1/N_mix V1=y1*V_mix "L" "Some Wrong Solutions with Common Mistakes:" W1_V1=V_mix*m1/(m1+m2) "Using mass fractions" W2_V1= V_mix "Assuming the volume to be the mixture volume"

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

13-91

13-112 An ideal gas mixture consists of 3 kg of Ar and 6 kg of CO2 gases. The mixture is now heated at constant volume from 250 K to 350 K. The amount of heat transfer is

(a) 374 kJ

(b) 436 kJ

(c) 488 kJ

(d) 525 kJ

(e) 664 kJ

Answer (c) 488 kJ Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T1=250 "K" T2=350 "K" Cv1=0.3122; Cp1=0.5203 "kJ/kg.K" Cv2=0.657; Cp2=0.846 "kJ/kg.K" m1=3 "kg" m2=6 "kg" MM1=39.95 "kg/kmol" MM2=44 "kg/kmol" "Applying Energy balance gives Q=DeltaU=DeltaU_Ar+DeltaU_CO2" Q=(m1*Cv1+m2*Cv2)*(T2-T1) "Some Wrong Solutions with Common Mistakes:" W1_Q = (m1+m2)*(Cv1+Cv2)/2*(T2-T1) "Using arithmetic average of properties" W2_Q = (m1*Cp1+m2*Cp2)*(T2-T1)"Using Cp instead of Cv" W3_Q = (m1*Cv1+m2*Cv2)*T2 "Using T2 instead of T2-T1"

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

13-92

13-113 An ideal gas mixture consists of 30% helium and 70% argon gases by mass. The mixture is now expanded isentropically in a turbine from 400°C and 1.2 MPa to a pressure of 200 kPa. The mixture temperature at turbine exit is

(a) 195°C

(b) 56°C

(c) 112°C

(d) 130°C

(e) 400°C

Answer (b) 56°C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T1=400+273"K" P1=1200 "kPa" P2=200 "kPa" mf_He=0.3 mf_Ar=0.7 k1=1.667 k2=1.667 "The specific heat ratio k of the mixture is also 1.667 since k=1.667 for all componet gases" k_mix=1.667 T2=T1*(P2/P1)^((k_mix-1)/k_mix)-273 "Some Wrong Solutions with Common Mistakes:" W1_T2 = (T1-273)*(P2/P1)^((k_mix-1)/k_mix) "Using C for T1 instead of K" W2_T2 = T1*(P2/P1)^((k_air-1)/k_air)-273; k_air=1.4 "Using k value for air" W3_T2 = T1*P2/P1 "Assuming T to be proportional to P"

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

13-93

13-114 One compartment of an insulated rigid tank contains 2 kmol of CO2 at 20°C and 150 kPa while the other compartment contains 5 kmol of H2 gas at 35°C and 300 kPa. Now the partition between the two gases is removed, and the two gases form a homogeneous ideal gas mixture. The temperature of the mixture is

(a) 25°C

(b) 29°C

(c) 22°C

(d) 32°C

(e) 34°C

Answer (b) 29°C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). N_H2=5 "kmol" T1_H2=35 "C" P1_H2=300 "kPa" N_CO2=2 "kmol" T1_CO2=20 "C" P1_CO2=150 "kPa" Cv_H2=10.183; Cp_H2=14.307 "kJ/kg.K" Cv_CO2=0.657; Cp_CO2=0.846 "kJ/kg.K" MM_H2=2 "kg/kmol" MM_CO2=44 "kg/kmol" m_H2=N_H2*MM_H2 m_CO2=N_CO2*MM_CO2 "Applying Energy balance gives 0=DeltaU=DeltaU_H2+DeltaU_CO2" 0=m_H2*Cv_H2*(T2-T1_H2)+m_CO2*Cv_CO2*(T2-T1_CO2) "Some Wrong Solutions with Common Mistakes:" 0=m_H2*Cp_H2*(W1_T2-T1_H2)+m_CO2*Cp_CO2*(W1_T2-T1_CO2) "Using Cp instead of Cv" 0=N_H2*Cv_H2*(W2_T2-T1_H2)+N_CO2*Cv_CO2*(W2_T2-T1_CO2) "Using N instead of mass" W3_T2 = (T1_H2+T1_CO2)/2 "Assuming averate temperature"

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

13-94

13-115 A piston-cylinder device contains an ideal gas mixture of 3 kmol of He gas and 7 kmol of Ar gas at 50°C and 400 kPa. Now the gas expands at constant pressure until its volume doubles. The amount of heat transfer to the gas mixture is

(a) 6.2 MJ

(b) 42 MJ

(c) 27 MJ

(d) 10 MJ

(e) 67 MJ

Answer (e) 67 MJ Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). N_He=3 "kmol" N_Ar=7 "kmol" T1=50+273 "C" P1=400 "kPa" P2=P1 "T2=2T1 since PV/T=const for ideal gases and it is given that P=constant" T2=2*T1 "K" MM_He=4 "kg/kmol" MM_Ar=39.95 "kg/kmol" m_He=N_He*MM_He m_Ar=N_Ar*MM_Ar Cp_Ar=0.5203; Cv_Ar = 3122 "kJ/kg.C" Cp_He=5.1926; Cv_He = 3.1156 "kJ/kg.K" "For a P=const process, Q=DeltaH since DeltaU+Wb is DeltaH" Q=m_Ar*Cp_Ar*(T2-T1)+m_He*Cp_He*(T2-T1) "Some Wrong Solutions with Common Mistakes:" W1_Q =m_Ar*Cv_Ar*(T2-T1)+m_He*Cv_He*(T2-T1) "Using Cv instead of Cp" W2_Q=N_Ar*Cp_Ar*(T2-T1)+N_He*Cp_He*(T2-T1) "Using N instead of mass" W3_Q=m_Ar*Cp_Ar*(T22-T1)+m_He*Cp_He*(T22-T1); T22=2*(T1-273)+273 "Using C for T1" W4_Q=(m_Ar+m_He)*0.5*(Cp_Ar+Cp_He)*(T2-T1) "Using arithmetic averate of Cp"

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13-95

13-116 An ideal gas mixture of helium and argon gases with identical mass fractions enters a turbine at 1200 K and 1 MPa at a rate of 0.3 kg/s, and expands isentropically to 100 kPa. The power output of the turbine is

(a) 478 kW

(b) 619 kW

(c) 926 kW

(d) 729 kW

(e) 564 kW

Answer (b) 619 kW Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m=0.3 "kg/s" T1=1200 "K" P1=1000 "kPa" P2=100 "kPa" mf_He=0.5 mf_Ar=0.5 k_He=1.667 k_Ar=1.667 Cp_Ar=0.5203 Cp_He=5.1926 Cp_mix=mf_He*Cp_He+mf_Ar*Cp_Ar "The specific heat ratio k of the mixture is also 1.667 since k=1.667 for all componet gases" k_mix=1.667 T2=T1*(P2/P1)^((k_mix-1)/k_mix) -W_out=m*Cp_mix*(T2-T1) "Some Wrong Solutions with Common Mistakes:" W1_Wout= - m*Cp_mix*(T22-T1); T22 = (T1-273)*(P2/P1)^((k_mix-1)/k_mix)+273 "Using C for T1 instead of K" W2_Wout= - m*Cp_mix*(T222-T1); T222 = T1*(P2/P1)^((k_air-1)/k_air)-273; k_air=1.4 "Using k value for air" W3_Wout= - m*Cp_mix*(T2222-T1); T2222 = T1*P2/P1 "Assuming T to be proportional to P" W4_Wout= - m*0.5*(Cp_Ar+Cp_He)*(T2-T1) "Using arithmetic average for Cp"

13-117 … 13-119 Design and Essay Problem

KJ

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14-1

Chapter 14 GAS-VAPOR MIXTURES AND AIR CONDITIONING Dry and Atmospheric Air, Specific and Relative Humidity

14-1C Yes; by cooling the air at constant pressure.

14-2C Yes.

14-3C Specific humidity will decrease but relative humidity will increase.

14-4C Dry air does not contain any water vapor, but atmospheric air does.

14-5C Yes, the water vapor in the air can be treated as an ideal gas because of its very low partial pressure.

14-6C The partial pressure of the water vapor in atmospheric air is called vapor pressure.

14-7C The same. This is because water vapor behaves as an ideal gas at low pressures, and the enthalpy of an ideal gas depends on temperature only.

14-8C Specific humidity is the amount of water vapor present in a unit mass of dry air. Relative humidity is the ratio of the actual amount of vapor in the air at a given temperature to the maximum amount of vapor air can hold at that temperature.

14-9C The specific humidity will remain constant, but the relative humidity will decrease as the temperature rises in a well-sealed room.

14-10C The specific humidity will remain constant, but the relative humidity will decrease as the temperature drops in a well-sealed room.

14-11C A tank that contains moist air at 3 atm is located in moist air that is at 1 atm. The driving force for moisture transfer is the vapor pressure difference, and thus it is possible for the water vapor to flow into the tank from surroundings if the vapor pressure in the surroundings is greater than the vapor pressure in the tank.

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14-2

14-12C Insulations on chilled water lines are always wrapped with vapor barrier jackets to eliminate the possibility of vapor entering the insulation. This is because moisture that migrates through the insulation to the cold surface will condense and remain there indefinitely with no possibility of vaporizing and moving back to the outside.

14-13C When the temperature, total pressure, and the relative humidity are given, the vapor pressure can be determined from the psychrometric chart or the relation Pv = φPsat where Psat is the saturation (or boiling) pressure of water at the specified temperature and φ is the relative humidity.

14-14E Humid air is expanded in an isentropic nozzle. The amount of water vapor that has condensed during the process is to be determined. Assumptions The air and the water vapor are ideal gases. Properties The specific heat ratio of air at room temperature is k = 1.4 (Table A-2a). The saturation properties of water are to be obtained from water tables. Analysis Since the mole fraction of the water vapor in this mixture is very small, ⎛P T2 = T1 ⎜⎜ 2 ⎝ P1

⎞ ⎟⎟ ⎠

( k −1) / k

⎛ 15 psia ⎞ ⎟⎟ = (860 R )⎜⎜ ⎝ 100 psia ⎠

0.4/1.4

= 500 R

We will assume that the air leaves the nozzle at a relative humidity of 100% (will be verified later). The vapor pressure and specific humidity at the outlet are then

100 psia 400°F ω1=0.025

AIR

15 psia

Pv , 2 = φ 2 Pg , 2 = φ 2 Psat @ 40°F = (1.0)(0.12173 psia) = 0.1217 psia

ω2 =

0.622 Pv , 2 P − Pv , 2

=

(0.622)(0.1217 psia) = 0.00509 lbm H 2 O/lbm dry air (15 − 0.1217) psia

This is less than the inlet specific humidity (0.025 lbm/lbm dry air), the relative humidity at the outlet must be 100% as originally assumed. The amount of liquid formation is then Δω = ω1 − ω 2 = 0.025 − 0.00509 = 0.0199 lbm H 2 O/lbm dry air

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14-3

14-15 Humid air is compressed in an isentropic compressor. The relative humidity of the air at the compressor outlet is to be determined. Assumptions The air and the water vapor are ideal gases. Properties The specific heat ratio of air at room temperature is k = 1.4 (Table A-2a). The saturation properties of water are to be obtained from water tables. Analysis At the inlet, Pv ,1 = φ1 Pg ,1 = φ1 Psat @ 20°C = (0.90)(2.3392 kPa) = 2.105 kPa

ω 2 = ω1 =

0.622 Pv,1 P − Pv ,1

=

(0.622)(2.105 kPa) = 0.0134 kg H 2 O/kg dry air (100 − 2.105) kPa

800 kPa

Humid air

Since the mole fraction of the water vapor in this mixture is very small, ⎛P T2 = T1 ⎜⎜ 2 ⎝ P1

⎞ ⎟⎟ ⎠

( k −1) / k

⎛ 800 kPa ⎞ = (293 K )⎜ ⎟ ⎝ 100 kPa ⎠

0.4/1.4

= 531 K

The saturation pressure at this temperature is

100 kPa 20°C 90% RH

Pg , 2 = Psat @ 258°C = 4542 kPa (from EES)

The vapor pressure at the exit is Pv , 2 =

ω 2 P2 ω 2 + 0.622

=

(0.0134)(800) = 16.87 kPa 0.0134 + 0.622

The relative humidity at the exit is then

φ2 =

Pv ,2 Pg , 2

=

16.87 = 0.0037 = 0.37% 4542

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14-4

14-16 A tank contains dry air and water vapor at specified conditions. The specific humidity, the relative humidity, and the volume of the tank are to be determined. Assumptions The air and the water vapor are ideal gases. Analysis (a) The specific humidity can be determined form its definition,

ω=

mv 0.3 kg = = 0.0143 kg H 2 O/kg dry air ma 21 kg

(b) The saturation pressure of water at 30°C is Pg = Psat @ 30°C = 4.2469 kPa

21 kg dry air 0.3 kg H2O vapor 30°C 100 kPa

Then the relative humidity can be determined from

φ=

(0.0143)(100 kPa) ωP = = 52.9% (0.622 + ω ) Pg (0.622 + 0.0143)(4.2469 kPa)

(c) The volume of the tank can be determined from the ideal gas relation for the dry air, Pv = φPg = (0.529)(4.2469 kPa) = 2.245 kPa Pa = P − Pv = 100 − 2.245 = 97.755 kPa

V =

m a R a T (21 kg)(0.287 kJ/kg ⋅ K)(303 K) = = 18.7 m 3 97.755 kPa Pa

14-17 A tank contains dry air and water vapor at specified conditions. The specific humidity, the relative humidity, and the volume of the tank are to be determined. Assumptions The air and the water vapor are ideal gases. Analysis (a) The specific humidity can be determined form its definition,

ω=

mv 0.3 kg = = 0.0143 kg H 2 O/kg dry air 21 kg ma

(b) The saturation pressure of water at 24°C is Pg = Psat @24°C = 2.986 kPa

21 kg dry air 0.3 kg H2O vapor 24°C 100 kPa

Then the relative humidity can be determined from

φ=

(0.0143)(100 kPa) ωP = = 75.2% (0.622 + ω ) Pg (0.622 + 0.0143)2.986 kPa

(c) The volume of the tank can be determined from the ideal gas relation for the dry air, Pv = φPg = (0.752)(2.986 kPa) = 2.245 kPa Pa = P − Pv = 100 − 2.245 = 97.755 kPa

V =

m a R a T (21 kg)(0.287 kJ/kg ⋅ K)(297 K) = = 18.3 m 3 97.755 kPa Pa

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14-5

14-18 A room contains air at specified conditions and relative humidity. The partial pressure of air, the specific humidity, and the enthalpy per unit mass of dry air are to be determined. Assumptions The air and the water vapor are ideal gases. Analysis (a) The partial pressure of dry air can be determined from Pv = φPg = φPsat @ 20°C = (0.85)(2.3392 kPa) = 1.988 kPa Pa = P − Pv = 98 − 1.988 = 96.01 kPa

(b) The specific humidity of air is determined from

ω=

AIR 20°C 98 kPa 85% RH

0.622 Pv (0.622)(1.988 kPa) = = 0.0129 kg H 2 O/kg dry air (98 − 1.988) kPa P − Pv

(c) The enthalpy of air per unit mass of dry air is determined from h = ha + ωhv ≅ c p T + ωh g = (1.005 kJ/kg ⋅ °C)(20°C) + (0.0129)(2537.4 kJ/kg) = 52.78 kJ/kg dry air

14-19 A room contains air at specified conditions and relative humidity. The partial pressure of air, the specific humidity, and the enthalpy per unit mass of dry air are to be determined. Assumptions The air and the water vapor are ideal gases. Analysis (a) The partial pressure of dry air can be determined from Pv = φPg = φPsat @ 20°C = (0.85)(2.3392 kPa) = 1.988 kPa Pa = P − Pv = 85 − 1.988 = 83.01 kPa

(b) The specific humidity of air is determined from

ω=

AIR 20°C 85 kPa 85% RH

0.622 Pv (0.622)(1.988 kPa) = = 0.0149 kg H 2 O/kg dry air (85 − 1.988) kPa P − Pv

(c) The enthalpy of air per unit mass of dry air is determined from h = ha + ωhv ≅ c p T + ωh g = (1.005 kJ/kg ⋅ °C)(20°C) + (0.0149)(2537.4 kJ/kg) = 57.90 kJ/kg dry air

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14-6

14-20E A room contains air at specified conditions and relative humidity. The partial pressure of air, the specific humidity, and the enthalpy per unit mass of dry air are to be determined. Assumptions The air and the water vapor are ideal gases. Analysis (a) The partial pressure of dry air can be determined from Pv = φPg = φPsat @ 70° F = (0.85)(0.36334 psia) = 0.309 psia Pa = P − Pv = 14.6 − 0.309 = 14.291 psia

(b) The specific humidity of air is determined from

ω=

AIR 70°F 14.6 psia 85% RH

0.622 Pv (0.622)(0.309 psia) = = 0.0134 lbm H 2 O/lbm dry air (14.6 − 0.309) psia P − Pv

(c) The enthalpy of air per unit mass of dry air is determined from h = ha + ωhv ≅ c p T + ωh g = (0.24 Btu/lbm ⋅ °F)(70°F) + (0.0134)(1091.8 Btu/lbm) = 31.43 Btu/lbm dry air

14-21 The masses of dry air and the water vapor contained in a room at specified conditions and relative humidity are to be determined. Assumptions The air and the water vapor are ideal gases. Analysis The partial pressure of water vapor and dry air are determined to be Pv = φPg = φPsat @ 23°C = (0.50)(2.811 kPa) = 1.41 kPa Pa = P − Pv = 98 − 1.41 = 96.59 kPa

The masses are determined to be ma =

PaV (96.59 kPa)(240 m 3 ) = = 272.9 kg R a T (0.287 kPa ⋅ m 3 /kg ⋅ K)(296 K)

mv =

PvV (1.41 kPa)(240 m 3 ) = = 2.47 kg Rv T (0.4615 kPa ⋅ m 3 /kg ⋅ K)(296 K)

ROOM 240 m3 23°C 98 kPa 50% RH

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14-7

Dew-point, Adiabatic Saturation, and Wet-bulb Temperatures

14-22C Dew-point temperature is the temperature at which condensation begins when air is cooled at constant pressure.

14-23C Andy’s. The temperature of his glasses may be below the dew-point temperature of the room, causing condensation on the surface of the glasses.

14-24C The outer surface temperature of the glass may drop below the dew-point temperature of the surrounding air, causing the moisture in the vicinity of the glass to condense. After a while, the condensate may start dripping down because of gravity.

14-25C When the temperature falls below the dew-point temperature, dew forms on the outer surfaces of the car. If the temperature is below 0°C, the dew will freeze. At very low temperatures, the moisture in the air will freeze directly on the car windows.

14-26C When the air is saturated (100% relative humidity).

14-27C These two are approximately equal at atmospheric temperatures and pressure.

14-28 A house contains air at a specified temperature and relative humidity. It is to be determined whether any moisture will condense on the inner surfaces of the windows when the temperature of the window drops to a specified value. Assumptions The air and the water vapor are ideal gases. Analysis The vapor pressure Pv is uniform throughout the house, and its value can be determined from Pv = φPg @ 25°C = (0.65)(3.1698 kPa) = 2.06 kPa

The dew-point temperature of the air in the house is

25°C

φ = 65% 10°C

Tdp = Tsat @ Pv = Tsat @ 2.06 kPa = 18.0°C

That is, the moisture in the house air will start condensing when the temperature drops below 18.0°C. Since the windows are at a lower temperature than the dew-point temperature, some moisture will condense on the window surfaces.

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14-8

14-29 A person wearing glasses enters a warm room at a specified temperature and relative humidity from the cold outdoors. It is to be determined whether the glasses will get fogged. Assumptions The air and the water vapor are ideal gases. Analysis The vapor pressure Pv of the air in the house is uniform throughout, and its value can be determined from Pv = φPg @ 25°C = (0.40)(3.1698 kPa) = 1.268 kPa

The dew-point temperature of the air in the house is

25°C φ = 40%

8°C

Tdp = Tsat @ Pv = Tsat @ 1.268 kPa = 10.5°C (from EES)

That is, the moisture in the house air will start condensing when the air temperature drops below 10.5°C. Since the glasses are at a lower temperature than the dew-point temperature, some moisture will condense on the glasses, and thus they will get fogged.

14-30 A person wearing glasses enters a warm room at a specified temperature and relative humidity from the cold outdoors. It is to be determined whether the glasses will get fogged. Assumptions The air and the water vapor are ideal gases. Analysis The vapor pressure Pv of the air in the house is uniform throughout, and its value can be determined from Pv = φPg @ 25°C = (0.30)(3.1698 kPa) = 0.95 kPa

The dew-point temperature of the air in the house is

25°C φ = 30%

8°C

Tdp = Tsat @ Pv = Tsat @ 0.95 kPa = 6.2°C (from EES)

That is, the moisture in the house air will start condensing when the air temperature drops below 6.2°C. Since the glasses are at a higher temperature than the dew-point temperature, moisture will not condense on the glasses, and thus they will not get fogged.

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14-9

14-31E A woman drinks a cool canned soda in a room at a specified temperature and relative humidity. It is to be determined whether the can will sweat. Assumptions The air and the water vapor are ideal gases.

80°F 50% RH

Analysis The vapor pressure Pv of the air in the house is uniform throughout, and its value can be determined from Pv = φPg @ 80° F = (0.50)(0.50745 psia) = 0.254 psia

The dew-point temperature of the air in the house is

Cola 40°F

Tdp = Tsat @ Pv = Tsat @ 0.254 psia = 59.7°F (from EES)

That is, the moisture in the house air will start condensing when the air temperature drops below 59.7°C. Since the canned drink is at a lower temperature than the dew-point temperature, some moisture will condense on the can, and thus it will sweat.

14-32 The dry- and wet-bulb temperatures of atmospheric air at a specified pressure are given. The specific humidity, the relative humidity, and the enthalpy of air are to be determined. Assumptions The air and the water vapor are ideal gases. Analysis (a) We obtain the properties of water vapor from EES. The specific humidity ω1 is determined from c p (T2 − T1 ) + ω 2 h fg 2

ω1 =

h g1 − h f 2

where T2 is the wet-bulb temperature, and ω2 is determined from

ω2 =

0.622 Pg 2 P2 − Pg 2

=

95 kPa 25°C Twb = 17°C

(0.622)(1.938 kPa) = 0.01295 kg H 2 O/kg dry air (95 − 1.938) kPa

Thus,

ω1 =

(1.005 kJ/kg ⋅ °C)(17 − 25)°C + (0.01295)(2460.6 kJ/kg) = 0.00963 kg H 2 O/kg dry air (2546.5 − 71.36) kJ/kg

(b) The relative humidity φ1 is determined from

φ1 =

ω 1 P1 (0.00963)(95 kPa) = = 0.457 or 45.7% (0.622 + ω 1 ) Pg1 (0.622 + 0.00963)(3.1698 kPa)

(c) The enthalpy of air per unit mass of dry air is determined from h1 = ha1 + ω1 hv1 ≅ c p T1 + ω1 h g1 = (1.005 kJ/kg ⋅ °C)(25°C) + (0.00963)(2546.5 kJ/kg) = 49.65 kJ/kg dry air

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14-10

14-33 The dry- and wet-bulb temperatures of air in room at a specified pressure are given. The specific humidity, the relative humidity, and the dew-point temperature are to be determined. Assumptions The air and the water vapor are ideal gases. Analysis (a) We obtain the properties of water vapor from EES. The specific humidity ω1 is determined from c p (T2 − T1 ) + ω 2 h fg 2

ω1 =

h g1 − h f 2

where T2 is the wet-bulb temperature, and ω2 is determined from

ω2 =

0.622 Pg 2 P2 − Pg 2

=

100 kPa 22°C Twb = 16°C

(0.622)(1.819 kPa) = 0.01152 kg H 2 O/kg dry air (100 − 1.819) kPa

Thus,

ω1 =

(1.005 kJ/kg ⋅ °C)(16 − 22)°C + (0.01152)(2463.0 kJ/kg) = 0.00903 kg H 2O/kg dry air (2541.1 − 67.17) kJ/kg

(b) The relative humidity φ1 is determined from

φ1 =

ω1 P1 (0.00903)(100 kPa) = = 0.541 or 54.1% (0.622 + ω 1 ) Pg1 (0.622 + 0.0091)(2.6452 kPa)

(c) The vapor pressure at the inlet conditions is Pv1 = φ1 Pg1 = φ1 Psat @ 22°C = (0.541)(2.6452 kPa) = 1.432 kPa

Thus the dew-point temperature of the air is Tdp = Tsat @ Pv = Tsat @ 1.432 kPa = 12.3°C

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14-11

14-34 EES Problem 14-33 is reconsidered. The required properties are to be determined using EES at 100 and 300 kPa pressures. Analysis The problem is solved using EES, and the solution is given below. Tdb=22 [C] Twb=16 [C] P1=100 [kPa] P2=300 [kPa] h1=enthalpy(AirH2O;T=Tdb;P=P1;B=Twb) v1=volume(AirH2O;T=Tdb;P=P1;B=Twb) Tdp1=dewpoint(AirH2O;T=Tdb;P=P1;B=Twb) w1=humrat(AirH2O;T=Tdb;P=P1;B=Twb) Rh1=relhum(AirH2O;T=Tdb;P=P1;B=Twb) h2=enthalpy(AirH2O;T=Tdb;P=P2;B=Twb) v2=volume(AirH2O;T=Tdb;P=P2;B=Twb) Tdp2=dewpoint(AirH2O;T=Tdb;P=P2;B=Twb) w2=humrat(AirH2O;T=Tdb;P=P2;B=Twb) Rh2=relhum(AirH2O;T=Tdb;P=P2;B=Twb) SOLUTION h1=45.09 [kJ/kga] h2=25.54 [kJ/kga] P1=100 [kPa] P2=300 [kPa] Rh1=0.541 Rh2=0.243 Tdb=22 [C] Tdp1=12.3 [C] Tdp2=0.6964 [C] Twb=16 [C] v1=0.8595 [m^3/kga] v2=0.283 [m^3/kga] w1=0.009029 [kgv/kga] w2=0.001336 [kgv/kga]

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14-12

14-35E The dry- and wet-bulb temperatures of air in room at a specified pressure are given. The specific humidity, the relative humidity, and the dew-point temperature are to be determined. Assumptions The air and the water vapor are ideal gases. Analysis (a) The specific humidity ω1 is determined from

14.7 psia 80°F Twb = 65°F

c p (T2 − T1 ) + ω 2 h fg 2

ω1 =

h g1 − h f 2

where T2 is the wet-bulb temperature, and ω2 is determined from

ω2 =

0.622 Pg 2 P2 − Pg 2

=

(0.622)(0.30578 psia) = 0.01321 lbm H 2 O/lbm dry air (14.7 − 0.30578) psia

Thus,

ω1 =

(0.24 Btu/lbm ⋅ °F)(65 − 80)°F + (0.01321)(1056.5 Btu/lbm) = 0.00974 lbm H 2O/lbm dry air (1096.1 − 33.08) Btu/lbm

(b) The relative humidity φ1 is determined from

φ1 =

ω1 P1 (0.00974)(14.7 psia) = = 0.447 or 44.7% (0.622 + ω1 ) Pg1 (0.622 + 0.00974)(0.50745 psia)

(c) The vapor pressure at the inlet conditions is Pv1 = φ1 Pg1 = φ1 Psat @ 70°F = (0.447)(0.50745 psia) = 0.2268 psia

Thus the dew-point temperature of the air is Tdp = Tsat @ Pv = Tsat @ 0.2268 psia = 56.6°F

(from EES)

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14-13

14-36 Atmospheric air flows steadily into an adiabatic saturation device and leaves as a saturated vapor. The relative humidity and specific humidity of air are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (m& a1 = m& a 2 = m& a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis The exit state of the air is completely specified, and the total pressure is 98 kPa. The properties of the moist air at the exit state may be determined from EES to be

h2 = 78.11 kJ/kg dry air

ω 2 = 0.02079 kg H 2 O/kg dry air

Water 25°C

The enthalpy of makeup water is h w 2 = h f@ 25°C = 104.83 kJ/kg

Humidifier

(Table A - 4)

An energy balance on the control volume gives h1 + (ω 2 − ω1 )h w = h2

35°C 98 kPa

AIR

25°C 98 kPa 100%

h1 + (0.02079 − ω1 )(104.83 kJ/kg) = 78.11 kJ/kg Pressure and temperature are known for inlet air. Other properties may be determined from this equation using EES. A hand solution would require a trial-error approach. The results are h1 = 77.66 kJ/kg dry air

ω1 = 0.01654 kg H 2 O/kg dry air φ1 = 0.4511

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14-14

Psychrometric Chart

14-37C They are very nearly parallel to each other.

14-38C The saturation states (located on the saturation curve).

14-39C By drawing a horizontal line until it intersects with the saturation curve. The corresponding temperature is the dew-point temperature. 14-40C No, they cannot. The enthalpy of moist air depends on ω, which depends on the total pressure.

14-41 [Also solved by EES on enclosed CD] The pressure, temperature, and relative humidity of air in a room are specified. Using the psychrometric chart, the specific humidity, the enthalpy, the wet-bulb temperature, the dew-point temperature, and the specific volume of the air are to be determined. Analysis From the psychrometric chart (Fig. A-31) we read

(a) ω = 0.0181 kg H 2 O / kg dry air (b) h = 78.4 kJ / kg dry air (c) Twb = 25.5°C (d) Tdp = 23.3°C (e) v = 0.890 m 3 / kg dry air

14-42 The pressure, temperature, and relative humidity of air in a room are specified. Using the psychrometric chart, the specific humidity, the enthalpy, the wet-bulb temperature, the dew-point temperature, and the specific volume of the air are to be determined. Analysis From the psychrometric chart (Fig. A-31) we read

(a) ω = 0.0148 kg H 2 O / kg dry air (b) h = 63.9 kJ / kg dry air (c) Twb = 21.9°C (d) Tdp = 20.1°C (e) v = 0.868 m 3 / kg dry air

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14-15

14-43 EES Problem 14-42 is reconsidered. The required properties are to be determined using EES. Also, the properties are to be obtained at an altitude of 2000 m. Analysis The problem is solved using EES, and the solution is given below. Tdb=26 [C] Rh=0.70 P1=101.325 [kPa] Z = 2000 [m] P2=101.325*(1-0.02256*Z*convert(m,km))^5.256 "Relation giving P as a function of altitude" h1=enthalpy(AirH2O,T=Tdb,P=P1,R=Rh) v1=volume(AirH2O,T=Tdb,P=P1,R=Rh) Tdp1=dewpoint(AirH2O,T=Tdb,P=P1,R=Rh) w1=humrat(AirH2O,T=Tdb,P=P1,R=Rh) Twb1=wetbulb(AirH2O,T=Tdb,P=P1,R=Rh) h2=enthalpy(AirH2O,T=Tdb,P=P2,R=Rh) v2=volume(AirH2O,T=Tdb,P=P2,R=Rh) Tdp2=dewpoint(AirH2O,T=Tdb,P=P2,R=Rh) w2=humrat(AirH2O,T=Tdb,P=P2,R=Rh) Twb2=wetbulb(AirH2O,T=Tdb,P=P2,R=Rh) SOLUTION h1=63.88 [kJ/kg] h2=74.55 [kJ/kg] P1=101.3 [kPa] P2=79.49 [kPa] Rh=0.7 Tdb=26 [C] Tdp1=20.11 [C] Tdp2=20.11 [C] Twb1=21.87 [C] Twb2=21.59 [C] v1=0.8676 [m^3/kg] v2=1.113 [m^3/kg] w1=0.0148 [kg/kg] w2=0.01899 [kg/kg] Z=2000 [m]

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14-16

14-44 The pressure and the dry- and wet-bulb temperatures of air in a room are specified. Using the psychrometric chart, the specific humidity, the enthalpy, the relative humidity, the dew-point temperature, and the specific volume of the air are to be determined. Analysis From the psychrometric chart (Fig. A-31) we read

(a) ω = 0.0092 kg H 2 O / kg dry air (b) h = 47.6 kJ / kg dry air (c) φ = 49.6% (d) Tdp = 12.8°C (e) v = 0.855 m3 / kg dry air

14-45 EES Problem 14-44 is reconsidered. The required properties are to be determined using EES. Also, the properties are to be obtained at an altitude of 3000 m. Analysis The problem is solved using EES, and the solution is given below. Tdb=24 [C] Twb=17 [C] P1=101.325 [kPa] Z = 3000 [m] P2=101.325*(1-0.02256*Z*convert(m,km))^5.256 "Relation giving P as function of altitude" h1=enthalpy(AirH2O,T=Tdb,P=P1,B=Twb) v1=volume(AirH2O,T=Tdb,P=P1,B=Twb) Tdp1=dewpoint(AirH2O,T=Tdb,P=P1,B=Twb) w1=humrat(AirH2O,T=Tdb,P=P1,B=Twb) Rh1=relhum(AirH2O,T=Tdb,P=P1,B=Twb) h2=enthalpy(AirH2O,T=Tdb,P=P2,B=Twb) v2=volume(AirH2O,T=Tdb,P=P2,B=Twb) Tdp2=dewpoint(AirH2O,T=Tdb,P=P2,B=Twb) w2=humrat(AirH2O,T=Tdb,P=P2,B=Twb) Rh2=relhum(AirH2O,T=Tdb,P=P2,B=Twb) SOLUTION h1=47.61 [kJ/kg] P1=101.3 [kPa] Rh1=0.4956 Tdb=24 [C] Tdp2=14.24 [C] v1=0.8542 [m^3/kg] w1=0.009219 [kg/kg] Z=3000 [m]

h2=61.68 [kJ/kg] P2=70.11 [kPa] Rh2=0.5438 Tdp1=12.81 [C] Twb=17 [C] v2=1.245 [m^3/kg] w2=0.01475 [kg/kg]

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14-17

14-46 The pressure, temperature, and relative humidity of air are specified. Using the psychrometric chart, the wet-bulb temperature, specific humidity, the enthalpy, the dew-point temperature, and the water vapor pressure are to be determined. Analysis From the psychrometric chart in Fig. A-31 or using EES psychrometric functions we obtain

(a) Twb = 27.1°C (b) ω = 0.0217 kg H 2 O / kg dry air (c) h = 85.5 kJ/kg dry air (d) Tdp = 26.2°C

Air 1 atm 30°C 80% RH

(e) Pv = φPg = φPsat @ 30°C = (0.80)(4.2469 kPa) = 3.40 kPa

14-47E The pressure, temperature, and wet-bulb temperature of air are specified. Using the psychrometric chart, the relative humidity, specific humidity, the enthalpy, the dew-point temperature, and the water vapor pressure are to be determined. Analysis From the psychrometric chart in Fig. A-31 or using EES psychrometric functions we obtain

(a) φ = 0.816 = 81.6% (b) ω = 0.0252 lbm H 2 O / lbm dry air (c) h = 49.4 Btu/lbm dry air (d) Tdp = 83.7°F

Air 1 atm 90°F Twb=85°F

(e) Pv = φPg = φPsat @ 90°F = (0.816)(0.69904 psia) = 0.570 psia

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14-18

14-48E The pressure, temperature, and wet-bulb temperature of air are specified. The adiabatic saturation temperature is to be determined. Analysis For an adiabatic saturation process, we obtained Eq. 14-14 in the text,

ω1 =

c p (T2 − T1 ) + ω 2 h fg 2

Water

h g1 − h f 2

This requires a trial-error solution for the adiabatic saturation temperature, T2. The inlet state properties are

ω1 = 0.0252 lbm H 2 O / lbm dry air

Humidifier 1 atm 90°F Twb=85°F

AIR 100%

h g1 = h g @ 90°F = 1100.4 Btu/lbm

As a first estimate, let us take T2 =85°F (the inlet wet-bulb temperature). Also, at the exit, the relative humidity is 100% ( φ 2 = 1 ) and the pressure is 1 atm. Other properties at the exit state are

ω 2 = 0.0264 lbm H 2 O / lbm dry air h f 2 = h f @ 85°F = 53.06 Btu/lbm (Table A - 4E) h fg 2 = h fg @ 85°F = 1045.2 Btu/lbm (Table A - 4E)

Substituting,

ω1 =

c p (T2 − T1 ) + ω 2 h fg 2 h g1 − h f 2

=

(0.240)(85 − 90) + (0.0264)(1045.2) = 0.0252 lbm H 2 O / lbm dry air 1100.4 − 53.06

which is equal to the inlet specific humidity. Therefore, the adiabatic saturation temperature is T2 = 85°F Discussion This result is not surprising since the wet-bulb and adiabatic saturation temperatures are approximately equal to each other for air-water mixtures at atmospheric pressure.

14-49 The pressure, temperature, and wet-bulb temperature of air are specified. Using the psychrometric chart, the relative humidity, specific humidity, the enthalpy, the dew-point temperature, and the water vapor pressure are to be determined. Analysis From the psychrometric chart in Fig. A-31 or using EES psychrometric functions we obtain

(a) φ = 0.618 = 61.8% (b) ω = 0.0148 kg H 2 O / kg dry air (c) h = 65.8 kJ/kg dry air (d) Twb = 22.4°C

Air 1 atm 28°C Tdp=20°C

(e) Pv = φPg = φPsat @ 28°C = (0.618)(3.780 kPa) = 2.34 kPa

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14-19

14-50 The pressure, temperature, and wet-bulb temperature of air are specified. The adiabatic saturation temperature is to be determined. Analysis For an adiabatic saturation process, we obtained Eq. 14-14 in the text,

ω1 =

c p (T2 − T1 ) + ω 2 h fg 2

Water

h g1 − h f 2

This requires a trial-error solution for the adiabatic saturation temperature, T2. The inlet state properties are

ω1 = 0.0148 kg H 2 O / kg dry air (Fig. A-31)

Humidifier 1 atm 28°C Tdp=20°C

AIR 100%

h g1 = h g @ 28°C = 2551.9 kJ/kg (Table A-4)

As a first estimate, let us take T2 =22°C (the inlet wet-bulb temperature). Also, at the exit, the relative humidity is 100% ( φ 2 = 1 ) and the pressure is 1 atm. Other properties at the exit state are

ω 2 = 0.0167 kg H 2 O / kg dry air h f 2 = h f @ 22°C = 92.28 kJ/kg (Table A - 4) h fg 2 = h fg @ 22°C = 2448.8 kJ/kg (Table A - 4)

Substituting,

ω1 =

c p (T2 − T1 ) + ω 2 h fg 2 h g1 − h f 2

=

(1.005)(22 − 28) + (0.0167)(2448.8) = 0.0142 kg H 2 O / kg dry air 2551.9 − 92.28

which is sufficiently close to the inlet specific humidity (0.0148). Therefore, the adiabatic saturation temperature is T2 ≅ 22°C Discussion This result is not surprising since the wet-bulb and adiabatic saturation temperatures are approximately equal to each other for air-water mixtures at atmospheric pressure.

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14-20

Human Comfort and Air-Conditioning

14-51C It humidifies, dehumidifies, cleans and even deodorizes the air.

14-52C (a) Perspires more, (b) cuts the blood circulation near the skin, and (c) sweats excessively.

14-53C It is the direct heat exchange between the body and the surrounding surfaces. It can make a person feel chilly in winter, and hot in summer.

14-54C It affects by removing the warm, moist air that builds up around the body and replacing it with fresh air.

14-55C The spectators. Because they have a lower level of activity, and thus a lower level of heat generation within their bodies.

14-56C Because they have a large skin area to volume ratio. That is, they have a smaller volume to generate heat but a larger area to lose it from.

14-57C It affects a body’s ability to perspire, and thus the amount of heat a body can dissipate through evaporation.

14-58C Humidification is to add moisture into an environment, dehumidification is to remove it.

14-59C The metabolism refers to the burning of foods such as carbohydrates, fat, and protein in order to perform the necessary bodily functions. The metabolic rate for an average man ranges from 108 W while reading, writing, typing, or listening to a lecture in a classroom in a seated position to 1250 W at age 20 (730 at age 70) during strenuous exercise. The corresponding rates for women are about 30 percent lower. Maximum metabolic rates of trained athletes can exceed 2000 W. We are interested in metabolic rate of the occupants of a building when we deal with heating and air conditioning because the metabolic rate represents the rate at which a body generates heat and dissipates it to the room. This body heat contributes to the heating in winter, but it adds to the cooling load of the building in summer.

14-60C The metabolic rate is proportional to the size of the body, and the metabolic rate of women, in general, is lower than that of men because of their smaller size. Clothing serves as insulation, and the thicker the clothing, the lower the environmental temperature that feels comfortable.

14-61C Sensible heat is the energy associated with a temperature change. The sensible heat loss from a human body increases as (a) the skin temperature increases, (b) the environment temperature decreases, and (c) the air motion (and thus the convection heat transfer coefficient) increases.

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14-21

14-62C Latent heat is the energy released as water vapor condenses on cold surfaces, or the energy absorbed from a warm surface as liquid water evaporates. The latent heat loss from a human body increases as (a) the skin wetness increases and (b) the relative humidity of the environment decreases. The rate of evaporation from the body is related to the rate of latent heat loss by Q& latent = m& vapor h fg where hfg is the latent heat of vaporization of water at the skin temperature.

14-63 An average person produces 0.25 kg of moisture while taking a shower. The contribution of showers of a family of four to the latent heat load of the air-conditioner per day is to be determined. Assumptions All the water vapor from the shower is condensed by the air-conditioning system. Properties The latent heat of vaporization of water is given to be 2450 kJ/kg. Analysis The amount of moisture produced per day is m& vapor = ( Moisture produced per person)(No. of persons) = (0.25 kg / person)(4 persons / day) = 1 kg / day

Then the latent heat load due to showers becomes Q& latent = m& vapor h fg = (1 kg / day)(2450 kJ / kg) = 2450 kJ / day

14-64 There are 100 chickens in a breeding room. The rate of total heat generation and the rate of moisture production in the room are to be determined. Assumptions All the moisture from the chickens is condensed by the air-conditioning system. Properties The latent heat of vaporization of water is given to be 2430 kJ/kg. The average metabolic rate of chicken during normal activity is 10.2 W (3.78 W sensible and 6.42 W latent). Analysis The total rate of heat generation of the chickens in the breeding room is Q& gen, total = q& gen, total (No. of chickens) = (10.2 W / chicken)(100 chickens) = 1020 W

The latent heat generated by the chicken and the rate of moisture production are Q& gen, latent = q& gen, latent (No. of chickens) = (6.42 W/chicken)(100 chickens) = 642 W = 0.642 kW m& moisture =

Q& gen, latent h fg

=

0.642 kJ / s = 0.000264 kg / s = 0.264 g / s 2430 kJ / kg

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14-22

14-65 A department store expects to have a specified number of people at peak times in summer. The contribution of people to the sensible, latent, and total cooling load of the store is to be determined. Assumptions There is a mix of men, women, and children in the classroom. Properties The average rate of heat generation from people doing light work is 115 W, and 70% of is in sensible form (see Sec. 14-6). Analysis The contribution of people to the sensible, latent, and total cooling load of the store are Q& people, total = (No. of people) × Q& person, total = 135 × (115 W) = 15,525 W Q& people, sensible = (No. of people) × Q& person, sensible = 135 × (0.7 × 115 W) = 10,868 W Q& people, latent = (No. of people) × Q& person, latent = 135 × (0.3 × 115 W) = 4658 W

14-66E There are a specified number of people in a movie theater in winter. It is to be determined if the theater needs to be heated or cooled. Assumptions There is a mix of men, women, and children in the classroom. Properties The average rate of heat generation from people in a movie theater is 105 W, and 70 W of it is in sensible form and 35 W in latent form. Analysis Noting that only the sensible heat from a person contributes to the heating load of a building, the contribution of people to the heating of the building is Q& people, sensible = (No. of people) × Q& person, sensible = 500 × (70 W) = 35,000 W = 119,420 Btu/h

since 1 W = 3.412 Btu/h. The building needs to be heated since the heat gain from people is less than the rate of heat loss of 130,000 Btu/h from the building.

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14-23

14-67 The infiltration rate of a building is estimated to be 1.2 ACH. The sensible, latent, and total infiltration heat loads of the building at sea level are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air infiltrates at the outdoor conditions, and exfiltrates at the indoor conditions. 3 Excess moisture condenses at room temperature of 24°C. 4 The effect of water vapor on air density is negligible. Properties The gas constant and the specific heat of air are R = 0.287 kPa.m3/kg.K and cp = 1.005 kJ/kg⋅°C (Table A-2). The heat of vaporization of water at 24°C is h fg = h fg @ 24°C = 2444.1 kJ/kg (Table A-4). The

properties of the ambient and room air are determined from the psychrometric chart (Fig. A-31) to be Tambient = 32º C⎫ = 0.0150 kg/kg dryair ⎬w φ ambient = 50% ⎭ ambient Troom = 24º C⎫ ⎬ wroom = 0.0093 kg/kg dryair

φ room = 50% ⎭

Analysis Noting that the infiltration of ambient air will cause the air in the cold storage room to be changed 1.2 times every hour, the air will enter the room at a mass flow rate of

ρ ambient =

P0 101.325 kPa = = 1.158 kg/m 3 RT0 (0.287 kPa.m 3 /kg.K)(32 + 273 K)

m& air = ρ ambientV room ACH = (1.158 kg/m 3 )(20 × 13 × 3 m 3 )(1.2 h -1 ) = 1084 kg/h = 0.301 kg/s

Then the sensible, latent, and total infiltration heat loads of the room are determined to be Q& infiltration, sensible = m& air c p (Tambient − Troom ) = (0.301 kg/s)(1.005 kJ/kg.°C)(32 − 24)°C = 2.42 kW Q& infiltration, latent = m& air ( wambient − wroom )h fg = (0.301 kg/s)(0.0150 − 0.0093)(2444.1 kJ/kg) = 4.16 kW Q& infiltration, total = Q& infiltration, sensible + Q& infiltration, latent = 2.42 + 4.16 = 6.58 kW

Discussion The specific volume of the dry air at the ambient conditions could also be determined from the psychrometric chart at ambient conditions.

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14-24

14-68 The infiltration rate of a building is estimated to be 1.8 ACH. The sensible, latent, and total infiltration heat loads of the building at sea level are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air infiltrates at the outdoor conditions, and exfiltrates at the indoor conditions. 3 Excess moisture condenses at room temperature of 24°C. 4 The effect of water vapor on air density is negligible. Properties The gas constant and the specific heat of air are R = 0.287 kPa.m3/kg.K and cp = 1.005 kJ/kg⋅°C (Table A-2). The heat of vaporization of water at 24°C is h fg = h fg @ 24°C = 2444.1 kJ/kg (Table A-4). The

properties of the ambient and room air are determined from the psychrometric chart (Fig. A-31) to be Tambient = 32º C⎫ = 0.0150 kg/kg dryair ⎬w φ ambient = 50% ⎭ ambient Troom = 24º C⎫ ⎬ wroom = 0.0093 kg/kg dryair

φ room = 50% ⎭

Analysis Noting that the infiltration of ambient air will cause the air in the cold storage room to be changed 1.8 times every hour, the air will enter the room at a mass flow rate of

ρ ambient =

P0 101.325 kPa = = 1.158 kg/m 3 RT0 (0.287 kPa.m 3 /kg.K)(32 + 273 K)

m& air = ρ ambientV room ACH = (1.158 kg/m 3 )(20 × 13 × 3 m 3 )(1.8 h -1 ) = 1084 kg/h = 0.4514 kg/s

Then the sensible, latent, and total infiltration heat loads of the room are determined to be Q& infiltration, sensible = m& air c p (Tambient − Troom ) = (0.4514 kg/s)(1.005 kJ/kg.°C)(32 − 24)°C = 3.63 kW Q& infiltration, latent = m& air ( wambient − wroom )h fg = (0.4514 kg/s)(0.0150 − 0.0093)(2444.1 kJ/kg) = 6.24 kW Q& infiltration, total = Q& infiltration, sensible + Q& infiltration, latent = 3.63 + 6.24 = 9.87 kW

Discussion The specific volume of the dry air at the ambient conditions could also be determined from the psychrometric chart at ambient conditions.

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14-25

Simple Heating and cooling

14-69C Relative humidity decreases during a simple heating process and increases during a simple cooling process. Specific humidity, on the other hand, remains constant in both cases. 14-70C Because a horizontal line on the psychrometric chart represents a ω = constant process, and the moisture content ω of air remains constant during these processes.

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14-26

14-71 Air enters a cooling section at a specified pressure, temperature, velocity, and relative humidity. The exit temperature, the exit relative humidity of the air, and the exit velocity are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (m& a1 = m& a 2 = m& a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis (a) The amount of moisture in the air remains constant (ω 1 = ω 2) as it flows through the cooling section since the process involves no humidification or dehumidification. The inlet state of the air is completely specified, and the total pressure is 1 atm. The properties of the air at the inlet state are determined from the psychrometric chart (Figure A-31) to be h1 = 55.0 kJ/kg dry air

ω1 = 0.0089 kg H 2 O/kg dry air (= ω 2 )

1200

v 1 = 0.877 m 3 / kg dry air The mass flow rate of dry air through the cooling section is m& a =

1

v1

1

32°C 30% 18 m/s

V1 A1 1

=

3

2 1 atm

AIR

(18 m/s)(π × 0.4 2 /4 m 2 )

(0.877 m / kg) = 2.58 kg/s

From the energy balance on air in the cooling section, & a ( h2 − h1 ) − Q& out = m −1200 / 60 kJ / s = (2.58 kg / s)( h2 − 55.0) kJ / kg h2 = 47.2 kJ / kg dry air

The exit state of the air is fixed now since we know both h2 and ω2. From the psychrometric chart at this state we read T2 = 24.4°C

(b)

φ 2 = 46.6% v 2 = 0.856 m 3 / kg dry air

(c) The exit velocity is determined from the conservation of mass of dry air, m& a1 = m& a 2 ⎯ ⎯→ V2 =

V A V A V&1 V&2 = ⎯ ⎯→ 1 = 2 v1 v 2 v1 v2

v2 0.856 V1 = (18 m/s) = 17.6 m/s v1 0.877

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14-27

14-72 Air enters a cooling section at a specified pressure, temperature, velocity, and relative humidity. The exit temperature, the exit relative humidity of the air, and the exit velocity are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (m& a1 = m& a 2 = m& a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis (a) The amount of moisture in the air remains constant (ω 1 = ω 2) as it flows through the cooling section since the process involves no humidification or dehumidification. The inlet state of the air is completely specified, and the total pressure is 1 atm. The properties of the air at the inlet state are determined from the psychrometric chart (Figure A-31) to be h1 = 55.0 kJ/kg dry air

ω1 = 0.0089 kg H 2 O/kg dry air (= ω 2 )

800 kJ/min

v 1 = 0.877 m 3 / kg dry air The mass flow rate of dry air through the cooling section is m& a =

1

v1

V1 A1 1

=

3

1

32°C 30% 18 m/s

2 1 atm

AIR

(18 m/s)(π × 0.4 2 /4 m 2 )

(0.877 m / kg) = 2.58 kg/s

From the energy balance on air in the cooling section, & a ( h2 − h1 ) − Q& out = m −800 / 60 kJ / s = (2.58 kg / s)( h2 − 55.0) kJ / kg h2 = 49.8 kJ / kg dry air

The exit state of the air is fixed now since we know both h2 and ω2. From the psychrometric chart at this state we read T2 = 26.9°C

(b)

φ 2 = 40.0% v 2 = 0.862 m 3 / kg dry air

(c) The exit velocity is determined from the conservation of mass of dry air, m& a1 = m& a 2 ⎯ ⎯→ V2 =

V A V A V&1 V&2 = ⎯ ⎯→ 1 = 2 v1 v 2 v1 v2

v2 0.862 V1 = (18 m/s) = 17.7 m/s v1 0.877

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14-28

14-73E Humid air at a specified state is cooled at constant pressure to the dew-point temperature. The cooling required for this process is to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (m& a1 = m& a 2 = m& a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis The amount of moisture in the air remains constant (ω 1 = ω 2) as it flows through the cooling section since the process involves no humidification or dehumidification. The inlet and exit states of the air are completely specified, and the total pressure is 1 atm. The properties of the air at the inlet state are determined from the psychrometric chart (Figure A-31E) to be h1 = 56.7 Btu/lbm dry air

ω1 = 0.0296 lbm H 2 O/lbm dry air (= ω 2 ) Tdp,1 = 88.4°F

The exit state enthalpy is P = 1 atm T2 = Tdp,1 = 88.4°F φ1 = 1

1 ⎫ ⎪ ⎬ h2 = 53.8 Btu/lbm dry air ⎪ ⎭

100°F 70% RH

1 atm

100% RH 2 AIR

From the energy balance on air in the cooling section, q out = h1 − h2 = 56.7 − 53.8 = 2.9 Btu/lbm dry air

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14-29

14-74 Humid air at a specified state is cooled at constant pressure to the dew-point temperature. The cooling required for this process is to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (m& a1 = m& a 2 = m& a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis The amount of moisture in the air remains constant (ω 1 = ω 2) as it flows through the cooling section since the process involves no humidification or dehumidification. The inlet state of the air is completely specified, and the total pressure is 150 kPa. The properties of the air at the inlet and exit states are determined to be Pv1 = φ1 Pg1 = φ1 Psat @ 40°C = (0.70)(7.3851 kPa) = 5.1696 kPa h g1 = h g @ 40°C = 2573.5 kJ/kg

ω1 =

0.622 Pv1 0.622(5.1696 kPa) = = 0.02220 kg H 2 O/kg dry air (150 − 5.1696) kPa P1 − Pv1

h1 = c p T1 + ω1 h g1 = (1.005 kJ/kg ⋅ °C)(40°C) + (0.02220)(2573.5 kJ/kg) = 97.33 kJ/kg dry air

Pv 2 = Pv1 = 5.1696 kPa Pg 2 =

Pv 2

φ2

=

5.1696 kPa = 5.1696 kPa 1

T2 = Tsat @ 5.1695 kPa = 33.5°C

1

40°C 70% RH

150 kPa

100% RH 2 AIR

h g 2 = h g @ 33.5°C = 2561.9 kJ/kg

ω 2 = ω1 h2 = c p T2 + ω 2 h g 2 = (1.005 kJ/kg ⋅ °C)(33.5°C) + (0.02220)(2561.9 kJ/kg) = 90.55 kJ/kg dry air

From the energy balance on air in the cooling section, q out = h1 − h2 = 97.33 − 90.55 = 6.78 kJ/kg dry air

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14-30

14-75 Saturated humid air at a specified state is heated to a specified temperature. The relative humidity at the exit and the rate of heat transfer are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (m& a1 = m& a 2 = m& a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis The amount of moisture in the air remains constant (ω 1 = ω 2) as it flows through the heating section since the process involves no humidification or dehumidification. The inlet state of the air is completely specified, and the total pressure is 200 kPa. The properties of the air at the inlet and exit states are determined to be Pv1 = φ1 Pg1 = φ1 Psat @ 15°C = (1.0)(1.7057 kPa) = 1.7057 kPa h g1 = h g @ 15°C = 2528.3 kJ/kg

Heating coils

Pa1 = P1 − Pv1 = 200 − 1.7057 = 198.29 kPa

v1 = =

R a T1 Pa1

1

(0.287 kPa ⋅ m 3 / kg ⋅ K)(288 K) 198.29 kPa

15°C 100% RH 20 m/s

200 kPa AIR

30°C 2

= 0.4168 m 3 / kg dry air

ω1 =

0.622 Pv1 0.622(1.7057 kPa) = = 0.005350 kg H 2 O/kg dry air P1 − Pv1 (200 − 1.7057) kPa

h1 = c p T1 + ω1 h g1 = (1.005 kJ/kg ⋅ °C)(15°C) + (0.005350)(2528.3 kJ/kg) = 28.60 kJ/kg dry air Pv 2 = Pv1 = 1.7057 kPa Pg 2 = Psat @ 30°C = 4.2469 kPa

φ2 =

Pv 2 1.7057 kPa = = 0.402 = 40.2% Pg 2 4.2469 kPa

h g 2 = h g @ 30°C = 2555.6 kJ/kg

ω 2 = ω1 h2 = c p T2 + ω 2 h g 2 = (1.005 kJ/kg ⋅ °C)(30°C) + (0.005350)(2555.6 kJ/kg) = 43.82 kJ/kg dry air

Then,

V&1 = V1 A1 = V1 m& a =

π D2 4

⎛ π (0.04 m) 2 = (20 m/s)⎜ ⎜ 4 ⎝

⎞ ⎟ = 0.02513 m 3 /s ⎟ ⎠

V&1 0.02513 m 3 / s = = 0.06029 kg/s v 1 0.4168 m 3 / kg dry air

From the energy balance on air in the heating section, Q& in = m& a (h2 − h1 ) = (0.06029 kg/s)(43.82 − 28.60)kJ/kg = 0.918 kW

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14-31

14-76 Saturated humid air at a specified state is heated to a specified temperature. The rate at which the exergy of the humid air is increased is to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (m& a1 = m& a 2 = m& a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis The amount of moisture in the air remains constant (ω 1 = ω 2) as it flows through the heating section since the process involves no humidification or dehumidification. The inlet state of the air is completely specified, and the total pressure is 200 kPa. The properties of the air at the inlet and exit states are determined to be Pv1 = φ1 Pg1 = φ1 Psat @ 15°C = (1.0)(1.7057 kPa) = 1.7057 kPa h g1 = h g @ 15°C = 2528.3 kJ/kg s g1 = s g @ 15°C = 8.7803 kJ/kg ⋅ K

1

Pa1 = P1 − Pv1 = 200 − 1.7057 = 198.29 kPa

15°C 100% RH

v1 =

R a T1 (0.287 kPa ⋅ m 3 / kg ⋅ K)(288 K) = = 0.4168 m 3 / kg dry air Pa1 198.29 kPa

ω1 =

0.622 Pv1 0.622(1.7057 kPa) = = 0.005350 kg H 2 O/kg dry air P1 − Pv1 (200 − 1.7057) kPa

200 kPa

30°C 2 AIR

h1 = c p T1 + ω1 h g1 = (1.005 kJ/kg ⋅ °C)(15°C) + (0.005350)(2528.3 kJ/kg) = 28.60 kJ/kg dry air Pv 2 = Pv1 = 1.7057 kPa Pg 2 = Psat @ 30°C = 4.2469 kPa

φ2 =

Pv 2 1.7057 kPa = = 0.402 = 40.2% Pg 2 4.2469 kPa

Pa 2 = P2 − Pv 2 = 200 − 1.7057 = 198.29 kPa h g 2 = h g @ 30°C = 2555.6 kJ/kg s g 2 = s g @ 30°C = 8.4520 kJ/kg ⋅ K

ω 2 = ω1 h2 = c p T2 + ω 2 h g 2 = (1.005 kJ/kg ⋅ °C)(30°C) + (0.005350)(2555.6 kJ/kg) = 43.82 kJ/kg dry air

The entropy change of the dry air is P T 303 198.29 ( s 2 − s1 ) dry air = c p ln 2 − R ln a 2 = (1.005) ln − (0.287) ln = 0.05103 kJ/kg ⋅ K 288 198.29 T1 Pa1 The entropy change of the air-water mixture is s 2 − s1 = ( s 2 − s1 ) dry air + ω ( s 2 − s1 ) water vapor = 0.05103 + (0.005350) (8.4520 − 8.7803) = 0.04927 kJ/kg ⋅ K The mass flow rate of the dry air is

V&1 = V1 A1 = V1 m& a =

π D2 4

⎛ π (0.04 m) 2 = (20 m/s)⎜ ⎜ 4 ⎝

⎞ ⎟ = 0.02513 m 3 /s ⎟ ⎠

V&1 0.02513 m 3 / s = = 0.06029 kg/s v 1 0.4168 m 3 / kg dry air

The exergy increase of the humid air during this process is then, ΔΦ = m& a (ψ 2 −ψ 1 ) = m& a [(h2 − h1 ) − T0 ( s 2 − s1 )]

= (0.06029 kg/s)[(43.82 − 28.60)kJ/kg - (288 K)(0.04927 kJ/kg ⋅ K)] = 0.062 kW/K

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14-32

Heating with Humidification

14-77C To achieve a higher level of comfort. Very dry air can cause dry skin, respiratory difficulties, and increased static electricity.

14-78 Air is first heated and then humidified by water vapor. The amount of steam added to the air and the amount of heat transfer to the air are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (m& a1 = m& a 2 = m& a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Properties The inlet and the exit states of the air are completely specified, and the total pressure is 1 atm. The properties of the air at various states are determined from the psychrometric chart (Figure A-31) to be h1 = 311 . kJ / kg dry air ω 1 = 0.0064 kg H 2 O / kg dry air ( = ω 2 ) h2 = 36.2 kJ / kg dry air h3 = 581 . kJ / kg dry air

Heating coils

ω 3 = 0.0129 kg H 2 O / kg dry air Analysis (a) The amount of moisture in the air remains constant it flows through the heating section (ω 1 = ω 2), but increases in the humidifying section (ω 3 > ω 2). The amount of steam added to the air in the heating section is

1 atm

T1 = 15°C φ 1 = 60%

AIR

T3 = 25°C φ 3 = 65%

T2 = 20°C

1

2

3

Δω = ω 3 − ω 2 = 0.0129 − 0.0064 = 0.0065 kg H 2 O / kg dry air

(b) The heat transfer to the air in the heating section per unit mass of air is qin = h2 − h1 = 36.2 − 311 . = 5.1 kJ / kg dry air

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

14-33

14-79E Air is first heated and then humidified by water vapor. The amount of steam added to the air and the amount of heat transfer to the air are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (m& a1 = m& a 2 = m& a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Properties The inlet and the exit states of the air are completely specified, and the total pressure is 1 atm. The properties of the air at various states are determined from the psychrometric chart (Figure A-31E) to be h1 = 17.0 Btu/lbm dry air

ω1 = 0.0046 lbm H 2 O/lbm dry air

Heating coils

h2 = 22.3 Btu/lbm dry air

ω 2 = ω1 = 0.0046 lbm H 2 O/lbm dry air h3 = 29.2 Btu/lbm dry air

ω 3 = 0.0102 lbm H 2 O/lbm dry air Analysis (a) The amount of moisture in the air remains constant it flows through the heating section (ω1 = ω2), but increases in the humidifying section (ω 3 > ω 2). The amount of steam added to the air in the heating section is

14.7 psia

T1 = 50°F φ 1 = 60%

AIR

T3 = 75°F φ 3 = 55%

T2 = 72°F

1

2

3

Δω = ω 3 − ω 2 = 0.0102 − 0.0046 = 0.0056 lbm H 2 O/lbm dry air (b) The heat transfer to the air in the heating section per unit mass of air is q in = h2 − h1 = 22.3 − 17.0 = 5.3 Btu/lbm dry air

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14-34

14-80 Air is first heated and then humidified by wet steam. The temperature and relative humidity of air at the exit of heating section, the rate of heat transfer, and the rate at which water is added to the air are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (m& a1 = m& a 2 = m& a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Properties The inlet and the exit states of the air are completely specified, and the total pressure is 1 atm. The properties of the air at various states are determined from the psychrometric chart (Figure A-31) to be h1 = 23.5 kJ/kg dry air

ω1 = 0.0053 kg H 2 O/kg dry air (= ω 2 ) v 1 = 0.809 m 3 /kg dry air

Sat. vapor 100°C

Heating coils

h3 = 42.3 kJ/kg dry air

Humidifier

ω 3 = 0.0087 kg H 2 O/kg dry air

m& a =

AIR

10°C 70% 35 m3/min

Analysis (a) The amount of moisture in the air remains constant it flows through the heating section (ω 1 = ω 2), but increases in the humidifying section (ω 3 > ω 2). The mass flow rate of dry air is

20°C 60%

1 atm 1

2

3

35 m3 / min V&1 = = 43.3 kg/min v1 0.809 m3 / kg

Noting that Q = W =0, the energy balance on the humidifying section can be expressed as E& in − E& out = ΔE& systemÊ0 (steady) = 0 E& in = E& out ∑ m& i hi = ∑ m& e he

⎯ ⎯→

m& w hw + m& a 2 h2 = m& a h3 (ω 3 − ω 2 )hw + h2 = h3

Solving for h2, h2 = h3 − (ω 3 − ω 2 )h g @ 100°C = 42.3 − (0.0087 − 0.0053)(2675.6) = 33.2 kJ/kg dry air

Thus at the exit of the heating section we have ω2 = 0.0053 kg H2O dry air and h2 = 33.2 kJ/kg dry air, which completely fixes the state. Then from the psychrometric chart we read T2 = 19.5°C

φ 2 = 37.8% (b) The rate of heat transfer to the air in the heating section is Q& in = m& a (h2 − h1 ) = (43.3 kg/min)(33.2 − 23.5) kJ/kg = 420 kJ/min

(c) The amount of water added to the air in the humidifying section is determined from the conservation of mass equation of water in the humidifying section, m& w = m& a (ω 3 − ω 2 ) = (43.3 kg/min)(0.0087 − 0.0053) = 0.15 kg/min

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14-35

14-81 Air is first heated and then humidified by wet steam. The temperature and relative humidity of air at the exit of heating section, the rate of heat transfer, and the rate at which water is added to the air are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (m& a1 = m& a 2 = m& a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Sat. vapor Analysis (a) The amount of moisture in the 100°C Heating air also remains constant it flows through coils Humidifier the heating section (ω 1 = ω 2), but increases in the humidifying section (ω 3 > AIR 10°C ω 2). The inlet and the exit states of the air 20°C 70% are completely specified, and the total 60% 3 95 kPa 35 m /min pressure is 95 kPa. The properties of the air at various states are determined to be 1 2 3 Pv1 = φ1 Pg1 = φ1 Psat @ 10°C = (0.70)(1.2281 kPa) = 0.860 kPa ( = Pv 2 ) Pa1 = P1 − Pv1 = 95 − 0.860 = 94.14 kPa

v1 =

R a T1 (0.287 kPa ⋅ m 3 / kg ⋅ K)(283 K) = = 0.863 m 3 / kg dry air Pa1 94.14 kPa

ω1 =

0.622 Pv1 0.622(0.86 kPa) = = 0.00568 kg H 2 O/kg dry air (= ω 2 ) P1 − Pv1 (95 − 0.86) kPa

h1 = c p T1 + ω1 h g1 = (1.005 kJ/kg ⋅ °C)(10°C) + (0.00568)(2519.2 kJ/kg) = 24.36 kJ/kg dry air Pv 3 = φ3 Pg 3 = φ3 Psat @ 20°C = (0.60)(2.3392 kPa) = 1.40 kPa

ω3 =

0.622 Pv3 0.622(1.40 kPa) = = 0.00930 kg H 2O/kg dry air P3 − Pv 3 (95 − 1.40) kPa

h3 = c pT3 + ω3hg 3 = (1.005 kJ/kg ⋅ °C)(20°C) + (0.0093)(2537.4 kJ/kg) = 43.70 kJ/kg dry air

Also, m& a =

V&1 35 m 3 / min = = 40.6 kg/min v 1 0.863 m 3 / kg

Noting that Q = W = 0, the energy balance on the humidifying section gives E& in − E& out = ΔE& system Ê0 (steady) = 0 ⎯ ⎯→ E& in = E& out ∑ m& e he = ∑ m& i hi ⎯ ⎯→ m& w hw + m& a 2 h2 = m& a h3 ⎯ ⎯→(ω 3 − ω 2 )h w + h2 = h3 h2 = h3 − (ω 3 − ω 2 )h g @ 100°C = 43.7 − (0.0093 − 0.00568) × 2675.6 = 34.0 kJ/kg dry air

Thus at the exit of the heating section we have ω = 0.00568 kg H2O dry air and h2 = 34.0 kJ/kg dry air, which completely fixes the state. The temperature of air at the exit of the heating section is determined from the definition of enthalpy, h2 = c p T2 + ω 2 h g 2 ≅ c p T2 + ω 2 (2500.9 + 1.82T2 ) 34.0 = (1.005)T2 + (0.00568)(2500.9 + 1.82T2 )

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14-36

Solving for h2, yields T2 = 19.5° C

The relative humidity at this state is

φ2 =

Pv 2 Pv 2 0.859 kPa = = = 0.377 or 37.7% Pg 2 Psat @ 19.5°C 2.2759 kPa

(b) The rate of heat transfer to the air in the heating section becomes Q& in = m& a (h2 − h1 ) = (40.6 kg/min)(34.0 − 24.36) kJ/kg = 391 kJ/min

(c) The amount of water added to the air in the humidifying section is determined from the conservation of mass equation of water in the humidifying section, &w = m & a (ω 3 − ω 2 ) = ( 40.6 kg / min)( 0.0093 − 0.00568) = 0.147 kg / min m

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14-37

Cooling with Dehumidification

14-82C To drop its relative humidity to more desirable levels.

14-83 Air is first cooled, then dehumidified, and finally heated. The temperature of air before it enters the heating section, the amount of heat removed in the cooling section, and the amount of heat supplied in the heating section are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (m& a1 = m& a 2 = m& a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis (a) The amount of moisture in the air decreases due to dehumidification (ω 3 < ω 1), and remains constant during heating (ω 3 = ω 2). The inlet and the exit states of the air are completely specified, and the total pressure is 1 atm. The intermediate state (state 2) is also known since φ2 = 100% and ω 2 = ω 3. Therefore, we can determine the properties of the air at all three states from the psychrometric chart (Fig. A-31) to be h1 = 95.2 kJ / kg dry air ω 1 = 0.0238 kg H 2 O / kg dry air

and

Heating section

Cooling section T1 = 34°C φ 1 = 70%

h3 = 431 . kJ / kg dry air ω 3 = 0.0082 kg H 2 O / kg dry air ( = ω 2 )

Also, hw ≅ h f @ 10°C = 42.02 kJ/kg (Table A - 4)

T2 1

T3 = 22°C

2 w

φ 3 = 50%

1 atm AIR 3

10°C

h2 = 31.8 kJ/kg dry air T2 = 11.1°C

(b) The amount of heat removed in the cooling section is determined from the energy balance equation applied to the cooling section, E& in − E& out = ΔE& systemÊ0 (steady) = 0 E& in = E& out ∑ m& i hi = ∑ m& e he + Q& out,cooling Q& out,cooling = m& a1h1 − (m& a 2 h2 + m& w hw ) = m& a (h1 − h2 ) − m& w hw

or, per unit mass of dry air, q out,cooling = (h1 − h2 ) − (ω 1 − ω 2 )hw = (95.2 − 31.8) − (0.0238 − 0.0082)42.02 = 62.7 kJ/kg dry air

(c) The amount of heat supplied in the heating section per unit mass of dry air is qin,heating = h3 − h2 = 431 . − 31.8 = 11.3 kJ / kg dry air

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14-38

14-84 [Also solved by EES on enclosed CD] Air is cooled by passing it over a cooling coil through which chilled water flows. The rate of heat transfer, the mass flow rate of water, and the exit velocity of airstream are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process. 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis (a) The saturation pressure of water at 35ºC is 5. 6291 kPa (Table A-4). Then the dew point temperature of the incoming air stream at 35°C becomes Tdp = Tsat @ Pv = Tsat @ 0.6×5.6291 kPa = 26°C (Table A-5) since air is cooled to 20°C, which is below its dew point temperature, some of the moisture in the air will condense. The amount of moisture in the air decreases due to dehumidification (ω 2 < ω 1 ) . The inlet and the exit states of the air are completely specified, and the total pressure is 1 atm. Then the properties of the air at both states are determined from the psychrometric chart (Fig. A-31) to be h1 = 90.3 kJ/kg dry air Water ω1 = 0.0215 kg H 2 O/kg dry air T + 8°C T v 1 = 0.904 m 3 /kg dry air and Cooling coils h2 = 57.5 kJ/kg dry air

ω 2 = 0.0147 kg H 2 O/kg dry air v 2 = 0.851 m 3 /kg dry air Also,

35°C 60% 120 m/min

1

hw ≅ h f @ 20°C = 83.93 kJ/kg (Table A-4)

AIR

20°C 2 Saturated

Then,

V&1 = V1 A1 = V1 m& a1 =

π D2 4

⎛ π (0.3 m) 2 = (120 m/min)⎜ ⎜ 4 ⎝

⎞ ⎟ = 8.48 m 3 / min ⎟ ⎠

V&1 8.48 m 3 / min = = 9.38 kg/min v 1 0.904 m 3 / kg dry air

Applying the water mass balance and the energy balance equations to the combined cooling and dehumidification section (excluding the water), Water Mass Balance: ∑ m& w,i = ∑ m& w,e ⎯ ⎯→ m& a1ω 1 = m& a 2ω 2 + m& w m& w = m& a ( ω1 − ω2 ) = ( 9.38 kg/min)(0.0215 − 0.0147 ) = 0.064 kg/min Energy Balance: Ê0 (steady) = 0⎯ ⎯→ E& = E& E& − E& = ΔE& in

out

system

in

out

∑ m& i hi = ∑ m& e he + Q& out ⎯ ⎯→ Qout = m& a1h1 − (m& a 2 h2 + m& w hw ) = m& a (h1 − h2 ) − m& w hw Q& out = (9.38 kg/min)(90.3 − 57.5)kJ/kg − (0.064 kg/min)(83.93 kJ/kg) = 302.3 kJ/min (b) Noting that the heat lost by the air is gained by the cooling water, the mass flow rate of the cooling water is determined from Q& cooling water = m& cooling water Δh = m& cooling water c p ΔT m& cooling water =

Q& w 302.3 kJ/min = = 9.04 kg/min c p ΔT (4.18 kJ/kg ⋅ °C)(8°C)

(c) The exit velocity is determined from the conservation of mass of dry air, V A V A V& V& m& a1 = m& a 2 ⎯ ⎯→ 1 = 2 ⎯ ⎯→ 1 = 2

v1

V2 =

v2

v1

v2

v2 0.851 V1 = (120 m/min) = 113 m/min 0.904 v1

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

14-39

14-85 EES Problem 14-84 is reconsidered. A general solution of the problem in which the input variables may be supplied and parametric studies performed is to be developed and the process is to be shown in the psychrometric chart for each set of input variables. Analysis The problem is solved using EES, and the solution is given below. "Input Data from the Diagram Window" {D=0.3 P[1] =101.32 [kPa] T[1] = 35 [C] RH[1] = 60/100 "%, relative humidity" Vel[1] = 120/60 "[m/s]" DELTAT_cw =8 [C] P[2] = 101.32 [kPa] T[2] = 20 [C]} RH[2] = 100/100 "%" "Dry air flow rate, m_dot_a, is constant" Vol_dot[1]= (pi * D^2)/4*Vel[1] v[1]=VOLUME(AirH2O,T=T[1],P=P[1],R=RH[1]) m_dot_a = Vol_dot[1]/v[1] "Exit vleocity" Vol_dot[2]= (pi * D^2)/4*Vel[2] v[2]=VOLUME(AirH2O,T=T[2],P=P[2],R=RH[2]) m_dot_a = Vol_dot[2]/v[2] "Mass flow rate of the condensed water" m_dot_v[1]=m_dot_v[2]+m_dot_w w[1]=HUMRAT(AirH2O,T=T[1],P=P[1],R=RH[1]) m_dot_v[1] = m_dot_a*w[1] w[2]=HUMRAT(AirH2O,T=T[2],P=P[2],R=RH[2]) m_dot_v[2] = m_dot_a*w[2] "SSSF conservation of energy for the air" m_dot_a *(h[1] + (1+w[1])*Vel[1]^2/2*Convert(m^2/s^2, kJ/kg)) + Q_dot = m_dot_a*(h[2] +(1+w[2])*Vel[2]^2/2*Convert(m^2/s^2, kJ/kg)) +m_dot_w*h_liq_2 h[1]=ENTHALPY(AirH2O,T=T[1],P=P[1],w=w[1]) h[2]=ENTHALPY(AirH2O,T=T[2],P=P[2],w=w[2]) h_liq_2=ENTHALPY(Water,T=T[2],P=P[2]) "SSSF conservation of energy for the cooling water" -Q_dot =m_dot_cw*Cp_cw*DELTAT_cw "Note: Q_netwater=-Q_netair" Cp_cw = SpecHeat(water,T=10,P=P[2])"kJ/kg-K"

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14-40

RH1

ma

mw

mcw

0.5 0.6 0.7 0.8 0.9

0.1574 0.1565 0.1556 0.1547 0.1538

0.0004834 0.001056 0.001629 0.002201 0.002774

0.1085 0.1505 0.1926 0.2346 0.2766

Q [kW] -3.632 -5.039 -6.445 -7.852 -9.258

Vel1 [m/s] 2 2 2 2 2

Vel2 [m/s] 1.894 1.883 1.872 1.861 1.85

T1 [C] 35 35 35 35 35

T2 [C] 20 20 20 20 20

w1

w2

0.01777 0.02144 0.02516 0.02892 0.03273

0.0147 0.0147 0.0147 0.0147 0.0147

0.050 0.045

Pressure = 101.0 [kPa]

0.040

Humidity Ratio

0.035

0.8

0.030

30 C

0.025

0.6

0.020 20 C

0.015 0.010

10 C

0.005 0.000 -10

0.4

0.2

0C

-5

-0

5

10

15

20

25

30

35

40

T [C]

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

14-41

14-86 Air is cooled by passing it over a cooling coil. The rate of heat transfer, the mass flow rate of water, and the exit velocity of airstream are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process. 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible.

Pv1 = φ1 Pg1 = φ1 Psat @ 35°C Tdp

T + 8°C

Water, T

Analysis (a) The dew point temperature of the incoming air stream at 35°C is

Cooling coils

= (0.6)(5.6291 kPa) = 3.38 kPa = Tsat @ Pv = Tsat @ 3.38 kPa = 25.9°C

1

Since air is cooled to 20°C, which is below its dew point temperature, some of the moisture in the air will condense.

35°C 60% 120 m/min

AIR

20°C 2 Saturated

The amount of moisture in the air decreases due to dehumidification (ω 2 < ω 1 ) . The inlet and the exit states of the air are completely specified, and the total pressure is 95 kPa. Then the properties of the air at both states are determined to be Pa1 = P1 − Pv1 = 95 − 3.38 = 91.62 kPa

v1 =

Ra T1 (0.287 kPa ⋅ m 3 / kg ⋅ K)(308 K) = = 0.965 m 3 / kg dry air 91.62 kPa Pa1

ω1 =

0.622 Pv1 0.622(3.38 kPa) = = 0.0229 kg H 2 O/kg dry air (95 − 3.38) kPa P1 − Pv1

h1 = c p T1 + ω1hg1 = (1.005 kJ/kg ⋅ °C)(35°C) + (0.0229)(2564.6 kJ/kg) = 93.90 kJ/kg dry air

and Pv 2 = φ 2 Pg 2 = (1.00) Psat @ 20°C = 2.3392 kPa

v2 =

Ra T2 (0.287 kPa ⋅ m 3 / kg ⋅ K)(293 K) = = 0.908 m 3 / kg dry air (95 − 2.339) kPa Pa 2

ω2 =

0.622 Pv 2 0.622(2.339 kPa) = = 0.0157 kg H 2 O/kg dry air (95 − 2.339) kPa P2 − Pv 2

h2 = c p T2 + ω 2 hg 2 = (1.005 kJ/kg ⋅ °C)(20°C) + (0.0157)(2537.4 kJ/kg) = 59.95 kJ/kg dry air

Also, hw ≅ h f @ 20°C = 83.915 kJ/kg

(Table A-4)

Then,

V&1 = V1 A1 = V1 m& a1 =

π D2 4

⎛ π (0.3 m) 2 = (120 m/min)⎜ ⎜ 4 ⎝

⎞ ⎟ = 8.48 m 3 / min ⎟ ⎠

V&1 8.48 m 3 / min = = 8.79 kg/min v 1 0.965 m 3 / kg dry air

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14-42

Applying the water mass balance and energy balance equations to the combined cooling and dehumidification section (excluding the water),

Water Mass Balance: ∑ m& w,i = ∑ m& w ,e

⎯ ⎯→

m& a1ω 1 = m& a 2ω 2 + m& w

m& w = m& a (ω 1 − ω 2 ) = (8.79 kg / min)(0.0229 − 0.0157) = 0.0633 kg / min

Energy Balance: E& in − E& out = ΔE& systemÊ0 (steady) = 0 E& in = E& out ∑ m& i hi = ∑ m& e he + Q& out → Q& out = m& a1h1 − (m& a 2 h2 + m& w hw ) = m& a (h1 − h2 ) − m& w hw Q& out = (8.79 kg/min)(93.90 − 59.94)kJ/kg − (0.0633 kg/min)(83.915 kJ/kg) = 293.2 kJ/min

(b) Noting that the heat lost by the air is gained by the cooling water, the mass flow rate of the cooling water is determined from Q& cooling water = m& cooling water Δh = m& cooling water c p ΔT m& cooling water =

Q& w 293.2 kJ/min = = 8.77 kg/min c p ΔT (4.18 kJ/kg ⋅ °C)(8°C)

(c) The exit velocity is determined from the conservation of mass of dry air, m& a1 = m& a 2 ⎯ ⎯→ V2 =

V A V A V&1 V&2 = ⎯ ⎯→ 1 = 2 v1 v 2 v1 v2

v2 0.908 V1 = (120 m/min) = 113 m/min 0.965 v1

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

14-43

14-87 Air is cooled and dehumidified at constant pressure. The amount of water removed from the air and the cooling requirement are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (m& a1 = m& a 2 = m& a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Properties The inlet and the exit states of the air are completely specified, and the total pressure is 1 atm. The properties of the air at various states are determined from the psychrometric chart (Figure A-31) to be

h1 = 85.5 kJ/kg dry air

ω1 = 0.0217 kg H 2 O/kg dry air

Cooling coils

and

T2 = 20°C φ 2 =100%

φ 2 = 1.0 h2 = 57.6 kJ/kg dry air

ω 2 = 0.0148 kg H 2 O/kg dry air

2

Also, h w ≅ h f @ 22°C = 92.28 kJ/kg

1 atm Condensate

T1 = 30°C φ 1 = 80%

1 22°C

(Table A-4)

Condensate removal

Analysis The amount of moisture in the air decreases due to dehumidification (ω 2 < ω 1). Applying the water mass balance and energy balance equations to the combined cooling and dehumidification section,

Water Mass Balance: ∑ m& w,i = ∑ m& w,e ⎯ ⎯→ m& a1ω1 = m& a 2 ω 2 + m& w Δω = ω1 − ω 2 = 0.0217 − 0.0148 = 0.0069 kg H 2 O/kg dry air

Energy Balance: E& in − E& out = ΔE& system ©0 (steady) = 0 E& in = E& out ∑ m& i hi = Q& out + ∑ m& e he Q& out = m& a1 h1 − (m& a 2 h2 + m& w h w ) = m& a (h1 − h2 ) − m& w hw q out = h1 − h2 − (ω1 − ω 2 )hw = (85.5 − 57.6)kJ/kg − (0.0069)(92.28) = 27.3 kJ/kg dry air

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14-44

14-88E Air is cooled and dehumidified at constant pressure. The amount of water removed from the air and the rate of cooling are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (m& a1 = m& a 2 = m& a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Properties The inlet and the exit states of the air are completely specified, and the total pressure is 1 atm. The properties of the air at various states are determined from the psychrometric chart (Figure A-31E) to be h1 = 37.8 Btu/lbm dry air

ω1 = 0.0158 lbm H 2 O/lbm dry air

Cooling coils

v 1 = 14.08 ft 3 /lbm dry air T2 = 60°F φ 2 =100%

and

φ 2 = 1.0 h2 = 26.5 Btu/lbm dry air

1 atm Condensate 2

1

ω 2 = 0.0111 lbm H 2 O/lbm dry air 65°F

Also, hw ≅ h f @ 65° F = 33.08 Btu/lbm

T1 = 85°F Tdp1=70°F

Condensate removal

(Table A-4E)

Analysis The amount of moisture in the air decreases due to dehumidification (ω 2 < ω 1). The mass flow rate of air is m& a1 =

V&1 (10,000 / 3600) ft 3 / s = = 0.1973 lbm/s v 1 14.08 ft 3 / lbm dry air

Applying the water mass balance and energy balance equations to the combined cooling and dehumidification section,

Water Mass Balance: ∑ m& w,i = ∑ m& w,e ⎯ ⎯→ m& a1ω1 = m& a 2 ω 2 + m& w m& w = m& a (ω1 − ω 2 ) = (0.1973 lbm/s)(0.0158 − 0.0111) = 0.000927 lbm/s

Energy Balance: E& in − E& out = ΔE& system ©0 (steady) = 0 E& in = E& out ∑ m& i hi = Q& out + ∑ m& e he Q& out = m& a1 h1 − (m& a 2 h2 + m& w h w ) = m& a (h1 − h2 ) − m& w h w Q& out = (0.1973 lbm/s)(37.8 − 26.5)Btu/lbm − (0.000927 lbm/s)(33.08 Btu/lbm) = 2.20 Btu/s

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

14-45

14-89 Air is cooled and dehumidified at constant pressure. The amount of water removed from the air and the rate of cooling are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (m& a1 = m& a 2 = m& a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Properties The inlet and the exit states of the air are completely specified, and the total pressure is 1 atm. The properties of the air at various states are determined from the psychrometric chart (Figure A-31) to be h1 = 106.8 kJ/kg dry air

ω1 = 0.0292 kg H 2 O/kg dry air

Cooling coils

v 1 = 0.905 m 3 /kg dry air T2 = 24°C φ 2 =60%

and h2 = 52.7 kJ/kg dry air

ω 2 = 0.0112 kg H 2 O/kg dry air

2

We assume that the condensate leaves this system at the average temperature of the air inlet and exit. Then, hw ≅ h f @ 28°C = 117.4 kJ/kg

1 atm Condensate

T1 = 32°C φ 1 =95%

1 28°C

Condensate removal

(Table A-4)

Analysis The amount of moisture in the air decreases due to dehumidification (ω 2 < ω 1). The mass of air is ma =

V1 1000 m 3 = = 1105 kg v 1 0.905 m 3 / kg dry air

Applying the water mass balance and energy balance equations to the combined cooling and dehumidification section,

Water Mass Balance: ∑ m& w,i = ∑ m& w,e ⎯ ⎯→ m& a1ω 1 = m& a 2ω 2 + m& w m w = m a (ω1 − ω 2 ) = (1105 kg)(0.0292 − 0.0112) = 19.89 kg

Energy Balance: E& in − E& out = ΔE& system ©0 (steady) = 0 E& in = E& out ∑ m& i hi = Q& out + ∑ m& e he Q& out = m& a1 h1 − (m& a 2 h2 + m& w h w ) = m& a (h1 − h2 ) − m& w h w Qout = m a (h1 − h2 ) − m w hw Qout = (1105 kg)(106.8 − 52.7)kJ/kg − (19.89 kg)(117.4 kJ/kg) = 57,450 kJ

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

14-46

14-90 The humid air of the previous problem is reconsidered. The exit temperature of the air to produce the desired dehumidification is to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (m& a1 = m& a 2 = m& a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Properties The inlet and the exit states of the air are completely specified, and the total pressure is 1 atm. The properties of the air at various states are determined from the psychrometric chart (Figure A-31) to be (from the data of the previous problem) h1 = 106.8 kJ/kg dry air

Cooling coils

ω1 = 0.0292 kg H 2 O/kg dry air v 1 = 0.905 m 3 /kg dry air and

T2 = 24°C φ 2 =60%

h2 = 52.7 kJ/kg dry air

ω 2 = 0.0112 kg H 2 O/kg dry air Analysis For the desired dehumidification, the air at the exit should be saturated with a specific humidity of 0.0112 kg water/kg dry air. That is,

1 atm Condensate 2

T1 = 32°C φ 1 =95%

1 28°C

Condensate removal

φ 2 = 1.0 ω 2 = 0.0112 kg H 2 O/kg dry air The temperature of the air at this state is T2 = 15.8°C

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

14-47

14-91 Air is cooled and dehumidified at constant pressure. The cooling required is provided by a simple ideal vapor-compression refrigeration system using refrigerant-134a as the working fluid. The exergy destruction in the total system per 1000 m3 of dry air is to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (m& a1 = m& a 2 = m& a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Properties The inlet and the exit states of the air are completely specified, and the total pressure is 1 atm. The properties of the air at various states are determined from the psychrometric chart (Figure A-31) to be h1 = 106.8 kJ/kg dry air

ω1 = 0.0292 kg H 2 O/kg dry air

Condenser

v 1 = 0.905 m 3 /kg dry air

3

and

2

E x

h2 = 52.7 kJ/kg dry air

Compressor

ω 2 = 0.0112 kg H 2 O/kg dry air 4

We assume that the condensate leaves this system at the average temperature of the air inlet and exit. Then, from Table A-4, hw ≅ h f @ 28°C = 117.4 kJ/kg

1 Evaporator

T2 = 24°C φ 2 = 60%

Analysis The amount of moisture in the air decreases due to dehumidification (ω 2 < ω 1). The mass of air is ma =

1 atm

T1 = 32°C φ 1 = 95%

Condensate

V1 1000 m 3 = = 1105 kg v 1 0.905 m 3 / kg dry air

Applying the water mass balance and energy balance equations to the combined cooling and dehumidification section,

Water Mass Balance: ∑ m& w,i = ∑ m& w,e ⎯ ⎯→ m& a1ω 1 = m& a 2ω 2 + m& w m w = m a (ω1 − ω 2 ) = (1105 kg)(0.0292 − 0.0112) = 19.89 kg

Energy Balance: E& in − E& out = ΔE& system ©0 (steady) = 0 E& in = E& out ∑ m& i hi = Q& out + ∑ m& e he Q& out = m& a1 h1 − (m& a 2 h2 + m& w h w ) = m& a (h1 − h2 ) − m& w h w Qout = m a (h1 − h2 ) − m w hw Qout = (1105 kg)(106.8 − 52.7)kJ/kg − (19.89 kg)(117.4 kJ/kg) = 57,450 kJ

We obtain the properties for the vapor-compression refrigeration cycle as follows (Tables A-11,through A13):

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14-48

T1 = 4°C ⎫ h1 = h g @ 4°C = 252.77 kJ/kg ⎬ sat. vapor ⎭ s1 = s g @ 4°C = 0.92927 kJ/kg ⋅ K

T

P2 = Psat @ 39.4°C = 1 MPa ⎫ ⎬ h2 = 275.29 kJ/kg s 2 = s1 ⎭

· QH

2

3 39.4°C

· Win

P3 = 1 MPa ⎫ h3 = h f @ 1 MPa = 107.32 kJ/kg ⎬ sat. liquid ⎭ s 3 = s f @ 1 MPa = 0.39189 kJ/kg ⋅ K

4 °C

h4 ≅ h3 = 107.32 kJ/kg ( throttling) T4 = 4°C ⎫ x 4 = 0.2561 ⎬ h4 = 107.32 kJ/kg ⎭ s 4 = 0.4045 kJ/kg ⋅ K

4s

4

· QL

1

The mass flow rate of refrigerant-134a is mR =

s

QL 57,450 kJ = = 395.0 kg h1 − h4 (252.77 − 107.32)kJ/kg

The amount of heat rejected from the condenser is Q H = m R (h2 − h3 ) = (395.0 kg)(275.29 − 107.32) kJ/kg = 66,350 kg

Next, we calculate the exergy destruction in the components of the refrigeration cycle: X destroyed,12 = m R T0 ( s 2 − s1 ) = 0 (since the process is isentropic) ⎛ Q ⎞ X destroyed, 23 = T0 ⎜⎜ m R ( s 3 − s 2 ) + H ⎟⎟ TH ⎠ ⎝ 66,350 kJ ⎞ ⎛ = (305 K )⎜ (395 kg)(0.39189 − 0.92927) kJ/kg ⋅ K + ⎟ = 1609 kJ 305 K ⎠ ⎝ X destroyed, 34 = m R T0 ( s 4 − s 3 ) = (395 kg)(305 K )(0.4045 − 0.39189) kJ/kg ⋅ K = 1519 kJ

The entropies of water vapor in the air stream are s g1 = s g @ 32°C = 8.4114 kJ/kg ⋅ K s g 2 = s g @ 24°C = 8.5782 kJ/kg ⋅ K

The entropy change of water vapor in the air stream is ΔS vapor = m a (ω 2 s g 2 − ω1 s g1 ) = (1105)(0.0112 × 8.5782 − 0.0292 × 8.4114) = −165.2 kJ/K

The entropy of water leaving the cooling section is S w = m w s f @ 28°C = (19.89 kg )(0.4091 kJ/kg ⋅ K) = 8.14 kJ/K

The partial pressures of water vapor and dry air for air streams are Pv1 = φ1 Pg1 = φ1 Psat @ 32°C = (0.95)(4.760 kPa) = 4.522 kPa Pa1 = P1 − Pv1 = 101.325 − 4.522 = 96.80 kPa Pv 2 = φ 2 Pg 2 = φ 2 Psat @ 24°C = (0.60)(2.986 kPa) = 1.792 kPa Pa 2 = P2 − Pv 2 = 101.325 − 1.792 = 99.53 kPa

The entropy change of dry air is

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14-49

⎛ P ⎞ T ΔS a = m a ( s 2 − s1 ) = m a ⎜⎜ c p ln 2 − R ln a 2 ⎟⎟ T1 Pa1 ⎠ ⎝ 297 99.53 ⎤ ⎡ = (1105) ⎢(1.005) ln − (0.287) ln = −38.34 kJ/kg dry air 305 96.80 ⎥⎦ ⎣

The entropy change of R-134a in the evaporator is ΔS R, 41 = m R ( s1 − s 4 ) = (395 kg )(0.92927 − 0.4045) = 207.3 kJ/K

An entropy balance on the evaporator gives S gen,evaporator = ΔS R,41 + ΔS vapor + ΔS a + S w = 207.3 + (−165.2) + (−38.34) + 8.14 = 11.90 kJ/K

Then, the exergy destruction in the evaporator is X dest = T0 S gen, evaporator = (305 K)(11.90 kJ/K) = 3630 kJ

Finally the total exergy destruction is X dest, total = X dest, compressor + X dest, condenser + X dest, throttle + X dest, evaporator = 0 + 1609 + 1519 + 3630 = 6758 kJ

The greatest exergy destruction occurs in the evaporator. Note that heat is absorbed from humid air and rejected to the ambient air at 32°C (305 K), which is also taken as the dead state temperature.

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

14-50

14-92 Atmospheric air enters the evaporator of an automobile air conditioner at a specified pressure, temperature, and relative humidity. The dew point and wet bulb temperatures at the inlet to the evaporator section, the required heat transfer rate from the atmospheric air to the evaporator fluid, and the rate of condensation of water vapor in the evaporator section are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (m& a1 = m& a 2 = m& a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis The inlet and exit states of the air are completely specified, and the total pressure is 1 atm. The properties of the air at the inlet and exit states may be determined from the psychrometric chart (Fig. A-31) or using EES psychrometric functions to be (we used EES) Tdp1 = 15.7°C

Cooling coils

Twb1 = 19.5°C h1 = 55.60 kJ/kg dry air

ω1 = 0.01115 kg H 2O/kg dry air

T2 =10°C φ 2 = 90%

1 atm Condensate

v1 = 0.8655 m3 / kg dry air 2

h2 = 27.35 kJ/kg dry air

ω2 = 0.00686 kg H 2O/kg dry air

m& a =

1 10°C

The mass flow rate of dry air is

T1 =27°C φ 1 = 50%

Condensate removal

V&1 V car ACH (2 m 3 /change)(5 changes/min) = = = 11.55 kg/min v1 v1 0.8655 m 3

The mass flow rates of vapor at the inlet and exit are m& v1 = ω1 m& a = (0.01115)(11.55 kg/min) = 0.1288 kg/min m& v 2 = ω 2 m& a = (0.00686)(11.55 kg/min) = 0.07926 kg/min

An energy balance on the control volume gives m& a h1 = Q& out + m& a h2 + m& w h w2

where the the enthalpy of condensate water is hw 2 = h f@ 10°C = 42.02 kJ/kg

(Table A - 4)

and the rate of condensation of water vapor is m& w = m& v1 − m& v 2 = 0.1288 − 0.07926 = 0.0495 kg/min

Substituting, m& a h1 = Q& out + m& a h2 + m& w hw 2 (11.55 kg/min)(55.60 kJ/kg) = Q& out + (11.55 kg/min)(27.35 kJ/kg) + (0.0495 kg/min)(42.02 kJ/kg) Q& = 324.4 kJ/min = 5.41 kW out

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14-51

AirH2O

0,050 Pressure = 101.3 [kPa] 0,045 0,040

35°C 0.8

Humidity Ratio

0,035 0,030 30°C

0.6

0,025 25°C

0,020

0.4 20°C

0,015 15°C

0,010

1 0.2

10°C

0,005 0,000 0

2 5

10

15

20

25

30

35

40

T [°C]

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

14-52

14-93 Atmospheric air flows into an air conditioner that uses chilled water as the cooling fluid. The mass flow rate of the condensate water and the volume flow rate of chilled water supplied to the air conditioner are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (m& a1 = m& a 2 = m& a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis We may assume that the exit relative humidity is 100 percent since the exit temperature (18°C) is below the dew-point temperature of the inlet air (25°C). The properties of the air at the exit state may be determined from the psychrometric chart (Fig. A-31) or using EES psychrometric functions to be (we used EES) h2 = 51.34 kJ/kg dry air

ω 2 = 0.01311 kg H 2 O/kg dry air The partial pressure of water vapor at the inlet state is (Table A-4)

Cooling coils

T2 = 18°C 100% RH 100 kPa

Pv1 = Psat@ 25°C = 3.17 kPa

T1 = 28°C Tdp1 = 25°C 2000 m3/h

Condensate 2

1

The saturation pressure at the inlet state is Pg1 = Psat@ 28°C = 3.783 kPa (Table A - 4)

18°C

Condensate removal

Then, the relative humidity at the inlet state becomes

φ1 =

Pv1 3.17 = = 0.8379 Pg1 3.783

Now, the inlet state is also fixed. The properties are obtained from EES to be h1 = 80.14 kJ/kg dry air

ω1 = 0.02036 kg H 2 O/kg dry air v 1 = 0.8927 m 3 /kg The mass flow rate of dry air is m& a =

V&1 (2000 / 60) m3/h = = 37.34 kg/min v1 0.8927 m3/kg

The mass flow rate of condensate water is m& w = m& a (ω1 − ω 2 ) = (37.34 kg/min)(0.02036 - 0.01311) = 0.2707 kg/min = 16.24 kg/h

The enthalpy of condensate water is hw 2 = h f@ 18°C = 75.54 kJ/kg

(Table A - 4)

An energy balance on the control volume gives

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14-53

m& a h1 = Q& out + m& a h2 + m& w hw2 (37.34 kg/min)(80.14 kJ/kg) = Q& out + (37.34 kg/min)(51.34 kJ/kg) + (0.2707 kg/min)(75.54 kJ/kg) Q& = 1055 kJ/min = 17.59 kW out

Noting that the rate of heat lost from the air is received by the cooling water, the mass flow rate of the cooling water is determined from ⎯→ m& cw = Q& in = m& cw c p ΔTcw ⎯

Q& in 1055 kJ/min = = 25.24 kg/min c p ΔTcw (4.18 kJ/kg.°C)(10°C)

where we used the specific heat of water value at room temperature. Assuming a density of 1000 kg/m3 for water, the volume flow rate is determined to be

V&cw =

m& cw

ρcw

=

25.24 kg/min = 0.0252 m3 /min 1000 kg/m3

AirH2O

0,050 Pressure = 100.0 [kPa] 0,045 0,040

0.8

0.

0,030

92

30°C

5

0,025

1

9

25°C

0,020

0.6

0.

Humidity Ratio

35°C

0,035

0.4 87

2

5

15°C 0 .8

0,010

3 /k

25

g

0 .8

0 .8

5

10

0.2

5m

10°C

0,005 0,000 0

0.

20°C

0,015

15

20

25

30

35

40

T [°C]

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

14-54

14-94 An automobile air conditioner using refrigerant 134a as the cooling fluid is considered. The inlet and exit states of moist air in the evaporator are specified. The volume flow rate of the air entering the evaporator of the air conditioner is to be determined. Assumptions 1 All processes are steady flow and the mass flow rate of dry air remains constant during the entire process (m& a1 = m& a 2 = m& a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis We assume that the total pressure of moist air is 100 kPa. Then, the inlet and exit states of the moist air for the evaporator are completely specified. The properties may be determined from the psychrometric chart (Fig. A-31) or using EES psychrometric functions to be (we used EES) h1 = 43.33 kJ/kg dry air

R-134a 375 kPa

ω1 = 0.008337 kg H 2 O/kg dry air v 1 = 0.8585 m 3 /kg dry air

Cooling coils

h2 = 23.31 kJ/kg dry air

ω 2 = 0.006065 kg H 2 O/kg dry air

T2 = 8°C φ 2 =90%

The mass flow rate of dry air is given by

2

V& V&1 m& a = 1 = v 1 0.8585 m 3 /kg

T1 =22°C φ 1 = 50%

AIR Condensate 1 8 °C

Condensate removal

The mass flow rate of condensate water is expressed as m& w = m& a (ω1 − ω 2 ) =

V&1 0.8585

(0.008337 - 0.006065) = 0.002646V&1

The enthalpy of condensate water is hw 2 = h f@ 8°C = 33.63 kJ/kg

(Table A - 4)

An energy balance on the control volume gives m& a h1 = Q& out + m& a h2 + m& w hw2 V&1 V&1 (43.05) = Q& out + (23.11) + 0.002646V&1 (33.63) 0.8585 0.8585

(1)

The properties of the R-134a at the inlet of the compressor and the enthalpy at the exit for the isentropic process are (R-134a tables) PR1 = 375 kPa ⎫ h R1 = 254.48 kJ/kg ⎬ x R1 = 1 ⎭ s R1 = 0.9278 kJ/kg.K PR 2 = 1800 kPa ⎫ ⎬h R 2, s = 286.90 kJ/kg s R 2 = s R1 ⎭

The enthalpies of R-134a at the condenser exit and the throttle exit are h R3 = h f@ 1800 kPa = 144.07 kJ/kg h R 4 = h R 3 = 144.07 kJ/kg

The mass flow rate of the refrigerant can be determined from the expression for the compressor power:

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14-55

W& C = m& R

h R 2, s − h R1

ηC

(286.90 − 254.48) kJ/kg 0.85 & m R = 0.1573 kg/s = 9.439 kg/min

6 kW = m& R

The rate of heat absorbed by the R-134a in the evaporator is Q& R,in = m& R (h R1 − h R 4 ) = (9.439 kg/min)(254.48 − 144.07) kJ/kg = 1042.1 kJ/min

The rate of heat lost from the air in the evaporator is absorbed by the refrigerant-134a. That is, Q& R ,in = Q& out . Then, the volume flow rate of the air at the inlet of the evaporator can be determined from Eq. (1) to be

V&1 0.8474

(43.05) = 1042.1 +

V&1 0.8474

(23.11) + 0.002646V1 (33.63) ⎯ ⎯→V&1 = 44.87 m 3 /min

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14-56

14-95 Air flows through an air conditioner unit. The inlet and exit states are specified. The rate of heat transfer and the mass flow rate of condensate water are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (m& a1 = m& a 2 = m& a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis The inlet state of the air is completely specified, and the total pressure is 98 kPa. The properties of the air at the inlet state may be determined from (Fig. A-31) or using EES psychrometric functions to be (we used EES)

Cooling coils

Tdb2 = 25°C Tdp2 = 6.5°C

h1 = 77.88 kJ/kg dry air

Condensate 2

ω1 = 0.01866 kg H 2 O/kg dry air φ1 = 0.6721

Tdb1 =30°C Twb1 =25°C P = 98 kPa

1 25°C

Condensate removal

The partial pressure of water vapor at the exit state is Pv 2 = Psat@ 6.5°C = 0.9682 kPa

(Table A - 4)

The saturation pressure at the exit state is Pg 2 = Psat@ 25°C = 3.17 kPa

(Table A - 4)

Then, the relative humidity at the exit state becomes

φ2 =

Pv 2 0.9682 = = 0.3054 3.17 Pg 2

Now, the exit state is also fixed. The properties are obtained from EES to be h2 = 40.97 kJ/kg dry air

ω 2 = 0.006206 kg H 2 O/kg dry air

v 2 = 0.8820 m 3 /kg The mass flow rate of dry air is m& a =

V&2 1000 m 3 /min = = 1133.8 kg/min v 2 0.8820 m 3 /kg

The mass flow rate of condensate water is m& w = m& a (ω1 − ω 2 ) = (1133.8 kg/min)(0.01866 - 0.006206) = 14.12 kg/min = 847.2 kg/h

The enthalpy of condensate water is h w 2 = h f@ 25°C = 104.83 kJ/kg

(Table A - 4)

An energy balance on the control volume gives m& a h1 = Q& out + m& a h2 + m& w h w2 (1133.8 kg/min)(77.88 kJ/kg) = Q& out + (1133.8 kg/min)(40.97 kJ/kg) + (14.12 kg/min)(104.83 kJ/kg) Q& = 40,377 kJ/min = 672.9 kW out

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14-57

Evaporative Cooling

14-96C In steady operation, the mass transfer process does not have to involve heat transfer. However, a mass transfer process that involves phase change (evaporation, sublimation, condensation, melting etc.) must involve heat transfer. For example, the evaporation of water from a lake into air (mass transfer) requires the transfer of latent heat of water at a specified temperature to the liquid water at the surface (heat transfer).

14-97C During evaporation from a water body to air, the latent heat of vaporization will be equal to convection heat transfer from the air when conduction from the lower parts of the water body to the surface is negligible, and temperature of the surrounding surfaces is at about the temperature of the water surface so that the radiation heat transfer is negligible.

14-98C Evaporative cooling is the cooling achieved when water evaporates in dry air. It will not work on humid climates.

14-99 Air is cooled by an evaporative cooler. The exit temperature of the air and the required rate of water supply are to be determined. Analysis (a) From the psychrometric chart (Fig. A-31) at 36°C and 20% relative humidity we read Twb1 = 19.5°C

ω1 = 0.0074 kg H 2 O/kg dry air

Water, m& ω

v 1 = 0.887 m /kg dry air

Humidifier

3

Assuming the liquid water is supplied at a temperature not much different than the exit temperature of the air stream, the evaporative cooling process follows a line of constant wet-bulb temperature. That is,

1 atm 36°C 20%

AIR 90%

Twb2 ≅ Twb1 = 19.5°C

At this wet-bulb temperature and 90% relative humidity we read T2 = 20.5° C ω 2 = 0.0137 kg H 2 O / kg dry air

Thus air will be cooled to 20.5°C in this evaporative cooler. (b) The mass flow rate of dry air is m& a =

V&1 4 m 3 / min = = 4.51 kg/min v 1 0.887 m 3 / kg dry air

Then the required rate of water supply to the evaporative cooler is determined from m& supply = m& w2 − m& w1 = m& a (ω 2 − ω1 ) = (4.51 kg/min)(0.0137 - 0.0074) = 0.028 kg/min

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14-58

14-100E Air is cooled by an evaporative cooler. The exit temperature of the air and the required rate of water supply are to be determined. Analysis (a) From the psychrometric chart (Fig. A-31E) at 90°F and 20% relative humidity we read Twb1 = 62.8°F

ω1 = 0.0060 lbm H 2O/lbm dry air

Water, m& ω

v1 = 14.0 ft 3/lbm dry air Assuming the liquid water is supplied at a temperature not much different than the exit temperature of the air stream, the evaporative cooling process follows a line of constant wet-bulb temperature. That is,

Humidifier 1 atm 90°F 20%

AIR 90%

Twb2 ≅ Twb1 = 62.8°F

At this wet-bulb temperature and 90% relative humidity we read

T2 = 65°F

ω 2 = 0.0116 lbm H 2 O/lbm dry air Thus air will be cooled to 64°F in this evaporative cooler. (b) The mass flow rate of dry air is m& a =

V&1 150 ft 3 / min = = 10.7 lbm/min v 1 14.0 ft 3 / lbm dry air

Then the required rate of water supply to the evaporative cooler is determined from m& supply = m& w2 − m& w1 = m& a (ω2 − ω1 ) = (10.7 lbm/min)(0.0116 - 0.0060) = 0.06 lbm/min

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14-59

14-101 Air is cooled by an evaporative cooler. The final relative humidity and the amount of water added are to be determined. Analysis (a) From the psychrometric chart (Fig. A-31) at 32°C and 30% relative humidity we read Twb1 = 19.4°C

ω1 = 0.0089 kg H 2O/kg dry air

Water

v1 = 0.877 m3/kg dry air Assuming the liquid water is supplied at a temperature not much different than the exit temperature of the air stream, the evaporative cooling process follows a line of constant wetbulb temperature. That is,

Humidifier 32°C 30% 2 m3/min

AIR

22°C

Twb2 ≅ Twb1 = 19.4°C

At this wet-bulb temperature and 22°C temperature we read

φ 2 = 79% ω 2 = 0.0130 kg H 2 O/kg dry air (b) The mass flow rate of dry air is m& a =

5 m3 / min V&1 = = 5.70 kg/min v1 0.877 m3 / kg dry air

Then the required rate of water supply to the evaporative cooler is determined from m& supply = m& w2 − m& w1 = m& a (ω 2 − ω1 ) = (5.70 kg/min)(0.0130 - 0.0089) = 0.0234 kg/min

14-102 Air enters an evaporative cooler at a specified state and relative humidity. The lowest temperature that air can attain is to be determined. Analysis From the psychrometric chart (Fig. A-31) at 29°C and 40% relative humidity we read Twb1 = 19.3°C

Assuming the liquid water is supplied at a temperature not much different than the exit temperature of the air stream, the evaporative cooling process follows a line of constant wet-bulb temperature, which is the lowest temperature that can be obtained in an evaporative cooler. That is, Tmin = Twb1 = 19.3°C

Water Humidifier 1 atm 29°C 40%

AIR 100%

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14-60

14-103 Air is first heated in a heating section and then passed through an evaporative cooler. The exit relative humidity and the amount of water added are to be determined. Analysis (a) From the psychrometric chart (Fig. A-31) at 15°C and 60% relative humidity we read

ω 1 = 0.00635 kg H 2 O / kg dry air The specific humidity ω remains constant during the heating process. Therefore, ω2 = ω1 = 0.00635 kg H2O / kg dry air. At this ω value and 30°C we read Twb2 = 16.7°C. Assuming the liquid water is supplied at a temperature not much different than the exit temperature of the air stream, the evaporative cooling process follows a line of constant wet-bulb temperature. That is, Twb3 ≅ Twb2 = 16.7°C. At this Twb value and 25°C we read

Water Heating coils 15°C 60%

Humidifier

1

25°C

30°C

AIR 1 atm 2

3

φ 3 = 42.6% ω 3 = 0.00840 kg H 2 O/kg dry air (b) The amount of water added to the air per unit mass of air is Δω 23 = ω 3 − ω 2 = 0.00840 − 0.00635 = 0.00205 kg H 2 O/kg dry air

14-104E Desert dwellers often wrap their heads with a water-soaked porous cloth. The temperature of this cloth on a desert with specified temperature and relative humidity is to be determined. Analysis Since the cloth behaves as the wick on a wet bulb thermometer, the temperature of the cloth will become the wet-bulb temperature. According to the pshchrometric chart, this temperature is T2 = Twb1 = 73.7°F

This process can be represented by an evaporative cooling process as shown in the figure.

Water Humidifier 1 atm 120°F 10%

AIR 100%

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14-61

Adiabatic Mixing of Airstreams

14-105C This will occur when the straight line connecting the states of the two streams on the psychrometric chart crosses the saturation line.

14-106C Yes.

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14-62

14-107 Two airstreams are mixed steadily. The specific humidity, the relative humidity, the dry-bulb temperature, and the volume flow rate of the mixture are to be determined. Assumptions 1 Steady operating conditions exist 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. 4 The mixing section is adiabatic. Properties Properties of each inlet stream are determined from the psychrometric chart (Fig. A-31) to be h1 = 62.7 kJ/kg dry air

ω1 = 0.0119 kg H 2 O/kg dry air

v 1 = 0.882 m 3 /kg dry air

1

and h2 = 31.9 kJ/kg dry air

32°C 40% 20 m3/min P = 1 atm AIR

ω 2 = 0.0079 kg H 2 O/kg dry air

v 2 = 0.819 m 3 /kg dry air Analysis The mass flow rate of dry air in each stream is m& a1 =

V&1 20 m 3 / min = = 22.7 kg/min v 1 0.882 m 3 / kg dry air

m& a 2 =

V&2 25 m 3 / min = = 30.5 kg/min v 2 0.819 m 3 / kg dry air

2

ω3 φ3

3

T3

25 m3/min 12°C 90%

From the conservation of mass, & a3 = m & a1 + m & a 2 = ( 22.7 + 30.5) kg / min = 53.2 kg / min m

The specific humidity and the enthalpy of the mixture can be determined from Eqs. 14-24, which are obtained by combining the conservation of mass and energy equations for the adiabatic mixing of two streams: & a1 ω 2 − ω 3 h2 − h3 m = = & ma 2 ω 3 − ω 1 h3 − h1

22.7 0.0079 − ω 3 319 . − h3 = = 30.5 ω 3 − 0.0119 h3 − 62.7

which yields,

ω 3 = 0.0096 kg H 2O / kg dry air h3 = 45.0 kJ / kg dry air

These two properties fix the state of the mixture. Other properties of the mixture are determined from the psychrometric chart: T3 = 20.6°C

φ3 = 63.4% v 3 = 0.845 m 3 /kg dry air Finally, the volume flow rate of the mixture is determined from

V&3 = m& a 3v 3 = (53.2 kg/min)(0.845 m 3 / kg) = 45.0 m 3 /min

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14-63

14-108 Two airstreams are mixed steadily. The specific humidity, the relative humidity, the dry-bulb temperature, and the volume flow rate of the mixture are to be determined. Assumptions 1 Steady operating conditions exist 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. 4 The mixing section is adiabatic. Analysis The properties of each inlet stream are determined to be Pv1 = φ1 Pg1 = φ1 Psat @ 32°C = (0.40)(4.760 kPa) = 1.90 kPa Pa1 = P1 − Pv1 = 90 − 1.90 = 88.10 kPa

v1 =

R a T1 (0.287 kPa ⋅ m 3 / kg ⋅ K)(305 K) = 88.10 kPa Pa1

1

= 0.994 m 3 / kg dry air 0.622 Pv1 0.622(1.90 kPa) ω1 = = P1 − Pv1 (90 − 1.90) kPa

P = 90 kPa AIR

= (1.005 kJ/kg ⋅ °C)(32°C) + (0.0134)(2559.2 kJ/kg) = 66.45 kJ/kg dry air

ω3 φ3 3 T3

3

= 0.0134 kg H 2 O/kg dry air h1 = c p T1 + ω1 h g1

32°C 40% 20 m3/min

2

25 m /min 12°C 90%

and Pv 2 = φ 2 Pg 2 = φ 2 Psat@12°C = (0.90)(1.403 kPa) = 1.26 kPa Pa 2 = P2 − Pv 2 = 90 − 1.26 = 88.74 kPa

v2 =

R a T2 (0.287 kPa ⋅ m 3 / kg ⋅ K)(285 K) = = 0.922 m 3 / kg dry air 88.74 kPa Pa 2

ω2 =

0.622 Pv 2 0.622(1.26 kPa) = = 0.00883 kg H 2 O/kg dry air (90 − 1.26) kPa P2 − Pv 2

h2 = c p T2 + ω 2 h g 2 = (1.005 kJ/kg ⋅ °C)(12°C) + (0.00883)(2522.9 kJ/kg) = 34.34 kJ/kg dry air

Then the mass flow rate of dry air in each stream is m& a1 =

V&1 20 m 3 / min = = 20.12 kg/min v 1 0.994 m 3 / kg dry air

m& a 2 =

V&2 25 m 3 / min = = 27.11 kg/min v 2 0.922 m 3 / kg dry air

From the conservation of mass, m& a 3 = m& a1 + m& a 2 = (20.12 + 27.11) kg/min = 47.23 kg/min

The specific humidity and the enthalpy of the mixture can be determined from Eqs. 14-24, which are obtained by combining the conservation of mass and energy equations for the adiabatic mixing of two streams: m& a1 ω2 − ω3 h2 − h3 20.12 0.00883 − ω3 34.34 − h3 = = ⎯ ⎯→ = = m& a 2 ω3 − ω1 h3 − h1 27.11 ω3 − 0.0134 h3 − 66.45

which yields

ω3 = 0.0108 kg H 2O/kg dry air h3 = 48.02 kJ/kg dry air PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

14-64

These two properties fix the state of the mixture. Other properties are determined from h3 = c pT3 + ω3hg 3 ≅ c pT3 + ω3 (2501.3 + 1.82T3 ) 48.02 kJ/kg = (1.005 kJ/kg ⋅ °C)T3 + (0.0108)(2500.9 + 1.82T3 ) kJ/kg ⎯⎯→ T3 = 20.5°C

ω3 = 0.0108 =

φ3 =

0.622 Pv 3 P3 − Pv 3 0.622 Pv 3 ⎯ ⎯→ Pv 3 = 1.54 kPa 90 − Pv 3 Pv 3 Pv 3 1.54 kPa = = = 0.639 or 63.9% Pg 3 Psat @ T3 2.41 kPa

Finally, Pa 3 = P3 − Pv3 = 90 − 1.54 = 88.46 kPa

v3 =

Ra T3 (0.287 kPa ⋅ m 3 / kg ⋅ K)(293.5 K) = = 0.952 m 3 /kg dry air Pa 3 88.46 kPa

V&3 = m& a 3v 3 = (47.23 kg/min)(0.952 m 3 / kg) = 45.0 m 3 /min

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14-65

14-109E Two airstreams are mixed steadily. The temperature and the relative humidity of the mixture are to be determined. Assumptions 1 Steady operating conditions exist 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. 4 The mixing section is adiabatic. Properties Properties of each inlet stream are determined from the psychrometric chart (Fig. A-31E or from EES) to be h1 = 66.7 Btu/lbm dry air

ω1 = 0.0386 lbm H 2 O/lbm dry air

1

v 1 = 14.98 ft 3 /lbm dry air and

100°F 90% 3 ft3/s

P = 1 atm AIR

h2 = 14.5 Btu/lbm dry air

ω 2 = 0.0023 lbm H 2 O/lbm dry air

v 2 = 12.90 ft 3 /lbm dry air Analysis The mass flow rate of dry air in each stream is m& a1 =

V&1 3 ft 3 / s = = 0.2002 lbm/s v 1 14.98 ft 3 / lbmdry air

m& a 2 =

V&2 1 ft 3 / s = = 0.07755 lbm/s v 2 12.90 ft 3 / lbm dry air

2

ω3 φ3 3 T3

1 ft3/s 50°F 30%

From the conservation of mass, m& a 3 = m& a1 + m& a 2 = (0.2002 + 0.07755) lbm/s = 0.2778 lbm/s

The specific humidity and the enthalpy of the mixture can be determined from Eqs. 14-24, which are obtained by combining the conservation of mass and energy equations for the adiabatic mixing of two streams: m& a1 ω 2 − ω 3 h2 − h3 = = m& a 2 ω 3 − ω1 h3 − h1 0.2002 0.0023 − ω 3 14.5 − h3 = = 0.07755 ω 3 − 0.0386 h3 − 66.7

which yields

ω 3 = 0.0284 lbm H 2 O/lbm dry air h3 = 52.1 Btu/lbm dry air These two properties fix the state of the mixture. Other properties of the mixture are determined from the psychrometric chart: T3 = 86.7°F

φ 3 = 1.0 = 100%

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14-66

14-110E Two airstreams are mixed steadily. The rate of entropy generation is to be determined. Assumptions 1 Steady operating conditions exist 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. 4 The mixing section is adiabatic. Properties Properties of each inlet stream are determined from the psychrometric chart (Fig. A-31 or from EES) to be h1 = 66.7 Btu/lbm dry air

ω1 = 0.0386 lbm H 2 O/lbm dry air

v 1 = 14.98 ft 3 /lbm dry air

1

and

100°F 90% 3 ft3/s

h2 = 14.5 Btu/lbm dry air

P = 1 atm AIR

ω 2 = 0.0023 lbm H 2 O/lbm dry air

v 2 = 12.90 ft /lbm dry air 3

T3

3

The entropies of water vapor in the air streams are s g1 = s g @ 100° F = 1.9819 Btu/lbm ⋅ R

ω3 φ3 3

2

1 ft /s 50°F 30%

s g 2 = s g @ 50° F = 2.1256 Btu/lbm ⋅ R

Analysis The mass flow rate of dry air in each stream is m& a1 =

V&1 3 ft 3 / s = = 0.2002 lbm/s v 1 14.98 ft 3 / lbmdry air

m& a 2 =

V&2 1 ft 3 / s = = 0.07755 lbm/s v 2 12.90 ft 3 / lbm dry air

From the conservation of mass, m& a 3 = m& a1 + m& a 2 = (0.2002 + 0.07755) lbm/s = 0.2778 lbm/s

The specific humidity and the enthalpy of the mixture can be determined from Eqs. 14-24, which are obtained by combining the conservation of mass and energy equations for the adiabatic mixing of two streams: m& a1 ω 2 − ω 3 h2 − h3 = = m& a 2 ω 3 − ω1 h3 − h1 0.2002 0.0023 − ω 3 14.5 − h3 = = 0.07755 ω 3 − 0.0386 h3 − 66.7

which yields

ω 3 = 0.0284 lbm H 2 O/lbm dry air h3 = 52.1 Btu/lbm dry air These two properties fix the state of the mixture. Other properties of the mixture are determined from the psychrometric chart: T3 = 86.7°F

φ 3 = 1.0 = 100% The entropy of water vapor in the mixture is

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14-67 s g 3 = s g @ 86.7° F = 2.0168 Btu/lbm ⋅ R

An entropy balance on the mixing chamber for the water gives ΔS& w = m& a 3ω 3 s 3 − m& a1ω1 s1 − m& a 2 ω 2 s 2 = 0.2778 × 0.0284 × 2.0168 − 0.2002 × 0.0386 × 1.9819 − 0.07755 × 0.0023 × 2.1256 = 2.169 × 10 − 4 Btu/s ⋅ R

The partial pressures of water vapor and dry air for all three air streams are Pv1 = φ1 Pg1 = φ1 Psat @ 100°F = (0.90)(0.95052 psia) = 0.8555 psia Pa1 = P1 − Pv1 = 14.696 − 0.8555 = 13.84 psia Pv 2 = φ 2 Pg 2 = φ 2 Psat @ 50° F = (0.30)(0.17812 psia) = 0.0534 psia Pa 2 = P2 − Pv 2 = 14.696 − 0.0534 = 14.64 psia Pv 3 = φ 3 Pg 3 = φ 3 Psat @ 86.7°C = (1.0)(0.6298 psia) = 0.6298 psia Pa 3 = P3 − Pv 3 = 14.696 − 0.6298 = 14.07 psia

An entropy balance on the mixing chamber for the dry air gives ΔS&a = m& a1( s3 − s1) + m& a 2 ( s3 − s2 ) ⎛ ⎛ T P ⎞ T P ⎞ = m& a1⎜⎜ c p ln 3 − R ln a3 ⎟⎟ + m& a 2 ⎜⎜ c p ln 3 − R ln a3 ⎟⎟ T P T P 1 a1 ⎠ 2 a2 ⎠ ⎝ ⎝ 14.07 ⎤ 546.7 14.07 ⎤ 546.7 ⎡ ⎡ = 0.2002⎢(0.240) ln − (0.06855) ln ⎥ + 0.07755⎢(0.240) ln 510 − (0.06855) ln 14.64 ⎥ 13 . 84 560 ⎣ ⎦ ⎣ ⎦ = (0.2002)(−0.006899) + (0.07755)(0.01940) = 1.233 × 10− 4 Btu/s ⋅ R

The rate of entropy generation is then S& gen = ΔS& a + ΔS& w = 1.233 × 10 −4 + 2.169 × 10 −4 = 3.402 × 10 −4 Btu/s ⋅ R

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

14-68

14-111 Two airstreams are mixed steadily. The mass flow ratio of the two streams for a specified mixture relative humidity and the temperature of the mixture are to be determined. Assumptions 1 Steady operating conditions exist 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. 4 The mixing section is adiabatic. Properties Properties of each inlet stream are determined from the psychrometric chart (Fig. A-31 or from EES) to be h1 = 29.4 kJ/kg dry air

ω1 = 0.0077 kg H 2 O/kg dry air and

1

h2 = 94.6 kJ/kg dry air

10°C 100%

ω 2 = 0.0244 kg H 2 O/kg dry air Analysis An application of Eq. 14-24, which are obtained by combining the conservation of mass and energy equations for the adiabatic mixing of two streams gives m& a1 ω 2 − ω 3 h2 − h3 = = m& a 2 ω 3 − ω1 h3 − h1 m& a1 0.0244 − ω 3 94.6 − h3 = = m& a 2 ω 3 − 0.0077 h3 − 29.4

P = 1 atm AIR

70%

ω3 3 T3

32°C 80% 2

This equation cannot be solved directly. An iterative solution is needed. A mixture temperature T3 is selected. At this temperature and given relative humidity (70%), specific humidity and enthalpy are read from the psychrometric chart. These values are substituted into the above equation. If the equation is not satisfied, a new value of T3 is selected. This procedure is repeated until the equation is satisfied. Alternatively, EES software can be used. We used the following EES program to get the results: "Given" P=101.325 [kPa] T_1=10 [C] phi_1=1.0 T_2=32 [C] phi_2=0.80 phi_3=0.70 "Analysis" Fluid$='AirH2O' "1st stream properties" h_1=enthalpy(Fluid$, T=T_1, P=P, R=phi_1) w_1=humrat(Fluid$, T=T_1, P=P, R=phi_1) "2nd stream properties" h_2=enthalpy(Fluid$, T=T_2, P=P, R=phi_2) w_2=humrat(Fluid$, T=T_2, P=P, R=phi_2) (w_2-w_3)/(w_3-w_1)=(h_2-h_3)/(h_3-h_1) Ratio=(w_2-w_3)/(w_3-w_1) "mixture properties" T_3=temperature(Fluid$, h=h_3, P=P, R=phi_3) h_3=enthalpy(Fluid$, T=T_3, P=P, R=phi_3)

The solution of this EES program is T3 = 24.0°C, ω 3 = 0.0149 kg H 2 O/kg dry air m& a1 = 1.31 h3 = 57.6 kJ/kg dry air, m& a 2

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14-69

14-112 A stream of warm air is mixed with a stream of saturated cool air. The temperature, the specific humidity, and the relative humidity of the mixture are to be determined. Assumptions 1 Steady operating conditions exist 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. 4 The mixing section is adiabatic. Properties The properties of each inlet stream are determined from the psychrometric chart (Fig. A-31) to be h1 = 110.2 kJ/kg dry air

ω1 = 0.0272 kg H 2 O/kg dry air

1

and

h2 = 50.9 kJ/kg dry air

40°C 8 kg/s Twb1 = 32°C

ω 2 = 0.0129 kg H 2 O/kg dry air

P = 1 atm AIR

Analysis The specific humidity and the enthalpy of the mixture can be determined from Eqs. 14-24, which are obtained by combining the conservation of mass and energy equations for the adiabatic mixing of two streams: m& a1 ω 2 − ω 3 h2 − h3 = = m& a 2 ω 3 − ω1 h3 − h1

2

ω3 φ3 3 T3

6 kg/s 18°C 100%

50.9 − h3 8.0 0.0129 − ω 3 = = 6.0 ω 3 − 0.0272 h3 − 110.2

which yields, (b)

ω 3 = 0.0211 kg H 2O / kg dry air h3 = 84.8 kJ / kg dry air

These two properties fix the state of the mixture. Other properties of the mixture are determined from the psychrometric chart: (a)

T3 = 30.7° C

(c)

φ 3 = 75.1%

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14-70

14-113 EES Problem 14-112 is reconsidered. The effect of the mass flow rate of saturated cool air stream on the mixture temperature, specific humidity, and relative humidity is to be investigated. Analysis The problem is solved using EES, and the solution is given below. P=101.325 [kPa] Tdb[1] =40 [C] Twb[1] =32 [C] m_dot[1] = 8 [kg/s] Tdb[2] =18 [C] Rh[2] = 1.0 m_dot[2] = 6 [kg/s] P[1]=P P[2]=P[1] P[3]=P[1] "Energy balance for the steady-flow mixing process:" "We neglect the PE of the flow. Since we don't know the cross sectional area of the flow streams, we also neglect theKE of the flow." E_dot_in - E_dot_out = DELTAE_dot_sys DELTAE_dot_sys = 0 [kW] E_dot_in = m_dot[1]*h[1]+m_dot[2]*h[2] E_dot_out = m_dot[3]*h[3] "Conservation of mass of dry air during mixing:" m_dot[1]+m_dot[2] = m_dot[3] "Conservation of mass of water vapor during mixing:" m_dot[1]*w[1]+m_dot[2]*w[2] = m_dot[3]*w[3] m_dot[1]=V_dot[1]/v[1]*convert(1/min,1/s) m_dot[2]=V_dot[2]/v[2]*convert(1/min,1/s) h[1]=ENTHALPY(AirH2O,T=Tdb[1],P=P[1],B=Twb[1]) Rh[1]=RELHUM(AirH2O,T=Tdb[1],P=P[1],B=Twb[1]) v[1]=VOLUME(AirH2O,T=Tdb[1],P=P[1],R=Rh[1]) w[1]=HUMRAT(AirH2O,T=Tdb[1],P=P[1],R=Rh[1]) h[2]=ENTHALPY(AirH2O,T=Tdb[2],P=P[2],R=Rh[2]) v[2]=VOLUME(AirH2O,T=Tdb[2],P=P[2],R=Rh[2]) w[2]=HUMRAT(AirH2O,T=Tdb[2],P=P[2],R=Rh[2]) Tdb[3]=TEMPERATURE(AirH2O,h=h[3],P=P[3],w=w[3]) Rh[3]=RELHUM(AirH2O,T=Tdb[3],P=P[3],w=w[3]) v[3]=VOLUME(AirH2O,T=Tdb[3],P=P[3],w=w[3]) Twb[2]=WETBULB(AirH2O,T=Tdb[2],P=P[2],R=RH[2]) Twb[3]=WETBULB(AirH2O,T=Tdb[3],P=P[3],R=RH[3]) m_dot[3]=V_dot[3]/v[3]*convert(1/min,1/s) m2 [kga/s] 0 2 4 6 8 10 12 14 16

Tdb3 [C] 40 35.69 32.79 30.7 29.13 27.91 26.93 26.13 25.45

Rh3 0.5743 0.6524 0.7088 0.751 0.7834 0.8089 0.8294 0.8462 0.8601

w3 [kgw/kga] 0.02717 0.02433 0.02243 0.02107 0.02005 0.01926 0.01863 0.01811 0.01768

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14-71

40 38

Tdb[3] [C]

36 34 32 30 28 26 24 0

2

4

6

8

10

12

14

16

12

14

16

m[2] [kga/s] 0.9 0.85

Rh[3]

0.8 0.75 0.7 0.65 0.6 0.55 0

2

4

6

8

10

m[2] [kga/s]

0.028

w [3] [kgw /kga]

0.026 0.024 0.022 0.02 0.018 0.016 0

2

4

6

8

10

12

14

16

m [2] [kga/s]

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14-72

Wet Cooling Towers

14-114C The working principle of a natural draft cooling tower is based on buoyancy. The air in the tower has a high moisture content, and thus is lighter than the outside air. This light moist air rises under the influence of buoyancy, inducing flow through the tower.

14-115C A spray pond cools the warm water by spraying it into the open atmosphere. They require 25 to 50 times the area of a wet cooling tower for the same cooling load.

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14-73

14-116 Water is cooled by air in a cooling tower. The volume flow rate of air and the mass flow rate of the required makeup water are to be determined. Assumptions 1 Steady operating conditions exist and thus mass flow rate of dry air remains constant during the entire process. 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. 4 The cooling tower is adiabatic. Analysis (a) The mass flow rate of dry air through the tower remains constant (m& a1 = m& a 2 = m& a ) , but the mass flow rate of liquid water decreases by an amount equal to the amount of water that vaporizes in the tower during the cooling process. The water lost through evaporation must be made up later in the cycle to maintain steady operation. Applying the mass and energy balances yields Dry Air Mass Balance: ∑ m& a ,i = ∑ m& a ,e

⎯ ⎯→

m& a1 = m& a 2 = m& a

Water Mass Balance: ∑ m& w,i = ∑ m& w,e → m& 3 + m& a1ω1 = m& 4 + m& a 2 ω 2

AIR 34°C 2 EXIT 90%

m& 3 − m& 4 = m& a (ω 2 − ω1 ) = m& makeup

Energy Balance: E& − E& = ΔE& in

out

system

Ê0 (steady)

=0

E& in = E& out & i hi = ∑ m & e he (since Q& = W& = 0) ∑m & e he − ∑ m & i hi 0= ∑m

WARM WATER

3

40°C 60 kg/s

& a 2 h2 + m & 4 h4 − m & a1h1 − m & 3h3 0=m & a ( h2 − h1 ) + ( m &3 − m & makeup )h4 − m & 3h3 0=m

Solving for m& a , m& 3 (h3 − h4 ) m& a = (h2 − h1 ) − (ω 2 − ω 1 )h4

From the psychrometric chart (Fig. A-31), h1 = 44.7 kJ/kg dry air

4 COOL WATER

1 AIR INLET 1 atm Tdb = 22°C Twb = 16°C

26°C Makeup water

ω1 = 0.0089 kg H 2 O/kg dry air

v 1 = 0.849 m 3 /kg dry air and h2 = 1135 . kJ / kg dry air

ω 2 = 0.0309 kg H 2 O / kg dry air From Table A-4, h3 ≅ h f @ 40°C = 167.53 kJ/kg H 2 O h4 ≅ h f @ 26°C = 109.01 kJ/kg H 2 O

Substituting, m& a =

(60 kg/s)(167.53 − 109.01)kJ/kg = 52.9 kg/s (113.5 − 44.7) kJ/kg − (0.0309 − 0.0089)(109.01) kJ/kg

Then the volume flow rate of air into the cooling tower becomes V& = m& v = (52.9 kg/s)(0.849 m 3 / kg ) = 44.9 m 3 /s 1

a 1

(b) The mass flow rate of the required makeup water is determined from & makeup = m & a (ω 2 − ω 1 ) = (52.9 kg / s)(0.0309 − 0.0089) = 1.16 kg / s m

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14-74

14-117 Water is cooled by air in a cooling tower. The mass flow rate of dry air is to be determined. Assumptions 1 Steady operating conditions exist and thus mass flow rate of dry air remains constant during the entire process. 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. 4 The cooling tower is adiabatic. Analysis The mass flow rate of dry air through the tower remains constant (m& a1 = m& a 2 = m& a ) , but the mass flow rate of liquid water decreases by an amount equal to the amount of water that vaporizes in the tower during the cooling process. The water lost through evaporation must be made up later in the cycle to maintain steady operation. Applying the mass and energy balances yields

Dry Air Mass Balance: ∑ m& a ,i = ∑ m& a ,e ⎯ ⎯→ m& a1 = m& a 2 = m& a Water Mass Balance:

AIR EXIT

∑ m& w,i = ∑ m& w,e → m& 3 + m& a1ω1 = m& 4 + m& a 2ω 2

2

18°C 95%

m& 3 − m& 4 = m& a (ω 2 − ω1 ) = m& makeup

Energy Balance: E& in − E& out = ΔE& systemÊ0 (steady) = 0 E&in = E& out

WARM WATER

3

30°C 5 kg/s

∑ m& i hi = ∑ m& e he (since Q& = W& = 0) 0 = ∑ m& e he − ∑ m& i hi 0 = m& a 2h2 + m& 4h4 − m& a1h1 − m& 3h3 0 = m& a (h2 − h1 ) + (m& 3 − m& makeup )h4 − m& 3h3

1 AIR 4

Solving for m& a , m& a =

m& 3 (h3 − h4 ) (h2 − h1 ) − (ω 2 − ω1 )h4

From the psychrometric chart (Fig. A-31),

COOL WATER

INLET 1 atm 15°C 25%

22°C Makeup water

h1 = 21.8 kJ/kg dry air

ω1 = 0.00264 kg H 2 O/kg dry air

v 1 = 0.820 m 3 /kg dry air and h2 = 49.3 kJ/kg dry air

ω 2 = 0.0123 kg H 2 O/kg dry air From Table A-4, h3 ≅ h f @ 30°C = 125.74 kJ/kg H 2 O h4 ≅ h f @ 22°C = 92.28 kJ/kg H 2 O

Substituting, m& a =

(5 kg/s)(125.74 − 92.28)kJ/kg = 6.29 kg/s (49.3 − 21.8) kJ/kg − (0.0123 − 0.00264)(92.28) kJ/kg

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14-75

14-118 Water is cooled by air in a cooling tower. The exergy lost in the cooling tower is to be determined. Assumptions 1 Steady operating conditions exist and thus mass flow rate of dry air remains constant during the entire process. 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. 4 The cooling tower is adiabatic. Analysis The mass flow rate of dry air through the tower remains constant (m& a1 = m& a 2 = m& a ) , but the mass flow rate of liquid water decreases by an amount equal to the amount of water that vaporizes in the tower during the cooling process. The water lost through evaporation must be made up later in the cycle to maintain steady operation. Applying the mass and energy balances yields

Dry Air Mass Balance: ∑ m& a ,i = ∑ m& a ,e ⎯ ⎯→ m& a1 = m& a 2 = m& a Water Mass Balance:

AIR EXIT

∑ m& w,i = ∑ m& w,e → m& 3 + m& a1ω1 = m& 4 + m& a 2ω 2

2

18°C 95%

m& 3 − m& 4 = m& a (ω 2 − ω1 ) = m& makeup

Energy Balance: E& in − E& out = ΔE& systemÊ0 (steady) = 0 E&in = E& out

WARM WATER

3

30°C 5 kg/s

∑ m& i hi = ∑ m& e he (since Q& = W& = 0) 0 = ∑ m& e he − ∑ m& i hi 0 = m& a 2h2 + m& 4h4 − m& a1h1 − m& 3h3 0 = m& a (h2 − h1 ) + (m& 3 − m& makeup )h4 − m& 3h3 4

Solving for m& a , m& a =

m& 3 (h3 − h4 ) (h2 − h1 ) − (ω 2 − ω1 )h4

From the psychrometric chart (Fig. A-31),

1 AIR

COOL WATER

INLET 1 atm 15°C 25%

22°C Makeup water

h1 = 21.8 kJ/kg dry air

ω1 = 0.00264 kg H 2 O/kg dry air

v 1 = 0.820 m 3 /kg dry air and h2 = 49.3 kJ/kg dry air

ω 2 = 0.0123 kg H 2 O/kg dry air From Table A-4, h3 ≅ h f @ 30°C = 125.74 kJ/kg H 2 O h4 ≅ h f @ 22°C = 92.28 kJ/kg H 2 O

Substituting, m& a =

(5 kg/s)(125.74 − 92.28)kJ/kg = 6.29 kg/s (49.3 − 21.8) kJ/kg − (0.0123 − 0.00264)(92.28) kJ/kg

The mass flow rate of water stream at state 3 per unit mass of dry air is PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

14-76

m3 =

m& 3 5 kg water/s = = 0.7949 kg water/kg dry air m& a 6.29 kg dry air/s

The mass flow rate of water stream at state 4 per unit mass of dry air is m 4 = m3 − (ω 2 − ω1 ) = 0.7949 − (0.0123 − 0.00264) = 0.7852 kg water/kg dry air

The entropies of water streams are s 3 = s f @ 30°C = 0.4368 kJ/kg ⋅ K s 4 = s f @ 22°C = 0.3249 kJ/kg ⋅ K

The entropy change of water stream is Δs water = m 4 s 4 − m3 s 3 = 0.7852 × 0.3249 − 0.7949 × 0.4368 = −0.09210 kJ/K ⋅ kg dry air

The entropies of water vapor in the air stream are s g1 = s g @ 15°C = 8.7803 kJ/kg ⋅ K s g 2 = s g @ 18°C = 8.7112 kJ/kg ⋅ K

The entropy change of water vapor in the air stream is Δs vapor = ω 2 s g 2 − ω1 s g1 = 0.0123 × 8.7112 − 0.00264 × 8.7803 = 0.08397 kJ/K ⋅ kg dry air

The partial pressures of water vapor and dry air for air streams are Pv1 = φ1 Pg1 = φ1 Psat @ 15°C = (0.25)(1.7057 kPa) = 0.4264 kPa Pa1 = P1 − Pv1 = 101.325 − 0.4264 = 100.90 kPa Pv 2 = φ 2 Pg 2 = φ 2 Psat @ 18°C = (0.95)(2.065 kPa) = 1.962 kPa Pa 2 = P2 − Pv 2 = 101.325 − 1.962 = 99.36 kPa

The entropy change of dry air is Δs a = s 2 − s1 = c p ln = (1.005) ln

P T2 − R ln a 2 Pa1 T1

291 99.36 − (0.287) ln = 0.01483 kJ/kg dry air 288 100.90

The entropy generation in the cooling tower is the total entropy change: s gen = Δs water + Δs vapor + Δs a = −0.09210 + 0.08397 + 0.01483 = 0.00670 kJ/K ⋅ kg dry air

Finally, the exergy destruction per unit mass of dry air is x dest = T0 s gen = (288 K)(0.00670 kJ/K ⋅ kg dry air) = 1.93 kJ/kg dry air

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14-77

14-119E Water is cooled by air in a cooling tower. The relative humidity of the air at the exit and the water’s exit temperature are to be determined. Assumptions 1 Steady operating conditions exist and thus mass flow rate of dry air remains constant during the entire process. 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. 4 The cooling tower is adiabatic. Analysis The mass flow rate of dry air through the tower remains constant (m& a1 = m& a 2 = m& a ) , but the mass flow rate of liquid water decreases by an amount equal to the amount of water that vaporizes in the tower during the cooling process. The water lost through evaporation must be made up later in the cycle to maintain steady operation. Applying the mass and energy balances yields Dry Air Mass Balance: ∑ m& a ,i = ∑ m& a ,e ⎯ ⎯→ m& a1 = m& a 2 = m& a

AIR EXIT

Water Mass Balance: ∑ m& w,i = ∑ m& w,e → m& 3 + m& a1ω1 = m& 4 + m& a 2ω 2

2

75°F

ω=0.018

m& 3 − m& 4 = m& a (ω 2 − ω1 ) = m& makeup

Energy Balance: E& − E& = ΔE& in

out

Ê0 (steady)

system

=0

E& in = E& out

WARM WATER

3

100°F 10,000 lbm/h

∑ m& i hi = ∑ m& e he (since Q& = W& = 0) 0 = ∑ m& e he − ∑ m& i hi 0 = m& a 2 h2 + m& 4 h4 − m& a1 h1 − m& 3 h3

1 AIR

0 = m& a (h2 − h1 ) + (m& 3 − m& makeup )h4 − m& 3 h3

Solving for h4, m& h − m& a (h2 − h1 ) h4 = 3 3 m& 3 − m& makeup

4 COOL WATER

From the psychrometric chart (Fig. A-31E), h1 = 16.8 Btu/lbm dry air

INLET 1 atm 60°F 20% 7000 lbm/h

Makeup water

ω1 = 0.00219 kg H 2 O/kg dry air

v 1 = 13.15 ft 3 /lbm dry air and h2 = 37.7 Btu/lbm dry air

φ 2 = 0.957 = 95.7% From Table A-4, h3 ≅ h f @ 100°F = 68.03 Btu/lbm H 2 O Also, m& makeup = m& a (ω 2 − ω1 ) = (7000 / 3600 lbm/s)(0.018 − 0.00219) = 0.03075 lbm/s

Substituting, m& h − m& a (h2 − h1 ) (10,000 / 3600)(68.03) − (7000 / 3600)(37.7 − 16.8) h4 = 3 3 = = 53.99 Btu/lbm (10,000 / 3600) − 0.03075 m& 3 − m& makeup The exit temperature of the water is then (Table A-4E) T4 = Tsat @ h f =53.99 Btu/lbm = 85.9°F

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

14-78

14-120 Water is cooled by air in a cooling tower. The volume flow rate of air and the mass flow rate of the required makeup water are to be determined. Assumptions 1 Steady operating conditions exist and thus mass flow rate of dry air remains constant during the entire process. 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. 4 The cooling tower is adiabatic. Analysis (a) The mass flow rate of dry air through the tower remains constant (m& a1 = m& a 2 = m& a ) , but the mass flow rate of liquid water decreases by an amount equal to the amount of water that vaporizes in the tower during the cooling process. The water lost through evaporation must be made up later in the cycle to maintain steady operation. Applying the mass and energy balances yields

Dry Air Mass Balance: ∑ m& a ,i = ∑ m& a ,e

2

35°C 100%

m& a1 = m& a 2 = m& a Water Mass Balance: ∑ m& w,i = ∑ m& w,e m& 3 + m& a1ω1 = m& 4 + m& a 2 ω 2 m& 3 − m& 4 = m& a (ω 2 − ω1 ) = m& makeup

Energy Balance: E& in − E& out = ΔE& systemÊ0 (steady) = 0 ⎯⎯→ E& in = E& out ∑ m& i hi = ∑ m& e he (since Q& = W& = 0) 0 = ∑ m& e he − ∑ m& i hi 0 = m& a 2 h2 + m& 4 h4 − m& a1h1 − m& 3h3 0 = m& a (h2 − h1 ) + (m& 3 − m& makeup )h4 − m& 3h3 m& a =

m& 3 (h3 − h4 ) (h2 − h1 ) − (ω 2 − ω 1 )h4

3

WATER 40°C 25 kg/s System boundary

1 4 30°C

AIR

96 kPa 20°C 70%

Makeup

The properties of air at the inlet and the exit are Pv1 = φ1Pg1 = φ1Psat @ 20°C = (0.70)(2.3392 kPa) = 1.637 kPa Pa1 = P1 − Pv1 = 96 − 1.637 = 94.363 kPa

v1 =

RaT1 (0.287 kPa ⋅ m3 / kg ⋅ K)(293 K) = = 0.891 m3 / kg dry air Pa1 94.363 kPa

ω1 =

0.622 Pv1 0.622(1.637 kPa) = = 0.0108 kg H 2O/kg dry air P1 − Pv1 (96 − 1.637) kPa

h1 = c pT1 + ω1hg1 = (1.005 kJ/kg ⋅ °C)(20°C) + (0.0108)(2537.4 kJ/kg) = 47.5 kJ/kg dry air

and Pv 2 = φ2 Pg 2 = φ2 Psat @ 35°C = (1.00)(5.6291 kPa) = 5.6291 kPa

ω2 =

0.622 Pv 2 0.622(5.6291 kPa) = = 0.0387 kg H 2O/kg dry air P2 − Pv 2 (96 − 5.6291) kPa

h2 = c pT2 + ω2 hg 2 = (1.005 kJ/kg ⋅ °C)(35°C) + (0.0387)(2564.6 kJ/kg) = 134.4 kJ/kg dry air

From Table A-4, PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

14-79 h3 ≅ h f @ 40°C = 167.53 kJ/kg H 2 O h4 ≅ h f @ 30°C = 125.74 kJ/kg H 2 O

Substituting, m& a =

(25 kg/s)(167.53 − 125.74)kJ/kg = 12.53 kg/s (134.4 − 47.5) kJ/kg − (0.0387 − 0.0108)(125.74) kJ/kg

Then the volume flow rate of air into the cooling tower becomes

V&1 = m& av 1 = (12.53 kg/s)(0.891 m 3 / kg ) = 11.2 m 3 /s (b) The mass flow rate of the required makeup water is determined from m& makeup = m& a (ω 2 − ω 1 ) = (12.53 kg/s)(0.0387 − 0.0108) = 0.35 kg/s

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14-80

14-121 A natural-draft cooling tower is used to remove waste heat from the cooling water flowing through the condenser of a steam power plant. The mass flow rate of the cooling water, the volume flow rate of air into the cooling tower, and the mass flow rate of the required makeup water are to be determined. Assumptions 1 All processes are steady-flow and the mass flow rate of dry air remains constant during the entire process (m& a1 = m& a 2 = m& a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis The inlet and exit states of the moist air for the tower are completely specified. The properties may be determined from the psychrometric chart (Fig. A-31) or using EES psychrometric functions to be (we used EES) h1 = 50.74 kJ/kg dry air

ω1 = 0.01085 kg H 2 O/kg dry air

v 1 = 0.8536 m 3 /kg dry air h2 = 142.83 kJ/kg dry air

T2 = 37°C φ 2 =100%

ω 2 = 0.04112 kg H 2 O/kg dry air The enthalpies of cooling water at the inlet and exit of the condenser are (Table A-4) hw3 = h f@ 40°C = 167.53 kJ/kg

T1 = 23°C Twb1 = 18°C

AIR 2

1 Makeup water

hw 4 = h f@ 26°C = 109.01 kJ/kg

The steam properties for the condenser are (Steam tables) Ps1 = 200 kPa ⎫ ⎬h s1 = 504.71 kJ/kg x s1 = 0 ⎭ Ps 2 = 10 kPa

⎫ ⎬hs 2 = 2524.3 kJ/kg s s 2 = 7.962 kJ/kg.K ⎭ Ps 3 = 10 kPa ⎫ ⎬h s 3 = 191.81 kJ/kg x s1 = 0 ⎭

The mass flow rate of dry air is given by V& V&1 m& a = 1 = v 1 0.8536 m 3 /kg The mass flow rates of vapor at the inlet and exit of the cooling tower are V&1 m& v1 = ω1m& a = (0.01085) = 0.01271V&1 0.8536 V&1 m& v 2 = ω 2 m& a = (0.04112) = 0.04817V&1 0.8536 Mass and energy balances on the cooling tower give m& v1 + m& cw3 = m& v 2 + m& cw4 m& a h1 + m& cw3 hw3 = m& a h2 + m& cw4 h w4 The mass flow rate of the makeup water is determined from m& makeup = m& v 2 − m& v1 = m& cw3 − m& cw4 An energy balance on the condenser gives 0.18m& s h s1 + 0.82m& s h s 2 + m& cw4 h w4 + m& makeup h w 4 = m& s h s 3 + m& cw3 h w3 Solving all the above equations simultaneously with known and determined values using EES, we obtain m& cw3 = 1413 kg/s V& = 47,700 m 3 /min 1

m& makeup = 28.19 kg/s

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14-81

Review Problems

14-122 Air is compressed by a compressor and then cooled to the ambient temperature at high pressure. It is to be determined if there will be any condensation in the compressed air lines. Assumptions The air and the water vapor are ideal gases. Properties The saturation pressure of water at 20°C is 2.3392 kPa (Table A-4).. Analysis The vapor pressure of air before compression is Pv1 = φ1 Pg = φ1 Psat @ 25°C = (0.50)(2.3392 kPa) = 1.17 kPa

The pressure ratio during the compression process is (800 kPa)/(92 kPa) = 8.70. That is, the pressure of air and any of its components increases by 8.70 times. Then the vapor pressure of air after compression becomes Pv 2 = Pv1 × (Pressure ratio) = (1.17 kPa)(8.70) = 10.2 kPa

The dew-point temperature of the air at this vapor pressure is Tdp = Tsat @ Pv 2 = Tsat @ 10.2 kPa = 46.1°C

which is greater than 20°C. Therefore, part of the moisture in the compressed air will condense when air is cooled to 20°C.

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14-82

14-123E The mole fraction of the water vapor at the surface of a lake and the mole fraction of water in the lake are to be determined and compared. Assumptions 1 Both the air and water vapor are ideal gases. 2 Air is weakly soluble in water and thus Henry’s law is applicable. Properties The saturation pressure of water at 60°F is 0.2564 psia (Table A-4E). Henry’s constant for air dissolved in water at 60ºF (289 K) is given in Table 16-2 to be H = 62,000 bar. Analysis The air at the water surface will be saturated. Therefore, the partial pressure of water vapor in the air at the lake surface will simply be the saturation pressure of water at 60°F, Pvapor = Psat @60° F = 0.2564 psia

Assuming both the air and vapor to be ideal gases, the mole fraction of water vapor in the air at the surface of the lake is determined to be y vapor =

Pvapor P

=

Air 13.8 psi

0.2564 psia = 0.0186 (or 1.86 percent) 13.8 psia

The partial pressure of dry air just above the lake surface is

60°F

Lake

Pdry air = P − Pvapor = 13.8 − 0.2564 = 13.54 psia

Then the mole fraction of air in the water becomes ydry air, liquid side =

Pdry air, gasside H

=

1354 . psia(1 atm / 14.696 psia ) = 151 . × 10 −5 62,000 bar (1 atm / 1.01325 bar)

which is very small, as expected. Therefore, the mole fraction of water in the lake near the surface is y water,liquid side = 1 − y dry air, liquid side = 1 − 1.51×10 −5 ≅ 1.0

Discussion The concentration of air in water just below the air-water interface is 1.51 moles per 100,000 moles. The amount of air dissolved in water will decrease with increasing depth.

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14-83

14-124 The mole fraction of the water vapor at the surface of a lake at a specified temperature is to be determined. Assumptions 1 Both the air and water vapor are ideal gases. 2 Air at the lake surface is saturated. Properties The saturation pressure of water at 18°C is 2.065 kPa (Table A-4). Analysis The air at the water surface will be saturated. Therefore, the partial pressure of water vapor in the air at the lake surface will simply be the saturation pressure of water at 18°C, Pvapor = Psat @18°C = 2.065 kPa

Assuming both the air and vapor to be ideal gases, the partial pressure and mole fraction of dry air in the air at the surface of the lake are determined to be Pdry air = P − Pvapor = 100 − 2.065 = 97.94 kPa y dry air =

Pdry air P

=

Air 100 kPa 18°C

Lake

97.94 kPa = 0.979 (or 97.9%) 100 kPa

Therefore, the mole fraction of dry air is 97.9 percent just above the air-water interface.

14-125E A room is cooled adequately by a 7500 Btu/h air-conditioning unit. If the room is to be cooled by an evaporative cooler, the amount of water that needs to be supplied to the cooler is to be determined. Assumptions 1 The evaporative cooler removes heat at the same rate as the air conditioning unit. 2 Water evaporates at an average temperature of 70°F. Properties The enthalpy of vaporization of water at 70°F is 1053.7 Btu/lbm (Table A-4E). Analysis Noting that 1 lbm of water removes 1053.7 Btu of heat as it evaporates, the amount of water that needs to evaporate to remove heat at a rate of 7500 Btu/h is determined from Q& = m& water h fg to be m& water =

Q& 7500 Btu/h = = 7.12 lbm/h h fg 1053.7 Btu/lbm

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14-84

14-126E The required size of an evaporative cooler in cfm (ft3/min) for an 8-ft high house is determined by multiplying the floor area of the house by 4. An equivalent rule is to be obtained in SI units. Analysis Noting that 1 ft = 0.3048 m and thus 1 ft2 = 0.0929 m2 and 1 ft3 = 0.0283 m3, and noting that a flow rate of 4 ft3/min is required per ft2 of floor area, the required flow rate in SI units per m2 of floor area is determined to 1 ft 2 ↔ 4 ft 3 / min 0.0929 m 2 ↔ 4 × 0.0283 m 3 / min 1 m 2 ↔ 1.22 m 3 / min

Therefore, a flow rate of 1.22 m3/min is required per m2 of floor area.

14-127 A cooling tower with a cooling capacity of 440 kW is claimed to evaporate 15,800 kg of water per day. It is to be determined if this is a reasonable claim. Assumptions 1 Water evaporates at an average temperature of 30°C. 2 The coefficient of performance of the air-conditioning unit is COP = 3. Properties The enthalpy of vaporization of water at 30°C is 2429.8 kJ/kg (Table A-4). Analysis Using the definition of COP, the electric power consumed by the air conditioning unit when running is W& in =

Q& cooling COP

=

440 kW = 146.7 kW 3

Then the rate of heat rejected at the cooling tower becomes

Q& rejected = Q& cooling + W& in = 440 + 146.7 = 586.7 kW Noting that 1 kg of water removes 2429.8 kJ of heat as it evaporates, the amount of water that needs to evaporate to remove heat at a rate of 586.7 kW is determined from Q& rejected = m& water h fg to be m& water =

Q& rejected h fg

=

586.7 kJ/s = 0.2415 kg/s = 869.3 kg/h = 20,860 kg/day 2429.8 kJ/kg

In practice, the air-conditioner will run intermittently rather than continuously at the rated power, and thus the water use will be less. Therefore, the claim amount of 15,800 kg per day is reasonable.

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14-85

14-128 Air is cooled by evaporating water into this air. The amount of water required and the cooling produced are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (m& a1 = m& a 2 = m& a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Properties The inlet and the exit states of the air are completely specified, and the total pressure is 1 atm. The properties of the air at various states are determined from the psychrometric chart (Figure A-31) to be

h1 = 64.0 kJ/kg dry air

ω1 = 0.0092 kg H 2 O/kg dry air

Water 20°C

and h2 = 66.0 kJ/kg dry air

1 atm 40°C 20%

ω 2 = 0.0160 kg H 2 O/kg dry air Also, hw ≅ h f @ 20°C = 83.92 kJ/kg

AIR

25°C 80%

(Table A-4)

Analysis The amount of moisture in the air increases due to humidification (ω 2 > ω 1). Applying the water mass balance and energy balance equations to the combined cooling and humidification section,

Water Mass Balance: ∑ m& w,i = ∑ m& w,e ⎯ ⎯→ m& a1ω1 = m& a 2 ω 2 + m& w Δω = ω 2 − ω1 = 0.0160 − 0.0092 = 0.0068 kg H 2 O/kg dry air

Energy Balance: E& in − E& out = ΔE& system ©0 (steady) = 0 E& in = E& out ∑ m& i hi = Q& out + ∑ m& e he Q& out = m& a1 h1 + m& w h w − m& a 2 h2 = m& a (h1 − h2 ) + m& w hw q out = h1 − h2 + (ω 2 − ω1 )h w = (64.0 − 66.0)kJ/kg + (0.0068)(83.92) = −1.43 kJ/kg dry air

The negative sign shows that the heat is actually transferred to the system.

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14-86

14-129 Air is humidified adiabatically by evaporating water into this air. The temperature of the air at the exit is to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (m& a1 = m& a 2 = m& a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Properties The inlet state of the air is completely specified, and the total pressure is 1 atm. The properties of the air at the inlet state are determined from the psychrometric chart (Figure A-31) to be

h1 = 64.0 kJ/kg dry air

ω1 = 0.0092 kg H 2 O/kg dry air and h w ≅ h f @ 20°C = 83.92 kJ/kg

(Table A-4)

Analysis The amount of moisture in the air increases due to humidification (ω 2 > ω 1). Applying the water mass balance and energy balance equations to the combined cooling and humidification section,

Water Mass Balance: ∑ m& w,i = ∑ m& w,e ⎯ ⎯→ m& a1ω1 = m& a 2 ω 2 + m& w

m& w = m& a (ω 2 − ω1 )

Water 20°C

Energy Balance: E& in − E& out = ΔE& system ©0 (steady) = 0 E& in = E& out ∑ m& i hi = ∑ m& e he & m a1 h1 + m& w hw = m& a 2 h2 m& w hw = m& a (h2 − h1 )

1 atm 40°C 20%

AIR

T2=? 80%

(ω 2 − ω1 )hw = h2 − h1

Substituting, (ω 2 − 0.0092)(83.92) = h2 − 64.0

The solution of this equation requires a trial-error method. An air exit temperature is assumed. At this temperature and given relative humidity, the enthalpy and specific humidity values are obtained from psychrometric chart and substituted into this equation. If the equation is not satisfied, a new value of exit temperature is assumed and this continues until the equation is satisfied. Alternatively, an equation solver such as EES may be used for the direct results. We used the following EES program. "Given" P=101.325 "[kPa]" T_1=40 "[C]" phi_1=0.20 phi_2=0.80 "Analysis" Fluid1$='AirH2O' Fluid2$='steam_iapws' h_1=enthalpy(Fluid1$, T=T_1, R=phi_1, P=P) w_1=humrat(Fluid1$, T=T_1, R=phi_1, P=P) h_2=enthalpy(Fluid1$, T=T_2, R=phi_2, P=P)

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14-87

w_2=humrat(Fluid1$, T=T_2, R=phi_2, P=P) h_w=enthalpy(Fluid2$, T=20, x=0) q=0 q=h_1-h_2+(w_2-w_1)*h_w

The results of these equations are T2 = 24.6°C h2 = 64.56 kJ/kg dry air

ω 2 = 0.01564 kg H 2 O/kg dry air

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14-88

14-130E Air is cooled and dehumidified at constant pressure. The rate of cooling and the minimum humid air temperature required to meet this cooling requirement are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (m& a1 = m& a 2 = m& a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Properties The inlet and the exit states of the air are completely specified, and the total pressure is 1 atm. The properties of the air at various states are determined from the psychrometric chart (Figure A-31) to be h1 = 50.6 Btu/lbm dry air

ω1 = 0.0263 lbm H 2 O/lbm dry air Cooling coils

v 1 = 14.44 ft 3 /lbm dry air and

T2 = 75°F φ 2 =50%

h2 = 28.2 Btu/lbm dry air

ω 2 = 0.0093 lbm H 2 O/lbm dry air

We assume that the condensate leaves this system at the average temperature of the air inlet and exit. Then, hw ≅ h f @ 82.5°F = 50.56 Btu/lbm

1 atm Condensate 2

T1 = 90°F φ 1 =85%

1 82.5°F

Condensate removal

(Table A-4)

Analysis The amount of moisture in the air decreases due to dehumidification (ω 2 < ω 1). The mass of air is ma =

V1 1000 ft 3 = = 69.25 lbm v 1 14.44 ft 3 / lbm dry air

Applying the water mass balance and energy balance equations to the combined cooling and dehumidification section, Water Mass Balance: ∑ m& w,i = ∑ m& w ,e ⎯ ⎯→ m& a1ω 1 = m& a 2ω 2 + m& w

m w = m a (ω1 − ω 2 ) = (69.25 kg)(0.0263 − 0.0093) = 1.177 lbm

Energy Balance: E& in − E& out = ΔE& system ©0 (steady) = 0 E& in = E& out ∑ m& i hi = Q& out + ∑ m& e he Q& out = m& a1 h1 − (m& a 2 h2 + m& w h w ) = m& a (h1 − h2 ) − m& w hw Qout = m a (h1 − h2 ) − m w hw Qout = (69.25 kg)(50.6 − 28.2)Btu/lbm − (1.177 lbm)(50.56 Btu/lbm) = 1492 Btu

For the desired dehumidification, the air at the exit should be saturated with a specific humidity of 0.0093 lbm water/lbm dry air. That is,

φ 2 = 1.0 ω 2 = 0.0093 lbm H 2 O/lbm dry air The temperature of the air at this state is the minimum air temperature required during this process: T2 = 55.2°F

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14-89

14-131E Air is cooled and dehumidified at constant pressure by a simple ideal vapor-compression refrigeration system. The system’s COP is to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (m& a1 = m& a 2 = m& a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis The inlet and the exit states of the air are completely specified, and the total pressure is 1 atm. The properties of the air at various states are determined from the psychrometric chart (Figure A-31) to be h1 = 50.6 Btu/lbm dry air

ω1 = 0.0263 lbm H 2 O/lbm dry air Cooling coils

v 1 = 14.44 ft 3 /lbm dry air and

T2 = 75°F φ 2 =50%

h2 = 28.2 Btu/lbm dry air

ω 2 = 0.0093 lbm H 2 O/lbm dry air

T1 = 90°F φ 1 =85%

1 atm Condensate 2

1

For the desired dehumidification, the air at the exit should be saturated with a specific humidity of 0.0093 lbm water/lbm dry air. That is,

82.5°F

Condensate removal

φ 2 = 1.0 ω 2 = 0.0093 lbm H 2 O/lbm dry air The temperature of the air at this state is the minimum air temperature required during this process: T2, min = 55.2°F

From the problem statement, the properties of R-134a at various states are (Tables A-11E through A-13E or from EES): T1 = 55.2 − 10 = 45.2°F ⎫ ⎪ h1 = h g @ 55 psia = 109.49 Btu/lbm P2 = Psat @ 45.2° F = 55 psia ⎬ ⎪ s1 = s g @ 55 psia = 0.22156 Btu/lbm ⋅ R sat. vapor ⎭ Tsat = 90 + 19.5 = 109.5°F

⎫ ⎪ P2 = Psat @ 109.5°F = 160 psia ⎬ h2 = 119.01 kJ/kg ⎪ s 2 = s1 ⎭

P3 = 160 psia ⎫ ⎬ h3 = h f sat. liquid ⎭

T · QH

@ 160 psia

· Win

= 48.52 Btu/lbm

h4 ≅ h3 = 48.52 Btu/lbm ( throttling)

The COP of this system is then COP =

2

3 109.5°F

qL h − h4 109.49 − 48.52 = 6.40 = = 1 win h2 − h1 119.01 − 109.49

45.2°F 4s

4

· QL

1

s

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14-90

14-132E Air at a specified state is heated to a specified temperature. The relative humidity after the heating is to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (m& a1 = m& a 2 = m& a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis There is no correspondence of inlet state from the psychrometric chart. Therefore, we have to use EES psychrometric functions to obtain the specific humidity:

ω1 = 0.0023 lbm H 2 O/lbm dry air As the outside air infiltrates into the dacha, it does not gain or lose any water. Therefore the humidity ratio inside the dacha is the same as that outside,

ω 2 = ω1 = 0.0023 lbm H 2 O/lbm dry air From EES or Fig. A-31E, at this humidity ratio and the temperature inside the dacha gives

70°F 1

32°F 60% RH

2 1 atm

AIR

φ 2 = 0.146 = 14.6%

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14-91

14-133E Air is humidified by evaporating water into this air. The amount of heating is to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (m& a1 = m& a 2 = m& a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Properties The inlet and the exit states of the air are completely specified, and the total pressure is 1 atm. The properties of the air at various states are determined from the psychrometric chart (Figure A-31) to be h1 = 19.3 Btu/lbm dry air

ω1 = 0.0023 kg H 2 O/kg dry air Water 60°F

v 1 = 13.40 m 3 /kg dry air and 1 atm 70°F 14.6%

h2 = 27.1 Btu/lbm dry air

ω 2 = 0.0094 lbm H 2 O/lbm dry air

AIR

70°F 60%

Also, hw ≅ h f @ 60°F = 28.08 Btu/lbm

(Table A-4E)

Analysis The amount of moisture in the air increases due to humidification (ω 2 > ω 1). Applying the water mass balance and energy balance equations to the combined cooling and humidification section,

Water Mass Balance: ∑ m& w,i = ∑ m& w,e ⎯ ⎯→ m& a1ω1 = m& a 2 ω 2 + m& w Energy Balance: E& in − E& out = ΔE& system ©0 (steady) = 0 E& in = E& out ∑ m& i hi + Q& in = ∑ m& e he Q& in = m& a 2 h2 − m& a1 h1 − m& w h w = m& a (h2 − h1 ) − m& w hw q in = h2 − h1 − (ω 2 − ω1 )h w = (27.1 − 19.3)Btu/lbm − (0.0094 − 0.0023)(28.08) = 7.59 Btu/lbm dry air

The mass of air that has to be humidified is ma =

V 16,000 ft 3 = = 1194 lbm dry air v 1 13.40 ft 3 /lbm dry air

The total heat requirement is then Qin = m a q in = (1194 lbm dry air )(7.59 Btu/lbm dry air ) = 9062 Btu

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14-92

14-134E It is estimated that 190,000 barrels of oil would be saved per day if the thermostat setting in residences in summer were raised by 6°F (3.3°C). The amount of money that would be saved per year is to be determined. Assumptions The average cooling season is given to be 120 days, and the cost of oil to be $20/barrel. Analysis The amount of money that would be saved per year is determined directly from (190,000 barrel/day)(120 days/year)($70/barrel) = $1,596,000 ,000

Therefore, the proposed measure will save more than one and half billion dollars a year.

14-135 Shading the condenser can reduce the air-conditioning costs by up to 10 percent. The amount of money shading can save a homeowner per year during its lifetime is to be determined. Assumptions It is given that the annual air-conditioning cost is $500 a year, and the life of the airconditioning system is 20 years. Analysis The amount of money that would be saved per year is determined directly from ($500 / year)(20 years)(0.10) = $1000

Therefore, the proposed measure will save about $1000 during the lifetime of the system.

14-136 A tank contains saturated air at a specified state. The mass of the dry air, the specific humidity, and the enthalpy of the air are to be determined. Assumptions The air and the water vapor are ideal gases. Analysis (a) The air is saturated, thus the partial pressure of water vapor is equal to the saturation pressure at the given temperature, Pv = Pg = Psat @ 25°C = 3.1698 kPa Pa = P − Pv = 97 − 3.1698 = 93.83 kPa

Treating air as an ideal gas, ma =

3 m3 25°C 97 kPa

PaV (93.83 kPa)(3 m3 ) = = 3.29 kg RaT (0.287 kPa ⋅ m3 / kg ⋅ K)(298 K)

(b) The specific humidity of air is determined from

ω=

0.622 Pv (0.622)(3.1698 kPa) = = 0.0210 kg H 2 O/kg dry air P − Pv (97 − 3.1698) kPa

(c) The enthalpy of air per unit mass of dry air is determined from h = ha + ωhv ≅ c p T + ωh g = (1.005 kJ/kg ⋅ °C)(25°C) + (0.0210)(2546.5 kJ/kg) = 78.6 kJ/kg dry air

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14-93

14-137 EES Problem 14-136 is reconsidered. The properties of the air at the initial state are to be determined and the effects of heating the air at constant volume until the pressure is 110 kPa is to be studied. Analysis The problem is solved using EES, and the solution is given below. "Input Data:" Tdb[1] = 25 [C] P[1]=97 [kPa] Rh[1]=1.0 P[2]=110 [kPa] Vol = 3 [m^3] w[1]=HUMRAT(AirH2O,T=Tdb[1],P=P[1],R=Rh[1]) v[1]=VOLUME(AirH2O,T=Tdb[1],P=P[1],R=Rh[1]) m_a=Vol/v[1] h[1]=ENTHALPY(AirH2O,T=Tdb[1],P=P[1],w=w[1]) "Energy Balance for the constant volume tank:" E_in - E_out = DELTAE_tank DELTAE_tank=m_a*(u[2] -u[1]) E_in = Q_in E_out = 0 [kJ] u[1]=INTENERGY(AirH2O,T=Tdb[1],P=P[1],w=w[1]) u[2]=INTENERGY(AirH2O,T=Tdb[2],P=P[2],w=w[2]) "The ideal gas mixture assumption applied to the constant volume process yields:" P[1]/(Tdb[1]+273)=P[2]/(Tdb[2]+273) "The mass of the water vapor and dry air are constant, thus:" w[2]=w[1] Rh[2]=RELHUM(AirH2O,T=Tdb[2],P=P[2],w=w[2]) h[2]=ENTHALPY(AirH2O,T=Tdb[2],P=P[2],w=w[2]) v[2]=VOLUME(AirH2O,T=Tdb[2],P=P[2],R=Rh[2]) PROPERTIES AT THE INITIAL STATE h[1]=78.67 [kJ/kga] m_a=3.289 [kga] v[1]=0.9121 [m^3/kga] w[1]=0.02101 [kgw/kga] 100 Qin [kJ] 0 15.12 30.23 45.34 60.45 75.55 90.65 98.2

75

Qin [kJ]

P2 [kPa] 97 99 101 103 105 107 109 110

50

25

0 96

98

100

102

104

106

108

110

P[2] [kPa]

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14-94

14-138E Air at a specified state and relative humidity flows through a circular duct. The dew-point temperature, the volume flow rate of air, and the mass flow rate of dry air are to be determined. Assumptions The air and the water vapor are ideal gases. Analysis (a) The vapor pressure of air is Pv = φPg = φPsat @ 60°F = (0.50)(0.2564 psia) = 0.128 psia

Thus the dew-point temperature of the air is

AIR 15 psia 50 f/s 60°F, 50%

Tdp = Tsat @ Pv = Tsat @ 0.128 psia = 41.3°F (from EES)

(b) The volume flow rate is determined from

V& = VA = V

πD 2 4

⎛ π × (8 / 12 ft ) 2 = (50 ft/s)⎜ ⎜ 4 ⎝

⎞ ⎟ = 17.45 ft 3 /s ⎟ ⎠

(c) To determine the mass flow rate of dry air, we first need to calculate its specific volume, Pa = P − Pv = 15 − 0.128 = 14.872 psia

v1 =

RaT1 (0.3704 psia ⋅ ft 3 / lbm ⋅ R)(520 R) = = 12.95 ft 3 / lbm dry air Pa1 14.872 psia

Thus, m& a1 =

V&1 17.45 ft 3 / s = = 1.35 lbm/s v1 12.95 ft 3 / lbm dry air

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

14-95

14-139 Air enters a cooling section at a specified pressure, temperature, and relative humidity. The temperature of the air at the exit and the rate of heat transfer are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (m& a1 = m& a 2 = m& a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis (a) The amount of moisture in the air also remains constant (ω 1 = ω 2 ) as it flows through the cooling section since the process involves no humidification or dehumidification. The total pressure is 97 kPa. The properties of the air at the inlet state are Pv1 = φ1Pg1 = φ1Psat @ 35°C = (0.3)(5.629 kPa) = 1.69 kPa

Cooling coils

Pa1 = P1 − Pv1 = 97 − 1.69 = 95.31 kPa

v1 =

RaT1 (0.287 kPa ⋅ m3 / kg ⋅ K)(308 K) = Pa1 95.31 kPa

= 0.927 m3 / kg dry air

ω1 =

1

35°C 30% 6 m3/min

2 97 kPa

AIR

0.622 Pv1 0.622(1.69 kPa) = = 0.0110 kg H 2O/kg dry air (= ω2 ) P1 − Pv1 (97 − 1.69) kPa

h1 = c pT1 + ω1hg1 = (1.005 kJ/kg°C)(35°C) + (0.0110)(2564.6 kJ/kg) = 63.44 kJ/kg dry air

The air at the final state is saturated and the vapor pressure during this process remains constant. Therefore, the exit temperature of the air must be the dew-point temperature, Tdp = Tsat @ Pv = Tsat @ 1.69 kPa = 14.8°C

(b) The enthalpy of the air at the exit is h2 = c pT2 + ω2 hg 2 = (1.005 kJ/kg ⋅ °C)(14.8°C) + (0.0110)(2528.1 kJ/kg) = 42.78 kJ/kg dry air

Also m& a =

V&1 6 m3 / s = = 6.47 kg/min v 1 0.927 m 3 / kg dry air

Then the rate of heat transfer from the air in the cooling section becomes Q& out = m& a (h1 − h2 ) = (6.47 kg/min)(63.44 − 42.78)kJ/kg = 134 kJ/min

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14-96

14-140 The outdoor air is first heated and then humidified by hot steam in an air-conditioning system. The rate of heat supply in the heating section and the mass flow rate of the steam required in the humidifying section are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (m& a1 = m& a 2 = m& a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Properties The amount of moisture in the air also remains constants it flows through the heating section (ω 1 = ω 2 ) , but increases in the humidifying section (ω 3 > ω 2 ) . The inlet and the exit states of the air are completely specified, and the total pressure is 1 atm. The properties of the air at various states are determined from the psychrometric chart (Fig. A-31) to be h1 = 17.7 kJ/kg dry air

ω1 = 0.0030 kg H 2 O/kg dry air (= ω 2 )

Heating coils

v 1 = 0.807 m 3 /kg dry air h2 = 29.8 kJ / kg dry air

ω 2 = ω 1 = 0.0030 kg H 2 O / kg dry air h3 = 52.9 kJ / kg dry air

1 atm

10°C 40% 22 m3/min

ω 3 = 0.0109 kg H 2 O / kg dry air

AIR 1

25°C 55%

22°C 2

3

Analysis (a) The mass flow rate of dry air is m& a =

V&1 22 m3 / min = = 27.3 kg/min v1 0.807 m3 / kg

Then the rate of heat transfer to the air in the heating section becomes & a ( h2 − h1 ) = ( 27.3 kg / min)(29.8 − 17.7 )kJ / kg = 330.3 kJ / min Q& in = m

(b) The conservation of mass equation for water in the humidifying section can be expressed as m& a 2ω 2 + m& w = m& a 3ω 3

or m& w = m& a (ω 3 − ω 2 )

Thus, & w = ( 27.3 kg / min)(0.0109 − 0.0030) = 0.216 kg / min m

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14-97

14-141 Air is cooled and dehumidified in an air-conditioning system with refrigerant-134a as the working fluid. The rate of dehumidification, the rate of heat transfer, and the mass flow rate of the refrigerant are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (m& a1 = m& a 2 = m& a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis (a) The saturation pressure of water at 30ºC is 4.2469 kPa. Then the dew point temperature of the incoming air stream at 30°C becomes Tdp = Tsat @ Pv = Tsat @ 0.7×4.2469 kPa = 24°C

Since air is cooled to 20°C, which is below its dew point temperature, some of the moisture in the air will condense. The mass flow rate of dry air remains constant during the entire process, but the amount of moisture in the air decreases due to dehumidification (ω 2 < ω 1 ) . The inlet and the exit states of the air are completely specified, and the total pressure is 1 atm. Then the properties of the air at both states are determined from the psychrometric chart (Fig. A-31) to be h1 = 78.3 kJ/kg dry air

4

3

ω1 = 0.0188 kg H 2 O/kg dry air

R-134a

v 1 = 0.885 m 3 /kg dry air

700 kPa x3 = 20%

700 kPa sat. vapor

and h2 = 57.5 kJ / kg dry air

1

ω 2 = 0.0147 kg H 2 O / kg dry air Also,

hw ≅ h f @ 20°C = 83.915 kJ/kg (Table A-4)

Then,

m& a1 =

30°C 70% 4 m3/min

AIR 1 atm

20°C

V&1 4 m 3 / min = = 4.52 kg/min v 1 0.885 m 3 / kg dry air

Applying the water mass balance and the energy balance equations to the combined cooling and dehumidification section (excluding the refrigerant), ∑ m& w,i = ∑ m& w,e

Water Mass Balance:

⎯ ⎯→

m& a1ω 1 = m& a 2ω 2 + m& w

m& w = m& a (ω 1 − ω 2 ) = (4.52 kg / min)(0.0188 − 0.0147) = 0.0185 kg / min

(b) Energy Balance: E& − E& = ΔE& in

out

Ê0 (steady)

system

E& in = E& out & i hi = Q& out + ∑ m & e he ∑m

=0 ⎯ ⎯→

& a1h1 − ( m & a 2 h2 + m & w hw ) = m & a ( h1 − h2 ) − m & w hw Q& out = m

Q& out = (4.52 kg/min)(78.3 − 57.5)kJ/kg − (0.0185 kg/min)(83.915 kJ/kg) = 92.5 kJ/min

(c) The inlet and exit enthalpies of the refrigerant are h3 = h g + x 3 h fg = 88.82 + 0.2 × 176.21 = 124.06 kJ/kg h4 = h g @ 700 kPa = 265.03 kJ/kg

Noting that the heat lost by the air is gained by the refrigerant, the mass flow rate of the refrigerant becomes Q& R 92.5 kJ/min = = 0.66 kg/min Q& R = m& R (h4 − h3 ) → m& R = h4 − h3 (265.03 − 124.06) kJ/kg

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2

14-98

14-142 Air is cooled and dehumidified in an air-conditioning system with refrigerant-134a as the working fluid. The rate of dehumidification, the rate of heat transfer, and the mass flow rate of the refrigerant are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process. 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis (a) The dew point temperature of the incoming air stream at 30°C is Pv1 = φ1Pg1 = φ1Psat @ 30°C = (0.7)(4.247 kPa) = 2.973 kPa Tdp = Tsat @ Pv = Tsat @ 2.973 kPa = 24°C

Since air is cooled to 20°C, which is below its dew point temperature, some of the moisture in the air will condense. The amount of moisture in the air decreases due to dehumidification (ω 2 < ω 1 ) . The inlet and the exit states of the air are completely specified, and the total pressure is 95 kPa. The properties of the air at both states are determined to be

4

3 R-134a

700 kPa x3 = 20%

30°C 70% 4 m3/min

1

700 kPa sat. vapor AIR

95 kPa

20°C

Pa1 = P1 − Pv1 = 95 − 2.97 = 92.03 kPa

v1 =

Ra T1 (0.287 kPa ⋅ m 3 / kg ⋅ K)(303 K) = = 0.945 m 3 / kg dry air Pa1 92.03 kPa

ω1 =

0.622 Pv1 0.622(2.97 kPa) = = 0.0201 kg H 2 O/kg dry air P1 − Pv1 (95 − 2.97) kPa

h1 = c p T1 + ω1hg1 = (1.005 kJ/kg ⋅ °C)(30°C) + (0.0201)(2555.6 kJ/kg) = 81.50 kJ/kg dry air

and Pv 2 = φ 2 Pg 2 = (1.00) Psat @ 20°C = 2.3392 kPa

ω2 =

0.622 Pv 2 0.622(2.3392 kPa) = = 0.0157 kg H 2 O/kg dry air P2 − Pv 2 (95 − 2.3392) kPa

h2 = c p T2 + ω 2 hg 2 = (1.005 kJ/kg ⋅ °C)(20°C) + (0.0157)(2537.4 kJ/kg) = 59.94 kJ/kg dry air

Also,

hw ≅ h f @ 20°C = 83.915 kJ/kg

(Table A-4)

Then, m& a1 =

4 m3 / min V&1 = = 4.23 kg/min v1 0.945 m3 / kg dry air

Applying the water mass balance and the energy balance equations to the combined cooling and dehumidification section (excluding the refrigerant), Water Mass Balance:

∑ m& w,i = ∑ m& w,e

⎯ ⎯→

m& a1ω 1 = m& a 2ω 2 + m& w

m& w = m& a (ω 1 − ω 2 ) = (4.23 kg / min)(0.0201 − 0.0157) = 0.0186 kg / min

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2

14-99

(b) Energy Balance: E& in − E& out = ΔE& system Ê0 (steady) = 0 E& in = E& out & i hi = Q& out + ∑ m & e he ∑m

⎯ ⎯→

& a1h1 − ( m & a 2 h2 + m & w hw ) = m & a ( h1 − h2 ) − m & w hw Q& out = m

Q& out = (4.23 kg/min)(81.50 − 59.94)kJ/kg − (0.0186 kg/min)(83.915 kJ/kg) = 89.7 kJ/min

(c) The inlet and exit enthalpies of the refrigerant are h3 = h g + x 3 h fg = 88.82 + 0.2 × 176.21 = 124.06 kJ/kg h4 = h g @ 700 kPa = 265.03 kJ/kg

Noting that the heat lost by the air is gained by the refrigerant, the mass flow rate of the refrigerant is determined from Q& R = m& R (h4 − h3 ) Q& R 89.7 kJ/min = = 0.636 kg/min m& R = h4 − h3 (265.03 − 124.06) kJ/kg

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14-100

14-143 Air is heated and dehumidified in an air-conditioning system consisting of a heating section and an evaporative cooler. The temperature and relative humidity of the air when it leaves the heating section, the rate of heat transfer in the heating section, and the rate of water added to the air in the evaporative cooler are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (m& a1 = m& a 2 = m& a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis (a) Assuming the wet-bulb temperature of the air remains constant during the evaporative cooling process, the properties of air at various states are determined from the psychrometric chart (Fig. A-31) to be h1 = 23.5 kJ/kg dry air T1 = 10°C ⎫ ⎬ ω = 0.00532 kg/ H 2 O/kg dry air φ1 = 70% ⎭ 1 v 1 = 0.810 m 3 / kg

ω 2 = ω1 Twb2 = Twb3

T2 = 28.3°C ⎫ ⎬ φ 2 = 22.3% ⎭ h ≅ h = 42.3 kJ / kg dry air 2 3

Water Heating coils 1 atm

10°C 70% 30 m3/min

h = 42.3 kJ/kg dry air T3 = 20°C ⎫ 3 ⎬ ω = 0.00875 kg/ H 2O/kg dry air φ3 = 60% ⎭ 3 Twb3 = 15.1°C

AIR 1

20°C 60%

T2 2

3

(b) The mass flow rate of dry air is m& a =

V&1 30 m 3 / min = = 37.0 kg/min v 1 0.810 m 3 / kg dry air

Then the rate of heat transfer to air in the heating section becomes Q& in = m& a (h2 − h1 ) = (37.0 kg/min)(42.3 − 23.5)kJ/kg = 696 kJ/min

(c) The rate of water added to the air in evaporative cooler is m& w, added = m& w3 − m& w2 = m& a (ω3 − ω2 ) = (37.0 kg/min)(0.00875 − 0.00532) = 0.127 kg/min

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14-101

14-144 EES Problem 14-143 is reconsidered. The effect of total pressure in the range 94 to 104 kPa on the results required in the problem is to be studied. Analysis The problem is solved using EES, and the solution is given below. P=101.325 [kPa] Tdb[1] =10 [C] Rh[1] = 0.70 Vol_dot[1]= 50 [m^3/min] Tdb[3] = 20 [C] Rh[3] = 0.60 P[1]=P P[2]=P[1] P[3]=P[1] "Energy balance for the steady-flow heating process 1 to 2:" "We neglect the PE of the flow. Since we don't know the cross sectional area of the flow streams, we also neglect theKE of the flow." E_dot_in - E_dot_out = DELTAE_dot_sys DELTAE_dot_sys = 0 [kJ/min] E_dot_in = m_dot_a*h[1]+Q_dot_in E_dot_out = m_dot_a*h[2] "Conservation of mass of dry air during mixing: m_dot_a = constant" m_dot_a = Vol_dot[1]/v[1] "Conservation of mass of water vapor during the heating process:" m_dot_a*w[1] = m_dot_a*w[2] "Conservation of mass of water vapor during the evaporative cooler process:" m_dot_a*w[2]+m_dot_w = m_dot_a*w[3] "During the evaporative cooler process:" Twb[2] = Twb[3] Twb[3] =WETBULB(AirH2O,T=Tdb[3],P=P[3],R=Rh[3]) h[1]=ENTHALPY(AirH2O,T=Tdb[1],P=P[1],R=Rh[1]) v[1]=VOLUME(AirH2O,T=Tdb[1],P=P[1],R=Rh[1]) w[1]=HUMRAT(AirH2O,T=Tdb[1],P=P[1],R=Rh[1]) {h[2]=ENTHALPY(AirH2O,T=Tdb[2],P=P[2],B=Twb[2])} h[2]=h[3] Tdb[2]=TEMPERATURE(AirH2O,h=h[2],P=P[2],w=w[2]) w[2]=HUMRAT(AirH2O,T=Tdb[2],P=P[2],R=Rh[2]) h[3]=ENTHALPY(AirH2O,T=Tdb[3],P=P[3],R=Rh[3]) w[3]=HUMRAT(AirH2O,T=Tdb[3],P=P[3],R=Rh[3]) mw [kg/min] 0.2112 0.2112 0.2111 0.2111 0.211 0.2109

Qin [kJ/min] 1119 1131 1143 1155 1168 1180

Rh2 0.212 0.2144 0.2167 0.219 0.2212 0.2233

Tdb2 [C] 29.2 29 28.82 28.64 28.47 28.3

P [kPa] 94 96 98 100 102 104

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14-102

0.224 0.222

Rh[2]

0.22 0.218 0.216 0.214 0.212 94

96

98

100

102

104

P [kPa]

29.2 29.1

Tdb[2] [C]

29 28.9 28.8 28.7 28.6 28.5 28.4 28.3 94

96

98

100

102

104

P [kPa]

1180

0.2113

0.2112

1160 1150

0.2111 1140 1130

0.211

mw [kg/min]

Qin [kJ/min]

1170

1120 1110 94

96

98

100

102

0.2109 104

P [kPa]

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14-103

14-145 Air is heated and dehumidified in an air-conditioning system consisting of a heating section and an evaporative cooler. The temperature and relative humidity of the air when it leaves the heating section, the rate of heat transfer in the heating section, and the rate at which water is added to the air in the evaporative cooler are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (m& a1 = m& a 2 = m& a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis (a) Assuming the wet-bulb temperature of the air remains constant during the evaporative cooling process, the properties of air at various states are determined to be Water Pv1 = φ1Pg1 = φ1Psat @ 10°C = (0.70)(1.2281 kPa) = 0.86 kPa Heating coils Pa1 = P1 − Pv1 = 96 − 0.86 = 95.14 kPa

v1 =

RaT1 (0.287 kPa ⋅ m 3 / kg ⋅ K)(283 K) = Pa1 95.14 kPa

= 0.854 m 3 / kg dry air

10°C 70% 30 m3/min

96 kPa AIR

1 0.622 Pv1 0.622(0.86 kPa) ω1 = = = 0.00562 kg H 2O/kg dry air P1 − Pv1 (96 − 0.86) kPa

20°C 60%

T2 2

3

h1 = c pT1 + ω1hg1 = (1.005 kJ/kg ⋅ °C)(10°C) + (0.00562)(2519.2 kJ/kg) = 24.21 kJ/kg dry air

and Pv 3 = φ3 Pg 3 = φ3 Psat @ 20°C = (0.60)(2.3392 kPa) = 1.40 kPa Pa 3 = P3 − Pv3 = 96 − 1.40 = 94.60 kPa

ω3 =

0.622 Pv3 0.622(1.40 kPa) = = 0.00923 kg H 2 O/kg dry air (96 − 1.40) kPa P3 − Pv 3

h3 = c p T3 + ω3 hg 3 = (1.005 kJ/kg ⋅ °C)(20°C) + (0.00921)(2537.4 kJ/kg) = 43.52 kJ/kg dry air

Also, h2 ≅ h3 = 43.52 kJ/kg

ω 2 = ω1 = 0.00562 kg H 2 O/kg dry air Thus, h2 = c pT2 + ω2 hg 2 ≅ c pT2 + ω2 (2500.9 + 1.82T2 ) = (1.005 kJ/kg ⋅ °C)T2 + (0.00562)(2500.9 + 1.82T2 )

Solving for T2, T2 = 29.0°C ⎯ ⎯→ Pg 2 = Psat@29°C = 4.013 kPa

Thus,

φ2 =

ω 2 P2 (0.00562)(96) = = 0.214 or 21.4% (0.622 + ω 2 ) Pg 2 (0.622 + 0.00562)(4.013)

(b) The mass flow rate of dry air is V& 30 m 3 / min m& a = 1 = = 35.1 kg/min v 1 0.854 m 3 / kg dry air Then the rate of heat transfer to air in the heating section becomes Q& = m& (h − h ) = (35.1 kg/min)(43.52 − 24.21)kJ/kg = 679 kJ/min in

a

2

1

(c) The rate of water addition to the air in evaporative cooler is m& w, added = m& w3 − m& w 2 = m& a (ω3 − ω2 ) = (35.1 kg/min)(0.00923 − 0.00562) = 0.127 kg/min

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

14-104

14-146 [Also solved by EES on enclosed CD] Waste heat from the cooling water is rejected to air in a natural-draft cooling tower. The mass flow rate of the cooling water, the volume flow rate of air, and the mass flow rate of the required makeup water are to be determined. Assumptions 1 Steady operating conditions exist. 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. 4 The cooling tower is adiabatic. Analysis (a) The mass flow rate of dry air through the tower remains constant (m& a1 = m& a 2 = m& a ) , but the mass flow rate of liquid water decreases by an amount equal to the amount of water that vaporizes in the tower during the cooling process. The water lost through evaporation is made up later in the cycle using water at 27°C. Applying the mass balance and the energy balance equations yields

Dry Air Mass Balance: ∑ m& a ,i = ∑ m& a ,e

⎯ ⎯→

m& a1 = m& a 2 = m& a

Water Mass Balance: ∑ m& w,i = ∑ m& w,e m& 3 + m& a1ω1 = m& 4 + m& a 2 ω 2 m& 3 − m& 4 = m& a (ω 2 − ω1 ) = m& makeup

Energy Balance: E& in − E& out = ΔE& systemÊ0 (steady) = 0 ⎯⎯→ E& in = E& out

AIR EXIT

∑ m& i hi = ∑ m& e he (since Q& = W& = 0) 0 = ∑ m& e he − ∑ m& i hi 0 = m& a 2 h2 + m& 4 h4 − m& a1h1 − m& 3h3

2 37°C saturated

0 = m& a (h2 − h1 ) + (m& 3 − m& makeup )h4 − m& 3h3

Solving for m& a , m& a =

m& 3 (h3 − h4 ) (h2 − h1 ) − (ω 2 − ω 1 )h4

3

ω1 = 0.0109 kg H 2O/kg dry air

42°C

WATER

From the psychrometric chart (Fig. A-31), h1 = 50.8 kJ/kg dry air

WARM

27°C 4

COOL WATER

1 AIR INLET Tdb = 23°C Twb = 18°C

v1 = 0.854 m3/kg dry air and h2 = 143.0 kJ / kg dry air ω 2 = 0.0412 kg H 2 O / kg dry air

From Table A-4, h3 ≅ h f @ 42°C = 175.90 kJ/kg H 2 O h4 ≅ h f @ 27°C = 113.19 kJ/kg H 2 O

Substituting m& a =

m& 3 (175.90 − 113.19)kJ/kg &3 = 0.706 m (143.0 − 50.8) kJ/kg − (0.0412 − 0.0109)(113.25) kJ/kg

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14-105

The mass flow rate of the cooling water is determined by applying the steady flow energy balance equation on the cooling water, & 3h3 − ( m &3 − m & makeup )h4 = m & 3h3 − [m &3 − m & a (ω 2 − ω 1 )]h4 Q& waste = m & 3h3 − m & 3[1 − 0.706( 0.0412 − 0.0109 )]h4 = m & 3 ( h3 − 0.9786h4 ) =m 50,000 kJ/s = m& 3 (175.90 − 0.9786 × 113.19) kJ/kg ⎯ ⎯→ m& 3 = 768.1 kg/s

and m& a = 0.706m& 3 = (0.706)(7681 . kg / s) = 542.3 kg / s

(b) Then the volume flow rate of air into the cooling tower becomes

V&1 = m& av 1 = (542.3 kg/s)(0.854 m 3 / kg ) = 463.1 m 3 /s (c) The mass flow rate of the required makeup water is determined from & makeup = m & a (ω 2 − ω 1 ) = (542.3 kg / s)(0.0412 − 0.0109) = 16.4 kg / s m

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14-106

14-147 EES Problem 14-146 is reconsidered. The effect of air inlet wet-bulb temperature on the required air volume flow rate and the makeup water flow rate is to be investigated. Analysis The problem is solved using EES, and the solution is given below. "Input Data" P_atm =101.325 [kPa] T_db_1 = 23 [C] T_wb_1 = 18 [C] T_db_2 = 37 [C] RH_2 = 100/100 "%. relative humidity at state 2, saturated condition" Q_dot_waste = 50 [MW]*Convert(MW, kW) T_cw_3 = 42 [C] "Cooling water temperature at state 3" T_cw_4 = 27 [C] "Cooling water temperature at state 4" "Dry air mass flow rates:" "RH_1 is the relative humidity at state 1 on a decimal basis" v_1=VOLUME(AirH2O,T=T_db_1,P=P_atm,R=RH_1) T_wb_1 = WETBULB(AirH2O,T=T_db_1,P=P_atm,R=RH_1) m_dot_a_1 = Vol_dot_1/v_1 "Conservaton of mass for the dry air (ma) in the SSSF mixing device:" m_dot_a_in - m_dot_a_out = DELTAm_dot_a_cv m_dot_a_in = m_dot_a_1 m_dot_a_out = m_dot_a_2 DELTAm_dot_a_cv = 0 "Steady flow requirement" "Conservation of mass for the water vapor (mv) and cooling water for the SSSF process:" m_dot_w_in - m_dot_w_out = DELTAm_dot_w_cv m_dot_w_in = m_dot_v_1 + m_dot_cw_3 m_dot_w_out = m_dot_v_2+m_dot_cw_4 DELTAm_dot_w_cv = 0 "Steady flow requirement" w_1=HUMRAT(AirH2O,T=T_db_1,P=P_atm,R=RH_1) m_dot_v_1 = m_dot_a_1*w_1 w_2=HUMRAT(AirH2O,T=T_db_2,P=P_atm,R=RH_2) m_dot_v_2 = m_dot_a_2*w_2 "Conservation of energy for the SSSF cooling tower process:" "The process is adiabatic and has no work done, ngelect ke and pe" E_dot_in_tower - E_dot_out_tower = DELTAE_dot_tower_cv E_dot_in_tower= m_dot_a_1 *h[1] + m_dot_cw_3*h_w[3] E_dot_out_tower = m_dot_a_2*h[2] + m_dot_cw_4*h_w[4] DELTAE_dot_tower_cv = 0 "Steady flow requirement" h[1]=ENTHALPY(AirH2O,T=T_db_1,P=P_atm,w=w_1) h[2]=ENTHALPY(AirH2O,T=T_db_2,P=P_atm,w=w_2) h_w[3]=ENTHALPY(steam,T=T_cw_3,x=0) h_w[4]=ENTHALPY(steam,T=T_cw_4,x=0) "Energy balance on the external heater determines the cooling water flow rate:" E_dot_in_heater - E_dot_out_heater = DELTAE_dot_heater_cv E_dot_in_heater = Q_dot_waste + m_dot_cw_4*h_w[4] E_dot_out_heater = m_dot_cw_3 * h_w[3] DELTAE_dot_heater_cv = 0 "Steady flow requirement"

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14-107 "Conservation of mass on the external heater gives the makeup water flow rate." "Note: The makeup water flow rate equals the amount of water vaporized in the cooling tower." m_dot_cw_in - m_dot_cw_out = DELTAm_dot_cw_cv m_dot_cw_in = m_dot_cw_4 + m_dot_makeup m_dot_cw_out = m_dot_cw_3 DELTAm_dot_cw_cv = 0 "Steady flow requirement"

Vol1 [m3/s] 408.3 420.1 433.2 447.5 463.4 481.2 501.1 523.7 549.3 578.7

mmakeup [kgw/s] 16.8 16.72 16.63 16.54 16.43 16.31 16.18 16.03 15.87 15.67

mcw3 [kgw/s] 766.6 766.7 766.8 767 767.2 767.4 767.7 767.9 768.2 768.6

ma1 [kga/s] 481.9 495 509.4 525.3 542.9 562.6 584.7 609.7 638.1 670.7

Twb1 [C] 14 15 16 17 18 19 20 21 22 23

Vol1 [m^3/s]

560

16.6

520

16.4 16.2

480

16.0 440 400 14.0

15.8

mmakeup [kgw/s]

16.8

15.6 16.0

18.0

20.0

22.0

Twb,1 [C]

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14-108

14-148 Atmospheric air enters an air-conditioning system at a specified pressure, temperature, and relative humidity. The heat transfer, the rate of condensation of water, and the mass flow rate of the refrigerant are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (m& a1 = m& a 2 = m& a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis The inlet and exit states of the air are completely specified, and the total pressure is 1 atm. The properties of the air at the inlet and exit states may be determined from the psychrometric chart (Figure A31) or using EES psychrometric functions to be (we used EES) R-134a h1 = 78.24 kJ/kg dry air 350 kPa 350 kPa ω1 = 0.01880 kg H 2O/kg dry air x = 1.0 x = 0.20

v1 = 0.8847 m3 / kg dry air h2 = 27.45 kJ/kg dry air

Cooling coils

ω2 = 0.002885 kg H 2O/kg dry air The mass flow rate of dry air is m& a =

T2 =20°C φ 2 = 20%

V&1 4 m3/min = = 4.521 kg/min v1 0.8847 m3

T1 =30°C φ 1 = 70% 4 m3/min

1 atm AIR Condensate 2

The mass flow rates of vapor at the inlet and exit are

1 20°C

Condensate removal

m& v1 = ω1 m& a = (0.01880)(4.521 kg/min) = 0.0850 kg/min m& v 2 = ω 2 m& a = (0.002885)(4.521 kg/min) = 0.01304 kg/min

An energy balance on the control volume gives m& a h1 = Q& out + m& a h2 + m& w h w2

where the the enthalpy of condensate water is h w 2 = h f@ 20°C = 83.91 kJ/kg

(Table A - 4)

and the rate of condensation of water vapor is m& w = m& v1 − m& v 2 = 0.0850 − 0.01304 = 0.07196 kg/min

Substituting, m& a h1 = Q& out + m& a h2 + m& w hw 2 (4.521 kg/min)(78.24 kJ/kg) = Q& out + (4.521 kg/min)(27.45 kJ/kg) + (0.07196 kg/min)(83.91 kJ/kg) Q& = 223.6 kJ/min = 3.727 kW out

The properties of the R-134a at the inlet and exit of the cooling section are PR1 = 350 kPa ⎫ ⎬h R1 = 97.56 kJ/kg x R1 = 0.20 ⎭ PR 2 = 350 kPa ⎫ ⎬h R 2 = 253.34 kJ/kg x R 2 = 1.0 ⎭

Noting that the rate of heat lost from the air is received by the refrigerant, the mass flow rate of the refrigerant is determined from

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14-109

m& R h R1 + Q& in = m& R h R 2 Q& in 223.6 kJ/min = = 1.435 kg/min m& R = h R 2 − h R1 (253.34 − 97.56) kJ/kg

AirH2O

0,050 Pressure = 101.3 [kPa] 0,045 0,040

35°C 0.8

Humidity Ratio

0,035 0,030 30°C

0.6

0,025 25°C

0,020

1

0.4

20°C

0,015 15°C

0,010

0.2

10°C

0,005 0,000 0

2 5

10

15

20

25

30

35

40

T [°C]

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14-110

14-149 An uninsulated tank contains moist air at a specified state. Water is sprayed into the tank until the relative humidity in the tank reaches a certain value. The amount of water supplied to the tank, the final pressure in the tank, and the heat transfer during the process are to be determined. Assumptions 1 Dry air and water vapor are ideal gases. 2 The kinetic and potential energy changes are negligible. Analysis The initial state of the moist air is completely specified. The properties of the air at the inlet state may be determined from the psychrometric chart (Figure A-31) or using EES psychrometric functions to be (we used EES) ω1 = 0.005433 kg H 2 O/kg dry air h1 = 49.16 kJ/kg dry air,

v 1 = 0.6863 m 3 / kg dry air The initial mass in the tank is V 0.5 m 3 ma = 1 = = 0.7285 kg v 1 0.6863 m 3 The partial pressure of dry air in the tank is Pa 2 =

ma Ra T2

=

(0.7285 kg)(0.287 kJ/kg.K)(35 + 273 K)

= 128.8 kPa (0.5 m 3 ) Then, the pressure of moist air in the tank is determined from ω ⎞ ω ⎞ ⎛ ⎛ P2 = Pa 2 ⎜⎜1 + 2 ⎟⎟ = (128.8 kPa)⎜⎜1 + 2 ⎟⎟ 0 . 622 0 .622 ⎠ ⎝ ⎠ ⎝ We cannot fix the final state explicitly by a hand-solution. However, using EES which has built-in functions for moist air properties, the final state properties are determined to be ω2 = 0.02446 kg H 2O/kg dry air P2 = 133.87 kPa

V

h2 = 97.97 kJ/kg dry air

v 2 = 0.6867 m3 / kg dry air

The partial pressures at the initial and final states are Pv1 = φ1 Psat@35°C = 0.20(5.6291 kPa) = 1.126 kPa Pa1 = P 1 − Pv1 = 130 − 1.126 = 128.87 kPa Pv 2 = P 2 − Pa 2 = 133.87 − 128.81 = 5.07 kPa The specific volume of water at 35ºC is v w1 = v w2 = v g @35°C = 25.205 m 3 /kg

The internal energies per unit mass of dry air in the tank are u1 = h1 − Pa1v 1 − w1 Pv1v w1 = 49.16 − 128.87 × 0.6863 − 0.005433 × 1.126 × 25.205 = −39.44 kJ/kg u 2 = h2 − Pa 2v 2 − w 2 Pv 2v w 2 = 97.97 − 128.81 × 0.6867 − 0.02446 × 5.07 × 25.205 = 6.396 kJ/kg The enthalpy of water entering the tank from the supply line is h w1 = hf @50°C = 209.34 kJ/kg The internal energy of water vapor at the final state is u w 2 = u g @35°C = 2422.7 kJ/kg The amount of water supplied to the tank is m w = m a (ω 2 − ω1 ) = (0.7285 kg)(0.02446 - 0.005433) = 0.01386 kg An energy balance on the system gives E in = ΔE tank Qin + m w h w1 = m a (u 2 − u1 ) + m w u w2

Qin + (0.01386 kg )(209.34 kJ/kg) = (0.7285 kg)[6.396 - (-39.44)kJ/kg ] + (0.01386 kg)(2422.7 kJ/kg) Q& = 64.1 kJ in

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14-111

Fundamentals of Engineering (FE) Exam Problems

14-150 A room is filled with saturated moist air at 25°C and a total pressure of 100 kPa. If the mass of dry air in the room is 100 kg, the mass of water vapor is

(a) 0.52 kg

(b) 1.97 kg

(c) 2.96 kg

(d) 2.04 kg

(e) 3.17 kg

Answer (d) 2.04 kg Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T1=25 "C" P=100 "kPa" m_air=100 "kg" RH=1 P_g=PRESSURE(Steam_IAPWS,T=T1,x=0) RH=P_v/P_g P_air=P-P_v w=0.622*P_v/(P-P_v) w=m_v/m_air "Some Wrong Solutions with Common Mistakes:" W1_vmass=m_air*w1; w1=0.622*P_v/P "Using P instead of P-Pv in w relation" W2_vmass=m_air "Taking m_vapor = m_air" W3_vmass=P_v/P*m_air "Using wrong relation"

14-151 A room contains 50 kg of dry air and 0.6 kg of water vapor at 25°C and 95 kPa total pressure. The relative humidity of air in the room is

(a) 1.2%

(b) 18.4%

(c) 56.7%

(d) 65.2%

(e) 78.0%

Answer (c) 56.7% Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T1=25 "C" P=95 "kPa" m_air=50 "kg" m_v=0.6 "kg" w=0.622*P_v/(P-P_v) w=m_v/m_air P_g=PRESSURE(Steam_IAPWS,T=T1,x=0) RH=P_v/P_g "Some Wrong Solutions with Common Mistakes:" W1_RH=m_v/(m_air+m_v) "Using wrong relation" W2_RH=P_g/P "Using wrong relation"

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14-112

14-152 A 40-m3 room contains air at 30°C and a total pressure of 90 kPa with a relative humidity of 75 percent. The mass of dry air in the room is

(a) 24.7 kg

(b) 29.9 kg

(c) 39.9 kg

(d) 41.4 kg

(e) 52.3 kg

Answer (c) 39.9 kg Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). V=40 "m^3" T1=30 "C" P=90 "kPa" RH=0.75 P_g=PRESSURE(Steam_IAPWS,T=T1,x=0) RH=P_v/P_g P_air=P-P_v R_air=0.287 "kJ/kg.K" m_air=P_air*V/(R_air*(T1+273)) "Some Wrong Solutions with Common Mistakes:" W1_mass=P_air*V/(R_air*T1) "Using C instead of K" W2_mass=P*V/(R_air*(T1+273)) "Using P instead of P_air" W3_mass=m_air*RH "Using wrong relation"

14-153 A room contains air at 30°C and a total pressure of 96.0 kPa with a relative humidity of 75 percent. The partial pressure of dry air is

(a) 82.0 kPa

(b) 85.8 kPa

(c) 92.8 kPa

(d) 90.6 kPa

(e) 72.0 kPa

Answer (c) 92.8 kPa Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T1=30 "C" P=96 "kPa" RH=0.75 P_g=PRESSURE(Steam_IAPWS,T=T1,x=0) RH=P_v/P_g P_air=P-P_v "Some Wrong Solutions with Common Mistakes:" W1_Pair=P_v "Using Pv as P_air" W2_Pair=P-P_g "Using wrong relation" W3_Pair=RH*P "Using wrong relation"

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14-113

14-154 The air in a house is at 20°C and 50 percent relative humidity. Now the air is cooled at constant pressure. The temperature at which the moisture in the air will start condensing is

(a) 8.7°C

(b) 11.3°C

(c) 13.8°C

(d) 9.3°C

(e) 10.0°C

Answer (d) 9.3°C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T1=20 "C" RH1=0.50 P_g=PRESSURE(Steam_IAPWS,T=T1,x=0) RH1=P_v/P_g T_dp=TEMPERATURE(Steam_IAPWS,x=0,P=P_v) "Some Wrong Solutions with Common Mistakes:" W1_Tdp=T1*RH1 "Using wrong relation" W2_Tdp=(T1+273)*RH1-273 "Using wrong relation" W3_Tdp=WETBULB(AirH2O,T=T1,P=P1,R=RH1); P1=100 "Using wet-bulb temperature"

14-155 On the psychrometric chart, a cooling and dehumidification process appears as a line that is

(a) horizontal to the left, (b) vertical downward, (c) diagonal upwards to the right (NE direction) (d) diagonal upwards to the left (NW direction) (e) diagonal downwards to the left (SW direction) Answer (e) diagonal downwards to the left (SW direction)

14-156 On the psychrometric chart, a heating and humidification process appears as a line that is

(a) horizontal to the right, (b) vertical upward, (c) diagonal upwards to the right (NE direction) (d) diagonal upwards to the left (NW direction) (e) diagonal downwards to the right (SE direction) Answer (c) diagonal upwards to the right (NE direction)

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14-114

14-157 An air stream at a specified temperature and relative humidity undergoes evaporative cooling by spraying water into it at about the same temperature. The lowest temperature the air stream can be cooled to is

(a) the dry bulb temperature at the given state (b) the wet bulb temperature at the given state (c) the dew point temperature at the given state (d) the saturation temperature corresponding to the humidity ratio at the given state (e) the triple point temperature of water Answer (a) the dry bulb temperature at the given state

14-158 Air is cooled and dehumidified as it flows over the coils of a refrigeration system at 85 kPa from 30°C and a humidity ratio of 0.023 kg/kg dry air to 15°C and a humidity ratio of 0.015 kg/kg dry air. If the mass flow rate of dry air is 0.7 kg/s, the rate of heat removal from the air is

(a) 5 kJ/s

(b) 10 kJ/s

(c) 15 kJ/s

(d) 20 kJ/s

(e) 25 kJ/s

Answer (e) 25 kJ/s Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P=85 "kPa" T1=30 "C" w1=0.023 T2=15 "C" w2=0.015 m_air=0.7 "kg/s" m_water=m_air*(w1-w2) h1=ENTHALPY(AirH2O,T=T1,P=P,w=w1) h2=ENTHALPY(AirH2O,T=T2,P=P,w=w2) h_w=ENTHALPY(Steam_IAPWS,T=T2,x=0) Q=m_air*(h1-h2)-m_water*h_w "Some Wrong Solutions with Common Mistakes:" W1_Q=m_air*(h1-h2) "Ignoring condensed water" W2_Q=m_air*Cp_air*(T1-T2)-m_water*h_w; Cp_air = 1.005 "Using dry air enthalpies" W3_Q=m_air*(h1-h2)+m_water*h_w "Using wrong sign"

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14-115

14-159 Air at a total pressure of 90 kPa, 15°C, and 75 percent relative humidity is heated and humidified to 25°C and 75 percent relative humidity by introducing water vapor. If the mass flow rate of dry air is 4 kg/s, the rate at which steam is added to the air is

(a) 0.032 kg/s

(b) 0.013 kg/s

(c) 0.019 kg/s

(d) 0.0079 kg/s

(e) 0 kg/s

Answer (a) 0.032 kg/s Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P=90 "kPa" T1=15 "C" RH1=0.75 T2=25 "C" RH2=0.75 m_air=4 "kg/s" w1=HUMRAT(AirH2O,T=T1,P=P,R=RH1) w2=HUMRAT(AirH2O,T=T2,P=P,R=RH2) m_water=m_air*(w2-w1) "Some Wrong Solutions with Common Mistakes:" W1_mv=0 "sine RH = constant" W2_mv=w2-w1 "Ignoring mass flow rate of air" W3_mv=RH1*m_air "Using wrong relation"

14-160 ··· 14-164 Design and Essay Problems

KJ

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15-1

Chapter 15 CHEMICAL REACTIONS Fuels and Combustion

15-1C Gasoline is C8H18, diesel fuel is C12H26, and natural gas is CH4.

15-2C Nitrogen, in general, does not react with other chemical species during a combustion process but its presence affects the outcome of the process because nitrogen absorbs a large proportion of the heat released during the chemical process.

15-3C Moisture, in general, does not react chemically with any of the species present in the combustion chamber, but it absorbs some of the energy released during combustion, and it raises the dew point temperature of the combustion gases.

15-4C The dew-point temperature of the product gases is the temperature at which the water vapor in the product gases starts to condense as the gases are cooled at constant pressure. It is the saturation temperature corresponding to the vapor pressure of the product gases.

15-5C The number of atoms are preserved during a chemical reaction, but the total mole numbers are not.

15-6C Air-fuel ratio is the ratio of the mass of air to the mass of fuel during a combustion process. Fuelair ratio is the inverse of the air-fuel ratio.

15-7C No. Because the molar mass of the fuel and the molar mass of the air, in general, are different.

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15-2

15-8 Sulfur is burned with oxygen to form sulfur dioxide. The minimum mass of oxygen required and the mass of sulfur dioxide in the products are to be determined when 1 kg of sulfur is burned. Properties The molar masses of sulfur and oxygen are 32.06 kg/kmol and 32.00 kg/kmol, respectively (Table A-1). Analysis The chemical reaction is given by S + O2 ⎯ ⎯→ SO 2

Hence, 1kmol of oxygen is required to burn 1 kmol of sulfur which produces 1 kmol of sulfur dioxide whose molecular weight is

S+ O2 ⎯ ⎯→ SO 2

M SO2 = M S + M O2 = 32.06 + 32.00 = 64.06 kg/kmol

Then, m O2 N O2 M O2 (1 kmol)(32 kg/kmol) = = = 0.998 kg O 2 /kg S (1 kmol)(32.06 kg/kmol) mS NSM S

and mSO2 N SO2 M SO2 (1 kmol)(64.06 kg/kmol) = = = 1.998 kg SO 2 /kg S (1 kmol)(32.06 kg/kmol) mS NSM S

15-9E Methane is burned with diatomic oxygen. The mass of water vapor in the products is to be determined when 1 lbm of methane is burned. Properties The molar masses of CH4, O2, CO2, and H2O are 16, 32, 44, and 18 lbm/lbmol, respectively (Table A-1E). Analysis The chemical reaction is given by CH 4 + 2O 2 ⎯ ⎯→ CO 2 + 2H 2 O

CH 4 + 2O 2 → CO 2 + 2H 2 O

Hence, for each lbmol of methane burned, 2 lbmol of water vapor are formed. Then, m H2O N H2O M H2O (2 lbmol)(18 lbm/lbmol) = = = 2.25 lbm H 2 O/lbm CH 4 (1 lbmol)(16 lbm/lbmol) m CH4 N CH4 M CH4

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15-3

Theoretical and Actual Combustion Processes

15-10C The causes of incomplete combustion are insufficient time, insufficient oxygen, insufficient mixing, and dissociation.

15-11C CO. Because oxygen is more strongly attracted to hydrogen than it is to carbon, and hydrogen is usually burned to completion even when there is a deficiency of oxygen.

15-12C It represent the amount of air that contains the exact amount of oxygen needed for complete combustion.

15-13C No. The theoretical combustion is also complete, but the products of theoretical combustion does not contain any uncombined oxygen.

15-14C Case (b).

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15-4

15-15 Propane is burned with theoretical amount of air. The mass fraction of carbon dioxide and the mole and mass fractions of the water vapor in the products are to be determined. Properties The molar masses of C3H8, O2, N2, CO2, and H2O are 44, 32, 28, 44, and 18 kg/kmol, respectively (Table A-1). Analysis (a) The reaction in terms of undetermined coefficients is C 3 H 8 + x(O 2 + 3.76 N 2 ) ⎯ ⎯→ yCO 2 + zH 2 O + pN 2

Balancing the carbon in this reaction gives

C3H8

y=3

Air

and the hydrogen balance gives

2z = 8 ⎯ ⎯→ z = 4

Combustion chamber

CO2, H2O, N2

100% theoretical

The oxygen balance produces 2x = 2 y + z ⎯ ⎯→ x = y + z / 2 = 3 + 4 / 2 = 5

A balance of the nitrogen in this reaction gives 2 × 3.76 x = 2 p ⎯ ⎯→ p = 3.76 x = 3.76 × 5 = 18.8

In balanced form, the reaction is C 3 H 8 + 5O 2 + 18.8 N 2 ⎯ ⎯→ 3CO 2 + 4H 2 O + 18.8N 2

The mass fraction of carbon dioxide is determined from mf CO2 =

m CO2 N CO2 M CO2 = m products N CO2 M CO2 + N H2O M H2O + N N2 M N2

(3 kmol)(44 kg/kmol) (3 kmol)(44 kg/kmol) + (4 kmol)(18 kg/kmol) + (18.8 kmol)(28 kg/kmol) 132 kg = = 0.181 730.4 kg

=

(b) The mole and mass fractions of water vapor are y H2O =

N H2O N H2O 4 kmol 4 kmol = = = = 0.155 N products N CO2 + N H2O + N N2 3 kmol + 4 kmol + 18.8 kmol 25.8 kmol

mf H2O =

m H2O N H2O M H2O = m products N CO2 M CO2 + N H2O M H2O + N N2 M N2

(4 kmol)(18 kg/kmol) (3 kmol)(44 kg/kmol) + (4 kmol)(18 kg/kmol) + (18.8 kmol)(28 kg/kmol) 72 kg = = 0.0986 730.4 kg

=

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15-5

15-16 Methane is burned with air. The mass flow rates at the two inlets are to be determined. Properties The molar masses of CH4, O2, N2, CO2, and H2O are 16, 32, 28, 44, and 18 kg/kmol, respectively (Table A-1). Analysis The stoichiometric combustion equation of CH4 is CH 4 + a th [O 2 + 3.76N 2 ] ⎯ ⎯→ CO 2 + 2H 2 O + 3.76a th N 2 a th = 1 + 1 ⎯ ⎯→ a th = 2

O2 balance: Substituting,

CH4 Products Air

CH 4 + 2[O 2 + 3.76N 2 ] ⎯ ⎯→ CO 2 + 2H 2 O + 7.52N 2

The masses of the reactants are m CH4 = N CH4 M CH4 = (1 kmol)(16 kg/kmol) = 16 kg m O2 = N O2 M O2 = (2 kmol)(32kg/kmol) = 64 kg m N2 = N N2 M N2 = (2 × 3.76 kmol)(28 kg/kmol) = 211 kg

The total mass is m total = m CH4 + m O2 + N N2 = 16 + 64 + 211 = 291 kg

Then the mass fractions are mf CH4 =

m CH4 16 kg = = 0.05498 m total 291 kg

mf O2 =

m O2 64 kg = = 0.2199 m total 291 kg

mf N2 =

m N2 211 kg = = 0.7251 m total 291 kg

For a mixture flow of 0.01 kg/s, the mass flow rates of the reactants are m& CH4 = mf CH4 m& = (0.05498)(0.01kg/s) = 0.0005498 kg/s m& air = m& − m& CH4 = 0.01 − 0.0005498 = 0.009450 kg/s

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15-6

15-17 n-Octane is burned with stoichiometric amount of oxygen. The mass fractions of each of the products and the mass of water in the products per unit mass of fuel burned are to be determined. Assumptions 1 Combustion is complete. 2 The combustion products contain CO2 and H2O. 3 Combustion gases are ideal gases. Properties The molar masses of C, H2, and O2 are 12 kg/kmol, 2 kg/kmol, and 32 kg/kmol, respectively (Table A-1). Analysis The combustion equation in this case is C 8 H 18 + 12.5O 2 ⎯ ⎯→ 8CO 2 + 9H 2 O

The mass of each product and the total mass are

C8H18 O2

Combustion chamber

Products

m CO2 = N CO2 M CO2 = (8 kmol)(44 kg/kmol) = 352 kg m H2O = N H2O M H2O = (9 kmol)(18 kg/kmol) = 162 kg m total = m CO2 + m H2O = 352 + 162 = 514 kg

Then the mass fractions are mf CO2 =

m CO2 352 kg = = 0.6848 m total 514 kg

mf H2O =

m H2O 162 kg = = 0.3152 m total 514 kg

The mass of water in the products per unit mass of fuel burned is determined from m H2O (9 × 18) kg = = 1.42 kg H 2 O/kg C 8 H18 m C8H18 (1× 114) kg

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15-7

15-18 Acetylene is burned with 10 percent excess oxygen. The mass fractions of each of the products and the mass of oxygen used per unit mass of fuel burned are to be determined. Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, and O2. 3 Combustion gases are ideal gases. Properties The molar masses of C, H2, and O2 are 12 kg/kmol, 2 kg/kmol, and 32 kg/kmol, respectively (Table A-1). Analysis The stoichiometric combustion equation is C 2 H 2 + 2.5O 2 ⎯ ⎯→ 2CO 2 + H 2 O

The combustion equation with 10% excess oxygen is

C2H2 O2

Combustion chamber

Products

C 2 H 2 + 2.75O 2 ⎯ ⎯→ 2CO 2 + H 2 O + 0.25O 2

The mass of each product and the total mass are m CO2 = N CO2 M CO2 = (2 kmol)(44 kg/kmol) = 88 kg m H2O = N H2O M H2O = (1 kmol)(18 kg/kmol) = 18 kg m O2 = N O2 M O2 = (0.25 kmol)(32 kg/kmol) = 8 kg m total = m CO2 + m H2O + m O2 = 88 + 18 + 8 = 114 kg

Then the mass fractions are mf CO2 =

m CO2 88 kg = = 0.7719 m total 114 kg

mf H2O =

m H2O 18 kg = = 0.1579 m total 114 kg

mf O2 =

m O2 8 kg = = 0.0702 m total 114 kg

The mass of oxygen per unit mass of fuel burned is determined from m O2 (2.75 × 32) kg = = 3.385 kg O 2 /kg C 2 H 2 (1× 26) kg m C2H2

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15-8

15-19 Coal whose mass percentages are specified is burned with 50 percent excess air. The fuel-air ratio is to be determined. Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, SO2, N2, and O2. 3 Combustion gases are ideal gases. Properties The molar masses of C, H2, O2, S, and air are 12, 2, 32, 32, and 29 kg/kmol, respectively (Table A-1). Analysis The mass fractions of the constituent of the coal when the ash is substituted are mf C =

mC 79.61 kg 79.61 kg = = = 0.8712 m total (100 - 8.62) kg 91.38 kg

mf H2 =

m H2 4.66 kg = = 0.05100 m total 91.38 kg

mf O2 =

m O2 4.76 kg = = 0.05209 m total 91.38 kg

mf N2 =

m N2 1.83 kg = = 0.02003 m total 91.38 kg

mf S =

mS 0.52 kg = = 0.00569 m total 91.38 kg

79.61% C 4.66% H2 4.76% O2 1.83% N2 0.52% S 8.62% ash (by mass)

We now consider 100 kg of this mixture. Then the mole numbers of each component are NC =

mC 87.12 kg = = 7.26 kmol M C 12 kg/kmol

N H2 =

m H2 5.10 kg = = 2.55 kmol M H2 2 kg/kmol

Coal

N O2

m 5.209 kg = O2 = = 0.1628 kmol M O2 32 kg/kmol

Air

N N2

m 2.003 kg = N2 = = 0.07154 kmol M N2 28 kg/kmol

NS =

Combustion chamber

CO2, H2O SO2, O2, N2

50% excess

mS 0.569 kg = = 0.01778 kmol M S 32 kg/kmol

The mole number of the mixture and the mole fractions are N m = 7.26 + 2.55 + 0.1628 + 0.07154 + 0.01778 = 10.06 kmol yC =

NC 7.26 kmol = = 0.7215 N m 10.06 kmol

y H2 =

N H2 2.55 kmol = = 0.2535 N m 10.06 kmol

y O2 =

N O2 0.1628 kmol = = 0.01618 10.06 kmol Nm

y N2 =

N N2 0.07154 kmol = = 0.00711 10.06 kmol Nm

yS =

N S 0.01778 kmol = = 0.00177 10.06 kmol Nm

Then, the combustion equation in this case may be written as

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15-9

0.7215C + 0.2535H 2 + 0.01618O 2 + 0.00711N 2 + 0.00177S + 1.5a th (O 2 + 3.76 N 2 ) ⎯ ⎯→ xCO 2 + yH 2 O + zSO 2 + kN 2 + 0.5a th O 2

According to the species balances, C balance : x = 0.7215 H 2 balance : y = 0.2535 S balance : z = 0.00177 O 2 balance : 0.01618 + 1.5a th = x + 0.5 y + z + 0.5a th 1.5a th − 0.5a th = 0.7215 + 0.5(0.2535) + 0.00177 − 0.01617 ⎯ ⎯→ a th = 0.8339 N 2 balance : 0.00711 + 1.5 × 3.76a th = k ⎯ ⎯→ k = 0.00711 + 1.5 × 3.76 × 0.8339 = 4.710

Substituting, 0.7215C + 0.2535H 2 + 0.01617O 2 + 0.00711N 2 + 0.00177S + 1.2509(O 2 + 3.76 N 2 ) ⎯ ⎯→ 0.7215CO 2 + 0.2535H 2 O + 0.0018SO 2 + 4.71N 2 + 0.4170O 2

The fuel-air mass ratio is then FA = =

m fuel (0.7215 × 12 + 0.2535 × 2 + 0.01617 × 32 + 0.00711× 28 + 0.00177 × 32) kg = (1.2509 × 4.76 × 29) kg m air 9.938 kg = 0.0576 kg fuel/kg air 172.5 kg

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15-10

15-20 Propane is burned with 75 percent excess air during a combustion process. The AF ratio is to be determined. Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, O2, and N2 only. Properties The molar masses of C, H2, and air are 12 kg/kmol, 2 kg/kmol, and 29 kg/kmol, respectively (Table A-1). Analysis The combustion equation in this case can be written as C 3 H 8 + 1.75a th [O 2 + 3.76N 2 ] ⎯ ⎯→ 3CO 2 + 4H 2 O + 0.75a th O 2 + (1.75 × 3.76)a th N 2

where ath is the stoichiometric coefficient for air. We have automatically accounted for the 75% excess air by using the factor 1.75ath instead of ath for air. The stoichiometric amount of oxygen (athO2) will be used to oxidize the fuel, and the remaining excess amount (0.75athO2) will appear in the products as free oxygen. The coefficient ath is determined from the O2 balance, O2 balance:

175 . a th = 3 + 2 + 0.75a th

⎯ ⎯→

C3H8 Products Air 75% excess a th = 5

Substituting,

C3H 8 + 8.75 O 2 + 3.76 N 2

⎯ ⎯→ 3CO 2 + 4H 2O + 3.75O 2 + 32.9 N 2

The air-fuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel, AF =

m air (8.75 × 4.76 kmol)(29 kg/kmol) = = 27.5 kg air/kg fuel m fuel (3 kmol)(12 kg/kmol) + (4 kmol)(2 kg/kmol)

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15-11

15-21 Acetylene is burned with the stoichiometric amount of air during a combustion process. The AF ratio is to be determined on a mass and on a mole basis. Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, and N2 only. Properties The molar masses of C, H2, and air are 12 kg/kmol, 2 kg/kmol, and 29 kg/kmol, respectively (Table A-1). Analysis This is a theoretical combustion process since C2H2 is burned completely with stoichiometric amount of air. The stoichiometric combustion equation of C2H2 is C 2 H 2 + a th [O 2 + 3.76N 2 ] ⎯ ⎯→ 2CO 2 + H 2 O + 3.76a th N 2

O2 balance:

a th = 2 + 0.5

⎯ ⎯→

a th = 2.5

Substituting,

C2H2 Products 100% theoretical air

C 2 H 2 + 2.5[O 2 + 3.76N 2 ] ⎯ ⎯→ 2CO 2 + H 2 O + 9.4N 2

The air-fuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel, AF =

m air (2.5 × 4.76 kmol)(29 kg/kmol) = = 13.3 kg air/kg fuel m fuel (2 kmol)(12 kg/kmol) + (1 kmol)(2 kg/kmol)

On a mole basis, the air-fuel ratio is expressed as the ratio of the mole numbers of the air to the mole numbers of the fuel, AFmole basis =

N air (2.5 × 4.76) kmol = = 11.9 kmol air/kmol fuel 1 kmol fuel N fuel

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15-12

15-22E Ethylene is burned with 200 percent theoretical air during a combustion process. The AF ratio and the dew-point temperature of the products are to be determined. Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, O2, and N2 only. 3 Combustion gases are ideal gases. Properties The molar masses of C, H2, and air are 12 lbm/lbmol, 2 lbm/lbmol, and 29 lbm/lbmol, respectively (Table A-1E). Analysis (a) The combustion equation in this case can be written as C 2 H 4 + 2a th O 2 + 3.76N 2

C2H4 Products 200% theoretical air

⎯ ⎯→ 2CO 2 + 2H 2O + a th O 2 + (2 × 3.76)a th N 2

where ath is the stoichiometric coefficient for air. It is determined from 2a th = 2 + 1 + a th

O2 balance:

⎯ ⎯→

a th = 3

Substituting, C 2 H 4 + 6 O 2 + 3.76N 2

⎯⎯→ 2CO 2 + 2 H 2 O + 3O 2 + 22.56N 2

The air-fuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel, AF =

m air (6 × 4.76 lbmol)(29 lbm/lbmol) = = 29.6 lbm air/lbm fuel m fuel (2 lbmol)(12 lbm/lbmol) + (2 lbmol)(2 lbm/lbmol)

(b) The dew-point temperature of a gas-vapor mixture is the saturation temperature of the water vapor in the product gases corresponding to its partial pressure. That is, ⎛ Nv Pv = ⎜ ⎜ N prod ⎝

⎞ ⎟ Pprod = ⎛⎜ 2 lbmol ⎞⎟(14.5 psia ) = 0.981 psia ⎜ 29.56 lbmol ⎟ ⎟ ⎠ ⎝ ⎠

Thus, Tdp = Tsat @ 0.981 psia = 101°F

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15-13

15-23 Propylene is burned with 50 percent excess air during a combustion process. The AF ratio and the temperature at which the water vapor in the products will start condensing are to be determined. Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, O2, and N2 only. 3 Combustion gases are ideal gases. Properties The molar masses of C, H2, and air are 12 kg/kmol, 2 kg/kmol, and 29 kg/kmol, respectively (Table A-1). Analysis (a) The combustion equation in this case can be written as

C3H6

Products

50% excess air

C 3 H 6 + 1.5a th [O 2 + 3.76N 2 ] ⎯ ⎯→ 3CO 2 + 3H 2 O + 0.5a th O 2 + (1.5 × 3.76)a th N 2

where ath is the stoichiometric coefficient for air. It is determined from O2 balance:

15 . a th = 3 + 15 . + 0.5a th

⎯ ⎯→

a th = 4.5

Substituting, C 3 H 6 + 6.75 O 2 + 3.76N 2

⎯⎯→ 3CO 2 + 3H 2 O + 2.25O 2 + 25.38N 2

The air-fuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel, AF =

m air (6.75 × 4.76 kmol)(29 kg/kmol) = = 22.2 kg air/kg fuel m fuel (3 kmol)(12 kg/kmol) + (3 kmol)(2 kg/kmol)

(b) The dew-point temperature of a gas-vapor mixture is the saturation temperature of the water vapor in the product gases corresponding to its partial pressure. That is, ⎛ Nv Pv = ⎜ ⎜ N prod ⎝

⎞ ⎟ Pprod = ⎛⎜ 3 kmol ⎞⎟(105 kPa ) = 9.367 kPa ⎜ 33.63 kmol ⎟ ⎟ ⎝ ⎠ ⎠

Thus, Tdp = Tsat @9.367 kPa = 44.5°C

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15-14

15-24 Butane C4H10 is burned with 200 percent theoretical air. The kmol of water that needs to be sprayed into the combustion chamber per kmol of fuel is to be determined. Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, O2, and N2 only. Properties The molar masses of C, H2, O2 and air are 12 kg/kmol, 2 kg/kmol, 32 kg/kmol, and 29 kg/kmol, respectively (Table A-1). Analysis The reaction equation for 200% theoretical air without the additional water is C 4 H 10 + 2a th [O 2 + 3.76N 2 ] ⎯ ⎯→ B CO 2 + D H 2 O + E O 2 + F N 2

where ath is the stoichiometric coefficient for air. We have automatically accounted for the 100% excess air by using the factor 2ath instead of ath for air. The coefficient ath and other coefficients are to be determined from the mass balances Carbon balance:

B=4

Hydrogen balance:

2 D = 10 ⎯ ⎯→ D = 5 2 × 2a th = 2 B + D + 2 E

Oxygen balance:

a th = E

C4H10 Products Air 200% theoretical

2a th × 3.76 = F

Nitrogen balance:

Solving the above equations, we find the coefficients (E = 6.5, F = 48.88, and ath = 6.5) and write the balanced reaction equation as C 4 H 10 + 13[O 2 + 3.76N 2 ] ⎯ ⎯→ 4 CO 2 + 5 H 2 O + 6.5 O 2 + 48.88 N 2

With the additional water sprayed into the combustion chamber, the balanced reaction equation is C 4 H 10 + 13[O 2 + 3.76N 2 ] + N v H 2 O ⎯ ⎯→ 4 CO 2 + (5 + N v ) H 2 O + 6.5 O 2 + 48.88 N 2

The partial pressure of water in the saturated product mixture at the dew point is Pv ,prod = Psat@60°C = 19.95 kPa

The vapor mole fraction is yv =

Pv ,prod Pprod

=

19.95 kPa = 0.1995 100 kPa

The amount of water that needs to be sprayed into the combustion chamber can be determined from yv =

N water N total,product

⎯ ⎯→ 0.1995 =

5 + Nv ⎯ ⎯→ N v = 9.796 kmol 4 + 5 + N v + 6.5 + 48.88

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15-15

15-25 A fuel mixture of 20% by mass methane, CH4, and 80% by mass ethanol, C2H6O, is burned completely with theoretical air. The required flow rate of air is to be determined. Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, and N2 only. Properties The molar masses of C, H2, O2 and air are 12 kg/kmol, 2 kg/kmol, 32 kg/kmol, and 29 kg/kmol, respectively (Table A-1). Analysis The combustion equation in this case can be written as x CH 4 + y C 2 H 6 O + a th [O 2 + 3.76N 2 ] ⎯ ⎯→ B CO 2 + D H 2 O + F N 2

where ath is the stoichiometric coefficient for air. The coefficient ath and other coefficients are to be determined from the mass balances Carbon balance:

x + 2y = B

Hydrogen balance:

4 x + 6 y = 2D

Oxygen balance:

2ath + y = 2 B + D

Nitrogen balance:

3.76a th = F

20% CH4 80% C2H6O Air

Products

100% theoretical

Solving the above equations, we find the coefficients as x = 0.4182

B = 1.582

y = 0.5818

D = 2.582

ath = 2.582

F = 9.708

Then, we write the balanced reaction equation as 0.4182 CH 4 + 0.5818 C 2 H 6O + 2.582 [O 2 + 3.76N 2 ] ⎯ ⎯→ 1.582 CO 2 + 2.582 H 2O + 9.708 N 2

The air-fuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel, AF =

m air m fuel

(2.582 × 4.76 kmol)(29 kg/kmol) (0.4182 kmol)(12 + 4 × 1)kg/kmol + (0.5818 kmol)(2 × 12 + 6 × 1 + 16)kg/kmol = 10.64 kg air/kg fuel =

Then, the required flow rate of air becomes m& air = AFm& fuel = (10.64)(31 kg/s) = 330 kg/s

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15-16

15-26 Octane is burned with 250 percent theoretical air during a combustion process. The AF ratio and the dew-pint temperature of the products are to be determined. Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, O2, and N2 only. 3 Combustion gases are ideal gases. Properties The molar masses of C, H2, and air are 12 kg/kmol, 2 kg/kmol, and 29 kg/kmol, respectively (Table A-1). Analysis (a) The combustion equation in this case can be written as

C8 H18 + 2.5a th O 2 + 3.76N 2

⎯ ⎯→ 8CO 2 + 9H 2O + 1.5a th O 2 + (2.5 × 3.76)a th N 2

where ath is the stoichiometric coefficient for air. It is determined from O2 balance:

C8H18

2.5a th = 8 + 4.5 + 1.5a th ⎯ ⎯→ a th = 12.5

Air 25°C

Substituting,

Combustion Products chamber P = 1 atm

C 8 H 18 + 31.25[O 2 + 3.76N 2 ] → 8CO 2 + 9H 2 O + 18.75O 2 + 117.5N 2

Thus, AF =

m air (31.25 × 4.76 kmol)(29 kg/kmol) = = 37.8 kg air/kg fuel m fuel (8 kmol)(12 kg/kmol) + (9 kmol)(2 kg/kmol)

(b) The dew-point temperature of a gas-vapor mixture is the saturation temperature of the water vapor in the product gases corresponding to its partial pressure. That is, ⎛ N ⎞ ⎛ 9 kmol ⎞ ⎟⎟(101.325 kPa ) = 5.951 kPa Pv = ⎜ v ⎟ Pprod = ⎜⎜ ⎜ N prod ⎟ ⎝ 153.25 kmol ⎠ ⎝ ⎠

Thus, Tdp = Tsat @5.951 kPa = 36.0°C

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15-17

15-27 Gasoline is burned steadily with air in a jet engine. The AF ratio is given. The percentage of excess air used is to be determined. Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, and N2 only. Properties The molar masses of C, H2, and air are 12 kg/kmol, 2 kg/kmol, and 29 kg/kmol, respectively (Table A-1). Analysis The theoretical combustion equation in this case can be written as C 8 H 18 + a th [O 2 + 3.76N 2 ] ⎯ ⎯→ 8CO 2 + 9H 2 O + 3.76a th N 2

Gasoline (C8H18)

where ath is the stoichiometric coefficient for air. It is determined from O2 balance:

a th = 8 + 4.5

⎯ ⎯→

Jet engine

Products

Air

a th = 12.5

The air-fuel ratio for the theoretical reaction is determined by taking the ratio of the mass of the air to the mass of the fuel for, AFth =

m air,th m fuel

=

(12.5 × 4.76 kmol)(29 kg/kmol) = 15.14 kg air/kg fuel (8 kmol)(12 kg/kmol) + (9 kmol)(2 kg/kmol)

Then the percent theoretical air used can be determined from Percent theoretical air =

AFact AFth

=

18 kg air/kg fuel = 119% 15.14 kg air/kg fuel

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

15-18

15-28 n-butane is burned with stoichiometric amount of air. The mass fraction of each product, the mass of CO2 and air per unit mass of fuel burned are to be determined. Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, O2, and N2 only. Properties The molar masses of C, H2, O2 and air are 12 kg/kmol, 2 kg/kmol, 32 kg/kmol, and 29 kg/kmol, respectively (Table A-1). Analysis The reaction equation for 100% theoretical air is C 4 H 10 + a th [O 2 + 3.76N 2 ] ⎯ ⎯→ B CO 2 + D H 2 O + E N 2

where ath is the stoichiometric coefficient for air. The coefficient ath and other coefficients are to be determined from the mass balances

C4H10 Products Air 100% theoretical

Carbon balance:

B=4

Hydrogen balance:

⎯→ D = 5 2 D = 10 ⎯

Oxygen balance:

2a th = 2 B + D ⎯ ⎯→ a th = 0.5(2 × 4 + 5) = 6.5

Nitrogen balance:

a th × 3.76 = E ⎯ ⎯→ E = 6.5 × 3.76 = 24.44

Substituting, the balanced reaction equation is C 4 H 10 + 6.5[O 2 + 3.76N 2 ] ⎯ ⎯→ 4 CO 2 + 5 H 2 O + 24.44 N 2

The mass of each product and the total mass are m CO2 = N CO2 M CO2 = (4 kmol)(44 kg/kmol) = 176 kg m H2O = N H2O M H2O = (5 kmol)(18 kg/kmol) = 90 kg m N2 = N N2 M N2 = (24.44 kmol)(28 kg/kmol) = 684 kg m total = m CO2 + m H2O = 176 + 90 + 684 = 950 kg

Then the mass fractions are mf CO2 =

m CO2 176 kg = = 0.1853 m total 950 kg

mf H2O =

m H2O 90 kg = = 0.0947 m total 950 kg

mf O2 =

m O2 684 kg = = 0.720 m total 950 kg

The mass of carbon dioxide per unit mass of fuel burned is m CO2 (4 × 44) kg = = 3.034 kg CO 2 /kg C 4 H10 m C4H10 (1× 58) kg

The mass of air required per unit mass of fuel burned is m air (6.5 × 4.76 × 29) kg = = 15.47 kg air/kg C 4 H10 (1× 58) kg m C4H10

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15-19

15-29 Coal whose mass percentages are specified is burned with stoichiometric amount of air. The mole fractions of the products, the apparent molecular weight of the product gas, and the air-fuel ratio are to be determined. Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, SO2, and N2. 3 Combustion gases are ideal gases. Properties The molar masses of C, H2, O2, S, and air are 12, 2, 32, 32, and 29 kg/kmol, respectively (Table A-1). Analysis The mass fractions of the constituent of the coal when the ash is substituted are mf C =

mC 84.36 kg 84.36 kg = = = 0.9153 m total (100 − 7.83) kg 92.17 kg

mf H2 =

m H2 1.89 kg = = 0.02051 m total 92.17 kg

mf O2 =

m O2 4.40 kg = = 0.04774 m total 92.17 kg

mf N2 =

m N2 0.63 kg = = 0.006835 m total 92.17 kg

mf S =

mS 0.89 kg = = 0.009656 m total 92.17 kg

84.36% C 1.89% H2 4.40% O2 0.63% N2 0.89% S 7.83% ash (by mass)

We now consider 100 kg of this mixture. Then the mole numbers of each component are NC =

mC 91.53 kg = = 7.628 kmol M C 12 kg/kmol

N H2

m 2.051 kg = H2 = = 1.026 kmol M H2 2 kg/kmol

N O2

m 4.774 kg = O2 = = 0.1492 kmol M O2 32 kg/kmol

N N2 =

m N2 0.6835 kg = = 0.02441 kmol M N2 28 kg/kmol

NS =

mS 0.9656 kg = = 0.03018 kmol M S 32 kg/kmol

Coal Air theoretical

Combustion chamber

Products

The mole number of the mixture and the mole fractions are N m = 7.628 + 1.026 + 0.1492 + 0.02441 + 0.03018 = 8.858 kmol yC =

N C 7.628 kmol = = 0.8611 N m 8.858 kmol

y H2 =

N H2 1.026 kmol = = 0.1158 Nm 8.858 kmol

y O2 =

N O2 0.1492 kmol = = 0.01684 Nm 8.858 kmol

y N2 =

N N2 0.02441 kmol = = 0.002756 Nm 8.858 kmol

yS =

N S 0.03018 kmol = = 0.003407 8.858 kmol Nm

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15-20

Then, the combustion equation in this case may be written as 0.8611C + 0.1158H 2 + 0.01684O 2 + 0.002756 N 2 + 0.003407S + a th (O 2 + 3.76 N 2 ) ⎯ ⎯→ xCO 2 + yH 2 O + zSO 2 + kN 2

According to the species balances, C balance : x = 0.8611 H 2 balance : y = 0.1158 S balance : z = 0.003407 O 2 balance : 0.01684 + a th = x + 0.5 y + z a th = 0.8611 + 0.5(0.1158) + 0.003407 − 0.01684 = 0.9056 N 2 balance : k = 0.002756 + 3.76a th = 0.002756 + 3.76 × 0.9056 = 3.408

Substituting, 0.8611C + 0.1158H 2 + 0.01684O 2 + 0.002756 N 2 + 0.003407S + 0.9056(O 2 + 3.76 N 2 ) ⎯ ⎯→ 0.8611CO 2 + 0.1158H 2 O + 0.00341SO 2 + 3.408N 2

The mole fractions of the products are N m = 0.8611 + 0.1158 + 0.00341 + 3.408 = 4.388 kmol y CO2 =

N CO2 0.8611 kmol = = 0.1962 4.388 kmol Nm

y H2O =

N H2O 0.1158 kmol = = 0.02639 4.388 kmol Nm

y SO2 =

N SO2 0.00341 kmol = = 0.00078 4.388 kmol Nm

y N2 =

N N2 3.408 kmol = = 0.7767 4.388 kmol Nm

The apparent molecular weight of the product gas is Mm =

m m (0.8611× 44 + 0.1158 × 18 + 0.00341× 64 + 3.408 × 28) kg = = 30.91 kg/kmol 4.388 kmol Nm

The air-fuel mass ratio is then AF = =

m air (0.9056 × 4.76 × 29) kg = m fuel (0.8611× 12 + 0.1158 × 2 + 0.01684 × 32 + 0.002756 × 28 + 0.003408 × 32) kg 125.0 kg = 11.07 kg air/kg fuel 11.29 kg

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15-21

15-30 Methyl alcohol is burned with stoichiometric amount of air. The mole fraction of each product, the apparent molar mass of the product gas, and the mass of water in the products per unit mass of fuel burned are to be determined. Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, and N2 only. Properties The molar masses of C, H2, O2 and air are 12 kg/kmol, 2 kg/kmol, 32 kg/kmol, and 29 kg/kmol, respectively (Table A-1). Analysis The balanced reaction equation for stoichiometric air is CH 3 OH + 1.5[O 2 + 3.76N 2 ] ⎯ ⎯→ CO 2 + 2 H 2 O + 5.64 N 2

The mole fractions of the products are N m = 1 + 2 + 5.64 = 8.64 kmol y CO2

N 1 kmol = CO2 = = 0.1157 8.64 kmol Nm

y H2O =

N H2O 2 kmol = = 0.2315 8.64 kmol Nm

y N2 =

N N2 5.64 kmol = = 0.6528 8.64 kmol Nm

CH3OH Air

Combustion Products chamber

100% theoretical

The apparent molecular weight of the product gas is Mm =

m m (1× 44 + 2 × 18 + 5.64 × 28) kg = = 27.54 kg/kmol 8.64 kmol Nm

The mass of water in the products per unit mass of fuel burned is m H2O (2 × 18) kg = = 1.125 kg H 2 O/kg CH 3 OH m CH3OH (1× 32) kg

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15-22

15-31 Coal whose mass percentages are specified is burned with stoichiometric amount of air. The combustion is incomplete. The mass fractions of the products, the apparent molecular weight of the product gas, and the air-fuel ratio are to be determined. Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, CO, H2O, SO2, and N2. 3 Combustion gases are ideal gases. Properties The molar masses of C, H2, O2, S, and air are 12, 2, 32, 32, and 29 kg/kmol, respectively (Table A-1). Analysis The mass fractions of the constituent of the coal when the ash is substituted are mf C =

mC 61.40 kg 61.40 kg = = = 0.6463 m total (100 − 5.00) kg 95.00 kg

mf H2 =

m H2 5.79 kg = = 0.06095 m total 95.00 kg

mf O2 =

m O2 25.31 kg = = 0.2664 m total 95.00 kg

mf N2 =

m N2 1.09 kg = = 0.01147 m total 95.00 kg

mf S =

mS 1.41 kg = = 0.01484 m total 95.00 kg

61.40% C 5.79% H2 25.31% O2 1.09% N2 1.41% S 5.00% ash (by mass)

We now consider 100 kg of this mixture. Then the mole numbers of each component are NC =

mC 64.63 kg = = 5.386 kmol M C 12 kg/kmol

N H2

m 6.095 kg = H2 = = 3.048 kmol M H2 2 kg/kmol

N O2

m 26.64 kg = O2 = = 0.8325 kmol M O2 32 kg/kmol

N N2 =

m N2 1.147 kg = = 0.04096 kmol M N2 28 kg/kmol

NS =

mS 1.484 kg = = 0.04638 kmol M S 32 kg/kmol

Coal Air theoretical

Combustion chamber

Products

The mole number of the mixture and the mole fractions are N m = 5.386 + 3.048 + 0.8325 + 0.04096 + 0.04638 = 9.354 kmol yC =

N C 5.386 kmol = = 0.5758 N m 9.354 kmol

y H2 =

N H2 3.048 kmol = = 0.3258 9.354 kmol Nm

y O2 =

N O2 0.8325 kmol = = 0.0890 9.354 kmol Nm

y N2 =

N N2 0.04096 kmol = = 0.00438 9.354 kmol Nm

yS =

N S 0.04638 kmol = = 0.00496 9.354 kmol Nm

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15-23

Then, the combustion equation in this case may be written as 0.5758C + 0.3258H 2 + 0.0890O 2 + 0.00438N 2 + 0.00496S + a th (O 2 + 3.76 N 2 ) ⎯ ⎯→ x(0.95CO 2 + 0.05CO) + yH 2 O + zSO 2 + kN 2

According to the species balances, C balance : x = 0.5758 H 2 balance : y = 0.3258 S balance : z = 0.00496 O 2 balance : 0.0890 + a th = 0.95 x + 0.5 × 0.05 x + 0.5 y + z a th = 0.95 × 0.5758 + 0.5 × 0.05 × 0.5758 + 0.5 × 0.3258 + 0.00496 − 0.0890 = 0.6403 N 2 balance : k = 0.00438 + 3.76a th = 0.00438 + 3.76 × 0.6403 = 2.412

Substituting, 0.5758C + 0.3258H 2 + 0.0890O 2 + 0.00438 N 2 + 0.00496S + 0.7293(O 2 + 3.76 N 2 ) ⎯ ⎯→ 0.5470CO 2 + 0.0288CO + 0.3258H 2 O + 0.00496SO 2 + 2.412 N 2

The mass fractions of the products are m total = 0.5470 × 44 + 0.0288 × 28 + 0.3258 × 18 + 0.00496 × 64 + 2.412 × 28 = 98.6 kg mf CO2 =

m CO2 (0.5470 × 44) kg = = 0.2441 m total 98.6 kg

mf CO =

m CO (0.0288 × 28) kg = = 0.0082 m total 98.6 kg

mf H2O =

m H2O (0.3258 × 18) kg = = 0.0595 m total 98.6 kg

mf SO2 =

mSO2 (0.00496 × 64) kg = = 0.0032 m total 98.6 kg

mf N2 =

m N2 (2.412 × 28) kg = = 0.6849 m total 98.6 kg

The total mole number of the products is N m = 0.5470 + 0.0288 + 0.3258 + 0.00496 + 2.712 = 3.319 kmol

The apparent molecular weight of the product gas is Mm =

mm 98.6 kg = = 29.71 kg/kmol N m 3.319 kmol

The air-fuel mass ratio is then AF =

m air (0.6403 × 4.76 × 29) kg = m fuel (0.5758 × 12 + 0.3258 × 2 + 0.0890 × 32 + 0.00438 × 28 + 0.00496 × 32) kg

88.39 kg 10.69 kg = 8.27 kg air/kg fuel =

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15-24

15-32 Propane is burned with 30 percent excess air. The mole fractions of each of the products, the mass of water in the products per unit mass of the fuel, and the air-fuel ratio are to be determined. Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, O2, and N2 only. Properties The molar masses of C, H2, O2, and air are 12 kg/kmol, 2 kg/kmol, 32 kg/kmol, and 29 kg/kmol, respectively (Table A-1). Analysis The combustion equation in this case can be written as C 3 H 8 + 1.3a th [O 2 + 3.76N 2 ] ⎯ ⎯→ 3CO 2 + 4H 2 O + 0.3a th O 2 + (1.3 × 3.76)a th N 2

where ath is the stoichiometric coefficient for air. We have automatically accounted for the 30% excess air by using the factor 1.3ath instead of ath for air. The stoichiometric amount of oxygen (athO2) will be used to oxidize the fuel, and the remaining excess amount (0.3athO2) will appear in the products as free oxygen. The coefficient ath is determined from the O2 balance, 1.3a th = 3 + 2 + 0.3a th ⎯ ⎯→ a th = 5

O2 balance:

C 3 H 8 + 6.5[O 2 + 3.76 N 2 ] ⎯ ⎯→ 3CO 2 + 4H 2 O + 1.5O 2 + 24.44 N 2

Substituting,

The mole fractions of the products are N m = 3 + 4 + 1.5 + 24.44 = 32.94 kmol y CO2 =

N CO2 3 kmol = = 0.0911 32.94 kmol Nm

y H2O =

N H2O 4 kmol = = 0.1214 Nm 32.94 kmol

y O2 =

N O2 1.5 kmol = = 0.0455 Nm 32.94 kmol

y N2 =

N N2 24.44 kmol = = 0.7420 Nm 32.94 kmol

C3H8 Products Air 75% excess

The mass of water in the products per unit mass of fuel burned is m H2O (4 × 18) kg = = 1.636 kg H 2 O/kg C 3 H 8 m C3H8 (1× 44) kg

The air-fuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel, AF =

m air (6.5 × 4.76 kmol)(29 kg/kmol) = = 20.39 kg air/kg fuel m fuel (3 kmol)(12 kg/ kmol) + (4 kmol)(2 kg/kmol)

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15-25

15-33 The volumetric fractions of the constituents of a certain natural gas are given. The AF ratio is to be determined if this gas is burned with the stoichiometric amount of dry air. Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, and N2 only. Properties The molar masses of C, H2, N2, O2, and air are 12 kg/kmol, 2 kg/kmol, 28 kg/kmol, 32 kg/kmol, and 29 kg/kmol, respectively (Table A-1). Analysis Considering 1 kmol of fuel, the combustion equation can be written as (0.65CH 4 + 0.08H 2 + 0.18N 2 + 0.03O 2 + 0.06CO 2 ) + a th (O 2 + 3.76N 2 ) ⎯ ⎯→ xCO 2 + yH 2 O + zN 2

The unknown coefficients in the above equation are determined from mass balances, C : 0.65 + 0.06 = x

⎯ ⎯→ x = 0.71

H : 0.65 × 4 + 0.08 × 2 = 2 y

O 2 : 0.03 + 0.06 + a th = x + y / 2 N 2 : 0.18 + 3.76a th = z

Natural gas

⎯ ⎯→ y = 1.38

Combustion Products chamber

⎯ ⎯→ a th = 1.31

⎯ ⎯→ z = 5.106

Dry air

Thus, (0.65CH 4 + 0.08H 2 + 018 . N 2 + 0.03O 2 + 0.06CO 2 ) + 131 . (O 2 + 3.76N 2 ) . H 2 O + 5106 . N2 ⎯⎯→ 0.71CO 2 + 138 The air-fuel ratio for the this reaction is determined by taking the ratio of the mass of the air to the mass of the fuel, m air = (1.31× 4.76 kmol)(29 kg/kmol) = 180.8 kg

m fuel = (0.65 × 16 + 0.08 × 2 + 0.18 × 28 + 0.03 × 32 + 0.06 × 44)kg = 19.2 kg and AFth =

m air, th m fuel

=

180.8 kg = 9.42 kg air/kg fuel 19.2 kg

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15-26

15-34 The composition of a certain natural gas is given. The gas is burned with stoichiometric amount of moist air. The AF ratio is to be determined. Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, and N2 only. Properties The molar masses of C, H2, N2, O2, and air are 12 kg/kmol, 2 kg/kmol, 28 kg/kmol, 32 kg/kmol, and 29 kg/kmol, respectively (Table A-1). Analysis The fuel is burned completely with the stoichiometric amount of air, and thus the products will contain only H2O, CO2 and N2, but no free O2. The moisture in the air does not react with anything; it simply shows up as additional H2O in the products. Therefore, we can simply balance the combustion equation using dry air, and then add the moisture to both sides of the equation. Considering 1 kmol of fuel, the combustion equation can be written as (0.65CH 4 + 0.08H 2 + 0.18N 2 + 0.03O 2 + 0.06CO 2 ) + a th (O 2 + 3.76N 2 ) ⎯ ⎯→ xCO 2 + yH 2O + zN 2

The unknown coefficients in the above equation are determined from mass balances, C : 0.65 + 0.06 = x

⎯ ⎯→ x = 0.71

H : 0.65 × 4 + 0.08 × 2 = 2 y

O 2 : 0.03 + 0.06 + a th = x + y / 2 N 2 : 0.18 + 3.76a th = z

Natural gas

⎯ ⎯→ y = 1.38

Combustion Products chamber

⎯ ⎯→ a th = 1.31

⎯ ⎯→ z = 5.106

Moist air

Thus, (0.65CH 4 + 0.08H 2 + 018 . N 2 + 0.03O 2 + 0.06CO 2 ) + 1.31(O 2 + 3.76N 2 ) ⎯⎯→ 0.71CO 2 + 138 . H 2 O + 5106 . N2

Next we determine the amount of moisture that accompanies 4.76ath = (4.76)(1.31) = 6.24 kmol of dry air. The partial pressure of the moisture in the air is

Pv,in = φ air Psat@ 25°C = (0.85)(3.1698 kPa) = 2.694 kPa Assuming ideal gas behavior, the number of moles of the moisture in the air (Nv, in) is determined to be ⎛ Pv ,in N v ,in = ⎜⎜ ⎝ Ptotal

⎞ ⎛ 2.694 kPa ⎞ ⎟ N total = ⎜ ⎯→ N v,air = 0.17 kmol ⎜ 101.325 kPa ⎟⎟ 6.24 + N v ,in ⎯ ⎟ ⎠ ⎝ ⎠

(

)

The balanced combustion equation is obtained by substituting the coefficients determined earlier and adding 0.17 kmol of H2O to both sides of the equation, ( 0.65CH 4 + 0.08H 2 + 018 . N 2 + 0.03O 2 + 0.06CO 2 ) + 131 . ( O 2 + 3.76N 2 ) + 017 . H 2O ⎯⎯→ 0.71CO 2 + 155 . H 2 O + 5106 . N2

The air-fuel ratio for the this reaction is determined by taking the ratio of the mass of the air to the mass of the fuel, m air = (1.31 × 4.76 kmol )(29 kg/kmol ) + (0.17 kmol × 18 kg/kmol ) = 183.9 kg

m fuel = (0.65 × 16 + 0.08 × 2 + 0.18 × 28 + 0.03 × 32 + 0.06 × 44 )kg = 19.2 kg

and AFth =

m air, th m fuel

=

183.9 kg = 9.58 kg air/kg fuel 19.2 kg

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15-27

15-35 The composition of a gaseous fuel is given. It is burned with 130 percent theoretical air. The AF ratio and the fraction of water vapor that would condense if the product gases were cooled are to be determined. Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, O2, and N2 only. Properties The molar masses of C, H2, N2, and air are 12 kg/kmol, 2 kg/kmol, 28 kg/kmol, and 29 kg/kmol, respectively (Table A-1). Analysis (a) The fuel is burned completely with excess air, and thus the products will contain H2O, CO2, N2, and some free O2. Considering 1 kmol of fuel, the combustion equation can be written as (0.60CH 4 + 0.30H 2 + 0.10N 2 ) + 1.3a th (O 2 + 3.76N 2 ) ⎯ ⎯→ xCO 2 + yH 2O + 0.3a th O 2 + zN 2

The unknown coefficients in the above equation are determined from mass balances, C : 0.60 = x

⎯ ⎯→ x = 0.60

H : 0.60 × 4 + 0.30 × 2 = 2 y O 2 : 1.3a th = x + y / 2 + 0.3a th N 2 : 0.10 + 3.76 × 1.3a th = z

⎯ ⎯→ y = 1.50 ⎯ ⎯→ a th = 1.35 ⎯ ⎯→ z = 6.70

Gaseous fuel Air

Combustion Products chamber

30% excess

Thus, (0.60CH 4 + 0.30H 2 + 0.10N 2 ) + 1.755(O 2 + 3.76N 2 ) ⎯ ⎯→ 0.6CO 2 + 1.5H 2O + 0.405O 2 + 6.7N 2

The air-fuel ratio for the this reaction is determined by taking the ratio of the mass of the air to the mass of the fuel, m air = (1.755 × 4.76 kmol )(29 kg/kmol ) = 242.3 kg

m fuel = (0.6 × 16 + 0.3 × 2 + 0.1× 28)kg = 13.0 kg

and AF =

mair 242.3 kg = = 18.6 kg air/kg fuel 13.0 kg mfuel

(b) For each kmol of fuel burned, 0.6 + 1.5 + 0.405 + 6.7 = 9.205 kmol of products are formed, including 1.5 kmol of H2O. Assuming that the dew-point temperature of the products is above 20°C, some of the water vapor will condense as the products are cooled to 20°C. If Nw kmol of H2O condenses, there will be 1.5 - Nw kmol of water vapor left in the products. The mole number of the products in the gas phase will also decrease to 9.205 - Nw as a result. Treating the product gases (including the remaining water vapor) as ideal gases, Nw is determined by equating the mole fraction of the water vapor to its pressure fraction, Nv N prod,gas

=

Pv 1.5 − N w 2.3392 kPa ⎯ ⎯→ = ⎯ ⎯→ N w = 1.32 kmol Pprod 9.205 − N w 101.325 kPa

since Pv = Psat @ 20°C = 2.3392 kPa. Thus the fraction of water vapor that condenses is 1.32/1.5 = 0.88 or 88%.

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15-28

15-36 EES Problem 15-35 is reconsidered. The effects of varying the percentages of CH4, H2 and N2 making up the fuel and the product gas temperature are to be studied. Analysis The problem is solved using EES, and the solution is given below. Let's modify this problem to include the fuels butane, ethane, methane, and propane in pull down menu. Adiabatic Combustion of fuel CnHm entering at T_fuel with Stoichiometric Air at T_air: Reaction: aCxHy+bH2+cN2 + (a*y/4 + a*x+b/2) (Theo_air/100) (O2 + 3.76 N2) a*xCO2 + ((a*y/2)+b) H2O + (c+3.76 (a*y/4 + a*x+b/2) (Theo_air/100)) N2 + (a*y/4 + a*x+b/2) (Theo_air/100 - 1) O2 T_prod is the product gas temperature. Theo_air is the % theoretical air. " Procedure H20Cond(P_prod,T_prod,Moles_H2O,M_other:T_DewPoint,Moles_H2O_vap,Moles_H2O_liq,Re sult$) P_v = Moles_H2O/(M_other+Moles_H2O)*P_prod T_DewPoint = temperature(steam,P=P_v,x=0) IF T_DewPoint = 100%, the solution assumes complete combustion.' {MolCO = 0 MolCO2 = x}

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15-132

w=0 MolO2 = A_th*(Th_air - 1) GOTO 10 ELSE w = 2*x + y/2 - z - 2*A_th*Th_air IF w > x then Call ERROR('The moles of CO2 are negative, the percent theoretical air must be >= xxxF3 %',ErrTh) Else SolMeth$ = '< 100%, the solution assumes incomplete combustion with no O_2 in products.' MolO2 = 0 endif; endif 10: END {"Input data from the diagram window" T_air = 298 [K] Theo_air = 200 "%" Fuel$='CH4(g)'} T_fuel = 298 [K] Call Fuel(Fuel$,T_fuel:x,y,z,h_fuel,Name$) A_th =x + y/4 - z/2 Th_air = Theo_air/100 Call Moles(x,y,z,Th_air,A_th:w,MolO2,SolMeth$) HR=h_fuel+ (x+y/4-z/2) *(Theo_air/100) *enthalpy(O2,T=T_air)+3.76*(x+y/4-z/2) *(Theo_air/100) *enthalpy(N2,T=T_air) HP=HR "Adiabatic" HP=(xw)*enthalpy(CO2,T=T_prod)+w*enthalpy(CO,T=T_prod)+(y/2)*enthalpy(H2O,T=T_prod)+3.76*(x +y/4-z/2)* (Theo_air/100)*enthalpy(N2,T=T_prod)+MolO2*enthalpy(O2,T=T_prod) Moles_O2=MolO2 Moles_N2=3.76*(x+y/4-z/2)* (Theo_air/100) Moles_CO2=x-w Moles_CO=w Moles_H2O=y/2

2500 Tprod [K] 2077 2287 2396 2122 1827 1506 1153 840.1 648.4

Adiabatic Flame Temp.

2100

for C8 H18 (liquid) Tprod [K]

Theoair [%] 75 90 100 120 150 200 300 500 800

1700 1300 900 500 0

100

200

300

400

500

600

700

800

Theoair [%]

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15-133

15-128 EES A general program is to be written to determine the adiabatic flame temperature during the complete combustion of a hydrocarbon fuel CnHm at 25°C in a steady-flow combustion chamber when the percent of excess air and its temperature are specified. Analysis The problem is solved using EES, and the solution is given below. Adiabatic Combustion of fuel CnHm entering at T_fuel with Stoichiometric Air at T_air: Reaction: CxHyOz + (y/4 + x-z/2) (Theo_air/100) (O2 + 3.76 N2) xCO2 + (y/2) H2O + 3.76 (y/4 + x-z/2) (Theo_air/100) N2 + (y/4 + x-z/2) (Theo_air/100 - 1) O2" "For theoretical oxygen, the complete combustion equation for CH3OH is" "CH3OH + A_th O2=1 CO2+2 H2O " "1+ 2*A_th=1*2+2*1""theoretical O balance" "Adiabatic, Incomplete Combustion of fuel CnHm entering at T_fuel with Stoichiometric Air at T_air: Reaction: CxHyOz + (y/4 + x-z/2) (Theo_air/100) (O2 + 3.76 N2) (x-w)CO2 +wCO + (y/2) H2O + 3.76 (y/4 + x-z/2) (Theo_air/100) N2 + ((y/4 + xz/2) (Theo_air/100 - 1) +w/2)O2" "T_prod is the adiabatic combustion temperature, assuming no dissociation. Theo_air is the % theoretical air. " "The initial guess value of T_prod = 450K ." Procedure Fuel(Fuel$,T_fuel:x,y,z,h_fuel,Name$) "This procedure takes the fuel name and returns the moles of C and moles of H" If fuel$='C2H2(g)' then x=2;y=2; z=0 Name$='acetylene' h_fuel = 226730 else If fuel$='C3H8(l)' then x=3; y=8; z=0 Name$='propane(liq)' h_fuel = -103850-15060 else If fuel$='C8H18(l)' then x=8; y=18; z=0 Name$='octane(liq)' h_fuel = -249950 else if fuel$='CH4(g)' then x=1; y=4; z=0 Name$='methane' h_fuel = enthalpy(CH4,T=T_fuel) else if fuel$='CH3OH(g)' then x=1; y=4; z=1 Name$='methyl alcohol' h_fuel = -200670 endif; endif; endif; endif; endif end Procedure Moles(x,y,z,Th_air,A_th:w,MolO2,SolMeth$) ErrTh =(2*x + y/2 - z - x)/(2*A_th)*100

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IF Th_air >= 1 then SolMeth$ = '>= 100%, the solution assumes complete combustion.' {MolCO = 0 MolCO2 = x} w=0 MolO2 = A_th*(Th_air - 1) GOTO 10 ELSE w = 2*x + y/2 - z - 2*A_th*Th_air IF w > x then Call ERROR('The moles of CO2 are negative, the percent theoretical air must be >= xxxF3 %',ErrTh) Else SolMeth$ = '< 100%, the solution assumes incomplete combustion with no O_2 in products.' MolO2 = 0 endif; endif 10: END {"Input data from the diagram window" T_air = 298 [K] Theo_air = 200 [%] Fuel$='CH4(g)'} T_fuel = 298 [K] Call Fuel(Fuel$,T_fuel:x,y,z,h_fuel,Name$) A_th =x + y/4 - z/2 Th_air = Theo_air/100 Call Moles(x,y,z,Th_air,A_th:w,MolO2,SolMeth$) HR=h_fuel+ (x+y/4-z/2) *(Theo_air/100) *enthalpy(O2,T=T_air)+3.76*(x+y/4-z/2) *(Theo_air/100) *enthalpy(N2,T=T_air) HP=HR "Adiabatic" HP=(xw)*enthalpy(CO2,T=T_prod)+w*enthalpy(CO,T=T_prod)+(y/2)*enthalpy(H2O,T=T_prod)+3.76*(x +y/4-z/2)* (Theo_air/100)*enthalpy(N2,T=T_prod)+MolO2*enthalpy(O2,T=T_prod) Moles_O2=MolO2 Moles_N2=3.76*(x+y/4-z/2)* (Theo_air/100) Moles_CO2=x-w Moles_CO=w Moles_H2O=y/2 SOLUTION for the sample calculation A_th=5 fuel$='C3H8(l)' HR=-119067 [kJ/kg] h_fuel=-118910 Moles_CO2=3.000 Moles_H2O=4 Moles_O2=2.500 MolO2=2.5 SolMeth$='>= 100%, the solution assumes complete combustion.' Theo_air=150 [%] Th_air=1.500 T_fuel=298 [K] T_prod=1820 [K] x=3 y=8

HP=-119067 [kJ/kg] Moles_CO=0.000 Moles_N2=28.200 Name$='propane(liq)' T_air=298 [K] w=0 z=0

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15-129 EES The minimum percent of excess air that needs to be used for the fuels CH4(g), C2H2(g), CH3OH(g), C3H8(g), and C8H18(l) if the adiabatic flame temperature is not to exceed 1500 K is to be determined. Analysis The problem is solved using EES, and the solution is given below. Adiabatic Combustion of fuel CnHm entering at T_fuel with Stoichiometric Air at T_air: Reaction: CxHyOz + (y/4 + x-z/2) (Theo_air/100) (O2 + 3.76 N2) xCO2 + (y/2) H2O + 3.76 (y/4 + x-z/2) (Theo_air/100) N2 + (y/4 + x-z/2) (Theo_air/100 - 1) O2" {"For theoretical oxygen, the complete combustion equation for CH3OH is" "CH3OH + A_th O2=1 CO2+2 H2O " 1+ 2*A_th=1*2+2*1"theoretical O balance"} "T_prod is the adiabatic combustion temperature, assuming no dissociation. Theo_air is the % theoretical air. " "The initial guess value of T_prod = 450K ." Procedure Fuel(Fuel$,T_fuel:x,y,z,h_fuel,Name$) "This procedure takes the fuel name and returns the moles of C and moles of H" If fuel$='C2H2(g)' then x=2;y=2; z=0 Name$='acetylene' h_fuel = 226730 else If fuel$='C3H8(g)' then x=3; y=8; z=0 Name$='propane' h_fuel = enthalpy(C3H8,T=T_fuel) else If fuel$='C8H18(l)' then x=8; y=18; z=0 Name$='octane' h_fuel = -249950 else if fuel$='CH4(g)' then x=1; y=4; z=0 Name$='methane' h_fuel = enthalpy(CH4,T=T_fuel) else if fuel$='CH3OH(g)' then x=1; y=4; z=1 Name$='methyl alcohol' h_fuel = -200670 endif; endif; endif; endif; endif end {"Input data from the diagram window" T_air = 298 [K] Fuel$='CH4(g)'} T_fuel = 298 [K] Excess_air=Theo_air - 100 "[%]" Call Fuel(Fuel$,T_fuel:x,y,z,h_fuel,Name$) A_th = y/4 + x-z/2

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Th_air = Theo_air/100 HR=h_fuel+ (y/4 + x-z/2) *(Theo_air/100) *enthalpy(O2,T=T_air)+3.76*(y/4 + x-z/2) *(Theo_air/100) *enthalpy(N2,T=T_air) HP=HR "Adiabatic" HP=x*enthalpy(CO2,T=T_prod)+(y/2)*enthalpy(H2O,T=T_prod)+3.76*(y/4 + x-z/2)* (Theo_air/100)*enthalpy(N2,T=T_prod)+(y/4 + x-z/2) *(Theo_air/100 - 1)*enthalpy(O2,T=T_prod) Moles_O2=(y/4 + x-z/2) *(Theo_air/100 - 1) Moles_N2=3.76*(y/4 + x-z/2)* (Theo_air/100) Moles_CO2=x Moles_H2O=y/2 T[1]=T_prod; xa[1]=Theo_air SOLUTION for a sample calculation A_th=2.5 fuel$='C2H2(g)' HR=226596 [kJ/kg] Moles_CO2=2 Moles_N2=24.09 Name$='acetylene' Th_air=2.563 T_air=298 [K] T_prod=1500 [K] xa[1]=256.3 z=0

Excess_air=156.251 [%] HP=226596 [kJ/kg] h_fuel=226730 Moles_H2O=1 Moles_O2=3.906 Theo_air=256.3 [%] T[1]=1500 [K] T_fuel=298 [K] x=2 y=2

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15-130 EES The minimum percentages of excess air that need to be used for the fuels CH4(g), C2H2(g), CH3OH(g), C3H8(g), and C8H18(l) AFOR adiabatic flame temperatures of 1200 K, 1750 K, and 2000 K are to be determined. Analysis The problem is solved using EES, and the solution is given below. Adiabatic Combustion of fuel CnHm entering at T_fuel with Stoichiometric Air at T_air: Reaction: CxHyOz + (y/4 + x-z/2) (Theo_air/100) (O2 + 3.76 N2) xCO2 + (y/2) H2O + 3.76 (y/4 + x-z/2) (Theo_air/100) N2 + (y/4 + x-z/2) (Theo_air/100 - 1) O2" {"For theoretical oxygen, the complete combustion equation for CH3OH is" "CH3OH + A_th O2=1 CO2+2 H2O " 1+ 2*A_th=1*2+2*1"theoretical O balance"} "T_prod is the adiabatic combustion temperature, assuming no dissociation. Theo_air is the % theoretical air. " "The initial guess value of T_prod = 450K ." Procedure Fuel(Fuel$,T_fuel:x,y,z,h_fuel,Name$) "This procedure takes the fuel name and returns the moles of C and moles of H" If fuel$='C2H2(g)' then x=2;y=2; z=0 Name$='acetylene' h_fuel = 226730 else If fuel$='C3H8(g)' then x=3; y=8; z=0 Name$='propane' h_fuel = enthalpy(C3H8,T=T_fuel) else If fuel$='C8H18(l)' then x=8; y=18; z=0 Name$='octane' h_fuel = -249950 else if fuel$='CH4(g)' then x=1; y=4; z=0 Name$='methane' h_fuel = enthalpy(CH4,T=T_fuel) else if fuel$='CH3OH(g)' then x=1; y=4; z=1 Name$='methyl alcohol' h_fuel = -200670 endif; endif; endif; endif; endif end {"Input data from the diagram window" T_air = 298 [K] Fuel$='CH4(g)'} T_fuel = 298 [K] Excess_air=Theo_air - 100 "[%]" Call Fuel(Fuel$,T_fuel:x,y,z,h_fuel,Name$) A_th = y/4 + x-z/2

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Th_air = Theo_air/100 HR=h_fuel+ (y/4 + x-z/2) *(Theo_air/100) *enthalpy(O2,T=T_air)+3.76*(y/4 + x-z/2) *(Theo_air/100) *enthalpy(N2,T=T_air) HP=HR "Adiabatic" HP=x*enthalpy(CO2,T=T_prod)+(y/2)*enthalpy(H2O,T=T_prod)+3.76*(y/4 + x-z/2)* (Theo_air/100)*enthalpy(N2,T=T_prod)+(y/4 + x-z/2) *(Theo_air/100 - 1)*enthalpy(O2,T=T_prod) Moles_O2=(y/4 + x-z/2) *(Theo_air/100 - 1) Moles_N2=3.76*(y/4 + x-z/2)* (Theo_air/100) Moles_CO2=x Moles_H2O=y/2 T[1]=T_prod; xa[1]=Theo_air SOLUTION for a sample calculation A_th=5 fuel$='C3H8(g)' HR=-103995 [kJ/kg] Moles_CO2=3 Moles_N2=24.7 Name$='propane' Th_air=1.314 T_air=298 [K] T_prod=2000 [K] xa[1]=131.4 z=0

Excess_air=31.395 [%] HP=-103995 [kJ/kg] h_fuel=-103858 Moles_H2O=4 Moles_O2=1.570 Theo_air=131.4 [%] T[1]=2000 [K] T_fuel=298 [K] x=3 y=8

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15-131 EES The adiabatic flame temperature of CH4(g) is to be determined when both the fuel and the air enter the combustion chamber at 25°C for the cases of 0, 20, 40, 60, 80, 100, 200, 500, and 1000 percent excess air. Analysis The problem is solved using EES, and the solution is given below. Adiabatic Combustion of fuel CnHm entering at T_fuel with Stoichiometric Air at T_air: Reaction: CxHyOz + (y/4 + x-z/2) (Theo_air/100) (O2 + 3.76 N2) xCO2 + (y/2) H2O + 3.76 (y/4 + x-z/2) (Theo_air/100) N2 + (y/4 + x-z/2) (Theo_air/100 - 1) O2" "For theoretical oxygen, the complete combustion equation for CH3OH is" "CH3OH + A_th O2=1 CO2+2 H2O " "1+ 2*A_th=1*2+2*1""theoretical O balance" "Adiabatic, Incomplete Combustion of fuel CnHm entering at T_fuel with Stoichiometric Air at T_air: Reaction: CxHyOz + (y/4 + x-z/2) (Theo_air/100) (O2 + 3.76 N2) (x-w)CO2 +wCO + (y/2) H2O + 3.76 (y/4 + x-z/2) (Theo_air/100) N2 + ((y/4 + xz/2) (Theo_air/100 - 1) +w/2)O2" "T_prod is the adiabatic combustion temperature, assuming no dissociation. Theo_air is the % theoretical air. " "The initial guess value of T_prod = 450K ." Procedure Fuel(Fuel$,T_fuel:x,y,z,h_fuel,Name$) "This procedure takes the fuel name and returns the moles of C and moles of H" If fuel$='C2H2(g)' then x=2;y=2; z=0 Name$='acetylene' h_fuel = 226730 else If fuel$='C3H8(g)' then x=3; y=8; z=0 Name$='propane' h_fuel = enthalpy(C3H8,T=T_fuel) else If fuel$='C8H18(l)' then x=8; y=18; z=0 Name$='octane' h_fuel = -249950 else if fuel$='CH4(g)' then x=1; y=4; z=0 Name$='methane' h_fuel = enthalpy(CH4,T=T_fuel) else if fuel$='CH3OH(g)' then x=1; y=4; z=1 Name$='methyl alcohol' h_fuel = -200670 endif; endif; endif; endif; endif end Procedure Moles(x,y,z,Th_air,A_th:w,MolO2,SolMeth$) ErrTh =(2*x + y/2 - z - x)/(2*A_th)*100

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IF Th_air >= 1 then SolMeth$ = '>= 100%, the solution assumes complete combustion.' {MolCO = 0 MolCO2 = x} w=0 MolO2 = A_th*(Th_air - 1) GOTO 10 ELSE w = 2*x + y/2 - z - 2*A_th*Th_air IF w > x then Call ERROR('The moles of CO2 are negative, the percent theoretical air must be >= xxxF3 %',ErrTh) Else SolMeth$ = '< 100%, the solution assumes incomplete combustion with no O_2 in products.' MolO2 = 0 endif; endif 10: END {"Input data from the diagram window" T_air = 298 [K] Theo_air = 200 [%] Fuel$='CH4(g)'} T_fuel = 298 [K] Call Fuel(Fuel$,T_fuel:x,y,z,h_fuel,Name$) A_th =x + y/4 - z/2 Th_air = Theo_air/100 Call Moles(x,y,z,Th_air,A_th:w,MolO2,SolMeth$) HR=h_fuel+ (x+y/4-z/2) *(Theo_air/100) *enthalpy(O2,T=T_air)+3.76*(x+y/4-z/2) *(Theo_air/100) *enthalpy(N2,T=T_air) HP=HR "Adiabatic" HP=(xw)*enthalpy(CO2,T=T_prod)+w*enthalpy(CO,T=T_prod)+(y/2)*enthalpy(H2O,T=T_prod)+3.76*(x +y/4-z/2)* (Theo_air/100)*enthalpy(N2,T=T_prod)+MolO2*enthalpy(O2,T=T_prod) Moles_O2=MolO2 Moles_N2=3.76*(x+y/4-z/2)* (Theo_air/100) Moles_CO2=x-w Product temperature vs % excess air for CH4 Moles_CO=w 3000 Moles_H2O=y/2 Tprod [K] 2329 2071 1872 1715 1587 1480 1137 749.5 553

2500 Tprod [K]

Theoair [%] 100 120 140 160 180 200 300 600 1100

2000 1500 1000 500 0 100 200 300 400 500 600 700 800 900 1000 1100 Theoair [%]

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15-132 EES The fuel among CH4(g), C2H2(g), C2H6(g), C3H8(g), and C8H18(l) that gives the highest temperature when burned completely in an adiabatic constant-volume chamber with the theoretical amount of air is to be determined. Analysis The problem is solved using EES, and the solution is given below. Adiabatic Combustion of fuel CnHm with Stoichiometric Air at T_fuel =T_air=T_reac in a constant volume, closed system: Reaction: CxHyOz + (x+y/4-z/2) (Theo_air/100) (O2 + 3.76 N2) --> xCO2 + (y/2) H2O + 3.76 (x+y/4-z/2) (Theo_air/100) N2 + (x+y/4-z/2) (Theo_air/100 - 1) O2" "For theoretical oxygen, the complete combustion equation for CH3OH is" "CH3OH + A_th O2=1 CO2+2 H2O " "1+ 2*A_th=1*2+2*1""theoretical O balance" "Adiabatic, Incomplete Combustion of fuel CnHm with Stoichiometric Air at T_fuel =T_air=T_reac in a constant volume, closed system: Reaction: CxHyOz + (x+y/4-z/2) (Theo_air/100) (O2 + 3.76 N2) --> (x-w)CO2 +wCO + (y/2) H2O + 3.76 (x+y/4-z/2) (Theo_air/100) N2 + ((x+y/4-z/2) (Theo_air/100 - 1) +w/2)O2" "T_prod is the adiabatic combustion temperature, assuming no dissociation. Theo_air is the % theoretical air. " "The initial guess value of T_prod = 450K ." Procedure Fuel(Fuel$,T_fuel:x,y,z,h_fuel,Name$) "This procedure takes the fuel name and returns the moles of C and moles of H" If fuel$='C2H2(g)' then x=2;y=2; z=0 Name$='acetylene' h_fuel = 226730"Table A.26" else If fuel$='C3H8(g)' then x=3; y=8; z=0 Name$='propane' h_fuel = enthalpy(C3H8,T=T_fuel) else If fuel$='C8H18(l)' then x=8; y=18; z=0 Name$='octane' h_fuel = -249950"Table A.26" else if fuel$='CH4(g)' then x=1; y=4; z=0 Name$='methane' h_fuel = enthalpy(CH4,T=T_fuel) else if fuel$='CH3OH(g)' then x=1; y=4; z=1 Name$='methyl alcohol' h_fuel = -200670"Table A.26" endif; endif; endif; endif; endif end

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Procedure Moles(x,y,z,Th_air,A_th:w,MolO2,SolMeth$) ErrTh =(2*x + y/2 - z - x)/(2*A_th)*100 IF Th_air >= 1 then SolMeth$ = '>= 100%, the solution assumes complete combustion.' w=0 MolO2 = A_th*(Th_air - 1) GOTO 10 ELSE w = 2*x + y/2 - z - 2*A_th*Th_air IF w > x then Call ERROR('The moles of CO2 are negative, the percent theoretical air must be >= xxxF3 %',ErrTh) Else SolMeth$ = '< 100%, the solution assumes incomplete combustion with no O_2 in products.' MolO2 = 0 endif; endif 10: END {"Input data from the diagram window" Theo_air = 200 [%] Fuel$='CH4(g)'} T_reac = 298 [K] T_air = T_reac T_fuel = T_reac R_u = 8.314 [kJ/kmol-K] Call Fuel(Fuel$,T_fuel:x,y,z,h_fuel,Name$) A_th =x + y/4 - z/2 Th_air = Theo_air/100 Call Moles(x,y,z,Th_air,A_th:w,MolO2,SolMeth$) UR=(h_fuel-R_u*T_fuel)+ (x+y/4-z/2) *(Theo_air/100) *(enthalpy(O2,T=T_air)R_u*T_air)+3.76*(x+y/4-z/2) *(Theo_air/100) *(enthalpy(N2,T=T_air)-R_u*T_air) UP=(x-w)*(enthalpy(CO2,T=T_prod)-R_u*T_prod)+w*(enthalpy(CO,T=T_prod)R_u*T_prod)+(y/2)*(enthalpy(H2O,T=T_prod)-R_u*T_prod)+3.76*(x+y/4-z/2)* (Theo_air/100)*(enthalpy(N2,T=T_prod)-R_u*T_prod)+MolO2*(enthalpy(O2,T=T_prod)R_u*T_prod) UR =UP "Adiabatic, constant volume conservation of energy" Moles_O2=MolO2 Moles_N2=3.76*(x+y/4-z/2)* (Theo_air/100) Moles_CO2=x-w Moles_CO=w Moles_H2O=y/2

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SOLUTION for CH4 A_th=2 fuel$='CH4(g)' h_fuel=-74875 Moles_CO=0.000 Moles_CO2=1.000 Moles_H2O=2 Moles_N2=7.520 Moles_O2=0.000 MolO2=0 Name$='methane' R_u=8.314 [kJ/kmol-K] SolMeth$='>= 100%, the solution assumes complete combustion.' Theo_air=100 [%] Th_air=1.000 T_air=298 [K] T_fuel=298 [K] T_prod=2824 [K] T_reac=298 [K] UP=-100981 UR=-100981 w=0 x=1 y=4 z=0 SOLUTION for C2H2 A_th=2.5 fuel$='C2H2(g)' h_fuel=226730 Moles_CO=0.000 Moles_CO2=2.000 Moles_H2O=1 Moles_N2=9.400 Moles_O2=0.000 MolO2=0 Name$='acetylene' R_u=8.314 [kJ/kmol-K] SolMeth$='>= 100%, the solution assumes complete combustion.' Theo_air=100 [%] Th_air=1.000 T_air=298 [K] T_fuel=298 [K] T_prod=3535 [K] T_reac=298 [K] UP=194717 UR=194717 w=0 x=2 y=2 z=0 SOLUTION for CH3OH A_th=1.5 fuel$='CH3OH(g)' h_fuel=-200670 Moles_CO=0.000 Moles_CO2=1.000 Moles_H2O=2 Moles_N2=5.640 Moles_O2=0.000 MolO2=0 Name$='methyl alcohol' R_u=8.314 [kJ/kmol-K] SolMeth$='>= 100%, the solution assumes complete combustion.' Theo_air=100 [%] Th_air=1.000 T_air=298 [K] T_fuel=298 [K] T_prod=2817 [K] T_reac=298 [K] UP=-220869 UR=-220869 w=0 x=1 y=4 z=1 SOLUTION for C3H8 A_th=5 fuel$='C3H8(g)' h_fuel=-103858 Moles_CO=0.000 Moles_CO2=3.000 Moles_H2O=4 Moles_N2=18.800 Moles_O2=0.000 MolO2=0 Name$='propane' R_u=8.314 [kJ/kmol-K] SolMeth$='>= 100%, the solution assumes complete combustion.' Theo_air=100 [%] Th_air=1.000 T_air=298 [K] T_fuel=298 [K] T_prod=2909 [K] T_reac=298 [K] UP=-165406 UR=-165406 w=0 x=3 y=8 z=0 SOLUTION for C8H18 A_th=12.5 fuel$='C8H18(l)' h_fuel=-249950 Moles_CO=0.000 Moles_CO2=8.000 Moles_H2O=9 Moles_N2=47.000 Moles_O2=0.000 MolO2=0 Name$='octane' R_u=8.314 [kJ/kmol-K] SolMeth$='>= 100%, the solution assumes complete combustion.' Theo_air=100 [%] Th_air=1.000 T_air=298 [K] T_fuel=298 [K] T_prod=2911 [K] T_reac=298 [K] UP=-400104 UR=-400104 w=0 x=8 y=18 z=0

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Fundamentals of Engineering (FE) Exam Problems

15-133 A fuel is burned with 90 percent theoretical air. This is equivalent to (a) 10% excess air (d) 90% deficiency of air

(b) 90% excess air (e) stoichiometric amount of air

(c) 10% deficiency of air

Answer (c) 10% deficiency of air

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). air_th=0.9 "air_th=air_access+1" air_th=1-air_deficiency

15-134 Propane C3H8 is burned with 150 percent theoretical air. The air-fuel mass ratio for this combustion process is (a) 5.3

(b) 10.5

(c) 15.7

(d) 23.4

(e) 39.3

Answer (d) 23.4

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). n_C=3 n_H=8 m_fuel=n_H*1+n_C*12 a_th=n_C+n_H/4 coeff=1.5 "coeff=1 for theoretical combustion, 1.5 for 50% excess air" n_O2=coeff*a_th n_N2=3.76*n_O2 m_air=n_O2*32+n_N2*28 AF=m_air/m_fuel

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15-145

15-135 One kmol of methane (CH4) is burned with an unknown amount of air during a combustion process. If the combustion is complete and there are 2 kmol of free O2 in the products, the air-fuel mass ratio is (a) 34.3

(b) 17.2

(c) 19.0

(d) 14.9

(e) 12.1

Answer (a) 34.3

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). n_C=1 n_H=4 m_fuel=n_H*1+n_C*12 a_th=n_C+n_H/4 (coeff-1)*a_th=2 "O2 balance: Coeff=1 for theoretical combustion, 1.5 for 50% excess air" n_O2=coeff*a_th n_N2=3.76*n_O2 m_air=n_O2*32+n_N2*28 AF=m_air/m_fuel "Some Wrong Solutions with Common Mistakes:" W1_AF=1/AF "Taking the inverse of AF" W2_AF=n_O2+n_N2 "Finding air-fuel mole ratio" W3_AF=AF/coeff "Ignoring excess air"

15-136 A fuel is burned steadily in a combustion chamber. The combustion temperature will be the highest except when (a) the fuel is preheated. (b) the fuel is burned with a deficiency of air. (c) the air is dry. (d) the combustion chamber is well insulated. (e) the combustion is complete. Answer (b) the fuel is burned with a deficiency of air.

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15-146

15-137 An equimolar mixture of carbon dioxide and water vapor at 1 atm and 60°C enter a dehumidifying section where the entire water vapor is condensed and removed from the mixture, and the carbon dioxide leaves at 1 atm and 60°C. The entropy change of carbon dioxide in the dehumidifying section is (a) –2.8 kJ/kg⋅K

(b) –0.13 kJ/kg⋅K

(c) 0

(d) 0.13 kJ/kg⋅K

(e) 2.8 kJ/kg⋅K

Answer (b) –0.13 kJ/kg⋅K

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). Cp_CO2=0.846 R_CO2=0.1889 T1=60+273 "K" T2=T1 P1= 1 "atm" P2=1 "atm" y1_CO2=0.5; P1_CO2=y1_CO2*P1 y2_CO2=1; P2_CO2=y2_CO2*P2 Ds_CO2=Cp_CO2*ln(T2/T1)-R_CO2*ln(P2_CO2/P1_CO2) "Some Wrong Solutions with Common Mistakes:" W1_Ds=0 "Assuming no entropy change" W2_Ds=Cp_CO2*ln(T2/T1)-R_CO2*ln(P1_CO2/P2_CO2) "Using pressure fractions backwards"

15-138 Methane (CH4) is burned completely with 80% excess air during a steady-flow combustion process. If both the reactants and the products are maintained at 25°C and 1 atm and the water in the products exists in the liquid form, the heat transfer from the combustion chamber per unit mass of methane is (a) 890 MJ/kg

(b) 802 MJ/kg

(c) 75 MJ/kg

(d) 56 MJ/kg

(e) 50 MJ/kg

Answer (d) 56 MJ/kg

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T= 25 "C" P=1 "atm" EXCESS=0.8 "Heat transfer in this case is the HHV at room temperature," HHV_CH4 =55.53 "MJ/kg" LHV_CH4 =50.05 "MJ/kg" "Some Wrong Solutions with Common Mistakes:" W1_Q=LHV_CH4 "Assuming lower heating value" W2_Q=EXCESS*hHV_CH4 "Assuming Q to be proportional to excess air"

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15-147

15-139 The higher heating value of a hydrocarbon fuel CnHm with m = 8 is given to be 1560 MJ/kmol of fuel. Then its lower heating value is (a) 1384 MJ/kmol

(b) 1208 MJ/kmol

(c) 1402 MJ/kmol

(d) 1540 MJ/kmol

(e) 1550 MJ/kmol

Answer (a) 1384 MJ/kmol

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). HHV=1560 "MJ/kmol fuel" h_fg=2.4423 "MJ/kg, Enthalpy of vaporization of water at 25C" n_H=8 n_water=n_H/2 m_water=n_water*18 LHV=HHV-h_fg*m_water "Some Wrong Solutions with Common Mistakes:" W1_LHV=HHV - h_fg*n_water "Using mole numbers instead of mass" W2_LHV= HHV - h_fg*m_water*2 "Taking mole numbers of H2O to be m instead of m/2" W3_LHV= HHV - h_fg*n_water*2 "Taking mole numbers of H2O to be m instead of m/2, and using mole numbers"

15-140 Acetylene gas (C2H2) is burned completely during a steady-flow combustion process. The fuel and the air enter the combustion chamber at 25°C, and the products leave at 1500 K. If the enthalpy of the products relative to the standard reference state is –404 MJ/kmol of fuel, the heat transfer from the combustion chamber is (a) 177 MJ/kmol

(b) 227 MJ/kmol

(c) 404 MJ/kmol

(d) 631 MJ/kmol

(e) 751 MJ/kmol

Answer (d) 631 MJ/kmol

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). hf_fuel=226730/1000 "MJ/kmol fuel" H_prod=-404 "MJ/kmol fuel" H_react=hf_fuel Q_out=H_react-H_prod "Some Wrong Solutions with Common Mistakes:" W1_Qout= -H_prod "Taking Qout to be H_prod" W2_Qout= H_react+H_prod "Adding enthalpies instead of subtracting them"

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15-148

15-141 Benzene gas (C6H6) is burned with 90 percent theoretical air during a steady-flow combustion process. The mole fraction of the CO in the products is (a) 1.6%

(b) 4.4%

(c) 2.5%

(d) 10.0%

(e) 16.7%

Answer (b) 4.4%

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). n_C=6 n_H=6 a_th=n_C+n_H/4 coeff=0.90 "coeff=1 for theoretical combustion, 1.5 for 50% excess air" "Assuming all the H burns to H2O, the combustion equation is C6H6+coeff*a_th(O2+3.76N2)----- (n_CO2) CO2+(n_CO)CO+(n_H2O) H2O+(n_N2) N2" n_O2=coeff*a_th n_N2=3.76*n_O2 n_H2O=n_H/2 n_CO2+n_CO=n_C 2*n_CO2+n_CO+n_H2O=2*n_O2 "Oxygen balance" n_prod=n_CO2+n_CO+n_H2O+n_N2 "Total mole numbers of product gases" y_CO=n_CO/n_prod "mole fraction of CO in product gases" "Some Wrong Solutions with Common Mistakes:" W1_yCO=n_CO/n1_prod; n1_prod=n_CO2+n_CO+n_H2O "Not including N2 in n_prod" W2_yCO=(n_CO2+n_CO)/n_prod "Using both CO and CO2 in calculations"

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15-149

15-142 A fuel is burned during a steady-flow combustion process. Heat is lost to the surroundings at 300 K at a rate of 1120 kW. The entropy of the reactants entering per unit time is 17 kW/K and that of the products is 15 kW/K. The total rate of exergy destruction during this combustion process is (a) 520 kW

(b) 600 kW

(c) 1120 kW

(d) 340 kW

(e) 739 kW

Answer (a) 520 kW

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). To=300 "K" Q_out=1120 "kW" S_react=17 "kW'K" S_prod= 15 "kW/K" S_react-S_prod-Q_out/To+S_gen=0 "Entropy balance for steady state operation, SinSout+Sgen=0" X_dest=To*S_gen "Some Wrong Solutions with Common Mistakes:" W1_Xdest=S_gen "Taking Sgen as exergy destruction" W2_Xdest=To*S_gen1; S_react-S_prod-S_gen1=0 "Ignoring Q_out/To"

15-143 ··· 15-147 Design and Essay Problems

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15-150

15-143 A certain industrial process generates a liquid solution of ethanol and water as the waste product. The solution is to be burned using methane. A combustion process is to be developed to accomplish this incineration process with minimum amount of methane. Analysis The mass flow rate of the liquid ethanol-water solution is given to be 10 kg/s. Considering that the mass fraction of ethanol in the solution is 0.2, m& ethanol = (0.2 )(10 kg/s ) = 2 kg/s m& water = (0.8)(10 kg/s ) = 8 kg/s Noting that the molar masses Methanol = 46 and Mwater = 18 kg/kmol and that mole numbers N = m/M, the mole flow rates become m& 2 kg/s N& ethanol = ethanol = = 0.04348 kmol/s M ethanol 46 kg/kmol m& 8 kg/s N& water = water = = 0.44444 kmol/s M water 18 kg/kmol Note that N& water 0.44444 = = 10.222 kmol H 2 O/kmol C 2 H 5 OH & N ethanol 0.04348

That is, 10.222 moles of liquid water is present in the solution for each mole of ethanol. Assuming complete combustion, the combustion equation of C2H5OH (l) with stoichiometric amount of air is C 2 H 5 OH(l ) + a th (O 2 + 3.76N 2 ) ⎯ ⎯→ 2CO 2 + 3H 2 O + 3.76a th N 2

where ath is the stoichiometric coefficient and is determined from the O2 balance, 1 + 2a th = 4 + 3 ⎯ ⎯→ a th = 3

Thus, C 2 H 5 OH(l ) + 3(O 2 + 3.76N 2 ) ⎯ ⎯→ 2CO 2 + 3H 2 O + 11.28N 2

Noting that 10.222 kmol of liquid water accompanies each kmol of ethanol, the actual combustion equation can be written as C 2 H 5 OH(l ) + 3(O 2 + 3.76N 2 ) + 10.222H 2 O(l ) ⎯ ⎯→ 2CO 2 + 3H 2 O(g ) + 11.28N 2 + 10.222H 2 O(l )

The heat transfer for this combustion process is determined from the steady-flow energy balance equation with W = 0, Q=

∑ N (h P

o f

+h −ho

) − ∑ N (h R

P

o f

+h −ho

)

R

Assuming the air and the combustion products to be ideal gases, we have h = h(T). We assume all the reactants to enter the combustion chamber at the standard reference temperature of 25°C. Furthermore, we assume the products to leave the combustion chamber at 1400 K which is a little over the required temperature of 1100°C. From the tables,

Substance C2H5OH (l) CH4 O2 N2 H2O (g) H2O (l) CO2

h fo

h 298 K

h1400 K

kJ/kmol

kJ/kmol ----8682 8669 9904 --9364

kJ/kmol ----45,648 43,605 53,351 --65,271

-277,690 -74,850 0 0 -241,820 -285,830 -393,520

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15-151

Thus,

Q = (2 )(−393,520 + 65,271 − 9364 ) + (3)(−241,820 + 53,351 − 9904 ) + (11.28)(0 + 43,605 − 8669) − (1)(− 277,690 ) − 0 − 0 + (10.222 )(− 241,820 + 53,351 − 9904) − (10.222)(− 285,830) = 295,409 kJ/kmol of C2H5OH

The positive sign indicates that 295,409 kJ of heat must be supplied to the combustion chamber from another source (such as burning methane) to ensure that the combustion products will leave at the desired temperature of 1400 K. Then the rate of heat transfer required for a mole flow rate of 0.04348 kmol C2H5OH/s CO becomes Q& = N& Q = (0.04348 kmol/s)(295,409 kJ/kmol) = 12,844 kJ/s Assuming complete combustion, the combustion equation of CH4(g) with stoichiometric amount of air is CH 4 + a th (O 2 + 3.76N 2 ) ⎯ ⎯→ CO 2 + 2H 2 O + 3.76a th N 2

where ath is the stoichiometric coefficient and is determined from the O2 balance, Thus, a th = 1 + 1 ⎯ ⎯→ a th = 2 CH 4 + 2(O 2 + 3.76N 2 ) ⎯ ⎯→ CO 2 + 2H 2 O + 7.52N 2

The heat transfer for this combustion process is determined from the steady-flow energy balance E in − E out = ΔE system equation as shown above under the same assumptions and using the same mini table: Q = (1)(−393,520 + 65,271 − 9364) + (2 )(−241,820 + 53,351 − 9904) + (7.52)(0 + 43,605 − 8669) − (1)(− 74,850) − 0 − 0 = −396,790 kJ/kmol of CH 4 That is, 396,790 kJ of heat is supplied to the combustion chamber for each kmol of methane burned. To supply heat at the required rate of 12,844 kJ/s, we must burn methane at a rate of 12,844 kJ/s Q& N& CH 4 = = = 0.03237 kmolCH 4 /s Q 396,790 kJ/kmol or, m& CH 4 = M CH 4 N& CH 4 = (16 kg/kmol)(0.03237 kmolCH 4 /s ) = 0.5179 kg/s Therefore, we must supply methane to the combustion chamber at a minimum rate 0.5179 kg/s in order to maintain the temperature of the combustion chamber above 1400 K.

KJ

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16-1

Chapter 16 CHEMICAL AND PHASE EQUILIBRIUM The Kp and Equilibrium Composition of Ideal Gases 16-1C Because when a reacting system involves heat transfer, the increase-in-entropy principle relation requires a knowledge of heat transfer between the system and its surroundings, which is impractical. The equilibrium criteria can be expressed in terms of the properties alone when the Gibbs function is used. 16-2C No, the wooden table is NOT in chemical equilibrium with the air. With proper catalyst, it will reach with the oxygen in the air and burn. 16-3C They are PC C PDν D v

Kp =

PAν A PBvB

, K p = e − ΔG*(T ) / RuT

ν

and K p =

N CC N νDD ⎛ P ⎜ ⎜ N νAA N νBB ⎝ N total

⎞ ⎟ ⎟ ⎠

Δν

where Δν = ν C + ν D −ν A −ν B . The first relation is useful in partial pressure calculations, the second in determining the Kp from gibbs functions, and the last one in equilibrium composition calculations. 16-4C (a) This reaction is the reverse of the known CO reaction. The equilibrium constant is then 1/ KP (b) This reaction is the reverse of the known CO reaction at a different pressure. Since pressure has no effect on the equilibrium constant, 1/ KP (c) This reaction is the same as the known CO reaction multiplied by 2. The quilibirium constant is then K P2

(d) This is the same as reaction (c) occurring at a different pressure. Since pressure has no effect on the equilibrium constant, K P2

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16-2

16-5C (a) This reaction is the reverse of the known H2O reaction. The equilibrium constant is then 1/ KP (b) This reaction is the reverse of the known H2O reaction at a different pressure. Since pressure has no effect on the equilibrium constant, 1/ KP (c) This reaction is the same as the known H2O reaction multiplied by 3. The quilibirium constant is then K P3

(d) This is the same as reaction (c) occurring at a different pressure. Since pressure has no effect on the equilibrium constant, K P3

16-6C (a) No, because Kp depends on temperature only. (b) In general, the total mixture pressure affects the mixture composition. The equilibrium constant for the reaction N 2 + O 2 ⇔ 2NO can be expressed as Kp =

NO N νNO

ν

ν

N NN 2 N OO 2 2

2

⎛ P ⎜ ⎜N ⎝ total

⎞ ⎟ ⎟ ⎠

(ν NO −ν N 2 −ν O 2 )

The value of the exponent in this case is 2-1-1 = 0. Therefore, changing the total mixture pressure will have no effect on the number of moles of N2, O2 and NO. 16-7C (a) The equilibrium constant for the reaction CO + 12 O 2 ⇔ CO 2 can be expressed as ν

Kp =

CO 2 N CO 2

ν

ν

CO N CO N OO 2 2

⎛ P ⎜ ⎜N ⎝ total

⎞ ⎟ ⎟ ⎠

(ν CO 2 −ν CO −ν O 2 )

Judging from the values in Table A-28, the Kp value for this reaction decreases as temperature increases. That is, the indicated reaction will be less complete at higher temperatures. Therefore, the number of moles of CO2 will decrease and the number moles of CO and O2 will increase as the temperature increases. (b) The value of the exponent in this case is 1-1-0.5=-0.5, which is negative. Thus as the pressure increases, the term in the brackets will decrease. The value of Kp depends on temperature only, and therefore it will not change with pressure. Then to keep the equation balanced, the number of moles of the products (CO2) must increase, and the number of moles of the reactants (CO, O2) must decrease.

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16-3

16-8C (a) The equilibrium constant for the reaction N 2 ⇔ 2N can be expressed as ν

Kp =

N NN ⎛ P ⎜ ⎜ ν N NN 2 ⎝ N total 2

⎞ ⎟ ⎟ ⎠

(ν N −ν N 2 )

Judging from the values in Table A-28, the Kp value for this reaction increases as the temperature increases. That is, the indicated reaction will be more complete at higher temperatures. Therefore, the number of moles of N will increase and the number moles of N2 will decrease as the temperature increases. (b) The value of the exponent in this case is 2-1 = 1, which is positive. Thus as the pressure increases, the term in the brackets also increases. The value of Kp depends on temperature only, and therefore it will not change with pressure. Then to keep the equation balanced, the number of moles of the products (N) must decrease, and the number of moles of the reactants (N2) must increase. 16-9C The equilibrium constant for the reaction CO + 12 O 2 ⇔ CO 2 can be expressed as ν

Kp =

CO 2 N CO 2

ν

ν

CO N CO N OO 2 2

⎛ P ⎜ ⎜N ⎝ total

⎞ ⎟ ⎟ ⎠

(ν CO 2 −ν CO −ν O 2 )

Adding more N2 (an inert gas) at constant temperature and pressure will increase Ntotal but will have no direct effect on other terms. Then to keep the equation balanced, the number of moles of the products (CO2) must increase, and the number of moles of the reactants (CO, O2) must decrease. 16-10C The values of the equilibrium constants for each dissociation reaction at 3000 K are, from Table A-28, N 2 ⇔ 2N ⇔ ln K p = −22.359 H 2 ⇔ 2H ⇔ ln K p = −3.685

(greater than - 22.359)

Thus H2 is more likely to dissociate than N2.

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16-4

16-11 The mole fractions of the constituents of an ideal gas mixture is given. The Gibbs function of the CO in this mixture at the given mixture pressure and temperature is to be determined. Analysis From Tables A-21 and A-26, at 1 atm pressure,

[

g * (800 K, 1 atm) = g of + Δ h (T ) − Ts o (T )

]

= −137,150 + (23,844 − 800 × 227.162) − (8669 − 298 × 197.543) = −244,837 kJ/kmol

The partial pressure of CO is PCO = y CO P = (0.30)(10 atm) = 3 atm

10% CO2 60% H2O 30% CO 10 atm 800 K

The Gibbs function of CO at 800 K and 3 atm is g (800 K, 3 atm) = g * (800 K, 1 atm) + Ru T ln PCO = −244,837 kJ/kmol + (8.314 kJ/kmol)(800 K)ln(3 atm) = −237,530 kJ/kmol

16-12 The partial pressures of the constituents of an ideal gas mixture is given. The Gibbs function of the nitrogen in this mixture at the given mixture pressure and temperature is to be determined. Analysis The partial pressure of nitrogen is PN2 = 130 kPa = (130 / 101.325) = 1.283 atm

The Gibbs function of nitrogen at 298 K and 3 atm is g (800 K, 3 atm) = g * (298 K, 1 atm) + Ru T ln PN2

N2 ,CO2, NO PN2 = 130 kPa 298 K

= 0 + (8.314 kJ/kmol)(298 K)ln(1.283 atm) = 617.4 kJ/kmol

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16-5

16-13E The equilibrium constant of the reaction H 2 O ⇔ H 2 + 12 O 2 is to be determined using Gibbs function. Analysis (a) The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data using

K p = e −ΔG*( T )/ Ru T or ln K p = − ΔG *(T ) / Ru T H2O ↔ H2 + ½O2

where ∗ ∗ ∗ ΔG * (T ) = ν H2 g H2 (T ) +ν O2 g O2 (T ) −ν H2O g H2O (T )

1440 R

At 1440 R, ∗ ∗ ∗ ΔG * (T ) = ν H2 g H2 (T ) + ν O2 g O2 (T ) −ν H2O g H2O (T )

= ν H2 (h − Ts ) H2 + ν O2 (h − Ts ) O2 −ν H2O (h − Ts ) H2O = ν H2 [(h f + h1440 − h537 ) − Ts ] H2 + ν O2 [(h f + h1440 − h537 ) − Ts ] O2 −ν H2O [(h f + h1440 − h537 ) − Ts ] H2O = 1× (0 + 9956.9 − 3640.3 − 1440 × 38.079) + 0.5 × (0 + 10,532.0 − 3725.1 − 1440 × 56.326) − 1× (−104,040 + 11,933.4 − 4258 − 1440 × 53.428) = 87,632 Btu/lbmol

Substituting, ln K p = −(87,632 Btu/lbmol)/[(1.986 Btu/lbmol.R)(1440 R)] = −30.64

or K p = 4.93 × 10 −14 (Table A - 28 : ln K p = −34.97 by interpolation ) At 3960 R, ∗ ∗ ∗ ΔG * (T ) = ν H2 g H2 (T ) +ν O2 g O2 (T ) −ν H2O g H2O (T )

= ν H2 (h − Ts ) H2 + ν O2 (h − Ts ) O 2 −ν H2O (h − Ts ) H2O = ν H2 [(h f + h3960 − h537 ) − Ts ] H2 + ν O2 [(h f + h3960 − h537 ) − Ts ] O2 −ν H2O [(h f + h3960 − h537 ) − Ts ] H2O = 1× (0 + 29,370.5 − 3640.3 − 3960 × 45.765) + 0.5 × (0 + 32,441 − 3725.1 − 3960 × 65.032) − 1× (−104,040 + 39,989 − 4258 − 3960 × 64.402) = 53,436 Btu/lbmol

Substituting, ln K p = −(53,436 Btu/lbmol)/[(1.986 Btu/lbmol.R)(3960 R)] = −6.79

or

K p = 1.125 × 10 −3 (Table A - 28 : ln K p = −6.768)

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16-6

16-14 The reaction 2H 2 O ⇔ 2H 2 + O 2 is considered. The mole fractions of the hydrogen and oxygen produced when this reaction occurs at 4000 K and 10 kPa are to be determined. Assumptions 1 The equilibrium composition consists of H2O, H2, and O2. 2 The constituents of the mixture are ideal gases. Analysis The stoichiometric and actual reactions in this case are

Stoichiometric:

2H 2 O ⇔ 2H 2 + O 2 (thus ν H2O = 2, ν H2 = 2, and ν O2 = 1)

Actual:

2H 2 O ⎯ ⎯→ xH 2 O + yH 2 + zO 2 123 14243 react.

products

4 = 2x + 2 y ⎯ ⎯→ y = 2 − x

2H2O

O balance:

2 = x + 2z ⎯ ⎯→ z = 1 − 0.5 x

Total number of moles:

N total = x + y + z = 3 − 0.5 x

4000 K 10 kPa

H balance:

The equilibrium constant relation can be expressed as ν

ν

Kp =

N H2H2 N O2O2 ⎛ P ⎜ ⎜N ν H2O N H2O ⎝ total

⎞ ⎟ ⎟ ⎠

(ν H2 +ν O2 −ν H2O )

From Table A-28, ln K p = −0.542 at 4000 K . Since the stoichiometric reaction being considered is double this reaction, K p = exp( −2 × 0.542) = 0.3382

Substituting, 0.3382 =

(2 − x) 2 (1 − 0.5 x) ⎛ 10 / 101.325 ⎞ ⎜ ⎟ x2 ⎝ 3 − 0.5 x ⎠

2 +1− 2

Solving for x, x = 0.4446

Then, y = 2 − x = 1.555 z = 1 − 0.5x = 0.7777

Therefore, the equilibrium composition of the mixture at 4000 K and 10 kPa is 0.4446 H 2 O + 1.555 H 2 + 0.7777 O 2

The mole fractions of hydrogen and oxygen produced are y H2 =

N H2 1.555 1.555 = = = 0.560 N total 3 − 0.5 × 0.4446 2.778

y O2 =

N O2 0.7777 = = 0.280 N total 2.778

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16-7

16-15 The reaction 2H 2 O ⇔ 2H 2 + O 2 is considered. The mole fractions of hydrogen gas produced is to be determined at 100 kPa and compared to that at 10 kPa. Assumptions 1 The equilibrium composition consists of H2O, H2, and O2. 2 The constituents of the mixture are ideal gases. Analysis The stoichiometric and actual reactions in this case are

Stoichiometric:

2H 2 O ⇔ 2H 2 + O 2 (thus ν H2O = 2, ν H2 = 2, and ν O2 = 1)

Actual:

2H 2 O ⎯ ⎯→ xH 2 O + yH 2 + zO 2 123 14243 react.

products

4 = 2x + 2 y ⎯ ⎯→ y = 2 − x

2H2O

O balance:

2 = x + 2z ⎯ ⎯→ z = 1 − 0.5 x

Total number of moles:

N total = x + y + z = 3 − 0.5 x

4000 K 100 kPa

H balance:

The equilibrium constant relation can be expressed as ν

ν

Kp =

N H2H2 N O2O2 ⎛ P ⎜ ⎜N ν H2O N H2O ⎝ total

⎞ ⎟ ⎟ ⎠

(ν H2 +ν O2 −ν H2O )

From Table A-28, ln K p = −0.542 at 4000 K . Since the stoichiometric reaction being considered is double this reaction, K p = exp( −2 × 0.542) = 0.3382

Substituting, 0.3382 =

(2 − x) 2 (1 − 0.5 x ) ⎛ 100 / 101.325 ⎞ ⎜ ⎟ x2 ⎝ 3 − 0.5 x ⎠

2 +1− 2

Solving for x, x = 0.8870

Then, y = 2 − x = 1.113 z = 1 − 0.5x = 0.5565

Therefore, the equilibrium composition of the mixture at 4000 K and 100 kPa is 0.8870 H 2 O + 1.113 H 2 + 0.5565 O 2

That is, there are 1.113 kmol of hydrogen gas. The mole number of hydrogen at 10 kPa reaction pressure was obtained in the previous problem to be 1.555 kmol. Therefore, the amount of hydrogen gas produced has decreased.

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16-8

16-16 The reaction 2H 2 O ⇔ 2H 2 + O 2 is considered. The mole number of hydrogen gas produced is to be determined if inert nitrogen is mixed with water vapor is to be determined and compared to the case with no inert nitrogen. Assumptions 1 The equilibrium composition consists of H2O, H2, O2, and N2. 2 The constituents of the mixture are ideal gases. Analysis The stoichiometric and actual reactions in this case are

Stoichiometric:

2H 2 O ⇔ 2H 2 + O 2 (thus ν H2O = 2, ν H2 = 2, and ν O2 = 1)

Actual:

2H 2 O + 0.5N 2 ⎯ ⎯→ xH 2 O + yH 2 + zO 2 + 0.5N 2 123 14243 123 react.

products

inert

H balance:

4 = 2x + 2 y ⎯ ⎯→ y = 2 − x

2H2O, 0.5N2

O balance:

2 = x + 2z ⎯ ⎯→ z = 1 − 0.5 x

Total number of moles:

N total = x + y + z + 0.5 = 3.5 − 0.5 x

4000 K 10 kPa

The equilibrium constant relation can be expressed as ν

ν

Kp =

N H2H2 N O2O2 ⎛ P ⎜ ⎜N ν H2O N H2O ⎝ total

⎞ ⎟ ⎟ ⎠

(ν H2 +ν O2 −ν H2O )

From Table A-28, ln K p = −0.542 at 4000 K . Since the stoichiometric reaction being considered is double this reaction, K p = exp(−2 × 0.542) = 0.3382

Substituting, 0.3382 =

(2 − x) 2 (1 − 0.5 x) ⎛ 10 / 101.325 ⎞ ⎜ ⎟ x2 ⎝ 3.5 − 0.5 x ⎠

2 +1− 2

Solving for x, x = 0.4187

Then, y = 2 − x = 1.581 z = 1 − 0.5x = 0.7907

Therefore, the equilibrium composition of the mixture at 4000 K and 10 kPa is 0.4187 H 2 O + 1.581 H 2 + 0.7907 O 2

That is, there are 1.581 kmol of hydrogen gas. The mole number of hydrogen without inert nitrogen case was obtained in Prob. 16-14 to be 1.555 kmol. Therefore, the amount of hydrogen gas produced has increased.

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16-9

16-17E The temperature at which 15 percent of diatomic oxygen dissociates into monatomic oxygen at two pressures is to be determined. Assumptions 1 The equilibrium composition consists of O2 and O. 2 The constituents of the mixture are ideal gases. Analysis (a) The stoichiometric and actual reactions can be written as

Stoichiometric:

O 2 ⇔ 2O (thus ν O2 = 1 and ν O = 2)

Actual:

O 2 ⇔ 0.85O 2 + 0{ .3O 1 424 3 react.

prod.

O2 ↔ 2O 15 % 3 psia

The equilibrium constant Kp can be determined from ν

Kp =

N OO ⎛ P ⎜ ⎜ ν N O2O2 ⎝ N total

ν O −ν O2

⎞ ⎟ ⎟ ⎠

=

0.3 2 ⎛ 3 / 14.696 ⎞ ⎜ ⎟ 0.85 ⎝ 0.85 + 0.3 ⎠

2 −1

= 0.01880

and ln K p = −3.974

From Table A-28, the temperature corresponding to this lnKp value is T = 3060 K = 5508 R

(b) At 100 psia, ν O −ν O2

Nν O ⎛ P ⎞ ⎟ K p = νO ⎜⎜ N O2O2 ⎝ N total ⎟⎠

0.32 ⎛ 100 / 14.696 ⎞ = ⎜ ⎟ 0.85 ⎝ 0.85 + 0.3 ⎠

2 −1

= 0.6265

ln K p = −0.4676

From Table A-28, the temperature corresponding to this lnKp value is T = 3701 K = 6662 R

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16-10

16-18 The dissociation reaction CO2 ⇔ CO + O is considered. The composition of the products at given pressure and temperature is to be determined. Assumptions 1 The equilibrium composition consists of CO2, CO, and O. 2 The constituents of the mixture are ideal gases. Analysis For the stoichiometric reaction CO 2 ⇔ CO + 12 O 2 , from Table A-28, at 2500 K ln K p = −3.331

For the oxygen dissociation reaction 0.5O 2 ⇔ O , from Table A-28, at 2500 K, ln K p = −8.509 / 2 = −4.255

For the desired stoichiometric reaction CO 2 ⇔ CO + O (thus ν CO2 = 1, ν CO = 1 and ν O = 1) , ln K p = −3.331 − 4.255 = −7.586

and K p = exp( −7.586) = 0.0005075

CO 2 ⎯ ⎯→ xCO 2 + yCO + zO 123 14243

Actual:

products

react.

C balance:

1= x+ y ⎯ ⎯→ y = 1 − x

O balance:

2 = 2x + y + z ⎯ ⎯→ z = 1 − x

Total number of moles:

N total = x + y + z = 2 − x

CO2 2500 K 1 atm

The equilibrium constant relation can be expressed as ν

ν

N CO N O ⎛ P K p = COν O ⎜⎜ CO2 N CO2 ⎝ N total

ν CO +ν O −ν CO2

⎞ ⎟ ⎟ ⎠

Substituting, 0.0005075 =

(1 − x)(1 − x) ⎛ 1 ⎞ ⎜ ⎟ x ⎝ 2− x⎠

1+1−1

Solving for x, x = 0.9775

Then, y = 1 − x = 0.0225 z = 1 − x = 0.0225

Therefore, the equilibrium composition of the mixture at 2500 K and 1 atm is 0.9775 CO 2 + 0.0225 CO + 0.0225 O

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16-11

16-19 The dissociation reaction CO2 ⇔ CO + O is considered. The composition of the products at given pressure and temperature is to be determined when nitrogen is added to carbon dioxide. Assumptions 1 The equilibrium composition consists of CO2, CO, O, and N2. 2 The constituents of the mixture are ideal gases. Analysis For the stoichiometric reaction CO 2 ⇔ CO + 12 O 2 , from Table A-28, at 2500 K ln K p = −3.331

For the oxygen dissociation reaction 0.5O 2 ⇔ O , from Table A-28, at 2500 K, ln K p = −8.509 / 2 = −4.255

For the desired stoichiometric reaction CO 2 ⇔ CO + O (thus ν CO2 = 1, ν CO = 1 and ν O = 1) , ln K p = −3.331 − 4.255 = −7.586

and K p = exp( −7.586) = 0.0005075

CO 2 + 3N 2 ⎯ ⎯→ xCO 2 + yCO + zO + 3N 123 14243 {2

Actual:

products

react.

C balance:

1= x+ y ⎯ ⎯→ y = 1 − x

O balance:

2 = 2x + y + z ⎯ ⎯→ z = 1 − x

Total number of moles:

N total = x + y + z + 3 = 5 − x

inert

CO2, 3N2 2500 K 1 atm

The equilibrium constant relation can be expressed as ν

ν

N CO N O ⎛ P K p = COν O ⎜⎜ CO2 N CO2 ⎝ N total

ν CO +ν O −ν CO2

⎞ ⎟ ⎟ ⎠

Substituting, 0.0005075 =

(1 − x)(1 − x) ⎛ 1 ⎞ ⎜ ⎟ x ⎝ 5− x ⎠

1+1−1

Solving for x, x = 0.9557

Then, y = 1 − x = 0.0443 z = 1 − x = 0.0443

Therefore, the equilibrium composition of the mixture at 2500 K and 1 atm is 0.9557 CO 2 + 0.0443 CO + 0.0443 O + 3N 2

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16-12

16-20 It is to be shown that as long as the extent of the reaction, α, for the disassociation reaction X2 ⇔ 2X KP 4+ KP

is smaller than one, α is given by α =

Assumptions The reaction occurs at the reference temperature. Analysis The stoichiometric and actual reactions can be written as

Stoichiometric:

X 2 ⇔ 2X (thus ν X2 = 1 and ν X = 2)

Actual:

X 2 ⇔ (1 − α )X 2 + 2{ αX 1424 3 react.

prod.

The equilibrium constant Kp is given by ν

Kp =

N XX ⎛ P ⎜ ⎜ ν N X2X2 ⎝ N total

ν X −ν X2

⎞ ⎟ ⎟ ⎠

=

(2α ) 2 ⎛ 1 ⎞ ⎜ ⎟ (1 − α ) ⎝ α + 1 ⎠

2 −1

=

4α 2 (1 − α )(1 + α )

Solving this expression for α gives

α=

KP 4+ KP

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16-13

16-21 A gaseous mixture consisting of methane and nitrogen is heated. The equilibrium composition (by mole fraction) of the resulting mixture is to be determined. Assumptions 1 The equilibrium composition consists of CH4, C, H2, and N2. 2 The constituents of the mixture are ideal gases. Analysis The stoichiometric and actual reactions in this case are

Stoichiometric:

CH 4 ⇔ C + 2H 2 (thus ν CH4 = 1, ν C = 1, and ν H2 = 2)

Actual:

CH 4 + N 2 ⎯ ⎯→ xCH 4 + yC + zH 2 + N 123 1424 3 {2 react.

products

C balance:

1= x+ y ⎯ ⎯→ y = 1 − x

H balance:

4 = 4x + 2z ⎯ ⎯→ z = 2 − 2 x

Total number of moles:

N total = x + y + z + 1 = 4 − 2 x

inert

CH4, N2 1000 K 1 atm

The equilibrium constant relation can be expressed as ν

Kp =

CH4 N CH4

ν

ν

N CC N H2H2

⎛ P ⎜ ⎜N ⎝ total

ν CH4 −ν C −ν H2

⎞ ⎟ ⎟ ⎠

From the problem statement at 1000 K, ln K p = 2.328 . Then, K p = exp(2.328) = 10.257

For the reverse reaction that we consider, K p = 1 / 10.257 = 0.09749

Substituting, 0.09749 =

⎛ 1 ⎞ ⎟ ⎜ 2 4 − 2x (1 − x)(2 − 2 x) ⎝ ⎠ x

1−1− 2

Solving for x, x = 0.02325 Then, y = 1 − x = 0.9768 z = 2 − 2x = 1.9535 Therefore, the equilibrium composition of the mixture at 1000 K and 1 atm is 0.02325 CH 4 + 0.9768 C + 1.9535 H 2 + 1 N 2

The mole fractions are y CH4 =

N CH4 0.02325 0.02325 = = = 0.0059 3.954 N total 4 − 2 × 0.02325

yC =

NC 0.9768 = = 0.2470 3.954 N total

y H2 =

N H2 1.9535 = = 0.4941 3.954 N total

y N2 =

N N2 1 = = 0.2529 N total 3.954

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16-14

16-22 The reaction N2 + O2 ⇔ 2NO is considered. The equilibrium mole fraction of NO 1000 K and 1 atm is to be determined. Assumptions 1 The equilibrium composition consists of N2, O2, and NO. 2 The constituents of the mixture are ideal gases. Analysis The stoichiometric and actual reactions in this case are

Stoichiometric:

N 2 + O 2 ⇔ 2 NO (thus ν N2 = 1, ν O2 = 1, and ν NO = 2)

Actual:

N2 + O2 ⎯ ⎯→ xN 2 + yO 2 + { zNO 14243 products react.

N balance:

2 = 2x + z ⎯ ⎯→ z = 2 − 2 x

O balance:

2 = 2y + z ⎯ ⎯→ y = x

Total number of moles:

N total = x + y + z = 2

N2, O2 1000 K 1 atm

The equilibrium constant relation can be expressed as ν

Kp =

NO N NO

ν

ν

N N2N2 N O2O2

⎛ P ⎜ ⎜N ⎝ total

⎞ ⎟ ⎟ ⎠

(ν NO −ν N2 −ν O2 )

From Table A-28, at 1000 K, ln K p = −9.388 . Since the stoichiometric reaction being considered is double this reaction, K p = exp(−2 × 9.388) = 7.009 × 10 −9

Substituting, 7.009 × 10 −9 =

(2 − 2 x) 2 ⎛ 1 ⎞ ⎜ ⎟ x2 ⎝2⎠

2 −1−1

Solving for x, x = 0.999958 Then, y = x = 0.999958 z = 2 − 2x = 8.4×10-5 Therefore, the equilibrium composition of the mixture at 1000 K and 1 atm is 0.999958 N 2 + 0.999958 O 2 + 8.4 × 10 −5 NO

The mole fraction of NO is then y NO =

N NO 8.4 × 10 −5 = = 4.2 × 10 −5 (42 parts per million) N total 2

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16-15

16-23 Oxygen is heated from a specified state to another state. The amount of heat required is to be determined without and with dissociation cases. Assumptions 1 The equilibrium composition consists of O2 and O. 2 The constituents of the mixture are ideal gases. Analysis (a) Obtaining oxygen properties from table A-19, an energy balance gives E −E 1in424out 3

=

Net energy transfer by heat, work, and mass

ΔE system 1 424 3

Change in internal, kinetic, potential, etc. energies

q in = u 2 − u1 = 57,192 − 6203 = 50,989 kJ/kmol

(b) The stoichiometric and actual reactions in this case are Stoichiometric:

O 2 ⇔ 2O (thus ν O2 = 1 and ν O = 2)

Actual:

O2 ⎯ ⎯→ { xO 2 + { yO react.

O2 2200 K 1 atm

products

O balance:

2 = 2x + y ⎯ ⎯→ y = 2 − 2 x

Total number of moles:

N total = x + y = 2 − x

The equilibrium constant relation can be expressed as ν

N O ⎛ P K p = νO ⎜⎜ N O2O2 ⎝ N total

ν O −ν O2

⎞ ⎟ ⎟ ⎠

From Table A-28, at 2200 K, ln K p = −11.827 . Then, K p = exp(−11.827) = 7.305 × 10 −6 Substituting, 7.305 × 10 − 6 =

(2 − 2 x) 2 x

⎛ 1 ⎞ ⎜ ⎟ ⎝ 2− x⎠

2 −1

Solving for x, x = 0.99865 Then, y = 2 − 2x = 0.0027 Therefore, the equilibrium composition of the mixture at 2200 K and 1 atm is 0.99865 O 2 + 0.0027 O

Hence, the oxygen ions are negligible and the result is same as that in part (a), q in = 50,989 kJ/kmol

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16-16

16-24 Air is heated from a specified state to another state. The amount of heat required is to be determined without and with dissociation cases. Assumptions 1 The equilibrium composition consists of O2 and O, and N2. 2 The constituents of the mixture are ideal gases. Analysis (a) Obtaining air properties from table A-17, an energy balance gives E −E 1in424out 3

=

Net energy transfer by heat, work, and mass

ΔE system 1 424 3

Change in internal, kinetic, potential, etc. energies

q in = u 2 − u1 = 1872.4 − 212.64 = 1660 kJ/kg

(b) The stoichiometric and actual reactions in this case are Stoichiometric:

O 2 ⇔ 2O (thus ν O2 = 1 and ν O = 2)

Actual:

O 2 + 3.76 N 2 ⎯ ⎯→ { xO 2 + { yO + 3.76 N 2 1 424 3 react.

products

O2, 3.76N2 2200 K 1 atm

inert

O balance:

2 = 2x + y ⎯ ⎯→ y = 2 − 2 x

Total number of moles:

N total = x + y + 3.76 = 5.76 − x

The equilibrium constant relation can be expressed as ν

N O ⎛ P K p = νO ⎜⎜ N O2O2 ⎝ N total

ν O −ν O2

⎞ ⎟ ⎟ ⎠

From Table A-28, at 2200 K, ln K p = −11.827 . Then, K p = exp(−11.827) = 7.305 × 10 −6 Substituting, 7.305 × 10 − 6 =

(2 − 2 x) 2 x

1 ⎛ ⎞ ⎜ ⎟ − x 5 . 76 ⎝ ⎠

2 −1

Solving for x, x = 0.99706 Then, y = 2 − 2x = 0.00588 Therefore, the equilibrium composition of the mixture at 2200 K and 1 atm is 0.99706 O 2 + 0.00588 O + 3.76 N 2

Hence, the atomic oxygen is negligible and the result is same as that in part (a), q in = 1660 kJ/kg

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16-17

16-25 The equilibrium constant of the reaction H2 + 1/2O2 ↔ H2O is listed in Table A-28 at different temperatures. The data are to be verified at two temperatures using Gibbs function data. Analysis (a) The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data using K p = e −ΔG*( T )/ Ru T or ln K p = − ΔG * (T ) / Ru T

H2 + ½O2 ↔ H2O

where ΔG * (T ) = ν H 2O g H∗ 2 O (T ) −ν H 2 g H∗ 2 (T ) −ν O 2 g O∗ 2 (T )

25ºC

At 25°C,

ΔG *(T ) = 1( −228,590) − 1(0) − 0.5(0) = −228,590 kJ / kmol Substituting, ln K p = −(−228,590 kJ/kmol)/[(8.314 kJ/kmol ⋅ K)(298 K)] = 92.26 or K p = 1.12 × 10 40 (Table A - 28: ln K p = 92.21)

(b) At 2000 K, ΔG * (T ) = ν H 2 O g H∗ 2 O (T ) −ν H 2 g H∗ 2 (T ) −ν O 2 g O∗ 2 (T ) = ν H 2 O ( h − T s ) H 2 O − ν H 2 ( h − Ts ) H 2 − ν O 2 ( h − Ts ) O 2 = ν H 2 O [(h f + h2000 − h298 ) − Ts ] H 2O −ν H 2 [(h f + h2000 − h298 ) − Ts ] H 2 −ν O 2 [(h f + h2000 − h298 ) − Ts ] O 2 = 1× (−241,820 + 82,593 − 9904 − 2000 × 264.571) − 1× (0 + 61,400 − 8468 − 2000 × 188.297) − 0.5 × (0 + 67,881 − 8682) − 2000 × 268.655) = −135,556 kJ/kmol

Substituting, ln K p = −(−135,556 kJ/kmol)/[(8.314 kJ/kmol ⋅ K)(2000 K)] = 8.152 or K p = 3471 (Table A - 28 : ln K p = 8.145)

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16-18

16-26E The equilibrium constant of the reaction H2 + 1/2O2 ↔ H2O is listed in Table A-28 at different temperatures. The data are to be verified at two temperatures using Gibbs function data. Analysis (a) The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data using K p = e −ΔG*( T )/ Ru T or ln K p = − ΔG * (T ) / Ru T

H2 + ½O2 ↔ H2O

where ΔG * (T ) = ν H 2O g H∗ 2 O (T ) −ν H 2 g H∗ 2 (T ) −ν O 2 g O∗ 2 (T )

537 R

At 537 R, ΔG *(T ) = 1( −98,350) − 1(0) − 0.5(0) = −98,350 Btu / lbmol

Substituting, ln K p = − ( −98,350 Btu / lbmol) / [(1.986 Btu / lbmol ⋅ R)(537 R)] = 92.22

or K p = 1.12 × 10 40 (Table A - 28: ln K p = 92.21)

(b) At 3240 R, ΔG * (T ) = ν H 2 O g H∗ 2 O (T ) −ν H 2 g H∗ 2 (T ) −ν O 2 g O∗ 2 (T ) = ν H 2 O ( h − T s ) H 2 O − ν H 2 ( h − Ts ) H 2 − ν O 2 ( h − Ts ) O 2 = ν H 2 O [(h f + h3240 − h537 ) − Ts ] H 2O −ν H 2 [(h f + h3240 − h298 ) − Ts ] H 2 −ν O 2 [(h f + h3240 − h298 ) − Ts ] O 2 = 1× (−104,040 + 31,204.5 − 4258 − 3240 × 61.948) − 1× (0 + 23,484.7 − 3640.3 − 3240 × 44.125) − 0.5 × (0 + 25,972 − 3725.1 − 3240 × 63.224) = −63,385 Btu/lbmol

Substituting, ln K p = −(−63,385 Btu/lbmol)/[(1.986 Btu/lbmol.R)(3240 R)] = 9.85 or K p = 1.90 × 104 (Table A - 28: ln K p = 9.83)

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16-19

16-27 The equilibrium constant of the reaction CO + 1/2O2 ↔ CO2 at 298 K and 2000 K are to be determined, and compared with the values listed in Table A-28. Analysis (a) The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data using K p = e −ΔG*( T )/ Ru T or ln K p = − ΔG * (T ) / Ru T

where ΔG * (T )

∗ ∗ ∗ = ν CO2 g CO2 (T ) −ν CO g CO (T ) −ν O2 g O2 (T )

CO + 1 O 2 ⇔ CO 2 2

298 K

At 298 K, ΔG * (T ) = 1(−394,360) − 1(−137,150) − 0.5(0) = −257,210 kJ/kmol

where the Gibbs functions are obtained from Table A-26. Substituting, ln K p = −

(−257,210 kJ/kmol) = 103.81 (8.314 kJ/kmol ⋅ K)(298 K) ln K p = 103.76

From Table A-28: (b) At 2000 K,

∗ ∗ ∗ ΔG * (T ) = ν CO2 g CO2 (T ) −ν CO g CO (T ) −ν O2 g O2 (T )

= ν CO2 (h − Ts ) CO2 −ν CO (h − Ts ) CO −ν O2 (h − Ts ) O2

= 1[(−302,128) − (2000)(309.00)] − 1[(−53,826) − (2000)(258.48)] − 0.5[(59,193) − (2000)(268.53)] = −110,409 kJ/kmol

The enthalpies at 2000 K and entropies at 2000 K and 101.3 kPa (1 atm) are obtained from EES. Substituting, ln K p = −

(−110,409 kJ/kmol) = 6.64 (8.314 kJ/kmol ⋅ K)(2000 K)

From Table A-28: ln K p = 6.635

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16-20

16-28 EES The effect of varying the percent excess air during the steady-flow combustion of hydrogen is to be studied. Analysis The combustion equation of hydrogen with stoichiometric amount of air is H 2 + 0.5[O 2 + 3.76N 2 ] ⎯ ⎯→ H 2 O + 0.5(3.76) N 2

For the incomplete combustion with 100% excess air, the combustion equation is H 2 + (1 + Ex)(0.5)[O 2 + 3.76N 2 ] ⎯ ⎯→ 0.97 H 2 O + a H 2 + b O 2 + c N 2

The coefficients are to be determined from the mass balances Hydrogen balance:

2 = 0.97 × 2 + a × 2 ⎯ ⎯→ a = 0.03

Oxygen balance:

(1 + Ex) × 0.5 × 2 = 0.97 + b × 2

Nitrogen balance: (1 + Ex) × 0.5 × 3.76 × 2 = c × 2 Solving the above equations, we find the coefficients (Ex = 1, a = 0.03 b = 0.515, c = 3.76) and write the balanced reaction equation as H 2 + [O 2 + 3.76N 2 ] ⎯ ⎯→ 0.97 H 2 O + 0.03 H 2 + 0.515 O 2 + 3.76 N 2

Total moles of products at equilibrium are N tot = 0.97 + 0.03 + 0.515 + 3.76 = 5.275

The assumed equilibrium reaction is H 2 O ←⎯→ H 2 + 0.5O 2

The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data using K p = e −ΔG*( T )/ Ru T or ln K p = − ΔG *(T ) / Ru T where ∗ ∗ ∗ ΔG * (T ) = ν H2 g H2 (Tprod ) + ν O2 g O2 (Tprod ) −ν H2O g H2O (Tprod )

and the Gibbs functions are defined as ∗ g H2 (Tprod ) = (h − Tprod s ) H2 ∗ g O2 (Tprod ) = (h − Tprod s ) O2 ∗ g H2O (Tprod ) = (h − Tprod s ) H2O

The equilibrium constant is also given by ⎛ P K p = ⎜⎜ ⎝ N tot

and

⎞ ⎟ ⎟ ⎠

1+ 0.5 −1

ab 0.5

⎛ 1 ⎞ =⎜ ⎟ 1 0.97 ⎝ 5.275 ⎠

0.5

(0.03)(0.515) 0.5 = 0.009664 0.97

ln K p = ln(0.009664) = −4.647

The corresponding temperature is obtained solving the above equations using EES to be Tprod = 2600 K

This is the temperature at which 97 percent of H2 will burn into H2O. The copy of EES solution is given next.

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16-21

"Input Data from parametric table:" {PercentEx = 10} Ex = PercentEx/100 "EX = % Excess air/100" P_prod =101.3"[kPa]" R_u=8.314 "[kJ/kmol-K]" "The combustion equation of H2 with stoichiometric amount of air is H2 + 0.5(O2 + 3.76N2)=H2O +0.5(3.76)N2" "For the incomplete combustion with 100% excess air, the combustion equation is H2 + (1+EX)(0.5)(O2 + 3.76N2)=0.97 H2O +aH2 + bO2+cN2" "Specie balance equations give the values of a, b, and c." "H, hydrogen" 2 = 0.97*2 + a*2 "O, oxygen" (1+Ex)*0.5*2=0.97 + b*2 "N, nitrogen" (1+Ex)*0.5*3.76 *2 = c*2 N_tot =0.97+a +b +c "Total kilomoles of products at equilibrium" "The assumed equilibrium reaction is H2O=H2+0.5O2" "The following equations provide the specific Gibbs function (g=h-Ts) for each H2mponent in the product gases as a function of its temperature, T_prod, at 1 atm pressure, 101.3 kPa" g_H2O=Enthalpy(H2O,T=T_prod )-T_prod *Entropy(H2O,T=T_prod ,P=101.3) g_H2=Enthalpy(H2,T=T_prod )-T_prod *Entropy(H2,T=T_prod ,P=101.3) g_O2=Enthalpy(O2,T=T_prod )-T_prod *Entropy(O2,T=T_prod ,P=101.3) "The standard-state Gibbs function is" DELTAG =1*g_H2+0.5*g_O2-1*g_H2O "The equilibrium constant is given by Eq. 15-14." K_P = exp(-DELTAG /(R_u*T_prod )) P=P_prod /101.3"atm" "The equilibrium constant is also given by Eq. 15-15." "K_ P = (P/N_tot)^(1+0.5-1)*(a^1*b^0.5)/(0.97^1)" sqrt(P/N_tot )*a *sqrt(b )=K_P *0.97 lnK_p = ln(k_P) 2625

-5.414 -5.165 -5.019 -4.918 -4.844 -4.786 -4.739 -4.7 -4.667 -4.639

PercentEx [%] 10 20 30 40 50 60 70 80 90 100

Tprod [K] 2440 2490 2520 2542 2557 2570 2580 2589 2596 2602

2585

2545 Tprod

ln Kp

2505

2465

2425 10

20

30

40

50 60 PercentEx

70

80

90

100

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16-22

16-29 The equilibrium constant of the reaction CH4 + 2O2 ↔ CO2 + 2H2O at 25°C is to be determined. Analysis The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data using

K p = e −ΔG*( T )/ Ru T or ln K p = − ΔG *(T ) / Ru T CH4 + 2O2 ↔ CO2 + 2H2O

where ∗ ∗ ΔG * (T ) = ν CO 2 g CO (T ) + ν H 2 O g H∗ 2O (T ) −ν CH 4 g CH (T ) −ν O 2 g O∗ 2 (T ) 2 4

25°C

At 25°C, ΔG *(T ) = 1( −394,360) + 2( −228,590) − 1( −50,790) − 2(0) = −800,750 kJ / kmol Substituting, ln K p = −(−800,750 kJ/kmol)/[(8.314 kJ/kmol ⋅ K)(298 K)] = 323.04

or K p = 1.96 × 10 140

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16-23

16-30 The equilibrium constant of the reaction CO2 ↔ CO + 1/2O2 is listed in Table A-28 at different temperatures. It is to be verified using Gibbs function data. Analysis (a) The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data using K p = e − ΔG*(T ) / RuT or ln K p = − ΔG * (T ) / Ru T

where

∗ ∗ ΔG * (T ) = ν CO g CO (T ) + ν O 2 g O∗ 2 (T ) −ν CO 2 g CO (T ) 2

CO2 ↔ CO + ½O2 298 K

At 298 K, ΔG * (T ) = 1(−137,150) + 0.5(0) − 1(−394,360) = 257,210 kJ/kmol

Substituting, ln K p = −(257,210 kJ/kmol)/[(8.314 kJ/kmol ⋅ K)(298 K)] = -103.81 or

K p = 8.24 × 10 -46 (Table A - 28 : ln K p = −103.76)

(b) At 1800 K, ∗ ∗ ΔG * (T ) = ν CO g CO (T ) + ν O 2 g O∗ 2 (T ) −ν CO 2 g CO (T ) 2

= ν CO ( h − Ts ) CO + ν O 2 ( h − Ts ) O 2 −ν CO 2 ( h − Ts ) CO 2 = ν CO [(h f + h1800 − h298 ) − Ts ] CO + ν O 2 [(h f + h1800 − h298 ) − Ts ] O 2 −ν CO 2 [(h f + h1800 − h298 ) − Ts ] CO 2 = 1× ( −110,530 + 58,191 − 8669 − 1800 × 254.797) + 0.5 × (0 + 60,371 − 8682 − 1800 × 264.701) − 1× ( −393,520 + 88,806 − 9364 − 1800 × 302.884) = 127,240.2 kJ/kmol

Substituting, or

ln K p = −(127,240.2 kJ/kmol)/[(8.314 kJ/kmol ⋅ K)(1800 K)] = −8.502

K p = 2.03 × 10 -4 (Table A - 28 : ln K p = −8.497)

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16-24

16-31 [Also solved by EES on enclosed CD] Carbon monoxide is burned with 100 percent excess air. The temperature at which 97 percent of CO burn to CO2 is to be determined. Assumptions 1 The equilibrium composition consists of CO2, CO, O2, and N2. 2 The constituents of the mixture are ideal gases. Analysis Assuming N2 to remain as an inert gas, the stoichiometric and actual reactions can be written as

Stoichiometric:

CO + 12 O 2 ⇔ CO 2 (thus ν CO 2 = 1, ν CO = 1, and ν O 2 = 12 )

Actual:

CO +1(O 2 + 3.76 N 2 )

⎯ ⎯→

0.97 CO 2 + 0.03 CO + 0.515O 2 + 3.76 N 2 1424 3 14442444 3 12 4 4 3 product

reactants

inert

The equilibrium constant Kp can be determined from ν

Kp =

N COCO2 2

νCO νO 2 N CO NO 2

⎛ P ⎞ ⎜⎜ ⎟⎟ ⎝ N total ⎠

( νCO 2 − νCO − νO 2 )

1−1.5

CO + ½O2 ↔ CO2 97 % 1 atm

0.97 1 ⎛ ⎞ ⎟ 0.5 ⎜ 0.03 × 0.515 ⎝ 0.97 + 0.03 + 0.515 + 3.76 ⎠ = 103.48 =

From Table A-28, the temperature corresponding to this Kp value is T = 2276 K

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16-25

16-32 EES Problem 16-31 is reconsidered. The effect of varying the percent excess air during the steadyflow process from 0 to 200 percent on the temperature at which 97 percent of CO burn into CO2 is to be studied. Analysis The problem is solved using EES, and the solution is given below. "To solve this problem, we need to give EES a guess value for T_prop other than the default value of 1. Set the guess value of T_prod to 1000 K by selecting Variable Infromation in the Options menu. Then press F2 or click the Calculator icon." "Input Data from the diagram window:" {PercentEx = 100} Ex = PercentEx/100 "EX = % Excess air/100" P_prod =101.3 [kPa] R_u=8.314 [kJ/kmol-K] "The combustion equation of CO with stoichiometric amount of air is CO + 0.5(O2 + 3.76N2)=CO2 +0.5(3.76)N2" "For the incomplete combustion with 100% excess air, the combustion equation is CO + (!+EX)(0.5)(O2 + 3.76N2)=0.97 CO2 +aCO + bO2+cN2" "Specie balance equations give the values of a, b, and c." "C, Carbon" 1 = 0.97 + a "O, oxygen" 1 +(1+Ex)*0.5*2=0.97*2 + a *1 + b*2 "N, nitrogen" (1+Ex)*0.5*3.76 *2 = c*2 N_tot =0.97+a +b +c "Total kilomoles of products at equilibrium" "The assumed equilibrium reaction is CO2=CO+0.5O2" "The following equations provide the specific Gibbs function (g=h-Ts) for each component in the product gases as a function of its temperature, T_prod, at 1 atm pressure, 101.3 kPa" g_CO2=Enthalpy(CO2,T=T_prod )-T_prod *Entropy(CO2,T=T_prod ,P=101.3) g_CO=Enthalpy(CO,T=T_prod )-T_prod *Entropy(CO,T=T_prod ,P=101.3) g_O2=Enthalpy(O2,T=T_prod )-T_prod *Entropy(O2,T=T_prod ,P=101.3) "The standard-state Gibbs function is" DELTAG =1*g_CO+0.5*g_O2-1*g_CO2 "The equilibrium constant is given by Eq. 15-14." K_P = exp(-DELTAG /(R_u*T_prod )) P=P_prod /101.3"atm" "The equilibrium constant is also given by Eq. 15-15." "K_ P = (P/N_tot)^(1+0.5-1)*(a^1*b^0.5)/(0.97^1)" sqrt(P/N_tot )*a *sqrt(b )=K_P *0.97 lnK_p = ln(k_P) "Compare the value of lnK_p calculated by EES with the value of lnK_p from table A-28 in the text."

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16-26

PercentEx [%] 0 20 40 60 80 100 120 140 160 180 200

Tprod [K] 2066 2194 2230 2250 2263 2273 2280 2285 2290 2294 2297

2350 2300

Tprod [K]

2250 2200 2150 2100 2050 0

40

80

120

160

200

PercentEx [%]

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16-27

16-33E Carbon monoxide is burned with 100 percent excess air. The temperature at which 97 percent of CO burn to CO2 is to be determined. Assumptions 1 The equilibrium composition consists of CO2, CO, O2, and N2. 2 The constituents of the mixture are ideal gases. Analysis Assuming N2 to remain as an inert gas, the stoichiometric and actual reactions can be written as

Stoichiometric:

CO + 12 O 2 ⇔ CO 2 (thus ν CO 2 = 1, ν CO = 1, and ν O 2 = 12 )

Actual:

CO +1(O 2 + 3.76 N 2 )

⎯ ⎯→

0.97 CO 2 + 0.03 CO + 0.515O 2 + 3.76 N 2 1424 3 14442444 3 12 4 4 3 product

reactants

inert

The equilibrium constant Kp can be determined from ν

Kp =

N COCO2 2

ν

νCO N CO N OO2 2

⎛ P ⎜⎜ ⎝ N total

0.97

=

0.03 × 0.5150.5 = 103.48

⎞ ⎟⎟ ⎠

( νCO 2 − νCO − νO 2 )

1 ⎛ ⎞ ⎜ ⎟ ⎝ 0.97 + 0.03 + 0.515 + 3.76 ⎠

CO + ½O2 ↔ CO2 97 % 1 atm

1−1.5

From Table A-28, the temperature corresponding to this Kp value is T = 2276 K = 4097 R

16-34 Hydrogen is burned with 150 percent theoretical air. The temperature at which 98 percent of H2 will burn to H2O is to be determined. Assumptions 1 The equilibrium composition consists of H2O, H2, O2, and N2. 2 The constituents of the mixture are ideal gases. Analysis Assuming N2 to remain as an inert gas, the stoichiometric and actual reactions can be written as

Stoichiometric:

H 2 + 21 O 2 ⇔ H 2 O (thus ν H 2 O = 1, ν H 2 = 1, and ν O 2 = 21 )

Actual:

H 2 + 0.75(O 2 + 3.76 N 2 )

⎯ ⎯→

0.98 H 2 O + 0.02 H 2 + 0.26O 2 + 2.82 N 2 1424 3 144 42444 3 12 4 4 3 product

reactants

inert

The equilibrium constant Kp can be determined from ν

Kp = =

N HHO2O 2

ν H2

ν O2

2

2

NH NO

⎛ P ⎜ ⎜N ⎝ total

0.98

0.02 × 0.26 0.5 = 194.11

⎞ ⎟ ⎟ ⎠

(ν H 2 O −ν H 2 −ν O 2 )

1 ⎛ ⎞ ⎜ ⎟ + + + 0 . 98 0 . 02 0 . 26 2 . 82 ⎝ ⎠

H2 Combustion chamber

1−1.5

H2O, H2 O2, N2

Air

From Table A-28, the temperature corresponding to this Kp value is T = 2472 K.

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16-28

16-35 Air is heated to a high temperature. The equilibrium composition at that temperature is to be determined. Assumptions 1 The equilibrium composition consists of N2, O2, and NO. 2 The constituents of the mixture are ideal gases. Analysis The stoichiometric and actual reactions in this case are N 2 + 12 O 2 ⇔ NO (thus ν NO = 1, ν N 2 = 12 , and ν O 2 = 12 )

Stoichiometric:

1 2

Actual:

3.76 N 2 + O 2

⎯ ⎯→

x NO + y N + z O 123 1422432 prod.

reactants

N balance:

7.52 = x + 2y or y = 3.76 - 0.5x

O balance:

2 = x + 2z or z = 1 - 0.5x

Total number of moles:

Ntotal = x + y + z = x + 4.76- x = 4.76

AIR 2000 K 2 atm

The equilibrium constant relation can be expressed as

Kp =

NO N νNO

ν

ν

N NN2 2 N OO2 2

⎛ P ⎜⎜ ⎝ N total

⎞ ⎟⎟ ⎠

(ν NO −ν N 2 −ν O 2 )

From Table A-28, ln Kp = -3.931 at 2000 K. Thus Kp = 0.01962. Substituting, 0.01962 =

x (3.76 − 0.5 x) 0.5 (1 − 0.5 x) 0.5

⎛ 2 ⎞ ⎜ ⎟ ⎝ 4.76 ⎠

1−1

Solving for x, x = 0.0376 Then, y = 3.76-0.5x = 3.7412 z = 1-0.5x = 0.9812 Therefore, the equilibrium composition of the mixture at 2000 K and 2 atm is 0.0376NO + 3.7412N 2 + 0.9812O 2

The equilibrium constant for the reactions O 2 ⇔ 2O (ln Kp = -14.622) and N 2 ⇔ 2 N (ln Kp = -41.645) are much smaller than that of the specified reaction (ln Kp = -3.931). Therefore, it is realistic to assume that no monatomic oxygen or nitrogen will be present in the equilibrium mixture. Also the equilibrium composition is in this case is independent of pressure since Δν = 1 − 0.5 − 0.5 = 0 .

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16-29

16-36 Hydrogen is heated to a high temperature at a constant pressure. The percentage of H2 that will dissociate into H is to be determined. Assumptions 1 The equilibrium composition consists of H2 and H. 2 The constituents of the mixture are ideal gases. Analysis The stoichiometric and actual reactions can be written as

Stoichiometric:

H 2 ⇔ 2H (thus ν H 2 = 1 and ν H = 2)

Actual:

H2 ⎯ ⎯→ { xH 2 + { yH react.

prod.

H balance:

2 = 2x + y or y = 2 - 2x

Total number of moles:

Ntotal = x + y = x + 2 - 2x = 2 - x

H2 3200 K 8 atm

The equilibrium constant relation can be expressed as ν H −ν H 2

Nν H ⎛ P ⎞ ⎟ K p = νH ⎜⎜ ⎟ N HH2 2 ⎝ N total ⎠

From Table A-28, ln Kp = -2.534 at 3200 K. Thus Kp = 0.07934. Substituting, 0.07934 =

Solving for x,

(2 − 2 x) 2 ⎛ 8 ⎞ ⎜ ⎟ x ⎝2− x⎠

2 −1

x = 0.95

Thus the percentage of H2 which dissociates to H at 3200 K and 8 atm is 1 - 0.95 = 0.05 or 5.0%

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16-30

16-37E A mixture of CO, O2, and N2 is heated to a high temperature at a constant pressure. The equilibrium composition is to be determined. Assumptions 1 The equilibrium composition consists of CO2, CO, O2, and N2. 2 The constituents of the mixture are ideal gases. Analysis The stoichiometric and actual reactions in this case are

Stoichiometric:

CO + 12 O 2 ⇔ CO 2 (thus ν CO 2 = 1, ν CO = 1, and ν O 2 = 12 )

Actual:

2 CO + 2 O 2 + 6 N 2

⎯ ⎯→

x CO + y CO + z O 2 + 6 N 2 1232 14244 3 products

C balance:

2= x+ y

⎯ ⎯→

O balance:

6 = 2x + y + 2z

Total number of moles:

N total = x + y + z + 6 = 10 − 0.5x

reactants

:

inert

2 CO 2 O2 6 N2 4320 R 3 atm

y = 2− x ⎯ ⎯→

z = 2 − 0.5x

The equilibrium constant relation can be expressed as ν

Kp =

CO 2 N CO 2

ν

ν

CO N CO N OO 2 2

⎛ P ⎜ ⎜N ⎝ total

⎞ ⎟ ⎟ ⎠

(ν CO 2 −ν CO −ν O 2 )

From Table A-28, ln K p = 3.860 at T = 4320 R = 2400 K. Thus K p = 47.465. Substituting, 47.465 =

x (2 − x)(2 − 0.5 x) 0.5

3 ⎛ ⎞ ⎜ ⎟ ⎝ 10 − 0.5 x ⎠

1−1.5

Solving for x, x = 1.930 Then, y = 2 - x = 0.070 z = 2 - 0.5x = 1.035 Therefore, the equilibrium composition of the mixture at 2400 K and 3 atm is 1.930CO 2 + 0.070CO + 1.035O 2 + 6N 2

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16-31

16-38 A mixture of N2, O2, and Ar is heated to a high temperature at a constant pressure. The equilibrium composition is to be determined. Assumptions 1 The equilibrium composition consists of N2, O2, Ar, and NO. 2 The constituents of the mixture are ideal gases. Analysis The stoichiometric and actual reactions in this case are N 2 + 12 O 2 ⇔ NO (thus ν NO = 1, ν N 2 = 12 , and ν O 2 = 12 )

Stoichiometric:

1 2

Actual:

3 N 2 + O 2 + 01 . Ar

⎯ ⎯→

x NO + y N + z O + 01 . Ar 123 1422432 123 prod.

reactants

N balance:

6 = x + 2y

⎯ ⎯→

y = 3 − 0.5x

O balance:

2 = x + 2z

⎯ ⎯→

z = 1 − 0.5x

Total number of moles:

N total = x + y + z + 01 . = 41 .

inert

3 N2 1 O2 0.1 Ar 2400 K 10 atm

The equilibrium constant relation becomes, Kp =

ν NO N NO ν

ν

N NN 2 N OO2 2

2

⎛ P ⎜⎜ ⎝ N total

⎞ ⎟⎟ ⎠

(ν NO − ν N 2 − νO2 )

=

x y 0.5 z 0.5

⎛ P ⎜⎜ ⎝ N total

⎞ ⎟⎟ ⎠

1−0 .5−0 .5

From Table A-28, ln K p = −3.019 at 2400 K. Thus K p = 0.04885. Substituting, 0.04885 =

x ×1 (3 − 0.5 x)0.5 (1 − 0.5 x)0.5

Solving for x, x = 0.0823 Then, y = 3 - 0.5x = 2.9589 z = 1 - 0.5x = 0.9589 Therefore, the equilibrium composition of the mixture at 2400 K and 10 atm is 0.0823NO + 2.9589N 2 + 0.9589O 2 + 0.1Ar

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16-32

16-39 The mole fraction of sodium that ionizes according to the reaction Na ⇔ Na+ + e- at 2000 K and 0.8 atm is to be determined. Assumptions All components behave as ideal gases. Analysis The stoichiometric and actual reactions can be written as

Stoichiometric:

Na ⇔ Na + + e - (thus ν Na = 1, ν Na + = 1 and ν e - = 1)

Actual:

Na ⎯ ⎯→ { x Na + y Na + + y e − 142 4 43 4 react.

Na ⇔ Na+ + e2000 K 0.8 atm

products

1 = x + y or y = 1 − x

Na balance:

N total = x + 2 y = 2 − x

Total number of moles:

The equilibrium constant relation becomes, Kp =

N νNaNa N

ν

e-

N νNaNa

⎛ P ⎜ ⎜N ⎝ total

(1 − x) 2 x

⎛ 0.8 ⎞ ⎜ ⎟ ⎝2− x⎠

e-

⎞ ⎟ ⎟ ⎠

(ν

Na +

+ν - −ν Na ) e

y2 = x

⎛ P ⎜ ⎜N ⎝ total

⎞ ⎟ ⎟ ⎠

1+1−1

Substituting, 0.668 =

Solving for x, x = 0.325 Thus the fraction of Na which dissociates into Na+ and e- is 1 - 0.325 = 0.675 or 67.5%

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16-33

16-40 Liquid propane enters a combustion chamber. The equilibrium composition of product gases and the rate of heat transfer from the combustion chamber are to be determined. Assumptions 1 The equilibrium composition consists of CO2, H2O, CO, N2, and O2. 2 The constituents of the mixture are ideal gases.

C3H8 25°C

Analysis (a) Considering 1 kmol of C3H8, the stoichiometric combustion equation can be written as

Air

C 3 H 8 (l) + a th (O 2 + 3.76 N 2 ) ⎯ ⎯→ 3CO 2 + 4H 2 O + 3.76a th N 2

Combustion chamber 2 atm

12°C

CO CO 1200 K 2 H2O O2 N2

where ath is the stoichiometric coefficient and is determined from the O2 balance, 2.5a th = 3 + 2 + 1.5a th

⎯ ⎯→

a th = 5

Then the actual combustion equation with 150% excess air and some CO in the products can be written as C3H 8 ( l ) + 12.5( O 2 + 3.76 N 2 )

⎯ ⎯→

xCO 2 + (3 − x )CO + (9 − 0.5 x )O 2 + 4H 2 O + 47N 2

After combustion, there will be no C3 H8 present in the combustion chamber, and H2O will act like an inert gas. The equilibrium equation among CO2, CO, and O2 can be expressed as CO 2 ⇔ CO + 12 O 2 (thus ν CO 2 = 1, ν CO = 1, and ν O 2 = 12 )

and ν

ν

CO N CO N OO 2 ⎛ P 2 ⎜ Kp = ⎜N ν CO 2 ⎝ total N CO 2

⎞ ⎟ ⎟ ⎠

(ν CO +ν O 2 −ν CO 2 )

where N total = x + (3 − x ) + (9 − 0.5x ) + 4 + 47 = 63 − 0.5x

From Table A-28, ln K p = −17.871 at 1200 K. Thus K p = 1.73 × 10 −8 . Substituting, 1.73 × 10 −8 =

(3 − x )(9 − 0.5 x ) 0.5 ⎛ 2 ⎞ ⎜ ⎟ x ⎝ 63 − 0.5 x ⎠

1.5−1

Solving for x, x = 2.9999999 ≅ 3.0

Therefore, the amount CO in the product gases is negligible, and it can be disregarded with no loss in accuracy. Then the combustion equation and the equilibrium composition can be expressed as C 3 H 8 ( l) + 12.5(O 2 + 3.76N 2 ) ⎯ ⎯→ 3CO 2 + 7.5O 2 + 4 H 2 O + 47N 2

and 3CO 2 + 7.5O 2 + 4H 2 O + 47N 2

(b) The heat transfer for this combustion process is determined from the steady-flow energy balance E in − E out = ΔE system on the combustion chamber with W = 0, − Qout =

∑ N (h P

o f

+h −ho

) − ∑ N (h P

R

o f

+ h −ho

)

R

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16-34

Assuming the air and the combustion products to be ideal gases, we have h = h(T). From the tables, (The h fo of liquid propane is obtained by adding the hfg at 25°C to h fo of gaseous propane).

h fo

h 285 K

h 298 K

h1200 K

kJ/kmol

kJ/kmol

kJ/kmol

kJ/kmol

C3H8 (l)

-118,910

---

---

---

O2

0

8696.5

8682

38,447

N2

0

8286.5

8669

36,777

H2O (g)

-241,820

---

9904

44,380

CO2

-393,520

---

9364

53,848

Substance

Substituting, − Qout = 3( −393,520 + 53,848 − 9364 ) + 4( −241,820 + 44,380 − 9904 ) + 7.5( 0 + 38,447 − 8682 ) + 47( 0 + 36,777 − 8669 ) − 1( −118,910 + h298 − h298 ) − 12.5( 0 + 8296.5 − 8682 ) − 47( 0 + 8186.5 − 8669 ) = −185,764 kJ / kmol of C3H 8

or Qout = 185,764 kJ / kmol of C3H 8

The mass flow rate of C3H8 can be expressed in terms of the mole numbers as & m . kg / min 12 N& = = = 0.02727 kmol / min M 44 kg / kmol

Thus the rate of heat transfer is Q& out = N& × Qout = (0.02727 kmol/min)(185,746 kJ/kmol) = 5066 kJ/min

The equilibrium constant for the reaction

1 2

N 2 + 12 O 2 ⇔ NO is ln Kp = -7.569, which is very small. This

indicates that the amount of NO formed during this process will be very small, and can be disregarded.

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16-35

16-41 EES Problem 16-40 is reconsidered. It is to be investigated if it is realistic to disregard the presence of NO in the product gases. Analysis The problem is solved using EES, and the solution is given below. "To solve this problem, the Gibbs function of the product gases is minimized. Click on the Min/Max icon." For this problem at 1200 K the moles of CO are 0.000 and moles of NO are 0.000, thus we can disregard both the CO and NO. However, try some product temperatures above 1286 K and observe the sign change on the Q_out and the amout of CO and NO present as the product temperature increases." "The reaction of C3H8(liq) with excess air can be written: C3H8(l) + (1+Ex)A_th (O2+3.76N2) = a C02 + b CO + c H2O + d N2 + e O2 + f NO The coefficients A_th and EX are the theoretical oxygen and the percent excess air on a decimal basis. Coefficients a, b, c, d, e, and f are found by minimiming the Gibbs Free Energy at a total pressure of the product gases P_Prod and the product temperature T_Prod. The equilibrium solution can be found by applying the Law of Mass Action or by minimizing the Gibbs function. In this problem, the Gibbs function is directly minimized using the optimization capabilities built into EES. To run this program, click on the Min/Max icon. There are six compounds present in the products subject to four specie balances, so there are two degrees of freedom. Minimize the Gibbs function of the product gases with respect to two molar quantities such as coefficients b and f. The equilibrium mole numbers a, b, c, d, e, and f will be determined and displayed in the Solution window." PercentEx = 150 [%] Ex = PercentEx/100 "EX = % Excess air/100" P_prod =2*P_atm T_Prod=1200 [K] m_dot_fuel = 0.5 [kg/s] Fuel$='C3H8' T_air = 12+273 "[K]" T_fuel = 25+273 "[K]" P_atm = 101.325 [kPa] R_u=8.314 [kJ/kmol-K] "Theoretical combustion of C3H8 with oxygen: C3H8 + A_th O2 = 3 C02 + 4 H2O " 2*A_th = 3*2 + 4*1 "Balance the reaction for 1 kmol of C3H8" "C3H8(l) + (1+Ex)A_th (O2+3.76N2) = a C02 + b CO + c H2O + d N2 + e O2 + f NO" b_max = 3 f_max = (1+Ex)*A_th*3.76*2 e_guess=Ex*A_th 1*3 = a*1+b*1 "Carbon balance" 1*8=c*2 "Hydrogen balance" (1+Ex)*A_th*2=a*2+b*1+c*1+e*2+f*1 "Oxygen balance" (1+Ex)*A_th*3.76*2=d*2+f*1 "Nitrogen balance" PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-36 "Total moles and mole fractions" N_Total=a+b+c+d+e+f y_CO2=a/N_Total; y_CO=b/N_Total; y_H2O=c/N_Total; y_N2=d/N_Total; y_O2=e/N_Total; y_NO=f/N_Total "The following equations provide the specific Gibbs function for each component as a function of its molar amount" g_CO2=Enthalpy(CO2,T=T_Prod)-T_Prod*Entropy(CO2,T=T_Prod,P=P_Prod*y_CO2) g_CO=Enthalpy(CO,T=T_Prod)-T_Prod*Entropy(CO,T=T_Prod,P=P_Prod*y_CO) g_H2O=Enthalpy(H2O,T=T_Prod)-T_Prod*Entropy(H2O,T=T_Prod,P=P_Prod*y_H2O) g_N2=Enthalpy(N2,T=T_Prod)-T_Prod*Entropy(N2,T=T_Prod,P=P_Prod*y_N2) g_O2=Enthalpy(O2,T=T_Prod)-T_Prod*Entropy(O2,T=T_Prod,P=P_Prod*y_O2) g_NO=Enthalpy(NO,T=T_Prod)-T_Prod*Entropy(NO,T=T_Prod,P=P_Prod*y_NO) "The extensive Gibbs function is the sum of the products of the specific Gibbs function and the molar amount of each substance" Gibbs=a*g_CO2+b*g_CO+c*g_H2O+d*g_N2+e*g_O2+f*g_NO "For the energy balance, we adjust the value of the enthalpy of gaseous propane given by EES:" h_fg_fuel = 15060"[kJ/kmol]" "Table A.27" h_fuel = enthalpy(Fuel$,T=T_fuel)-h_fg_fuel "Energy balance for the combustion process:" "C3H8(l) + (1+Ex)A_th (O2+3.76N2) = a C02 + b CO + c H2O + d N2 + e O2 + f NO" HR =Q_out+HP HR=h_fuel+ (1+Ex)*A_th*(enthalpy(O2,T=T_air)+3.76*enthalpy(N2,T=T_air)) HP=a*enthalpy(CO2,T=T_prod)+b*enthalpy(CO,T=T_prod)+c*enthalpy(H2O,T=T_prod)+d*enthal py(N2,T=T_prod)+e*enthalpy(O2,T=T_prod)+f*enthalpy(NO,T=T_prod) "The heat transfer rate is:" Q_dot_out=Q_out/molarmass(Fuel$)*m_dot_fuel "[kW]" SOLUTION a=3.000 [kmol] A_th=5 b=0.000 [kmol] b_max=3 c=4.000 [kmol] d=47.000 [kmol] e=7.500 [kmol] Ex=1.5 e_guess=7.5 f=0.000 [kmol] Fuel$='C3H8' f_max=94 Gibbs=-17994897 [kJ] g_CO=-703496 [kJ/kmol]

g_CO2=-707231 [kJ/kmol] g_H2O=-515974 [kJ/kmol] g_N2=-248486 [kJ/kmol] g_NO=-342270 [kJ/kmol] g_O2=-284065 [kJ/kmol] HP=-330516.747 [kJ/kmol] HR=-141784.529 [kJ/kmol] h_fg_fuel=15060 [kJ/kmol] h_fuel=-118918 [kJ/kmol] m_dot_fuel=0.5 [kg/s] N_Total=61.5 [kmol/kmol_fuel] PercentEx=150 [%] P_atm=101.3 [kPa] P_prod=202.7 [kPa]

Q_dot_out=2140 [kW] Q_out=188732 [kJ/kmol_fuel] R_u=8.314 [kJ/kmol-K] T_air=285 [K] T_fuel=298 [K] T_Prod=1200.00 [K] y_CO=1.626E-15 y_CO2=0.04878 y_H2O=0.06504 y_N2=0.7642 y_NO=7.857E-08 y_O2=0.122

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-37

16-42 Oxygen is heated during a steady-flow process. The rate of heat supply needed during this process is to be determined for two cases. Assumptions 1 The equilibrium composition consists of O2 and O. 2 All components behave as ideal gases. Analysis (a) Assuming some O2 dissociates into O, the dissociation equation can be written as O2

⎯ ⎯→

x O 2 + 2(1 − x )O

The equilibrium equation among O2 and O can be expressed as O 2 ⇔ 2O (thus ν O 2 = 1 and ν O = 2)

Q&

Assuming ideal gas behavior for all components, the equilibrium constant relation can be expressed as ν

N O ⎛ P K p = νO ⎜⎜ N OO 2 ⎝ N total 2

where

ν O −ν O 2

⎞ ⎟ ⎟ ⎠

O2

O2, O

298 K

3000 K

N total = x + 2(1 − x ) = 2 − x

From Table A-28, ln K p = −4.357 at 3000 K. Thus K p = 0.01282. Substituting, 2 −1

(2 − 2 x) 2 ⎛ 1 ⎞ ⎜ ⎟ x ⎝2− x⎠ Solving for x gives x = 0.943 Then the dissociation equation becomes 0.01282 =

O2

⎯ ⎯→

0.943 O 2 + 0114 . O

The heat transfer for this combustion process is determined from the steady-flow energy balance E in − E out = ΔE system on the combustion chamber with W = 0, Qin =

∑ N (h P

o f

+ h −ho

) − ∑ N (h R

P

o f

+ h −ho

)

R

Assuming the O2 and O to be ideal gases, we have h = h(T). From the tables,

Substance

O O2

h fo

h 298 K

h 3000 K

kJ/kmol

kJ/kmol

kJ/kmol

249,190 0

6852 8682

63,425 106,780

Substituting, Qin = 0.943( 0 + 106,780 − 8682) + 0114 . ( 249,190 + 63,425 − 6852 ) − 0 = 127,363 kJ / kmol O 2

The mass flow rate of O2 can be expressed in terms of the mole numbers as 0.5 kg/min m& N& = = = 0.01563 kmol/min M 32 kg/kmol

Thus the rate of heat transfer is Q& in = N& × Qin = (0.01563 kmol/min)(127,363 kJ/kmol) = 1990 kJ/min (b) If no O2 dissociates into O, then the process involves no chemical reactions and the heat transfer can be determined from the steady-flow energy balance for nonreacting systems to be Q& = m& (h − h ) = N& (h − h ) = (0.01563 kmol/min)(106,780 - 8682) kJ/kmol = 1533 kJ/min in

2

1

2

1

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16-38

16-43 The equilibrium constant, Kp is to be estimated at 2500 K for the reaction CO + H2O = CO2 + H2. Analysis (a) The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data using K p = e − ΔG*(T ) / RuT or ln K p = −ΔG * (T ) / Ru T

where ∗ ∗ ∗ ∗ ΔG * (T ) = ν CO2 g CO2 (T ) + ν H2 g H2 (T ) −ν CO g CO (T ) −ν H2O g H2O (T )

At 2500 K, ∗ ∗ ∗ ∗ ΔG * (T ) = ν CO2 g CO2 (T ) + ν H2 g H2 (T ) −ν CO g CO (T ) −ν H2O g H2O (T )

= ν CO2 (h − Ts ) CO2 + ν H2 (h − Ts ) H2 −ν CO (h − Ts ) CO −ν H2O (h − Ts ) H2O = 1[(−271,641) − (2500)(322.60)] + 1[(70,452) − (2500)(196.10)]

− 1[(−35,510) − (2500)(266.65)] − 1[(−142,891) − (2500)(276.18)]

= 37,525 kJ/kmol

The enthalpies at 2500 K and entropies at 2500 K and 101.3 kPa (1 atm) are obtained from EES. Substituting, ln K p = −

37,525 kJ/kmol = −1.8054 ⎯ ⎯→ K p = 0.1644 (8.314 kJ/kmol ⋅ K)(2500 K)

The equilibrium constant may be estimated using the integrated van't Hoff equation: ⎛ K p ,est ⎞ ⎟= ln⎜ ⎜ K p1 ⎟ ⎝ ⎠ ⎛ K p ,est ⎞ ⎟= ln⎜⎜ ⎟ ⎝ 0.2209 ⎠

hR Ru

⎛ 1 1⎞ ⎜⎜ − ⎟⎟ ⎝ TR T ⎠

− 26,176 kJ/kmol ⎛ 1 1 ⎞ − ⎯→ K p ,est = 0.1612 ⎜ ⎟⎯ 8.314 kJ/kmol.K ⎝ 2000 K 2500 K ⎠

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16-39

16-44 A constant volume tank contains a mixture of H2 and O2. The contents are ignited. The final temperature and pressure in the tank are to be determined. Analysis The reaction equation with products in equilibrium is H2 + O2 ⎯ ⎯→ a H 2 + b H 2 O + c O 2

The coefficients are determined from the mass balances Hydrogen balance:

2 = 2a + 2b

Oxygen balance:

2 = b + 2c

The assumed equilibrium reaction is H 2 O ←⎯→ H 2 + 0.5O 2

The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data using K p = e −ΔG *( T )/ Ru T or ln K p = − ΔG * (T ) / Ru T

where ∗ ∗ ∗ ΔG * (T ) = ν H2 g H2 (Tprod ) + ν O2 g O2 (Tprod ) −ν H2O g H2O (Tprod )

and the Gibbs functions are given by ∗ (Tprod ) = (h − Tprod s ) H2 g H2 ∗ (Tprod ) = (h − Tprod s ) O2 g O2 ∗ (Tprod ) = (h − Tprod s ) H2O g H2O

The equilibrium constant is also given by a 1c 0.5 ⎛ P ⎜ Kp = b 1 ⎜⎝ N tot

⎞ ⎟ ⎟ ⎠

1+ 0.5 −1

=

ac 0.5 b

⎛ P2 / 101.3 ⎞ ⎟⎟ ⎜⎜ ⎝ a+b+c ⎠

0.5

An energy balance on the tank under adiabatic conditions gives UR =UP

where U R = 1(hH2@25°C − Ru Treac ) + 1(hO2@25°C − Ru Treac ) = 0 − (8.314 kJ/kmol.K)(298.15 K) + 0 − (8.314 kJ/kmol.K)(298.15 K) = −4958 kJ/kmol U P = a (hH2@ Tprod − Ru Tprod ) + b(hH2O@ Tprod − Ru Tprod ) + c(hO2@ Tprod − Ru Tprod )

The relation for the final pressure is P2 =

N tot Tprod ⎛ a + b + c ⎞⎛⎜ Tprod ⎞⎟ (101.3 kPa) P1 = ⎜ ⎟⎜ 2 N 1 Treac ⎝ ⎠⎝ 298.15 K ⎟⎠

Solving all the equations simultaneously using EES, we obtain the final temperature and pressure in the tank to be Tprod = 3857 K P2 = 1043 kPa

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16-40

Simultaneous Reactions

16-45C It can be expresses as “(dG)T,P = 0 for each reaction.” Or as “the Kp relation for each reaction must be satisfied.” 16-46C The number of Kp relations needed to determine the equilibrium composition of a reacting mixture is equal to the difference between the number of species present in the equilibrium mixture and the number of elements.

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16-41

16-47 Two chemical reactions are occurring in a mixture. The equilibrium composition at a specified temperature is to be determined. Assumptions 1 The equilibrium composition consists of H2O, OH, O2, and H2. 2 The constituents of the mixture are ideal gases. Analysis The reaction equation during this process can be expressed as H 2O

⎯ ⎯→

H 2O ⇒

x H 2O + y H 2 + z O 2 + w OH

2 = 2x + 2 y + w

(1)

O balance:

1 = x + 2z + w

(2)

O2 ,H 2

3400 K 1 atm

Mass balances for hydrogen and oxygen yield H balance:

H 2O,OH

The mass balances provide us with only two equations with four unknowns, and thus we need to have two more equations (to be obtained from the Kp relations) to determine the equilibrium composition of the mixture. They are

H 2 O ⇔ H 2 + 12 O 2

(reaction 1)

H 2 O ⇔ 12 H 2 + OH

(reaction 2)

The equilibrium constant for these two reactions at 3400 K are determined from Table A-28 to be

ln K P1 = −1891 .

⎯ ⎯→

K P1 = 015092 .

ln K P 2 = −1576 .

⎯ ⎯→

K P 2 = 0.20680

The Kp relations for these two simultaneous reactions are ν

K P1

ν

N HH 2 N OO 2 ⎛ P 2 2 ⎜ = ⎜ ν N HH O2 O ⎝ N total 2

where

⎞ ⎟ ⎟ ⎠

ν

(ν H 2 +ν O 2 −ν H 2 O )

and

K P2

ν

OH N HH 2 N OH ⎛ P 2 ⎜ = ⎜N ν H 2O ⎝ total NH O 2

⎞ ⎟ ⎟ ⎠

(ν H 2 +ν OH −ν H 2 O )

N total = N H2O + N H2 + N O2 + N OH = x + y + z + w

Substituting, 1/ 2

( y )( z )1 / 2 x

⎛ ⎞ 1 ⎜⎜ ⎟⎟ x + y + z + w ⎝ ⎠

( w)( y )1/ 2 0.20680 = x

⎛ ⎞ 1 ⎜⎜ ⎟⎟ x + y + z + w ⎝ ⎠

0.15092 =

(3) 1/ 2

(4)

Solving Eqs. (1), (2), (3), and (4) simultaneously for the four unknowns x, y, z, and w yields x = 0.574

y = 0.308

z = 0.095

w = 0.236

Therefore, the equilibrium composition becomes 0.574H 2 O + 0.308H 2 + 0.095O 2 + 0.236OH

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16-42

16-48 Two chemical reactions are occurring in a mixture. The equilibrium composition at a specified temperature is to be determined. Assumptions 1 The equilibrium composition consists of CO2, CO, O2, and O. 2 The constituents of the mixture are ideal gases. Analysis The reaction equation during this process can be expressed as ⎯ ⎯→

2 CO 2 + O 2

x CO 2 + y CO + z O 2 + w O

CO2, CO, O2, O 3200 K 2 atm

Mass balances for carbon and oxygen yield C balance:

2= x+ y

(1)

O balance:

6 = 2x + y + 2z + w

(2)

The mass balances provide us with only two equations with four unknowns, and thus we need to have two more equations (to be obtained from the KP relations) to determine the equilibrium composition of the mixture. They are

CO 2 ⇔ CO + 12 O 2

(reaction 1)

O 2 ⇔ 2O

(reaction 2)

The equilibrium constant for these two reactions at 3200 K are determined from Table A-28 to be

ln K P1 = −0.429

⎯ ⎯→

K P1 = 0.65116

ln K P 2 = −3.072

⎯ ⎯→

K P 2 = 0.04633

The KP relations for these two simultaneous reactions are ν

K P1

CO N νCO N OO 2 ⎛ P 2 ⎜ = ⎜N ν CO 2 ⎝ total N CO 2

νO

K P2

N ⎛ P = νO ⎜⎜ O2 N O ⎝ N total 2

⎞ ⎟ ⎟ ⎠

(ν CO +ν O 2 −ν CO 2 )

ν O −ν O 2

⎞ ⎟ ⎟ ⎠

where

N total = N CO2 + N O2 + N CO + N O = x + y + z + w Substituting, 0.65116 =

( y )( z )1 / 2 x

0.04633 =

w2 z

⎛ ⎞ 2 ⎜⎜ ⎟⎟ ⎝ x + y + z + w⎠

⎛ ⎞ 2 ⎜⎜ ⎟⎟ ⎝ x + y + z + w⎠

1/ 2

(3)

2 −1

(4)

Solving Eqs. (1), (2), (3), and (4) simultaneously for the four unknowns x, y, z, and w yields x = 1.127

y = 0.873

z = 1.273

w = 0.326

Thus the equilibrium composition is 1.127CO 2 + 0.873CO + 1.273O 2 + 0.326O

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16-43

16-49 Two chemical reactions are occurring at high-temperature air. The equilibrium composition at a specified temperature is to be determined. Assumptions 1 The equilibrium composition consists of O2, N2, O, and NO. 2 The constituents of the mixture are ideal gases. Analysis The reaction equation during this process can be expressed as ⎯ ⎯→

O 2 + 3.76 N 2

Heat

x N 2 + y NO + z O 2 + w O

Mass balances for nitrogen and oxygen yield

AIR

N balance:

7.52 = 2 x + y

(1)

O balance:

2 = y + 2z + w

(2)

Reaction chamber, 2 atm

O2, N2, O, NO 3000 K

The mass balances provide us with only two equations with four unknowns, and thus we need to have two more equations (to be obtained from the Kp relations) to determine the equilibrium composition of the mixture. They are 1 2

N 2 + 21 O 2 ⇔ NO

(reaction 1)

O 2 ⇔ 2O

(reaction 2)

The equilibrium constant for these two reactions at 3000 K are determined from Table A-28 to be

ln K P1 = −2.114

⎯ ⎯→

K P1 = 012075 .

ln K P 2 = −4.357

⎯ ⎯→

K P 2 = 0.01282

The KP relations for these two simultaneous reactions are ν

K P1 =

ν

ν

N NN 2 N OO 2 2

K P2 =

⎛ P ⎜ ⎜N ⎝ total

NO N NO

2

νO

NO ⎛ P ⎜ ⎜ ν N OO 2 ⎝ N total 2

where

⎞ ⎟ ⎟ ⎠

(ν NO −ν N 2 −ν O 2 )

ν O −ν O 2

⎞ ⎟ ⎟ ⎠

N total = N N 2 + N NO + N O 2 + N O = x + y + z + w

Substituting, 0.12075 =

0.01282 =

y x 0 .5 z 0 .5 w2 z

⎛ ⎞ 2 ⎜⎜ ⎟⎟ ⎝ x+ y+ z+w⎠

⎛ ⎞ 2 ⎜⎜ ⎟⎟ ⎝ x+ y+ z+w⎠

1− 0.5 − 0.5

(3)

2 −1

(4)

Solving Eqs. (1), (2), (3), and (4) simultaneously for the four unknowns x, y, z, and w yields x = 3.656

y = 0.2086

z = 0.8162

w = 0.1591

Thus the equilibrium composition is 3.656N 2 + 0.2086NO + 0.8162O 2 + 0.1591O

The equilibrium constant of the reaction N 2 ⇔ 2N at 3000 K is lnKP = -22.359, which is much smaller than the KP values of the reactions considered. Therefore, it is reasonable to assume that no N will be present in the equilibrium mixture.

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16-44

16-50E [Also solved by EES on enclosed CD] Two chemical reactions are occurring in air. The equilibrium composition at a specified temperature is to be determined. Assumptions 1 The equilibrium composition consists of O2, N2, O, and NO. 2 The constituents of the mixture are ideal gases. Analysis The reaction equation during this process can be expressed as ⎯ ⎯→

O 2 + 3.76 N 2

Heat

x N 2 + y NO + z O 2 + w O

Mass balances for nitrogen and oxygen yield

AIR

N balance:

7.52 = 2 x + y

(1)

O balance:

2 = y + 2z + w

(2)

Reaction chamber, 1 atm

O2, N2, O, NO 5400 R

The mass balances provide us with only two equations with four unknowns, and thus we need to have two more equations (to be obtained from the Kp relations) to determine the equilibrium composition of the mixture. They are 1 2

N 2 + 12 O 2 ⇔ NO

(reaction 1)

O 2 ⇔ 2O

(reaction 2)

The equilibrium constant for these two reactions at T = 5400 R = 3000 K are determined from Table A-28 to be ln K P1 = −2.114

⎯ ⎯→

K P1 = 012075 .

ln K P 2 = −4.357

⎯ ⎯→

K P 2 = 0.01282

The KP relations for these two simultaneous reactions are ν

K P1 =

ν

ν

N NN 2 N OO 2 2

K P2 =

⎛ P ⎜ ⎜N ⎝ total

NO N NO

2

νO

NO ⎛ P ⎜ ⎜ ν N OO 2 ⎝ N total 2

where

⎞ ⎟ ⎟ ⎠

(ν NO −ν N 2 −ν O 2 )

ν O −ν O 2

⎞ ⎟ ⎟ ⎠

N total = N N 2 + N NO + N O2 + N O = x + y + z + w

Substituting, 0.12075 =

0.01282 =

y x 0 . 5 z 0 .5 w2 z

⎛ ⎞ 1 ⎜⎜ ⎟⎟ ⎝ x + y + z + w⎠

⎛ ⎞ 1 ⎜⎜ ⎟⎟ ⎝ x + y + z + w⎠

1− 0.5− 0.5

(3)

2 −1

(4)

Solving Eqs. (1), (2), (3), and (4) simultaneously for the four unknowns x, y, z, and w yields x = 3.658

y = 0.2048

z = 0.7868

w = 0.2216

Thus the equilibrium composition is 3.658N 2 + 0.2048NO + 0.7868O 2 + 0.2216O

The equilibrium constant of the reaction N 2 ⇔ 2N at 5400 R is lnKP = -22.359, which is much smaller than the KP values of the reactions considered. Therefore, it is reasonable to assume that no N will be present in the equilibrium mixture.

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-45

14-51E EES Problem 16-50E is reconsidered. Using EES (or other) software, the equilibrium solution is to be obtained by minimizing the Gibbs function by using the optimization capabilities built into EES. This solution technique is to be compared with that used in the previous problem. Analysis The problem is solved using EES, and the solution is given below. "This example illustrates how EES can be used to solve multi-reaction chemical equilibria problems by directly minimizing the Gibbs function. 0.21 O2+0.79 N2 = a O2+b O + c N2 + d NO Two of the four coefficients, a, b, c, and d, are found by minimiming the Gibbs function at a total pressure of 1 atm and a temperature of 5400 R. The other two are found from mass balances. The equilibrium solution can be found by applying the Law of Mass Action to two simultaneous equilibrium reactions or by minimizing the Gibbs function. In this problem, the Gibbs function is directly minimized using the optimization capabilities built into EES. To run this program, select MinMax from the Calculate menu. There are four compounds present in the products subject to two elemental balances, so there are two degrees of freedom. Minimize Gibbs with respect to two molar quantities such as coefficients b and d. The equilibrium mole numbers of each specie will be determined and displayed in the Solution window. Minimizing the Gibbs function to find the equilibrium composition requires good initial guesses." "Data from Data Input Window" {T=5400 "R" P=1 "atm" } AO2=0.21; BN2=0.79 "Composition of air" AO2*2=a*2+b+d "Oxygen balance" BN2*2=c*2+d "Nitrogen balance" "The total moles at equilibrium are" N_tot=a+b+c+d y_O2=a/N_tot; y_O=b/N_tot; y_N2=c/N_tot; y_NO=d/N_tot "The following equations provide the specific Gibbs function for three of the components." g_O2=Enthalpy(O2,T=T)-T*Entropy(O2,T=T,P=P*y_O2) g_N2=Enthalpy(N2,T=T)-T*Entropy(N2,T=T,P=P*y_N2) g_NO=Enthalpy(NO,T=T)-T*Entropy(NO,T=T,P=P*y_NO) "EES does not have a built-in property function for monatomic oxygen so we will use the JANAF procedure, found under Options/Function Info/External Procedures. The units for the JANAF procedure are kgmole, K, and kJ so we must convert h and s to English units." T_K=T*Convert(R,K) "Convert R to K" Call JANAF('O',T_K:Cp`,h`,S`) "Units from JANAF are SI" S_O=S`*Convert(kJ/kgmole-K, Btu/lbmole-R) h_O=h`*Convert(kJ/kgmole, Btu/lbmole) "The entropy from JANAF is for one atmosphere so it must be corrected for partial pressure." g_O=h_O-T*(S_O-R_u*ln(Y_O)) R_u=1.9858 "The universal gas constant in Btu/mole-R " "The extensive Gibbs function is the sum of the products of the specific Gibbs function and the molar amount of each substance." Gibbs=a*g_O2+b*g_O+c*g_N2+d*g_NO

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16-46

d [lbmol] 0.002698 0.004616 0.007239 0.01063 0.01481 0.01972 0.02527 0.03132 0.03751 0.04361

b [lbmol] 0.00001424 0.00006354 0.0002268 0.000677 0.001748 0.004009 0.008321 0.01596 0.02807 0.04641

Gibbs [Btu/lbmol] -162121 -178354 -194782 -211395 -228188 -245157 -262306 -279641 -297179 -314941

yO2

yO

yNO

yN2

0.2086 0.2077 0.2062 0.2043 0.2015 0.1977 0.1924 0.1849 0.1748 0.1613

0.0000 0.0001 0.0002 0.0007 0.0017 0.0040 0.0083 0.0158 0.0277 0.0454

0.0027 0.0046 0.0072 0.0106 0.0148 0.0197 0.0252 0.0311 0.0370 0.0426

0.7886 0.7877 0.7863 0.7844 0.7819 0.7786 0.7741 0.7682 0.7606 0.7508

T [R] 3000 3267 3533 3800 4067 4333 4600 4867 5133 5400

Mole fraction of NO and O

0.050

0.040

0.030

NO

0.020

O 0.010

0.000 3000

3500

4000

4500

5000

5500

T [R] Discussion The equilibrium composition in the above table are based on the reaction in which the reactants are 0.21 kmol O2 and 0.79 kmol N2. If you multiply the equilibrium composition mole numbers above with 4.76, you will obtain equilibrium composition for the reaction in which the reactants are 1 kmol O2 and 3.76 kmol N2.This is the case in problem 16-43E.

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16-47

16-52 Water vapor is heated during a steady-flow process. The rate of heat supply for a specified exit temperature is to be determined for two cases. Assumptions 1 The equilibrium composition consists of H2O, OH, O2, and H2. 2 The constituents of the mixture are ideal gases.

Q

Analysis (a) Assuming some H2O dissociates into H2, O2, and O, the dissociation equation can be written as ⎯ ⎯→

H 2O

x H 2O + y H 2 + z O 2 + w OH

H2O

Mass balances for hydrogen and oxygen yield

H2O, H2, O2, OH

298 K

H balance:

2 = 2x + 2 y + w

(1)

O balance:

1 = x + 2z + w

(2)

3000 K

The mass balances provide us with only two equations with four unknowns, and thus we need to have two more equations (to be obtained from the KP relations) to determine the equilibrium composition of the mixture. They are

H 2 O ⇔ H 2 + 12 O 2

(reaction 1)

H 2 O ⇔ 12 H 2 + OH

(reaction 2)

The equilibrium constant for these two reactions at 3000 K are determined from Table A-28 to be ln K P1 = −3.086

⎯ ⎯→

K P1 = 0.04568

ln K P 2 = −2.937

⎯ ⎯→

K P 2 = 0.05302

The KP relations for these three simultaneous reactions are ν

K P1

ν

N HH 2 N OO 2 ⎛ P 2 2 ⎜ = ⎜ ν N HH O2 O ⎝ N total 2

ν H2

K P2

ν

OH N H N OH ⎛ P 2 ⎜ = ⎜N ν H 2O ⎝ total NH O 2

⎞ ⎟ ⎟ ⎠

(ν H 2 +ν O 2 −ν H 2 O )

⎞ ⎟ ⎟ ⎠

(ν H 2 +ν O 2 −ν H 2 O )

where

N total = N H 2O + N H 2 + N O 2 + N OH = x + y + z + w Substituting, 0.04568 =

( y )( z )1/ 2 x

0.05302 =

( w)( y )1 / 2 x

⎛ ⎞ 1 ⎜⎜ ⎟⎟ ⎝ x + y + z + w⎠

1/ 2

⎞ ⎛ 1 ⎟⎟ ⎜⎜ ⎝ x + y + z + w⎠

(3) 1/ 2

(4)

Solving Eqs. (1), (2), (3), and (4) simultaneously for the four unknowns x, y, z, and w yields x = 0.784

y = 0.162

z = 0.054

w = 0.108

Thus the balanced equation for the dissociation reaction is H 2O

⎯ ⎯→

0.784H 2 O + 0.162H 2 + 0.054O 2 + 0.108OH

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16-48

The heat transfer for this dissociation process is determined from the steady-flow energy balance E in − E out = ΔE system with W = 0, Qin =

∑ N (h P

o f

+h −ho

) − ∑ N (h P

R

o f

+h −ho

)

R

Assuming the O2 and O to be ideal gases, we have h = h(T). From the tables,

hfo

h 298 K

h 3000 K

kJ/kmol

kJ/kmol

kJ/kmol

H2O

-241,820

9904

136,264

H2

0

8468

97,211

O2

0

8682

106,780

OH

39,460

9188

98,763

Substance

Substituting, Qin = 0.784( −241,820 + 136,264 − 9904) + 0162 . ( 0 + 97,211 − 8468) + 0.054( 0 + 106,780 − 8682) + 0108 . (39,460 + 98,763 − 9188) − ( −241,820) = 184,909 kJ / kmol H 2 O

The mass flow rate of H2O can be expressed in terms of the mole numbers as 0.2 kg / min m& = = 0.01111 kmol / min N& = M 18 kg / kmol

Thus,

Q& in = N& × Qin = (0.01111 kmol/min)(184,909 kJ/kmol) = 2055 kJ/min (b) If no dissociates takes place, then the process involves no chemical reactions and the heat transfer can be determined from the steady-flow energy balance for nonreacting systems to be

& ( h2 − h1 ) = N& ( h2 − h1 ) Q& in = m = ( 0.01111 kmol / min)(136,264 − 9904) kJ / kmol = 1404 kJ / min

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16-49

16-53 EES Problem 16-52 is reconsidered. The effect of the final temperature on the rate of heat supplied for the two cases is to be studied. Analysis The problem is solved using EES, and the solution is given below. "This example illustrates how EES can be used to solve multi-reaction chemical equilibria problems by directly minimizing the Gibbs function. H2O = x H2O+y H2+z O2 + w OH Two of the four coefficients, x, y, z, and w are found by minimiming the Gibbs function at a total pressure of 1 atm and a temperature of 3000 K. The other two are found from mass balances. The equilibrium solution can be found by applying the Law of Mass Action (Eq. 15-15) to two simultaneous equilibrium reactions or by minimizing the Gibbs function. In this problem, the Gibbs function is directly minimized using the optimization capabilities built into EES. To run this program, click on the Min/Max icon. There are four compounds present in the products subject to two elemental balances, so there are two degrees of freedom. Minimize Gibbs with respect to two molar quantities such as coefficient z and w. The equilibrium mole numbers of each specie will be determined and displayed in the Solution window. Minimizing the Gibbs function to find the equilibrium composition requires good initial guesses." "T_Prod=3000 [K]" P=101.325 [kPa] m_dot_H2O = 0.2 [kg/min] T_reac = 298 [K] T = T_prod P_atm=101.325 [kPa] "H2O = x H2O+y H2+z O2 + w OH" AH2O=1 "Solution for 1 mole of water" AH2O=x+z*2+w "Oxygen balance" AH2O*2=x*2+y*2+w "Hydrogen balance" "The total moles at equilibrium are" N_tot=x+y+z+w y_H2O=x/N_tot; y_H2=y/N_tot; y_O2=z/N_tot; y_OH=w/N_tot "EES does not have a built-in property function for monatomic oxygen so we will use the JANAF procedure, found under Options/Function Info/External Procedures. The units for the JANAF procedure are kgmole, K, and kJ." Call JANAF('OH',T_prod:Cp`,h`,S`) "Units from JANAF are SI" S_OH=S` h_OH=h` "The entropy from JANAF is for one atmosphere so it must be corrected for partial pressure." g_OH=h_OH-T_prod*(S_OH-R_u*ln(y_OH*P/P_atm)) R_u=8.314 "The universal gas constant in kJ/kmol-K " "The following equations provide the specific Gibbs function for three of the components." g_O2=Enthalpy(O2,T=T_prod)-T_prod*Entropy(O2,T=T_prod,P=P*y_O2) g_H2=Enthalpy(H2,T=T_prod)-T_prod*Entropy(H2,T=T_prod,P=P*y_H2) PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-50 g_H2O=Enthalpy(H2O,T=T_prod)-T_prod*Entropy(H2O,T=T_prod,P=P*y_H2O) "The extensive Gibbs function is the sum of the products of the specific Gibbs function and the molar amount of each substance." Gibbs=x*g_H2O+y*g_H2+z*g_O2+w*g_OH "H2O = x H2O+y H2+z O2 + w OH" 1*Enthalpy(H2O,T=T_reac)+Q_in=x*Enthalpy(H2O,T=T_prod)+y*Enthalpy(H2,T=T_prod)+z*Enth alpy(O2,T=T_prod)+w*h_OH N_dot_H2O = m_dot_H2O/molarmass(H2O) Q_dot_in_Dissoc = N_dot_H2O*Q_in Q_dot_in_NoDissoc = N_dot_H2O*(Enthalpy(H2O,T=T_prod) - Enthalpy(H2O,T=T_reac)) Tprod [K] 2500 2600 2700 2800 2900 3000 3100 3200 3300 3400 3500

Qin,Dissoc [kJ/min] 1266 1326 1529 1687 1862 2053 2260 2480 2710 2944 3178

Qin,NoDissoc [kJ/min] 1098 1158 1219 1280 1341 1403 1465 1528 1590 1653 1716

3000

Qin [kJ/min]

2500

Qin,Dissoc 2000

Qin,NoDissoc 1500

1000 2500

2700

2900

3100

3300

3500

TProd [K]

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16-51

16-54 EES Ethyl alcohol C2H5OH (gas) is burned in a steady-flow adiabatic combustion chamber with 40 percent excess air. The adiabatic flame temperature of the products is to be determined and the adiabatic flame temperature as a function of the percent excess air is to be plotted. Analysis The complete combustion reaction in this case can be written as C 2 H 5 OH (gas) + (1 + Ex)a th [O 2 + 3.76N 2 ] ⎯ ⎯→ 2 CO 2 + 3 H 2 O + ( Ex)(a th ) O 2 + f N 2

where ath is the stoichiometric coefficient for air. The oxygen balance gives 1 + (1 + Ex)a th × 2 = 2 × 2 + 3 × 1 + ( Ex)(a th ) × 2

The reaction equation with products in equilibrium is C 2 H 5 OH (gas) + (1 + Ex)a th [O 2 + 3.76N 2 ] ⎯ ⎯→ a CO 2 + b CO + d H 2 O + e O 2 + f N 2 + g NO

The coefficients are determined from the mass balances Carbon balance:

2 = a+b

Hydrogen balance:

6 = 2d ⎯ ⎯→ d = 3

Oxygen balance:

1 + (1 + Ex)a th × 2 = a × 2 + b + d + e × 2 + g

Nitrogen balance: (1 + Ex)a th × 3.76 × 2 = f × 2 + g Solving the above equations, we find the coefficients to be Ex = 0.4, ath = 3, a = 1.995, b = 0.004712, d = 3, e = 1.17, f = 15.76, g = 0.06428 Then, we write the balanced reaction equation as C 2 H 5 OH (gas) + 4.2[O 2 + 3.76N 2 ] ⎯ ⎯→ 1.995 CO 2 + 0.004712 CO + 3 H 2 O + 1.17 O 2 + 15.76 N 2 + 0.06428 NO

Total moles of products at equilibrium are N tot = 1.995 + 0.004712 + 3 + 1.17 + 15.76 = 21.99

The first assumed equilibrium reaction is CO 2 ←⎯→ CO + 0.5O 2

The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data using ⎛ − ΔG1 * (Tprod ) ⎞ ⎟ K p1 = exp⎜ ⎜ ⎟ Ru Tprod ⎝ ⎠

Where

∗ ∗ ∗ ΔG1 * (Tprod ) = ν CO g CO (Tprod ) + ν O2 g O2 (Tprod ) −ν CO2 g CO2 (Tprod )

and the Gibbs functions are defined as ∗ (Tprod ) = (h − Tprod s ) CO g CO ∗ (Tprod ) = (h − Tprod s ) O2 g O2 ∗ g CO2 (Tprod ) = (h − Tprod s ) CO2

The equilibrium constant is also given by

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16-52

K p1 =

be 0.5 a

⎛ P ⎜ ⎜N ⎝ tot

⎞ ⎟ ⎟ ⎠

1+ 0.5 −1

(0.004712)(1.17) 0.5 ⎛ 1 ⎞ ⎜ ⎟ 1.995 ⎝ 21.99 ⎠

=

0.5

= 0.0005447

The second assumed equilibrium reaction is 0.5N 2 + 0.5O 2 ←⎯→ NO

Also, for this reaction, we have ∗ (Tprod ) = (h − Tprod s ) NO g NO ∗ g N2 (Tprod ) = (h − Tprod s ) N2 ∗ g O2 (Tprod ) = (h − Tprod s ) O2

∗ ∗ ∗ ΔG 2 * (Tprod ) = ν NO g NO (Tprod ) −ν N2 g N2 (Tprod ) −ν O2 g O2 (Tprod )

⎛ − ΔG 2 * (Tprod ) ⎞ ⎟ K p 2 = exp⎜ ⎜ ⎟ R T u prod ⎝ ⎠ K p2

⎛ P = ⎜⎜ ⎝ N tot

⎞ ⎟ ⎟ ⎠

1− 0.5 − 0.5

0

g e o.5 f

0.5

0.06428 ⎛ 1 ⎞ = 0.01497 =⎜ ⎟ ⎝ 21.99 ⎠ (1.17) 0.5 (15.76) 0.5

A steady flow energy balance gives HR = HP

where H R = h fo fuel@25°C + 4.2hO2@25°C + 15.79h N2@25°C = (−235,310 kJ/kmol) + 4.2(0) + 15.79(0) = −235,310 kJ/kmol H P = 1.995hCO2@Tprod + 0.004712hCO@Tprod + 3hH2O@Tprod + 1.17 hO2@Tprod + 15.76h N2@Tprod + 0.06428h NO@Tprod

Solving the energy balance equation using EES, we obtain the adiabatic flame temperature Tprod = 1901 K

The copy of entire EES solution including parametric studies is given next: "The reactant temperature is:" T_reac= 25+273 "[K]" "For adiabatic combustion of 1 kmol of fuel: " Q_out = 0 "[kJ]" PercentEx = 40 "Percent excess air" Ex = PercentEx/100 "EX = % Excess air/100" P_prod =101.3"[kPa]" R_u=8.314 "[kJ/kmol-K]" "The complete combustion reaction equation for excess air is:" "C2H5OH(gas)+ (1+Ex)*A_th (O2 +3.76N2)=2 CO2 + 3 H2O + Ex*A_th O2 + f N2 " "Oxygen Balance for complete combustion:" 1 + (1+Ex)*A_th*2=2*2+3*1 + Ex*A_th*2 "The reaction equation for excess air and products in equilibrium is:" "C2H5OH(gas)+ (1+Ex)*A_th (O2 +3.76N2)=a CO2 + b CO+ d H2O + e O2 + f N2 + g NO" PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-53

"Carbon Balance:" 2=a + b "Hydrogen Balance:" 6=2*d "Oxygen Balance:" 1 + (1+Ex)*A_th*2=a*2+b + d + e*2 +g "Nitrogen Balance:" (1+Ex)*A_th*3.76 *2= f*2 + g N_tot =a +b + d + e + f +g "Total kilomoles of products at equilibrium" "The first assumed equilibrium reaction is CO2=CO+0.5O2" "The following equations provide the specific Gibbs function (g=h-Ts) for each component in the product gases as a function of its temperature, T_prod, at 1 atm pressure, 101.3 kPa" g_CO2=Enthalpy(CO2,T=T_prod )-T_prod *Entropy(CO2,T=T_prod ,P=101.3) g_CO=Enthalpy(CO,T=T_prod )-T_prod *Entropy(CO,T=T_prod ,P=101.3) g_O2=Enthalpy(O2,T=T_prod )-T_prod *Entropy(O2,T=T_prod ,P=101.3) "The standard-state Gibbs function is" DELTAG_1 =1*g_CO+0.5*g_O2-1*g_CO2 "The equilibrium constant is given by Eq. 15-14." K_P_1 = exp(-DELTAG_1 /(R_u*T_prod )) P=P_prod /101.3"atm" "The equilibrium constant is also given by Eq. 15-15." "K_ P_1 = (P/N_tot)^(1+0.5-1)*(b^1*e^0.5)/(a^1)" sqrt(P/N_tot) *b *sqrt(e) =K_P_1*a "The econd assumed equilibrium reaction is 0.5N2+0.5O2=NO" g_NO=Enthalpy(NO,T=T_prod )-T_prod *Entropy(NO,T=T_prod ,P=101.3) g_N2=Enthalpy(N2,T=T_prod )-T_prod *Entropy(N2,T=T_prod ,P=101.3) "The standard-state Gibbs function is" DELTAG_2 =1*g_NO-0.5*g_O2-0.5*g_N2 "The equilibrium constant is given by Eq. 15-14." K_P_2 = exp(-DELTAG_2 /(R_u*T_prod )) "The equilibrium constant is also given by Eq. 15-15." "K_ P_2 = (P/N_tot)^(1-0.5-0.5)*(g^1)/(e^0.5*f^0.5)" g=K_P_2 *sqrt(e*f) "The steady-flow energy balance is:" H_R = Q_out+H_P h_bar_f_C2H5OHgas=-235310 "[kJ/kmol]" H_R=1*(h_bar_f_C2H5OHgas ) +(1+Ex)*A_th*ENTHALPY(O2,T=T_reac)+(1+Ex)*A_th*3.76*ENTHALPY(N2,T=T_reac) "[kJ/kmol]" H_P=a*ENTHALPY(CO2,T=T_prod)+b*ENTHALPY(CO,T=T_prod)+d*ENTHALPY(H2O,T=T_pro d)+e*ENTHALPY(O2,T=T_prod)+f*ENTHALPY(N2,T=T_prod)+g*ENTHALPY(NO,T=T_prod) "[kJ/kmol]"

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16-54

ath

a

b

d

e

f

g

3 3 3 3 3 3 3 3 3 3

1.922 1.971 1.988 1.995 1.998 1.999 2 2 2 2

0.07779 0.0293 0.01151 0.004708 0.001993 0.0008688 0.0003884 0.0001774 0.00008262 0.00003914

3 3 3 3 3 3 3 3 3 3

0.3081 0.5798 0.8713 1.17 1.472 1.775 2.078 2.381 2.683 2.986

12.38 13.5 14.63 15.76 16.89 18.02 19.15 20.28 21.42 22.55

0.0616 0.06965 0.06899 0.06426 0.05791 0.05118 0.04467 0.03867 0.0333 0.02856

PercentEx [%] 10 20 30 40 50 60 70 80 90 100

Tprod [K] 2184 2085 1989 1901 1820 1747 1682 1621 1566 1516

2200

2100

Tprod (K)

2000

1900

1800

1700

1600

1500 10

20

30

40

50

60

70

80

90

100

PercentEx

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16-55

Variations of Kp with Temperature

16-55C It enables us to determine the enthalpy of reaction hR from a knowledge of equilibrium constant KP. 16-56C At 2000 K since combustion processes are exothermic, and exothermic reactions are more complete at lower temperatures.

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-56

16-57 The hR at a specified temperature is to be determined using the enthalpy and KP data. Assumptions Both the reactants and products are ideal gases. Analysis (a) The complete combustion equation of CO can be expressed as

CO + 12 O 2 ⇔ CO 2 The hR of the combustion process of CO at 2200 K is the amount of energy released as one kmol of CO is burned in a steady-flow combustion chamber at a temperature of 2200 K, and can be determined from hR =

∑ N (h P

o f

+h −ho

) − ∑ N (h P

R

o f

+h −ho

)

R

Assuming the CO, O2 and CO2 to be ideal gases, we have h = h(T). From the tables,

hfo

h 298 K

h 2200 K

kJ/kmol

kJ/kmol

kJ/kmol

CO2

-393,520

9364

112,939

CO

-110,530

8669

72,688

O2

0

8682

75,484

Substance

Substituting, hR = 1( −393,520 + 112,939 − 9364) − 1( −110,530 + 72,688 − 8669) − 0.5(0 + 75,484 − 8682) = −276,835 kJ / kmol

(b) The hR value at 2200 K can be estimated by using KP values at 2000 K and 2400 K (the closest two temperatures to 2200 K for which KP data are available) from Table A-28, ln

K P 2 hR ≅ K P1 Ru

⎛1 1 ⎜⎜ − ⎝ T1 T2

3.860 − 6.635 ≅

⎞ h ⎟⎟ or ln K P 2 − ln K P1 ≅ R Ru ⎠

⎛1 1 ⎜⎜ − ⎝ T1 T2

⎞ ⎟⎟ ⎠

hR 1 ⎞ ⎛ 1 − ⎜ ⎟ 8.314 kJ/kmol ⋅ K ⎝ 2000 K 2400 K ⎠

h R ≅ −276,856 kJ/kmol

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16-57

16-58E The hR at a specified temperature is to be determined using the enthalpy and KP data. Assumptions Both the reactants and products are ideal gases. Analysis (a) The complete combustion equation of CO can be expressed as

CO + 12 O 2 ⇔ CO 2 The hR of the combustion process of CO at 3960 R is the amount of energy released as one kmol of H2 is burned in a steady-flow combustion chamber at a temperature of 3960 R, and can be determined from hR =

∑ N (h P

o f

+h −ho

) − ∑ N (h P

R

o f

+h −ho

)

R

Assuming the CO, O2 and CO2 to be ideal gases, we have h = h (T). From the tables,

hfo

h 537 R

h 3960 R

Btu/lbmol

Btu/lbmol

Btu/lbmol

CO2

-169,300

4027.5

48,647

CO

-47,540

3725.1

31,256.5

O2

0

3725.1

32,440.5

Substance

Substituting, hR = 1( −169,300 + 48,647 − 4027.5) .) − 1( −47,540 + 31,256.5 − 37251 .) − 0.5(0 + 32,440.5 − 37251 = −119,030 Btu / lbmol

(b) The hR value at 3960 R can be estimated by using KP values at 3600 R and 4320 R (the closest two temperatures to 3960 R for which KP data are available) from Table A-28, ln

K P 2 hR ≅ K P1 Ru

⎛1 1 ⎜⎜ − ⎝ T1 T2

3.860 − 6.635 ≅

⎞ h ⎟⎟ or ln K P 2 − ln K P1 ≅ R Ru ⎠

⎛1 1 ⎜⎜ − ⎝ T1 T2

⎞ ⎟⎟ ⎠

hR 1 ⎞ ⎛ 1 − ⎜ ⎟ 1.986 Btu/lbmol ⋅ R ⎝ 3600 R 4320 R ⎠

h R ≅ −119,041 Btu/lbmol

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16-58

16-59 The KP value of the combustion process H2 + 1/2O2 ⇔ H2O is to be determined at a specified temperature using hR data and KP value . Assumptions Both the reactants and products are ideal gases. Analysis The hR and KP data are related to each other by ln

h ⎛ 1 K P 2 hR ⎛ 1 1 ⎞ 1 ⎞ ⎜⎜ − ⎟⎟ or ln K P 2 − ln K P1 ≅ R ⎜⎜ − ⎟⎟ ≅ Ru ⎝ T1 T2 ⎠ K P1 Ru ⎝ T1 T2 ⎠

The hR of the specified reaction at 2400 K is the amount of energy released as one kmol of H2 is burned in a steady-flow combustion chamber at a temperature of 2400 K, and can be determined from hR =

∑ N (h P

o f

+h −ho

) − ∑ N (h P

R

o f

+h −ho

)

R

Assuming the H2O, H2 and O2 to be ideal gases, we have h = h (T). From the tables, h fo

h 298 K

h 2400 K

kJ/kmol

kJ/kmol

kJ/kmol

H2O

-241,820

9904

103,508

H2

0

8468

75,383

O2

0

8682

83,174

Substance

Substituting, hR = 1( −241,820 + 103,508 − 9904) − 1(0 + 75,383 − 8468) − 0.5(0 + 83,174 − 8682) = −252,377 kJ / kmol

The KP value at 2600 K can be estimated from the equation above by using this hR value and the KP value at 2200 K which is ln KP1 = 6.768, ln K P 2 − 6.768 ≅

− 252,377 kJ/kmol ⎛ 1 1 ⎞ − ⎜ ⎟ 8.314 kJ/kmol ⋅ K ⎝ 2200 K 2600 K ⎠

ln K P 2 = 4.645

(Table A - 28: lnK P 2 = 4.648)

or K P 2 = 104.1

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16-59

16-60 The hR value for the dissociation process CO2 ⇔ CO + 1/2O2 at a specified temperature is to be determined using enthalpy and Kp data. Assumptions Both the reactants and products are ideal gases. Analysis (a) The dissociation equation of CO2 can be expressed as CO 2 ⇔ CO + 12 O 2

The hR of the dissociation process of CO2 at 2200 K is the amount of energy absorbed or released as one kmol of CO2 dissociates in a steady-flow combustion chamber at a temperature of 2200 K, and can be determined from hR =

∑ N (h P

o f

+h −ho

) − ∑ N (h P

R

o f

+h −ho

)

R

Assuming the CO, O2 and CO2 to be ideal gases, we have h = h (T). From the tables, h fo

h 298 K

h 2200 K

kJ/kmol

kJ/kmol

kJ/kmol

CO2

-393,520

9364

112,939

CO

-110,530

8669

72,688

O2

0

8682

75,484

Substance

Substituting, hR = 1( −110,530 + 72,688 − 8669 ) + 0.5( 0 + 75,484 − 8682 ) − 1( −393,520 + 112,939 − 9364) = 276,835 kJ / kmol

(b) The hR value at 2200 K can be estimated by using KP values at 2000 K and 2400 K (the closest two temperatures to 2200 K for which KP data are available) from Table A-28, ln

K P 2 hR ⎛ 1 h ⎛1 1 ⎞ 1 ⎞ ⎟ ⎜ − ⎟ or ln K P 2 − ln K P1 ≅ R ⎜⎜ − ≅ Ru ⎝ T1 T2 ⎟⎠ K P1 Ru ⎜⎝ T1 T2 ⎟⎠

− 3.860 − (−6.635) ≅

hR 1 ⎞ ⎛ 1 − ⎜ ⎟ 8.314 kJ/kmol ⋅ K ⎝ 2000 K 2400 K ⎠

h R ≅ 276,856 kJ/kmol

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16-60

16-61 The hR value for the dissociation process O2 ⇔ 2O at a specified temperature is to be determined using enthalpy and KP data. Assumptions Both the reactants and products are ideal gases. Analysis (a) The dissociation equation of O2 can be expressed as O 2 ⇔ 2O

The hR of the dissociation process of O2 at 3100 K is the amount of energy absorbed or released as one kmol of O2 dissociates in a steady-flow combustion chamber at a temperature of 3100 K, and can be determined from hR =

∑ N (h P

o f

+h −ho

) − ∑ N (h P

R

o f

+h −ho

)

R

Assuming the O2 and O to be ideal gases, we have h = h (T). From the tables, h fo

h 298 K

h 2900 K

kJ/kmol

kJ/kmol

kJ/kmol

O

249,190

6852

65,520

O2

0

8682

110,784

Substance

Substituting, hR = 2(249,190 + 65,520 − 6852) − 1(0 + 110,784 − 8682) = 513,614 kJ/kmol

(b) The hR value at 3100 K can be estimated by using KP values at 3000 K and 3200 K (the closest two temperatures to 3100 K for which KP data are available) from Table A-28, ln

K P 2 hR ⎛ 1 h ⎛1 1 ⎞ 1 ⎞ ⎟ ⎟⎟ or ln K P 2 − ln K P1 ≅ R ⎜⎜ − ⎜⎜ − ≅ K P1 Ru ⎝ T1 T2 ⎠ Ru ⎝ T1 T2 ⎟⎠

− 3.072 − (−4.357) ≅

hR 1 ⎞ ⎛ 1 − ⎜ ⎟ 8.314 kJ/kmol ⋅ K ⎝ 3000 K 3200 K ⎠

h R ≅ 512,808 kJ/kmol

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16-61

16-62 The enthalpy of reaction for the equilibrium reaction CH4 + 2O2 = CO2 + 2H2O at 2500 K is to be estimated using enthalpy data and equilibrium constant, Kp data. Analysis The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data using K p = e − ΔG*(T ) / RuT or ln K p = −ΔG * (T ) / Ru T

where ∗ ∗ ∗ ∗ ΔG * (T ) = ν CO2 g CO2 (T ) + ν H2O g H2O (T ) −ν CH4 g CH4 (T ) −ν O2 g O2 (T )

At T1 = 2500 - 10 = 2490 K: ∗ ∗ ∗ ∗ ΔG1 * (T ) = ν CO2 g CO2 (T1 ) + ν H2O g H2O (T1 ) −ν CH4 g CH4 (T1 ) −ν O2 g O2 (T1 )

= 1(−1.075 ×10 6 ) + 2( −830,577) − 1( −717,973) − 2(−611,582) = −794,929 kJ/kmol

At T2 = 2500 + 10 = 2510 K: ∗ ∗ ∗ ∗ ΔG 2 * (T ) = ν CO2 g CO2 (T2 ) + ν H2O g H2O (T2 ) −ν CH4 g CH4 (T2 ) −ν O2 g O2 (T2 )

= 1(−1.081× 10 6 ) + 2(−836,100) − 1(−724,516) − 2(−617,124) = −794,801 kJ/kmol

The Gibbs functions are obtained from enthalpy and entropy properties using EES. Substituting, ⎞ ⎛ − 794,929 kJ/kmol ⎟⎟ = 4.747 × 1016 K p1 = exp⎜⎜ − ⎝ (8.314 kJ/kmol ⋅ K)(2490 K) ⎠ ⎞ ⎛ − 794,801 kJ/kmol ⎟⎟ = 3.475 × 1016 K p 2 = exp⎜⎜ − (8.314 kJ/kmol K)(2510 K) ⋅ ⎠ ⎝

The enthalpy of reaction is determined by using the integrated van't Hoff equation: ⎛ K p2 ln⎜ ⎜ K p1 ⎝ ⎛ 3.475 × 1016 ln⎜⎜ 16 ⎝ 4.747 × 10

⎞ hR ⎛ 1 1 ⎞ ⎟= ⎜ − ⎟ ⎟ Ru ⎜⎝ T1 T2 ⎟⎠ ⎠ ⎞ hR 1 ⎞ ⎛ 1 ⎟= ⎯→ h R = −810,845 kJ/kmol ⎟ 8.314 kJ/kmol.K ⎜⎝ 2490 K − 2510 K ⎟⎠ ⎯ ⎠

The enthalpy of reaction can also be determined from an energy balance to be hR = H P − H R

where H R = 1hCH4 @ 2500 K + 2hO2 @ 2500 K = 96,668 + 2(78,377) = 253,422 kJ/kmol H P = 1hCO2 @ 2500 K + 2hH2O @ 2500 K = (−271,641) + 2(−142,891) = −557,423 kJ/kmol

The enthalpies are obtained from EES. Substituting, h R = H P − H R = ( −557,423) − ( 253,422) = −810,845 kJ/kmol

which is identical to the value obtained using Kp data.

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16-62

Phase Equilibrium

16-63C No. Because the specific gibbs function of each phase will not be affected by this process; i.e., we will still have gf = gg. 16-64C Yes. Because the number of independent variables for a two-phase (PH=2), two-component (C=2) mixture is, from the phase rule, IV = C - PH + 2 = 2 - 2 + 2 = 2 Therefore, two properties can be changed independently for this mixture. In other words, we can hold the temperature constant and vary the pressure and still be in the two-phase region. Notice that if we had a single component (C=1) two phase system, we would have IV=1, which means that fixing one independent property automatically fixes all the other properties. 11-65C Using solubility data of a solid in a specified liquid, the mass fraction w of the solid A in the liquid at the interface at a specified temperature can be determined from mf A =

msolid msolid + m liquid

where msolid is the maximum amount of solid dissolved in the liquid of mass mliquid at the specified temperature. 11-66C The molar concentration Ci of the gas species i in the solid at the interface Ci, solid side (0) is proportional to the partial pressure of the species i in the gas Pi, gas side(0) on the gas side of the interface, and is determined from C i, solid side (0) = S × Pi, gas side (0)

(kmol/m3)

where S is the solubility of the gas in that solid at the specified temperature. 11-67C Using Henry’s constant data for a gas dissolved in a liquid, the mole fraction of the gas dissolved in the liquid at the interface at a specified temperature can be determined from Henry’s law expressed as yi, liquid side (0) =

Pi, gas side (0) H

where H is Henry’s constant and Pi,gas side(0) is the partial pressure of the gas i at the gas side of the interface. This relation is applicable for dilute solutions (gases that are weakly soluble in liquids).

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16-63

16-68 It is to be shown that a mixture of saturated liquid water and saturated water vapor at 100°C satisfies the criterion for phase equilibrium. Analysis Using the definition of Gibbs function and enthalpy and entropy data from Table A-4, g f = h f − Ts f = (419.17 kJ/kg) − (373.15 K)(1.3072 kJ/kg ⋅ K) = −68.61 kJ/kg g g = h g − Ts g = (2675.6 kJ/kg) − (373.15 K)(7.3542 kJ/kg ⋅ K) = −68.62 kJ/kg

which are practically same. Therefore, the criterion for phase equilibrium is satisfied.

16-69 It is to be shown that a mixture of saturated liquid water and saturated water vapor at 300 kPa satisfies the criterion for phase equilibrium. Analysis The saturation temperature at 300 kPa is 406.7 K. Using the definition of Gibbs function and enthalpy and entropy data from Table A-5, g f = h f − Ts f = (561.43 kJ/kg) − (406.7 K)(1.6717 kJ/kg ⋅ K) = −118.5 kJ/kg g g = h g − Ts g = (2724.9 kJ/kg) − (406.7 K)(6.9917 kJ/kg ⋅ K) = −118.6 kJ/kg

which are practically same. Therefore, the criterion for phase equilibrium is satisfied.

16-70 It is to be shown that a saturated liquid-vapor mixture of refrigerant-134a at -10°C satisfies the criterion for phase equilibrium. Analysis Using the definition of Gibbs function and enthalpy and entropy data from Table A-11, g f = h f − Ts f = (38.55 kJ/kg) − (263.15 K)(0.15504 kJ/kg ⋅ K) = −2.249 kJ/kg g g = h g − Ts g = (244.51 kJ/kg) − (263.15 K)(0.93766 kJ/kg ⋅ K) = −2.235 kJ/kg

which are sufficiently close. Therefore, the criterion for phase equilibrium is satisfied.

16-71 The number of independent properties needed to fix the state of a mixture of oxygen and nitrogen in the gas phase is to be determined. Analysis In this case the number of components is C = 2 and the number of phases is PH = 1. Then the number of independent variables is determined from the phase rule to be

IV = C - PH + 2 = 2 - 1 + 2 = 3 Therefore, three independent properties need to be specified to fix the state. They can be temperature, the pressure, and the mole fraction of one of the gases.

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16-64

16-72 The values of the Gibbs function for saturated refrigerant-134a at 0°C as a saturated liquid, saturated vapor, and a mixture of liquid and vapor are to be calculated. Analysis Obtaining properties from Table A-11, the Gibbs function for the liquid phase is, g f = h f − Ts f = 51.86 kJ/kg − (273.15 K )(0.20439 kJ/kg ⋅ K ) = −3.97 kJ/kg

For the vapor phase, g g = h g − Ts g = 250.45 kJ/kg − (273.15 K )(0.93139 kJ/kg ⋅ K ) = −3.96 kJ/kg

R-134a 0°C

For the saturated mixture with a quality of 30%, h = h f + xh fg = 51.86 kJ/kg + (0.30)(198.60 kJ/kg ) = 111.44 kJ/kg s = s f + xs fg = 0.20439 kJ/kg ⋅ K + (0.30)(0.72701 kJ/kg ⋅ K ) = 0.42249 kJ/kg ⋅ K g = h − Ts = 111.44 kJ/kg − (273.15 kJ/kg )(0.42249 kJ/kg ⋅ K ) = −3.96 kJ/kg

The results agree and demonstrate that phase equilibrium exists.

16-73 The values of the Gibbs function for saturated refrigerant-134a at -10°C are to be calculated. Analysis Obtaining properties from Table A-11, the Gibbs function for the liquid phase is, g f = h f − Ts f = 38.55 kJ/kg − (263.15 K )(0.15504 kJ/kg ⋅ K ) = −2.25 kJ/kg

For the vapor phase, g g = h g − Ts g = 244.51 kJ/kg − (263.15 K )(0.93766 kJ/kg ⋅ K ) = −2.24 kJ/kg

R-134a −10°C x = 0.4

The results agree and demonstrate that phase equilibrium exists.

16-74 Using the liquid-vapor equilibrium diagram of an oxygen-nitrogen mixture at a specified pressure, the temperature is to be determined for a specified composition of the nitrogen and the mass fraction of the oxygen at this temperature is to be determined. Analysis From the equilibrium diagram (Fig. 16-21) we read T = 88 K.

For the liquid phase, from the same figure, y f ,O2 = 0.90 and

y f , N2 = 0.10

Then, mf f ,O2 =

m f ,O2 m f , total

=

y f ,O2 M O2 y f ,O2 M O2 + y f , N2 M N2

=

(0.90)(32) = 0.911 (0.90)(32) + (0.10)(28)

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16-65

16-75 A liquid-vapor mixture of ammonia and water in equilibrium at a specified temperature is considered. The pressure of ammonia is to be determined for two compositions of the liquid phase. Assumptions The mixture is ideal and thus Raoult’s law is applicable. Analysis According to Raoults’s law, when the mole fraction of the ammonia liquid is 20%, PNH3 = y f , NH3 Psat, NH3 (T ) = 0.20(615.3 kPa) = 123.1 kPa

H2O + NH3 10°C

When the mole fraction of the ammonia liquid is 80%, PNH3 = y f , NH3 Psat, NH3 (T ) = 0.80(615.3 kPa) = 492.2 kPa

16-76 A liquid-vapor mixture of ammonia and water in equilibrium at a specified temperature is considered. The composition of the liquid phase is given. The composition of the vapor phase is to be determined. Assumptions The mixture is ideal and thus Raoult’s law is applicable. Properties At 25°C, Psat, H 2 O = 3.170 kPa and

Psat, NH 3 = 1003.5 kPa .

Analysis The vapor pressures are PH 2O = y f ,H 2O Psat,H 2O (T ) = 0.50(3.170 kPa) = 1.585 kPa

H2O + NH3 25°C

PNH3 = y f , NH 3 Psat, NH3 (T ) = 0.50(1003.5 kPa) = 501.74 kPa

Thus the total pressure of the mixture is Ptotal = PH 2O + PNH 3 = (1.585 + 501.74) kPa = 503.33 kPa

Then the mole fractions in the vapor phase become y g , H 2O = y g , NH3 =

PH 2O Ptotal PNH 3 Ptotal

=

1.585 kPa = 0.0031 or 0.31% 503.33 kPa

=

501.74 kPa = 0.9969 or 99.69% 503.33 kPa

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16-66

16-77 A liquid-vapor mixture of ammonia and water in equilibrium at a specified temperature is considered. The composition of the vapor phase is given. The composition of the liquid phase is to be determined. Assumptions The mixture is ideal and thus Raoult’s law is applicable. Properties At 50°C, Psat, H 2O = 12.352 kPa and Psat, NH 3 = 2033.5 kPa. Analysis We have y g ,H 2O = 1% and y g , NH3 = 99% . For an ideal two-phase mixture we have y g , H 2 O Pm = y f , H 2 O Psat, H 2 O (T ) y g , NH 3 Pm = y f , NH 3 Psat, NH 3 (T )

H2O + NH3

y f , H 2 O + y f , NH 3 = 1

50°C

Solving for y f ,H 2O, y f ,H 2O =

y g ,H 2O Psat, NH3 y g , NH3 Psat,H 2O

(1 − y f ,H 2O ) =

(0.01)(2033.5 kPa) (1 − y f ,H 2O ) (0.99)(12.352 kPa)

It yields y f ,H 2O = 0.624 and y f , NH3 = 0.376

16-78 Using the liquid-vapor equilibrium diagram of an oxygen-nitrogen mixture, the composition of each phase at a specified temperature and pressure is to be determined. Analysis From the equilibrium diagram (Fig. 16-21) we read

Liquid: 30% N 2 and 70% O 2 Vapor: 66% N 2 and 34% O 2

16-79 Using the liquid-vapor equilibrium diagram of an oxygen-nitrogen mixture at a specified pressure, the temperature is to be determined for a specified composition of the vapor phase. Analysis From the equilibrium diagram (Fig. 16-21) we read T = 82 K.

16-80 Using the liquid-vapor equilibrium diagram of an oxygen-nitrogen mixture at a specified pressure, the temperature is to be determined for a specified composition of the liquid phase. Analysis From the equilibrium diagram (Fig. 16-21) we read T = 84 K.

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16-67

16-81 A rubber wall separates O2 and N2 gases. The molar concentrations of O2 and N2 in the wall are to be determined. Assumptions The O2 and N2 gases are in phase equilibrium with the rubber wall. Properties The molar mass of oxygen and nitrogen are 32.0 and 28.0 kg/kmol, respectively (Table A-1). The solubility of oxygen and nitrogen in rubber at 298 K are 0.00312 and 0.00156 kmol/m3⋅bar, respectively (Table 16-3). Analysis Noting that 500 kPa = 5 bar, the molar densities of oxygen and nitrogen in the rubber wall are determined to be

Rubber plate

C O 2 , solid side (0) = S × PO 2 , gas side = (0.00312 kmol/m 3 .bar )(5 bar) = 0.0156 kmol/m

3

CN 2 , solid side (0) = S × PN 2 , gas side

O2 25°C 500 kPa

CO2 CN2

N2 25°C 500 kPa

= (0.00156 kmol / m3 . bar )(5 bar) = 0.0078 kmol / m 3

That is, there will be 0.0156 kmol of O2 and 0.0078 kmol of N2 gas in each m3 volume of the rubber wall.

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16-68

16-82 An ammonia-water absorption refrigeration unit is considered. The operating pressures in the generator and absorber, and the mole fractions of the ammonia in the strong liquid mixture being pumped from the absorber and the weak liquid solution being drained from the generator are to be determined. Assumptions The mixture is ideal and thus Raoult’s law is applicable. Properties At 0°C, Psat,H2O = 0.6112 kPa and at 46°C, Psat,H2O = 10.10 kPa (Table A-4). The saturation

pressures of ammonia at the same temperatures are given to be 430.6 kPa and 1830.2 kPa, respectively. Analysis According to Raoults’s law, the partial pressures of ammonia and water are given by Pg, NH3 = y f , NH3 Psat, NH3 Pg, H2O = y f , H2O Psat,H2O = (1 − y f , NH3 ) Psat,H2O

Using Dalton’s partial pressure model for ideal gas mixtures, the mole fraction of the ammonia in the vapor mixture is y g , NH3 = 0.96 =

y f , NH3 Psat, NH3 y f , NH3 Psat, NH3 + (1 − y f , NH3 Psat, H2O ) 430.6 y f , NH3 430.6 y f , NH3 + 0.6112(1 − y f , NH3 )

⎯ ⎯→ y f , NH3 = 0.03294

Then, P = y f , NH3 Psat, NH3 + (1 − y f , NH3 ) Psat, H2O = (0.03294)(430.6) + (1 − 0.03294)(0.6112) = 14.78 kPa

Performing the similar calculations for the regenerator, 0.96 =

1830.2 y f , NH3 1830.2 y f , NH3 + 10.10(1 − y f , NH3 )

⎯ ⎯→ y f , NH3 = 0.1170

P = (0.1170)(1830.2) + (1 − 0.1170)(10.10) = 223.1 kPa

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16-69

16-83 An ammonia-water absorption refrigeration unit is considered. The operating pressures in the generator and absorber, and the mole fractions of the ammonia in the strong liquid mixture being pumped from the absorber and the weak liquid solution being drained from the generator are to be determined. Assumptions The mixture is ideal and thus Raoult’s law is applicable. Properties At 6°C, Psat,H2O = 0.9353 kPa and at 40°C, Psat,H2O = 7.3851 kPa (Table A-4 or EES). The

saturation pressures of ammonia at the same temperatures are given to be 534.8 kPa and 1556.7 kPa, respectively. Analysis According to Raoults’s law, the partial pressures of ammonia and water are given by Pg, NH3 = y f , NH3 Psat, NH3 Pg, H2O = y f , H2O Psat,H2O = (1 − y f , NH3 ) Psat,H2O

Using Dalton’s partial pressure model for ideal gas mixtures, the mole fraction of the ammonia in the vapor mixture is y g , NH3 = 0.96 =

y f , NH3 Psat, NH3 y f , NH3 Psat, NH3 + (1 − y f , NH3 Psat,H2O ) 534.8 y f , NH3 534.8 y f , NH3 + 0.9353(1 − y f , NH3 )

⎯ ⎯→ y f , NH3 = 0.04028

Then, P = y f , NH3 Psat, NH3 + (1 − y f , NH3 ) Psat, H2O = (0.04028)(534.8) + (1 − 0.04028)(0.9353) = 22.44 kPa Performing the similar calculations for the regenerator, 0.96 =

1556.7 y f , NH3 1556.7 y f , NH3 + 7.3851(1 − y f , NH3 )

⎯ ⎯→ y f , NH3 = 0.1022

P = (0.1022)(1556.7) + (1 − 0.1022)(7.3851) = 165.7 kPa

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16-70

16-84 A liquid mixture of water and R-134a is considered. The mole fraction of the water and R-134a vapor are to be determined. Assumptions The mixture is ideal and thus Raoult’s law is applicable. Properties At 20°C, Psat, H2O = 2.3392 kPa and Psat,R = 572.07 kPa (Tables A-4, A-11). The molar masses

of water and R-134a are 18.015 and 102.03 kg/kmol, respectively (Table A-1). Analysis The mole fraction of the water in the liquid mixture is y f , H2O = =

N f ,H2O N total

=

mf f ,H2O / M H2O (mf f ,H2O / M H2O ) + (mf f ,R / M R )

0.9 / 18.015 = 0.9808 (0.9 / 18.015) + (0.1 / 102.03)

H2O + R-134a 20°C

According to Raoults’s law, the partial pressures of R-134a and water in the vapor mixture are Pg ,R = y f ,R Psat,R = (1 − 0.9808)(572.07 kPa) = 10.98 kPa

Pg , H2O = y f , H2O Psat, H2O = (0.9808)(2.3392 kPa) = 2.294 kPa

The total pressure of the vapor mixture is then Ptotal = Pg ,R + Pg ,H2O = 10.98 + 2.294 = 13.274 kPa

Based on Dalton’s partial pressure model for ideal gases, the mole fractions in the vapor phase are y g ,H2O = y g ,R =

Pg , H2O Ptotal

Pg ,R Ptotal

=

=

2.294 kPa = 0.1728 13.274 kPa

10.98 kPa = 0.8272 13.274 kPa

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16-71

16-85E A mixture of water and R-134a is considered. The mole fractions of the R-134a in the liquid and vapor phases are to be determined. Assumptions The mixture is ideal and thus Raoult’s law is applicable. Properties At 77°F, Psat,H2O = 0.4597 psia and Psat, R = 96.56 psia (Tables A-4E, A-11E or EES). Analysis According to Raoults’s law, the partial pressures of R-134a and water in the vapor phase are given by Pg ,R = y f , R Psat,R =

N f ,R N f , R + N f , H2O

Pg , H2O = y f , H2O Psat, H2O =

(96.56 psia)

N f ,H2O N f ,H2O + N f ,H2O

H2O + R-134a 14.7 psia, 77°F

(0.4597 psia)

The sum of these two partial pressures must equal the total pressure of the vapor mixture. In terms of N f ,H2O , this sum is x= N f ,R 96.56 0.4597 x + = 14.7 x +1 x +1

Solving this expression for x gives x = 5.748 kmol H2O/kmol R-134a In the vapor phase, the partial pressure of the R-134a vapor is Pg , R =

96.56 96.56 = = 14.31 psia x + 1 5.748 + 1

The mole fraction of R-134a in the vapor phase is then y g ,R =

Pg , R P

=

14.31 psia = 0.9735 14.7 psia

According to Raoult’s law, y f ,R =

Pg , R Psat,R

=

14.31 psia = 0.1482 96.56 psia

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16-72

16-86 A glass of water is left in a room. The mole fraction of the water vapor in the air and the mole fraction of air in the water are to be determined when the water and the air are in thermal and phase equilibrium. Assumptions 1 Both the air and water vapor are ideal gases. 2 Air is saturated since the humidity is 100 percent. 3 Air is weakly soluble in water and thus Henry’s law is applicable. Properties The saturation pressure of water at 27°C is 3.568 kPa (Table A-4). Henry’s constant for air dissolved in water at 27ºC (300 K) is given in Table 16-2 to be H = 74,000 bar. Molar masses of dry air and water are 29 and 18 kg/kmol, respectively (Table A-1). Analysis (a) Noting that air is saturated, the partial pressure of water vapor in the air will simply be the saturation pressure of water at 27°C,

Pvapor = Psat @ 27°C = 3.600 kPa (Table A-4) Assuming both the air and vapor to be ideal gases, the mole fraction of water vapor in the air is determined to be y vapor =

Pvapor P

=

Air 27ºC 97 kPa φ = 100%

3.600 kPa = 0.0371 97 kPa

(b) Noting that the total pressure is 97 kPa, the partial pressure of dry air is

Pdry air = P − Pvapor = 97 − 3.600 = 93.4 kPa = 0.934 bar

Water 27ºC

From Henry’s law, the mole fraction of air in the water is determined to be y dry air,liquid side =

Pdry air,gas side H

=

0.934 bar = 1.26 ×10 − 5 74,000 bar

Discussion The amount of air dissolved in water is very small, as expected.

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16-73

16-87 A carbonated drink in a bottle is considered. Assuming the gas space above the liquid consists of a saturated mixture of CO2 and water vapor and treating the drink as a water, determine the mole fraction of the water vapor in the CO2 gas and the mass of dissolved CO2 in a 300 ml drink are to be determined when the water and the CO2 gas are in thermal and phase equilibrium. Assumptions 1 The liquid drink can be treated as water. 2 Both the CO2 and the water vapor are ideal gases. 3 The CO2 gas and water vapor in the bottle from a saturated mixture. 4 The CO2 is weakly soluble in water and thus Henry’s law is applicable. Properties The saturation pressure of water at 27°C is 3.568 kPa (Table A-4). Henry’s constant for CO2 dissolved in water at 27ºC (300 K) is given in Table 16-2 to be H = 1710 bar. Molar masses of CO2 and water are 44 and 18 kg/kmol, respectively (Table A-1). Analysis (a) Noting that the CO2 gas in the bottle is saturated, the partial pressure of water vapor in the air will simply be the saturation pressure of water at 27°C,

Pvapor = Psat @ 27°C = 3.568 kPa

(more accurate EES value compared to interpolation value from Table A-4)

Assuming both CO2 and vapor to be ideal gases, the mole fraction of water vapor in the CO2 gas becomes y vapor =

Pvapor P

=

3.568 kPa = 0.0274 130 kPa

(b) Noting that the total pressure is 130 kPa, the partial pressure of CO2 is

PCO 2 gas = P − Pvapor = 130 − 3.568 = 126.4 kPa = 1.264 bar From Henry’s law, the mole fraction of CO2 in the drink is determined to be yCO 2 ,liquid side =

PCO 2 ,gas side H

=

1264 . bar = 7.39 × 10 −4 1710 bar

Then the mole fraction of water in the drink becomes y water, liquid side = 1 − y CO 2 , liquid side = 1 − 7.39 × 10 −4 = 0.9993

The mass and mole fractions of a mixture are related to each other by mf i =

mi N M Mi = i i = yi mm N m M m Mm

where the apparent molar mass of the drink (liquid water - CO2 mixture) is Mm =

∑y M i

i

= yliquid water M water + y CO2 M CO2 = 0.9993 × 18.0 + ( 7.39 × 10 −4 ) × 44 = 18.02 kg / kmol

Then the mass fraction of dissolved CO2 gas in liquid water becomes mf CO 2 , liquid side = y CO 2 , liquid side (0)

M CO 2 Mm

= 7.39 ×10 − 4

44 = 0.00180 18.02

Therefore, the mass of dissolved CO2 in a 300 ml ≈ 300 g drink is

mCO 2 = mf CO 2 m m = (0.00180)(300 g) = 0.54 g

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16-74

Review Problems

16-88 The equilibrium constant of the dissociation process O2 ↔ 2O is given in Table A-28 at different temperatures. The value at a given temperature is to be verified using Gibbs function data. Analysis The KP value of a reaction at a specified temperature can be determined from the Gibbs function data using K p = e −ΔG *( T )/ Ru T or ln K p = − ΔG * (T ) / Ru T

O2 ↔ 2O

where

2000 K

ΔG * (T ) = ν O g O* (T ) −ν O 2 g O* 2 (T ) = ν O ( h − Ts ) O − ν O 2 ( h − Ts ) O 2 = ν O [(h f + h2000 − h298 ) − Ts ] O −ν O 2 [(h f + h2000 − h298 ) − Ts ] O 2 = 2 × (249,190 + 42,564 − 6852 − 2000 × 201.135) − 1× (0 + 67,881 − 8682 − 2000 × 268.655) = 243,375 kJ/kmol

Substituting,

ln K p = −(243,375 kJ/kmol)/[(8.314 kJ/kmol ⋅ K)(2000 K)] = −14.636 or K p = 4.4 × 10 −7

(Table A-28: ln KP = -14.622)

16-89 A mixture of H2 and Ar is heated is heated until 15% of H2 is dissociated. The final temperature of mixture is to be determined. Assumptions 1 The constituents of the mixture are ideal gases. 2 Ar in the mixture remains an inert gas. Analysis The stoichiometric and actual reactions can be written as

Stoichiometric:

H 2 ⇔ 2H (thus ν H 2 = 1 and ν H = 2)

Actual:

H 2 + Ar ⎯ ⎯→ 0{ .3H + 0.85H 2 + Ar { 1 424 3 inert prod

H 2 ⇔ 2H

Ar 1 atm

react.

The equilibrium constant KP can be determined from

Kp =

ν N HH ⎛ P ⎜ ⎜ ν N HH 2 ⎝ N total 2

ν H −ν H 2

⎞ ⎟ ⎟ ⎠

=

0.3 2 ⎛ 1 ⎞ ⎜ ⎟ 0.85 ⎝ 0.85 + 0.3 + 1 ⎠

2 −1

= 0.0492

From Table A-28, the temperature corresponding to this KP value is T = 3117 K.

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16-75

16-90 A mixture of H2O, O2, and N2 is heated to a high temperature at a constant pressure. The equilibrium composition is to be determined. Assumptions 1 The equilibrium composition consists of H2O, O2, N2 and H2. 2 The constituents of the mixture are ideal gases. Analysis The stoichiometric and actual reactions in this case are

Stoichiometric:

H 2 O ⇔ H 2 + 12 O 2 (thus ν H 2 O = 1, ν H 2 = 1, and ν O 2 = 12 )

Actual:

H 2 O + 2O 2 + 5 N 2 ⎯ ⎯→ x H 2 O + y H 2 + z O 2 + 5 N 2 123 14243 { react.

H balance:

2 = 2x + 2 y

O balance:

5 = x + 2z ⎯ ⎯→ z = 2.5 − 0.5 x

Total number of moles:

⎯ ⎯→

products

inert

y = 1− x

1 H2O 2 O2 5 N2 2200 K 5 atm

N total = x + y + z + 5 = 8.5 − 0.5 x

The equilibrium constant relation can be expressed as ν

Kp =

ν

N HH 2 N OO 2 ⎛ P ⎞ 2 2 ⎜ ⎟ ν H 2O ⎜ ⎟ N H O ⎝ N total ⎠ 2

(ν H 2 −ν O 2 −ν H 2 O )

=

1+ 0.5 −1

z 0.5 ⎛ P ⎞ ⎜ ⎟ ⎜N ⎟ x ⎝ total ⎠

y

From Table A-28, lnKP = -6.768 at 2200 K. Thus KP = 0.00115. Substituting,

(1 − x)(1.5 − 0.5 x) 0.5 ⎛ 5 ⎞ 0.00115 = ⎜ ⎟ x ⎝ 8.5 − 0.5 x ⎠

0.5

Solving for x, x = 0.9981 Then, y = 1 - x = 0.0019 z = 2.5 - 0.5x = 2.00095 Therefore, the equilibrium composition of the mixture at 2200 K and 5 atm is 0.9981H 2 O + 0.0019H 2 + 2.00095O 2 + 5N 2

The equilibrium constant for the reaction H 2 O ⇔ OH + 21 H 2 is lnKP = -7.148, which is very close to the KP value of the reaction considered. Therefore, it is not realistic to assume that no OH will be present in equilibrium mixture.

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16-76

16-91 [Also solved by EES on enclosed CD] Methane gas is burned with stoichiometric amount of air during a combustion process. The equilibrium composition and the exit temperature are to be determined. Assumptions 1 The product gases consist of CO2, H2O, CO, N2, and O2. 2 The constituents of the mixture are ideal gases. 3 This is an adiabatic and steady-flow combustion process. Analysis (a) The combustion equation of CH4 with stoichiometric amount of O2 can be written as CH 4 + 2( O 2 + 3.76 N 2 )

⎯ ⎯→

xCO 2 + (1 − x )CO + (0.5 − 0.5x )O 2 + 2H 2O + 7.52N 2

After combustion, there will be no CH4 present in the combustion chamber, and H2O will act like an inert gas. The equilibrium equation among CO2, CO, and O2 can be expressed as CO 2 ⇔ CO + 12 O 2 (thus ν CO 2 = 1, ν CO = 1, and ν O 2 = 12 )

and

CH4 ν

CO N νCO N OO2 2 ⎛ P ⎞ ⎜ ⎟ Kp = ν ⎜ ⎟ N CO 2 ⎝ N total ⎠

(ν CO +ν O 2 −ν CO 2 )

25°C

CO 2

Air

where N total = x + (1 − x ) + (15 . − 0.5 x ) + 2 + 7.52 = 12.02 − 0.5x

CO CO2 H2O O2 N2

Combustion chamber 1 atm

25°C

Substituting, 1.5−1

Kp =

(1 − x)(0.5 − 0.5 x)0.5 ⎛ 1 ⎞ ⎜ ⎟ x ⎝ 12.02 − 0.5 x ⎠

The value of KP depends on temperature of the products, which is yet to be determined. A second relation to determine KP and x is obtained from the steady-flow energy balance expressed as 0=

∑ N (h P

o f

+h −ho

) − ∑ N (h R

P

o f

+h −ho

)

R

⎯ ⎯→ 0 =

∑ N (h P

o f

+h −ho

) −∑ N P

o Rhf R

since the combustion is adiabatic and the reactants enter the combustion chamber at 25°C. Assuming the air and the combustion products to be ideal gases, we have h = h (T). From the tables,

hfo

h 298 K

kJ/kmol

kJ/kmol

CH4(g)

-74,850

--

N2

0

8669

O2

0

8682

H2O(g)

-241,820

9904

CO

-110,530

8669

CO2

-393,520

9364

Substance

Substituting, 0 = x ( −393,520 + hCO 2 − 9364) + (1 − x )( −110,530 + hCO − 8669) + 2( −241,820 + hH 2 O − 9904) + ( 0.5 − 0.5x )( 0 + hO 2 − 8682) + 7.52(0 + hN 2 − 8669) − 1( −74,850 + h298 − h298 ) − 0 − 0

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16-77

which yields x hCO 2 + (1 − x )hCO + 2 hH 2O + ( 0.5 − 0.5 x )hO 2 + 7.52 hN 2 − 279,344 x = 617,329

Now we have two equations with two unknowns, TP and x. The solution is obtained by trial and error by assuming a temperature TP, calculating the equilibrium composition from the first equation, and then checking to see if the second equation is satisfied. A first guess is obtained by assuming there is no CO in the products, i.e., x = 1. It yields TP = 2328 K. The adiabatic combustion temperature with incomplete combustion will be less. Take Tp = 2300 K

⎯ ⎯→

ln K p = −4.49

Take Tp = 2250 K

⎯ ⎯→

ln K p = −4.805

⎯ ⎯→ ⎯ ⎯→

x = 0.870

⎯ ⎯→

RHS = 641,093

x = 0.893

⎯ ⎯→

RHS = 612,755

By interpolation,

Tp = 2258 K and x = 0.889 Thus the composition of the equilibrium mixture is 0.889CO 2 + 0.111CO + 0.0555O 2 + 2H 2 O + 7.52N 2

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16-78

16-92 EES Problem 16-91 is reconsidered. The effect of excess air on the equilibrium composition and the exit temperature by varying the percent excess air from 0 to 200 percent is to be studied. Analysis The problem is solved using EES, and the solution is given below. "Often, for nonlinear problems such as this one, good gusses are required to start the solution. First, run the program with zero percent excess air to determine the net heat transfer as a function of T_prod. Just press F3 or click on the Solve Table icon. From Plot Window 1, where Q_net is plotted vs T_prod, determnine the value of T_prod for Q_net=0 by holding down the Shift key and move the cross hairs by moving the mouse. Q_net is approximately zero at T_prod = 2269 K. From Plot Window 2 at T_prod = 2269 K, a, b, and c are approximately 0.89, 0.10, and 0.056, respectively." "For EES to calculate a, b, c, and T_prod directly for the adiabatic case, remove the '{ }' in the last line of this window to set Q_net = 0.0. Then from the Options menu select Variable Info and set the Guess Values of a, b, c, and T_prod to the guess values selected from the Plot Windows. Then press F2 or click on the Calculator icon." "Input Data" {PercentEx = 0} Ex = PercentEX/100 P_prod =101.3 [kPa] R_u=8.314 [kJ/kmol-K] T_fuel=298 [K] T_air=298 [K] "The combustion equation of CH4 with stoichiometric amount of air is CH4 + (1+Ex)(2)(O2 + 3.76N2)=CO2 +2H2O+(1+Ex)(2)(3.76)N2" "For the incomplete combustion process in this problem, the combustion equation is CH4 + (1+Ex)(2)(O2 + 3.76N2)=aCO2 +bCO + cO2+2H2O+(1+Ex)(2)(3.76)N2" "Specie balance equations" "O" 4=a *2+b +c *2+2 "C" 1=a +b N_tot =a +b +c +2+(1+Ex)*(2)*3.76 "Total kilomoles of products at equilibrium" "We assume the equilibrium reaction is CO2=CO+0.5O2" "The following equations provide the specific Gibbs function (g=h-Ts) for each component as a function of its temperature at 1 atm pressure, 101.3 kPa" g_CO2=Enthalpy(CO2,T=T_prod )-T_prod *Entropy(CO2,T=T_prod ,P=101.3) g_CO=Enthalpy(CO,T=T_prod )-T_prod *Entropy(CO,T=T_prod ,P=101.3) g_O2=Enthalpy(O2,T=T_prod )-T_prod *Entropy(O2,T=T_prod ,P=101.3) "The standard-state Gibbs function is" DELTAG =1*g_CO+0.5*g_O2-1*g_CO2 "The equilibrium constant is given by Eq. 16-14." K_P = exp(-DELTAG /(R_u*T_prod )) P=P_prod /101.3"atm" "The equilibrium constant is also given by Eq. 16-15." "K_ P = (P/N_tot)^(1+0.5-1)*(b^1*c^0.5)/(a^1)" sqrt(P/N_tot )*b *sqrt(c )=K_P *a "Conservation of energy for the reaction, assuming SSSF, neglecting work , ke, and pe:" E_in - E_out = DELTAE_cv E_in = Q_net + HR "The enthalpy of the reactant gases is" HR=enthalpy(CH4,T=T_fuel)+ (1+Ex)*(2) *enthalpy(O2,T=T_air)+(1+Ex)*(2)*3.76 *enthalpy(N2,T=T_air) E_out = HP "The enthalpy of the product gases is"

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16-79

HP=a *enthalpy(CO2,T=T_prod )+b *enthalpy(CO,T=T_prod ) +2*enthalpy(H2O,T=T_prod )+(1+Ex)*(2)*3.76*enthalpy(N2,T=T_prod ) + c *enthalpy(O2,T=T_prod ) DELTAE_cv = 0 "Steady-flow requirement" Q_net=0 "For an adiabatic reaction the net heat added is zero." Tprod [K] 2260 2091 1940 1809 1695 1597 1511 1437 1370 1312 1259

PercentEx 0 20 40 60 80 100 120 140 160 180 200 2400

Tprod [K]

2200 2000 1800 1600 1400 1200 0

40

80

120

160

200

Percent Excess Air [%] Coefficients for CO2, CO, and O2 vs Tprod

Coefficients: a, b, c

1.10 0.90

a CO2 b CO c O2

0.70 0.50 0.30 0.10 -0.10 1200

1400

1600

1800

2000

2200

2400

2600

Tprod, K

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16-80

16-93 The equilibrium constant for the reaction CH4 + 2O2 ⇔ CO2 + 2H2O at 100 kPa and 2000 K is to be determined. Assumptions 1 The constituents of the mixture are ideal gases. Analysis This is a simultaneous reaction. We can begin with the dissociation of methane and carbon dioxide, CH 4 ⇔ C + 2H 2

K P = e −7.847

C + O 2 ⇔ CO 2

K P = e 23.839

When these two reactions are summed and the common carbon term cancelled, the result is CH 4 + O 2 ⇔ CO 2 + 2H 2

K P = e ( 23.839 − 7.847 ) = e15.992

CH4+2O2 ⇔ CO2+2H2O 2000 K 100 kPa

Next, we include the water dissociation reaction (Table A-28), 2H 2 + O 2 ⇔ 2H 2 O

K P = e 2(8.145) = e16.29

which when summed with the previous reaction and the common hydrogen term is cancelled yields CH 4 + 2O 2 ⇔ CO 2 + 2H 2 O

K P = e15.992 +16.29 = e 32.282

Then, ln K P = 32.282

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16-81

16-94 The equilibrium mole fraction of the water vapor for the reaction CH4 + 2O2 ⇔ CO2 + 2H2O at 100 kPa and 2000 K is to be determined. Assumptions 1 The equilibrium composition consists of CH4, O2, CO2, and H2O. 2 The constituents of the mixture are ideal gases. Analysis This is a simultaneous reaction. We can begin with the dissociation of methane and carbon dioxide, CH 4 ⇔ C + 2H 2

K P = e −7.847

C + O 2 ⇔ CO 2

K P = e 23.839

When these two reactions are summed and the common carbon term cancelled, the result is K P = e ( 23.839 − 7.847 ) = e15.992

CH 4 + O 2 ⇔ CO 2 + 2H 2

Next, we include the water dissociation reaction, 2H 2 + O 2 ⇔ 2H 2 O

CH4+2O2 ⇔ CO2+2H2O 2000 K 100 kPa

K P = e 2(8.145) = e16.29

which when summed with the previous reaction and the common hydrogen term is cancelled yields CH 4 + 2O 2 ⇔ CO 2 + 2H 2 O

K P = e15.992 +16.29 = e 32.282

Then, ln K P = 32.282 CH 4 + 2O 2 ⎯ ⎯→ xCH 4 + yO 2 + zCO 2 + mH 2 O 14 4244 3 1442443

Actual reeaction:

react.

C balance:

1= x+ z ⎯ ⎯→ z = 1 − x

H balance:

4 = 4 x + 2m ⎯ ⎯→ m = 2 − 2 x

O balance:

4 = 2 y + 2z + m ⎯ ⎯→ y = 2 x

Total number of moles:

N total = x + y + z + m = 3

products

The equilibrium constant relation can be expressed as ν

Kp =

ν

CO2 H2O N CO2 N H2O ⎛ P ⎜ ν CH4 ν O2 ⎜ N N CH4 N O2 ⎝ total

ν CO2 +ν H2O −ν CH4 −ν O2

⎞ ⎟ ⎟ ⎠

Substituting, e 32.282 =

(1 − x)(2 − 2 x) 2 ⎛ 100 / 101.325 ⎞ ⎟ ⎜ 3 x(2 x) 2 ⎠ ⎝

1+ 2 −1− 2

Solving for x, x = 0.00002122 Then, y = 2x = 0.00004244 z = 1 − x = 0.99997878 m = 2 − 2x = 1.99995756 Therefore, the equilibrium composition of the mixture at 2000 K and 100 kPa is 0.00002122 CH 4 + 0.00004244 O 2 + 0.99997878 CO 2 + 1.99995756 H 2 O The mole fraction of water vapor is then 1.99995756 y H2O = = 0.6667 3

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16-82

16-95 The equilibrium partial pressure of the carbon dioxide for the reaction CH4 + 2O2 ⇔ CO2 + 2H2O at 700 kPa and 3000 K is to be determined. Assumptions 1 The equilibrium composition consists of CH4, O2, CO2, and H2O. 2 The constituents of the mixture are ideal gases. Analysis This is a simultaneous reaction. We can begin with the dissociation of methane and carbon dioxide, CH 4 ⇔ C + 2H 2

K P = e −9.685

C + O 2 ⇔ CO 2

K P = e15.869

When these two reactions are summed and the common carbon term cancelled, the result is K P = e (15.869 −9.685) = e 6.184

CH 4 + O 2 ⇔ CO 2 + 2H 2

Next, we include the water dissociation reaction, 2H 2 + O 2 ⇔ 2H 2 O

CH4+2O2 ⇔ CO2+2H2O 3000 K 700 kPa

K P = e 2(3.086) = e 6.172

which when summed with the previous reaction and the common hydrogen term is cancelled yields CH 4 + 2O 2 ⇔ CO 2 + 2H 2 O

K P = e 6.184 + 6.172 = e 12.356

Then, ln K P = 12.356 CH 4 + 2O 2 ⎯ ⎯→ xCH 4 + yO 2 + zCO 2 + mH 2 O 14 4244 3 1442443

Actual reeaction:

react.

C balance:

1= x+ z ⎯ ⎯→ z = 1 − x

H balance:

4 = 4 x + 2m ⎯ ⎯→ m = 2 − 2 x

O balance:

4 = 2 y + 2z + m ⎯ ⎯→ y = 2 x

Total number of moles:

N total = x + y + z + m = 3

products

The equilibrium constant relation can be expressed as ν

Kp =

ν

CO2 H2O N CO2 N H2O ⎛ P ⎜ ν CH4 ν O2 ⎜ N N CH4 N O2 ⎝ total

ν CO2 +ν H2O −ν CH4 −ν O2

⎞ ⎟ ⎟ ⎠

Substituting, e12.356 =

(1 − x)(2 − 2 x) 2 ⎛ 700 / 101.325 ⎞ ⎜ ⎟ 3 x(2 x) 2 ⎠ ⎝

1+ 2 −1− 2

Solving for x, x = 0.01601 Then, y = 2x = 0.03202 z = 1 − x = 0.98399 m = 2 − 2x = 1.96798 Therefore, the equilibrium composition of the mixture at 3000 K and 700 kPa is 0.01601 CH 4 + 0.03202 O 2 + 0.98399 CO 2 + 1.96798 H 2 O The mole fraction of carbon dioxide is 0.98399 y CO2 = = 0.3280 3 and the partial pressure of the carbon dioxide in the product mixture is PCO2 = y CO2 P = (0.3280)(700 kPa ) = 230 kPa

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16-83

16-96 Methane is heated from a specified state to another state. The amount of heat required is to be determined without and with dissociation cases. Properties The molar mass and gas constant of methane are 16.043 kg/kmol and 0.5182 kJ/kg⋅K (Table A1). Assumptions 1 The equilibrium composition consists of O2 and O. 2 The constituents of the mixture are ideal gases. Analysis (a) An energy balance for the process gives E −E 1in424out 3

Net energy transfer by heat, work, and mass

=

ΔE system 1 424 3

Change in internal, kinetic, potential, etc. energies

Qin = N (u 2 − u1 )

[

= N h2 − h1 − Ru (T2 − T1 )

]

Using the empirical coefficients of Table A-2c, 2

∫

h2 − h1 = c p dT = a (T2 − T1 ) + 1

= 19.89(1000 − 298) +

b 2 c d (T2 − T12 ) + (T23 − T13 ) + (T24 − T14 ) 2 3 4

0.05024 1.269 × 10 −5 (1000 2 − 298 2 ) + (1000 3 − 298 3 ) 2 3

− 11.01× 10 −9 (1000 4 − 298 4 ) 4 = 38,239 kJ/kmol +

Substituting, Qin = (10 kmol)[38,239 kJ/kmol − (8.314 kJ/kmol ⋅ K)(1000 − 298)K ] = 324,000 kJ

(b) The stoichiometric and actual reactions in this case are CH 4 ⇔ C + 2H 2

Stoichiometric:

(thus ν CH4 = 1, ν C = 1 and ν H2 = 2)

CH 4 ⎯ ⎯→ xCH 4 + yC + zH 2 123 1424 3

Actual:

react.

products

C balance:

1= x+ y ⎯ ⎯→ y = 1 − x

H balance:

4 = 4x + 2z ⎯ ⎯→ z = 2 − 2 x

Total number of moles:

N total = x + y + z = 3 − 2 x

CH4 1000 K 1 atm

The equilibrium constant relation can be expressed as ν

N C N ν H2 ⎛ P K p = C ν H2 ⎜⎜ CH4 N CH4 ⎝ N total

ν C +ν H2 −ν CH4

⎞ ⎟ ⎟ ⎠

From the problem statement, at 1000 K, ln K p = −2.328 . Then, K P = e −2.328 = 0.09749

Substituting,

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16-84

0.09749 =

(1 − x )(2 − 2 x) 2 x

⎛ 1 ⎞ ⎜ ⎟ ⎝ 3 − 2x ⎠

1+ 2 −1

Solving for x, x = 0.6414 Then, y = 1 − x = 0.3586 z = 2 − 2x = 0.7172 Therefore, the equilibrium composition of the mixture at 1000 K and 1 atm is 0.6414 CH 4 + 0.3586 C + 0.7172 H 2

The mole fractions are y CH4 =

N CH4 0.6414 0.6414 = = = 0.3735 N total 0.6414 + 0.3586 + 0.7172 1.7172

yC =

NC 0.3586 = = 0.2088 N total 1.7172

y H2 =

N H2 0.7172 = = 0.4177 N total 1.7172

The heat transfer can be determined from Qin = N ( y CH4 cv ,CH4 T2 + y H2 cv ,H2 T2 + y C cv ,C T2 ) − Ncv ,CH4 T1

= (10)[(0.3735)(63.3)(1000) + (0.4177)(21.7)(1000) + (0.2088)(0.711)(1000)] − (10)(27.8)(298) = 245,700 kJ

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16-85

16-97 Solid carbon is burned with a stoichiometric amount of air. The number of moles of CO2 formed per mole of carbon is to be determined. Assumptions 1 The equilibrium composition consists of CO2, CO, O2, and N2. 2 The constituents of the mixture are ideal gases. Analysis Inspection of Table A-28 reveals that the dissociation equilibrium constants of CO2, O2, and N2 are quite small and therefore may be neglected. (We learned from another source that the equilibrium constant for CO is also small). The combustion is then complete and the reaction is described by

Carbon + Air 25°C

C + (O 2 + 3.76N 2 ) ⎯ ⎯→ CO 2 + 3.76 N 2

The number of moles of CO2 in the products is then N CO2 =1 NC

16-98 Solid carbon is burned with a stoichiometric amount of air. The amount of heat released per kilogram of carbon is to be determined. Assumptions 1 The equilibrium composition consists of CO2, CO, O2, and N2. 2 The constituents of the mixture are ideal gases. Analysis Inspection of Table A-28 reveals that the dissociation equilibrium constants of CO2, O2, and N2 are quite small and therefore may be neglected. (We learned from another source that the equilibrium constant for CO is also small). The combustion is then complete and the reaction is described by

Carbon + Air 25°C

C + (O 2 + 3.76N 2 ) ⎯ ⎯→ CO 2 + 3.76 N 2

The heat transfer for this combustion process is determined from the energy balance E in − E out = ΔE system applied on the combustion chamber with W = 0. It reduces to − Qout =

∑ N (h P

o f

+h −ho

) − ∑ N (h P

R

o f

+h −ho

)

R

Assuming the air and the combustion products to be ideal gases, we have h = h(T). From the tables, h fo

h298K

h1000 K

kJ/kmol

kJ/kmol

kJ/kmol

N2

0

8669

30,129

CO2

-393,520

9364

42,769

Substance

Substituting, −Qout = (1)(−393,520 + 42,769 − 9364) + (3.76)(0 + 30,129 − 8669) = −279,400 kJ/kmol C or

Qout =

279,400 kJ/kmol = 23,280 kJ/kg C 12 kg/kmol

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16-86

16-99 Methane gas is burned with 30 percent excess air. The equilibrium composition of the products of combustion and the amount of heat released by this combustion are to be determined. Assumptions 1 The equilibrium composition consists of CO2, H2O, O2, NO, and N2. 2 The constituents of the mixture are ideal gases. Analysis Inspection of the equilibrium constants of the possible reactions indicate that only the formation of NO need to be considered in addition to other complete combustion products. Then, the stoichiometric and actual reactions in this case are Stoichiometric: N 2 + O 2 ⇔ 2 NO (thus ν N2 = 1, ν O2 = 1, and ν NO = 2) CH 4 + 2.6(O 2 + 3.76 N 2 ) ⎯ ⎯→ CO 2 + 2H 2 O + xNO + yO 2 + zN 2

Actual:

9.776 = x + 2 z ⎯ ⎯→ z = 4.888 − 0.5 x

N balance:

O balance: 5.2 = 2 + 2 + x + 2 y ⎯ ⎯→ y = 0.6 − 0.5 x N total = 1 + 2 + x + y + z = 8.488 Total number of moles: The equilibrium constant relation can be expressed as (ν

ν

−ν

−ν

Since the stoichiometric reaction being considered is double this reaction, K p = exp(−2 × 5.294) = 2.522 × 10 −5

CH4 25°C 30% excess air

x2 ⎛ 1 ⎞ ⎜ ⎟ (0.6 − 0.5 x)(4.888 − 0.5 x) ⎝ 8.488 ⎠

Combustion chamber 1 atm

CO2, H2O NO, O2, N2 1600 K

25°C

Substituting, 2.522 × 10 −5 =

Qout

)

⎛ P ⎞ NO N2 O2 ⎜ ⎟ Kp = ν ⎜ ⎟ ν N N2N2 N O2O2 ⎝ N total ⎠ From Table A-28, at 1600 K, ln K p = −5.294 . NO N NO

2 −1−1

Solving for x, x = 0.008566 Then, y = 0.6 − 0.5x = 0.5957 z = 4.888 − 0.5x =4.884 Therefore, the equilibrium composition of the products mixture at 1600 K and 1 atm is CH 4 + 2.6(O 2 + 3.76N 2 ) ⎯ ⎯→ CO 2 + 2H 2 O + 0.008566NO + 0.5957O 2 + 4.884N 2 The heat transfer for this combustion process is determined from the energy balance E in − E out = ΔE system

applied on the combustion chamber with W = 0. It reduces to − Qout = N P h fo + h − h o P − N R h fo + h − h o

∑ (

) ∑ (

)

R

Assuming the air and the combustion products to be ideal gases, we have h = h(T). From the tables, h fo h298K h1600 K Substance kJ/kmol kJ/kmol kJ/kmol CH4 -74,850 ----O2 0 8682 52,961 0 8669 50,571 N2 -241,820 9904 62,748 H2O -393,520 9364 76,944 CO2 Neglecting the effect of NO in the energy balance and substituting, −Qout = (1)(−393,520 + 76,944 − 9364 ) + (2)(−241,820 + 62,748 − 9904) + 0.5957(52,961 − 8682) + (4.884)(50,571 − 8669) − (−74,850) = −472,900 kJ/kmol CH 4

or

Qout = 472,900 kJ/kmol CH 4

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16-87

16-100 Propane gas is burned with 30% excess air. The equilibrium composition of the products of combustion and the amount of heat released by this combustion are to be determined. Assumptions 1 The equilibrium composition consists of CO2, H2O, O2, NO, and N2. 2 The constituents of the mixture are ideal gases. Analysis (a) The stoichiometric and actual reactions in this case are Stoichiometric: N 2 + O 2 ⇔ 2 NO (thus ν N2 = 1, ν O2 = 1, and ν NO = 2) C 3 H 8 + 1.3 × 5(O 2 + 3.76 N 2 ) ⎯ ⎯→ 3CO 2 + 4H 2 O + xNO + yO 2 + zN 2

Actual:

48.88 = x + 2 z ⎯ ⎯→ z = 24.44 − 0.5 x

N balance:

O balance: 13 = 6 + 4 + x + 2 y ⎯ ⎯→ y = 1.5 − 0.5 x N total = 3 + 4 + x + y + z = 32.94 Total number of moles: The equilibrium constant relation can be expressed as (ν

ν

−ν

−ν

C3H8 25°C

Since the stoichiometric reaction being considered is double this reaction, K p = exp(−2 × 5.294) = 2.522 × 10 −5

30% excess air

Combustion chamber 1 atm

Products 1600 K

25°C

Substituting, 2.522 × 10 −5 =

Qout

)

⎛ P ⎞ NO N2 O2 ⎟ ⎜ Kp = ν ⎟ ⎜ ν N N2N2 N O2O2 ⎝ N total ⎠ From Table A-28, at 1600 K, ln K p = −5.294 . NO N NO

x2 ⎛ 1 ⎞ ⎜ ⎟ (1.5 − 0.5 x)(24.44 − 0.5 x) ⎝ 32.94 ⎠

2 −1−1

Solving for x, x = 0.03024 Then, y = 1.5 − 0.5x = 1.485 z = 24.44 − 0.5x = 24.19 Therefore, the equilibrium composition of the products mixture at 1600 K and 1 atm is C 3 H 8 + 6.5(O 2 + 3.76N 2 ) ⎯ ⎯→ 3CO 2 + 4H 2 O + 0.03024NO + 1.485O 2 + 24.19N 2 (b) The heat transfer for this combustion process is determined from the energy balance E in − E out = ΔE system applied on the combustion chamber with W = 0. It reduces to − Qout =

∑ N (h P

o f

+h −ho

) − ∑ N (h P

R

o f

+h −ho

)

R

Assuming the air and the combustion products to be ideal gases, we have h = h(T). From the tables, Substance C3H8 O2 N2 H2O CO2

h fo

h298K

h1600 K

kJ/kmol -103,850 0 0 -241,820 -393,520

kJ/kmol --8682 8669 9904 9364

kJ/kmol --52,961 50,571 62,748 76,944

Neglecting the effect of NO in the energy balance and substituting, −Qout = (3)(−393,520 + 76,944 − 9364 ) + (4)(−241,820 + 62,748 − 9904) + 1.485(52,961 − 8682) + (24.19)(50,571 − 8669) − (−103,850) = −654,360 kJ/kmol C 3 H 8

or

Qout =

654,360 kJ/kmol = 14,870 kJ/kg C 3 H 8 44 kg/kmol

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16-88

16-101E Gaseous octane gas is burned with 40% excess air. The equilibrium composition of the products of combustion is to be determined. Assumptions 1 The equilibrium composition consists of CO2, H2O, O2, NO, and N2. 2 The constituents of the mixture are ideal gases. Analysis The stoichiometric and actual reactions in this case are

Stoichiometric:

N 2 + O 2 ⇔ 2 NO (thus ν N2 = 1, ν O2 = 1, and ν NO = 2)

Actual:

C 8 H 18 + 1.4 × 12.5(O 2 + 3.76 N 2 ) ⎯ ⎯→ 8CO 2 + 9H 2 O + xNO + yO 2 + zN 2

N balance:

131.6 = x + 2 z ⎯ ⎯→ z = 65.8 − 0.5 x

O balance:

35 = 16 + 9 + x + 2 y ⎯ ⎯→ y = 5 − 0.5 x

Total number of moles:

N total = 8 + 9 + x + y + z = 87.8

The equilibrium constant relation can be expressed as ν

Kp =

NO N NO

ν

ν

N N2N2 N O2O2

⎛ P ⎜ ⎜N ⎝ total

⎞ ⎟ ⎟ ⎠

(ν NO −ν N2 −ν O2 )

From Table A-28, at 2000 K (3600 R), ln K p = −3.931 . Since the stoichiometric

C8H18 Combustion chamber 40% excess air

600 psia

CO2, H2O NO, O2, N2 3600 R

reaction being considered is double this reaction, K p = exp(−2 × 3.931) = 3.851× 10 −4 Substituting, 3.851× 10 − 4 =

x2 ⎛ 600 / 14.7 ⎞ ⎜ ⎟ (5 − 0.5 x)(65.8 − 0.5 x) ⎝ 87.8 ⎠

2 −1−1

Solving for x, x = 0.1119 Then, y = 5 − 0.5x = 4.944 z = 65.8 − 0.5x = 65.74 Therefore, the equilibrium composition of the products mixture at 3600 R and 600 psia is C 8 H18 + 17.5(O 2 + 3.76N 2 ) ⎯ ⎯→ 8CO 2 + 9H 2 O + 0.1119NO + 4.944O 2 + 65.74N 2

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16-89

16-102 A mixture of H2O and O2 is heated to a high temperature. The equilibrium composition is to be determined. Assumptions 1 The equilibrium composition consists of H2O, OH, O2, and H2. 2 The constituents of the mixture are ideal gases. Analysis The reaction equation during this process can be expressed as ⎯ ⎯→

2H 2 O + 3O 2

H2O, OH, H2, O2 3600 K 8 atm

x H 2 O + y H 2 + z O 2 + w OH

Mass balances for hydrogen and oxygen yield H balance:

4 = 2x + 2 y + w

(1)

O balance:

8 = x + 2z + w

(2)

The mass balances provide us with only two equations with four unknowns, and thus we need to have two more equations (to be obtained from the KP relations) to determine the equilibrium composition of the mixture. They are H 2 O ⇔ H 2 + 21 O 2

(reaction 1)

H 2 O ⇔ 21 H 2 + OH

(reaction 2)

The equilibrium constant for these two reactions at 3600 K are determined from Table A-28 to be ln K P1 = −1392 .

⎯ ⎯→

K P1 = 0.24858

ln K P 2 = −1088 .

⎯ ⎯→

K P 2 = 0.33689

The KP relations for these two simultaneous reactions are ν

K P1

ν

N HH 2 N OO 2 ⎛ P 2 2 ⎜ = ⎜ ν N HH O2 O ⎝ N total 2

ν H2

K P2 =

ν

OH N H N OH ⎛ P 2 ⎜ ⎜N ν H 2O ⎝ total NH O 2

⎞ ⎟ ⎟ ⎠

(ν H 2 +ν O 2 −ν H 2 O )

⎞ ⎟ ⎟ ⎠

(ν H 2 +ν OH −ν H 2 O )

where N total = N H 2 O + N H 2 + N O 2 + N OH = x + y + z + w

Substituting, 1/ 2

0.24858 =

( y )( z )1 / 2 x

⎞ ⎛ 8 ⎟⎟ ⎜⎜ x + y + z + w ⎠ ⎝

0.33689 =

( w)( y )1 / 2 x

⎛ ⎞ 8 ⎜⎜ ⎟⎟ x + y + z + w ⎝ ⎠

(3) 1/ 2

(4)

Solving Eqs. (1), (2), (3), and (4) simultaneously for the four unknowns x, y, z, and w yields x = 1.371

y = 0.1646

z = 2.85

w = 0.928

Therefore, the equilibrium composition becomes 1.371H 2O + 0.165H 2 + 2.85O 2 + 0.928OH

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16-90

16-103 A mixture of CO2 and O2 is heated to a high temperature. The equilibrium composition is to be determined. Assumptions 1 The equilibrium composition consists of CO2, CO, O2, and O. 2 The constituents of the mixture are ideal gases. Analysis The reaction equation during this process can be expressed as ⎯ ⎯→

3C 2 O + 3O 2

CO2, CO, O2, O 3400 K 2 atm

x CO 2 + y CO + z O 2 + w O

Mass balances for carbon and oxygen yield C balance:

3= x+ y

(1)

O balance:

12 = 2 x + y + 2 z + w

(2)

The mass balances provide us with only two equations with four unknowns, and thus we need to have two more equations (to be obtained from the KP relations) to determine the equilibrium composition of the mixture. They are CO 2 ⇔ CO + 12 O 2

(reaction 1)

O 2 ⇔ 2O

(reaction 2)

The equilibrium constant for these two reactions at 3400 K are determined from Table A-28 to be ln K P1 = 0169 .

⎯ ⎯→

K P1 = 11841 .

ln K P 2 = −1935 .

⎯ ⎯→

K P 2 = 01444 .

The KP relations for these two simultaneous reactions are ν

ν

K P1

CO N CO N OO 2 ⎛ P 2 ⎜ = ⎜N ν CO 2 ⎝ total N CO 2

νO

K P2

N ⎛ P = νO ⎜⎜ O2 N O ⎝ N total 2

⎞ ⎟ ⎟ ⎠

(ν CO +ν O 2 −ν CO 2 )

ν O −ν O 2

⎞ ⎟ ⎟ ⎠

where N total = N CO 2 + N O 2 + N CO + N O = x + y + z + w

Substituting, ( y )( z )1 / 2 1.1841 = x 0.1444 =

w2 z

⎛ ⎞ 2 ⎜⎜ ⎟⎟ ⎝ x+ y+ z + w⎠

⎛ ⎞ 2 ⎜⎜ ⎟⎟ ⎝ x + y + z + w⎠

1/ 2

(3)

2 −1

(4)

Solving Eqs. (1), (2), (3), and (4) simultaneously for the four unknowns x, y, z, and w yields x = 1.313

y = 1.687

z = 3.187

w = 1.314

Thus the equilibrium composition is 1.313CO 2 + 1.687CO + 3.187O 2 + 1.314O

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16-91

16-104 EES Problem 16-103 is reconsidered. The effect of pressure on the equilibrium composition by varying pressure from 1 atm to 10 atm is to be studied. Analysis The problem is solved using EES, and the solution is given below. "For EES to calculate a, b, c, and d at T_prod and P_prod press F2 or click on the Calculator icon. The EES results using the built in function data is not the same as the anwers provided with the problem. However, if we supply the K_P's from Table A-28 to ESS, the results are equal to the answer provided. The plot of moles CO vs. P_atm was done with the EES property data." "Input Data" P_atm = 2 [atm] P_prod =P_atm*101.3 R_u=8.314 [kJ/kmol-K] T_prod=3400 [K] P=P_atm "For the incomplete combustion process in this problem, the combustion equation is 3 CO2 + 3 O2=aCO2 +bCO + cO2+dO" "Specie balance equations" "O" 3*2+3*2=a *2+b +c *2+d*1 "C" 3*1=a*1 +b*1 N_tot =a +b +c +d "Total kilomoles of products at equilibrium" "We assume the equilibrium reactions are CO2=CO+0.5O2 O2=2O" "The following equations provide the specific Gibbs function (g=h-Ts) for each component as a function of its temperature at 1 atm pressure, 101.3 kPa" g_CO2=Enthalpy(CO2,T=T_prod )-T_prod *Entropy(CO2,T=T_prod ,P=101.3) g_CO=Enthalpy(CO,T=T_prod )-T_prod *Entropy(CO,T=T_prod ,P=101.3) g_O2=Enthalpy(O2,T=T_prod )-T_prod *Entropy(O2,T=T_prod ,P=101.3) "EES does not have a built-in property function for monatomic oxygen so we will use the JANAF procedure, found under Options/Function Info/External Procedures. The units for the JANAF procedure are kmol, K, and kJ. The values are calculated for 1 atm. The entropy must be corrected for other pressrues." Call JANAF('O',T_prod:Cp,h_O,s_O) "Units from JANAF are SI" "The entropy from JANAF is for one atmosphere and that's what we need for this approach." g_O=h_O-T_prod*s_O "The standard-state (at 1 atm) Gibbs functions are" DELTAG_1 =1*g_CO+0.5*g_O2-1*g_CO2 DELTAG_2 =2*g_O-1*g_O2 "The equilibrium constants are given by Eq. 15-14." {K_P_2=0.1444 "From Table A-28" K_P_1 = 0.8445}"From Table A-28" K_p_1 = exp(-DELTAG_1/(R_u*T_prod)) "From EES data" K_P_2 = exp(-DELTAG_2/(R_u*T_prod)) "From EES data"

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16-92

"The equilibrium constant is also given by Eq. 15-15." "Write the equilibrium constant for the following system of equations: 3 CO2 + 3 O2=aCO2 +bCO + cO2+dO CO2=CO+0.5O2 O2=2O" "K_ P_1 = (P/N_tot)^(1+0.5-1)*(b^1*c^0.5)/(a^1)" sqrt(P/N_tot )*b *sqrt(c )/a=K_P_1 "K_ P_2 = (P/N_tot)^(2-1)*(d^2)/(c^1)" P/N_tot *d^2/c =K_P_2 b [kmolCO] 1.968 1.687 1.52 1.404 1.315 1.244 1.186 1.136 1.093 1.055

Patm [atm] 1 2 3 4 5 6 7 8 9 10

2

b [kmolCO]

1.8

1.6

1.4

1.2

1 1

2

3

4

5

6

7

8

9

10

Patm [atm]

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16-93

16-105 The hR at a specified temperature is to be determined using enthalpy and Kp data. Assumptions Both the reactants and products are ideal gases. Analysis (a) The complete combustion equation of H2 can be expressed as H 2 + 12 O 2 ⇔ H 2 O

The hR of the combustion process of H 2 at 2400 K is the amount of energy released as one kmol of H2 is burned in a steady-flow combustion chamber at a temperature of 2400 K, and can be determined from hR =

∑ N (h P

o f

+h −ho

) − ∑ N (h P

R

o f

+h −ho

)

R

Assuming the H2O, H2, and O2 to be ideal gases, we have h = h (T). From the tables, h fo

h 298 K

h 2400 K

kJ/kmol

kJ/kmol

kJ/kmol

H2O

-241,820

9904

103,508

H2

0

8468

75,383

O2

0

8682

83,174

Substance

Substituting, hR = 1( −241,820 + 103,508 − 9904) − 1(0 + 75,383 − 8468) − 0.5(0 + 83,174 − 8682) = −252,377 kJ / kmol

(b) The hR value at 2400 K can be estimated by using KP values at 2200 K and 2600 K (the closest two temperatures to 2400 K for which KP data are available) from Table A-28, ln

K P 2 hR ≅ K P1 Ru

⎛1 1 ⎜⎜ − ⎝ T1 T2

4.648 − 6.768 ≅

⎞ h ⎟⎟ or ln K P 2 − ln K P1 ≅ R Ru ⎠

⎛1 1 ⎜⎜ − ⎝ T1 T2

⎞ ⎟⎟ ⎠

hR 1 ⎞ ⎛ 1 − ⎜ ⎟ 8.314 kJ/kmol ⋅ K ⎝ 2200 K 2600 K ⎠

h R ≅ -252,047 kJ/kmol

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16-94

16-106 EES Problem 16-105 is reconsidered. The effect of temperature on the enthalpy of reaction using both methods by varying the temperature from 2000 to 3000 K is to be investigated. Analysis The problem is solved using EES, and the solution is given below. "Input Data" T_prod=2400 [K] DELTAT_prod =25 [K] R_u=8.314 [kJ/kmol-K] T_prod_1 = T_prod - DELTAT_prod T_prod_2 = T_prod + DELTAT_prod "The combustion equation is 1 H2 + 0.5 O2 =>1 H2O" "The enthalpy of reaction H_bar_R using enthalpy data is:" h_bar_R_Enthalpy = HP - HR HP = 1*Enthalpy(H2O,T=T_prod ) HR = 1*Enthalpy(H2,T=T_prod ) + 0.5*Enthalpy(O2,T=T_prod ) "The enthalpy of reaction H_bar_R using enthalpy data is found using the following equilibruim data:" "The following equations provide the specific Gibbs function (g=h-Ts) for each component as a function of its temperature at 1 atm pressure, 101.3 kPa" g_H2O_1=Enthalpy(H2O,T=T_prod_1 )-T_prod_1 *Entropy(H2O,T=T_prod_1 ,P=101.3) g_H2_1=Enthalpy(H2,T=T_prod_1 )-T_prod_1 *Entropy(H2,T=T_prod_1 ,P=101.3) g_O2_1=Enthalpy(O2,T=T_prod_1 )-T_prod_1 *Entropy(O2,T=T_prod_1 ,P=101.3) g_H2O_2=Enthalpy(H2O,T=T_prod_2 )-T_prod_2 *Entropy(H2O,T=T_prod_2 ,P=101.3) g_H2_2=Enthalpy(H2,T=T_prod_2 )-T_prod_2 *Entropy(H2,T=T_prod_2 ,P=101.3) g_O2_2=Enthalpy(O2,T=T_prod_2 )-T_prod_2 *Entropy(O2,T=T_prod_2 ,P=101.3) "The standard-state (at 1 atm) Gibbs functions are" DELTAG_1 =1*g_H2O_1-0.5*g_O2_1-1*g_H2_1 DELTAG_2 =1*g_H2O_2-0.5*g_O2_2-1*g_H2_2 "The equilibrium constants are given by Eq. 15-14." K_p_1 = exp(-DELTAG_1/(R_u*T_prod_1)) "From EES data" K_P_2 = exp(-DELTAG_2/(R_u*T_prod_2)) "From EES data" "the entahlpy of reaction is estimated from the equilibrium constant K_p by using EQ 15-18 as:" ln(K_P_2/K_P_1)=h_bar_R_Kp/R_u*(1/T_prod_1 - 1/T_prod_2) PercentError = ABS((h_bar_R_enthalpy - h_bar_R_Kp)/h_bar_R_enthalpy)*Convert(, %)

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16-95

Percent Error [%] 0.0002739 0.0002333 0.000198 0.0001673 0.0001405 0.0001173 0.00009706 0.00007957 0.00006448 0.00005154 0.0000405

Tprod [K] 2000 2100 2200 2300 2400 2500 2600 2700 2800 2900 3000

hREnthalpy [kJ/kmol] -251723 -251920 -252096 -252254 -252398 -252532 -252657 -252778 -252897 -253017 -253142

hRKp [kJ/kmol] -251722 -251919 -252095 -252254 -252398 -252531 -252657 -252777 -252896 -253017 -253142

-251500

DELTATprod = 25 K -251850

hR [kJ/kmol]

Enthalpy Data Kp Data

-252200 -252550 -252900 -253250 2000

2200

2400

2600

2800

3000

Tprod [k]

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16-96

16-107 The KP value of the dissociation process O2 ⇔ 2O at a specified temperature is to be determined using the hR data and KP value at a specified temperature. Assumptions Both the reactants and products are ideal gases. Analysis The hR and KP data are related to each other by ln

K P 2 hR ⎛ 1 h ⎛1 1 ⎞ 1 ⎞ ⎜⎜ − ⎟⎟ or ln K P 2 − ln K P1 ≅ R ⎜⎜ − ⎟⎟ ≅ K P1 Ru ⎝ T1 T2 ⎠ Ru ⎝ T1 T2 ⎠

The hR of the specified reaction at 2800 K is the amount of energy released as one kmol of O2 dissociates in a steady-flow combustion chamber at a temperature of 2800 K, and can be determined from hR =

∑ N (h P

o f

+h −ho

) − ∑ N (h P

R

o f

+h −ho

)

R

Assuming the O2 and O to be ideal gases, we have h = h (T). From the tables, h fo

h 298 K

h 2800 K

kJ/kmol

kJ/kmol

kJ/kmol

O

249,190

6852

59,241

O2

0

8682

98,826

Substance

Substituting, hR = 2(249,190 + 59,241 − 6852) − 1(0 + 98,826 − 8682) = 513,014 kJ / kmol The KP value at 3000 K can be estimated from the equation above by using this hR value and the KP value at 2600 K which is ln KP1 = -7.521, ln K P 2 − (−7.521) = ln K P 2 = −4.357

513,014 kJ/kmol ⎛ 1 1 ⎞ − ⎜ ⎟ 8.314 kJ/kmol ⋅ K ⎝ 2600 K 3000 K ⎠

(Table A - 28 : ln K P 2 = −4.357)

or K P 2 = 0.0128

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16-97

16-108 It is to be shown that when the three phases of a pure substance are in equilibrium, the specific Gibbs function of each phase is the same. Analysis The total Gibbs function of the three phase mixture of a pure substance can be expressed as G = m s g s + ml g l + m g g g

where the subscripts s, l, and g indicate solid, liquid and gaseous phases. Differentiating by holding the temperature and pressure (thus the Gibbs functions, g) constant yields dG = g s dm s + g l dm l + g g dm g

From conservation of mass, dms + dml + dmg = 0

mg ⎯ ⎯→

dms = −dml − dmg

Substituting, dG = − g s (dml + dmg ) + g l dml + g g dmg

ml ms

Rearranging, dG = ( g l − g s )dml + ( g g − g s )dmg

For equilibrium, dG = 0. Also dml and dmg can be varied independently. Thus each term on the right hand side must be zero to satisfy the equilibrium criteria. It yields g l = g s and g g = g s

Combining these two conditions gives the desired result, gl = g s = g s

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16-98

16-109 It is to be shown that when the two phases of a two-component system are in equilibrium, the specific Gibbs function of each phase of each component is the same. Analysis The total Gibbs function of the two phase mixture can be expressed as G = (ml1 g l1 + mg1 g g1 ) + (ml2 g l 2 + mg 2 g g 2 )

where the subscripts l and g indicate liquid and gaseous phases. Differentiating by holding the temperature and pressure (thus the Gibbs functions) constant yields dG = g l1dml1 + g g1dmg1 + g l2 dml2 + g g 2 dmg 2

From conservation of mass, dmg1 = − dml1 and dmg 2 = − dml 2

mg1 mg2 ml 1 ml 2

Substituting, dG = ( g l1 − g g1 )dml1 + ( g l2 − g g 2 )dml 2

For equilibrium, dG = 0. Also dml1 and dml2 can be varied independently. Thus each term on the right hand side must be zero to satisfy the equilibrium criteria. Then we have g l1 = g g1 and g l2 = g g 2

which is the desired result.

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16-99

16-110 A mixture of CO and O2 contained in a tank is ignited. The final pressure in the tank and the amount of heat transfer are to be determined. Assumptions 1 The equilibrium composition consists of CO2 and O2. 2 Both the reactants and the products are ideal gases. Analysis The combustion equation can be written as CO + 3 O 2

⎯ ⎯→

∑ N (h P

o f

25°C 2 atm

CO 2 + 2.5 O 2

The heat transfer can be determined from − Qout =

CO2, CO, O2

+ h − h o − Pv

) − ∑ N (h R

P

o f

+ h − h o − Pv

)

R

Both the reactants and the products are assumed to be ideal gases, and thus all the internal energy and enthalpies depend on temperature only, and the Pv terms in this equation can be replaced by RuT. It yields − Qout =

∑ N (h P

o f

+ h500 K − h298 K − Ru T

) − ∑ N (h P

R

o f

− Ru T

)

R

since reactants are at the standard reference temperature of 25°C. From the tables,

h fo

h 298 K

h 500 K

kJ/kmol

kJ/kmol

kJ/kmol

CO

-110,530

8669

14,600

O2

0

8682

14,770

CO2

-393,520

9364

17,678

Substance

Substituting, −Qout = 1(−393,520 + 17,678 − 9364 − 8.314 × 500) + 2.5(0 + 14,770 − 8682 − 8.314 × 500) − 3(0 − 8.314 × 298) − 1(−110,530 − 8.314 × 298) = −264,095 kJ/kmol CO

or Qout = 264,095 kJ/kmol CO

The final pressure in the tank is determined from N R T P1V N T 3.5 500 K = 1 u 1 ⎯ ⎯→ P2 = 2 2 P1 = × (2 atm) = 2.94 atm P2V N 2 Ru T2 N 1T1 4 298 K

The equilibrium constant for the reaction CO + 21 O 2 ⇔ CO 2 is ln KP = 57.62, which is much greater than 7. Therefore, it is not realistic to assume that no CO will be present in equilibrium mixture.

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16-100

16-111 Using Henry’s law, it is to be shown that the dissolved gases in a liquid can be driven off by heating the liquid. Analysis Henry’s law is expressed as yi, liquid side (0) =

Pi, gas side (0) H

Henry’s constant H increases with temperature, and thus the fraction of gas i in the liquid yi,liquid side decreases. Therefore, heating a liquid will drive off the dissolved gases in a liquid.

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16-101

16-112 A 2-L bottle is filled with carbonated drink that is fully charged (saturated) with CO2 gas. The volume that the CO2 gas would occupy if it is released and stored in a container at room conditions is to be determined. Assumptions 1 The liquid drink can be treated as water. 2 Both the CO2 gas and the water vapor are ideal gases. 3 The CO2 gas is weakly soluble in water and thus Henry’s law is applicable. Properties The saturation pressure of water at 17°C is 1.938 kPa (Table A-4). Henry’s constant for CO2 dissolved in water at 17ºC (290 K) is H = 1280 bar (Table 16-2). Molar masses of CO2 and water are 44.01 and 18.015 kg/kmol, respectively (Table A-1). The gas constant of CO2 is 0.1889 kPa.m3/kg.K. Also, 1 bar = 100 kPa. Analysis In the charging station, the CO2 gas and water vapor mixture above the liquid will form a saturated mixture. Noting that the saturation pressure of water at 17°C is 1.938 kPa, the partial pressure of the CO2 gas is PCO 2 , gas side = P − Pvapor = P − Psat @ 17°C = 600 − 1.938 = 598.06 kPa = 5.9806 bar

From Henry’s law, the mole fraction of CO2 in the liquid drink is determined to be y CO 2 ,liquid side =

PCO 2 ,gas side H

=

5.9806 bar = 0.00467 1280 bar

Then the mole fraction of water in the drink becomes y water, liquid side = 1 − yCO 2 , liquid side = 1 − 0.00467 = 0.99533

The mass and mole fractions of a mixture are related to each other by wi =

mi N M Mi = i i = yi mm N m M m Mm

where the apparent molar mass of the drink (liquid water - CO2 mixture) is Mm =

∑y M i

i

= yliquid water Mwater + yCO 2 MCO 2

= 0.99533 × 18.015 + 0.00467 × 44.01 = 1814 . kg / kmol

Then the mass fraction of dissolved CO2 in liquid drink becomes wCO 2 , liquid side = yCO 2 , liquid side (0)

M CO 2 Mm

= 0.00467

44.01 = 0.0113 1814 .

Therefore, the mass of dissolved CO2 in a 2 L ≈ 2 kg drink is mCO 2 = wCO 2 mm = 0.0113(2 kg) = 0.0226 kg

Then the volume occupied by this CO2 at the room conditions of 20°C and 100 kPa becomes

V =

mRT (0.0226 kg)(0.1889 kPa ⋅ m 3 / kg ⋅ K )(293 K) = = 0.0125 m 3 = 12.5 L 100 kPa P

Discussion Note that the amount of dissolved CO2 in a 2-L pressurized drink is large enough to fill 6 such bottles at room temperature and pressure. Also, we could simplify the calculations by assuming the molar mass of carbonated drink to be the same as that of water, and take it to be 18 kg/kmol because of the very low mole fraction of CO2 in the drink.

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16-102

16-113 EES Ethyl alcohol C2H5OH (gas) is burned in a steady-flow adiabatic combustion chamber with 40 percent excess air. The adiabatic flame temperature of the products is to be determined and the adiabatic flame temperature as a function of the percent excess air is to be plotted. Analysis The complete combustion reaction in this case can be written as C 2 H 5 OH (gas) + (1 + Ex)a th [O 2 + 3.76N 2 ] ⎯ ⎯→ 2 CO 2 + 3 H 2 O + ( Ex)(a th ) O 2 + f N 2

where ath is the stoichiometric coefficient for air. The oxygen balance gives 1 + (1 + Ex)a th × 2 = 2 × 2 + 3 × 1 + ( Ex)(a th ) × 2

The reaction equation with products in equilibrium is C 2 H 5 OH (gas) + (1 + Ex) a th [O 2 + 3.76N 2 ] ⎯ ⎯→ a CO 2 + b CO + d H 2 O + e O 2 + f N 2

The coefficients are determined from the mass balances Carbon balance:

2 = a+b

Hydrogen balance:

6 = 2d ⎯ ⎯→ d = 3

Oxygen balance:

1 + (1 + Ex)a th × 2 = a × 2 + b + d + e × 2

Nitrogen balance: (1 + Ex)a th × 3.76 = f Solving the above equations, we find the coefficients to be Ex = 0.4, ath = 3, a = 1.995, b = 0.004938, d = 3, e = 1.202, f = 15.79 Then, we write the balanced reaction equation as C 2 H 5 OH (gas) + 4.2[O 2 + 3.76N 2 ] ⎯ ⎯→ 1.995 CO 2 + 0.004938 CO + 3 H 2 O + 1.202 O 2 + 15.79 N 2

Total moles of products at equilibrium are N tot = 1.995 + 0.004938 + 3 + 1.202 + 15.79 = 21.99

The assumed equilibrium reaction is CO 2 ←⎯→ CO + 0.5O 2

The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data using K p = e −ΔG*( T )/ Ru T or ln K p = − ΔG * (T ) / Ru T

where ∗ ∗ ∗ ΔG * (T ) = ν CO g CO (Tprod ) + ν O2 g O2 (Tprod ) −ν CO2 g CO2 (Tprod )

and the Gibbs functions are defined as ∗ g CO (Tprod ) = ( h − Tprod s ) CO ∗ g O2 (Tprod ) = ( h − Tprod s ) O2 ∗ g CO2 (Tprod ) = ( h − Tprod s ) CO2

The equilibrium constant is also given by Kp =

be 0.5 a

⎛ P ⎜ ⎜N ⎝ tot

⎞ ⎟ ⎟ ⎠

1+ 0.5 −1

=

(0.004938)(1.202) 0.5 ⎛ 1 ⎞ ⎜ ⎟ 1.995 ⎝ 21.99 ⎠

0.5

= 0.0005787

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16-103

A steady flow energy balance gives HR = HP

where H R = h fo fuel@25°C + 4.2hO2@25°C + 15.79h N2@25°C = (−235,310 kJ/kmol) + 4.2(0) + 15.79(0) = −235,310 kJ/kmol H P = 1.995hCO2@Tprod + 0.004938hCO@Tprod + 3hH2O@Tprod + 1.202hO2@Tprod + 15.79h N2@Tprod

Solving the energy balance equation using EES, we obtain the adiabatic flame temperature to be Tprod = 1907 K

The copy of entire EES solution including parametric studies is given next: "The product temperature isT_prod" "The reactant temperature is:" T_reac= 25+273.15 "[K]" "For adiabatic combustion of 1 kmol of fuel: " Q_out = 0 "[kJ]" PercentEx = 40 "Percent excess air" Ex = PercentEx/100 "EX = % Excess air/100" P_prod =101.3"[kPa]" R_u=8.314 "[kJ/kmol-K]" "The complete combustion reaction equation for excess air is:" "C2H5OH(gas)+ (1+Ex)*A_th (O2 +3.76N2)=2 CO2 + 3 H2O +Ex*A_th O2 + f N2" "Oxygen Balance for complete combustion:" 1 + (1+Ex)*A_th*2=2*2+3*1 + Ex*A_th*2 "The reaction equation for excess air and products in equilibrium is:" "C2H5OH(gas)+ (1+Ex)*A_th (O2 +3.76N2)=a CO2 + b CO+ d H2O + e O2 + f N2" "Carbon Balance:" 2=a + b "Hydrogen Balance:" 6=2*d "Oxygen Balance:" 1 + (1+Ex)*A_th*2=a*2+b + d + e*2 "Nitrogen Balance:" (1+Ex)*A_th*3.76 = f N_tot =a +b + d + e + f "Total kilomoles of products at equilibrium" "The assumed equilibrium reaction is CO2=CO+0.5O2" "The following equations provide the specific Gibbs function (g=h-Ts) for each component in the product gases as a function of its temperature, T_prod, at 1 atm pressure, 101.3 kPa" g_CO2=Enthalpy(CO2,T=T_prod )-T_prod *Entropy(CO2,T=T_prod ,P=101.3) g_CO=Enthalpy(CO,T=T_prod )-T_prod *Entropy(CO,T=T_prod ,P=101.3) g_O2=Enthalpy(O2,T=T_prod )-T_prod *Entropy(O2,T=T_prod ,P=101.3) "The standard-state Gibbs function is" DELTAG =1*g_CO+0.5*g_O2-1*g_CO2 "The equilibrium constant is given by Eq. 15-14." K_P = exp(-DELTAG /(R_u*T_prod )) P=P_prod /101.3"atm" "The equilibrium constant is also given by Eq. 15-15." PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-104

"K_ P = (P/N_tot)^(1+0.5-1)*(b^1*e^0.5)/(a^1)" sqrt(P/N_tot )*b *sqrt(e )=K_P *a "The steady-flow energy balance is:" H_R = Q_out+H_P h_bar_f_C2H5OHgas=-235310 "[kJ/kmol]" H_R=1*(h_bar_f_C2H5OHgas ) +(1+Ex)*A_th*ENTHALPY(O2,T=T_reac)+(1+Ex)*A_th*3.76*ENTHALPY(N2,T=T_reac) "[kJ/kmol]" H_P=a*ENTHALPY(CO2,T=T_prod)+b*ENTHALPY(CO,T=T_prod)+d*ENTHALPY(H2O,T=T_prod) +e*ENTHALPY(O2,T=T_prod)+f*ENTHALPY(N2,T=T_prod) "[kJ/kmol]"

a

ath

b

d

e

f

1.922 1.97 1.988 1.995 1.998 1.999 2 2 2 2

3 3 3 3 3 3 3 3 3 3

0.07809 0.03017 0.01201 0.004933 0.002089 0.0009089 0.000405 0.0001843 0.0000855 0.00004036

3 3 3 3 3 3 3 3 3 3

0.339 0.6151 0.906 1.202 1.501 1.8 2.1 2.4 2.7 3

12.41 13.54 14.66 15.79 16.92 18.05 19.18 20.3 21.43 22.56

PercentEx [%] 10 20 30 40 50 60 70 80 90 100

Tprod [K] 2191 2093 1996 1907 1826 1752 1685 1625 1569 1518

2200 2100

Tprod (K)

2000 1900 1800 1700 1600 1500 10

20

30

40

50

60

70

80

90

100

PercentEx

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16-105

16-114 EES The percent theoretical air required for the combustion of octane such that the volume fraction of CO in the products is less than 0.1% and the heat transfer are to be determined. Also, the percent theoretical air required for 0.1% CO in the products as a function of product pressure is to be plotted. Analysis The complete combustion reaction equation for excess air is C 8 H18 + Pth a th [O 2 + 3.76N 2 ] ⎯ ⎯→ 8 CO 2 + 9 H 2 O + ( Pth − 1)a th O 2 + f N 2

The oxygen balance is Pth a th × 2 = 8 × 2 + 9 × 1 + ( Pth − 1)a th × 2

The reaction equation for excess air and products in equilibrium is C 8 H18 + Pth a th [O 2 + 3.76N 2 ] ⎯ ⎯→ a CO 2 + b CO + d H 2 O + e O 2 + f N 2

The coefficients are to be determined from the mass balances Carbon balance:

8=a+b

Hydrogen balance:

18 = 2d ⎯ ⎯→ d = 9

Oxygen balance:

Pth a th × 2 = a × 2 + b + d + e × 2

Nitrogen balance:

Pth a th × 3.76 = f

Volume fraction of CO must be less than 0.1%. That is, y CO =

b b = = 0.001 N tot a + b + d + e + f

The assumed equilibrium reaction is CO 2 ←⎯→ CO + 0.5O 2

The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data: ∗ g CO (Tprod ) = (h − Tprod s ) CO = (−53,826) − (2000)(258.48) = −570,781 kJ/kmol ∗ g O2 (Tprod ) = (h − Tprod s ) O2 = (59,193) − (2000)(268.53) = −477,876 kJ/kmol ∗ g CO2 (Tprod ) = (h − Tprod s ) CO2 = (−302,128) − (2000)(309.00) = −920,121 kJ/kmol

The enthalpies at 2000 K and entropies at 2000 K and 101.3 kPa are obtained from EES. Substituting, ∗ ∗ ∗ ΔG * (Tprod ) = ν CO g CO (Tprod ) + ν O2 g O2 (Tprod ) −ν CO2 g CO2 (Tprod )

= 1(−570,781) + 0.5(−477,876) − (−920,121) = 110,402 kJ/kmol ⎛ − ΔG * (Tprod ) ⎞ ⎟ = exp⎛⎜ − 110,402 ⎞⎟ = 0.001308 K p = exp⎜ ⎜ (8.314)(2000) ⎟ ⎜ Ru Tprod ⎟ ⎝ ⎠ ⎝ ⎠

The equilibrium constant is also given by be 0.5 Kp = a

⎛ P ⎜ ⎜N ⎝ tot

⎞ ⎟ ⎟ ⎠

1+ 0.5 −1

be 0.5 = a

⎛ Pprod / 101.3 ⎜ ⎜ a+b+d +e+ f ⎝

⎞ ⎟ ⎟ ⎠

1+ 0.5 −1

The steady flow energy balance gives H R = Qout + H P

where PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-106

H R = 1hC8H18 @ 298 K + Pth a th hO2 @ 298 K + ( Pth a th × 3.76) h N2 @ 298 K = ( −208,459) + Pth a th (0) + ( Pth a th × 3.76)(0) = −208,459 kJ/kmol H P = ahCO2 @ 2000 K + bhCO @ 2000 K + dhH2O @ 2000 K + ehO2 @ 2000 K + f h N2 @ 2000 K = a( −302,128) + b( −53,826) + d (−169,171) + e(59,193) + f (56,115)

The enthalpies are obtained from EES. Solving all the equations simultaneously using EES, we obtain Pth = 1.024, a th = 12.5, a = 7.935, b = 0.06544, d = 9, e = 0.3289, f = 48.11 PercentTh = Pth × 100 = 1.024 × 100 = 102.4% Qout = 995,500 kJ/kmol C 8 H18

The copy of entire EES solution including parametric studies is given next: "The product temperature is:" T_prod = 2000 "[K]" "The reactant temperature is:" T_reac= 25+273 "[K]" "PercentTH is Percent theoretical air" Pth= PercentTh/100 "Pth = % theoretical air/100" P_prod = 5 "[atm]" *convert(atm,kPa)"[kPa]" R_u=8.314 "[kJ/kmol-K]" "The complete combustion reaction equation for excess air is:" "C8H18+ Pth*A_th (O2 +3.76N2)=8 CO2 + 9 H2O +(Pth-1)*A_th O2 + f N2" "Oxygen Balance for complete combustion:" Pth*A_th*2=8*2+9*1 + (Pth-1)*A_th*2 "The reaction equation for excess air and products in equilibrium is:" "C8H18+ Pth*A_th (O2 +3.76N2)=a CO2 + b CO+ d H2O + e O2 + f N2" "Carbon Balance:" 8=a + b "Hydrogen Balance:" 18=2*d "Oxygen Balance:" Pth*A_th*2=a*2+b + d + e*2 "Nitrogen Balance:" Pth*A_th*3.76 = f N_tot =a +b + d + e + f "Total kilomoles of products at equilibrium" "The volume faction of CO in the products is to be less than 0.1%. For ideal gas mixtures volume fractions equal mole fractions." "The mole fraction of CO in the product gases is:" y_CO = 0.001 y_CO = b/N_tot "The assumed equilibrium reaction is CO2=CO+0.5O2" "The following equations provide the specific Gibbs function (g=h-Ts) for each component in the product gases as a function of its temperature, T_prod, at 1 atm pressure, 101.3 kPa" g_CO2=Enthalpy(CO2,T=T_prod )-T_prod *Entropy(CO2,T=T_prod ,P=101.3) g_CO=Enthalpy(CO,T=T_prod )-T_prod *Entropy(CO,T=T_prod ,P=101.3) g_O2=Enthalpy(O2,T=T_prod )-T_prod *Entropy(O2,T=T_prod ,P=101.3) "The standard-state Gibbs function is" DELTAG =1*g_CO+0.5*g_O2-1*g_CO2 "The equilibrium constant is given by Eq. 15-14." K_P = exp(-DELTAG /(R_u*T_prod ))

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16-107

P=P_prod /101.3"atm" "The equilibrium constant is also given by Eq. 15-15." "K_ P = (P/N_tot)^(1+0.5-1)*(b^1*e^0.5)/(a^1)" sqrt(P/N_tot )*b *sqrt(e )=K_P *a "The steady-flow energy balance is:" H_R = Q_out+H_P H_R=1*ENTHALPY(C8H18,T=T_reac)+Pth*A_th*ENTHALPY(O2,T=T_reac)+Pth*A_th*3.76*EN THALPY(N2,T=T_reac) "[kJ/kmol]" H_P=a*ENTHALPY(CO2,T=T_prod)+b*ENTHALPY(CO,T=T_prod)+d*ENTHALPY(H2O,T=T_pro d) +e*ENTHALPY(O2,T=T_prod)+f*ENTHALPY(N2,T=T_prod) "[kJ/kmol]" Pprod [kPa] 100 300 500 700 900 1100 1300 1500 1700 1900 2100 2300

PercentTh [%] 112 104.1 102.4 101.7 101.2 101 100.8 100.6 100.5 100.5 100.4 100.3

112

PercentTh %

110 108 106 104 102 100 0

500

1000

1500

2000

2500

Pprod [kPa]

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16-108

Fundamentals of Engineering (FE) Exam Problems

16-115 If the equilibrium constant for the reaction H2 + ½O2 → H2O is K, the equilibrium constant for the reaction 2H2O → 2H2 + O2 at the same temperature is

(a) 1/K

(b) 1/(2K)

(c) 2K

(d) K2

(e) 1/K2

Answer (e) 1/K2

16-116 If the equilibrium constant for the reaction CO + ½O2 → CO2 is K, the equilibrium constant for the reaction CO2 + 3N2 → CO + ½O2 + 3N2 at the same temperature is

(a) 1/K

(b) 1/(K + 3)

(c) 4K

(d) K

(e) 1/K2

Answer (a) 1/K

16-117 The equilibrium constant for the reaction H2 + ½O2 → H2O at 1 atm and 1500°C is given to be K. Of the reactions given below, all at 1500°C, the reaction that has a different equilibrium constant is

(a) H2 + ½O2 → H2O at 5 atm, (b) 2H2 + O2 → 2H2O at 1 atm, (c) H2 + O2 → H2O+ ½O2 at 2 atm, (d) H2 + ½O2 + 3N2 → H2O+ 3N2 at 5 atm, (e) H2 + ½O2 + 3N2 → H2O+ 3N2 at 1 atm, Answer (b) 2H2 + O2 → 2H2O at 1 atm,

16-118 Of the reactions given below, the reaction whose equilibrium composition at a specified temperature is not affected by pressure is

(a) H2 + ½O2 → H2O (b) CO + ½O2 → CO2 (c) N2 + O2 → 2NO (d) N2 → 2N (e) all of the above. Answer (c) N2 + O2 → 2NO

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16-109

16-119 Of the reactions given below, the reaction whose number of moles of products increases by the addition of inert gases into the reaction chamber at constant pressure and temperature is

(a) H2 + ½O2 → H2O (b) CO + ½O2 → CO2 (c) N2 + O2 → 2NO (d) N2 → 2N (e) none of the above. Answer (d) N2 → 2N

16-120 Moist air is heated to a very high temperature. If the equilibrium composition consists of H2O, O2, N2, OH, H2, and NO, the number of equilibrium constant relations needed to determine the equilibrium composition of the mixture is

(a) 1

(b) 2

(c) 3

(d) 4

(e) 5

Answer (c) 3

16-121 Propane C3H8 is burned with air, and the combustion products consist of CO2, CO, H2O, O2, N2, OH, H2, and NO. The number of equilibrium constant relations needed to determine the equilibrium composition of the mixture is

(a) 1

(b) 2

(c) 3

(d) 4

(e) 5

Answer (d) 4

16-122 Consider a gas mixture that consists of three components. The number of independent variables that need to be specified to fix the state of the mixture is

(a) 1

(b) 2

(c) 3

(d) 4

(e) 5

Answer (d) 4

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16-123 The value of Henry’s constant for CO2 gas dissolved in water at 290 K is 12.8 MPa. Consider water exposed to air at 100 kPa that contains 3 percent CO2 by volume. Under phase equilibrium conditions, the mole fraction of CO2 gas dissolved in water at 290 K is

(a) 2.3×10-4

(b) 3.0×10-4

(c) 0.80×10-4

(d) 2.2×10-4

(e) 5.6×10-4

Answer (a) 2.3×10-4 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). H=12.8 "MPa" P=0.1 "MPa" y_CO2_air=0.03 P_CO2_air=y_CO2_air*P y_CO2_liquid=P_CO2_air/H "Some Wrong Solutions with Common Mistakes:" W1_yCO2=P_CO2_air*H "Multiplying by H instead of dividing by it" W2_yCO2=P_CO2_air "Taking partial pressure in air"

16-124 The solubility of nitrogen gas in rubber at 25°C is 0.00156 kmol/m3⋅bar. When phase equilibrium is established, the density of nitrogen in a rubber piece placed in a nitrogen gas chamber at 800 kPa is

(a) 0.012 kg/m3

(b) 0.35 kg/m3

(c) 0.42 kg/m3

(d) 0.56 kg/m3 (e) 0.078 kg/m3

Answer (b) 0.35 kg/m3 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T=25 "C" S=0.00156 "kmol/bar.m^3" MM_N2=28 "kg/kmol" S_mass=S*MM_N2 "kg/bar.m^3" P_N2=8 "bar" rho_solid=S_mass*P_N2 "Some Wrong Solutions with Common Mistakes:" W1_density=S*P_N2 "Using solubility per kmol"

16-125 … 16-128 Design and Essay Problems

KJ

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17-1

Chapter 17 COMPRESSIBLE FLOW Stagnation Properties 17-1C The temperature of the air will rise as it approaches the nozzle because of the stagnation process. 17-2C Stagnation enthalpy combines the ordinary enthalpy and the kinetic energy of a fluid, and offers convenience when analyzing high-speed flows. It differs from the ordinary enthalpy by the kinetic energy term. 17-3C Dynamic temperature is the temperature rise of a fluid during a stagnation process. 17-4C No. Because the velocities encountered in air-conditioning applications are very low, and thus the static and the stagnation temperatures are practically identical.

17-5 The state of air and its velocity are specified. The stagnation temperature and stagnation pressure of air are to be determined. Assumptions 1 The stagnation process is isentropic. 2 Air is an ideal gas. Properties The properties of air at room temperature are cp = 1.005 kJ/kg⋅K and k = 1.4 (Table A-2a). Analysis The stagnation temperature of air is determined from T0 = T +

(470 m/s) 2 V2 ⎛ 1 kJ/kg = 245.9 K + ⎜ 2c p 2 × 1.005 kJ/kg ⋅ K ⎝ 1000 m 2 /s 2

⎞ ⎟ = 355.8 K ⎠

Other stagnation properties at the specified state are determined by considering an isentropic process between the specified state and the stagnation state, ⎛T ⎞ P0 = P⎜ 0 ⎟ ⎝T ⎠

k /( k −1)

355.8 K ⎞ = (44 kPa) ⎛⎜ ⎟ ⎝ 245.9 K ⎠

1.4 /(1.4 −1)

= 160.3 kPa

Discussion Note that the stagnation properties can be significantly different than thermodynamic properties.

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17-6 Air at 300 K is flowing in a duct. The temperature that a stationary probe inserted into the duct will read is to be determined for different air velocities. Assumptions The stagnation process is isentropic. Properties The specific heat of air at room temperature is cp = 1.005 kJ/kg⋅K (Table A-2a). Analysis The air which strikes the probe will be brought to a complete stop, and thus it will undergo a stagnation process. The thermometer will sense the temperature of this stagnated air, which is the stagnation temperature, T0. It is determined from T0 = T +

V2 2c p

(a)

⎛ 1 kJ/kg (1 m/s) 2 ⎜ T0 = 300 K + 2 × 1.005 kJ/kg ⋅ K ⎜⎝ 1000 m 2 / s 2

(b)

T0 = 300 K +

(10 m/s) 2 ⎛ 1 kJ/kg ⎞ ⎜ ⎟ = 300.1 K 2 × 1.005 kJ/kg ⋅ K ⎝ 1000 m 2 / s 2 ⎠

(c)

T0 = 300 K +

(100 m/s) 2 ⎛ 1 kJ/kg ⎞ ⎜ ⎟ = 305.0 K 2 × 1.005 kJ/kg ⋅ K ⎝ 1000 m 2 / s 2 ⎠

(d)

T0 = 300 K +

(1000 m/s) 2 ⎛ 1 kJ/kg ⎞ ⎜ ⎟ = 797.5 K 2 × 1.005 kJ/kg ⋅ K ⎝ 1000 m 2 / s 2 ⎠

⎞ ⎟ = 300.0 K ⎟ ⎠

AIR 300 K V

Discussion Note that the stagnation temperature is nearly identical to the thermodynamic temperature at low velocities, but the difference between the two is very significant at high velocities,

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17-7 The states of different substances and their velocities are specified. The stagnation temperature and stagnation pressures are to be determined. Assumptions 1 The stagnation process is isentropic. 2 Helium and nitrogen are ideal gases. Analysis (a) Helium can be treated as an ideal gas with cp = 5.1926 kJ/kg·K and k = 1.667 (Table A-2a). Then the stagnation temperature and pressure of helium are determined from T0 = T +

(240 m/s) 2 V2 ⎛ 1 kJ/kg ⎞ = 50°C + ⎜ ⎟ = 55.5°C 2c p 2 × 5.1926 kJ/kg ⋅ °C ⎝ 1000 m 2 / s 2 ⎠

⎛T ⎞ P0 = P⎜ 0 ⎟ ⎝T ⎠

k / ( k −1)

⎛ 328.7 K ⎞ = (0.25 MPa)⎜ ⎟ ⎝ 323.2 K ⎠

1.667 / (1.667 −1)

= 0.261 MPa

(b) Nitrogen can be treated as an ideal gas with cp = 1.039 kJ/kg·K and k =1.400. Then the stagnation temperature and pressure of nitrogen are determined from T0 = T +

(300 m/s) 2 V2 ⎛ 1 kJ/kg ⎞ = 50°C + ⎜ ⎟ = 93.3°C 2c p 2 × 1.039 kJ/kg ⋅ °C ⎝ 1000 m 2 / s 2 ⎠

⎛T ⎞ P0 = P⎜ 0 ⎟ ⎝T ⎠

k /( k −1)

⎛ 366.5 K ⎞ = (0.15 MPa)⎜ ⎟ ⎝ 323.2 K ⎠

1.4 /(1.4 −1)

= 0.233 MPa

(c) Steam can be treated as an ideal gas with cp = 1.865 kJ/kg·K and k =1.329. Then the stagnation temperature and pressure of steam are determined from T0 = T +

(480 m/s) 2 V2 ⎛ 1 kJ/kg ⎞ = 350°C + ⎜ ⎟ = 411.8°C = 685 K 2c p 2 × 1.865 kJ/kg ⋅ °C ⎝ 1000 m 2 / s 2 ⎠

⎛T ⎞ P0 = P⎜⎜ 0 ⎟⎟ ⎝T ⎠

k /( k −1)

⎛ 685 K ⎞ = (0.1 MPa)⎜ ⎟ ⎝ 623.2 K ⎠

1.329 /(1.329 −1)

= 0.147 MPa

Discussion Note that the stagnation properties can be significantly different than thermodynamic properties.

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17-4

17-8 The inlet stagnation temperature and pressure and the exit stagnation pressure of air flowing through a compressor are specified. The power input to the compressor is to be determined. Assumptions 1 The compressor is isentropic. 2 Air is an ideal gas. Properties The properties of air at room temperature are cp = 1.005 kJ/kg⋅K and k = 1.4 (Table A-2a).

900 kPa

Analysis The exit stagnation temperature of air T02 is determined from ⎛P T02 = T01 ⎜⎜ 02 ⎝ P01

⎞ ⎟ ⎟ ⎠

( k −1) / k

900 ⎞ = (300.2 K)⎛⎜ ⎟ ⎝ 100 ⎠

(1.4 −1) / 1.4

AIR 0.02 kg/s

= 562.4 K

From the energy balance on the compressor,

& W

P2= 270 kPa V2= 620 m/s

W& in = m& (h20 − h01 )

or,

W& in = m& c p (T02 − T01 ) = (0.02 kg/s)(1.005 kJ/kg ⋅ K)(562.4 − 300.2)K = 5.27 kW Discussion Note that the stagnation properties can be used conveniently in the energy equation.

17-9E Steam flows through a device. The stagnation temperature and pressure of steam and its velocity are specified. The static pressure and temperature of the steam are to be determined. Assumptions 1 The stagnation process is isentropic. 2 Steam is an ideal gas. Properties Steam can be treated as an ideal gas with cp = 0.445 Btu/lbm·R and k =1.329 (Table A-2Ea). Analysis The static temperature and pressure of steam are determined from T = T0 −

⎛ 1 Btu/lbm (900 ft/s) 2 V2 ⎜ = 700°F − 2c p 2 × 0.445 Btu/lbm ⋅ °F ⎜⎝ 25,037 ft 2 / s 2

⎛T P = P0 ⎜⎜ ⎝ T0

⎞ ⎟ ⎟ ⎠

k /( k −1)

⎛ 1123.6 R ⎞ = (120 psia)⎜ ⎟ ⎝ 1160 R ⎠

⎞ ⎟ = 663.6°F ⎟ ⎠

1.329 /(1.329 −1)

= 105.5 psia

Discussion Note that the stagnation properties can be significantly different than thermodynamic properties.

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17-5

17-10 The inlet stagnation temperature and pressure and the exit stagnation pressure of products of combustion flowing through a gas turbine are specified. The power output of the turbine is to be determined. Assumptions 1 The expansion process is isentropic. 2 Products of combustion are ideal gases.

1 MPa 750°C

Properties The properties of products of combustion are given to be cp = 1.157 kJ/kg⋅K, R = 0.287 kJ/kg⋅K, and k = 1.33. Analysis The exit stagnation temperature T02 is determined to be T02

⎛P = T01 ⎜⎜ 02 ⎝ P01

⎞ ⎟ ⎟ ⎠

( k −1) / k

0.1 ⎞ = (1023.2 K)⎛⎜ ⎟ ⎝ 1 ⎠

(1.33−1) / 1.33

STEAM

W

= 577.9 K

100 kPa

Also, c p = kc v = k (c p − R) ⎯ ⎯→ c p =

kR 1.33(0.287 kJ/kg ⋅ K) = = 1.157 kJ/kg ⋅ K k −1 1.33 − 1

From the energy balance on the turbine, − wout = (h20 − h01 )

or, wout = c p (T01 − T02 ) = (1.157 kJ/kg ⋅ K)(1023.2 − 577.9) K = 515.2 kJ/kg

Discussion Note that the stagnation properties can be used conveniently in the energy equation.

17-11 Air flows through a device. The stagnation temperature and pressure of air and its velocity are specified. The static pressure and temperature of air are to be determined. Assumptions 1 The stagnation process is isentropic. 2 Air is an ideal gas. Properties The properties of air at an anticipated average temperature of 600 K are cp = 1.051 kJ/kg⋅K and k = 1.376 (Table A-2b). Analysis The static temperature and pressure of air are determined from T = T0 −

(570 m/s) 2 V2 ⎛ 1 kJ/kg = 673.2 − ⎜ 2c p 2 × 1.051 kJ/kg ⋅ K ⎝ 1000 m 2 / s 2

⎞ ⎟ = 518.6 K ⎠

and ⎛T P2 = P02 ⎜⎜ 2 ⎝ T02

⎞ ⎟ ⎟ ⎠

k /( k −1)

518.6 K ⎞ = (0.6 MPa)⎛⎜ ⎟ ⎝ 673.2 K ⎠

1.376 /(1.376 −1)

= 0.23 MPa

Discussion Note that the stagnation properties can be significantly different than thermodynamic properties.

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17-6

Speed of sound and Mach Number

17-12C Sound is an infinitesimally small pressure wave. It is generated by a small disturbance in a medium. It travels by wave propagation. Sound waves cannot travel in a vacuum. 17-13C Yes, it is. Because the amplitude of an ordinary sound wave is very small, and it does not cause any significant change in temperature and pressure. 17-14C The sonic speed in a medium depends on the properties of the medium, and it changes as the properties of the medium change.

17-15C In warm (higher temperature) air since c = kRT

17-16C Helium, since c = kRT and helium has the highest kR value. It is about 0.40 for air, 0.35 for argon and 3.46 for helium. 17-17C Air at specified conditions will behave like an ideal gas, and the speed of sound in an ideal gas depends on temperature only. Therefore, the speed of sound will be the same in both mediums. 17-18C In general, no. Because the Mach number also depends on the speed of sound in gas, which depends on the temperature of the gas. The Mach number will remain constant if the temperature is maintained constant.

17-19 The Mach number of scramjet and the air temperature are given. The speed of the engine is to be determined. Assumptions Air is an ideal gas with constant specific heats at room temperature. Properties The gas constant of air is R = 0.287 kJ/kg·K. Its specific heat ratio at room temperature is k = 1.4 (Table A-2a). Analysis The speed of sound is ⎛ 1000 m 2 / s 2 c = kRT = (1.4)(0.287 kJ/kg ⋅ K)(253 K)⎜⎜ ⎝ 1 kJ/kg

⎞ ⎟ = 318.8 m/s ⎟ ⎠

and ⎛ 3.6 km/h ⎞ V = cMa = (318.83 m/s)(7)⎜ ⎟ = 8035 km/h ⎝ 1 m/s ⎠

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17-20E The Mach number of scramjet and the air temperature are given. The speed of the engine is to be determined. Assumptions Air is an ideal gas with constant specific heats at room temperature. Properties The gas constant of air is R = 0.06855 Btu/lbm·R. Its specific heat ratio at room temperature is k = 1.4 (Table A-2Ea). Analysis The speed of sound is ⎛ 25,037 ft 2 / s 2 c = kRT = (1.4)(0.06855 Btu/lbm ⋅ R)(460 R)⎜⎜ ⎝ 1 Btu/lbm

⎞ ⎟ = 1051.3 ft/s ⎟ ⎠

and ⎛ 1 mi/h ⎞ V = cMa = (1051.3 ft/s)(7)⎜ ⎟ = 5018 mi/h ⎝ 1.46667 ft/s ⎠

17-21 The speed of an airplane and the air temperature are give. It is to be determined if the speed of this airplane is subsonic or supersonic. Assumptions Air is an ideal gas with constant specific heats at room temperature. Properties The gas constant of air is R = 0.287 kJ/kg·K. Its specific heat ratio at room temperature is k = 1.4 (Table A-2a). Analysis The speed of sound is ⎛ 1000 m 2 / s 2 c = kRT = (1.4)(0.287 kJ/kg ⋅ K)(223 K)⎜⎜ ⎝ 1 kJ/kg

⎞ ⎛ 3.6 km/h ⎞ ⎟⎜ ⎟ ⎝ 1 m/s ⎟⎠ = 1077.6 km/h ⎠

and Ma =

V 920 km/h = = 0.854 c 1077.6 km/h

The speed of the airplane is subsonic since the Mach number is less than 1.

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17-22 The Mach number of an aircraft and the velocity of sound in air are to be determined at two specified temperatures. Assumptions Air is an ideal gas with constant specific heats at room temperature. Analysis (a) At 300 K air can be treated as an ideal gas with R = 0.287 kJ/kg·K and k = 1.4 (Table A-2a). Thus ⎛ 1000 m 2 / s 2 ⎞ ⎟ = 347.2 m/s c = kRT = (1.4)(0.287 kJ/kg ⋅ K)(300 K)⎜ ⎜ 1 kJ/kg ⎟ ⎠ ⎝

and Ma =

V 280 m/s = = 0.81 c 347.2 m/s

(b) At 1000 K, ⎛ 1000 m 2 / s 2 c = kRT = (1.4)(0.287 kJ/kg ⋅ K)(1000 K)⎜⎜ ⎝ 1 kJ/kg

⎞ ⎟ = 634 m/s ⎟ ⎠

and Ma =

V 280 m/s = = 0.442 c 634 m/s

Discussion Note that a constant Mach number does not necessarily indicate constant speed. The Mach number of a rocket, for example, will be increasing even when it ascends at constant speed. Also, the specific heat ratio k changes with temperature, and the accuracy of the result at 1000 K can be improved by using the k value at that temperature (it would give k = 1.336, c = 619 m/s, and Ma = 0.452).

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17-23 Carbon dioxide flows through a nozzle. The inlet temperature and velocity and the exit temperature of CO2 are specified. The Mach number is to be determined at the inlet and exit of the nozzle. Assumptions 1 CO2 is an ideal gas with constant specific heats at room temperature. 2 This is a steady-flow process. Properties The gas constant of carbon dioxide is R = 0.1889 kJ/kg·K. Its constant pressure specific heat and specific heat ratio at room temperature are cp = 0.8439 kJ/kg⋅K and k = 1.288 (Table A-2a).

1200 K 50 m/s

Carbon dioxide

400 K

Analysis (a) At the inlet

⎛ 1000 m 2 / s 2 c1 = k1 RT1 = (1.288)(0.1889 kJ/kg ⋅ K)(1200 K)⎜⎜ ⎝ 1 kJ/kg

⎞ ⎟ = 540.4 m/s ⎟ ⎠

Thus, Ma 1 =

V1 50 m/s = = 0.0925 c1 540.4 m/s

(b) At the exit, ⎛ 1000 m 2 / s 2 c 2 = k 2 RT2 = (1.288)(0.1889 kJ/kg ⋅ K)(400 K)⎜⎜ ⎝ 1 kJ/kg

⎞ ⎟ = 312 m/s ⎟ ⎠

The nozzle exit velocity is determined from the steady-flow energy balance relation, 0 = h2 − h1 +

V 2 2 − V1 2 2

→

0 = c p (T2 − T1 ) +

0 = (0.8439 kJ/kg ⋅ K)(1200 − 400 K) +

V 2 2 − V1 2 2

V 2 2 − (50 m/s) 2 ⎛ 1 kJ/kg ⎞ ⎯→ V 2 = 1163 m/s ⎟⎯ ⎜ 2 ⎝ 1000 m 2 / s 2 ⎠

Thus, Ma 2 =

V 2 1163 m/s = = 3.73 c2 312 m/s

Discussion The specific heats and their ratio k change with temperature, and the accuracy of the results can be improved by accounting for this variation. Using EES (or another property database):

→

c1 = 516 m/s,

V1 = 50 m/s,

At 400 K: cp = 0.9383 kJ/kg⋅K, k = 1.252 →

c2 = 308 m/s,

V2 = 1356 m/s,

At 1200 K: cp = 1.278 kJ/kg⋅K, k = 1.173

Ma1 = 0.0969 Ma2 = 4.41

Therefore, the constant specific heat assumption results in an error of 4.5% at the inlet and 15.5% at the exit in the Mach number, which are significant.

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17-10

17-24 Nitrogen flows through a heat exchanger. The inlet temperature, pressure, and velocity and the exit pressure and velocity are specified. The Mach number is to be determined at the inlet and exit of the heat exchanger. Assumptions 1 N2 is an ideal gas. 2 This is a steady-flow process. 3 The potential energy change is negligible. Properties The gas constant of N2 is R = 0.2968 kJ/kg·K. Its constant pressure specific heat and specific heat ratio at room temperature are cp = 1.040 kJ/kg⋅K and k = 1.4 (Table A-2a).

120 kJ/kg 150 kPa 10°C 100 m/s

100 kPa 200 m/s

Nitrogen

Analysis At the inlet, the speed of sound is

⎛ 1000 m 2 / s 2 c1 = k1 RT1 = (1.400)(0.2968 kJ/kg ⋅ K)(283 K)⎜⎜ ⎝ 1 kJ/kg

⎞ ⎟ = 342.9 m/s ⎟ ⎠

Thus, Ma 1 =

V1 100 m/s = = 0.292 c1 342.9 m/s

From the energy balance on the heat exchanger, qin = c p (T2 − T1 ) +

V2 2 − V12 2

120 kJ/kg = (1.040 kJ/kg.°C)(T2 − 10°C) +

(200 m/s) 2 − (100 m/s) 2 ⎛ 1 kJ/kg ⎞ ⎟ ⎜ 2 ⎝ 1000 m 2 / s 2 ⎠

It yields T2 = 111°C = 384 K ⎛ 1000 m 2 / s 2 c 2 = k 2 RT2 = (1.4 )(0.2968 kJ/kg ⋅ K)(384 K)⎜⎜ ⎝ 1 kJ/kg

⎞ ⎟ = 399 m/s ⎟ ⎠

Thus, Ma 2 =

V 2 200 m/s = = 0.501 c 2 399 m/s

Discussion The specific heats and their ratio k change with temperature, and the accuracy of the results can be improved by accounting for this variation. Using EES (or another property database):

→

c1 = 343 m/s,

V1 = 100 m/s,

Ma1 = 0.292

At 111°C cp = 1.041 kJ/kg⋅K, k = 1.399 →

c2 = 399 m/s,

V2 = 200 m/s,

Ma2 = 0.501

At 10°C : cp = 1.038 kJ/kg⋅K, k = 1.400

Therefore, the constant specific heat assumption results in no error at the inlet and at the exit in the Mach number.

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17-11

17-25 The speed of sound in refrigerant-134a at a specified state is to be determined. Assumptions R-134a is an ideal gas with constant specific heats at room temperature. Properties The gas constant of R-134a is R = 0.08149 kJ/kg·K. Its specific heat ratio at room temperature is k = 1.108. Analysis From the ideal-gas speed of sound relation,

⎛ 1000 m 2 / s 2 c = kRT = (1.108)(0.08149 kJ/kg ⋅ K)(60 + 273 K)⎜⎜ ⎝ 1 kJ/kg

⎞ ⎟ = 173 m/s ⎟ ⎠

Discussion Note that the speed of sound is independent of pressure for ideal gases.

17-26 The Mach number of a passenger plane for specified limiting operating conditions is to be determined. Assumptions Air is an ideal gas with constant specific heats at room temperature. Properties The gas constant of air is R = 0.287 kJ/kg·K. Its specific heat ratio at room temperature is k = 1.4 (Table A-2a). Analysis From the speed of sound relation ⎛ 1000 m 2 / s 2 c = kRT = (1.4)(0.287 kJ/kg ⋅ K)(-60 + 273 K)⎜⎜ ⎝ 1 kJ/kg

⎞ ⎟ = 293 m/s ⎟ ⎠

Thus, the Mach number corresponding to the maximum cruising speed of the plane is Ma =

V max (945 / 3.6) m/s = = 0.897 c 293 m/s

Discussion Note that this is a subsonic flight since Ma < 1. Also, using a k value at -60°C would give practically the same result.

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17-12

17-27E Steam flows through a device at a specified state and velocity. The Mach number of steam is to be determined assuming ideal gas behavior. Assumptions Steam is an ideal gas with constant specific heats. Properties The gas constant of steam is R = 0.1102 Btu/lbm·R. Its specific heat ratio is given to be k = 1.3. Analysis From the ideal-gas speed of sound relation,

⎛ 25,037 ft 2 / s 2 c = kRT = (1.3)(0.1102 Btu/lbm ⋅ R)(1160 R)⎜⎜ ⎝ 1 Btu/lbm

⎞ ⎟ = 2040.8 ft/s ⎟ ⎠

Thus, Ma =

900 ft/s V = = 0.441 c 2040 ft/s

Discussion Using property data from steam tables and not assuming ideal gas behavior, it can be shown that the Mach number in steam at the specified state is 0.446, which is sufficiently close to the ideal-gas value of 0.441. Therefore, the ideal gas approximation is a reasonable one in this case.

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17-13

17-28E EES Problem 17-27E is reconsidered. The variation of Mach number with temperature as the temperature changes between 350 and 700°F is to be investigated, and the results are to be plotted. Analysis Using EES, this problem can be solved as follows: T=Temperature+460 R=0.1102 V=900 k=1.3 c=SQRT(k*R*T*25037) Ma=V/c

Mach number Ma 0.528 0.520 0.512 0.505 0.498 0.491 0.485 0.479 0.473 0.467 0.462 0.456 0.451 0.446 0.441

0.53 0.52 0.51 0.5 0.49

Ma

Temperature, T, °F 350 375 400 425 450 475 500 525 550 575 600 625 650 675 700

0.48 0.47 0.46 0.45 0.44 350

400

450

500

550

600

650

700

Temperature, °F

Discussion Note that for a specified flow speed, the Mach number decreases with increasing temperature, as expected.

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17-14

17-29 The expression for the speed of sound for an ideal gas is to be obtained using the isentropic process equation and the definition of the speed of sound. Analysis The isentropic relation Pvk = A where A is a constant can also be expressed as k

⎛1⎞ P = A⎜ ⎟ = Aρ k ⎝v ⎠

Substituting it into the relation for the speed of sound, ⎛ ∂ ( Aρ ) k ⎛ ∂P ⎞ c 2 = ⎜⎜ ⎟⎟ = ⎜ ⎝ ∂ρ ⎠ s ⎜⎝ ∂ρ

⎞ ⎟ = kAρ k −1 = k ( Aρ k ) / ρ = k ( P / ρ ) = kRT ⎟ ⎠s

since for an ideal gas P = ρRT or RT = P/ρ. Therefore, c = kRT

which is the desired relation.

17-30 The inlet state and the exit pressure of air are given for an isentropic expansion process. The ratio of the initial to the final speed of sound is to be determined. Assumptions Air is an ideal gas with constant specific heats at room temperature. Properties The properties of air are R = 0.287 kJ/kg·K and k = 1.4 (Table A-2a). The specific heat ratio k varies with temperature, but in our case this change is very small and can be disregarded. Analysis The final temperature of air is determined from the isentropic relation of ideal gases, ⎛P T2 = T1 ⎜⎜ 2 ⎝ P1

⎞ ⎟⎟ ⎠

( k −1) / k

0.4 MPa ⎞ = (333.2 K)⎛⎜ ⎟ ⎝ 1.5 MPa ⎠

(1.4 −1) / 1.4

= 228.4 K

Treating k as a constant, the ratio of the initial to the final speed of sound can be expressed as Ratio =

c2 = c1

k1RT1 k2 RT2

=

T1 T2

=

333.2 = 1.21 228.4

Discussion Note that the speed of sound is proportional to the square root of thermodynamic temperature.

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17-15

17-31 The inlet state and the exit pressure of helium are given for an isentropic expansion process. The ratio of the initial to the final speed of sound is to be determined. Assumptions Helium is an ideal gas with constant specific heats at room temperature. Properties The properties of helium are R = 2.0769 kJ/kg·K and k = 1.667 (Table A-2a). Analysis The final temperature of helium is determined from the isentropic relation of ideal gases, ⎛P T2 = T1 ⎜⎜ 2 ⎝ P1

⎞ ⎟⎟ ⎠

( k −1) / k

0.4 ⎞ = (333.2 K)⎛⎜ ⎟ ⎝ 1.5 ⎠

(1.667 −1) / 1.667

= 196.3 K

The ratio of the initial to the final speed of sound can be expressed as Ratio =

c2 = c1

k1RT1 k2 RT2

=

T1 T2

=

333.2 = 1.30 196.3

Discussion Note that the speed of sound is proportional to the square root of thermodynamic temperature.

17-32E The inlet state and the exit pressure of air are given for an isentropic expansion process. The ratio of the initial to the final speed of sound is to be determined. Assumptions Air is an ideal gas with constant specific heats at room temperature. Properties The properties of air are R = 0.06855 Btu/lbm·R and k = 1.4 (Table A-2Ea). The specific heat ratio k varies with temperature, but in our case this change is very small and can be disregarded. Analysis The final temperature of air is determined from the isentropic relation of ideal gases, ⎛P T2 = T1 ⎜⎜ 2 ⎝ P1

⎞ ⎟⎟ ⎠

( k −1) / k

⎛ 60 ⎞ = (659.7 R)⎜ ⎟ ⎝ 170 ⎠

(1.4 −1) / 1.4

= 489.9 R

Treating k as a constant, the ratio of the initial to the final speed of sound can be expressed as Ratio =

c2 = c1

k1RT1 k2 RT2

=

T1 T2

=

659.7 = 1.16 489.9

Discussion Note that the speed of sound is proportional to the square root of thermodynamic temperature.

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17-16

One Dimensional Isentropic Flow

17-33C (a) The exit velocity remain constant at sonic speed, (b) the mass flow rate through the nozzle decreases because of the reduced flow area. 17-34C (a) The velocity will decrease, (b), (c), (d) the temperature, the pressure, and the density of the fluid will increase. 17-35C (a) The velocity will increase, (b), (c), (d) the temperature, the pressure, and the density of the fluid will decrease. 17-36C (a) The velocity will increase, (b), (c), (d) the temperature, the pressure, and the density of the fluid will decrease. 17-37C (a) The velocity will decrease, (b), (c), (d) the temperature, the pressure and the density of the fluid will increase. 17-38C They will be identical. 17-39C No, it is not possible.

17-40 Air enters a converging-diverging nozzle at specified conditions. The lowest pressure that can be obtained at the throat of the nozzle is to be determined. Assumptions 1 Air is an ideal gas with constant specific heats at room temperature. 2 Flow through the nozzle is steady, one-dimensional, and isentropic. Properties The specific heat ratio of air at room temperature is k = 1.4 (Table A-2a). Analysis The lowest pressure that can be obtained at the throat is the critical pressure P*, which is determined from 2 ⎞ P* = P0 ⎛⎜ ⎟ ⎝ k + 1⎠

k /( k −1)

2 ⎞ = (1.2 MPa)⎛⎜ ⎟ ⎝ 1.4 + 1 ⎠

1.4 /(1.4 −1)

= 0.634 MPa

Discussion This is the pressure that occurs at the throat when the flow past the throat is supersonic.

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17-17

17-41 Helium enters a converging-diverging nozzle at specified conditions. The lowest temperature and pressure that can be obtained at the throat of the nozzle are to be determined. Assumptions 1 Helium is an ideal gas with constant specific heats. 2 Flow through the nozzle is steady, one-dimensional, and isentropic. Properties The properties of helium are k = 1.667 and cp = 5.1926 kJ/kg·K (Table A-2a). Analysis The lowest temperature and pressure that can be obtained at the throat are the critical temperature T* and critical pressure P*. First we determine the stagnation temperature T0 and stagnation pressure P0, T0 = T +

(100 m/s) 2 V2 ⎛ 1 kJ/kg ⎞ = 800 K + ⎟ = 801 K ⎜ 2 × 5.1926 kJ/kg ⋅ °C ⎝ 1000 m 2 / s 2 ⎠ 2c p

⎛T ⎞ P0 = P ⎜ 0 ⎟ ⎝T ⎠

k /( k −1)

801 K ⎞ = (0.7 MPa)⎛⎜ ⎟ ⎝ 800 K ⎠

1.667 /(1.667 −1)

= 0.702 MPa

Helium

Thus, 2 ⎞ ⎛ ⎛ 2 ⎞ T * = T0 ⎜ ⎟ = 601 K ⎟ = (801 K)⎜ ⎝ 1.667 + 1 ⎠ ⎝ k +1⎠

and ⎛ 2 ⎞ P* = P0 ⎜ ⎟ ⎝ k + 1⎠

k /( k −1)

2 ⎛ ⎞ = (0.702 MPa)⎜ ⎟ ⎝ 1.667 + 1 ⎠

1.667 /(1.667 −1)

= 0.342 MPa

Discussion These are the temperature and pressure that will occur at the throat when the flow past the throat is supersonic.

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17-18

17-42 The critical temperature, pressure, and density of air and helium are to be determined at specified conditions. Assumptions Air and Helium are ideal gases with constant specific heats at room temperature. Properties The properties of air at room temperature are R = 0.287 kJ/kg·K, k = 1.4, and cp = 1.005 kJ/kg·K. The properties of helium at room temperature are R = 2.0769 kJ/kg·K, k = 1.667, and cp = 5.1926 kJ/kg·K (Table A-2a). Analysis (a) Before we calculate the critical temperature T*, pressure P*, and density ρ*, we need to determine the stagnation temperature T0, pressure P0, and density ρ0. T0 = 100°C +

⎛T ⎞ P0 = P ⎜ 0 ⎟ ⎝T ⎠

ρ0 =

(250 m/s) 2 V2 ⎛ 1 kJ/kg ⎞ = 100 + ⎟ = 131.1°C ⎜ 2 × 1.005 kJ/kg ⋅ °C ⎝ 1000 m 2 / s 2 ⎠ 2c p k /( k −1)

⎛ 404.3 K ⎞ = (200 kPa)⎜ ⎟ ⎝ 373.2 K ⎠

1.4 /(1.4 −1)

= 264.7 kPa

P0 264.7 kPa = = 2.281 kg/m 3 RT0 (0.287 kPa ⋅ m 3 /kg ⋅ K)(404.3 K)

Thus, ⎛ 2 ⎞ ⎛ 2 ⎞ T * = T0 ⎜ ⎟ = (404.3 K)⎜ ⎟ = 337 K ⎝ k +1⎠ ⎝ 1.4 + 1 ⎠

⎛ 2 ⎞ P* = P0 ⎜ ⎟ ⎝ k +1⎠

k /( k −1)

⎛ 2 ⎞ = (264.7 kPa)⎜ ⎟ ⎝ 1.4 + 1 ⎠

1 /( k −1)

⎛ 2 ⎞ = (2.281 kg/m 3 )⎜ ⎟ ⎝ 1.4 + 1 ⎠

2 ⎞ ⎟ ⎝ k +1⎠

ρ * = ρ 0 ⎛⎜

1.4 /(1.4 −1)

= 140 kPa

1 /(1.4 −1)

= 1.45 kg/m 3

(b) For helium, T0 = T +

(300 m/s) 2 V2 ⎛ 1 kJ/kg ⎞ = 40 + ⎜ ⎟ = 48.7°C 2c p 2 × 5.1926 kJ/kg ⋅ °C ⎝ 1000 m 2 / s 2 ⎠

⎛T ⎞ P0 = P ⎜ 0 ⎟ ⎝T ⎠

ρ0 =

k /( k −1)

⎛ 321.9 K ⎞ = (200 kPa)⎜ ⎟ ⎝ 313.2 K ⎠

1.667 /(1.667 −1)

= 214.2 kPa

P0 214.2 kPa = = 0.320 kg/m 3 RT0 (2.0769 kPa ⋅ m 3 /kg ⋅ K)(321.9 K)

Thus, 2 ⎛ 2 ⎞ ⎛ ⎞ T * = T0 ⎜ ⎟ = (321.9 K)⎜ ⎟ = 241 K ⎝ k +1⎠ ⎝ 1.667 + 1 ⎠ ⎛ 2 ⎞ P* = P0 ⎜ ⎟ ⎝ k +1⎠

k /( k −1)

2 ⎞ ⎟ ⎝ k +1⎠

ρ * = ρ 0 ⎛⎜

1 /( k −1)

2 ⎛ ⎞ = (214.2 kPa)⎜ ⎟ 1.667 + 1 ⎝ ⎠

1.667 /(1.667 −1)

2 ⎛ ⎞ = (0.320 kg/m 3 )⎜ ⎟ ⎝ 1.667 + 1 ⎠

= 104.3 kPa 1 /(1.667 −1)

= 0.208 kg/m 3

Discussion These are the temperature, pressure, and density values that will occur at the throat when the flow past the throat is supersonic.

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17-19

17-43 Stationary carbon dioxide at a given state is accelerated isentropically to a specified Mach number. The temperature and pressure of the carbon dioxide after acceleration are to be determined. Assumptions Carbon dioxide is an ideal gas with constant specific heats. Properties The specific heat ratio of the carbon dioxide at 400 K is k = 1.252 (Table A-2b). Analysis The inlet temperature and pressure in this case is equivalent to the stagnation temperature and pressure since the inlet velocity of the carbon dioxide said to be negligible. That is, T0 = Ti = 400 K and P0 = Pi = 600 kPa. Then, ⎛ 2 T = T0 ⎜ ⎜ 2 + (k − 1)M 2 ⎝

⎞ ⎛ 2 ⎟ = (400 K)⎜ ⎟ ⎜ 2 + (1.252 - 1)(0.5) 2 ⎠ ⎝

⎞ ⎟ = 387.8 K ⎟ ⎠

and ⎛T P = P0 ⎜⎜ ⎝ T0

⎞ ⎟ ⎟ ⎠

k /( k −1)

⎛ 387.8 K ⎞ = (600 kPa)⎜ ⎟ ⎝ 400 K ⎠

1.252 /(1.252 −1)

= 514.3 kPa

Discussion Note that both the pressure and temperature drop as the gas is accelerated as part of the internal energy of the gas is converted to kinetic energy.

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17-20

17-44 Air flows through a duct. The state of the air and its Mach number are specified. The velocity and the stagnation pressure, temperature, and density of the air are to be determined. Assumptions Air is an ideal gas with constant specific heats at room temperature. Properties The properties of air at room temperature are R = 0.287 kPa.m3/kg.K and k = 1.4 (Table A-2a). Analysis The speed of sound in air at the specified conditions is

⎛ 1000 m 2 / s 2 c = kRT = (1.4)(0.287 kJ/kg ⋅ K)(373.2 K)⎜⎜ ⎝ 1 kJ/kg

⎞ ⎟ = 387.2 m/s ⎟ ⎠

Thus, V = Ma × c = (0.8)(387.2 m/s) = 310 m/s

AIR

Also,

ρ=

P 200 kPa = = 1.867 kg/m 3 RT (0.287 kPa ⋅ m 3 /kg ⋅ K)(373.2 K)

Then the stagnation properties are determined from ⎛ (k − 1)Ma 2 T0 = T ⎜ 1 + ⎜ 2 ⎝

⎛T ⎞ P0 = P⎜ 0 ⎟ ⎝T ⎠

k /( k −1)

⎛ T0 ⎞ ⎟ ⎝T ⎠

ρ0 = ρ⎜

1 /( k −1)

2 ⎞ ⎛ ⎟ = (373.2 K)⎜1 + (1.4 - 1)(0.8) ⎟ ⎜ 2 ⎠ ⎝

⎛ 421.0 K ⎞ = (200 kPa)⎜ ⎟ ⎝ 373.2 K ⎠

1.4 /(1.4 −1)

⎛ 421.0 K ⎞ = (1.867 kg/m 3 )⎜ ⎟ ⎝ 373.2 K ⎠

⎞ ⎟ = 421 K ⎟ ⎠

= 305 kPa

1 /(1.4 −1)

= 2.52 kg/m 3

Discussion Note that both the pressure and temperature drop as the gas is accelerated as part of the internal energy of the gas is converted to kinetic energy.

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17-21

17-45 EES Problem 17-44 is reconsidered. The effect of Mach number on the velocity and stagnation properties as the Ma is varied from 0.1 to 2 are to be investigated, and the results are to be plotted. Analysis Using EES, the problem is solved as follows:

"Stagnation properties" T0=T*(1+(k-1)*Ma^2/2) P0=P*(T0/T)^(k/(k-1)) rho0=rho*(T0/T)^(1/(k-1))

1600 1400

V, T0, P0, and 100•ρ0

P=200 T=100+273.15 R=0.287 k=1.4 c=SQRT(k*R*T*1000) Ma=V/c rho=P/(R*T)

P0 1200 1000 800 600

T0

400

ρ0

200

V 0 0

0.4

0.8

1.2

1.6

Ma

Mach num. Ma 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0

Velocity, V, m/s 38.7 77.4 116.2 154.9 193.6 232.3 271.0 309.8 348.5 387.2 425.9 464.7 503.4 542.1 580.8 619.5 658.3 697.0 735.7 774.4

Stag. Temp, T0, K 373.9 376.1 379.9 385.1 391.8 400.0 409.7 420.9 433.6 447.8 463.5 480.6 499.3 519.4 541.1 564.2 588.8 615.0 642.6 671.7

Stag. Press, P0, kPa 201.4 205.7 212.9 223.3 237.2 255.1 277.4 304.9 338.3 378.6 427.0 485.0 554.1 636.5 734.2 850.1 987.2 1149.2 1340.1 1564.9

Stag. Density, ρ0, kg/m3 1.877 1.905 1.953 2.021 2.110 2.222 2.359 2.524 2.718 2.946 3.210 3.516 3.867 4.269 4.728 5.250 5.842 6.511 7.267 8.118

Discussion Note that as Mach number increases, so does the flow velocity and stagnation temperature, pressure, and density.

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2

17-22

17-46E Air flows through a duct at a specified state and Mach number. The velocity and the stagnation pressure, temperature, and density of the air are to be determined. Assumptions Air is an ideal gas with constant specific heats at room temperature. Properties The properties of air are R = 0.06855 Btu/lbm.R = 0.3704 psia⋅ft3/lbm.R and k = 1.4 (Table A2Ea). Analysis The speed of sound in air at the specified conditions is

⎛ 25,037 ft 2 / s 2 c = kRT = (1.4)(0.06855 Btu/1bm ⋅ R)(671.7 R)⎜⎜ ⎝ 1 Btu/1bm

⎞ ⎟ = 1270.4 ft/s ⎟ ⎠

Thus, V = Ma × c = (0.8)(1270.4 ft/s) = 1016 ft/s

Also,

ρ=

30 psia P = = 0.1206 1bm/ft 3 RT (0.3704 psia ⋅ ft 3 /lbm ⋅ R)(671.7 R)

Then the stagnation properties are determined from ⎛ (k − 1)Ma 2 T0 = T ⎜⎜1 + 2 ⎝ ⎛T ⎞ P0 = P⎜ 0 ⎟ ⎝T ⎠

k /( k −1)

⎛ T0 ⎞ ⎟ ⎝T ⎠

ρ0 = ρ⎜

1 /( k −1)

⎞ ⎛ (1.4 - 1)(0.8) 2 ⎟ = (671.7 R)⎜1 + ⎟ ⎜ 2 ⎠ ⎝

⎛ 757.7 R ⎞ = (30 psia)⎜ ⎟ ⎝ 671.7 R ⎠

1.4 /(1.4 −1)

⎛ 757.7 R ⎞ = (0.1206 1bm/ft 3 )⎜ ⎟ ⎝ 671.7 R ⎠

⎞ ⎟ = 758 R ⎟ ⎠

= 45.7 psia 1 /(1.4 −1)

= 0.163 1bm/ft 3

Discussion Note that the temperature, pressure, and density of a gas increases during a stagnation process.

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17-23

17-47 An aircraft is designed to cruise at a given Mach number, elevation, and the atmospheric temperature. The stagnation temperature on the leading edge of the wing is to be determined. Assumptions Air is an ideal gas. Properties The properties of air are R = 0.287 kPa.m3/kg.K, cp = 1.005 kJ/kg·K, and k = 1.4 (Table A-2a). Analysis The speed of sound in air at the specified conditions is ⎛ 1000 m 2 / s 2 ⎞ ⎟ = 308.0 m/s c = kRT = (1.4)(0.287 kJ/kg ⋅ K)(236.15 K)⎜ ⎜ 1 kJ/kg ⎟ ⎝ ⎠

Thus, V = Ma × c = (1.2)(308.0 m/s) = 369.6 m/s

Then, T0 = T +

V2 (369.6 m/s) 2 ⎛ 1 kJ/kg ⎞ = 236.15 + ⎜ ⎟ = 304.1 K 2c p 2 × 1.005 kJ/kg ⋅ K ⎝ 1000 m 2 / s 2 ⎠

Discussion Note that the temperature of a gas increases during a stagnation process as the kinetic energy is converted to enthalpy.

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17-24

Isentropic Flow Through Nozzles

17-48C (a) The exit velocity will reach the sonic speed, (b) the exit pressure will equal the critical pressure, and (c) the mass flow rate will reach the maximum value. 17-49C (a) None, (b) None, and (c) None. 17-50C They will be the same. 17-51C Maximum flow rate through a nozzle is achieved when Ma = 1 at the exit of a subsonic nozzle. For all other Ma values the mass flow rate decreases. Therefore, the mass flow rate would decrease if hypersonic velocities were achieved at the throat of a converging nozzle. 17-52C Ma* is the local velocity non-dimensionalized with respect to the sonic speed at the throat, whereas Ma is the local velocity non-dimensionalized with respect to the local sonic speed. 17-53C The fluid would accelerate even further instead of decelerating. 17-54C The fluid would decelerate instead of accelerating. 17-55C (a) The velocity will decrease, (b) the pressure will increase, and (c) the mass flow rate will remain the same. 17-56C No. If the velocity at the throat is subsonic, the diverging section will act like a diffuser and decelerate the flow.

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17-25

17-57 It is to be explained why the maximum flow rate per unit area for a given ideal gas depends only on & max / A * = a P0 / T0 . P0 / T0 . Also for an ideal gas, a relation is to be obtained for the constant a in m

(

)

Properties The properties of the ideal gas considered are R = 0.287 kPa.m3/kg⋅K and k = 1.4 (Table A-2a). Analysis The maximum flow rate is given by 2 ⎞ m& max = A * P0 k / RT0 ⎛⎜ ⎟ ⎝ k + 1⎠

( k +1) / 2 ( k −1)

or

(

m& max / A* = P0 / T0

)

⎛ 2 ⎞ k / R⎜ ⎟ ⎝ k + 1⎠

( k +1) / 2 ( k −1)

For a given gas, k and R are fixed, and thus the mass flow rate depends on the parameter P0 / T0 .

(

)

m& max / A * can be expressed as a m& max / A* = a P0 / T0 where

2 ⎞ a = k / R ⎛⎜ ⎟ ⎝ k + 1⎠

( k +1) / 2 ( k −1)

=

1.4 ⎛ 1000 m 2 / s 2 (0.287 kJ/kg.K)⎜⎜ ⎝ 1 kJ/kg

⎛ 2 ⎞ ⎜ ⎟ ⎞ ⎝ 1.4 + 1 ⎠ ⎟ ⎟ ⎠

2.4 / 0.8

= 0.0404 (m/s) K

Discussion Note that when sonic conditions exist at a throat of known cross-sectional area, the mass flow rate is fixed by the stagnation conditions.

17-58 For an ideal gas, an expression is to be obtained for the ratio of the speed of sound where Ma = 1 to the speed of sound based on the stagnation temperature, c*/c0. Analysis For an ideal gas the speed of sound is expressed as c = kRT . Thus, c* = c0

kRT * kRT0

⎛T *⎞ ⎟ = ⎜⎜ ⎟ ⎝ T0 ⎠

1/ 2

⎛ 2 ⎞ =⎜ ⎟ ⎝ k +1⎠

1/ 2

Discussion Note that a speed of sound changes the flow as the temperature changes.

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17-26

17-59 For subsonic flow at the inlet, the variation of pressure, velocity, and Mach number along the length of the nozzle are to be sketched for an ideal gas under specified conditions. Analysis Using EES and CO2 as the gas, we calculate and plot flow area A, velocity V, and Mach number Ma as the pressure drops from a stagnation value of 1400 kPa to 200 kPa. Note that the curve for A represents the shape of the nozzle, with horizontal axis serving as the centerline.

Normalized A, Ma, P, V

k=1.289 Cp=0.846 "kJ/kg.K" R=0.1889 "kJ/kg.K" P0=1400 "kPa" T0=473 "K" m=3 "kg/s" rho_0=P0/(R*T0) rho=P/(R*T) rho_norm=rho/rho_0 "Normalized density" T=T0*(P/P0)^((k-1)/k) Tnorm=T/T0 "Normalized temperature" V=SQRT(2*Cp*(T0-T)*1000) V_norm=V/500 A=m/(rho*V)*500 C=SQRT(k*R*T*1000) Ma=V/C

Mai < 1

P A

Ma V

200

400

600

800

P, kPa

1000

1200

1400

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17-27

17-60 For supersonic flow at the inlet, the variation of pressure, velocity, and Mach number along the length of the nozzle are to be sketched for an ideal gas under specified conditions. Analysis Using EES and CO2 as the gas, we calculate and plot flow area A, velocity V, and Mach number Ma as the pressure rises from 200 kPa at a very high velocity to the stagnation value of 1400 kPa. Note that the curve for A represents the shape of the nozzle, with horizontal axis serving as the centerline. k=1.289 Cp=0.846 "kJ/kg.K" R=0.1889 "kJ/kg.K" P0=1400 "kPa"

Mai > 1

Normalized A, Ma, P, V

T0=473 "K" m=3 "kg/s" rho_0=P0/(R*T0) rho=P/(R*T) rho_norm=rho/rho_0 "Normalized density" T=T0*(P/P0)^((k-1)/k) Tnorm=T/T0 "Normalized temperature" V=SQRT(2*Cp*(T0-T)*1000) V_norm=V/500 A=m/(rho*V)*500 C=SQRT(k*R*T*1000) Ma=V/C

P A

Ma V

200

400

600

800

P, kPa

1000

1200

1400

Discussion Note that this problem is identical to the proceeding one, except the flow direction is reversed.

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17-28

17-61 Air enters a nozzle at specified temperature, pressure, and velocity. The exit pressure, exit temperature, and exit-to-inlet area ratio are to be determined for a Mach number of Ma = 1 at the exit. Assumptions 1 Air is an ideal gas with constant specific heats at room temperature. 2 Flow through the nozzle is steady, one-dimensional, and isentropic. Properties The properties of air are k = 1.4 and cp = 1.005 kJ/kg·K (Table A-2a). Analysis The properties of the fluid at the location where Ma = 1 are the critical properties, denoted by superscript *. We first determine the stagnation temperature and pressure, which remain constant throughout the nozzle since the flow is isentropic. T0 = Ti +

Vi 2 (150 m/s) 2 ⎛ 1 kJ/kg = 350 K + ⎜ 2c p 2 × 1.005 kJ/kg ⋅ K ⎝ 1000 m 2 / s 2

i 150 m/s

AIR

* Ma = 1

⎞ ⎟ = 361.2 K ⎠

and ⎛T P0 = Pi ⎜⎜ 0 ⎝ Ti

⎞ ⎟ ⎟ ⎠

k /( k −1)

361.2 K ⎞ = (0.2 MPa)⎛⎜ ⎟ ⎝ 350 K ⎠

1.4 /(1.4 −1)

= 0.223 MPa

From Table A-32 (or from Eqs. 17-18 and 17-19) at Ma = 1, we read T/T0 = 0.8333, P/P0 = 0.5283. Thus, T = 0.8333T0 = 0.8333(361.2 K) = 301 K and P = 0.5283P0 = 0.5283(0.223 MPa) = 0.118 MPa Also, ⎛ 1000 m 2 / s 2 c i = kRT i = (1.4 )(0.287 kJ/kg ⋅ K)(350 K)⎜⎜ ⎝ 1 kJ/kg

⎞ ⎟ = 375 m/s ⎟ ⎠

and Ma i =

Vi 150 m/s = = 0.40 c i 375 m/s

From Table A-32 at this Mach number we read Ai /A* = 1.5901. Thus the ratio of the throat area to the nozzle inlet area is A* 1 = = 0.629 Ai 15901 .

Discussion We can also solve this problem using the relations for compressible isentropic flow. The results would be identical.

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17-29

17-62 Air enters a nozzle at specified temperature and pressure with low velocity. The exit pressure, exit temperature, and exit-to-inlet area ratio are to be determined for a Mach number of Ma = 1 at the exit. Assumptions 1 Air is an ideal gas. 2 Flow through the nozzle is steady, one-dimensional, and isentropic. Properties The specific heat ratio of air is k = 1.4 (Table A-2a). Analysis The properties of the fluid at the location where Ma = 1 are the critical properties, denoted by superscript *. The stagnation temperature and pressure in this case are identical to the inlet temperature and pressure since the inlet velocity is negligible. They remain constant throughout the nozzle since the flow is isentropic.

T0 = Ti = 350 K P0 = Pi = 0.2 MPa From Table A-32 (or from Eqs. 17-18 and 17-19) at Ma =1, we read T/T0 =0.8333, P/P0 = 0.5283. Thus, T = 0.8333T0 = 0.83333(350 K) = 292 K

i Vi ≈ 0

AIR

* Ma = 1

and P = 0.5283P0 = 0.5283(0.2 MPa) = 0.106 MPa The Mach number at the nozzle inlet is Ma = 0 since Vi ≅ 0. From Table A-32 at this Mach number we read Ai/A* = ∞. Thus the ratio of the throat area to the nozzle inlet area is A* 1 = =0 Ai ∞

Discussion We can also solve this problem using the relations for compressible isentropic flow. The results would be identical.

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17-30

17-63E Air enters a nozzle at specified temperature, pressure, and velocity. The exit pressure, exit temperature, and exit-to-inlet area ratio are to be determined for a Mach number of Ma = 1 at the exit. Assumptions 1 Air is an ideal gas with constant specific heats at room temperature. 2 Flow through the nozzle is steady, one-dimensional, and isentropic. Properties The properties of air are k = 1.4 and cp = 0.240 Btu/lbm·R (Table A-2Ea). Analysis The properties of the fluid at the location where Ma =1 are the critical properties, denoted by superscript *. We first determine the stagnation temperature and pressure, which remain constant throughout the nozzle since the flow is isentropic. T0 = T +

i 450 ft/s

AIR

* Ma = 1

⎛ 1 Btu/1bm ⎞ (450 ft/s) 2 Vi 2 ⎜ ⎟ = 646.9 R = 630 R + 2c p 2 × 0.240 Btu/lbm ⋅ R ⎜⎝ 25,037 ft 2 / s 2 ⎟⎠

and

⎛T P0 = Pi ⎜⎜ 0 ⎝ Ti

⎞ ⎟⎟ ⎠

k /( k −1)

⎛ 646.9 K ⎞ = (30 psia)⎜ ⎟ ⎝ 630 K ⎠

1.4 /(1.4 −1)

= 32.9 psia

From Table A-32 (or from Eqs. 17-18 and 17-19) at Ma =1, we read T/T0 =0.8333, P/P0 = 0.5283. Thus, T = 0.8333T0 = 0.8333(646.9 R) = 539 R and P = 0.5283P0 = 0.5283(32.9 psia) = 17.4 psia Also, ⎛ 25,037 ft 2 / s 2 c i = kRT i = (1.4 )(0.06855 Btu/1bm ⋅ R)(630 R)⎜⎜ ⎝ 1 Btu/1bm

⎞ ⎟ = 1230 ft/s ⎟ ⎠

and Ma i =

Vi 450 ft/s = = 0.3657 c i 1230 ft/s

From Table A-32 at this Mach number we read Ai/A* = 1.7426. Thus the ratio of the throat area to the nozzle inlet area is 1 A* = = 0.574 Ai 1.7426

Discussion We can also solve this problem using the relations for compressible isentropic flow. The results would be identical.

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17-31

17-64 Air enters a converging-diverging nozzle at a specified pressure. The back pressure that will result in a specified exit Mach number is to be determined. Assumptions 1 Air is an ideal gas. 2 Flow through the nozzle is steady, one-dimensional, and isentropic. Properties The specific heat ratio of air is k = 1.4 (Table A-2a). Analysis The stagnation pressure in this case is identical to the inlet pressure since the inlet velocity is negligible. It remains constant throughout the nozzle since the flow is isentropic,

P0 = Pi = 0.5 MPa From Table A-32 at Mae =1.8, we read Pe /P0 = 0.1740. Thus, P = 0.1740P0 = 0.1740(0.5 MPa) = 0.087 MPa Discussion We can also solve this problem using the relations for compressible isentropic flow. The results would be identical.

i

AIR

e Mae = 1.8

Vi ≈ 0

17-65 Nitrogen enters a converging-diverging nozzle at a given pressure. The critical velocity, pressure, temperature, and density in the nozzle are to be determined. Assumptions 1 Nitrogen is an ideal gas. 2 Flow through the nozzle is steady, one-dimensional, and isentropic. Properties The properties of nitrogen are k = 1.4 and R = 0.2968 kJ/kg·K (Table A-2a). Analysis The stagnation properties in this case are identical to the inlet properties since the inlet velocity is negligible. They remain constant throughout the nozzle,

P0 = Pi = 700 kPa T0 = Ti = 450 K

ρ0 =

i

N2

Vi ≈ 0

P0 700 kPa = = 5.241 kg/m 3 RT0 (0.2968 kPa ⋅ m 3 /kg ⋅ K)(450 K)

*

Critical properties are those at a location where the Mach number is Ma= 1. From Table A-32 at Ma =1, we read T/T0 =0.8333, P/P0 = 0.5283, and ρ/ρ0 = 0.6339. Then the critical properties become T* = 0.8333T0 = 0.8333(450 K) = 375 K P* = 0.52828P0 = 0.5283(700 kPa) = 370 MPa

ρ* = 0.63394ρ0 = 0.6339(5.241 kg/m3) = 3.32 kg/m3 Also, ⎛ 1000 m 2 /s 2 V * = c* = kRT * = (1.4)(0.2968 kJ/kg ⋅ K)(375.0 K)⎜⎜ ⎝ 1 kJ/kg

⎞ ⎟ = 395 m/s ⎟ ⎠

Discussion We can also solve this problem using the relations for compressible isentropic flow. The results would be identical.

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17-32

17-66 An ideal gas is flowing through a nozzle. The flow area at a location where Ma = 2.4 is specified. The flow area where Ma = 1.2 is to be determined. Assumptions Flow through the nozzle is steady, one-dimensional, and isentropic. Properties The specific heat ratio is given to be k = 1.4. Analysis The flow is assumed to be isentropic, and thus the stagnation and critical properties remain constant throughout the nozzle. The flow area at a location where Ma2 = 1.2 is determined using A /A* data from Table A-32 to be Ma 1 = 2.4 :

A1 A1 25 cm 2 = 2.4031 ⎯ ⎯→ A* = = = 10.40 cm 2 A* 2.4031 2.4031

Ma 2 = 1.2 :

A2 = 1.0304 ⎯ ⎯→ A2 = (1.0304) A* = (1.0304)(10.40 cm 2 ) = 10.7 cm 2 A*

Discussion We can also solve this problem using the relations for compressible isentropic flow. The results would be identical.

17-67 An ideal gas is flowing through a nozzle. The flow area at a location where Ma = 2.4 is specified. The flow area where Ma = 1.2 is to be determined. Assumptions Flow through the nozzle is steady, one-dimensional, and isentropic. Analysis The flow is assumed to be isentropic, and thus the stagnation and critical properties remain constant throughout the nozzle. The flow area at a location where Ma2 = 1.2 is determined using the A /A* relation, A 1 ⎧⎛ 2 ⎞⎛ k − 1 ⎞⎫ Ma 2 ⎟⎬ = ⎟⎜1 + ⎨⎜ A * Ma ⎩⎝ k + 1 ⎠⎝ 2 ⎠⎭

( k +1) / 2 ( k −1)

For k = 1.33 and Ma1 = 2.4: A1 1 ⎧⎛ 2 ⎞⎛ 1.33 − 1 ⎫ 2.4 2 ⎞⎟⎬ = ⎟⎜1 + ⎨⎜ A * 2.4 ⎩⎝ 1.33 + 1 ⎠⎝ 2 ⎠⎭

2.33 / 2×0.33

= 2.570

and, A* =

A1 25 cm 2 = = 9.729 cm 2 2.570 2.570

For k = 1.33 and Ma2 = 1.2: A2 1 ⎧⎛ 2 ⎞⎛ 1.33 − 1 2 ⎞⎫ 1.2 ⎟⎬ = ⎟⎜1 + ⎨⎜ A * 1.2 ⎩⎝ 1.33 + 1 ⎠⎝ 2 ⎠⎭

2.33 / 2×0.33

= 1.0316

and A2 = (1.0316) A* = (1.0316)(9.729 cm 2 ) = 10.0 cm 2

Discussion Note that the compressible flow functions in Table A-32 are prepared for k = 1.4, and thus they cannot be used to solve this problem.

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17-33

17-68 [Also solved by EES on enclosed CD] Air enters a converging nozzle at a specified temperature and pressure with low velocity. The exit pressure, the exit velocity, and the mass flow rate versus the back pressure are to be calculated and plotted. Assumptions 1 Air is an ideal gas with constant specific heats at room temperature. 2 Flow through the nozzle is steady, one-dimensional, and isentropic. Properties The properties of air are k = 1.4, R = 0.287 kJ/kg·K, and cp = 1.005 kJ/kg·K (Table A-2a). Analysis The stagnation properties in this case are identical to the inlet properties since the inlet velocity is negligible. They remain constant throughout the nozzle since the flow is isentropic. P0 = Pi = 900 kPa T0 = Ti = 400 K The critical pressure is determined to be k /( k −1)

1.4 / 0.4

2 ⎞ 2 ⎞ P* = P0 ⎛⎜ = 475.5 kPa = (900 kPa)⎛⎜ i AIR e ⎟ ⎟ ⎝ 1.4 + 1 ⎠ ⎝ k + 1⎠ Vi ≈ 0 Then the pressure at the exit plane (throat) will be Pe = Pb for Pb ≥ 475.5 kPa for Pb < 475.5 kPa (choked flow) Pe = P* = 475.5 kPa Thus the back pressure will not affect the flow when 100 < Pb < 475.5 kPa. For a specified exit pressure Pe, the temperature, the velocity and the mass flow rate can be determined from ⎛P Te = T0 ⎜⎜ e ⎝ P0

Temperature

⎞ ⎟ ⎟ ⎠

( k −1) / k

⎛ P ⎞ = (400 K)⎜⎜ e ⎟⎟ ⎝ 900 ⎠

0.4 / 1.4

⎛ 1000 m 2 /s 2 Velocity V = 2c p (T0 − Te ) = 2(1.005 kJ/kg ⋅ K)(400 - Te )⎜⎜ ⎝ 1 kJ/kg

⎞ ⎟ ⎟ ⎠

Pe Pe = RTe (0.287 kPa ⋅ m3/kg ⋅ K )Te

Density

ρe =

Mass flow rate

m& = ρeVe Ae = ρeVe (0.001 m 2 )

The results of the calculations can be tabulated as Pb, kPa

Pe, kPa

Te, K

Ve, m/s

ρe, kg/m3

900 800 700 600 500 475.5 400 300 200 100

900 800 700 600 500 475.5 475.5 475.5 475.5 475.5

400 386.8 372.3 356.2 338.2 333.3 333.3 333.3 333.3 333.3

0 162.9 236.0 296.7 352.4 366.2 366.2 366.2 366.2 366.2

7.840 7.206 6.551 5.869 5.151 4.971 4.971 4.971 4.971 4.971

& , kg/s m 0 1.174 1.546 1.741 1.815 1.820 1.820 1.820 1.820 1.820 & m

Ve

Pe

c

Pb

& max m

Pb

100

475.5

900

Pb, kPa

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17-34

17-69 EES Problem 17-68 is reconsidered. Using EES (or other) software, The problem is to be solved for the inlet conditions of 1 MPa and 1000 K. Analysis Using EES, the problem is solved as follows: Procedure ExitPress(P_back,P_crit : P_exit, Condition$) If (P_back>=P_crit) then P_exit:=P_back "Unchoked Flow Condition" Condition$:='unchoked' else P_exit:=P_crit "Choked Flow Condition" Condition$:='choked' Endif End "Input data from Diagram Window" {Gas$='Air' A_cm2=10 P_inlet = 900"kPa" T_inlet= 400"K"} {P_back =475.5 "kPa"}

"Throat area, cm2"

A_exit = A_cm2*Convert(cm^2,m^2) C_p=specheat(Gas$,T=T_inlet) C_p-C_v=R k=C_p/C_v M=MOLARMASS(Gas$) "Molar mass of Gas$" R= 8.314/M "Gas constant for Gas$" "Since the inlet velocity is negligible, the stagnation temperature = T_inlet; and, since the nozzle is isentropic, the stagnation pressure = P_inlet." P_o=P_inlet "Stagnation pressure" T_o=T_inlet "Stagnation temperature" P_crit /P_o=(2/(k+1))^(k/(k-1)) "Critical pressure from Eq. 16-22" Call ExitPress(P_back,P_crit : P_exit, Condition$) T_exit /T_o=(P_exit/P_o)^((k-1)/k)

"Exit temperature for isentopic flow, K"

V_exit ^2/2=C_p*(T_o-T_exit)*1000 "Exit velocity, m/s" Rho_exit=P_exit/(R*T_exit)

"Exit density, kg/m3"

m_dot=Rho_exit*V_exit*A_exit

"Nozzle mass flow rate, kg/s"

"If you wish to redo the plots, hide the diagram window and remove the { } from the first 4 variables just under the procedure. Next set the desired range of back pressure in the parametric table. Finally, solve the table (F3). "

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17-35

SOLUTION A_cm2=10 [cm^2] A_exit=0.001 [m^2] Condition$='choked' C_p=1.14 [kJ/kg-K] C_v=0.8532 [kJ/kg-K] Gas$='Air' k=1.336 M=28.97 [kg/kmol] m_dot=1.258 [kg/s] P_back=300 [kPA]

m [kg/s] 1.819 1.819 1.819 1.819 1.819 1.74 1.546 1.176 0

P_crit=539.2 [kPA] P_exit=539.2 [kPA] P_inlet=1000 [kPA] P_o=1000 [kPA] R=0.287 [kJ/kg-K] Rho_exit=2.195 [m^3/kg] T_exit=856 [K] T_inlet=1000 [K] T_o=1000 [K] V_exit=573 [m/s]

Pexit [kPa] 475.5 475.5 475.5 475.5 475.5 600 700 800 900

Texit [K] 333.3 333.3 333.3 333.3 333.3 356.2 372.3 386.8 400

Vexit [m/s] 366.1 366.1 366.1 366.1 366 296.6 236 163.1 0

ρexit [kg/m3] Pback [kPa] 4.97 100 4.97 200 4.97 300 4.97 400 4.97 475.5 5.868 600 6.551 700 7.207 800 7.839 900

2.0

m, kg/s

1.6

Air, A=10 cm2

1.2

0.8

0.4

0.0 100

200

300

400

500

600

Pback, kPa

700

800

900

.

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17-36

17-70E Air enters a converging-diverging nozzle at a specified temperature and pressure with low velocity. The pressure, temperature, velocity, and mass flow rate are to be calculated in the specified test section. Assumptions 1 Air is an ideal gas. 2 Flow through the nozzle is steady, one-dimensional, and isentropic. Properties The properties of air are k = 1.4 and R = 0.06855 Btu/lbm·R = 0.3704 psia·ft3/lbm·R (Table A2Ea). Analysis The stagnation properties in this case are identical to the inlet properties since the inlet velocity is negligible. They remain constant throughout the nozzle since the flow is isentropic.

P0 = Pi = 150 psia T0 = Ti = 100°F = 560 R Then,

i ⎛ 2 T e = T0 ⎜ ⎜ 2 + (k − 1)Ma 2 ⎝

⎞ ⎛ 2 ⎟ = (560 R)⎜ ⎟ ⎜ 2 + (1.4 - 1)2 2 ⎠ ⎝

⎞ ⎟ = 311 R ⎟ ⎠

AIR

e

Vi ≈ 0

and ⎛T Pe = P0 ⎜⎜ ⎝ T0

⎞ ⎟ ⎟ ⎠

k /( k −1)

311 ⎞ = (150 psia)⎛⎜ ⎟ ⎝ 560 ⎠

1.4 / 0.4

= 19.1 psia

Also,

ρe =

Pe 19.1 psia = = 0.166 1bm/ft 3 RTe (0.3704 psia.ft 3 /1bm.R)(311 R)

The nozzle exit velocity can be determined from Ve = Maece , where ce is the speed of sound at the exit conditions, ⎛ 25,037 ft 2 / s 2 Ve = Maece = Ma e kRTe = (2) (1.4)(0.06855 Btu/1bm.R)(311 R)⎜⎜ ⎝ 1 Btu/1bm

⎞ ⎟ = 1729 ft/s ⎟ ⎠

Finally, m& = ρe AeVe = (0. 166 1bm/ft3)(5 ft2 )(1729 ft/s) = 1435 1bm/s

Discussion Air must be very dry in this application because the exit temperature of air is extremely low, and any moisture in the air will turn to ice particles.

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17-37

Shock Waves and Expansion Waves

17-71C No, because the flow must be supersonic before a shock wave can occur. The flow in the converging section of a nozzle is always subsonic. 17-72C The Fanno line represents the states which satisfy the conservation of mass and energy equations. The Rayleigh line represents the states which satisfy the conservation of mass and momentum equations. The intersections points of these lines represents the states which satisfy the conservation of mass, energy, and momentum equations. 17-73C No, the second law of thermodynamics requires the flow after the shock to be subsonic.. 17-74C (a) decreases, (b) increases, (c) remains the same, (d) increases, and (e) decreases. 17-75C Oblique shocks occur when a gas flowing at supersonic speeds strikes a flat or inclined surface. Normal shock waves are perpendicular to flow whereas inclined shock waves, as the name implies, are typically inclined relative to the flow direction. Also, normal shocks form a straight line whereas oblique shocks can be straight or curved, depending on the surface geometry. 17-76C Yes, the upstream flow have to be supersonic for an oblique shock to occur. No, the flow downstream of an oblique shock can be subsonic, sonic, and even supersonic. 17-77C Yes. Conversely, normal shocks can be thought of as special oblique shocks in which the shock angle is β = π/2, or 90o. 17-78C When the wedge half-angle δ is greater than the maximum deflection angle θmax, the shock becomes curved and detaches from the nose of the wedge, forming what is called a detached oblique shock or a bow wave. The numerical value of the shock angle at the nose is be β = 90o. 17-79C When supersonic flow impinges on a blunt body like the rounded nose of an aircraft, the wedge half-angle δ at the nose is 90o, and an attached oblique shock cannot exist, regardless of Mach number. Therefore, a detached oblique shock must occur in front of all such blunt-nosed bodies, whether twodimensional, axisymmetric, or fully three-dimensional. 17-80C Isentropic relations of ideal gases are not applicable for flows across (a) normal shock waves and (b) oblique shock waves, but they are applicable for flows across (c) Prandtl-Meyer expansion waves.

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17-38

17-81 For an ideal gas flowing through a normal shock, a relation for V2/V1 in terms of k, Ma1, and Ma2 is to be developed. Analysis The conservation of mass relation across the shock is ρ1V1 = ρ 2V2 and it can be expressed as V2 ρ1 P / RT1 ⎛ P1 ⎞⎛ T2 = = 1 = ⎜ ⎟⎜ V1 ρ 2 P2 / RT2 ⎜⎝ P2 ⎟⎠⎜⎝ T1

⎞ ⎟⎟ ⎠

From Eqs. 17-35 and 17-38, V 2 ⎛ 1 + kMa 22 =⎜ V1 ⎜⎝ 1 + kMa 12

⎞⎛ 1 + Ma 12 (k − 1) / 2 ⎞ ⎟⎜ ⎟ ⎟⎜ 1 + Ma 2 (k − 1) / 2 ⎟ 2 ⎠⎝ ⎠

Discussion This is an important relation as it enables us to determine the velocity ratio across a normal shock when the Mach numbers before and after the shock are known.

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17-39

17-82 Air flowing through a converging-diverging nozzle experiences a normal shock at the exit. The effect of the shock wave on various properties is to be determined. Assumptions 1 Air is an ideal gas. 2 Flow through the nozzle is steady, one-dimensional, and isentropic before the shock occurs. 3 The shock wave occurs at the exit plane. Properties The properties of air are k = 1.4 and R = 0.287 kJ/kg·K (Table A-2a). Analysis The inlet stagnation properties in this case are identical to the inlet properties since the inlet velocity is negligible. Then,

Shock wave

P01 = Pi = 1.5 MPa

i

T01 = Ti = 350 K

AIR

1

2

Vi ≈ 0

Then, ⎛ 2 T1 = T01 ⎜ ⎜ 2 + (k − 1)Ma 2 1 ⎝

⎞ ⎛ 2 ⎟ = (350 K)⎜ ⎜ 2 + (1.4 - 1)2 2 ⎟ ⎝ ⎠

⎞ ⎟ = 194.4 K ⎟ ⎠

and ⎛T P1 = P01 ⎜⎜ 1 ⎝ T0

⎞ ⎟ ⎟ ⎠

k /( k −1)

⎛ 194.4 ⎞ = (1.5 MPa)⎜ ⎟ ⎝ 300 ⎠

1.4 / 0.4

= 0.1917 MPa

The fluid properties after the shock (denoted by subscript 2) are related to those before the shock through the functions listed in Table A-33. For Ma1 = 2.0 we read Ma 2 = 0.5774 ,

P02 P T = 0.7209, 2 = 4.5000, and 2 = 1.6875 P01 P1 T1

Then the stagnation pressure P02, static pressure P2, and static temperature T2, are determined to be P02 = 0.7209P01 = (0.7209)(1.5 MPa) = 1.081 MPa P2= 4.5000P1 = (4.5000)(0.1917 MPa) = 0.863 MPa T2 = 1.6875T1 = (1.6875)(194.4 K) = 328.1 K The air velocity after the shock can be determined from V2 = Ma2c2, where c2 is the velocity of sound at the exit conditions after the shock, ⎛ 1000 m 2 / s 2 V2 = Ma2c2 = Ma 2 kRT2 = (0.5774) (1.4)(0.287 kJ/kg ⋅ K)(328.1 K)⎜ ⎜ 1 kJ/kg ⎝

⎞ ⎟ = 209.6 m/s ⎟ ⎠

Discussion We can also solve this problem using the relations for normal shock functions. The results would be identical.

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17-40

17-83 Air enters a converging-diverging nozzle at a specified state. The required back pressure that produces a normal shock at the exit plane is to be determined for the specified nozzle geometry. Assumptions 1 Air is an ideal gas. 2 Flow through the nozzle is steady, one-dimensional, and isentropic before the shock occurs. 3 The shock wave occurs at the exit plane. Analysis The inlet stagnation pressure in this case is identical to the inlet pressure since the inlet velocity is negligible. Since the flow before the shock to be isentropic,

P01 = Pi = 2 MPa It is specified that A/A* =3.5. From Table A-32, Mach number and the pressure ratio which corresponds to this area ratio are the Ma1 =2.80 and P1/P01 = 0.0368. The pressure ratio across the shock for this Ma1 value is, from Table A-33, P2/P1 = 8.98. Thus the back pressure, which is equal to the static pressure at the nozzle exit, must be

shock wave i

AIR

1

2

Vi ≈ 0 Pb

P2 =8.98P1 = 8.98×0.0368P01 = 8.98×0.0368×(2 MPa) = 0.661 MPa Discussion We can also solve this problem using the relations for compressible flow and normal shock functions. The results would be identical.

17-84 Air enters a converging-diverging nozzle at a specified state. The required back pressure that produces a normal shock at the exit plane is to be determined for the specified nozzle geometry. Assumptions 1 Air is an ideal gas. 2 Flow through the nozzle is steady, one-dimensional, and isentropic before the shock occurs. Analysis The inlet stagnation pressure in this case is identical to the inlet pressure since the inlet velocity is negligible. Since the flow before the shock to be isentropic,

shock wave

P01= Pi = 2 MPa It is specified that A/A* = 2. From Table A-32, the Mach number and the pressure ratio which corresponds to this area ratio are the Ma1 =2.20 and P1/P01 = 0.0935. The pressure ratio across the shock for this Ma1 value is, from Table A-33, P2/P1 = 5.48. Thus the back pressure, which is equal to the static pressure at the nozzle exit, must be

i

AIR

1

2

Vi ≈ 0 Pb

P2 =5.48P1 = 5.48×0.0935P01 = 5.48×0.0935×(2 MPa) = 1.02 MPa Discussion We can also solve this problem using the relations for compressible flow and normal shock functions. The results would be identical.

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17-41

17-85 Air flowing through a nozzle experiences a normal shock. The effect of the shock wave on various properties is to be determined. Analysis is to be repeated for helium under the same conditions. Assumptions 1 Air and helium are ideal gases with constant specific heats. 2 Flow through the nozzle is steady, one-dimensional, and isentropic before the shock occurs. Properties The properties of air are k = 1.4 and R = 0.287 kJ/kg·K, and the properties of helium are k = 1.667 and R = 2.0769 kJ/kg·K (Table A-2a). Analysis The air properties upstream the shock are

shock wave

Ma1 = 2.5, P1 = 61.64 kPa, and T1 = 262.15 K Fluid properties after the shock (denoted by subscript 2) are related to those before the shock through the functions in Table A-33. For Ma1 = 2.5, Ma 2 = 0.513,

i

AIR

1

2 Ma1 = 2.5

P02 P T = 8.5262, 2 = 7.125, and 2 = 2.1375 P1 P1 T1

Then the stagnation pressure P02, static pressure P2, and static temperature T2, are determined to be P02 = 8.5261P1 = (8.5261)(61.64 kPa) = 526 kPa P2 = 7.125P1 = (7.125)(61.64 kPa) = 439 kPa T2 = 2.1375T1 = (2.1375)(262.15 K) = 560 K The air velocity after the shock can be determined from V2 = Ma2c2, where c2 is the speed of sound at the exit conditions after the shock, ⎛ 1000 m 2 / s 2 V 2 = Ma 2 c 2 = Ma 2 kRT2 = (0.513) (1.4)(0.287 kJ/kg ⋅ K)(560.3 K)⎜⎜ ⎝ 1 kJ/kg

⎞ ⎟ = 243 m/s ⎟ ⎠

We now repeat the analysis for helium. This time we cannot use the tabulated values in Table A-33 since k is not 1.4. Therefore, we have to calculate the desired quantities using the analytical relations, ⎛ Ma 12 + 2 /( k − 1) ⎞ ⎟ Ma 2 = ⎜⎜ 2 ⎟ ⎝ 2Ma 1 k /( k − 1) − 1 ⎠

1/ 2

⎞ ⎛ 2.5 2 + 2 /(1.667 − 1) ⎟ = ⎜⎜ 2 ⎟ ⎝ 2 × 2.5 × 1.667 /(1.667 − 1) − 1 ⎠

1/ 2

= 0.553

P2 1 + kMa 12 1 + 1.667 × 2.5 2 = = 7.5632 = P1 1 + kMa 22 1 + 1.667 × 0.553 2 T2 1 + Ma 12 (k − 1) / 2 1 + 2.5 2 (1.667 − 1) / 2 = 2.7989 = = 2 T1 1 + Ma 2 (k − 1) / 2 1 + 0.553 2 (1.667 − 1) / 2 P02 ⎛ 1 + kMa 12 =⎜ P1 ⎜⎝ 1 + kMa 22

⎞ ⎟ 1 + (k − 1)Ma 22 / 2 ⎟ ⎠

(

⎛ 1 + 1.667 × 2.5 2 = ⎜⎜ 2 ⎝ 1 + 1.667 × 0.553

Thus,

)k /(k −1)

(

)

⎞ 1.667 / 0.667 ⎟ 1 + (1.667 − 1) × 0.553 2 / 2 = 9.641 ⎟ ⎠

P02 = 11.546P1 = (9.641)(61.64 kPa) = 594 kPa P2 = 7.5632P1 = (7.5632)(61.64 kPa) = 466 kPa T2 = 2.7989T1 = (2.7989)(262.15 K) = 734 K

⎛ 1000 m 2 / s 2 V 2 = Ma 2 c 2 = Ma 2 kRT y = (0.553) (1.667)(2.0769 kJ/kg ⋅ K)(733.7 K)⎜⎜ ⎝ 1 kJ/kg

⎞ ⎟ = 881 m/s ⎟ ⎠

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17-42

17-86 Air flowing through a nozzle experiences a normal shock. The entropy change of air across the normal shock wave is to be determined. Assumptions 1 Air and helium are ideal gases with constant specific heats. 2 Flow through the nozzle is steady, one-dimensional, and isentropic before the shock occurs. Properties The properties of air are R = 0.287 kJ/kg·K and cp = 1.005 kJ/kg·K, and the properties of helium are R = 2.0769 kJ/kg·K and cp = 5.1926 kJ/kg·K (Table A-2a). Analysis The entropy change across the shock is determined to be s 2 − s1 = c p ln

P T2 − R ln 2 = (1.005 kJ/kg ⋅ K)ln(2.1375) - (0.287 kJ/kg ⋅ K)ln(7.125) = 0.200 kJ/kg ⋅ K P1 T1

For helium, the entropy change across the shock is determined to be s 2 − s1 = c p ln

P T2 − R ln 2 = (5.1926 kJ/kg ⋅ K)ln(2.7989) - (2.0769 kJ/kg ⋅ K)ln(7.5632) = 1.14 kJ/kg ⋅ K P1 T1

Discussion Note that shock wave is a highly dissipative process, and the entropy generation is large during shock waves.

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17-43

17-87E [Also solved by EES on enclosed CD] Air flowing through a nozzle experiences a normal shock. Effect of the shock wave on various properties is to be determined. Analysis is to be repeated for helium. Assumptions 1 Air and helium are ideal gases with constant specific heats. 2 Flow through the nozzle is steady, one-dimensional, and isentropic before the shock occurs. Properties The properties of air are k = 1.4 and R = 0.06855 Btu/lbm·R, and the properties of helium are k = 1.667 and R = 0.4961 Btu/lbm·R. shock Analysis The air properties upstream the shock are wave Ma = 2.5, P = 10 psia, and T = 440.5 R 1

1

1

Fluid properties after the shock (denoted by subscript i 2) are related to those before the shock through the functions listed in Table A-33. For Ma1 = 2.5, P P T Ma 2 = 0.513, 02 = 8.5262, 2 = 7.125, and 2 = 2.1375 P1 P1 T1

AIR

2

1

Ma1 = 2.5

Then the stagnation pressure P02, static pressure P2, and static temperature T2, are determined to be P02 = 8.5262P1 = (8.5262)(10 psia) = 85.3 psia P2 = 7.125P1 = (7.125)(10 psia) = 71.3 psia T2 = 2.1375T1 = (2.1375)(440.5 R) = 942 R The air velocity after the shock can be determined from V2 = Ma2c2, where c2 is the speed of sound at the exit conditions after the shock, ⎛ 25,037 ft 2 / s 2 V 2 = Ma 2 c 2 = Ma 2 kRT2 = (0.513) (1.4)(0.06855 Btu/1bm ⋅ R)(941.6 R)⎜⎜ ⎝ 1 Btu/1bm

⎞ ⎟ = 772 ft/s ⎟ ⎠

We now repeat the analysis for helium. This time we cannot use the tabulated values in Table A-33 since k is not 1.4. Therefore, we have to calculate the desired quantities using the analytical relations, ⎛ Ma 12 + 2 /( k − 1) ⎞ ⎟ Ma 2 = ⎜⎜ 2 ⎟ ⎝ 2Ma 1 k /( k − 1) − 1 ⎠

1/ 2

⎞ ⎛ 2.5 2 + 2 /(1.667 − 1) ⎟ = ⎜⎜ 2 ⎟ ⎝ 2 × 2.5 × 1.667 /(1.667 − 1) − 1 ⎠

1/ 2

= 0.553

P2 1 + kMa 12 1 + 1.667 × 2.5 2 = = 7.5632 = P1 1 + kMa 22 1 + 1.667 × 0.553 2 T2 1 + Ma 12 (k − 1) / 2 1 + 2.5 2 (1.667 − 1) / 2 = 2.7989 = = T1 1 + Ma 22 (k − 1) / 2 1 + 0.553 2 (1.667 − 1) / 2 P02 ⎛ 1 + kMa 12 =⎜ P1 ⎜⎝ 1 + kMa 22

⎞ ⎟ 1 + (k − 1)Ma 22 / 2 ⎟ ⎠

(

⎛ 1 + 1.667 × 2.5 2 = ⎜⎜ 2 ⎝ 1 + 1.667 × 0.553 Thus,

)k /(k −1)

⎞ 1.667 / 0.667 ⎟ 1 + (1.667 − 1) × 0.553 2 / 2 = 9.641 ⎟ ⎠

(

)

P02 = 11.546P1 = (9.641)(10 psia) = 594 psia P2 = 7.5632P1 = (7.5632)(10 psia) = 75.6 psia T2 = 2.7989T1 = (2.7989)(440.5 R) = 1233 R

⎛ 25,037 ft 2 / s 2 V 2 = Ma 2 c 2 = Ma 2 kRT2 = (0.553) (1.667)(0.4961 Btu/1bm.R)(1232.9 R)⎜⎜ ⎝ 1 Btu/1bm

⎞ ⎟ = 2794 ft/s ⎟ ⎠

Discussion This problem could also be solved using the relations for compressible flow and normal shock functions. The results would be identical.

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17-44

17-88E EES Problem 17-87E is reconsidered. The effects of both air and helium flowing steadily in a nozzle when there is a normal shock at a Mach number in the range 2 < Ma1 < 3.5 are to be studied. Also, the entropy change of the air and helium across the normal shock is to be calculated and the results are to be tabulated. Analysis Using EES, the problem is solved as follows: Procedure NormalShock(M_x,k:M_y,PyOPx, TyOTx,RhoyORhox, PoyOPox, PoyOPx) If M_x < 1 Then M_y = -1000;PyOPx=-1000;TyOTx=-1000;RhoyORhox=-1000 PoyOPox=-1000;PoyOPx=-1000 else M_y=sqrt( (M_x^2+2/(k-1)) / (2*M_x^2*k/(k-1)-1) ) PyOPx=(1+k*M_x^2)/(1+k*M_y^2) TyOTx=( 1+M_x^2*(k-1)/2 )/(1+M_y^2*(k-1)/2 ) RhoyORhox=PyOPx/TyOTx PoyOPox=M_x/M_y*( (1+M_y^2*(k-1)/2)/ (1+M_x^2*(k-1)/2) )^((k+1)/(2*(k-1))) PoyOPx=(1+k*M_x^2)*(1+M_y^2*(k-1)/2)^(k/(k-1))/(1+k*M_y^2) Endif End Function ExitPress(P_back,P_crit) If P_back>=P_crit then ExitPress:=P_back If P_back 94 kPa

Therefore, the flow is choked, and the velocity at the exit of the hole is the sonic speed. Then the flow properties at the exit becomes

ρ0 =

P0 314 kPa = = 3.671 kg/m 3 RT0 (0.287 kPa ⋅ m 3 / kg ⋅ K )(298 K) 2 ⎞ ⎟ ⎝ k + 1⎠

ρ * = ρ 0 ⎛⎜ T* =

1 /( k −1)

2 ⎞ = (3.671 kg/m 3 )⎛⎜ ⎟ ⎝ 1.4 + 1 ⎠

1 /(1.4 −1)

= 2.327 kg/m 3

2 2 T0 = (298 K) = 248.3 K k +1 1.4 + 1

⎛ 1000 m 2 / s 2 V = c = kRT * = (1.4 )(0.287 kJ/kg ⋅ K)⎜⎜ ⎝ 1 kJ/kg

⎞ ⎟(248.3 K) = 315.9 m/s ⎟ ⎠

Then the initial mass flow rate through the hole becomes m& = ρAV = (2.327 kg/m 3 )[π (0.004 m) 2 /4](315.9 m/s) = 0.00924 kg/s = 0.554 kg/min

Discussion The mass flow rate will decrease with time as the pressure inside the tire drops.

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17-80

17-122 The thrust developed by the engine of a Boeing 777 is about 380 kN. The mass flow rate of air through the nozzle is to be determined. Assumptions 1 Air is an ideal gas with constant specific properties. 2 Flow of combustion gases through the nozzle is isentropic. 3 Choked flow conditions exist at the nozzle exit. 4 The velocity of gases at the nozzle inlet is negligible. Properties The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A-1), and it can also be used for combustion gases. The specific heat ratio of combustion gases is k = 1.33 (Table 17-2). Analysis The velocity at the nozzle exit is the sonic velocity, which is determined to be ⎛ 1000 m 2 / s 2 V = c = kRT = (1.33)(0.287 kJ/kg ⋅ K)⎜ ⎜ 1 kJ/kg ⎝

⎞ ⎟(265 K) = 318.0 m/s ⎟ ⎠

Noting that thrust F is related to velocity by F = m& V , the mass flow rate of combustion gases is determined to be m& =

F 380,000 N ⎛⎜ 1 kg.m/s 2 = 318.0 m/s ⎜⎝ 1 N V

⎞ ⎟ = 1194.8 kg/s ⎟ ⎠

Discussion The combustion gases are mostly nitrogen (due to the 78% of N2 in air), and thus they can be treated as air with a good degree of approximation.

17-123 A stationary temperature probe is inserted into an air duct reads 85°C. The actual temperature of air is to be determined. Assumptions 1 Air is an ideal gas with constant specific heats at room temperature. 2 The stagnation process is isentropic. Properties The specific heat of air at room temperature is cp = 1.005 kJ/kg⋅K (Table A-2a). Analysis The air that strikes the probe will be brought to a complete stop, and thus it will undergo a stagnation process. The thermometer will sense the temperature of this stagnated air, which is the stagnation temperature. The actual air temperature is determined from T = T0 −

(250 m/s) 2 V2 ⎛ 1 kJ/kg ⎞ = 85°C − ⎜ ⎟ = 53.9°C 2c p 2 × 1.005 kJ/kg ⋅ K ⎝ 1000 m 2 / s 2 ⎠

Discussion Temperature rise due to stagnation is very significant in high-speed flows, and should always be considered when compressibility effects are not negligible.

T 250 m/s

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17-81

17-124 Nitrogen flows through a heat exchanger. The stagnation pressure and temperature of the nitrogen at the inlet and the exit states are to be determined. Assumptions 1 Nitrogen is an ideal gas with constant specific properties. 2 Flow of nitrogen through the heat exchanger is isentropic. Properties The properties of nitrogen are cp = 1.039 kJ/kg.K and k = 1.4 (Table A-2a).

Qin 150 kPa 10°C 100 m/s

Nitrogen

100 kPa 180 m/s

Analysis The stagnation temperature and pressure of nitrogen at the inlet and the exit states are determined from T01 = T1 +

V1 2 (100 m/s) 2 ⎛ 1 kJ/kg ⎞ = 10°C + ⎜ ⎟ = 14.8°C 2c p 2 × 1.039 kJ/kg ⋅ °C ⎝ 1000 m 2 / s 2 ⎠

⎛T ⎞ P01 = P1 ⎜⎜ 01 ⎟⎟ ⎝ T1 ⎠

k /( k −1)

⎛ 288.0 K ⎞ = (150 kPa)⎜ ⎟ ⎝ 283.2 K ⎠

1.4 /(1.4 −1)

= 159.1 kPa

From the energy balance relation E in − E out = ΔE system with w = 0 V 22 − V1 2 + Δpe ©0 2 (180 m/s) 2 − (100 m/s) 2 ⎛ 1 kJ/kg ⎞ 125 kJ/kg = (1.039 kJ/kg ⋅ °C)(T2 − 10°C) + ⎜ ⎟ 2 ⎝ 1000 m 2 / s 2 ⎠ T2 = 119.5°C q in = c p (T2 − T1 ) +

and, T02 = T2 +

V2 2 (180 m/s) 2 ⎛ 1 kJ/kg ⎞ = 119.5°C + ⎜ ⎟ = 135.1°C 2c p 2 × 1.039 kJ/kg ⋅ °C ⎝ 1000 m 2 / s 2 ⎠

⎛T P02 = P2 ⎜⎜ 02 ⎝ T2

⎞ ⎟⎟ ⎠

k /( k −1)

⎛ 408.3 K ⎞ = (100 kPa)⎜ ⎟ ⎝ 392.7 K ⎠

1.4 /(1.4 −1)

= 114.6 kPa

Discussion Note that the stagnation temperature and pressure can be very different than their thermodynamic counterparts when dealing with compressible flow.

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17-82

17-125 An expression for the speed of sound based on van der Waals equation of state is to be derived. Using this relation, the speed of sound in carbon dioxide is to be determined and compared to that obtained by ideal gas behavior. Properties The properties of CO2 are R = 0.1889 kJ/kg·K and k = 1.279 at T = 50°C = 323.2 K (Table A2b). Analysis Van der Waals equation of state can be expressed as P=

RT

−

a

v −b v 2

Differentiating, RT 2a ⎛ ∂P ⎞ + 3 ⎜ ⎟ = 2 v ⎝ ∂v ⎠ T (v − b) ⎯→ dρ = −dv / v 2 , the speed of sound relation becomes Noting that ρ = 1 / v ⎯ ⎛ ∂P ⎞ ⎛ ∂P ⎞ c 2 = k⎜ ⎟ = v 2 k⎜ ⎟ ∂ r ⎝ ⎠T ⎝ ∂v ⎠ T

Substituting, c2 =

v 2 kRT 2ak − v (v − b) 2

Using the molar mass of CO2 (M = 44 kg/kmol), the constant a and b can be expressed per unit mass as a = 0.1882 kPa ⋅ m 6 /kg 2 and

b = 9.70 × 10 −4 m 3 / kg

The specific volume of CO2 is determined to be 200 kPa =

(0.1889 kPa ⋅ m 3 / kg ⋅ K)(323.2 K)

v − 0.000970 m 3 /kg

−

2 × 0.1882 kPa ⋅ m 6 / kg 2

v2

→ v = 0.300 m 3 / kg

Substituting, ⎛ (0.300 m 3 / kg) 2 (1.279)(0.1889 kJ/kg ⋅ K)(323.2 K) 1000 m 2 / s 2 ⎜ ⎜ 1 kJ/kg (0.300 − 0.000970 m 3 / kg ) 2 c=⎜ 6 3 ⎜ 2(0.1882 kPa.m / kg )(1.279) 1000 m 2 / s 2 ⎜− (0.300 m 3 / kg) 2 1 kPa ⋅ m 3 /kg ⎝

⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠

1/ 2

= 271 m/s

If we treat CO2 as an ideal gas, the speed of sound becomes ⎛ 1000 m 2 / s 2 c = kRT = (1.279)(0.1889 kJ/kg ⋅ K)(323.2 K)⎜⎜ ⎝ 1 kJ/kg

⎞ ⎟ = 279 m/s ⎟ ⎠

Discussion Note that the ideal gas relation is the simplest equation of state, and it is very accurate for most gases encountered in practice. At high pressures and/or low temperatures, however, the gases deviate from ideal gas behavior, and it becomes necessary to use more complicated equations of state.

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17-83

17-126 The equivalent relation for the speed of sound is to be verified using thermodynamic relations. ⎛ ∂P ⎞ ⎛ ∂P ⎞ Analysis The two relations are c 2 = ⎜⎜ ⎟⎟ and c 2 = k ⎜⎜ ⎟⎟ ⎝ ∂ρ ⎠ s ⎝ ∂ρ ⎠ T ⎯→ dr = −dv / v 2 . Thus, From r = 1 / v ⎯ ⎛ ∂P ⎞ ⎛ ∂P ⎞ 2 ⎛ ∂P ∂T ⎞ 2 ⎛ ∂P ⎞ ⎛ ∂T ⎞ c 2 = ⎜ ⎟ = −v 2 ⎜ ⎟ = −v ⎜ ⎟ = −v ⎜ ⎟ ⎜ ⎟ ⎝ ∂r ⎠ s ⎝ ∂v ⎠ s ⎝ ∂T ∂v ⎠ s ⎝ ∂T ⎠ s ⎝ ∂v ⎠ s

From the cyclic rule, ⎛ ∂P ⎞ ⎛ ∂T ⎞ ⎛ ∂s ⎞ ⎛ ∂P ⎞ ⎛ ∂s ⎞ ⎛ ∂P ⎞ ( P, T , s ) : ⎜ ⎯→ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = −1 ⎯ ⎟ = −⎜ ⎟ ⎜ ⎟ ∂ ∂ ∂ ∂ T s P T ⎝ ⎠s ⎝ ⎠P ⎝ ⎠T ⎝ ⎠s ⎝ ∂T ⎠ P ⎝ ∂s ⎠ T ⎛ ∂T ⎞ ⎛ ∂v ⎞ ⎛ ∂s ⎞ ⎛ ∂T ⎞ ⎛ ∂s ⎞ ⎛ ∂T ⎞ ⎯→ ⎜ (T , v , s ) : ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = −1 ⎯ ⎟ = −⎜ ⎟ ⎜ ⎟ ∂ ∂ s ∂ T ∂ v v ⎠v ⎝ ⎠s ⎝ ∂v ⎠ T ⎝ ∂s ⎠ v ⎝ ⎠s ⎝ ⎠T ⎝

Substituting, ⎛ ∂s ⎞ ⎛ ∂P ⎞ ⎛ ∂s ⎞ ⎛ ∂T ⎞ 2 ⎛ ∂s ⎞ ⎛ ∂T ⎞ ⎛ ∂P ⎞ c 2 = −v 2 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = −v ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠ P ⎝ ∂s ⎠T ⎝ ∂v ⎠T ⎝ ∂s ⎠v ⎝ ∂T ⎠ P ⎝ ∂s ⎠v ⎝ ∂s ⎠T Recall that ⎛ ∂s ⎞ =⎜ ⎟ T ⎝ ∂T ⎠ P

cp

and

cv ⎛ ∂s ⎞ =⎜ ⎟ T ⎝ ∂T ⎠ v

Substituting, ⎛ c p ⎞⎛ T c 2 = −v 2 ⎜⎜ ⎟⎟⎜⎜ ⎝ T ⎠⎝ cv

⎞⎛ ∂P ⎞ 2 ⎛ ∂P ⎞ ⎟⎟⎜ ⎟ = −v k ⎜ ⎟ ⎝ ∂v ⎠T ⎠⎝ ∂v ⎠T

Replacing − dv / v 2 by dρ, ⎛ ∂P ⎞ ⎟⎟ c 2 = k ⎜⎜ ⎝ ∂ρ ⎠ T

Discussion Note that the differential thermodynamic property relations are very useful in the derivation of other property relations in differential form.

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17-84

17-127 For ideal gases undergoing isentropic flows, expressions for P/P*, T/T*, and ρ/ρ* as functions of k and Ma are to be obtained. Analysis Equations 17-18 and 17-21 are given to be

T0 2 + (k − 1)Ma 2 = T 2 and T* 2 = T0 k + 1

Multiplying the two, ⎛ T0 T * ⎞ ⎛ 2 + (k − 1)Ma 2 ⎟ ⎜ ⎜ ⎜ T T ⎟=⎜ 2 0 ⎠ ⎝ ⎝

⎞⎛ 2 ⎞ ⎟⎜ ⎟⎝ k + 1 ⎟⎠ ⎠

Simplifying and inverting, T k +1 = T * 2 + (k − 1)Ma 2

(1)

From P ⎛ T ⎞ =⎜ ⎟ P* ⎝T *⎠

k /( k −1)

ρ ⎛ ρ ⎞ ⎟ =⎜ ρ * ⎜⎝ ρ * ⎟⎠

k /( k −1)

P ⎛ k +1 ⎯ ⎯→ = ⎜⎜ P * ⎝ 2 + (k − 1)Ma 2

⎞ ⎟ ⎟ ⎠

k /( k −1)

ρ ⎛⎜ k +1 ⎯ ⎯→ =⎜ ρ * ⎝ 2 + (k − 1)Ma 2

⎞ ⎟ ⎟ ⎠

k /( k −1)

(2)

From (3)

Discussion Note that some very useful relations can be obtained by very simple manipulations.

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17-85

17-128 It is to be verified that for the steady flow of ideal gases dT0/T = dA/A + (1-Ma2) dV/V. The effect of heating and area changes on the velocity of an ideal gas in steady flow for subsonic flow and supersonic flow are to be explained. Analysis We start with the relation

V2 = c p (T0 − T ) , 2

V dV = c p (dT0 − dT )

Differentiating,

dρ

We also have

ρ dP

and

ρ

+

(1) (2)

dA dV + =0 A V

(3)

+ V dV = 0

(4) dP dρ dT = + =0 P T ρ

Differentiating the ideal gas relation P = ρRT,

(5)

From the speed of sound relation,

c 2 = kRT = (k − 1)c p T = kP / ρ

(6)

Combining Eqs. (3) and (5),

dP dT dA dV − + + =0 P T A V

(7)

Combining Eqs. (4) and (6),

dP

ρ

=

dP kP / c 2

= −VdV

or,

dP k V 2 dV dV = − 2 V dV = −k 2 = − kMa 2 P V V c c

Combining Eqs. (2) and (6),

dT = dT0 − V

or,

dV cp

dT dT0 V 2 dV dT dT0 V2 dV dT0 dV = − = = − 2 = − (k − 1)Ma 2 T T c pT V T T T V c /(k − 1) V

Combining Eqs. (7), (8), and (9),

[

− (k − 1)Ma 2

(8)

(9)

dV dT0 dV dA dV − + (k − 1)Ma 2 + + =0 V T V A V

]

or,

dT0 dA dV = + − kMa 2 + (k − 1)Ma 2 + 1 T A V

Thus,

dT0 dA dV = + (1 − Ma 2 ) T A V

(10)

Differentiating the steady-flow energy equation q = h02 − h01 = c p (T02 − T01 )

δq = c p dT0

(11)

Eq. (11) relates the stagnation temperature change dT0 to the net heat transferred to the fluid. Eq. (10) relates the velocity changes to area changes dA, and the stagnation temperature change dT0 or the heat transferred. (a) When Ma < 1 (subsonic flow), the fluid will accelerate if the duck converges (dA < 0) or the fluid is heated (dT0 > 0 or δq > 0). The fluid will decelerate if the duck converges (dA < 0) or the fluid is cooled (dT0 < 0 or δq < 0). (b) When Ma > 1 (supersonic flow), the fluid will accelerate if the duck diverges (dA > 0) or the fluid is cooled (dT0 < 0 or δq < 0). The fluid will decelerate if the duck converges (dA < 0) or the fluid is heated (dT0 > 0 or δq > 0).

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17-86

17-129 A pitot tube measures the difference between the static and stagnation pressures for a subsonic airplane. The speed of the airplane and the flight Mach number are to be determined.

Assumptions 1 Air is an ideal gas with constant specific heat ratio. 2 The stagnation process is isentropic. Properties The properties of air are R = 0.287 kJ/kg.K and k = 1.4 (Table A-2a). Analysis The stagnation pressure of air at the specified conditions is P0 = P + ΔP = 70.109 + 35 = 105.109 kPa

Then, P0 ⎛ (k − 1)Ma 2 = ⎜1 + P ⎜⎝ 2

⎞ ⎟ ⎟ ⎠

k / k −1

105.109 ⎛⎜ (1.4 − 1)Ma 2 ⎯ ⎯→ = 1+ 70.109 ⎜⎝ 2

⎞ ⎟ ⎟ ⎠

1.4 / 0.4

It yields Ma = 0.783 The speed of sound in air at the specified conditions is ⎛ 1000 m 2 / s 2 ⎞ ⎟ = 328.5 m/s c = kRT = (1.4)(0.287 kJ/kg ⋅ K)(268.65 K)⎜ ⎜ 1 kJ/kg ⎟ ⎝ ⎠

Thus, V = Ma × c = (0.783)(328.5 m/s) = 257.3 m/s

Discussion Note that the flow velocity can be measured in a simple and accurate way by simply measuring pressure.

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17-87

17-130 The mass flow parameter m& RT0 / ( AP0 ) versus the Mach number for k = 1.2, 1.4, and 1.6 in the range of 0 ≤ Ma ≤ 1 is to be plotted.

Analysis The mass flow rate parameter (m& RT0 ) / P0 A can be expressed as m& RT0 P0 A

⎛ 2 = Ma k ⎜⎜ 2 ⎝ 2 + (k − 1) M

⎞ ⎟ ⎟ ⎠

( k +1) / 2 ( k −1)

Thus, Ma

k = 1.2

k = 1.4

k = 1.6

0.0

0

0

0

0.1

0.1089

0.1176

0.1257

0.2

0.2143

0.2311

0.2465

0.3

0.3128

0.3365

0.3582

0.4

0.4015

0.4306

0.4571

0.5

0.4782

0.5111

0.5407

0.6

0.5411

0.5763

0.6077

0.7

0.5894

0.6257

0.6578

0.8

0.6230

0.6595

0.6916

0.9

0.6424

0.6787

0.7106

1.0

0.6485

0.6847

0.7164

0.75 k = 1.6 1.4 1.2

0.60

0.45

0.30

0.15 Ma 0

0.2

0.4

0.6

0.8

1.0

Discussion Note that the mass flow rate increases with increasing Mach number and specific heat ratio. It levels off at Ma = 1, and remains constant (choked flow).

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17-88

17-131 Helium gas is accelerated in a nozzle. The pressure and temperature of helium at the location where Ma = 1 and the ratio of the flow area at this location to the inlet flow area are to be determined.

Assumptions 1 Helium is an ideal gas with constant specific heats. 2 Flow through the nozzle is steady, one-dimensional, and isentropic. Properties The properties of helium are R = 2.0769 kJ/kg⋅K, cp = 5.1926 kJ/kg⋅K, and k = 1.667 (Table A2a). Analysis The properties of the fluid at the location where Ma = 1 are the critical properties, denoted by superscript *. We first determine the stagnation temperature and pressure, which remain constant throughout the nozzle since the flow is isentropic. T0 = Ti +

Vi 2 (120 m/s) 2 ⎛ 1 kJ/kg ⎞ = 500 K + ⎜ ⎟ = 501.4 K 2c p 2 × 5.1926 kJ/kg ⋅ K ⎝ 1000 m 2 / s 2 ⎠ i 120 m/s

and ⎛T P0 = Pi ⎜⎜ 0 ⎝ Ti

⎞ ⎟ ⎟ ⎠

k /( k −1)

501.4 K ⎞ = (0.8 MPa)⎛⎜ ⎟ ⎝ 500 K ⎠

1.667 /(1.667 −1)

He

* Ma = 1

= 0.806 MPa

The Mach number at the nozzle exit is given to be Ma = 1. Therefore, the properties at the nozzle exit are the critical properties determined from 2 ⎛ 2 ⎞ ⎛ ⎞ T * = T0 ⎜ ⎟ = (501.4 K)⎜ ⎟ = 376 K + 1 1.667 + 1 k ⎝ ⎠ ⎝ ⎠ 2 ⎞ P* = P0 ⎛⎜ ⎟ ⎝ k + 1⎠

k /( k −1)

2 ⎞ = (0.806 MPa)⎛⎜ ⎟ ⎝ 1.667 + 1 ⎠

1.667 /(1.667 −1)

= 0.393 MPa

The speed of sound and the Mach number at the nozzle inlet are ⎛ 1000 m 2 / s 2 c i = kRT i = (1.667)(2.0769 kJ/kg ⋅ K)(500 K)⎜⎜ ⎝ 1 kJ/kg Ma i =

⎞ ⎟ = 1316 m/s ⎟ ⎠

Vi 120 m/s = = 0.0912 c i 1316 m/s

The ratio of the entrance-to-throat area is 1 ⎡⎛ 2 ⎞⎛ k − 1 ⎞⎤ = Ma i2 ⎟⎥ ⎟⎜1 + ⎢⎜ * Ma i ⎣⎝ k + 1 ⎠⎝ 2 ⎠⎦ A Ai

=

( k +1) /[ 2 ( k −1)]

1 ⎡⎛ 2 ⎞⎛ 1.667 − 1 ⎞⎤ (0.0912) 2 ⎟⎥ ⎟⎜1 + ⎢⎜ 0.0912 ⎣⎝ 1.667 + 1 ⎠⎝ 2 ⎠⎦

2.667 /( 2×0.667 )

= 6.20

Then the ratio of the throat area to the entrance area becomes 1 A* = = 0.161 Ai 6.20

Discussion The compressible flow functions are essential tools when determining the proper shape of the compressible flow duct.

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17-89

17-132 Helium gas enters a nozzle with negligible velocity, and is accelerated in a nozzle. The pressure and temperature of helium at the location where Ma = 1 and the ratio of the flow area at this location to the inlet flow area are to be determined.

Assumptions 1 Helium is an ideal gas with constant specific heats. 2 Flow through the nozzle is steady, one-dimensional, and isentropic. 3 The entrance velocity is negligible. Properties The properties of helium are R = 2.0769 kJ/kg⋅K, cp = 5.1926 kJ/kg⋅K, and k = 1.667 (Table A2a). Analysis We treat helium as an ideal gas with k = 1.667. The properties of the fluid at the location where Ma = 1 are the critical properties, denoted by superscript *. The stagnation temperature and pressure in this case are identical to the inlet temperature and pressure since the inlet velocity is negligible. They remain constant throughout the nozzle since the flow is isentropic. T0 = Ti = 500 K P0 = Pi = 0.8 MPa The Mach number at the nozzle exit is given to be Ma = 1. Therefore, the properties at the nozzle exit are the critical properties determined from

i Vi ≈ 0

He

* Ma = 1

2 ⎛ 2 ⎞ ⎛ ⎞ T * = T0 ⎜ ⎟ = (500 K)⎜ ⎟ = 375 K ⎝ k +1⎠ ⎝ 1.667 + 1 ⎠ 2 ⎞ P* = P0 ⎛⎜ ⎟ ⎝ k + 1⎠

k /( k −1)

2 ⎞ = (0.8 MPa)⎛⎜ ⎟ ⎝ 1.667 + 1 ⎠

1.667 /(1.667 −1)

= 0.390 MPa

The ratio of the nozzle inlet area to the throat area is determined from Ai A*

=

1 Ma i

⎡⎛ 2 ⎞⎛ k − 1 2 ⎞⎤ ⎢⎜ k + 1 ⎟⎜1 + 2 Ma i ⎟⎥ ⎠⎝ ⎠⎦ ⎣⎝

( k +1) /[ 2 ( k −1)]

But the Mach number at the nozzle inlet is Ma = 0 since Vi ≅ 0. Thus the ratio of the throat area to the nozzle inlet area is A* 1 = =0 Ai ∞

Discussion The compressible flow functions are essential tools when determining the proper shape of the compressible flow duct.

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17-90

17-133 EES Air enters a converging nozzle. The mass flow rate, the exit velocity, the exit Mach number, and the exit pressure-stagnation pressure ratio versus the back pressure-stagnation pressure ratio for a specified back pressure range are to be calculated and plotted.

Assumptions 1 Air is an ideal gas with constant specific heats at room temperature. 2 Flow through the nozzle is steady, one-dimensional, and isentropic. Properties The properties of air at room temperature are R = 0.287 kJ/kg⋅K, cp = 1.005 kJ/kg⋅K, and k = 1.4 (Table A-2a). Analysis The stagnation properties remain constant throughout the nozzle since the flow is isentropic. They are determined from T0 = Ti +

Vi 2 (180 m/s) 2 ⎛ 1 kJ/kg ⎞ = 400 K + ⎜ ⎟ = 416.1 K 2c p 2 × 1.005 kJ/kg ⋅ K ⎝ 1000 m 2 / s 2 ⎠

and

Air

i

⎛T P0 = Pi ⎜⎜ 0 ⎝ Ti

⎞ ⎟ ⎟ ⎠

k /( k −1)

416.1 K ⎞ = (900 kPa)⎛⎜ ⎟ ⎝ 400 K ⎠

1.4 /(1.4 −1)

= 1033.3 kPa

180 m/s

The critical pressure is determined to be ⎛ 2 ⎞ P* = P0 ⎜ ⎟ ⎝ k + 1⎠

k /( k −1)

⎛ 2 ⎞ = (1033.3 kPa)⎜ ⎟ ⎝ 1.4 + 1 ⎠

1.4 / 0.4

= 545.9 kPa

Then the pressure at the exit plane (throat) will be Pe = Pb

for

Pb ≥ 545.9 kPa

Pe = P* = 545.9 kPa

for

Pb < 545.9 kPa

(choked flow)

Thus the back pressure will not affect the flow when 100 < Pb < 545.9 kPa. For a specified exit pressure Pe, the temperature, the velocity and the mass flow rate can be determined from ⎞ ⎟ ⎟ ⎠

( k −1) / k

0.4 / 1.4

Temperature

⎛P Te = T0 ⎜⎜ e ⎝ P0

Velocity

⎛ 1000 m 2 /s 2 V = 2c p (T0 − Te ) = 2(1.005 kJ/kg ⋅ K)(416.1 − Te )⎜⎜ ⎝ 1 kJ/kg

Speed of sound

c e = kRT

Mach number

Ma e = Ve / c e

Density

ρe =

Mass flow rate

m& = ρ eV e Ae = ρ eVe (0.001 m 2 )

e

⎛ Pe ⎞ = (416.1 K)⎜ ⎟ ⎝ 1033.3 ⎠

⎛ 1000 m 2 / s 2 = (1.4)(0.287 kJ/kg ⋅ K)⎜⎜ ⎝ 1 kJ/kg

⎞ ⎟ ⎟ ⎠

⎞ ⎟ ⎟ ⎠

Pe Pe = RTe (0.287 kPa ⋅ m 3 / kg ⋅ K )Te

The results of the calculations can be tabulated as

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

e

17-91

Ma

ρe, kg/m3

& kg / s m,

Pb, kPa

Pb, P0

Pe, kPa

Pb, P0

Te, K

Ve, m/s

900

0.871

900

0.871

400.0

180.0

0.45

7.840

0

800

0.774

800

0.774

386.8

162.9

0.41

7.206

1.174

700

0.677

700

0.677

372.3

236.0

0.61

6.551

1.546

600

0.581

600

0.581

356.2

296.7

0.78

5.869

1.741

545.9

0.528

545.9

0.528

333.3

366.2

1.00

4.971

1.820

500

0.484

545.9

0.528

333.2

366.2

1.00

4.971

1.820

400

0.387

545.9

0.528

333.3

366.2

1.00

4.971

1.820

300

0.290

545.9

0.528

333.3

366.2

1.00

4.971

1.820

200

0.194

545.9

0.528

333.3

366.2

1.00

4.971

1.820

100

0.097

545.9

0.528

333.3

366.2

1.00

4.971

1.820

475.5

P 900 kPa

Pe

Pb

Ve

& m

c

& max m

Pb

100

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17-92

17-134 EES Steam enters a converging nozzle. The exit pressure, the exit velocity, and the mass flow rate versus the back pressure for a specified back pressure range are to be plotted. Assumptions 1 Steam is to be treated as an ideal gas with constant specific heats. 2 Flow through the nozzle is steady, one-dimensional, and isentropic. 3 The nozzle is adiabatic. Properties The ideal gas properties of steam are given to be R = 0.462 kJ/kg.K, cp = 1.872 kJ/kg.K, and k = 1.3. Analysis The stagnation properties in this case are identical to the inlet properties since the inlet velocity is negligible. Since the flow is isentropic, they remain constant throughout the nozzle, P0 = Pi = 6 MPa T0 = Ti = 700 K The critical pressure is determined from to be i e STEAM k /( k −1) 1.3 / 0.3 2 ⎞ 2 ⎞ ⎛ ⎛ P* = P0 ⎜ = (6 MPa)⎜ = 3.274 MPa Vi ≈ 0 ⎟ ⎟ ⎝ k + 1⎠ ⎝ 1.3 + 1 ⎠ Then the pressure at the exit plane (throat) will be Pe = Pb

for

Pb ≥ 3.274 MPa

Pe = P* = 3.274 MPa

for

Pb < 3.274 MPa (choked flow)

Pe

Thus the back pressure will not affect the flow when 3 < Pb < 3.274 MPa. For a specified exit pressure Pe, the temperature, the velocity and the mass flow rate can be determined from Temperature ⎛P Te = T0 ⎜⎜ e ⎝ P0

⎞ ⎟ ⎟ ⎠

( k −1) / k

⎛P ⎞ = (700 K)⎜ e ⎟ ⎝ 6 ⎠

P

0.3 / 1.3

Ve

C

Velocity ⎛ 1000 m 2 /s 2 V = 2c p (T0 − Te ) = 2(1.872 kJ/kg ⋅ K)(700 − Te )⎜⎜ ⎝ 1 kJ/kg

⎞ ⎟ ⎟ ⎠

P & m

Density

ρe =

Pe Pe = RTe (0.462 kPa ⋅ m 3 / kg ⋅ K )Te

& max m

Mass flow rate m& = ρ eV e Ae = ρ eVe (0.0008 m 2 )

3

The results of the calculations can be tabulated as follows:

3.274

Pb, MPa

Pe, MPa

Te, K

Ve, m/s

ρe, kg/m3

& kg / s m,

6.0 5.5 5.0 4.5 4.0 3.5 3.274 3.0

6.0 5.5 5.0 4.5 4.0 3.5 3.274 3.274

700 686.1 671.2 655.0 637.5 618.1 608.7 608.7

0 228.1 328.4 410.5 483.7 553.7 584.7 584.7

18.55 17.35 16.12 14.87 13.58 12.26 11.64 11.64

0 3.166 4.235 4.883 5.255 5.431 5.445 5.445

6

P MPa

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17-93

17-135 An expression for the ratio of the stagnation pressure after a shock wave to the static pressure before the shock wave as a function of k and the Mach number upstream of the shock wave is to be found.

Analysis The relation between P1 and P2 is ⎛ 1 + kMa 12 P2 1 + kMa 22 ⎜ P P = ⎯ ⎯→ = 2 1 ⎜ 1 + kMa 2 P1 1 + kMa 12 2 ⎝

⎞ ⎟ ⎟ ⎠

Substituting this into the isentropic relation

(

P02 = 1 + (k − 1)Ma 22 / 2 P1

)k /(k −1)

Then, P02 ⎛ 1 + kMa 12 =⎜ P1 ⎜⎝ 1 + kMa 22

⎞ ⎟ 1 + (k − 1)Ma 22 / 2 ⎟ ⎠

(

)k /(k −1)

where Ma 22 =

Ma 12 + 2 /( k − 1) 2kMa 22 /( k − 1) − 1

Substituting, P02 ⎛ (1 + kMa 12 )(2kMa 12 − k + 1) ⎞⎛ (k − 1)Ma 12 / 2 + 1 ⎞ ⎟⎜1 + ⎟ =⎜ ⎟⎜ 2kMa 2 /( k − 1) − 1 ⎟ P1 ⎜⎝ kMa 12 (k + 1) − k + 3 1 ⎠⎝ ⎠

k /( k −1)

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17-94

17-136 Nitrogen entering a converging-diverging nozzle experiences a normal shock. The pressure, temperature, velocity, Mach number, and stagnation pressure downstream of the shock are to be determined. The results are to be compared to those of air under the same conditions.

Assumptions 1 Nitrogen is an ideal gas with constant specific heats. 2 Flow through the nozzle is steady, one-dimensional, and isentropic. 3 The nozzle is adiabatic. Properties The properties of nitrogen are R = 0.2968 kJ/kg⋅K and k = 1.4 (Table A-2a). Analysis The inlet stagnation properties in this case are identical to the inlet properties since the inlet velocity is negligible. Assuming the flow before the shock to be isentropic,

shock wave

P01 = Pi = 700 kPa T01 = Ti = 300 K

i

N2

1

2

Vi ≈ 0 Ma1 = 3.0

Then, ⎛ 2 T1 = T01 ⎜ ⎜ 2 + (k − 1)Ma 2 1 ⎝

⎞ ⎛ 2 ⎟ = (300 K)⎜ ⎜ 2 + (1.4 - 1)3 2 ⎟ ⎝ ⎠

⎞ ⎟ = 107.1 K ⎟ ⎠

and ⎛T ⎞ P1 = P01 ⎜⎜ 1 ⎟⎟ ⎝ T01 ⎠

k /( k −1)

⎛ 107.1 ⎞ = (700 kPa)⎜ ⎟ ⎝ 300 ⎠

1.4 / 0.4

= 19.06 kPa

The fluid properties after the shock (denoted by subscript 2) are related to those before the shock through the functions listed in Table A-14. For Ma1 = 3.0 we read Ma 2 = 0.4752,

P02 P T = 0.32834, 2 = 10.333, and 2 = 2.679 P01 P1 T1

Then the stagnation pressure P02, static pressure P2, and static temperature T2, are determined to be P02 = 0.32834 P01 = (0.32834 )(700 kPa) = 230 kPa P2 = 10.333P1 = (10.333)(19.06 kPa) = 197 kPa T2 = 2.679T1 = (2.679)(107.1 K) = 287 K

The velocity after the shock can be determined from V2 = Ma2c2, where c2 is the speed of sound at the exit conditions after the shock, ⎛ 1000 m 2 / s 2 V 2 = Ma 2 c 2 = Ma 2 kRT2 = (0.4752) (1.4)(0.2968 kJ/kg ⋅ K)(287 K)⎜ ⎜ 1 kJ/kg ⎝

⎞ ⎟ = 164 m/s ⎟ ⎠

Discussion For air at specified conditions k = 1.4 (same as nitrogen) and R = 0.287 kJ/kg·K. Thus the only quantity which will be different in the case of air is the velocity after the normal shock, which happens to be 161.3 m/s.

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

17-95

17-137 The diffuser of an aircraft is considered. The static pressure rise across the diffuser and the exit area are to be determined. Assumptions 1 Air is an ideal gas with constant specific heats at room temperature. 2 Flow through the diffuser is steady, oneAIR dimensional, and isentropic. 3 The diffuser is adiabatic. 1 2 Properties Air properties at room temperature are R = 0.287 Diffuser Ma1 = 0.8 Ma2 = 0.3 kJ/kg⋅K, cp = 1.005 kJ/kg·K, and k = 1.4 (Table A-2a). Analysis The inlet velocity is ⎛ 1000 m 2 / s 2 V1 = Ma 1c1 = M 1 kRT1 = (0.8) (1.4)(0.287 kJ/kg ⋅ K)(242.7 K)⎜⎜ ⎝ 1 kJ/kg

⎞ ⎟ = 249.8 m/s ⎟ ⎠

Then the stagnation temperature and pressure at the diffuser inlet become T01 = T1 +

V1 2 (249.8 m/s) 2 ⎛ 1 kJ/kg ⎞ = 242.7 + ⎜ ⎟ = 273.7 K 2c p 2(1.005 kJ/kg ⋅ K) ⎝ 1000 m 2 / s 2 ⎠

⎛T ⎞ P01 = P1 ⎜⎜ 01 ⎟⎟ ⎝ T1 ⎠

k /( k −1)

273.7 K ⎞ = (41.1 kPa)⎛⎜ ⎟ ⎝ 242.7 K ⎠

1.4 /(1.4 −1)

= 62.6 kPa

For an adiabatic diffuser, the energy equation reduces to h01 = h02. Noting that h = cpT and the specific heats are assumed to be constant, we have T01 = T02 = T0 = 273.7 K

The isentropic relation between states 1 and 02 gives ⎛T P02 = P02 = P1 ⎜⎜ 02 ⎝ T1

⎞ ⎟⎟ ⎠

k /( k −1)

⎛ 273.72 K ⎞ = (41.1 kPa)⎜ ⎟ ⎝ 242.7 K ⎠

1.4 /(1.4 −1)

= 62.61 kPa

The exit velocity can be expressed as ⎛ 1000 m 2 / s 2 V 2 = Ma 2 c 2 = Ma 2 kRT2 = (0.3) (1.4)(0.287 kJ/kg ⋅ K) T2 ⎜⎜ ⎝ 1 kJ/kg

Thus ,

T2 = T02 −

⎞ ⎟ = 6.01 T2 ⎟ ⎠

6.012 T2 m 2 /s 2 ⎛ 1 kJ/kg ⎞ V2 2 = (273.7) − ⎜ ⎟ = 268.9 K 2c p 2(1.005 kJ/kg ⋅ K) ⎝ 1000 m 2 / s 2 ⎠

Then the static exit pressure becomes ⎛T P2 = P02 ⎜⎜ 2 ⎝ T02

⎞ ⎟ ⎟ ⎠

k /( k −1)

⎛ 268.9 K ⎞ = (62.61 kPa)⎜ ⎟ ⎝ 273.7 K ⎠

1.4 /(1.4 −1)

= 58.85 kPa

Thus the static pressure rise across the diffuser is ΔP = P2 − P1 = 58.85 − 41.1 = 17.8 kPa

Also,

ρ2 =

P2 58.85 kPa = = 0.7626 kg/m 3 RT2 (0.287 kPa ⋅ m 3 /kg ⋅ K)(268.9 K)

V 2 = 6.01 T2 = 6.01 268.9 = 98.6 m/s

Thus

A2 =

65 kg/s m& = = 0.864 m 2 ρ 2V 2 (0.7626 kg/m 3 )(98.6 m/s)

Discussion The pressure rise in actual diffusers will be lower because of the irreversibilities. However, flow through well-designed diffusers is very nearly isentropic.

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17-96

17-138 Helium gas is accelerated in a nozzle isentropically. For a specified mass flow rate, the throat and exit areas of the nozzle are to be determined. Assumptions 1 Helium is an ideal gas with constant specific heats. 2 Flow through the nozzle is steady, one-dimensional, and isentropic. 3 The nozzle is adiabatic. Properties The properties of helium are R = 2.0769 kJ/kg.K, cp = 5.1926 kJ/kg.K, k = 1.667 (Table A-2a). Analysis The inlet stagnation properties in this case are identical to the inlet properties since the inlet velocity is negligible, T01 = T1 = 500 K P01 = P1 = 1.0 MPa

The flow is assumed to be isentropic, thus the stagnation temperature and pressure remain constant throughout the nozzle, T02 = T01 = 500 K

He

1

*

2

Vi ≈ 0

P02 = P01 = 1.0 MPa The critical pressure and temperature are determined from 2 ⎞ 2 ⎛ ⎞ = 375.0 K T * = T0 ⎛⎜ ⎟ = (500 K)⎜ ⎟ ⎝ k + 1⎠ ⎝ 1.667 + 1 ⎠ ⎛ 2 ⎞ P* = P0 ⎜ ⎟ ⎝ k + 1⎠

ρ* =

k /( k −1)

2 ⎞ = (1.0 MPa)⎛⎜ ⎟ ⎝ 1.667 + 1 ⎠

1.667 /(1.667 −1)

= 0.487 MPa

P* 487 kPa = = 0.625 kg/m 3 RT * (2.0769 kPa ⋅ m 3 /kg ⋅ K)(375 K)

⎛ 1000 m 2 / s 2 V * = c* = kRT * = (1.667)(2.0769 kJ/kg ⋅ K)(375 K)⎜⎜ ⎝ 1 kJ/kg

⎞ ⎟ = 1139.4 m/s ⎟ ⎠

Thus the throat area is 0.25 kg/s m& A* = = = 3.51× 10 − 4 m 2 = 3.51 cm 2 ρ * V * (0.625 kg/m 3 )(1139.4 m/s) At the nozzle exit the pressure is P2 = 0.1 MPa. Then the other properties at the nozzle exit are determined to be k /( k −1)

1.667 / 0.667

P0 ⎛ k − 1 1.0 MPa ⎛ 1.667 − 1 ⎞ ⎞ = ⎜1 + ⎯ ⎯→ = ⎜1 + Ma 22 ⎟ Ma 22 ⎟ 2 0.1 MPa ⎝ 2 P2 ⎝ ⎠ ⎠ It yields Ma2 = 2.130, which is greater than 1. Therefore, the nozzle must be converging-diverging. ⎞ ⎛ ⎞ ⎛ 2 2 ⎟ = (500 K )⎜ ⎟ = 199.0 K T2 = T0 ⎜⎜ 2 ⎟ 2 ⎟ ⎜ ⎝ 2 + (1.667 − 1) × 2.13 ⎠ ⎝ 2 + (k − 1)Ma 2 ⎠ P 100 kPa = 0.242 kg/m 3 ρ2 = 2 = RT2 (2.0769 kPa ⋅ m 3 /kg ⋅ K)(199 K) ⎛ 1000 m 2 / s 2 V 2 = Ma 2 c 2 = Ma 2 kRT2 = (2.13) (1.667)(2.0769 kJ/kg ⋅ K)(199 K)⎜⎜ ⎝ 1 kJ/kg

⎞ ⎟ = 1768.0 m/s ⎟ ⎠

Thus the exit area is 0.25 kg/s m& A2 = = = 5.84 × 10 − 4 m 2 = 5.84 cm 2 ρ 2V 2 (0.242 kg/m 3 )(1768 m/s)

Discussion Flow areas in actual nozzles would be somewhat larger to accommodate the irreversibilities.

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17-97

17-139E Helium gas is accelerated in a nozzle. For a specified mass flow rate, the throat and exit areas of the nozzle are to be determined for the cases of isentropic and 97% efficient nozzles. Assumptions 1 Helium is an ideal gas with constant specific heats. 2 Flow through the nozzle is steady, one-dimensional, and isentropic. 3 The nozzle is adiabatic. Properties The properties of helium are R = 0.4961 Btu/lbm·R = 2.6809 psia·ft3/lbm·R, cp = 1.25 Btu/lbm·R, and k = 1.667 (Table A-2Ea). Analysis The inlet stagnation properties in this case are identical to the inlet properties since the inlet velocity is negligible, T01 = T1 = 900 R P01 = P1 = 150 psia

The flow is assumed to be isentropic, thus the stagnation temperature and pressure remain constant throughout the nozzle, T02 = T01 = 900 R

1

He

*

2

Vi ≈ 0

P02 = P01 = 150 psia

The critical pressure and temperature are determined from 2 ⎛ 2 ⎞ ⎛ ⎞ T * = T0 ⎜ ⎟ = (900 R)⎜ ⎟ = 674.9 R ⎝ k + 1⎠ ⎝ 1.667 + 1 ⎠ ⎛ 2 ⎞ P* = P0 ⎜ ⎟ ⎝ k + 1⎠

ρ* =

and

k /( k −1)

2 ⎞ = (150 psia)⎛⎜ ⎟ 1.667 + 1 ⎝ ⎠ 73.1 psia

1.667 /(1.667 −1)

= 73.1 psia

P* = = 0.0404 1bm/ft 3 RT * (2.6809 psia ⋅ ft 3 /lbm ⋅ R)(674.9 R)

⎛ 25,037 ft 2 / s 2 V* = c* = kRT * = (1.667)(0.4961 Btu/lbm ⋅ R)(674.9 R)⎜⎜ ⎝ 1 Btu/1bm 0.2 1bm/s m& A* = = = 0.00132 ft 2 ρ * V * (0.0404 1bm/ft 3 )(3738 ft/s)

⎞ ⎟ = 3738 ft/s ⎟ ⎠

At the nozzle exit the pressure is P2 = 15 psia. Then the other properties at the nozzle exit are determined to be k /( k −1) 1.667 / 0.667 p0 ⎛ k − 1 150 psia ⎛ 1.667 − 1 ⎞ ⎞ = ⎜1 + ⎯ ⎯→ = ⎜1 + Ma 22 ⎟ Ma 22 ⎟ 2 15 psia ⎝ 2 p2 ⎝ ⎠ ⎠

It yields Ma2 = 2.130, which is greater than 1. Therefore, the nozzle must be converging-diverging. ⎞ ⎛ ⎞ ⎛ 2 2 ⎟ = (900 R )⎜ ⎟ T2 = T0 ⎜⎜ 2 ⎟ 2 ⎜ 2 + (1.667 − 1) × 2.13 ⎟ = 358.1 R ⎠ ⎝ ⎝ 2 + (k − 1)Ma 2 ⎠ P 15 psia = 0.0156 1bm/ft 3 ρ2 = 2 = 3 RT2 (2.6809 psia ⋅ ft /lbm ⋅ R)(358.1 R) ⎛ 25,037 ft 2 / s 2 V 2 = Ma 2 c 2 = Ma 2 kRT2 = (2.13) (1.667)(0.4961 Btu/lbm ⋅ R)(358.1 R)⎜⎜ ⎝ 1 Btu/1bm

⎞ ⎟ = 5800 ft/s ⎟ ⎠

Thus the exit area is 0.2 lbm/s m& A2 = = = 0.00221 ft 2 ρ 2V 2 (0.0156 lbm/ft 3 )(5800 ft/s) Discussion Flow areas in actual nozzles would be somewhat larger to accommodate the irreversibilities.

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

17-98

17-140 [Also solved by EES on enclosed CD] Using the compressible flow relations, the one-dimensional compressible flow functions are to be evaluated and tabulated as in Table A-32 for an ideal gas with k = 1.667. Properties The specific heat ratio of the ideal gas is given to be k = 1.667. Analysis The compressible flow functions listed below are expressed in EES and the results are tabulated. Ma * = Ma A A*

=

1 Ma

k +1 2 + (k − 1)Ma 2

⎡⎛ 2 ⎞⎛ k − 1 2 ⎞⎤ ⎢⎜ k + 1 ⎟⎜1 + 2 Ma ⎟⎥ ⎝ ⎠ ⎝ ⎠⎦ ⎣

P ⎛ k −1 ⎞ = ⎜1 + Ma 2 ⎟ 2 P0 ⎝ ⎠

0.5( k +1) /( k −1)

− k /( k −1)

−1 /( k −1) ρ ⎛ k −1 ⎞ = ⎜1 + Ma 2 ⎟ 2 ρ0 ⎝ ⎠

T ⎛ k −1 ⎞ = ⎜1 + Ma 2 ⎟ T0 ⎝ 2 ⎠

−1

k=1.667 PP0=(1+(k-1)*M^2/2)^(-k/(k-1)) TT0=1/(1+(k-1)*M^2/2) DD0=(1+(k-1)*M^2/2)^(-1/(k-1)) Mcr=M*SQRT((k+1)/(2+(k-1)*M^2)) AAcr=((2/(k+1))*(1+0.5*(k-1)*M^2))^(0.5*(k+1)/(k-1))/M

Ma 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0 5.0 ∝

Ma* 0 0.1153 0.2294 0.3413 0.4501 0.5547 0.6547 0.7494 0.8386 0.9222 1.0000 1.1390 1.2572 1.3570 1.4411 1.5117 1.5713 1.6216 1.6643 1.7007 1.7318 1.8895 1.9996

A/A* ∞ 5.6624 2.8879 1.9891 1.5602 1.3203 1.1760 1.0875 1.0351 1.0081 1.0000 1.0267 1.0983 1.2075 1.3519 1.5311 1.7459 1.9980 2.2893 2.6222 2.9990 9.7920 ∞

P/P0 1.0000 0.9917 0.9674 0.9288 0.8782 0.8186 0.7532 0.6850 0.6166 0.5501 0.4871 0.3752 0.2845 0.2138 0.1603 0.1202 0.0906 0.0686 0.0524 0.0403 0.0313 0.0038 0

ρ/ρ0 1.0000 0.9950 0.9803 0.9566 0.9250 0.8869 0.8437 0.7970 0.7482 0.6987 0.6495 0.5554 0.4704 0.3964 0.3334 0.2806 0.2368 0.2005 0.1705 0.1457 0.1251 0.0351 0

T/T0 1.0000 0.9967 0.9868 0.9709 0.9493 0.9230 0.8928 0.8595 0.8241 0.7873 0.7499 0.6756 0.6047 0.5394 0.4806 0.4284 0.3825 0.3424 0.3073 0.2767 0.2499 0.1071 0

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17-99

17-141 [Also solved by EES on enclosed CD] Using the normal shock relations, the normal shock functions are to be evaluated and tabulated as in Table A-33 for an ideal gas with k = 1.667. Properties The specific heat ratio of the ideal gas is given to be k = 1.667. Analysis The normal shock relations listed below are expressed in EES and the results are tabulated. Ma 2 =

P2 1 + kMa 12 2kMa 12 − k + 1 = = P1 1 + kMa 22 k +1

(k − 1)Ma 12 + 2 2kMa 12 − k + 1

(k + 1)Ma 12 V ρ 2 P2 / P1 = = = 1 , ρ 1 T2 / T1 2 + (k − 1)Ma 12 V 2

T2 2 + Ma 12 (k − 1) = T1 2 + Ma 22 (k − 1) k +1

P02 Ma 1 ⎡1 + Ma 22 (k − 1) / 2 ⎤ 2( k −1) = ⎢ ⎥ P01 Ma 2 ⎢⎣1 + Ma 12 (k − 1) / 2 ⎥⎦

P02 (1 + kMa 12 )[1 + Ma 22 (k − 1) / 2] k /( k −1) = P1 1 + kMa 22

k=1.667 My=SQRT((Mx^2+2/(k-1))/(2*Mx^2*k/(k-1)-1)) PyPx=(1+k*Mx^2)/(1+k*My^2) TyTx=(1+Mx^2*(k-1)/2)/(1+My^2*(k-1)/2) RyRx=PyPx/TyTx P0yP0x=(Mx/My)*((1+My^2*(k-1)/2)/(1+Mx^2*(k-1)/2))^(0.5*(k+1)/(k-1)) P0yPx=(1+k*Mx^2)*(1+My^2*(k-1)/2)^(k/(k-1))/(1+k*My^2)

Ma1 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.0 4.0 5.0 ∞

Ma2 1.0000 0.9131 0.8462 0.7934 0.7508 0.7157 0.6864 0.6618 0.6407 0.6227 0.6070 0.5933 0.5814 0.5708 0.5614 0.5530 0.5455 0.5388 0.5327 0.5273 0.5223 0.4905 0.4753 0.4473

P2/P1 1.0000 1.2625 1.5500 1.8626 2.2001 2.5626 2.9501 3.3627 3.8002 4.2627 4.7503 5.2628 5.8004 6.3629 6.9504 7.5630 8.2005 8.8631 9.5506 10.2632 11.0007 19.7514 31.0022 ∞

ρ2/ρ1 1.0000 1.1496 1.2972 1.4413 1.5805 1.7141 1.8415 1.9624 2.0766 2.1842 2.2853 2.3802 2.4689 2.5520 2.6296 2.7021 2.7699 2.8332 2.8923 2.9476 2.9993 3.3674 3.5703 3.9985

T2/T1 1.0000 1.0982 1.1949 1.2923 1.3920 1.4950 1.6020 1.7135 1.8300 1.9516 2.0786 2.2111 2.3493 2.4933 2.6432 2.7989 2.9606 3.1283 3.3021 3.4819 3.6678 5.8654 8.6834 ∞

P02/P01 1 0.999 0.9933 0.9813 0.9626 0.938 0.9085 0.8752 0.8392 0.8016 0.763 0.7243 0.6861 0.6486 0.6124 0.5775 0.5442 0.5125 0.4824 0.4541 0.4274 0.2374 0.1398 0

P02/P1 2.0530 2.3308 2.6473 2.9990 3.3838 3.8007 4.2488 4.7278 5.2371 5.7767 6.3462 6.9457 7.5749 8.2339 8.9225 9.6407 10.3885 11.1659 11.9728 12.8091 13.6750 23.9530 37.1723 ∞

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17-100

17-142 The critical temperature, pressure, and density of an equimolar mixture of oxygen and nitrogen for specified stagnation properties are to be determined. Assumptions Both oxygen and nitrogen are ideal gases with constant specific heats at room temperature. Properties The specific heat ratio and molar mass are k = 1.395 and M = 32 kg/kmol for oxygen, and k = 1.4 and M = 28 kg/kmol for nitrogen (Tables A-1 and A-2). Analysis The gas constant of the mixture is M m = y O 2 M O 2 + y N 2 M N 2 = 0.5 × 32 + 0.5 × 28 = 30 kg/kmol Rm =

8.314 kJ/kmol ⋅ K Ru = = 0.2771 kJ/kg ⋅ K 30 kg/kmol Mm

The specific heat ratio is 1.4 for nitrogen, and nearly 1.4 for oxygen. Therefore, the specific heat ratio of the mixture is also 1.4. Then the critical temperature, pressure, and density of the mixture become ⎛ 2 ⎞ = (800 K)⎛ 2 ⎞ T * = T0 ⎜ ⎟ ⎜ ⎟ = 667 K ⎝ k + 1⎠ ⎝ 1.4 + 1 ⎠ ⎛ 2 ⎞ P* = P0 ⎜ ⎟ ⎝ k + 1⎠

ρ* =

k /( k −1)

2 ⎞ = (500 kPa)⎛⎜ ⎟ ⎝ 1.4 + 1 ⎠

1.4 /(1.4 −1)

= 264 kPa

264 kPa P* = = 1.43 kg/m 3 RT * (0.2771 kPa ⋅ m 3 /kg ⋅ K)(667 K)

Discussion If the specific heat ratios k of the two gases were different, then we would need to determine the k of the mixture from k = cp,m/cv,m where the specific heats of the mixture are determined from c p ,m = mf O 2 c p ,O 2 + mf N 2 c p , N 2 = ( y O 2 M O 2 / M m )c p ,O 2 + ( y N 2 M N 2 / M m )c p , N 2

cv , m = mf O 2 cv ,O 2 + mf N 2 cv , N 2 = ( y O 2 M O 2 / M m )cv ,O 2 + ( y N 2 M N 2 / M m )cv , N 2

where mf is the mass fraction and y is the mole fraction. In this case it would give c p ,m = (0.5 × 32 / 30) × 0.918 + (0.5 × 28 / 30) × 1.039 = 0.974 kJ/kg.K c p ,m = (0.5 × 32 / 30) × 0.658 + (0.5 × 28 / 30) × 0.743 = 0.698 kJ/kg.K

and k = 0.974/0.698 = 1.40

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17-101

17-143 EES Using EES (or other) software, the shape of a converging-diverging nozzle is to be determined for specified flow rate and stagnation conditions. The nozzle and the Mach number are to be plotted. Assumptions 1 Air is an ideal gas with constant specific heats. 2 Flow through the nozzle is steady, onedimensional, and isentropic. 3 The nozzle is adiabatic. Properties The specific heat ratio of air at room temperature is 1.4 (Table A-2a). Analysis The problem is solved using EES, and the results are tabulated and plotted below. k=1.4 Cp=1.005 "kJ/kg.K" R=0.287 "kJ/kg.K" P0=1400 "kPa" T0=200+273 "K" m=3 "kg/s" rho_0=P0/(R*T0) rho=P/(R*T) T=T0*(P/P0)^((k-1)/k) V=SQRT(2*Cp*(T0-T)*1000) A=m/(rho*V)*10000 "cm2" C=SQRT(k*R*T*1000) Ma=V/C

Pressure P, kPa

Flow area A, cm2

Mach number Ma

1400 1350 1300 1250 1200 1150 1100 1050 1000 950 900 850 800 750 700 650 600 550 500 450 400 350 300 250 200 150 100

∞ 30.1 21.7 18.1 16.0 14.7 13.7 13.0 12.5 12.2 11.9 11.7 11.6 11.5 11.5 11.6 11.8 12.0 12.3 12.8 13.3 14.0 15.0 16.4 18.3 21.4 27.0

0 0.229 0.327 0.406 0.475 0.538 0.597 0.655 0.710 0.766 0.820 0.876 0.931 0.988 1.047 1.107 1.171 1.237 1.308 1.384 1.467 1.559 1.663 1.784 1.929 2.114 2.373

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17-102

2.5

2

Ma

1.5

1

0.5

0 0

200

400

600

800

1000

1200

1400

1000

1200

1400

P, kPa

50 45 40

Flow area A, cm

2

35 30 25 20 15 10 0

200

400

600

800

P, kPa

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17-103

17-144 EES Using the compressible flow relations, the one-dimensional compressible flow functions are to be evaluated and tabulated as in Table A-32 for air. Properties The specific heat ratio is given to be k = 1.4 for air Analysis The compressible flow functions listed below are expressed in EES and the results are tabulated. Ma * = Ma A A*

=

1 Ma

k +1 2 + (k − 1)Ma 2

⎡⎛ 2 ⎞⎛ k − 1 2 ⎞⎤ ⎢⎜ k + 1 ⎟⎜1 + 2 Ma ⎟⎥ ⎠⎝ ⎠⎦ ⎣⎝

P ⎛ k −1 ⎞ = ⎜1 + Ma 2 ⎟ 2 P0 ⎝ ⎠

0.5( k +1) /( k −1)

− k /( k −1)

−1 /( k −1) ρ ⎛ k −1 ⎞ = ⎜1 + Ma 2 ⎟ 2 ρ0 ⎝ ⎠

T ⎛ k −1 ⎞ = ⎜1 + Ma 2 ⎟ T0 ⎝ 2 ⎠

−1

Air: k=1.4 PP0=(1+(k-1)*M^2/2)^(-k/(k-1)) TT0=1/(1+(k-1)*M^2/2) DD0=(1+(k-1)*M^2/2)^(-1/(k-1)) Mcr=M*SQRT((k+1)/(2+(k-1)*M^2)) AAcr=((2/(k+1))*(1+0.5*(k-1)*M^2))^(0.5*(k+1)/(k-1))/M Ma 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0 8.5 9.0 9.5 10.0

Ma* 1.0000 1.3646 1.6330 1.8257 1.9640 2.0642 2.1381 2.1936 2.2361 2.2691 2.2953 2.3163 2.3333 2.3474 2.3591 2.3689 2.3772 2.3843 2.3905

A/A* 1.0000 1.1762 1.6875 2.6367 4.2346 6.7896 10.7188 16.5622 25.0000 36.8690 53.1798 75.1343 104.1429 141.8415 190.1094 251.0862 327.1893 421.1314 535.9375

P/P0 0.5283 0.2724 0.1278 0.0585 0.0272 0.0131 0.0066 0.0035 0.0019 0.0011 0.0006 0.0004 0.0002 0.0002 0.0001 0.0001 0.0000 0.0000 0.0000

ρ/ρ0 0.6339 0.3950 0.2300 0.1317 0.0762 0.0452 0.0277 0.0174 0.0113 0.0076 0.0052 0.0036 0.0026 0.0019 0.0014 0.0011 0.0008 0.0006 0.0005

T/T0 0.8333 0.6897 0.5556 0.4444 0.3571 0.2899 0.2381 0.1980 0.1667 0.1418 0.1220 0.1058 0.0926 0.0816 0.0725 0.0647 0.0581 0.0525 0.0476

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17-104

17-145 EES Using the compressible flow relations, the one-dimensional compressible flow functions are to be evaluated and tabulated as in Table A-32 for methane. Properties The specific heat ratio is given to be k = 1.3 for methane. Analysis The compressible flow functions listed below are expressed in EES and the results are tabulated. Ma * = Ma A A*

=

1 Ma

k +1 2 + (k − 1)Ma 2

⎡⎛ 2 ⎞⎛ k − 1 2 ⎞⎤ ⎢⎜ k + 1 ⎟⎜1 + 2 Ma ⎟⎥ ⎠⎝ ⎠⎦ ⎣⎝

P ⎛ k −1 ⎞ = ⎜1 + Ma 2 ⎟ 2 P0 ⎝ ⎠

0.5( k +1) /( k −1)

− k /( k −1)

−1 /( k −1) ρ ⎛ k −1 ⎞ = ⎜1 + Ma 2 ⎟ 2 ρ0 ⎝ ⎠

T ⎛ k −1 ⎞ = ⎜1 + Ma 2 ⎟ T0 ⎝ 2 ⎠

−1

Methane: k=1.3 PP0=(1+(k-1)*M^2/2)^(-k/(k-1)) TT0=1/(1+(k-1)*M^2/2) DD0=(1+(k-1)*M^2/2)^(-1/(k-1)) Mcr=M*SQRT((k+1)/(2+(k-1)*M^2)) AAcr=((2/(k+1))*(1+0.5*(k-1)*M^2))^(0.5*(k+1)/(k-1))/M Ma 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0 8.5 9.0 9.5 10.0

Ma* 1.0000 1.3909 1.6956 1.9261 2.0986 2.2282 2.3263 2.4016 2.4602 2.5064 2.5434 2.5733 2.5978 2.6181 2.6350 2.6493 2.6615 2.6719 2.6810

A/A* 1.0000 1.1895 1.7732 2.9545 5.1598 9.1098 15.9441 27.3870 45.9565 75.2197 120.0965 187.2173 285.3372 425.8095 623.1235 895.5077 1265.6040 1761.2133 2416.1184

P/P0 0.5457 0.2836 0.1305 0.0569 0.0247 0.0109 0.0050 0.0024 0.0012 0.0006 0.0003 0.0002 0.0001 0.0001 0.0000 0.0000 0.0000 0.0000 0.0000

ρ/ρ0 0.6276 0.3793 0.2087 0.1103 0.0580 0.0309 0.0169 0.0095 0.0056 0.0033 0.0021 0.0013 0.0008 0.0006 0.0004 0.0003 0.0002 0.0001 0.0001

T/T0 0.8696 0.7477 0.6250 0.5161 0.4255 0.3524 0.2941 0.2477 0.2105 0.1806 0.1563 0.1363 0.1198 0.1060 0.0943 0.0845 0.0760 0.0688 0.0625

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17-105

17-146 EES Using the normal shock relations, the normal shock functions are to be evaluated and tabulated as in Table A-33 for air. Properties The specific heat ratio is given to be k = 1.4 for air. Analysis The normal shock relations listed below are expressed in EES and the results are tabulated. Ma 2 =

P2 1 + kMa 12 2kMa 12 − k + 1 = = P1 1 + kMa 22 k +1

(k − 1)Ma 12 + 2 2kMa 12 − k + 1

(k + 1)Ma 12 V ρ 2 P2 / P1 = = = 1 , ρ 1 T2 / T1 2 + (k − 1)Ma 12 V 2

T2 2 + Ma 12 (k − 1) = T1 2 + Ma 22 (k − 1) k +1

P02 Ma 1 ⎡1 + Ma 22 (k − 1) / 2 ⎤ 2( k −1) = ⎢ ⎥ P01 Ma 2 ⎢⎣1 + Ma 12 (k − 1) / 2 ⎥⎦

P02 (1 + kMa 12 )[1 + Ma 22 (k − 1) / 2] k /( k −1) = P1 1 + kMa 22

Air: k=1.4 My=SQRT((Mx^2+2/(k-1))/(2*Mx^2*k/(k-1)-1)) PyPx=(1+k*Mx^2)/(1+k*My^2) TyTx=(1+Mx^2*(k-1)/2)/(1+My^2*(k-1)/2) RyRx=PyPx/TyTx P0yP0x=(Mx/My)*((1+My^2*(k-1)/2)/(1+Mx^2*(k-1)/2))^(0.5*(k+1)/(k-1)) P0yPx=(1+k*Mx^2)*(1+My^2*(k-1)/2)^(k/(k-1))/(1+k*My^2)

Ma1 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0 8.5 9.0 9.5 10.0

Ma2 1.0000 0.7011 0.5774 0.5130 0.4752 0.4512 0.4350 0.4236 0.4152 0.4090 0.4042 0.4004 0.3974 0.3949 0.3929 0.3912 0.3898 0.3886 0.3876

P2/P1 1.0000 2.4583 4.5000 7.1250 10.3333 14.1250 18.5000 23.4583 29.0000 35.1250 41.8333 49.1250 57.0000 65.4583 74.5000 84.1250 94.3333 105.1250 116.5000

ρ2/ρ1 1.0000 1.8621 2.6667 3.3333 3.8571 4.2609 4.5714 4.8119 5.0000 5.1489 5.2683 5.3651 5.4444 5.5102 5.5652 5.6117 5.6512 5.6850 5.7143

T2/T1 1.0000 1.3202 1.6875 2.1375 2.6790 3.3151 4.0469 4.8751 5.8000 6.8218 7.9406 9.1564 10.4694 11.8795 13.3867 14.9911 16.6927 18.4915 20.3875

P02/P01 1 0.9298 0.7209 0.499 0.3283 0.2129 0.1388 0.0917 0.06172 0.04236 0.02965 0.02115 0.01535 0.01133 0.008488 0.006449 0.004964 0.003866 0.003045

P02/P1 1.8929 3.4133 5.6404 8.5261 12.0610 16.2420 21.0681 26.5387 32.6535 39.4124 46.8152 54.8620 63.5526 72.8871 82.8655 93.4876 104.7536 116.6634 129.2170

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17-106

17-147 EES Using the normal shock relations, the normal shock functions are to be evaluated and tabulated as in Table A-14 for methane. Properties The specific heat ratio is given to be k = 1.3 for methane. Analysis The normal shock relations listed below are expressed in EES and the results are tabulated. Ma 2 =

P2 1 + kMa 12 2kMa 12 − k + 1 = = P1 1 + kMa 22 k +1

(k − 1)Ma 12 + 2 2kMa 12 − k + 1

(k + 1)Ma 12 V ρ 2 P2 / P1 = = = 1 , ρ 1 T2 / T1 2 + (k − 1)Ma 12 V 2

T2 2 + Ma 12 (k − 1) = T1 2 + Ma 22 (k − 1) k +1

P02 Ma 1 ⎡1 + Ma 22 (k − 1) / 2 ⎤ 2( k −1) = ⎢ ⎥ P01 Ma 2 ⎢⎣1 + Ma 12 (k − 1) / 2 ⎥⎦

P02 (1 + kMa 12 )[1 + Ma 22 (k − 1) / 2] k /( k −1) = P1 1 + kMa 22

Methane: k=1.3 My=SQRT((Mx^2+2/(k-1))/(2*Mx^2*k/(k-1)-1)) PyPx=(1+k*Mx^2)/(1+k*My^2) TyTx=(1+Mx^2*(k-1)/2)/(1+My^2*(k-1)/2) RyRx=PyPx/TyTx P0yP0x=(Mx/My)*((1+My^2*(k-1)/2)/(1+Mx^2*(k-1)/2))^(0.5*(k+1)/(k-1)) P0yPx=(1+k*Mx^2)*(1+My^2*(k-1)/2)^(k/(k-1))/(1+k*My^2)

Ma1 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0 8.5 9.0 9.5 10.0

Ma2 1.0000 0.6942 0.5629 0.4929 0.4511 0.4241 0.4058 0.3927 0.3832 0.3760 0.3704 0.3660 0.3625 0.3596 0.3573 0.3553 0.3536 0.3522 0.3510

P2/P1 1.0000 2.4130 4.3913 6.9348 10.0435 13.7174 17.9565 22.7609 28.1304 34.0652 40.5652 47.6304 55.2609 63.4565 72.2174 81.5435 91.4348 101.8913 112.9130

ρ2/ρ1 1.0000 1.9346 2.8750 3.7097 4.4043 4.9648 5.4118 5.7678 6.0526 6.2822 6.4688 6.6218 6.7485 6.8543 6.9434 7.0190 7.0837 7.1393 7.1875

T2/T1 1.0000 1.2473 1.5274 1.8694 2.2804 2.7630 3.3181 3.9462 4.6476 5.4225 6.2710 7.1930 8.1886 9.2579 10.4009 11.6175 12.9079 14.2719 15.7096

P02/P01 1 0.9261 0.7006 0.461 0.2822 0.1677 0.09933 0.05939 0.03613 0.02243 0.01422 0.009218 0.006098 0.004114 0.002827 0.001977 0.001404 0.001012 0.000740

P02/P1 1.8324 3.2654 5.3700 8.0983 11.4409 15.3948 19.9589 25.1325 30.9155 37.3076 44.3087 51.9188 60.1379 68.9658 78.4027 88.4485 99.1032 110.367 122.239

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17-107

17-148 Air flowing at a supersonic velocity in a duct is accelerated by cooling. For a specified exit Mach number, the rate of heat transfer is to be determined. Assumptions The assumptions associated with Rayleigh flow (i.e., steady one-dimensional flow of an ideal gas with constant properties through a constant cross-sectional area duct with negligible frictional effects) are valid.

Q& P01 = 240 kPa T01 = 350 K

Properties We take the properties of air to be k = 1.4, cp = 1.005 kJ/kg⋅K, and R = 0.287 kJ/kg⋅K (Table A-2a).

Ma2 = 2

Ma1 = 1.2

Analysis Knowing stagnation properties, the static properties are determined to be ⎛ k −1 ⎞ Ma 12 ⎟ T1 = T01 ⎜1 + 2 ⎝ ⎠

−1

⎛ k −1 ⎞ Ma 12 ⎟ P1 = P01 ⎜1 + 2 ⎝ ⎠

− k /( k −1)

ρ1 =

⎛ 1.4 - 1 2 ⎞ = (350 K)⎜1 + 1.2 ⎟ 2 ⎝ ⎠

−1

= 271.7 K

⎛ 1.4 - 1 2 ⎞ = (240 kPa)⎜1 + 1.2 ⎟ 2 ⎝ ⎠

−1.4 / 0.4

= 98.97 kPa

P1 98.97 kPa = = 1.269 kg/m 3 RT1 (0.287 kJ/kgK)(271.7 K)

Then the inlet velocity and the mass flow rate become ⎛ 1000 m 2 / s 2 c1 = kRT1 = (1.4)(0.287 kJ/kg ⋅ K)(271.7 K)⎜⎜ ⎝ 1 kJ/kg

⎞ ⎟ = 330.4 m/s ⎟ ⎠

V1 = Ma 1c1 = 1.2(330.4 m/s ) = 396.5 m/s m& air = ρ1 Ac1V1 = (1.269 kg/m 3 )[π (0.20 m) 2 / 4](330.4 m/s) = 15.81 kg/s The Rayleigh flow functions T0/T0* corresponding to the inlet and exit Mach numbers are (Table A-34):

Ma1 = 1.8:

T01/T0* = 0.9787

Ma2 = 2:

T02/T0* = 0.7934

Then the exit stagnation temperature is determined to be T0 2 T02 / T0* 0.7934 = = = 0.8107 T0 1 T01 / T0* 0.9787

→ T02 = 0.8107T01 = 0.8107(350 K ) = 283.7 K

Finally, the rate of heat transfer is Q& = m& air c p (T02 − T01 ) = (15.81 kg/s )(1.005 kJ/kg ⋅ K )(283.7 − 350) K = -1053 kW

Discussion The negative sign confirms that the gas needs to be cooled in order to be accelerated. Also, it can be shown that the thermodynamic temperature drops to 158 K at the exit, which is extremely low. Therefore, the duct may need to be heavily insulated to maintain indicated flow conditions.

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17-108

17-149 Air flowing at a subsonic velocity in a duct is accelerated by heating. The highest rate of heat transfer without affecting the inlet conditions is to be determined. Assumptions 1 The assumptions associated with Rayleigh flow (i.e., steady one-dimensional flow of an ideal gas with constant properties through a constant cross-sectional area duct with negligible frictional effects) are valid. 2 Inlet conditions (and thus the mass flow rate) remain constant. Properties We take the properties of air to be k = 1.4, cp = 1.005 kJ/kg⋅K, and R = 0.287 kJ/kg⋅K (Table A-2a).

Q&

Analysis Heat transfer will stop when the flow is choked, and thus Ma2 = V2/c2 = 1. The inlet density and stagnation temperature are

P1 = 400 kPa T1 = 360 K

ρ1 =

P1 400 kPa = = 3.872 kg/m 3 RT1 (0.287 kJ/kgK)(360 K)

Ma2 = 1

Ma1 = 0.4

⎛ k −1 ⎞ ⎛ 1.4 - 1 ⎞ T01 = T1 ⎜1 + Ma 12 ⎟ = (360 K)⎜1 + 0.4 2 ⎟ = 371.5 K 2 2 ⎝ ⎠ ⎝ ⎠

Then the inlet velocity and the mass flow rate become ⎛ 1000 m 2 / s 2 c1 = kRT1 = (1.4 )(0.287 kJ/kg ⋅ K)(360 K)⎜⎜ ⎝ 1 kJ/kg

⎞ ⎟ = 380.3 m/s ⎟ ⎠

V1 = Ma 1c1 = 0.4(380.3 m/s ) = 152.1 m/s

m& air = ρ 1 Ac1V1 = (3.872 kg/m 3 )(0.1× 0.1 m 2 )(152.1 m/s) = 5.890 kg/s

The Rayleigh flow functions corresponding to the inlet and exit Mach numbers are T02/T0* = 1 (since Ma2 = 1) T01 T0*

=

(k + 1)Ma 12 [2 + (k − 1)Ma 12 ] (1 + kMa 12 ) 2

=

(1.4 + 1)0.4 2 [2 + (1.4 − 1)0.4 2 ] (1 + 1.4 × 0.4 2 ) 2

= 0.5290

Therefore, T0 2 T02 / T0* 1 = = T0 1 T01 / T0* 0.5290

→ T02 = T01 / 0.5290 = (371.5 K ) / 0.5290 = 702.3 K

Then the rate of heat transfer becomes Q& = m& air c p (T02 − T01 ) = (5.890 kg/s )(1.005 kJ/kg ⋅ K )(702.3 − 371.5) K = 1958 kW

Discussion It can also be shown that T2 = 585 K, which is the highest thermodynamic temperature that can be attained under stated conditions. If more heat is transferred, the additional temperature rise will cause the mass flow rate to decrease. We can also solve this problem using the Rayleigh function values listed in Table A-34.

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17-109

17-150 Helium flowing at a subsonic velocity in a duct is accelerated by heating. The highest rate of heat transfer without affecting the inlet conditions is to be determined. Assumptions 1 The assumptions associated with Rayleigh flow (i.e., steady one-dimensional flow of an ideal gas with constant properties through a constant cross-sectional area duct with negligible frictional effects) are valid. 2 Inlet conditions (and thus the mass flow rate) remain constant. Properties We take the properties of helium to be k = 1.667, cp = 5.193 kJ/kg⋅K, and R = 2.077 kJ/kg⋅K (Table A-2a).

Q&

Analysis Heat transfer will stop when the flow is choked, and thus Ma2 = V2/c2 = 1. The inlet density and stagnation temperature are

ρ1 =

P1 400 kPa = = 0.5350 kg/m 3 RT1 (2.077 kJ/kgK)(360 K)

P1 = 400 kPa T1 = 360 K

Ma2 = 1

Ma1 = 0.4

⎛ k −1 ⎞ ⎛ 1.667 - 1 ⎞ T01 = T1 ⎜1 + Ma 12 ⎟ = (360 K)⎜1 + 0.4 2 ⎟ = 379.2 K 2 2 ⎝ ⎠ ⎝ ⎠

Then the inlet velocity and the mass flow rate become ⎛ 1000 m 2 / s 2 c1 = kRT1 = (1.667)(2.077 kJ/kg ⋅ K)(360 K)⎜⎜ ⎝ 1 kJ/kg

⎞ ⎟ = 1116 m/s ⎟ ⎠

V1 = Ma 1c1 = 0.4(1116 m/s ) = 446.6 m/s m& air = ρ 1 Ac1V1 = (0.5350 kg/m 3 )(0.1× 0.1 m 2 )(446.6 m/s) = 2.389 kg/s The Rayleigh flow functions corresponding to the inlet and exit Mach numbers are T02/T0* = 1 (since Ma2 = 1)

T01 T0*

=

(k + 1)Ma 12 [2 + (k − 1)Ma 12 ] (1 + kMa 12 ) 2

=

(1.667 + 1)0.4 2 [2 + (1.667 − 1)0.4 2 ] (1 + 1.667 × 0.4 2 ) 2

= 0.5603

Therefore,

T0 2 T02 / T0* 1 = = * T0 1 T01 / T0 0.5603

→ T02 = T01 / 0.5603 = (379.2 K ) / 0.5603 = 676.8 K

Then the rate of heat transfer becomes Q& = m& air c p (T02 − T01 ) = (2.389 kg/s)(5.193 kJ/kg ⋅ K )(676.8 − 379.2) K = 3693 kW

Discussion It can also be shown that T2 = 508 K, which is the highest thermodynamic temperature that can be attained under stated conditions. If more heat is transferred, the additional temperature rise will cause the mass flow rate to decrease. Also, in the solution of this problem, we cannot use the values of Table A34 since they are based on k = 1.4.

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17-110

17-151 Air flowing at a subsonic velocity in a duct is accelerated by heating. For a specified exit Mach number, the heat transfer for a specified exit Mach number as well as the maximum heat transfer are to be determined. Assumptions 1 The assumptions associated with Rayleigh flow (i.e., steady one-dimensional flow of an ideal gas with constant properties through a constant cross-sectional area duct with negligible frictional effects) are valid. 2 Inlet conditions (and thus the mass flow rate) remain constant. Properties We take the properties of air to be k = 1.4, cp = 1.005 kJ/kg⋅K, and R = 0.287 kJ/kg⋅K (Table A2a). Analysis The inlet Mach number and stagnation temperature are ⎛ 1000 m 2 / s 2 c1 = kRT1 = (1.4)(0.287 kJ/kg ⋅ K)(400 K)⎜⎜ ⎝ 1 kJ/kg Ma 1 =

V1 100 m/s = = 0.2494 c1 400.9 m/s

⎛ k −1 ⎞ T01 = T1⎜1 + Ma12 ⎟ 2 ⎝ ⎠ ⎛ 1.4 - 1 ⎞ = (400 K)⎜1 + 0.24942 ⎟ 2 ⎝ ⎠ = 405.0 K

⎞ ⎟ = 400.9 m/s ⎟ ⎠

q P1 = 35 kPa T1 = 400 K

Ma2 = 0.8

V1 = 100 m/s

The Rayleigh flow functions corresponding to the inlet and exit Mach numbers are (Table A-34): Ma1 = 0.2494:

T01/T* = 0.2559

Ma2 = 0.8:

T02/T* = 0.9639

Then the exit stagnation temperature and the heat transfer are determined to be T0 2 T02 / T * 0.9639 = = = 3.7667 → T0 2 = 3.7667T01 = 3.7667(405.0 K ) = 1526 K T0 1 T01 / T * 0.2559 q = c p (T02 − T01 ) = (1.005 kJ/kg ⋅ K )(1526 − 405) K = 1126 kJ/kg

Maximum heat transfer will occur when the flow is choked, and thus Ma2 = 1 and thus T02/T* = 1. Then, T0 2 T02 / T * 1 = = → T0 2 = T01 / 0.2559 = 405.0 K ) / 0.2559 = 1583 K T0 1 T01 / T * 0.2559 q max = c p (T02 − T01 ) = (1.005 kJ/kg ⋅ K )(1583 − 405) K = 1184 kJ/kg

Discussion This is the maximum heat that can be transferred to the gas without affecting the mass flow rate. If more heat is transferred, the additional temperature rise will cause the mass flow rate to decrease.

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17-111

17-152 Air flowing at sonic conditions in a duct is accelerated by cooling. For a specified exit Mach number, the amount of heat transfer per unit mass is to be determined. Assumptions The assumptions associated with Rayleigh flow (i.e., steady one-dimensional flow of an ideal gas with constant properties through a constant cross-sectional area duct with negligible frictional effects) are valid. Properties We take the properties of air to be k = 1.4, cp = 1.005 kJ/kg⋅K, and R = 0.287 kJ/kg⋅K (Table A2a). Analysis Noting that Ma1 = 1, the inlet stagnation temperature is ⎛ k −1 ⎞ T01 = T1 ⎜1 + Ma 12 ⎟ 2 ⎝ ⎠ ⎛ 1.4 - 1 2 ⎞ = (500 K)⎜1 + 1 ⎟ = 600 K 2 ⎝ ⎠ *

The Rayleigh flow functions T0/T0 corresponding to the inlet and exit Mach numbers are (Table A-34): Ma1 = 1:

T01/T0* = 1

Ma2 = 1.6:

T02/T0* = 0.8842

q P01 = 420 kPa T01 = 500 K

Ma2 = 1.6

Ma1 = 1

Then the exit stagnation temperature and heat transfer are determined to be T0 2 T02 / T0* 0.8842 = = = 0.8842 T0 1 T01 / T0* 1

→ T02 = 0.8842T01 = 0.8842(600 K ) = 530.5 K

q = c p (T02 − T01 ) = (1.005 kJ/kg ⋅ K )(530.5 − 600) K = - 69.8 kJ/kg

Discussion The negative sign confirms that the gas needs to be cooled in order to be accelerated. Also, it can be shown that the thermodynamic temperature drops to 351 K at the exit

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17-112

17-153 Saturated steam enters a converging-diverging nozzle with a low velocity. The throat area, exit velocity, mass flow rate, and exit Mach number are to be determined for isentropic and 90 percent efficient nozzle cases. Assumptions 1 Flow through the nozzle is steady and onedimensional. 2 The nozzle is adiabatic. Analysis (a) The inlet stagnation properties in this case are identical to the inlet properties since the inlet velocity is negligible. Thus h10 = h1. At the inlet,

1

Steam

Vi ≈ 0

h1 = (h f + x1 h fg ) @ 3 MPa = 1008.3 + 0.95 × 1794.9 = 2713.4 kJ/kg

t

a) ηN = 100% b) ηN = 90%

s1 = ( s f + x1 s fg ) @ 3 MPa = 2.6454 + 0.95 × 3.5402 = 6.0086 kJ/kg ⋅ K

At the exit, P2 = 1.2 MPa and s2 = s2s = s1 = 6.0086 kJ/kg·K. Thus, s2 = s f + x2 s fg → 6.0086 = 2.2159 + x2 (4.3058) → x2 = 0.8808 h2 = h f + x2 h fg = 798.33 + 0.8808 × 1985.4 = 2547.2 kJ/kg

v 2 = v f + x2v fg = 0.001138 + 0.8808 × (0.16326 − 0.001138) = 0.1439 m3 / kg Then the exit velocity is determined from the steady-flow energy balance to be h1 +

V12 V2 V 2 − V12 = h2 + 2 → 0 = h2 − h1 + 2 2 2 2

Solving for V2, ⎛ 1000 m 2 / s 2 ⎞ ⎟ = 576.7 m/s V2 = 2(h1 − h2 ) = 2(2713.4 - 2547.2)kJ/kg⎜ ⎜ 1 kJ/kg ⎟ ⎝ ⎠

The mass flow rate is determined from m& =

1

v2

A2V2 =

1 (16 × 10− 4 m 2 )(576.7 m/s) = 6.41 kg/s 3 0.1439 m / kg

The velocity of sound at the exit of the nozzle is determined from 1/ 2

⎛ ∂P ⎞ c=⎜ ⎟ ⎝ ∂r ⎠ s

1/ 2

⎛ ΔP ⎞ ⎟⎟ ≅ ⎜⎜ ⎝ Δ (1 / v ) ⎠ s

The specific volume of steam at s2 = 6.0086 kJ/kg·K and at pressures just below and just above the specified pressure (1.1 and 1.3 MPa) are determined to be 0.1555 and 0.1340 m3/kg. Substituting, c2 =

(1300 − 1100) kPa

⎛ 1000 m 2 / s 2 ⎜ 3 ⎜ 1 ⎞ ⎛ 1 3 ⎝ 1 kPa ⋅ m − ⎜ ⎟ kg/m ⎝ 0.1340 0.1555 ⎠

⎞ ⎟ = 440.3 m/s ⎟ ⎠

Then the exit Mach number becomes Ma 2 =

V2 576.7 m/s = = 1.310 c2 440.3 m/s

The steam is saturated, and thus the critical pressure which occurs at the throat is taken to be Pt = P* = 0.576 × P01 = 0.576 × 3 = 1.728 MPa

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17-113

Pt = 1.728 MPa and s t = s1 = 6.0086 kJ/kg ⋅ K

Thus, ht = 2611.4 kJ/kg v t = 0.1040 m3 / kg Then the throat velocity is determined from the steady-flow energy balance, Ê0

h1 +

V12 2

= ht +

Vt 2 V2 → 0 = ht − h1 + t 2 2

Solving for Vt, ⎛ 1000 m 2 / s 2 ⎞ ⎟ = 451.7 m/s Vt = 2(h1 − ht ) = 2(2713.4 − 2611.4)kJ/kg⎜ ⎜ 1 kJ/kg ⎟ ⎝ ⎠

Thus the throat area is At =

m& v t (6.41 kg/s)(0.1040 m3 / kg) = = 14.75 × 10− 4 m 2 = 14.75 cm 2 (451.7 m/s) Vt

(b) The inlet stagnation properties in this case are identical to the inlet properties since the inlet velocity is negligible. Thus h10 = h1. At the inlet, h1 = (h f + x1 h fg ) @ 3 MPa = 1008.3 + 0.95 × 1794.9 = 2713.4 kJ/kg

1

Steam

Vi ≈ 0

s1 = ( s f + x1 s fg ) @ 3 MPa = 2.6454 + 0.95 × 3.5402 = 6.0086 kJ/kg ⋅ K

t

a) ηN = 100% b) ηN = 90%

At state 2s, P2 = 1.2 MPa and s2 = s2s = s1 = 6.0086 kJ/kg·K. Thus, s 2 s = s f + x 2 s s fg ⎯ ⎯→ 6.0086 = 2.2159 + x 2 s (4.3058) ⎯ ⎯→ x 2 s = 0.8808 h2 s = h f + x 2 s h fg = 798.33 + 0.8808 × 1985.4 = 2547.2 kJ/kg

The enthalpy of steam at the actual exit state is determined from

ηN =

h01 − h2 2713.4 − h2 ⎯ ⎯→ 0.90 = ⎯ ⎯→ h2 = 2563.8 kJ/kg h01 − h2 s 2713.4 − 2547.2

Therefore at the exit, P2 = 1.2 MPa and h2 = 2563.8 kJ/kg·K. Thus, h2 = h f + x2 h fg ⎯ ⎯→ 2563.8 = 798.33 + x2 (1985.4) ⎯ ⎯→ x2 = 0.8892 s2 = s f + x2 s fg = 2.2159 + 0.8892 × 4.3058 = 6.0447 v 2 = v f + x2v fg = 0.001138 + 0.8892 × (0.16326 − 0.001138) = 0.1453 kJ / kg

Then the exit velocity is determined from the steady-flow energy balance to be h1 +

V12 V2 V 2 − V12 = h2 + 2 → 0 = h2 − h1 + 2 2 2 2

Solving for V2, ⎛ 1000 m 2 / s 2 ⎞ ⎟ = 547.1 m/s V2 = 2(h1 − h2 ) = 2(2713.4 − 2563.8)kJ/kg⎜ ⎜ 1 kJ/kg ⎟ ⎝ ⎠

The mass flow rate is determined from

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17-114

m& =

1

A2V2 =

v2

1 0.1453 m3 / kg

(16 × 10− 4 m 2 )(547.1 m/s) = 6.02 kg/s

The velocity of sound at the exit of the nozzle is determined from 1/ 2

⎛ ∂P ⎞ c = ⎜⎜ ⎟⎟ ⎝ ∂ρ ⎠ s

1/ 2

⎛ ΔP ⎞ ⎟⎟ ≅ ⎜⎜ ⎝ Δ (1 / v ) ⎠ s

The specific volume of steam at s2 = 6.0447 kJ/kg·K and at pressures just below and just above the specified pressure (1.1 and 1.3 MPa) are determined to be 0.1570 and 0.1353 m3/kg. Substituting,

(1300 − 1100) kPa

⎛ 1000 m 2 / s 2 ⎞ ⎜ ⎟ = 442.6 m/s 3 ⎟ ⎜ 1 ⎞ ⎛ 1 3 ⎝ 1 kPa ⋅ m ⎠ − ⎜ ⎟ kg/m ⎝ 0.1353 0.1570 ⎠

c2 =

Then the exit Mach number becomes Ma 2 =

V2 547.1 m/s = = 1.236 c2 442.6 m/s

The steam is saturated, and thus the critical pressure which occurs at the throat is taken to be Pt = P* = 0.576 × P01 = 0.576 × 3 = 1.728 MPa

At state 2ts, Pts = 1.728 MPa and sts = s1 = 6.0086 kJ/kg·K. Thus, hts = 2611.4 kJ/kg. The actual enthalpy of steam at the throat is

ηN =

2713.4 − ht h01 − ht ⎯ ⎯→ 0.90 = ⎯ ⎯→ ht = 2621.6 kJ/kg 2713.4 − 2611.4 h01 − hts

Therefore at the throat, P2 = 1.728 MPa and ht = 2621.6 kJ/kg. Thus, vt = 0.1046 m3/kg. Then the throat velocity is determined from the steady-flow energy balance, Ê0

V2 h1 + 1 2

= ht +

Vt 2 V2 → 0 = ht − h1 + t 2 2

Solving for Vt, ⎛ 1000 m 2 / s 2 ⎞ ⎟ = 428.5 m/s Vt = 2(h1 − ht ) = 2(2713.4 − 2621.6)kJ/kg⎜ ⎜ 1 kJ/kg ⎟ ⎝ ⎠

Thus the throat area is At =

m& v t (6.02 kg/s)(0.1046 m3 / kg) = = 14.70 × 10− 4 m 2 = 14.70 cm 2 (428.5 m/s) Vt

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Fundamentals of Engineering (FE) Exam Problems 17-154 An aircraft is cruising in still air at 5°C at a velocity of 400 m/s. The air temperature at the nose of the aircraft where stagnation occurs is (a) 5°C

(b) 25°C

(c) 55°C

(d) 80°C

(e) 85°C

Answer (e) 85°C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). k=1.4 Cp=1.005 "kJ/kg.K" T1=5 "C" Vel1= 400 "m/s" T1_stag=T1+Vel1^2/(2*Cp*1000) "Some Wrong Solutions with Common Mistakes:" W1_Tstag=T1 "Assuming temperature rise" W2_Tstag=Vel1^2/(2*Cp*1000) "Using just the dynamic temperature" W3_Tstag=T1+Vel1^2/(Cp*1000) "Not using the factor 2"

17-155 Air is flowing in a wind tunnel at 15°C, 80 kPa, and 200 m/s. The stagnation pressure at a probe inserted into the flow stream is (a) 82 kPa

(b) 91 kPa

(c) 96 kPa

(d) 101 kPa

(e) 114 kPa

Answer (d) 101 kPa Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). k=1.4 Cp=1.005 "kJ/kg.K" T1=15 "K" P1=80 "kPa" Vel1= 200 "m/s" T1_stag=(T1+273)+Vel1^2/(2*Cp*1000) "C" T1_stag/(T1+273)=(P1_stag/P1)^((k-1)/k) "Some Wrong Solutions with Common Mistakes:" T11_stag/T1=(W1_P1stag/P1)^((k-1)/k); T11_stag=T1+Vel1^2/(2*Cp*1000) "Using deg. C for temperatures" T12_stag/(T1+273)=(W2_P1stag/P1)^((k-1)/k); T12_stag=(T1+273)+Vel1^2/(Cp*1000) "Not using the factor 2" T13_stag/(T1+273)=(W3_P1stag/P1)^(k-1); T13_stag=(T1+273)+Vel1^2/(2*Cp*1000) "Using wrong isentropic relation"

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17-116

17-156 An aircraft is reported to be cruising in still air at -20°C and 40 kPa at a Mach number of 0.86. The velocity of the aircraft is (a) 91 m/s

(b) 220 m/s

(c) 186 m/s

(d) 280 m/s

(e) 378 m/s

Answer (d) 280 m/s Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). k=1.4 Cp=1.005 "kJ/kg.K" R=0.287 "kJ/kg.K" T1=-10+273 "K" P1=40 "kPa" Mach=0.86 VS1=SQRT(k*R*T1*1000) Mach=Vel1/VS1 "Some Wrong Solutions with Common Mistakes:" W1_vel=Mach*VS2; VS2=SQRT(k*R*T1) "Not using the factor 1000" W2_vel=VS1/Mach "Using Mach number relation backwards" W3_vel=Mach*VS3; VS3=k*R*T1 "Using wrong relation"

17-157 Air is flowing in a wind tunnel at 12°C and 66 kPa at a velocity of 230 m/s. The Mach number of the flow is (a) 0.54

(b) 0.87

(c) 3.3

(d) 0.36

(e) 0.68

Answer (e) 0.68 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). k=1.4 Cp=1.005 "kJ/kg.K" R=0.287 "kJ/kg.K" T1=12+273 "K" P1=66 "kPa" Vel1=230 "m/s" VS1=SQRT(k*R*T1*1000) Mach=Vel1/VS1 "Some Wrong Solutions with Common Mistakes:" W1_Mach=Vel1/VS2; VS2=SQRT(k*R*(T1-273)*1000) "Using C for temperature" W2_Mach=VS1/Vel1 "Using Mach number relation backwards" W3_Mach=Vel1/VS3; VS3=k*R*T1 "Using wrong relation"

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17-117

17-158 Consider a converging nozzle with a low velocity at the inlet and sonic velocity at the exit plane. Now the nozzle exit diameter is reduced by half while the nozzle inlet temperature and pressure are maintained the same. The nozzle exit velocity will (a) remain the same. (b) double. (e) go down to one-fourth.

(c) quadruple.

(d) go down by half.

Answer (a) remain the same.

17-159 Air is approaching a converging-diverging nozzle with a low velocity at 20°C and 300 kPa, and it leaves the nozzle at a supersonic velocity. The velocity of air at the throat of the nozzle is (a) 290 m/s

(b) 98 m/s

(c) 313 m/s

(d) 343 m/s

(e) 412 m/s

Answer (c) 313 m/s Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). k=1.4 Cp=1.005 "kJ/kg.K" R=0.287 "kJ/kg.K" "Properties at the inlet" T1=20+273 "K" P1=300 "kPa" Vel1=0 "m/s" To=T1 "since velocity is zero" Po=P1 "Throat properties" T_throat=2*To/(k+1) P_throat=Po*(2/(k+1))^(k/(k-1)) "The velocity at the throat is the velocity of sound," V_throat=SQRT(k*R*T_throat*1000) "Some Wrong Solutions with Common Mistakes:" W1_Vthroat=SQRT(k*R*T1*1000) "Using T1 for temperature" W2_Vthroat=SQRT(k*R*T2_throat*1000); T2_throat=2*(To-273)/(k+1) "Using C for temperature" W3_Vthroat=k*R*T_throat "Using wrong relation"

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17-118

17-160 Argon gas is approaching a converging-diverging nozzle with a low velocity at 20°C and 120 kPa, and it leaves the nozzle at a supersonic velocity. If the cross-sectional area of the throat is 0.015 m2, the mass flow rate of argon through the nozzle is (a) 0.41 kg/s

(b) 3.4 kg/s

(c) 5.3 kg/s

(d) 17 kg/s

(e) 22 kg/s

Answer (c) 5.3 kg/s Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). k=1.667 Cp=0.5203 "kJ/kg.K" R=0.2081 "kJ/kg.K" A=0.015 "m^2" "Properties at the inlet" T1=20+273 "K" P1=120 "kPa" Vel1=0 "m/s" To=T1 "since velocity is zero" Po=P1 "Throat properties" T_throat=2*To/(k+1) P_throat=Po*(2/(k+1))^(k/(k-1)) rho_throat=P_throat/(R*T_throat) "The velocity at the throat is the velocity of sound," V_throat=SQRT(k*R*T_throat*1000) m=rho_throat*A*V_throat "Some Wrong Solutions with Common Mistakes:" W1_mass=rho_throat*A*V1_throat; V1_throat=SQRT(k*R*T1_throat*1000); T1_throat=2*(To273)/(k+1) "Using C for temp" W2_mass=rho2_throat*A*V_throat; rho2_throat=P1/(R*T1) "Using density at inlet"

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17-161 Carbon dioxide enters a converging-diverging nozzle at 60 m/s, 310°C, and 300 kPa, and it leaves the nozzle at a supersonic velocity. The velocity of carbon dioxide at the throat of the nozzle is (a) 125 m/s

(b) 225 m/s

(c) 312 m/s

(d) 353 m/s

(e) 377 m/s

Answer (d) 353 m/s Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). k=1.289 Cp=0.846 "kJ/kg.K" R=0.1889 "kJ/kg.K" "Properties at the inlet" T1=310+273 "K" P1=300 "kPa" Vel1=60 "m/s" To=T1+Vel1^2/(2*Cp*1000) To/T1=(Po/P1)^((k-1)/k) "Throat properties" T_throat=2*To/(k+1) P_throat=Po*(2/(k+1))^(k/(k-1)) "The velocity at the throat is the velocity of sound," V_throat=SQRT(k*R*T_throat*1000) "Some Wrong Solutions with Common Mistakes:" W1_Vthroat=SQRT(k*R*T1*1000) "Using T1 for temperature" W2_Vthroat=SQRT(k*R*T2_throat*1000); T2_throat=2*(T_throat-273)/(k+1) "Using C for temperature" W3_Vthroat=k*R*T_throat "Using wrong relation"

17-162 Consider gas flow through a converging-diverging nozzle. Of the five statements below, select the one that is incorrect: (a) The fluid velocity at the throat can never exceed the speed of sound. (b) If the fluid velocity at the throat is below the speed of sound, the diversion section will act like a diffuser. (c) If the fluid enters the diverging section with a Mach number greater than one, the flow at the nozzle exit will be supersonic. (d) There will be no flow through the nozzle if the back pressure equals the stagnation pressure. (e) The fluid velocity decreases, the entropy increases, and stagnation enthalpy remains constant during flow through a normal shock. Answer (c) If the fluid enters the diverging section with a Mach number greater than one, the flow at the nozzle exit will be supersonic.

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17-163 Combustion gases with k = 1.33 enter a converging nozzle at stagnation temperature and pressure of 400°C and 800 kPa, and are discharged into the atmospheric air at 20°C and 100 kPa. The lowest pressure that will occur within the nozzle is (a) 26 kPa

(b) 100 kPa

(c) 321 kPa

(d) 432 kPa

(e) 272 kPa

Answer (d) 432 kPa Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). k=1.33 Po=800 "kPa" "The critical pressure is" P_throat=Po*(2/(k+1))^(k/(k-1)) "The lowest pressure that will occur in the nozzle is the higher of the critical or atmospheric pressure." "Some Wrong Solutions with Common Mistakes:" W2_Pthroat=Po*(1/(k+1))^(k/(k-1)) "Using wrong relation" W3_Pthroat=100 "Assuming atmospheric pressure"

17-164 ··· 17-166 Design and Essay Problems

KJ

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