chapter 9 gas power cycles - به نام یگانه ... · pdf file9-2 9-14 the four processes...
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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) all the processes are internally reversible, (3) the combustion process is replaced by the heat addition process, and (4) the exhaust process is replaced by the heat rejection process which returns the working fluid to its original state. 9-7C The clearance volume is the minimum volume formed in the cylinder whereas the displacement volume is the volume displaced by the piston as the piston moves between the top dead center and the bottom dead center. 9-8C 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 remain the same (otherwise it will increase). The mean effective pressure, on the other hand, will decrease as a car gets older as a result of wear and tear. 9-12C 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.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-2
9-14 The four processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-s diagrams, and the net work output and the thermal efficiency are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The properties of air are given in Table A-17. Analysis (b) The properties of air at various states are
P
qout
qin 3
4
2
1
qout
qin 3
4
2
1
( )
( )
( ) kJ/kg 828.149.101310kPa 2668
kPa 100
kPa 2668kPa 800K 539.8K 1800
1310kJ/kg 1487.2
K 1800
K 539.8kJ/kg 389.22
088.111.386kPa 100kPa 800
386.1kJ/kg 300.19
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
T v
T
From energy balances,
kJ/kg570.1 9.5270.1098
kJ/kg 527.919.3001.828
kJ/kg 1098.02.3892.1487
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
wη
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-3
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 be plotted. 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"
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-4
T3 [K] ηth qin,total [kJ/kg] Wnet [kJ/kg] 1500 50.91 815.4 415.1 1700 51.58 1002 516.8 1900 52.17 1192 621.7 2100 52.69 1384 729.2 2300 53.16 1579 839.1 2500 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
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-5
1500 1700 1900 2100 2300 250050.5
51
51.5
52
52.5
53
53.5
54
T[3] [K]
ηth
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]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-6
9-16 The four processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-s diagrams, and the 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
( )
( ) K 579.2kPa 100kPa 1000K 300
0.4/1.4/1
1
212 =
=
=
− kk
PPTT
( )
( )( ) K 3360==→−⋅=
−=−=
3max3
2323in
579.2KkJ/kg 1.005kJ/kg 2800 TTT
TTchhq p
v (c) ( ) K 336K 3360
kPa 1000kPa 100
33
44
4
44
3
33 ===→= TPPT
TP
TP vv
( ) ( )( ) ( )
( )( ) ( )( )
21.0%=−=−=
=−⋅+−⋅=
−+−=
−+−=+=
kJ/kg 2800kJ/kg 2212
11
kJ/kg 2212K300336KkJ/kg 1.005K3363360KkJ/kg 0.718
in
outth
1443
1443out41,out34,out
TTcTTchhuuqqq
p
η
v
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 the temperature 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-s diagrams, and the total heat input and the thermal efficiency are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The properties of air are given in Table A-17E. Analysis (b) The properties of air at various states are
P T Btu/lbm 129.06 Btu/lbm, 92.04R 540 111 ==→= hu
q12
qout1
2 q23
4
3
s
qout
q23
q12
1
2
4
3
v
( )
( ) Btu/lbm 593.22317.01242psia 57.6psia 14.7
1242Btu/lbm 48.849
R 3200
psia 57.6psia 14.7R 540R 2116
Btu/lbm 1.537,R2116Btu/lbm 04.39230004.92
43
4
33
11
22
1
11
2
22
22
in,121212in,12
34
3
=→===
==
→=
===→=
==
=+=+=→−=
hPPP
P
Ph
T
PTT
PT
PT
P
hTquu
uuq
rr
r
vv
T
From energy balance,
Btu/lbm 464.1606.12922.593 38.312300
Btu/lbm 312.381.53748.849
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
qη
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-7
9-18E The four processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-s diagrams, and the total heat input and the thermal efficiency are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 0.240 Btu/lbm.R, cv = 0.171 Btu/lbm.R, and k = 1.4 (Table A-2E).
PAnalysis (b) ( )
( )( )
( )
( ) ( )( ) Btu/lbm 217.4R22943200RBtu/lbm 0.24
psia 62.46psia 14.7R 540R 2294
R 2294R540Btu/lbm.R 0.171Btu/lbm 300
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/lbm 378.55402117Btu/lbm.R 0.240
4.217300
R 2117psia 62.46
psia 14.7R 3200
1414out
in,23in,12in
0.4/1.4/1
3
434
=−=−=−=
=+=+=
=
=
=
−
TTchhq
qqq
PP
TT
p
kk
Btu/lbm517.4
T
q12
qout1
2 q23
4
3
(c) 26.8%=−=−=Btu/lbm 517.4Btu/lbm 378.5
11in
outth q
qη s
9-19 The three processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-s diagrams, and the heat rejected and the thermal efficiency are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-2).
Analysis (b) ( )
( ) K 579.2kPa 100kPa 1000
K 3000.4/1.4/1
1
212 =
=
=
− kk
PP
TT P
qout
qin
1
2 3
v
( ) ( )( )( )( ) K 12662.579KkJ/kg 1.005kg 0.004kJ 2.76 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/kg 7.273KkJ/kg 0.287kPa 1000K 1266
kPa 100K 300
2kPa 1001000
22area
3
3
1
11331
1331
=⋅
−
+=
−
+=−
+==
PRT
PRTPPPP
w vv
T
Energy balance for process 3-1 gives
( )[ ]( ) ( )( )[ ] kJ 1.679=⋅+−=
−+−=−−−=
−=−−→∆=−
K1266-300KkJ/kg 0.718273.7kg 0.004)(
)(
31out31,31out31,out31,
31out31,out31,systemoutin
TTcwmTTmcmwQ
uumWQEEE
vv
qin
qout 1
2 3
s
(c) 39.2%=−=−=kJ 2.76kJ 1.679
11in
outth Q
Qη
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-8
9-20 The three processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-s diagrams, and the 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
( )
( )
( ) ( )( )
( ) ( )( )
kJ0.422 651.1073.2
kJ 1.651kJ/kg290.16840.38kg 0.003
kJ 2.073kJ/kg206.91897.91kg 0.003
kJ/kg 840.3851.8207.2kPa 380
kPa 95
2.207kJ/kg, 897.91
K 1160K 290kPa 95kPa 380
kJ/kg 290.16kJ/kg 206.91
K 029
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
3 1
s(c) 20.4%===
kJ 2.073kJ 0.422
in
outnet,th Q
Wη
9-21 The three processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-s diagrams, and the 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,
( )
( )( )
( ) ( )
( )( )( )
( ) ( )
( )( )( )
kJ0.39 48.187.1
kJ 1.48K290780.6KkJ/kg 1.005kg 0.003
kJ 1.87K2901160KkJ/kg 0.718kg 0.003
K 780.6kPa 380
kPa 95K 1160
K 1160K 290kPa 95kPa 380
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
3 1
s
qout
qin
1
2
3
v
T
(c) 20.9%===kJ1.87kJ0.39
in
netth Q
Wη
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-9
9-22 A Carnot cycle with the specified temperature limits is considered. The net work output per cycle is to be determined. Assumptions Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K, cv = 0.718 kJ/kg·K, R = 0.287 kJ/kg.K, and k = 1.4 (Table A-2). Analysis The minimum pressure in the cycle is P3 and the maximum pressure is P1. Then,
or,
( )
( )( ) kPa 3.935
K 300K 900
kPa 201.4/0.41/
3
232
/1
3
2
3
2
=
=
=
=
−
−
kk
kk
TT
PP
PP
TT
T
The heat input is determined from
Then,
( )
( ) ( )( )( )
( )( ) kJ 0.393===
=−=−=
=⋅=−=
⋅=⋅−=−=−
kJ 0.58890.667
.7%66K 900K 300
11
kJ 0.5889KkJ/kg 0.2181K 900kg 0.003
KkJ/kg 0.2181kPa 2000kPa 935.3
lnKkJ/kg 0.287lnln
inthoutnet,
th
12in
1
20
1
212
QW
TT
ssmTQ
PP
RTT
css
H
L
H
p
η
η
1
4 qout
qin 2
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
Qin 2
3
1200
( ) max41 kPa 300379.2
238
4
1 PPP
PP
r
r==== kPa30,013
350(b) The heat input is determined from
s
( ) ( ) kJ0.706 0.7083/kJ 0.5/
%83.70K 1200K 350
11
thoutnet,in
th
===
=−=−=
η
η
WQ
TT
H
L
(c) The mass of air is
( ) ( )
( )( ) ( )( )
kg0.00296 kJ/kg 169.15kJ 0.5
kJ/kg169.15K3501200KkJ/kg0.199
KkJ/kg 0.199
kPa 150kPa 300
lnKkJ/kg 0.287ln
outnet,
outnet,
12outnet,
21
3
403434
===
=−⋅=−−=
−=⋅−=
⋅−=−−=−
wW
m
TTssw
ss
PP
Rssss
LH
oo
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-10
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,
( )
( )( ) kPa 6524=
=
=
=
−
−
71.667/0.661/
4
141
/1
4
1
4
1
K 350K 1200
kPa 300kk
kk
TT
PP
PP
TT
Wnet = 0.5 kJ
350
1200 1
4
Qin 2
3
T
(b) The heat input is determined from s
( ) ( ) kJ0.706 0.7083/kJ 0.5/
70.83%K 1200K 350
11
thoutnet,in
th
===
=−=−=
η
η
WQ
TT
H
L
(c) The mass of helium is determined from
( )
( )( ) ( )( )
kg 0.000409===
=−⋅=−−=
−=⋅−=
⋅−=−=−
kJ/kg 1223.7kJ 0.5
kJ/kg 1223.7K3501200KkJ/kg 1.4396
KkJ/kg 1.4396
kPa 150kPa 300
lnKkJ/kg 2.0769lnln
outnet,
outnet,
12outnet,
21
3
40
3
434
wW
m
TTssw
ss
PP
RTT
css
LH
p
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-11
9-25 A Carnot cycle executed in a closed system with air as the working fluid is considered. The minimum pressure in the cycle, the heat rejection from the cycle, the thermal efficiency of the cycle, and the second-law efficiency of an actual cycle operating between the same temperature limits are to be determined. Assumptions Air is an ideal gas with constant specific heats. Properties The properties of air at room temperatures are R = 0.287 kJ/kg.K and k = 1.4 (Table A-2). Analysis (a) The minimum temperature is determined from
( )( ) ( )( ) K 350K750KkJ/kg 0.25kJ/kg 10012net =→−⋅=→−−= LLLH TTTTssw
The pressure at state 4 is determined from
or
( )
( )
kPa 1.110K 350K 750kPa 800 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
qin 2
3
T
s
The minimum pressure in the cycle is determined from
( ) kPa 46.1=→⋅−=⋅−
−=∆−=∆
33
3
40
3
43412
kPa 110.1lnKkJ/kg 0.287KkJ/kg 25.0
lnln
PP
PP
RTT
css p
(b) The heat rejection from the cycle is kgkJ/ 87.5==∆= kJ/kg.K) K)(0.25 350(12out sTq L
(c) The thermal efficiency is determined from
0.533=−=−=K 750K 350
11thH
L
TT
η
(d) The power output for the Carnot cycle is
kW 9000kJ/kg) kg/s)(100 90(netCarnot === wmW &&
Then, the second-law efficiency of the actual cycle becomes
0.578===kW 9000kW 5200
Carnot
actualII W
W&
&η
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-12
Otto Cycle 9-26C The four processes that make up the Otto cycle are (1) isentropic compression, (2) v = constant heat addition, (3) isentropic expansion, and (4) v = constant heat rejection. 9-27C 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 gasoline engines.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-13
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
v
P
750 kJ/kg 2
3
1
4
v
( )
( ) ( ) kPa 1705kPa 95K 300K 673.1
8
kJ/kg 491.2K 673.1
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.
588.6
kJ/kg 241.217502.4913
3in23,2323in23, =
=→=+=+=→−=
r
Tquuuuq
v
K1539
( ) kPa3898 kPa 1705K 673.1K 1539
22
33
2
22
3
33 =
==→= P
TT
PT
PT
P vv
(b) Process 3-4: isentropic expansion.
( )( )kJ/kg 571.69
K 774.570.52588.68
4
4
2
1334 =
=→====
uT
r rrr vvv
vv
Process 4-1: v = constant heat rejection. q u uout 571.69 214.07 357.62 kJ / kg= − = − =4 1
kJ/kg 392.4=−=−= 62.357750outinoutnet, qqw
(c) 52.3%===kJ/kg 750kJ/kg 392.4
in
outnet,th q
wη
(d) ( )( )
( )( )kPa495.0
kJmkPa
1/81/kgm 0.906kJ/kg 392.4
)/11(MEP
/kgm0.906kPa95
K300K/kgmkPa0.287
3
31
outnet,
21
outnet,
max2min
max3
3
1
11
=
⋅
−=
−=
−=
==
==⋅⋅
==
rww
r
PRT
vvv
vvv
vv
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-14
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_23 q_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]"
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-15
rcomp ηth MEP [kPa] wnet [kJ/kg]
5 43.78 452.9 328.4 6 47.29 469.6 354.7 7 50.08 483.5 375.6 8 52.36 495.2 392.7 9 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
]
95
kPa
3900 kPa
0.9
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
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-16
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
ηth
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-17
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.
( )( )
( ) ( ) kPa 1745kPa 95K 300K 689
8
K 6898K300
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.
( )
( )(K1734
K689KkJ/kg 0.718kJ/kg 750
3
3
2323in23,
=−⋅=
−=−=
TTTTcuuq v
)
( ) kPa4392 kPa 1745K 689K 1734
22
33
2
22
3
33 =
==→= P
TT
PT
PT
P vv
(b) Process 3-4: isentropic expansion.
( ) K 75581K 1734
0.41
4
334 =
=
=
−k
TTv
v
Process 4-1: v = constant heat rejection. ( ) ( )( ) kJ/kg 327K300755KkJ/kg 0.7181414out =−⋅=−=−= TTcuuq v
kJ/kg423 327750outinoutnet, =−=−= qqw
(c) 56.4%===kJ/kg750kJ/kg423
in
outnet,th q
wη
(d) ( )( )
( )( )kPa534
kJmkPa
1/81/kgm 0.906kJ/kg 423
)/11(MEP
/kgm 0.906kPa 95
K 300K/kgmkPa 0.287
3
31
outnet,
21
outnet,
max2min
max3
3
1
11
=
⋅
−=
−=
−=
==
==⋅⋅
==
rww
r
PRT
vvv
vvv
vv
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-18
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 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
( )( )
( ) ( ) kPa 2338kPa 100K 308K 757.9
9.5
K 757.99.5K 308
11
2
2
12
1
11
2
22
0.41
2
112
=
==→=
==
=
−
PTT
PT
PT
P
TTk
v
vvv
v
v
Qout Qin
2
3
1
4
vProcess 3-4: isentropic expansion.
( )( ) K 1969==
=
−0.4
1
3
443 9.5K 800
k
TTv
v
Process 2-3: v = constant heat addition.
( ) kPa 6072=
==→= kPa 2338
K 757.9K 1969
22
33
2
22
3
33 PTT
PT
PT
P vv
(b) ( )( )
( )( )kg10788.6
K 308K/kgmkPa 0.287m 0.0006kPa 100 4
3
3
1
11 −×=⋅⋅
==RTP
mV
( ) ( ) ( )( )( ) kJ 0.590=−⋅×=−=−= − K757.91969KkJ/kg 0.718kg106.788 42323in TTmcuumQ v
(c) Process 4-1: v = constant heat rejection.
( ) ( )( )( ) kJ0.2 40K308800KkJ/kg 0.718kg106.788)( 41414out =−⋅×−=−=−= −TTmcuumQ v
kJ 0.350240.0590.0outinnet =−=−= QQW
59.4%===kJ 0.590kJ 0.350
in
outnet,th Q
Wη
(d)
( )( )kPa 652=
⋅
−=
−=
−=
==
kJmkPa
1/9.51m 0.0006kJ 0.350
)/11(MEP
3
31
outnet,
21
outnet,
max2min
rWW
r
VVV
VVV
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-19
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 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
( )( )
( ) ( ) kPa 2338kPa 100K 308K 757.9
9.5
K 757.99.5K 308
11
2
2
12
1
11
2
22
0.41
2
112
=
==→=
==
=
−
PTT
PT
PT
P
TTk
v
vvv
v
v
800 K
Polytropic
Qout
Qin
2
3
1308 K
4
vProcess 3-4: polytropic expansion.
( )( )( )( )
kg10788.6K 308K/kgmkPa 0.287
m 0.0006kPa 100 43
3
1
11 −×=⋅⋅
==RTP
mV
( )( )
( ) ( )( )( )kJ 0.5338
1.351K1759800KkJ/kg 0.287106.788
1
9.5K 800
434
34
35.01
3
443
=−
−⋅×=
−−
=
==
=
−
−
nTTmR
W
TTn
K 1759v
v
Then energy balance for process 3-4 gives
( )( ) ( )
( )( )( ) kJ 0.0664kJ 0.5338K1759800KkJ/kg 0.718kg106.788 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 probably is due to assuming constant specific heats at room temperature). (b) Process 2-3: v = constant heat addition.
( ) kPa 5426=
==→= kPa 2338
K 757.9K 1759
22
33
2
22
3
33 PTT
PT
PT
P vv
( ) ( )( )( )( ) kJ 0.4879K757.91759KkJ/kg 0.718kg106.788 4
in23,
2323in23,
=−⋅×=
−=−=−Q
TTmcuumQ v
Therefore, kJ 0.5543=+=+= 0664.04879.0in34,in23,in QQQ (c) Process 4-1: v = constant heat rejection. ( ) ( ) ( )( )( ) kJ 0.2398K308800KkJ/kg 0.718kg 106.788 4
1414out =−⋅×=−=−= −TTmcuumQ v kJ 0.31452398.05543.0outinoutnet, =−=−= QQW
56.7%===kJ 0.5543kJ 0.3145
in
outnet,th Q
Wη
(d)
( )( )kPa 586=
⋅
−=
−=
−=
==
kJmkPa
1/9.51m 0.0006kJ 0.3145
)/11(MEP
3
31
outnet,
21
outnet,
max2min
rWW
rV
VVV
VV
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-20
9-39E An ideal Otto cycle with air as the working fluid has a compression ratio of 8. The amount of heat transferred to the air during the heat addition process, the thermal efficiency, and the thermal efficiency of a Carnot cycle operating between the same temperature limits are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. 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/lbm 11.28204.1832.144811
21
2222
=→==== ur rrr vv
v
vv
Process 2-3: v = constant heat addition.
Btu/lbm 241.42=−=−=
==
→=
28.21170.452
419.2Btu/lbm 452.70
R2400
23
33
3
uuq
uT
in
rv
(b) Process 3-4: isentropic expansion.
( )( ) Btu/lbm 205.5435.19419.28 43
4334
=→==== ur rrr vvv
vv
Process 4-1: v = constant heat rejection. Btu/lbm 50.11304.9254.20514out =−=−= uuq
53.0%=−=−=Btu/lbm 241.42Btu/lbm 113.50
11in
outth q
qη
(c) 77.5%=−=−=R 2400
R 54011Cth,
H
L
TT
η
9-40E An ideal Otto cycle with argon as the working fluid has a compression ratio of 8. The amount of heat transferred to the argon during the heat addition process, the thermal efficiency, and the thermal efficiency of a Carnot cycle operating between the same temperature limits are to be determined. Assumptions 1 The air-standard assumptions are applicable with argon as the working fluid. 2 Kinetic and potential energy changes are negligible. 3 Argon is an ideal gas with constant specific heats. Properties The properties of argon are cp = 0.1253 Btu/lbm.R, cv = 0.0756 Btu/lbm.R, and k = 1.667 (Table A-2E). Analysis (a) Process 1-2: isentropic compression.
P
( )( ) R 21618R 540 0.6671
2
112 ==
=
−k
TTv
v
Process 2-3: v = constant heat addition. ( ) ( )( ) Btu/lbm 18.07=−=−=−= R 21612400Btu/lbm.R 0.07562323in TTcuuq v
qout
qin 2
3
1
4
v(b) Process 3-4: isentropic expansion.
( ) R 60081R 2400
0.6671
4
334 =
=
=
−k
TTv
v
Process 4-1: v = constant heat rejection. ( ) ( )( ) Btu/lbm 4.536R540600Btu/lbm.R 0.07561414out =−=−=−= TTcuuq v
74.9%=−=−=Btu/lbm 18.07Btu/lbm 4.536
11in
outth q
qη
(c) 77.5%=−=−=R 2400R 540
11Cth,H
L
TT
η
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-21
9-41 A gasoline engine operates on an Otto cycle. The compression and expansion processes are modeled as polytropic. The temperature at the end of expansion process, the net work output, the thermal efficiency, the mean effective pressure, the engine speed for a given net power, and the specific fuel consumption are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at 850 K are cp = 1.110 kJ/kg·K, cv = 0.823 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.349 (Table A-2b). Analysis (a) Process 1-2: polytropic compression
Qout
4
3
2
Qin
1
( )( )
( )( ) kPa 199510kPa 100
K 4.66410K 333
1.3
2
112
1-1.31
2
112
==
=
==
=
−
n
n
PP
TT
v
v
v
v
Process 2-3: constant volume heat addition
( ) K 2664kPa 1995kPa 8000K 664.4
2
323 =
=
=
PP
TT
( )
( )( ) kJ/kg 1646K664.42664KkJ/kg 0.823 2323in
=−⋅=−=−= TTcuuq v
Process 3-4: polytropic expansion.
( ) K 1335=
=
=
− 1-1.31
4
334 10
1K 2664n
TTv
v
( ) kPa 9.400101kPa 8000
1.3
1
234 =
=
=
n
PPv
v
Process 4-1: constant voume heat rejection. ( ) ( )( ) kJ/kg 8.824K3331335KkJ/kg 0.8231414out =−⋅=−=−= TTcuuq v
(b) The net work output and the thermal efficiency are kJ/kg 820.9=−=−= 8.8241646outinoutnet, qqw
0.499===kJ/kg 1646kJ/kg 820.9
in
outnet,th q
wη
(c) The mean effective pressure is determined as follows
( )( )
( )( )kPa 954.3=
⋅
−=
−=
−=
==
==⋅⋅
==
kJmkPa
1/101/kgm 0.9557kJ/kg 820.9
)/11(MEP
/kgm 0.9557kPa 100
K 333K/kgmkPa 0.287
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 (state 1) are
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-22
33
m 0002444.0m 0022.0
10 =→+
=→+
= cc
c
c
dc VV
VV
VVr
31 m 002444.00022.00002444.0 =+=+= dc VVV
The total mass contained in the cylinder is
( )( )kg 0.002558
K 333K/kgmkPa 0.287)m 444kPa)/0.002 100(
3
3
1
11 =⋅⋅
==RT
VPmt
The engine speed for a net power output of 70 kW is
rev/min 4001=
⋅==
min 1s 60
cycle)kJ/kg kg)(820.9 002558.0(kJ/s 70rev/cycle) 2(2
net
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
kg 0001505.0kg) 002558.0(
16AF =→−
=→−
== ff
f
f
ft
f
a mm
mm
mmmm
Finally, the specific fuel consumption is
g/kWh 258.0=
==
kWh 1kJ 3600
kg 1g 1000
kJ/kg) kg)(820.9 002558.0(kg 0001505.0
sfcnetwm
m
t
f
Diesel Cycle 9-42C A diesel engine differs from the gasoline engine in the way combustion is initiated. In diesel engines combustion is initiated by compressing the air above the self-ignition temperature of the fuel whereas it is initiated by a spark plug in a gasoline engine. 9-43C 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 the cutoff ratio decreases, the efficiency of the diesel cycle increases.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-23
9-47 An air-standard Diesel cycle with a compression ratio of 16 and a cutoff ratio of 2 is considered. The temperature after the heat addition process, the thermal efficiency, and the mean effective pressure are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The gas constant of air is R = 0.287 kJ/kg.K. The properties of air are given in Table A-17.
P
qout
qin 3 2
1
4
v
Analysis (a) Process 1-2: isentropic compression.
2.621kJ/kg214.07
K3001
11 =
=→=
r
uT
v
( )kJ/kg 90.98K 862.4
825.382.6211611
2
2
1
2112 =
=→====
hT
r rrr vvv
vv
Process 2-3: P = constant heat addition.
( )( ) 546.4kJ/kg 1910.6
K 862.4223
322
2
33
2
22
3
33==
→====→=r
hTTT
TP
TP
vv
vvv K 1724.8
(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/kg 1019.7kJ/kg 445.63
11
kJ/kg 445.6307.2147.659
in
outth
14out
uuq
η
(c)
( )( )
( ) ( )( )kPa675.9
kJmkPa
1/161/kgm 0.906kJ/kg 574.07
/11MEP
/kgm 0.906kPa 95
K 300K/kgmkPa 0.287
kJ/kg 574.0763.4457.1019
3
31
outnet,
21
outnet,
max2min
max3
3
1
11
outinoutnet,
=
⋅
−=
−=
−=
==
==⋅⋅
==
=−=−=
rww
r
PRT
qqw
vvv
vvv
vv
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-24
9-48 An air-standard Diesel cycle with a compression ratio of 16 and a cutoff ratio of 2 is considered. The temperature after the heat addition process, the thermal efficiency, and the mean effective pressure are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with 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 3 2
1
4
v
( )( ) K909.416K300 0.41
2
112 ==
=
−k
TTv
v
Process 2-3: P = constant heat addition.
( )( ) K1818.8====→= K909.422 222
33
2
22
3
33 TTTT
PT
Pv
vvv
(b) ( ) ( )( ) kJ/kg 913.9K909.41818.8KkJ/kg 1.0052323in =−⋅=−=−= 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/kg 913.9kJ/kg 353
11
kJ/kg353K300791.7KkJ/kg0.718
in
outth
1414out
TTcuuq
η
v
(c)
( )( )
( ) ( )( )kPa660.4
kJmkPa
1/161/kgm 0.906kJ/kg 560.9
/11MEP
/kgm 0.906kPa 95
K 300K/kgmkPa 0.287
kJ/kg560.93539.913
3
31
outnet,
21
outnet,
max2min
max3
3
1
11
outinnet.out
=
⋅
−=
−=
−=
==
==⋅⋅
==
=−=−=
rww
r
PRT
qqw
vvv
vvv
vv
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-25
9-49E An air-standard Diesel cycle with a compression ratio of 18.2 is considered. The cutoff ratio, the heat rejection per unit mass, and the thermal efficiency are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The properties of air are given in Table A-17E.
PAnalysis (a) Process 1-2: isentropic compression.
3000 R
qout
qin 32
1
4
v
T 32.144Btu/lbm 40.92
R 5401
11 =
=→=
r
uv
( )Btu/lbm 402.05R 1623.6
93.732.1442.18
11
2
2
1
2112 =
=→====
hT
r rrr vvv
vv
Process 2-3: P = constant heat addition.
1.848===→=R 1623.6
R 3000
2
3
2
3
2
22
3
33
TT
TP
TP
v
vvv
(b)
Btu/lbm 388.6305.40268.790
180.1Btu/lbm 790.68
R 3000
23in
33
3
=−=−=
==
→=
hhq
hT
rv
Process 3-4: isentropic expansion.
( ) Btu/lbm 91.250621.11180.1848.1
2.18848.1848.1 4
2
4
3
43334
=→===== urrrrr vv
v
vv
v
vv
Process 4-1: v = constant heat rejection.
(c) 59.1%
Btu/lbm 158.87
=−=−=
=−=−=
Btu/lbm 388.63Btu/lbm 158.87
11
04.9291.250
in
outth
14out
uuq
η
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-26
9-50E An air-standard Diesel cycle with a compression ratio of 18.2 is considered. The cutoff ratio, the heat rejection per unit mass, and the thermal efficiency are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 0.240 Btu/lbm.R, cv = 0.171 Btu/lbm.R, and k = 1.4 (Table A-2E). Analysis (a) Process 1-2: isentropic compression. P
qin 32
1
4
v
( )( ) R172418.2R540 0.41
2
112 ==
=
−k
TTv
v
Process 2-3: P = constant heat addition.
1.741===→=R 1724R 3000
2
3
2
3
2
22
3
33
TT
TP
TP
v
vvv
(b) ( ) ( )( ) Btu/lbm 306R17243000Btu/lbm.R 0.2402323in =−=−=−= TTchhq p
Process 3-4: isentropic expansion.
( ) R 117318.21.741R 3000
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/lbm 108
=−=−=
=−=−=−=
Btu/lbm 306Btu/lbm 108
11
R0541173Btu/lbm.R 0.171
in
outth
1414out
TTcuuq
η
v
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-27
9-51 An ideal diesel engine with air as the working fluid has a compression ratio of 20. The thermal efficiency and the mean effective pressure are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4 (Table A-2). Analysis (a) Process 1-2: isentropic compression.
P
qout
qin 32
1
4
v
( )( ) K 971.120K 293 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/kg 1235kJ/kg 784.4
kJ/kg 784.46.4501235
kJ/kg 450.6K293920.6KkJ/kg 0.718
kJ/kg 1235K971.12200KkJ/kg 1.005
K 920.620
2.265K 2200265.2265.2
in
outnet,th
outinoutnet,
1414out
2323in
0.41
3
1
4
23
1
4
334
qw
qqw
TTcuuq
TTchhq
rTTTT
p
kkk
η
v
V
V
V
V
(b) ( )( )
( ) ( )( )kPa933
kJmkPa
1/201/kgm 0.885kJ/kg 784.4
/11MEP
/kgm 0.885kPa 95
K 293K/kgmkPa 0.287
3
31
outnet,
21
outnet,
max2min
max3
3
1
11
=
⋅
−=
−=
−=
==
==⋅⋅
==
rww
r
PRT
vvv
vvv
vv
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-28
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
( )( ) K 971.120K 293 0.41
2
112 ==
=
−k
TTV
V
Process 2-3: P = constant heat addition.
2.265K 971.1K 2200
2
3
2
3
2
22
3
33 ===→=TT
TP
TP
V
VVV
Process 3-4: polytropic expansion.
( )
( ) ( )( )( ) ( )( ) kJ/kg 526.3K 2931026KkJ/kg 0.718
kJ/kg 1235K 971.12200KkJ/kg 1.005
K 102620
2.265K 2200r
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/kg 120.1K 22001026KkJ/kg 0.718kJ/kg 963
kJ/kg 9631.351
K 22001026KkJ/kg 0.2871
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 loses heat during polytropic expansion. The cause of this unrealistic result is the constant specific heat assumption. If we were to use u data from the air table, we would obtain ( ) kJ/kg 1.128)4.18723.781(96334out34,in34, −=−+=−+= uuwq
which is a heat loss as expected. Then qout becomes kJ/kg 654.43.5261.128out41,out34,out =+=+= qqq
and
47.0%===
=−=−=
kJ/kg 1235kJ/kg 580.6
kJ/kg 580.64.6541235
in
outnet,th
outinoutnet,
qw
qqw
η
(b) ( )( )
( ) ( )( )kPa 691=
⋅
−=
−=
−=
==
==⋅⋅
==
kJmkPa 1
1/201/kgm 0.885kJ/kg 580.6
/11MEP
/kgm 0.885kPa 95
K 293K/kgmkPa 0.287
3
31
outnet,
21
outnet,
max2min
max3
3
1
11
rww
r
PRT
vvv
vvv
vv
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-29
9-53 EES Problem 9-52 is reconsidered. The effect of the compression ratio on the net work output, mean effective pressure, 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: 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])^n 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 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])
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-30
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 ηth MEP [kPa] wnet [kJ/kg] 14 47.69 970.8 797.9 16 50.14 985 817.4 18 52.16 992.6 829.8 20 53.85 995.4 837.0 22 55.29 994.9 840.6 24 56.54 992 841.5
4.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5200400600800
10001200140016001800200022002400
s [kJ/kg-K]
T[K]
95 kP
a
340
.1 kP
a
592
0 kPa
0.044
0.1 0
.88 m3/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
6.74 kJ/kg-K
Air
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-31
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
ηth
14 16 18 20 22 24970
975
980
985
990
995
1000
rcomp
MEP[kPa]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-32
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
( )( ) K 101917K 328 0.41
2
112 ==
=
−k
TTV
V
Process 2-3: P = constant heat addition.
( )( ) K 2241K 10192.22.2 222
33
2
22
3
33 ====→= TTTT
PT
Pv
vvv
Process 3-4: isentropic expansion.
( )
( ) ( )
( ) ( )( )( )( )
( )( ) kW 46.6===
=−=−=
=−⋅×=−=−=
=−⋅×=
−=−=
×=⋅⋅
==
=
=
=
=
=
−
−
−
−−−
kJ/rev 1.864rev/s 1500/60
kJ/rev 864.1174.1038.3
kJ 174.1K328989.2KkJ/kg 0.718kg10473.2
kJ 3.038K)10192241)(KkJ/kg 1.005)(kg10.4732(
kg10473.2)K 328)(K/kgmkPa 0.287(
)m 0.0024)(kPa 97(
K 989.217
2.2K 22412.22.2
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).
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-33
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 Kinetic and potential energy changes are negligible. 3 Nitrogen is an ideal gas with constant specific heats. Properties The properties of nitrogen at room temperature are cp = 1.039 kJ/kg·K, cv = 0.743 kJ/kg·K, R = 0.2968 kJ/kg·K, and k = 1.4 (Table A-2).
PAnalysis Process 1-2: isentropic compression.
Qout
Qin 32
1
4
v
( )( ) K 101917K 328 0.41
2
112 ==
=
−k
TTV
V
Process 2-3: P = constant heat addition.
( )( ) K 2241K 10192.22.2 222
33
2
22
3
33 ====→= TTTT
PT
Pv
vvv
Process 3-4: isentropic expansion.
( )
( )( )( )( )
( ) ( )( )( )( )( ) ( )
( )( )( )
( )( ) kW 46.6===
=−=−=
=−⋅×=−=−=
=−⋅×=
−=−=
×=⋅⋅
==
=
=
=
=
=
−
−
−
−−−
kJ/rev 1.863rev/s 1500/60
kJ/rev .8631175.1037.3
kJ 1.175K328989.2KkJ/kg 0.743kg102.391
kJ 3.037K10192241KkJ/kg 1.039kg102.391
kg10391.2K 328K/kgmkPa 0.2968
m 0.0024kPa 97
K 989.2172.2K 22412.22.2
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).
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-34
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
1520.4 kJ/kg
3
2
1
4
v
Analysis (a) Process 1-2: isentropic compression.
621.2kJ/kg 214.07
K 3001
11 =
=→=
r
uT
v
( )kJ/kg 611.2K 823.1
37.442.6211411
2
2
1
2112 =
=→====
uT
r rrr vvv
vv
Process 2-x, x-3: heat addition,
( ) (
( ) ( )xx
xx
r
huhhuuqqq
hT
−+−=
−+−=+=
==
→=
−−
2.25032.6114.1520
012.2kJ/kg 2503.2
K 2200
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/kg 1520.4kJ/kg 411.62
ratio
kJ/kg 411.622.61182.1022
in
in,2
2in,2
uuq
x
xx
(b) cxxx
xx rTT
TP
TP
====→= 1.692K 1300K 220033
3
33
v
vvv
( ) kJ/kg 886.3648.16012.2692.114
692.1692.1 42
4
3
43334
=→===== urrrrr vv
v
vv
v
vv
Process 4-1: v = constant heat rejection.
55.8%=−=−=
=−=−=
kJ/kg 1520.4kJ/kg 672.23
11
kJ/kg 72.23607.2143.886
in
outth
14out
uuq
η
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-35
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
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-36
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 ηth [%] wnet [kJ/kg] 10 52.33 795.4 11 53.43 812.1 12 54.34 826 13 55.09 837.4 14 55.72 846.9 15 56.22 854.6 16 56.63 860.7 17 56.94 865.5 18 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
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-37
10 11 12 13 14 15 16 17 1852
53
54
55
56
57
58
rv
ηth[%]
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 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 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.
( )( ) K 86214K 300 0.41
2
112 ==
=
−k
TTv
v
Process 2-x, x-3: heat addition,
( ) ( )
( ) ( )( )( ) ( )( xx
xpx
xxxx
TT
TTcTTchhuuqqq
−⋅+−⋅=
−+−=
)
−+−=+= −−
2200KkJ/kg 1.005862KkJ/kg 0.718kJ/kg 1520.432
32in,3in,2in
v
P
4
1
2
3
1520.4 kJ/kg
x
Qout
v
Solving for Tx we get Tx = 250 K which is impossible. Therefore, constant specific heats at room temperature turned out to be an unreasonable assumption in this case because of the very high temperatures involved.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-38
9-59 A six-cylinder compression ignition engine operates on the ideal Diesel cycle. The maximum temperature in the cycle, the cutoff ratio, the net work output per cycle, the thermal efficiency, the mean effective pressure, the net power output, and the specific fuel consumption are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at 850 K are cp = 1.110 kJ/kg·K, cv = 0.823 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.349 (Table A-2b). Analysis (a) Process 1-2: Isentropic compression
Qout
4
32
Qin
1
( )( )
( )( ) kPa 434117kPa 95
K 7.88117K 328
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
m 0002813.0
m 0045.017
=
+=→
+=
c
c
c
c
dcr
V
V
V
V
VV
31 m 004781.00045.00002813.0 =+=+= dc VVV
The total mass contained in the cylinder is
( )( )kg 0.004825
K 328K/kgmkPa 0.287)m 781kPa)(0.004 95(
3
3
1
11 =⋅⋅
==RTP
mV
The mass of fuel burned during one cycle is
kg 000193.0kg) 004825.0(
24 =→−
=→−
== ff
f
f
f
f
a mm
mm
mmmm
AF
Process 2-3: constant pressure heat addition kJ 039.88)kJ/kg)(0.9 kg)(42,500 000193.0(HVin === cf qmQ η
K 2383=→−=→−= 3323in K)7.881(kJ/kg.K) kg)(0.823 004825.0(kJ 039.8)( TTTTmcQ v
The cutoff ratio is
2.7===K 881.7K 2383
2
3
TT
β
(b) 33
12 m 0002813.0
17m 0.004781
===r
VV
23
14
3323 m 00076.0)m 0002813.0)(70.2(
PP ==
===VV
VV β
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-39
Process 3-4: isentropic expansion.
( )
( ) kPa 2.363m 0.004781m 0.00076kPa 4341
K 1254m 0.004781m 0.00076K 2383
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 677.3K3281254KkJ/kg 0.823kg) 004825.0(14out =−⋅=−= TTmcQ v
The net work output and the thermal efficiency are kJ 4.361=−=−= 677.3039.8outinoutnet, QQW
0.543===kJ 8.039kJ 4.361
in
outnet,th Q
Wη
(c) The mean effective pressure is determined to be
kPa 969.2=
⋅
−=
−=
kJmkPa
m)0002813.0004781.0(kJ 4.361
MEP3
321
outnet,
VV
W
(d) The power for engine speed of 2000 rpm is
kW 72.7=
==
s 60min 1
rev/cycle) 2((rev/min) 2000kJ/cycle) (4.361
2netnetnWW&&
Note that there are two revolutions in one cycle in four-stroke engines. (e) Finally, the specific fuel consumption is
g/kWh 159.3=
==
kWh 1kJ 3600
kg 1g 1000
kJ/kg 4.361kg 000193.0sfc
netWmf
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.