Solution Manual for Advanced Thermodynamics for Engineers [2 ed.]

This Solution Manual contains the solutions to the even and odds problems of the text.This manual cover the chapters 1,

283 22 66MB

English Pages 222 Year 2015

Report DMCA / Copyright

DOWNLOAD PDF FILE

Table of contents :
Chapter_01
Chapter_02
Chapter_03
Chapter_04
Chapter_05
Chapter_06
Chapter_07
Chapter_08
Chapter_09
Chapter_10
Chapter_12
Chapter_14
Chapter_15
Chapter_16
Chapter_17
Chapter_18
Chapter_19
Chapter_20
Chapter_21
Recommend Papers

Solution Manual for Advanced Thermodynamics for Engineers [2 ed.]

  • 0 0 0
  • Like this paper and download? You can publish your own PDF file online for free in a few minutes! Sign Up
File loading please wait...
Citation preview

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 12: dU12  U2  U1   Q12   W12  10  18  28J Process 23: dU23  U3  U2   Q23   W23  100  0  100J Process 34: dU34  U4  U3   Q34   W34  20  70  90J Process 41: 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

Page 1 Current edition: 13/02/2015

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 p-V diagram 2. A more general one based on the p-V 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.

Solutions Chapter 1  D E Winterbone

Page 2 Current edition: 13/02/2015

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 non-linear 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 = 212.4C

Initial conditions:

ug  2600kJ / kg ; hg  2799kJ / kg; vg  0.0995m3 / kg p  20bar; t  500C

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

Solutions Chapter 1  D E Winterbone

Page 3 Current edition: 13/02/2015

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

_________________________________________________________________________________________

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  229C 3

Using the saturated water tables, vf = 0.001209m3/kg Hence, x 

0.00317  0.001209  0.0274  2.74% 0.072788  0.001209

_________________________________________________________________________________________

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 102 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 102 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

Solutions Chapter 1  D E Winterbone

Page 4 Current edition: 13/02/2015

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  pV  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

_________________________________________________________________________________________

P1.9

t A  20C; ps  5.673bar; hg  195.78kJ / kg ; h f  54.87kJ / kg

Tank A: 1kg Freon-12

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 

___________________________________________________________________________________

Solutions Chapter 1  D E Winterbone

Page 5 Current edition: 13/02/2015

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 .

Solutions Chapter 1  D E Winterbone

Page 6 Current edition: 13/02/2015

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.

Solutions Chapter 1  D E Winterbone

Page 7 Current edition: 13/02/2015

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.

Solutions Chapter 1  D E Winterbone

Page 8 Current edition: 13/02/2015

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. __________________________________________________________________________________________

Solutions Chapter 1  D E Winterbone

Page 9 Current edition: 13/02/2015

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(

-- Q-l"t

$€q"1 oddzd , q,l^

(trn-tnr) =. A1W -2Q{

H",rt- \Lt -- q,t- : 1+r :

Aa.g+%

-

2B-lt

4/3

#*r^i=i= lue I - !p.+ *rt^ Wen flur* l- W, =! t- W a Li.E'A 25lt

C^*,rd areo

%t^

.fi-oienqy b4b^t€rr Wu tt^r6 b^get-h^tcs \rla-(arncf %qV ( would ; 2*n* - ztr*.+ - B3.g

P3.to.

$

bns .**eu." ve"staan ts er^Tedranua b .Aeo"c .n

qa effinf b ncrease- Vra- {$uss -b pnt ca. G"dt^$!ans ab- sta}re- podrts l, L, c.na 3 oft-s,a.nt,z- as tn P3." C"arratttans

olt 5

or