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Chapter 1 Solutions
P1.1
This example shows how the First Law can be applied to individual processes and how these can make up a cycle. A cycle of events is shown in Fig A.10. It is made up of four processes, and the heat and work associated with those processes is as given.
Fig A.10: Cycle of events made up of four processes. Process 12: Process 23: Process 34: Process 41:
Q = +10J Q = +100J Q = 20J Q = 10J
W = 18J W = 0J W = +70J W = +28J
Calculate the values of Un  U1, the net work, the net heat transfer and the heat supplied for the cycle. Solution This problem can be solved by applying the First Law to each of the processes in turn, when the change in internal energy is dU Q W . Process 12: dU12 U2 U1 Q12 W12 10 18 28J Process 23: dU23 U3 U2 Q23 W23 100 0 100J Process 34: dU34 U4 U3 Q34 W34 20 70 90J Process 41: dU41 U1 U4 Q41 W41 10 28 38J The change in internal energy relative to the internal energy at point 1, U1, is given by dU1n U n U 1 U n U n 1 U n 1 U n 2 ......U 2 U1
dU n 1,n dU n 2,n 1 ...... dU 12 Hence: dU12 U 2 U1 28J dU13 dU12 dU 23 28 100 128J dU14 dU12 dU 23 dU 34 28 100 (90) 38J
dU11 dU12 dU 23 dU 34 dU 41 28 100 (90) (38) 0J The result dU11 = 0 confirms that the four processes constitute a cycle, because the net change of state is zero. The net work done in the cycle is W W12 W23 W34 W41 cycle
18 0 70 28 80J The positive sign indicates that the system does work on the surroundings. The net heat supplied in the cycle is
Solutions Chapter 1 D E Winterbone
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Chapter 1 Solutions
Q Q
12
cycle
Q23 Q34 Q41
10 100 20 10 80J
Hence the net work done and net heat supplied are both 80J, as would be expected because the state of the system has not changed between both ends of the cycle. It is also possible to differentiate between the heat supplied to the system, Q > 0, and heat rejected from the system, Q < 0. In this case the total heat supplied is Q Q12 Q23 cycle
10 100 110J while the heat rejected is the sum of the negative heat transfer terms, viz.: Q Q34 Q41 . cycle 20 10 30J It should also be noted that the work done, W, is the difference between the heat supplied and that rejected. This is an important point when considering the conversion of heat into work, which is dealt with by the Second Law of Thermodynamics. __________________________________________________________________________________
P1.2 There are two ways to solve this problem: 1. A simple one based on the pV diagram 2. A more general one based on the pV relationship
Work done = Area under lines. Assuming a linear spring, then a linear (straight) line relationship joins the points. 1 105 0.5 4 3 100kJ 2 10 Work done by gas = Area adeca = Area abca + Area adbca
Hence, Work done by spring = Area abca = Ws
105 150kJ 103 More general method is to evaluate the work as W pdV . = 100 1.0 0.5
First find the relationship for the spring. If linear pg1 kV1 k ,and pg 2 kV2 k Thus
k
p g 2 p g1 V2 V1
8bar/m3 , and k 7bar
giving pg 8V 7.
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Chapter 1 Solutions
δWg pg dV , 2
2 8V 2 giving Wg 8V 7 dV 7V Work done by gas, 2 1 1
4 2.25 1 7 1.5 1 102 150kJ δWs ps dV 2
V2 Ws 8V 8 dV 8 V 100kJ 2 1 1 The benefit of the latter approach is that it can be applied in the case of a nonlinear spring: it is more difficult to use the simpler approach. _________________________________________________________________________________________
Work done by spring,
2
P1.3 Pressure of gas is proportional to diameter, i.e. p d , giving p1 kd1, and hence k Volume V
p1 1.5 5bar/m d1 0.3
13
d3
6V , thus d 6
Work done during process, W pdV Working in terms of d dV 2
W kd 1
3 d 2 d2 dd dd 6 2
d2 2
dd
2 k k d 2 4 d14 3 d .d d 2 1 2 4 4
5 0.334 0.34 105 738J 4 2 This problem can also be solved in terms of V; however, it cannot be solved using a linear approximation. You might have been close to the correct solution for P1.3, but it does not work for P1.4. _________________________________________________________________________________________
P1.4 The problem is the same as P1.3, but the final diameter is 1m. 2 2 d2 k k d 2 4 d14 3 W kd dd d .d d 2 2 1 2 4 4 1 The work done is 5 4 1.0 0.34 105 194759J 4 2 Using the WRONG APPROACH, assuming a linear relationship, gives 1.5 5 W d 23 d13 105 165580J 2 6 _________________________________________________________________________________________
P1.5 ts at 20bar = 212.4C
Initial conditions:
ug 2600kJ / kg ; hg 2799kJ / kg; vg 0.0995m3 / kg p 20bar; t 500C
Final conditions:
ug 3116kJ / kg ; hg 3467kJ / kg; vg 0.1756m3 / kg
To evaluate the energy added use 1st Law for closed system
Q dU W dU pdV
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Chapter 1 Solutions
20 105 Q m 3116 2600 0.1756 0.09957 3 10 Hence energy added, 3 516 152.06 2004kJ Alternative method Constant pressure process, hence enthalpy can be used
Q dh m h2 h1 3 3467 2799 2004kJ W
20 105 0.1756 0.9957 3 456kJ 103
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P1.6 This is a constant volume (isochoric) process. Hence
v1 0.00317m3 /kg at critical point, and v1 0.00317m3 /kg. But v2 xvg 1 x v f v f xv fg
At p2 27.5bar, interpolation vg 0.072788m /kg, and t2 229C 3
Using the saturated water tables, vf = 0.001209m3/kg Hence, x
0.00317 0.001209 0.0274 2.74% 0.072788 0.001209
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P1.7 Initial and final conditions:
p1 A 3.5bar; m1 A 1.0kg ; x1 A 1.0; u g1 A 2549kJ / kg ; u1 A 2549kJ / kg p1B 7.0bar; m1B 2.0kg ; x1B 0.8; u g1B 2573kJ / kg ; u f 1B 696kJ / kg ; u1B 2197.6kJ / kg
p2 5bar; m2 3.0kg Total U1 2549 4395.2 6944.2kJ / kg Volumes
vg1 A 0.5241m3 / kg ;V1 A 0.5241m3 / kg vg1B 0.2728m3 / kg ; v f 1B 0.1107 102 m3 / kg (interp); V1B 0.21846m3 / kg Total volume, V1 0.5241 2 0.21846 0.9610m3 / kg Conditions at 2
V2 0.9610m3 ; v2
V2 0.3203m3 / kg m2
At 5bar, vg 0.3748m3 / kg ; v f 01093 102 m3 / kg x2
v2 v f 2 vg 2 v f 2
0.3192 0.8541 0.3737
Thus U 2 3 0.8541 2562 0.1459 639 6846kJ Total heat transfer, Q U 2 U1 98kJ _________________________________________________________________________________________
P1.8
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Chapter 1 Solutions
m 5kg ; p1 14bar; x1 0.8 vg1 0.1408m3 / kg ; v f 1 0.001149m3 / kg u g1 2593kJ / kg ; u f 1 828kJ / kg v1 x1vg1 1 x1 v f 1 0.1129m3 / kg : V1 5 0.1129 0.5645m3 u1 x1u g1 1 x1 u f 1 2240kJ / kg : U1 5 2240 11200kJ Spring equation, p = kV, because p = 0 when V = 0. When volume is 150% of initial value,
p2 1.5 p1 21bar : V2 1.5 0.5645 0.8468m3
Work done, W pV kVdV
W pdV kVdV Hence,
p1 p pV VdV 1 V22 V12 1 1 1.25 V1 2V1 2
14 0.5645 105 1.25 3 494kJ 2 10 3 3 Conditions at 2: V2 0.8468m ; v2 0.16936m / kg : p2 21bar
Based on steam tables condition 2 is in the superheat region: volumes listed below
p = 20bar p = 21bar p = 30bar
0.1634 0.1578 0.1078
0.1756 0.1696 (interpolated) 0.1161
Hence t2 = 500C
u2 3116 0.1 3108 3116 3115kJ / kg
Thus
U 2 5 3115 15576kJ Heat transfer from 1st Law Q dU W 15576 11200 494 4870kJ
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P1.9
t A 20C; ps 5.673bar; hg 195.78kJ / kg ; h f 54.87kJ / kg
Tank A: 1kg Freon12
ug hg pg vg 195.78 5.673
105 0.0308 178.3kJ / kg 103
Cylinder: isobaric expansion at 2bar Heat transferred so temperature constant at 20°C: calculate heat transfer. Flow across valve is isenthalpic:
Vp1 0m3 ;Vp 2 ?
Initial energy,
U1 178.3kJ ; initial volume, V1 0.0308m3
Final volume,
V2 0.0969m3
2 105 0.0969 0.0308 13.22kJ 103 2 105 Final energy, U 2 202.27 0.0969 182.9kJ 103 Hence, heat transfer, Q dU W 182.9 178.3 13.22 17.8kJ Hence work done, W
pdV
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Solutions Chapter 1 D E Winterbone
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Chapter 1 Solutions
Problems P1.10 to P1.14 require the USFEE These examples demonstrate that it is possible to obtain solutions for problems that have flows across the system boundaries by closed system methods.
P1.10 An insulated bottle is initially evacuated, i.e. it contains a vacuum, and then the stopper is removed allowing atmospheric air to fill the bottle. Evaluate the final conditions in a bottle when the air has just filled it. Comment: This is a relatively complex problem in unsteady gas dynamics if the processes between the removal of the stopper and the quiescent end state are considered in detail. In the flow processes pressure waves will travel into the bottle and be reflected; these will cause the atmospheric air to start flowing into the bottle. The waves will ultimately die out due to the flow interactions at the neck of the bottle and fluid friction, and finally a steady state will be reached. The great strength of the approaches of thermodynamics are that they allow the final state to be evaluated without any knowledge of gas dynamics. The bottle is shown in Fig A. 11; it has a volume of V. At t = 0, referred to as state 1, there is no gas in the bottle and its pressure is zero. Hence, p1 = 0 and m1 = 0. Gas is admitted to the bottle until the pressure of the gas in the bottle is atmospheric pressure; p2 = patm. It is possible to treat this as a closed system problem if the system boundaries are drawn in such a way that no flow occurs across them during the filling process. This has been done by drawing a boundary around the gas which flows into the bottle during the process: in this case it could be an actual boundary such as an extremely flexible balloon which has been filled sufficiently to hold all the air required to fill the bottle.
Fig A.11: Bottle being filled with atmospheric air. The bottle is initially evacuated: ma = 0 at t = 0. Considering the total system at time t = 0, i.e. system a + system b. The First Law can be applied to the combined system, giving Q dU W . In this case Q = 0, because the bottle is insulated. The total work done during the time interval is the pdV work done on the gas in system b, because system a does not change volume and no stirring work is done. Hence dU U 2 U 1 W , where U 1 U 1a U 1b 0 minuin minuin and U 2 U 2a U 2b minu2 0 minu2. Rearranging the equation gives U 2 U1 W where W patmVin min patmvin .
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Chapter 1 Solutions Thus U 2 min u2 min u1 min patm vin min uin patm vin , giving the surprising result that the specific internal energy of the gas in the bottle is not the same as that of the atmosphere, but is u2 uin patm vin hin . i.e. the internal energy of the gas in the bottle is equal to the enthalpy of the gas that was forced into the bottle. The reason for this is that the atmosphere did work on pushing the gas from system b into system a. ________________________________________________________________________________
P1.11
Filling a bottle which already contains some fluid. The analysis adopted above may be applied to a bottle which is not initially evacuated, i.e. pa1 0. This is shown in Fig P1.11. The approach to solving this problem is similar to that adopted above. The same technique is used to make the problem a closed system one. As before, Q = 0, because the bottle is insulated. The total work done during the time interval is the pdV work done on the gas in system b, because system a does not change volume and no stirring work is done. The mass of air in the bottle at the beginning of the process will be denoted by m1, and its specific internal energy will be u1. Hence, again dU U 2 U 1 W , but now U 1 U 1a U 1b m1u1 minuin and U 2 U 2 a U 2b minu2 0 minu2.
Fig P1.11(a): Filling of a bottle which is not initially evacuated Rearranging the equation gives U 2 U1 W where W patmVin min patmvin . U 2 min u2 m1u1 min uin min patm vin m1u1 min uin patm vin , Thus where min m2 m1. Substituting this value in the equation for u2 gives
min uin pin vin min hin m2 m1 hin m2 u2 m1u1 which can be rearranged to give m2 hin u2 m1 hin u1 . This equation has two unknowns, m2 and u2, and, in general, has to be solved iteratively, i.e. by trial R R and cp and error. In the special case of a perfect gas, where pv RT , with cv , it is 1 1 possible to get the following solution.
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Chapter 1 Solutions
Using the equation m2 hin u2 m1 hin u1 and substituting the following values for the initial and final conditions RT1 pV u1 cv T1 , m1 1 1 , 1 RT1 RT2 pV u2 cv T2 , m2 2 2 1 RT2 and the value for the inflowing enthalpy RTa hin 1 gives paV1 RTa RT2 p1V1 RTa RT1 . RT1 1 1 RT1 1 1 This equation can be rearranged to give Ta T2 p1 Ta 1 1 pa T1 If p1 = 0, then the result reduces to that for filling an evacuated bottle, viz., T2 = Ta. Fig P1.11(b) shows how the value of T2 varies with the ratios p1/pa and Ta/T1.
Temperature ratio, T 2/T a
1.4
1.2
1
T1/Ta=1.0 T1/Ta=0.8 0.8
T1/Ta=1.2
0.6 0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Pressure ratio, p 1/p a
Fig P1.11(b): Variation of temperature ratio, T2/Ta, with pressure ratio, p1/pa, for different levels of temperature ratio, T1/Ta. In the case of gases which are not perfect it is not possible to derive a simple equation for the variation of the final temperature with initial conditions, and an iteration must be performed to evaluate it. _________________________________________________________________________________
P1.12
Discharge from a bottle The two questions above have dealt with filling a bottle, and shown how this can be considered to be a closed system problem. A similar approach can be applied to a bottle discharging to the surroundings. The system is shown in Fig P1.12.
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Chapter 1 Solutions
Fig P1.12: Bottle discharging to atmosphere. The initial mass in the bottle (system a) is m1 with a specific internal energy of u1, while the final mass in the bottle is m2 with a specific internal energy of u2. The mass flow out of the cylinder, into system b, is mexh, and the mean specific internal energy of the fluid passing from system a to system b is uexh. If the systems are insulated, then Q = 0. During the discharge process work is done by the gas on the surroundings as the fluid "inflates" system b. This work is W mexh pexhvexh . Applying the First Law to the process Q W dE U 2 U1 , which gives on substitution 0 mexh pexhvexh m2 u2 mexhuexh m1u1, which can be rearranged to give m1 m2 pexh vexh uexh m2 u2 m1u1 m1 m2 hexh . Hence m2 hexh u2 m1 hexh u1 , giving
m2 m1
hexh u1 . hexh u2
The final mass has to be evaluated by trial and error in most cases because the value of hexh is an average value during the exhausting process. __________________________________________________________________________________________
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Note: Solutions are not provided for problems P2.6 and P2.7: these are left for the reader to consider. Also, a solution has not been provided for P2.15: this can be solved using a spreadsheet. See also Section 8.3
Solutions Manual Chapter 2
Pz.l
A
10 kg of water at 0"C is brought into contact with a large heat reservoir at 100"C.
of
mass
(a)
When the water has the change the change the change
(r) (ii) (iii)
reached l00oC what has been: of entropy of the water; of antropy of the reservoir; of entropy of the universe?
( 13. 1 lkJ/K; lt.26kJ lK:' 1.85kJ/K) the water had been heated from OoC to 100oC by first bringing it into contact with a reservoir at 30oC and then a reservoir of 100"C, what would have been the change in the
If
O)
entropy of the universe? (1.070kJ/K)
(c)
Explain how the water could have been heated to give no change in the entropy of the
universe'
flnfinite number of reservoirs
P2.2,
A system contains
a

reversible heat transfer)
fluid at a temperature of 70"C and I bar. It undergoes a reversible process during
which the temp€rature of the system remains constant. Given that the heat transfer to the fluid during the process is 100 kJ, evaluate: (a) the increase in entropy. (b) if the system has a mass of 2.31kg. Evaluate the increase in specific enfopy of the system. a second fluid system, identical to the first one urdergoes an irreversible isothermal process from the same initial state to the same final state as above; and the heat transfer to the fluid in this irreversible process is 180 kJ; evaluate the increase in entropy ofthe fluid. (0.29 t SkJ/K; 0.1262kJ lk gK; 0.29 I skJ/kgK)
(c) If
P2.3
Calculate the gain in entropy when 1 kg of water at 30"C is converted into steam at 150"C and then superheated to 300"C, with the process taking place at constant pressue. :2.1 kJ/kg K hg:2600  1.5 t, where temperature in oC. Take cp(wate )= 4.2 kJ/kg K cp(steam) (7.6533kJ/K)
/:
P2.4 A mass of a liquid, m, at temperature,
21, is mixed with an equal mass of the same liquid at
tempereture, 72. The system is therrnally insulated. Show that the change of entropy of the Universe is:
zmc,tn(
 Ql"t
$€q"1 oddzd , q,l^
(trntnr) =. A1W 2Q{
H",rt \Lt  q,t : 1+r :
Aa.g+%

2Blt
4/3
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