# FTFS Chap08 P001 ```Chapter 8 Power and Refrigeration Cycles
Chapter 8
POWER AND REFRIGERATION CYCLES
Actual and Ideal Cycles, Carnot cycle, Air-Standard Assumptions
8-1C The Carnot cycle is not suitable as an ideal cycle for all power producing devices because it cannot
be approximated using the hardware of actual power producing devices.
8-2C It is less than the thermal efficiency of a Carnot cycle.
8-3C It represents the net work on both diagrams.
8-4C The cold air standard assumptions involves the additional assumption that air can be treated as an
ideal gas with constant specific heats at room temperature.
8-5C Under the air standard assumptions, the combustion process is modeled as a heat addition process,
and the exhaust process as a heat rejection process.
8-6C The air standard assumptions are: (1) the working fluid is air which behaves as an ideal gas, (2) all
the processes are internally reversible, (3) the combustion process is replaced by the heat addition process,
and (4) the exhaust process is replaced by the heat rejection process which returns the working fluid to its
original state.
8-7C The clearance volume is the minimum volume formed in the cylinder whereas the displacement
volume is the volume displaced by the piston as the piston moves between the top dead center and the
8-8C It is the ratio of the maximum to minimum volumes in the cylinder.
8-9C The MEP is the fictitious pressure which, if acted on the piston during the entire power stroke, would
produce the same amount of net work as that produced during the actual cycle.
8-10C Yes.
8-11C Assuming no accumulation of carbon deposits on the piston face, the compression ratio will remain
the same (otherwise it will increase). The mean effective pressure, on the other hand, will decrease as a car
gets older as a result of wear and tear.
8-12C The SI and CI engines differ from each other in the way combustion is initiated; by a spark in SI
engines, and by compressing the air above the self-ignition temperature of the fuel in CI engines.
8-13C Stroke is the distance between the TDC and the BDC, bore is the diameter of the cylinder, TDC is
the position of the piston when it forms the smallest volume in the cylinder, and clearance volume is the
minimum volume formed in the cylinder.
8-1
Chapter 8 Power and Refrigeration Cycles
8-14 The four processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-s
diagrams, and the net work output 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-21.
Analysis (b) The properties of air at various states are
T1  300 K 

Pr2 
P
3
h1  300.19 kJ/kg
qin
Pr1  1.386
2
u  389.22 kJ/kg
P2
800 kPa
1.386   11 .088  2
Pr1 
T2  539.8 K
P1
100 kPa
T3  1800 K 

1
u 3  1487.2 kJ/kg
qout 4
Pr3  1310
P3 v3 P2 v 2
T
1800 K
800 kPa   2668 kPa


 P3  3 P2 
T3
T2
T2
539.8 K
P
100 kPa
1310   49.10  h4  828.1 kJ/kg
Pr4  4 Pr3 
P3
2668 kPa
T
3
qin
2
From energy balances,
4
qin  u3  u2  1487.2  389.2  1098.0 kJ / kg
qout  h4  h1  8281
.  30019
.  527.9 kJ / kg
wnet ,out  qin  qout  1098.0  527.9  570.1 kJ / kg
(c) Then the thermal efficiency becomes
 th 
wnet ,out
qin
v

570.1 kJ / kg
 51.9%
1098.0 kJ / kg
8-2
1
qout
s
Chapter 8 Power and Refrigeration Cycles
8-15 Problem 8-14 is reconsidered. The effect of varying the temperature after the constant
volume heat addition from 1500 K to 2500 K is to be investigated. The net work output and
thermal efficiency are to be plotted as a function of the maximum temperature of the cycle as well
as the T-s and P-v diagrams for the cycle when the maximum temperature of the cycle is 1800 K.
&quot;We assume that this ideal gas cycle takes place in a piston-cylinder device;
therefore, we will use a closed system analysis.&quot;
&quot;See the T-s diagram in Plot Window1 and the P-v diagram in Plot Window2&quot;
&quot;Input Data&quot;
T=300&quot;K&quot;
P=100&quot;kPa&quot;
P = 800&quot;[kPa]&quot;
T=1800&quot;K&quot;
P = 100 &quot;[kPa]&quot;
&quot;Process 1-2 is isentropic compression&quot;
s=entropy(air,T=T,P=P)
s=s
T=temperature(air, s=s, P=P)
P*v/T=P*v/T
P*v=0.287*T
&quot;Conservation of energy for process 1 to 2&quot;
q_12 -w_12 = DELTAu_12
q_12 =0&quot;isentropic process&quot;
DELTAu_12=intenergy(air,T=T)-intenergy(air,T=T)
&quot;Process 2-3 is constant volume heat addition&quot;
s=entropy(air, T=T, P=P)
{P*v/T=P*v/T}
P*v=0.287*T
v=v
&quot;Conservation of energy for process 2 to 3&quot;
q_23 -w_23 = DELTAu_23
w_23 =0&quot;constant volume process&quot;
DELTAu_23=intenergy(air,T=T)-intenergy(air,T=T)
&quot;Process 3-4 is isentropic expansion&quot;
s=entropy(air,T=T,P=P)
s=s
P*v/T=P*v/T
{P*v=0.287*T}
&quot;Conservation of energy for process 3 to 4&quot;
q_34 -w_34 = DELTAu_34
q_34 =0&quot;isentropic process&quot;
DELTAu_34=intenergy(air,T=T)-intenergy(air,T=T)
&quot;Process 4-1 is constant pressure heat rejection&quot;
{P*v/T=P*v/T}
&quot;Conservation of energy for process 4 to 1&quot;
q_41 -w_41 = DELTAu_41
w_41 =P*(v-v)
&quot;constant pressure process&quot;
DELTAu_41=intenergy(air,T=T)-intenergy(air,T=T)
q_in_total=q_23
w_net = w_12+w_23+w_34+w_41
Eta_th=w_net/q_in_total*100 &quot;Thermal efficiency, in percent&quot;
8-3
Chapter 8 Power and Refrigeration Cycles

50.91
51.58
52.17
52.69
53.16
53.58
q
[kJ/kg]
815.4
1002
1192
1384
1579
1775
w
[kJ/kg]
415.1
516.8
621.7
729.2
839.1
951.2
T
[K]
1500
1700
1900
2100
2300
2500
Air
2000
3
1800
1600
T [K]
1400
1200
800 kPa
1000
100 kPa
800
4
600
2
400
200
5.0
1
5.3
5.5
5.8
6.0
6.3
6.5
6.8
7.0
7.3
7.5
s [kJ/kg-K]
Air
4x103
3
P [kPa]
103
2
102
4
1
1800 K
300 K
101
10-2
10-1
100
v [m3/kg]
8-4
101
102
Chapter 8 Power and Refrigeration Cycles
54
53.5
53
52.5
th
52
51.5
51
50.5
1500
1700
1900
2100
2300
2500
2300
2500
T [K]
1800
qin,total [kJ/kg]
1600
1400
1200
1000
800
1500
1700
1900
2100
T [K]
1000
wnet [kJ/kg]
900
800
700
600
500
400
1500
1700
1900
2100
T [K]
8-5
2300
2500
Chapter 8 Power and Refrigeration Cycles
8-16 The four 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.
Properties The properties of air at room temperature are Cp = 1.005 kJ/kg.K, Cv = 0.718 kJ/kg&middot;K, and k
= 1.4 (Table A-2).
Analysis (b) From the ideal gas isentropic relations and energy balance,
P
T2  T1  2
 P1



k 1 / k
 1000 kPa
 300 K 
 100 kPa



P
0.4/1.4
qin
 579.2 K
3
2
q in  h3  h2  C p T3  T2 
or,
q34
1 q41
2800 kJ/kg  1.005 kJ/kg  K T3  579.2 
4
v
Tmax  T3  3360 K
100 kPa
P3 v 3 P4 v 4
P
3360 K   336 K


 T4  4 T3 
T3
T4
P3
1000 kPa
(c)
T
3
q out  q 34,out  q 41,out  u 3  u 4   h4  h1 
qin
 C v T3  T4   C p T4  T1 
2
 0.718 kJ/kg  K 3360  336 K  1.005 kJ/kg  K 336  300 K
 2212 kJ/kg
 th
2212 kJ/kg
q
 1  out  1 
 21.0%
q in
2800 kJ/kg
8-6
q34
4
1
q41
s
Chapter 8 Power and Refrigeration Cycles
8-17E The four processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-s
diagrams, and the total heat input 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-21E.
Analysis (b) The properties of air at various states are
T1  540 R


q in,12  u 2  u1 

u1  92.04 Btu / lbm
q12
T2  2116 R , h2  537 .1Btu/lbm
h3  849 .48 Btu/lbm
3
2
u 2  u1  q in,12  92 .04  300  392 .04 Btu/lbm
4
1
qout
v
T
3
q23
Pr3  1242
2
P
14.7 psia
1242   317.0  h4  593.22 Btu/lbm
Pr4  4 Pr3 
P3
57.6 psia
q12
1
4
qout
s
From energy balance,
q23,in  h3  h2  849.48  5371
.  312.38 Btu / lbm
qin  q12,in  q23,in  300  312.38  612.38 Btu / lbm
qout  h4  h1  593.22  129.06  464.16 Btu / lbm
(c) Then the thermal efficiency becomes
 th  1 
q23
h1  129.06 Btu / lbm
P2 v 2 P1v1
T
2116 R
14.7 psia   57.6 psia


 P2  2 P1 
T2
T1
T1
540 R
T3  3200 R 

P
qout
464.16 Btu / lbm
 1
 24.2%
qin
612.38 Btu / lbm
8-7
Chapter 8 Power and Refrigeration Cycles
8-18E The four processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-s
diagrams, and the total heat input 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).
P
Analysis (b)
q23
q in,12  u 2  u1  C v T2  T1 
3
2
300 Btu/lbm  0.171 Btu/lbm.R T2  540 R
q12
T2  2294 R
4
1
qout
v
P2 v 2 P1v1
T
2294 R
14.7 psia   62.46 psia


 P2  2 P1 
T2
T1
T1
540 R
q in, 23  h3  h2  C P T3  T2   0.24 Btu/lbm  R 3200  2294 R  217.4 Btu/lbm
Process 3-4 is isentropic:
P
T4  T3  4
 P3




k 1 / k
T
 14.7 psia 

 3200 R 
 62.46 psia 
0.4/1.4
2
q12
q in  q in,12  q in, 23  300  217 .4  517.4Btu/lbm
q out  h4  h1  C p T4  T1   0.240 Btu/lbm.R 2117  540 
 378.5 Btu/lbm
(c)
 th  1 
3
q23
 2117 R
1
4
qout
s
qout
378.5 Btu / lbm
 1
 26.8%
qin
517.4 Btu / lbm
8-8
Chapter 8 Power and Refrigeration Cycles
8-19 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 heat rejected 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 = 1.005 kJ/kg.K, Cv = 0.718 kJ/kg&middot;K, and k
= 1.4 (Table A-2).
Analysis (b)
P
T2  T1  2
 P1



k 1 / k
 1000 kPa
 300 K 
 100 kPa



P
0.4/1.4
 579.2 K
qin
2
3
Qin  mh3  h2   mC p T3  T2 
or,
qout
1
2.76 kJ  0.0015 kg 1.005 kJ/kg  K T3  579 .2 
 T3  2410 K
v
Process 3-1 is a straight line on the P-v diagram, thus the w31 is simply
the area under the process curve,
w31  area 
P3  P1
P P
v1  v3   3 1
2
2
 1000  100 kPa

2

 RT1 RT3 


 P  P 
3 
 1
2410 K
 300 K


100
kPa
1000
kPa

T

0.287 kJ/kg  K   93.1 kJ/kg

qin
3
2
Energy balance for process 3-1 gives
1
qout
s
E in  E out  E system
 Q31,out  W31,out  m(u1  u 3 )


Q31,out   mw 31,out  mC v (T1  T3 )   m w31,out  C v T1  T3 
 0.0015 kg 93.1  0.718 kJ/kg  K 300 - 2410 K   2.133 kJ
(c)
th  1 
Qout
2.133 kJ
1
 22.7%
Qin
2.76 kJ
8-9
Chapter 8 Power and Refrigeration Cycles
8-20 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 net work per 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-21.
Analysis (b) The properties of air at various states are
T1  290 K 

u1  206.91 kJ/kg
P
h1  290.16 kJ/kg
2
P2 v 2 P1v1
P
380 kPa
290 K   1160 K


 T2  2 T1 
T2
T1
P1
95 kPa
qin

 u 2  897.91 kJ/kg, Pr2  207 .2
Pr3 
3
1
qout
v
P3
95 kPa
207.2   51.8  h3  840.38 kJ/kg
Pr2 
P2
380 kPa
T
Qin  mu 2  u1   0.003 kg 897.91  206.91 kJ/kg  2.073 kJ
2
qin
Qout  mh3  h1   0.003 kg 840.38  290.16 kJ/kg  1.651 kJ
Wnet ,out  Qin  Qout  2.073  1.651  0.422kJ
3
1
qout
s
(c)
 th 
Wnet ,out
Qin

0.422 kJ
 20.4%
2.073 kJ
8-10
Chapter 8 Power and Refrigeration Cycles
8-21 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 net work per 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.
Properties The properties of air at room temperature are Cp = 1.005 kJ/kg.K, Cv = 0.718 kJ/kg&middot;K, and k
= 1.4 (Table A-2).
P
Analysis (b) From the isentropic relations and energy balance,
P2 v 2 P1v1
P
380 kPa
290 K   1160 K


 T2  2 T1 
T2
T1
P1
95 kPa
P
T3  T2  3
 P2



k 1 / k
 95 kPa
 1160 K 
 380 kPa



2
qin
0.4/1.4
 780.6 K
3
1
qout
v
T
2
Qin  mu 2  u1   mC v T2  T1 
qin
 0.003 kg 0.718 kJ/kg  K 1160  290 K  1.87 kJ
Qout  mh3  h1   mC p T3  T1 
 0.003 kg 1.005 kJ/kg  K 780.6  290 K  1.48 kJ
Wnet ,out  Qin  Qout  1.87  1.48  0.39kJ
(c)
 th 
3
1
Wnet 0.39 kJ

 20.9%
Qin
1.87 kJ
8-11
qout
s
Chapter 8 Power and Refrigeration Cycles
8-22 A Carnot cycle with the specified temperature limits is considered. The net work output per cycle is 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&middot;K, and k
= 1.4 (Table A-2).
Analysis The minimum pressure in the cycle is P3 and the maximum pressure is P1. Then,
T2  P2 
 
T3  P3 
k 1 / k
T
or,
T
P2  P3  2
 T3




k / k 1
 1000 K 

 20 kPa 
 300 K 
qin
1000
1
300
4
1.4/0.4
 1352 kPa
The heat input is determined from
s 2  s1  C p ln
T2
T1
3
qout
s
0
 Rln
P2
1352 kPa
 0.287 kJ/kg  K ln
 0.08205 kJ/kg  K
P1
1800 kPa
Qin  mT H s 2  s1   0.003 kg 1000 K 0.08205 kJ/kg  K   0.246 kJ
Then,
 th  1 
2
TL
300 K
1
 70.0%
TH
1000 K
Wnet ,out   th Qin  0.70 0.246 kJ   0.172 kJ
8-12
Chapter 8 Power and Refrigeration Cycles
8-23 A Carnot cycle with specified temperature limits is considered. The maximum pressure in the cycle,
the heat transfer to the working fluid, and the mass of the working fluid are to be determined. 
Assumptions Air is an ideal gas with variable specific heats.
Analysis (a) In a Carnot cycle, the maximum pressure occurs at the beginning of the expansion process,
which is state 1.
T1  1200 K


T4  350 K


P1 
Pr1
P4 
Pr4
Pr1  238
Pr 4  2.379
T
(Table A-21)
238
300 kPa   30,013kPa  Pmax
2.379
1200
Qin
Wnet = 0.5 kJ
350
4
TL
350 K
1
 70 .83 %
TH
1200 K
Qin  Wnet ,out /  th  0.5 kJ / 0.7083   0.706kJ
(c) The mass of air is

s 4  s 3  s 4  s 3

0
 Rln
P4
300 kPa
 0.287 kJ/kg  K ln
P3
150 kPa
 0.199 kJ/kg  K  s1  s 2
wnet ,out  s 2  s1 TH  TL   0.199 kJ/kg  K 1200  350 K  169.15 kJ/kg
m
Wnet ,out
wnet ,out

3
Qout
(b) The heat input is determined from
 th  1 
2
1
0.5 kJ
 0.00296kg
169.15 kJ/kg
8-13
s
Chapter 8 Power and Refrigeration Cycles
8-24 A Carnot cycle with specified temperature limits is considered. The maximum pressure in the cycle,
the heat transfer to the working fluid, and the mass of the working fluid are to be determined.
Assumptions Helium is an ideal gas with constant specific heats.
Properties The properties of helium at room temperature are R = 2.0769 kJ/kg.K and k = 1.667 (Table A2).
Analysis (a) In a Carnot cycle, the maximum pressure occurs at the beginning of the expansion process,
which is state 1.
T1  P1 
 
T4  P4 
k 1 / k
or,
T
P1  P4  1
 T4



k / k 1
 1200 K 

 300 kPa 
 350 K 
1.667/0.667
 6524 kPa
Qin
1200
(b) The heat input is determined from
 th  1 
T
TL
350 K
1
 70.83%
TH
1200 K
Wnet = 400 kJ
Qin  Wnet ,out /  th  0.5 kJ / 0.7083   0.706kJ
350
(c) The mass of helium is determined from
s 4  s 3  C p ln
T4
T3
0
 Rln
P4
300 kPa
 2.0769 kJ/kg  K ln
P3
150 kPa
 1.4396 kJ/kg  K  s1  s 2
wnet ,out  s 2  s1 TH  TL   1.4396 kJ/kg  K 1200  350 K  1223.7 kJ/kg
m
Wnet ,out
wnet ,out

2
1
0.5 kJ
 0.000409kg
1223.7 kJ/kg
8-14
4
3
s
Chapter 8 Power and Refrigeration Cycles
Otto Cycle
8-25C The four processes that make up the Otto cycle are (1) isentropic compression, (2) v = constant heat
addition, (3) isentropic expansion, and (4) v = constant heat rejection.
8-26C The ideal Otto cycle involves external irreversibilities, and thus it has a lower thermal efficiency.
8-27C For actual four-stroke engines, the rpm is twice the number of thermodynamic cycles; for two-stroke
engines, it is equal to the number of thermodynamic cycles.
8-28C They are analyzed as closed system processes because no mass crosses the system boundaries
during any of the processes.
8-29C It increases with both of them.
8-30C Because high compression ratios cause engine knock.
8-31C The thermal efficiency will be the highest for argon because it has the highest specific heat ratio, k
= 1.667.
8-32C The fuel is injected into the cylinder in both engines, but it is ignited with a spark plug in gasoline
engines.
8-15
Chapter 8 Power and Refrigeration Cycles
8-33 An ideal Otto cycle with air as the working fluid has a compression ratio of 8. The pressure and
temperature at the end of the heat addition process, the net work output, the thermal efficiency, and the
mean effective pressure for the cycle 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-21.
P
Analysis (a) Process 1-2: isentropic compression.
T1  300 K
v r2


3
u1  214.07 kJ / kg
vr1  621.2
750 kJ/kg
v
1
1
 2 v r1  v r1  621 .2  77 .65
v1
r
8


4
2
1
T2  673.1 K
v
u 2  491.2 kJ/kg
 673.1 K 
P2 v 2 P1v1
v T
95kPa   1705 kPa


 P2  1 2 P1  8
T2
T1
v 2 T1
 300 K 
Process 2-3: v = constant heat addition.
q23,in  u3  u2 
 u3  u2  q23,in  491.2  750  1241.2 kJ / kg 

T3  1539 K
vr 3  6.588
 1539 K 
P3 v3 P2 v 2
T
1705 kPa   3898kPa


 P3  3 P2  
T3
T2
T2
 673.1 K 
(b) Process 3-4: isentropic expansion.
v r4 
T4  774.5K
v1
v r3  rvr3  86.588   52.70 

u 4  571.69 kJ/kg
v2
Process 4-1: v = constant heat rejection.
qout  u4  u1  571.69  214.07  357.62 kJ / kg
wnet ,out  qin  qout  750  357.62  392.38 kJ / kg
(c)
 th 
(d)
v1 
wnet,out
q in
392.38 kJ/kg
 52.3%
750 kJ/kg



RT1
0.287 kPa  m 3 /kg  K 300 K 

 0.906 m 3 /kg  v max
P1
95 kPa
v min  v 2 
MEP 
v max
r
wnet,out
v1  v 2

wnet,out
v1 (1  1 / r )

 kPa  m 3

kJ
0.906 m 3 /kg 1  1/8  

392.38 kJ/kg

8-16

  495.0kPa


Chapter 8 Power and Refrigeration Cycles
8-34 Problem 8-33 is reconsidered. The effect of varying the compression ratio from 5 to 10 is to
be investigated. The net work output and thermal efficiency are to be plotted as a function of the
compression ratio. Also, the T-s and P-v diagrams for the cycle are to be plotted when the
compression ratio is 8.
&quot;We assume that this ideal gas cycle takes place in a piston-cylinder device;
therefore, we will use a closed system analysis.&quot;
&quot;See the T-s diagram in Plot Window1 and the P-v diagram in Plot Window2&quot;
&quot;Input Data&quot;
T=300&quot;K&quot;
P=95&quot;kPa&quot;
q_23 = 750 &quot;[kJ/kg]&quot;
{r_comp = 8}
&quot;Process 1-2 is isentropic compression&quot;
s=entropy(air,T=T,P=P)
s=s
T=temperature(air, s=s, P=P)
P*v/T=P*v/T
P*v=0.287*T
V = V/ r_comp
&quot;Conservation of energy for process 1 to 2&quot;
q_12 - w_12 = DELTAu_12
q_12 =0&quot;isentropic process&quot;
DELTAu_12=intenergy(air,T=T)-intenergy(air,T=T)
&quot;Process 2-3 is constant volume heat addition&quot;
v=v
s=entropy(air, T=T, P=P)
P*v=0.287*T
&quot;Conservation of energy for process 2 to 3&quot;
q_23 - w_23 = DELTAu_23
w_23 =0&quot;constant volume process&quot;
DELTAu_23=intenergy(air,T=T)-intenergy(air,T=T)
&quot;Process 3-4 is isentropic expansion&quot;
s=s
s=entropy(air,T=T,P=P)
P*v=0.287*T
&quot;Conservation of energy for process 3 to 4&quot;
q_34 -w_34 = DELTAu_34
q_34 =0&quot;isentropic process&quot;
DELTAu_34=intenergy(air,T=T)-intenergy(air,T=T)
&quot;Process 4-1 is constant volume heat rejection&quot;
V = V
&quot;Conservation of energy for process 4 to 1&quot;
q_41 - w_41 = DELTAu_41
w_41 =0
&quot;constant volume process&quot;
DELTAu_41=intenergy(air,T=T)-intenergy(air,T=T)
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*100 &quot;Thermal efficiency, in percent&quot;
&quot;The mean effective pressure is:&quot;
MEP = w_net/(V-V)&quot;[kPa]&quot;
8-17
Chapter 8 Power and Refrigeration Cycles

MEP
[kPa]
452.9
469.6
483.5
495.2
505.3
514.2
43.78
47.29
50.08
52.36
54.28
55.93
r
w
[kJ/kg]
328.4
354.7
375.6
392.7
407.1
419.5
5
6
7
8
9
10
Air
104
3
2
103
s4 = 33 = 6.424 kJ/kg-K
P [kPa]
300 K
4
102
s2 = s1 = 5.716 kJ/kg-K
101
10-2
1
1500 K
102
101
100
10-1
v [m3/kg]
Air
1600
3
0.9
1400
kP
1000
95
T [K]
a
1200
800
600
0
39
400
200
4.5
0
11
0.
5.0
Pa
2
4
k
kg
3/
m
1
5.5
6.0
s [kJ/kg-K]
8-18
6.5
7.0
7.5
Chapter 8 Power and Refrigeration Cycles
420
wnet [kJ/kg]
400
380
360
340
320
5
6
7
8
9
10
8
9
10
rcomp
520
510
MEP [kPa]
500
490
480
470
460
450
5
6
7
rcomp
56
54
52
th
50
48
46
44
42
5
6
7
8
rcomp
8-19
9
10
Chapter 8 Power and Refrigeration Cycles
8-35 An ideal Otto cycle with air as the working fluid has a compression ratio of 8. The pressure and
temperature at the end of the heat addition process, the net work output, the thermal efficiency, and the
mean effective pressure for the cycle 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&middot;K, Cv = 0.718 kJ/kg&middot;K, and k
= 1.4 (Table A-2).
P
Analysis (a) Process 1-2: isentropic compression.
v
T2  T1  1
 v2



k 1
 300 K 8
0.4
3
 689 K
750 kJ/kg
 689 K 
P2 v 2 P1v1
v T
95 kPa   1745 kPa


 P2  1 2 P1  8
T2
T1
v 2 T1
 300 K 
1
v
Process 2-3: v = constant heat addition.
q 23,in  u 3  u 2  C v T3  T2 
750 kJ/kg  0.718 kJ/kg  K T3  689 K
T3  1734K
 1734 K 
P3 v3 P2 v 2
T
1745 kPa   4392kPa


 P3  3 P2  
T3
T2
T2
 689 K 
(b) Process 3-4: isentropic expansion.
v
T4  T3  3
 v4



k 1
1
 1734 K  
8
0.4
 755 K
Process 4-1: v = constant heat rejection.
qout  u 4  u1  Cv T4  T1   0.718 kJ/kg  K 755  300 K  327 kJ/kg
wnet,out  q in  q out  750  327  423kJ/kg
(c)
(d)
 th 
wnet,out
v1 
qin
423 kJ / kg
 56.4%
750 kJ / kg


RT1
0.287 kPa  m 3 /kg  K 300 K 

 0.906 m 3 /kg  v max
P1
95 kPa
v min  v 2 
MEP 

v max
r
wnet,out
v1  v 2

wnet,out
v1 (1  1 / r )
4
2

 kPa  m 3

kJ
0.906 m 3 /kg 1  1/8  

423 kJ/kg

8-20

  534kPa


Chapter 8 Power and Refrigeration Cycles
8-36 An ideal Otto cycle with air as the working fluid has a compression ratio of 9.5. The highest pressure
and temperature in the cycle, the amount of heat transferred, 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&middot;K, Cv = 0.718 kJ/kg&middot;K, and k
= 1.4 (Table A-2).
Analysis (a) Process 1-2: isentropic compression.
v
T2  T1  1
 v2



k 1
P
 290 K 9.5 0.4  713.7 K
3
Qin
 713.7 K 
P2 v 2 P1v1
v T
100 kPa   2338 kPa


 P2  1 2 P1  9.5 
T2
T1
v 2 T1
 290 K 
2
v
T3  T4  4
 v3




v
 800 K 9.5 0.4  1969K
Process 2-3: v = constant heat addition.
 1969 K 
P3 v3 P2 v 2
T
2338 kPa   6449kPa


 P3  3 P2  
T3
T2
T2
 713.7 K 
(b)
m


P1V1
100 kPa  0.0006 m 3

 7.21  10  4 kg
RT1
0.287 kPa  m 3 /kg  K 290 K 




Qin  mu3  u 2   mCv T3  T2   7.21 104 kg 0.718kJ/kg  K1969  713.7K  0.650kJ
(c) Process 4-1: v = constant heat rejection.


Qout  m(u 4  u1 )  mCv T4  T1    7.21 104 kg 0.718kJ/kg  K800  290K  0.264kJ
Wnet  Qin  Qout  0.650  0.264  0.386 kJ
 th 
(d)
Wnet,out
Qin
Vmin  V2 
MEP 

0.386 kJ
 59.4%
0.650 kJ
Vmax
r
Wnet ,out
V1  V2

4
1
Process 3-4: isentropic expansion.
k 1
Qout
Wnet ,out
V1 (1  1 / r )

 kPa  m 3

0.0006 m 3 1  1/9.5   kJ

0.386 kJ

8-21

  719kPa


Chapter 8 Power and Refrigeration Cycles
8-37 An Otto cycle with air as the working fluid has a compression ratio of 9.5. The highest pressure and
temperature in the cycle, the amount of heat transferred, 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&middot;K, Cv = 0.718 kJ/kg&middot;K, and k
= 1.4 (Table A-2).
Analysis (a) Process 1-2: isentropic compression.
v
T2  T1  1
 v2



k 1
 290 K 9.5 0.4  713.7 K
P
 713.7 K 
P2 v 2 P1v1
v T
100 kPa   2338 kPa


 P2  1 2 P1  9.5
T2
T1
v 2 T1
 290 K 
3
Polytropic
Process 3-4: polytropic expansion.
PV
(100 kPa)(0.0006 m 3 )
m 1 1 
 7.209  10  4 kg
RT1 (0.287 kPa  m 3 /kg  K )(290 K )
2
n 1
v 
0.35
T3  T4  4   800 K 9.5  1759 K
v
 3
mR T4  T3  7.209  10  4 0.287 kJ/kg  K 800  1759 K
W34 

 0.567 kJ
1 n
1  1.35


Then energy balance for process 3-4 gives
E in  E out  E system
Q34,in  W34,out  mu 4  u 3 
Q34,in  mu 4  u 3   W34,out  mC v T4  T3   W34,out


Q34,in  7.209  10  4 kg 0.718 kJ/kg  K 800  1759 K  0.567 kJ  0.071 kJ
That is, 0.071 kJ of heat is added to the air during the expansion process (This is not realistic, and probably
is due to assuming constant specific heats at room temperature).
(b) Process 2-3: v = constant heat addition.
 1759 K 
P3 v3 P2 v 2
T
2338 kPa   5762kPa


 P3  3 P2  
T3
T2
T2
 713.7 K 
Q23,in  mu 3  u 2   mC v T3  T2 


Q23,in  7.209  10  4 kg 0.718 kJ/kg  K 1759  713.7 K  0.541kJ
Therefore,
Qin  Q23,in  Q34,in  0.541  0.071  0.612 kJ
(c) Process 4-1: v = constant heat rejection.


Qout  mu 4  u1   mCv T4  T1   7.209 104 kg 0.718kJ/kg  K800  290K  0.264kJ
Wnet ,out  Qin  Qout  0.612  0.264  0.348 kJ
 th 
Wnet ,out
Qin

4 800 K
Qin
0.348 kJ
 56.9%
0.612 kJ
8-22
Qout
1
290 K
v
Chapter 8 Power and Refrigeration Cycles
(d)
Vmin  V2 
MEP 
Vmax
r
Wnet ,out
V1  V2

Wnet ,out
V1 (1  1 / r )

 kPa  m 3

0.0006 m 3 1  1/9.5   kJ

0.348 kJ

8-23

  648kPa


Chapter 8 Power and Refrigeration Cycles
8-38E An ideal Otto cycle with air as the working fluid has a compression ratio of 8. The amount of heat
transferred to the air during the heat addition process, the thermal efficiency, and the thermal efficiency of a
Carnot cycle operating between the same temperature limits 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.
P
2400 R
Properties The properties of air are given in Table A-21E.
3
Analysis (a) Process 1-2: isentropic compression.
T1  540 R
v r2 


u1  92.04 Btu / lbm
vr1  144.32
qin
2
Process 2-3: v = constant heat addition.
u 3  452.70 Btu/lbm
v r3  2.419
qin  u 3  u 2  452 .70  211 .28  241 .42 Btu/lbm
(b) Process 3-4: isentropic expansion.
v r4 
v4
v r  rvr3  82.419   19.35 
 u 4  205.54 Btu/lbm
v3 3
Process 4-1: v = constant heat rejection.
q out  u 4  u1  205 .54  92.04  113 .50 Btu/lbm
th  1 
(c)
q out
113.50 Btu/lbm
1
 47.0%
qin
241.42 Btu/lbm
th,C  1 
4
1 540 R
v2
1
1
v r2  v r2  144 .32   18 .04 
 u 2  211.28 Btu/lbm
v1
r
8
T3  2400 R 

qout
TH
540 R
1
 77.5%
TL
2400 R
8-24
v
Chapter 8 Power and Refrigeration Cycles
8-39E An ideal Otto cycle with argon as the working fluid has a compression ratio of 8. The amount of
heat transferred to the argon during the heat addition process, the thermal efficiency, and the thermal
efficiency of a Carnot cycle operating between the same temperature limits are to be determined. 
Assumptions 1 The air-standard assumptions are applicable with argon as the working fluid. 2 Kinetic and
potential energy changes are negligible. 3 Argon is an ideal gas with constant specific heats.
Properties The properties of argon are Cp = 0.1253 Btu/lbm.R, Cv = 0.0756 Btu/lbm.R, and k = 1.667
P
(Table A-2E).
3
Analysis (a) Process 1-2: isentropic compression.
v
T2  T1  1
 v2



k 1
 540 R 8
0.667
 2161 R
qin
2
qout
4
1
Process 2-3: v = constant heat addition.
qin  u 3  u 2  C v T3  T2   0.0756 Btu/lbm.R 2400  2161 R  18.07 Btu/lbm.R
(b) Process 3-4: isentropic expansion.
v
T4  T3  3
 v4



k 1
1
 2400 R  
8
0.667
 600 R
Process 4-1: v = constant heat rejection.
q out  u 4  u1  C v T4  T1   0.0756 Btu/lbm.R 600  540 R  4.536 Btu/lbm
 th  1 
(c)
4.536 Btu/lbm
q out
1
 74.9%
qin
18.07 Btu/lbm
th,C  1 
TH
540 R
1
 77.5%
TL
2400 R
8-25
v
Chapter 8 Power and Refrigeration Cycles
Diesel Cycle
8-40C 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.
8-41C 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.
8-42C The gasoline engine.
8-43C Diesel engines operate at high compression ratios because the diesel engines do not have the engine
knock problem.
8-44C 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.
8-26
Chapter 8 Power and Refrigeration Cycles
8-45 An air-standard Diesel cycle with a compression ratio of 16 and a cutoff ratio of 2 is considered. The
temperature after the heat addition process, 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 properties of air are given in Table A-21.
P
2
qin
3
Analysis (a) Process 1-2: isentropic compression.
T1  300 K
v r2 


u1  214.07 kJ / kg
4
vr1  621.2
1
qout
T2  862.4 K
v2
1
1
v r1  v r1  621 .2  38.825 

h2  890.9 kJ/kg
v1
r
16
Process 2-3: P = constant heat addition.
P3 v3 P2 v 2
v


 T3  3 T2  2T2  2862.4 K   1724.8 K
T3
T2
v2


(b)
h3  1910.6 kJ/kg
v r3  4.546
qin  h3  h2  1910.6  890.9  1019.7 kJ / kg
Process 3-4: isentropic expansion.
v r4 
v4
v
r
16
v r  4 v r  v r  4.546   36 .37 
 u 4  659.7 kJ/kg
v3 3 2v 2 3 2 3 2
Process 4-1: v = constant heat rejection.
qout  u4  u1  659.7  214.07  445.63 kJ / kg
 th  1 
(c)
qout
445.63 kJ / kg
 1
 56.3%
qin
1019.7 kJ / kg
wnet ,out  q in  q out  1019 .7  445 .63  574.07 kJ/kg
v1 


RT1
0.287kPa  m 3 /kg  K 300 K 

 0.906 m 3 /kg  v max
P1
95 kPa
v min  v 2 
MEP 
v max
r
wnet ,out
v1  v 2

wnet ,out
v1 1  1 / r 

 kPa  m 3

0.906 m 3 /kg 1  1/16   kJ

574.07 kJ/kg

8-27

  675.9kPa


v
Chapter 8 Power and Refrigeration Cycles
8-46 An air-standard Diesel cycle with a compression ratio of 16 and a cutoff ratio of 2 is considered. The
temperature after the heat addition process, 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&middot;K, Cv = 0.718 kJ/kg&middot;K, and k
= 1.4 (Table A-2).
P
Analysis (a) Process 1-2: isentropic compression.
v
T2  T1  1
 v2



k 1
2
4
qout
1
P3 v3 P2 v 2
v


 T3  3 T2  2T2  2909.4 K   1818.8K
T3
T2
v2
v
q in  h3  h2  C p T3  T2   1.005 kJ/kg  K 1818.8  909.4 K  913.9 kJ/kg
Process 3-4: isentropic expansion.
v
T4  T3  3
 v4



k 1
 2v
 T3  2
 v4



k 1
2
 1818.8 K  
 16 
0.4
 791.7 K
Process 4-1: v = constant heat rejection.
q out  u 4  u1  C v T4  T1   0.718 kJ/kg  K 791.7  300 K  353 kJ/kg
 th  1 
(c)
q out
353 kJ/kg
1
 61.4%
qin
913.9 kJ/kg
wnet.out  q in  q out  913 .9  353  560.9 kJ/kg
v1 


RT1
0.287kPa  m 3 /kg  K 300 K 

 0.906 m 3 /kg  v max
P1
95 kPa
v min  v 2 
MEP 
3
 300 K 16 0.4  909.4 K
Process 2-3: P = constant heat addition.
(b)
qin
v max
r
wnet ,out
v1  v 2

wnet ,out
v1 1  1 / r 

 kPa  m 3

0.906 m 3 /kg 1  1/16   kJ

560.9 kJ/kg

8-28

  660.4kPa


Chapter 8 Power and Refrigeration Cycles
8-47E 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-21E.
Analysis (a) Process 1-2: isentropic compression.
T1  540 R
v r2 


u1  92.04 Btu / lbm
vr1  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.
P
P3 v3 P2 v 2
v
T
3000 R


 3  3 
 1.848
T3
T2
v 2 T2 1623.6 R
(b)
T3  3000 R 

2
qin
3 3000 R
h3  790.68 Btu/lbm
4
v r3  1.180
1
qout
q in  h3  h2  790 .68  402 .05  388.63 Btu/lbm
Process 3-4: isentropic expansion.
v r4 
v4
v4
r
18.2
1.180   11.621  u 4  250.91Btu/lbm
v r3 
v r3 
v r3 
v3
1.848v 2
1.848
1.848
Process 4-1: v = constant heat rejection.
q out  u 4  u1  250 .91  92.04  158.87 Btu/lbm
(c)
 th  1 
q out
158.87 Btu/lbm
1
 59.1%
q in
388.63 Btu/lbm
8-29
v
Chapter 8 Power and Refrigeration Cycles
8-48E 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).
P
Analysis (a) Process 1-2: isentropic compression.
v
T2  T1  1
 v2



k 1
2
qin
 540 R 18.2 0.4  1724 R
4
Process 2-3: P = constant heat addition.
1
P3 v3 P2 v 2
v
T
3000 R


 3  3 
 1.741
T3
T2
v 2 T2 1724 R
(b)
v
qin  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
Process 4-1: v = constant heat rejection.
q out  u 4  u1  C v T4  T1 
 0.171 Btu/lbm.R 1173  540R  108 Btu/lbm
(c)
 th  1 
3
q out
108 Btu/lbm
1
 64.6%
qin
306 Btu/lbm
8-30
 1173 R
Chapter 8 Power and Refrigeration Cycles
8-49 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&middot;K, Cv = 0.718 kJ/kg&middot;K, and k
= 1.4 (Table A-2).
Analysis (a) Process 1-2: isentropic compression.
V
T2  T1  1
 V2



P
2
k 1
 293 K 20 
qin
3
 971.1 K
0.4
4
Process 2-3: P = constant heat addition.
P3V3 P2 V2

T3
T2


qout
1
V3 T3 2200 K


 2.265
V2 T2 971.1 K
v
Process 3-4: isentropic expansion.
V
T4  T3  3
 V4



k 1
 2.265V2
 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  C v T4  T1   0.718 kJ/kg  K 920.6  293 K  450.6 kJ/kg
wnet ,out  qin  q out  1235  450 .6  784.4 kJ/kg
 th 
(b)
v1 
wnet ,out
qin

784.4 kJ/kg
 63.5%
1235 kJ/kg


RT1
0.287kPa  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
v1 1  1 / r 

 kPa  m 3

kJ
0.885 m 3 /kg 1  1/20  

784.4 kJ/kg

8-31

  933kPa


Chapter 8 Power and Refrigeration Cycles
8-50 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 = 1.005 kJ/kg&middot;K, Cv = 0.718 kJ/kg&middot;K, and k
= 1.4 (Table A-2).
Analysis (a) Process 1-2: isentropic compression.
V
T2  T1  1
 V2



P
k 1
2
 293 K 20 0.4  971.1 K
qin
3
Polytropic
4
Process 2-3: P = constant heat addition.
P3V3 P2 V2

T3
T2


qout
V3 T3 2200 K


 2.265
V2 T2 971.1 K
1
v
Process 3-4: polytropic expansion.
V
T4  T3  3
 V4



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  C v 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

 963 kJ/kg
1 n
1  1.35
 E system
w34,out 
E in  E out
q 34,in  w34,out  u 4  u 3 
 q 34,in  w34,out  C v T4  T3 
 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
 th 
wnet ,out
q in

580.6 kJ/kg
 47 .0%
1235 kJ/kg
8-32
Chapter 8 Power and Refrigeration Cycles
(c)
v1 


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
v1 1  1 / r 

 1 kPa  m 3

kJ
0.885 m 3 /kg 1  1/20  

580.6 kJ/kg

8-33

  691 kPa


Chapter 8 Power and Refrigeration Cycles
8-51 Problem 8-50 is reconsidered. The effect of varying the compression ratio from 14 to 24 is to
be investigated. The net work output, mean effective pressure and thermal efficiency as to be
plotted as a function of the compression ratio. The T-s and P-v diagrams for the cycle are also to
be plotted when the compression ratio is 20.
&quot;Let's take advantage of the capabilities of EES and do this for variable specific heats.&quot;
&quot;We assume that this ideal gas cycle takes place in a piston-cylinder device;
therefore, we will use a closed system analysis.&quot;
&quot;See the T-s diagram in Plot Window1 and the P-v diagram in Plot Window2&quot;
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 &gt; 0) THEN q_in_total = q_12 ELSE q_out_total = - q_12
If q_23 &gt; 0 then q_in_total = q_in_total + q_23 else q_out_total = q_out_total - q_23
If q_34 &gt; 0 then q_in_total = q_in_total + q_34 else q_out_total = q_out_total - q_34
If q_41 &gt; 0 then q_in_total = q_in_total + q_41 else q_out_total = q_out_total - q_41
END
&quot;Input Data&quot;
T=293&quot;K&quot;
P=95&quot;kPa&quot;
T = 2200&quot;[K]&quot;
n=1.35
{r_comp = 20}
&quot;Process 1-2 is isentropic compression&quot;
s=entropy(air,T=T,P=P)
s=s
T=temperature(air, s=s, P=P)
P*v/T=P*v/T
P*v=0.287*T
V = V/ r_comp
&quot;Conservation of energy for process 1 to 2&quot;
q_12 - w_12 = DELTAu_12
q_12 =0&quot;isentropic process&quot;
DELTAu_12=intenergy(air,T=T)-intenergy(air,T=T)
&quot;Process 2-3 is constant pressure heat addition&quot;
P=P
s=entropy(air, T=T, P=P)
P*v=0.287*T
&quot;Conservation of energy for process 2 to 3&quot;
q_23 - w_23 = DELTAu_23
w_23 =P*(V - V)&quot;constant pressure process&quot;
DELTAu_23=intenergy(air,T=T)-intenergy(air,T=T)
&quot;Process 3-4 is polytropic expansion&quot;
P/P =(V/V)^n
s=entropy(air,T=T,P=P)
P*v=0.287*T
&quot;Conservation of energy for process 3 to 4&quot;
q_34 - w_34 = DELTAu_34 &quot;q_34 is not 0 for the polytropic process&quot;
DELTAu_34=intenergy(air,T=T)-intenergy(air,T=T)
P*V^n = Const
w_34=(P*V-P*V)/(1-n)
&quot;Process 4-1 is constant volume heat rejection&quot;
V = V
&quot;Conservation of energy for process 4 to 1&quot;
q_41 - w_41 = DELTAu_41
w_41 =0
&quot;constant volume process&quot;
DELTAu_41=intenergy(air,T=T)-intenergy(air,T=T)
8-34
Chapter 8 Power and Refrigeration Cycles
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 &quot;Thermal efficiency, in percent&quot;
&quot;The mean effective pressure is:&quot;
MEP = w_net/(V-V)&quot;[kPa]&quot;

47.69
50.14
52.16
53.85
55.29
56.54
MEP
[kPa]
970.8
985
992.6
995.4
994.9
992
r
w
[kJ/kg]
797.9
817.4
829.8
837.0
840.6
841.5
14
16
18
20
22
24
Air
104
104
P [kPa]
103
103
293 K
2200 K
102
102
1049 K
74
6.
9
kJ
5.6
/kg
-K
101
10-2
10-1
100
101
102
101
v [m3/kg]
Air
2400
2200
3
3/k
a
2
1000
4
kP
a
800
95
kP
1200
0.1
600
400
200
4.0
0.8
44
0.0
0.1
1400
1
4.5
5.0
5.5
34
T [K]
1600
8m
1800
g
59
20
k
Pa
2000
6.0
s [kJ/kg-K]
8-35
6.5
7.0
7.5
Chapter 8 Power and Refrigeration Cycles
850
840
wnet [kJ/kg]
830
820
810
800
790
14
16
18
20
22
24
rcomp
57
55
th
53
51
49
47
14
16
18
20
22
24
rcomp
1000
MEP [kPa]
995
990
985
980
975
970
14
16
18
20
rcomp
8-36
22
24
Chapter 8 Power and Refrigeration Cycles
8-52 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 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&middot;K, Cv = 0.718 kJ/kg&middot;K, and k
= 1.4 (Table A-2).
P
2
Analysis Process 1-2: isentropic compression.
V
T2  T1  1
 V2



k 1
 300 K 17 0.4  931.8 K
Qin
3
4
Qout
1
Process 2-3: P = constant heat addition.
v
P3 v3 P2 v 2
v


 T3  3 T2  2.2T2  2.2 931.8 K   2050 K
T3
T2
v2
Process 3-4: isentropic expansion.
V
T4  T3  3
 V4



n 1
 2.2V 2
 T3 
 V4



n 1
 2.2 
 T3 

 r 
n 1
 2.2 
 2050 K 

 17 
0.4
 904.8 K
V1  V disp  V 2  V disp  V1 / r  V1  V disp /(1  1 / r )  3 /(1  1 / 17 )  3.188 L
m
(97 kPa )( 0.003188 m 3 )
P1V1

 3.592  10 3 kg
RT1 (0.287 kPa  m 3 /kg  K )( 300 K )
Qin  mh3  h2   mC p T3  T2 
 (3.592  10 3 kg )(1.005 kJ/kg  K )( 2050  931.8 )K  4.036 kJ
Qout  mu 4  u1   mC v T4  T1 
 3.592  10 3 kg 0.718 kJ/kg  K 904.8  300 K  1.560 kJ


W net ,out  Qin  Qout  4.036  1.560  2.476 kJ/rev
W net ,out  n W net ,out  1500/60 rev/s2.476 kJ/rev   61.9 kW
Discussion Note that for 2-stroke engines, 1 thermodynamic cycle is equivalent to 1 mechanical cycle (and
thus revolutions).
8-37
Chapter 8 Power and Refrigeration Cycles
8-53 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&middot;K, Cv = 0.743 kJ/kg&middot;K,
and k = 1.4 (Table A-2).
P
Analysis Process 1-2: isentropic compression.
V
T2  T1  1
 V2



k 1
2
Qin
3
 300 K 17 0.4  931.8 K
4
Qout
Process 2-3: P = constant heat addition.
1
P3 v3 P2 v 2
v


 T3  3 T2  2.2T2  2.2 931.8 K   2050 K
T3
T2
v2
v
Process 3-4: isentropic expansion.
V
T4  T3  3
 V4



n 1
 2.2V 2
 T3 
 V4



n 1
 2.2 
 T3 

 r 
n 1
 2.2 
 2050 K 

 17 
0.4
 904.8 K
V1  V disp  V 2  V disp  V1 / r  V1  V disp /(1  1 / r )  3 /(1  1 / 17 )  3.188 L
m


Qin  mh3  h2   mC p T3  T2 


97 kPa  0.003188 m 3
P1V1

 3.473  10 3 kg
RT1
0.2968 kPa  m 3 /kg  K 300 K 


 3.473  10 3 kg 1.039 kJ/kg  K 2050  931.8 K  4.035 kJ
Qout  mu 4  u1   mC v T4  T1 
 3.473  10 3 kg 0.743 kJ/kg  K 904.8  300 K  1.561 kJ


W net ,out  Qin  Qout  4.035  1.561  2.474 kJ/rev
W net ,out  n W net ,out  1500/60 rev/s2.474 kJ/rev   61.8 kW
Discussion Note that for 2-stroke engines, 1 thermodynamic cycle is equivalent to 1 mechanical cycle (and
thus revolutions).
8-38
Chapter 8 Power and Refrigeration Cycles
8-54 [Also solved by EES on enclosed CD] An ideal dual cycle with air as the working fluid has a
compression ratio of 14. The fraction of heat transferred at constant volume and the thermal efficiency of
the cycle 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-21.
Analysis (a) Process 1-2: isentropic compression.
T1  300 K
v r2 


u1  214.07 kJ / kg
vr1  621.2
T  823.1 K
v2
1
1
v r1  v r1  621 .2  44 .37 
 2
u 2  611.2 kJ/kg
v1
r
14
P
T3  2200 K 

3
x
h3  2503.2 kJ/kg
v r3  2.012
2
1520.4
kJ/kg
q in  q x  2,in  q 3 x,in  u x  u 2   h3  h x 
1520 .4  u x  611 .2  2503 .2  h x 
Thus,
q2  x ,in  u x  u2  1022.82  611.2  411.62 kJ / kg
and
(b)
q2  x ,in
qin
P3 v3 Px v x

T3
Tx
v r4 

411.62 kJ / kg
 27.1%
1520.4 kJ / kg


v3 T3 2200 K


 1.692  rc
v x Tx 1300 K
v4
v4
r
14
2.012   16.648  u 4  886.3 kJ/kg
v r3 
vr 
vr 
v3
1.692 v 2 3 1.692 3 1.692
Process 4-1: v = constant heat rejection.
qout  u4  u1  886.3  214.07  672.23 kJ / kg
 th  1 
Qout
1
v
By trial and error, we get T x = 1300 K and hx = 1395.97, ux = 1022.82 kJ /kg.
ratio 
4
qout
672.23 kJ / kg
 1
 55.8%
qin
1520.4 kJ / kg
8-39
Chapter 8 Power and Refrigeration Cycles
8-55 Problem 8-54 is reconsidered. The effect of varying the compression ratio from 10 to 18 is to
be investigated. For a compression ratio of 14, the T-s and P-v diagrams for the cycle are to be
plotted.
&quot;We assume that this ideal dual cycle takes place in a piston-cylinder device;
therefore, we will use a closed system analysis.&quot;
&quot;See Figure 8-23 for the P-v diagram for the cycle. See the T-s diagram in
Plot Window1 and the P-v diagram in Plot Window2&quot;
&quot;Input Data&quot;
T=300&quot;[K]&quot;
P=100&quot;[kPa]&quot;
T=2200&quot;[K]&quot;
q_in_total=1520&quot;[kJ/kg]&quot;
r_v = 14
v/v=r_v &quot;Compression ratio&quot;
&quot;Process 1-2 is isentropic compression&quot;
s=entropy(air,T=T,P=P)&quot;[kJ/kg-K]&quot;
s=s&quot;[kJ/kg-K]&quot;
s=entropy(air, T=T, v=v)&quot;[kJ/kg-K]&quot;
P*v/T=P*v/T
P*v=0.287*T
&quot;Conservation of energy for process 1 to 2&quot;
q_12 -w_12 = DELTAu_12
q_12 =0&quot;[kJ/kg]&quot;&quot;isentropic process&quot;
DELTAu_12=intenergy(air,T=T)-intenergy(air,T=T)&quot;[kJ/kg]&quot;
&quot;Process 2-3 is constant volume heat addition&quot;
s=entropy(air, T=T, P=P)&quot;[kJ/kg-K]&quot;
{P*v/T=P*v/T}
P*v=0.287*T
v=v&quot;[m^3/kg]&quot;
&quot;Conservation of energy for process 2 to 3&quot;
q_23 -w_23 = DELTAu_23
w_23 =0&quot;constant volume process&quot;
DELTAu_23=intenergy(air,T=T)-intenergy(air,T=T)&quot;[kJ/kg]&quot;
&quot;Process 3-4 is constant pressure heat addition&quot;
s=entropy(air, T=T, P=P)&quot;[kJ/kg-K]&quot;
{P*v/T=P*v/T}
P*v=0.287*T
P=P&quot;[kPa]&quot;
&quot;Conservation of energy for process 3 to4&quot;
q_34 -w_34 = DELTAu_34
w_34 =P*(v-v)
&quot;constant pressure process&quot;
DELTAu_34=intenergy(air,T=T)-intenergy(air,T=T)
q_in_total=q_23+q_34
&quot;Process 4-5 is isentropic expansion&quot;
s=entropy(air,T=T,P=P)&quot;[kJ/kg-K]&quot;
s=s&quot;[kJ/kg-K]&quot;
P*v/T=P*v/T
{P*v=0.287*T}
&quot;Conservation of energy for process 4 to 5&quot;
q_45 -w_45 = DELTAu_45
q_45 =0&quot;[kJ/kg]&quot;&quot;isentropic process&quot;
DELTAu_45=intenergy(air,T=T)-intenergy(air,T=T)&quot;[kJ/kg]&quot;
8-40
Chapter 8 Power and Refrigeration Cycles
&quot;Process 5-1 is constant volume heat rejection&quot;
v=v&quot;[m^3/kg]&quot;
&quot;Conservation of energy for process 2 to 3&quot;
q_51 -w_51 = DELTAu_51
w_51 =0&quot;[kJ/kg]&quot;&quot;constant volume process&quot;
DELTAu_51=intenergy(air,T=T)-intenergy(air,T=T)&quot;[kJ/kg]&quot;
w_net = w_12+w_23+w_34+w_45+w_51
Eta_th=w_net/q_in_total*100 &quot;Thermal efficiency, in percent&quot;
 [%]
r
52.33
53.43
54.34
55.09
55.72
56.22
56.63
56.94
57.17
10
11
12
13
14
15
16
17
18
w
[kJ/kg]
795.4
812.1
826
837.4
846.9
854.6
860.7
865.5
869
T-s Diagram for Air Dual Cycle
3500
3000
6025 kPa
2500
3842 kPa
T [K]
4
p=const
2000
382.7 kPa
1500
3
5
1000
100 kPa
2
v=const
500
1
0
4.0
4.5
5.0
5.5
6.0
6.5
s [kJ/kg-K]
8-41
7.0
7.5
8.0
8.5
Chapter 8 Power and Refrigeration Cycles
P-v Diagram for Air Dual Cycle
8x103
4
3
2
103
2200 K
P [kPa]
s=const
5
102
1
300 K
101
10-2
10-1
100
101
102
3
v [m /kg]
58
57
th [%]
56
55
54
53
52
10
11
12
13
14
15
16
17
18
rv
870
860
wnet [kJ/kg]
850
840
830
820
810
800
790
10
11
12
13
14
rv
8-42
15
16
17
18
Chapter 8 Power and Refrigeration Cycles
8-56 An ideal dual cycle with air as the working fluid has a compression ratio of 14. The fraction of heat
transferred at constant volume and the thermal efficiency of the cycle 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&middot;K, Cv = 0.718 kJ/kg&middot;K, and k
= 1.4 (Table A-2).
P
Analysis (a) Process 1-2: isentropic compression.
v
T2  T1  1
 v2



k 1
3
x
 300 K 14 0.4  862 K
2
1520.4
kJ/kg
4
Qout