9-70
Brayton Cycle with Intercooling, Reheating, and Regeneration 9-101C As the number of compression and expansion stages are increased and regeneration is employed, the ideal Brayton cycle will approach the Ericsson cycle. 9-102C (a) decrease, (b) decrease, and (c) decrease. 9-103C (a) increase, (b) decrease, and (c) decrease. 9-104C (a) increase, (b) decrease, (c) decrease, and (d) increase. 9-105C (a) increase, (b) decrease, (c) increase, and (d) decrease. 9-106C Because the steady-flow work is proportional to the specific volume of the gas. Intercooling decreases the average specific volume of the gas during compression, and thus the compressor work. Reheating increases the average specific volume of the gas, and thus the turbine work output. 9-107C (c) The Carnot (or Ericsson) cycle efficiency.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-71
9-108 An ideal gas-turbine cycle with two stages of compression and two stages of expansion is considered. The back work ratio and the thermal efficiency of the cycle are to be determined for the cases of with and without a regenerator. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of air are given in Table A-17. Analysis (a) The work inputs to each stage of compressor are identical, so are the work outputs of each stage of the turbine since this is an ideal cycle. Then,
( )( )
( )
( ) ( )( ) ( ) kJ/kg 62.86636.94679.127722
kJ/kg 2.142219.30026.41122
kJ/kg 946.3633.7923831
238kJ/kg 77.7912
K 2001
kJ/kg 411.26158.4386.13
386.1kJ/kg 300.19
K 300
65outT,
12inC,
865
6
755
421
2
11
56
5
12
1
=−=−=
=−=−=
==→=
==
===
→=
==→===
==
→=
hhw
hhw
hhPPP
P
Phh
T
hhPPP
P
Ph
T
rr
r
rr
r
2
9
10
7
6 8
300 K
1200 K qin 5
1
4
3
T
s
Thus, 33.5%===kJ/kg 662.86kJ/kg 222.14
outT,
inC,bw w
wr
( ) ( ) ( ) ( )
36.8%===
=−=−=
=−+−=−+−=
kJ/kg 1197.96kJ/kg 440.72
kJ/kg 440.72222.1486.662
kJ/kg 1197.9636.94679.127726.41179.1277
in
netth
inC,outT,net
6745in
qw
www
hhhhq
η
(b) When a regenerator is used, rbw remains the same. The thermal efficiency in this case becomes
( ) ( )( )
55.3%===
=−=−=
=−=−=
kJ/kg 796.63kJ/kg 440.72
kJ/kg 796.6333.40196.1197
kJ/kg 33.40126.41136.94675.0
in
netth
regenoldin,in
48regen
qw
qqq
hhq
η
ε
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-72
9-109 A gas-turbine cycle with two stages of compression and two stages of expansion is considered. The back work ratio and the thermal efficiency of the cycle are to be determined for the cases of with and without a regenerator. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of air are given in Table A-17. Analysis (a) The work inputs to each stage of compressor are identical, so are the work outputs of each stage of the turbine. Then,
( )( )
( )( ) ( )
( )
( )( )(
( ) ( )( ) ( ) kJ/kg 63.44507.99679.127722
kJ/kg 77.68219.30003.43922
kJ/kg 96.07936.94679.127785.079.1277
kJ/kg 946.3633.7923831
238kJ/kg 1277.79K 1200
kJ/kg 9.034380.0/19.30026.41119.300
/
kJ/kg 411.26158.4386.13
386.1kJ/kg 300.19K 300
65outT,
12inC,
6558665
65
865
6
755
1214212
12
421
2
11
56
5
12
1
=−=−=
=−=−=
=−−=
−−==→−−
=
==→=
==
===→=
=−+=
−+==→−−
=
==→===
==→=
hhw
hhw
hhhhhhhhh
hhPPP
P
PhhT
hhhhhhhhh
hhPPP
P
PhT
sTs
T
rr
r
Css
C
ssrr
r
ηη
ηη
)
T
4
8
22s
9
10
7
66s
8qin
5
1
4
3s
Thus, 49.3%===kJ/kg 563.44kJ/kg 277.68
outT,
inC,bw w
wr
( ) ( ) ( ) ( )
25.5%===
=−=−=
=−+−=−+−=
kJ/kg 1120.48kJ/kg 285.76
kJ/kg 5.762868.27744.563
kJ/kg 1120.48996.071277.79439.031277.79
in
netth
inC,outT,net
6745in
qw
www
hhhhq
η
(b) When a regenerator is used, rbw remains the same. The thermal efficiency in this case becomes
( ) ( )( )
40.7%===
=−=−=
=−=−=
kJ/kg 702.70kJ/kg 285.76
kJ/kg .7070278.41748.1120
kJ/kg 17.78403.43907.99675.0
in
netth
regenoldin,in
48regen
qw
qqq
hhq
η
ε
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-73
9-110 A regenerative gas-turbine cycle with two stages of compression and two stages of expansion is considered. The minimum mass flow rate of air needed to develop a specified net power output is to be determined. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of air are given in Table A-17. Analysis The mass flow rate will be a minimum when the cycle is ideal. That is, the turbine and the compressors are isentropic, the regenerator has an effectiveness of 100%, and the compression ratios across each compression or expansion stage are identical. In our case it is rp = 9 = 3. Then the work inputs to each stage of compressor are identical, so are the work outputs of each stage of the turbine.
( )( )
( )
( ) ( )( ) ( )
kg/s 249.6===
=−=−=
=−=−=
=−=−=
==→=
==
===→=
==→===
==→=
kJ/kg 440.72kJ/s 110,000
kJ/kg 440.72222.1468.662
kJ/kg 662.86946.361277.7922
kJ/kg 222.1419.30026.41122
kJ/kg 6.369433.7923831
238 kJ/kg, 1277.79K 0120
kJ/kg 11.264158.4386.13
386.1 kJ/kg, 00.193K 003
net
net
inC,outT,net
65outT,
12inC,
865
6
755
421
2
11
56
5
12
1
wW
m
www
hhw
hhw
hhPPP
P
PhhT
hhPPP
P
PhT
rr
r
rr
r
&&
2
7
6 8
300 K
1200 K 5
1
4
3
T
s
9-111 A regenerative gas-turbine cycle with two stages of compression and two stages of expansion is considered. The minimum mass flow rate of air needed to develop a specified net power output is to be determined. Assumptions 1 Argon is an ideal gas with constant specific heats. 2 Kinetic and potential energy changes are negligible. Properties The properties of argon at room temperature are cp = 0.5203 kJ/kg.K and k = 1.667 (Table A-2a). Analysis The mass flow rate will be a minimum when the cycle is ideal. That is, the turbine and the compressors are isentropic, the regenerator has an effectiveness of 100%, and the compression ratios across each compression or expansion stage are identical. In our case it is rp = 9 = 3. Then the work inputs to each stage of compressor are identical, so are the work outputs of each stage of the turbine.
( )( )( )
( )( )
( ) ( ) ( )( )
( ) ( ) ( )( )
kg/s 404.7===
=−=−=
=−⋅=−=−=
=−⋅=−=−=
=
=
=
==
=
−
−
kJ/kg 271.8kJ/s 110,000
kJ/kg 271.83.1721.444
kJ/kg 444.1K773.21200KkJ/kg 0.5203222
kJ/kg 172.3K300465.6KkJ/kg 0.5203222
K 773.231K 1200
K 465.63K 300
net
net
inC,outT,net
6565outT,
1212inC,
70.667/1.66/1
5
656
70.667/1.66/1
1
212
wW
m
www
TTchhw
TTchhw
PP
TT
PP
TT
p
p
kk
kk
&&
2
7
6 8
300 K
1200 K 5
1
4
3
T
s
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-74
Jet-Propulsion Cycles 9-112C The power developed from the thrust of the engine is called the propulsive power. It is equal to thrust times the aircraft velocity. 9-113C The ratio of the propulsive power developed and the rate of heat input is called the propulsive efficiency. It is determined by calculating these two quantities separately, and taking their ratio. 9-114C It reduces the exit velocity, and thus the thrust. 9-115E A turbojet engine operating on an ideal cycle is flying at an altitude of 20,000 ft. The pressure at the turbine exit, the velocity of the exhaust gases, and the propulsive efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with constant specific heats at room temperature. 4 Kinetic and potential energies are negligible, except at the diffuser inlet and the nozzle exit. 5 The turbine work output is equal to the compressor work input. Properties The properties of air at room temperature are cp = 0.24 Btu/lbm.R and k = 1.4 (Table A-2Ea). Analysis (a) For convenience, we assume the aircraft is stationary and the air is moving towards the aircraft at a velocity of V1 = 900 ft/s. Ideally, the air will leave the diffuser with a negligible velocity (V2 ≅ 0). Diffuser:
( )
( )( )( )
( )( ) psia 11.19
R 470R 537.3
psia 7
R 537.4/sft 25,037
Btu/lbm 1RBtu/lbm 0.242
ft/s 900470
2
2/02
0
2/2/
1.4/0.41/
1
212
22
221
12
2112
21
022
12
222
211
outin
(steady) 0systemoutin
=
=
=
=
⋅+=+=
−−=
−+−=
+=+
=
∆=−
−kk
p
p
TT
PP
cV
TT
VTTc
VVhh
VhVh
EE
EEE&&
&&&
T
6
qout
5
qin
3
4
2
1s
Compressor:
( )( ) ( )( )( )
( )( ) R 1118.313R 537.4
psia 145.5psia 11.1913
0.4/1.4/1
2
323
243
==
=
====
− kk
p
PP
TT
PrPP
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-75
Turbine:
( ) ( )54235423outturb,incomp, TTcTTchhhhww pp −=−→−=−→=
or,
( )( ) psia 55.2=
=
=
=+−=+−=
− 1.4/0.41/
4
545
2345
R 2400R 1819.1
psia 145.5
R 1819.14.5371118.34002kk
TT
PP
TTTT
(b) Nozzle:
( )( )
( ) 2/02
0
2/2/
R 1008.6psia 55.2
psia 7R 1819.1
2656
025
26
56
266
255
outin
(steady) 0systemoutin
0.4/1.4/1
5
656
VTTc
VVhh
VhVh
EE
EEE
PP
TT
p
kk
+−=
−+−=
+=+
=
∆=−
=
=
=
−
&&
&&&
or,
( )( )( ) s/ft 3121=
−⋅=
Btu/lbm 1/sft 25,037
R1008.61819.1RBtu/lbm 0.240222
6V
(c) The propulsive efficiency is the ratio of the propulsive work to the heat input,
( )
( )[ ]( )
( ) ( )( )
%25.9===
=−⋅=−=−=
=
−=
−=
Btu/lbm 307.6Btu/lbm 79.8
Btu/lbm 307.6R1118.32400RBtu/lbm 0.24
Btu/lbm 79.8/sft 25,037
Btu/lbm 1ft/s 900ft/s 9003121
in
3434in
22
aircraftinletexit
qw
TTchhq
VVVw
pp
p
p
η
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-76
9-116E A turbojet engine operating on an ideal cycle is flying at an altitude of 20,000 ft. The pressure at the turbine exit, the velocity of the exhaust gases, and the propulsive efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with variable specific heats. 4 Kinetic and potential energies are negligible, except at the diffuser inlet and the nozzle exit. 5 The turbine work output is equal to the compressor work input. Properties The properties of air are given in Table A-17E. Analysis (a) For convenience, we assume the aircraft is stationary and the air is moving towards the aircraft at a velocity of V1= 900 ft/s. Ideally, the air will leave the diffuser with a negligible velocity (V 2 ≅ 0).
Diffuser: 8548.0
Btu/lbm 112.20R 470
1
11
==→=
rPhT
T
( )
( ) psia 11.220.85481.3698psia 7
3698.1Btu/lbm 128.48/sft 25,037
Btu/lbm 12ft/s 900
20.1122
20
2/2/
1
2
2
12
r22
221
12
21
022
12
222
211
outin
(steady) 0systemoutin
=
=
=
=→=
+=+=
−+−=
+=+
=
∆=−
r
r
P
PPP
PV
hh
VVhh
VhVh
EE
EEE&&
&&&
6
qout
5
qin
3
4
2
1s
Compressor:
( )( ) ( )( )
( ) Btu/lbm 267.5680.171.36811.22145.8
psia 145.8psia 11.2213
32
3
243
23=→=
=
=
====
hPPP
P
PrPP
rr
p
Turbine: T 6.367Btu/lbm 22.617
R 24004
44 =
=→=
rPh
5423
outturb,incomp,
hhhh
ww
−=−
=
or,
( ) psia 56.6=
=
=
=→=+−=+−=
367.6142.7psia 145.8
7.142 Btu/lbm14.47848.12856.26722.617
4
5
5
45
2345
r
r
r
P
PPP
Phhhh
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-77
(b) Nozzle:
20
2/2/
Btu/lbm 266.9317.66psia 56.6
psia 7)7.142(
025
26
56
266
255
outin
(steady) 0systemoutin
65
656
VVhh
VhVh
EE
EEE
hPP
PP rr
−+−=
+=+
=
∆=−
=→=
=
=
&&
&&&
or,
( ) ( ) ft/s 3252=
−=−=
Btu/lbm 1/sft 25,037
Btu/lbm)93.26614.478(2222
656 hhV
(c) The propulsive efficiency is the ratio of the propulsive work to the heat input,
( )
[ ]
24.2%===
=−=−=
=
−=
−=
Btu/lbm 349.66Btu/lbm 84.55
Btu/lbm 349.6656.26722.617
Btu/lbm 84.55/sft 25,037
Btu/lbm 1ft/s) (900ft/s) 900)(3252
in
34
22
aircraftinletexit
qw
hhq
VVVw
pp
in
p
η
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-78
9-117 A turbojet aircraft flying at an altitude of 9150 m is operating on the ideal jet propulsion cycle. The velocity of exhaust gases, the propulsive power developed, and the rate of fuel consumption are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with constant specific heats at room temperature. 4 Kinetic and potential energies are negligible, except at the diffuser inlet and the nozzle exit. 5 The turbine work output is equal to the compressor work input. Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K and k = 1.4 (Table A-2a). Analysis (a) We assume the aircraft is stationary and the air is moving towards the aircraft at a velocity of V 1 = 320 m/s. Ideally, the air will leave the diffuser with a negligible velocity (V 2 ≅ 0). Diffuser:
( )( )
( )( )( )
( ) kPa 62.6K 241K 291.9
kPa 32
K 291.9/sm 1000
kJ/kg 1KkJ/kg 1.0052
m/s 320K 241
2
2/02
02/2/
1.4/0.41/
1
212
22
221
12
2112
21
022
122
222
11
outin(steady) 0
systemoutin
=
=
=
=
⋅+=+=
−−=
−+−=→+=+
=→∆=−
−kk
p
p
TT
PP
cV
TT
VTTc
VVhhVhVh
EEEEE &&&&&
T
6
5
Qi·
3
4
2
1s
Compressor:
( )( ) ( )( )( )
( )( ) K 593.712K 291.9
kPa 751.2kPa 62.612
0.4/1.4/1
2
323
243
==
=
====
− kk
p
PP
TT
PrPP
Turbine:
or, ( ) ( )
K1098.2291.9593.714002345
54235423outturb,incomp,
=+−=+−=
−=−→−=−→=
TTTT
TTcTTchhhhww pp
Nozzle:
( )( )
( ) 2/02
0
2/2/
K 568.2kPa 751.2
kPa 32K 1400
2656
025
26
56
266
255
outin(steady) 0
systemoutin
0.4/1.4/1
4
646
VTTcVV
hh
VhVh
EEEEE
PP
TT
p
kk
+−=→−
+−=
+=+
=→∆=−
=
=
=
−
&&&&&
or, ( )( )( ) m/s1032 kJ/kg 1
/sm 1000K568.21098.2KkJ/kg 1.0052
22
6 =
−⋅=V
(b) ( ) ( )( ) ( ) kW 13,670=
−=−=
22aircraftinletexit/sm 1000
kJ/kg 1m/s 320m/s3201032kg/s 60VVVmW p &&
(c) ( ) ( ) ( )( )( )
kg/s 1.14===
=−⋅=−=−=
kJ/kg 42,700kJ/s 48,620
HV
kJ/s 48,620K593.71400KkJ/kg 1.005kg/s 60
infuel
3434in
Qm
TTcmhhmQ p
&&
&&&
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-79
9-118 A turbojet aircraft is flying at an altitude of 9150 m. The velocity of exhaust gases, the propulsive power developed, and the rate of fuel consumption are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with constant specific heats at room temperature. 4 Kinetic and potential energies are negligible, except at the diffuser inlet and the nozzle exit. Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K and k = 1.4 (Table A-2a). Analysis (a) For convenience, we assume the aircraft is stationary and the air is moving towards the aircraft at a velocity of V 1 = 320 m/s. Ideally, the air will leave the diffuser with a negligible velocity (V 2 ≅ 0). Diffuser:
( )
( )( )( )
( )( ) kPa 62.6
K 241K 291.9
kPa 32
K 291.9/sm 1000
kJ/kg 1KkJ/kg 1.0052
m/s 320K 241
2
2/02
0
2/2/
1.4/0.41/
1
212
22
221
12
2112
21
022
12
222
211
outin
(steady) 0systemoutin
=
=
=
=
⋅+=+=
−−=
−+−=
+=+
=
∆=−
−kk
p
p
TT
PP
cV
TT
VTTc
VVhh
VhVh
EE
EEE&&
&&&
T
5
6
5s
Qi·
3
4
2
1s
Compressor:
( )( ) ( )( )( )
( )( ) K 593.712K 291.9
kPa 751.2kPa 62.612
0.4/1.4/1
2
323
243
==
=
====
− kk
s
p
PP
TT
PrPP
( )( )
( ) ( ) ( ) K 669.20.80/291.9593.71.929/2323
23
23
23
23
=−+=−+=
−
−=
−−
=
Cs
p
spsC
TTTT
TTcTTc
hhhh
η
η
Turbine:
or, ( ) ( )
K 1022.7291.9669.214002345
54235423outturb,incomp,
=+−=+−=
−=−→−=−→=
TTTT
TTcTTchhhhww pp
( )( )
( ) ( )( )
( ) kPa 197.7K 1400K 956.1
kPa 751.2
K 956.185.0/1022.714001400/1.4/0.41/
4
545
5445
54
54
54
54
=
=
=
=−−=−−=
−
−=
−−
=
−kks
Ts
sp
p
sT
TT
PP
TTTT
TTcTTc
hhhh
η
η
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-80
Nozzle:
( )( )
( ) 2/02
0
2/2/
K 607.8kPa 197.7
kPa 32K 1022.7
2656
025
26
56
266
255
outin
(steady) 0systemoutin
0.4/1.4/1
5
656
VTTc
VVhh
VhVh
EE
EEE
PP
TT
p
kk
+−=
−+−=
+=+
=
∆=−
=
=
=
−
&&
&&&
or,
( )( )( ) m/s 913.2=
−⋅=
kJ/kg 1/sm 1000
K607.81022.7KkJ/kg 1.005222
6V
(b) ( )( )( ) ( )
kW 11,390=
−=
−=
22
aircraftinletexit
/sm 1000kJ/kg 1
m/s 320m/s320913.2kg/s 60
VVVmW p &&
(c) ( ) ( ) ( )( )( )
kg/s 1.03===
=−⋅=−=−=
kJ/kg 42,700kJ/s 44,067
HV
kJ/s 44,067K669.21400KkJ/kg 1.005kg/s 60
infuel
3434in
Qm
TTcmhhmQ p
&&
&&&
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-81
9-119 A turbojet aircraft that has a pressure rate of 12 is stationary on the ground. The force that must be applied on the brakes to hold the plane stationary is to be determined. Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with variable specific heats. 4 Kinetic and potential energies are negligible, except at the nozzle exit. Properties The properties of air are given in Table A17.
TAnalysis (a) Using variable specific heats for air,
5
4
qin
2
3
1
Compressor: T 386.1
kJ/kg 300.19K 300
1
11
==→=
rPh
( )( )
( )( )
27.396 kJ/kg 1464.65854610.65
kJ/kg 854kg/s 10
kJ/s 8540
kJ/s 8540kJ/kg 42,700kg/s 0.2HV
kJ/kg 610.6563.16386.112
3
12
2323in
inin
fuelin
21
2
=→
=+=+=→−=
===
==×=
=→===
r
in
rr
Pqhhhhq
mQ
q
mQ
hPPP
P
&
&
&&
s
Turbine:
or, kJ/kg 741.17300.19610.65464.6511234
4312outturb,incomp,
=+−=+−=
−=−→=
hhhh
hhhhww
Nozzle:
( )
20
2/2/
kJ/kg 41.79702.3312127.396
024
25
45
255
244
outin
(steady) 0systemoutin
53
535
VVhh
VhVh
EE
EEE
hPP
PP rr
−+−=
+=+
=
∆=−
=→=
=
=
&&
&&&
or,
( ) ( )( ) m/s 908.9kJ/kg 1
/sm 1000kJ/kg741.171154.1922
22
545 =
−=−= hhV
Brake force = Thrust = ( ) ( )( ) N 9089=
⋅−=−
2inletexitm/skg 1N 1
m/s0908.9kg/s 10VVm&
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-82
9-120 EES Problem 9-119 is reconsidered. The effect of compressor inlet temperature on the force that must be applied to the brakes to hold the plane stationary is to be investigated. Analysis Using EES, the problem is solved as follows: P_ratio = 12 T_1 = 27 [C] T[1] = T_1+273 "[K]" P[1]= 95 [kPa] P[5]=P[1] Vel[1]=0 [m/s] V_dot[1] = 9.063 [m^3/s] HV_fuel = 42700 [kJ/kg] m_dot_fuel = 0.2 [kg/s] Eta_c = 1.0 Eta_t = 1.0 Eta_N = 1.0 "Inlet conditions" h[1]=ENTHALPY(Air,T=T[1]) s[1]=ENTROPY(Air,T=T[1],P=P[1]) v[1]=volume(Air,T=T[1],P=P[1]) m_dot = V_dot[1]/v[1] "Compressor anaysis" s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor" P_ratio=P[2]/P[1]"Definition of pressure ratio - to find P[2]" T_s[2]=TEMPERATURE(Air,s=s_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exit" h_s[2]=ENTHALPY(Air,T=T_s[2]) Eta_c =(h_s[2]-h[1])/(h[2]-h[1]) "Compressor adiabatic efficiency; Eta_c = W_dot_c_ideal/W_dot_c_actual. " m_dot*h[1] +W_dot_c=m_dot*h[2] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" "External heat exchanger analysis" P[3]=P[2]"process 2-3 is SSSF constant pressure" h[3]=ENTHALPY(Air,T=T[3]) Q_dot_in = m_dot_fuel*HV_fuel m_dot*h[2] + Q_dot_in= m_dot*h[3]"SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0" "Turbine analysis" s[3]=ENTROPY(Air,T=T[3],P=P[3]) s_s[4]=s[3] "For the ideal case the entropies are constant across the turbine" {P_ratio= P[3] /P[4]} T_s[4]=TEMPERATURE(Air,h=h_s[4]) "Ts[4] is the isentropic value of T[4] at turbine exit" {h_s[4]=ENTHALPY(Air,T=T_s[4])} "Eta_t = W_dot_t /Wts_dot turbine adiabatic efficiency, Wts_dot > W_dot_t" Eta_t=(h[3]-h[4])/(h[3]-h_s[4]) m_dot*h[3] = W_dot_t + m_dot*h[4] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" T[4]=TEMPERATURE(Air,h=h[4]) P[4]=pressure(Air,s=s_s[4],h=h_s[4]) "Cycle analysis" W_dot_net=W_dot_t-W_dot_c"Definition of the net cycle work, kW" W_dot_net = 0 [kW]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-83
"Exit nozzle analysis:" s[4]=entropy('air',T=T[4],P=P[4]) s_s[5]=s[4] "For the ideal case the entropies are constant across the nozzle" T_s[5]=TEMPERATURE(Air,s=s_s[5], P=P[5]) "T_s[5] is the isentropic value of T[5] at nozzle exit" h_s[5]=ENTHALPY(Air,T=T_s[5]) Eta_N=(h[4]-h[5])/(h[4]-h_s[5]) m_dot*h[4] = m_dot*(h_s[5] + Vel_s[5]^2/2*convert(m^2/s^2,kJ/kg)) m_dot*h[4] = m_dot*(h[5] + Vel[5]^2/2*convert(m^2/s^2,kJ/kg)) T[5]=TEMPERATURE(Air,h=h[5]) s[5]=entropy('air',T=T[5],P=P[5]) "Brake Force to hold the aircraft:" Thrust = m_dot*(Vel[5] - Vel[1]) "[N]" BrakeForce = Thrust "[N]" "The following state points are determined only to produce a T-s plot" T[2]=temperature('air',h=h[2]) s[2]=entropy('air',T=T[2],P=P[2])
BrakeForce
[N]
m [kg/s]
T3 [K]
T1 [C]
9971 11.86 1164 -20 9764 11.41 1206 -10 9568 10.99 1247 0 9383 10.6 1289 10 9207 10.24 1330 20 9040 9.9 1371 30
4.5 5.0 5.5 6.0 6.5 7.0 7.5
200
400
600
800
1000
1200
1400
1600
s [kJ/kg-K]
T [K
]
95
kPa
114
0 kP
a
Air
1
2s
3
4s
5s
-20 -10 0 10 20 309000
9200
9400
9600
9800
10000
T1 [C]
Bra
keFo
rce
[N]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-84
9-121 Air enters a turbojet engine. The thrust produced by this turbojet engine is to be determined. Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with variable specific heats. 4 Kinetic and potential energies are negligible, except at the diffuser inlet and the nozzle exit. Properties The properties of air are given in Table A-17. Analysis We assume the aircraft is stationary and the air is moving towards the aircraft at a velocity of V 1 = 300 m/s. Taking the entire engine as our control volume and writing the steady-flow energy balance yield
T h
T h1 1
2 2
280 28013
700 713 27
= → =
= → =
K k
K k
.
.
J / kg
J / kg 15,000 kJ/s
( ) ( )
−+−=
−+−=
+=++
=
∆=−
22
222
21
22
12in
222
211in
outin
(steady) 0systemoutin
/sm 1000kJ/kg 1
2m/s 300
13.28027.713kg/s 16kJ/s 000,15
2
)2/()2/(
V
VVhhmQ
VhmVhmQ
EE
EEE
&&
&&&
&&
&&&
7°C 300 m/s16 kg/s
427°C
1 2
It gives V 2 = 1048 m/s Thus, ( ) ( )( ) N 11,968=−=−= m/s3001048kg/s 1612 VVmFp &
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-85
Second-Law Analysis of Gas Power Cycles 9-122 The total exergy destruction associated with the Otto cycle described in Prob. 9-34 and the exergy at the end of the power stroke are to be determined. Analysis From Prob. 9-34, qin = 750, qout = 357.62 kJ/kg, T1 = 300 K, and T4 = 774.5 K. The total exergy destruction associated with this Otto cycle is determined from
( ) kJ/kg 245.12=
−=
−=
K 2000kJ/kg 750
K 300kJ/kg 357.62
K 300inout0destroyed
HL Tq
Tq
Tx
Noting that state 4 is identical to the state of the surroundings, the exergy at the end of the power stroke (state 4) is determined from ( ) ( ) ( )040040044 vv −+−−−= PssTuuφ
where
( ) KkJ/kg 0.7081K 300K 774.5ln KkJ/kg 0.2871.70203.68232
lnlnln
0kJ/kg357.62
1
414
41
1414
1
4141404
1404
out1404
⋅=⋅−−=
−−=−−=−−=−=−
=−=−==−=−
TT
RssTT
RssPP
Rssssss
quuuu
oooooo
v
v
vvvv
Thus, ( ) ( )( ) kJ/kg 145.2=+⋅−= 0KkJ/kg 0.7081K 300kJ/kg 357.624φ
9-123 The total exergy destruction associated with the Diesel cycle described in Prob. 9-47 and the exergy at the end of the compression stroke are to be determined. Analysis From Prob. 9-47, qin = 1019.7, qout = 445.63 kJ/kg, T1 = 300 K, v1 = 0.906 m3/kg, and v 2 = v 1 / r = 0.906 / 12 = 0.0566 m3/kg. The total exergy destruction associated with this Otto cycle is determined from
( ) kJ/kg 292.7=
−=
−=
K 2000kJ/kg 1019.7
K 300kJ/kg 445.63
K 300inout0destroyed
HL Tq
Tq
Tx
Noting that state 1 is identical to the state of the surroundings, the exergy at the end of the compression stroke (state 2) is determined from
( ) ( ) ( )( ) ( ) ( )( ) ( )( )
kJ/kg 348.6=
⋅−+−−=
−+−−−=−+−−−=
33
12012012
020020022
mkPa 1kJ 1
/kgm0.9060.0566kPa 950214.07643.3
vv
vv
PssTuuPssTuuφ
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-86
9-124E The exergy destruction associated with the heat rejection process of the Diesel cycle described in Prob. 9-49E and the exergy at the end of the expansion stroke are to be determined. Analysis From Prob. 9-49E, qout = 158.9 Btu/lbm, T1 = 540 R, T4 = 1420.6 R, and v 4 = v 1. At Tavg = (T4 + T1)/2 = (1420.6 + 540)/2 = 980.3 R, we have cv,avg = 0.180 Btu/lbm·R. The entropy change during process 4-1 is
( ) RBtu/lbm 0.1741R 1420.6
R 540lnRBtu/lbm 0.180lnln0
4
1
4
141 ⋅−=⋅=+=−
vvR
TTcss v
Thus,
( ) Btu/lbm 64.9=
+⋅−=
+−=
R 540Btu/lbm 158.9RBtu/lbm 0.1741R54041,
41041 destroyed,R
R
Tq
ssTx
Noting that state 4 is identical to the state of the surroundings, the exergy at the end of the power stroke (state 4) is determined from ( ) ( ) ( )040040044 vv −+−−−= PssTuuφ
where
RBtu/lbm 0.1741
0RBtu/lbm 9.581
1404
1404
out1404
⋅=−=−=−=−
⋅==−=−
ssss
quuuuvvvv
Thus, ( ) ( )( ) Btu/lbm 64.9=+⋅−= 0RBtu/lbm 0.1741R540Btu/lbm 158.94φ
Discussion Note that the exergy at state 4 is identical to the exergy destruction for the process 4-1 since state 1 is identical to the dead state, and the entire exergy at state 4 is wasted during process 4-1.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-87
9-125 The exergy destruction associated with each of the processes of the Brayton cycle described in Prob. 9-73 is to be determined. Analysis From Prob. 9-73, qin = 584.62 kJ/kg, qout = 478.92 kJ/kg, and
KkJ/kg2.67602kJ/kg789.16
KkJ/kg3.13916K1160
KkJ/kg2.47256kJ/kg6.364
KkJ/kg.734981K310
44
33
22
11
⋅=→=
⋅=→=
⋅=→=
⋅=→=
o
o
o
o
sh
sT
sh
sT
Thus,
( )
( ) ( ) ( )( )
( )
( )
( ) ( ) ( )( )
( ) kJ/kg 206.0
kJ/kg 38.76
kJ/kg 87.35
kJ/kg 40.83
=
+−=
+−−=
+−==
=⋅−−=
=
−−=−==
=
−−=
−+−−=
+−==
=⋅−−=
=
−−=−==
K 310kJ/kg 478.92
2.676021.73498K 290
ln
1/8lnKkJ/kg 0.2873.139162.67602K 290
ln
K 1600kJ/kg 584.62
2.472563.13916K 290
ln
8lnKkJ/kg 0.2871.734982.47256K 290
ln
out0
4
1410
41,41041gen,041 destroyed,
3
434034034gen,034 destroyed,
in0
2
3230
23,23023gen,023 destroyed,
1
212012012gen,012 destroyed,
LR
R
HR
R
Tq
PP
RssTT
qssTsTx
PP
RssTssTsTx
Tq
PP
RssTT
qssTsTx
PP
RssTssTsTx
oo
oo
oo
oo
9-126 The total exergy destruction associated with the Brayton cycle described in Prob. 9-93 and the exergy at the exhaust gases at the turbine exit are to be determined. Analysis From Prob. 9-93, qin = 601.94, qout = 279.68 kJ/kg, and h6 = 579.87 kJ/kg. The total exergy destruction associated with this Otto cycle is determined from
( ) kJ/kg 179.4=
−=
−=
K 1800kJ/kg 601.94
K 300kJ/kg 279.68
K 300inout0destroyed
HL Tq
Tq
Tx
Noting that h0 = h@ 300 K = 300.19 kJ/kg, the stream exergy at the exit of the regenerator (state 6) is determined from
( ) ( ) 06
026
060066 2gz
VssThh ++−−−=φ
where s s s s s s RPP6 0 6 1 6 1
6
1
0
− = − = − − = − = ⋅o o ln 2.36275 1.70203 0.66072 kJ / kg K
Thus, ( )( ) kJ/kg 81.5=⋅−−= KkJ/kg 0.66072K 300300.1979.8756φ
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-88
9-127 EES Problem 9-126 is reconsidered. The effect of the cycle pressure on the total irreversibility for the cycle and the exergy of the exhaust gas leaving the regenerator is to be investigated. Analysis Using EES, the problem is solved as follows: "Input data" T_o = 300 [K] T_L = 300 [K] T_H = 1400 [K] T[3] = 1200 [K] {Pratio = 10} T[1] = 300 [K] C_P=1.005 [kJ/kg-K] P[1]= 100 [kPa] P_o=P[1] Eta_reg = 1.0 Eta_c =1.0"Compressor isentorpic efficiency" Eta_t =1.0"Turbien isentropic efficiency" MM=MOLARMASS(Air) R=R_u/MM R_u=8.314 [kJ/kmol-K] C_V=C_P - R k=C_P/C_V "Isentropic Compressor anaysis" "For the ideal case the entropies are constant across the compressor" P[2] = Pratio*P[1] "T_s[2] is the isentropic value of T[2] at compressor exit" T_s[2]=T[1]*(Pratio)^((k-1)/k) Eta_c = w_compisen/w_comp "compressor adiabatic efficiency, W_comp > W_compisen" "Conservation of energy for the compressor for the isentropic case: e_in - e_out = DELTAe=0 for steady-flow" w_compisen = C_P*(T_s[2]-T[1]) "Actual compressor analysis:" w_comp = C_P*(T[2]-T[1]) "External heat exchanger analysis" "SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0 e_in - e_out =DELTAe_cv =0 for steady flow" q_in_noreg = C_P*(T[3]-T[2]) P[3]=P[2]"process 2-3 is SSSF constant pressure" "Turbine analysis" "For the ideal case the entropies are constant across the turbine" P[4] = P[3] /Pratio T_s[4]=T[3]*(1/Pratio)^((k-1)/k) "T_s[4] is the isentropic value of T[4] at turbine exit" Eta_t = w_turb /w_turbisen "turbine adiabatic efficiency, w_turbisen > w_turb" "SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0 e_in -e_out = DELTAe_cv = 0 for steady-flow" w_turbisen=C_P*(T[3] - T_s[4]) "Actual Turbine analysis:" w_turb= C_P*(T[3]-T[4]) "Cycle analysis" w_net=w_turb-w_comp "[kJ/kg]" Eta_th_noreg=w_net/q_in_noreg*Convert(, %) "[%]" "Cycle thermal efficiency" Bwr=w_comp/w_turb "Back work ratio"
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-89
"With the regenerator the heat added in the external heat exchanger is" q_in_withreg = C_P*(T[3]-T[5]) P[5]=P[2] "The regenerator effectiveness gives h[5] and thus T[5] as:" Eta_reg = (T[5]-T[2])/(T[4]-T[2]) "Energy balance on regenerator gives h[6] and thus T[6] as:" "h[2] + h[4]=h[5] + h[6]" T[2] + T[4]=T[5] + T[6] P[6]=P[4] "Cycle thermal efficiency with regenerator" Eta_th_withreg=w_net/q_in_withreg*Convert(, %) "[%]" "Irreversibility associated with the Brayton cycle is determined from:" q_out_withreg = q_in_withreg - w_net i_withreg = T_o*(q_out_withreg/T_L - q_in_withreg/T_H) q_out_noreg = q_in_noreg - w_net i_noreg = T_o*(q_out_noreg/T_L - q_in_noreg/T_H) "Neglecting the ke and pe of the exhaust gases, the exergy of the exhaust gases at the exit of the regenerator is:" "Psi_6 = (h[6] - h_o) - T_o(s[6] - s_o)" Psi_exit_withreg = C_P*(T[6] - T_o) - T_o*(C_P*ln(T[6]/T_o)-R*ln(P[6]/P_o)) Psi_exit_noreg = C_P*(T[4] - T_o) - T_o*(C_P*ln(T[4]/T_o)-R*ln(P[4]/P_o))
inoreg iwithreg Pratio Ψexit,noreg [kJ/kg]
Ψexit,withreg [kJ/kg]
ηth,noreg [%]
ηth,withreg [%]
270.8 97.94 6 157.8 47.16 40.05 58.3 244.5 113.9 7 139.9 56.53 42.63 56.42 223 128.8 8 125.5 65.46 44.78 54.73 205 142.7 9 113.6 74 46.61 53.18
189.6 155.9 10 103.6 82.17 48.19 51.75 176.3 168.4 11 95.05 90.02 49.58 50.41 164.6 180.2 12 87.62 97.58 50.82 49.17 154.2 191.5 13 81.11 104.9 51.93 47.99 144.9 202.3 14 75.35 111.9 52.94 46.88
4.5 5.0 5.5 6.0 6.5 7.0 7.5200
400
600
800
1000
1200
1400
1600
s [kJ/kg-K]
T [K
]
100 kPa
1000 kPa
Air
2s
1
2 5
3
44s
6
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-90
6 7 8 9 10 11 12 13 1475
115
155
195
235
275
Pratio
i [kJ
/kg]
No regeneration
With regeneration
6 7 8 9 10 11 12 13 1440
60
80
100
120
140
160
Pratio
Psi
exit
[kJ/
kg]
No regeneration
With regeneration
6 7 8 9 10 11 12 13 1440
44
48
52
56
60
Pratio
η th
[%
]
With regeneration
No regeneration
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-91
9-128 The exergy destruction associated with each of the processes of the Brayton cycle described in Prob. 9-98 and the exergy at the end of the exhaust gases at the exit of the regenerator are to be determined. Properties The gas constant of air is R = 0.287 kJ/kg·K (Table A-1). Analysis From Prob. 9-98, qin = 539.23 kJ/kg, qout = 345.17 kJ/kg, and
KkJ/kg 08336.2kJ/kg 738.56
KkJ/kg 68737.2kJ/kg 88.797
KkJ/kg 17888.3K 1200
KkJ/kg .373482kJ/kg 04.586
KkJ/kg 1.70203K 300
55
44
33
22
11
⋅=→=
⋅=→=
⋅=→=
⋅=→=
⋅=→=
o
o
o
o
o
sh
sh
sT
sh
sT
and, from an energy balance on the heat exchanger,
KkJ/kg 2.47108
kJ/kg 645.36.04586738.5688.797
6
66425
⋅=→
=+−=→−=−os
hhhhh
Thus,
( )
( ) ( ) ( )( )
( )
( ) ( ) ( )( )( ) ( )[ ] ( ) ( )[ ]
( )( )
( )
( ) kJ/kg 114.5
kJ/kg 42.78
kJ/kg 5.57
kJ/kg 31.59
kJ/kg 22.40
=
+−=
+−−=
+−==
=
−−=
−−−=
−−==
=−+−=
−+−=−+−==
=⋅−−=
−−=−==
=⋅−−=
−−=−==
K 300kJ/kg 345.17
2.471081.70203K 300
ln
K 1260kJ/kg 539.23
2.608333.17888K 300
ln
2.687372.471082.373482.60833K 300
1/8lnKkJ/kg0.2873.178882.68737K 300
ln
8lnKkJ/kg 0.2871.702032.37348K 300
ln
out0
6
1610
61,61061gen,061 destroyed,
0
5
3530
53,53053gen,053 destroyed,
4625046250regengen,0regen destroyed,
3
434034034gen,034 destroyed,
1
212012012gen,012 destroyed,
LR
R
H
in
R
R
Tq
PP
RssTT
qssTsTx
Tq
PP
RssTT
qssTsTx
ssssTssssTsTx
PP
RssTssTsTx
PP
RssTssTsTx
oo
oo
oooo
oo
oo
Noting that h0 = h@ 300 K = 300.19 kJ/kg, the stream exergy at the exit of the regenerator (state 6) is determined from
( ) ( ) 06
026
060066 2gzVssThh ++−−−=φ
where KkJ/kg 0.769051.702032.47108ln0
1
6161606 ⋅=−=−−=−=−
PP
Rssssss oo
Thus, ( )( ) kJ/kg 114.5=⋅−−= KkJ/kg 0.76905K 300300.1936.6456φ
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-92
9-129 A gas-turbine plant uses diesel fuel and operates on simple Brayton cycle. The isentropic efficiency of the compressor, the net power output, the back work ratio, the thermal efficiency, and the second-law 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.
700 kPa
Diesel fuel
100 kPa
Compres
4
3 2
Turbine
Combustion chamber
1
Properties The properties of air at 500ºC = 773 K are cp = 1.093 kJ/kg·K, cv = 0.806 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.357 (Table A-2b). Analysis (a) The isentropic efficiency of the compressor may be determined if we first calculate the exit temperature for the isentropic case
( ) K 6.505kPa 100kPa 700K 303
1)/1.357-(1.357/)1(
1
212 =
=
=
− kk
s PP
TT
0.881=−−
=−−
=K)303533(K)3036.505(
12
12
TTTT s
Cη
(b) The total mass flowing through the turbine and the rate of heat input are
kg/s 81.12kg/s 21.0kg/s 6.1260
kg/s 6.12kg/s 6.12AF
=+=+=+=+= aafat
mmmmm
&&&&&
kW 85557)kJ/kg)(0.9 00kg/s)(42,0 21.0(HVin === cf qmQ η&& The temperature at the exit of combustion chamber is K 1144)K533kJ/kg.K)( 3kg/s)(1.09 81.12(kJ/s 8555)( 3323in =→−=→−= TTTTcmQ p&&
The temperature at the turbine exit is determined using isentropic efficiency relation
( ) K 7.685kPa 700kPa 100K 1144
1)/1.357-(1.357/)1(
3
434 =
=
=
− kk
s PP
TT
K 4.754K)7.6851144(
K)1144(85.0 4
4
43
43 =→−−
=→−−
= TT
TTTT
sTη
The net power and the back work ratio are kW 3168)K30333kJ/kg.K)(5 3kg/s)(1.09 6.12()( 12inC, =−=−= TTcmW pa&
&
kW 5455)K4.754144kJ/kg.K)(1 3kg/s)(1.09 81.12()( 43outT, =−=−= TTcmW p&&
kW 2287=−=−= 31685455inC,outT,net WWW &&&
0.581===kW 5455kW 3168
outT,
inC,bw W
Wr &
&
(c) The thermal efficiency is 0.267===kW 8555kW 2287
in
netth Q
W&
&η
The second-law efficieny of the cycle is defined as the ratio of actual thermal efficiency to the maximum possible thermal efficiency (Carnot efficiency). The maximum temperature for the cycle can be taken to be the turbine inlet temperature. That is,
735.0K 1144K 30311
3
1max =−=−=
TT
η
and 0.364===735.0267.0
max
thII η
ηη
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-93
9-130 A modern compression ignition engine operates on the ideal dual cycle. The maximum temperature in the cycle, the net work output, the thermal efficiency, the mean effective pressure, the net power output, the second-law efficiency of the cycle, and the rate of exergy of the exhaust gases 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 850 K are cp = 1.110 kJ/kg·K, cv = 0.823 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.349 (Table A-2b). Analysis (a) The clearance volume and the total volume of the engine at the beginning of compression process (state 1) are
xcc
c
c
dcr VVVV
VV
VV===→
+=→
+= 2
33
m 0002154.0m 0028.014
43
1 m 003015.00028.00002154.0 VVVV ==+=+= dc
Process 1-2: Isentropic compression
( )( )
( )( ) kPa 334114kPa 95
K 9.82314K 328
1.349
2
112
1-1.3491
2
112
==
=
==
=
−
k
k
PP
TT
v
v
v
v
Process 2-x and x-3: Constant-volume and constant pressure heat addition processes:
K 2220kPa 3341kPa 9000K) 9.823(
22 ===
PP
TT xx
kJ/kg 1149K)9.8232220(kJ/kg.K) (0.823)( 2-2 =−=−= TTcq xx v
x
Qout
V
P
4
3
2Qin
1
K 3254=→−=→−==− 3333-2 K)2220(kJ/kg.K) (0.823kJ/kg 1149)( TTTTcqq xpxx
(b) kJ/kg 2298114911493-2in =+=+= − xx qqq
3333 m 0003158.0
K 2220K 3254)m 0002154.0( ===
xx T
TVV
Process 3-4: isentropic expansion.
( )
( ) kPa 9.428m 0.003015m 0.0003158kPa 9000
K 1481m 0.003015m 0.0003158K 3254
1.349
3
3
4
334
1-1.349
3
31
4
334
=
=
=
=
=
=
−
k
k
PP
TT
V
V
V
V
Process 4-1: constant voume heat rejection. ( ) ( )( ) kJ/kg 7.948K3281481KkJ/kg 0.82314out =−⋅=−= TTcq v
The net work output and the thermal efficiency are kJ/kg 1349=−=−= 7.9482298outinoutnet, qqw
0.587===kJ/kg 2298kJ/kg 1349
in
outnet,th q
wη
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-94
(c) The mean effective pressure is determined to be
( )( )kg 0.003043
K 328K/kgmkPa 0.287)m 015kPa)(0.003 95(
3
3
1
11 =⋅⋅
==RTP
mV
kPa 1466=
⋅
−=
−=
kJmkPa
m)0002154.0003015.0(kJ/kg) kg)(1349 (0.003043
MEP3
321
outnet,
VV
mw
(d) The power for engine speed of 3500 rpm is
kW 120=
==
s 60min 1
rev/cycle) 2((rev/min) 3500kJ/kg) kg)(1349 003043.0(
2netnetnmwW&&
Note that there are two revolutions in one cycle in four-stroke engines. (e) The second-law efficieny of the cycle is defined as the ratio of actual thermal efficiency to the maximum possible thermal efficiency (Carnot efficiency). We take the dead state temperature and pressure to be 25ºC and 100 kPa.
908.0K 3254
K )27325(113
0max =
+−=−=
TT
η
and 0.646===908.0587.0
max
thII η
ηη
The rate of exergy of the exhaust gases is determined as follows
( )
( )( ) kJ/kg 6.567100
9.428ln287.0298
1481kJ/kg.Kln 110.1()298(29814810.823
lnln)(0
4
0
4004040044
=
−−−=
−−−=−−−=
PP
RTT
cTTTcssTuux pv
kW 50.4=
==
s 60min 1
rev/cycle) 2((rev/min) 3500kJ/kg) kg)(567.6 003043.0(
244nmxX&&
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-95
9-131 A gas-turbine plant operates on the regenerative Brayton cycle. The isentropic efficiency of the compressor, the effectiveness of the regenerator, the air-fuel ratio in the combustion chamber, the net power output, the back work ratio, the thermal efficiency, the second law efficiency, the exergy efficiencies of the compressor, the turbine, and the regenerator, and the rate of the exergy of the combustion gases at the regenerator exit 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 500ºC = 773 K are cp = 1.093 kJ/kg·K, cv = 0.806 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.357 (Table A-2b). Analysis (a) For the compressor and the turbine:
( ) K 6.505kPa 100kPa 700K 303
1.3571-1.3571
1
212 =
=
=
−k
k
s PPTT
65
400°C 700 kPa 260°C
Regenerator
871°C
Compress.
432
Turbine
Combustion chamber
1100 kPa 30°C
0.881=−−
=−−
=K)303533(K)3036.505(
12
12
TTTT s
Cη
( ) K 6.685kPa 700kPa 100K 1144
1)/1.357-(1.357/)1(
3
434 =
=
=
− kk
s PP
TT
K 4.754K)6.6851144(
K)1144(85.0 4
4
43
43 =→−−
=→−−
= TT
TTTT
sTη
(b) The effectiveness of the regenerator is
0.632=−−
=−−
=K)5334.754(
K)533673(
24
25regen TT
TTε
(c) The fuel rate and air-fuel ratio are
kg/s 1613.0673)K144kJ/kg.K)(1 093.1)(6.12(7)kJ/kg)(0.9 000,42(
)()( 53HVin
=→−+=
−+==
fff
pafcf
mmm
TTcmmqmQ
&&&
&&&& η
78.14===1613.0
6.12AFf
a
mm&
&
Also, kg/s 76.121613.06.12 =+=+= fa mmm &&&
kW 65707)kJ/kg)(0.9 00kg/s)(42,0 1613.0(HVin === cf qmQ η&& (d) The net power and the back work ratio are
kW 3168)K30333kJ/kg.K)(5 3kg/s)(1.09 6.12()( 12inC, =−=−= TTcmW pa&&
kW 5434)K4.754144kJ/kg.K)(1 3kg/s)(1.09 76.12()( 43outT, =−=−= TTcmW p&&
kW 2267=−=−= 31685434inC,outT,net WWW &&&
0.583===kW 5434kW 3168
outT,
inC,bw W
Wr
&
&
(e) The thermal efficiency is
0.345===kW 6570kW 2267
in
net
QW
th &
&η
(f) The second-law efficieny of the cycle is defined as the ratio of actual thermal efficiency to the maximum possible thermal efficiency (Carnot efficiency). The maximum temperature for the cycle can be taken to be the turbine inlet temperature. That is,
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-96
735.0K 1144K 30311
3
1max =−=−=
TT
η
and 0.469===735.0345.0
max
thII η
ηη
(g) The exergy efficiency for the compressor is defined as the ratio of stream exergy difference between the inlet and exit of the compressor to the actual power input:
[ ] ( )
kW 2943100700ln287.0
303533ln)093.1()303()303533)(093.1()6.12(
lnln)(1
2
1
201212012C
=
−
−−=
−−−=−−−=∆
PP
RTT
cTTTcmssThhmX ppaa &&&
0.929==∆
=kW 3168kW 2943
inC,
CCII, W
X&
&η
The exergy efficiency for the turbine is defined as the ratio of actual turbine power to the stream exergy difference between the inlet and exit of the turbine:
( )
kW 5834100700ln287.0
754.41144ln)093.1()303()4.7541144)(093.1()76.12(
lnln4
3
4
3043T
=
−
−−=
−−−=∆
PP
RTT
cTTTcmX pp&&
0.932==∆
=kW 5834kW 5434
T
inT,, X
WTII &
&η
An energy balance on the regenerator gives
K 2.616)4.754)(093.1)(76.12()53373)(1.093)(66.12(
)()(
66
6425
=→−=−
−=−
TT
TTcmTTcm ppa &&
The exergy efficiency for the regenerator is defined as the ratio of the exergy increase of the cold fluid to the exergy decrease of the hot fluid:
( )
kW 10730616.2754.4ln)093.1()303()2.6164.754)(093.1()76.12(
0ln6
4064hotregen,
=
−
−−=
−−−=∆
TT
cTTTcmX pp&&
( )
kW 8.9540533673ln)093.1()303()533673)(093.1()76.12(
0ln2
5025coldregen,
=
−
−−=
−−−=∆
TT
cTTTcmX pp&&
0.890==∆
∆=
kW 1073kW 8.954
hotregen,
coldregen,, X
XTII &
&η
The exergy of the combustion gases at the regenerator exit:
( )
kW 1351=
−
−−=
−−−=
0303
616.2ln)093.1()303()3032.616)(093.1()76.12(
0ln0
60066 T
TcTTTcmX pp&&
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.