Download - ThermoSolutions CHAPTER09
9-1
Chapter 9 GAS POWER CYCLES
Actual and Ideal Cycles, Carnot cycle, Air-Standard Assumptions
9-1C The Carnot cycle is not suitable as an ideal cycle for all power producing devices because it cannot be approximated using the hardware of actual power producing devices.
9-2C It is less than the thermal efficiency of a Carnot cycle.
9-3C It represents the net work on both diagrams.
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 Under the air standard assumptions, the combustion process is modeled as a heat addition process, and the exhaust process as a heat rejection process.
9-6C The air standard assumptions are: (1) the working fluid is air which behaves as an ideal gas, (2) allthe 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.
9-7C The clearance volume is the minimum volume formed in the cylinder whereas the displacementvolume is the volume displaced by the piston as the piston moves between the top dead center and the bottom dead center.
9-8C It is the ratio of the maximum to minimum volumes in the cylinder.
9-9C The MEP is the fictitious pressure which, if acted on the piston during the entire power stroke,would produce the same amount of net work as that produced during the actual cycle.
9-10C Yes.
9-11C Assuming no accumulation of carbon deposits on the piston face, the compression ratio will remainthe same (otherwise it will increase). The mean effective pressure, on the other hand, will decrease as a cargets older as a result of wear and tear.
9-12C The SI and CI engines differ from each other in the way combustion is initiated; by a spark in SI engines, and by compressing the air above the self-ignition temperature of the fuel in CI engines.
9-13C Stroke is the distance between the TDC and the BDC, bore is the diameter of the cylinder, TDC is the position of the piston when it forms the smallest volume in the cylinder, and clearance volume is the minimum volume formed in the cylinder.
9-2
9-14 The four processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-sdiagrams, 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
P
qout
qin
3
4
2
1
qout
qin
3
4
2
1
kJ/kg828.149.101310kPa2668
kPa100
kPa2668kPa800K539.8K1800
1310kJ/kg1487.2
K1800
K539.8kJ/kg389.22
088.111.386kPa100kPa800
386.1kJ/kg300.19
K300
43
4
22
33
2
22
3
33
33
2
2
1
2
11
34
3
12
1
hPPP
P
PTT
PT
PT
P
Pu
Tu
PPP
P
Ph
T
rr
r
rr
r
vv
Tv
T
From energy balances,
kJ/kg570.19.5270.1098
kJ/kg527.919.3001.828
kJ/kg1098.02.3892.1487
outinoutnet,
14out
23in
qqw
hhq
uuq s
(c) Then the thermal efficiency becomes
51.9%kJ/kg1098.0
kJ/kg570.1
in
outnet,th q
w
9-3
9-15 EES Problem 9-14 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 beplotted.Analysis Using EES, the problem is solved as follows:
"Input Data"T[1]=300 [K] P[1]=100 [kPa]P[2] = 800 [kPa] T[3]=1800 [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"
9-4
T3 [K] th qin,total [kJ/kg] Wnet [kJ/kg]1500 50.91 815.4 415.11700 51.58 1002 516.81900 52.17 1192 621.72100 52.69 1384 729.22300 53.16 1579 839.12500 53.58 1775 951.2
5.0 5.3 5.5 5.8 6.0 6.3 6.5 6.8 7.0 7.3 7.5200
400
600
800
1000
1200
1400
1600
1800
2000
s [kJ/kg-K]
T[K]
100 kPa
800 kPa
Air
1
2
3
4
10-2 10-1 100 101 102101
102
103
4x103
v [m3/kg]
P[kPa]
300 K
1800 K
Air
1
2
3
4
9-5
1500 1700 1900 2100 2300 250050.5
51
51.5
52
52.5
53
53.5
54
T[3] [K]
th
1500 1700 1900 2100 2300 2500800
1000
1200
1400
1600
1800
T[3] [K]
qin,total
[kJ/kg]
1500 1700 1900 2100 2300 2500400
500
600
700
800
900
1000
T[3] [K]
wnet
[kJ/kg]
9-6
9-16 The four processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-sdiagrams, and the maximum temperature in the cycle and the thermal efficiency are to be determined.Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K, cv = 0.718 kJ/kg·K, and k= 1.4 (Table A-2). Analysis (b) From the ideal gas isentropic relations and energy balance, P
qin
q41
q34
3
4
2
1
K579.2kPa100kPa1000K300
0.4/1.4/1
1
212
kk
PPTT
K33603max3
2323in
579.2KkJ/kg1.005kJ/kg2800 TTT
TTchhq p
v (c) K336K3360
kPa1000kPa100
33
44
4
44
3
33 TPPT
TP
TP vv
21.0%kJ/kg2800kJ/kg2212
11
kJ/kg2212K300336KkJ/kg1.005K3363360KkJ/kg0.718
in
outth
1443
1443out41,out34,out
TTcTTchhuuqqq
pv
T
q34
q41
qin
3
4
2
1
s
Discussion The assumption of constant specific heats at room temperature is not realistic in this case thetemperature changes involved are too large.
9-17E The four processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-sdiagrams, 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. Analysis (b) The properties of air at various states are
PT Btu/lbm129.06Btu/lbm,92.04R540 111 hu
q12
qout1
2q23
4
3
s
qout
q23
q12
1
2
4
3
v
Btu/lbm593.22317.01242psia57.6psia14.7
1242Btu/lbm48.849
R3200
psia57.6psia14.7R540R2116
Btu/lbm1.537,R2116Btu/lbm04.39230004.92
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
T
From energy balance,
Btu/lbm464.1606.12922.59338.312300
Btu/lbm312.381.53748.849
14out
in23,in12,in
23in23,
hhqqqq
hhqBtu/lbm612.38
(c) Then the thermal efficiency becomes
24.2%Btu/lbm612.38Btu/lbm464.16
11in
outth q
q
9-7
9-18E The four processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-sdiagrams, 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).
PAnalysis (b)
Btu/lbm217.4R22943200RBtu/lbm0.24
psia62.46psia14.7R540R2294
R2294R540Btu/lbm.R0.171Btu/lbm300
2323in,23
11
22
1
11
2
22
2
2
1212in,12
TTchhq
PTT
PT
PT
PT
TTTcuuq
P
vv
v
qout
q23
q12
1
2
4
3
v
Process 3-4 is isentropic:
Btu/lbm378.55402117Btu/lbm.R0.240
4.217300
R2117psia62.46
psia14.7R3200
1414out
in,23in,12in
0.4/1.4/1
3
434
TTchhq
qqq
PP
TT
p
kk
Btu/lbm517.4
T
q12
qout1
2q23
4
3
(c) 26.8%Btu/lbm517.4Btu/lbm378.5
11in
outth q
q s
9-19 The three processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-sdiagrams, 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) K579.2kPa100kPa1000
K3000.4/1.4/1
1
212
kk
PP
TTP
qout
qin
1
2 3
v
K12662.579KkJ/kg1.005kg0.004kJ2.76 33
2323in
TT
TTmchhmQ p
Process 3-1 is a straight line on the P-v diagram, thus the w31 is simply the area under the process curve,
kJ/kg7.273KkJ/kg0.287kPa1000K1266
kPa100K300
2kPa1001000
22area
3
3
1
11331
1331 P
RTP
RTPPPPw vv T
Energy balance for process 3-1 gives
kJ1.679K1266-300KkJ/kg0.718273.7kg0.004)(
)(
31out31,31out31,out31,
31out31,out31,systemoutin
TTcwmTTmcmwQ
uumWQEEE
vv
qin
qout1
23
s
(c) 39.2%kJ2.76kJ1.679
11in
outth Q
Q
9-8
9-20 The three processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-sdiagrams, and the net work per cycle and the thermal efficiency are to be determined.Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The properties of air are given in Table A-17. Analysis (b) The properties of air at various states are P
qout
qin
1
2
3
kJ0.422651.1073.2
kJ1.651kJ/kg290.16840.38kg0.003
kJ2.073kJ/kg206.91897.91kg0.003
kJ/kg840.3851.8207.2kPa380
kPa95
2.207kJ/kg,897.91
K1160K290kPa95kPa380
kJ/kg290.16kJ/kg206.91
K029
outinoutnet,
13out
12in
32
3
2
11
22
1
11
2
22
1
11
23
2
QQW
hhmQ
uumQ
hPPP
P
Pu
TPP
TT
PT
P
hu
T
rr
r
vv
v
T
qout
qin
2
31
s(c) 20.4%
kJ2.073kJ0.422
in
outnet,th Q
W
9-21 The three processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-sdiagrams, and the net work per cycle and the thermal efficiency are to be determined.Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K, cv = 0.718 kJ/kg·K, and k= 1.4 (Table A-2). Analysis (b) From the isentropic relations and energy balance,
kJ0.3948.187.1
kJ1.48K290780.6KkJ/kg1.005kg0.003
kJ1.87K2901160KkJ/kg0.718kg0.003
K780.6kPa380
kPa95K1160
K1160K290kPa95kPa380
outinoutnet,
1313out
1212in
0.4/1.4/1
2
323
11
22
1
11
2
22
QQW
TTmchhmQ
TTmcuumQ
PP
TT
TPP
TT
PT
P
p
kk
v
vv P
qout
qin
2
31
s
qout
qin
1
2
3
v
T
(c) 20.9%kJ1.87kJ0.39
in
netth Q
W
9-9
9-22 A Carnot cycle with the specified temperature limits is considered. The net work output per cycle isto 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,
kPa3.935K300K900
kPa201.4/0.41/
3
232
/1
3
2
3
2
kk
kk
TT
PP
PP
TT T
The heat input is determined from
Then,
kJ0.393kJ0.58890.667
.7%66K900K300
11
kJ0.5889KkJ/kg0.2181K900kg0.003
KkJ/kg0.2181kPa2000kPa935.3
lnKkJ/kg0.287lnln
inthoutnet,
th
12in
1
20
1
212
QW
TT
ssmTQ
PP
RTT
css
H
L
H
p
1
4qout
qin2
3
900
300
s
9-23 A Carnot cycle with specified temperature limits is considered. The maximum pressure in the cycle, the heat transfer to the working fluid, and the mass of the working fluid are to be determined.Assumptions Air is an ideal gas with variable specific heats. Analysis (a) In a Carnot cycle, the maximum pressure occurs at the beginning of the expansion process,which is state 1.
(Table A-17)T P
T P
r
r
1
4
1
4
238
2 379
1200 K
350 K .T
Wnet = 0.5 kJ
1
4Qout
Qin2
3
1200
max41 kPa300379.2
238
4
1 PPP
PP
r
rkPa30,013
350(b) The heat input is determined from
s
kJ0.7060.7083/kJ0.5/
%83.70K1200K350
11
thoutnet,in
th
WQ
TT
H
L
(c) The mass of air is
kg0.00296kJ/kg169.15kJ0.5
kJ/kg169.15K3501200KkJ/kg0.199
KkJ/kg0.199
kPa150kPa300
lnKkJ/kg0.287ln
outnet,
outnet,
12outnet,
21
3
403434
wW
m
TTssw
ss
PP
Rssss
LH
9-10
9-24 A Carnot cycle with specified temperature limits is considered. The maximum pressure in the cycle, the heat transfer to the working fluid, and the mass of the working fluid are to be determined.Assumptions Helium is an ideal gas with constant specific heats. Properties The properties of helium at room temperature are R = 2.0769 kJ/kg.K and k = 1.667 (Table A-2).Analysis (a) In a Carnot cycle, the maximum pressure occurs at the beginning of the expansion process,which is state 1.
or,
kPa652471.667/0.661/
4
141
/1
4
1
4
1
K350K1200
kPa300kk
kk
TT
PP
PP
TT
Wnet = 0.5 kJ
350
1200 1
4
Qin2
3
T
(b) The heat input is determined from s
kJ0.7060.7083/kJ0.5/
70.83%K1200K350
11
thoutnet,in
th
WQ
TT
H
L
(c) The mass of helium is determined from
kg0.000409kJ/kg1223.7kJ0.5
kJ/kg1223.7K3501200KkJ/kg1.4396
KkJ/kg1.4396
kPa150kPa300
lnKkJ/kg2.0769lnln
outnet,
outnet,
12outnet,
21
3
40
3
434
wW
m
TTssw
ss
PP
RTT
css
LH
p
9-11
9-25 A Carnot cycle executed in a closed system with air as the working fluid is considered. The minimumpressure 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
K350K750KkJ/kg0.25kJ/kg10012net LLLH TTTTssw
The pressure at state 4 is determined from
or
kPa1.110K350K750kPa800 4
1.4/0.4
4
1/
4
141
/1
4
1
4
1
PP
TT
PP
PP
TT
kk
kk
qout
wnet=100 kJ/kg
750 K 1
4
qin2
3
T
s
The minimum pressure in the cycle is determined from
kPa46.133
3
40
3
43412
kPa110.1lnKkJ/kg0.287KkJ/kg25.0
lnln
PP
PP
RTT
css p
(b) The heat rejection from the cycle is kgkJ/87.5kJ/kg.K)K)(0.25350(12out sTq L
(c) The thermal efficiency is determined from
0.533K750K350
11thH
L
TT
(d) The power output for the Carnot cycle is
kW9000kJ/kg)kg/s)(10090(netCarnot wmW
Then, the second-law efficiency of the actual cycle becomes
0.578kW9000kW5200
Carnot
actualII W
W
9-12
Otto Cycle
9-26C The four processes that make up the Otto cycle are (1) isentropic compression, (2) v = constant heataddition, (3) isentropic expansion, and (4) v = constant heat rejection.
9-27C The ideal Otto cycle involves external irreversibilities, and thus it has a lower thermal efficiency.
9-28C 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.
9-29C They are analyzed as closed system processes because no mass crosses the system boundaries during any of the processes.
9-30C It increases with both of them.
9-31C Because high compression ratios cause engine knock.
9-32C The thermal efficiency will be the highest for argon because it has the highest specific heat ratio, k= 1.667.
9-33C The fuel is injected into the cylinder in both engines, but it is ignited with a spark plug in gasolineengines.
9-13
9-34 An ideal Otto cycle with air as the working fluid has a compression ratio of 8. The pressure and temperature at the end of the heat addition process, the net work output, the thermal efficiency, and the mean effective pressure for the cycle are to be determined.Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The gas constant of air is R = 0.287 kJ/kg.K. The properties of air are given in Table A-17. Analysis (a) Process 1-2: isentropic compression.
621.2kJ/kg214.07
K3001
11
r
uT
vP
750 kJ/kg 2
3
1
4
vkPa1705kPa95K300K673.1
8
kJ/kg491.2K673.1
65.772.621811
11
2
2
12
1
11
2
22
2
2
1
2112
PTT
PT
PT
P
uT
r rrr
v
vvv
vvv
vv
Process 2-3: v = constant heat addition.
588.6kJ/kg241.217502.4913
3in23,2323in23,
r
Tquuuuq
v
K1539
kPa3898kPa1705K673.1K1539
22
33
2
22
3
33 PTT
PT
PT
P vv
(b) Process 3-4: isentropic expansion.
kJ/kg571.69K774.5
70.52588.684
4
2
1334 u
Tr rrr vv
v
vv
Process 4-1: v = constant heat rejection.q u uout 571.69 214.07 357.62 kJ / kg4 1
kJ/kg392.462.357750outinoutnet, qqw
(c) 52.3%kJ/kg750kJ/kg392.4
in
outnet,th q
w
(d)
kPa495.0kJ
mkPa1/81/kgm0.906
kJ/kg392.4)/11(
MEP
/kgm0.906kPa95
K300K/kgmkPa0.287
3
31
outnet,
21
outnet,
max2min
max3
3
1
11
rww
r
PRT
vvv
vvv
vv
9-14
9-35 EES Problem 9-34 is reconsidered. The effect of the compression ratio 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]=300 [K] P[1]=95 [kPa]q_23 = 750 [kJ/kg] {r_comp = 8}
"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_23q_out_total = -q_41 w_net = w_12+w_23+w_34+w_41 Eta_th=w_net/q_in_total*100 "Thermal efficiency, in percent""The mean effective pressure is:"MEP = w_net/(V[1]-V[2])"[kPa]"
9-15
rcomp th MEP [kPa] wnet [kJ/kg]5 43.78 452.9 328.46 47.29 469.6 354.77 50.08 483.5 375.68 52.36 495.2 392.79 54.28 505.3 407.1
10 55.93 514.2 419.5
4.5 5.0 5.5 6.0 6.5 7.0 7.5200
400
600
800
1000
1200
1400
1600
s [kJ/kg-K]
T [K
]
95kP
a
3900 kPa0.
9
0.11 m3/kg
Air
1
2
3
4
10-2 10-1 100 101 102101
102
103
104
v [m3/kg]
P [k
Pa]
300 K
1500 K
Air
1
2
3
4
s2 = s1 = 5.716 kJ/kg-K
s4 = 33 = 6.424 kJ/kg-K
9-16
5 6 7 8 9 10320
340
360
380
400
420
rcomp
wnet
[kJ/kg]
5 6 7 8 9 10450
460
470
480
490
500
510
520
rcomp
MEP[kPa]
5 6 7 8 9 1042
44
46
48
50
52
54
56
rcomp
th
9-17
9-36 An ideal Otto cycle with air as the working fluid has a compression ratio of 8. The pressure and temperature at the end of the heat addition process, the net work output, the thermal efficiency, and the mean effective pressure for the cycle are to be determined.Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 1.005 kJ/kg·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.
kPa1745kPa95K300K689
8
K6898K300
11
2
2
12
1
11
2
22
0.41
2
112
PTT
PT
PT
P
TTk
v
vvv
v
v P
750 kJ/kg 2
3
1
4
vProcess 2-3: v = constant heat addition.
K1734K689KkJ/kg0.718kJ/kg750
3
3
2323in23,
TTTTcuuq v
kPa4392kPa1745K689K1734
22
33
2
22
3
33 PTT
PT
PT
P vv
(b) Process 3-4: isentropic expansion.
K75581K1734
0.41
4
334
k
TTv
v
Process 4-1: v = constant heat rejection.kJ/kg327K300755KkJ/kg0.7181414out TTcuuq v
kJ/kg423327750outinoutnet, qqw
(c) 56.4%kJ/kg750kJ/kg423
in
outnet,th q
w
(d)
kPa534kJ
mkPa1/81/kgm0.906
kJ/kg423)/11(
MEP
/kgm0.906kPa95
K300K/kgmkPa0.287
3
31
outnet,
21
outnet,
max2min
max3
3
1
11
rww
r
PRT
vvv
vvv
vv
9-18
9-37 An ideal Otto cycle with air as the working fluid has a compression ratio of 9.5. The highest pressure and temperature in the cycle, the amount of heat transferred, the thermal efficiency, and the mean effectivepressure 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.
P
kPa2338kPa100K308K757.9
9.5
K757.99.5K308
11
2
2
12
1
11
2
22
0.41
2
112
PTT
PT
PT
P
TTk
v
vvv
v
v
QoutQin
2
3
1
4
vProcess 3-4: isentropic expansion.
K19690.41
3
443 9.5K800
k
TTv
v
Process 2-3: v = constant heat addition.
kPa6072kPa2338K757.9K1969
22
33
2
22
3
33 PTT
PT
PT
P vv
(b) kg10788.6K308K/kgmkPa0.287
m0.0006kPa100 43
3
1
11
RTP
mV
kJ0.590K757.91969KkJ/kg0.718kg106.788 42323in TTmcuumQ v
(c) Process 4-1: v = constant heat rejection.
kJ0.240K308800KkJ/kg0.718kg106.788)( 41414out TTmcuumQ v
kJ0.350240.0590.0outinnet QQW
59.4%kJ0.590kJ0.350
in
outnet,th Q
W
(d)
kPa652kJ
mkPa1/9.51m0.0006
kJ0.350)/11(
MEP3
31
outnet,
21
outnet,
max2min
rWW
r
VVV
VVV
9-19
9-38 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 effectivepressure 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.
P
kPa2338kPa100K308K757.9
9.5
K757.99.5K308
11
2
2
12
1
11
2
22
0.41
2
112
PTT
PT
PT
P
TTk
v
vvv
v
v
800 K
Polytropic
Qout
Qin
2
3
1308 K
4
vProcess 3-4: polytropic expansion.
kg10788.6K308K/kgmkPa0.287
m0.0006kPa100 43
3
1
11
RTP
mV
kJ0.53381.351
K1759800KkJ/kg0.287106.7881
9.5K800
434
34
35.01
3
443
nTTmR
W
TTn
K1759v
v
Then energy balance for process 3-4 gives
kJ0.0664kJ0.5338K1759800KkJ/kg0.718kg106.788 4in34,
out34,34out34,34in34,
34out34,in34,
outin
QWTTmcWuumQ
uumWQ
EEE system
v
That is, 0.066 kJ of heat is added to the air during the expansion process (This is not realistic, and probablyis due to assuming constant specific heats at room temperature).(b) Process 2-3: v = constant heat addition.
kPa5426kPa2338K757.9K1759
22
33
2
22
3
33 PTT
PT
PT
P vv
kJ0.4879K757.91759KkJ/kg0.718kg106.788 4in23,
2323in23,
QTTmcuumQ v
Therefore,kJ0.55430664.04879.0in34,in23,in QQQ
(c) Process 4-1: v = constant heat rejection.kJ0.2398K308800KkJ/kg0.718kg106.788 4
1414out TTmcuumQ v
kJ0.31452398.05543.0outinoutnet, QQW
56.7%kJ0.5543kJ0.3145
in
outnet,th Q
W
(d)
kPa586kJ
mkPa1/9.51m0.0006
kJ0.3145)/11(
MEP3
31
outnet,
21
outnet,
max2min
rWW
rV
VVV
VV
9-20
9-39E An ideal Otto cycle with air as the working fluid has a compression ratio of 8. The amount of heattransferred 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. Properties The properties of air are given in Table A-17E. Analysis (a) Process 1-2: isentropic compression. P
540
2400 R
qout
qin
2
3
1
4
v
T 32.144Btu/lbm92.04
R5401
11
r
uv
Btu/lbm11.28204.1832.144811
21
2222
ur rrr vv
v
vv
Process 2-3: v = constant heat addition.
Btu/lbm241.4228.21170.452
419.2Btu/lbm452.70
R2400
23
33
3
uuq
uT
in
rv
(b) Process 3-4: isentropic expansion.
Btu/lbm205.5435.19419.28 43
4334
ur rrr vvv
vv
Process 4-1: v = constant heat rejection.Btu/lbm50.11304.9254.20514out uuq
53.0%Btu/lbm241.42Btu/lbm113.50
11in
outth q
q
(c) 77.5%R2400
R54011Cth,
H
L
TT
9-40E An ideal Otto cycle with argon as the working fluid has a compression ratio of 8. The amount ofheat transferred to the argon during the heat addition process, the thermal efficiency, and the thermalefficiency 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 andpotential 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.
P
R21618R540 0.6671
2
112
k
TTv
v
Process 2-3: v = constant heat addition.Btu/lbm18.07R21612400Btu/lbm.R0.07562323in TTcuuq v
qout
qin
2
3
1
4
v(b) Process 3-4: isentropic expansion.
R60081R2400
0.6671
4
334
k
TTv
v
Process 4-1: v = constant heat rejection.Btu/lbm4.536R540600Btu/lbm.R0.07561414out TTcuuq v
74.9%Btu/lbm18.07Btu/lbm4.536
11in
outth q
q
(c) 77.5%R2400R540
11Cth,H
L
TT
9-21
9-41 A gasoline engine operates on an Otto cycle. The compression and expansion processes are modeledas 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
Qout
4
3
2
Qin
1
kPa199510kPa100
K4.66410K333
1.3
2
112
1-1.31
2
112
n
n
PP
TT
v
v
v
v
Process 2-3: constant volume heat addition
K2664kPa1995kPa8000K664.4
2
323 P
PTT
kJ/kg1646K664.42664KkJ/kg0.8232323in TTcuuq v
Process 3-4: polytropic expansion.
K13351-1.31
4
334 10
1K2664n
TTv
v
kPa9.400101kPa8000
1.3
1
234
n
PPv
v
Process 4-1: constant voume heat rejection.kJ/kg8.824K3331335KkJ/kg0.8231414out TTcuuq v
(b) The net work output and the thermal efficiency are kJ/kg820.98.8241646outinoutnet, qqw
0.499kJ/kg1646kJ/kg820.9
in
outnet,th q
w
(c) The mean effective pressure is determined as follows
kPa954.3kJ
mkPa1/101/kgm0.9557
kJ/kg820.9)/11(
MEP
/kgm0.9557kPa100
K333K/kgmkPa0.287
3
31
outnet,
21
outnet,
max2min
max3
3
1
11
rww
r
PRT
vvv
vvv
vv
(d) The clearance volume and the total volume of the engine at the beginning of compression process (state1) are
9-22
33
m0002444.0m0022.0
10 cc
c
c
dc VV
VV
VVr
31 m002444.00022.00002444.0dc VVV
The total mass contained in the cylinder is
kg0.002558K333K/kgmkPa0.287
)m444kPa)/0.002100(3
3
1
11
RTVP
mt
The engine speed for a net power output of 70 kW is
rev/min4001min1
s60cycle)kJ/kgkg)(820.9002558.0(
kJ/s70rev/cycle)2(2net
net
wmW
nt
Note that there are two revolutions in one cycle in four-stroke engines.
(e) The mass of fuel burned during one cycle is
kg0001505.0kg)002558.0(
16AF ff
f
f
ft
f
a mm
mm
mmmm
Finally, the specific fuel consumption is
g/kWh258.0kWh1
kJ3600kg1
g1000kJ/kg)kg)(820.9002558.0(
kg0001505.0sfc
netwmm
t
f
Diesel Cycle
9-42C A diesel engine differs from the gasoline engine in the way combustion is initiated. In dieselengines 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.
9-43C 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.
9-44C The gasoline engine.
9-45C Diesel engines operate at high compression ratios because the diesel engines do not have the engine knock problem.
9-46C Cutoff ratio is the ratio of the cylinder volumes after and before the combustion process. As thecutoff ratio decreases, the efficiency of the diesel cycle increases.
9-23
9-47 An air-standard Diesel cycle with a compression ratio of 16 and a cutoff ratio of 2 is considered. Thetemperature after the heat addition process, the thermal efficiency, and the mean effective pressure are to be determined.Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The gas constant of air is R = 0.287 kJ/kg.K.The properties of air are given in Table A-17.
P
qout
qin 32
1
4
v
Analysis (a) Process 1-2: isentropic compression.
2.621kJ/kg214.07
K3001
11
r
uT
v
kJ/kg90.98K862.4
825.382.6211611
2
2
1
2112 h
Tr rrr vv
v
vv
Process 2-3: P = constant heat addition.
546.4kJ/kg1910.6
K862.4223
322
2
33
2
22
3
33
r
hTTT
TP
TP
vv
vvvK1724.8
(b) kJ/kg019.719.8906.191023in hhq
Process 3-4: isentropic expansion.
kJ/kg659.737.36546.42
1622 4
2
4
3
43334
urrrrr vv
v
vv
v
vv
Process 4-1: v = constant heat rejection.
56.3%kJ/kg1019.7kJ/kg445.63
11
kJ/kg445.6307.2147.659
in
outth
14out
uuq
(c)
kPa675.9kJ
mkPa1/161/kgm0.906
kJ/kg574.07/11
MEP
/kgm0.906kPa95
K300K/kgmkPa0.287
kJ/kg574.0763.4457.1019
3
31
outnet,
21
outnet,
max2min
max3
3
1
11
outinoutnet,
rww
r
PRT
qqw
vvv
vvv
vv
9-24
9-48 An air-standard Diesel cycle with a compression ratio of 16 and a cutoff ratio of 2 is considered. Thetemperature after the heat addition process, the thermal efficiency, and the mean effective pressure are to be determined.Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 1.005 kJ/kg·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. P
qout
qin 32
1
4
v
K909.416K300 0.41
2
112
k
TTv
v
Process 2-3: P = constant heat addition.
K1818.8K909.422 222
33
2
22
3
33 TTTT
PT
Pv
vvv
(b) kJ/kg913.9K909.41818.8KkJ/kg1.0052323in TTchhq p
Process 3-4: isentropic expansion.
K791.7162K1818.8
2 0.41
4
23
1
4
334
kk
TTTv
v
v
v
Process 4-1: v = constant heat rejection.
61.4%kJ/kg913.9
kJ/kg35311
kJ/kg353K300791.7KkJ/kg0.718
in
outth
1414out
TTcuuq v
(c)
kPa660.4kJ
mkPa1/161/kgm0.906
kJ/kg560.9/11
MEP
/kgm0.906kPa95
K300K/kgmkPa0.287
kJ/kg560.93539.913
3
31
outnet,
21
outnet,
max2min
max3
3
1
11
outinnet.out
rww
r
PRT
qqw
vvv
vvv
vv
9-25
9-49E An air-standard Diesel cycle with a compression ratio of 18.2 is considered. The cutoff ratio, theheat 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.
PAnalysis (a) Process 1-2: isentropic compression.3000 R
qout
qin 32
1
4
v
T 32.144Btu/lbm40.92
R5401
11
r
uv
Btu/lbm402.05R1623.6
93.732.1442.18
11
2
2
1
2112 h
Tr rrr vv
v
vv
Process 2-3: P = constant heat addition.
1.848R1623.6
R3000
2
3
2
3
2
22
3
33
TT
TP
TP
v
vvv
(b)
Btu/lbm388.6305.40268.790
180.1Btu/lbm790.68
R3000
23in
33
3
hhq
hT
rv
Process 3-4: isentropic expansion.
Btu/lbm91.250621.11180.1848.1
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/lbm158.87
Btu/lbm388.63Btu/lbm158.87
11
04.9291.250
in
outth
14out
uuq
9-26
9-50E An air-standard Diesel cycle with a compression ratio of 18.2 is considered. The cutoff ratio, theheat 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. P qin 32
1
4
v
R172418.2R540 0.41
2
112
k
TTv
v
Process 2-3: P = constant heat addition.
1.741R1724R3000
2
3
2
3
2
22
3
33
TT
TP
TP
v
vvv
(b) Btu/lbm306R17243000Btu/lbm.R0.2402323in TTchhq p
Process 3-4: isentropic expansion.
R117318.21.741R3000
741.1 0.41
4
23
1
4
334
kk
vTTT
v
v
v
Process 4-1: v = constant heat rejection.
(c) 64.6%
Btu/lbm108
Btu/lbm306Btu/lbm108
11
R0541173Btu/lbm.R0.171
in
outth
1414out
TTcuuq v
9-27
9-51 An ideal diesel engine with air as the working fluid has a compression ratio of 20. The thermalefficiency 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.
P
qout
qin 32
1
4
v
K971.120K293 0.41
2
112
k
TTV
V
Process 2-3: P = constant heat addition.
2.265K971.1K2200
2
3
2
3
2
22
3
33
TT
TP
TP
V
VVV
Process 3-4: isentropic expansion.
63.5%kJ/kg1235kJ/kg784.4
kJ/kg784.46.4501235
kJ/kg450.6K293920.6KkJ/kg0.718
kJ/kg1235K971.12200KkJ/kg1.005
K920.620
2.265K2200265.2265.2
in
outnet,th
outinoutnet,
1414out
2323in
0.41
3
1
4
23
1
4
334
qw
qqw
TTcuuq
TTchhq
rTTTT
p
kkk
v
V
V
V
V
(b)
kPa933kJ
mkPa1/201/kgm0.885
kJ/kg784.4/11
MEP
/kgm0.885kPa95
K293K/kgmkPa0.287
3
31
outnet,
21
outnet,
max2min
max3
3
1
11
rww
r
PRT
vvv
vvv
vv
9-28
9-52 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, and k= 1.4 (Table A-2).Analysis (a) Process 1-2: isentropic compression. P
Polytropic
qout
qin 32
1
4
v
K971.120K293 0.41
2
112
k
TTV
V
Process 2-3: P = constant heat addition.
2.265K971.1K2200
2
3
2
3
2
22
3
33
TT
TP
TP
V
VVV
Process 3-4: polytropic expansion.
kJ/kg526.3K2931026KkJ/kg0.718
kJ/kg1235K971.12200KkJ/kg1.005
K102620
2.265K2200r
2.2652.265
1414out
2323in
0.351n
3
1n
4
23
1
4
334
TTcuuq
TTchhq
TTTT
p
n
v
V
V
V
V
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:
kJ/kg120.1K22001026KkJ/kg0.718kJ/kg963
kJ/kg9631.351
K22001026KkJ/kg0.2871
34out34,in34,34out34,in34,
systemoutin
34out34,
TTcwquuwq
EEEnTTR
w
v
which means that 120.1 kJ/kg of heat is transferred to the combustion gases during the expansion process.This is unrealistic since the gas is at a much higher temperature than the surroundings, and a hot gas losesheat during polytropic expansion. The cause of this unrealistic result is the constant specific heatassumption. If we were to use u data from the air table, we would obtain
kJ/kg1.128)4.18723.781(96334out34,in34, uuwq
which is a heat loss as expected. Then qout becomeskJ/kg654.43.5261.128out41,out34,out qqq
and
47.0%kJ/kg1235kJ/kg580.6
kJ/kg580.64.6541235
in
outnet,th
outinoutnet,
qw
qqw
(b)
kPa691kJ
mkPa11/201/kgm0.885
kJ/kg580.6/11
MEP
/kgm0.885kPa95
K293K/kgmkPa0.287
3
31
outnet,
21
outnet,
max2min
max3
3
1
11
rww
r
PRT
vvv
vvv
vv
9-29
9-53 EES Problem 9-52 is reconsidered. The effect of the compression ratio on the net work output, meaneffective pressure, and thermal efficiency is to be investigated. Also, T-s and P-v diagrams for the cycle areto be plotted.Analysis Using EES, the problem is solved as follows:
Procedure QTotal(q_12,q_23,q_34,q_41: q_in_total,q_out_total)q_in_total = 0 q_out_total = 0 IF (q_12 > 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_41 > 0 then q_in_total = q_in_total + q_41 else q_out_total = q_out_total - q_41 END
"Input Data"T[1]=293 [K] P[1]=95 [kPa]T[3] = 2200 [K] n=1.35{r_comp = 20}
"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 pressure heat addition"P[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 w_23 =P[2]*(V[3] - V[2])"constant pressure process"DELTAu_23=intenergy(air,T=T[3])-intenergy(air,T=T[2])"Process 3-4 is polytropic expansion"P[3]/P[4] =(V[4]/V[3])^ns[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 ploytropic process"DELTAu_34=intenergy(air,T=T[4])-intenergy(air,T=T[3])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 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])
9-30
Call QTotal(q_12,q_23,q_34,q_41: q_in_total,q_out_total)w_net = w_12+w_23+w_34+w_41 Eta_th=w_net/q_in_total*100 "Thermal efficiency, in percent""The mean effective pressure is:"MEP = w_net/(V[1]-V[2])
rcomp th MEP [kPa] wnet [kJ/kg]14 47.69 970.8 797.916 50.14 985 817.418 52.16 992.6 829.820 53.85 995.4 837.022 55.29 994.9 840.624 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]
95kP
a
340.1
kPa
5920
kPa
0.044
0.1
0.88m3/kg
Air
2
1
3
4
10-2 10-1 100 101 102101
102
103
104
101
102
103
104
v [m3/kg]
P[kPa] 293 K
1049 K
2200 K
5.69
6.74kJ/kg-K
Air
9-31
14 16 18 20 22 24790
800
810
820
830
840
850
rcomp
wnet
[kJ/kg]
14 16 18 20 22 2447
49
51
53
55
57
rcomp
th
14 16 18 20 22 24970
975
980
985
990
995
1000
rcomp
MEP[kPa]
9-32
9-54 A four-cylinder ideal diesel engine with air as the working fluid has a compression ratio of 17 and a cutoff ratio of 2.2. The power the engine will deliver at 1500 rpm is to be determined.Assumptions 1 The 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.
P
Qout
Qin 32
1
4
v
K101917K328 0.41
2
112
k
TTV
V
Process 2-3: P = constant heat addition.
K2241K10192.22.2 222
33
2
22
3
33 TTTT
PT
Pv
vvv
Process 3-4: isentropic expansion.
kW46.6kJ/rev1.864rev/s1500/60
kJ/rev864.1174.1038.3
kJ174.1K328989.2KkJ/kg0.718kg10473.2
kJ3.038K)10192241)(KkJ/kg1.005)(kg10.4732(
kg10473.2)K328)(K/kgmkPa0.287(
)m0.0024)(kPa97(
K989.217
2.2K22412.22.2
outnet,outnet,
outinoutnet,
31414out
32323in
33
3
1
11
4.01
3
1
4
23
1
4
334
WnW
QQW
TTmcuumQ
TTmchhmQ
RTP
m
rTTTT
v
p
kkk
V
V
V
V
V
Discussion Note that for 2-stroke engines, 1 thermodynamic cycle is equivalent to 1 mechanical cycle (and thus revolutions).
9-33
9-55 A four-cylinder ideal diesel engine with nitrogen as the working fluid has a compression ratio of 17 and a cutoff ratio of 2.2. The power the engine will deliver at 1500 rpm is to be determined.Assumptions 1 The air-standard assumptions are applicable with nitrogen as the working fluid. 2 Kineticand 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).
PAnalysis Process 1-2: isentropic compression.
Qout
Qin 32
1
4
v
K101917K328 0.41
2
112
k
TTV
V
Process 2-3: P = constant heat addition.
K2241K10192.22.2 222
33
2
22
3
33 TTTT
PT
Pv
vvv
Process 3-4: isentropic expansion.
kW46.6kJ/rev1.863rev/s1500/60
kJ/rev.8631175.1037.3
kJ1.175K328989.2KkJ/kg0.743kg102.391
kJ3.037K10192241KkJ/kg1.039kg102.391
kg10391.2K328K/kgmkPa0.2968
m0.0024kPa97
K989.2172.2K22412.22.2
outnet,outnet,
outinoutnet,
31414out
32323in
33
3
1
11
0.41
3
1
4
23
1
4
334
WnW
QQW
TTmcuumQ
TTmchhmQ
RTP
m
rTTTT
p
kkk
v
V
V
V
V
V
Discussion Note that for 2-stroke engines, 1 thermodynamic cycle is equivalent to 1 mechanical cycle (and thus revolutions).
9-34
9-56 [Also solved by EES on enclosed CD] An ideal dual cycle with air as the working fluid has a compression ratio of 14. The fraction of heat transferred at constant volume and the thermal efficiency of the cycle are to be determined.Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The properties of air are given in Table A-17. P
x
Qout
1520.4kJ/kg
3
2
1
4
v
Analysis (a) Process 1-2: isentropic compression.
621.2kJ/kg214.07
K3001
11
r
uT
v
kJ/kg611.2K823.1
37.442.6211411
2
2
1
2112 u
Tr rrr vv
v
vv
Process 2-x, x-3: heat addition,
xx
xx
r
huhhuuqqq
hT
2.25032.6114.1520
012.2kJ/kg2503.2
K2200
32inx,3in2,xin
33
3v
By trial and error, we get Tx = 1300 K and hx = 1395.97, ux = 1022.82 kJ /kg.Thus,
and
27.1%kJ/kg1520.4kJ/kg411.62
ratio
kJ/kg411.622.61182.1022
in
in,2
2in,2
uuq
x
xx
(b) cxxx
xx rTT
TP
TP
1.692K1300K220033
3
33
v
vvv
kJ/kg886.3648.16012.2692.114
692.1692.1 42
4
3
43334
urrrrr vv
v
vv
v
vv
Process 4-1: v = constant heat rejection.
55.8%kJ/kg1520.4kJ/kg672.23
11
kJ/kg72.23607.2143.886
in
outth
14out
uuq
9-35
9-57 EES Problem 9-56 is reconsidered. The effect of the compression ratio 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]=300 [K] P[1]=100 [kPa] T[4]=2200 [K] q_in_total=1520 [kJ/kg] r_v = 14 v[1]/v[2]=r_v "Compression ratio"
"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"q_12 -w_12 = DELTAu_12 q_12 =0 [kJ/kg]"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 constant pressure heat addition"s[4]=entropy(air, T=T[4], P=P[4]) {P[4]*v[4]/T[4]=P[3]*v[3]/T[3]}P[4]*v[4]=R*T[4]P[4]=P[3]"Conservation of energy for process 3 to4"q_34 -w_34 = DELTAu_34 w_34 =P[3]*(v[4]-v[3]) "constant pressure process"DELTAu_34=intenergy(air,T=T[4])-intenergy(air,T=T[3])q_in_total=q_23+q_34"Process 4-5 is isentropic expansion"s[5]=entropy(air,T=T[5],P=P[5])s[5]=s[4]P[5]*v[5]/T[5]=P[4]*v[4]/T[4]{P[5]*v[5]=0.287*T[5]}"Conservation of energy for process 4 to 5"q_45 -w_45 = DELTAu_45 q_45 =0 [kJ/kg] "isentropic process"DELTAu_45=intenergy(air,T=T[5])-intenergy(air,T=T[4])"Process 5-1 is constant volume heat rejection"v[5]=v[1]"Conservation of energy for process 2 to 3"q_51 -w_51 = DELTAu_51
9-36
w_51 =0 [kJ/kg] "constant volume process"DELTAu_51=intenergy(air,T=T[1])-intenergy(air,T=T[5])
w_net = w_12+w_23+w_34+w_45+w_51 Eta_th=w_net/q_in_total*Convert(, %) "Thermal efficiency, in percent"
rv th [%] wnet [kJ/kg]10 52.33 795.411 53.43 812.112 54.34 82613 55.09 837.414 55.72 846.915 56.22 854.616 56.63 860.717 56.94 865.518 57.17 869
4.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0 8.50
500
1000
1500
2000
2500
3000
3500
s [kJ/kg-K]
T[K]
100 kPa
382.7 kPa
3842 kPa
6025 kPa
T-s Diagram for Air Dual Cycle
1
2
3
4
5
v=const
p=const
1
2
3 4
5
s=const
10-2 10-1 100 101 102101
102
103
8x103
v [m3/kg]
P[kPa]
300 K
2200 K
P-v Diagram for Air Dual Cycle
9-37
10 11 12 13 14 15 16 17 1852
53
54
55
56
57
58
rv
th[%]
10 11 12 13 14 15 16 17 18790
800
810
820
830
840
850
860
870
rv
wnet
[kJ/kg]
9-58 An ideal dual cycle with air as the working fluid has a compression ratio of 14. The fraction of heattransferred at constant volume and the thermal efficiency of the cycle are to be determined.Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, and k= 1.4 (Table A-2).Analysis (a) Process 1-2: isentropic compression.
K86214K300 0.41
2
112
k
TTv
v
Process 2-x, x-3: heat addition,
xx
xpx
xxxx
TT
TTcTTchhuuqqq
2200KkJ/kg1.005862KkJ/kg0.718kJ/kg1520.432
32in,3in,2in
v
P
4
1
2
3
1520.4kJ/kg
x
Qout
v
Solving for Tx we get Tx = 250 K which is impossible. Therefore, constant specific heats at roomtemperature turned out to be an unreasonable assumption in this case because of the very high temperaturesinvolved.
9-38
9-59 A six-cylinder compression ignition engine operates on the ideal Diesel cycle. The maximumtemperature in the cycle, the cutoff ratio, the net work output per cycle, the thermal efficiency, the meaneffective 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
Qout
4
32
Qin
1
kPa434117kPa95
K7.88117K328
1.349
2
112
1-1.3491
2
112
k
k
PP
TT
v
v
v
v
The clearance volume and the total volume of the engine at the beginning of compression process (state 1) are
3
3
m0002813.0
m0045.017
c
c
c
c
dcr
V
V
V
V
VV
31 m004781.00045.00002813.0dc VVV
The total mass contained in the cylinder is
kg0.004825K328K/kgmkPa0.287
)m781kPa)(0.00495(3
3
1
11
RTP
mV
The mass of fuel burned during one cycle is
kg000193.0kg)004825.0(
24 ff
f
f
f
f
a mm
mm
mmmm
AF
Process 2-3: constant pressure heat additionkJ039.88)kJ/kg)(0.9kg)(42,500000193.0(HVin cf qmQ
K23833323in K)7.881(kJ/kg.K)kg)(0.823004825.0(kJ039.8)( TTTTmcQ v
The cutoff ratio is
2.7K881.7K2383
2
3
TT
(b) 33
12 m0002813.0
17m0.004781
rV
V
23
14
3323 m00076.0)m0002813.0)(70.2(
PPVV
VV
9-39
Process 3-4: isentropic expansion.
kPa2.363m0.004781m0.00076kPa4341
K1254m0.004781m0.00076K2383
1.349
3
3
4
334
1-1.349
3
31
4
334
k
k
PP
TT
V
V
V
V
Process 4-1: constant voume heat rejection.kJ677.3K3281254KkJ/kg0.823kg)004825.0(14out TTmcQ v
The net work output and the thermal efficiency are kJ4.361677.3039.8outinoutnet, QQW
0.543kJ8.039kJ4.361
in
outnet,th Q
W
(c) The mean effective pressure is determined to be
kPa969.2kJ
mkPam)0002813.0004781.0(
kJ4.361MEP
3
321
outnet,
VV
W
(d) The power for engine speed of 2000 rpm is
kW72.7s60
min1rev/cycle)2(
(rev/min)2000kJ/cycle)(4.3612netnetnWW
Note that there are two revolutions in one cycle in four-stroke engines.(e) Finally, the specific fuel consumption is
g/kWh159.3kWh1
kJ3600kg1
g1000kJ/kg4.361
kg000193.0sfcnetW
mf
9-40
Stirling and Ericsson Cycles
9-60C The efficiencies of the Carnot and the Stirling cycles would be the same, the efficiency of the Otto cycle would be less.
9-61C The efficiencies of the Carnot and the Ericsson cycles would be the same, the efficiency of the Diesel cycle would be less.
9-62C The Stirling cycle.
9-63C The two isentropic processes of the Carnot cycle are replaced by two constant pressure regeneration processes in the Ericsson cycle.
9-64E An ideal Ericsson engine with helium as the working fluid operates between the specifiedtemperature and pressure limits. The thermal efficiency of the cycle, the heat transfer rate in theregenerator, and the power delivered are to be determined.Assumptions Helium is an ideal gas with constant specific heats. Properties The gas constant and the specific heat of helium at room temperature are R = 0.4961 Btu/lbm.Rand cp = 1.25 Btu/lbm·R (Table A-2E).Analysis (a) The thermal efficiency of this totally reversible cycle is determined from
81.67%R3000
R55011th
H
L
TT
(b) The amount of heat transferred in the regenerator is
Btu/s42,875R5503000RBtu/lbm1.25lbm/s14
4141in41,regen TTcmhhmQQ p
(c) The net power output is determined from
2
T
550 R
3000 R 1
4Qout·
Qin·
3
s
Btu/s35,384328,438167.0
Btu/s,32843RBtu/lbm1.0316R3000lbm/s14
RBtu/lbm1.0316psia200
psia25lnRBtu/lbm0.4961lnln
inthoutnet,
12in
1
20
1
212
QW
ssTmQ
PP
RTT
css
H
p
9-41
9-65 An ideal steady-flow Ericsson engine with air as the working fluid is considered. The maximumpressure 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
KkJ/kg0.5K300
kJ/kg150
0
out34,34 T
qss
300 K
1200 K 1
4qout
qin2
3
T
andKkJ/kg0.5
kPa120P
lnKkJ/kg0.287
lnln
4
3
40
3
434 P
PR
TT
css p
s
It yields P4 = 685.2 kPa
(b) For reversible cycles, kJ/kg600kJ/kg150K300K1200
outinin
out qTT
qTT
L
H
H
L
Thus, kJ/kg450150600outinoutnet, qqw
(c) The thermal efficiency of this totally reversible cycle is determined from
75.0%K1200K300
11thH
L
TT
9-66 An ideal Stirling engine with helium as the working fluid operates between the specified temperatureand pressure limits. The thermal efficiency of the cycle, the amount of heat transfer in the regenerator, andthe work output per cycle are to be determined.Assumptions Helium is an ideal gas with constant specific heats. Properties The gas constant and the specific heat of helium at room temperature are R = 2.0769 kJ/kg.K,cv = 3.1156 kJ/kg.K and cp = 5.1926 kJ/kg.K (Table A-2).Analysis (a) The thermal efficiency of this totally reversible cycle is determined from
85.0%K2000
K30011th
H
L
TT
300 K
2000 K 1
4qout
qin2
3
T(b) The amount of heat transferred in the regenerator is
kJ635.6K3002000KkJ/kg3.1156kg0.12
4141in41,regen TTmcuumQQ v
(c) The net work output is determined from s
kJ465.5kJ6.54785.0
kJ547.6KkJ/kg2.282K2000kg0.12
KkJ/kg282.23lnKkJ/kg2.0769lnln
3kPa150K2000kPa3000K300
inthoutnet,
12in
1
20
1
212
1
2
31
13
1
3
1
11
3
33
QW
ssmTQ
RTT
css
PTPT
TP
TP
H
v
v
v
v
v
vvv
v
9-42
Ideal and Actual Gas-Turbine (Brayton) Cycles
9-67C In gas turbine engines a gas is compressed, and thus the compression work requirements are verylarge since the steady-flow work is proportional to the specific volume.
9-68C They are (1) isentropic compression (in a compressor), (2) P = constant heat addition, (3) isentropicexpansion (in a turbine), and (4) P = constant heat rejection.
9-69C 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.
9-70C Back work ratio is the ratio of the compressor (or pump) work input to the turbine work output. It isusually between 0.40 and 0.6 for gas turbine engines.
9-71C As a result of turbine and compressor inefficiencies, (a) the back work ratio increases, and (b) the thermal efficiency decreases.
9-72E A simple ideal Brayton cycle with air as the working fluid has a pressure ratio of 10. The airtemperature 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 Kineticand 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,
520 R
2000 R
qout
qin
3
4
2
1
T
ThPr
11
112147520 R
124.27 Btu / lbm.
Btu/lbm240.11147.122147.110
2
2
1
212 h
TP
PP
P rrR996.5
(b) Process 3-4 is isentropic, and thuss
Btu/lbm38.88283.26571.504
Btu/lbm115.8427.12411.240
Btu/lbm265.834.170.174101
0.174Btu/lbm504.71
R2000
43outT,
12inC,
43
4
33
34
3
hhw
hhw
hPPP
P
Ph
T
rr
r
Then the back-work ratio becomes
48.5%Btu/lbm238.88Btu/lbm115.84
outT,
inC,bw w
wr
(c)
46.5%Btu/lbm264.60Btu/lbm123.04
Btu/lbm123.0484.11588.238
Btu/lbm264.6011.24071.504
in
outnet,th
inC,outT,outnet,
23in
qw
www
hhq
9-43
9-73 [Also solved by EES on enclosed CD] A simple Brayton cycle with air as the working fluid has a pressure ratio of 8. 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 andpotential energy changes are negligible. 4 Air is an ideal gas with variable specific heats.
4
2
310 K
1160 K
qout
qin
3
4s
2s
1
T
Properties The properties of air are given in Table A-17. Analysis (a) Noting that process 1-2s is isentropic,
ThPr
11
115546310 K
310.24 kJ / kg.
s
kJ/kg89.16719.69292.123082.092.1230
K3.680andkJ/kg692.1990.252.20781
2.207kJ/kg1230.92
K1160
kJ/kg646.775.0
24.31058.56224.310
K557.25andkJ/kg562.5844.125546.18
433443
43
443
4
33
1212
12
12
221
2
34
3
12
sTs
T
ssrr
r
C
ssC
ssrr
hhhhhhhh
ThPPP
P
Ph
T
hhhh
hhhh
ThPPP
P
Thus, T4 = 770.1 K(b)
kJ/kg105.392.4782.584
kJ/kg478.9224.31016.789
kJ/kg584.27.64692.1230
outinoutnet,
14out
23in
qqw
hhq
hhq
(c) 18.0%kJ/kg584.2kJ/kg105.3
in
outnet,th q
w
9-44
9-74 EES Problem 9-73 is reconsidered. The mass flow rate, pressure ratio, turbine inlet temperature, andthe isentropic efficiencies of the turbine and compressor are to be varied and a general solution for theproblem 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 = 8}{T[1] = 310 [K] P[1]= 100 [kPa] T[3] = 1160 [K] m_dot = 20 [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])
9-45
Bwr Pratio Wc [kW] Wnet [kW] Wt [kW] Qin [kW]0.5229 0.1 2 1818 1659 3477 165870.6305 0.1644 4 4033 2364 6396 143730.7038 0.1814 6 5543 2333 7876 128620.7611 0.1806 8 6723 2110 8833 116820.8088 0.1702 10 7705 1822 9527 10700
0.85 0.1533 12 8553 1510 10063 98520.8864 0.131 14 9304 1192 10496 91020.9192 0.1041 16 9980 877.2 10857 84260.9491 0.07272 18 10596 567.9 11164 78090.9767 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
Cycleefficiency,
Wnet
[kW]
Wnet
Tmax=1160 K
Note Pratio for maximum work and
c = 0.75t = 0.82
9-46
9-75 A simple Brayton cycle with air as the working fluid has a pressure ratio of 8. 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 Kineticand 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,
K733.94.640116082.01160
K645.375.0
3105.561310
K640.481K1160
K561.58K310
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
spsC
kk
s
kk
s
TTTTTTcTTc
hhhh
TTTT
TTcTTc
hhhh
PP
TT
PP
TT
4
2
310 K
1160 K
qout
qin
3
4s
2s
1
T
s
(b)
kJ/kg91.30.4263.517
kJ/kg426.0K310733.9KkJ/kg1.005
kJ/kg517.3K645.31160KkJ/kg1.005
outinoutnet,
1414out
2323in
qqw
TTchhq
TTchhq
p
p
(c) 17.6%kJ/kg517.3
kJ/kg91.3
in
outnet,th q
w
9-47
9-76 A gas turbine power plant that operates on the simple Brayton cycle with air as the working fluid hasa 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 Kineticand potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature arecp = 1.005 kJ/kg·K and k = 1.4 (Table A-2).Analysis (a) Using the isentropic relations,
kg/s352kJ/kg199.1
kJ/s70,000
kJ/kg99.1175.31184.510
kJ/kg510.84K491.71000KkJ/kg1.005
kJ/kg311.75K300610.2KkJ/kg1.005
K491.7121K1000
K610.212K300
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
1
212
wW
m
www
TTchhw
TTchhw
PP
TT
PP
TT
s
sps
sps
kk
s
kk
s
4
2
300 K
1000 K 3
4s
2s
1
T
s
(b) The net work output is determined to be
kg/s1037kJ/kg67.5
kJ/s70,000
kJ/kg67.50.85311.7584.51085.0
/
outnet,a,
outnet,
inC,s,outT,s,inC,a,outT,a,outnet,a,
wW
m
wwwww
a
CT
9-48
9-77 A stationary gas-turbine power plant operates on a simple ideal Brayton cycle with air as the working fluid. The power delivered by this plant is to be determined assuming constant and variable specific heats.Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kineticand potential energy changes are negligible. 4 Air is an ideal gas. Analysis (a) Assuming constant specific heats,
kW15,680kW35,0000.448
448.03.5251100
2902.6071111
K607.281K1100
K525.38K290
inthoutnet,
23
14
23
14
in
outth
0.4/1.4/1
3
434
0.4/1.4/1
1
212
QW
TTTT
TTcTTc
PP
TT
PP
TT
p
p
kk
s
kk
s
290 K
1100 K
qout
qin
3
4
2
1
T
s
(b) Assuming variable specific heats (Table A-17),
kW15,085kW35,0000.431
431.011.52607.116116.29037.651111
kJ/kg651.3789.201.16781
1.167kJ/kg1161.07
K1100
kJ/kg526.128488.92311.18
2311.1kJ/kg290.16
K290
inoutnet,
23
14
in
outth
43
4
33
21
2
11
34
3
12
1
QW
hhhh
hPPP
P
Ph
T
hPPP
P
Ph
T
T
rr
r
rr
r
9-49
9-78 An actual gas-turbine power plant operates at specified conditions. The fraction of the turbine workoutput 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 Kineticand 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,
T h
T h1 1
2 2
300 K 300.19 kJ / kg
580 K 586.04 kJ / kg
kJ/kg542.0905.831536.0486.0
kJ/kg285.8519.30004.586
kJ/kg05.83973.6711.47471
11.474
kJ/kg1536.0404.586950
7100700
43outT,
12inC,
43
4
323in
1
2
34
3
sT
srr
r
p
hhw
hhw
hPPP
P
P
hhhq
PP
r4
2
300 K
580 K
950 kJ/kg 3
4s
2s
1
T
s
Thus, 52.7%kJ/kg542.0kJ/kg285.85
outT,
inC,bw w
wr
(b)
27.0%kJ/kg950
kJ/kg256.15
kJ/kg256.15285.85.0542
in
outnet,th
inC,outT,net.out
qw
www
9-50
9-79 A gas-turbine power plant operates at specified conditions. The fraction of the turbine work outputused 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. 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/kg562.2K874.81525.3KkJ/kg1.00586.0
kJ/kg281.4K300580KkJ/kg1.005
K874.871K1525.3
K1525.3KkJ/kg1.005/kJ/kg950K580
/
7100700
4343outT,
1212inC,
0.4/1.4/k1k
3
434s
in232323in
1
2
spTsT
p
pp
p
TTchhw
TTchhw
PP
T
cqTTTTchhq
PP
r
T
4
2
300 K
580 K
950 kJ/kg 3
4s
2s
1
T
s
Thus, 50.1%kJ/kg562.2kJ/kg281.4
outT,
inC,bw w
wr
(b)
29.6%kJ/kg950kJ/kg280.8
kJ/kg280.8281.42.562
in
outnet,th
inC,outT,outnet,
qw
www
9-80E A gas-turbine power plant operates on a simple Brayton cycle with air as the working fluid. The netpower output of the plant is to be determined.Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kineticand 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 Using variable specific heats for air,
T h
T h3 3
4 4
2000
1200 291.30
R 504.71 Btu / lbm
R Btu / lbm
kW3373Btu/s197353398536
Btu/s8536Btu/lbm291.30504.71lbm/s40
Btu/s53390.80/131.30238.07lbm/s40/
Btu/lbm8.072379.11474.18
474.1Btu/lbm131.306400/40291.30
815
120
inC,outT,outnet,
43outT,
12inC,
21
2
114out
1
2
12
1
WWW
hhmW
hhmW
hPPP
P
PhhhmQ
PP
r
Cs
srr
r
p
4
21200 R
2000 R
6400 Btu/s
3
4s
2s
1
T
s
9-51
9-81E A gas-turbine power plant operates on a simple Brayton cycle with air as the working fluid. The compressor efficiency for which the power plant produces zero net work is to be determined.Assumptions 1 Steady operating conditions exist. 2 The air-standardassumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with variable specific heats.
TProperties The properties of air are given in Table A-17E. Analysis Using variable specific heats,
T h
T h3 3
4 4
2000
1200 291.30
R 504.71 Btu / lbm
R Btu / lbm
Btu/lbm8.072379.11474.18
474.1Btu/lbm131.306400/40291.30
815
120
21
2
114out
1
2
12
1
srr
r
p
hPPP
P
PhhhmQ
PP
r
4
2Wnet = 0 ·
3
4s
2s
1
2000 R
1200 R
s
Then,
50.0%30.29171.50430.13107.238
/
43
12
4312outT,inC,
hhhh
hhmhhmWW
sC
Cs
9-82 A 32-MW gas-turbine power plant operates on a simple Brayton cycle with air as the working fluid.The mass flow rate of air through the cycle is to be determined.Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kineticand 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 Using variable specific heats,
T 5546.1kJ/kg310.24
K3101
11
rPh
kJ/kg40.6880.0/24.31026.56232.51993.93286.0
/
kJ/kg519.32411.929.7581
29.75kJ/kg32.939
K900
kJ/kg62.26544.125546.18
1243inC,outT,outnet,
43
4
33
21
2
34
3
12
CssT
srr
r
srr
hhhhwww
hPPP
P
Ph
T
hPPP
P4
2
310 K
900 K
Wnet =32 MW
·
3
4s
2s
1
T
s
and kg/s786.6kJ/kg40.68
kJ/s32,000
outnet,
outnet,
wW
m
9-52
9-83 A 32-MW gas-turbine power plant operates on a simple Brayton cycle with air as the working fluid.The mass flow rate of air through the cycle is to be determined.Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kineticand 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 Using constant specific heats,
kJ/kg32.5
K0.80/310561.5496.89000.86KkJ/kg1.005
/
K496.881K900
K561.58K310
1243CoutT,outnet,
0.4/1.4/1
3
434
0.4/1.4/1
1
212
in, CspspT
kk
s
kk
s
TTcTTcwww
PP
TT
PP
TT
4
2
310 K
900 K
Wnet =32 MW
·
3
4s
2s
1
T
s
and
kg/s984.6kJ/kg32.5
kJ/s32,000
outnet,
outnet,
wW
m
9-53
9-84 A gas-turbine plant operates on the simple Brayton cycle. The net power output, the back work ratio,and the thermal efficiency are to be determined.Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The gas constant of air is R = 0.287 kJ/kg·K (Table A-1).Analysis (a) For this problem, we use theproperties from EES software. Remember that for an ideal gas, enthalpy is a function of temperature only whereas entropy is functions ofboth temperature and pressure.
1.2 MPa
500°C100 kPa 30°C
Compress.
4
32
Turbine
Combustionchamber
1
Process 1-2: Compression
KkJ/kg7159.5kPa100
C30kJ/kg60.303C30
11
1
11
sPT
hT
kJ/kg37.617kJ/kg.K7159.5
kPa12002
12
2sh
ssP
kJ/kg24.68660.303
60.30337.61782.0 2212
12C h
hhhhh s
Process 3-4: Expansion
kJ/kg62.792C500 44 hT
ss hhh
hhhh
43
3
43
43T
62.79288.0
We cannot find the enthalpy at state 3 directly. However, using the following lines in EES together withthe isentropic efficiency relation, we find h3 = 1404.7 kJ/kg, T3 = 1034ºC, s3 = 6.5699 kJ/kg.K. The solution by hand would require a trial-error approach.
h_3=enthalpy(Air, T=T_3)s_3=entropy(Air, T=T_3, P=P_2)h_4s=enthalpy(Air, P=P_1, s=s_3)
The mass flow rate is determined from
kg/s875.2K27330K/kgmkPa0.287
)s/m0kPa)(150/6100(3
3
1
11
RTP
mV
The net power output iskW1100)kJ/kg60.30324kg/s)(686.875.2()( 12inC, hhmW
W kW1759)kJ/kg62.792.7kg/s)(1404875.2()( 43outT, hhm
kW65911001759inC,outT,net WWW
(b) The back work ratio is
0.625kW1759kW1100
outT,
inC,bw W
Wr
(c) The rate of heat input and the thermal efficiency are kW2065)kJ/kg24.686.7kg/s)(1404875.2()( 23in hhmQ
0.319kW2065
kW659
in
net
QW
th
9-54
Brayton Cycle with Regeneration
9-85C Regeneration increases the thermal efficiency of a Brayton cycle by capturing some of the wasteheat from the exhaust gases and preheating the air before it enters the combustion chamber.
9-86C Yes. At very high compression ratios, the gas temperature at the turbine exit may be lower than thetemperature at the compressor exit. Therefore, if these two streams are brought into thermal contact in aregenerator, heat will flow to the exhaust gases instead of from the exhaust gases. As a result, the thermalefficiency will decrease.
9-87C The extent to which a regenerator approaches an ideal regenerator is called the effectiveness , and is defined as = qregen, act /qregen, max.
9-88C (b) turbine exit.
9-89C 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 and W increases while Qin remainsconstant. Steam can be obtained by utilizing the hot exhaust gases.
9-55
9-90E 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 Theambient air is 540 R and 14.5 psia. 4 The effectiveness of the regenerator is 0.9, and the isentropicefficiencies 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 fromTable A-17E. Analysis The gas turbine cycle with regeneration can be analyzed as follows:
Btu/lbm372.253.5712.23041
12.230Btu/lbm549.35
R2160
Btu/lbm0.192544.5386.14
386.1Btu/lbm129.06
R054
43
4
33
21
2
11
34
3
12
1
srr
r
srr
r
hPPP
P
Ph
T
hPPP
P
Ph
T
24
5
540 R
2160 R qin 3
4s2s
1
T
s
and
Btu/lbm63.4072.37235.549
35.5490.80
Btu/lbm74.20706.129
06.1290.1920.80
44
43
43turb
2212
12comp
hh
hhhh
hhhh
hh
s
s
Then the thermal efficiency of the gas turbine cycle becomesq Btu/lbm179.9)74.20763.407(9.0)( 24regen hh
Btu/lbm63.0)06.12974.207()63.40735.549()()(
Btu/lbm161.7=9.179)74.20735.549()(
1243inC,outT,outnet,
regen23in
hhhhwww
qhhq
0.39Btu/lbm161.7Btu/lbm63.0
in
outnet,th %39
qw
Finally, the mass flow rate of air through the turbine becomes
lbm/s1.07hp1
Btu/s7068.0Btu/lbm63.0
hp95
net
netair w
Wm
9-56
9-91 [Also solved by EES on enclosed CD] The thermal efficiency and power output of an actual gas turbine are given. The isentropic efficiency of the turbine and of the compressor, and the thermal efficiency of the gas turbine modified with a regenerator are to be determined.Assumptions 1 Air is an ideal gas with variable specific heats. 2 Kinetic and potential energy changes are negligible. 3 The mass flow rates of air and of the combustion gases are the same, and the properties of combustion gases are the same as those of air. Properties The properties of air are given in Table A-17. Analysis The properties at various states are
24
5
293 K
1561 K qin 3
4s2s
1
T
kJ/kg23.82547.485.7127.14
1
5.712kJ/kg1710.0
K1561C1288
kJ/kg3.643765.182765.17.14
2765.1kJ/kg293.2
K293=C20
43
4
33
21
2
11
34
3
12
1
srr
r
srr
r
hPPP
P
Ph
T
hPPP
P
Ph
T
s
The net work output and the heat input per unit mass are
C650kJ/kg54.9582.29334.665kJ/kg34.66566.3720.1038
kJ/kg0.67210381710
kJ/kg0.10380.359
kJ/kg372.66
kJ/kg66.372h1
s3600kg/h1,536,000kW159,000
41out414out
netinout
3223in
th
netin
netnet
Thqhhhqwqq
qhhhhq
wq
mW
w
in
Then the compressor and turbine efficiencies become
0.924
0.849
2.2936722.2933.643
23.825171054.9581710
12
12
43
43
hhhh
hhhh
sC
sT
When a regenerator is added, the new heat input and the thermal efficiency become
0.496kJ/kg751.46kJ/kg372.66
kJ/kg751.46=54.2861038
kJ/kg286.54=672.0)-.54(0.80)(958=)(
newin,
netnewth,
regeninnewin,
24regen
qw
qqq
hhq
Discussion Note an 80% efficient regenerator would increase the thermal efficiency of this gas turbinefrom 35.9% to 49.6%.
9-57
9-92 EES Problem 9-91 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] = 20 [C] P[1]= 100 [kPa] {T[4]=589 [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 = 0.80 Eta_c = 0.892 "Compressor isentorpic efficiency"Eta_t = 0.926 "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_doth[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] /Pratios_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"
9-58
m_dot*h[3] = W_dot_turbisen + m_dot*h_s[4] h_s[4]=ENTHALPY(Air,T=T_s[4])"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]
t c th,noreg th,withreg Qinnoreg
[kW]Qinwithreg
[kW]Wnet [kW]
0.7 0.892 0.2309 0.3405 442063 299766 102.10.75 0.892 0.2736 0.3841 442063 314863 120.90.8 0.892 0.3163 0.4237 442063 329960 139.8
0.85 0.892 0.359 0.4599 442063 345056 158.70.9 0.892 0.4016 0.493 442063 360153 177.6
0.95 0.892 0.4443 0.5234 442063 375250 196.41 0.892 0.487 0.5515 442063 390346 215.3
9-59
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 1100
120
140
160
180
200
220
t
Wnet
[kW]
0.7 0.75 0.8 0.85 0.9 0.95 1275000
310000
345000
380000
415000
450000
t
Qdot,in
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
0.6
t
Etath
with regeneration
no regeneration
9-60
9-93 An ideal Brayton cycle with regeneration is considered. The effectiveness of the regenerator is 100%. The net work output and the thermal efficiency of the cycle 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 Noting that this is an ideal cycle and thus the compression and expansion processes are isentropic,we have
kJ/kg601.9485.67579.1277
kJ/kg279.6819.30087.579
kJ/kg675.858.23238101
238kJ/kg1277.79
K1200
kJ/kg579.8786.13386.110
386.1kJ/kg300.19
K300
43outT,
12inC,
43
4
33
21
2
11
34
3
12
1
hhw
hhw
hPPP
P
Ph
T
hPPP
P
Ph
T
rr
r
rr
r
5
300 K
1200 K qin3
42
1
T
s
Thus,kJ/kg322.2668.27994.601inC,outT,net www
Also, 100% 5 4h h 675.85 kJ / kg
and
53.5%kJ/kg601.94kJ/kg322.26
kJ/kg601.9485.67579.1277
in
netth
53in
qw
hhq
9-61
9-94 EES Problem 9-93 is reconsidered. The effects of the isentropic efficiencies for the compressor and turbine and regenerator effectiveness on net work done and the heat supplied to the cycle are 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] = 1200 [K] Pratio = 10 T[1] = 300 [K] P[1]= 100 [kPa] Eta_reg = 1.0 Eta_c =0.8 "Compressor isentorpic efficiency"Eta_t =0.9 "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_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"h[1] + w_compisen = 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 = 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_in - e_out =DELTAe_cv =0 for steady flow"h[2] + q_in_noreg = h[3] 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] /Pratios_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_turb /w_turbisen "turbine adiabatic efficiency, w_turbisen > w_turb""SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0 e_in -e_out = DELTAe_cv = 0 for steady-flow"h[3] = w_turbisen + h_s[4] h_s[4]=ENTHALPY(Air,T=T_s[4])"Actual Turbine analysis:"h[3] = w_turb + h[4] h[4]=ENTHALPY(Air,T=T[4])s[4]=ENTROPY(Air,T=T[4], P=P[4])"Cycle analysis"w_net=w_turb-w_compEta_th_noreg=w_net/q_in_noreg*Convert(, %) "[%]" "Cycle thermal efficiency"Bwr=w_comp/w_turb"Back work ratio"
9-62
"With the regenerator the heat added in the external heat exchanger is"h[5] + q_in_withreg = h[3] 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:"h[2] + h[4]=h[5] + 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_net/q_in_withreg*Convert(, %) "[%]""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]
c t th,noreg th,withreg qinnoreg
[kJ/kg]qinwithreg
[kJ/kg]wnet
[kJ/kg]0.6 0.9 14.76 13.92 510.9 541.6 75.4
0.65 0.9 20.35 20.54 546.8 541.6 111.30.7 0.9 24.59 26.22 577.5 541.6 142
0.75 0.9 27.91 31.14 604.2 541.6 168.60.8 0.9 30.59 35.44 627.5 541.6 192
0.85 0.9 32.79 39.24 648 541.6 212.50.9 0.9 34.64 42.61 666.3 541.6 230.8
4.5 5.0 5.5 6.0 6.5 7.0 7.5200
400
600
800
1000
1200
1400
1600
s [kJ/kg-K]
T[K] 10
0kPa
1000
kPa
Air
2s
1
2 5
3
44s
6
9-63
0.7 0.75 0.8 0.85 0.9 0.95 110
15
20
25
30
35
40
45
t
th
c = 0.8
With regeneration
No regeneration
0.7 0.75 0.8 0.85 0.9 0.95 150
95
140
185
230
275
t
wnet
[kJ/kg]
c = 0.8
0.7 0.75 0.8 0.85 0.9 0.95 1400
450
500
550
600
650
t
qin
c = 0.8
No regeneration
With regeneration
9-64
0.6 0.65 0.7 0.75 0.8 0.85 0.910
15
20
25
30
35
40
45
c
th
t = 0.9
With regeneration
No regeneration
0.6 0.65 0.7 0.75 0.8 0.85 0.975
110
145
180
215
250
c
wnet
[kJ/kg]
t = 0.9
0.6 0.65 0.7 0.75 0.8 0.85 0.9500
520
540
560
580
600
620
640
660
680
c
qin
t = 0.9
No regeneration
With regeneration
9-65
9-95 An ideal Brayton cycle with regeneration is considered. The effectiveness of the regenerator is 100%. The net work output and the thermal efficiency of the cycle 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 Noting that this is an ideal cycle and thus the compression and expansion processes are isentropic, we have
517.0)10(120030011or
5.62112003002.5791111
K579.2andK621.5%100
K621.5101K1200
K579.210K300
1)/1.4(1.4/)1(
3
1th
53
16
53
16
in
outth
2645
0.4/1.4/1
3
434
0.4/1.4/1
1
212
kkp
p
p
kk
kk
rTT
TTTT
TTcTTc
TTTT
PP
TT
PP
TT
0.517
5
300 K
1200 K qin3
42
1
T
s
Then,
kJ/kg300.8300)]K-(579.2-621.5)-1200kJ/kg.K)[(005.1(
)]()[(
)()(
1243
1243incomp,outturb,net
TTTTc
hhhhwww
p
or,
kgkJ6300=621.5)-200kJ/kg.K)(1005.1)(517.0(
)()(
53th
53th
inthnet
/.
TTchh
qw
p
9-66
9-96 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 areapplicable. 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 of air at various states are
kJ/kg803.14711.801219.250.821219.25
kJ/kg711.8059.2815.20071
15.200kJ/kg1219.25
K1501
kJ/kg618.260.75/310.24541.2610.243/
kJ/kg541.2688.105546.17
5546.1kJ/kg310.24
K310
433443
43
43
4
33
121212
12
21
2
11
34
3
12
1
sTs
T
srr
r
Css
C
srr
r
hhhhhhhh
hPPP
P
Ph
T
hhhhhhhh
hPPP
P
Ph
T2
4
6
5
310 K
1150 K qin 3
4s2s
1
T
s
Thus, T4 = 782.8 K(b)
kJ/kg108.0924.31026.61814.80325.1219
1243inC,outT,net hhhhwww
(c)
kJ/kg738.43618.26803.140.6526.618
242524
25 hhhhhhhh
Then,
22.5%kJ/kg480.82kJ/kg108.09
kJ/kg480.82738.4319.2512
in
netth
53in
qw
hhq
9-67
9-97 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 potentialenergy 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 obtainedfrom Table A-17. Analysis (a) Assuming constant specific heats,
kW39,188kW75,0000.5225
5225.02.6071100
2903.5251111
K525.3andK607.2%100
K607.281K1100
K525.38K290
innet
53
16
53
16
in
outth
2645
0.4/1.4/1
3
434
0.4/1.4/1
1
212
QW
TTTT
TTcTTc
TTTT
PP
TT
PP
TT
T
p
p
kk
kk
6
qout
5
290 K
1100 K 75,000 kW 3
42
1
T
s
(b) Assuming variable specific heats,
kW40,283kW75,0000.5371
5371.037.65107.116116.29012.526111
kJ/kg526.12andkJ/kg651.37%100
kJ/kg651.3789.201.16781
1.167kJ/kg1161.07
K1100
kJ/kg526.128488.92311.18
2311.1kJ/kg90.162
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
9-68
9-98 A regenerative gas-turbine engine using air as the working fluid is considered. The amount of heattransfer 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
kJ/kg152.504.58688.79772.0kJ/kg797.88
75.71979.127786.079.1277
kJ/kg719.7575.290.23881
238.0kJ/kg1277.79K2001
kJ/kg86.045K580kJ/kg00.193K300
8100/800/
24regen
433443
43
43
4
33
22
11
12
34
3
hhq
hhhhhhhh
hPPP
P
PhThThT
PPr
sTs
T
srr
r
p
2580 K
4
6
5
300 K
1200 K qin 3
4s2s
1
T
s
(b)
36.0%kJ/kg539.23kJ/kg194.06
kJ/kg539.23152.52586.041277.79
kJ/kg194.06300.19586.0488.79779.1277
in
netth
regen23in
1243inC,outT,net
qw
qhhq
hhhhwww
9-69
9-99 A regenerative gas-turbine engine using air as the working fluid is considered. The amount of heattransfer 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).Analysis (a) Using the isentropic relations and turbine efficiency,
kJ/kg114.2K580737.8KkJ/kg1.0050.72K737.8
5.662120086.01200
K662.581K1200
8100/800/
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
2580 K
4
6
5
300 K
1200 K qin 3
4s2s
1
T
s
(b)
36.0%kJ/kg508.9kJ/kg183.1
kJ/kg508.9114.2K5801200KkJ/kg1.005
kJ/kg183.1K300580737.81200KkJ/kg1.005
in
netth
regen23regen23in
1243inC,outT,net
qw
qTTcqhhq
TTcTTcwww
p
pp
9-100 A regenerative gas-turbine engine using air as the working fluid is considered. The amount of heattransfer 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 of air at various states are
kJ/kg148.304.58688.79770.0
kJ/kg797.8875.71979.127786.079.1277
kJ/kg719.7575.290.23881
238.0kJ/kg1277.79K2001
kJ/kg86.045K580kJ/kg00.193K300
8100/800/
23regen
433443
43
43
4
33
22
11
12
34
3
hhq
hhhhhhhh
hPPP
P
PhThThT
PPr
sTs
T
srr
r
p
2580 K
4
6
5
300 K
1200 K qin 3
4s2s
1
T
s
(b)
35.7%kJ/kg5543.kJ/kg194.06
kJ/kg543.5148.3586.041277.79
kJ/kg194.06300.19586.0488.79779.1277
in
netth
regen23in
1243inC,outT,net
qw
qhhq
hhhhwww
9-70
Brayton Cycle with Intercooling, Reheating, and Regeneration
9-101C As the number of compression and expansion stages are increased and regeneration is employed,the ideal Brayton cycle will approach the Ericsson cycle.
9-102C (a) decrease, (b) decrease, and (c) decrease.
9-103C (a) increase, (b) decrease, and (c) decrease.
9-104C (a) increase, (b) decrease, (c) decrease, and (d) increase.
9-105C (a) increase, (b) decrease, (c) increase, and (d) decrease.
9-106C Because the steady-flow work is proportional to the specific volume of the gas. Intercooling decreases the average specific volume of the gas during compression, and thus the compressor work. Reheating increases the average specific volume of the gas, and thus the turbine work output.
9-107C (c) The Carnot (or Ericsson) cycle efficiency.
9-71
9-108 An ideal gas-turbine cycle with two stages of compression and two stages of expansion isconsidered. The back work ratio and the thermal efficiency of the cycle are to be determined for the cases of with and without a regenerator.Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible.Properties The properties of air are given in Table A-17. Analysis (a) The work inputs to each stage of compressor are identical, so are the work outputs of each stage of the turbine since this is an ideal cycle. Then,
kJ/kg62.86636.94679.127722
kJ/kg2.142219.30026.41122
kJ/kg946.3633.7923831
238kJ/kg77.7912
K2001
kJ/kg411.26158.4386.13
386.1kJ/kg300.19
K300
65outT,
12inC,
865
6
755
421
2
11
56
5
12
1
hhw
hhw
hhPPP
P
Phh
T
hhPPP
P
Ph
T
rr
r
rr
r
2
9
10
7
6 8
300 K
1200 K qin5
1
4
3
T
s
Thus, 33.5%kJ/kg662.86kJ/kg222.14
outT,
inC,bw w
wr
36.8%kJ/kg1197.96kJ/kg440.72
kJ/kg440.72222.1486.662
kJ/kg1197.9636.94679.127726.41179.1277
in
netth
inC,outT,net
6745in
qw
www
hhhhq
(b) When a regenerator is used, rbw remains the same. The thermal efficiency in this case becomes
55.3%kJ/kg796.63kJ/kg440.72
kJ/kg796.6333.40196.1197
kJ/kg33.40126.41136.94675.0
in
netth
regenoldin,in
48regen
qw
qqq
hhq
9-72
9-109 A gas-turbine cycle with two stages of compression and two stages of expansion is considered. The back work ratio and the thermal efficiency of the cycle are to be determined for the cases of with and without a regenerator.Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible.Properties The properties of air are given in Table A-17. Analysis (a) The work inputs to each stage of compressor are identical, so are the work outputs of each stage of the turbine. Then,
kJ/kg63.44507.99679.127722
kJ/kg77.68219.30003.43922
kJ/kg96.07936.94679.127785.079.1277
kJ/kg946.3633.7923831
238kJ/kg1277.79K1200
kJ/kg9.034380.0/19.30026.41119.300
/
kJ/kg411.26158.4386.13
386.1kJ/kg300.19K300
65outT,
12inC,
6558665
65
865
6
755
1214212
12
421
2
11
56
5
12
1
hhw
hhw
hhhhhhhhh
hhPPP
P
PhhT
hhhhhhhhh
hhPPP
P
PhT
sTs
T
rr
r
Css
C
ssrr
r T
4
8
22s
9
10
7
66s
8qin
5
1
4
3s
Thus, 49.3%kJ/kg563.44kJ/kg277.68
outT,
inC,bw w
wr
25.5%kJ/kg1120.48kJ/kg285.76
kJ/kg5.762868.27744.563
kJ/kg1120.48996.071277.79439.031277.79
in
netth
inC,outT,net
6745in
qw
www
hhhhq
(b) When a regenerator is used, rbw remains the same. The thermal efficiency in this case becomes
40.7%kJ/kg702.70kJ/kg285.76
kJ/kg.7070278.41748.1120
kJ/kg17.78403.43907.99675.0
in
netth
regenoldin,in
48regen
qw
qqq
hhq
9-73
9-110 A regenerative gas-turbine cycle with two stages of compression and two stages of expansion isconsidered. The minimum mass flow rate of air needed to develop a specified net power output is to be determined.Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible.Properties The properties of air are given in Table A-17. Analysis The mass flow rate will be a minimum when the cycle is ideal. That is, the turbine and thecompressors are isentropic, the regenerator has an effectiveness of 100%, and the compression ratios across each compression or expansion stage are identical. In our case it is rp = 9 = 3. Then the work inputs toeach stage of compressor are identical, so are the work outputs of each stage of the turbine.
kg/s249.6kJ/kg440.72
kJ/s110,000
kJ/kg440.72222.1468.662
kJ/kg662.86946.361277.7922
kJ/kg222.1419.30026.41122
kJ/kg6.369433.7923831
238kJ/kg,1277.79K0120
kJ/kg11.264158.4386.13
386.1kJ/kg,00.193K003
net
net
inC,outT,net
65outT,
12inC,
865
6
755
421
2
11
56
5
12
1
wW
m
www
hhw
hhw
hhPPP
P
PhhT
hhPPP
P
PhT
rr
r
rr
r
2
7
6 8
300 K
1200 K 5
1
4
3
T
s
9-111 A regenerative gas-turbine cycle with two stages of compression and two stages of expansion isconsidered. The minimum mass flow rate of air needed to develop a specified net power output is to be determined.Assumptions 1 Argon is an ideal gas with constant specific heats. 2 Kinetic and potential energy changes are negligible.Properties The properties of argon at room temperature are cp = 0.5203 kJ/kg.K and k = 1.667 (Table A-2a).Analysis The mass flow rate will be a minimum when the cycle is ideal. That is, the turbine and thecompressors are isentropic, the regenerator has an effectiveness of 100%, and the compression ratios across each compression or expansion stage are identical. In our case it is rp = 9 = 3. Then the work inputs toeach stage of compressor are identical, so are the work outputs of each stage of the turbine.
kg/s404.7kJ/kg271.8
kJ/s110,000
kJ/kg271.83.1721.444
kJ/kg444.1K773.21200KkJ/kg0.5203222
kJ/kg172.3K300465.6KkJ/kg0.5203222
K773.231K1200
K465.63K300
net
net
inC,outT,net
6565outT,
1212inC,
70.667/1.66/1
5
656
70.667/1.66/1
1
212
wW
m
www
TTchhw
TTchhw
PP
TT
PP
TT
p
p
kk
kk
2
7
6 8
300 K
1200 K 5
1
4
3
T
s
9-74
Jet-Propulsion Cycles
9-112C The power developed from the thrust of the engine is called the propulsive power. It is equal tothrust times the aircraft velocity.
9-113C The ratio of the propulsive power developed and the rate of heat input is called the propulsiveefficiency. It is determined by calculating these two quantities separately, and taking their ratio.
9-114C It reduces the exit velocity, and thus the thrust.
9-115E A turbojet engine operating on an ideal cycle is flying at an altitude of 20,000 ft. The pressure atthe turbine exit, the velocity of the exhaust gases, and the propulsive efficiency are to be determined.Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with constant specific heats at room temperature. 4 Kinetic and potential energies are negligible,except at the diffuser inlet and the nozzle exit. 5 The turbine work output is equal to the compressor workinput.Properties The properties of air at room temperature are cp = 0.24 Btu/lbm.R and k = 1.4 (Table A-2Ea). Analysis (a) For convenience, we assume the aircraft is stationary and the air is moving towards theaircraft at a velocity of V1 = 900 ft/s. Ideally, the air will leave the diffuser with a negligible velocity (V20).Diffuser:
psia11.19R470R537.3
psia7
R537.4/sft25,037
Btu/lbm1RBtu/lbm0.242
ft/s900470
2
2/02
0
2/2/
1.4/0.41/
1
212
22
221
12
2112
21
022
12
222
211
outin
(steady)0systemoutin
kk
p
p
TT
PP
cV
TT
VTTc
VVhh
VhVh
EE
EEE T
6
qout
5
qin
3
4
2
1s
Compressor:
R1118.313R537.4
psia145.5psia11.1913
0.4/1.4/1
2
323
243
kk
p
PP
TT
PrPP
9-75
Turbine:
54235423outturb,incomp, TTcTTchhhhww pp
or,
psia55.21.4/0.41/
4
545
2345
R2400R1819.1
psia145.5
R1819.14.5371118.34002kk
TT
PP
TTTT
(b) Nozzle:
2/02
0
2/2/
R1008.6psia55.2
psia7R1819.1
2656
025
26
56
266
255
outin
(steady)0systemoutin
0.4/1.4/1
5
656
VTTc
VVhh
VhVh
EE
EEE
PP
TT
p
kk
or,
s/ft3121Btu/lbm1
/sft25,037R1008.61819.1RBtu/lbm0.2402
22
6V
(c) The propulsive efficiency is the ratio of the propulsive work to the heat input,
%25.9Btu/lbm307.6
Btu/lbm79.8
Btu/lbm307.6R1118.32400RBtu/lbm0.24
Btu/lbm79.8/sft25,037
Btu/lbm1ft/s900ft/s9003121
in
3434in
22
aircraftinletexit
qw
TTchhq
VVVw
pp
p
p
9-76
9-116E A turbojet engine operating on an ideal cycle is flying at an altitude of 20,000 ft. The pressure atthe turbine exit, the velocity of the exhaust gases, and the propulsive efficiency are to be determined.Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with variable specific heats. 4 Kinetic and potential energies are negligible, except at the diffuserinlet and the nozzle exit. 5 The turbine work output is equal to the compressor work input.Properties The properties of air are given in Table A-17E. Analysis (a) For convenience, we assume the aircraft is stationary and the air is moving towards the aircraftat a velocity of V1= 900 ft/s. Ideally, the air will leave the diffuser with a negligible velocity (V 2 0).
Diffuser:8548.0
Btu/lbm112.20R470
1
11
rPhT
T
psia11.220.85481.3698psia7
3698.1Btu/lbm128.48/sft25,037
Btu/lbm12ft/s900
20.1122
20
2/2/
1
2
2
12
r22
221
12
21
022
12
222
211
outin
(steady)0systemoutin
r
r
P
PPP
PV
hh
VVhh
VhVh
EE
EEE
6
qout
5
qin
3
4
2
1s
Compressor:
Btu/lbm267.5680.171.36811.22145.8
psia145.8psia11.2213
32
3
243
23hP
PP
P
PrPP
rr
p
Turbine: T 6.367Btu/lbm22.617
R24004
44
rPh
5423
outturb,incomp,
hhhh
ww
or,
psia56.6367.6142.7psia145.8
7.142Btu/lbm14.47848.12856.26722.617
4
5
5
45
2345
r
r
r
P
PPP
Phhhh
9-77
(b) Nozzle:
20
2/2/
Btu/lbm266.9317.66psia56.6
psia7)7.142(
025
26
56
266
255
outin
(steady)0systemoutin
65
656
VVhh
VhVh
EE
EEE
hPP
PP rr
or,
ft/s3252Btu/lbm1
/sft25,037Btu/lbm)93.26614.478(22
22
656 hhV
(c) The propulsive efficiency is the ratio of the propulsive work to the heat input,
24.2%Btu/lbm349.66
Btu/lbm84.55
Btu/lbm349.6656.26722.617
Btu/lbm84.55/sft25,037
Btu/lbm1ft/s)(900ft/s)900)(3252
in
34
22
aircraftinletexit
qw
hhq
VVVw
pp
in
p
9-78
9-117 A turbojet aircraft flying at an altitude of 9150 m is operating on the ideal jet propulsion cycle. The velocity of exhaust gases, the propulsive power developed, and the rate of fuel consumption are to be determined.Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with constant specific heats at room temperature. 4 Kinetic and potential energies are negligible,except at the diffuser inlet and the nozzle exit. 5 The turbine work output is equal to the compressor workinput.Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K and k = 1.4 (Table A-2a). Analysis (a) We assume the aircraft is stationary and the air is moving towards the aircraft at a velocity of V 1 = 320 m/s. Ideally, the air will leave the diffuser with a negligible velocity (V 2 0). Diffuser:
kPa62.6K241K291.9
kPa32
K291.9/sm1000
kJ/kg1KkJ/kg1.0052
m/s320K241
2
2/02
02/2/
1.4/0.41/
1
212
22
221
12
2112
21
022
122
222
11
outin(steady)0
systemoutin
kk
p
p
TT
PP
cV
TT
VTTc
VVhhVhVh
EEEEE
T
6
5
Qi·
3
4
2
1s
Compressor:
K593.712K291.9
kPa751.2kPa62.612
0.4/1.4/1
2
323
243
kk
p
PP
TT
PrPP
Turbine:
or,K1098.2291.9593.714002345
54235423outturb,incomp,
TTTT
TTcTTchhhhww pp
Nozzle:
2/02
0
2/2/
K568.2kPa751.2
kPa32K1400
2656
025
26
56
266
255
outin(steady)0
systemoutin
0.4/1.4/1
4
646
VTTcVV
hh
VhVh
EEEEE
PP
TT
p
kk
or, m/s1032kJ/kg1
/sm1000K568.21098.2KkJ/kg1.0052
22
6V
(b) kW13,67022aircraftinletexit
/sm1000kJ/kg1
m/s320m/s3201032kg/s60VVVmW p
(c)
kg/s1.14kJ/kg42,700kJ/s48,620
HV
kJ/s48,620K593.71400KkJ/kg1.005kg/s60
infuel
3434in
Qm
TTcmhhmQ p
9-79
9-118 A turbojet aircraft is flying at an altitude of 9150 m. The velocity of exhaust gases, the propulsive power developed, and the rate of fuel consumption are to be determined.Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with constant specific heats at room temperature. 4 Kinetic and potential energies are negligible,except at the diffuser inlet and the nozzle exit.Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K and k = 1.4 (Table A-2a). Analysis (a) For convenience, we assume the aircraft is stationary and the air is moving towards theaircraft at a velocity of V 1 = 320 m/s. Ideally, the air will leave the diffuser with a negligible velocity (V 2
0). Diffuser:
kPa62.6K241K291.9
kPa32
K291.9/sm1000
kJ/kg1KkJ/kg1.0052
m/s320K241
2
2/02
0
2/2/
1.4/0.41/
1
212
22
221
12
2112
21
022
12
222
211
outin
(steady)0systemoutin
kk
p
p
TT
PP
cV
TT
VTTc
VVhh
VhVh
EE
EEE T
5
6
5s
Qi·
3
4
2
1s
Compressor:
K593.712K291.9
kPa751.2kPa62.612
0.4/1.4/1
2
323
243
kk
s
p
PP
TT
PrPP
K669.20.80/291.9593.71.929/2323
23
23
23
23
Cs
p
spsC
TTTT
TTcTTc
hhhh
Turbine:
or,K1022.7291.9669.214002345
54235423outturb,incomp,
TTTT
TTcTTchhhhww pp
kPa197.7K1400K956.1
kPa751.2
K956.185.0/1022.714001400/1.4/0.41/
4
545
5445
54
54
54
54
kks
Ts
sp
p
sT
TT
PP
TTTT
TTcTTc
hhhh
9-80
Nozzle:
2/02
0
2/2/
K607.8kPa197.7
kPa32K1022.7
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/s913.2kJ/kg1
/sm1000K607.81022.7KkJ/kg1.0052
22
6V
(b)
kW11,39022
aircraftinletexit
/sm1000kJ/kg1
m/s320m/s320913.2kg/s60
VVVmW p
(c)
kg/s1.03kJ/kg42,700kJ/s44,067
HV
kJ/s44,067K669.21400KkJ/kg1.005kg/s60
infuel
3434in
Qm
TTcmhhmQ p
9-81
9-119 A turbojet aircraft that has a pressure rate of 12 is stationary on the ground. The force that must be applied on the brakes to hold the plane stationary is to be determined.Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with variable specific heats. 4 Kinetic and potential energies are negligible, except at the nozzle exit.Properties The properties of air are given in Table A17.
TAnalysis (a) Using variable specific heats for air,
5
4
qin
2
3
1
Compressor: T386.1
kJ/kg300.19K300
1
11
rPh
27.396kJ/kg1464.65854610.65
kJ/kg854kg/s10
kJ/s8540
kJ/s8540kJ/kg42,700kg/s0.2HV
kJ/kg610.6563.16386.112
3
12
2323in
inin
fuelin
21
2
r
in
rr
Pqhhhhq
mQ
q
mQ
hPPP
P
s
Turbine:
or,kJ/kg741.17300.19610.65464.6511234
4312outturb,incomp,
hhhh
hhhhww
Nozzle:
20
2/2/
kJ/kg41.79702.3312127.396
024
25
45
255
244
outin
(steady)0systemoutin
53
535
VVhh
VhVh
EE
EEE
hPP
PP rr
or,
m/s908.9kJ/kg1
/sm1000kJ/kg741.171154.1922
22
545 hhV
Brake force = Thrust = N90892inletexit
m/skg1N1
m/s0908.9kg/s10VVm
9-82
9-120 EES Problem 9-119 is reconsidered. The effect of compressor inlet temperature on the force thatmust be applied to the brakes to hold the plane stationary is to be investigated.Analysis Using EES, the problem is solved as follows:
P_ratio = 12 T_1 = 27 [C] T[1] = T_1+273 "[K]"P[1]= 95 [kPa] P[5]=P[1]Vel[1]=0 [m/s]V_dot[1] = 9.063 [m^3/s] HV_fuel = 42700 [kJ/kg] m_dot_fuel = 0.2 [kg/s] Eta_c = 1.0 Eta_t = 1.0 Eta_N = 1.0
"Inlet conditions"h[1]=ENTHALPY(Air,T=T[1])s[1]=ENTROPY(Air,T=T[1],P=P[1])v[1]=volume(Air,T=T[1],P=P[1])m_dot = V_dot[1]/v[1]"Compressor anaysis"s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor"P_ratio=P[2]/P[1]"Definition of pressure ratio - to find P[2]"T_s[2]=TEMPERATURE(Air,s=s_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exit"h_s[2]=ENTHALPY(Air,T=T_s[2])Eta_c =(h_s[2]-h[1])/(h[2]-h[1]) "Compressor adiabatic efficiency; Eta_c = W_dot_c_ideal/W_dot_c_actual. "m_dot*h[1] +W_dot_c=m_dot*h[2] "SSSF First Law for the actual compressor, assuming:adiabatic, ke=pe=0"
"External heat exchanger analysis"P[3]=P[2]"process 2-3 is SSSF constant pressure"h[3]=ENTHALPY(Air,T=T[3])Q_dot_in = m_dot_fuel*HV_fuelm_dot*h[2] + Q_dot_in= m_dot*h[3]"SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0"
"Turbine analysis"s[3]=ENTROPY(Air,T=T[3],P=P[3])s_s[4]=s[3] "For the ideal case the entropies are constant across the turbine"{P_ratio= P[3] /P[4]}T_s[4]=TEMPERATURE(Air,h=h_s[4]) "Ts[4] is the isentropic value of T[4] at turbine exit"{h_s[4]=ENTHALPY(Air,T=T_s[4])} "Eta_t = W_dot_t /Wts_dot turbine adiabatic efficiency,Wts_dot > W_dot_t"Eta_t=(h[3]-h[4])/(h[3]-h_s[4])m_dot*h[3] = W_dot_t + m_dot*h[4] "SSSF First Law for the actual compressor, assuming:adiabatic, ke=pe=0"T[4]=TEMPERATURE(Air,h=h[4])P[4]=pressure(Air,s=s_s[4],h=h_s[4])"Cycle analysis"W_dot_net=W_dot_t-W_dot_c"Definition of the net cycle work, kW"W_dot_net = 0 [kW]
9-83
"Exit nozzle analysis:"s[4]=entropy('air',T=T[4],P=P[4])s_s[5]=s[4] "For the ideal case the entropies are constant across the nozzle"
T_s[5]=TEMPERATURE(Air,s=s_s[5], P=P[5]) "T_s[5] is the isentropic value of T[5] at nozzleexit"h_s[5]=ENTHALPY(Air,T=T_s[5])Eta_N=(h[4]-h[5])/(h[4]-h_s[5])m_dot*h[4] = m_dot*(h_s[5] + Vel_s[5]^2/2*convert(m^2/s^2,kJ/kg))m_dot*h[4] = m_dot*(h[5] + Vel[5]^2/2*convert(m^2/s^2,kJ/kg))T[5]=TEMPERATURE(Air,h=h[5])s[5]=entropy('air',T=T[5],P=P[5])
"Brake Force to hold the aircraft:"Thrust = m_dot*(Vel[5] - Vel[1]) "[N]"BrakeForce = Thrust "[N]""The following state points are determined only to produce a T-s plot"T[2]=temperature('air',h=h[2])s[2]=entropy('air',T=T[2],P=P[2])
BrakeForce
[N]
m[kg/s]
T3
[K]T1
[C]
9971 11.86 1164 -209764 11.41 1206 -109568 10.99 1247 09383 10.6 1289 109207 10.24 1330 209040 9.9 1371 30
4.5 5.0 5.5 6.0 6.5 7.0 7.5200
400
600
800
1000
1200
1400
1600
s [kJ/kg-K]
T [K
]
95kP
a
114
0kP
a
Air
1
2s
3
4s
5s
-20 -10 0 10 20 309000
9200
9400
9600
9800
10000
T1 [C]
Bra
keFo
rce
[N]
9-84
9-121 Air enters a turbojet engine. The thrust produced by this turbojet engine is to be determined.Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with variable specific heats. 4 Kinetic and potential energies are negligible, except at the diffuserinlet and the nozzle exit.Properties The properties of air are given in Table A-17. AnalysisWe assume the aircraft is stationary and the air is moving towards the aircraft at a velocity of V 1= 300 m/s. Taking the entire engine as our control volume and writing the steady-flow energy balance yield
T h
T h1 1
2 2
280 28013
700 713 27
K k
K k
.
.
J / kg
J / kg 15,000 kJ/s
22
222
21
22
12in
222
211in
outin
(steady)0systemoutin
/sm1000kJ/kg1
2m/s300
13.28027.713kg/s16kJ/s000,15
2
)2/()2/(
V
VVhhmQ
VhmVhmQ
EE
EEE 7 C300 m/s16 kg/s
427 C
1 2
It gives V 2 = 1048 m/sThus,
N11,968m/s3001048kg/s1612 VVmFp
9-85
Second-Law Analysis of Gas Power Cycles
9-122 The total exergy destruction associated with the Otto cycle described in Prob. 9-34 and the exergy atthe end of the power stroke are to be determined.Analysis From Prob. 9-34, qin = 750, qout = 357.62 kJ/kg, T1 = 300 K, and T4 = 774.5 K. The total exergy destruction associated with this Otto cycle is determined from
kJ/kg245.12K2000
kJ/kg750K300kJ/kg357.62
K300inout0destroyed
HL Tq
Tq
Tx
Noting that state 4 is identical to the state of the surroundings, the exergy at the end of the power stroke(state 4) is determined from
040040044 vvPssTuu
where
KkJ/kg0.7081K300K774.5lnKkJ/kg0.2871.70203.68232
lnlnln
0kJ/kg357.62
1
414
41
1414
1
4141404
1404
out1404
TT
RssTT
RssPP
Rssssss
quuuu
v
v
vvvv
Thus,kJ/kg145.20KkJ/kg0.7081K300kJ/kg357.624
9-123 The total exergy destruction associated with the Diesel cycle described in Prob. 9-47 and the exergyat the end of the compression stroke are to be determined.Analysis From Prob. 9-47, qin = 1019.7, qout = 445.63 kJ/kg, T1 = 300 K, v1 = 0.906 m3/kg, and v 2 = v 1 / r= 0.906 / 12 = 0.0566 m3/kg.The total exergy destruction associated with this Otto cycle is determined from
kJ/kg292.7K2000
kJ/kg1019.7K300kJ/kg445.63
K300inout0destroyed
HL Tq
Tq
Tx
Noting that state 1 is identical to the state of the surroundings, the exergy at the end of the compressionstroke (state 2) is determined from
kJ/kg348.63
312012012
020020022
mkPa1kJ1
/kgm0.9060.0566kPa950214.07643.3
vv
vv
PssTuuPssTuu
9-86
9-124E The exergy destruction associated with the heat rejection process of the Diesel cycle described in Prob. 9-49E and the exergy at the end of the expansion stroke are to be determined.Analysis From Prob. 9-49E, qout = 158.9 Btu/lbm, T1 = 540 R, T4 = 1420.6 R, and v 4 = v 1. At Tavg = (T4 + T1)/2 = (1420.6 + 540)/2 = 980.3 R, we have cv,avg = 0.180 Btu/lbm·R. The entropy change during process 4-1 is
RBtu/lbm0.1741R1420.6
R540lnRBtu/lbm0.180lnln0
4
1
4
141 v
vRTTcss v
Thus,
Btu/lbm64.9R540
Btu/lbm158.9RBtu/lbm0.1741R54041,41041destroyed,
R
R
Tq
ssTx
Noting that state 4 is identical to the state of the surroundings, the exergy at the end of the power stroke(state 4) is determined from
040040044 vvPssTuu
where
RBtu/lbm0.17410
RBtu/lbm9.581
1404
1404
out1404
ssss
quuuuvvvv
Thus,Btu/lbm64.90RBtu/lbm0.1741R540Btu/lbm158.94
Discussion Note that the exergy at state 4 is identical to the exergy destruction for the process 4-1 sincestate 1 is identical to the dead state, and the entire exergy at state 4 is wasted during process 4-1.
9-87
9-125 The exergy destruction associated with each of the processes of the Brayton cycle described in Prob. 9-73 is to be determined.Analysis From Prob. 9-73, qin = 584.62 kJ/kg, qout = 478.92 kJ/kg, and
KkJ/kg2.67602kJ/kg789.16
KkJ/kg3.13916K1160
KkJ/kg2.47256kJ/kg6.364
KkJ/kg.734981K310
44
33
22
11
sh
sT
sh
sT
Thus,
kJ/kg206.0
kJ/kg38.76
kJ/kg87.35
kJ/kg40.83
K310kJ/kg478.92
2.676021.73498K290
ln
1/8lnKkJ/kg0.2873.139162.67602K290
ln
K1600kJ/kg584.62
2.472563.13916K290
ln
8lnKkJ/kg0.2871.734982.47256K290
ln
out0
4
1410
41,41041gen,041destroyed,
3
434034034gen,034destroyed,
in0
2
3230
23,23023gen,023destroyed,
1
212012012gen,012destroyed,
LR
R
HR
R
Tq
PP
RssTT
qssTsTx
PP
RssTssTsTx
Tq
PP
RssTT
qssTsTx
PP
RssTssTsTx
9-126 The total exergy destruction associated with the Brayton cycle described in Prob. 9-93 and theexergy at the exhaust gases at the turbine exit are to be determined.Analysis From Prob. 9-93, qin = 601.94, qout = 279.68 kJ/kg, and h6 = 579.87 kJ/kg.The total exergy destruction associated with this Otto cycle is determined from
kJ/kg179.4K1800
kJ/kg601.94K300kJ/kg279.68
K300inout0destroyed
HL Tq
Tq
Tx
Noting that h0 = h@ 300 K = 300.19 kJ/kg, the stream exergy at the exit of the regenerator (state 6) isdetermined from
06
026
060066 2gz
VssThh
where s s s s s s R PP6 0 6 1 6 1
6
1
0
ln 2.36275 1.70203 0.66072 kJ / kg K
Thus, kJ/kg81.5KkJ/kg0.66072K300300.1979.8756
9-88
9-127 EES Problem 9-126 is reconsidered. The effect of the cycle pressure on the total irreversibility forthe cycle and the exergy of the exhaust gas leaving the regenerator is to be investigated.Analysis Using EES, the problem is solved as follows:
"Input data"T_o = 300 [K] T_L = 300 [K] T_H = 1400 [K] T[3] = 1200 [K] {Pratio = 10}T[1] = 300 [K] C_P=1.005 [kJ/kg-K] P[1]= 100 [kPa] P_o=P[1]Eta_reg = 1.0 Eta_c =1.0"Compressor isentorpic efficiency"Eta_t =1.0"Turbien isentropic efficiency"MM=MOLARMASS(Air)R=R_u/MMR_u=8.314 [kJ/kmol-K]C_V=C_P - R k=C_P/C_V"Isentropic Compressor anaysis""For the ideal case the entropies are constant across the compressor"P[2] = Pratio*P[1] "T_s[2] is the isentropic value of T[2] at compressor exit"T_s[2]=T[1]*(Pratio)^((k-1)/k)Eta_c = w_compisen/w_comp"compressor adiabatic efficiency, W_comp > W_compisen""Conservation of energy for the compressor for the isentropic case: e_in - e_out = DELTAe=0 for steady-flow"w_compisen = C_P*(T_s[2]-T[1]) "Actual compressor analysis:"w_comp = C_P*(T[2]-T[1]) "External heat exchanger analysis""SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0 e_in - e_out =DELTAe_cv =0 for steady flow"q_in_noreg = C_P*(T[3]-T[2]) P[3]=P[2]"process 2-3 is SSSF constant pressure""Turbine analysis""For the ideal case the entropies are constant across the turbine"P[4] = P[3] /PratioT_s[4]=T[3]*(1/Pratio)^((k-1)/k)"T_s[4] is the isentropic value of T[4] at turbine exit"Eta_t = w_turb /w_turbisen "turbine adiabatic efficiency, w_turbisen > w_turb""SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0 e_in -e_out = DELTAe_cv = 0 for steady-flow"w_turbisen=C_P*(T[3] - T_s[4]) "Actual Turbine analysis:"w_turb= C_P*(T[3]-T[4])
"Cycle analysis"w_net=w_turb-w_comp "[kJ/kg]"Eta_th_noreg=w_net/q_in_noreg*Convert(, %) "[%]" "Cycle thermal efficiency"Bwr=w_comp/w_turb "Back work ratio"
9-89
"With the regenerator the heat added in the external heat exchanger is"q_in_withreg = C_P*(T[3]-T[5]) P[5]=P[2]"The regenerator effectiveness gives h[5] and thus T[5] as:"Eta_reg = (T[5]-T[2])/(T[4]-T[2]) "Energy balance on regenerator gives h[6] and thus T[6] as:""h[2] + h[4]=h[5] + h[6]"T[2] + T[4]=T[5] + T[6]P[6]=P[4]"Cycle thermal efficiency with regenerator"Eta_th_withreg=w_net/q_in_withreg*Convert(, %) "[%]"
"Irreversibility associated with the Brayton cycle is determined from:"q_out_withreg = q_in_withreg - w_net i_withreg = T_o*(q_out_withreg/T_L - q_in_withreg/T_H)q_out_noreg = q_in_noreg - w_net i_noreg = T_o*(q_out_noreg/T_L - q_in_noreg/T_H)"Neglecting the ke and pe of the exhaust gases, the exergy of the exhaust gases at the exit of the regenerator is:""Psi_6 = (h[6] - h_o) - T_o(s[6] - s_o)"Psi_exit_withreg = C_P*(T[6] - T_o) - T_o*(C_P*ln(T[6]/T_o)-R*ln(P[6]/P_o))Psi_exit_noreg = C_P*(T[4] - T_o) - T_o*(C_P*ln(T[4]/T_o)-R*ln(P[4]/P_o))
inoreg iwithreg Pratio exit,noreg
[kJ/kg]exit,withreg
[kJ/kg]th,noreg
[%]th,withreg
[%]270.8 97.94 6 157.8 47.16 40.05 58.3244.5 113.9 7 139.9 56.53 42.63 56.42223 128.8 8 125.5 65.46 44.78 54.73205 142.7 9 113.6 74 46.61 53.18
189.6 155.9 10 103.6 82.17 48.19 51.75176.3 168.4 11 95.05 90.02 49.58 50.41164.6 180.2 12 87.62 97.58 50.82 49.17154.2 191.5 13 81.11 104.9 51.93 47.99144.9 202.3 14 75.35 111.9 52.94 46.88
4.5 5.0 5.5 6.0 6.5 7.0 7.5200
400
600
800
1000
1200
1400
1600
s [kJ/kg-K]
T [K
]
100 kPa
1000 kPa
Air
2s
1
2 5
3
44s
6
9-90
6 7 8 9 10 11 12 13 1475
115
155
195
235
275
Pratio
i [kJ
/kg]
No regeneration
With regeneration
6 7 8 9 10 11 12 13 1440
60
80
100
120
140
160
Pratio
Psi
exit
[kJ/
kg]
No regeneration
With regeneration
6 7 8 9 10 11 12 13 1440
44
48
52
56
60
Pratio
th[%
]
With regeneration
No regeneration
9-91
9-128 The exergy destruction associated with each of the processes of the Brayton cycle described in Prob. 9-98 and the exergy at the end of the exhaust gases at the exit of the regenerator are to bedetermined.Properties The gas constant of air is R = 0.287 kJ/kg·K (Table A-1).Analysis From Prob. 9-98, qin = 539.23 kJ/kg, qout = 345.17 kJ/kg, and
KkJ/kg08336.2kJ/kg738.56
KkJ/kg68737.2kJ/kg88.797
KkJ/kg17888.3K1200
KkJ/kg.373482kJ/kg04.586
KkJ/kg1.70203K300
55
44
33
22
11
sh
sh
sT
sh
sT
and, from an energy balance on the heat exchanger,
KkJ/kg2.47108
kJ/kg645.36.04586738.5688.797
6
66425
s
hhhhh
Thus,
kJ/kg114.5
kJ/kg42.78
kJ/kg5.57
kJ/kg31.59
kJ/kg22.40
K300kJ/kg345.17
2.471081.70203K300
ln
K1260kJ/kg539.23
2.608333.17888K300
ln
2.687372.471082.373482.60833K300
1/8lnKkJ/kg0.2873.178882.68737K300
ln
8lnKkJ/kg0.2871.702032.37348K300
ln
out0
6
1610
61,61061gen,061destroyed,
0
5
3530
53,53053gen,053destroyed,
4625046250regengen,0regendestroyed,
3
434034034gen,034destroyed,
1
212012012gen,012destroyed,
LR
R
H
in
R
R
Tq
PP
RssTT
qssTsTx
Tq
PP
RssTT
qssTsTx
ssssTssssTsTx
PP
RssTssTsTx
PP
RssTssTsTx
Noting that h0 = h@ 300 K = 300.19 kJ/kg, the stream exergy at the exit of the regenerator (state 6) isdetermined from
06
026
060066 2gzVssThh
where KkJ/kg0.769051.702032.47108ln0
1
6161606 P
PRssssss
Thus, kJ/kg114.5KkJ/kg0.76905K300300.1936.6456
9-92
9-129 A gas-turbine plant uses diesel fuel and operates on simple Brayton cycle. The isentropic efficiency of the compressor, the net power output, the back work ratio, the thermal efficiency, and the second-lawefficiency are to be determined.Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energychanges are negligible. 3 Air is an ideal gas withconstant specific heats.
700kPa
Diesel fuel
100kPa
Compres
4
32
Turbine
Combustionchamber
1
Properties The properties of air at 500ºC = 773 K are cp = 1.093 kJ/kg·K, cv = 0.806 kJ/kg·K,R = 0.287 kJ/kg·K, and k = 1.357 (Table A-2b).Analysis (a) The isentropic efficiency of thecompressor may be determined if we firstcalculate the exit temperature for the isentropic case
K6.505kPa100kPa700K303
1)/1.357-(1.357/)1(
1
212
kk
s PP
TT
0.881K)303533(K)3036.505(
12
12
TTTT s
C
(b) The total mass flowing through the turbine and the rate of heat input are
kg/s81.12kg/s21.0kg/s6.1260
kg/s6.12kg/s6.12AF
aafat
mmmmm
kW85557)kJ/kg)(0.900kg/s)(42,021.0(HVin cf qmQThe temperature at the exit of combustion chamber is
K1144)K533kJ/kg.K)(3kg/s)(1.0981.12(kJ/s8555)( 3323in TTTTcmQ p
The temperature at the turbine exit is determined using isentropic efficiency relation
K7.685kPa700kPa100K1144
1)/1.357-(1.357/)1(
3
434
kk
s PP
TT
K4.754K)7.6851144(
K)1144(85.0 4
4
43
43 TT
TTTT
sT
The net power and the back work ratio are kW3168)K30333kJ/kg.K)(53kg/s)(1.096.12()( 12inC, TTcmW pa
kW5455)K4.754144kJ/kg.K)(13kg/s)(1.0981.12()( 43outT, TTcmW p
kW228731685455inC,outT,net WWW
0.581kW5455kW3168
outT,
inC,bw W
Wr
(c) The thermal efficiency is 0.267kW8555kW2287
in
netth Q
W
The second-law efficieny of the cycle is defined as the ratio of actual thermal efficiency to the maximumpossible thermal efficiency (Carnot efficiency). The maximum temperature for the cycle can be taken to be the turbine inlet temperature. That is,
735.0K1144K30311
3
1max T
T
and 0.364735.0267.0
max
thII
9-93
9-130 A modern compression ignition engine operates on the ideal dual cycle. The maximum temperaturein the cycle, the net work output, the thermal efficiency, the mean effective pressure, the net power output,the second-law efficiency of the cycle, and the rate of exergy of the exhaust gases are to be determined.Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at 850 K are cp = 1.110 kJ/kg·K, cv = 0.823 kJ/kg·K, R = 0.287 kJ/kg·K,and k = 1.349 (Table A-2b).Analysis (a) The clearance volume and the total volume of the engine at the beginning of compressionprocess (state 1) are
xcc
c
c
dcr VVVV
VV
VV2
33
m0002154.0m0028.014
43
1 m003015.00028.00002154.0 VVVV dc
Process 1-2: Isentropic compression
kPa334114kPa95
K9.82314K328
1.349
2
112
1-1.3491
2
112
k
k
PP
TT
v
v
v
v
Process 2-x and x-3: Constant-volume and constant pressure heat addition processes:
K2220kPa3341kPa9000K)9.823(
22 P
PTT x
x
kJ/kg1149K)9.8232220(kJ/kg.K)(0.823)( 2-2 TTcq xx v
x
Qout
V
P
4
3
2Qin
1
K32543333-2 K)2220(kJ/kg.K)(0.823kJ/kg1149)( TTTTcqq xpxx
(b) kJ/kg2298114911493-2in xx qqq
3333 m0003158.0
K2220K3254)m0002154.0(
xx T
TVV
Process 3-4: isentropic expansion.
kPa9.428m0.003015m0.0003158kPa9000
K1481m0.003015m0.0003158K3254
1.349
3
3
4
334
1-1.349
3
31
4
334
k
k
PP
TT
V
V
V
V
Process 4-1: constant voume heat rejection.kJ/kg7.948K3281481KkJ/kg0.82314out TTcq v
The net work output and the thermal efficiency are kJ/kg13497.9482298outinoutnet, qqw
0.587kJ/kg2298kJ/kg1349
in
outnet,th q
w
9-94
(c) The mean effective pressure is determined to be
kg0.003043K328K/kgmkPa0.287
)m015kPa)(0.00395(3
3
1
11
RTP
mV
kPa1466kJ
mkPam)0002154.0003015.0(
kJ/kg)kg)(1349(0.003043MEP
3
321
outnet,
VV
mw
(d) The power for engine speed of 3500 rpm is
kW120s60
min1rev/cycle)2(
(rev/min)3500kJ/kg)kg)(1349003043.0(2netnetnmwW
Note that there are two revolutions in one cycle in four-stroke engines.(e) The second-law efficieny of the cycle is defined as the ratio of actual thermal efficiency to themaximum possible thermal efficiency (Carnot efficiency). We take the dead state temperature and pressureto be 25ºC and 100 kPa.
908.0K3254
K)27325(113
0max T
T
and 0.646908.0587.0
max
thII
The rate of exergy of the exhaust gases is determined as follows
kJ/kg6.567100
9.428ln287.0298
1481kJ/kg.Kln110.1()298(29814810.823
lnln)(0
4
0
4004040044 P
PR
TT
cTTTcssTuux pv
kW50.4s60
min1rev/cycle)2(
(rev/min)3500kJ/kg)kg)(567.6003043.0(244nmxX
9-95
9-131 A gas-turbine plant operates on the regenerative Brayton cycle. The isentropic efficiency of the compressor, the effectiveness of the regenerator, the air-fuel ratio in the combustion chamber, the net power output, the back work ratio, the thermal efficiency, the second law efficiency, the exergy efficienciesof the compressor, the turbine, and the regenerator, and the rate of the exergy of the combustion gases atthe regenerator exit are to be determined.Assumptions 1 The air-standard assumptions areapplicable. 2 Kinetic and potential energychanges are negligible. 3 Air is an ideal gas withconstant specific heats. Properties The properties of air at 500ºC = 773 K are cp = 1.093 kJ/kg·K, cv = 0.806 kJ/kg·K,R = 0.287 kJ/kg·K, and k = 1.357 (Table A-2b).Analysis (a) For the compressor and the turbine:
K6.505kPa100kPa700K303
1.3571-1.3571
1
212
kk
s PPTT
65
400°C700 kPa 260°C
Regenerator
871°C
Compress.
432
Turbine
Combustionchamber
1100 kPa 30°C
0.881K)303533(K)3036.505(
12
12
TTTT s
C
K6.685kPa700kPa100K1144
1)/1.357-(1.357/)1(
3
434
kk
s PP
TT
K4.754K)6.6851144(
K)1144(85.0 4
4
43
43 TT
TTTT
sT
(b) The effectiveness of the regenerator is
0.632K)5334.754(
K)533673(
24
25regen TT
TT
(c) The fuel rate and air-fuel ratio are
kg/s1613.0673)K144kJ/kg.K)(1093.1)(6.12(7)kJ/kg)(0.9000,42(
)()( 53HVin
fff
pafcf
mmm
TTcmmqmQ
78.141613.0
6.12AFf
a
mm
Also, kg/s76.121613.06.12fa mmm
kW65707)kJ/kg)(0.900kg/s)(42,01613.0(HVin cf qmQ(d) The net power and the back work ratio are
kW3168)K30333kJ/kg.K)(53kg/s)(1.096.12()( 12inC, TTcmW pa
kW5434)K4.754144kJ/kg.K)(13kg/s)(1.0976.12()( 43outT, TTcmW p
kW226731685434inC,outT,net WWW
0.583kW5434kW3168
outT,
inC,bw W
Wr
(e) The thermal efficiency is
0.345kW6570kW2267
in
net
QW
th
(f) The second-law efficieny of the cycle is defined as the ratio of actual thermal efficiency to themaximum possible thermal efficiency (Carnot efficiency). The maximum temperature for the cycle can be taken to be the turbine inlet temperature. That is,
9-96
735.0K1144K30311
3
1max T
T
and 0.469735.0345.0
max
thII
(g) The exergy efficiency for the compressor is defined as the ratio of stream exergy difference between the inlet and exit of the compressor to the actual power input:
kW2943100700ln287.0
303533ln)093.1()303()303533)(093.1()6.12(
lnln)(1
2
1
201212012C P
PR
TT
cTTTcmssThhmX ppaa
0.929kW3168kW2943
inC,
CCII, W
X
The exergy efficiency for the turbine is defined as the ratio of actual turbine power to the stream exergy difference between the inlet and exit of the turbine:
kW5834100700ln287.0
754.41144ln)093.1()303()4.7541144)(093.1()76.12(
lnln4
3
4
3043T P
PR
TT
cTTTcmX pp
0.932kW5834kW5434
T
inT,, X
WTII
An energy balance on the regenerator gives
K2.616)4.754)(093.1)(76.12()53373)(1.093)(66.12(
)()(
66
6425
TT
TTcmTTcm ppa
The exergy efficiency for the regenerator is defined as the ratio of the exergy increase of the cold fluid to the exergy decrease of the hot fluid:
kW10730616.2754.4ln)093.1()303()2.6164.754)(093.1()76.12(
0ln6
4064hotregen, T
TcTTTcmX pp
kW8.9540533673ln)093.1()303()533673)(093.1()76.12(
0ln2
5025coldregen, T
TcTTTcmX pp
0.890kW1073kW8.954
hotregen,
coldregen,, X
XTII
The exergy of the combustion gases at the regenerator exit:
kW13510303
616.2ln)093.1()303()3032.616)(093.1()76.12(
0ln0
60066 T
TcTTTcmX pp
9-97
Review Problems
9-132 A turbocharged four-stroke V-16 diesel engine produces 3500 hp at 1200 rpm. The amount ofpower 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 mechanicalcycles, we have (a)
cycle)mechkJ/cyl8.16(=hp1Btu/min42.41
rev/min)(1200cylinders)(16hp3500
cycles)mechanicalof(No.cylinders)of(No.producedpowerTotal
mechanical
cyclemechBtu/cyl7.73
w
(b)
cycle)thermkJ/cyl16.31(=hp1Btu/min42.41
rev/min)(1200/2cylinders)(16hp3500
cycles)amic thermodynof(No.cylinders)of(No.producedpowerTotal
micthermodyna
cyclethermBtu/cyl15.46
w
9-133 A simple ideal Brayton cycle operating between the specified temperature limits is considered. The pressure ratio for which the compressor and the turbine exit temperature of air are equal is to bedetermined.Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kineticand 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).Analysis 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
kk
p
kk
kkp
kk
rT
PPTT
rTPPTT
/1
3
/1
3
434
/11
/1
1
212
1
T1
T3
qout
qin
3
4
2
1
T
Setting T2 = T4 and solving for rp givess
16.71.4/0.812/
1
3
K300K1500
kk
p TT
r
Therefore, the compressor and turbine exit temperatures will be equal when the compression ratio is 16.7.
9-98
9-134 The four processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-sdiagrams, 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.
PProperties The properties of air are given in Table A-17.
qout
q23
q12
1
2
4
3Analysis (b) We treat air as an ideal gas with variable specific heats,
kJ/kg187.2827.73655.923
kJ/kg736.2719.30046.1036
kJ/kg923.5593.93297.139507.21458.674
kJ/kg1036.46
3.110330.9kPa300kPa100
9.330kJ/kg,1395.97kJ/kg1022.82K0130
kJ/kg932.93kJ/kg674.58K900
K300kPa100kPa300
kJ/kg300.19kJ/kg214.07K300
outinnet
14out
2312in23,in12,in
4
3
4
33
33
2
2
11
22
1
11
2
22
1
11
34
qqw
hhq
hhuuqqq
h
PPP
P
PhuT
hu
TPP
TT
PT
P
huT
rr
r
vvv
T
q12
qout1
2q23
4
3
s
(c) 20.3%kJ/kg923.55kJ/kg187.28
in
netth q
w
9-99
9-135 All four processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-sdiagrams, 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 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).
PAnalysis (b) Process 3-4 is isentropic:
6538.832
kJ/kg653K300949.8KkJ/kg1.005
kJ/kg832.8K9001300KkJ/kg1.005K300900KkJ/kg0.718
K900K300kPa100kPa300
K949.831K1300
outinnet
1414out
23122312in23,in12,in
11
22
1
11
2
22
0.4/1.4/1
3
434
kJ/kg179.8qqw
TTchhq
TTcTTchhuuqqq
TPP
TT
PT
P
PP
TT
p
pv
kk
vvqout
q23
q12
1
2
4
3
v
T
q12
qout1
2q23
4
3
(c) 21.6%kJ/kg832.8kJ/kg179.8
in
netth q
w s
9-136 The three processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-sdiagrams, and the maximum temperature in the cycle and the thermal efficiency are to be determined.Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The properties of air are given in Table A-17. P
qout
qin
1
2 3Analysis (b) We treat air as an ideal gas with variable specific heats,
kJ/kg2377.7kJ/kg1775.3K2100
K300kPa100kPa700
kJ/kg523.90702.9386.1kPa100kPa700
386.1kJ/kg214.07K300
3
33
11
33max
1
11
3
33
21
2
11
12
1
huT
TPP
TTT
PT
P
hPPP
P
PuT
rr
r
K2100vv
v
T
qin
qout1
23
(c)
kJ/kg1853.8kJ/kg1561.23
11
kJ/kg1561.2307.2143.1775
kJ/kg1853.89.5237.2377
in
outth
13out
23in
15.8%qq
uuq
hhqs
9-100
9-137 All three processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-sdiagrams, and the maximum temperature in the cycle and the thermal efficiency are to be determined.Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K, cv = 0.718 kJ/kg·K, and k= 1.4 (Table A-2).
P
qout
qin
1
2 3
Analysis (b) We treat air as an ideal gas with constant specific heats. Process 1-2 is isentropic:
K2100K300kPa100kPa700
K523.1kPa100kPa700K300
11
33max
1
11
3
33
0.4/1.4/1
1
212
TPPTT
TP
TP
PPTT
kk
vv vT
qin
qout1
23(c)
18.5%kJ/kg1584.8kJ/kg1292.4
11
kJ/kg1292.4K3002100KkJ/kg0.718
kJ/kg1584.8K523.12100KkJ/kg1.005
in
outth
1313out
2323in
TTcuuq
TTchhq
v
p
s
9-138 A Carnot cycle executed in a closed system uses air as the working fluid. The net work output per cycle is to be determined.Assumptions 1 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).
th = 60%
300 K
1 MPa1
4
2700 kPa
3
T
s
Analysis (a) The maximum temperature is determined from
kJ0.115K300750KkJ/kg0.1024kg0.0025
KkJ/kg0.1204kPa1000kPa700
lnKkJ/kg0.287ln
K750K300
160.01
12
1
21
0
212
LHnet
HHH
Lth
TTssmW
PP
Rssss
TTT
T
9-101
9-139 [Also solved by EES on enclosed CD] A four-cylinder spark-ignition engine with a compression ratioof 8 is considered. The amount of heat supplied per cylinder, the thermal efficiency, and the rpm for a netpower 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 inTable A-17.Analysis (a) Process 1-2: isentropic compression.
P
1800 K
Qout
Qin
2
3
1
4
v
kJ/kg475.11
51.841.676811
1.676kJ/kg206.91K029
2
1
2
11
112
1
ur
uT
rrr
r
vvv
vv
v
Process 2-3: v = constant heat addition.
kJ0.715kJ/kg475.111487.2kg107.065
kg107.065K290K/kgmkPa0.287
m0.0006kPa98
994.3kJ/kg1487.2K1800
423in
43
3
1
11
33
3
uumQ
RTP
m
uT
r
V
v
(b) Process 3-4: isentropic expansion.
kJ/kg693.2395.31994.38 43
4334
ur rrr vvv
vv
Process 4-1: v = constant heat rejection.
51.9%kJ0.715kJ0.371
kJ0.371344.0715.0
kJ/kg206.91693.23kg107.065
in
netth
outinnet
-414out
QW
QQW
uumQ kJ0.344
(c) rpm4852min1
s60kJ/cycle)0.371(4
kJ/s60rev/cycle)2(2
cylnet,cyl
net
WnW
n
Note that for four-stroke cycles, there are two revolutions per cycle.
9-102
9-140 EES Problem 9-139 is reconsidered. The effect of the compression ratio net work done and theefficiency 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]=(17+273) [K] P[1]=98 [kPa]T[3]=1800 [K] V_cyl=0.6 [L]*Convert(L, m^3) r_v=8 "Compression ratio"W_dot_net = 60 [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"thermalcycle"/2"mechanical cycles"
9-103
th[%]
rv wnet[kJ/kg]
42.81 5 467.146.39 6 492.549.26 7 509.851.63 8 521.753.63 9 529.855.35 10 535.256.85 11 538.5
10-2 10-1 100 101 10250
100
1000
8000
v [m3/kg]
P[kPa]
290 K
1800 K
Air Otto Cycle P-v Diagram
s = const
1
2
3
4
1
2
3
4
v = const
4.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0 8.50
500
1000
1500
2000
2500
s [kJ/kg-K]
T[K] 98 kPa
4866 kPa
Air Otto Cycle T-s Diagram
9-104
5 6 7 8 9 10 11460
470
480
490
500
510
520
530
540
rv
wne
t [k
J/kg
]
5 6 7 8 9 10 1142
44
46
48
50
52
54
56
58
rv
th [
%]
9-105
9-141 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 effectivepressure 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 inTable A-17. Analysis (a) Process 1-2: isentropic compression.
kPa2129kPa98K300K708.3
9.2
kJ/kg9.518K3.70852.672.621
2.911
2.621kJ/kg214.07K300
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
P
qout
qin
2
3
1
4
v
Process 2-3: v = constant heat addition.
kJ/kg609.89.5187.1128
593.8kJ/kg1128.7K1416.63.70822
23
3222
33
2
22
3
33
3
uuq
uTTPP
TT
PT
P
in
rv
vv
(b) Process 3-4: isentropic expansion.
kJ/kg487.7506.79593.82.9 43
4334
ur rrr vvv
vv
Process 4-1: v = constant heat rejection.
kJ/kg336.17.2738.609kJ/kg273.707.21475.487
outinnet
14out
qqwuuq
(c) 55.1%kJ/kg609.8kJ/kg336.1
in
netth q
w
(d)
kPa429kJ1
mkPa11/9.21/kgm0.879
kJ/kg336.1/11
MEP
/kgm0.879kPa98
K300K/kgmkPa0.287
3
31
net
21
net
max2min
33
1
11max
rww
r
PRT
vvv
vvv
vv
9-106
9-142 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 effectivepressure 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 (a) Process 1-2 is isentropic compression:
kPa2190kPa98K300K728.8
9.2
K728.89.2K300
11
2
2
12
1
11
2
22
0.41
2
112
PTT
PT
PT
P
TTk
v
vvv
v
v P
Process 2-3: v = constant heat addition.
kJ/kg523.3K728.81457.6KkJ/kg0.718
K457.618.72822
2323
222
33
2
22
3
33
TTcuuq
TTPP
TT
PT
P
in v
vv
qout
qin
2
3
1
4
v
(b) Process 3-4: isentropic expansion.
K600.09.21K1457.6
0.41
4
334
k
TTv
v
Process 4-1: v = constant heat rejection.
kJ/kg307.94.2153.523
kJ/kg215.4K300600KkJ/kg0.718
outinnet
1414out
qqw
TTcuuq v
(c) 58.8%kJ/kg523.3kJ/kg307.9
in
netth q
w
(d)
kPa393kJ1
mkPa11/9.21/kgm0.879
kJ/kg307.9/11
MEP
/kgm0.879kPa98
K300K/kgmkPa0.287
3
31
net
21
net
max2min
33
1
11max
rww
r
PRT
vvv
vvv
vv
9-107
9-143 An engine operating on the ideal diesel cycle with air as the working fluid is considered. The pressure at the beginning of the heat-rejection process, the net work per cycle, 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 inTable A-17.Analysis (a) The compression and the cutoff ratios are
P2
cm75cm15016
cm75cm1200
3
3
2
33
3
2
1
VV
VV
crr
Process 1-2: isentropic compression.
kJ/kg863.03K837.3256.421.676
1611
1.676kJ/kg206.91K029
2
21
2
11
112
1
hT
r
uT
rrr
r
vvv
vv
v
qout
qin 32
1
4
v
Process 2-3: P = constant heat addition.
002.5kJ/kg1848.9
K1674.63.83722
3
3
222
33
2
22
3
33
r
h
TTTT
PT
P
v
v
vvv
Process 3-4: isentropic expansion.
kJ/kg636.00K853.4016.40002.5
216
224
42
4
3
43334
uTr
rrrr vvv
vv
vv
v
Process 4-1: v = constant heat rejection.
kPa294.3kPa100K290K853.4
11
44
1
11
4
44 PTTP
TP
TP vv
(b) kg101.442K290K/kgmkPa0.287
m0.0012kPa100 33
3
1
11
RTP
mV
kJ0.803619.0422.1
kJ0.619kJ/kg206.91636.00kg101.442
kJ1.422863.081848.9kg101.442
outinnet
3-14out
-323in
QQW
uumQ
hhmQ
(c) kPa714kJ1
mkPa11/161m0.0012
kJ0.803/11
MEP3
31
net
21
net
rWW
VVV
9-108
9-144 An engine operating on the ideal diesel cycle with argon as the working fluid is considered. The pressure at the beginning of the heat-rejection process, the net work per cycle, 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 Argon is an ideal gas with constant specific heats.Properties The properties of argon at room temperature are cp = 0.5203 kJ/kg.K, cv = 0.3122 kJ/kg·K, R= 0.2081 kJ/kg·K and k = 1.667 (Table A-2). Analysis (a) The compression and the cutoff ratios are
2cm75cm15016
cm75cm1200
3
3
2
33
3
2
1
VV
VV
crr P
Qout
Qin 32
1
4
v
Process 1-2: isentropic compression.
K184316K290 0.6671
1
212
k
TTVV
Process 2-3: P = constant heat addition.
K3686184322 222
33
2
22
3
33 TTTT
PT
Pvvvv
Process 3-4: isentropic expansion.
K920.9162K368622 0.6671
3
1
4
23
1
4
334
kkk
rTTTT
VV
VV
Process 4-1: v = constant heat rejection.
kPa317.6kPa100K290K920.9
11
44
1
11
4
44 PTTP
TP
TP vv
(b) kg101.988K290K/kgmkPa0.2081
m0.0012kPa100 33
3
1
11
RTPm V
kJ1.514392.0906.1
kJ0.392K290920.9KkJ/kg0.3122kg101.988
kJ1.906K18433686KkJ/kg0.5203kg101.988
outinnet
3-1414out
-32323in
QQW
TTmcuum
TTmchhmQ p
vQ
(c) kPa1346kJ1
mkPa11/161m0.0012
kJ1.514/11
MEP3
31
net
21
net
rWW
VVV
9-109
9-145E An ideal dual cycle with air as the working fluid with a compression ratio of 12 is considered. Thethermal 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
lbm103.132R550R/lbmftpsia0.3704
ft75/1728psia14.7 33
3
1
11
RTPm V P
0.3 Btu
x
Qout
1.1 Btu 3
2
1
4
v
Process 1-2: isentropic compression.
R148612R550 0.41
2
112
k
TTVV
Process 2-x: v = constant heat addition,
R2046R1486RBtu/lbm0.171lbm103.132Btu0.3 x3
22in,2
x
xxx
TT
TTmcuumQ v
Process x-3: P = constant heat addition.
1.715R2046R3509
R3509R2046RBtu/lbm0.240lbm103.132Btu1.1
33
3
33
333
33in,3
xxc
x
xx
xpxx
TT
rT
PT
P
TT
TTmchhmQ
V
VVV
Process 3-4: isentropic expansion.
R161112
1.715R3509715.1715.1 0.41
3
1
4
13
1
4
334
kkk
rTTTT
V
V
V
V
Process 4-1: v = constant heat rejection.
59.4%Btu1.4
Btu0.56811
Btu0.568R5501611RBtu/lbm0.171lbm103.132
in
outth
31414out
TTmcuumQ v
9-110
9-146 An ideal Stirling cycle with air as the working fluid is considered. The maximum pressure in thecycle and the net work output are to be determined.Assumptions 1 The air-standard assumptions are applicable. 2Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats.
350 K
1800 K 1
4qout
qin = 900 kJ/kg
2
3
T
Properties The properties of air at room temperature areR = 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
sKkJ/kg0.5
K1800kJ/kg90012
12HT
qss
and
kPa5873K350K1800
5.710kPa200
710.5lnKkJ/kg0.287KkJ/kg0.5lnln
3
1
1
23
3
1
1
331
1
11
3
33
1
2
1
2
1
20
1
212
TT
PTT
PPT
PT
P
RTT
css
v
v
v
vvv
v
v
v
v
v
vv
(b) kJ/kg725kJ/kg900K1800K350
11 ininthnet qTT
qwH
L
9-111
9-147 A simple ideal Brayton cycle with air as the working fluid is considered. The changes in the network 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. Analysis The properties at various states are T
3
2qin 3
4qout
2
1
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
.
.
For rp = 6, s
%9.37kJ/kg894.57kJ/kg339.46
kJ/kg339.4611.55557.894kJ/kg555.1119.3003.855
kJ/kg894.5740.50197.1395
kJ/kg855.315.559.33061
kJ/kg501.40316.8386.16
in
netth
outinnet
14out
23in
43
4
21
2
34
12
qw
qqwhhqhhq
hPPP
P
hPPP
P
rr
rr
For rp = 12,
%5.48kJ/kg785.37kJ/kg380.96
kJ/kg380.9641.40437.785kJ/kg404.4119.3006.704
kJ/kg785.3760.61097.1395
kJ/kg704.658.279.330121
kJ/kg610.663.16386.112
in
netth
outinnet
14out
23in
43
4
21
2
34
12
qw
qqwhhqhhq
hPPP
P
hPPP
P
rr
rr
Thus,(a) increase46.33996.380net kJ/kg41.5w
(b) increase%9.37%5.48th 10.6%
9-112
9-148 A simple ideal Brayton cycle with air as the working fluid is considered. The changes in the network 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/kg803.4kJ/kg321.9
kJ/kg321.95.4814.803
kJ/kg481.5K300779.1KkJ/kg1.005
kJ/kg803.4K500.61300KkJ/kg1.005
K779.161K1300
K500.66K300
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
T3
2qin 3
4qout
2
1s
For rp = 12,
%8.50kJ/kg693.2kJ/kg352.3
kJ/kg352.39.3402.693
kJ/kg340.9K300639.2KkJ/kg1.005
kJ/kg693.2K610.21300KkJ/kg1.005
K639.2121K1300
K610.212K300
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
Thus,(a) increase9.3213.352net kJ/kg30.4w
(b) increase%1.40%8.50th 10.7%
9-113
9-149 A regenerative Brayton cycle with helium as the working fluid is considered. The thermal efficiency and the required mass flow rate of helium are to be determined for 100 percent and 80 percent isentropicefficiencies for both the compressor and the turbine. 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. Properties The properties of helium are cp = 5.1926 kJ/kg.Kand k = 1.667 (Table A-2). T
Analysis (a) Assuming T = C = 100%,
K759.84.6893.78375.04.689
K783.381K1800
K689.48K300
242524
25
24
25
70.667/1.66/1
3
434
70.667/1.66/1
1
212
TTTTTTcTTc
hhhh
PP
T
PP
TT
p
p
kk
s
kk
s
T
24
6
5
qin 3
4s2s
1
1800 K
300 Ks
60.3%kJ/kg5401.3kJ/kg3257.3
kJ/kg5401.3K759.81800KkJ/kg5.1926
kJ/kg3257.3kJ/s60,000
kJ/kg3257.3K300689.4783.31800KkJ/kg5.1926
in
netth
5353in
net
net
12431243inC,outT,net
qw
TTchhq
wW
m
TTTTchhhhwww
p
p
kg/s18.42
(b) Assuming T = C = 80%,
K986.63.783180080.01800
K783.381K1800
K786.880.0/3004.689300/
K689.48K300
433443
43
43
43
70.667/1.66/1
3
434
121212
12
12
12
70.667/1.66/1
1
212
sTsp
p
sT
kk
s
Csp
spsC
kk
s
TTTTTTcTTc
hhhh
PP
TT
TTTTTTcTTc
hhhh
PP
TT
37.8%kJ/kg4482.8kJ/kg1695.9
kJ/kg4482.8K936.71800KkJ/kg5.1926
kJ/kg1695.9kJ/s60,000
kJ/kg1695.9K300786.8986.61800KkJ/kg5.1926
K936.78.7866.98675.08.786
in
netth
5353in
net
net
12431243inC,outT,net
242524
25
24
25
qw
TTchhq
wWm
TTTTchhhhwww
TTTTTTcTTc
hhhh
p
p
p
p
kg/s35.4
9-114
9-150 A regenerative gas-turbine engine operating with two stages of compression and two stages ofexpansion 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. 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/kg624.1K889.51200KkJ/kg1.005222
kJ/kg332.7K300465.5KkJ/kg1.005222
K770.85.4655.88972.05.465
K889.59.838120086.01200
K838.93.51K1200
K465.578.0/3001.429300
/
K429.13.5K300
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 T
4
9
22s
5
8
77s
9
6
1
4
3s
Thus, 53.3%kJ/kg624.1kJ/kg332.7
outT,
inC,bw w
wr
39.2%kJ/kg743.4kJ/kg291.4
kJ/kg291.47.3321.624
kJ/kg743.4K889.51200770.81200KkJ/kg1.005
in
netth
inC,outT,net
78567856in
qw
www
TTTTchhhhq p
9-115
9-151 EES Problem 9-150 is reconsidered. The effect of the isentropic efficiencies for the compressor andturbine and regenerator effectiveness on net work done and the heat supplied to the cycle is to be investigated. Also, the T-s diagram for the cycle is to be plotted. Analysis Using EES, the problem is solved as follows:
"Input data"T[6] = 1200 [K] T[8] = T[6]Pratio = 3.5 T[1] = 300 [K] P[1]= 100 [kPa] T[3] = T[1]Eta_reg = 0.72 "Regenerator effectiveness"Eta_c =0.78 "Compressor isentorpic efficiency"Eta_t =0.86 "Turbien isentropic efficiency"
"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 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_compisen_LP/w_comp_LP"compressor adiabatic efficiency, W_comp > W_compisen"
"Conservation of energy for the LP compressor for the isentropic case: e_in - e_out = DELTAe=0 for steady-flow"h[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] h[2]=ENTHALPY(Air,T=T[2])s[2]=ENTROPY(Air,T=T[2], P=P[2])
"HP Compressor:"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 HP compressor"P[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 compressor for the isentropic case: e_in - e_out = DELTAe=0 for steady-flow"h[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]
9-116
h[4]=ENTHALPY(Air,T=T[4])s[4]=ENTROPY(Air,T=T[4], P=P[4])
"Intercooling heat loss:"h[2] = q_out_intercool + h[3]
"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"h[4] + q_in_noreg = h[6]
h[6]=ENTHALPY(Air,T=T[6])P[6]=P[4]"process 4-6 is SSSF constant pressure"
"HP Turbine analysis"
s[6]=ENTROPY(Air,T=T[6],P=P[6])s_s[7]=s[6] "For the ideal case the entropies are constant across the turbine"P[7] = P[6] /Pratios_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"Eta_t = w_turb_HP /w_turbisen_HP "turbine adiabatic efficiency, w_turbisen > w_turb"
"SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0 e_in -e_out = DELTAe_cv = 0 for steady-flow"h[6] = w_turbisen_HP + h_s[7] h_s[7]=ENTHALPY(Air,T=T_s[7])"Actual Turbine analysis:"h[6] = w_turb_HP + h[7] h[7]=ENTHALPY(Air,T=T[7])s[7]=ENTROPY(Air,T=T[7], P=P[7])
"Reheat Q_in:"h[7] + q_in_reheat = h[8] h[8]=ENTHALPY(Air,T=T[8])
"HL Turbine analysis"
P[8]=P[7]s[8]=ENTROPY(Air,T=T[8],P=P[8])s_s[9]=s[8] "For the ideal case the entropies are constant across the turbine"P[9] = P[8] /Pratios_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"Eta_t = w_turb_LP /w_turbisen_LP "turbine adiabatic efficiency, w_turbisen > w_turb"
"SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0 e_in -e_out = DELTAe_cv = 0 for steady-flow"h[8] = w_turbisen_LP + h_s[9] h_s[9]=ENTHALPY(Air,T=T_s[9])"Actual Turbine analysis:"h[8] = w_turb_LP + h[9] h[9]=ENTHALPY(Air,T=T[9])s[9]=ENTROPY(Air,T=T[9], P=P[9])
"Cycle analysis"w_net=w_turb_HP+w_turb_LP - w_comp_HP - w_comp_LPq_in_total_noreg=q_in_noreg+q_in_reheatEta_th_noreg=w_net/(q_in_total_noreg)*Convert(, %) "[%]" "Cycle thermal efficiency"
9-117
Bwr=(w_comp_HP + w_comp_LP)/(w_turb_HP+w_turb_LP)"Back work ratio"
"With the regenerator, the heat added in the external heat exchanger is"h[5] + q_in_withreg = h[6] h[5]=ENTHALPY(Air, T=T[5]) s[5]=ENTROPY(Air,T=T[5], P=P[5])P[5]=P[4]
"The regenerator effectiveness gives h[5] and thus T[5] as:"Eta_reg = (h[5]-h[4])/(h[9]-h[4]) "Energy balance on regenerator gives h[10] and thus T[10] as:"h[4] + h[9]=h[5] + h[10] h[10]=ENTHALPY(Air, T=T[10]) s[10]=ENTROPY(Air,T=T[10], P=P[10])P[10]=P[9]
"Cycle thermal efficiency with regenerator"q_in_total_withreg=q_in_withreg+q_in_reheatEta_th_withreg=w_net/(q_in_total_withreg)*Convert(, %) "[%]"
"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]s_s[8]=s[8]T_s[8]=T[8]s_s[10]=s[10]T_s[10]=T[10]
C reg t th,noreg
[%]th,withreg
[%]qin,total,noreg
[kJ/kg]qin,total,withreg
[kJ/kg]wnet
[kJ/kg]0.78 0.6 0.86 27.03 36.59 1130 834.6 305.40.78 0.65 0.86 27.03 37.7 1130 810 305.40.78 0.7 0.86 27.03 38.88 1130 785.4 305.40.78 0.75 0.86 27.03 40.14 1130 760.8 305.40.78 0.8 0.86 27.03 41.48 1130 736.2 305.40.78 0.85 0.86 27.03 42.92 1130 711.6 305.40.78 0.9 0.86 27.03 44.45 1130 687 305.40.78 0.95 0.86 27.03 46.11 1130 662.4 305.40.78 1 0.86 27.03 47.88 1130 637.8 305.4
9-118
4.5 5.0 5.5 6.0 6.5 7.0 7.5200
400
600
800
1000
1200
1400
1600
s [kJ/kg-K]
T[K] 10
0kPa
350kP
a
1225
kPa
Air
1
2
3
4
5
6
7
8
9
10
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
t
th
reg = 0.72
c = 0.78
With regeneration
No regeneration
0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1600
700
800
900
1000
1100
1200
t
q in,
tota
l
reg = 0.72
c = 0.78No regenerationWith regeneration
9-119
0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1100
150
200
250
300
350
400
450
t
wne
t [k
J/kg
]reg = 0.72
c = 0.78
0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 125
30
35
40
45
50
reg
th
c = 0.78
t = 0.86
With regenertion
No regeneration
0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 115
20
25
30
35
40
45
50
c
th
reg = 0.72
t = 0.86
With regeneration
No regeneration
9-120
9-152 A regenerative gas-turbine engine operating with two stages of compression and two stages ofexpansion 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. Properties The properties of helium at room temperature are cp = 5.1926 kJ/kg.K and k = 1.667 (TableA-2).Analysis The work inputs to each stage of compressor are identical, so are the work outputs of each stage of the turbine.
kJ/kg4225.7K793.11200KkJ/kg5.1926222
kJ/kg2599.4K300550.3KkJ/kg5.1926222
K725.13.5501.79372.03.550
K793.19.726120086.01200
K726.93.51K1200
K550.378.0/3002.495300
/
K495.23.5K300
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 T
4
9
22s
5
8
77s
9qin
6
1
4
3s
Thus, 61.5%kJ/kg4225.7kJ/kg2599.4
outT,
inC,bw w
wr
35.5%kJ/kg4578.8kJ/kg1626.3
kJ/kg1626.34.25997.4225
kJ/kg4578.8K793.11200725.11200KkJ/kg5.1926
in
netth
inC,outT,net
78567856in
qw
www
TTTTchhhhq p
9-121
9-153 An ideal regenerative Brayton cycle is considered. The pressure ratio that maximizes the thermalefficiency of the cycle is to be determined, and to be compared with the pressure ratio that maximizes the cycle net work. Analysis Using the isentropic relations, the temperatures at the compressor and turbine exit can be expressed as
kkp
kk
p
kk
kkp
kk
rTr
TPP
TT
rTPP
TT
/13
/1
3
/1
3
434
/11
/1
1
212
1
T
Then,
1/1
1/1
33outinnet
/11121616out
/13435353in
1
1
TrTrTTcqqw
rTcTTcTTchhq
rTcTTcTTchhq
kkp
kkpp
kkpppp
kkpppp
6
5
qin3
42
1s
To maximize the net work, we must have
01111 /11
/13
net kkp
kkpp
prT
kkrT
kkc
rw
Solving for rp giveskk
p TTr
12/
3
1
Similarly,
kkpp
kkpp
rTc
rTcqq
/13
/11
in
outth
1
111
which simplifies to
kkpr
TT /1
3
1th 1
When rp = 1, the thermal efficiency becomes th = 1 - T1/T3, which is the Carnot efficiency. Therefore, the efficiency is a maximum when rp = 1, and must decrease as rp increases for the fixed values of T1 and T3.Note that the compression ratio cannot be less than 1, and the factor
kkpr /1
is always greater than 1 for rp > 1. Also note that the net work wnet = 0 for rp = 1. This being the case, thepressure ratio for maximum thermal efficiency, which is rp = 1, is always less than the pressure ratio formaximum work.
9-122
9-154 An ideal gas-turbine cycle with one stage of compression and two stages of expansion andregeneration is considered. The thermal efficiency of the cycle as a function of the compressor pressureratio and the high-pressure turbine to compressor inlet temperature ratio is to be determined, and to becompared with the efficiency of the standard regenerative cycle.Analysis The T-s diagram of the cycle is as shown in thefigure. If the overall pressure ratio of the cycle is rp, which is the pressure ratio across the compressor, then the pressure ratioacross each turbine stage in the ideal case becomes rp. Usingthe isentropic relations, the temperatures at the compressor and turbine exit can be expressed as
T
kkp
kkp
kkp
kkp
kk
p
kk
kkp
kk
p
kk
kkp
kk
rTrrTrTr
TPPTT
rTr
TPPTTT
rTPPTTT
2/11
2/1/11
2/12
/1
5
/1
5
656
2/13
/1
3
/1
3
4347
/11
/1
1
2125
1
1 qout
3
4
6
5
qin
7
2
1s
Then,
1
1
2/111616out
2/137373in
kkppp
kkppp
rTcTTchhq
rTcTTchhq
and thus
kkpp
kkpp
rTc
rTcqq
2/13
2/11
in
outth
1
111
which simplifies to
kkpr
TT 2/1
3
1th 1
The thermal efficiency of the single stage ideal regenerative cycle is given as
kkpr
TT /1
3
1th 1
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.
9-123
9-155 A spark-ignition engine operates on an ideal Otto cycle with a compression ratio of 11. The maximum temperature and pressure in the cycle, the net work per cycle and per cylinder, the thermalefficiency, the mean effective pressure, and the power output for a specified engine speed 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) First the mass in one cylinder is determined as follows
cylinderoneform000045.04/)0018.0(
11 3c
c
c
c
dcr VV
V
V
VV
31 m000495.000045.0000045.0dc VVV
kg0.0004805K323K/kgmkPa0.287
)m495kPa)(0.00090(3
3
1
11
RTPm V
For this problem, we use the properties from EES software. Remember that for an ideal gas, enthalpy is afunction of temperature only whereas entropy is functions of both temperature and pressure. Process 1-2: Isentropic compression
KkJ/kg8100.5kPa90
K233C50kJ/kg88.230K233C50
11
1
11
sPT
uT
Qout
V
P
4
3
2
Qin
1/kgm09364.0kg0004805.0
m000045.0
m000045.0
/kgm0302.1kg0004805.0
m000495.0
33
22
32
33
11
m
m
c
Vv
VV
Vv
K3.807kg/m09364.0
kJ/kg.K8100.523
2
12 Tss
v
kJ/kg33.598K3.807 22 uT
kPa2474K323K807.3
m0.000045m0.000495kPa)90(
3
3
1
2
2
112 T
TPP
V
V
Process 2-3: constant volume heat addition
kJ/kg8.3719kJ/kg)33.598kg)(0004805.0(kJ5.1)( 3323in uuuumQ
K403732 kJ/kg33.598 Tu
kPa12,375K807.3K4037kPa)2474(
2
323 T
TPP
KkJ/kg3218.7kPa375,12
K40373
3
3 sPT
9-124
(b) Process 3-4: isentropic expansion.
K2028kg/m0302.1
kJ/kg.K3218.743
14
34 Tssvv
kJ/kg6.1703K2028 44 uT
kPa565K4037K2028
111kPa)375,12(
3
4
4
334 T
TPP
V
V
Process 4-1: constant voume heat rejectionkJ7077.0kJ/kg88.2301703.6kg)0004805.0()( 14out uumQ
The net work output and the thermal efficiency are cylinder)percycle(per7077.05.1outinoutnet, kJ0.792QQW
0.528kJ1.5
kJ0.792
in
outnet,th Q
W
(c) The mean effective pressure is determined to be
kPa1761kJ
mkPam)000045.0000495.0(
kJ0.7923MEP
3
321
outnet,
VV
W
(d) The power for engine speed of 3000 rpm is
kW79.2s60
min1rev/cycle)2(
rev/min)3000(cycle)-rkJ/cylinde0.792cylinder)(4(2netcylnetnWnW
Note that there are two revolutions in one cycle in four-stroke engines.
9-125
9-156 A gas-turbine plant operates on the regenerative Brayton cycle with reheating and intercooling. Theback work ratio, the net work output, the thermal efficiency, the second-law efficiency, and the exergies atthe 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: kPa4.346)1200)(100(412 PPP
Process 1-2, 3-4: Compression
T
KkJ/kg7054.5kPa100K300
kJ/kg43.300K300
11
1
11
sPT
hT
4
8
22s
9
10
7
66s
8qin
5
1
4
3
kJ/kg79.428kJ/kg.K7054.5
kPa4.3462
12
2sh
ssP
kJ/kg88.46043.300
43.30079.42880.0 2212
12C h
hhhhh s s
KkJ/kg5040.5kPa4.346
K350kJ/kg78.350K350
33
3
33
sPT
hT
kJ/kg42.500kJ/kg.K5040.5
kPa12004
34
4sh
ssP
kJ/kg83.53778.350
78.35042.50080.0 4434
34C h
hhhhh s
Process 6-7, 8-9: Expansion
KkJ/kg6514.6kPa1200K1400
kJ/kg9.1514K1400
66
6
66
sPT
hT
kJ/kg9.1083kJ/kg.K6514.6
kPa4.3467
67
7sh
ssP
kJ/kg1.11709.10839.1514
9.151480.0 7
7
76
76T h
hhhhh
s
KkJ/kg9196.6kPa4.346
K1300kJ/kg6.1395K1300
88
8
88
sPT
hT
kJ/kg00.996kJ/kg.K9196.6
kPa1009
89
9sh
ssP
9-126
kJ/kg9.107500.9966.1395
6.139580.0 9
9
98
98T h
hhhhh
s
Cycle analysis: kJ/kg50.34778.35083.53743.30088.4603412inC, hhhhw
kJ/kg50.6649.10756.13951.11709.15149876outT, hhhhw
0.52350.66450.347
outT,
inC,bw w
wr
kJ/kg317.050.34750.664inC,outT,net www
Regenerator analysis:
kJ/kg36.67283.5379.1075
9.107575.0 10
10
49
109regen h
hhhhh
KkJ/kg5157.6kPa100
K36.67210
10
10 sPh
kJ/kg40.94183.53736.6729.1075 5545109regen hhhhhhq
(b) kJ/kg54.57340.9419.151456in hhq
0.55354.5730.317
in
netth q
w
(c) The second-law efficieny of the cycle is defined as the ratio of actual thermal efficiency to themaximum possible thermal efficiency (Carnot efficiency). The maximum temperature for the cycle can be taken to be the turbine inlet temperature. That is,
786.0K1400K30011
6
1max T
T
and 0.704786.0553.0
max
thII
(d) The exergies at the combustion chamber exit and the regenerator exit are kJ/kg930.7kJ/kg.K)7054.56514.6)(K300(kJ/kg)43.3009.1514()( 060066 ssThhx
kJ/kg128.8kJ/kg.K)7054.55157.6)(K300(kJ/kg)43.30036.672()( 010001010 ssThhx
9-127
9-157 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 flowrate 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. Analysis (a) For this problem, we use theproperties of air from EES software. Rememberthat for an ideal gas, enthalpy is a function oftemperature only whereas entropy is functions ofboth temperature and pressure.
5Heatexchanger
Sat. vap.200°C
25°C350°C
1.2 MPa 500°C
100 kPa30°C
Compress.
43
2
Turbine
Combustionchamber
1
Process 1-2: Compression
KkJ/kg7159.5kPa100
C30kJ/kg60.303C30
11
1
11
sPT
hT
kJ/kg37.617kJ/kg.K7159.5
kPa12002
12
2sh
ssP
kJ/kg24.68660.303
60.30337.61782.0 2212
12C h
hhhhh s
Process 3-4: Expansion
kJ/kg62.792C500 44 hT
ss hhh
hhhh
43
3
43
43T
62.79282.0
We cannot find the enthalpy at state 3 directly. However, using the following lines in EES together withthe isentropic efficiency relation, we find h3 = 1404.7 kJ/kg, T3 = 1034ºC, s3 = 6.5699 kJ/kg.K. The solution by hand would require a trial-error approach.
h_3=enthalpy(Air, T=T_3)s_3=entropy(Air, T=T_3, P=P_2)h_4s=enthalpy(Air, P=P_1, s=s_3)
Also, kJ/kg44.631C350 55 hT
The inlet water is compressed liquid at 25ºC 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 liquidenthalpy at the given temperature.
kJ/kg0.27921
C200
kJ/kg27.106kPa1555
C25
22
2
11
1
ww
ww
hx
T
hP
T
The net work output is
kJ/kg03.61262.7927.1404kJ/kg64.38260.30324.686
43outT,
12inC,
hhwhhw
kJ/kg39.22964.38203.612inC,outT,net www
9-128
The mass flow rate of air is
kg/s3.487kJ/kg39.229
kJ/s800
net
net
wW
ma
(b) The back work ratio is
0.62503.61264.382
outT,
inC,bw w
wr
The rate of heat input and the thermal efficiency are
kW2505)kJ/kg24.686.7kg/s)(1404487.3()( 23in hhmQ a
0.319kW2505
kW800
in
netth Q
W
(c) An energy balance on the heat exchanger gives
kg/s0.2093ww
wwwa
mm
hhmhhm
)kJ/kg27.106(2792.0)kJ/kg44.63162kg/s)(792.487.3(
)()( 1254
(d) The heat supplied to the water in the heat exchanger (process heat) and the utilization efficiency are
kW1.562)kJ/kg27.106.0kg/s)(27922093.0()( 12p www hhmQ
0.544kW2505
1.562800
in
pnet
Q
QWu
9-129
9-158 A turbojet aircraft flying is considered. The pressure of the gases at the turbine exit, the mass flowrate 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. 3Air is an ideal gas with variable specific heats. Properties The gas constant of air is R = 0.287 kJ/kg·K (Table A-1). TAnalysis (a) For this problem, we use the properties from EES software.Remember that for an ideal gas, enthalpy is a function of temperatureonly whereas entropy is functions of both temperature and pressure. Diffuser, Process 1-2:
kJ/kg23.238C35 11 hT
kJ/kg37.269/sm1000
kJ/kg12m/s)(15
/sm1000kJ/kg1
2m/s)(900/3.6kJ/kg)23.238(
22
222
2
222
2
22
2
21
1
hh
VhVh
6
qout
5
qin
3
4
2
1s
KkJ/kg7951.5kPa50
kJ/kg37.2692
2
2 sPh
Compressor, Process 2-3:
kJ/kg19.505kJ/kg.K7951.5
kPa4503
23
3sh
ssP
kJ/kg50.55337.269
37.26919.50583.0 3323
23C h
hhhhh s
Turbine, Process 3-4:
kJ/kg8.1304C950 44 hT
kJ/kg6.10208.130437.26950.553 555423 hhhhhh
where the mass flow rates through the compressor and the turbine are assumed equal.
kJ/kg45.9628.1304
6.10208.130483.0 5554
54T s
ssh
hhhhh
KkJ/kg7725.6kPa450C950
44
4 sPT
kPa147.4545
5
KkJ/kg7725.6kJ/kg45.962
Pss
h s
(b) The mass flow rate of the air through the compressor is
kg/s1.760kJ/kg)37.26950.553(
kJ/s500
23
C
hhW
m
(c) Nozzle, Process 5-6:
KkJ/kg8336.6kPa4.147kJ/kg6.1020
55
5 sPh
9-130
kJ/kg66.709kJ/kg.K8336.6
kPa406
56
6sh
ssP
kJ/kg52.76266.7096.1020
6.102083.0 6
6
65
65N h
hhhhh
s
m/s718.5622
26
26
6
25
5
/sm1000kJ/kg1
2kJ/kg52.7620kJ/kg)6.1020(
22
VV
VhVh
where the velocity at nozzle inlet is assumed zero.(d) The propulsive power and the propulsive efficiency are
kW206.122116 /sm1000kJ/kg1m/s)250)(m/s250m/s5.718(kg/s)76.1()( VVVmWp
kW1322kJ/kg)50.5538.1304(kg/s)76.1()( 34in hhmQ
0.156kW1322kW1.206
inQ
W pp
9-131
9-159 EES The effect of variable specific heats on the thermal efficiency of the ideal Otto cycle using air as the working fluid is to be investigated. The percentage of error involved in using constant specific heatvalues at room temperature for different combinations of compression ratio and maximum cycletemperature is to be determined.Analysis Using EES, the problem is solved as follows:
Procedure ConstPropResult(T[1],P[1],r_comp,T[3]:Eta_th_ConstProp,Eta_th_easy)"For Air:"C_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(, %) "[%]"
"The Easy Way to calculate the constant property Otto cycle efficiency is:"Eta_th_easy = (1 - 1/r_comp^(k-1))*Convert(, %) "[%]"
END
"Input Data"T[1]=300 [K] P[1]=100 [kPa]"T[3] = 1000 [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"
9-132
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_23q_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"
Call ConstPropResult(T[1],P[1],r_comp,T[3]:Eta_th_ConstProp,Eta_th_easy)PerCentError = ABS(Eta_th - Eta_th_ConstProp)/Eta_th*Convert(, %) "[%]"
PerCentError
[%]
rcomp th
[%]th,ConstProp
[%]th,easy
[%]T3
[K
3.604 12 60.8 62.99 62.99 10006.681 12 59.04 62.99 62.99 15009.421 12 57.57 62.99 62.99 200011.64 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
rcomp
Per
Cen
tErr
or [
%]
Percent Error = | th - th,ConstProp | / th
Tmax = 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
9-133
9-160 EES 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_23q_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"
9-134
th[%]
rcomp wnet[kJ/kg]
45.83 6 567.448.67 7 589.351.03 8 604.953.02 9 616.254.74 10 624.356.24 11 63057.57 12 633.858.75 13 636.359.83 14 637.560.8 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
th [
%]
9-135
9-161 The effects of pressure ratio on the net work output and the thermal efficiency of a simple Braytoncycle is to be investigated. The pressure ratios at which the net work output and the thermal efficiency are maximum are to be determined.Analysis Using EES, the problem is solved as follows:
P_ratio = 8 T[1] = 300 [K] P[1]= 100 [kPa] T[3] = 1800 [K] m_dot = 1 [kg/s] Eta_c = 100/100 Eta_t = 100/100
"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])
9-136
Bwr Pratio Wc[kW]
Wnet[kW]
Wt[kW]
Qin[kW]
0.254 0.3383 5 175.8 516.3 692.1 15260.2665 0.3689 6 201.2 553.7 754.9 15010.2776 0.3938 7 223.7 582.2 805.9 14780.2876 0.4146 8 244.1 604.5 848.5 14580.2968 0.4324 9 262.6 622.4 885 14390.3052 0.4478 10 279.7 637 916.7 14220.313 0.4615 11 295.7 649 944.7 14060.3203 0.4736 12 310.6 659.1 969.6 13920.3272 0.4846 13 324.6 667.5 992.1 13780.3337 0.4945 14 337.8 674.7 1013 13640.3398 0.5036 15 350.4 680.8 1031 13520.3457 0.512 16 362.4 685.9 1048 13400.3513 0.5197 17 373.9 690.3 1064 13280.3567 0.5269 18 384.8 694.1 1079 13170.3618 0.5336 19 395.4 697.3 1093 13070.3668 0.5399 20 405.5 700 1106 12970.3716 0.5458 21 415.3 702.3 1118 12870.3762 0.5513 22 424.7 704.3 1129 12770.3806 0.5566 23 433.8 705.9 1140 12680.385 0.5615 24 442.7 707.2 1150 12590.3892 0.5663 25 451.2 708.3 1160 12510.3932 0.5707 26 459.6 709.2 1169 12430.3972 0.575 27 467.7 709.8 1177 12340.401 0.5791 28 475.5 710.3 1186 12270.4048 0.583 29 483.2 710.6 1194 12190.4084 0.5867 30 490.7 710.7 1201 12110.412 0.5903 31 498 710.8 1209 12040.4155 0.5937 32 505.1 710.7 1216 11970.4189 0.597 33 512.1 710.4 1223 11900.4222 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
Wnet
[kW]
th
9-137
9-162 EES The effects of pressure ratio on the net work output and the thermal efficiency of a simpleBrayton cycle is to be investigated assuming adiabatic efficiencies of 85 percent for both the turbine and the compressor. The pressure ratios at which the net work output and the thermal efficiency are maximumare to be determined.Analysis Using EES, the problem is solved as follows:
P_ratio = 8 T[1] = 300 [K] P[1]= 100 [kPa] T[3] = 1800 [K] m_dot = 1 [kg/s] Eta_c = 85/100 Eta_t = 85/100
"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])
9-138
s[4]=entropy('air',T=T[4],P=P[4])
9-139
Bwr Pratio Wc[kW]
Wnet[kW]
Wt[kW]
Qin[kW]
0.3515 0.2551 5 206.8 381.5 588.3 14950.3689 0.2764 6 236.7 405 641.7 14650.3843 0.2931 7 263.2 421.8 685 14390.3981 0.3068 8 287.1 434.1 721.3 14150.4107 0.3182 9 309 443.3 752.2 13930.4224 0.3278 10 329.1 450.1 779.2 13730.4332 0.3361 11 347.8 455.1 803 13540.4433 0.3432 12 365.4 458.8 824.2 13370.4528 0.3495 13 381.9 461.4 843.3 13200.4618 0.355 14 397.5 463.2 860.6 13050.4704 0.3599 15 412.3 464.2 876.5 12900.4785 0.3643 16 426.4 464.7 891.1 12760.4862 0.3682 17 439.8 464.7 904.6 12620.4937 0.3717 18 452.7 464.4 917.1 12490.5008 0.3748 19 465.1 463.6 928.8 12370.5077 0.3777 20 477.1 462.6 939.7 12250.5143 0.3802 21 488.6 461.4 950 12140.5207 0.3825 22 499.7 460 959.6 12020.5268 0.3846 23 510.4 458.4 968.8 11920.5328 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]
thPratio for
W net,max
W netth
9-140
9-163 EES The effects of pressure ratio, maximum cycle temperature, and compressor and turbine inefficiencies on the net work output per unit mass and the thermal efficiency of a simple Brayton cyclewith air as the working fluid is to be investigated. Constant specific heats at room temperature are to be used.Analysis Using EES, the problem is solved as follows:
Procedure ConstPropResult(T[1],P[1],r_comp,T[3]:Eta_th_ConstProp,Eta_th_easy)"For Air:"C_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(, %) "[%]"
"The Easy Way to calculate the constant property Otto cycle efficiency is:"Eta_th_easy = (1 - 1/r_comp^(k-1))*Convert(, %) "[%]"END
"Input Data"T[1]=300 [K] P[1]=100 [kPa]{T[3] = 1000 [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])
9-141
"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_23q_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"Call ConstPropResult(T[1],P[1],r_comp,T[3]:Eta_th_ConstProp,Eta_th_easy)PerCentError = ABS(Eta_th - Eta_th_ConstProp)/Eta_th*Convert(, %) "[%]"
PerCentError[%]
rcomp th[%]
th,ConstProp[%]
th,easy[%]
T3[K]
3.604 12 60.8 62.99 62.99 10006.681 12 59.04 62.99 62.99 15009.421 12 57.57 62.99 62.99 200011.64 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 Error = | th - th ,ConstProp | / th
Tm ax = 1000 K
9-142
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
9-143
9-164 EES 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])
9-144
s[4]=entropy('air',T=T[4],P=P[4])
9-145
Bwr Pratio Wc[kW]
Wnet[kW]
Wt[kW]
Qin[kW]
0.5229 0.1 2 1818 1659 3477 165870.6305 0.1644 4 4033 2364 6396 143730.7038 0.1814 6 5543 2333 7876 128620.7611 0.1806 8 6723 2110 8833 116820.8088 0.1702 10 7705 1822 9527 10700
0.85 0.1533 12 8553 1510 10063 98520.8864 0.131 14 9304 1192 10496 91020.9192 0.1041 16 9980 877.2 10857 84260.9491 0.07272 18 10596 567.9 11164 78090.9767 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]
W net
Tmax=1160 K
Note Pratio for maximum work and
c = 0.75t = 0.82
9-146
9-165 EES 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 fluid is to be investigated.Analysis Using EES, the problem is solved as follows:
Function hFunc(WorkFluid$,T,P) "The EES functions teat helium as a real gas; thus, T and P are needed for helium's enthalpy."IF WorkFluid$ = 'Air' then hFunc:=enthalpy(Air,T=T) ELSE
hFunc: = enthalpy(Helium,T=T,P=P) endifENDProcedure EtaCheck(Eta_th:EtaError$) If Eta_th < 0 then EtaError$ = 'Why are the net work done and efficiency < 0?' Else EtaError$ = '' END"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 = 0.8 Eta_t = 0.8 WorkFluid$ = 'Helium'}"Inlet conditions"h[1]=hFunc(WorkFluid$,T[1],P[1])s[1]=ENTROPY(WorkFluid$,T=T[1],P=P[1])"Compressor anaysis"s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor"P_ratio=P[2]/P[1]"Definition of pressure ratio - to find P[2]"T_s[2]=TEMPERATURE(WorkFluid$,s=s_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exit"h_s[2]=hFunc(WorkFluid$,T_s[2],P[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]=hFunc(WorkFluid$,T[3],P[3])m_dot*h[2] + Q_dot_in= m_dot*h[3]"SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0""Turbine analysis"s[3]=ENTROPY(WorkFluid$,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(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_s[4],P[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_th=W_dot_net/Q_dot_in"Cycle thermal efficiency"
9-147
Call EtaCheck(Eta_th:EtaError$) 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])
Bwr Pratio Wc
[kW]Wnet
[kW]Wt
[kW]Qin
[kW]0.5229 0.1 2 1818 1659 3477 165870.6305 0.1644 4 4033 2364 6396 143730.7038 0.1814 6 5543 2333 7876 128620.7611 0.1806 8 6723 2110 8833 116820.8088 0.1702 10 7705 1822 9527 10700
0.85 0.1533 12 8553 1510 10063 98520.8864 0.131 14 9304 1192 10496 91020.9192 0.1041 16 9980 877.2 10857 84260.9491 0.07272 18 10596 567.9 11164 78090.9767 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
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
Cycleefficiency,
Wnet
[kW]
Wnet
Tmax=1160 K
Note Pratio for maximum work and
c = 0.75t = 0.82
Brayton Cycle using Airm air = 20 kg/s
9-148
9-166 EES The effects of pressure ratio, maximum cycle temperature, regenerator effectiveness, and compressor and turbine efficiencies on the net work output per unit mass and the thermalefficiency of a regenerative Brayton cycle with air as the working fluid is to be investigated. Constant specific heats at room temperature are to be used. Analysis Using EES, the problem is solved as follows:
"Input data for air"C_P = 1.005 [kJ/kg-K] k = 1.4 "Other Input data from the diagram window"{T[3] = 1200 [K] Pratio = 10 T[1] = 300 [K] P[1]= 100 [kPa] Eta_reg = 1.0 Eta_c =0.8"Compressor isentorpic efficiency" Eta_t =0.9"Turbien isentropic efficiency"}
"Isentropic Compressor anaysis"T_s[2] = T[1]*Pratio^((k-1)/k)P[2] = Pratio*P[1] "T_s[2] 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_s[2]-T[1])
"Actual compressor analysis:" w_comp = C_P*(T[2]-T[1])
"External heat exchanger analysis""SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0 e_in - e_out =DELTAe_cv =0 for steady flow"q_in_noreg = C_P*(T[3] - T[2]) P[3]=P[2]"process 2-3 is SSSF constant pressure"
"Turbine analysis"P[4] = P[3] /Pratio T_s[4] = T[3]*(1/Pratio)^((k-1)/k)"T_s[4] is the isentropic value of T[4] at turbine exit"Eta_t = w_turb /w_turbisen "turbine adiabatic efficiency, w_turbisen > w_turb"
"SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0 e_in -e_out = DELTAe_cv = 0 for steady-flow"w_turbisen = C_P*(T[3] - T_s[4])
"Actual Turbine analysis:"w_turb = C_P*(T[3] - T[4])
"Cycle analysis"w_net=w_turb-w_compEta_th_noreg=w_net/q_in_noreg*Convert(, %) "[%]" "Cycle thermal efficiency"Bwr=w_comp/w_turb "Back work ratio"
9-149
"With the regenerator the heat added in the external heat exchanger is"q_in_withreg = C_P*(T[3] - T[5]) P[5]=P[2]
"The regenerator effectiveness gives h[5] and thus T[5] as:"Eta_reg = (T[5]-T[2])/(T[4]-T[2]) "Energy balance on regenerator gives h[6] and thus T[6] as:"T[2] + T[4]=T[5] + T[6] P[6]=P[4]
"Cycle thermal efficiency with regenerator"Eta_th_withreg=w_net/q_in_withreg*Convert(, %) "[%]"
c t th,noreg[%]
th,withreg[%]
qin,noreg[kJ/kg]
qin,withreg[kJ/kg]
wnet[kJ/kg]
0.6 0.9 14.76 13.92 510.9 541.6 75.40.65 0.9 20.35 20.54 546.8 541.6 111.30.7 0.9 24.59 26.22 577.5 541.6 1420.75 0.9 27.91 31.14 604.2 541.6 168.60.8 0.9 30.59 35.44 627.5 541.6 1920.85 0.9 32.79 39.24 648 541.6 212.50.9 0.9 34.64 42.61 666.3 541.6 230.8
9-150
9-167 EES The effects of pressure ratio, maximum cycle temperature, regenerator effectiveness, and compressor and turbine efficiencies on the net work output per unit mass and the thermal efficiency of a regenerative Brayton cycle with air as the working fluid is to be investigated. Variable specific heats are tobe used. Analysis Using EES, the problem is solved as follows:
"Input data""Input data from the diagram window"{T[3] = 1200 [K] Pratio = 10 T[1] = 300 [K] P[1]= 100 [kPa] Eta_reg = 1.0 Eta_c =0.8 "Compressor isentorpic efficiency" Eta_t =0.9 "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_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"h[1] + w_compisen = 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 = 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_in - e_out =DELTAe_cv =0 for steady flow"h[2] + q_in_noreg = h[3]
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_turb /w_turbisen "turbine adiabatic efficiency, w_turbisen > w_turb"
"SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0 e_in -e_out = DELTAe_cv = 0 for steady-flow"
9-151
h[3] = w_turbisen + h_s[4] h_s[4]=ENTHALPY(Air,T=T_s[4])"Actual Turbine analysis:"h[3] = w_turb + h[4] h[4]=ENTHALPY(Air,T=T[4])s[4]=ENTROPY(Air,T=T[4], P=P[4])"Cycle analysis"w_net=w_turb-w_compEta_th_noreg=w_net/q_in_noreg*Convert(, %) "[%]" "Cycle thermal efficiency"Bwr=w_comp/w_turb"Back work ratio"
"With the regenerator the heat added in the external heat exchanger is"h[5] + q_in_withreg = h[3]
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:"h[2] + h[4]=h[5] + 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_net/q_in_withreg*Convert(, %) "[%]"
4.5 5.0 5.5 6.0 6.5 7.0 7.5200
400
600
800
1000
1200
1400
1600
s [kJ/kg-K]
T [K
]
100 kPa
1000 kPa
Air
2s
1
2 5
3
44s
6
"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]
c t th,noreg[%]
th,withreg[%]
qin,noreg[kJ/kg]
qin,withreg[kJ/kg]
wnet[kJ/kg]
0.6 0.9 14.76 13.92 510.9 541.6 75.40.65 0.9 20.35 20.54 546.8 541.6 111.30.7 0.9 24.59 26.22 577.5 541.6 1420.75 0.9 27.91 31.14 604.2 541.6 168.60.8 0.9 30.59 35.44 627.5 541.6 1920.85 0.9 32.79 39.24 648 541.6 212.50.9 0.9 34.64 42.61 666.3 541.6 230.8
9-152
0.7 0.75 0.8 0.85 0.9 0.95 110
15
20
25
30
35
40
45
t
thc = 0.8
With regeneration
No regeneration
0.7 0.75 0.8 0.85 0.9 0.95 150
95
140
185
230
275
t
wne
t [k
J/kg
]
c = 0.8
0.7 0.75 0.8 0.85 0.9 0.95 1400
450
500
550
600
650
t
q in
c = 0.8
No regeneration
With regeneration
9-153
0.6 0.65 0.7 0.75 0.8 0.85 0.910
15
20
25
30
35
40
45
c
tht = 0.9
With regeneration
No regeneration
0.6 0.65 0.7 0.75 0.8 0.85 0.975
110
145
180
215
250
c
wne
t[k
J/kg
]
t = 0.9
0.6 0.65 0.7 0.75 0.8 0.85 0.9500
520
540
560
580
600
620
640
660
680
c
q in
t = 0.9
No regeneration
With regeneration
9-154
9-168 EES The effects of pressure ratio, maximum cycle temperature, regenerator effectiveness, and compressor and turbine efficiencies on the net work output per unit mass and the thermal efficiency of a regenerative Brayton cycle with 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
"Other Input data from the diagram window"{T[3] = 1200 [K] Pratio = 10 T[1] = 300 [K] P[1]= 100 [kPa] Eta_reg = 1.0 Eta_c =0.8 "Compressor isentorpic efficiency" Eta_t =0.9 "Turbien isentropic efficiency"}
"Isentropic Compressor anaysis"
T_s[2] = T[1]*Pratio^((k-1)/k)P[2] = Pratio*P[1] "T_s[2] 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_s[2]-T[1])
"Actual compressor analysis:" w_comp = C_P*(T[2]-T[1])
"External heat exchanger analysis""SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0 e_in - e_out =DELTAe_cv =0 for steady flow"q_in_noreg = C_P*(T[3] - T[2]) P[3]=P[2]"process 2-3 is SSSF constant pressure"
"Turbine analysis"P[4] = P[3] /Pratio
T_s[4] = T[3]*(1/Pratio)^((k-1)/k)"T_s[4] is the isentropic value of T[4] at turbine exit"Eta_t = w_turb /w_turbisen "turbine adiabatic efficiency, w_turbisen > w_turb"
"SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0 e_in -e_out = DELTAe_cv = 0 for steady-flow"w_turbisen = C_P*(T[3] - T_s[4])
"Actual Turbine analysis:"w_turb = C_P*(T[3] - T[4])
"Cycle analysis"w_net=w_turb-w_comp
9-155
Eta_th_noreg=w_net/q_in_noreg*Convert(, %) "[%]" "Cycle thermal efficiency"Bwr=w_comp/w_turb "Back work ratio"
"With the regenerator the heat added in the external heat exchanger is"q_in_withreg = C_P*(T[3] - T[5]) P[5]=P[2]
"The regenerator effectiveness gives h[5] and thus T[5] as:"Eta_reg = (T[5]-T[2])/(T[4]-T[2]) "Energy balance on regenerator gives h[6] and thus T[6] as:"T[2] + T[4]=T[5] + T[6] P[6]=P[4]
"Cycle thermal efficiency with regenerator"Eta_th_withreg=w_net/q_in_withreg*Convert(, %) "[%]"
c t th,noreg[%]
th,withreg[%]
qin,noreg[kJ/kg]
qin,withreg[kJ/kg]
wnet[kJ/kg]
0.6 0.9 14.76 13.92 510.9 541.6 75.40.65 0.9 20.35 20.54 546.8 541.6 111.30.7 0.9 24.59 26.22 577.5 541.6 1420.75 0.9 27.91 31.14 604.2 541.6 168.60.8 0.9 30.59 35.44 627.5 541.6 1920.85 0.9 32.79 39.24 648 541.6 212.50.9 0.9 34.64 42.61 666.3 541.6 230.8
9-156
9-169 EES The effect of the number of compression and expansion stages on the thermal efficiency of anideal 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"
9-157
w_net=w_turb_total-w_comp_total "[kJ/kg]"Bwr=w_comp/w_turb "Back work ratio"P_4=P_2P_5=P_4P_6=P_5T_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(, %) "[%]"
th,Ericksson[%]
th,Regenerative[%]
Nstages
75 49.15 175 64.35 275 68.32 375 70.14 475 72.33 775 73.79 1575 74.05 1975 74.18 22
0 2 4 6 8 10 12 14 16 18 20 22 2440
50
60
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Nstages
th [%
]
Ericsson
Ideal Regenerative Brayton
9-158
9-170 EES 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 workingfluid 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
9-159
"Cycle analysis"w_net=w_turb_total-w_comp_totalBwr=w_comp/w_turb "Back work ratio"
P_4=P_2P_5=P_4P_6=P_5T_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(, %) "[%]"
th,Ericksson[%]
th,Regenerative[%]
Nstages
75 32.43 175 58.9 275 65.18 375 67.95 475 71.18 775 73.29 1575 73.66 1975 73.84 22
0 2 4 6 8 10 12 14 16 18 20 22 2430
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Nstages
th [%
]
EricssonIdeal Regenerative Brayton
9-160
Fundamentals of Engineering (FE) Exam Problems
9-171 An Otto cycle with air as the working fluid has a compression ratio of 8.2. Under cold air standardconditions, the thermal efficiency of this cycle is(a) 24% (b) 43% (c) 52% (d) 57% (e) 75%
Answer (d) 57%
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 numericalvalues).
r=8.2k=1.4Eta_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"
9-172 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
9-173 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 numericalvalues).
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"
9-161
9-174 Air in an ideal Diesel cycle is compressed from 3 L to 0.15 L, and then it expands during theconstant pressure heat addition process to 0.30 L. Under cold air standard conditions, the thermalefficiency of this cycle is(a) 35% (b) 44% (c) 65% (d) 70% (e) 82%
Answer (c) 65%
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 numericalvalues).
V1=3 "L"V2= 0.15 "L"V3= 0.30 "L"r=V1/V2rc=V3/V2k=1.4Eta_Diesel=1-(1/r^(k-1))*(rc^k-1)/k/(rc-1)
"Some Wrong Solutions with Common Mistakes:"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"
9-175 Helium gas in an ideal Otto cycle is compressed from 20 C and 2.5 L to 0.25 L, and its temperatureincreases by an additional 700 C during the heat addition process. The temperature of helium before theexpansion 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 numericalvalues).
k=1.667V1=2.5V2=0.25r=V1/V2T1=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"
9-162
9-176 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 numericalvalues).
rho1=1.20 "kg/m^3"k=1.4V1=2.2V2=0.26m=rho1*V1/1000 "kg"w_net=440 "kJ/kg"Wtotal=m*w_netMEP=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"
9-177 In an ideal Brayton cycle, air is compressed from 95 kPa and 25 C to 800 kPa. Under cold airstandard conditions, the thermal efficiency of this cycle is(a) 46% (b) 54% (c) 57% (d) 39% (e) 61%
Answer (a) 46%
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 numericalvalues).
P1=95 "kPa"P2=800 "kPa"T1=25+273 "K"rp=P2/P1k=1.4Eta_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"
9-163
9-178 Consider an ideal Brayton cycle executed between the pressure limits of 1200 kPa and 100 kPa andtemperature 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 numericalvalues).
P1=100 "kPa"P2=1200 "kPa"T1=20+273 "K"T3=1000+273 "K"rp=P2/P1k=1.667Cp=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"
9-179 An ideal Brayton cycle has a net work output of 150 kJ/kg and a backwork ratio of 0.4. If both theturbine 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 numericalvalues).
wcomp/wturb=0.4wturb-wcomp=150 "kJ/kg"Eff=0.85w_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"
9-164
9-180 In an ideal Brayton cycle, air is compressed from 100 kPa and 25 C to 1 MPa, and then heated to1200 C before entering the turbine. Under cold air standard conditions, the air temperature at the turbine exit is(a) 490 C (b) 515 C (c) 622 C (d) 763 C (e) 895 C
Answer (a) 490 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 numericalvalues).
P1=100 "kPa"P2=1000 "kPa"T1=25+273 "K"T3=1200+273 "K"rp=P2/P1k=1.4T4=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"
9-181 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 numericalvalues).
k=1.667Cp=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/P1T2=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"
9-165
9-182 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 numericalvalues).
k=1.4Cp=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/P1T2check=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"
9-183 Consider a gas turbine that has a pressure ratio of 6 and operates on the Brayton cycle withregeneration between the temperature limits of 20 C and 900 C. If the specific heat ratio of the workingfluid 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 numericalvalues).
k=1.3rp=6T1=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)"
9-166
9-184 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 numericalvalues).
k=1.4rp=10T1=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"
9-185 Air enters a turbojet engine at 260 m/s at a rate of 30 kg/s, and exits at 800 m/s relative to theaircraft. The thrust developed by the engine is(a) 8 kN (b) 16 kN (c) 24 kN (d) 20 kN (e) 32 kN
Answer (b) 16 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 numericalvalues).
Vel1=260 "m/s"Vel2=800 "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"
9-186 ··· 9-191 Design and Essay Problems.