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Chapter 9
9–14 An air-standard cycle with variable specific heats is executed in a closed system
and is composed of the following four processes:
1-2 Isentropic compression from 100 kPa and 27°C to 800 kPa
2-3 v = constant heat addition to 1800 K
3-4 Isentropic expansion to 100 kPa
4-1 P = constant heat rejection to initial state
( a ) Show the cycle on P-v and T - s diagrams.
( b ) Calculate the net work output per unit mass.
( c ) Determine the thermal efficiency.
9-14 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-17.
Analysis ( b ) The properties of air at various states are
( 1310 ) 49.10 828.1kJ/kg
2668 kPa
100 kPa
800 kPa 2668 kPa 539.8K
1800 K
1487.2kJ/kg 1800 K
539.8K
389.22kJ/kg 1.386 11. 088 100 kPa
800 kPa
300.19kJ/kg 300 K
4 3
4
2 2
3 3 2
2 2
3
3 3
3 3
2
2
1
2
1 1
4 3
3
2 1
1
P h P
P
P
P
T
T
P
T
P
T
P
P
u T
T
u P P
P
P
P
h T
r r
r
r r
r
v v
From energy balances,
0 527. 9 570.1 kJ/kg
1 300. 19 527.9kJ/kg
2 389. 2 1098.0kJ/kg
net,out in out
out 4 1
in 3 2
w q q
q h h
q u u
( c ) Then the thermal efficiency becomes
1098.0kJ/kg
570.1kJ/kg
in
net,out th q
w
v
P
1
2
4
3
qin
q (^) out
s
T
1
2
4
3
qin
qout
9–22 Consider a Carnot cycle executed in a closed system with 0.003 kg of air. The
temperature limits of the cycle are 300 and 900 K, and the minimum and maximum
pressures that occur during the cycle are 20 and 2000 kPa. Assuming constant specific
heats, determine the net work output per cycle.
9-22 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·K, R = 0.287 kJ/kg.K, and k = 1.4 (Table A-2).
Analysis The minimum pressure in the cycle is P 3 and the maximum pressure is P 1.
Then,
or,
( )
( )
( ) 935. 3 kPa 300 K
900 K
20 kPa
/ (^1) 1.4/0.
3
2 2 3
1 /
3
2
3
2
−
−
k k
k k
T
T
P P
P
P
T
T
The heat input is determined from
Then,
( )
( ) ( )( )( )
= =( )( ) = 0.393kJ
0.667 0.5889 kJ
900 K
300 K
0.003kg 900 K 0.2181kJ/kgK 0.5889kJ
0.2181kJ/kgK 2000 kPa
935.3kPa ln ln 0.287kJ/kgKln
net,out th in
th
in 2 1
1
2
0
1
2 2 1
W Q
T
T
Q mT s s
P
P
R
T
T
s s c
H
L
H
p
©
s
T
3
2
qin
q (^) out
4
900 1
300
( c ) = = = 52.3% 750 kJ/kg
392.4kJ/kg
in
net,out th q
w
( d )
( )( )
( )( )
495.0kPa kJ
kPam
0.906m/kg 1 1/
392.4kJ/kg
( 1 1 / )
MEP
0.906m/kg 95 kPa
0.287kPa m/kgK 300 K
3
3 1
net,out
1 2
net,out
max min 2
max
3
3
1
1 1
r
w w
r
P
RT
v v v
v v v
v v
9–47 An air-standard Diesel cycle has a compression ratio of 16 and a cutoff ratio of 2.
At the beginning of the compression process, air is at 95 kPa and 27°C. Accounting for
the variation of specific heats with temperature, determine ( a ) the temperature after the
heat-addition process, ( b ) the thermal efficiency, and ( c ) the mean effective pressure.
Answers: ( a ) 1724.8 K, ( b ) 56.3 percent, ( c ) 675.9 kPa
9-47 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 gas constant of air is R =
0.287 kJ/kg.K. The properties of air are given
in Table A-17.
Analysis ( a ) Process 1-2: isentropic compression.
214.07kJ/kg 300 K 1
1 1 =
r
u T v
( ) 8 90.9kJ/kg
862.4K
2
2
1
2 2 1 1 =
h
T
r
r v r^ v r
v
v
v
Process 2-3: P = constant heat addition.
( )( )
- 546
1910.6kJ/kg 2 2 862.4K 3
3 2 2 2
3 3 2
2 2
3
3 3
r
h T T T T
P
T
P
v^ v
v v v
1724.8K
( b ) q in = h 3 − h 2 = 1910. 6 − 890. 9 = 1 019.7kJ/kg
Process 3-4: isentropic expansion.
( 4. 546 ) 36. 37 659.7kJ/kg 2
4 2
4
3
4 4 3 3 3 = = = = = ⎯⎯→ u =
r
r r v r^ v r
v
v
v
v
v
v
Process 4-1: v = constant heat rejection.
v
P
4
1
2 3
qin
qout
1019.7 kJ/kg
445.63kJ/kg 1 1
- 7 214. 07 445.63kJ/kg
in
out th
out 4 1
q
q
q u u
( c )
( )( )
( ) (^) ( )( )
675.9kPa kJ
kPa m
0.906m/kg 1 1/
574.07kJ/kg
1 1 /
MEP
0.906m/kg 95 kPa
0.287kPam/kg K 300 K
- 7 445. 63 574.07kJ/kg
3
3 1
net,out
1 2
net,out
max min 2
max
3
3
1
1 1
net,out in out
r
w w
r
P
RT
w q q
v v v
v
v v
v v
9–84 A gas-turbine power plant operates on the simple Brayton cycle between the
pressure limits of 100 and 1200 kPa. The working fluid is air, which enters the
compressor at 30°C at a rate of 150 m3/min and leaves the turbine at 500°C. Using
variable specific heats for air and assuming a compressor isentropic efficiency of 82
percent and a turbine isentropic efficiency of 88 percent, determine ( a ) the net power
output, ( b ) the back work ratio, and ( c ) the thermal efficiency. Answers: ( a ) 659 kW, ( b )
0.625, ( c ) 0.
9-84 A gas-turbine plant operates on the simple Brayton cycle. The net power output,
the back work ratio, and the thermal efficiency are to be determined.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential
energy changes are negligible. 3 Air is an ideal gas with variable specific heats.
Properties The gas constant of air is R = 0.287 kJ/kg·K (Table A-1).
Analysis ( a ) For this problem, we use
the properties from EES software.
Remember that for an ideal gas,
enthalpy is a function of temperature
only whereas entropy is functions of
both temperature and pressure.
Process 1-2: Compression
- 7159 kJ/kg K 100 kPa
30 C
30 C 303. 60 kJ/kg
1 1
1
1 1
s P
T
T h
- 37 kJ/kg
- 7159 kJ/kg.K
1200 kPa 2 2 1
2
⎭
hs s s
P
1
Combustion chamber
Turbin
2
3
4
Compress.
100 kPa
30°C
500°C
