ME 470 Gas Power Cycle HW Solutions
Instructor: Shoeleh Di Julio
Chapter 9, Solution 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·K, and k
= 1.4 (Table A-2).
Analysis (b) From the ideal gas isentropic relations and energy balance,
P
k 1 / k
P 
 1000 kPa
T2  T1 2 
 300 K 
P
 100 kPa
 1
q in  h3  h2  c p T3  T2 



0.4/1.4
qin
 579.2 K
3
2
q34
2800 kJ/kg  1.005 kJ/kg  K T3  579.2  
 Tmax  T3  3360 K
(c)
P3v 3 P4v 4
P
100 kPa
3360 K   336 K


 T4  4 T3 
T3
T4
P3
1000 kPa
q out  q 34,out  q 41,out  u 3  u 4   h4  h1 
1 q41
4
v
T
3
 cv T3  T4   c p T4  T1 
qin
 0.718 kJ/kg  K 3360  336 K  1.005 kJ/kg  K 336  300 K
 2212 kJ/kg
2
q34
4
1
q41
q
2212 kJ/kg
s
 th  1  out  1 
 21.0%
q in
2800 kJ/kg
Discussion The assumption of constant specific heats at room temperature is not realistic in this case the
temperature changes involved are too large.
Chapter 9, Solution 37.
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·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 
v2 
k 1
P
 308 K 9.5
0.4
 757.9 K
 757.9 K 
P2v 2 P1v1
v T
100 kPa   2338 kPa


 P2  1 2 P1  9.5
T2
T1
v 2 T1
 308 K 
Process 3-4: isentropic expansion.
v 
T3  T4  4 
 v3 
k 1
 800 K 9.50.4  1969 K
Process 2-3: v = constant heat addition.
3
Qin
2
Qout
4
1
v
v
 1969 K 
P3v 3 P2v 2
T
2338 kPa   6072 kPa


 P3  3 P2  
T3
T2
T2
 757.9 K 
(b)
m


P1V1
100 kPa  0.0006 m3

 6.788 10  4 kg
RT1
0.287 kPa  m 3 /kg  K 308 K 




Qin  mu3  u 2   mcv T3  T2   6.78810 4 kg 0.718 kJ/kg K1969  757.9K  0.590 kJ
(c) Process 4-1: v = constant heat rejection.


Qout  m(u 4  u1 )  mcv T4  T1    6.78810 4 kg 0.718 kJ/kg K800  308K  0.240 kJ
Wnet  Qin  Qout  0.590  0.240  0.350 kJ
 th 
(d)
Wnet,out
Vmin  V2 
MEP 

Qin
0.350 kJ
 59.4%
0.590 kJ
Vmax
r
Wnet,out
V1  V2

Wnet,out
V1 (1  1 / r )

 kPa  m 3 

  652 kPa
0.0006 m 1  1/9.5   kJ 

0.350 kJ
3

Chapter 9, Solution 65.
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
s 4  s 3  c p ln
and
T4
T3

150 kJ/kg
 0.5 kJ/kg  K
300 K
0
 Rln
P4
P3
T
300 K
P4
 0.287 kJ/kg  K ln
 0.5 kJ/kg  K
120 kPa
It yields
qin
4
3
q out TL
T
1200 K
150 kJ/kg   600 kJ/kg


 q in  H q out 
q in
TH
TL
300 K
wnet,out  q in  q out  600  150  450 kJ/kg
(c) The thermal efficiency of this totally reversible cycle is determined from
 th  1 
TL
300 K
 1
 75.0%
TH
1200 K
2
qout
P4 = 685.2 kPa
(b) For reversible cycles,
Thus,
1
1200 K
s
Chapter 9, Solution 66.
An ideal Stirling engine with helium as the working fluid operates between the specified temperature and
pressure limits. The thermal efficiency of the cycle, the amount of heat transfer in the regenerator, and the
work output per cycle are to be determined.
Assumptions Helium is an ideal gas with constant specific heats.
Properties The gas constant and the specific heat of helium at room temperature are R = 2.0769 kJ/kg.K, cv
= 3.1156 kJ/kg.K and cp = 5.1926 kJ/kg.K (Table A-2).
Analysis (a) The thermal efficiency of this totally reversible cycle is determined from
 th  1 
TL
300 K
 1
 85.0%
TH
2000 K
(b) The amount of heat transferred in the regenerator is
Qregen  Q41,in  mu1  u4   mcv T1  T4 
T
1
2000 K
qin
2
 0.12 kg 3.1156 kJ/kg  K 2000  300 K
 635.6 kJ
300 K
P3v 3 P1v 1
v
T P 300 K 3000 kPa 
v


 3  3 1 
3 2
T3
T1
v 1 T1 P3 2000 K 150 kPa 
v1
T2
T1
3
qout
(c) The net work output is determined from
s2  s1  cv ln
4
0
 Rln
v2
 2.0769 kJ/kg  K ln 3  2.282 kJ/kg  K
v1
Qin  mTH s2  s1   0.12 kg 2000 K 2.282 kJ/kg  K   547.6 kJ
Wnet,out   th Qin  0.85 547 .6 kJ   465.5 kJ
s