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9-1
Solutions Manual for
Thermodynamics: An Engineering Approach Seventh Edition
Yunus A. Cengel, Michael A. Boles McGraw-Hill, 2011
Chapter 9 GAS POWER CYCLES
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9-2
Actual and Ideal Cycles, Carnot cycle, Air-Standard Assumptions, Reciprocating Engines
9-1C It is less than the thermal efficiency of a Carnot cycle.
9-2C It represents the net work on both diagrams.
9-3C 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.
9-4C 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.
9-5C 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 bottom dead center.
9-6C It is the ratio of the maximum to minimum volumes in the cylinder.
9-7C 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.
9-8C Yes.
9-9C 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.
9-10C 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.
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9-3
9-11C 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.
9-12E The maximum possible thermal efficiency of a gas power cycle with specified reservoirs is to be determined.
Analysis The maximum efficiency this cycle can have is
0.643=+
+−=−= 1 LT
R )460(940R )460(401Carnotth,
HTη
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9-4
9-13 An air-standard cycle executed in a piston-cylinder system is composed of three specified processes. The cycle is to be sketcehed on the P-v and T-s diagrams and the back work ratio 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 are given as R = 0.287 kPa·m3/kg·K, cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, and k = 1.4.
Analysis (a) The P-v and T-s diagrams of the cycle are shown in the figures.
(b) Process 1-2: Isentropic compression
)( 12,21 TTmcw in −=− v
s
T
3 2
1
v
P
32
1
11
212 =⎟⎟
⎠⎜⎜⎝
= rTTTv
1
1 −−
⎞⎛ kk
v
rocess 2-3: Constant pressure heat addition
TTmRP −=−= VVv
he back w rk ratio is
P
2
,32 Pdw out = ∫− )()( 23232
3
T o
)()(
23
12
,32
,21
TTmRTTmc
ww
rout
inbw −
−==
−
− v
Noting that
1
thus,and and −
==−=k
Rccc
kccR pp vv
v
From ideal gas relation,
rTT
===2
1
2
3
2
3
vv
vv
Substituting these into back work relation,
( )
( )0.256=
−−
−=
−−
−=
−
⎟⎠⎞
⎜⎝⎛ −
−=
−−
−=
−
−−
1661
14.11
11
11
1
11
11
)1/()/1(1
1
4.0
11
23
21
2
2
rr
krr
k
TTTT
TT
RkRr
kk
bw
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9-5
9-14 The three processes of an air-standard cycle are described. The cycle is to be shown on the P-v and T-s diagrams, and 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 constant specific heats.
Properties The properties of air are given as R = 0.287 kJ/kg.K, cp = 1.005 kJ/kg.K, cv = 0.718 kJ/kg·K, and
k = 1.4.
Analysis (a) The P-v and T-s diagrams of the cycle are shown in the figures.
(b) The temperature at state 2 is
K 2100kPa 100kPa 700K) 300(
1
212 ===
PP
TT
K 210023 == TT
During process 1-3, we have
3
,13
=−⋅−=
−−=−−=−= ∫− TTRPPdw in VVv
During process 2-3, we have
kJ/kg 516.600)K21K)(300kJ/kg 287.0(
)()( 31311
1
s
T
32
1
v
P
3
2
1
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kJ/kg 8.1172n7K)(2100)KlkJ/kg 287.0(22
=⋅=
7ln7
lnln2
2
2
33
,3 ===== ∫∫− RTRTRTdRTPdw out V
V
V
Vv
Vv
The back work ratio is then
3
2
0.440===−
−
kJ/kg 8.1172kJ/kg 6.516
,32
,13
out
inbw w
wr
Heat input is determined from an energy balance on the cycle during process 1-3,
) ,321
,3231,31
31
=+−⋅
+−
+∆=−
∆=
−
−−−
−
out
outin
ut
wTwuq
u
The net work output is
,32,31 − −− oin wq
718.0(( 3
=
= v Tc
kJ/kg 2465kJ/kg 1172.8300)K)(2100kJ/kg
kJ/kg 2.6566.5168.1172,13,32 =−=−= −− inoutnet www
(c) The thermal efficiency is then
26.6%==== 266.0kJ 2465kJ 656.2
in
netth q
wη
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9-6
9-15 The three processes of an ideal gas power cycle are described. The cycle is to be shown on the P-v and T-s diagrams, and the maximum temperature, expansion and compression works, and thermal efficiency are to be determined.
Assumptions 1 Kinetic and potential energy changes are negligible. 2 The ideal gas has constant specific heats.
Properties The properties of ideal gas are given as R = 0.3 kJ/kg.K, cp = 0.9 kJ/kg.K, cv = 0.6 kJ/kg·K, and
k = 1.5.
Analysis (a) The P-v and T-s diagrams of the cycle are shown in the figures.
(b) The maximum temperature is determined from
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
K 734.8=+==⎟⎟⎠
⎜⎜⎝
== 11
2
112max 27(rTTTT
V
⎞ −−−
11.51
K)(6) 273kk
V
) An energy balance during process 2-3 gives
voutin TTTTcuwq 232332,32,32 since 0)(−−− ==−=∆=−
Then, the work of compression is
s
T
32
1
v
P
3
2
1
⎛
(c
ouin wq ,32,32 −− = t
kJ/kg 395.0=⋅=
=== ∫K)ln6 K)(734.8kJ/kg 3.0(
lnln 22
32
22,32,32 rRTRTdRTPdoutin V
Vv
Vv
(d) The work during isentropic compression is determined from an energy 1-2:
−=∆= −
300))( 1221,2 TTcuw vin
) Net work output is
== ∫−−
33
wq
balance during process
−1
kJ/kg 260.9=
⋅= K)(734.8kJ/kg 6.0( −
(e
kJ/kg 1.1349.2600.395,21,32 =−=−= −− inoutnet www
The thermal efficiency is then
33.9%==== 339.0kJ 395.0kJ 134.1
in
netth q
wη
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9-7
9-16 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 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
v
P
1
2
4
3qin
qout
s
T
1
24
3qin
qout
( )
( )
( ) kJ/kg 71.73979.32601.9kPa 1835.6
kPa 100
kPa 6.1835kPa 600K 490.3K 1500
9.601kJ/kg 41.1205
K 1500
K 3.490kJ/kg 352.29
841.71.3068kPa 100kPa 600
3068.1kJ/kg .17295
K 2951 ⎯→⎯=T
43
4
22
323
33
2
2
1
2
1
34
3
12
1
=⎯→⎯===
=
==
⎯→⎯=
==
⎯→⎯===
==
hPPP
P
TT
Pu
T
Tu
PPP
P
Ph
rr
r
rr
r
From energy balances,
(c) Then the thermal efficiency becomes
32233 ==⎯→⎯= PT
PPP vv
T
kJ/kg 408.6=−=−=
=−=−=
=−=−=
5.4441.853
kJ/kg .544417.29571.739
kJ/kg 1.85329.35241.1205
outinoutnet,
14out
23in
qqw
hhq
uuq
47.9%==== 479.0kJ/kg 853.1kJ/kg 408.6
in
outnet,th q
wη
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9-8
9-17 Problem 9-16 is reconsidered. The effect of the maximum temperature of the cycle on the net work output and thermal efficiency is to be investigated. Also, T-s and P-v diagrams for the cycle are to be plotted.
Analysis Using EES, the problem is solved as follows:
"Input Data" T[1]=295 [K] P[1]=100 [kPa] P[2] = 600 [kPa] T[3]=1500 [K] P[4] = 100 [kPa] "Process 1-2 is isentropic compression" s[1]=entropy(air,T=T[1],P=P[1]) s[2]=s[1] T[2]=temperature(air, s=s[2], P=P[2]) P[2]*v[2]/T[2]=P[1]*v[1]/T[1] P[1]*v[1]=R*T[1] R=0.287 [kJ/kg-K] "Conservation of energy for process 1 to 2" q_12 -w_12 = DELTAu_12 q_12 =0"isentropic process" DELTAu_12=intenergy(air,T=T[2])-intenergy(air,T=T[1]) "Process 2-3 is constant volume heat addition" s[3]=entropy(air, T=T[3], P=P[3]) {P[3]*v[3]/T[3]=P[2]*v[2]/T[2]} P[3]*v[3]=R*T[3] v[3]=v[2] "Conservation of energy for process 2 to 3" q_23 -w_23 = DELTAu_23 w_23 =0"constant volume process" DELTAu_23=intenergy(air,T=T[3])-intenergy(air,T=T[2]) "Process 3-4 is isentropic expansion" s[4]=entropy(air,T=T[4],P=P[4]) s[4]=s[3] P[4]*v[4]/T[4]=P[3]*v[3]/T[3] {P[4]*v[4]=0.287*T[4]} "Conservation of energy for process 3 to 4" q_34 -w_34 = DELTAu_34 q_34 =0"isentropic process" DELTAu_34=intenergy(air,T=T[4])-intenergy(air,T=T[3]) "Process 4-1 is constant pressure heat rejection" {P[4]*v[4]/T[4]=P[1]*v[1]/T[1]} "Conservation of energy for process 4 to 1" q_41 -w_41 = DELTAu_41 w_41 =P[1]*(v[1]-v[4]) "constant pressure process" DELTAu_41=intenergy(air,T=T[1])-intenergy(air,T=T[4]) q_in_total=q_23 w_net = w_12+w_23+w_34+w_41 Eta_th=w_net/q_in_total*100 "Thermal efficiency, in percent"
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9-9
T3 [K]
ηth qin,total [kJ/kg]
Wnet [kJ/kg]
1500 1600 1700 1800 1900 2000 2100 2200 2300 2400 2500
47.91 48.31 48.68 49.03 49.35 49.66 49.95 50.22 50.48 50.72 50.95
852.9 945.7 1040 1134 1229 1325 1422 1519 1617 1715 1813
408.6 456.9 506.1 556
606.7 658.1 710.5 763
816.1 869.8 924
1500 1700 1900 2100 2300 250047.5
48
48.5
49
49.5
50
50.5
51
T[3] [K]
ηth
1500 1700 1900 2100 2300 2500800
1020
1240
1460
1680
1900
T[3
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] [K]
q in,
tota
l [k
J/k
g]
preparation. If you are a student using this Manual, you are using it without permission.
9-10
1500 1700 1900 2100 2300 2500400
500
600
700
800
900
1000
T[3] [K]
wne
t [k
J/kg
]
5.0 5.5 6.0 6.5 7.0 7.5 8.0 8.50
500
1000
1500
2000
s [kJ/kg-K]
T [K
]
100 kPa
600 kPa
Air
1
2
3
4
10-2 10-1 100 101 102101
102
103
104
v [m3/kg]
P [k
Pa]
295 K 1500 K
Air
1
2
3
4
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9-11
9-18 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·K, and k = 1.4 (Table A-2).
Analysis (b) The temperature at state 2 and the heat input are
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( )( ) K 579.2
kPa 100kPa 1000
K 3000.4/1.4/1
1
212 =⎟⎟
⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
− kk
PP
TT
( ) ( )( )( )( )2.579KkJ/kg 1.005kg 0.004kJ 2.76 3 ⎯→⎯−⋅= T
P
K 12663
2323in
=
−=−=
T
TTmchhmQ p
Process 3-1 is a straight line on the -v diagram, thus the w31 is simply the area under the process curve,
( )
( )
v
P
32
1
qin
qout
s
T
32
1 qout
qin
kJ/kg 7.273=
KkJ/kg 0.287kPa 1000K 1266
kPa 100K 300
2kPa 1001000
22area
3
3
1
11331
1331
⋅⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟
⎠
⎞⎜⎝
⎛ +=
⎟⎟⎠
⎞⎜⎜⎝
⎛−
+=−
+==
PRT
PRTPPPP
w vv
Energy balance for process 3-1 gives
−−−=
−=−−→
K1266-300KkJ/kg 0.718273.7kg 0.004)(
)(
31out31,31out31,out31,
31out31,out31,systemoutin
TTcwmTTmcmwQ
uumWQ
vv
) The thermal efficiency is then
( )[ ]( ) ( )( )[ ] kJ 1.679=⋅+−=
−+−
⎯⎯∆=− EEE
=
(c
39.2%=−=−=kJ 2.76kJ 1.679
11in
outth Q
Qη
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9-12
9-19E 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-17E.
v
P
3
4
2
1
q12
q23
qout
s
T3
42
1 qout
q23
q12
Analysis (b) The properties of air at various states are
Btu/lbm 129.06 Btu/lbm, 92.04R 540 111 ==⎯→⎯= huT
( )
( ) Btu/lbm 593.22317.01242psia 57.6psia 14.7
1242Btu/lbm 48.849
R 3200
psia 57.6psia 14.7R 540R 2116
Btu/lbm 1.537,R2116Btu/lbm 04.39230004.92
43
4
33
11
22
1
11
2
22
22
in,121212in,12
34
3
=⎯→⎯===
==
⎯→⎯=
===⎯→⎯=
==
=+=+=⎯→⎯−=
hPPP
P
Ph
T
PTT
PT
PT
P
hTquu
uuq
rr
r
vv
From energy balance,
Btu/lbm 464.1606.12922.593 38.31300in23,in12,in
23in23,
+=+= qqq 2Btu/lbm 312.381.53748.849
14out =−=−=
=
=−=−=
hhq
hhqBtu/lbm612.38
(c) Then the thermal efficiency becomes
24.2%=−=−=Btu/lbm612.38Btu/lbm464.16
11in
outth q
qη
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9-13
9-20E 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).
Analysis (b) The temperature at state 2 and the heat input are
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
)( )( )
(
( )
( ) ( )( ) Btu/lbm 217.4R22943200RBtu/lbm 0.24
psia 62.46psia 14.7R 540R 2294
R 2294R540Btu/lbm.R .171
2323in,23
11
22
1
11
2
22
2
2
1212in,12
=−⋅=−=−=
===⎯→⎯=
=−
−=−=
TTchhq
PTT
PT
PT
PT
TTTcuuq
P
vv
v
Process 3-4 is isentropic:
0Btu/lbm 300 =
( )( )
( ) ( )( ) Btu/lbm 378.55402117Btu/lbm.R 0.240
4.217300
R 2117psia 62.46
psia 14.7R 3200
1414out
in,23in,12in
0.4/1.4
3
434
=−=−=−=
=+=+=
=⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
TTchhq
qqq
PP
T
p
Btu/lbm517.4
(c) The thermal efficiency is then
/1− kk
T
26.8%=−=−=Btu/lbm 517.4Btu/lbm 378.5
11in
outth q
qη
v
P
3
4
2
1
q12
q23
qout
s
T
3
42
1 qout
q23
q12
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9-14
9-21 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 P3 and the maximum pressure is P1. Then,
or
( )
( )( ) kPa 1888
K 300K 1100
kPa 201.4/0.41/
3
232
/1
3
2
3
2
=⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
⎟⎟⎠
⎞⎜⎜⎝
⎛=
−
−
kk
kk
TT
PP
PP
TT
s
T
3
2qin
qout4
11100
300 The heat input is determined from
Then,
( )
( ) ( )( )( )
( )( ) kJ 63.8===
==−=−=
=⋅=−=
⋅=⋅−=−=−
kJ 87.730.7273
.7%727273.0K 1100K 30011
kJ 73.87KkJ/kg 0.1329K 1100kg 0.6
KkJ/kg 0.1329kPa 3000kPa 1888lnKkJ/kg 0.287lnln
inthoutnet,
th
12in
1
20
1
212
QW
TT
ssmTQ
PP
RTT
css
H
L
H
p
η
η
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
9-15
9-22 A Carnot cycle executed in a closed system with air as the working fluid is considered. The minimum pressure in the cycle, the heat rejection from the cycle, the thermal efficiency of the cycle, and the second-law efficiency of an actual cycle operating between the same temperature limits are to be determined.
Assumptions Air is an ideal gas with constant specific heats.
Properties The properties of air at room temperatures are R = 0.287 kJ/kg.K and k = 1.4 (Table A-2).
Analysis (a) The minimum temperature is determined from
( )( ) ( )( ) K 350K750KkJ/kg 0.25kJ/kg 10012net =⎯→⎯−⋅=⎯→⎯−−= LLLH TTTTssw
The pressure at state 4 is determined from ( )
( )
kPa 1.110
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
or
K 350 ⎠⎝
K 750kPa 800 4
1.4/0.4
4
1/
4
141
/1
=⎯→⎯⎟⎟⎞
⎜⎜⎛
=
⎟⎟⎠
⎞⎜⎜⎝
⎛=
⎞⎛
−
−
PP
TT
PP
PT
kk
kk
he minim m pressure in the cycle is determined from s
T
3
2qin
4
1750 K
wnet=100 kJ/kg
qout
4
1
4
1⎟⎟⎠
⎜⎜⎝
=PT
T u
( ) kPa 46.1=⎯→⎯⋅−=⋅−
−=∆−=∆0
4lnT
css
33
3
43412
kPa 110.1lnKkJ/kg 0.287KkJ/kg 25.0
ln
PP
PP
RTp
) The heat rejection from the cycle is
3
(b
kgkJ/ 87.5==∆= kJ/kg.K) K)(0.25 350(12out sTq L
(c) The thermal efficiency is determined from
0.533=−=−=K 750K 350
11thH
L
TT
η
(d) The power output for the Carnot cycle is
Then, the second-law efficiency of the actual cycle becomes
kW 9000kJ/kg) kg/s)(100 90(netCarnot === wmW &&
0.578===kW 9000kW 5200
Carnot
actualII W
W&
&η
preparation. If you are a student using this Manual, you are using it without permission.
9-16
9-23 An ideal gas Carnot cycle with air as the working fluid is considered. The maximum temperature of the low-temperature energy reservoir, the cycle's thermal efficiency, and the amount of heat that must be supplied per 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.K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-2a).
Analysis The temperature of the low-temperature reservoir can be found by applying the isentropic expansion process relation
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
K 481.1=⎟⎠
⎜⎝
+=⎟⎟⎠
⎜⎜⎝
=1
21 12K) 2731027(TT
v ⎞⎛⎞⎛ −− 14.11
2 1k
v
ince the Carnot engine is completely reversible, its efficiency is
s
T
3
2qin
qout4
1 1300 K S
0.630=+
−=−= 11Carnotth,H
L
Tη
K 273)(1027K 481.1T
he work tput per cycle is
T ou
kJ/cycle 20min 1
s 60cycle/min 1500
kJ/s 500netnet =⎟
⎠⎞
⎜⎝⎛==
nW&
W&
According to the definition of the cycle efficiency,
kJ/cycle 31.75===⎯→⎯=0.63kJ/cycle 20
Carnotth,
netin
in
netCarnotth, η
ηW
QQW
preparation. If you are a student using this Manual, you are using it without permission.
9-17
9-24 An air-standard cycle executed in a piston-cylinder system is composed of three specified processes. The cycle is to be sketcehed on the P-v and T-s diagrams; the heat and work interactions and the thermal efficiency of the cycle are to be determined; and an expression for thermal efficiency as functions of compression ratio and specific heat ratio is to be obtained.
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 are given as R = 0.3 kJ/kg·K and cv = 0.3 kJ/kg·K.
Analysis (a) The P-v and T-s diagrams of the cycle are shown in the figures.
(b) Noting that
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
1.4290.1
===7.0
K kJ/kg 0.13.07.0 ⋅=+=+=
v
v
c
Rcc
Process 1-2: Isentropic compression
s
T
32
1
v
P
32
1
p
ck p
K 4.584)5)(K 293( 429.011
1
2
112 ===⎟⎟
⎠
⎞⎜⎜⎝
⎛= −
−k
k
rTTTv
v
=− ( 2in2,1 Tcw v kJ/kg 204.0=−⋅=− K )2934.584)(KkJ/kg 7.0()1T
2q 0=−1
From ideal gas relation,
2922)5)(4.584(32
1
2
3
2
3 ==⎯→⎯=== TrTT
v
v
v
v
Process 2-3: Constant pressure heat addition
K )4.5842922)(KkJ/kg 3.0(
)23
3
2
TT
K )4.5842922)(KkJ/kg 1()( 23
3232,32in3,
TTchuwq
p
out
Process 3-1: Constant volume heat rejection
=−== ∫− ()( 232out3,2 RPPdw vvv
kJ/kg 701.3=−⋅=
−
kJ/kg 2338=−⋅=−=
∆=∆+= −−−−2
=∆= −− (31out1,3 Tcuq v kJ/kg 1840.3=⋅=− K 293)-K)(2922kJ/kg 7.0()13 T
w
(c) Net work is
0=−13
KkJ/kg 3.4970.2043.701in2,1out3,2net ⋅=−=−= −− www
The thermal efficiency is then
21.3%==== 213.0kJ 2338kJ 497.3
in
netth q
wη
preparation. If you are a student using this Manual, you are using it without permission.
9-18
(d) The expression for the cycle thermal efficiency is obtained as follows:
⎟⎠
⎞⎜⎝
⎛ −−
−⎟⎠⎞
⎜⎝⎛ −=
⎟⎠
⎞⎜⎝
⎛ −−
−=
⎟⎟⎠
⎞⎜⎜⎝
⎛−
−−=
−
⎟⎟⎠
⎞⎜⎜⎝
⎛−
−=
−
−−=
−−−−
=
−==
−
−
−
−
−−
−−
−
−−
1
1
11
1
11
11
111
11
11
11
1
23
1223
in
in2,1out3,2
in
netth
11)1(
111
11)1(
1
1)1(
1
)1(
1
)()(
)()()(
k
kp
kp
kp
kk
v
p
kkp
kv
p
p
v
rrkk
rrkcR
rTT
rkcR
rrTc
rTT
rTc
cR
rTrrTcTrTc
cR
TTcTTcTTR
qww
qw
η
since
111kc
cc
cccR
p
v
p
vp
p−=−=
−=
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
9-19
Otto Cycle
9-25C 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.
9-26C The ideal Otto cycle involves external irreversibilities, and thus it has a lower thermal efficiency.
9-27C 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.
9-28C They are analyzed as closed system processes because no mass crosses the system boundaries during any of the processes.
9-29C It increases with both of them.
9-30C Because high compression ratios cause engine knock.
9-31C The thermal efficiency will be the highest for argon because it has the highest specific heat ratio, k = 1.667.
9-32C The fuel is injected into the cylinder in both engines, but it is ignited with a spark plug in gasoline engines.
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
9-20
9-33 An ideal Otto cycle is considered. The thermal efficiency and the rate of heat input 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-2a).
Analysis The definition of cycle thermal efficiency reduces to
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
61.0%==−=−=−−
6096.010.5
1111 11.41th krη
The rate of heat addition is then
kW 148===0.6096
kW 90
th
netin η
WQ
&&
2 Kinetic and potential energy changes are negligible. 3 Air is
ies The properties of air at room temperature are c = 1.005 kJ/kg.K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-
nalysis The definition of cycle thermal efficiency reduces to
v
P
4
1
3
2
qin
qout
9-34 An ideal Otto cycle is considered. The thermal efficiency and the rate of heat input are to be determined.
Assumptions 1 The air-standard assumptions are applicable. an ideal gas with constant specific heats.
Propert p2a).
A
v
P
4
1
3
2
qin
qout
57.5%==−=−=− 8.5
11 1.1th krη
−5752.011
14
he rate o eat addition is then T f h
kW 157===0.5752
kW 90
th
netin η
WQ
&&
preparation. If you are a student using this Manual, you are using it without permission.
9-21
9-35 The two isentropic processes in an Otto cycle are replaced with polytropic processes. The heat added to and rejected from this cycle, and the cycle’s 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 R = 0.287 kPa·m3/kg·K, cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-2a).
Analysis The temperature at the end of the compression is
v
P
4
1
3
2
K 4.537K)(8) 288( 13.111
1
2
112 ===⎟⎟
⎠
⎞⎜⎜⎝
⎛= −−
−n
n
rTTTv
v
And the temperature at the end of the expansion is
K 4.7898
K) 1473(34
34 =⎟⎠
⎜⎝
=⎟⎠
⎜⎝
=⎟⎟⎠
⎜⎜⎝
=r
TTTv
11 13.1113 ⎞⎛⎞⎛⎞⎛ −−− nn
v
for the polytropic compression gives The integral of the work expression
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
kJ/kg 6.238)18(13.1
11 2
21 −⎥⎦⎢⎣⎟⎠
⎜⎝−− n
wv
K) K)(288kJ/kg 287.0( 13.11
11 =−⋅
=⎥⎤
⎢⎡
−⎟⎞
⎜⎛
= −−n
RT v
imilarly, the work produced during the expansion is S
kJ/kg 0.654181
13.1K) K)(1473kJ/kg 287.0(
11
13.11 ⎤⎡ −n
4
3343 =
⎥⎥⎦
⎤
⎢⎢⎣
⎡−⎟
⎠⎞
⎜⎝⎛
−⋅
−=⎥⎥
⎦⎢⎢
⎣−⎟⎟
⎠
⎞⎜⎜⎝
⎛−
−=−
− nRT
wv
v
pplication of the first law to each of the four processes gives
A
1 kJ/kg 53.59K)2884.537)(KkJ/kg 718.0(kJ/kg 6.238)( 12212 =−⋅−=−−= − TTcwq v −
1473)(KkJ/kg 718.0()( 2332 ⋅=−=− TTcq v kJ/kg 8.671K)4.537 =−
kJ/kg 2.163K)4.7891473)(KkJ/kg 718.0(kJ/kg 0.654)( 434343 =−⋅−=−−= −− TTcwq v
kJ/kg 0.360K)2884.789)(KkJ/kg 718.0()( 1414 =−⋅=−=− TTcq v
The head added and rejected from the cycle are
The thermal efficiency of this cycle is then
kJ/kg 419.5kJ/kg 835.0
=+=+=
=+=+=
−−
−−
0.36053.592.1638.671
1421out
4332in
qqqqqq
0.498=−=−=0.8355.41911
in
outth q
qη
preparation. If you are a student using this Manual, you are using it without permission.
9-22
9-36 An ideal Otto cycle is considered. The heat added to and rejected from this cycle, and the cycle’s 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 R = 0.287 kPa·m3/kg·K, cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-2a).
Analysis The temperature at the end of the compression is
v
P
4
1
3
2
K 7.661K)(8) 288( 14.111
1
2
112 ===⎟⎟
⎠
⎞⎜⎜⎝
⎛= −−
−k
k
rTTTv
v
and the temperature at the end of the expansion is
K 2.648
K) 1473(34
34 =⎟⎠
⎜⎝
=⎟⎠
⎜⎝
=⎟⎟⎠
⎜⎜⎝
=r
TTTv
111 14.1113 ⎞⎛⎞⎛⎞⎛ −−− kk
v
pplication of the first law to the heat addition process gives A
=−= ()( 23in TTcq v
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
kJ/kg 582.5=−⋅ K)7.6611473)(KkJ/kg 718.0
imilarly, the heat rejected is S
⋅=−= KkJ/kg 718.0()( TTcq kJ/kg 253.6=− K)2882.641)(14out v
he thermal efficiency of this cycle is then T
0.565=−=−=5.5826.25311
in
outth q
qη
preparation. If you are a student using this Manual, you are using it without permission.
9-23
9-37E A six-cylinder, four-stroke, spark-ignition engine operating on the ideal Otto cycle is considered. The power produced by the engine 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 R = 0.3704 psia·ft3/lbm.R (Table A-1E), cp = 0.240 Btu/lbm·R, cv = 0.171 Btu/lbm·R, and k = 1.4 (Table A-2Ea).
Analysis From the data specified in the problem statement,
v
P
4
1
3
2
143.7
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
140 12 vv .11 ==
vvr
Since the compression and expansion processes are isentropic,
=
( ) R 1153143.7R) 525( 14.111
1
2
112 ===⎟⎟
⎠
⎞⎜⎜⎝
⎛= −−
−k
k
rTTTvv
R 2.9387.143
11 14.1113 ⎞⎛⎞⎛⎞⎛ −−− kk
vR) 2060(3
434 =⎟
⎠⎜⎝
=⎟⎠
⎜⎝
=⎟⎟⎠
⎜⎜⎝
=r
TTTv
mpression and expansion processes gives
=−⋅
Application of the first law to the co
171.0(R)2.9382060)(RBtu/lbm 171.0()()( 1243net
−−⋅=−−−= TTcTTcw vv
Btu/lbm 44.84R)5251153)(RBtu/lbm
When each cylinder is charged with the air-fuel mixture,
/lbmft 89.13psia 14
R) 525)(R/lbmftpsia 3704.0( 33
1
11 =
⋅⋅==
PRT
v
The total air mass taken by all 6 cylinders when they are charged is
lbm 009380.0/lbmft 89.13
ft)/4 12/9.3(ft) 12/5.3()6(4/cylcyl ==
∆=
πvv
V SBNNm 3
2
1
2
1=
π
he net work produced per cycle is
T
Btu/cycle 7920.0Btu/lbm) lbm)(84.44 (0.009380netnet === mwW
he power produced is determined from
T
hp 23.3=⎟⎠
⎞⎜⎝
⎛==Btu/s 7068.0
hp 1rev/cycle 2
rev/s) (2500/60Btu/cycle) (0.7920
rev
netnet N
nWW
&&
since there are two revolutions per cycle in a four-stroke engine.
preparation. If you are a student using this Manual, you are using it without permission.
9-24
9-38E An Otto cycle with non-isentropic compression and expansion processes is considered. The thermal efficiency, the heat addition, 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 R = 0.3704 psia·ft3/lbm.R (Table A-1E), cp = 0.240 Btu/lbm·R, cv = 0.171 Btu/lbm·R, and k = 1.4 (Table A-2Ea).
Analysis We begin by determining the temperatures of the cycle states using the process equations and component efficiencies. The ideal temperature at the end of the compression is then
v
P
4
1
3
2 qout
qin
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
( ) R 11958R) 520(12
12 ===⎟⎟⎠
⎜⎜⎝
=s rTTTv
14.111
1 ⎞⎛ −−−
kk
v
With the isentropic compression efficiency, the actual temperature at e end of the compression is th
R 13140.85
R 520)(1195R) 520(1212
12
12 ⎯→⎯−
=η TTT
s =−
+=−
+=−
ηTT
TTT s
imilarly for the expansion,
S
R 120181R) 4602300(1 14.11
3
1
4 ⎠⎝3
34 =⎟⎠⎞
⎜⎝⎛+=⎟
⎠⎞
⎜⎝⎛=⎟⎟
⎞⎜⎜⎛
=−−− kk
s rTTT
v
v
R 1279R )1201(2760)95.0(R) 2760()( 433443 sTT43 =−−=−−=⎯→⎯
−−
sTTTTTT
ηη
The specific heat addition is that of process 2-3,
=
Btu/lbm 247.3=−⋅=−= R)13142760)(RBtu/lbm 171.0()( 23in TTcq v
The net work production is the difference between the work produced by the expansion and that used by the compression,
−⋅−−⋅=
The thermal efficiency of this cycle is then
)()( 1243net −−−= TTcTTcw vv
R)5201314)(RBtu/lbm 171.0(R)12792760)(RBtu/lbm 171.0(Btu/lbm 5.117=
0.475===Btu/lbm 247.3inqBtu/lbm 117.5net
thw
η
t the beg ning of compression, the maximum specific volume of this cycle is A in
/lbmft 82.14psia 13
R) 520)(R/lbmftpsia 3704.0( 31
1⋅⋅
==RT
v 3
1=
P
hile the minimum specific volume of the cycle occurs at the end of the compression
w
/lbmft 852.18
/lbmft 82.14 33
12 ===
rv
v
The engine’s mean effective pressure is then
psia 49.0=⎟⎟⎠
⎞⎜⎜⎝
⎛ ⋅
−=
−=
Btu 1ftpsia 404.5
/lbmft )852.182.14(Btu/lbm 5.117MEP
3
321
net
vv
w
preparation. If you are a student using this Manual, you are using it without permission.
9-25
9-39 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
P
4
1
3
2Qin Qout
( )( )
( ) ( ) kPa 2338kPa 100K
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
308K 757.9
9.5
K 757.99.5K
11
2
2
12
1
11
2
22
0.41
=⎟⎟⎠
⎞⎜⎜⎝
⎛==⎯→⎯=
=⎞⎛
−
PTT
PT
PT
P
k
v
vvv
v
308
2
112 =⎟⎟
⎠⎜⎜⎝
= TTv
Process 3-4: isentropic expansion.
( )( ) K 1969==⎟⎟⎠
⎞⎜⎜⎝
⎛=
−0.4
1
3
443 9.5K 800
k
TTv
v
Process 2-3: v = constant heat addition.
( ) kPa 6072=⎟⎟⎠
⎞⎜⎜⎝
⎛==⎯→⎯= kPa 2338
K 757.9K 1969
22
323 TTT
3223 PT
PPP vv
(b)
3
( )( )( )( )
kg10788.6K 308K/kgmkPa 0.287
m 0.0006kPa 100 43
3
1
11 −×=⋅⋅
==RTP
mV
( ) ( ) ( )( )( ) kJ 0.590=−⋅×=−=−= − K757.91969KkJ/kg 0.718kg106.788 42323in TTmcuumQ v
(c) Process 4-1: v = constant heat rejection.
( ) ( )( )( ) kJ 0.240K308800KkJ/kg 0.718kg106.788)( 41414out =−⋅×−=−=−= −TTmcuumQ v
kJ 0.350240.0590.0outinnet =−=−= QQW
59.4%===kJ 0.590kJ 0.350
in
outnet,th Q
Wη
( )( )kPa 652=⎟
⎟⎠
⎞⎜⎜⎝
⎛ ⋅
−=
−=
−=
==
kJmkPa
1/9.51m 0.0006kJ 0.350
)/11(MEP
3
31
outnet,
21
outnet,
max2min
rWW
r
VVV
VVV (d)
preparation. If you are a student using this Manual, you are using it without permission.
9-26
9-40 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.
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
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.
( )( )
( ) ( ) kPa 2338kPa 100K 308K 757.9
9.5
K 757.99.5K 308
11
2
2
12
1
11
2
22
0.41
2
112
=⎟⎟⎠
⎞⎜⎜⎝
⎛==⎯→⎯=
==⎟⎟⎠
⎞⎜⎜⎝
⎛=
−
PTT
PT
PT
P
TTk
v
vvv
v
v
v
P
4
1
3
2
Qin
Qout
Polytropic
800 K
308 K
Process 3-4: polytropic expansion.
( )( )( )( )
kg10788.6K 308K/kgmkPa 0.287
m 0.0006kPa 100 43
3
1
11 −×=⋅⋅
==RTP
mV
( )( )
( ) ( )( )( )kJ 0.5338
1.351K1759800KkJ/kg 0.287
106.788
134
34×
=−
−=
nTTmR
W
9.5K 800
4
35.01
3
443
=−
−⋅
==⎟⎟⎠
⎞⎜⎜⎝
⎛=
−
−
TTn
K 1759v
v
=+−⋅×=
+−=+−=−Q
WTTmcWuumQ v
That is, 0.066 kJ of heat is added to the air during the expansion process (This is not realistic, and probably is due to nstant specific heats at room temperature).
v = constant heat addition.
T gyhen ener balance for process 3-4 gives
outin ∆=− EEE system
( )34out34,in34, −=− uumWQ( ) ( )
( )( )( ) kJ 0.0664kJ 0.5338K1759800KkJ/kg 0.718kg106.788 4in34,
out34,34out34,34in34,
assuming co(b) Process 2-3:
( ) kPa 5426=⎟⎟⎠
⎞⎜⎜⎝
⎛TP vv==⎯→⎯= kPa 2338
K 757.9K 1759
22
33
2
22
3
3 PT
PT
P
=−⋅×=
−=−=−Q
TTmcuumQ v
efore,
3
T
( ) ( )( )( )( ) kJ 0.4879K757.91759KkJ/kg 0.718kg106.788 4
in23,
2323in23,
kJ 0.5543=+=+= 0664.04879.0in34,in23,in QQQ Ther
(c) Process 4-1: v = constant heat rejection.
( ) ( ) ( )( )( ) kJ 0.2398K308800KkJ/kg 0.718kg 106.788 41414out =−⋅×=−=−= −TTmcuumQ v
kJ 0.31452398.05543.0outinoutnet, =−=−= QQW
56.7%===kJ 0.5543kJ 0.3145
in
outnet,th Q
Wη
( )( )kPa 586=⎟
⎟⎠
⎞⎜⎜⎝
⎛ ⋅
−=
−=
−=
==
kJmkPa
1/9.51m 0.0006kJ 0.3145
)/11(MEP
3
31
outnet,
21
outnet,
max2min
rWW
r
VVV
VVV (d)
preparation. If you are a student using this Manual, you are using it without permission.
9-27
9-41E 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.
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
v
P
4
1
3
2
qinqout
2400 R
540 R
Properties The properties of air are given in Table A-17E.
Analysis (a) Process 1-2: isentropic compression.
32.144Btu/lbm92.04
R5401
11 =
=⎯→⎯=
r
uT
v
( ) Btu/lbm 11.28204.1832.14481
222===
r rrr vvv
v11
22 =⎯→⎯= u
v
rocess 2- v = constant heat addition.
⎯→⎯=
28.21170.452
419.2Btu/lbm 0
R2400
23
33
uuq
T
in
rv
P 3:
= 452.73u
Btu/lbm 241.42=−=−=
=
(b) Process 3-4: isentropic expansion.
( )( ) Btu/lbm 205.5435.19419.28 43
4334
=⎯→⎯==== ur rrr vvv
vv
Process 4-1: v = constant heat rejection.
Btu/lbm 50.11304.9254.20514out =−=−= uuq
53.0%=−=−=Btu/lbm 241.42Btu/lbm 113.50
11in
outth q
qη
(c) The thermal efficiency of a Carnot cycle operating between the same temperature limits is
77.5%=−=−=R 2400
R 54011Cth,
H
L
TT
η
preparation. If you are a student using this Manual, you are using it without permission.
9-28
9-42E 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 (Table A-2E).
Analysis (a) Process 1-2: isentropic compression.
( )( ) R 21618R 540 0.6671
2
112 ==⎟⎟
⎠
⎞⎜⎜⎝
⎛=
−k
TTv
v
v
P
4
1
3
2
qinqout
Process 2-3: v = constant heat addition.
( )
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
( )(= 2400Btu/lbm.R 0.0756 )Btu/lbm 18.07=
−−=−=
R 21612323 TTcuuq v
) Process 3-4: isentropic expansion.
in
(b
( ) R 6008
0.667
4=
⎠⎝⎠⎝v
Process 4-1: v = constant heat rejection.
1R 24001
334 ⎟
⎞⎜⎛=⎟⎟
⎞⎜⎜⎛
=−k
TTv
( ) ( )( ) Btu/lbm 4.536R540600Btu/lbm.R 0.07561414out =−=−=−= TTcuuq v
74.9%=−=−=Btu/lbm 4.536
11 outqη
Btu/lbm 18.07inth q
(c) The thermal efficiency of a Carnot cycle operating between the same temperature limits is
77.5%=−=−=R 2400R 540
11Cth,H
L
TT
η
preparation. If you are a student using this Manual, you are using it without permission.
9-29
9-43 A gasoline engine operates on an Otto cycle. The compression and expansion processes are modeled as polytropic. The temperature at the end of expansion process, the net work output, the thermal efficiency, the mean effective pressure, the engine speed for a given net power, and the specific fuel consumption 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) Process 1-2: polytropic compression
1
Qin
2
3
4
P
V
Qout
( )( )
( )( ) kPa 225811kPa 100
K 5.63611K 310
1.3
2
112
1-1.31
2
112
==⎟⎟⎠
⎞⎜⎜⎝
⎛=
==⎟⎟⎠
⎞⎜⎜⎝
⎛=
−
n
n
PP
TT
v
v
v
v
kJ/kg 3.3121.31112 −− n
310)KK)(636.5kJ/kg (0.287)( 12 −=−⋅
=−
=TTR
w
Process 2-3: constant volume heat addition
( ) K 2255kPa 2258kPa 8000K 636.5
2
323 =⎟
⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
PP
TT
( 2323in −=−= TTcuuq v )( )( ) kJ/kg 1332K636.52255KkJ/kg 0.823 =−⋅=
rocess 3-4: polytropic expansion.
P
( ) K 1098=⎟⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
− 1-1.31
4
334 11
1K 2255n
TTv
v
( ) kPa 2.3541 1.3⎞⎛⎞⎛
nv
11
kPa 80001
234 =⎟
⎠⎜⎝
=⎟⎟⎠
⎜⎜⎝
= PPv
kJ/kg 11061.31
2255)KK)(1098kJ/kg (0.287)( −⋅− TTR1
34 =−
=−
=n
w
Process 4-1: constant volume heat rejection.
(b) The net work output and the thermal efficiency are
34
kJ/kg 794=−=−= 3.31211061234outnet, www
59.6%==== 596.0kJ/kg 1332kJ/kg 794
in
outnet,th q
wη
(c) The mean effective pressure is determined as follows
( )( )
( )( )kPa 982=⎟
⎟⎠
⎞⎜⎜⎝
⎛ ⋅
−=
−=
−=
==
==⋅⋅
==
kJmkPa
1/111/kgm 0.8897kJ/kg 794
)/11(MEP
/kgm 0.8897kPa 100
K 310K/kgmkPa 0.287
3
31
outnet,
21
outnet,
max2min
max3
3
1
11
rww
r
PRT
vvv
vvv
vv
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. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
9-30
(d) The clearance volume and the total volume of the engine at the beginning of compression process (state 1) are
33
m 00016.0m 0016.0
11 =⎯→⎯+
=⎯→⎯+
= cc
c
c
dcr VV
V
V
VV
31 m 00176.00016.000016.0 =+=+= dc VVV
The total mass contained in the cylinder is
( )( )kg 0.001978
K 310K/kgmkPa 0.287)m 76kPa)/0.001 100(
3
3
1
11 =⋅⋅
==RTP
mtV
The engine speed for a net power output of 50 kW is
rev/min 3820=⎟⎠⎞
⎜⎝⋅
==min 1cycle)kJ/kg kg)(794 001978.0(
rev/cycle) 2(2net
net
wmn
t& ⎛ s 60kJ/s 50W&
n four-stroke engines.
) The mass of fuel burned during one cycle is
Note that there are two revolutions in one cycle i
(e
kg 0001164.0kg) 001978.0(
=⎯→⎯−
16AF =⎯→⎯==ff
a
mm
−f
f
fft mm
mmmm
inally, the specific fuel consumption is
F
g/kWh 267=⎟⎠⎞
⎜⎝⎛
⎟⎟⎠
⎞⎜⎜⎝
⎛==
kWh 1kJ 3600
kg 1g 1000
kJ/kg) kg)(794 001978.0(kg 0001164.0sfc
netwmm
t
f
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
9-31
9-44 The expressions for the maximum gas temperature and pressure of an ideal Otto cycle are to be determined when the compression ratio is doubled.
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.
Analysis The temperature at the end of the compression varies with the compression ratio as
11
1
2
112
−−
=⎟⎟⎠
⎞⎜⎜⎝
⎛= k
k
rTTTvv
v
P
4
1
3
2 qout
qinsince T1 is fixed. The temperature rise during the combustion remains constant since the amount of heat addition is fixed. Then, the maximum cycle temperature is given by
11in2in3 // −+=+= krTcqTcqT vv
The smallest gas specific volume during the cycle is
r1
3v
v =
When this is combined with the maximum temperature, the maximum pressure is given by
( )11in
13
33 / −+== krTcqRrRT
P vvv
9-45 It is to be determined if the polytropic exponent to be used in an Otto cycle model will be greater than or less than the isentropic exponent.
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.
nalysis During a polytropic process,
=T
=Pv
is lost during the expansion of the gas,
>
here T4s is the temperature that would occur if the expansion were reversible and adiabatic (n=k). This can only occur when
A
constant/)1(
=− nn
nPv P
4
1
3
2 qout
qin
P constant
and for an isentropic process,
k constant
constant/)1( =− kkTP
If heat v
4 sTT 4
w
kn ≤
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
9-32
Diesel Cycle
9-46C 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.
9-47C 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.
9-48C The gasoline engine.
9-49C Diesel engines operate at high compression ratios because the diesel engines do not have the engine knock problem.
9-50C 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.
9-51 An ideal diesel cycle has a compression ratio of 20 and a cutoff ratio of 1.3. The maximum temperature of the air and the rate of heat addition 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-2a).
Analysis We begin by using the process types to fix the temperatures of the states.
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
( ) K 6.95420K) 288( 14.111
1
2
112 ===⎟⎟
⎠
⎞⎜⎜⎝
⎛= −−
−k
k
rTTTv
v
K 1241===⎟⎠
⎜⎝
= K)(1.3) 6.954(22
23 crTTTv
⎟⎞
⎜⎛ 3v
ombining the first law as applied to the various processes with the process equations gives v
P
4
1
2 3qin
qout
C
6812.0)13.1(4.120
1)1(
111.41th −
−=−
−=−−
ck rkr
η 13.1111 4.1=
−−kcr
ccording the definition of the thermal efficiency,
A to
kW 367===0.6812
kW 250
th
netin η
WQ
&&
preparation. If you are a student using this Manual, you are using it without permission.
9-33
9-52E An ideal diesel cycle has a a cutoff ratio of 1.4. The power produced 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 R = 0.3704 psia·ft3/lbm.R (Table A-1E), cp = 0.240 Btu/lbm·R, cv = 0.171 Btu/lbm·R, and k = 1.4 (Table A-2Ea).
Analysis The specific volume of the air at the start of the compression is
v
P
4
1
2 3qin
qout
/lbmft 12.13psia 4.14
R) 510)(R/lbmftpsia 3704.0( 33
1
11 =
⋅⋅==
PRT
v
The total air mass taken by all 8 cylinders when they are charged is
lbm 01774.0/lbmft 12.13
)8(3
1cyl
1cyl ===
vvNNm ft)/4 12/4(ft) 12/4(4/ 22
=∆ ππV SB
he rate at which air is processed by the engine is determined from T
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
lbm/h 0.958lbm/s 2661.0rev/cycle 2rev
====N
m& rev/s) (1800/60lbm/cycle) (0.01774nm &
nce there re two revolutions per cycle in a four-stroke engine. The compression ratio is
si a
22.22045.01
==r
=== −−krTT
work integral to the constant pressure heat addition gives
At the end of the compression, the air temperature is
( ) R 176322.22R) 510( 14.111 2
Application of the first law and
Btu/lbm 239.3R)17632760)(RBtu/lbm 240.0()( 23in =−⋅=−= TTcq p
while the thermal efficiency is
6892.0)14.1(4.1
14.122.22
111 −kr 1
)1(1
4.1
11.41=
−−
−=−
−=−−
c
ck rkr
η
he power produced by this engine is then
th
T
hp 62.1=
⎟⎠
⎞⎜⎝
⎛=
==
Btu/h 5.2544hp 1
Btu/lbm) 892)(239.3lbm/h)(0.6 (958.0
inthnetnet qmwmW η&&&
preparation. If you are a student using this Manual, you are using it without permission.
9-34
9-53 An ideal dual cycle has a compression ratio of 14 and cutoff ratio of 1.2. The thermal efficiency, amount of heat added, and the maximum gas pressure and temperature 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-2a).
Analysis The specific volume of the air at the start of the compression is
v
P
4
1
2
3
qout
x
qin/kgm 051.1
kPa 80K) 293)(K/kgmkPa 287.0( 3
3
1
11 =
⋅⋅==
PRT
v
and the specific volume at the end of the compression is
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
/kgm 07508.0142 ===
/kgm 051.1 33
1vv
r
The pressure at the end of the compression is
( ) kPa 321914kPa) 80( 4.11
212 ⎟
⎠⎜⎝v
1 ===⎟⎞
⎜⎛
= kk
rPPPv
nd the ma imum pressure is
he temperature at the end of the compression is
a x
4829==== kPa) 3219)(5.1(23 PrPP px kPa
T
( ) K 0.84214K) 293( 14.111
1
2
11 2 ===⎟⎟
⎠
⎞⎜⎜⎝
⎛= −−
−k
k
rTTTv
v
nd K 1263kPa 3219kPa 4829K) 0.842(
2
32 =⎟
⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎛ P⎜⎜⎝
=P
TTx
he remaining state temperatures are then
a
From the definition of cutoff ratio
/kgm 09010.0)/kgm 07508.0)(2.1( 3323 ==== vvv cxc rr
T
K 1516=⎟⎠
⎜⎝
=⎟⎟⎠
⎜⎜⎝
=0.07508
K) 1263(33
xxTT
v ⎞⎛⎞⎛ 0.09010v
K 5.5671.051
0.09010K) 1516(14.11
4
334 ⎜⎜⎛
= TTv
=⎟⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞
⎝
−−k
v
ng the first law and work expression to the heat addition processes gives
The heat rejected is
Applyi
kJ/kg 556.5=−⋅+−⋅=
−+−=
K)12631516)(KkJ/kg 005.1(K)0.8421263)(KkJ/kg 718.0(
)()( 32in xpx TTcTTcq v
kJ/kg 1.197K)2935.567)(KkJ/kg 718.0()( 14out =−⋅=−= TTcq v
0.646=−=−=kJ/kg 556.5kJ/kg 197.1
11in
outth q
qη Then,
preparation. If you are a student using this Manual, you are using it without permission.
9-35
9-54 An ideal dual cycle has a compression ratio of 14 and cutoff ratio of 1.2. The thermal efficiency, amount of heat added, and the maximum gas pressure and temperature 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 The specific volume of the air at the start of the compression is
/kgm 9076.0kPa 80
K) 253)(K/kgmkPa 287.0( 33
1
11 =
⋅⋅==
PRT
v
v
P
4
1
2
3
qout
x
qinand the specific volume at the end of the compression is
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
/kgm 06483.0142 ===
/kgm 9076.0 33
1vv
r
The pressure at the end of the compression is
( ) kPa 321914kPa) 80( 4.11
212 ⎟
⎠⎜⎝v
1 ===⎟⎞
⎜⎛
= kk
rPPPv
nd the ma imum pressure is
he temperature at the end of the compression is
a x
4829==== kPa) 3219)(5.1(23 PrPP px kPa
T
( ) K 1.72714K) 253( 14.111
1
2
11 2 ===⎟⎟
⎠
⎞⎜⎜⎝
⎛= −−
−k
k
rTTTv
v
nd K 1091kPa 3219kPa 4829K) 1.727(
2
32 =⎟
⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎛ P⎜⎜⎝
=P
TTx
he remaining state temperatures are then
a
From the definition of cutoff ratio
/kgm 07780.0)/kgm 06483.0)(2.1( 3323 ==== vvv cxc rr
T
K 1309=⎟⎠
⎜⎝
=⎟⎟⎠
⎜⎜⎝
=0.06483
K) 1091(33
xxTT
v ⎞⎛⎞⎛ 0.07780v
K 0.4900.9076
0.07780K) 1309(14.11
4
334 ⎜⎜⎛
= TTv
=⎟⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞
⎝
−−k
v
ng the first law and work expression to the heat addition processes gives
The heat rejected is
Applyi
kJ/kg 480.4=−⋅+−⋅=
−+−=
K)10911309)(KkJ/kg 005.1(K)1.7271091)(KkJ/kg 718.0(
)()( 32in xpx TTcTTcq v
kJ/kg 2.170K)2530.490)(KkJ/kg 718.0()( 14out =−⋅=−= TTcq v
0.646=−=−=kJ/kg 480.4kJ/kg 170.2
11in
outth q
qη Then,
preparation. If you are a student using this Manual, you are using it without permission.
9-36
9-55E 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-17E.
v
P
4
1
2 3qin
qout
3000 RAnalysis (a) Process 1-2: isentropic compression.
32.144Btu/lbm 40.92
R 5401
11 =
=⎯→⎯=
r
uT
v
( )Btu/lbm 402.05R 1623.6
93.732.1442.181
112 r rrr v11
2
22
==
⎯→⎯====hT
vvv
v
rocess 2-3: P = constant heat addition. P
1.848===⎯→⎯=R 1623.6
R 3000
2
3
2
3
2
22
3
33
TT
TP
T
Pv
vvv
(b) R 3000
23in
33
=−=−=
=⎯→⎯=
hhq
Trv
Btu/lbm 388.6305.40268.790
180.1Btu/lbm 790.683 =h
Process 3-4: isentropic expansion.
( ) Btu/lbm 91.250621.11180.1848.1
2.18848.1848.1 4
2
4
3
43334
=⎯→⎯===== urrrrr vv
v
vv
v
vv
Process 4-1: v = constant heat rejection.
(c) 59.1%
Btu/lbm 158.87
=−=−=
=−=−=
Btu/lbm 388.63Btu/lbm 158.87
11
04.9291.250
in
outth
14out
uuq
η
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
9-37
9-56E 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).
Analysis (a) Process 1-2: isentropic compression.
v
P
4
1
2 3qin
( )( ) R172418.2R540 0.41
2
112 ==⎟⎟
⎠
⎞⎜⎜⎝
⎛=
−k
TTv
v
Process 2-3: P = constant heat addition.
1.741===⎯→⎯=R 1724R 3000
2
3
2
3
2
22
3
33
TT
TP
TP
v
vvv
(b) ( ) ( )( ) Btu/lbm 306R17243000Btu/lbm.R 0.2402323in =−=
rocess 3- isentropic expansion.
−=−= TTchhq p
P 4:
( ) R 117318.21.741R 3000
741.11
3 ⎞⎛⎞⎛−k
vv 0.41
4
23
434 =⎟
⎠⎞
⎜⎝⎛=⎟⎟
⎠⎜⎜⎝
=⎟⎟⎠
⎜⎜⎝
=−k
vTTT
v
Process 4-1: v = constant heat rejection.
(c)
( )( )( )
64.6%
Btu/lbm 108
=−=−=
=−=
−=−=
Btu/lbm 306Btu/lbm 108
11
R0541173Btu/lbm.R 0.171
in
outth
1414out
TTcuuq
η
v
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
9-38
9-57 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·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
P
4
1
2 3 qin
qout
( )( ) K 971.120K 2932
112 =⎟⎟
⎠⎜⎜⎝
= TTV
0.41
=⎞⎛
−kV
rocess 2-3: P = constant heat addition.
P
2.265K971.1K2200
2
3
2
3
2
22
3
33 =⎯→⎯=TT V
==TTPP VVV
rocess 3- isentropic expansion.
P 4:
( )
( ) ( )( )( ) ( )( )
63.5%===
=−=−=
=−⋅=−=−=
=−⋅=−=−=
=⎟⎠⎞
⎜⎝⎛=⎟
⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
−−−
kJ/kg 1235kJ/kg 784.4
kJ/kg 784.46.4501235
kJ/kg 450.6K293920.6KkJ/kg 0.718
kJ/kg 1235K971.12200KkJ/kg 1.005
K 920.620
2.265K 2200265.2265.2
in
outnet,th
outinoutnet,
1414out
2323in
0.41
3
1
4
23
1
4
334
qw
qqw
TTcuuq
TTchhq
rTTTT
p
kkk
η
v
V
V
V
V
(b) ( )( )
( ) ( )( )kPa933
kJmkPa
1/201/kgm 0.885kJ/kg 784.4
/11MEP
/kgm 0.885kPa 95
K 293K/kgmkPa 0.287
3
31
outnet,
21
outnet,
max2min
max3
3
1
11
=⎟⎟⎠
⎞⎜⎜⎝
⎛ ⋅
−=
−=
−=
==
==⋅⋅
==
rww
r
PRT
vvv
vvv
vv
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
9-39
9-58 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·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
P
4
1
2 3 qin
qout
Polytropic
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
( )( )20K 293 0.4
2
112 =⎟⎟
⎠⎜⎜⎝
= TTV
K 971.11
=⎞⎛
−kV
rocess 2-3: P = constant heat addition. P
2.265K 971.1K 2200
2
3
2
32233 =⎯→⎯=T
PT
PV
VVV
23==
TT
Process 3-4: polytropic expansion.
( )
( ) ( )( )( ) ( )( ) kJ/kg 526.3K 2931026KkJ/kg 0.718
kJ/kg 1235K 971.12200KkJ/kg 1.005
K 102620
2.265K 2200r
1414out
2323in
0.35
34
34
34
=−⋅=−=−=
=−⋅=−=−=
=⎟⎠⎞
⎜⎝⎛=⎟
⎠⎜⎝
=⎟⎟⎠
⎜⎝
=⎟⎠
⎜⎝
=
TTcuuq
TTchhq
TTTT
p
v
VV
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:
2.2652.265 1n1n2
13 ⎞⎛⎞
⎜⎛
⎟⎞
⎜⎛ −−−n
VV
( ) ( )( )
( )( )(
kJ/kg )
1K 22001026KkJ/kg 0.718kJ/kg 963
K 22001026KkJ/kg 0.287
34out34,in34,34
−⋅+=
−+=⎯→⎯−
−⋅−
TTcwqu
TTR
v
hich means that 120.1 kJ/kg of heat is transferred to the comlistic since the gas is at a much higher temperature than the surroundings, and a hot gas loses heat during polytropic
xpansion The cause of this unrealistic result is the constant specific heat assumption. If we were to use u data from the ir table, w would obtain
kJ/kg 9631.3511
out34,in34,
systemoutin
34out34,
=−
∆=−
=−
=−
=
uwq
EEEn
w
120.=
w bustion gases during the expansion process. This is unreae . a e
( ) kJ/kg 1.128)4.18723.781(96334out34,in34, −=−+=−+= uuwq
which is a heat loss as expected. Then qout becomes kJ/kg 654.43.5261.128out41,out34,out =+=+= qqq
and
47.0%===
=−=−=
kJ/kg 1235kJ/kg 580.6
kJ/kg 580.64.6541235
in
outnet,th
outinoutnet,
qw
qqw
η
( )( )
( ) ( )( )kPa 691=⎟
⎟⎠
⎞⎜⎜⎝
⎛ ⋅
−=
−=
−=
==
==⋅⋅
==
kJmkPa 1
1/201/kgm 0.885kJ/kg 580.6
/11MEP
/kgm 0.885kPa 95
K 293K/kgmkPa 0.287
3
31
outnet,
21
outnet,
max2min
max3
3
1
11
rww
r
PRT
vvv
vvv
vv (b)
preparation. If you are a student using this Manual, you are using it without permission.
9-40
9-59 Problem 9-58 is reconsidered. The effect of the compression ratio on the net work output, mean effective lso, T-s and P-v diagrams for the cycle are to be plotted.
nalysis Using EES, the problem is solved as follows:
tal(q_12,q_23,q_34,q_41: q_in_total,q_out_total)
> 0 then q_in_total = q_in_total + q_41 else q_out_total = q_out_total - q_41
ssion"
P=P[2]) v[1]/T[1]
rocess 1 to 2"
=T[1]) heat addition"
nergy(air,T=T[2]) ion"
pic process" )-intenergy(air,T=T[3])
4 to 1"
pressure, and thermal efficiency is to be investigated. A
A
Procedure QToq_in_total = 0 q_out_total = 0 IF (q_12 > 0) THEN q_in_total = q_12 ELSE q_out_total = - q_12 If q_23 > 0 then q_in_total = q_in_total + q_23 else q_out_total = q_out_total - q_23 If q_34 > 0 then q_in_total = q_in_total + q_34 else q_out_total = q_out_total - q_34 If q_41END "Input Data" T[1]=293 [K] P[1]=95 [kPa] T[3] = 2200 [K]
=1.35 n{r_comp = 20}
1-2 is isentropic compre"Process s[1]=entropy(air,T=T[1],P=P[1]) s[2]=s[1]
(air, s=s[2],T[2]=temperatureP[2]*v[2]/T[2]=P[1]*P[1]*v[1]=R*T[1] R=0.287 [kJ/kg-K] V[2] = V[1]/ r_comp "Conservation of energy for pq_12 - w_12 = DELTAu_12 q_12 =0"isentropic process" DELTAu_12=intenergy(air,T=T[2])-intenergy(air,T"Process 2-3 is constant pressureP[3]=P[2] s[3]=entropy(air, T=T[3], P=P[3]) P[3]*v[3]=R*T[3] "Conservation of energy for process 2 to 3" q_23 - w_23 = DELTAu_23
sure process" w_23 =P[2]*(V[3] - V[2])"constant presDELTAu_23=intenergy(air,T=T[3])-inte"Process 3-4 is polytropic expans
[3])^n P[3]/P[4] =(V[4]/Vs[4]=entropy(air,T=T[4],P=P[4]) P[4]*v[4]=R*T[4] "Conservation of energy for process 3 to 4" q_34 - w_34 = DELTAu_34 "q_34 is not 0 for the ploytroDELTAu_34=intenergy(air,T=T[4]P[3]*V[3]^n = Const w_34=(P[4]*V[4]-P[3]*V[3])/(1-n) "Process 4-1 is constant volume heat rejection"V[4] = V[1] "Conservation of energy for process q_41 - w_41 = DELTAu_41
_41 =0 "constant volume process" wDELTAu_41=intenergy(air,T=T[1])-intenergy(air,T=T[4]) 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
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
9-41
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
"Thermal efficiency, in percent" he mean effective pressure is:" EP = w_net/(V[1]-V[2])
r
Eta_th=w_net/q_in_total*100 "TM
comp ηth MEP [kPa]
wnet [kJ/kg]
14 47.69 970.8 797.9 16 50.14 985 817.4 18 52.16 992.6 829.8 20 53.85 995.4 837.0 22 55.29 994.9 840.6 24 56.54 992 841.5
4.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5200400600800
10001200140016001800200022002400
s [kJ/kg-K]
T [K
]
95
kPa
340
.1 kP
a
592
0 kP
a
0.0
44
0.1
0.8
8 m
3/kg
Air
2
1
3
4
10-2 10-1 100 101 102101
102
103
104
101
102
103
104
v [m3/kg]
P [k
Pa] 293 K
1049 K
2200 K
5.69 6.74 kJ/kg-K
Air
preparation. If you are a student using this Manual, you are using it without permission.
9-42
14 16 18 20 22 24790
800
810
820
830
840
850
rcomp
wne
t [k
J/kg
]
14 16 18 20 22 2447
49
51
53
55
57
rcomp
ηth
14 16 18 20 22 24970
975
980
985
990
995
1000
rcomp
MEP
[kP
a]
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
9-43
9-60 A four-cylinder ideal diesel engine with air as the working fluid has a compression ratio of 22 and a cutoff ratio of 1.8. The power the engine will deliver at 2300 rpm is to be determined.
Assumptions 1 The cold 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 Process 1-2: isentropic compression.
v
P
4
1
2 3 Qin
Qout
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
( )( )22K 3432
112 =⎟⎟
⎠⎜⎜⎝
= TTV
K 11810.41
=⎞⎛
−kV
Process 2-3: P = constant heat addition.
( )( ) K 2126K 11811.88.1 222
33
2
22
3
33 =⎯→⎯= TTT
=== TTPP
v
vvv
rocess 3- isentropic expansion.
P 4:
( )
( ) ( )
( ) ( )( )( )( )
( )( ) kW 48.0===
=−=−=
=−⋅=−=−=
=−⋅=−=−=
=⋅⋅
==
=⎟⎠⎞
⎜⎝⎛=⎟
⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
−−− 4.0111
Tkkk
kJ/rev 1.251rev/s 2300/60
kJ/rev 251.16198.0871.1
kJ 6198.0K343781KkJ/kg 0.718kg 001971.0
kJ .8711K)11812216)(KkJ/kg 1.005)(kg 001971.0(
kg 001971.0)K 343)(K/kgmkPa 0.287(
)m 0.0020)(kPa 97(
K 78122
8.1K 22162.22.2
outnet,outnet,
outinoutnet,
1414out
2323in
3
3
1
11
34
23
4
334
WnW
QQW
TTmcuumQ
TTmchhmQ
RTP
m
rTTT
v
p
&&
V
VV
V
V
Discussion Note that for 2-stroke engines, 1 thermodynamic cycle is equivalent to 1 mechanical cycle (and thus revolutions).
preparation. If you are a student using this Manual, you are using it without permission.
9-44
9-61 A four-cylinder ideal diesel engine with nitrogen as the working fluid has a compression ratio of 22 and a cutoff ratio of 1.8. The power the engine will deliver at 2300 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·K, cv = 0.743 kJ/kg·K, R = 0.2968 kJ/kg·K, and k = 1.4 (Table A-2).
Analysis Process 1-2: isentropic compression.
v
P
4
1
2 3 Qin
Qout
( )( ) K 118122K 343 0.41
2
112 ==⎟⎟
⎠
⎞⎜⎜⎝
⎛=
−k
TTV
V
Process 2-3: P = constant heat addition.
( )( ) K 2126K 11811.88.1 222
33
23 TT223 ====⎯→⎯= TTT
PPv
vvv
Process 3-4: isentropic expansion.
3
( )
( ) ( )
( ) ( )( )( )( )
( )( ) kW 47.9=== kJ/rev 1.251rev/s 2300/60outnet,outnet, WnW &&
Discussion No
=−=−=
=−⋅=−=−=
=−⋅=−=−=
=⋅⋅
==
=⎟⎠⎞
⎜⎝⎛=⎟
⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
−−−
kJ/rev 251.16202.0871.1
kJ 6202.0K343781KkJ/kg 0.743kg 001906.0
kJ .8711K)11812216)(KkJ/kg 1.039)(kg 001906.0(
kg 001906.0)K 343)(K/kgmkPa 0.2968(
)m 0.0020)(kPa 97(
K 78122
8.1K 22162.22.2
outinoutnet,
1414out
2323in
3
3
1
11
4.01
3
1
4
23
1
4
334
QQW
TTmcuumQ
TTmchhmQ
RTP
m
rTTTT
v
p
kkk
V
VV
V
V
te that for 2-stroke engines, 1 thermodynamic cycle is equivalent to 1 mechanical cycle (and thus revolutions).
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
9-45
9-62 An ideal dual cycle has a compression ratio of 18 and cutoff ratio of 1.1. The power produced by the cycle 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·K, cv = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4 (Table A-2a).
Analysis We begin by fixing the temperatures at all states.
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
( ) K 7.92418K) 291( 14.111
1
2
112 ===⎟⎟
⎠
⎞⎜⎜⎝
⎛= −−
−k
k
rTTTv
v
v
P
4
1
2
3
qout
x
qin
( ) kPa 514818kPa) 90( 4.11
2
11 2 ===⎟⎟
⎠
⎞⎜⎜⎝
⎛= k
k
rPPPv
v
kPa 5663kPa) 5148)(1.1(23 ==== PrPP px
K 1017kPa 5148kPa 5663K) 7.924(
22 =⎟
⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
PP
TT x x
K 1119K) 1017)(1.1(3 === xcTrT
K 8.365181.K) 1119(
1
33
34 ⎜⎛=⎟⎟
⎞⎜⎜⎛
=⎟⎟⎞
⎜⎜⎛
=k
crTTTv 1 14.11
4=⎟
⎠⎞
⎝⎠⎝⎠⎝
−−−k
rv
pplying the first law to each of the processes gives
A
1 kJ/kg 0.455K)2917.924)(KkJ/kg 718.0()( 122 =−⋅=−= TTcw v −
kJ/kg 5.102K)10171119)(KkJ/kg 005.1()( 33 =−⋅=−=− xpx TTcq
( 333 −= −− xx Tcqw v kJ/kg 26.29K)10171119)(KkJ/kg 718.0(5.102) =−⋅−=− xT
3 kJ/kg 8.540K)8.3651119)(KkJ/kg 718.0()( 434 =−⋅=−=− TTcw v
1.1150.45526.298.54021343net
The net work of the cycle is
k J/kg=−+=−+= −−− wwww x
The mass in the device is given by
kg 003233.0K) 291)(K/kgmkPa 287.0(
)m kPa)(0.003 90( 3PV3
1
11 =⋅⋅
=RT
m
he net power produced by this engine is then
=
T
kW 24.8=== cycle/s) 0/60kJ/kg)(400 115.1kg/cycle)( 003233.0(netnet nmwW &&
preparation. If you are a student using this Manual, you are using it without permission.
9-46
9-63 A dual cycle with non-isentropic compression and expansion processes is considered. The power produced by the cycle 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·K, cv = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4 (Table A-2a).
Analysis We begin by fixing the temperatures at all states.
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
( ) K 7.92418K) 291( 14.111
1
2
112 ===⎟⎟
⎠
⎞⎜⎜⎝
⎛= −−
−k
k
s rTTTv
v
K 10370.85
K )291(924.7K) 291(1212
12
12 ⎯→⎯−
=TT s =
−+=
−+=
− ηη
TTTT
TTs
v
P
4
1
2
3
qout
x
qin
( ) kPa 514818kPa) 90( 4.11
2
112 ===⎟⎟
⎠
⎞⎜⎜⎝
⎛= k
k
rPPPv
v
kPa 5663kPa) 5148)(1.1(23 ==== PrPP px
K 1141kPa 5148kPa 5663K) 1037(
22 =⎟
⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
PP
TT x x
K 1255K) 1141)(1.1(3 === xcTrT
K 3.410181.1K) 1255(
14.11
3
1
4
33 =⎟
⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
−−− kc
k
rr
TTTv
v 4s
K 8.494K )3.410(1255)90.0(K) 1255()( 433443
43 = −−=−−=⎯→⎯−−
= ss
TTTTTTTT
ηη
pplying e first law to each of the processes gives A th
kJ/kg 6.535K)2911037)(KkJ/kg 718.0()( 1221 =−⋅=−=− TTcw v
kJ/kg 6.114K)11411255)(KkJ/kg 005.1()( 33 =−⋅=−= TTcq
11411255)(KkJ/kg 718.0(6.114)( 333
− xpx
) kJ/kg 75.32K =−⋅−=−−= −− xxx TTcqw v
8 71.0()( 4343 =−=− TTcw v kJ/kg 8.545K)8.4941255)(KkJ/kg =−⋅
he net w k of the cycle is T or
328.54521343net +=−+= −−− wwww x kJ/kg 95.426.53575. =−
he mass in the device is given by
T
kg 003233.0K) 291)(K/kgmkPa 287.0(
)m kPa)(0.003 90(3
3
1
11 =⋅⋅
==RTP
mV
The net power produced by this engine is then
kW 9.26=== cycle/s) 0/60kJ/kg)(400 42.95kg/cycle)( 003233.0(netnet nmwW &&
preparation. If you are a student using this Manual, you are using it without permission.
9-47
9-64E An ideal dual cycle has a compression ratio of 15 and cutoff ratio of 1.4. The net work, heat addition, 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 R = 0.3704 psia·ft3/lbm.R (Table A-1E), cp = 0.240 Btu/lbm·R, cv = 0.171 Btu/lbm·R, and k = 1.4 (Table A-2Ea).
Analysis Working around the cycle, the germane properties at the various states are
( ) R 158015R) 535( 14.111
1
2
112 ===⎟⎟
⎠
⎞⎜⎜⎝
⎛= −−
−k
k
rTTTv
v
v
P
4
1
2
3
qout
x
qin( ) psia 2.62915psia) 2.14( 4.11
2
11
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
2 ===⎟⎟⎠
⎞⎜⎜⎝
⎛= k
k
rPPPv
v
psia 1.692psia) 2.629)(1.1(23 ==== PrPP px
R 1738psia 629.2psia 692.1
R) 1580(2
2 =⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
PP
TT xx
R 2433R)(1.4) 1738(33 ===⎟⎟
⎠
⎞⎜⎜⎝
⎛= cx
xx rTTT
v
v
R 2.942151.4R) 2433(
14.11
3
1 −−k
4
33 =⎟
⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
−kc
rr
TTTv
v
pplying e first law to each of the processes gives
4
A th
Btu/lbm 7.178R)5351580)(RBtu/lbm 171.0()( 1221 =−⋅=−=− TTcw v
2 Btu/lbm 02.27R)15801738)(RBtu/lbm 171.0()( 2 =−⋅=−= TTcq xx v −
)( 33 Btu/lbm 8.166R)17382433)(RBtu/lbm 240.0( =−⋅=
2433)(RBtu/lbm 171.0(Btu/lbm 8.166)( 333
−=− xpx TTcq
8 Btu/lbm 96.47R)173 =−⋅−=−−= −− xxx TTcqw v
)( 4343 Btu/lbm 9.254R)2.9422433)(RBtu/lbm 171.0( =−⋅=
he net w k of the cycle is
−=− TTcw v
T or
−+= wwww Btu/lbm 124.2=−+=−−− 7.17896.479.25421343net x
nd the net heat addition is
Hence, the thermal efficiency is
a
Btu/lbm 193.8=+=+= −− 8.16602.2732in xx qqq
0.641===Btu/lbm 193.8Btu/lbm 124.2
in
netth q
wη
preparation. If you are a student using this Manual, you are using it without permission.
9-48
9-65 A six-cylinder compression ignition engine operates on the ideal Diesel cycle. The maximum temperature in the cycle, the cutoff ratio, the net work output per cycle, the thermal efficiency, the mean effective pressure, the net power output, and the specific fuel consumption 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) Process 1-2: Isentropic compression
1
Qin
2 3
4
Qout
( )( )
( )( ) kPa 504419kPa 95 1.349
2
112
1-1.3491
==⎟⎟⎠
⎞⎜⎜⎝
⎛=
⎞⎛−k
PPv
v
v
The clearance volume and the total volume of the engine at the beginning of compression process (state 1) are
K 1.95019K 3402
112 ==⎟⎟
⎠⎜⎜⎝
=
k
TTv
3
3
m 0001778.0
m 0045.019 =⎯→⎯=
c
c
c
dcrVV
=
++
cV
VVV
m 003378.00032.00001778.0 =+=+= VVV 31 dc
ass contained in the cylinder is The total m
( )( )kg 0.003288
K 340K/kgmkPa 0.287)3
1=
⋅⋅
The mass of fuel burned during one cycle is
m 378kPa)(0.003 95(3
11 ==RTP
mV
kg 0001134.0kg) 003288.0(
28 =⎯→⎯−
=⎯→⎯−
== a
mm
AF ff
f
f
f
fm
mm
mmm
rocess 2-3: constant pressure heat a
P ddition
kJ 723.48)kJ/kg)(0.9 kg)(42,500 0001134.0(HVin === cf qmQ η
The cutoff ratio is
K 2244=⎯→⎯−=⎯→⎯−= 3323in K)1.950(kJ/kg.K) kg)(0.823 003288.0(kJ 723.4)( TTTTmcQ v
2.362===K 950.1K 2244
2
3
TT
β
(b) 33
12 m 0001778.0
19m 0.003378
===r
VV
23
14
3323 m 0004199.0)m 0001778.0)(362.2(
PP ==
===VV
VV β
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. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
9-49
Process 3-4: isentropic expansion.
( )
( ) kPa 9.302
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
m 0.003378kPa 5044 3
434 ⎜
⎜⎝
=⎟⎟⎠
⎜⎜⎝
= PPV
m 0.0004199
K 1084m 78m 0.0004199
1.34933
1-1.349
3
31
=⎟⎟⎠
⎞⎛⎞⎛
=⎟⎟⎠
⎞⎛⎞⎛−
k
k
V
V
rocess 4- constant voume heat rejection.
0.0033K 2244
4
334 ⎜
⎜⎝
=⎟⎟⎠
⎜⎜⎝
= TTV
P 1:
( ) ( )( ) kJ 013.2K3401084KkJ/kg 0.823kg) 003288.0(14out =−⋅=−= TTmcQ v
The net work output and the thermal efficiency are
=−=−= 013.2723.4QQW kJ 2.710outinout net,
57==== 5737.0kJ 2.710outnet,th
Wη .4%
kJ 4.723inQ
) The mean effective pressure is determined to be (c
kPa 847=⎟⎟⎞
⎜⎜⎛ ⋅
==mkPakJ 2.710MEP
3outnet,W
⎠⎝ kJm)1778 3
) The power for engine speed of 1750 rpm is
−− 000.0003378.0(21 VV
(d
kW 39.5=⎟⎠⎞
⎜⎝⎛==
s 60min 1
rev/cycle) 2((rev/min) 1750kJ/cycle) (2.710
2netnetnWW&&
Note that there are two revolutions in one cycle in four-stroke engines.
(e) Finally, the specific fuel consumption is
g/kWh 151=⎟⎠⎞
⎜⎝⎛
⎟⎟⎠
⎞⎜⎜⎝
⎛==
kWh 1kJ 3600
kg 1g 1000
kJ/kg 2.710kg 0001134.0sfc
netWm f
preparation. If you are a student using this Manual, you are using it without permission.
9-50
9-66 An expression for cutoff ratio of an ideal diesel cycle is to be developed.
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.
Analysis Employing the isentropic process equations,
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
TrrrTT kcc
−==
v
P
4
1
2 3 qin
qout
112
−= krTT
while the ideal gas law gives
3 11
2
When the first law and the closed system work integral is applied to the constant pressure heat addition, the result is
)()( 11
11
23in TrTrrcTTcq kkcpp
−− −=−=
When this is solved for cutoff ratio, the result is
11
in1Trc
qr kp
c −+=
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9-51
9-67 An expression for the thermal efficiency of a dual cycle is to be developed and the thermal efficiency for a given case is to be calculated.
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 The thermal efficiency of a dual cycle may be expressed as
)()(
)(11
32
14
in
outth
xpx TTcTTcTTc
−+−−
−=−=v
vη
By applying the isentropic process relations for ideal gases with constant specific heats to the processes 1-2 and 3-4, as well as the ideal gas equation of state, the temperatures may be eliminated from the thermal efficiency expression. This yields the result
⎥⎥⎦
⎤
⎢⎢⎣
⎡
−+−
−−=
− 1)1(111
1thpcp
kcp
k rrkrrr
rη
v
P
4
1
2
3
qout
x
qinwhere
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
2P
Pr x
p = and x
cr v
v 3=
hen rc = rp, we obtain W
⎟⎟
⎠
⎞ ⎜
⎝ −−=
− (1
21thp
k rkrη ⎜
⎛
−+
−+
1)
11 1
pp
kp
rr
r
For the case r = 20 and rp = 2,
0.660=⎟⎟⎠
⎞⎜⎜⎝
⎛
−+−
−−=
+
− 12)22(4.112
2011
2
14.1
14.1thη
preparation. If you are a student using this Manual, you are using it without permission.
9-52
9-68 An expression regarding the thermal efficiency of a dual cycle for a special case is to be obtained.
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.
Analysis The thermal efficiency of a dual cycle may be expressed as
)()(
)(11
32
14
in
outth
xpx TTcTTcTTc
−+−−
−=−=v
vη
By applying the isentropic process relations for ideal gases with constant specific heats to the processes 1-2 and 3-4, as well as the ideal gas equation of state, the temperatures may be eliminated from the thermal efficiency expression. This yields the result
⎥⎥⎦
⎤
⎢⎢⎣
⎡
−+−
−−=
− 1)1(111
1thpcp
kcp
k rrkrrr
rη
v
P
4
1
2
3
qout
x
qinwhere
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
2P
Pr x
p = and x
cr v
v 3=
hen rc = rp, we obtain W
⎟⎟
⎠
⎞ ⎜
⎝ −−=
− )(1
21thpp
k rrkrη ⎜
⎛
−+
−+
1
11 1
p
kp
r
r
earrangement of this result gives
R
1th2
1
)1(1)(
1 −+
−=−+−
− k
ppp
kp r
rrrk
rη
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9-53
9-69 The five processes of the dual cycle is described. The P-v and T-s diagrams for this cycle is to be sketched. An expression for the cycle thermal efficiency is to be obtained and the limit of the efficiency is to be evaluated for certain cases.
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.
Analysis (a) The P-v and T-s diagrams for this cycle are as shown.
s
T
52
1
4
3
v
P
5
1
2
4
qout
3
qin
(b) Apply first law to the closed system for processes 2-3, 3-4, and 5-1 to show:
( ) ( )( )
3 2 4 3
5 1
in v p
out v
q C T T C T T
q C T T
= − + −
= −
The cycle thermal efficiency is given by
( )( ) ( )
( )( ) ( )
( )( ) ( )
5 1 1 5 1
4 3 2 3 2 3 4 3
5 1
323 2 4 3
/ 11
/ 1 / 1
/ 11
/ 1 / 1
vout
p
th
C T T T T TqT C T T T T T kT T T
TT T T k T Tη
− −= −
− + − − + −
−= −
− + −
Process 1-2 is isentropic; therefore,
3 2
1 1thin vq C T
η = − = −
1 1T T
T T
112 1
kkT V
−
−⎛ ⎞
1 2
rT V
= =⎜ ⎟⎝ ⎠
Process 2-3 is constant volume; therefore,
3 3
2 2
3 3
2 2pr
T PV P= = =
T PV P
Process 3-4 is constant pressure; therefore,
3 34 4 4 4
4 3 3 3c
PVPV T V rT T T V
= ⇒ = =
Process 4-5 is isentropic; therefore, 1 1 1 1 1
5 4T rV VT V
3 24
4 5 1 1 1
k k k k kc c cV r V r
V V V r
− − − − −⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠
Process 5-1 is constant volume; however T5/T1 is found from the following.
= ⎜ ⎟
15 5 34 2 1
cc k k
c p pT T T T T rr r r r r−= = =⎜ ⎟
⎝ ⎠
1 4 3 2 1
kT T T rT T −⎛ ⎞
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9-54
The ratio T3/T1 is found from the following.
3 3 2
1 2 1
1kp
T T TT T T
r r −= =
The efficiency becomes
)( ) (1 1
11
1 1
kc p
th kp c
kp
r rr r k rr r
η− −
−= −
− + −
(c) In the limit as rp approaches unity, the cycle thermal efficiency becomes
( ) ( )
( )
11 1
1
1
1
1lim 1 lim
1 1
11lim 11
p p
p
kc p
th kr rp c
kc
th th Dieselrc
kp
k
r rr r k r
rk r
r r
r
η
η η
−→ →
→
−
−
⎧ ⎫−⎪ ⎪= − ⎨ ⎬− + −⎪ ⎪⎩ ⎭
⎡ ⎤−⎢ ⎥= − =−⎢ ⎥⎣ ⎦
(d) In the limit as rc approaches unity, the cycle thermal efficiency becomes
( ) ( ) ( )1 11 1
1
1
1
1 1lim 1 lim 1
1 1
1lim 1
c p
c
kc p p
th k kr rp c
th th Ottor
kp
k
r r rr r k r r rr r
r
η
η η
− −→ →
→
−
−
⎧ ⎫− −⎪ ⎪= − = −⎨ ⎬− + − −⎪ ⎪⎩ ⎭
= − =
1p
⎧ ⎫⎪ ⎪⎨ ⎬⎪ ⎪⎩ ⎭
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
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9-55
Stirling and Ericsson Cycles
9-70C The Stirling cycle.
9-71C The two isentropic processes of the Carnot cycle are replaced by two constant pressure regeneration processes in the Ericsson cycle.
9-72C The efficiencies of the Carnot and the Stirling cycles would be the same, the efficiency of the Otto cycle would be less.
9-73C The efficiencies of the Carnot and the Ericsson cycles would be the same, the efficiency of the Diesel cycle would be less.
9-74 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
T
3
2qin
qout
4
1 1200 K
300 K
KkJ/kg 0.5K 300
kJ/kg 150
0
out34,34 ⋅−=−=−=−
Tq
ss
and
( ) KkJ/kg 0.5kPa 120
PlnKkJ/kg 0.287
lnln
4
3
40
3
434
⋅−=⋅−=
−=−PPR
TTcss p
(b) For reversible cycles,
It yields P4 = 685.2 kPa
( ) kJ/kg 600kJ/kg 150K 300K 1200
outinin
outq===⎯→⎯= q
TT
qTT
q L
H
H
L
et, qqw
(c) The thermal efficiency of this totally reversible cycle is determined from
Thus,
kJ/kg 450=−=−= 150600outinoutn
75.0%=−=−=K1200K300
11thH
L
TT
η
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9-56
9-75 An ideal Stirling engine with air as the working fluid operates between the specified temperature and pressure limits. The net work produced per cycle and the thermal efficiency of the cycle are 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-2a).
Analysis Since the specific volume is constant during process 2-3,
s
T
3
2qin
qout
4
1 800 K
300 K
kPa 7.266K 300K 800kPa) 100(
3
232 =⎟
⎠⎞
⎜⎝⎛==
TT
PP
Heat is only added to the system during reversible process 1-2. Then,
kJ/kg .6462)KkJ/kg 0.5782)( K800()(
KkJ/kg 5782.0kPa 2000kPa 266.7ln)KkJ/kg 0.287(0
lnln
121in
1112 −=−
PR
Tcss p
20
2
=⋅=−=
⋅=
⎟⎠⎞
⎜⎝⎛⋅−=
ssTq
PT
he thermal efficiency of this totally reversible cycle is determined from
T
0.625=−=−=K 800K 300
11thH
L
TT
η
Then,
kJ 289.1=== kJ/kg) kg)(462.6 (0.625)(1inthnet mqW η
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
9-57
9-76 An ideal Stirling engine with air as the working fluid operates between the specified temperature and pressure limits. The power produced and the rate of heat input are 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-2a).
Analysis Since the specific volume is constant during process 2-3,
s
T
3
2qin
qout
4
1 800 K
300 K
kPa 7.266K 300K 800kPa) 100(
3
232 =⎟
⎠⎞
⎜⎝⎛==
TT
PP
Heat is only added to the system during reversible process 1-2. Then,
kJ/kg .6462)KkJ/kg 0.5782)( K800()(
KkJ/kg 5782.0kPa
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
2266.7
000kPa ln)KkJ/kg 0.287(0
121in
20
2
=⋅=−=
⋅=
⎟⎠⎞
⎜⎝⎛⋅−=
ssTq
PT
ycle is determined from
lnln11
12 −=−P
RT
css p
The thermal efficiency of this totally reversible c
0.625K 800K 300
11th =−=−=H
L
TT
η
Then,
kJ 289.1kJ/kg) kg)(462.6 (0.625)(1inthnet === mqW η
The rate at which heat is added to this engine is
while the power produced by the engine is
kW 10,020=== cycle/s) 0/60kJ/kg)(130 462.6kg/cycle)( 1(inin nmqQ &&
kW 6264=== cycle/s) 1300/60kJ/cycle)( 1.289()netnet nWW &&
preparation. If you are a student using this Manual, you are using it without permission.
9-58
9-77 An ideal Ericsson cycle operates between the specified temperature limits. The rate of heat addition is to be determined.
Analysis The thermal efficiency of this totally reversible cycle is determined from
s
T
3
2qin
qout
4
1 900 K
280 K
0.6889K 900K 280
11th =−=−=H
L
TT
η
According to the general definition of the thermal efficiency, the rate of heat addition is
kW 726===0.6889
kW 500
th
netin η
WQ
&&
9-78 An ideal Ericsson cycle operates between the specified temperature limits. The power produceddetermined.
Analysis The power output is 500 kW when t
by the cycle is to be
he cycle is repeated 2000 times inute. Then the work per cycle is
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
per m
kJ/cycle 15cycle/s (2000/60)net ===
nW
& kJ/s 500netW&
hen the cycle is repeated 3000 times per minute, the power output will be
considered. The temperature of the source-energy reservoir, r pressure during the cycle are to be determined.
ature are R = 0.3704 psia·ft3/lbm.R (Table A-1E), cp = 0.240 Btu/lbm·R, v
Analysis From the thermal efficiency relation,
s
T
3
2qin
qout
4
1 900 K
280 KW
kW 750=== kJ/cycle) 5cycle/s)(1 000/603(netnet WnW &&
9-79E An ideal Stirling engine with air as the working fluid isthe amount of air contained in the engine, and the maximum ai
Assumptions Air is an ideal gas with constant specific heats.
Properties The properties of air at room temperc = 0.171 Btu/lbm·R, and k = 1.4 (Table A-2Ea).
s
T
3
2qin
qout
4
1
510 R
R 765=⎯→⎯−=⎯→⎯−== HHH
L TTT
TQ
W R 5101Btu 6Btu 21
in
netthη
State 3 ma ine the mass of air in the system, y be used to determ
lbm 0.02647=⋅⋅
==R) 510)(R/lbmftpsia 3704.0(
)ft psia)(0.5 10(3
3
3
33
RTP
mV
The maximum pressure occurs at state 1,
psia 125=⋅⋅
==3
3
1
11
ft 0.06R) 765)(R/lbmftpsia 3704.0(lbm) 02647.0(
V
mRTP
preparation. If you are a student using this Manual, you are using it without permission.
9-59
9-80E An ideal Stirling engine with air as the working fluid is considered. The temperature of the source-energy reservoir, the amount of air contained in the engine, and the maximum air pressure during the cycle are to be determined.
Assumptions Air is an ideal gas with constant specific heats.
Properties The properties of air at room temperature are R = 0.3704 psia·ft3/lbm.R, cp = 0.240 Btu/lbm·R, cv = 0.171 Btu/lbm·R, and k = 1.4 (Table A-2E).
Analysis From the thermal efficiency relation,
s
T
3
2qin
qout
4
1
510 R
R 874=⎯→⎯−=⎯→⎯−== HHH TTQ Btu 6in
thL T
TW R 5101Btu 2.51netη
State 3 may be used to determine the mass of air in the system,
lbm 0.02647=⋅⋅ /lbmftpsia 3704.0( 3
3RT==
R) 510)(R)ft psia)(0.5 10( 3
33Pm
V
he maxim m pressure occurs at state 1, T u
psia 143=⋅⋅
==3
3
1
11
ft 0.06R) 874)(R/lbmftpsia 3704.0(lbm) 02647.0(
V
mRTP
9-81 An ideal Stirling engine with air as the working fluid operates between specifi imits. The heat added to the cycle and the net work produced by the cycle are to be determined.
ed pressure l
Properties The properties of air at room temperature are R = 0.287 kPa·m /kg.K, cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, nd k = 1.4 (Table A-2a).
Analysis Applying the ideal gas equation to the isothermal process 3-4 gives
Assumptions Air is an ideal gas with constant specific heats. 3
a
kPa 600kPa)(12) 50(4
s
T
3
2qin
qout
3v3 ===
vPP
4
1
298 K
4
Since process 4-1 is one of constant volume,
K 1788kPa 600kPa 3600K) 298(
4
141 =⎟
⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
PP
TT
Adapting the first law and work integral to the heat addition process gives
== −21in wq kJ/kg 1275=⋅= K)ln(12) 1788)(KkJ/kg 0.287(ln1
21 v
vRT
imilarly,
S
kJ/kg 212.5=⎟⎠⎞
⎜⎝⎛⋅=== − 12
1K)ln 298)(KkJ/kg 0.287(ln3
4343out v
vRTwq
The net work is then
kJ/kg 1063=−=−= 5.2121275outinnet qqw
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
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9-60
9-82 An ideal Stirling engine with air as the working fluid operates between specified pressure limits. The heat transfer in the regenerator is to be determined.
Assumptions Air is an ideal gas with constant specific heats.
Properties The properties of air at room temperature are R = 0.287 kPa·m3/kg.K, cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-2a).
Analysis Applying the ideal gas equation to the isothermal process 3-4 gives
s
T
3
2qin
qout
4
1
298 K
kPa 600kPa)(12) 50(4
34 v3 ===
vPP
Since process 4-1 is one of constant volume,
K 1788kPa 6004
41⎠⎝⎟
⎠⎜⎝ P
kPa 3600K) 298(1 =⎟⎞
⎜⎛=⎟
⎞⎜⎛
=P
TT
pplication of the first law to process 4-1 gives A
kJ/kg 1070=−⋅=−= K)2981788)(KkJ/kg 718.0()( 41regen TTcq v
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9-61
Ideal and Actual Gas-Turbine (Brayton) Cycles
9-83C They are (1) isentropic compression (in a compressor), (2) P = constant heat addition, (3) isentropic expansion (in a turbine), and (4) P = constant heat rejection.
9-84C For fixed maximum and minimum temperatures, (a) the thermal efficiency increases with pressure ratio, (b) the net work first increases with pressure ratio, reaches a maximum, and then decreases.
9-85C Back work ratio is the ratio of the compressor (or pump) work input to the turbine work output. It is usually between 0.40 and 0.6 for gas turbine engines.
9-86C In gas turbine engines a gas is compressed, and thus the compression work requirements are very large since the steady-flow work is proportional to the specific volume.
9-87C As a result of turbine and compressor inefficiencies, (a) the back work ratio increases, and (b) the thermal efficiency decreases.
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9-62
9-88E A simple ideal Brayton cycle with air as the working fluid has a pressure ratio of 10. The air temperature at the compressor exit, the back work ratio, and the thermal efficiency are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with variable specific heats.
Properties The properties of air are given in Table A-17E.
Analysis (a) Noting that process 1-2 is isentropic,
s
T
1
24
3qin
qout
2000
520
ThPr
11
11 2147= ⎯ →⎯
=
=520 R
124.27 Btu / lbm.
( )( )Btu/lbm 240.11
147.122147.110
2
2
112 P rr
2
==
⎯→⎯===hT
PP
PR996.5
(b) Process 3-4 is isentropic, and thus
( )
Btu/lbm 38.88283.26571.50443outT, −=−= hhw
Btu/lbm 115.8427.12411.240
Btu/lbm 265.834.170.174101
0.174Btu/lbm 504.71
R 2000
12inC,
43
4
33
34
3
=
=−=−=
=⎯→⎯=⎟⎠⎞
⎜⎝⎛==
==
⎯→⎯=
hhw
hPPP
P
Ph
T
rr
r
Then the back-work ratio becomes
48.5%===Btu/lbm 238.88Btu/lbm 115.84
outT,
inC,bw w
wr
46.5%===
=−=−=
=−=−=
Btu/lbm 264.60Btu/lbm 123.04
Btu/lbm 123.0484.11588.238
Btu/lbm 264.6011.24071.504
in
outnet,th
inC,outT,outnet,
23in
qw
www
hhq
η
(c)
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
PROPRIETARY MATERIAL. © 2011 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.
9-63
9-89 A simple Brayton cycle with air as the working fluid has a pressure ratio of 10. The air temperature at the etermined.
anges are s.
-17.
Analysis (a) Noting that process 1-2s is isentropic,
turbine exit, the net work output, and the thermal efficiency are to be d
Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy ch
s
T
1
2s
4s
3qin
qout
1240 K
295 K
2
4
negligible. 4 Air is an ideal gas with variable specific heat
Properties The properties of air are given in Table A
3068.1kJ/kg .17295
K 2951
11 =
=⎯→⎯=
rPh
T
( )( )
( )
( )( )( )
kJ/kg 83.04707.70293.132487.093.1324
K 6.689 and kJ/kg .0770223.273.272101
3.272kJ/kg 1324.93
K 1240
kJ/kg .6062683.0
17.29526.57017.295
K 564.9 and kJ/kg 570.2607.133068.110
433443
43
443
4
3 3T
1212
12
12
222
34
3
=−−=
−−=⎯→⎯−−
=
==⎯→⎯=⎟⎠⎞
⎜⎝⎛==
==
⎯→⎯=
=−
+=
−+=⎯→⎯
−=
==⎯→⎯===
sTs
T
ssrr
r
C
ssC
ssr
hhhhhhhh
ThPPP
P
Ph
hhhh
hhhh
ThPP
ηη
ηη
Thus,
T4 = 764.4 K
(b)
(c)
rP1
12
−P
kJ/kg 210.4=−=−=
=−=−=
=−=−=
9.4873.698
kJ/kg 487.917.29504.783
kJ/kg .369860.62693.1324
outinoutnet,
14out
23in
qqw
hhq
hhq
30.1%==== 3013.0kJ/kg 698.3kJ/kg 210.4
in
outnet,th q
wη
9-64
9-90 Problem 9-89 is reconsidered. The mass flow rate, pressure ratio, turbine inlet temperature, and the isentropic efficiencies of the turbine and compressor are to be varied and a general solution for the problem by taking advantage of the diagram window method for supplying data to EES is to be developed.
Analysis Using EES, the problem is solved as follows:
"Input data - from diagram window" {P_ratio = 10} {T[1] = 295 [K] P[1]= 100 [kPa] T[3] = 1240 [K] m_dot = 20 [kg/s] Eta_c = 83/100 Eta_t = 87/100} "Inlet conditions" h[1]=ENTHALPY(Air,T=T[1]) s[1]=ENTROPY(Air,T=T[1],P=P[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]) 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,s=s_s[4],P=P[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" "Cycle analysis" W_dot_net=W_dot_t-W_dot_c"Definition of the net cycle work, kW" Eta=W_dot_net/Q_dot_in"Cycle thermal efficiency" Bwr=W_dot_c/W_dot_t "Back work ratio" "The following state points are determined only to produce a T-s plot" T[2]=temperature(air,h=h[2]) T[4]=temperature(air,h=h[4]) s[2]=entropy(air,T=T[2],P=P[2]) s[4]=entropy(air,T=T[4],P=P[4])
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9-65
Bwr η Pratio Wc
[kW] Wnet [kW]
Wt [kW]
Qin [kW]
0.5229 0.1 2 1818 1659 3477 16587 0.6305 0.1644 4 4033 2364 6396 14373 0.7038 0.1814 6 5543 2333 7876 12862 0.7611 0.1806 8 6723 2110 8833 11682 0.8088 0.1702 10 7705 1822 9527 10700
0.85 0.1533 12 8553 1510 10063 9852 0.8864 0.131 14 9304 1192 10496 9102 0.9192 0.1041 16 9980 877.2 10857 8426 0.9491 0.07272 18 10596 567.9 11164 7809 0.9767 0.03675 20 11165 266.1 11431 7241
2 4 6 8 10 12 14 16 18 200.12
0.16
0.2
0.24
0.28
0.32
0.36
2250
2700
3150
3600
4050
4500
Pratio
η
Wne
t [k
W]
4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0 8.50
500
1000
1500
s [kJ/kg-K]
T [K
]
100 kPa
1000 kPa
Air
1
2
3
42s 4s
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
9-66
9-91 A simple Brayton cycle with air as the working fluid has a pressure ratio of 10. The air temperature at the turbine exit, the net work output, and the thermal efficiency are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2).
Analysis (a) Using the compressor and turbine efficiency relations, ( )
( )( )
( )( )
( )( )
( )( ) ( )
( )( )K 720=
−−=−−=⎯→⎯
−
−=
−−
=
=−
+=
−+=⎯→⎯
−
−=
−−
=
=⎟⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
==⎟⎟⎠
⎞⎜⎜⎝
⎛=
−
−
3.642124087.01240
K 625.883.0
2956.569295
K 642.3101K 1240
K 6.56910K 295
433443
43
43
43
1212
12
12
12
12
0.4/1.4/1
3
434
0.4/1.4/1
1
212
sTsp
p
sT
C
s
p
sps
s
T
1
2s
4s
3qin
qout
1240 K
295 K
2
4Cη
kk
s
kk
s
TTTTTTcTTc
hhhh
TTTT
TTcTTc
hhhh
PP
TT
PP
TT
ηη
η
(b) ( ) ( )( )( ) ( )( )
kJ/kg 190.2=−=−=
=−⋅=−=−=
=−⋅=−=−=
1.4273.617
kJ/kg 427.1K295720KkJ/kg 1.005
kJ/kg 17.36K625.81240KkJ/kg 1.005
outinoutnet,
1414out
2323in
qqw
TTchhq
TTchhq
p
p
30.8%==== 3081.0kJ/kg 617.3kJ/kg 190.2
in
outnet,th q
wη (c)
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9-67
9-92E A simple ideal Brayton cycle with helium has a pressure ratio of 14. The power output is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Helium is an ideal gas with constant specific heats.
Properties The properties of helium are cp = 1.25 Btu/lbm·R and k = 1.667 (Table A-2Ea).
Analysis Using the isentropic relations for an ideal gas,
s
T
1
2
4
3 qin
qout
1760 R
520 R
R 1495R)(14) 520( 70.667/1.66/)1(1
/)1(
1
212 ===⎟⎟
⎠
⎞⎜⎜⎝
⎛= −
−kk
p
kk
rTPP
TT
Similarly,
K 2.61214
R) 1760(33
434 =⎟
⎠⎜⎝
=⎟⎟⎠
⎜⎜⎝
=⎟⎟⎠
⎜⎜⎝
=pr
TP
TT 11 70.667/1.66/)1(/)1(⎞⎛⎞⎛⎞⎛
−− kkkkP
ssure heat addition process 2-3 produces Applying the first law to the constant-pre
q Btu/lbm 3.331R)14951760)(RBtu/lbm 1.25()( 23in =−⋅=−= TTc p
Similarly,
Btu/lbm 3.115R)5202.612)(RBtu/lbm 25.1()( 14out =−⋅=−= TTcq p
net work production is then
et =−=−= qqw
nd
The
n Btu/lbm 216.03.1153.331outin
a
hp 509.3=⎟⎠
⎞⎜⎝
⎛==Btu/min 42.41hp 1Btu/lbm) 0.216(lbm/min) 100(netnet wmW &&
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
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9-68
9-93E A simple Brayton cycle with helium has a pressure ratio of 14. The power output is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Helium is an ideal gas with constant specific heats.
Properties The properties of helium at room temperature are cp = 1.25 Btu/lbm·R and k = 1.667 (Table A-2Ea).
Analysis For the compression process,
R 1495R)(14) 520( 70.667/1.66/)1(1
/)1(
1
212 ===⎟⎟
⎠
⎞⎜⎜⎝
⎛= −
−kk
p
kk
s rTPP
TT
4
s
T
1
2s
3 qin
qout
1760 R
520 R
2
R 154695.0
5201495520
)( 1212 −=
−=
pC TTchh
η)( 12
121212
=−
+=
−+=⎯→⎯
−−
C
ssps TTTT
TTchhη
For the isentropic expansion process,
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
R 2.612141R) 1760(1 70.667/1.66/)1(/)1( ⎞⎛⎞
−− kkkk
33
4 =⎟⎠⎞
⎜⎝⎛=⎟
⎟⎠
⎜⎜⎝
=⎟⎟⎠⎝ pr
TPP
pplying the first law to the constant-pressure heat addition process 2-3 produces
34 ⎜⎜⎛
= TT
A
Btu/lbm 5.267R)15461760)(RBtu/lbm 1.25()( 23in =−⋅=−= TTcq
imilarly,
p
S
Btu/lbm 3.115R)5202.612)(RBtu/lbm 25.1()( 14out =−⋅=−= TTcq p
The net work production is then
Btu/lbm 152.23.1155.267outinnet =−=−= qqw
nd
a
hp 358.9=⎟⎠
⎞⎜⎝
⎛==Btu/min 42.41hp 1
Btu/lbm) 2.152(lbm/min) 100(netnet wmW &&
preparation. If you are a student using this Manual, you are using it without permission.
9-69
9-94 A simple Brayton cycle with air as the working fluid operates between the specified temperature and pressure limits. The effects of non-isentropic compressor and turbine on the back-work ratio is to be compared.
Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2a).
Analysis For the compression process,
4s
s
T
1
2s
4
3 qin
qout
873 K
288 K
2
K 8.585K)(12) 288( 0.4/1.4/)1(
1
212 ==⎟⎟
⎠
⎞⎜⎜⎝
⎛=
− kk
s PP
TT
K 9.61890.0
2888.585288
)()( 12
1212
12
12
12
=−
+=
−+=⎯→⎯
−
−=
−−
=C
s
p
spsC
TTTT
TTcTTc
hhhh
ηη
F paor the ex nsion process,
K 2.429121K) 873(
0.4/1.4/)1(
3
434 =⎟
⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
− kk
s PP
TT
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
K 6.473)2.429873)(90.0(873
)()( 334
43
43
43
43 −=⎯→⎯−
=−
= Tsp
p
sT TTT
TTchhηη
)(4
=−−=
−−−
sTTTchh
The isentropic and actual work of compressor and turbine are
kJ/kg 3.299K)2888.585)(KkJ/kg 1.005()( 12Comp, =−⋅=−= TTcW sps
kJ/kg 6.332K)2889.618)(KkJ/kg 1.005()( 12Comp =−⋅=−= TTcW
4
p
( 3Turb, kJ/kg 0.446K)2.429873)(KkJ/kg 1.005() =−⋅=− sTW = ps Tc
kJ/kg 4.401K)6.473873)(KkJ/kg 1.005()( 43Turb =−⋅=−= TTcW p
The back work ratio for 90% efficient compressor and isentropic turbine case is
0.7457===kJ/kg 446.0Turb,
bwsW
rkJ/kg 332.6CompW
The back work ratio for 90% efficient turbine and isentropic compressor case is
0.7456===kJ/kg 401.4kJ/kg 299.3
Turb
Comp,bw W
Wr s
The two results are almost identical.
preparation. If you are a student using this Manual, you are using it without permission.
9-70
9-95 A gas turbine power plant that operates on the simple Brayton cycle with air as the working fluid has a specified pressure ratio. The required mass flow rate of air is to be determined for two cases.
Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2).
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
Analysis (a) Using the isentropic relations,
( )( )( )
( )( )
( ) ( )( )( ) ( )( )
kg/s 352===
=−=−=
=−⋅=−=−=
=−⋅=−=−=
=⎟⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
==⎟⎟⎠
⎜⎜⎝
= K 3001
12 PTT s
⎞⎛
−
−
kJ/kg 199.1kJ/s 70,000
kJ/kg 99.1175.31184.510
kJ/kg 510.84K491.71000KkJ/kg 1.005
kJ/kg 311.75K300610.2KkJ/kg 1.005
K 491.7121K 1000
K 610.212
outnet,s,
outnet,
inC,s,outT,s,outnet,s,
4343outT,s,
1212inC,s,
0.4/1.4/1
3
434
0.4/1.4/1
2
wW
m
www
TTchhw
TTchhw
PP
TT
P
s
sps
sps
kk
s
kk
&&
(b) The net work output is determined to be
s
T
1
2s
4s
31000 K
300 K
2
4
( )( )
kg/s 1037===
=−=
−=−=
kJ/kg 67.5kJ/s 70,000
kJ/kg 67.50.85311.7584.51085.0
/
outnet,a,
outnet,
inC,s,outT,s,inC,a,outT,a,outnet,a,
wW
m
wwwww
a
CT
&&
ηη
preparation. If you are a student using this Manual, you are using it without permission.
9-71
9-96 An actual gas-turbine power plant operates at specified conditions. The fraction of the turbine work output used to drive the compressor and the thermal efficiency are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with variable specific heats.
Properties The properties of air are given in Table A-17.
Analysis (a) Using the isentropic relations,
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
→
= ⎯ →⎯ =
300.19 kJ / kg
580 K 586.04 kJ / kg
T h
T h1 1
2 2
= ⎯⎯ =300 K
( )
( ) ( )( ) kJ/kg 542.0905.831536.0486.0
kJ/kg 285.8519.30004.586
kJ/kg 05.83973.6711.47471
11.474
kJ/kg1536.0404.586950
7100700
43outT,w
12inC,
43
4
323in
1
2
34
3
=−=−=
=−=−=
=⎯→⎯=⎟⎠⎞
⎜⎝⎛==
=→
=+=⎯→⎯−=
===
sT
srr
r
p
hh
hhw
hPPP
P
P
hhhq
PP
r
η
Thus,
s
T
1
2s
4s
3950 kJ/kg
580 K
300 K
2
4
52.7%===kJ/kg 542.0kJ/kg 285.85
outT,
inC,bw w
wr
27.0%===
=−=−=
kJ/kg 950kJ/kg 256.15
kJ/kg 256.15285.85.0542
in
outnet,th
inC,outT,net.out
qw
www
η
(b)
preparation. If you are a student using this Manual, you are using it without permission.
9-72
9-97 A gas-turbine power plant operates at specified conditions. The fraction of the turbine work output used to drive the compressor and the thermal efficiency are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats.
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2).
Analysis (a) Using constant specific heats,
( )( ) ( )
( )( )
( ) ( )( )( ) ( ) ( )( )( ) kJ/kg 562.2K874.81525.3KkJ/kg 1.00586.0
kJ/kg 281.4K300580KkJ/kg1.005
K 874.871K 1525.3
K 1525.3KkJ/kg 1.005/kJ/kg 950K 580
/
7100700
4343outT,
1212inC,
0.4/1.4/k1k
3
434s
in232323in
1
2
=−⋅=−=−=
=−⋅=−=−=
=⎟⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
=⋅+=
+=⎯→⎯−=−=
===
−
spTsT
p
pp
p
TTchhw
TTchhw
PP
TT
cqTTTTchhq
PP
r
ηη
Thus,
s
T
1
2s
4s
3950 kJ/kg
580 K
300 K
2
4
50.1%===kJ/kg 562.2kJ/kg 281.4
outT,
inC,bw w
wr
29.6%===
=−=−=
kJ/kg 950kJ/kg 280.8
kJ/kg 280.8281.42.562
in
outnet,th
inC,outT,outnet,
qw
www
η
(b)
preparation. If you are a student using this Manual, you are using it without permission.
9-73
9-98 An aircraft engine operates as a simple ideal Brayton cycle with air as the working fluid. The pressure ratio and the rate of heat input are given. The net power and the thermal efficiency are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2a).
Analysis For the isentropic compression process,
K .1527K)(10) 273( 0.4/1.4/)1(12 === − kk
prTT
s
T
1
2
4
3 qin
qout273 K
The heat addition is
kJ/kg 500kg/s 1in ===
mq
& kW 500inQ&
Applying the first law to the heat addition process,
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
K 1025
KkJ 1.00523pc /kg
kJ/kg 500K 1.527
)(
in
23in
=⋅
+=+=
−p
qTT
TTc
he temperature at the exit of the turbine is
=q
T
K 9.530101K) 1025(1 0.4/1.4/)1(
⎞⎛− kk
34 =⎟⎠⎞
⎜⎝⎛=⎟
⎟⎠
⎜⎜⎝
=pr
TT
pplying the first law to the adiabatic turbine and the compressor produce
43T
A
1025)(KkJ/kg 1.005()( ⋅=−= TTcw kJ/kg 6.496K)9.530 =−p
kJ/kg 4.255K)2731.527)(KkJ/kg 1.005()( 12C =−⋅=−= TTcw
he net po er produced by the engine is then
Finally the thermal efficiency is
p
T w
kW 241.2=−=−= kJ/kg)4.2556kg/s)(496. 1()( CTnet wwmW &&
0.482===kW 500kW 241.2
in
netth Q
W&
&η
preparation. If you are a student using this Manual, you are using it without permission.
9-74
9-99 An aircraft engine operates as a simple ideal Brayton cycle with air as the working fluid. The pressure ratio and the rate of heat input are given. The net power and the thermal efficiency are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2a).
Analysis For the isentropic compression process,
K .8591K)(15) 273( 0.4/1.4/)1(12 === − kk
prTT
s
T
1
2
4
3 qin
qout273 K
The heat addition is
kJ/kg 500kg/s 1in ===
mq
& kW 500inQ&
Applying the first law to the heat addition process,
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
K 1089
Kk 1.00523pc J/kg
kJ/kg 500K 8.591
)(
in
23in
=⋅
+=+=
−p
qTT
TTc
he temperature at the exit of the turbine is
=q
T
K 3.502151K) 1089(1 0.4/1.4/)1(
⎞⎛− kk
34 =⎟⎠⎞
⎜⎝⎛=⎟
⎟⎠
⎜⎜⎝
=pr
TT
pplying the first law to the adiabatic turbine and the compressor produce
43T
A
1089)(KkJ/kg 1.005()( ⋅=−= TTcw kJ/kg 6.589K)3.502 =−p
kJ/kg 4.320K)2738.591)(KkJ/kg 1.005()( 12C =−⋅=−= TTcw
he net po er produced by the engine is then
Finally the thermal efficiency is
p
T w
kW 269.2=−=−= kJ/kg)4.3206kg/s)(589. 1()( CTnet wwmW &&
0.538===kW 500kW 269.2
in
netth Q
W&
&η
preparation. If you are a student using this Manual, you are using it without permission.
9-75
9-100 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.
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
=⎭⎬⎫
===
shss
P 1
Combustion chamber
Turbine
23
4
Compress.
100 kPa 40°C
650°C
2 MPa
Process 1-2: Compression
KkJ/kg 749.5kPa 100
C40kJ/kg 6.313C40
11
1
11
⋅=⎭⎬⎫
=°=
=⎯→⎯°=
sPT
hT
kJ/kg 7.736kJ/kg.K 749.5
kPa 20002
12
2
kJ/kg 4.8116.313
6.3137.73685.0 2212
12C ⎯
−=
hhsη =⎯→⎯
−−
=⎯→−
hh
hh
Process 3-4: Expansion
kJ/kg 2.959C650 44 =⎯→⎯°= hT
ss hhh
hhhh
43
3
43
43T
2.95988.0
−−
=⎯→⎯−−
=η
We cannot find the enthalpy at state 3 directly. However, using the following lines in EES together with the isentropic efficienc 73 kJ/kg, T3 = 1421ºC, s3 = 6.736 kJ/kg.K. The solution by hand would require a trial-error app
) h_4s=enthalpy(Air, P=P_1, s=s_3)
ermined from
y relation, we find h3 = 18roach. h_3=enthalpy(Air, T=T_3) s_3=entropy(Air, T=T_3, P=P_2
The mass flow rate is det
( )( )kg/s 99.12
K 27340K/kgmkPa 0.287)s/m 0kPa)(700/6 100(
3
3
1
11 =+⋅⋅
==RTP
mV&
&
The net power output is
=−=&
=−=−= hhmW &
(b) The back work ratio is
( 12inC, −= hhmW & kW 6464)kJ/kg6.3134kg/s)(811. 99.12()
kW 868,11)kJ/kg2.959kg/s)(1873 99.12()( 43outT,&
kW 5404=−=−= 6464868,11inC,outT,net WWW &&&
0.545===kW 868,11
kW 6464
outT,
inC,bw W
Wr
&
&
(c) The rate of heat input and the thermal efficiency are
kW 788,13)kJ/kg4.811kg/s)(1873 99.12()( 23in =−=−= hhmQ &&
39.2%==== 392.0kW 788,13
kW 5404
in
net
QW
th &
&η
preparation. If you are a student using this Manual, you are using it without permission.
9-76
9-101 A simple Brayton cycle with air as the working fluid operates between the specified temperature and pressure limits. The cycle is to be sketched on the T-s cycle and the isentropic efficiency of the turbine and the cycle thermal efficiency are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats.
Properties The properties of air are given as cv = 0.718 kJ/kg·K, cp = 1.005 kJ/kg·K, R = 0.287 kJ/kg·K, k = 1.4.
Analysis (b) For the compression process,
4s
s
T
1
2s
4
3 873 K
303 K
2 kW 300,60
K)30330)(KkJ/kg 1.005(kg/s) 200(
)( 12Comp
=−⋅=
−= TTcmW p&&
For the turbine during the isentropic process,
K 9.772kPa 800kPa 100K) 1400(
0.4/1.4/)1(
3
434 =⎟
⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
− kk
s PP
TT
kg/s 200()( 43sTurb, =−= sp TTcmW && kW 050,126K)9.7721400)(KkJ/kg 1.005() =−⋅
The actual power output from the turbine is
,120300,60000,60TurbnetTurb
CompTurbnet
=+=+= WWW &&&
The isentropic efficiency of the turbine is then
−= WWW &&&
kW 300
95.4%==== 954.0kW 050,126kW 300,120
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
sTurb,
Turbrb W
W&
η
(c) The rate of heat input is
=−= TTcmQ p&&
he thermal efficiency is then
&Tu
kW 200,160K)]273330(1400)[(KkJ/kg 1.005(kg/s) 200()( 23in =+−⋅
T
37.5%==== 375.0kW 200,160kW 000,60
in
netth Q
W&
&η
preparation. If you are a student using this Manual, you are using it without permission.
9-77
9-102 A modified Brayton cycle with air as the working fluid operates at a specified pressure ratio. The T-s diagram is to be sketched and the temperature and pressure at the exit of the high-pressure turbine and the mass flow rate of air are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats.
Properties The properties of air are given as cv = 0.718 kJ/kg·K, cp = 1.005 kJ/kg·K, R = 0.287 kJ/kg·K, k = 1.4.
Analysis (b) For the compression process,
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
K 5.494K)(8) 273(1
12 ==⎟⎟⎠
⎜⎜⎝
=P
TT 0.4/1.4/)1(
2 ⎞⎛− kk
P
The power input to the compressor is equal to the power output from the high-pressure turbine. Then,
−=−
−=−
=
5.4942731500
)()(
21
4
4312
outTurb, HPinComp,
TTTTTT
TTcmTTcm
WW
pp &&
&&
The pressure at this state is
s
T
1
4
5
3
P4
1500 K
273 K
2
P5
P3 = P2
= 34
312
TT K 1278.5=−+=−+
kPa 457.3=⎟⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=⎯→⎯⎟⎟
⎠
⎞⎜⎜⎝
⎛=
−− /()1/( kkk 4.0/4.1)1
3
414
3
4
3
4
K 1500K 5.1278kPa) 100(8
k
TT
rPPTT
PP
(c) The temperature at state 5 is determined from
K 1.828kPa 3.457
kPa 100K) 5.1278(0.4/1.4/)1(
5 ⎜⎛=⎟
⎞⎜⎛
=− kk
PTT
44 =⎟
⎠⎞
⎝⎟⎠
⎜⎝ P
he net power is that generated by the low-pressure turbine since the power output from the high-pressure turbine is equal the power input to the compressor. Then,
5
Tto
kg/s 441.8=−⋅
=−
=
−=
K)1.8285.1278)(KkJ/kg 1.005(kW 000,200
)(
)(
54
Turb LP
54Turb LP
TTcW
m
TTcmW
p
p
&&
&&
preparation. If you are a student using this Manual, you are using it without permission.
9-78
9-103 A simple Brayton cycle with air as the working fluid operates at a specified pressure ratio and between the specified temperature and pressure limits. The cycle is to be sketched on the T-s cycle and the volume flow rate of the air into the compressor is to be determined. Also, the effect of compressor inlet temperature on the mass flow rate and the net power output are to be investigated.
Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats.
Properties The properties of air are given as cv = 0.718 kJ/kg·K, cp = 1.005 kJ/kg·K, R = 0.287 kJ/kg·K, k = 1.4.
Analysis (b) For the compression process,
( ) K 0.4767K) 273( 0.4/1.4/)1(
1
212 ==⎟⎟
⎠
⎞⎜⎜⎝
⎛=
− kk
s PP
TT
4s
s
T
1
2s
4
3 1500 K
273 K
2
K 8.526273
2730.47680.0
)()(
22
12
12
12
12
CompComp ==
W&η sComp,
=⎯→⎯−
−=
−−
=−
−
TT
TTTT
TTcmTTcmW s
p
sp
&
&&
or the ex nsion process,
F pa
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
K 3.8601K) 1500(0.4/1.4/)1(
4 =⎟⎞
⎜⎛=⎟
⎞⎜⎛
=− kk
PTT
7334
⎠⎝⎟⎠
⎜⎝
s P
K 3.9243.8601500
150090.0
)()(
44
43
43
43
43
sTurb,
TurbTurb
=⎯→⎯−
−=
−−
=−
−==
TT
TTTT
TTcmTTcm
WW
ssp
p
&
&
&
&η
Given the net power, the mass flow rate is determined from
[ ]
[ ]
[ ]kg/s 7.463
)2738.526()3.9241500()KkJ/kg 1.005(kW 000,150
=−−−⋅
=
The specific volume and the volume flow rate at the inlet of the c mpressor are
)()(
)()(
1243
net
1243CompTurbnet
−−−=
−−−=−=
TTTTcW
m
TTcmTTcmWWW
p
pp
&&
&&&&&
o
)()( 1243net −−−= TTTTcmW p&&
/kgm 7835.0kPa 100
K) 273)(KkJ/kg 287.0( 3
1
11 =
PRT
v =⋅
=
(c) For a fixed compressor inlet velocity and flow area, when the compressor inlet temperature increases, the specific
volume increases since
/sm 363.2 3=== )/kgm 7835.0)(kg/s 7.463( 311 vV m&&
PRT
=v . When specific volume increases, the mass flow rate decreases since vV&
& =m . Note that
volume flow rate is the same since inlet velocity and flow area are fixed ( ). When mass flow rate decreases, the net power decreases since . Therefore, when the inlet temperature increases, both mass flow rate and the net power decrease.
AV=V&
)( CompTurbnet wwmW −= &&
preparation. If you are a student using this Manual, you are using it without permission.
9-79
Brayton Cycle with Regeneration
9-104C Regeneration increases the thermal efficiency of a Brayton cycle by capturing some of the waste heat from the exhaust gases and preheating the air before it enters the combustion chamber.
9-105C Yes. At very high compression ratios, the gas temperature at the turbine exit may be lower than the temperature at the compressor exit. Therefore, if these two streams are brought into thermal contact in a regenerator, heat will flow to the exhaust gases instead of from the exhaust gases. As a result, the thermal efficiency will decrease.
9-106C The extent to which a regenerator approaches an ideal regenerator is called the effectiveness ε, and is defined as ε = qregen, act /qregen, max.
9-107C (b) turbine exit.
9-108C The steam injected increases the mass flow rate through the turbine and thus the power output. This, in turn, increases the thermal efficiency since in/ QW=η and W increases while Qin remains constant. Steam can be obtained by utilizing the hot exhaust gases.
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
9-80
9-109 A Brayton cycle with regeneration produces 150 kW power. The rates of heat addition and rejection are to be determined.
Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K and k = 1.4 (Table A-2a).
Analysis According to the isentropic process expressions for an ideal gas,
K 8.530K)(8) 293( 0.4/1.4/)1(12 === − kk
prTT
s
T
1
2 5
4 qin1073 K
293 K
3
6
qout
K 3.5928
K) 1073(45 =⎟⎠
⎜⎝
=⎟⎟⎠
⎜⎜⎝
=pr
TT 11 0.4/1.4/)1(⎞⎛⎞⎛
− kk
hen the first law is applied to the heat exchanger, the result is
TT
The simultaneous solution of these two results gives
W
TTTT −=− 3 652
while the regenerator temperature specification gives
=−=−= K 3.582103.5921053
6 K 8.540)8.5303.582(3.592)( 235 =−−=−−= TTTT
ation of the first law to the turbine and compressor gives
)3.5921073)(KkJ/kg 005.1(
)()( 1254net
=−⋅−−⋅=
−−−= TTcTTcw pp
Applic
KkJ/kg 244.1
K )2938.530)(KkJ/kg 005.1(
Then,
kg/s 6145.0kJ/kg 244.1kW 150
net
net ===wW
m& &
Applying the first law to the combustion chamber produces
Similarly,
kW 303.0=−⋅=−= K)3.5821073)(KkJ/kg 005.1(kg/s) 6145.0()( 34in TTcmQ p&&
kW 153.0=−⋅=−= K)2938.540)(KkJ/kg 005.1(kg/s) 6145.0()( 16out TTcmQ p&&
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
9-81
9-110 A Brayton cycle with regeneration produces 150 kW power. The rates of heat addition and rejection are to be determined.
Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K and k = 1.4 (Table A-2a).
Analysis For the compression and expansion processes we have
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
=== − kkprTT K 8.530K)(8) 293( 0.4/1.4/)1(
12s
s
T
1
2 5s
4qin1073 K
293 K
3 6
qout
52s K 3.566
87.02938.530293
)()( 12
1212
12
=−
+=
−+=⎯→⎯
−
−=
C
s
p
spC
TTTT
TTcTTc
ηη
K 3.59281K) 1073(1 0.4/1.4/)1(
45 =⎟⎠⎞
⎜⎝⎛=⎟
⎟⎠
⎞⎜⎜⎝
⎛=
− kk
ps r
TT
K 9.625)3.5921073)(93.0(107354 −−=sp
)()()(
544554
=
−−=⎯→⎯−
−= sT
pT TTTT
TTcTTc
ηη
result is
=−=−= TT
When the first law is applied to the heat exchanger, the
6523 TTTT −=−
while the regenerator temperature specification gives
3 K 9.615109.625105
The simultaneous solution of these two results gives
K 3.576)3.5669.615(9.625)( 2356 =−−=−−= TTTT
Application of the first law to the turbine and compressor gives
kJ/kg 7.174K )2933.566)(KkJ/kg 005.1(K )9.6251073)(KkJ/kg 005.1(
)()( 1254net
=−⋅−−⋅=
−−−= TTcTTcw pp
Then,
kg/s 8586.0kJ/kg 174.7kW 150
net
net ===wW
m&
&
pplying the first law to the combustion chamber produces
Similarly,
A
kW 394.4=−⋅=−= K)9.6151073)(KkJ/kg 005.1(kg/s) 8586.0()( 34in TTcmQ p&&
kW 244.5=−⋅=−= K)2933.576)(KkJ/kg 005.1(kg/s) 8586.0()( 16out TTcmQ p&&
preparation. If you are a student using this Manual, you are using it without permission.
9-82
9-111 A Brayton cycle with regeneration is considered. The thermal efficiencies of the cycle for parallel-flow and counter-flow arrangements of the regenerator are to be compared.
Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg⋅K and k = 1.4 (Table A-2a).
Analysis According to the isentropic process expressions for an ideal gas,
K 9.510K)(7) 293( 0.4/1.4/)1(12 === − kk
prTT
s
T
1
2 5
4qin1000 K
293 K
3
6
qout
K 5.5737
K) 1000(45 =⎟⎠
⎜⎝
=⎟⎟⎠
⎜⎜⎝
=pr
TT 11 0.4/1.4/)1(⎞⎛⎞⎛
− kk
hen the first law is applied to the heat exchanger as originally Warranged, the result is
TTTT −=− 3 652
while the regenerator temperature specification gives
K 5.56765.573653 =−=−= TT
The simultaneous solution of these two results gives
2356
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
K 9.5169.5105.5675.573 =+T
n
+−= TTT −=
The thermal efficiency of the cycle is the
0.482=−
−−=
−−
−=−=5.5671000
293516.9111 34in
16th TT
TTqq
η
cle,
An energy balance on the heat exchanger gives
TTT −=−
6 =TT
The thermal efficiency of the cycle is then
out
For the rearranged version of this cy
663 −= TT
s
T
1
2 5
4qin1000 K
293 K
3 6
qout
T6523
The solution of these two equations is
K 2.5393 =
K 2.545
0.453=−
−−=
−−
−=−=2.5391000
293545.211134
16
in
outth TT
TTqq
η
preparation. If you are a student using this Manual, you are using it without permission.
9-83
9-112E An ideal Brayton cycle with regeneration has a pressure ratio of 11. The thermal efficiency of the cycle is to be determined with and without regenerator cases.
Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible.
Properties The properties of air at room temperature are cp = 0.24 Btu/lbm⋅R and k = 1.4 (Table A-2Ea).
Analysis According to the isentropic process expressions for an ideal gas,
R 1111R)(11) 560( 0.4/1.4/)1(12 === − kk
prTT
s
T
1
2 5
4qin2400 R
560 R
3
6 qout
R 121011
R) 2400(45 =⎟⎠
⎜⎝
=⎟⎟⎠
⎜⎜⎝
=pr
TT 11 0.4/1.4/)1(⎞⎛⎞⎛
− kk
he regenerator is ideal (i. , the effectiveness is 100%) and thus,
== TT
T e.
R 1111
53
== TT
R 1210
26
The thermal efficiency of the cycle is then
53.7%==−−
−=−−
−=−= 537.0121024005601111111
34
16
in
outth TT
TTqq
η
The solution without a regenerator is as follows:
s
T
1
2
4
3 qin
qout
2400 R
560 R
R 1111R)(11) 560( 0.4/1.4/)1(12 === − kk
prTT
R 1210111R) 2400(1 0.4/1.4/)1(
34 =⎟⎠⎞
⎜⎝⎛=⎟
⎟⎠
⎞⎜⎜⎝
⎛=
− kk
prTT
49.6%==−−
−=−−
−=−= 496.0111124005601210111
23
14
in
outth TT
TTqq
η
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
9-84
9-113E A car is powered by a gas turbine with a pressure ratio of 4. The thermal efficiency of the car and the mass flow rate of air for a net power output of 95 hp are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with variable specific heats. 3 The ambient air is 540 R and 14.5 psia. 4 The effectiveness of the regenerator is 0.9, and the isentropic efficiencies for both the compressor and the turbine are 80%. 5 The combustion gases can be treated as air.
Properties The properties of air at the compressor and turbine inlet temperatures can be obtained from Table A-17E.
Analysis The gas turbine cycle with regeneration can be analyzed as follows:
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
( )( )
( ) 37253.5712.23014
434
=⎯→⎯=⎟⎞
⎜⎛== srr hP
PP Btu/lbm .2
4
12.230Btu/lbm 549.35
R 2160
Btu/lbm 0.192544.5386.14
386.1Btu/lbm 129.06
R 054
3
33
21
2
11
3
12
1
⎠⎝
==
⎯→⎯=
=⎯→⎯===
==
⎯→⎯=
r
srr
r
P
Ph
T
hPPP
P
Ph
T
and
s
T
1
2s 4s
3qin2160 R
540 R
5
4
2
Btu/lbm 63.407 2.37235.549
35.549 443
212
−− hhh0.80
Btu/lbm 74.207 06.129
06.1290.1920.80
443
turb
212
comp
=→−
=→−
=
=→−
−=→
−−
=
hhh
hhhh
hh
s
s
η
η
Then the thermal efficiency of the gas turbine cycle becomes
Btu/lbm 179.9)74.20763.407(9.0)( 24regen =−=−= hhq ε
Btu/lbm 63.0)06.12974.207()63.40735.549()()(
Btu/lbm 161.7=9.179)74.20735.549()(
1243inC,outT,outnet,
regen23in
=−−−=−−−=−=
−−=−−=
hhhhwww
qhhq
0.39Btu/lbm 161.7Btu/lbm 63.0
in
outnet,th %39====
qw
η
Finally, the mass flow rate of air through the turbine becomes
lbm/s 1.07=⎟⎟⎠
⎞⎜⎜⎝
⎛==
hp 1Btu/s 7068.0
Btu/lbm 63.0hp 95
net
netair w
Wm
&&
preparation. If you are a student using this Manual, you are using it without permission.
9-85
9-114 The thermal efficiency and power output of an actual gas turbine are given. The isentropic efficiency of the
1 Air is an ideal gas with variable specific heats. 2 Kinetic and potential energy changes are negligible. 3 The same, and the properties of combustion gases are the same as
Analysis The properties at various states are
turbine and of the compressor, and the thermal efficiency of the gas turbine modified with a regenerator are to be determined.
Assumptionsmass flow rates of air and of the combustion gases are the those of air.
Properties The properties of air are given in Table A-17.
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
( )( )
( ) 47.485.7127.14
14
3
434
=⎯→⎯=⎟⎠⎞
⎜⎝⎛== srr hP
PP
P kJ/kg 23.825
5.712kJ/kg 1710.0
K 1561C1288
kJ/kg 25.65310.214356.17.14
4356.1kJ/kg .21303
K 303=C30
33
21
2
11
3
12
1
==
⎯→⎯=°=
=⎯→⎯===
==
⎯→⎯°=
r
srr
r
Ph
T
hPPP
P
Ph
T
ass are
s
T
1
2s 4s
3qin1561 K
303 K
5
4
2
The net work output and the heat input per unit m
C7.658K 7.931 kJ/kg 55.96821.30334.665 kJ/kg 34.66566.3720.1038netinout =−=−= wqq
kJ/kg 0.67210381710
kJ/kg 0.10380.359
kJ/kg 372.66
kJ/kg 66.372h 1
s 3600kg/h 1,536,000kW 159,000
41out414t
3223in
th
netin
netnet
°==→=+=+=→−=
=−=−=→−=
===
=⎟⎠⎞
⎜⎝⎛==
Thqhhhq
qhhhhq
wq
mW
w
in
η
&
&
ou
Then the compressor and turbine efficiencies become
94.9%
83.8%
==−
−=
−−
=
==−−
=−−
=
949.021.303672
21.30325.653
838.023.825171055.9681710
12
12
43
43
hhhh
hhhh
sC
sT
η
η
When a regenerator is added, the new heat input and the thermal efficiency become
44.1%====
−=−=
−=
441.0kJ/kg 845.2kJ/kg 372.66
kJ/kg 845.2=8.1921038
kJ/kg 192.8=672.0)-.55(0.65)(968=)(
newin,
netnewth,
regeninnewin,
24regen
qw
qqq
hhq
η
ε
Discussion Note a 65% efficient regenerator would increase the thermal efficiency of this gas turbine from 35.9% to 44.1%.
preparation. If you are a student using this Manual, you are using it without permission.
9-86
9-115 Problem 9-114 is reconsidered. A solution that allows different isentropic efficiencies for the compressor and turbine is to be developed and the effect of the isentropic efficiencies on net work done and the heat supplied to the cycle is to be studied. Also, the T-s diagram for the cycle is to be plotted.
Analysis Using EES, the problem is solved as follows:
"Input data" T[3] = 1288 [C] Pratio = 14.7 T[1] = 30 [C] P[1]= 100 [kPa] {T[4]=659 [C]} {W_dot_net=159 [MW] }"We omit the information about the cycle net work" m_dot = 1536000 [kg/h]*Convert(kg/h,kg/s) {Eta_th_noreg=0.359} "We omit the information about the cycle efficiency." Eta_reg = 0.65 Eta_c = 0.84 "Compressor isentropic efficiency" Eta_t = 0.95 "Turbien isentropic efficiency" "Isentropic Compressor anaysis" s[1]=ENTROPY(Air,T=T[1],P=P[1]) s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor" P[2] = Pratio*P[1] s_s[2]=ENTROPY(Air,T=T_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exit" Eta_c = W_dot_compisen/W_dot_comp "compressor adiabatic efficiency, W_dot_comp > W_dot_compisen" "Conservation of energy for the compressor for the isentropic case: E_dot_in - E_dot_out = DELTAE_dot=0 for steady-flow" m_dot*h[1] + W_dot_compisen = m_dot*h_s[2] h[1]=ENTHALPY(Air,T=T[1]) h_s[2]=ENTHALPY(Air,T=T_s[2]) "Actual compressor analysis:" m_dot*h[1] + W_dot_comp = m_dot*h[2] h[2]=ENTHALPY(Air,T=T[2]) s[2]=ENTROPY(Air,T=T[2], P=P[2]) "External heat exchanger analysis" "SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0 E_dot_in - E_dot_out =DELTAE_dot_cv =0 for steady flow" m_dot*h[2] + Q_dot_in_noreg = m_dot*h[3] q_in_noreg=Q_dot_in_noreg/m_dot h[3]=ENTHALPY(Air,T=T[3]) P[3]=P[2]"process 2-3 is SSSF constant pressure" "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[4] = P[3] /Pratio s_s[4]=ENTROPY(Air,T=T_s[4],P=P[4])"T_s[4] is the isentropic value of T[4] at turbine exit" Eta_t = W_dot_turb /W_dot_turbisen "turbine adiabatic efficiency, W_dot_turbisen > W_dot_turb" "SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0 E_dot_in -E_dot_out = DELTAE_dot_cv = 0 for steady-flow" m_dot*h[3] = W_dot_turbisen + m_dot*h_s[4] h_s[4]=ENTHALPY(Air,T=T_s[4])
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
9-87
"Actual Turbine analysis:" m_dot*h[3] = W_dot_turb + m_dot*h[4] h[4]=ENTHALPY(Air,T=T[4]) s[4]=ENTROPY(Air,T=T[4], P=P[4]) "Cycle analysis" "Using the definition of the net cycle work and 1 MW = 1000 kW:" W_dot_net*1000=W_dot_turb-W_dot_comp "kJ/s" Eta_th_noreg=W_dot_net*1000/Q_dot_in_noreg"Cycle thermal efficiency" Bwr=W_dot_comp/W_dot_turb"Back work ratio" "With the regenerator the heat added in the external heat exchanger is" m_dot*h[5] + Q_dot_in_withreg = m_dot*h[3] q_in_withreg=Q_dot_in_withreg/m_dot h[5]=ENTHALPY(Air, T=T[5]) s[5]=ENTROPY(Air,T=T[5], P=P[5]) P[5]=P[2] "The regenerator effectiveness gives h[5] and thus T[5] as:" Eta_reg = (h[5]-h[2])/(h[4]-h[2]) "Energy balance on regenerator gives h[6] and thus T[6] as:" m_dot*h[2] + m_dot*h[4]=m_dot*h[5] + m_dot*h[6] h[6]=ENTHALPY(Air, T=T[6]) s[6]=ENTROPY(Air,T=T[6], P=P[6]) P[6]=P[4] "Cycle thermal efficiency with regenerator" Eta_th_withreg=W_dot_net*1000/Q_dot_in_withreg "The following data is used to complete the Array Table for plotting purposes." s_s[1]=s[1] T_s[1]=T[1] s_s[3]=s[3] T_s[3]=T[3] s_s[5]=ENTROPY(Air,T=T[5],P=P[5]) T_s[5]=T[5] s_s[6]=s[6] T_s[6]=T[6]
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
9-88
ηt ηc ηth,noreg ηth,withreg Qinnoreg
[kW] Qinwithreg
[kW] Wnet [kW]
0.7 0.75 0.8
0.85 0.9
0.95 1
0.84 0.84 0.84 0.84 0.84 0.84 0.84
0.2044 0.2491 0.2939 0.3386 0.3833 0.4281 0.4728
0.27 0.3169 0.3605 0.4011 0.4389 0.4744 0.5076
422152 422152 422152 422152 422152 422152 422152
319582 331856 344129 356403 368676 380950 393223
86.3 105.2 124.1 142.9 161.8 180.7 199.6
4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0 8.5-100
100
300
500
700
900
1100
1300
1500
1700
s [kJ/kg-K]
T [C
] 100 kPa
1470 kPa
T-s Diagram for Gas Turbine with Regeneration
1
2s2
5
3
4s
4
6
0.7 0.75 0.8 0.85 0.9 0.95 180
100
120
140
160
180
200
η t
Wne
t [k
W]
0.7 0.75 0.8 0.85 0.9 0.95 1310000
330000
350000
370000
390000
410000
430000
η t
Qin
[kW
]
no regeneration
with regeneration
0.7 0.75 0.8 0.85 0.9 0.95 10.2
0.25
0.3
0.35
0.4
0.45
0.5
0.55
ηt
ηth
no regeneration
with regeneration
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
9-89
9-116 A Brayton cycle with regeneration using air as the working fluid is considered. The air temperature at the turbine exit, the net work output, and the thermal efficiency are 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.
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
Properties The properties of air are given in Table A-17.
Analysis (a) The properties of air at various states are
( )( )
( ) ( ) ( )
( )
( ) ( )( ) kJ/kg 803.14711.801219.250.821219.25
kJ/kg 711.8059.2815.20071
15.200kJ/kg 1219.25
K 1501
kJ/kg 618.260.75/310.24541.2610.243/
kJ/kg 541.2688.105546.17
5546.1kJ/kg 310.24
K 310
433443
43
43
4
33T
121212
12
21
1
34
3
12
1
=−−=−−=⎯→⎯−−
=
=⎯→⎯=⎟⎠⎞
⎜⎝⎛==
==
⎯→⎯=
=−+=−+=⎯→⎯−−
=
=⎯→⎯=
==
⎯→⎯=
sTs
T
srr
r
Css
C
srr
r
hhhhhhhh
hPPP
P
Ph
hhhhhhhh
hP
Ph
ηη
ηη
Thus,
T4 = 782.8 K
(b)
)
s
T
1
2s4s
3qin1150 K
310 K
5
6
42
1T
2 == PP
P
( ) ( )( ) (
kJ/kg 108.09=−−−=
−−−=−=
24.31026.61814.80325.12191243inC,outT,net hhhhwww
( )( )(
kJ/kg 738.43618.26803.140.6526.618
242524
25
=−+=
−+=⎯→⎯−−
= hhhhhhhh
εε (c) )
Then,
22.5%===
=−=−=
kJ/kg 480.82kJ/kg 108.09
kJ/kg 480.82738.4319.2512
in
netth
53in
qw
hhq
η
preparation. If you are a student using this Manual, you are using it without permission.
9-90
9-117 A stationary gas-turbine power plant operating on an ideal regenerative Brayton cycle with air as the working fluid is considered. The power delivered by this plant is to be determined for two cases.
Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas. 3 Kinetic and potential energy changes are negligible.
Properties When assuming constant specific heats, the properties of air at room temperature are cp = 1.005 kJ/kg.K and k = 1.4 (Table A-2a). When assuming variable specific heats, the properties of air are obtained from Table A-17.
Analysis (a) Assuming constant specific heats,
( )( )( )
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
( )( )
( )( )
( )( ) kW 39,188===
=−
−−=
−−
−=−
−−=−=
====⎯→⎯=
=⎟⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
=⎞⎛
−
−
kW 75,0000.5225
5225.02.6071100
2903.5251111
K 525.3andK 607.2%100
K 607.281K 1100
K 525.3
innet
53
16
53
16
in
outth
2645
0.4/1.4/1
3
434
0.4/1.4/1
QW
TTTT
TTcTTc
TTTT
PP
TT
P
T
p
p
kk
kk
&& η
η
ε
) Assuming variable specific heats,
=⎟⎟⎠
⎜⎜⎝
= 8K 2901
212 P
TT
(b
( )( )
( )
( )( ) kW 40,283===
=−−
−=−−
−=−=
====⎯→⎯=
=⎯→⎯=⎟⎠⎞
⎜⎝⎛==
==
⎯→⎯=
=⎯→⎯===
==
⎯→⎯=
kW 75,0000.5371
5371.037.65107.116116.29012.526111
kJ/kg 526.12andkJ/kg 651.37%100
kJ/kg 651.3789.201.16781
1.167kJ/kg 1161.07
K1100
kJ/kg 526.128488.92311.18
2311.1kJ/kg 90.162
K029
innet
53
16
in
outth
2645
43
4
33
21
2
11
34
3
12
1
QW
hhhh
hhhh
hPPP
P
Ph
T
hPPP
P
Ph
T
T
rr
r
rr
r
&& η
η
ε
s
T
1
2 4
375,000 kW1100 K
290 K
5
qout
6
preparation. If you are a student using this Manual, you are using it without permission.
9-91
9-118 A regenerative gas-turbine engine using air as the working fluid is considered. The amount of heat transfer in the regenerator and the thermal efficiency are 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 (a) The properties at various states are
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
( )
( )( )(
( ) ( )( ) kJ/kg 192.7=−=−==
−−=−−=⎯→⎯
−−
=
=⎯→⎯=⎟⎠⎞
⎜⎝⎛==
==⎯→⎯==⎯→⎯=
=⎯→⎯=
===
84.65974.90080.0kJ/kg .74900
44.83242.151590.042.1515
kJ/kg .4483206.505.45091
.5450kJ/kg 1515.42K 0014
kJ/kg .84659K 065kJ/kg 10.243K 310
9100/900/
24regen
433443
43
43
4
33
22
11
12
34
3
hhq
hhhhhhhh
hPPP
P
PhThThT
PPr
sTs
T
srr
r
)
ε
ηη
(b)
p
( ) ( )( ) ( )( ) ( )
40.0%====
=−−=−−=
=−−−=−−−=−=
400.0kJ/kg 662.88kJ/kg 265.08
kJ/kg .886627.192.846591515.42
kJ/kg .08265.24310659.8474.90042.1515
in
netth
regen23in
1243inC,outT,net
qw
qhhq
hhhhwww
η
s
T
1
2s4s
3 qin1400 K
310 K
5
6
4650 K 2
preparation. If you are a student using this Manual, you are using it without permission.
9-92
9-119 A regenerative gas-turbine engine using air as the working fluid is considered. The amount of heat transfer in the regenerator and the thermal efficiency are to be determined.
Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats. 3 Kinetic and potential energy changes are negligible.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K and k = 1.4 (Table A-2a).
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
Analysis (a) Using the isentropic relations and turbine efficiency,
( )( )
( )( ) ( )
( )( )
( ) ( ) ( )( )( ) kJ/kg 130.7=−⋅=−=−==
−−=−−=⎯→⎯
−
−=
−−
=
=⎟⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
===
−
K065812.6KkJ/kg 1.0050.80K .6812
3.747140090.01400
K .374791K 1400
9100/900/
2424regen
433443
43
43
43
4.1/4.0/1
3
434
12
TTchhq
TTTTTTcTTc
hhhh
PP
TT
PPr
p
sTsp
p
sT
kk
s
p
εε
ηη s
T
1
2s 4s
3qin1400 K
310 K
5
6
4650 K 2
( ) ( )( ) ( ) ( )[ ]( ) ( )( )( )
39.9%====
=−−⋅=
−−=−−=
=−−−⋅=
−−−=−=
399.0kJ/kg 623.1kJ/kg 248.7
kJ/kg .1623.7130K0651400KkJ/kg 1.005
kJ/kg .7248K310650.68121400KkJ/kg 1.005
in
netth
regen23regen23in
1243inC,outT,net
qw
qTTcqhhq
TTcTTcwww
p
pp
η
(b)
preparation. If you are a student using this Manual, you are using it without permission.
9-93
9-120 A regenerative gas-turbine engine using air as the working fluid is considered. The amount of heat transfer in the regenerator and the thermal efficiency are 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.
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
Analysis (a) The properties at various states are
( )
( )( )(
( ) ( )( ) kJ/kg 168.6=−=−==
−−=−−=⎯→⎯
−−
=
=⎯→⎯=⎟⎠⎞
⎜⎝⎛==
==⎯→⎯==⎯→⎯=
=⎯→⎯=
===
84.65974.90070.0kJ/kg .74900
44.83242.151590.042.1515
kJ/kg .4483206.505.45091
.5450kJ/kg 1515.42K 0014
kJ/kg .84659K 065kJ/kg 10.243K 310
9100/900/
24regen
433443
43
43
4
33
22
11
12
34
3
hhq
hhhhhhhh
hPPP
P
PhThThT
PPr
sTs
T
srr
r
)
ε
ηη
(b)
p
s
T
1
2s4s
3 qin1400 K
310 K
5
6
4650 K 2
( ) ( )( ) ( )( ) ( )
38.6%====
=−−=−−=
=−−−=−−−=−=
386.0kJ/kg 687.18kJ/kg 265.08
kJ/kg 18.6876.168.846591515.42
kJ/kg .08265.24310659.8474.90042.1515
in
netth
regen23in
1243inC,outT,net
qw
qhhq
hhhhwww
η
preparation. If you are a student using this Manual, you are using it without permission.
9-94
9-121 An expression for the thermal efficiency of an ideal Brayton cycle with an ideal regenerator is to be developed.
Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible.
Analysis The expressions for the isentropic compression and expansion processes are
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
−= 2kk
prTT /)1(1
s
T
1
2 4
3qin
5
6qout
kk
prTT
/)1(
341
−
⎟⎟⎠
⎞⎜⎜⎝
⎛=
For an ideal regenerator,
45
TT =
26
TT =
The thermal efficiency of the cycle is
kkp
kkp
kkp
rTT
r
rTT
TTTT
TT
TTTT
TT
TTTT
/)1(
3
1
/)1(
/)1(
3
1
34
12
3
1
35
16
3
1
53
16
in
outth
1
1
11
)/(11)/(
1
)/(11)/(
111
−
−−
−
−=
−
−−=
−−
−=
−−
−=−−
−=−=η
preparation. If you are a student using this Manual, you are using it without permission.
9-95
Brayton Cycle with Intercooling, Reheating, and Regeneration
9-122C As the number of compression and expansion stages are increased and regeneration is employed, the ideal Brayton cycle will approach the Ericsson cycle.
9-123C 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.
9-124C (a) decrease, (b) decrease, and (c) decrease.
9-125C (a) increase, (b) decrease, and (c) decrease.
9-126C (a) increase, (b) decrease, (c) decrease, and (d) increase.
9-127C (a) increase, (b) decrease, (c) increase, and (d) decrease.
9-128C (c) The Carnot (or Ericsson) cycle efficiency.
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
9-96
9-129 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.
s
T
3
4
1
5qin1200 K
300 K
86
7
10
9
2
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
5
⎞⎛P
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
12
1
=−=−=
=−=−=
==⎯→⎯=⎟⎠
⎜⎝
==
===
⎯→⎯=
==⎯→⎯===
==
⎯→⎯=
hhw
hhw
hhPP
P
Phh
T
hhPPP
P
Ph
T
rr
r
rr
r
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 MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
9-97
9-130 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,
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
( )( )
( )( ) ( )
( )
( )( )( )
( ) ( )( ) ( ) kJ/kg 3.325813.98679.127722
kJ/kg .4626419.30042.43222
kJ/kg .1398636.94679.127788.079.1277
kJ/kg 946.3633.7923831
238kJ/kg 1277.79K 1200
kJ/kg .42432
/
kJ/kg 411.26158.4386.13
386.1kJ/kg 300.19K 300
65outT,
12inC,
6558665
65
865
6
755
1214212
421
2
11
56
5
12
1
=−=−=
=−=−=
=−−=
−−==⎯→⎯−−
=
==⎯→⎯=⎟⎠⎞
⎜⎝⎛==
===⎯→⎯=
=
−+==⎯→⎯−−
=
=⎯→⎯===
==⎯→⎯=
hhw
hhw
hhhhhhhhh
hhPPP
P
PhhT
hhhhhhhhh
hhPPP
P
PhT
sTs
T
ssrr
r
Css
C
ssrr
r
ηη
ηη
Thus,
s
T
3
4
1
5qin
86
7
1
9
2
86s
84.0/19.30026.41119.30012 −+=
=
45.3%==== 453.0kJ/kg 583.32kJ/kg 264.46
outT,
inC,bw w
wr
4 2s
( ) ( ) ( ) ( )
28.0%====
=−=−=
=−+−=−+−=
280.0kJ/kg 1137.03kJ/kg 318.86
kJ/kg .8631846.26432.583
kJ/kg 1137.03986.131277.79432.421277.79
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
( ) ( )( )
44.2%====
=−=−=
=−=−=
442.0kJ/kg 721.75kJ/kg 318.86
kJ/kg .7572128.41503.1137
kJ/kg 15.28442.43213.98675.0
in
netth
regenoldin,in
48regen
qw
qqq
hhq
η
ε
preparation. If you are a student using this Manual, you are using it without permission.
9-98
9-131E An ideal regenerative gas-turbine cycle with two stages of compression and two stages of expansion is considered. The power produced and consumed by each compression and expansion stage, and the rate of heat rejected are to be determined.
Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible.
Properties The properties of air at room temperature are cp = 0.24 Btu/lbm⋅R and k = 1.4 (Table A-2Ea).
Analysis The pressure ratio for each stage is
464.312 ==pr
s
T
3
4
1
6
520 R
97
8
2
5
1
According to the isentropic process expressions for an ideal gas,
R 6.741R)(3.464) 520( 0.4/1.4/)1(142 ==== − kk
prTTT
Since this is an ideal cycle,
6.741504975
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
R 6.79150 =+
For the isentropic expansion processes,
==== − kkprTTT
he heat i is
=+=== TTTT
R 1129R)(3.464) 6.791( 0.4/1.4/)1(786
T nput
(2 56in Btu/lbm 162.0R )6.791R)(1129Btu/lbm (0.242) =−⋅=
The mass flow rate is then
−= TTcq p
lbm/s 3.086Btu/lbm 0.162inqBtu/s 500in ===
Qm
&&
pplication of the first law to the expansion process 6-7 gives
A
kW 263.6=
⎟⎠⎞
⎜⎝⎛−⋅=
−= )( 76out7,-6 TTcmW p&&
Btu/s 94782.0kW 1R )6.791R)(1129Btu/lbm 4lbm/s)(0.2 086.3(
The same amount of power is produced in process 8-9. When the first law is adapted to the compression process 1-2 it becomes
−= )( 12in12, TTcmW p&&
kW 173.2=
⎟⎠⎞
⎜⎝⎛−⋅=
Btu/s 94782.0kW 1R )520R)(741.6Btu/lbm 4lbm/s)(0.2 086.3(
Compression process 3-4 uses the same amount of power. The rate of heat rejection from the cycle is
Btu/s 328.3=
−⋅=
−=
R )520R)(741.6Btu/lbm 4lbm/s)(0.2 086.3(2
)(2 32out TTcmQ p&&
preparation. If you are a student using this Manual, you are using it without permission.
9-99
9-132E An ideal regenerative gas-turbine cycle with two stages of compression and two stages of expansion is considered. The power produced and consumed by each compression and expansion stage, and the rate of heat rejected are to be determined.
Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible.
Properties The properties of air at room temperature are cp = 0.24 Btu/lbm⋅R and k = 1.4 (Table A-2Ea).
Analysis The pressure ratio for each stage is
464.312 ==pr
s
T
3
4s
1
6
520 R
97
8
2
5
14 2s
7s 9s
For the compression processes,
2s R 6.741R)(3.464) 520( 0.4/1.4/)1(14 ==== − kk
ps rTTT
R 7.78085.0
5206.741520
)()( 12
14212
12
=−
+=
−+==⎯→⎯
−
−=
C
s
p
spC
TTTTT
TTcTTc
ηη
Since the regenerator is ideal,
R 7.830507.780504975 =+=+=== TTTT
For the expansion processes,
R 1185R)(3.464) 7.830( 0.4/1.4/)1(786 ==== − kk
pss rTTT
R 1150)7.8301185)(90.0(7.830)()()(
7678676
76 =−+=−+==⎯→⎯−
−=
c pTη TTTTT
TTcTT
sTsp
η
The heat input is
Btu/lbm 153.3R )7.830R)(1150Btu/lbm (0.242)(2 56
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
in =−⋅=TTcq p
−=
The mass flow rate is then
lbm/s 3.262Btu/lbm 3.153Btu/s 500
in
in ===qQ
m&
&
Application of the first law to the expansion process 6-7 gives
−= )( 76out7,-6 TTcmW p&&
kW 263.7=⎟
⎠⎞
⎜⎝⎛−⋅=
Btu/s 94782.0kW 1R )7.830R)(1150Btu/lbm 4lbm/s)(0.2 262.3(
t becomes
The same amount of power is produced in process 8-9. When the first law is adapted to the compression process 1-2 i
kW 215.3=⎟⎠⎞
⎜⎝⎛−⋅=
−=
Btu/s 94782.0kW 1R )520R)(780.7Btu/lbm 4lbm/s)(0.2 262.3(
)( 12in12, TTcmW p&&
Compression process 3-4 uses the same amount of power. The rate of heat rejection from the cycle is
Btu/s 408.2=−⋅=
−=
R )520R)(780.7Btu/lbm 4lbm/s)(0.2 262.3(2
)(2 32out TTcmQ p&&
preparation. If you are a student using this Manual, you are using it without permission.
9-100
9-133 A regenerative gas-turbine cycle with two stages of compression and two stages of expansion is considered. The thermal efficiency of the cycle is to be determined.
Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg⋅K and k = 1.4 (Table A-2a).
Analysis The temperatures at various states are obtained as follows
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
==== − kkprTTT
2 K 9.430K)(4) 290( 0.4/1.4/)1(14
K 9.450209.4302045 =+=+= TT
s
T
3
4
1
6
290 K
97
8
2 5
10 K 4.749
KkJ/kg 1.005kJ/kg 300
K 9.450
)(
in56
56in
=⋅
+=+=
−=
p
p
cq
TT
TTcq
K 3.50441K) 4.749(1 0.4/1.4/)1(
67 =⎟⎠⎞
⎜⎝⎛=⎟
⎟⎠
⎞⎜⎜⎝
⎛=
− kk
prTT
K 8.802KkJ/kg 1.005
kJ/kg 300K 3.504in
78 =⋅
+=+=pc
qTT
K 2.54041K) 8.802(1 0.4/1.4/)1(
89 =⎟⎠⎞
⎜⎝⎛=⎟
⎟⎠
⎞⎜⎜⎝
=− kk
prTT
9 =−=
e heat rejected is
The thermal efficiency of the cycle is then
⎛
20−= TT K 2.520202.54010
The heat input is
kJ/kg 600300300in =+=q
Th
kJ/kg 373.0R )2909.430290K)(520.2kJ/kg (1.005
)()( 32110out
=−+−⋅=
−+−= TTcTTcq pp
0.378=−=−=600
0.37311in
outth q
qη
preparation. If you are a student using this Manual, you are using it without permission.
9-101
9-134 A regenerative gas-turbine cycle with three stages of compression and three stages of expansion is considered. The thermal efficiency of the cycle is to be determined.
Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg⋅K and k = 1.4 (Table A-2a).
Analysis The temperatures at various states are obtained as follows
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
===== − kkprTTTT
2 K 9.430K)(4) 290( 0.4/1.4/)1(164
s
T
3
4
1290 K
9
8
2
5
1
11
67
12
13
14
K 9.450209.4302067 =+=+= TT
K 4.749KkJ/kg 1.005
kJ/kg 300K 9.450
)(
in78
78in
=⋅
+=+=
−=
p
p
cq
TT
TTcq
K 3.50441K) 4.749(1 0.4/1.4/)1(
89 =⎟⎠⎞
⎜⎝⎛=⎟
⎟⎠
⎞⎜⎜⎝
⎛=
− kk
prTT
K 8.802KkJ/kg 1.005
kJ/kg 300K 3.504in
910 =⋅
+=+=pc
qTT
K 2.54041K) 8.802(1 0.4/1.4/)1(
1011 =⎟⎠⎞
⎜⎝⎛=⎟
⎟⎠
⎞⎜⎜⎝
⎛=
− kk
prTT
K 7.838KkJ/kg 1.005
kJ/kg 300K 2.540in =
⋅+=+
pcq
1112 = TT
K 4.56441K) 7.838(1 0.4/1.4/)1(
1213 ⎜⎜=
rTT =⎟
⎠⎞
⎜⎝⎛=⎟
⎟⎠
⎞
⎝
⎛− kk
p
The heat input is
he heat rejected is
The thermal efficiency of the cycle is then
K 4.544204.564201314 =−=−= TT
kJ/kg 900300300300in =++=q
T
kJ/kg .9538R )2909.4302909.430290K)(544.4kJ/kg (1.005
)()()( 5432114out
=−+−+−⋅=
−+−+−= TTcTTcTTcq ppp
40.1%==−=−= 0.401900
9.53811in
outth q
qη
preparation. If you are a student using this Manual, you are using it without permission.
9-102
9-135 A regenerative gas-turbine cycle with three stages of compression and three stages of expansion is considered. The thermal efficiency of the cycle is to be determined.
Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg⋅K and k = 1.4 (Table A-2a).
Analysis Since all compressors share the same compression ratio and begin at the same temperature,
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
===== − kkprTTTT
2 K 9.430K)(4) 290( 0.4/1.4/)1(164
s
T
3
4
1290 K
9
8
2
5
1
11
67
12
13
14
From the problem statement,
65137 −= TT
The relations for heat input and expansion processes are
pp c
qTTTTcq in
7878in )( +=⎯→⎯−=
kk
prTT
/)1(
891
−
⎟⎟⎠
⎞⎜⎜⎝
⎛=
pcq
TT in910 += ,
kk
prTT
/)1(
10111
−
⎟⎟⎠
⎞⎜⎜⎝
⎛=
kk
ppcTT in
1112 += , q
rTT 1213
1⎟⎟
⎜⎜=
/)1( −
⎠
⎞
⎝
⎛
ing EES software gives the following results
ance on the regenerator,
61414136
141367
=+=+=⎯→⎯−=
−=−
TTTTT
TTT
The heat input is
The heat rejected is
54
=−+−+−⋅=
− TTc p
The thermal efficiency of the cycle is then
The simultaneous solution of above equations us
K 7.585 K, 4.870 K, 9.571 K, 8.849K 3.551 K, 2.819 K, 7.520
13121110
987
=======
TTTTTTT
From an energy bal
T
)65( 13 −−T K 9.495659.43065
kJ/kg 900300300300in =++=q
kJ/kg 1.490R )2909.4302909.430290K)(495.9kJ/kg (1.005
)()()( 32114out +−+−= TTcTTcq pp
45.5%==−=−= 455.0900
1.49011in
outth q
qη
preparation. If you are a student using this Manual, you are using it without permission.
9-103
Jet-Propulsion Cycles
9-136C The power developed from the thrust of the engine is called the propulsive power. It is equal to thrust times the aircraft velocity.
9-137C 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.
9-138C It reduces the exit velocity, and thus the thrust.
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
9-104
9-139E A turboprop engine operating on an ideal cycle is considered. The thrust force generated is 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 The turbine work output is equal to the compressor work input.
Properties The properties of air at room temperature are R = 0.3704 psia⋅ft3/lbm⋅R (Table A-1E), cp = 0.24 Btu/lbm⋅R and k = 1.4 (Table A-2Ea).
Analysis Working across the two isentropic processes of the cycle yields
s
T
1
2 4
3qin
5qout
R 8.868R)(10) 450( 0.4/1.4/)1(12 === − kk
prTT
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
R 1.72510
R) 1400(35 =⎟⎠
⎜⎝
=⎟⎟⎠
⎜⎜⎝
=pr
TT 11 0.4/1.4/)1(⎞⎛⎞⎛
− kk
ince the work produced by expansion 3-4 equals that used b pression 1-2, an energy balance gives
S ycom
R 2.981)4508.868(1400)(4 123 =−−=−−= TTTT
The excess enthalpy generated by expansion 4-5 is used to increase the kinetic energy of the flow through the propeller,
2
)(22
inletexit54
VVmTTcm ppe
−=− &&
which when solved for the velocity at which the air leaves the propeller gives
ft/s 9.716
)ft/s 600(Btu/lbm 1
/sft 25,037R)1.7252.981)(RBtu/lbm 24.0(
2012
⎢⎢ ⋅=
)(2
2/12
22
2/12
inlet54exit
=
⎥⎥⎦
⎤
⎣
⎡+⎟
⎟⎠
⎞⎜⎜⎝
⎛−
⎥⎥⎦
⎤
⎢⎢⎣
⎡+−= VTTc
mm
V pp
e
&
&
The mass flow rate through the propeller is
lbm/s 2261/lbmft 84.20
ft/s 6004
ft) 10(4
/lbmft 84.20psia 8
R) 450()ftpsia 3704.0(
3
2
1
12
1
1
33
1
11
====
=⋅
==
ππvv
v
VDAVm
PRT
p&
The thrust force generated by this propeller is then
lbf 8215=⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=−=
2inletexitft/slbm 32.174
lbf 1600)ft/s.9lbm/s)(716 (2261)( VVmF p&
preparation. If you are a student using this Manual, you are using it without permission.
9-105
9-140E A turboprop engine operating on an ideal cycle is considered. The thrust force generated is 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 The turbine work output is equal to the compressor work input.
Properties The properties of air at room temperature are R = 0.3704 psia⋅ft3/lbm⋅R (Table A-1E), cp = 0.24 Btu/lbm⋅R and k = 1.4 (Table A-2Ea).
Analysis Working across the two isentropic processes of the cycle yields
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
=== − kkprTT
s
T
1
2 4
3qin
5qout
2 R 8.868R)(10) 450( 0.4/1.4/)1(1
R 1.725101R) 1400(1 0.4/1.4/)1(
35 =⎟⎠⎞
⎜⎝⎛=⎟
⎟⎠
⎞⎜⎜⎝
⎛=
− kk
prTT
Since the work produced by expansion 3-4 equals that used by compression 1-2, an energy balance gives
R 2.981)4508.868(1400)( 1234 =−−=−−= TTTT
The mass flow rate through the propeller is
lbm/s 1447
/lbmft 84.20ft/s 600
4ft) 8(
4
/lbmft 84.20psia 8
R) 450()ftpsia 3704.0(
3
2
11
12
1
33
1
====
=⋅
==
ππvv
v
VDAVm
PRT
p&
ccording the previous problem, A to
lbm/s 1.113lbm/s 2261=== p
em
m&
& 2020
he excess enthalpy generated by expansion 4-5 is used to increase the kinetic energy of the flow through the propeller, T
2
)(2
inlet2
exit54
VVmTTcm ppe
−=− &&
which when solved for the velocity at which the air leaves the propeller gives
ft/s 0.775
)ft/s 600(Btu/lbm 1
/sft 25,037R)1.7252.981)(RBtu/lbm 24.0(
lbm/s 1447lbm/s 1.1132
)(2
2/12
22
2/12
inlet54exit
=
⎥⎥⎦
⎤
⎢⎢⎣
⎡+⎟
⎟⎠
⎞⎜⎜⎝
⎛−⋅=
⎥⎥⎦
⎤
⎢⎢⎣
⎡+−= VTTc
mm
V pp
e
&
&
The thrust force generated by this propeller is then
lbf 7870=⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=−=
2inletexitft/slbm 32.174
lbf 1600)ft/slbm/s)(775 (1447)( VVmF p&
preparation. If you are a student using this Manual, you are using it without permission.
9-106
9-141 A turbofan engine operating on an ideal cycle produces 50,000 N of thrust. The air temperature at the fan outlet needed to produce this thrust is 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 The turbine work output is equal to the compressor work input.
Properties The properties of air at room temperature are R = 0.287 kPa⋅m3/kg⋅K, cp = 1.005 kJ/kg⋅K and k = 1.4 (Table A-2a).
Analysis The total mass flow rate is
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
kg/s 1.676
/kgm 452.1m/s 200
4m) 5.2(
4
/kgm 452.1kPa 50
K) 253()mkPa 287.0(
3
2
1
12
1
1
33
1
====
=⋅
==
ππvv
v
VDAVm
PRT
&
s
T
1
2 4
3qin
5
qoutNow,
kg/s 51.8488e
kg/s 1.676===
mm&
The mass flow rate through the fan is
In order to produce the specified thrust force, the velocity at the fan exit will be
&
kg/s 6.59151.841.676 =−=−= ef mmm &&&
)( inletexit −= f VVmF &
m/s 284.5
N 1m/skg 1
kg/s 591.6N 50,000m/s) (200
2
inletexit =⎟⎟⎠
⎞⎜⎜⎝
⎛ ⋅+=+=
fmFVV&
An energy balance on the stream passing through the fan gives
K 232.6=
⎟⎠
⎞⎜⎝
⎛⋅
−−=
−−=
−=−
22
22
2inlet
2exit
45
2inlet
2exit
54
/sm 1000kJ/kg 1
)KkJ/kg 005.1(2)m/s 200()m/s 5.284(K 253
2
2)(
p
p
cVV
TT
VVTTc
preparation. If you are a student using this Manual, you are using it without permission.
9-107
9-142 A pure jet engine operating on an ideal cycle is considered. The velocity at the nozzle exit and the thrust produced 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 The turbine work output is equal to the compressor work input. Properties The properties of air at room temperature are R = 0.287 kPa⋅m3/kg⋅K, 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 = 240 m/s. Ideally, the air will leave the diffuser with a negligible velocity (V 2 ≅ 0).
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
Diffuser:
( )( )
( )( )( )
( ) kPa 88.64K 260K 288.7
kPa 45
K 7.288/sm 1000
kJ/kg 1KkJ/kg 1.0052
m/s 240K 2602
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 &&&&&
qin
qouts
T
1
2
4
3 5
6
Compressor: ( )( ) ( )( )
( )( )( ) K 7.60013K 288.7
kPa 5.843kPa 64.8813
0.4/1.4/1
2
323
243
==⎟⎟
⎠
⎞⎜⎜⎝
⎛=
====
− kk
p
PP
TT
PrPP
Turbine:
( ) ( )54235423outturb,incomp, TTcTTchhhhww pp −=−⎯→⎯−=−⎯→⎯=
K 0.5187.2887.6008302345 =+−=+−= TTTT or Nozzle:
( )( )
( ) 2/02
0
2/2/
K 3.359kPa 843.5
kPa 45K 830
2656
022 VV − 5656
266
25
outin(steady) 0
systemoutin
0.4/1.4/1
4
646
VTTchh
VhVh
EEEEE
PP
TT
p
kk
+−=⎯→⎯+−=
+=+
=⎯→⎯∆=−
=⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
−
&&&&&
or
5
( )( )( ) m/s 564.8=⎟⎟⎠
⎞⎜⎜⎝
⎛−⋅==
kJ/kg 1/s2m 1000
K3.359518.0KkJ/kg 1.00522
6 exitVV
ine is
The mass flow rate through the eng
kg/s 0.291/kgm 658.1
m/s 2404
m) 6.1(4
/kgm 658.1kPa 45
K) 260()mkPa 28.0(==
RT 7
3
2
1
12
1
1
33
1
====
=⋅
ππvv
v
VDAVm
P
&
The thrust force generated is then
N 94,520=⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=−=
2inletexitm/skg 1N 10)m/s248kg/s)(564. (291.0)( VVmF &
preparation. If you are a student using this Manual, you are using it without permission.
9-108
9-143 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).
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
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 &&&&&
s
T
1
2
4
3
Qi
5
6
·
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:
( ) ( )54235423outturb,incomp, TTcTTchhhhww pp −=−⎯→⎯−=−⎯→⎯=
K1098.2291.9593.714002345 =+−=+−= TTTT or
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
+−=⎯→⎯−
+−=
+=+
=⎯→⎯∆=−
=⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
−
&&&&&
( )( )( ) m/s1032 kJ/kg 1
/sm 1000K568.21098.2KkJ/kg 1.0052
22
6 =⎟⎟⎠
⎞⎜⎜⎝
⎛−⋅=V or
( ) ( )( ) ( ) kW 13,670=⎟⎟⎠
⎞⎜⎜⎝
⎛−=−=
22aircraftinletexit/sm 1000
kJ/kg 1m/s 320m/s3201032kg/s 60VVVmW p && (b)
( ) ( ) ( )( )( )
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
&&
&&& (c)
preparation. If you are a student using this Manual, you are using it without permission.
9-109
9-144 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&&
&&&
s
T
1
2
4
3
Qin
5s
6
·
5
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
TTc pTTchhhhww p
( )( )
( ) ( )( )
( ) 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 MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
9-110
Nozzle:
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
( )( )
( ) 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
( )( )( ) ( )
kW 11,390=⎟⎟⎠
⎞⎜⎜⎝
⎛−=
−=
22
aircraftinletexit
/sm 1000kJ/kg 1
m/s 320m/s320913.2kg/s 60
VVVmW p && (b)
(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
&&
&&&
preparation. If you are a student using this Manual, you are using it without permission.
9-111
9-145 A turbojet aircraft that has a pressure rate of 9 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 A-17.
Analysis (a) Using variable specific heats for air,
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
Compressor:
2311.1kJ/kg 0.1629K 029
1
11
==⎯→⎯=
rPhT
( )( )
( )( )
5.568kJ/kg 1611.65.1067.07544
kJ/kg 5.1067kg/s 20
kJ/s 21,350
kJ/s 350,21kJ/kg 42,700kg/s 0.5HV
kJ/kg .0754408.112311.19
3
12
2323in
inin
fuelin
21
2
=⎯→⎯=+=+=⎯→⎯−=
===
==×=
=⎯→⎯===
rin
rr
Pqhhhhq
mQ
q
mQ
hPPP
P
&
&
&&
bine:
or
Nozzle:
s
T
1
3
2
qin
4
5
Tur
kJ/kg 7.1357.16290.07544.616111234
4312outturb,incomp,
=+−=+−=
−=−⎯→⎯=
hhhh
hhhhww
( )
20
2/2/
kJ/kg .5688817.63915.568
024
25
45
255
244
outin
(steady) 0systemoutin
53
535
VVhh
VhVh
EE
EEE
hPP
PP rr
−+−=
+=+
=
∆=−
=⎯→⎯=⎟⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
&&
&&&
or
( ) ( )( ) m/s .6968kJ/kg 1
/sm 1000kJ/kg.568881357.722
22
545 =⎟⎟⎠
⎞⎜⎜⎝
⎛−=−= hhV
( ) ( )( ) N 19,370=⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=−
2inletexit m/skg 1N 1m/s0968.6kg/s 20VVm& Brake force = Thrust =
preparation. If you are a student using this Manual, you are using it without permission.
9-112
9-146 Problem 9-145 is reconsidered. The effect of compressor inlet temperature on the force that must be applied to ed.
nalysis Using EES, the problem is solved as follows:
"[K]"
0.5 [kg/s]
P=P[1])
"SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0"
tant pressure"
_in= m_dot*h[3]"SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0"
SSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0"
W_dot_c"Definition of the net cycle work, kW"
[5], P=P[5]) "T_s[5] is the isentropic value of T[5] at nozzle exit"
g))
the brakes to hold the plane stationary is to be investigat
A
P_ratio =9 T_1 = 7 [C]
273T[1] = T_1+P[1]= 95 [kPa] P[5]=P[1] Vel[1]=0 [m/s] V_dot[1] = 18.1 [m^3/s]
700 [kJ/kg]HV_fuel = 42=m_dot_fuel
Eta_c = 1.0 ta_t = 1.0 E
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],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] "External heat exchanger analysis" P[3]=P[2]"process 2-3 is SSSF consh[3]=ENTHALPY(Air,T=T[3]) Q_dot_in = m_dot_fuel*HV_fuel m_dot*h[2] + Q_dot "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] "ST[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_net = 0 [kW] "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_sh_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/k
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
9-113
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
2/2*convert(m^2/s^2,kJ/kg))
[5]=entropy('air',T=T[5],P=P[5])
l[1]) "[N]"
determined only to produce a T-s plot"
[2]=entropy('air',T=T[2],P=P[2])
Force [kg/s] [K] [C]
m_dot*h[4] = m_dot*(h[5] + Vel[5]^T[5]=TEMPERATURE(Air,h=h[5])s "Brake Force to hold the aircraft:" Thrust = m_dot*(Vel[5] - VeBrakeForce = Thrust "[N]" "The following state points are T[2]=temperature('air',h=h[2]) s
Brake
[N]
m T3 T1
21232 21007 20788 20576 20369 20168 19972 19782 19596 19415 19238 19.77 1510 30
23.68 23.22 22.78 22.35 21.94 21.55 21.17 20.8
20.45 20.1
1284 1307 1330 1352 1375 1398 1420 1443 1466 1488
-20 -15 -10 -5 0 5
10 15 20 25
-20 -10 0 10 20 3019000
19450
19900
20350
20800
21250
T1 [C]
Bra
keFo
rce
[N]
4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0 8.50
500
1000
1500
0x100
5x102
103
103
s [kJ/kg-K]
T [K
]
Air
855 kPa
95 kPa
1
2s
3
4s
5
preparation. If you are a student using this Manual, you are using it without permission.
9-114
9-147 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
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
J / kg
J / kg
T h
T h1 1
2 2
280 28013
700 713 27
= ⎯ →⎯ =
= ⎯ →⎯ =
K k
K k
.
.
( ) ( )⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛−+−=
⎟⎟⎠
⎞⎜⎜⎝
⎛ −+−=
+=++
=
∆=−
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
&&
&&&
&&
&&&
15,000 kJ/s
1
7°C 300 m/s16 kg/s
2
427°C
It gives
V = 1048 m/s
hus,
2
T
( ) ( )( ) N 11,968=−=−= m/s3001048kg/s 1612 VVmFp &
preparation. If you are a student using this Manual, you are using it without permission.
9-115
Second-Law Analysis of Gas Power Cycles
9-148 The process with the highest exergy destruction for an ideal Otto cycle described in Prob. 9-36 is to be determined.
Analysis From Prob. 9-36, qin = 582.5 kJ/kg, qout = 253.6 kJ/kg, T1 = 288 K, T2 = 661.7 K, T3 = 1473 K, and T4 = 641.2 K. The exergy destruction during a process of the cycle is
⎟⎟⎠
⎞⎜⎜⎝
⎛+−∆==
sink
out
source
in0gen0dest T
qT
qsTsTx
v
P
4
1
3
2
s
T
1
2
4
3 qin
qout
Application of this equation for each process of the cycle gives
0=2-dest,1 (isentropicx process)
KkJ/kg 5746.00
K 7.661K 1473ln)KkJ/kg 718.0(
lnln2
3
2
31423
⋅=+⋅=
+=−=−v
vv R
TT
cssss
kJ/kg 51.59=
⎟
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
⎠⎝ K 1473⎞
⎜⎛ −⋅=
⎞⎛=
kJ/kg 582.5KkJ/kg 0.5746K) 288(
3-dest,2q
Tx
(isentropic process)
⎟⎟⎠
⎜⎜⎝
−−source
in230 T
ss
0=4-dest,3x
kJ/kg 88.12=
⎟⎠
⎞⎜⎝
⎛ +⋅−=
⎟⎟⎠
⎞⎜⎜⎝
⎛+−=
K 288kJ/kg 253.6KkJ/kg 0.5746K) 288(
sink
out4101-dest,4 T
qssTx
The largest exergy destruction in the cycle occurs during the heat-rejection process.
preparation. If you are a student using this Manual, you are using it without permission.
9-116
9-149E The exergy destruction associated with the heat rejection process of the Diesel cycle described in Prob. 9-55E and the exergy at the end of the expansion stroke are to be determined.
Analysis From Prob. 9-55E, qout = 158.9 Btu/lbm, T1 = 540 R, P1 = 14.7 psia, T4 = 1420.6 R, P4 = 38.62 psia and v 4 = v 1.
The entropy change during process 4-1 is
RBtu/lbm 0.1728)62.38/7.14ln()06855.0(83984.060078.0
)/ln( 41R @1420.6o4@540R
o141
⋅−=−−=
−−=− PPRssss
Thus,
( ) Btu/lbm 65.6=⎟⎟⎠
⎜⎜⎝
+⋅−=⎟⎟⎠
⎜⎜⎝
+−=R 540
RBtu/lbm 0.1728R54041041 destroyed,RT
ssTx ⎞⎛⎞⎛ Btu/lbm 158.941,Rq
oting that state 4 is identical to the state of the surroundings, the exergy at the end of the power stroke (state 4) is etermined from
Nd
( ) ( ) ( 040040044 vv )−+−−−= PssTuuφ
where
1404
out1404
=−=− 0RBtu/lbm 9.581 ⋅==−=− quuuu
vvvv RBtu/lbm 0.17411404 ⋅=−=− ssss
Thus,
( ) ( )( ) Btu/lbm 65.6=+⋅−= 0RBtu/lbm 0.1728R540Btu/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 MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
9-117
9-150 The exergy loss of each process for an ideal dual cycle described in Prob. 9-63 is to be determined.
Analysis From Prob. 9-63, qin,x-3 = 114.6 kJ/kg, T1 = 291 K, T2 = 1037 K, Tx = 1141 K, T3 = 1255 K, and T4 = 494.8 K. Also,
kJ/kg 74.67K)10371141)(KkJ/kg 718.0()( 22,in =−⋅=−=− TTcq xx v
kJ/kg 3.146K)2918.494)(KkJ/kg 718.0()( 14out =−⋅=−= TTcq v
v
P
4
1
2
3
qout
x
qinThe exergy destruction during a process of the cycle is
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
⎟⎟⎠
⎜⎜⎝
+−∆==sink
out
source
in0gen0dest TT
sTsTx ⎞⎛ qq
Application of this equation for each process of the cycle gives
KkJ/kg 1158.0kPa 90kPa 5148ln)KkJ/kg 287.0(
K 291K 1037ln)KkJ/kg 005.1(
lnln1
2
1
212
⋅=
⋅−⋅=
−=−PP
RTT
css p
kJ/kg 33.7=⋅=−= K)kJ/kg K)(0.1158 291()( 1202-1 dest, ssTx
KkJ/kg 06862.00K 1037K 1141ln)KkJ/kg 718.0(
lnln22
2
⋅=+⋅=
+=−v
vv
xxx R
TT
css
kJ/kg 2.65=⎟⎠
⎞⎜⎝
⎛ −⋅=⎟⎟⎠
⎞⎜⎜⎝
⎛−−=
K 1255kJ/kg 74.67
KkJ/kg 0.06862K) 291(source
-2 in,20-2 dest, T
qssTx x
xx
KkJ/kg 09571.00K 1141K 1255ln)KkJ/kg 005.1(lnln 33
3 ⋅=−⋅=−=−xx
px PP
RTT
css
kJ/kg 1.28=⎟⎠
⎞⎜⎝
⎛ −⋅=⎟⎟⎠
⎞⎜⎜⎝
⎛−−=
K 1255kJ/kg 114.6
KkJ/kg 0.09571K) 291(source
3- in,303- dest, T
qssTx x
xx
KkJ/kg 1339.01.1
18ln)KkJ/kg 287.0(K 1255K 8.494ln)KkJ/kg 718.0(
lnlnlnln3
4
3
4
3
434
⋅=⋅+⋅=
+=+=−crrR
TT
cRTT
css vv v
v
kJ/kg 39.0=⋅=−= K)kJ/kg K)(0.1339 291()( 3404-3 dest, ssTx
KkJ/kg 3811.00K 8.494
K 291ln)KkJ/kg 718.0(lnln4
1
4
141 ⋅−=+⋅=+=−
v
vv R
TT
css
kJ/kg 35.4=⎟⎠
⎞⎜⎝
⎛ +⋅−=⎟⎟⎠
⎞⎜⎜⎝
⎛+−=
K 291kJ/kg 146.3
KkJ/kg 0.3811K) 291(sink
out4101-dest,4 T
qssTx
preparation. If you are a student using this Manual, you are using it without permission.
9-118
9-151 The exergy loss of each process for an air-standard Stirling cycle described in Prob. 9-81 is to be determined.
Analysis From Prob. 9-81, qin = 1275 kJ/kg, qout = 212.5 kJ/kg, T1 = T2 = 1788 K, T3 = T4 = 298 K. The exergy destruction during a process of the cycle is
⎟⎟⎠
⎞⎜⎜⎝
⎛+−∆==
sink
out
source
in0gen0dest T
qT
qsTsTx
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
Application of this equation for each process of the cycle gives
KkJ/kg 7132.0)12ln()KkJ/kg 287.0(0
lnln1
2
1
212
⋅=⋅+=
+=−v
vv R
TT
css
0kJ/kg 0.034 ≈=⎟⎠
⎞⎜⎝
⎛ −⋅=
⎟⎟⎠
⎞⎜⎜⎝
⎛−−=
K 1788kJ/kg 1275KkJ/kg 0.7132K) 298(
source
in1202-1 dest, T
qssTx
s
T
3
2 qin
qout
4
1
KkJ/kg 7132.012 ⎠⎝
1ln)KkJ/kg 287.0(0
lnln3
4
3
434
⋅−=⎟⎞
⎜⎛⋅+=
+=−v
vv R
TT
css
0kJ/kg 0.034 ≈−=⎟⎠
⎞⎜⎝
⎛ +⋅−=
⎟⎟⎠
⎞⎛ outq⎜⎜⎝
+−=
K 298kJ/kg 212.5KkJ/kg 0.7132K) 298(
sink3404-3 dest, T
ssTx
These results are not surprising since Stirling cycle is totally reversible. Exergy destructions are not calculated for processes 2-3 and 4-1 because there is no interaction with the surroundings during these processes to alter the exergy destruction.
preparation. If you are a student using this Manual, you are using it without permission.
9-119
9-152 The exergy destruction associated with each of the processes of the Brayton cycle described in Prob. 9-89 is to be determined.
Analysis From Prob. 9-89, qin = 698.3 kJ/kg, qout = 487.9 kJ/kg, and
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
Thus,
KkJ/kg 2.66807kJ/kg 783.04
KkJ/kg 3.21751K 1240
KkJ/kg 2.44117kJ/kg .60626
KkJ/kg .685151K 295
44
33
22
11
⋅=⎯→⎯=
⋅=⎯→⎯=
⋅=⎯→⎯=
⋅=⎯→⎯=
o
o
o
o
sh
sT
sh
sT
( )
( ) ( ) ( )( )
( )
( )
( ) ( ) ( )( )
( ) kJ/kg 183.2
kJ/kg 34.53
kJ/kg 105.4
kJ/kg 29.51
=⎟⎟⎠
⎞⎜⎜⎝
⎛+−=
⎟⎟⎠
⎞⎜⎜⎝
⎛+−−=⎟
⎟⎠
⎞⎜⎜⎝
⎛+−==
=⋅−−=
=⎟⎟⎠
⎞⎜⎜⎝
⎛−−=−==
=⎟⎟⎠
⎞⎜⎜⎝
⎛−−=
⎟⎟⎠
⎞⎜⎜⎝
⎛ −+−−=⎟
⎟⎠
⎞⎜⎜⎝
⎛+−==
=⋅−−=
=⎟⎟⎠
⎞⎜⎜⎝
⎛−−=−==
K 310kJ/kg 487.92.668071.68515K 310
ln
1/10lnKkJ/kg 0.2873.217512.66807K 310
ln
K 1600kJ/kg 698.3
2.441173.21751K 310
ln
10lnKkJ/kg 0.2871.685152.44117K 310
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
preparation. If you are a student using this Manual, you are using it without permission.
9-120
9-153 Exergy analysis is to be used to answer the question in Prob. 9-94.
Analysis From Prob. 9-94, T1 = 288 K, T2s = 585.8 K, T2 = 618.9 K, T3 = 873 K, T4s = 429.2 K, T4 = 473.6 K, rp = 12. The exergy change of a flow stream between an inlet and exit state is given by
)(0 ieie ssThh −−−=∆ψ
4s
s
T
1
2s
4
3 qin
qout
873 K
288 K
2
This is also the expression for reversible work. Application of this equation for isentropic and actual compression processes gives
KkJ/kg 0003998.0
)12ln()KkJ/kg 287.0(K 288K 8.585ln)KkJ/kg 005.1(
lnln1
2
1
212
⋅=
⋅−⋅=
−=−PP
RTT
css sps
kJ/kg 2.299)KkJ/kg 0003998.0)(K 288(K)2888.585)(KkJ/kg 005.1(
)()( 12s01221 rev,
=⋅−−⋅=
−−−=− ssTTTcw sps
KkJ/kg 05564.0)12ln()KkJ/kg 287.0(K 288K 9.618ln)KkJ/kg 005.1(
lnln1
2
1
212
⋅=⋅−⋅=
−=−PP
RTT
css p
)()( 1201221 rev, −−−=− ssTTTcw p
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
kJ/kg 5.316)KkJ/kg 05564.0)(K 288(K)2889.618)(KkJ/kg 005.1( =⋅−−⋅=
inimum work that must be supplied to the compressor by The irreversibilities therefore increase the m
kJ/kg 17.3=−=−=∆ −− 2.2995.31621 rev,21 rev,Crev, swww
Repeating the calculations for the turbine,
KkJ/kg 0003944.0)12ln()KkJ/kg 287.0(K2.429
K 873ln)KkJ/kg 005.1(
lnln4
3
4
343
⋅=⋅−⋅=
−=−PP
RTT
csss
ps
v, −−−=− ssps ssTTTcw
)()( 4304343 re
kJ/kg 9.445)KkJ/kg 0003944.0)(K 288(K)2.429873)(KkJ/kg 005.1( =⋅−−⋅=
KkJ/kg 09854.0)12ln()KkJ/kg 287.0(K6.473
K 873ln)KkJ/kg 005.1(
lnln4
3
4
343
⋅−=⋅−⋅=
−=−PP
RTT
css p
kJ/kg 8.429)KkJ/kg 09854.0)(K 288(K)6.473873)(KkJ/kg 005.1(
)()( 4304343 rev,
=⋅−−−⋅=
−−−=− ssp ssTTTcw
kJ/kg 16.1=−=−=∆ −− 8.4299.44543 rev,43 rev,Trev, www s
Hence, it is clear that the compressor is a little more sensitive to the irreversibilities than the turbine.
preparation. If you are a student using this Manual, you are using it without permission.
9-121
9-154 The total exergy destruction associated with the Brayton cycle described in Prob. 9-116 and the exergy at the exhaust gases at the turbine exit are to be determined.
Properties The gas constant of air is R = 0.287 kJ/kg·K (Table A-1).
Analysis From Prob. 9-116, qin = 480.82, qout = 372.73 kJ/kg, and
s
T
1
2s 4s
3qin1150 K
310 K
5
6
42
KkJ/kg 08156.2kJ/kg 738.43
KkJ/kg 69407.2kJ/kg 14.803
KkJ/kg 12900.3K 1150
KkJ/kg 42763.2kJ/kg 26.618
KkJ/kg 1.73498K 310
55
44
33
22
11
⋅=⎯→⎯=
⋅=⎯→⎯=
⋅=⎯→⎯=
⋅=⎯→⎯=
⋅=⎯→⎯=
o
o
o
o
o
sh
sh
sT
sh
sT
and, from an energy balance on the heat exchanger,
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
⋅=⎯→⎯
=−−=⎯→⎯−=−os
hhhhh
Thus,
KkJ/kg 2.52861
kJ/kg 97.682)26.61843.738(14.803
6
66425
( )
( ) ( ) ( )( )
( )
( ) ( ) ( )( )( ) ( )[ ] ( ) ( )[ ]
( )( )
( )
( ) kJ/kg 142.6=⎟⎟⎠
⎜⎜⎝
+−=K 290
2.528611.73498K 290
kJ/kg 58.09
kJ/kg 4.37
kJ/kg 35.83
kJ/kg 38.91
⎞⎛
⎟⎟⎠
⎞⎜⎜⎝
⎛+−−=⎟
⎟⎠
⎞⎜⎜⎝
⎛+−==
=⎟⎟⎠
⎞⎜⎜⎝
⎛−−=
⎟⎟⎠
⎞⎜⎜⎝
⎛−−−=⎟
⎟⎠
⎞⎜⎜⎝
⎛−−==
=−+−=
−+−=−+−==
=⋅−−=
⎟⎟⎠
⎞⎜⎜⎝
⎛−−=−==
=⋅−−=
⎟⎟⎠
⎞⎜⎜⎝
⎛−−=−==
kJ/kg 372.73
ln
K 1500kJ/kg 480.822.608153.12900K 290
ln
2.694072.528612.427632.60815K 290
1/7lnKkJ/kg0.2873.129002.69407K 290
ln
7lnKkJ/kg 0.2871.734982.42763K 290
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
oting that h0 = h@ 290 K = 290.16 kJ/kg and 29 10 ⋅=⎯→⎯= osT , the stream exergy at the exit of the regenerator (state 6) is determined from
N 0 KkJ/kg 1.66802K
( ) ( ) 06
026
060066 2gzVssThh ++−−−=φ
where
KkJ/kg 0.860591.668022.52861ln0
1
6161606 ⋅=−=−−=−=−
PP
Rssssss oo
Thus,
( )( ) kJ/kg 143.2=⋅−−= KkJ/kg 0.86059K 2900.162997.6826φ
preparation. If you are a student using this Manual, you are using it without permission.
9-122
9-155 Prob. 9-154 is reconsidered. The effect of the cycle pressure on the total irreversibility for the cycle and the e investigated.
nalysis Using EES, the problem is solved as follows:
[1] ]
blem 9-154"
=P[6])
0)
])
estroyed_53+x_destroyed_61 ce state 0 and state 1 are identical"
Problem 9-116"
, s=s_s[2])
)
, s=s_s[4])
) en=epsilon*(h[4]-h[2])
exergy of the exhaust gas leaving the regenerator is to b
A
"Given"
] T[1]=310 [KP[1]=100 [kPa] Ratio_P=7
*PP[2]=Ratio_PT[3]=1150 [Keta_C=0.75 eta_T=0.82
5 epsilon=0.6T_H=1500 [K]
0=290 [K] TP0=100 [kPa] "Analysis for Proq_in=h[3]-h[5] q_out=h[6]-h[1] h[5]-h[2]=h[4]-h[6] s[2]=entropy(Fluid$, P=P[2], h=h[2])
py(Fluid$, h=h[4], P=P[4]) s[4]=entros[5]=entropy(Fluid$, h=h[5], P=P[5]) P[5]=P[2] s[6]=entropy(Fluid$, h=h[6], PP[6]=P[1] h[0]=enthalpy(Fluid$, T=T0) s[0]=entropy(Fluid$, T=T0, P=Px_destroyed_12=T0*(s[2]-s[1]) x_destroyed_34=T0*(s[4]-s[3]) x_destroyed_regen=T0*(s[5]-s[2]+s[6]-s[4x_destroyed_53=T0*(s[3]-s[5]-q_in/T_H) x_destroyed_61=T0*(s[1]-s[6]+q_out/T0) _total=x_destroyed_12+x_destroyed_34+x_destroyed_regen+x_dx
x6=h[6]-h[0]-T0*(s[6]-s[0]) "sin
ysis for"AnalFluid$='air' "(a)" h[1]=enthalpy(Fluid$, T=T[1]) s[1]=entropy(Fluid$, T=T[1], P=P[1])
sion" s_s[2]=s[1] "isentropic compres]h_s[2]=enthalpy(Fluid$, P=P[2
eta_C=(h_s[2]-h[1])/(h[2]-h[1]) lpy(Fluid$, T=T[3]) h[3]=entha
s[3]=entropy(Fluid$, T=T[3], P=P[3]P[3]=P[2]
] "isentropic expansion" s_s[4]=s[3h_s[4]=enthalpy(Fluid$, P=P[4]P[4]=P[1]
ta_T=(h[3]-h[4])/(h[3]-h_s[4]eq_reg "(b)" w_C_in=(h[2]-h[1]) w_T_out=h[3]-h[4]
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
9-123
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
C_in _in=(h[3]-h[2])-q_regen ta_th=w_net_out/q_in
Ra P [k
w_net_out=w_T_out-w_qe
tio_ xtotal
J/kg]x6
[kJ/kg] 6 7 8 9
10 11 12 13 14
335.1 343.3
177.1 182.1
270.1 280
289.9 299.5 308.8 317.8 326.6
137.2 143.5 149.6 155.5 161.1 166.6 171.9
6 7 8 9 10 11 12 13 14130
150
170
190
210
230
250
270
290
310
330
350
RatioP
x [k
J/kg
]
xtotal,dest
x6
preparation. If you are a student using this Manual, you are using it without permission.
9-124
9-156 The exergy loss of each process for a regenerative Brayton cycle with three stages of reheating and intercooling described in Prob. 9-135 is to be determined.
Analysis From Prob. 9-135,
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
K 3.551 K, 2.819 K, 7.520
1413
98
==
rp = 4, qin,7-8 = qin,9-10 = qin,11-12 = 300 kJ/kg,
qout,14-1 = 206.9 kJ/kg, qout,2-3 = qout,4-5 = 141.6 kJ/kg,
T1 = T3 = T5 = 290 K , T2 = T4 = T6 = 430.9 K
K 9.495 K, 7.585 K, 4.870 K, 9.571 K, 8.849 121110 ===
7 ===
TT
TTT
he exergy destruction during a process of a stream f m an inlet state to exit state is given by
s
T
TTT
T ro
⎟⎟⎞
⎜⎜⎛
+−−== outin0gen0dest
qqssTsTx ie
⎠⎝ sinksource TT
Application of this equation for each process of the cycle gives
0kJ/kg 0.03 ≈=⎥⎦⎤
⎢⎣⎡ −=
⎟⎟⎠
⎞⎜⎜⎝
⎛−===
)4ln()287.0(290
430.9(1.005)ln)290(
lnln1
2
1
206-5 dest,4-3 dest,2-1 dest, P
PR
TT
cTxxx p
kJ/kg 32.1=⎥⎦⎤
⎢⎣⎡ −−=⎟⎟
⎠
⎞⎜⎜⎝
⎛−−=
4.8703000
520.7819.2(1.005)ln)290(lnln
source
8-in,7
7
8
7
808-7 dest, T
qPP
RTT
cTx p
kJ/kg 26.2=⎥⎦⎤
⎢⎣⎡ −−=⎟⎟
⎠
⎞⎜⎜⎝
⎛−−=
4.8703000
551.3849.8(1.005)ln)290(lnln
source
10-in,9
7
8
9
10010-9 dest, T
qPP
RTT
cTx p
kJ/kg 22.5=⎥⎦⎤
⎢⎣⎡ −−=⎟⎟
⎠
⎞⎜⎜⎝
⎛−−=
4.8703000
571.9870.4(1.005)ln)290(lnln
source
12-in,11
11
12
11
12012-11 dest, T
qPP
RTT
cTx p
0kJ/kg 0.05 ≈−=⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛−=⎟⎟
⎠
⎞⎜⎜⎝
⎛−=
41ln)287.0(
819.2551.3(1.005)ln)290(lnln
8
9
8
9098- dest, P
PR
TT
cTx p
0kJ/kg 0.04 ≈−=⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛−=⎟⎟
⎠
⎞⎜⎜⎝
⎛−=
41ln)287.0(
849.8571.9(1.005)ln)290(lnln
10
11
10
11011-10 dest, P
PR
TT
cTx p
0kJ/kg 0.08 ≈−=⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛−=⎟⎟
⎠
⎞⎜⎜⎝
⎛−=
41ln)287.0(
870.4585.7(1.005)ln)290(lnln
12
13
12
13013-12 dest, P
PR
TT
cTx p
kJ/kg 50.6=⎥⎦⎤
⎢⎣⎡ +−=⎟⎟
⎠
⎞⎜⎜⎝
⎛+−=
2909.2060
495.9290(1.005)ln)290(lnln
sink
1-out,14
14
1
14
101-14 dest, T
qPP
RTT
cTx p
kJ/kg 26.2=⎥⎦⎤
⎢⎣⎡ +−=⎟⎟
⎠
⎞⎜⎜⎝
⎛+−==
2906.1410
430.9290(1.005)ln)290(lnln
sink
3-out,2
2
3
2
305-4 dest,3-2 dest, T
qPP
RTT
cTxx p
kJ/kg 6.66=⎥⎦⎤
⎢⎣⎡ +=
⎟⎟⎠
⎞⎜⎜⎝
⎛+=∆+∆= −−
585.7495.9(1.005)ln
430.9520.7(1.005)ln)290(
lnln)(13
14
6
701413760regendest, T
Tc
TT
cTssTx pp
3
4
1
9
8
2
5
10
11
67
12
13
14
preparation. If you are a student using this Manual, you are using it without permission.
9-125
9-157 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.
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
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) 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
1
Combustion chamber
Turbine
23
4
Compress.
100 kPa30°C
Diesel fuel
700 kPa260°C
0.881=−−
=−−
=K)303533(K)3036.505(12
TTTT s
Cη 12
(b) The total mass flowing through the turbine and the rate of heat inp are ut
kg/s 81.12kg/s 21.0kg/s 6.1260
kg/s 6.12AF
+=+=+= aafat mmmm &&&&
kg/s 6.12=+=
m&
The temperature at the exit of combustion chamber is
3kg/s)(1.09 81.12(kJ/s 8555)( 323in =⎯→⎯−=⎯→⎯−= TTTcmQ p&&
The temperature at the turbine exit is determined using isentropic efficiency relation
kW 85557)kJ/kg)(0.9 00kg/s)(42,0 21.0(HVin === cf qmQ η&&
K 1144)K533kJ/kg.K)( 3T
( ) K 7.685kPa 700kPa 0 1)/1.357-(1.357
334 =⎟
⎠⎞
⎝⎟⎠
⎜⎝
s P 10K 1144
/)1(4 ⎜
⎛=⎟⎞
⎜⎛
=− kk
PTT
K 4.754K)7.6851144(
K)1144(85.0 4
4
43
43 =⎯→⎯−
−=⎯→⎯
−−
= TT
TTTT
sTη
T whe net po er and the back work ratio are =−= )( 12inC, −= TTcmW pa&
& kW 3168)K30333kJ/kg.K)(5 3kg/s)(1.09 6.12(
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 3168inC,W
r&
kW 5455outT,
bw W&
(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
0.364===735.0267.0
max
thII η
ηη and
preparation. If you are a student using this Manual, you are using it without permission.
9-126
9-158 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 1000 K are cp = 1.142 kJ/kg·K, cv = 0.855 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.336 (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
V
V
VV===⎯→⎯
+=⎯→⎯
+= 2
33
m 00012.0m 0018.0
16
43
1 0018.000012.0VVV +=+= dc m 00192.0 V==
Process 1- ntropic compression
1
Qin2
3
4
P
V
Qout
x2: Ise
( )( )
( )( ) kPa 385916kPa 95
K 7.87016
1.336
2
112
1-1.336
2
==⎟⎟⎠
⎞⎜⎜⎝
⎛=
=⎠⎝
k
PPv
v
v
Process 2-x and x-3: Constant-volume and constant pressure heat addition processes:
K 3431
112 =⎟⎟
⎞⎜⎜⎛
=−k
TTv
K 1692kPa 3859kPa 7500K) 7.870(
22 ===
PP
TT xx
kJ/kg 6.702K)7.8701692(kJ/kg.K) (0.855)( 2-2 =−=−= TTcq xx v
−== )( TTcqq K 2308=⎯→⎯−=⎯→⎯ 3333- K)1692(kJ/kg.K) (0.855kJ/kg 6.702 TTxpx
−2 x
(b) 6.7026.7023-2in =+=+= − xx qqq kJ/kg 1405
3333 m 0001636.0
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
K 1692xTK 2308)m 00012.0( === x
TVV
rocess 3- isentropic expansion. P 4:
( )
( ) kPa 4.279m 0.00192
m 0.0001636kPa 7500
K 1009m 0.00192m 0.0001636K 2308
1.336
3
3
4
334
1-1.336
3
31
4
334
=⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
=⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
−
k
k
PP
TT
V
V
V
V
Process 4-1: constant voume heat rejection.
( ) ( )( ) kJ/kg 3.569K3431009KkJ/kg 0.85514out =−⋅=−= TTcq v
The net work output and the thermal efficiency are
kJ/kg 835.8=−=−= 3.5691405outinoutnet, qqw
59.5%==== 5948.0kJ/kg 1405kJ/kg 835.8
in
outnet,th q
wη
preparation. If you are a student using this Manual, you are using it without permission.
9-127
(c) The mean effective pressure is determined to be
( )( )kg 0.001853
K 343K/kgmkPa 0.287)m 92kPa)(0.001 95(
3
3
1
11 =⋅⋅
==RTP
mV
kPa 860.4=⎟⎟⎠
⎞⎜⎜⎝
⎛ ⋅
−− 0001.000192.0(21 VV==
kJmkPa
m)2kJ/kg)kg)(835.8 (0.001853
MEP3
3outnet,mw
d) The power for engine speed of 3500 rpm is (
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
kW 28.39=⎟⎠
⎜⎝
==s 60rev/cycle) 2(
kJ/kg) kg)(835.8 001853.0(2netnet mwW& ⎞
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.
⎛ min 1(rev/min) 2200n&
8709.0K 2308
K )27325(113
0max =
+−=−=
TT
η
and
68.3%==== 683.08709.05948.0
max
thII η
ηη
The rate of exergy of the exhaust gases is determined as follows
( )
( )( ) kJ/kg 0.285100
4.279ln)kJ/kg.K 287.0(298
1009kJ/kg.K)ln 142.1()298(29810090.855
lnln)(0
4
0
400404004
4
=⎥⎦⎤
⎢⎣⎡ −−−=
⎥⎦
⎤⎢⎣
⎡−−−=−−−=
PP
RTT
cTTTcssTuux pv
kW 9.683=⎟⎠⎞
⎜⎝⎛==
s 60min 1
rev/cycle) 2((rev/min) 2200kJ/kg) kg)(285.0 001853.0(
244nmxX&&
preparation. If you are a student using this Manual, you are using it without permission.
9-128
Review Problems
9-159 An Otto cycle with a compression ratio of 7 is considered. The thermal efficiency is to be determined using constant and variable specific heats.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible.
Properties The properties of air at room temperature are R = 0.287 kPa·m3/kg·K, cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-2a).
Analysis (a) Constant specific heats:
54.1%==−=−=−−
0.54087
1111 11.41th krη
v
P
4
1
3
2(b) Variable specific heats: (using air properties from Table A-17)
Process 1-2: isentropic compression.
1.688kJ/kg 48.205
K 2881
11 =
=⎯→⎯=
r
uT
v
kJ/kg 62.4473.98)1.688(722
12 r rrr vv
vv
112
2 =⎯→⎯==== uv
rocess 2- v = constant heat addition.
8K 1273
23
33
=−=−==
uuq
T
in
rv
P 3:
kJ/kg .8955062.44751.998045.12
kJ/kg 51.993 =⎯→⎯=
u
Process 3-4: isentropic expansion.
kJ/kg 54.47532.84)045.12)(7( 4333
44 =⎯→⎯==== ur rrr vv
v
vv
Process 4-1: v = constant heat rejection.
kJ/kg 06.27048.20554.47514out =−=−= uuq
51.0%==−=−= 0.5098kJ/kg 550.89kJ/kg 270.0611
in
outth q
qη
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
9-129
9-160E An ideal diesel engine with air as the working fluid has a compression ratio of 20. The thermal efficiency is to be determined using constant and variable specific heats. 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, R = 0.06855 Btu/lbm·R, and k = 1.4 (Table A-2Ea). Analysis (a) Constant specific heats:
v
P
4
1
2 3qin
qout
Process 1-2: isentropic compression.
R 8.1673R)(20) 505( 0.4
2
112 =⎟⎟
⎠⎜⎜⎝
= TTV
1
=⎞⎛
−kV
rocess 2-3: P = constant heat addition.
P
350.1R 1673.8
R 2260
2
3
2
3
2
2233 =⎯→⎯=TT V
3
==TTPP VVV
Process 3-4: isentropic expansion.
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
( ) ( )( )( ) ( )( )
67.9%====
=
=−⋅=−=−=
==−=−=
=⎟⎠⎞
⎜⎝⎛=⎟
⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞
⎝⎠⎝
−
0.6794Btu/lbm 140.7Btu/lbm 95.59
Btu/lbm 59.9511.
Btu/lbm 11.45R505768.8RBtu/lbm 0.171
Btu/lbm 7.140R8.0.240
R 8.76820
1.350R) 2260(350.1350
in
outnet,th
1414out
2323in
0.41
34
2
4
qw
TTcuuq
TTchhq
rT
p
k
η
v
V
V
V
) Variable specific heats: (using air properties from Table A-17) .
11 =
=⎯→⎯=
uT
v
⎜⎜⎛
=⎟⎟⎞
⎜⎜⎛
=−
.13
13
34 TTTk
V−1k
−⋅ 16732260RBtu/lbm
−=−= 457.140outinoutnet, qqw
(bProcess 1-2: isentropic compression
82.170Btu/lbm 06.86
R 5051r
Btu/lbm 01.391R 3.1582
541.8)82.170(2011v
2
211
1
2
==
⎯→⎯====hT
r rr vvv
v
s 2-3: P = constant heat addition.
2r
Proces
428.1R 1582.3
R 2260
2
3
2
3
2
22
3
33 ===⎯→⎯=TT
TP
TP
v
vvv
Process 3-4: isentropic expansion.
Btu/lbm 51.18601.39152.577
922.2Btu/lbm 52.577
R 2260
23in
3
33
=−=−=
=
=⎯→⎯=
hhq
hT
rv
Btu/lbm 65.15292.40)922.2(428.120
428.1428.1 4332
43
3
44 =⎯→⎯===== ur
rrrr vvv
vv
v
vv
Process 4-1: v = constant heat rejection.
Then
Btu/lbm 59.6606.8665.15214out =−=−= uuq
64.3%==−=−= 0.6430Btu/lbm 186.51Btu/lbm 66.5911
in
outth q
qη
preparation. If you are a student using this Manual, you are using it without permission.
9-130
9-161E A simple ideal Brayton cycle with air as the working fluid operates between the specified temperature limits. The net work is to be determined using constant and variable specific heats.
Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible.
Properties The properties of air at room temperature are cp = 0.240 Btu/lbm·R and k = 1.4 (Table A-2Ea).
Analysis (a) Constant specific heats:
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
=== − kkprTT R 3.976R)(12) 480( 0.4/1.4/)1(
12
s
T
1
2
4
3 qin
qout
1460 R
480 R
R 8.717121R) 1460(1 0.4/1.4/)1(
34 =⎟⎠⎞
⎜⎝⎛=⎟
⎟⎠
⎞⎜⎜⎝
⎛=
− kk
prTT
−=
R)3.9764808.7171460)(RBtu/lbm 240.0(
)( 2143
1243
TTTTc
www
p
pp
(b) Variable specific heats: (using air properties from Table A-17E)
−−−= )()(compturbnet
TTcTTc
Btu/lbm 59.0=−+−⋅=
−+−=
9182.0Btu/lbm 69.114
R 4801
11 =
=⎯→⎯=
rPh
T
Btu/lbm 63.23302.11)9182.0)(12( 211
22r =⎯→⎯=== hP
PP
P r
Btu/lbm 32.17612.4)40.50(121
40.50Btu/lbm 63.358
R 1460
433
44
3
33
=⎯→⎯=⎟⎠⎞
⎜⎝⎛==
==
⎯→⎯=
hPPP
P
Ph
T
rr
r
Btu/lbm 63.4=−−−=
−−−=
−=
)69.11463.233()32.17663.358()()( 1243
compturbnet
hhhh
www
preparation. If you are a student using this Manual, you are using it without permission.
9-131
9-162 A turbocharged four-stroke V-16 diesel engine produces 3500 hp at 1200 rpm. The amount of power produced per cylinder per mechanical and per thermodynamic cycle is to be determined.
Analysis Noting that there are 16 cylinders and each thermodynamic cycle corresponds to 2 mechanical cycles (revolutions), we have
(a)
cycle)mech kJ/cyl 8.16(= hp 1Btu/min 42.41
rev/min) (1200cylinders) (16hp 3500
cycles) mechanical of (No.cylinders) of (No.producedpower Total
mechanical
⋅⋅=
⎟⎟⎠
⎞⎜⎜⎝
⎛=
=
cycle mechBtu/cyl 7.73
w
(b)
cycle) thermkJ/cyl 16.31(= hp 1Btu/min 42.41
rev/min) (1200/2cylinders) (16hp 3500
cycles) amic thermodynof (No.cylinders) of (No.producedpower Total
micthermodyna
⋅⋅=
⎟⎟⎠
⎞⎜⎜⎝
⎛=
=
cycle thermBtu/cyl 15.46
w
ified temperature limits is considered. The pressure ratio for
2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats.
Properties The specific heat ratio of air is k =1.4 (Table A-2).
nalysis We treat air as an ideal gas with constant specific heats. Using the isentropic relations, the temperatures at the compressor and turbine exit can be expressed as
9-163 A simple ideal Brayton cycle operating between the specwhich the compressor and the turbine exit temperature of air are equal is to be determined.
Assumptions 1 Steady operating conditions exist.
A
( )( )( )
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
( ) ( ) kk
p
kk
kkk
P
/1/1
3
//1
−−
−
⎠
⎞
⎝
⎛⎞⎛
⎞⎛
Setting T2 = T4 and solving for rp gives
s
T
1
2
4
3 q
kprT
PTT 1
11
212
−=⎟⎟⎠
⎜⎜⎝
=
rT
PPTT 3
434
1⎟⎟
⎜⎜=⎟⎟
⎠⎜⎜⎝
=in
qout
T3
T1( )16.7=⎟⎟
⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
− 1.4/0.812/
1
3
K 300K 1500
kk
p TT
r
Therefore, the compressor and turbine exit temperatures will be equal when the compression ratio is 16.7.
preparation. If you are a student using this Manual, you are using it without permission.
9-132
9-164 A four-cylinder spark-ignition engine with a compression ratio of 8 is considered. The amount of heat supplied per cylinder, the thermal efficiency, and the rpm for a net power output of 60 kW 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). The properties of air are given in Table A-17.
Analysis (a) Process 1-2: isentropic compression.
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
( )
kJ/kg .29564
50.543.5725.10
11
3.5721
=rv
kJ/kg 221.25K 031
2
1
2
1
112
=⎯→⎯
====
=⎯→⎯=
ur
uT
rrr vvv
vv
rocess 2-3: v = constant heat addition.
1
P
( )( )
( )( )
( ) ( )( ) kJ 0.5336=−×=−=
×=⋅⋅ /kgmkPa 0.2871RT
=
==⎯→⎯=
−
−
kJ/kg.295641775.3kg 104.406
kg 10.4064K 310K
m 0.0004kPa 98
356.2kJ/kg .31775K 0021
423in
43
311
33
uumQ
Pm
uT
r
V
v
) Process 3-4: isentropic expansion.
v
P
4
1
3
2
Qin
Qout
1800 K
3
=
(b
( )( ) kJ/kg 05.76474.24356.25.10 43
4334
=⎯→⎯==== ur rrr vvv
vv
Process 4-1: v = constant heat rejection.
( ) ( )( )
55.2%====
=−=−=
=−×=−= kJ/kg.25221764.05kg 104.406 -414out uumQ
5517.0kJ 0.5336kJ 0.2944
kJ 0.29442392.05336.0
kJ 0.2392
in
netth
outinnet
QW
QQW
η
(c) rpm 4586=⎟⎟⎠
⎞⎜⎜⎝
⎛×
==min 1
s 60kJ/cycle) 0.2944(4
kJ/s 45rev/cycle) 2(2
cylnet,cyl
net
WnW
n&
&
Note that for four-stroke cycles, there are two revolutions per cycle.
preparation. If you are a student using this Manual, you are using it without permission.
9-133
9-165 Problem 9-164 is reconsidered. The effect of the compression ratio net work done and the efficiency of the cycle is to be investigated. Also, the T-s and P-v diagrams for the cycle are to be plotted.
Analysis Using EES, the problem is solved as follows:
"Input Data" T[1]=(37+273) [K] P[1]=98 [kPa] T[3]= 2100 [K] V_cyl=0.4 [L]*Convert(L, m^3) r_v=10.5 "Compression ratio" W_dot_net = 45 [kW] N_cyl=4 "number of cyclinders" v[1]/v[2]=r_v "The first part of the solution is done per unit mass." "Process 1-2 is isentropic compression" s[1]=entropy(air,T=T[1],P=P[1]) s[2]=s[1] s[2]=entropy(air, T=T[2], v=v[2]) P[2]*v[2]/T[2]=P[1]*v[1]/T[1] P[1]*v[1]=R*T[1] R=0.287 [kJ/kg-K] "Conservation of energy for process 1 to 2: no heat transfer (s=const.) with work input" w_in = DELTAu_12 DELTAu_12=intenergy(air,T=T[2])-intenergy(air,T=T[1]) "Process 2-3 is constant volume heat addition" s[3]=entropy(air, T=T[3], P=P[3]) {P[3]*v[3]/T[3]=P[2]*v[2]/T[2]} P[3]*v[3]=R*T[3] v[3]=v[2] "Conservation of energy for process 2 to 3: the work is zero for v=const, heat is added" q_in = DELTAu_23 DELTAu_23=intenergy(air,T=T[3])-intenergy(air,T=T[2]) "Process 3-4 is isentropic expansion" s[4]=entropy(air,T=T[4],P=P[4]) s[4]=s[3] P[4]*v[4]/T[4]=P[3]*v[3]/T[3] {P[4]*v[4]=R*T[4]} "Conservation of energy for process 3 to 4: no heat transfer (s=const) with work output" - w_out = DELTAu_34 DELTAu_34=intenergy(air,T=T[4])-intenergy(air,T=T[3]) "Process 4-1 is constant volume heat rejection" v[4]=v[1] "Conservation of energy for process 2 to 3: the work is zero for v=const; heat is rejected" - q_out = DELTAu_41 DELTAu_41=intenergy(air,T=T[1])-intenergy(air,T=T[4]) w_net = w_out - w_in Eta_th=w_net/q_in*Convert(, %) "Thermal efficiency, in percent" "The mass contained in each cylinder is found from the volume of the cylinder:" V_cyl=m*v[1] "The net work done per cycle is:" W_dot_net=m*w_net"kJ/cyl"*N_cyl*N_dot"mechanical cycles/min"*1"min"/60"s"*1"thermal cycle"/2"mechanical cycles"
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
9-134
rv ηth
[%] wnet
[kJ/kg] 5 6 7 8 9 10 11
42 45.55 48.39 50.74 52.73 54.44 55.94
568.3 601.9 625.7 642.9 655.5 664.6 671.2
10-2 10-1 100 101 102101
102
103
104
v [m3/kg]P
[kPa
]
310 K 2100 K
1
2
3
4
s = const
Air Otto Cycle P-v Diagram
4.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
8.50
500
1000
1500
2000
2500
3000
s [kJ/kg-K]
T [K
]
98 kPa
6971 kPa
Air Otto Cycle T-s Diagram
1
2
3
4
v = const
5 6 7 8 9 10 1142
44
46
48
50
52
54
56
rv
ηth
[%
]
5 6 7 8 9 10 11
560
580
600
620
640
660
680
rv
wne
t [k
J/kg
]
preparation. If you are a student using this Manual, you are using it without permission.
9-135
9-166 An ideal gas Carnot cycle with helium as the working fluid is considered. The pressure ratio, compression ratio, and minimum temperature of the energy source are to be determined.
Assumptions 1 Kinetic and potential energy changes are negligible. 2 Helium is an ideal gas with constant specific heats.
Properties The specific heat ratio of helium is k = 1.667 (Table A-2a).
Analysis From the definition of the thermal efficiency of a Carnot heat engine,
s
T
3
2qin
qout4
1TH
288 K
K 576=−− 0.5011 Carnotth,
Carnotth, ηHHT
An isentropic process for an i
+==⎯→⎯−=
K 273)(151η LL TT
T
deal gas is one in which Pvk remains constant. Then, the pressure ratio is
5.65=⎟⎠⎝⎟
⎠⎜⎝ 11 K 288TP
⎞⎜⎛=⎟
⎞⎜⎛
=−− )1667.1/(667.1)1/(
22 K 576kk
TP
ased on t ion, the compression ratio is B he process equat
2.83==⎟⎟⎠
⎞⎜⎜⎝
⎛= 667.1/1
/1
1
2
2
1 )65.5(k
PP
v
v
9-167E An ideal gas Carnot cycle with helium as the working fluid is conand minimum temperature of the energy-source reservoir are to be determ
sidered. The pressure ratio, compression ratio, ined.
an ideal gas with constant specific heats.
roperties The specific heat ratio of helium is k = 1.667 (Table A-2Ea).
Analysis From the definition of the thermal efficiency of a Carnot heat engine,
Assumptions 1 Kinetic and potential energy changes are negligible. 2 Helium is
P
s
T
3
2qin
qout4
1TH
520 R
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
R 1300=−+
−=R )460(601Carnotth,η L TT
=−
=⎯→⎯0.6011 Carnotth,η
LH
HT
T
k remains constant. An isentropic process for an ideal gas is one in which PvThen, the pressure ratio is
9.88=⎟⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
−− )1667
.1/(667.1)1/(
1
2
1
2
R 520R 1300
kk
TT
PP
ased on the process equation, the compression ratio is
B
3.95==⎟⎟⎠
⎞⎜⎜⎝
⎛= 667.1/1
/1
1
2
2
1 )88.9(k
PP
v
v
preparation. If you are a student using this Manual, you are using it without permission.
9-136
9-168 The compression ratio required for an ideal Otto cycle to produce certain amount of work when consuming a given amount of fuel 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. 4 The combustion efficiency is 100 percent.
Properties The properties of air at room temperature are k = 1.4 (Table A-2).
Analysis The heat input to the cycle for 0.043 grams of fuel consumption is
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
=×== −qmQ
v
P
4
1
3
2 qout
qinkJ 505.1kJ/kg) kg)(43,000 10035.0( 3
HVfuelin
The thermal efficiency is then
6645.0kJ 1.505
kJ 1
in
netth ===
QW
η
From the definition of thermal efficiency, we obtain the required compression ratio to be
15.3=−
=−
=⎯→⎯−=−−− )14.1/(1)1/(1
th1th
)6645.01(1
)1(111
kkr
r ηη
preparation. If you are a student using this Manual, you are using it without permission.
9-137
9-169 An ideal Otto cycle with air as the working fluid with a compression ratio of 9.2 is considered. The amount of heat transferred to the air, the net work output, 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 (Table A-1). The properties of air are given in Table A-17.
Analysis (a) Process 1-2: isentropic compression.
v
P
4
1
3
2
qin
qout
( )
( ) ( ) kPa 2129kPa 98K 300K 708.3
9.2
kJ/kg 9.518K 3.70852.672.621
2.911
2.621kJ/kg 214.07K 300
11
2
2
12
1
11
2
22
2
21
2
11
112
1
=⎟⎟⎠
⎞⎜⎜⎝
⎛==⎯→⎯=
==⎯→⎯====
==⎯→⎯=
PTT
PT
PT
P
uT
r
uT
rrr
r
v
vvv
vvv
vv
v
Process 2-3: v = constant heat addition.
( )( )
kJ/kg609.8 9.5187.1128
593.8kJ/kg 1128.7K 1416.63.70822
23
3222
323 PTT
3223
3
=−=−=
==⎯→⎯====⎯→⎯=
uuq
uTTP
TPP
in
rv
vv
(b) Process 3-4: isentropic expansion.
3
( )( ) kJ/kg 487.7506.79593.82.9 43
433
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
4r =⎯→⎯==== ur rr vv
vv
ess 4-1: v = constant heat rejection.
14t
=−=−==−=−=
qqwuuq
(c)
v
Proc
kJ/kg336.1 7.2738.609kJ/kg 273.707.21475.487
outin
net
ou
55.1%===kJ/kg 609.8kJ/kg 336.1
in
netth q
wη
( )( )
( ) ( )( )kPa429
kJ 1mkPa 1
1/9.21/kgm 0.879kJ/kg 336.1
/11MEP
/kgm 0.879kPa 98
K 300K/kgmkPa 0.287
3
31
net
21
net
max2min
33
1
11max
=⎟⎟⎠
⎞⎜⎜⎝
⎛ ⋅
−=
−=
−=
==
=⋅⋅
===
rww
r
PRT
vvv
vvv
vv (d)
preparation. If you are a student using this Manual, you are using it without permission.
9-138
9-170 An ideal Otto cycle with air as the working fluid with a compression ratio of 9.2 is considered. The amount of heat transferred to the air, the net work output, 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.
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
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-2a).
Analysis (a) Process 1-2 is isentropic compression:
( )( )
( ) ( ) kPa 2190kPa 98K 300K 728.8
9.2
K 728.89.2K 300
11
2
2
12
1
11
2
22
0.41
2
112
=⎟⎟⎠
⎞⎜⎜⎝
⎛==⎯→⎯=
==⎟⎟⎠
⎞⎜⎜⎝
⎛=
−
PTT
PT
PT
P
TTk
v
vvv
v
v
v
P
4
1
3
2
qin
qout
Process 2-3: v = constant heat addition.
( )( )
( ) ( )( ) kJ/kg523.3 K728.81457.6KkJ/kg 0.718
K 457.618.72822
2323
222
323 PTT
3223
=−⋅=−=−=
====⎯→⎯=
TTcuuq
TTP
TPP
in v
vv
(b) Process 3-4: isentropic expansion.
3
( ) K 600.09.21K 1457.6
0.413 ⎛⎞⎛
−kv
43 =⎟
⎠⎞
⎜⎝
=⎟⎟⎠
⎜⎜⎝
= TTv
ess 4-1: v = constant heat rejection.
4
Proc
( ) ( )( )
kJ/kg307.9 4.2153.523
kJ/kg 215.4K300600KkJ/kg 0.718
outinnet
1414out
=−=−=
=−⋅=−=−=
qqw
TTcuuq v
(c) 58.8%===kJ/kg 523.3kJ/kg 307.9
in
netth q
wη
( )( )
( ) ( )( )kPa393
kJ 1mkPa 1
1/9.21/kgm 0.879kJ/kg 307.9
/11MEP
/kgm 0.879kPa 98
K 300K/kgmkPa 0.287
3
31
net
21
net
max2min
33
1
11max
=⎟⎟⎠
⎞⎜⎜⎝
⎛ ⋅
−=
−=
−=
==
=⋅⋅
===
rww
r
PRT
vvv
vvv
vv (d)
preparation. If you are a student using this Manual, you are using it without permission.
9-139
9-171E An ideal dual cycle with air as the working fluid with a compression ratio of 12 is considered. The thermal efficiency of the cycle 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 = 0.240 Btu/lbm.R, cv = 0.171 Btu/lbm.R, and k = 1.4 (Table A-2E).
Analysis The mass of air is
( )( )( )( )
lbm 003881.0R 580R/lbmftpsia 0.3704
ft 98/1728psia 14.73
3
1
11 =⋅⋅
==RTP
mV
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
Process 1-2: isentropic compression.
( )( ) R 166714R 5802
12 =⎟⎟⎠
⎜⎜⎝
= TTV
0.41
1 =⎞⎛
−kV
rocess 2-x: v = constant heat addition,
)( ) R 2571R1667RBtu/lbm x
22in,2
=⎯→⎯−⋅
−=−=−
x
xxx
TT
TTmcuumQ v
rocess x- P = constant heat addition.
v
P
4
1
2
3 1.1 Btu
Qout
x
0.6 Btu
P
( )( 0.171lbm 0.003881Btu 0.6 =
( ) ( )
P 3:
( ) ( )( )( )( )
1.459R 2571R 3752
R 3752R2571RBtu/lbm 0.240lbm 0.003881Btu 1.1
3
3
3
33 ==⎯→⎯=x
cx
xx rTT V
33
33in,3
==
=⎯→⎯−⋅=
−−
x
xpxx
TTPP
TT
TT
VVV
rocess 3- isentropic expansion.
=−= mchhmQ
P 4:
( ) R 151914
1.459R 3752459.1459.1 0.41
3
1
4
13
1
4
334 =⎟
⎠⎞
⎜⎝⎛=⎟
⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
−−− kkk
rTTTT
V
V
V
V
Process 4-1: v = constant heat rejection.
( ) ( )
( )( )( )
63.4%==−=−=
=−⋅=−=−=
6336.0Btu 1.7
Btu 0.622911
Btu 0.6229R5801519RBtu/lbm 0.171lbm 0.003881
in
outth
1414out
TTmcuumQ
η
v
preparation. If you are a student using this Manual, you are using it without permission.
9-140
9-172 An ideal Stirling cycle with air as the working fluid is considered. The maximum pressure in the cycle and the net work output 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.
s
T
3
2
qin = 900 kJ/kg
qout
4
1 1800 K
350 K
Properties The properties of air at room temperature are R = 0.287 kJ/kg.K, cp = 1.005 kJ/kg.K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-2).
Analysis (a) The entropy change during process 1-2 is
KkJ/kg 0.5K 1800
kJ/kg 9001212 ⋅===−
HTq
ss
and
( )
( )( ) kPa 5873=⎟⎟⎠
⎞⎜⎜⎝
⎛===⎯→⎯=
=⎯→⎯⋅=⋅⎯→⎯+=− ln1
12 Tcss v
K 350K 800
5.710kPa 200
710.5lnKkJ/kg 0.287KkJ/kg 0.5ln
3
1
1
23
3
1
1
331
1
11
3
33
1
2
1
2
1
20
2
TT
PTT
PPT
PT
P
RT
v
v
v
vvv
v
v
v
v
v
v
(b) The net work output is
1
( ) kJ/kg 725=⎟⎟⎠
⎞⎜⎜⎝
⎛−=⎟⎟
⎠
⎞⎜⎜⎝
⎛−== kJ/kg 900
K 1800K 350
11 ininthnet qTT
qwH
Lη
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
9-141
9-173 A simple ideal Brayton cycle with air as the working fluid is considered. The changes in the net work output 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-17.
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
For rp = 6,
s
T
1
2
4
3 qin
qout
2′
3′ Analysis The properties at various states are
T hP
T hP
r
r
1 1
3 3
1
3
1 386
330 9
= ⎯ →⎯ =
=
= ⎯ →⎯ ==
300 K 300.19 kJ / kg
1300 K 1395.97 kJ / kg
.
.
( )( )
( )
%9.37kJ/kg 894.57kJ/kg 339.46
kJ/kg 339.4611.55557.894kJ/kg 555.1119.3003.855
kJ/kg 894.5740.50197.1395
kJ/kg 855.315.559.33061
kJ/kg 501.40316.8386.16
in
netth
outinnet
outw
14
23in
43
4
21
2
34
12
===
=−=−==−=−=
=−=−=
=⎯→⎯=⎟⎠⎞
⎜⎝⎛==
=⎯→⎯===
qw
qqhhqhhq
hPPP
P
hPPP
P
rr
rr
η
rp = 12,
For
( )( )
( )
%5.48kJ/kg 785.37kJ/kg 380.96
kJ/kg 380.9641.40437.785kJ/kg 404.4119.3006.704
kJ/kg 785.3760.61097.1395
kJ/kg 704.658.279.330121
kJ/kg 610.663.16386.112
in
netth
outinnet
14out
23in
43
4
21
2
34
12
===
=−=−==−=−=
=−=−=
=⎯→⎯=⎟⎠⎞
⎜⎝⎛==
=⎯→⎯===
qw
qqwhhqhhq
hPPP
P
hPPP
P
rr
rr
η
Thus,
(a) ( )increase46.33996.380net kJ/kg 41.5=−=∆w
( )increase%9.37%5.48th 10.6%=−=∆η (b)
preparation. If you are a student using this Manual, you are using it without permission.
9-142
9-174 A simple ideal Brayton cycle with air as the working fluid is considered. The changes in the net work output 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 R = 0.287 kJ/kg.K, cp = 1.005 kJ/kg.K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-2).
Analysis Processes 1-2 and 3-4 are isentropic. Therefore, For rp = 6,
( )( )( )
( )( )
( )( )( )
( )( )( )
%1.40kJ/kg 803.4kJ/kg 321.9
kJ/kg 321.95.4814.803
kJ/kg 481.5K300779.1KkJ/kg 1.005
kJ/kg 803.4K500.61300KkJ/kg 1.005
K 779.161K 1300
K 500.66K 300
in
netth
outinnet
1414out
2323in
0.4/1.4/1
3
434
0.4/1.4/1
1
212
===
=−=−=
=−⋅=−=−=
=−⋅=−=−=
=⎟⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
==⎟⎟⎠
⎞⎜⎜⎝
⎛=
−
−
qw
qqw
TTchhq
TTchhq
PP
TT
PP
TT
p
p
kk
kk
η
s
T
1
24
3 qin
qout
2
3′
For rp = 12, ( )
( )( )
( )( )
( )( )( )
( )( )( )
%8.50kJ/kg 693.2kJ/kg 352.3
kJ/kg 352.39.3402.693
kJ/kg 340.9K300639.2KkJ/kg 1.005
kJ/kg 693.2K610.21300KkJ/kg 1.005
K 639.2121K 1300
K 610.212K 300
in
netth
outinnet
1414out
2323in
0.4/1.4/1
3
434
0.4/1.4/1
1
212
===
=−=−=
=−⋅=−=−=
=−⋅=−=−=
=⎟⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
==⎟⎟⎠
⎞⎜⎜⎝
⎛=
−
−
qw
qqw
TTchhq
TTchhq
PP
TT
PP
TT
p
p
kk
kk
η
Thus,
(a) ( )increase9.3213.352net kJ/kg 30.4=−=∆w
( )increase%1.40%8.50th 10.7%=−=∆η (b)
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
9-143
9-175 A regenerative gas-turbine engine operating with two stages of compression and two stages of expansion is considered. 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 constant specific heats.
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
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 The work inputs to each stage of compressor are identical, so are the work outputs of each stage of the turbine.
( )( )( )
( )( ) ( )
( ) ( )
( )( )
( )( ) ( )
( )( )
( )( ) ( )
( )( )
( ) ( ) ( )( )
( ) ( ) ( )( ) kJ/kg .5791K10061400KkJ/kg 1.005222
kJ/kg 375.7K300486.9KkJ/kg 1.005222
K .48769.486100675.09.486
K 10061.942140086.01400
K .194241K 1400
K 486.978.0/3008.445300
/
K 445.84K 300
7676outT,
1212inC,
494549
45
49
45
7667976
76
76
76
0.4/1.4/1
6
7679
1212412
12
12
12
0.4/1.4/1
1
2124
=−⋅=−=−=
=−⋅=−=−=
=−+=
−+=⎯→⎯−
−=
−−
=
=−−=
−−==⎯→⎯−
−=
−−
=
=⎟⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛==
=−+=
−+==⎯→⎯−
−=
−−
=
==⎟⎟⎠
⎞⎜⎜⎝
⎛==
−
−
TTchhw
TTchhw
TTTTTTcTTc
hhhh
TTTTTTTcTTc
hhhh
PP
TTT
TTTTTTTcTTc
hhhh
PP
TTT
p
p
p
p
sTsp
p
sT
kk
ss
Csp
spsC
kk
ss
εε
ηη
ηη
s
T
3
4
1
6
97
8
5
2
97s
4 2s
Thus,
0.475===kJ/kg 791.5kJ/kg 375.7
outT,
inC,bw w
wr
( ) ( ) ( ) ( )[ ]( ) ( ) ( )[ ]
45.1%====
=−=−=
=−+−⋅=−+−=−+−=
451.0kJ/kg 922.0kJ/kg 415.8
kJ/kg .84157.3755.791
kJ/kg .0922K10061400.48761400KkJ/kg 1.005
in
netth
inC,outT,net
78567856in
qw
www
TTTTchhhhq p
η
preparation. If you are a student using this Manual, you are using it without permission.
9-144
9-176 Problem 9-175 is reconsidered. The effect of the isentropic efficiencies for the compressor and turbine and regenerator effectiveness on net work done and the heat supplied to the cycle is to be investigated. Also, the T-s diagram for
nalysis Using EES, the problem is solved as follows:
[K]
cy" sentropic efficiency"
ies are constant across the compressor"
ssor for the isentropic case: dy-flow"
])
T[2], P=P[2])
ideal case the entropies are constant across the HP compressor"
r for the isentropic case: dy-flow"
])
4], P=P[4])
the cycle is to be plotted.
A
"Input data" T[6] = 1400 T[8] = T[6] Pratio = 4 T[1] = 300 [K] P[1]= 100 [kPa] T[3] = T[1] Eta_reg = 0.75 "Regenerator effectiveness
ta_c =0.78 "Compressor isentorpic efficien"
EEta_t =0.86 "Turbine i "LP Compressor:" "Isentropic Compressor anaysis" s[1]=ENTROPY(Air,T=T[1],P=P[1]) s_s[2]=s[1] "For the ideal case the entropP[2] = Pratio*P[1] s_s[2]=ENTROPY(Air,T=T_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exit" Eta_c = w_compisen_LP/w_comp_LP "compressor adiabatic efficiency, W_comp > W_compisen" "Conservation of energy for the LP compree_in - e_out = DELTAe=0 for steah[1] + w_compisen_LP = h_s[2] h[1]=ENTHALPY(Air,T=T[1]) h_s[2]=ENTHALPY(Air,T=T_s[2 "Actual compressor analysis:"h[1] + w_comp_LP = h[2]
[2]=ENTHALPY(Air,T=T[2]) hs[2]=ENTROPY(Air,T= "HP Compressor:" s[3]=ENTROPY(Air,T=T[3],P=P[3]) s_s[4]=s[3] "For theP[4] = Pratio*P[3] P[3] = P[2] s_s[4]=ENTROPY(Air,T=T_s[4],P=P[4]) "T_s[4] is the isentropic value of T[4] at compressor exit" Eta_c = w_compisen_HP/w_comp_HP "compressor adiabatic efficiency, W_comp > W_compisen" "Conservation of energy for the compressoe_in - e_out = DELTAe=0 for steah[3] + w_compisen_HP = h_s[4] h[3]=ENTHALPY(Air,T=T[3]) h_s[4]=ENTHALPY(Air,T=T_s[4 "Actual compressor analysis:"h[3] + w_comp_HP = h[4] h[4]=ENTHALPY(Air,T=T[4]) s[4]=ENTROPY(Air,T=T[
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
9-145
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
ntercooling heat loss:"
eat exchanger, assuming W=0, ke=pe=0 _in - e_out =DELTAe_cv =0 for steady flow"
[6]=ENTHALPY(Air,T=T[6]) s SSSF constant pressure"
r,T=T[6],P=P[6])
_s[7]=ENTROPY(Air,T=T_s[7],P=P[7])"T_s[7] is the isentropic value of T[7] at HP turbine exit"
ic turbine, assuming: adiabatic, ke=pe=0 teady-flow"
s[7] =T_s[7])
[7]=ENTHALPY(Air,T=T[7]) (Air,T=T[7], P=P[7])
[7] + q_in_reheat = h[8] T[8])
e analysis"
r,T=T[8],P=P[8])
_s[9]=ENTROPY(Air,T=T_s[9],P=P[9])"T_s[9] is the isentropic value of T[9] at LP turbine exit" n > w_turb"
ic turbine, assuming: adiabatic, ke=pe=0 teady-flow"
[9] =T_s[9])
[9]=ENTHALPY(Air,T=T[9]) ir,T=T[9], P=P[9])
ta_th_noreg=w_net/(q_in_total_noreg)*Convert(, %) "[%]" "Cycle thermal efficiency" atio"
t added in the external heat exchanger is"
ALPY(Air, T=T[5]) [5]=ENTROPY(Air,T=T[5], P=P[5])
"Ih[2] = q_out_intercool + h[3] "External heat exchanger analysis" "SSSF First Law for the heh[4] + q_in_noreg = h[6] hP[6]=P[4]"process 4-6 i "HP Turbine analysis" s[6]=ENTROPY(Ais_s[7]=s[6] "For the ideal case the entropies are constant across the turbine" P[7] = P[6] /Pratio sEta_t = w_turb_HP /w_turbisen_HP "turbine adiabatic efficiency, w_turbisen > w_turb" "SSSF First Law for the isentrope_in -e_out = DELTAe_cv = 0 for sh[6] = w_turbisen_HP + h_h_s[7]=ENTHALPY(Air,T"Actual Turbine analysis:" h[6] = w_turb_HP + h[7] hs[7]=ENTROPY "Reheat Q_in:" hh[8]=ENTHALPY(Air,T= "HL Turbin P[8]=P[7] s[8]=ENTROPY(Ais_s[9]=s[8] "For the ideal case the entropies are constant across the turbine" P[9] = P[8] /Pratio sEta_t = w_turb_LP /w_turbisen_LP "turbine adiabatic efficiency, w_turbise "SSSF First Law for the isentrope_in -e_out = DELTAe_cv = 0 for sh[8] = w_turbisen_LP + h_sh_s[9]=ENTHALPY(Air,T"Actual Turbine analysis:" h[8] = w_turb_LP + h[9] hs[9]=ENTROPY(A "Cycle analysis" w_net=w_turb_HP+w_turb_LP - w_comp_HP - w_comp_LP q_in_total_noreg=q_in_noreg+q_in_reheat EBwr=(w_comp_HP + w_comp_LP)/(w_turb_HP+w_turb_LP)"Back work r "With the regenerator, the heah[5] + q_in_withreg = h[6] h[5]=ENTHsP[5]=P[4]
preparation. If you are a student using this Manual, you are using it without permission.
9-146
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
]-h[4]) ives h[10] and thus T[10] as:"
ALPY(Air, T=T[10]) [10]=ENTROPY(Air,T=T[10], P=P[10])
_in_total_withreg=q_in_withreg+q_in_reheat
g data is used to complete the Array Table for plotting purposes."
OPY(Air,T=T[5],P=P[5])
_s[10]=s[10] _s[10]=T[10]
η qi g qin eg
"The regenerator effectiveness gives h[5] and thus T[5] as:" Eta_reg = (h[5]-h[4])/(h[9"Energy balance on regenerator gh[4] + h[9]=h[5] + h[10] h[10]=ENTHsP[10]=P[9] "Cycle thermal efficiency with regenerator" qEta_th_withreg=w_net/(q_in_total_withreg)*Convert(, %) "[%]" "The followins_s[1]=s[1] T_s[1]=T[1] s_s[3]=s[3] T_s[3]=T[3] s_s[5]=ENTRT_s[5]=T[5] s_s[6]=s[6] T_s[6]=T[6] s_s[8]=s[8] T_s[8]=T[8] sT
ηreg ηC ηt ηth,noreg [%]
th,withreg
[%] n,total,nore
[kJ/kg] ,total,withr
[kJ/kg] wnet
[kJ/kg] 0.6
0.65 0.7
0.75 0.8
0.85 0.9
0.95 1
0.78 0.78
0.86 0.86
30.57 30.57
51.89 53.87
1434 1434
845.1 814.1
438.5 438.5
0.78 0.78 0.78 0.78 0.78 0.78 0.78
0.86 0.86 0.86 0.86 0.86 0.86 0.86
30.57 30.57 30.57 30.57 30.57 30.57 30.57
41.29 42.53 43.85 45.25 46.75 48.34 50.06
1434 1434 1434 1434 1434 1434 1434
1062 1031 1000 969.1 938.1 907.1 876.1
438.5 438.5 438.5 438.5 438.5 438.5 438.5
4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0 8.50
400
800
1200
1600
s [kJ/kg-K]
T [K
]
100 k
Pa
400 k
Pa
160
0 kP
a
Air
1
2
3
4
5
6
7
8
9
10
preparation. If you are a student using this Manual, you are using it without permission.
9-147
0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 128
32
36
40
44
48
52
ηreg
ηth
[%
]
No regeneration
With regeneration
0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 110
15
20
25
30
35
40
45
50
55
η t
ηth
[%
]
With regeneration
No regeneration
0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1150
200
250
300
350
400
450
500
550
600
η t
wne
t [k
J/kg
]
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
9-148
0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1
800
900
1000
1100
1200
1300
1400
1500
ηt
q in,
tota
l [kJ
/kg]
With regeneration
No regeneration
0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1
25
30
35
40
45
50
55
ηc
ηth
[%
] With regeneration
No regeneration
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
9-149
9-177 A regenerative gas-turbine engine operating with two stages of compression and two stages of expansion is considered. 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 Helium is an ideal gas with constant specific heats.
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
Properties The properties of helium at room temperature are cp = 5.1926 kJ/kg.K and k = 1.667 (Table A-2).
Analysis The work inputs to each stage of compressor are identical, so are the work outputs of each stage of the turbine.
( )( )( )
( )( ) ( )
( ) ( )
( )( )
( )( ) ( )
( )( )
( )( ) ( )
( )( )
( ) ( ) ( )( )
( ) ( ) ( )( ) kJ/kg 5323K.48871400KkJ/kg 5.1926222
kJ/kg 2961K300585.2KkJ/kg 5.1926222
K .88112.5854.88775.02.585
K .48870.804140086.01400
K 0.80441K 1400
K .258578.0/3004.522300
/
K .45224K 300
7676outT,
1212inC,
494549
45
49
45
7667976
76
76
76
70.667/1.66/1
6
7679
1212412
12
12
12
70.667/1.66/1
1
2124
=−⋅=−=−=
=−⋅=−=−=
=−+=
−+=⎯→⎯−
−=
−−
=
=−−=
−−==⎯→⎯−
−=
−−
=
=⎟⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛==
=−+=
−+==⎯→⎯−
−=
−−
=
==⎟⎟⎠
⎞⎜⎜⎝
⎛==
−
−
TTchhw
TTchhw
TTTTTTcTTc
hhhh
TTTTTTTcTTc
hhhh
PP
TTT
TTTTTTTcTTc
hhhh
PP
TTT
p
p
p
p
sTsp
p
sT
kk
ss
Csp
spsC
kk
ss
εε
ηη
ηη
s
T
3
4
1
6qin
97
8
5
2
97s
4 2s
Thus,
0.556===kJ/kg 5323kJ/kg 2961
outT,
inC,bw w
wr
( ) ( ) ( ) ( )[ ]( ) ( ) ( )[ ]
41.3%====
=−=−=
=−+−⋅=−+−=−+−=
4133.0kJ/kg 5716kJ/kg 2362
kJ/kg 236229615323
kJ/kg 5716K.48871400.88111400KkJ/kg 5.1926
in
netth
inC,outT,net
78567856in
qw
www
TTTTchhhhq p
η
preparation. If you are a student using this Manual, you are using it without permission.
9-150
9-178 An ideal gas-turbine cycle with one stage of compression and two stages of expansion and regeneration is considered. The thermal efficiency of the cycle as a function of the compressor pressure ratio and the high-pressure turbine to compressor inlet temperature ratio is to be determined, and to be compared with the efficiency of the standard regenerative cycle.
Analysis The T-s diagram of the cycle is as shown in the figure. If the overall pressure ratio of the cycle is rp, which is the pressure ratio across the compressor, then the pressure ratio across each turbine stage in the ideal case becomes √ rp. Using the isentropic relations, the temperatures at the compressor and turbine exit can be expressed as
( )( )( )
( ) ( )( )
( ) ( )( ) ( ) ( ) ( ) kk
pkk
pkk
pkk
p
kk
p
kk
kkp
kk
p
kk
kkp
kk
rTrrTrTr
TPTT
rTr
TPPTTT
rTPPTTT
2/11
2/1/11
2/12
/1
5
/16
5
2/13
/1
3
/1
3
4347
/11
/1
1
2125
1
1
−−−−
−−
−
−−
−−
===⎟⎟
⎠
⎞
⎜⎜
⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎛
=
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛==
=⎟⎟⎠
⎞⎜⎜⎝
⎛==
6 P5⎝
Then,
( ) ( )( )
( ) ( )( )12/1 −=−=−= − kkrTcTTchhq
1
1161
2/137373in −=−=−= −
ppp
kkppp rTcTTchhq
6out
and thus ( )( )
( )( )kkpp rTcq 2/1
3inth
111
−−−=−=η
which simplifies to
kkpp rTcq 2/1
1out 1− −
( ) kkpr
TT 2/1
3
1th 1 −−=η
The thermal efficiency of the single stage ideal regenerative cycle is given as
( ) kkpr
TT /1
3
1th 1 −−=η
Therefore, the regenerative cycle with two stages of expansion has a higher thermal efficiency than the standard regenerative cycle with a single stage of expansion for any given value of the pressure ratio rp.
s
T
1
2
7
qin
5
6
4
3
qout
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
9-151
9-179 A gas-turbine plant operates on the regenerative Brayton cycle with reheating and intercooling. The back work ratio, the net work output, the thermal efficiency, the second-law efficiency, and the exergies at the exits of the combustion chamber and the regenerator 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.
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.
Optimum intercooling and reheating pressure is
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
kPa 4.346)1200)(100(412 === PPP
Process 1-2, 3-4: Compression
KkJ/kg 7054.5kPa 100K 300
kJ/kg 43.300K 300
11
1
11
⋅=⎭⎬⎫
==
=⎯→⎯=
sPT
hT
=⎭⎬⎫
===
shss
P kJ/kg 79.428
kJ/kg.K 7054.5kPa 4.346
212
2
kJ/kg 88.46043.300
43.30079.42880.0 2212
12C =⎯→⎯
−−
=⎯→⎯−−
= hhhh
hh sη
s
T
3
4
1
5 qin
8 6
7
10
9
2
8s 6s
4s 2s
KkJ/kg 5040.5kPa 4.346
K 350kJ/kg 78.350K 350
33
3
33
⋅=⎭⎬⎫
==
=⎯→⎯=
sPT
hT
43
=⎭⎬⎫
== shss
kJ/kg 42.500kJ/kg.K 5040.5
kPa 12004 =P
4
kJ/kg 83.53778.350
78.35042.50080.0 4434
34C =⎯→⎯
−−
=⎯→⎯−−
= hhhh
hh sη
Process 6-7, 8-9: Expansion
=⎭⎬⎫
===
shss
P
KkJ/kg 6514.6kPa 1200K 1400
kJ/kg 9.1514K 1400
66
6
66
⋅=⎭⎬⎫
==
=⎯→⎯=
sPT
hT
kJ/kg 9.1083kJ/kg.K 6514.6
kPa 4.3467
67
7
kJ/kg 1.11709.10839.1514
9.151480.0 7
7
76T =
hη 76 =⎯→⎯
−−
=⎯→⎯−−
hh
hhh
s
8
88
⋅=⎭⎬⎫
==
=⎯→⎯=
sPT
hT
96kPa 100
99 =
⎭⎬⎫=
shP
KkJ/kg 9196.6kPa 4.346
K 1300kJ/kg 6.1395K 1300
88
kJ/kg 00.996kJ/kg.K 91.689 == ss
preparation. If you are a student using this Manual, you are using it without permission.
9-152
kJ/kg 9.107500.9966.1395
6.139580.0 9
9
98
98T =⎯→⎯
−−
=⎯→⎯−−
= hh
hhhh
sη
C ysycle anal is:
kJ/kg 50.34778.35083.53743.30088.4603412inC, =−+−=−+−= hhhhw
kJ/kg 50.6649.10756.13951.11709.15149876outT, =−+−=−+−= hhhhw
0.523===50.66450.347
outT,
inC,bw w
wr
kJ/kg 317.0=−=−= 50.34750.664inC,outT,net www
Regenerator analysis:
kJ/kg 36.67283.5379.1075
9.107575.0 10
10
49
109regen =⎯→⎯
−−
=⎯→⎯−−
= hh
hhhh
ε
1010
10 ⋅=⎭⎬=
sP
(b) =−=−= hhq
KkJ/kg 5157.6kPa 100
K 36.672 ⎫=h
kJ/kg 40.94183.53736.6729.1075 5545109regen =⎯→⎯−=−⎯→⎯−=−= hhhhhhq
kJ/kg 54.57340.9419.151456in
0.553===54.5730.317
in
netth q
wη
(c) 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,
786.0K 1400K 30011
6
1max =−=−=
TT
η
and
0.704===786.0553.0
maxηη
η thII
(d) The exergies at the combustion chamber exit and the regenerator exit are
kJ/kg 930.7=−−−=
−−−=kJ/kg.K)7054.56514.6)(K 300(kJ/kg)43.3009.1514(
)( 060066 ssThhx
kJ/kg 128.8=−−−=
−−−=kJ/kg.K)7054.55157.6)(K 300(kJ/kg)43.30036.672(
)( 010001010 ssThhx
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
9-153
9-180 The thermal efficiency of a two-stage gas turbine with regeneration, reheating and intercooling to that of a three-stage gas turbine is to be compared.
Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg⋅K and k = 1.4 (Table A-2a).
Analysis
Two Stages:
s
T
3
4
1
6 873 K
283 K
97
8
2
5
10
The pressure ratio across each stage is
416 ==pr
The temperatures at the end of compression and expansion are
c K 5.420K)(4) 283( 0.4/1.4/)1(min === − kk
prTT
K 5.58741K) 873(1 0.4/1.4/)1(
max =⎟⎠⎞
⎜⎝⎛=⎟
⎟⎠
⎞⎜⎜⎝
⎛=
− kk
pe r
TT
The heat input and heat output are
kJ/kg 9.573K )5.587K)(873kJ/kg 2(1.005)(2 maxin =−⋅=−= ep TTcq
kJ/kg 4.276K )283K)(420.5kJ/kg 2(1.005)(2
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
minout =−⋅=−= TTcq
e therm of the cycle is then
cp
Th al efficiency
0.518=−=−= 1thη9.573inq4.2761outq
hree Stages:
The pressure ratio across each stage is
==r
The temperatures at the end of compression and expansion are
T
s
T
3
4
1
9
8
2
5
10
11
6
7
12
13
14
p 520.216 3/1
K 5.368K)(2.520) 283( 0.4/1.4/)1(min === − kk
pc rTT
K 4.6702.520
1K) 873(1/)1(
=⎟⎞
⎜⎛
=− kk
TT0.4/1.4
max =⎟⎠⎞
⎜⎝⎛
⎟⎠
⎜⎝ pr
he heat input and heat output are
kg
e
T
kJ/ 3(1.005)(3 maxin =−= ep TTcq kJ/kg 8.610K )4.670K)(873 =−⋅
kJ/kg 8.257K )283K)(368.5kJ/kg (1.0053)(3 min out =−⋅=−= TTcq cp
The thermal efficiency of the cycle is then
0.578=−=−=8.6108.25711
in
outth q
qη
preparation. If you are a student using this Manual, you are using it without permission.
9-154
9-181E A pure jet engine operating on an ideal cycle is considered. The thrust force produced per unit mass flow rate is 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 The turbine work output is equal to the compressor work input.
Properties The properties of air at room temperature are R = 0.3704 psia⋅ft3/lbm⋅R (Table A-1E), cp = 0.24 Btu/lbm⋅R and k = 1.4 (Table A-2Ea).
Analysis (a) We assume the aircraft is stationary and the air is moving towards the aircraft at a velocity of V 1 = 1200 ft/s. Ideally, the air will leave the diffuser with a negligible velocity (V 2 ≅ 0).
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
Diffuser:
( )( )
( )( )( )
( ) psia 5.21 R490 R609.8psia 10
R8.609/sft 25,037
Btu/lbm1R BtuJlbm0.242
ft/s 1200 R4902
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
TTPP
cVTT
VTTc
VVhhVhVh
EEEEE &&&&&
qin
qout
s
T
1
2
4
3 5
6
Compressor:
( )( ) ( )( )( )
( )( ) R4.11429 R609.8
psia 5.193psia 21.59
0.4/1.4/1
2
323
243
==⎟⎟⎠
⎞⎜⎜⎝
⎛=
====
− kk
p
PP
TT
PrPP
bine:
Tur
( ) ( )54235423outturb,incomp, TTcTTchhhhww pp −=−⎯→⎯−=−⎯→⎯=
R4.6278.6094.114211602345 =+−=+−= TTTT or
Nozzle: ( )
( ) R5.497psia 193.5
psia 10R 11600.4/1.4/1
4
646 =⎟⎟
⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
− kk
PPTT
( ) 2/02
0
2/2/2
systemoutin
VhVh +=+
2656
025
26
56
26
outin(steady) 0
VTTcVV
hh
EEEEE
p +−=⎯→⎯−
+−=
=⎯→⎯∆=− &&&&&
or,
655
( )( )( ) ft/s 1249Btu/lbm 1
/sft 25,037R5.497627.4RBtu/lbm 0.24222
6 =⎟⎟⎠
⎞⎜⎜⎝
⎛−⋅== exitVV
The specific impulse is then
m/s 49=−=−= 12001249inletexit VVmF&
preparation. If you are a student using this Manual, you are using it without permission.
9-155
9-182 The electricity and the process heat requirements of a manufacturing facility are to be met by a cogeneration plant consisting of a gas-turbine and a heat exchanger for steam production. The mass flow rate of the air in the cycle, the back work ratio, the thermal efficiency, the rate at which steam is produced in the heat exchanger, and the utilization efficiency of the cogeneration plant 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.
1
Combustion chamber
Turbine
23
4
Compress.
100 kPa20°C
450°C1 MPa
325°C 15°C
Sat. vap. 200°C
Heat exchanger
5
Analysis (a) For this problem, we use the properties of air 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
KkJ/kg 682.5kPa 100
C20kJ/kg 5.293C20
11
1
11
⋅=⎭⎬⎫
=°=
=⎯→⎯°=
sPT
hT
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
=⎭⎬⎫
===
shss
P kJ/kg 2.567
kJ/kg.K 682.5kPa 1000
212
2
kJ/kg 8.6115.293
5.2932.56786.0 2212
C ⎯−
=hh
η 12 =⎯→⎯−
−=⎯→
−h
hhh s
Process 3-4: Expansion
kJ/kg 5.738C450 44 =⎯→⎯°= hT
ss hhh
hhhh
43
3
43
43T
5.73888.0
−−
=⎯→⎯−−
=η
We cannot find the enthalpy at state 3 directly. However, using the following lines in EES together with the isentropic efficienc 62 kJ/kg, T3 = 913.2ºC, s3 = 6.507 kJ/kg.K. The solution by hand would require a trial-error app
3=entropy(Air, T=T_3, P=P_2)
h_4s=enthalpy(Air, P=P_1, s=s_3)
The inlet water is compressed liquid at 15ºC and at the saturation pressure of steam at 200ºC (1555 kPa). This is not available in the tables but we can obtain it in EES. The alternative is to use saturated liquid enthalpy at the given temperature.
kJ/kg 47.64kPa 1555
C15
22
11
1
=⎭=
=⎭⎬⎫
=°=
w
ww hP
T
The net work output is
1
43out =−=−=
=−
hhw
y relation, we find h3 = 12roach.
h_3=enthalpy(Air, T=T_3)
s_
Also,
kJ/kg 4.605C325 55 =⎯→⎯°= hT
kJ/kg 27921
C2002⎬⎫°=w h
Tx
kJ/kg 4.5235.7381262kJ/kg 2.3185.2938.6112inC, =−= hhw
T,
preparation. If you are a student using this Manual, you are using it without permission.
9-156
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
, CoutT,net kJ/kg 2.2052.3184.523in =−=
The mass flow rate of air is
−= www
kg/s 7.311===kJ/kg 2.205netw
ma& kJ/s 1500netW&
(b) The back work ratio is
0.608===4.5232.318
outT,
inC,bw w
wr
The rate of heat input and the thermal efficiency are
kW 4753)kJ/kg8.611kg/s)(1262 311.7()( 23in =−=−= hhmQ a&&
31.6%==== 3156.0kW 4753kW 1500
in
netth Q
W&
&η
(c) An energy balance on the heat exchanger gives
(d) The heat supplied to the water in the heat exchanger (process heat) and the utilization efficiency are
kg/s 0.3569=⎯→⎯−=−
−=−
ww
wwwa
mm
hhmhhm
&&
&&
)kJ/kg47.64(2792)kJ/kg4.6055kg/s)(738. 311.7(
)()( 1254
kW 5.973)kJ/kg47.64kg/s)(2792 3569.0()( 12p =−=−= www hhmQ &&
52.0%==+
=+
= 5204.0kW 4753
5.9731500
in
pnet
Q
QWu &
&&ε
preparation. If you are a student using this Manual, you are using it without permission.
9-157
9-183 A turbojet aircraft flying is considered. The pressure of the gases at the turbine exit, the mass flow rate of the air through the compressor, the velocity of the gases at the nozzle exit, the propulsive power, and the propulsive efficiency of the cycle are to be determined.
Assumptions 1 The air-standard assumptions are applicable. 2 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).
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
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.
Diffuser, Process 1-2:
kJ/kg 23.238C35 11 =⎯→⎯°−= hT
kJ/kg 37.269/sm 1000
kJ/kg 12m/s) (15
/sm 1000kJ/kg 1
2m/s) (900/3.6kJ/kg) 23.238(
22
222
2
222
2
22
2
21
1
=⎯→⎯⎟⎠⎞
⎜⎝⎛+=⎟
⎠⎞
⎜⎝⎛+
+=+
hh
VhVh
s
T
1
2
4
3
qin
5
qout
6
KkJ/kg 7951.5kPa 50
kJ/k 37.2692 = g2
2⋅=
⎭⎬⎫
=s
Ph
Compressor, Process 2-3:
32
=⎭⎬⎫
== shss
kJ/kg 19.505kJ/kg.K 7951.5
kPa 4503 =P
3
kJ/kg 50.55337.269
37.26919.50583.0 3323
23C =⎯→⎯
−−
=⎯→⎯−−
= hhhh
hh sη
urbine, P cess 3-4:
here the ass flow rates through the compressor and the turbine are assumed equal.
T ro
kJ/kg 8.1304C950 44 =⎯→⎯°= hT
kJ/kg 6.10208.130437.26950.553 555423 =⎯→⎯−=−⎯→⎯−=− hhhhhh
w m
kJ/kg 45.9628.1304
6.10208.130454 ⎯→⎯−
=⎯→⎯−
=hh
η 83.0 5554
T =−− s
ssh
hhh
C9504
4 ⋅=⎫°=T
(b) The mass flow rate of the air through the compressor is
kPa 4504 ⎭
⎬=s
PKkJ/kg 7725.6
kPa 147.4=⎭⎬⎫
⋅===
545
5
KkJ/kg 7725.6kJ/kg 45.962
Pss
h s
kg/s 1.760=−
=−
=kJ/kg )37.26950.553(
kJ/s 500
23
C
hhW
m&
&
(c) Nozzle, Process 5-6:
kJ/kg 6.1020
55 ⋅=⎬
⎫=s
h KkJ/kg 8336.6
kPa 4.1475 ⎭=P
preparation. If you are a student using this Manual, you are using it without permission.
9-158
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
=⎭⎬⎫
===
shss
P kJ/kg 66.709
kJ/kg.K 8336.6kPa 40
656
6
kJ/kg 52.76266.7096.1020
6.102083.0 6
6
65
65N =⎯→⎯
−−
=⎯→⎯−−
= hh
hhhh
sη
m/s 718.5=⎯→⎯⎟⎠⎞
⎜⎝⎛+=+
+=+
622
26
26
6
25
5
/sm 1000kJ/kg 1
2kJ/kg 52.7620kJ/kg) 6.1020(
22
VV
VhVh
where the velocity at nozzle inlet is assumed zero.
(d) The propulsive power and the propulsive efficiency are
kW 206.1=⎟⎠⎞
⎜⎝⎛−=−= 22116 /sm 1000
kJ/kg 1m/s) 250)(m/s 250m/s 5.718(kg/s) 76.1()( VVVmWp &&
kW 1322kJ/kg)50.5538.1304(kg/s) 76.1()( 34in =−=−= hhmQ &&
0.156===kW 1322kW 1.206
inQ
W pp &
&η
preparation. If you are a student using this Manual, you are using it without permission.
9-159
9-184 The three processes of an air standard cycle are described. The cycle is to be shown on the P-v and T-s diagrams, and the expressions for back work ratio and the thermal efficiency are to be obtained.
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.
Analysis (a) The P-v and T-s diagrams for this cycle are as shown.
(b) The work of compression is found by the first law for process 1-2:
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
T
)3 2 3 22
Pdv P v v R T T= − = −∫
The back work ratio is
( )( )
1 2 1 2 1 2
1 2
1 2 1 2 2 1
1 2 2 1
0( )
v
comp v
q w uq isentropic processw u C T
w w C T T
− − −
−
− −
−
− = ∆=
= −∆ = − −
= − = −
The expansion work is found by
( ) (3
exp 2 3w w −= =
( )( )
( )( )
3 1 3 11
2 3 2
/ 1/ 1
comp v vw C T T T TC TT T T
− −
exp 3 2w R T T R= =
−
Process 1-2 is isentropic; therefore,
−
s
T
3 2
1
v
P
32
1
1
1 21
2 1kT V r −= =1
kT V
−⎛ ⎞⎜ ⎟⎝ ⎠
2 1
1 2P= = and
kkP V r
V⎛ ⎞⎜ ⎟⎝ ⎠
Process 2-3 is constant pressure; therefore,
3 3 3 32 2 1
3 2 2 2 2
PV T VPV V rT T T V V
= ⇒ = = =
Process 3-1 is constant volume; therefore,
3
T T3 3 31 1 2 kr
T P P= ⇒ = = =
The back work ratio becomes (Cv=R/(k-1))
PV T PPV P
3 1 1 1 1
1
1exp
1 11 1
kcomp
k
w 1w k r r
−
−
−=
− −
2
3 1out v
T
q C T T= −
The cycle thermal efficiency is given by
r
(c) Apply first law to the closed system for processes 2-3 and 3-1 to show:
( )3in pq C T= −
( )
( )( )
( )( )
3 1 1 3 1/ 111 1 1vout C T T T T Tqη− −
= − = − = −3 2 2 3 2/ 1th
in pq C T T k T T T− −
The efficiency becomes
preparation. If you are a student using this Manual, you are using it without permission.
9-160
1
1 1 111
k
th k
rk r r
η −
−= −
−
(d) Determine the value of the back work ratio and efficiency as r goes to unity.
( )( )
1
1exp
21 1
1 1 1 21 1 1exp 1
lim lim lim lim1 1 1 1 1k k k k kr r r
rw k r r k r r k kr k r− − − −→ → →
→
1exp
1 1 11 1
11 1 1 1 1 1
1 1 1 1lim 11 1 1 1
kcomp
k
kk kcomp
comp
r
w rw k r r
w k rr r
w k kw k k k k
−
−
−− −
→
−=
− −
⎧ ⎫ ⎧ ⎫−⎧ ⎫− −⎪ ⎪ ⎪ ⎪= = =⎨ ⎬ ⎨ ⎬ ⎨ ⎬− − − − − − −⎪ ⎪⎩ ⎭⎪ ⎪ ⎩ ⎭⎩
⎭− −⎧ ⎫ ⎧ ⎫= = =⎨ ⎬ ⎨ ⎬− − + −⎩ ⎭ ⎩ ⎭
( )
1
1
1 1 11 1 1 1
1
1 1 111
1 1 1 1 1 1lim 1 lim 1 lim 1 lim1 1
1 1lim 1 1 01 1
k
th k
k k k
th k k k kr r r r
thr
rk r r
r r krk r r k r r k kr k r
k kk k k k
η
η
η
−
−
− − −→ → → →
→
−= −
−
2k −
⎧ ⎫⎧ ⎫ ⎧ ⎫− − ⎪ ⎪= − = − = −⎨ ⎬ ⎨ ⎬ ⎨− − − − ⎬⎪ ⎪⎩ ⎭ ⎩ ⎭ ⎩ ⎭
⎧ ⎫ ⎧ ⎫= − = − =⎨ ⎬ ⎨ ⎬− +⎩ ⎭ ⎩ ⎭
These results show that if there is no compression (i.e. r = 1), there can be no expansion and no net work will be done even though heat may be added to the system.
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
9-161
9-185 The three processes of an air standard cycle are described. The cycle is to be shown on the P-v and T-s diagrams, and the expressions for back work ratio and the thermal efficiency are to be obtained.
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.
Analysis (a) The P-v and T-s diagrams for this cycle are as shown.
(b) The work of expansion is found by the first law for process 2-3:
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
3 2 3 2 3
2 3 2 3 3 2
exp 2 3 2 3
)
v
v
q w ucess
w u C T T
w w C T T
−
− −
−
− = ∆
= −∆ = − −
= = −
s found by
)1 3 3 13
v R T T− = −
The back work ratio is
( )( )
2− −
2 3 0(q isentropic pro− =
The compression work i
(3 1compw w Pdv P v−= − = − = −∫ ) (1
( )( )
( )( )
( )( )
3 1comp vw C T T C−= = 1 3 1 33
exp 2 3 3 2 3 2 3
1 / 1 // 1 / 1
v vT T T TT Cw R T T R T T T R T T
− −=
− −
Process 3-1 is constant pressure; therefore,
−
3 3 1 1 1 1 2
T T T V V= ⇒ = = =
3 1 3 3 3
1PV PV T V Vr
Process 2-3 is isentropic; therefore, 1k − k
12 2
3 2
kT V rT V
−⎛ ⎞= =⎜ ⎟
⎝ ⎠ and 32
3 2
kVP rP V
⎛ ⎞= =⎜ ⎟
⎝ ⎠
The back work ratio becomes (Cv=R/(k-1))
( ) 111 1
compk
rkw r r−= − =
− 1exp
11 1 1k
w k rr −
− − −−
(c) Apply first law to the closed system for processes 1-2 and 3-1 to show:
1
3 1t p
T
q C T T= −
( )( )
2in vq C T= −
ou
The cycle thermal efficiency is given by
( )( )
( )( )
3 1 1 3 1
2 1 1 2 1
/ 11 1 1
/ 1pout
thin v
C T T T T Tq kq C T T T T T
η− −
= − = − = −− −
Process 1-2 is constant volume; therefore,
2 2 1 1 2 2 2
2 1 1 1 3
kPV PV T P P rT T T P P
= ⇒ = = =
The efficiency becomes
v
P
3
2
1
s
T
3
2
1
preparation. If you are a student using this Manual, you are using it without permission.
9-162
111th k
rkr
η −= −
−
(d) Determine the value of the back work ratio and efficiency as r goes to unity.
( )
( ) ( )
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
( )
1 1exp
11 1 1exp
11 1 111 1
1 1lim 1 lim 1 lim1
compk k
compk kr r r
w k rrkw r r r
w rk kw r r kr
− −
−→
− − −= − =
− −
⎧ ⎫ ⎧ ⎫−= − = −⎨ ⎬ ⎨ ⎬− −⎩ ⎭ ⎩ ⎭→ →
1exp
1lim 1 11
comp
r
wk
w k→
⎧ ⎫= −
=⎨ ⎬−⎩ ⎭
11 1 1
1
111
1 1lim 1 lim 1 lim1
1lim 1 0
th k
th k kr r r
thr
rkr
rk kr k
kk
η
η
η
−→ → →
→
r
−= −
−⎧ ⎫ ⎧−
= − = −⎨ ⎬ ⎨−⎩ ⎭ ⎩⎧ ⎫= − =⎨ ⎬⎩ ⎭
⎫⎬⎭
These results show that if there is no compression (i.e. r = 1), there can be no expansion and no net work will be done even though heat may be added to the system.
preparation. If you are a student using this Manual, you are using it without permission.
9-163
9-186 The four processes of an air-standard cycle are described. The cycle is to be shown on the P-v and T-s diagrams; an expression for the cycle thermal efficiency is to be obtained; and the limit of the efficiency as the volume ratio during heat rejection approaches unity is to be evaluated.
Analysis (a) The P-v and T-s diagrams of the cycle are shown in the figures.
(b) Apply first law to the closed system for processes 2-3 and 4-1 to show:
( )( )
3 2
4 1
in v
out p
q C T T
q C T T
= −
= −
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
The cycle thermal efficiency is given by
( )( )
( )( )
4 1 1 4 1
3in v 2 2 3 2
/ 11 1 1
/ 1pout
th
C T T T T Tq kq C T T T T T
η− −
= − = − = −− −
s
T
42
1
3
v
P3
2
1
4
1
1 2
2 1
Process 1-2 is isentropic; therefore, 1
1k
k
T VT V r
−
−
⎛ ⎞= =⎜ ⎟
⎝ ⎠
Process 3-4 is isentropic; therefore, 1
13 4
4 3
kk
eT V rT V
−
−⎛ ⎞= =⎜ ⎟
⎝ ⎠
Process 4-1 is constant pressure; therefore,
4 4 1 1 4 4PV PV T V r= ⇒ = = 4 1 1 1
pT T T V
113 3 4 1
12 4 1 2
1 kk e
e p pk
T T rT T r r rT T T T r r
−−
−
⎛ ⎞= = = ⎜ ⎟⎝ ⎠
tant volume and V3 = V2, Since process 2-3 is cons
4 4 4 1
3 2 1 2e p
V V V Vr rV V V V
= = = r =
1kr rT−
⎛ ⎞3
2
p kp pr r
T r= =⎜ ⎟
⎝ ⎠
The efficiency becomes
1
1111
pth k k
p
rk
r rη −
−= −
−
(c) In the limit as rp approaches unity, the cycle thermal efficiency becomes
1 11 1
1 11
11 1lim 1 lim 1 lim1
1 1 1lim 1 1
p p p
p
pth k k k kr r r
p p
th th Ottok kr
rk k
r r r kr
kr k r
η
η η
− −→ → →
− −→
⎧ ⎫ ⎧−⎪ ⎪ ⎪= − = −⎨ ⎬ ⎨−⎪ ⎪ ⎪⎩ ⎭ ⎩⎧ ⎫= − = − =⎨ ⎬⎩ ⎭
11
1−
⎫⎪⎬⎪⎭
preparation. If you are a student using this Manual, you are using it without permission.
9-164
9-187 The four processes of an air-standard cycle are described. The back work ratio and its limit as rp goes to unity are to be determined, and the result is to be compared to the expression for the Otto cycle.
Analysis The work of compression for process 1-2 is found by the first law:
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
T
)14
T
( )( )
1 2 1 2 1 2
1 2
1 2 1 2 2 1
,1 2 1 2 2 1
0( )
v
comp v
q w uq isentropic processw u C T T
w w C T
− − −
−
− −
− −
− = ∆=
= −∆ = − −
= − = −
The work of compression for process 4-1 is found by
( ) (,4 1 4 1 1 4 4compw w Pdv P v v R T− −
1
= − = − −
The work of expansion for process 3-4 is found by the first law:
4 3 4
0( )
v
v
q w uq isentropic processw u C T T
C T T
− − −
−
− −
−
− = ∆=
= −∆ = − −
= −
The back work ratio is
= − − =∫
( )( )
3 4 3 4 3 4
3 4
3 4 3 4 4 3
exp,3 4 3w w− = −
( ) ( )( )
( ) ( )
( )4 1 2 1
4 1 2 1 1
exp 3 4 3 4 3
/ 1 / 1
1 /comp v v
v
T T T Tw R T T C T T CTw C T T T T T
R− + −
− + −= =
− −
Using data from the previous problem and Cv = R/(k-1)
( ) ( )11
exp 31 1
( 1) 1 1
1w T=
⎛ ⎞
1
kpcomp
k kp
k r rw T
r r
−
− −
− − + −
−⎜ ⎟⎜ ⎟⎝ ⎠
( ) ( ) ( )( ) ( )1 11 1
1 1exp 3 3
11 1
11
1exp 3
1
( 1) 1 1 1 0 1lim lim 11 11
1lim 11
p p
p
k kpcomp
r r
kk kp
kcomp
r
k
k r r k rw T Tw T T
rr r
w T rw T
r
− −
→ →
−− −
−
→
−
⎧ ⎫⎧ ⎫⎪ ⎪⎪ ⎪⎪ ⎪− − + − − + −⎪ ⎪ ⎪= =⎨ ⎬ ⎨
⎛ ⎞⎪ ⎪ ⎪ −−⎜ ⎟⎪ ⎪ ⎪⎜ ⎟ ⎩ ⎭⎝ ⎠⎪ ⎪⎩ ⎭⎧ ⎫⎪ ⎪−
= ⎨ ⎬⎪ ⎪−⎩ ⎭
⎪⎬⎪⎪
This result is the same expression for the back work ratio for the Otto cycle.
preparation. If you are a student using this Manual, you are using it without permission.
9-165
9-188 The effects of compression ratio on the net work output and the thermal efficiency of the Otto cycle for given operating conditions is to be investigated.
Analysis Using EES, the problem is solved as follows:
"Input Data" T[1]=300 [K] P[1]=100 [kPa] T[3] = 2000 [K] r_comp = 12 "Process 1-2 is isentropic compression" s[1]=entropy(air,T=T[1],P=P[1]) s[2]=s[1] T[2]=temperature(air, s=s[2], P=P[2]) P[2]*v[2]/T[2]=P[1]*v[1]/T[1] P[1]*v[1]=R*T[1] R=0.287 [kJ/kg-K] V[2] = V[1]/ r_comp "Conservation of energy for process 1 to 2" q_12 - w_12 = DELTAu_12 q_12 =0"isentropic process" DELTAu_12=intenergy(air,T=T[2])-intenergy(air,T=T[1]) "Process 2-3 is constant volume heat addition" v[3]=v[2] s[3]=entropy(air, T=T[3], P=P[3]) P[3]*v[3]=R*T[3] "Conservation of energy for process 2 to 3" q_23 - w_23 = DELTAu_23 w_23 =0"constant volume process" DELTAu_23=intenergy(air,T=T[3])-intenergy(air,T=T[2]) "Process 3-4 is isentropic expansion" s[4]=s[3] s[4]=entropy(air,T=T[4],P=P[4]) P[4]*v[4]=R*T[4] "Conservation of energy for process 3 to 4" q_34 -w_34 = DELTAu_34 q_34 =0"isentropic process" DELTAu_34=intenergy(air,T=T[4])-intenergy(air,T=T[3]) "Process 4-1 is constant volume heat rejection" V[4] = V[1] "Conservation of energy for process 4 to 1" q_41 - w_41 = DELTAu_41 w_41 =0 "constant volume process" DELTAu_41=intenergy(air,T=T[1])-intenergy(air,T=T[4]) 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*Convert(, %) "Thermal efficiency, in percent"
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
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9-166
ηth[%]
rcomp wnet[kJ/kg]
45.83 6 567.4 48.67 7 589.3 51.03 8 604.9 53.02 9 616.2 54.74 10 624.3 56.24 11 630 57.57 12 633.8 58.75 13 636.3 59.83 14 637.5 60.8 15 637.9
6 7 8 9 10 11 12 13 14 15560
570
580
590
600
610
620
630
640
rcomp
wne
t [k
J/kg
]
6 7 8 9 10 11 12 13 14 1545
48.5
52
55.5
59
62.5
rcomp
ηth
[%
]
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
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9-167
9-189 The effects of pressure ratio on the net work output and the thermal efficiency of a simple Brayton cbe investigate
ycle is to d. The pressure ratios at which the net work output and the thermal efficiency are maximum are to be
nalysis Using EES, the problem is solved as follows:
P[1])
"SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0"
ot_t + m_dot*h[4] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0"
ycle work, kW"
determined only to produce a T-s plot"
s[4]=entropy(air,T=T[4],P=P[4])
determined.
A
P_ratio = 8 T[1] = 300 [K] P[1]= 100 [kPa] T[3] = 1800 [K] m_dot = 1 [kg/s]
ta_c = 100/100 EEta_t = 100/100 "Inlet conditions" [1]=ENTHALPY(Air,T=T[1]) h
s[1]=ENTROPY(Air,T=T[1],P= "Compressor anaysis" s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor"
sure ratio - to find P[2]" P_ratio=P[2]/P[1]"Definition of presT_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])
ta_c =(h_s[2]-h[1])/(h[2]-h[1]) "Compressor adiabatic efficiency; Eta_c = W_dot_c_ideal/W_dot_c_actual. " Em_dot*h[1] +W_dot_c=m_dot*h[2]
lysis" "External heat exchanger anaP[3]=P[2]"process 2-3 is SSSF constant pressure"
ir,T=T[3]) h[3]=ENTHALPY(Am_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"
,T=T[3],P=P[3]) s[3]=ENTROPY(Airs_s[4]=s[3] "For the ideal case the entropies are constant across the turbine" P_ratio= P[3] /P[4]
=s_s[4],P=P[4]) "Ts[4] is the isentropic value of T[4] at turbine exit" T_s[4]=TEMPERATURE(Air,sh_s[4]=ENTHALPY(Air,T=T_s[4]) "Eta_t = W_dot_t /Wts_dot turbine adiabatic efficiency, Wts_dot > W_dot_t"
ta_t=(h[3]-h[4])/(h[3]-h_s[4]) Em_dot*h[3] = W_d "Cycle analysis"
f the net cW_dot_net=W_dot_t-W_dot_c"Definition oEta=W_dot_net/Q_dot_in"Cycle thermal efficiency"
work ratio" Bwr=W_dot_c/W_dot_t "Back"The following state points areT[2]=temperature(air,h=h[2]) T[4]=temperature(air,h=h[4]) s[2]=entropy(air,T=T[2],P=P[2])
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
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9-168
Bwr η Pratio Wc
[kW] Wnet[kW]
Wt[kW]
Qin[kW]
0.254 0.3383 5 175.8 516.3 692.1 1526 0.2665 0.3689 6 201.2 553.7 754.9 1501 0.2776 0.3938 7 223.7 582.2 805.9 1478 0.2876 0.4146 8 244.1 604.5 848.5 1458 0.2968 0.4324 9 262.6 622.4 885 1439 0.3052 0.4478 10 279.7 637 916.7 1422 0.313 0.4615 11 295.7 649 944.7 1406 0.3203 0.4736 12 310.6 659.1 969.6 1392 0.3272 0.4846 13 324.6 667.5 992.1 1378 0.3337 0.4945 14 337.8 674.7 1013 1364 0.3398 0.5036 15 350.4 680.8 1031 1352 0.3457 0.512 16 362.4 685.9 1048 1340 0.3513 0.5197 17 373.9 690.3 1064 1328 0.3567 0.5269 18 384.8 694.1 1079 1317 0.3618 0.5336 19 395.4 697.3 1093 1307 0.3668 0.5399 20 405.5 700 1106 1297 0.3716 0.5458 21 415.3 702.3 1118 1287 0.3762 0.5513 22 424.7 704.3 1129 1277 0.3806 0.5566 23 433.8 705.9 1140 1268 0.385 0.5615 24 442.7 707.2 1150 1259 0.3892 0.5663 25 451.2 708.3 1160 1251 0.3932 0.5707 26 459.6 709.2 1169 1243 0.3972 0.575 27 467.7 709.8 1177 1234 0.401 0.5791 28 475.5 710.3 1186 1227 0.4048 0.583 29 483.2 710.6 1194 1219 0.4084 0.5867 30 490.7 710.7 1201 1211 0.412 0.5903 31 498 710.8 1209 1204 0.4155 0.5937 32 505.1 710.7 1216 1197 0.4189 0.597 33 512.1 710.4 1223 1190 0.4222 0.6002 34 518.9 710.1 1229 1183
5 10 15 20 25 30 35500
525
550
575
600
625
650
675
700
725
0.3
0.35
0.4
0.45
0.5
0.55
0.6
0.65
Pratio
Wne
t [k
W]
η th
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
9-169
9-190 The effects of pressure ratio on the net work output and the thermal efficiency of a simple Brayton cycis to be investigated assuming adiabatic efficiencies of 85 percent for both the turbine and the comp
le ressor. The
mal efficiency are maximum are to be determined.
nalysis Using EES, the problem is solved as follows:
w for the actual compressor, assuming: adiabatic, ke=pe=0"
ot_t + m_dot*h[4] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0"
ycle work, kW" fficiency"
determined only to produce a T-s plot"
s[4]=entropy(air,T=T[4],P=P[4])
pressure ratios at which the net work output and the ther
A
P_ratio = 8 T[1] = 300 [K] P[1]= 100 [kPa] T[3] = 1800 [K]
] m_dot = 1 [kg/sta_c = 80/100 E
Eta_t = 80/100 "Inlet conditions"
T[1]) h[1]=ENTHALPY(Air,T=s[1]=ENTROPY(Air,T=T[1],P=P[1]) "Compressor anaysis" s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor"
sure ratio - to find P[2]" P_ratio=P[2]/P[1]"Definition of presT_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])
mpressor adiabatic efficiency; Eta_c = W_dot_c_ideal/W_dot_c_actual. " Eta_c =(h_s[2]-h[1])/(h[2]-h[1]) "Com_dot*h[1] +W_dot_c=m_dot*h[2] "SSSF First La
lysis" "External heat exchanger anaP[3]=P[2]"process 2-3 is SSSF constant pressure"
ir,T=T[3]) h[3]=ENTHALPY(Am_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"
,T=T[3],P=P[3]) s[3]=ENTROPY(Airs_s[4]=s[3] "For the ideal case the entropies are constant across the turbine" P_ratio= P[3] /P[4]
=s_s[4],P=P[4]) "Ts[4] is the isentropic value of T[4] at turbine exit" T_s[4]=TEMPERATURE(Air,sh_s[4]=ENTHALPY(Air,T=T_s[4]) "Eta_t = W_dot_t /Wts_dot turbine adiabatic efficiency, Wts_dot > W_dot_t"
ta_t=(h[3]-h[4])/(h[3]-h_s[4]) Em_dot*h[3] = W_d "Cycle analysis"
f the net cW_dot_net=W_dot_t-W_dot_c"Definition ota=W_dot_net/Q_dot_in"Cycle thermal eE
Bwr=W_dot_c/W_dot_t "Back work ratio" "The following state points areT[2]=temperature(air,h=h[2]) T[4]=temperature(air,h=h[4]) s[2]=entropy(air,T=T[2],P=P[2])
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
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9-170
Bwr η Pratio Wc
[kW] Wnet[kW]
Wt[kW]
Qin[kW]
0.3515 0.2551 5 206.8 381.5 588.3 1495 0.3689 0.2764 6 236.7 405 641.7 1465 0.3843 0.2931 7 263.2 421.8 685 1439 0.3981 0.3068 8 287.1 434.1 721.3 1415 0.4107 0.3182 9 309 443.3 752.2 1393 0.4224 0.3278 10 329.1 450.1 779.2 1373 0.4332 0.3361 11 347.8 455.1 803 1354 0.4433 0.3432 12 365.4 458.8 824.2 1337 0.4528 0.3495 13 381.9 461.4 843.3 1320 0.4618 0.355 14 397.5 463.2 860.6 1305 0.4704 0.3599 15 412.3 464.2 876.5 1290 0.4785 0.3643 16 426.4 464.7 891.1 1276 0.4862 0.3682 17 439.8 464.7 904.6 1262 0.4937 0.3717 18 452.7 464.4 917.1 1249 0.5008 0.3748 19 465.1 463.6 928.8 1237 0.5077 0.3777 20 477.1 462.6 939.7 1225 0.5143 0.3802 21 488.6 461.4 950 1214 0.5207 0.3825 22 499.7 460 959.6 1202 0.5268 0.3846 23 510.4 458.4 968.8 1192 0.5328 0.3865 24 520.8 456.6 977.4 1181
5 9 13 17 21 25380
390
400
410
420
430
440
450
460
470
0.24
0.26
0.28
0.3
0.32
0.34
0.36
0.38
0.4
Pratio
Wne
t [k
W]
η thPratio for
W net,max
W net η th
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
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9-171
9-191 The effects of pressure ratio, maximum cycle temperature, and compressor and turbine inefficiencies onnet work output per unit mass and the thermal efficiency of a simple Bray
the ton cycle with air as the working fluid is to be
e are to be used.
nalysis Using EES, the problem is solved as follows:
esult(T[1],P[1],r_comp,T[3]:Eta_th_ConstProp,Eta_th_easy)
ycle efficiency is:" _easy = (1 - 1/r_comp^(k-1))*Convert(, %) "[%]"
]}
ession"
P=P[2]) v[1]/T[1]
rocess 1 to 2"
T=T[1]) eat addition"
2 to 3"
nergy(air,T=T[2]) ion"
rocess 3 to 4"
T=T[3]) n"
to 1"
investigated. Constant specific heats at room temperatur
A
Procedure ConstPropR
:" "For AirC_V = 0.718 [kJ/kg-K] k = 1.4 T2 = T[1]*r_comp^(k-1) P2 = P[1]*r_comp^k q_in_23 = C_V*(T[3]-T2) T4 = T[3]*(1/r_comp)^(k-1) q_out_41 = C_V*(T4-T[1])
) "[%]" Eta_th_ConstProp = (1-q_out_41/q_in_23)*Convert(, %Easy Way to calculate the constant property Otto c"The
ta_thEEND "Input Data" T[1]=300 [K]
] P[1]=100 [kPaT[3] = 1000 [K{
r_comp = 12
1-2 is isentropic compr"Process s[1]=entropy(air,T=T[1],P=P[1]) s[2]=s[1]
(air, s=s[2],T[2]=temperatureP[2]*v[2]/T[2]=P[1]*P[1]*v[1]=R*T[1] R=0.287 [kJ/kg-K] V[2] = V[1]/ r_comp "Conservation of energy for pq_12 - w_12 = DELTAu_12 q_12 =0"isentropic process"
12=intenergy(air,T=T[2])-intenergy(air,DELTAu_"Process 2-3 is constant volume hv[3]=v[2] s[3]=entropy(air, T=T[3], P=P[3]) P[3]*v[3]=R*T[3] "Conservation of energy for process q_23 - w_23 = DELTAu_23 w_23 =0"constant volume process"
23=intenergy(air,T=T[3])-inteDELTAu_"Process 3-4 is isentropic expanss[4]=s[3] s[4]=entropy(air,T=T[4],P=P[4]) P[4]*v[4]=R*T[4] "Conservation of energy for pq_34 -w_34 = DELTAu_34 q_34 =0"isentropic process"
=intenergy(air,T=T[4])-intenergy(air,DELTAu_34"Process 4-1 is constant volume heat rejectioV[4] = V[1] "Conservation of energy for process 4q_41 - w_41 = DELTAu_41
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PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
t volume process" y(air,T=T[1])-intenergy(air,T=T[4])
all ConstPropResult(T[1],P[1],r_comp,T[3]:Eta_th_ConstProp,Eta_th_easy) erCentError = ABS(Eta_th - Eta_th_ConstProp)/Eta_th*Convert(, %) "[%]"
PerCentError r η
w_41 =0 "constanDELTAu_41=intenergq_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*Convert(, %) "Thermal efficiency, in percent" CP
[%] comp ηth
[%] th,ConstProp
[%] ηth,easy
[%] T3[K]
3.604 12 60.8 62.99 62.99 1000 6.681 12 59.04 62.99 62.99 1500 9.421 12 57.57 62.99 62.99 2000 11.64 12 56.42 62.99 62.99 2500
6 7 8 9 10 11 123.6
3.7
3.8
3.9
4
4.1
4.2
4.3
rcom p
Per
Cen
tErr
or [
%]
P ercent E rror = |η th - η th ,ConstProp | / η th
Tm ax = 1000 K
1000 1200 1400 1600 1800 2000 2200 2400 26004
6.2
8.4
10.6
12.8
15
T[3] [K]
Per
Cen
tErr
or [
%]
rcomp = 6
=12
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9-173
9-192 The effects of pressure ratio, maximum cycle temperature, and compressor and turbine efficiencies on the net work output per unit mass and the thermal efficiency of a simple Brayton cycle with air as the working fluid is to be investigated. Variable specific heats are to be used.
Analysis Using EES, the problem is solved as follows:
"Input data - from diagram window" {P_ratio = 8} {T[1] = 300 [K] P[1]= 100 [kPa] T[3] = 800 [K] m_dot = 1 [kg/s] Eta_c = 75/100 Eta_t = 82/100} "Inlet conditions" h[1]=ENTHALPY(Air,T=T[1]) s[1]=ENTROPY(Air,T=T[1],P=P[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]) 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,s=s_s[4],P=P[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" "Cycle analysis" W_dot_net=W_dot_t-W_dot_c"Definition of the net cycle work, kW" Eta=W_dot_net/Q_dot_in"Cycle thermal efficiency" Bwr=W_dot_c/W_dot_t "Back work ratio" "The following state points are determined only to produce a T-s plot" T[2]=temperature('air',h=h[2]) T[4]=temperature('air',h=h[4]) s[2]=entropy(air,T=T[2],P=P[2]) s[4]=entropy(air,T=T[4],P=P[4])
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9-174
Bwr η Pratio Wc
[kW] Wnet[kW]
Wt[kW]
Qin[kW]
0.5229 0.1 2 1818 1659 3477 16587 0.6305 0.1644 4 4033 2364 6396 14373 0.7038 0.1814 6 5543 2333 7876 12862 0.7611 0.1806 8 6723 2110 8833 11682 0.8088 0.1702 10 7705 1822 9527 10700
0.85 0.1533 12 8553 1510 10063 9852 0.8864 0.131 14 9304 1192 10496 9102 0.9192 0.1041 16 9980 877.2 10857 8426 0.9491 0.07272 18 10596 567.9 11164 7809 0.9767 0.03675 20 11165 266.1 11431 7241
5.0 5.5 6.0 6.5 7.0 7.50
500
1000
1500
s [kJ/kg-K]
T [K
]
100 kPa
800 kPa
1
2s
2
3
4
4s
Air Standard Brayton Cycle
Pressure ratio = 8 and Tmax = 1160K
2 4 6 8 10 12 14 16 18 200.00
0.05
0.10
0.15
0.20
0.25
0
500
1000
1500
2000
2500
Pratio
Cyc
le e
ffic
ienc
y,
Wne
t [k
W]
η Wnet
Tmax=1160 K
Note Pratio for maximum work and η
ηc = 0.75
ηt = 0.82
η
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
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9-175
9-193 The effects of pressure ratio, maximum cycle temperature, and compressor and turbine efficiencies on the net work output per unit mass and the thermal efficiency of a simple Brayton cycle with helium as the working
nalysis Using EES, the problem is solved as follows:
are needed for helium's enthalpy." r ir,T=T) ELSE hFunc: = enthalpy(Helium,T=T,P=P)
re the net work done and efficiency < 0?' Else EtaError$ = ''
m diagram window"
g/s]
t"
w for the actual compressor, assuming: adiabatic, ke=pe=0"
rst Law for the heat exchanger, assuming W=0, ke=pe=0"
[4],P[4]) "Eta_t = W_dot_t /Wts_dot turbine adiabatic efficiency, Wts_dot >
mpressor, assuming: adiabatic, ke=pe=0"
e work, kW" l efficiency"
determined only to produce a T-s plot"
s[4]=entropy(air,T=T[4],P=P[4])
fluid is to be investigated.
A
Function hFunc(WorkFluid$,T,P) The E ; thus, T and P " ES functions teat helium as a real gas
kFluid$ = 'Air' then hFunc:=enthalpy(AIF Wo endif END
dure EtaCheck(Eta_th:EtaError$) ProceIf Eta_th < 0 then EtaError$ = 'Why aEND "Input data - fro{P_ratio = 8}
{T[1] = 300 [K]P[1]= 100 [kPa]
] T[3] = 800 [Km_dot = 1 [kEta_c = 0.8 Eta_t = 0.8 WorkFluid$ = 'Helium'} "Inlet conditions"
,T[1],P[1]) h[1]=hFunc(WorkFluid$s[1]=ENTROPY(WorkFluid$,T=T[1],P=P[1]) "Compressor anaysis" s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor"
ratio - to find P[2]" P_ratio=P[2]/P[1]"Definition of pressureT_s[2]=TEMPERATURE(WorkFluid$,s=s_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exih_s[2]=hFunc(WorkFluid$,T_s[2],P[2])
mpressor adiabatic efficiency; Eta_c = W_dot_c_ideal/W_dot_c_actual. " Eta_c =(h_s[2]-h[1])/(h[2]-h[1]) "Com_dot*h[1] +W_dot_c=m_dot*h[2] "SSSF First La
"External heat exchanger analysis"P[3]=P[2]"process 2-3 is SSSF constant pressure"
luid$,T[3],P[3]) h[3]=hFunc(WorkFm_dot*h[2] + Q_dot_in= m_dot*h[3]"SSSF Fi"Turbine analysis"
rkFluid$,T=T[3],P=P[3]) s[3]=ENTROPY(Wos_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(WorkFluid$,s=s_s[4],P=P[4]) "Ts[4] is the isentropic value of T[4] at turbine exit" h_s[4]=hFunc(WorkFluid$,T_sW_dot_t"
(h[3]-h_s[4]) Eta_t=(h[3]-h[4])/m_dot*h[3] = W_dot_t + m_dot*h[4] "SSSF First Law for the actual co"Cycle analysis"
finition of the net cyclW_dot_net=W_dot_t-W_dot_c"DeEta_th=W_dot_net/Q_dot_in"Cycle thermaCall EtaCheck(Eta_th:EtaError$)
work ratio" Bwr=W_dot_c/W_dot_t "Back"The following state points areT[2]=temperature(air,h=h[2]) T[4]=temperature(air,h=h[4]) s[2]=entropy(air,T=T[2],P=P[2])
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9-176
Bwr η Pratio Wc
[kW] Wnet
[kW] Wt
[kW] Qin
[kW] 0.5229 0.1 2 1818 1659 3477 16587 0.6305 0.1644 4 4033 2364 6396 14373 0.7038 0.1814 6 5543 2333 7876 12862 0.7611 0.1806 8 6723 2110 8833 11682 0.8088 0.1702 10 7705 1822 9527 10700
0.85 0.1533 12 8553 1510 10063 9852 0.8864 0.131 14 9304 1192 10496 9102 0.9192 0.1041 16 9980 877.2 10857 8426 0.9491 0.07272 18 10596 567.9 11164 7809 0.9767 0.03675 20 11165 266.1 11431 7241
5.0 5 .5 6 .0 6 .5 7 .0 7 .50
500
1000
1500
s [kJ /kg -K ]
T [K
]
100 kP a
800 kP a
1
2 s
2
3
4
4 s
B rayto n C yc le
P ressu re ra tio = 8 an d T m ax = 1160K
2 4 6 8 10 12 14 16 18 200.00
0.05
0.10
0.15
0.20
0.25
0
500
1000
1500
2000
2500
Pratio
Cyc
le e
ffici
ency
,
Wne
t [k
W]
ηWnet
Tmax=1160 K
Note Pratio for maximum work and η
η c = 0.75η t = 0.82
η
Brayton Cycle using Airm air = 20 kg/s
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9-177
9-194 The effect of the number of compression and expansion stages on the thermal efficiency of an ideal regenerative Brayton cycle with multistage compression and expansion and air as the working fluid is to be investigated.
Analysis Using EES, the problem is solved as follows:
"Input data for air" C_P = 1.005 [kJ/kg-K] k = 1.4 "Nstages is the number of compression and expansion stages" Nstages = 1 T_6 = 1200 [K] Pratio = 12 T_1 = 300 [K] P_1= 100 [kPa] Eta_reg = 1.0 "regenerator effectiveness" Eta_c =1.0 "Compressor isentorpic efficiency" Eta_t =1.0 "Turbine isentropic efficiency" R_p = Pratio^(1/Nstages) "Isentropic Compressor anaysis" T_2s = T_1*R_p^((k-1)/k) P_2 = R_p*P_1 "T_2s is the isentropic value of T_2 at compressor exit" 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_2s-T_1) "Actual compressor analysis:" w_comp = C_P*(T_2 - T_1) "Since intercooling is assumed to occur such that T_3 = T_1 and the compressors have the same pressure ratio, the work input to each compressor is the same. The total compressor work is:" w_comp_total = Nstages*w_comp "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" "The heat added in the external heat exchanger + the reheat between turbines is" q_in_total = C_P*(T_6 - T_5) +(Nstages - 1)*C_P*(T_8 - T_7) "Reheat is assumed to occur until:" T_8 = T_6 "Turbine analysis" P_7 = P_6 /R_p "T_7s is the isentropic value of T_7 at turbine exit" T_7s = T_6*(1/R_p)^((k-1)/k) "Turbine adiabatic efficiency, w_turbisen > w_turb" Eta_t = w_turb /w_turbisen "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_6 - T_7s) "Actual Turbine analysis:" w_turb = C_P*(T_6 - T_7) w_turb_total = Nstages*w_turb "Cycle analysis" w_net=w_turb_total-w_comp_total "[kJ/kg]" Bwr=w_comp/w_turb "Back work ratio"
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9-178
P_4=P_2 P_5=P_4 P_6=P_5 T_4 = T_2 "The regenerator effectiveness gives T_5 as:" Eta_reg = (T_5 - T_4)/(T_9 - T_4) T_9 = T_7 "Energy balance on regenerator gives T_10 as:" T_4 + T_9=T_5 + T_10 "Cycle thermal efficiency with regenerator" Eta_th_regenerative=w_net/q_in_total*Convert(, %) "[%]" "The efficiency of the Ericsson cycle is the same as the Carnot cycle operating between the same max and min temperatures, T_6 and T_1 for this problem." Eta_th_Ericsson = (1 - T_1/T_6)*Convert(, %) "[%]"
ηth,Ericksson[%]
ηth,Regenerative[%]
Nstages
75 49.15 1 75 64.35 2 75 68.32 3 75 70.14 4 75 72.33 7 75 73.79 15 75 74.05 19 75 74.18 22
0 2 4 6 8 10 12 14 16 18 20 22 2440
50
60
70
80
Nstages
ηth
[%]
Ericsson
Ideal Regenerative Brayton
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9-179
9-195 The effect of the number of compression and expansion stages on the thermal efficiency of an ideal regenerative Brayton cycle with multistage compression and expansion and helium as the working fluid is to be investigated.
Analysis Using EES, the problem is solved as follows:
"Input data for Helium" C_P = 5.1926 [kJ/kg-K] k = 1.667 "Nstages is the number of compression and expansion stages" {Nstages = 1} T_6 = 1200 [K] Pratio = 12 T_1 = 300 [K] P_1= 100 [kPa] Eta_reg = 1.0 "regenerator effectiveness" Eta_c =1.0 "Compressor isentorpic efficiency" Eta_t =1.0 "Turbine isentropic efficiency" R_p = Pratio^(1/Nstages) "Isentropic Compressor anaysis" T_2s = T_1*R_p^((k-1)/k) P_2 = R_p*P_1 "T_2s is the isentropic value of T_2 at compressor exit" 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_2s-T_1) "Actual compressor analysis:" w_comp = C_P*(T_2 - T_1) "Since intercooling is assumed to occur such that T_3 = T_1 and the compressors have the same pressure ratio, the work input to each compressor is the same. The total compressor work is:" w_comp_total = Nstages*w_comp "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" "The heat added in the external heat exchanger + the reheat between turbines is" q_in_total = C_P*(T_6 - T_5) +(Nstages - 1)*C_P*(T_8 - T_7) "Reheat is assumed to occur until:" T_8 = T_6 "Turbine analysis" P_7 = P_6 /R_p "T_7s is the isentropic value of T_7 at turbine exit" T_7s = T_6*(1/R_p)^((k-1)/k) "Turbine adiabatic efficiency, w_turbisen > w_turb" Eta_t = w_turb /w_turbisen "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_6 - T_7s) "Actual Turbine analysis:" w_turb = C_P*(T_6 - T_7) w_turb_total = Nstages*w_turb "Cycle analysis" w_net=w_turb_total-w_comp_total
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9-180
Bwr=w_comp/w_turb "Back work ratio" P_4=P_2 P_5=P_4 P_6=P_5 T_4 = T_2 "The regenerator effectiveness gives T_5 as:" Eta_reg = (T_5 - T_4)/(T_9 - T_4) T_9 = T_7 "Energy balance on regenerator gives T_10 as:" T_4 + T_9=T_5 + T_10 "Cycle thermal efficiency with regenerator" Eta_th_regenerative=w_net/q_in_total*Convert(, %) "[%]" "The efficiency of the Ericsson cycle is the same as the Carnot cycle operating between the same max and min temperatures, T_6 and T_1 for this problem." Eta_th_Ericsson = (1 - T_1/T_6)*Convert(, %) "[%]"
ηth,Ericksson[%]
ηth,Regenerative[%]
Nstages
75 32.43 1 75 58.9 2 75 65.18 3 75 67.95 4 75 71.18 7 75 73.29 15 75 73.66 19 75 73.84 22
0 2 4 6 8 10 12 14 16 18 20 22 2430
40
50
60
70
80
Nstages
ηth
[%]
EricssonIdeal Regenerative Brayton
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Fundamentals of Engineering (FE) Exam Problems
9-196 An Otto cycle with air as the working fluid has a compression ratio of 10.4. Under cold air standard conditions, the thermal efficiency of this cycle is
(a) 10% (b) 39% (c) 61% (d) 79% (e) 82%
Answer (c) 61%
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).
r=10.4 k=1.4 Eta_Otto=1-1/r^(k-1) "Some Wrong Solutions with Common Mistakes:" W1_Eta = 1/r "Taking efficiency to be 1/r" W2_Eta = 1/r^(k-1) "Using incorrect relation" W3_Eta = 1-1/r^(k1-1); k1=1.667 "Using wrong k value"
9-197 For specified limits for the maximum and minimum temperatures, the ideal cycle with the lowest thermal efficiency is
(a) Carnot (b) Stirling (c) Ericsson (d) Otto (e) All are the same
Answer (d) Otto
9-198 A Carnot cycle operates between the temperatures limits of 300 K and 2000 K, and produces 600 kW of net power. The rate of entropy change of the working fluid during the heat addition process is
(a) 0 (b) 0.300 kW/K (c) 0.353 kW/K (d) 0.261 kW/K (e) 2.0 kW/K
Answer (c) 0.353 kW/K
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).
TL=300 "K" TH=2000 "K" Wnet=600 "kJ/s" Wnet= (TH-TL)*DS "Some Wrong Solutions with Common Mistakes:" W1_DS = Wnet/TH "Using TH instead of TH-TL" W2_DS = Wnet/TL "Using TL instead of TH-TL" W3_DS = Wnet/(TH+TL) "Using TH+TL instead of TH-TL"
9-182
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9-199 Air in an ideal Diesel cycle is compressed from 2 L to 0.13 L, and then it expands during the constant pressure heat addition process to 0.30 L. Under cold air standard conditions, the thermal efficiency of this cycle is
(a) 41% (b) 59% (c) 66% (d) 70% (e) 78%
Answer (b) 59%
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).
V1=2 "L" V2= 0.13 "L" V3= 0.30 "L"
W1_Eta = 1-(1/r1^(k-1))*(rc^k-1)/k/(rc-1); r1=V1/V3 "Wrong r value" W2_Eta = 1-Eta_Diesel "Using incorrect relation" W3_Eta = 1-(1/r^(k1-1))*(rc^k1-1)/k1/(rc-1); k1=1.667 "Using wrong k value" W4_Eta = 1-1/r^(k-1) "Using Otto cycle efficiency"
9-200 Helium gas in an ideal Otto cycle is compressed from 20°C and 2.5 L to 0.25 L, and its temperature increases by an additional 700°C during the heat addition process. The temperature of helium before the expansion process is
(a) 1790°C (b) 2060°C (c) 1240°C (d) 620°C (e) 820°C
Answer (a) 1790°C
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).
k=1.667 V1=2.5 V2=0.25 r=V1/V2 T1=20+273 "K" T2=T1*r^(k-1) T3=T2+700-273 "C" "Some Wrong Solutions with Common Mistakes:" W1_T3 =T22+700-273; T22=T1*r^(k1-1); k1=1.4 "Using wrong k value" W2_T3 = T3+273 "Using K instead of C" W3_T3 = T1+700-273 "Disregarding temp rise during compression" W4_T3 = T222+700-273; T222=(T1-273)*r^(k-1) "Using C for T1 instead of K"
r=V1/V2 rc=V3/V2 k=1.4 Eta_Diesel=1-(1/r^(k-1))*(rc^k-1)/k/(rc-1) "Some Wrong Solutions with Common Mistakes:"
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9-183
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9-201 In an ideal Otto cycle, air is compressed from 1.20 kg/m3 and 2.2 L to 0.26 L, and the net work output of the cycle is 440 kJ/kg. The mean effective pressure (MEP) for this cycle is
(a) 612 kPa (b) 599 kPa (c) 528 kPa (d) 416 kPa (e) 367 kPa
Answer (b) 599 kPa
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).
rho1=1.20 "kg/m^3" k=1.4 V1=2.2 V2=0.26 m=rho1*V1/1000 "kg" w_net=440 "kJ/kg" Wtotal=m*w_net MEP=Wtotal/((V1-V2)/1000) "Some Wrong Solutions with Common Mistakes:" W1_MEP = w_net/((V1-V2)/1000) "Disregarding mass" W2_MEP = Wtotal/(V1/1000) "Using V1 instead of V1-V2" W3_MEP = (rho1*V2/1000)*w_net/((V1-V2)/1000); "Finding mass using V2 instead of V1" W4_MEP = Wtotal/((V1+V2)/1000) "Adding V1 and V2 instead of subtracting"
9-202 In an ideal Brayton cycle, air is compressed from 95 kPa and 25°C to 1100 kPa. Under cold air standard conditions, the thermal efficiency of this cycle is
(a) 45% (b) 50% (c) 62% (d) 73% (e) 86%
Answer (b) 50%
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).
P1=95 "kPa" P2=1100 "kPa" T1=25+273 "K" rp=P2/P1 k=1.4 Eta_Brayton=1-1/rp^((k-1)/k) "Some Wrong Solutions with Common Mistakes:" W1_Eta = 1/rp "Taking efficiency to be 1/rp" W2_Eta = 1/rp^((k-1)/k) "Using incorrect relation" W3_Eta = 1-1/rp^((k1-1)/k1); k1=1.667 "Using wrong k value"
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9-203 Consider an ideal Brayton cycle executed between the pressure limits of 1200 kPa and 100 kPa and temperature limits of 20°C and 1000°C with argon as the working fluid. The net work output of the cycle is
(a) 68 kJ/kg (b) 93 kJ/kg (c) 158 kJ/kg (d) 186 kJ/kg (e) 310 kJ/kg
Answer (c) 158 kJ/kg
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).
P1=100 "kPa" P2=1200 "kPa" T1=20+273 "K" T3=1000+273 "K" rp=P2/P1 k=1.667 Cp=0.5203 "kJ/kg.K" Cv=0.3122 "kJ/kg.K" T2=T1*rp^((k-1)/k) q_in=Cp*(T3-T2) Eta_Brayton=1-1/rp^((k-1)/k) w_net=Eta_Brayton*q_in "Some Wrong Solutions with Common Mistakes:" W1_wnet = (1-1/rp^((k-1)/k))*qin1; qin1=Cv*(T3-T2) "Using Cv instead of Cp" W2_wnet = (1-1/rp^((k-1)/k))*qin2; qin2=1.005*(T3-T2) "Using Cp of air instead of argon" W3_wnet = (1-1/rp^((k1-1)/k1))*Cp*(T3-T22); T22=T1*rp^((k1-1)/k1); k1=1.4 "Using k of air instead of argon" W4_wnet = (1-1/rp^((k-1)/k))*Cp*(T3-T222); T222=(T1-273)*rp^((k-1)/k) "Using C for T1 instead of K"
9-204 An ideal Brayton cycle has a net work output of 150 kJ/kg and a backwork ratio of 0.4. If both the turbine and the compressor had an isentropic efficiency of 85%, the net work output of the cycle would be
(a) 74 kJ/kg (b) 95 kJ/kg (c) 109 kJ/kg (d) 128 kJ/kg (e) 177 kJ/kg
Answer (b) 95 kJ/kg
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).
wcomp/wturb=0.4 wturb-wcomp=150 "kJ/kg" Eff=0.85 w_net=Eff*wturb-wcomp/Eff "Some Wrong Solutions with Common Mistakes:" W1_wnet = Eff*wturb-wcomp*Eff "Making a mistake in Wnet relation" W2_wnet = (wturb-wcomp)/Eff "Using a wrong relation" W3_wnet = wturb/eff-wcomp*Eff "Using a wrong relation"
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9-205 In an ideal Brayton cycle, air is compressed from 100 kPa and 25°C to 1 MPa, and then heated to 927°C before entering the turbine. Under cold air standard conditions, the air temperature at the turbine exit is
(a) 349°C (b) 426°C (c) 622°C (d) 733°C (e) 825°C
Answer (a) 349°C
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).
P1=100 "kPa" P2=1000 "kPa" T1=25+273 "K" T3=900+273 "K" rp=P2/P1 k=1.4 T4=T3*(1/rp)^((k-1)/k)-273 "Some Wrong Solutions with Common Mistakes:" W1_T4 = T3/rp "Using wrong relation" W2_T4 = (T3-273)/rp "Using wrong relation" W3_T4 = T4+273 "Using K instead of C" W4_T4 = T1+800-273 "Disregarding temp rise during compression"
9-206 In an ideal Brayton cycle with regeneration, argon gas is compressed from 100 kPa and 25°C to 400 kPa, and then heated to 1200°C before entering the turbine. The highest temperature that argon can be heated in the regenerator is
(a) 246°C (b) 846°C (c) 689°C (d) 368°C (e) 573°C
Answer (e) 573°C
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). k=1.667 Cp=0.5203 "kJ/kg.K" P1=100 "kPa" P2=400 "kPa" T1=25+273 "K" T3=1200+273 "K" "The highest temperature that argon can be heated in the regenerator is the turbine exit temperature," rp=P2/P1 T2=T1*rp^((k-1)/k) T4=T3/rp^((k-1)/k)-273 "Some Wrong Solutions with Common Mistakes:" W1_T4 = T3/rp "Using wrong relation" W2_T4 = (T3-273)/rp^((k-1)/k) "Using C instead of K for T3" W3_T4 = T4+273 "Using K instead of C" W4_T4 = T2-273 "Taking compressor exit temp as the answer"
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9-207 In an ideal Brayton cycle with regeneration, air is compressed from 80 kPa and 10°C to 400 kPa and 175°C, is heated to 450°C in the regenerator, and then further heated to 1000°C before entering the turbine. Under cold air standard conditions, the effectiveness of the regenerator is
(a) 33% (b) 44% (c) 62% (d) 77% (e) 89%
Answer (d) 77%
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).
k=1.4 Cp=1.005 "kJ/kg.K" P1=80 "kPa" P2=400 "kPa" T1=10+273 "K" T2=175+273 "K" T3=1000+273 "K" T5=450+273 "K" "The highest temperature that the gas can be heated in the regenerator is the turbine exit temperature," rp=P2/P1 T2check=T1*rp^((k-1)/k) "Checking the given value of T2. It checks." T4=T3/rp^((k-1)/k) Effective=(T5-T2)/(T4-T2) "Some Wrong Solutions with Common Mistakes:" W1_eff = (T5-T2)/(T3-T2) "Using wrong relation" W2_eff = (T5-T2)/(T44-T2); T44=(T3-273)/rp^((k-1)/k) "Using C instead of K for T3" W3_eff = (T5-T2)/(T444-T2); T444=T3/rp "Using wrong relation for T4"
9-208 Consider a gas turbine that has a pressure ratio of 6 and operates on the Brayton cycle with regeneration between the temperature limits of 20°C and 900°C. If the specific heat ratio of the working fluid is 1.3, the highest thermal efficiency this gas turbine can have is
(a) 38% (b) 46% (c) 62% (d) 58% (e) 97%
Answer (c) 62%
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).
k=1.3 rp=6 T1=20+273 "K" T3=900+273 "K" Eta_regen=1-(T1/T3)*rp^((k-1)/k) "Some Wrong Solutions with Common Mistakes:" W1_Eta = 1-((T1-273)/(T3-273))*rp^((k-1)/k) "Using C for temperatures instead of K" W2_Eta = (T1/T3)*rp^((k-1)/k) "Using incorrect relation" W3_Eta = 1-(T1/T3)*rp^((k1-1)/k1); k1=1.4 "Using wrong k value (the one for air)"
preparation. If you are a student using this Manual, you are using it without permission.
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PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
9-209 An ideal gas turbine cycle with many stages of compression and expansion and a regenerator of 100 percent effectiveness has an overall pressure ratio of 10. Air enters every stage of compressor at 290 K, and every stage of turbine at 1200 K. The thermal efficiency of this gas-turbine cycle is
(a) 36% (b) 40% (c) 52% (d) 64% (e) 76%
Answer (e) 76%
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).
k=1.4 rp=10 T1=290 "K" T3=1200 "K" Eff=1-T1/T3 "Some Wrong Solutions with Common Mistakes:" W1_Eta = 100 W2_Eta = 1-1/rp^((k-1)/k) "Using incorrect relation" W3_Eta = 1-(T1/T3)*rp^((k-1)/k) "Using wrong relation" W4_Eta = T1/T3 "Using wrong relation"
9-210 Air enters a turbojet engine at 320 m/s at a rate of 30 kg/s, and exits at 650 m/s relative to the aircraft. The thrust developed by the engine is
(a) 5 kN (b) 10 kN (c) 15 kN (d) 20 kN (e) 26 kN
Answer (b) 10 kN
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).
Vel1=320 "m/s" Vel2=650 "m/s" Thrust=m*(Vel2-Vel1)/1000 "kN" m= 30 "kg/s" "Some Wrong Solutions with Common Mistakes:" W1_thrust = (Vel2-Vel1)/1000 "Disregarding mass flow rate" W2_thrust = m*Vel2/1000 "Using incorrect relation"
9-211 ··· 9-219 Design and Essay Problems.
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