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9-1
Solutions Manual for
Thermodynamics: An Engineering Approach Seventh Edition in SI Units
Yunus A. Cengel, Michael A. Boles McGraw-Hill, 2011
Chapter 9 GAS POWER CYCLES
PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The McGraw-Hill Companies, Inc. (“McGraw-Hill”) and protected by copyright and other state and federal laws. By opening and using this Manual the user agrees to the following restrictions, and if the recipient does not agree to these restrictions, the Manual should be promptly returned unopened to McGraw-Hill: This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the affiliated textbook. No other use or distribution of this Manual is permitted. This Manual may not be sold and may not be distributed to or used by any student or other third party. No part of this Manual may be reproduced, displayed or distributed in any form or by any means, electronic or otherwise, without the prior written permission of McGraw-Hill.
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
Actual and Ideal Cycles, Carnot cycle, Air-Standard Assumptions, Reciprocating Engines
9-1C It represents the net work on both diagrams.
9-2C The air standard assumptions are: (1) the working fluid is air which behaves as an ideal gas, (2) all the processes are internally reversible, (3) the combustion process is replaced by the heat addition process, and (4) the exhaust process is replaced by the heat rejection process which returns the working fluid to its original state.
9-4C The cold air standard assumptions involves the additional assumption that air can be treated as an ideal gas with constant specific heats at room temperature.
9-5C The clearance volume is the minimum volume formed in the cylinder whereas the displacement volume is the volume displaced by the piston as the piston moves between the top dead center and the bottom dead center.
9-6C It is the ratio of the maximum to minimum volumes in the cylinder.
9-6C 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.
9-7C Yes.
9-8C 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.
9-9C 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.
9-10C 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 MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
9-11 The maximum possible thermal efficiency of a gas power cycle with specified reservoirs is to be determined.
Analysis The maximum efficiency this cycle can have is
64.0%==+
+−=−= 1
T0.640
K )273(500K )273(51Carnotth,
H
L
Tη
specified processes. The cycle is to be
2 Kinetic and potential energy changes are negligible. 3 Air is
erties The properties of air are given as R = 0.287 kPa·m3/kg·K, c = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, and k =
the cycle are shown in the figures.
(b) Process 1-2: Isentropic com
TTmcw in −= v
9-12 An air-standard cycle executed in a piston-cylinder system is composed of three sketcehed on the P-v and T-s diagrams and the back work ratio are to be determined.
Assumptions 1 The air-standard assumptions are applicable. an ideal gas with constant specific heats.
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9-4
9-13 The three processes of an air-standard cycle are described. The cycle is to be shown on the P-v and T-s diagrams, and the back work ratio and the thermal efficiency are to be determined.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats.
Properties The properties of air are given as R = 0.287 kJ/kg.K, cp = 1.005 kJ/kg.K, cv = 0.718 kJ/kg·K, and
k = 1.4.
Analysis (a) The P-v and T-s diagrams of the cycle are shown in the figures.
preparation. If you are a student using this Manual, you are using it without permission.
9-5
9-14 The three processes of an ideal gas power cycle are described. The cycle is to be shown on the P-v and T-s diagrams, and the maximum temperature, expansion and compression works, and thermal efficiency are to be determined.
Assumptions 1 Kinetic and potential energy changes are negligible. 2 The ideal gas has constant specific heats.
Properties The properties of ideal gas are given as R = 0.3 kJ/kg.K, cp = 0.9 kJ/kg.K, cv = 0.6 kJ/kg·K, and
k = 1.5.
Analysis (a) The P-v and T-s diagrams of the cycle are shown in the figures.
(d) The work during isentropic compression is determined from an energy
−=∆= −
300))( 1221,2 TTcuw vin
) Net work output is
== ∫−−
33
,32,32 wq outin
balance during process 1-2:
−1
kJ/kg 260.9=
⋅= K)(734.8kJ/kg 6.0( −
(e
kJ/kg 1.1349.2600.395,21,32 =−=−= −− inoutnet www
The thermal efficiency is then
33.9%==== 339.0kJ 395.0kJ 134.1
in
netth q
wη
preparation. If you are a student using this Manual, you are using it without permission.
9-6
9-15 The four processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-s diagrams, and the net work output and the thermal efficiency are to be determined.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats.
Properties The properties of air are given in Table A-17.
Analysis (b) The properties of air at various states are
v
P
1
2
4
3qin
qout
s
T
1
24
3qin
qout
( )
( )
( ) kJ/kg 71.73979.32601.9kPa 1835.6
kPa 100
kPa 6.1835kPa 600K 490.3K 1500
9.601kJ/kg 41.1205
K 1500
K 3.490kJ/kg 352.29
841.71.3068kPa 100kPa 600
3068.1kJ/kg .17295
K 2951 ⎯→⎯=T
43
4
22
323
33
2
2
1
2
1
34
3
12
1
=⎯→⎯===
=
==
⎯→⎯=
==
⎯→⎯===
==
hPPP
P
TT
Pu
T
Tu
PPP
P
Ph
rr
r
rr
r
From energy balances,
(c) Then the thermal efficiency becomes
32233 ==⎯→⎯= PT
PPP vv
T
kJ/kg 408.6=−=−=
=−=−=
=−=−=
5.4441.853
kJ/kg .544417.29571.739
kJ/kg 1.85329.35241.1205
outinoutnet,
14out
23in
qqw
hhq
uuq
47.9%==== 479.0kJ/kg 853.1kJ/kg 408.6
in
outnet,th q
wη
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
9-16 Problem 9-15 is reconsidered. The effect of the maximum temperature of the cycle on the net work output and thermal efficiency is to be investigated. Also, T-s and P-v diagrams for the cycle are to be plotted.
Analysis Using EES, the problem is solved as follows:
"Input Data" T[1]=295 [K] P[1]=100 [kPa] P[2] = 600 [kPa] T[3]=1500 [K] P[4] = 100 [kPa] "Process 1-2 is isentropic compression" s[1]=entropy(air,T=T[1],P=P[1]) s[2]=s[1] T[2]=temperature(air, s=s[2], P=P[2]) P[2]*v[2]/T[2]=P[1]*v[1]/T[1] P[1]*v[1]=R*T[1] R=0.287 [kJ/kg-K] "Conservation of energy for process 1 to 2" q_12 -w_12 = DELTAu_12 q_12 =0"isentropic process" DELTAu_12=intenergy(air,T=T[2])-intenergy(air,T=T[1]) "Process 2-3 is constant volume heat addition" s[3]=entropy(air, T=T[3], P=P[3]) {P[3]*v[3]/T[3]=P[2]*v[2]/T[2]} P[3]*v[3]=R*T[3] v[3]=v[2] "Conservation of energy for process 2 to 3" q_23 -w_23 = DELTAu_23 w_23 =0"constant volume process" DELTAu_23=intenergy(air,T=T[3])-intenergy(air,T=T[2]) "Process 3-4 is isentropic expansion" s[4]=entropy(air,T=T[4],P=P[4]) s[4]=s[3] P[4]*v[4]/T[4]=P[3]*v[3]/T[3] {P[4]*v[4]=0.287*T[4]} "Conservation of energy for process 3 to 4" q_34 -w_34 = DELTAu_34 q_34 =0"isentropic process" DELTAu_34=intenergy(air,T=T[4])-intenergy(air,T=T[3]) "Process 4-1 is constant pressure heat rejection" {P[4]*v[4]/T[4]=P[1]*v[1]/T[1]} "Conservation of energy for process 4 to 1" q_41 -w_41 = DELTAu_41 w_41 =P[1]*(v[1]-v[4]) "constant pressure process" DELTAu_41=intenergy(air,T=T[1])-intenergy(air,T=T[4]) q_in_total=q_23 w_net = w_12+w_23+w_34+w_41 Eta_th=w_net/q_in_total*100 "Thermal efficiency, in percent"
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
9-17 The three processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-s diagrams, and the heat rejected and the thermal efficiency are to be determined.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-2).
Analysis (b) The temperature at state 2 and the heat input are
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9-12
9-19 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
Then, the second-law efficiency of the actual cycle becomes
th
kW 9000kJ/kg) kg/s)(100 90(netCarnot === wmW &&
0.578===kW 9000kW 5200
Carnot
actualII W
W&
&η
preparation. If you are a student using this Manual, you are using it without permission.
9-13
9-20 An ideal gas Carnot cycle with air as the working fluid is considered. The maximum temperature of the low-temperature energy reservoir, the cycle's thermal efficiency, and the amount of heat that must be supplied per cycle are to be determined.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-2a).
Analysis The temperature of the low-temperature reservoir can be found by applying the isentropic expansion process relation
K 481.1=⎟⎠⎞
⎜⎝⎛+=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
−− 14.11
1
221 12
1K) 2731027(k
TTv
v
s
T
3
2qin
qout4
11300 K Since the Carnot engine is completely reversible, its efficiency is
0.630=+
−=−=K 273)(1027
K 481.111Carnotth,HTLT
η
The work output per cycle is
kJ/cycle 20min 1cycle/min 1500net
⎠⎝n&s 60kJ/s 500net =⎟
⎞⎜⎛==
WW
&
ccording the definition of the cycle efficiency,
A to
kJ/cycle 31.75===⎯→⎯=0.63kJ/cycle 20
Carnotth,
netin
in
netCarnotth, η
ηW
QQW
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
9-21 An air-standard cycle executed in a piston-cylinder system is composed of three specified processes. The cycle is to be sketcehed on the P-v and T-s diagrams; the heat and work interactions and the thermal efficiency of the cycle are to be determined; and an expression for thermal efficiency as functions of compression ratio and specific heat ratio is to be obtained.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats.
Properties The properties of air are given as R = 0.3 kJ/kg·K and cv = 0.3 kJ/kg·K.
Analysis (a) The P-v and T-s diagrams of the cycle are shown in the figures.
(b) Noting that
1.429
7.00.1
K kJ/kg 0.13.07.0
===
⋅=+=+=
v
v
cc
k
Rcc
p
p
s
T
32
1
v
P
32
1Process 1-2: Isentropic compression
K 4.584)5)(K 293( 429.011
1
2
112 ===⎟⎟
⎠
⎞⎜⎜⎝
⎛= −
−k
k
rTTTvv
kJ/kg 204.0=−⋅=−=−1 K )2934.584)(KkJ/kg 7.0()( 12in2, TTcw v
rom ideal gas relation,
0=q −21
F
2922)5)(4.584(3133 =⎯→⎯=== Tr
T v
v
v
v
22=
T
Constant pressure heat addition
−=−== ∫−
K )4.
)()( 23232
3
out3,2 TTmRPPdw VVv
huq
p
out
ant volume heat rejection
2
Process 2-3:
kJ/kg 701.3=−⋅= 5842922)(KkJ/kg 3.0(2
kJ/kg 2338=−⋅=−=
∆=∆+= −−−−
K )4.5842922)(KkJ/kg 1()( 23
3232,32in3,2
TTcw
Process 3-1: Const
=∆= −− (31out1,3 cuq v kJ/kg 1840.3=⋅=− K 293)-K)(2922kJ/kg 7.0()13 TT
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9-16
Otto Cycle
9-22C 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.
9-23C The ideal Otto cycle involves external irreversibilities, and thus it has a lower thermal efficiency.
9-24C 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.
9-25C They are analyzed as closed system processes because no mass crosses the system boundaries during any of the processes.
9-26C It increases with both of them.
9-27C Because high compression ratios cause engine knock.
9-28C The thermal efficiency will be the highest for argon because it has the highest specific heat ratio, k = 1.667.
9-29C The fuel is injected into the cylinder in both engines, but it is ignited with a spark plug in gasoline engines.
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
9-30 An ideal Otto cycle is considered. The thermal efficiency and the rate of heat input are to be determined.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-2a).
Analysis The definition of cycle thermal efficiency reduces to
v
P
4
1
3
2
qin
qout
61.0%==−=−=−− 10.51th kr
6096.01111 11.4η
The rate of heat addition is then
kW 148===0.6096
kW 90
th
netin η
WQ
&&
9-31 An ideal Otto cycle is considered. The thermal efficiency and the rate of heat input are to be determined.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is
erties 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-
nalysis The definition of cycle thermal efficiency reduces to
an ideal gas with constant specific heats.
Prop2a).
A
v
P
4
1
3
2
qin
qout
57.5%==−=−=1111thη
−−5752.0
8.5 11.41kr
he rate of heat addition is then
T
kW 157===0.5752
kW 90
th
netin η
WQ
&&
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
9-32 The two isentropic processes in an Otto cycle are replaced with polytropic processes. The heat added to and rejected from this cycle, and the cycle’s thermal efficiency are to be determined.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats.
Properties The properties of air at room temperature are R = 0.287 kPa·m3/kg·K, cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-2a).
Analysis The temperature at the end of the compression is
kJ/kg 718.0()( 1414 ⋅=−=− TTcq v kJ/kg 0.360K)2884.789)(K =−
he head added and rejected from the cycle are
The thermal efficiency of this cycle is then
T
kJ/kg 419.5kJ/kg 835.0
=+=+==+=+=
−−
−−
0.36053.592.1638.671
1421out
4332in
qqqqqq
0.498=−=−=0.8355.41911
in
outth q
qη
preparation. If you are a student using this Manual, you are using it without permission.
9-19
9-33 An ideal Otto cycle is considered. The heat added to and rejected from this cycle, and the cycle’s thermal efficiency are to be determined.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats.
Properties The properties of air at room temperature are R = 0.287 kPa·m3/kg·K, cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-2a).
Analysis The temperature at the end of the compression is
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9-20
9-34 A six-cylinder, four-stroke, spark-ignition engine operating on the ideal Otto cycle is considered. The power produced by the engine is to be determined.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats.
Properties The properties of air at room temperature are R = 0.287 kPa·m3/kg.K (Table A-1), cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-2a).
Analysis From the data specified in the problem statement,
v
P
4
1
3
2
143.7140 1
1
2
1 ===v
v
v
v
.r
Since the compression and expansion processes are isentropic,
( ) K 7.636143.7K) 290( 14.111
1
2
112 ===⎟⎟
⎠
⎞⎜⎜⎝
⎛= −−
−k
k
rTTTvv
K 6.5207.143
K) 1143(34
34 =⎟⎠
⎜⎝
=⎟⎠
⎜⎝
=⎟⎟⎠
⎜⎜⎝
=r
TTTv
11 14.1113 ⎞⎛⎞⎛⎞⎛ −−− kk
v
pplicatio of the first law to the compression and expansion processes gives
since there are two revolutions per cycle in a four-stroke engine.
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9-21
9-35 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).
preparation. If you are a student using this Manual, you are using it without permission.
9-22
9-36 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.
( )( )
( ) ( ) 2338kPa 100K 308K 757.9
9.5
K 757.99.5
11
2
2
12
1
11
2
22
0.41
=⎟⎟⎠
⎞⎜⎜⎝
⎛==⎯→⎯=
=⎞⎛
−
PTT
PT
PT
P
k
v
vvv
v
rocess 3- polytropic expansion.
kPa v
P
4
1
3
2
Qin
Qout
Polytropic
800 K
308 KK 3082
112 =⎟⎟
⎠⎜⎜⎝
= TTv
P 4:
( )( )( )( )
kg10788.6K 308K/kgmkPa 0.287
m 43
3
1
11 −×=⋅⋅
==RT
m
0.0006kPa 100PV
( )( )
( ) ( )( )( )kJ 0.5338
1.351K1759800KkJ/kg 0.287106.788
1
434
34 =−
−⋅×=
−−
=−
nTTmR
W
9.5K 800 35.01
3
443 ==⎟⎟
⎠
⎞⎜⎜⎝
⎛=
−
TTn
K 1759v
v
ring the expansion process (This is not realistic, and probably is due to ssuming constant specific heats at room temperature). ) Process 2-3: v = constant heat addition.
preparation. If you are a student using this Manual, you are using it without permission.
9-23
9-37 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-17.
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9-24
9-38 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 = 0.5203 kJ/kg.K, cv = 0.3122 kJ/kg.K, and k = 1.667 (Table A-2).
Analysis (a) Process 1-2: isentropic compression.
v
P
4
1
3
2
qinqout
( )( )8K 3002
12 =⎟⎟⎠
⎜⎜⎝
= TTv
K 12010.6671
1 =⎞⎛
−kv
Process 2-3: v = constant heat addition.
( ) ( )( ) kJ/kg 43.45=−=−=−= k 0.31222323in TTcuuq v K12011340J/kg.K
9-39 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).
9-40 The expressions for the maximum gas temperature and pressure of an ideal Otto cycle are to be determined when the compression ratio is doubled.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats.
Analysis The temperature at the end of the compression varies with the compression ratio as
11
212 =⎟⎟
⎠⎜⎜⎝
= rTTTv
1
1 −−
⎞⎛ kk
v
T1 is fixed. The temperature rise during the combustion remains constant since the mount of heat addition is fixed. Then, the maximum cycle temperature is given by
he smallest gas specific volume during the cycle is
v
P
4
1
3
2 qout
qinsince a
11in2in3 // −+=+= krTcqTcqT vv
T
r1
3v
v =
When this is combined with the maximum temperature, the maximum pressure is given by
( )11in
13
33 / −+== krTcqRrRTP vvv
9-41 It is to be determined if the polytropic exponent to be used in an Otto cycle model will be greater than or less than the isentropic exponent.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is n ideal ga with constant specific heats.
rocess,
=kPv
heat is lost during the expansion of the gas,
where T4s is the temperature that would occur if the expansion were reversible and adiabatic (n=k). This can only occur when
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9-28
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.
9-47 An ideal diesel cycle has a compression ratio of 20 and a cutoff ratio of 1.3. The maximum temperature of the air and the rate of heat addition are to be determined.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4 (Table A-2a).
Analysis We begin by using the process types to fix the temperatures of the states.
ombining e first law as applied to the various processes with the process equations gi
C th ves v
P
4
1
2 3qin
qout
6812.0)13.1(4.1
13.1111 4.1 −−kr20
1)1(
11.41th =
−−=
−−=
−c
ck rkr
η
According to the definition of the thermal efficiency,
1−
kW 367===0.6812
kW 250
th
netin η
WQ
&&
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9-29
9-48 An ideal dual cycle has a compression ratio of 14 and cutoff ratio of 1.2. The thermal efficiency, amount of heat added, and the maximum gas pressure and temperature are to be determined.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4 (Table A-2a).
Analysis The specific volume of the air at the start of the compression is
v
P
4
1
2
3
qout
x
qin/kgm 051.1
kPa 8011 ===
Pv
K) 293)(K/kgmkPa 287.0( 33
1 ⋅⋅RT
olume at the end of the compression is and the specific v
kJ/kg 1.197K)2935.567)(KkJ/kg 718.0()( 14out =−⋅=−= TTcq v
0.646=−=−=kJ/kg 556.5kJ/kg 197.1
11in
outth q
qη Then,
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9-30
9-49 An ideal dual cycle has a compression ratio of 14 and cutoff ratio of 1.2. The thermal efficiency, amount of heat added, and the maximum gas pressure and temperature are to be determined.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4 (Table A-2).
Analysis The specific volume of the air at the start of the compression is
kJ/kg 2.170K)2530.490)(KkJ/kg 718.0()( 14out =−⋅=−= TTcq v
0.646=−=−=kJ/kg 480.4kJ/kg 170.2
11in
outth q
qη Then,
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9-31
9-50 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-17.
1700 KAnalysis (a) Process 1-2: isentropic compression.
2.621kJ/kg 07.214
K 3001
11 =
=⎯→⎯=
r
uT
v
( )kJ/kg 83.934
K 7.90113.34 2.621
2.181112
====r rrr vv
vv
11
2
22
==
⎯→⎯hTv
rocess 2- P = constant heat addition. P 3:
1.885===⎯→⎯=K 901.7K 1700
2
3
2
3
2
22
3
33
TT
TP
TP
v
vvv
(b) K 001733
=⎯→⎯=T
rv
kJ/kg 27.94583.9341.1880
761.4kJ/kg 1.1880
23
3
=−=−=
=
hhq
h
in
Process 3-4: isentropic expansion.
( ) kJ/kg 91.60297.45761.4885.1
2.18885.1885.1 4
2
4
3
43334
=⎯→⎯===== urrrrr vv
v
vv
v
vv
Process 4-1: v = constant heat rejection.
(c) 58.9%
kJ/kg 388.84
==−=−=
=−=−=
589.0kJ/kg 945.27kJ/kg 388.84
11
07.21491.602
in
outth
14out
qq
uuq
η
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9-32
9-51 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 = 1.005 kJ/kg.K, cv = 0.718 kJ/kg.K, and k = 1.4 (Table A-2).
preparation. If you are a student using this Manual, you are using it without permission.
9-33
9-52 An ideal diesel engine with air as the working fluid has a compression ratio of 20. The thermal efficiency and the mean effective pressure are to be determined.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4 (Table A-2).
Analysis (a) Process 1-2: isentropic compression.
v
P
4
1
2 3 qin
qout
( )( ) K 971.120K 293 0.41
2
112 ==⎟⎟
⎠
⎞⎜⎜⎝
⎛=
−k
TTV
V
Process 2-3: P = constant heat addition.
2.265K971.1K2200
2
3
223 TT V3223 ===⎯→⎯=
TTPP VVV
Process 3-4: isentropic expansion.
3
( )
( ) ( )( )( ) ( )( )
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
( )( )
( ) ( )( )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 (b)
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
9-53 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.
t some heat is also rejected during the polytropic rocess, w ch is determined from an energy balance on process 3-4:
No oup hi
in this case does not represent the entire heat rejected since
( ) ( )( )
( )( )( )
kJ/kg 120.1K 22001026KkJ/kg 0.718kJ/kg
34out34,in34,34out34,in34,
system
=−⋅+
−+=⎯→⎯−=− TTcwquuwq v
h 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
pansion. The cause of this unrealistic result is the constant specific heat assumption. If we were to use u data from the ir table, we would obtain
whic hh is a eat loss as expected. Then qout becomes kJ/kg 654.43.5261.128out41,out34,out =+=+= qqq
and
47.0%===
=−=−=
kJ/kg 1235kJ/kg 580.6
kJ/kg 580.64.6541235
in
outnet,th
outinoutnet,
qw
qqw
η
( )( )
( ) ( )( )kPa 691=⎟
⎟⎠
⎞⎜⎜⎝
⎛ ⋅
−=
−=
−=
==
==⋅⋅
==
kJmkPa 1
1/201/kgm 0.885kJ/kg 580.6
/11MEP
/kgm 0.885kPa 95
K 293K/kgmkPa 0.287
3
31
outnet,
21
outnet,
max2min
max3
3
1
11
rww
r
PRT
vvv
vvv
vv (b)
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9-35
9-54 Problem 9-53 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]) Call QTotal(q_12,q_23,q_34,q_41: q_in_total,q_out_total) w_net = w_12+w_23+w_34+w_41
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
9-55 A four-cylinder ideal diesel engine with air as the working fluid has a compression ratio of 22 and a cutoff ratio of 1.8. The power the engine will deliver at 2300 rpm is to be determined.
Assumptions 1 The cold air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4 (Table A-2).
9-56 A four-cylinder ideal diesel engine with nitrogen as the working fluid has a compression ratio of 22 and a cutoff ratio of 1.8. The power the engine will deliver at 2300 rpm is to be determined.
Assumptions 1 The air-standard assumptions are applicable with nitrogen as the working fluid. 2 Kinetic and potential energy changes are negligible. 3 Nitrogen is an ideal gas with constant specific heats.
Properties The properties of nitrogen at room temperature are cp = 1.039 kJ/kg·K, cv = 0.743 kJ/kg·K, R = 0.2968 kJ/kg·K, and k = 1.4 (Table A-2).
Discussion Note that for 2-stroke engines, 1 thermodynamic cycle is equivalent to 1 mechanical cycle (and thus revolutions).
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9-40
9-57 An ideal dual cycle has a compression ratio of 18 and cutoff ratio of 1.1. The power produced by the cycle is to be determined.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4 (Table A-2a).
Analysis We begin by fixing the temperatures at all states.
( ) K 7.92418K) 291( 14.111
1
2
112 ===⎟⎟
⎠
⎞⎜⎜⎝
⎛= −−
−k
k
rTTTv
v
v
P
4
1
2
3
qout
x
qin
( ) kPa 514818kPa) 90( 4.11
2
112 ===⎟⎟
⎠
⎞⎜⎜⎝
⎛= k
k
rPPPv
v
x kPa 5663kPa) 5148)(1.1(23 ==== PrPP p
K 1017kPa 5148kPa 5663K) 7.924(
22 =⎟
⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
PP
TT x
xc
x
K 1119K) 1017)(1.1( === TrT 3
K 8.365181.1K) 1119(
14.11
3
1
4
33 4 =⎟
⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
−−− kc
k
rr
TTTv
v
Applying the first law to each of the processes gives
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9-41
9-58 A dual cycle with non-isentropic compression and expansion processes is considered. The power produced by the cycle is to be determined.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4 (Table A-2a).
Analysis We begin by fixing the temperatures at all states.
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9-42
9-59 An ideal dual cycle has a compression ratio of 15 and cutoff ratio of 1.4. The net work, heat addition, and the thermal efficiency are to be determined.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats.
Properties The properties of air at room temperature are R = 0.287 kPa·m3/kg.K (Table A-1), cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-2a).
Analysis Working around the cycle, the germane properties at the various states are
( ) K 4.87715K) 297( 14.111
1
2
112 ===⎟⎟
⎠
⎞⎜⎜⎝
⎛= −−
−k
k
rTTTv
v
v
P
4
1
2
3
qout
x
qin( ) kPa 434315kPa) 98( 4.11
2
112 ===⎟⎟
⎠
⎞⎜⎜⎝
⎛= k
k
rPPPv
v
23 === PrP px kPa 4777kPa) 4343)(1.1( =P
K 1.965kPa 4343kPa 4777K) 4.877(
22 =⎟
⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
PP
TT xx
K 1351K)(1.4) 1.965(33 ===⎟⎟
⎠⎜⎜⎝
= cxx
x rTTTv
⎞⎛ v
K 2.523151.4K) 1351(
14.11
3
1
4
33 4 =⎟
⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
−−− kc
k
rr
TTTv
v
Applying the first law to each of the processes gives
kJ/kg 7.416K)2974.877)(KkJ/kg 718.0()( 1221 =−⋅=−=− TTcw v
kJ/kg 289=−+=−+= −−− 7.4168.1104.59421343 net wwww x
and the net heat addition is
Hence, the thermal efficiency is
kJ/kg 451=+=+= −− 8.3870.6332in xx qqq
0.641===kJ/kg 451kJ/kg 289
in
netth q
wη
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9-43
9-60 A six-cylinder compression ignition engine operates on the ideal Diesel cycle. The maximum temperature in the cycle, the cutoff ratio, the net work output per cycle, the thermal efficiency, the mean effective pressure, the net power output, and the specific fuel consumption are to be determined.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats.
Properties The properties of air at 850 K are cp = 1.110 kJ/kg·K, cv = 0.823 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.349 (Table A-2b).
Analysis (a) Process 1-2: Isentropic compression
1
Qin
2 3
4
Qout
( )( )
( )( ) kPa 504419kPa 95
K 1.95019K 340
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
ote that there are two revolutions in one cycle in four-stroke engines
(e) Finally, the specific fuel consumption is
==2(
kJ/cycle) (2.7102netnet WW&
N .
g/kWh 151=⎟⎠⎞
⎜⎝⎛
⎟⎟⎠
⎞⎜⎜⎝
⎛==
kWh 1kJ 3600
kg 1g 1000
kJ/kg 2.710kg 0001134.0sfc
netWm f
preparation. If you are a student using this Manual, you are using it without permission.
9-45
9-61 An expression for cutoff ratio of an ideal diesel cycle is to be developed.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats.
Analysis Employing the isentropic process equations,
When this is solved for cutoff ratio, the result is
v
P
4
1
2 3 qin
qout
21
1
while the ideal gas law gives
11
23 TrrrTT kcc
−==
When the first law and the closed system work integral is applied to the constant pressure heat addition, the result is
)()( 11
123in cpp
11
in1Trc
qr kp
c −+=
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9-46
9-62 An expression for the thermal efficiency of a dual cycle is to be developed and the thermal efficiency for a given case is to be calculated.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4 (Table A-2)
Analysis The thermal efficiency of a dual cycle may be expressed as
)()( 32in
thxpx TTcTTcq −+−v
By applying the isentropic process relations for ideal gases with constant specific heats to the processes 1-2 and 3-4, as wellas the ideal gas equation of state, the tem
)(11 14out TTcq −
−=−= vη
peratures may be eliminated from the thermal efficiency expression. This yields
preparation. If you are a student using this Manual, you are using it without permission.
9-47
9-63 An expression regarding the thermal efficiency of a dual cycle for a special case is to be obtained.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats.
Analysis The thermal efficiency of a dual cycle may be expressed as
)()( 32in
thxpx TTcTTcq −+−v
By applying the isentropic process relations for ideal gases with constant specific heats to the processes 1-2 and 3-4, as wellas the ideal gas equation of state, the tem
)(11 14out TTcq −
−=−= vη
peratures may be eliminated from the thermal efficiency expression. This yields
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9-48
9-64 The five processes of the dual cycle is described. The P-v and T-s diagrams for this cycle is to be sketched. An expression for the cycle thermal efficiency is to be obtained and the limit of the efficiency is to be evaluated for certain cases.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats.
Analysis (a) The P-v and T-s diagrams for this cycle are as shown.
s
T
52
1
4
3
v
P
5
1
2
4
qout
3
qin
(b) Apply first law to the closed system for processes 2-3, 3-4, and 5-1 to show:
(c) In the limit as rp approaches unity, the cycle thermal efficiency becomes
( ) ( )
( )
11 1
1
1
1
1lim 1 lim
1 1
11lim 11
p p
p
kc p
th kr rp c
kc
th th Dieselrc
kp
k
r rr r k r
rk r
r r
r
η
η η
−→ →
→
−
−
⎧ ⎫−⎪ ⎪= − ⎨ ⎬− + −⎪ ⎪⎩ ⎭
⎡ ⎤−⎢ ⎥= − =−⎢ ⎥⎣ ⎦
(d) In the limit as rc approaches unity, the cycle thermal efficiency becomes
( ) ( ) ( )1 11 1
1
1
1
1 1lim 1 lim 1
1 1
1lim 1
c p
c
kc p p
th k kr rp c
th th Ottor
kp
k
r r rr r k r r rr r
r
η
η η
− −→ →
→
−
−
⎧ ⎫− −⎪ ⎪= − = −⎨ ⎬− + − −⎪ ⎪⎩ ⎭
= − =
1p
⎧ ⎫⎪ ⎪⎨ ⎬⎪ ⎪⎩ ⎭
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9-50
Stirling and Ericsson Cycles
9-65C The Stirling cycle.
9-66C The two isentropic processes of the Carnot cycle are replaced by two constant pressure regeneration processes in the Ericsson cycle.
9-67C The efficiencies of the Carnot and the Stirling cycles would be the same, the efficiency of the Otto cycle would be less.
9-68C The efficiencies of the Carnot and the Ericsson cycles would be the same, the efficiency of the Diesel cycle would be less.
9-69 An ideal steady-flow Ericsson engine with air as the working fluid is considered. The maximum pressure in the cycle, the net work output, and the thermal efficiency of the cycle are to be determined.
Assumptions Air is an ideal gas.
Properties The gas constant of air is R = 0.287 kJ/kg.K (Table A-1).
Analysis (a) The entropy change during process 3-4 is
(c) The thermal efficiency of this totally reversible cycle is determined from
ou
kJ/kg 450=−=−= 150600outinoutnet, qqw
75.0%=−=−=K1200K300
11thH
L
TT
η
preparation. If you are a student using this Manual, you are using it without permission.
9-51
9-70 An ideal Stirling engine with air as the working fluid operates between the specified temperature and pressure limits. The net work produced per cycle and the thermal efficiency of the cycle are to be determined.
Assumptions Air is an ideal gas with constant specific heats.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4 (Table A-2a).
Analysis Since the specific volume is constant during process 2-3,
s
T
3
2qin
qout
4
1 800 K
300 K
kPa 7.266K 3003
32⎠⎝T
K 800kPa) 100(2 =⎟⎞
⎜⎛==
TPP
Heat is only added to the system during reversible process 1-2. Then,
9-71 An ideal Stirling engine with air as the working fluid operates between the specified temperature and pressure limits. The power produced and the rate of heat input are to be determined.
Assumptions Air is an ideal gas with constant specific heats.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4 (Table A-2a).
Analysis Since the specific volume is constant during process 2-3,
s
T
3
2qin
qout
4
1 800 K
300 K
kPa 7.266K 3003
32⎠⎝T
K 800kPa) 100(2 =⎟⎞
⎜⎛==
TPP
Heat is only added to the system during reversible process 1-2. Then,
9-74 An ideal Stirling engine with air as the working fluid operates between specified pressure limits. The heat added to the cycle and the net work produced by the cycle are to be determined.
Assumptions Air is an ideal gas with constant specific heats.
Properties The properties of air at room temperature are R = 0.287 kPa·m3/kg.K, cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-2a).
Analysis Applying the ideal gas equation to the isothermal process 3-4 gives
s
T
3
2qin
qout
4
1
298 K
kPa 600kPa)(12) 50(4
34 v3 ===
vPP
Since process 4-1 is one of constant volume,
K 1788kPa 6004
41⎠⎝⎟
⎠⎜⎝ P
kPa 3600K) 298(1 =⎟⎞
⎜⎛=⎟
⎞⎜⎛
=P
TT
dapting t work integral to the heat addition process gives A he first law and
kJ/kg 1275=⋅=== − K)ln(12) 1788)(KkJ/kg 0.287(ln1
2121in v
vRTwq
Similarly,
kJ/kg 212.5=⎟⎠⎞
⎜⎝⎛⋅== = wq − 12
1K)ln 298)(KkJ/kg 0.287(ln3
4343t v
vRT
he net work is then
Properties The properties of air at room temperature are R = 0.287 kPa·m /kg.K, cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, nd k = 1.4 (Table A-2a).
Analysis Applying the ideal gas equation to the isothermal process 3-4 gives
ou
T
kJ/kg 1063=−=−= 5.2121275outinnet qqw
9-75 An ideal Stirling engine with air as the working fluid operates between specified pressure limits. The heat transfer in the regenerator is to be determined.
Assumptions Air is an ideal gas with constant specific heats. 3
a
kPa 600kPa)(12) 50(4
s
T
3
2qin
qout
4
1
298 K
3v3 ===
vPP 4
Since process 4-1 is one of constant volume,
K 1788kPa 600kPa 3600K) 298(
4
141 =⎟
⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
PP
TT
Application of the first law to process 4-1 gives
kJ/kg 1070=−⋅=−= K)2981788)(KkJ/kg 718.0()( 41regen TTcq v
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
9-76C They are (1) isentropic compression (in a compressor), (2) P = constant heat addition, (3) isentropic expansion (in a turbine), and (4) P = constant heat rejection.
9-77C 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.
9-78C Back work ratio is the ratio of the compressor (or pump) work input to the turbine work output. It is usually between 0.40 and 0.6 for gas turbine engines.
9-79C In gas turbine engines a gas is compressed, and thus the compression work requirements are very large since the steady-flow work is proportional to the specific volume.
9-80C As a result of turbine and compressor inefficiencies, (a) the back work ratio increases, and (b) the thermal efficiency decreases.
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
9-81 A simple ideal Brayton cycle with air as the working fluid has a pressure ratio of 10. The air temperature at the compressor exit, the back work ratio, and the thermal efficiency are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with variable specific heats.
Properties The properties of air are given in Table A-17.
Analysis (a) Noting that process 1-2 is isentropic,
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9-57
9-82 A simple Brayton cycle with air as the working fluid has a pressure ratio of 10. The air temperature at the turbine exit, the net work output, and the thermal efficiency are to be determined.
s
T
1
2s
4s
3qin
qout
1240 K
295 K
2
4
Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with variable specific heats.
Properties The properties of air are given in Table A-17.
Analysis (a) Noting that process 1-2s is isentropic,
3068.1kJ/kg .17295
K 2951
11 =
=⎯→⎯=
rPh
T
( )( )
( )
( )( )( )
kJ/kg 83.04707.70293.132487.093.1324
K 6.689 and kJ/kg .0770223.273.272101
3.272kJ/kg 1324.93
K 1240
kJ/kg .6062683.0
17.29526.57017.295
K 564.9 and kJ/kg 570.2607.133068.110
433443
43
443
4
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
ηη
ηη
Thus,
T = 764.4 K
(c)
4
(b)
kJ/kg 210.4=−=−=
=−=−=
=−=−=
9.4873.698
kJ/kg 487.917.29504.783
kJ/kg .369860.62693.1324
outinoutnet,
14out
23in
qqw
hhq
hhq
30.1%==== 3013.0kJ/kg 698.3kJ/kg 210.4
in
outnet,th q
wη
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
9-83 Problem 9-82 is reconsidered. The mass flow rate, pressure ratio, turbine inlet temperature, and the isentropic efficiencies of the turbine and compressor are to be varied and a general solution for the problem by taking advantage of the
be developed.
nalysis Using EES, the problem is solved as follows:
m diagram window"
]
P[1])
compressor, assuming: adiabatic, ke=pe=0"
_in= m_dot*h[3]"SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0"
ot_t + m_dot*h[4] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0"
ycle work, kW" fficiency"
determined only to produce a T-s plot"
[4]=entropy(air,T=T[4],P=P[4])
diagram window method for supplying data to EES is to
ta_c = 83/100 EEta_t = 87/100} "Inlet conditions" h[1]=ENTHALPY(Air,T=T[1]) s[1]=ENTROPY(Air,T=T[1],P= "Compressor anaysis" s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor"
sure ratio - to find P[2]" P_ratio=P[2]/P[1]"Definition of presT_s[2]=TEMPERATURE(Air,s=s_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exit" h_s[2]=ENTHALPY(Air,T=T_s[2])
mpressor adiabatic efficiency; Eta_c = W_dot_c_ideal/W_dot_c_actual. " Eta_c =(h_s[2]-h[1])/(h[2]-h[1]) "Com_dot*h[1] +W_dot_c=m_dot*h[2] "SSSF First Law for the actual
lysis" "External heat exchanger anaP[3]=P[2]"process 2-3 is SSSF constant pressure" h[3]=ENTHALPY(Air,T=T[3]) m_dot*h[2] + Q_dot "Turbine analysis" s[3]=ENTROPY(Air,T=T[3],P=P[3]) s_s[4]=s[3] "For the ideal case the entropies are constant across the turbine" P_ratio= P[3] /P[4] T_s[4]=TEMPERATURE(Air,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_d "Cycle analysis" W_dot_net=W_dot_t-W_dot_c"Definition oEta=W_dot_net/Q_dot_in"Cycle thermal e
f the net c
Bwr=W_dot_c/W_dot_t "Back work ratio" "The following state points areT[2]=temperature(air,h=h[2]) T[4]=temperature(air,h=h[4]) s[2]=entropy(air,T=T[2],P=P[2]) s
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
9-84 A simple Brayton cycle with air as the working fluid has a pressure ratio of 10. The air temperature at the turbine exit, the net work output, and the thermal efficiency are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2).
Analysis (a) Using the compressor and turbine efficiency relations, ( )
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9-61
9-85 A simple ideal Brayton cycle with helium has a pressure ratio of 14. The power output is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Helium is an ideal gas with constant specific heats.
Properties The properties of helium are cp = 5.1926 kJ/kg·K and k = 1.667 (Table A-2a).
Analysis Using the isentropic relations for an ideal gas,
s
T
1
2
4
3 qin
qout
973 K
288 K
K 9.827K)(14) 288( 70.667/1.66/)1(1
/)1(
1
212T ===⎟⎟
⎠
⎞⎜⎜⎝
⎛= −
−kk
p
kk
rTPP
T
Similarly,
K 5.33814
K) 973(33
34 =⎟⎠
⎜⎝
=⎟⎠
⎜⎝
=⎟⎟⎠
⎜⎜⎝
=pr
TP
TT 11 70.667/1.66/)1(/)1(4 ⎞⎛⎟
⎞⎜⎛⎞⎛
−− kkkkP
t law to the constant-pressure heat addition process 2-3 produces
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9-62
9-86 A simple Brayton cycle with helium has a pressure ratio of 14. The power output is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Helium is an ideal gas with constant specific heats.
Properties The properties of helium are cp = 5.1926 kJ/kg·K and k = 1.667 (Table A-2a).
.19265()( 23in =−= TTcq p kJ/kg 0.606K)3.856973)(kJ/kg·K
S
kJ/kg 2.262K)2885.338)(kJ/kg·K 1926.5()( 14out =−=−= TTcq p
T orhe net w k production is then
and
kJ/kg 8.3432.2620.606outinnet =−=−= qqw
kW 287=⎟⎠⎞
⎜⎝⎛==
s 60min 1kJ/kg) 8.343(kg/min) 50(netnet wmW &&
preparation. If you are a student using this Manual, you are using it without permission.
9-63
9-87 A simple Brayton cycle with air as the working fluid operates between the specified temperature and pressure limits. The effects of non-isentropic compressor and turbine on the back-work ratio is to be compared.
Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2a).
The back work ratio for 90% efficient turbine and isentropic compressor case is
0.7456===kJ/kg 401.4kJ/kg 299.3
Turb
Comp,bw W
Wr s
The two results are almost identical.
preparation. If you are a student using this Manual, you are using it without permission.
9-64
9-88 A gas turbine power plant that operates on the simple Brayton cycle with air as the working fluid has a specified pressure ratio. The required mass flow rate of air is to be determined for two cases.
Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2).
preparation. If you are a student using this Manual, you are using it without permission.
9-65
9-89 An actual gas-turbine power plant operates at specified conditions. The fraction of the turbine work output used to drive the compressor and the thermal efficiency are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with variable specific heats.
Properties The properties of air are given in Table A-17.
preparation. If you are a student using this Manual, you are using it without permission.
9-66
9-90 A gas-turbine power plant operates at specified conditions. The fraction of the turbine work output used to drive the compressor and the thermal efficiency are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats.
preparation. If you are a student using this Manual, you are using it without permission.
9-67
9-91 An aircraft engine operates as a simple ideal Brayton cycle with air as the working fluid. The pressure ratio and the rate of heat input are given. The net power and the thermal efficiency are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2a).
preparation. If you are a student using this Manual, you are using it without permission.
9-68
9-92 An aircraft engine operates as a simple ideal Brayton cycle with air as the working fluid. The pressure ratio and the rate of heat input are given. The net power and the thermal efficiency are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2a).
preparation. If you are a student using this Manual, you are using it without permission.
9-69
9-93 A gas-turbine plant operates on the simple Brayton cycle. The net power output, the back work ratio, and the thermal efficiency are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The gas constant of air is R = 0.287 kJ/kg·K (Table A-1). Analysis (a) For this problem, we use the properties from EES software. Remember that for an ideal gas, enthalpy is a function of temperature only whereas entropy is functions of both temperature and pressure.
. However, using the following lines in EES together with the isentropic J/kg, T3 = 1421ºC, s3 = 6.736 kJ/kg.K. The solution by hand would require a trial-
error approach. h_3=enthalpy(Air, T=T_3)
=T_3, P=P_2) P_1, s=s_3)
The mass flow rate is determined from
hh 43 −
We cannot find the enthalpy at state 3 directlyefficiency relation, we find h3 = 1873 k
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9-70
9-94 A simple Brayton cycle with air as the working fluid operates between the specified temperature and pressure limits. The cycle is to be sketched on the T-s cycle and the isentropic efficiency of the turbine and the cycle thermal efficiency are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats.
Properties The properties of air are given as cv = 0.718 kJ/kg·K, cp = 1.005 kJ/kg·K, R = 0.287 kJ/kg·K, k = 1.4.
preparation. If you are a student using this Manual, you are using it without permission.
9-71
9-95 A modified Brayton cycle with air as the working fluid operates at a specified pressure ratio. The T-s diagram is to be sketched and the temperature and pressure at the exit of the high-pressure turbine and the mass flow rate of air are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats.
Properties The properties of air are given as cv = 0.718 kJ/kg·K, cp = 1.005 kJ/kg·K, R = 0.287 kJ/kg·K, k = 1.4.
Analysis (b) For the compression process,
K 5.494K)(8) 273( 0.4/1.4/)1(
1
212 ==⎟⎟
⎠
⎞⎜⎜⎝
⎛=
− kk
PP
TT
s
T
1
4
5
3
P4
1500 K
273 K
2
P5
P3 = P2
The power input to the compressor is equal to the power output from the high-pressure turbine. Then,
K 1278.5=−+=−+=−=−
−=− )( 12 cmTTcm p &&
=
5.4942731500
)(
2134
4312
43
outTurb, HPinComp,
TTTTTTTT
TT
WW
p
&&
The pressure at this state is
kPa 457.3=⎟⎠⎝⎟
⎠⎜⎝
⎟⎠
⎜⎝ 3
1433 K 1500TTP
(c) The temperature at state 5 is determined f
⎞⎜⎛=⎟
⎞⎜⎛
=⎯→⎯⎟⎞
⎜⎛
=−− 4.0/4.1)1/(
4)1/(
4 K 5.1278kPa) 100(8kkkk
TrPP
TP
rom
4
K 1.828kPa 3.457
kPa 100K) 5.1278(0.4/1.4/)1(
4
545 =⎟
⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
− kk
PP
TT
The net power is that generated by the low-pressure turbine since the power output from the high-pressure turbine is equal the power input to the compressor. Then,
to
kg/s 441.8=−⋅
=−
=
−=
K)1.8285.1278)(KkJ/kg 1.005(kW 000,200
)(
)(
54
Turb LP
54Turb LP
TTcW
m
TTcmW
p
p
&&
&&
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
9-96 A simple Brayton cycle with air as the working fluid operates at a specified pressure ratio and between the specified temperature and pressure limits. The cycle is to be sketched on the T-s cycle and the volume flow rate of the air into the compressor is to be determined. Also, the effect of compressor inlet temperature on the mass flow rate and the net power output are to be investigated.
Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats.
Properties The properties of air are given as cv = 0.718 kJ/kg·K, cp = 1.005 kJ/kg·K, R = 0.287 kJ/kg·K, k = 1.4.
Given the net power, the mass flow rate is determined from
)( 4343sTurb,Turb −− TTTTcmW ssp&&
.0
[ ]
[ ]
[ ]kg/s 7.63
)2738.526()3.9241500()KkJ/kg 1.005(
)( 123
net
=−−−⋅
=
−=
TTTcW
mp
&&
4
kW 000,150
()
)()(
()(
4
1243
1243CompTurbnet
−−
−−−=
−−−=−=
T
TTTTcmW
TTcmTTcmWWW
p
pp
&
&&&&&
The specific volume and the volume flow rate at the inlet of the compressor are
)
net&
/kgm 7835.0kPa 100
K) 273)(KkJ/kg 287.0( 3
1
11 =
⋅==
PRT
v
(c) For a fixed compressor inlet velocity and flow area, when the compressor inlet temperature increases, the specific
volume increases since
3m 342.2=== )/kgm 7835.0)(kg/s 7.436( 311 vV m&&
PRT
=v . When specific volume increases, the mass flow rate decreases since vV&
& =m . Note that
volume flow rate is the same since inlet velocity and flow area are fixed ( ). When mass flow rate decreases, the net power decreases since . Therefore, when the inlet temperature increases, both mass flow rate and
the net power decrease.
AV=V&
)( CompTurbnet wwmW −= &&
preparation. If you are a student using this Manual, you are using it without permission.
9-73
Brayton Cycle with Regeneration
9-97C Regeneration increases the thermal efficiency of a Brayton cycle by capturing some of the waste heat from the exhaust gases and preheating the air before it enters the combustion chamber.
9-98C Yes. At very high compression ratios, the gas temperature at the turbine exit may be lower than the temperature at the compressor exit. Therefore, if these two streams are brought into thermal contact in a regenerator, heat will flow to the exhaust gases instead of from the exhaust gases. As a result, the thermal efficiency will decrease.
9-99C The extent to which a regenerator approaches an ideal regenerator is called the effectiveness ε, and is defined as ε = qregen, act /qregen, max.
9-100C (b) turbine exit.
9-101C The steam injected increases the mass flow rate through the turbine and thus the power output. This, in turn, increases the thermal efficiency since in/ QW=η and W increases while Qin remains constant. Steam can be obtained by utilizing the hot exhaust gases.
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
9-102 A Brayton cycle with regeneration produces 150 kW power. The rates of heat addition and rejection are to be determined.
Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K and k = 1.4 (Table A-2a).
Analysis According to the isentropic process expressions for an ideal gas,
preparation. If you are a student using this Manual, you are using it without permission.
9-75
9-103 A Brayton cycle with regeneration produces 150 kW power. The rates of heat addition and rejection are to be determined.
Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K and k = 1.4 (Table A-2a).
Analysis For the compression and expansion processes we have
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9-76
9-104 A Brayton cycle with regeneration is considered. The thermal efficiencies of the cycle for parallel-flow and counter-flow arrangements of the regenerator are to be compared.
Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg⋅K and k = 1.4 (Table A-2a).
Analysis According to the isentropic process expressions for an ideal gas,
When the first law is applied to the heat exchanger as originally arranged, the result is
6523 TTTT −=−
while the regenerator temperature specification gives
=−=−= TT
he simul ution of these two results gives
K 5.56765.573653
T taneous sol
5.5675.5732356 +−=+−= TTTT K 9.5169.510 =
The thermal efficiency of the cycle is then
0.482=−
−−=
− 34in TTq−
−=−=5.5671000
293516.9111 16outth
TTqη
or the rearranged version of this cycle,
n energy balance on the eat exchanger gives
he soluti hese two equations is
6 =T
The thermal efficiency of the cycle is then
F
663 −= TT
s
T
1
2 5
4qin1000 K
293 K
3 6
qout
A h
6523 TTTT −=−
T on of t
K 2.5393 =T
K 2.545
0.453=−
−−=
−−
−=−=2.5391000
293545.211134
16
in
outth TT
TTqq
η
preparation. If you are a student using this Manual, you are using it without permission.
9-77
9-105 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 70 kW are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with variable specific heats. 3 The ambient air is 300 K and 100 kPa. 4 The effectiveness of the regenerator is 0.9, and the isentropic efficiencies for both the compressor and the turbine are 80%. 5 The combustion gases can be treated as air.
Properties The properties of air at the compressor and turbine inlet temperatures can be obtained from Table A-17.
Analysis The gas turbine cycle with regeneration can be analyzed as follows:
Finally, the mass flow rate of air through the turbine becomes
kg/s 0.499===kJ/kg 140.34
kW 70
net
netair w
Wm
&&
preparation. If you are a student using this Manual, you are using it without permission.
9-78
9-106 The thermal efficiency and power output of an actual gas turbine are given. The isentropic efficiency of the
1 Air is an ideal gas with variable specific heats. 2 Kinetic and potential energy changes are negligible. 3 The same, and the properties of combustion gases are the same as
e A-17.
Analysis The properties at various states are
turbine and of the compressor, and the thermal efficiency of the gas turbine modified with a regenerator are to be determined.
Assumptionsmass flow rates of air and of the combustion gases are the those of air.
Properties The properties of air are given in Tabl
Then the compressor and turbine efficiencies become
94.9%
83.8%
==−
−=
−−
=
==−−
=−−
=
949.021.303672
21.30325.653
838.023.825171055.9681710
12
12
43
43
hhhh
hhhh
sC
sT
η
η
When a regenerator is added, the new heat input and the thermal efficiency become
44.1%====
−=−=
−=
441.0kJ/kg 845.2kJ/kg 372.66
kJ/kg 845.2=8.1921038
kJ/kg 192.8=672.0)-.55(0.65)(968=)(
newin,
netnewth,
regeninnewin,
24regen
qw
qqq
hhq
η
ε
Discussion Note a 65% efficient regenerator would increase the thermal efficiency of this gas turbine from 35.9% to 44.1%.
preparation. If you are a student using this Manual, you are using it without permission.
9-79
9-107 Problem 9-106 is reconsidered. A solution that allows different isentropic efficiencies for the compressor and turbine is to be developed and the effect of the isentropic efficiencies on net work done and the heat supplied to the cycle is to be studied. Also, the T-s diagram for the cycle is to be plotted.
Analysis Using EES, the problem is solved as follows:
"Input data" T[3] = 1288 [C] Pratio = 14.7 T[1] = 30 [C] P[1]= 100 [kPa] {T[4]=659 [C]} {W_dot_net=159 [MW] }"We omit the information about the cycle net work" m_dot = 1536000 [kg/h]*Convert(kg/h,kg/s) {Eta_th_noreg=0.359} "We omit the information about the cycle efficiency." Eta_reg = 0.65 Eta_c = 0.84 "Compressor isentropic efficiency" Eta_t = 0.95 "Turbien isentropic efficiency" "Isentropic Compressor anaysis" s[1]=ENTROPY(Air,T=T[1],P=P[1]) s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor" P[2] = Pratio*P[1] s_s[2]=ENTROPY(Air,T=T_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exit" Eta_c = W_dot_compisen/W_dot_comp "compressor adiabatic efficiency, W_dot_comp > W_dot_compisen" "Conservation of energy for the compressor for the isentropic case: E_dot_in - E_dot_out = DELTAE_dot=0 for steady-flow" m_dot*h[1] + W_dot_compisen = m_dot*h_s[2] h[1]=ENTHALPY(Air,T=T[1]) h_s[2]=ENTHALPY(Air,T=T_s[2]) "Actual compressor analysis:" m_dot*h[1] + W_dot_comp = m_dot*h[2] h[2]=ENTHALPY(Air,T=T[2]) s[2]=ENTROPY(Air,T=T[2], P=P[2]) "External heat exchanger analysis" "SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0 E_dot_in - E_dot_out =DELTAE_dot_cv =0 for steady flow" m_dot*h[2] + Q_dot_in_noreg = m_dot*h[3] q_in_noreg=Q_dot_in_noreg/m_dot h[3]=ENTHALPY(Air,T=T[3]) P[3]=P[2]"process 2-3 is SSSF constant pressure" "Turbine analysis" s[3]=ENTROPY(Air,T=T[3],P=P[3]) s_s[4]=s[3] "For the ideal case the entropies are constant across the turbine" P[4] = P[3] /Pratio s_s[4]=ENTROPY(Air,T=T_s[4],P=P[4])"T_s[4] is the isentropic value of T[4] at turbine exit" Eta_t = W_dot_turb /W_dot_turbisen "turbine adiabatic efficiency, W_dot_turbisen > W_dot_turb" "SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0 E_dot_in -E_dot_out = DELTAE_dot_cv = 0 for steady-flow" m_dot*h[3] = W_dot_turbisen + m_dot*h_s[4] h_s[4]=ENTHALPY(Air,T=T_s[4])
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
"Actual Turbine analysis:" m_dot*h[3] = W_dot_turb + m_dot*h[4] h[4]=ENTHALPY(Air,T=T[4]) s[4]=ENTROPY(Air,T=T[4], P=P[4]) "Cycle analysis" "Using the definition of the net cycle work and 1 MW = 1000 kW:" W_dot_net*1000=W_dot_turb-W_dot_comp "kJ/s" Eta_th_noreg=W_dot_net*1000/Q_dot_in_noreg"Cycle thermal efficiency" Bwr=W_dot_comp/W_dot_turb"Back work ratio" "With the regenerator the heat added in the external heat exchanger is" m_dot*h[5] + Q_dot_in_withreg = m_dot*h[3] q_in_withreg=Q_dot_in_withreg/m_dot h[5]=ENTHALPY(Air, T=T[5]) s[5]=ENTROPY(Air,T=T[5], P=P[5]) P[5]=P[2] "The regenerator effectiveness gives h[5] and thus T[5] as:" Eta_reg = (h[5]-h[2])/(h[4]-h[2]) "Energy balance on regenerator gives h[6] and thus T[6] as:" m_dot*h[2] + m_dot*h[4]=m_dot*h[5] + m_dot*h[6] h[6]=ENTHALPY(Air, T=T[6]) s[6]=ENTROPY(Air,T=T[6], P=P[6]) P[6]=P[4] "Cycle thermal efficiency with regenerator" Eta_th_withreg=W_dot_net*1000/Q_dot_in_withreg "The following data is used to complete the Array Table for plotting purposes." s_s[1]=s[1] T_s[1]=T[1] s_s[3]=s[3] T_s[3]=T[3] s_s[5]=ENTROPY(Air,T=T[5],P=P[5]) T_s[5]=T[5] s_s[6]=s[6] T_s[6]=T[6]
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
9-108 A Brayton cycle with regeneration using air as the working fluid is considered. The air temperature at the turbine exit, the net work output, and the thermal efficiency are to be determined.
Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible.
Properties The properties of air are given in Table A-17.
Analysis (a) The properties of air at various states are
( )( )
( ) ( ) ( )
( )
( ) ( )( ) kJ/kg 803.14711.801219.250.821219.25
kJ/kg 711.8059.2815.20071
15.200kJ/kg 1219.25
K 1501
kJ/kg 618.260.75/310.24541.2610.243/
kJ/kg 541.2688.105546.17
5546.1kJ/kg 310.24
K 310
433443
43
43
4
33
121212
12
21
2 == PP
P
1
34
3
12
1
=−−=−−=⎯→⎯−−
=
=⎯→⎯=⎟⎠⎞
⎜⎝⎛==
==
⎯→⎯=
=−+=−+=⎯→⎯−−
=
=⎯→⎯=
==
⎯→⎯=
sTs
T
srr
r
Css
C
srr
r
hhhhhhhh
hPPP
P
Ph
hhhhhhhh
hP
Ph
ηη
ηη
Thus,
T4 = 782.8 K
(b)
)
s
T
1
2s4s
3qin1150 K
310 K
5
6
42
1T
T
( ) ( )( ) (
kJ/kg 108.09=−−−=
−−−=−=
24.31026.61814.80325.12191243inC,outT,net hhhhwww
( )( )(
kJ/kg 738.43618.26803.140.6526.618
242524
25
=−+=
−+=⎯→⎯−−
= hhhhhhhh
εε (c) )
Then,
22.5%===
=−=−=
kJ/kg 480.82kJ/kg 108.09
kJ/kg 480.82738.4319.2512
in
netth
53in
qw
hhq
η
preparation. If you are a student using this Manual, you are using it without permission.
9-83
9-109 A stationary gas-turbine power plant operating on an ideal regenerative Brayton cycle with air as the working fluid is considered. The power delivered by this plant is to be determined for two cases.
Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas. 3 Kinetic and potential energy changes are negligible.
Properties When assuming constant specific heats, the properties of air at room temperature are cp = 1.005 kJ/kg.K and k = 1.4 (Table A-2a). When assuming variable specific heats, the properties of air are obtained from Table A-17.
preparation. If you are a student using this Manual, you are using it without permission.
9-84
9-110 A regenerative gas-turbine engine using air as the working fluid is considered. The amount of heat transfer in the regenerator and the thermal efficiency are to be determined.
Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible.
Properties The properties of air are given in Table A-17.
preparation. If you are a student using this Manual, you are using it without permission.
9-85
9-111 A regenerative gas-turbine engine using air as the working fluid is considered. The amount of heat transfer in the regenerator and the thermal efficiency are to be determined.
Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats. 3 Kinetic and potential energy changes are negligible.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K and k = 1.4 (Table A-2a).
Analysis (a) Using the isentropic relations and turbine efficiency,
( )( )
( )( ) ( )
( )( )
( ) ( ) ( )( )( ) kJ/kg 130.7=−⋅=−=−==
−−=−−=⎯→⎯
−
−=
−−
=
=⎟⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
===
−
K065812.6KkJ/kg 1.0050.80K .6812
3.747140090.01400
K .374791K 1400
9100/900/
2424regen
433443
43
43
43
4.1/4.0/1
3
434
12
TTchhq
TTTTTTcTTc
hhhh
PP
TT
PPr
p
sTsp
p
sT
kk
s
p
εε
ηη s
T
1
2s 4s
3qin1400 K
310 K
5
6
4650 K 2
( ) ( )( ) ( ) ( )[ ]( ) ( )( )( )
39.9%====
=−−⋅=
−−=−−=
=−−−⋅=
−−−=−=
399.0kJ/kg 623.1kJ/kg 248.7
kJ/kg .1623.7130K0651400KkJ/kg 1.005
kJ/kg .7248K310650.68121400KkJ/kg 1.005
in
netth
regen23regen23in
1243inC,outT,net
qw
qTTcqhhq
TTcTTcwww
p
pp
η
(b)
preparation. If you are a student using this Manual, you are using it without permission.
9-86
9-112 A regenerative gas-turbine engine using air as the working fluid is considered. The amount of heat transfer in the regenerator and the thermal efficiency are to be determined.
Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible.
Properties The properties of air are given in Table A-17.
preparation. If you are a student using this Manual, you are using it without permission.
9-87
9-113 An expression for the thermal efficiency of an ideal Brayton cycle with an ideal regenerator is to be developed.
Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible.
Analysis The expressions for the isentropic compression and expansion processes are
preparation. If you are a student using this Manual, you are using it without permission.
9-88
Brayton Cycle with Intercooling, Reheating, and Regeneration
9-114C As the number of compression and expansion stages are increased and regeneration is employed, the ideal Brayton cycle will approach the Ericsson cycle.
9-115C 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.
9-116C (a) decrease, (b) decrease, and (c) decrease.
9-117C (a) increase, (b) decrease, and (c) decrease.
9-118C (a) increase, (b) decrease, (c) decrease, and (d) increase.
9-119C (a) increase, (b) decrease, (c) increase, and (d) decrease.
9-120C (c) The Carnot (or Ericsson) cycle efficiency.
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
9-121 An ideal gas-turbine cycle with two stages of compression and two stages of expansion is considered. The back work ratio and the thermal efficiency of the cycle are to be determined for the cases of with and without a regenerator.
Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible.
s
T
3
4
1
5qin1200 K
300 K
86
7
10
9
2
Properties The properties of air are given in Table A-17.
Analysis (a) The work inputs to each stage of compressor are identical, so are the work outputs of each stage of the turbine since this is an ideal cycle. Then,
( )( )
( )
( ) ( )( ) ( ) kJ/kg 62.86636.94679.127722
kJ/kg 2.142219.30026.41122
kJ/kg 946.3633.7923831
5
⎞⎛P
238kJ/kg 77.7912
K 2001
kJ/kg 411.26158.4386.13
386.1kJ/kg 300.19
K 300
65outT,
12inC,
865
6
755
421
2
11
56
12
1
=−=−=
=−=−=
==⎯→⎯=⎟⎠
⎜⎝
==
===
⎯→⎯=
==⎯→⎯===
==
⎯→⎯=
hhw
hhw
hhPP
P
Phh
T
hhPPP
P
Ph
T
rr
r
rr
r
Thus,
33.5%===kJ/kg 662.86kJ/kg 222.14
outT,
inC,bw w
wr
( ) ( ) ( ) ( )
36.8%===
=−=−=
=−+−=−+−=
kJ/kg 1197.96kJ/kg 440.72
kJ/kg 440.72222.1486.662
kJ/kg 1197.9636.94679.127726.41179.1277
in
netth
inC,outT,net
6745in
qw
www
hhhhq
η
(b) When a regenerator is used, rbw remains the same. The thermal efficiency in this case becomes
( ) ( )( )
55.3%===
=−=−=
=−=−=
kJ/kg 796.63kJ/kg 440.72
kJ/kg 796.6333.40196.1197
kJ/kg 33.40126.41136.94675.0
in
netth
regenoldin,in
48regen
qw
qqq
hhq
η
ε
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
9-122 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,
(b) When a regenerator is used, rbw remains the same. The thermal efficiency in this case becomes
( ) ( )( )
44.2%====
=−=−=
=−=−=
442.0kJ/kg 721.75kJ/kg 318.86
kJ/kg .7572128.41503.1137
kJ/kg 15.28442.43213.98675.0
in
netth
regenoldin,in
48regen
qw
qqq
hhq
η
ε
preparation. If you are a student using this Manual, you are using it without permission.
9-91
9-123 An ideal regenerative gas-turbine cycle with two stages of compression and two stages of expansion is considered. The power produced and consumed by each compression and expansion stage, and the rate of heat rejected are to be determined.
Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg⋅K and k = 1.4 (Table A-2a).
Analysis The pressure ratio for each stage is
464.312 ==pr
s
T
3
4
1
6
288 K
97
8
2
5
1
According to the isentropic process expressions for an ideal gas,
pplication of the first law to the expansion process 6-7 gives
K )7.438K)(625.7kJ/kg 5kg/s)(1.00 331.1(
he same amount of power is produced in process 8-9. When e first law is adapted to the compression process 1-2 it becomes
K )288K)(410.7kJ/kg 5kg/s)(1.00 331.1(
Compression process 3-4 uses the same amount of power. The rate of heat rejection from the cycle is
A
−= )( 76out7,-6 TTcmW p&&
kW 250.1=−⋅=
T th
−= )( 1212,in TTcmW p&&
kW 164.1=
−⋅=
kW 328.3=−⋅=
−=
K )288K)(410.7kJ/kg 5kg/s)(1.00 331.1(2
)(2 32out TTcmQ p&&
preparation. If you are a student using this Manual, you are using it without permission.
9-92
9-124 A regenerative gas-turbine cycle with two stages of compression and two stages of expansion is considered. The thermal efficiency of the cycle is to be determined.
Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg⋅K and k = 1.4 (Table A-2a).
Analysis The temperatures at various states are obtained as follows
preparation. If you are a student using this Manual, you are using it without permission.
9-93
9-125 A regenerative gas-turbine cycle with three stages of compression and three stages of expansion is considered. The thermal efficiency of the cycle is to be determined.
Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg⋅K and k = 1.4 (Table A-2a).
Analysis The temperatures at various states are obtained as follows
preparation. If you are a student using this Manual, you are using it without permission.
9-94
9-126 A regenerative gas-turbine cycle with three stages of compression and three stages of expansion is considered. The thermal efficiency of the cycle is to be determined.
Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg⋅K and k = 1.4 (Table A-2a).
Analysis Since all compressors share the same compression ratio and begin at the same temperature,
preparation. If you are a student using this Manual, you are using it without permission.
9-95
Jet-Propulsion Cycles
9-127C The power developed from the thrust of the engine is called the propulsive power. It is equal to thrust times the aircraft velocity.
9-128C The ratio of the propulsive power developed and the rate of heat input is called the propulsive efficiency. It is determined by calculating these two quantities separately, and taking their ratio.
9-129C It reduces the exit velocity, and thus the thrust.
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
9-130 A turboprop engine operating on an ideal cycle is considered. The thrust force generated is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with constant specific heats at room temperature. 4 The turbine work output is equal to the compressor work input.
Properties The properties of air at room temperature are R = 0.287 kPa⋅m3/kg⋅K (Table A-1), cp = 1.005 kJ/kg⋅K and k = 1.4 (Table A-2a).
Analysis Working across the two isentropic processes of the cycle yields
Since the work produced by expansion 3-4 equals that used by compression 1-2, an energy balance gives
K 3.545)2507.482(778)( 1234 =−−=−−= TTTT
The excess enthalpy generated by expansion 4-5 is used to increase the kinetic energy of the flow through the propeller,
2
)(2
inlet2
exit54
VVmTTcm ppe
−=− &&
ich when solved for the velocity at which the air leaves the propeller gives wh
m/s 1.216
)m/s 180(kJ/kg 1
/sm 1000K)0.4033.545)(KkJ/kg 005.1(2012
)(21
2 ⎥⎤
⎢⎡
+−= VTTcm
V e&
2/12
22
2/
inlet54exit
=
⎥⎥⎦
⎤
⎢⎢⎣
⎡+
⎠⎜⎜⎝
⎛−⋅=
⎥⎦⎢⎣ m pp&
⎟⎟⎞
The mass flow rate through the propeller is
kg/s 0.975/kgm 305.1
m/s 1804m) 3(
4
/kgm 305.1kPa 55
K) 250()mkPa
287.0(RT
3
2
1
12
1
1
33
1
====
=⋅
==
ππvv
v
VDAVm
P
p&
The thrust force generated by this propeller is then
kN 35.2=⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=−=
2inletexitm/skg 1000
kN 10)m/s181kg/s)(216. (975.0)( VVmF p&
preparation. If you are a student using this Manual, you are using it without permission.
9-97
9-131 A turboprop engine operating on an ideal cycle is considered. The thrust force generated is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with constant specific heats at room temperature. 4 The turbine work output is equal to the compressor work input.
Properties The properties of air at room temperature are R = 0.287 kPa⋅m3/kg⋅K (Table A-1), cp = 1.005 kJ/kg⋅K and k = 1.4 (Table A-2a).
Analysis Working across the two isentropic processes of the cycle yields
The thrust force generated by this propeller is then
ex
kN 33.7=⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=−=
2inletexitm/skg 1000
kN 10)m/s180kg/s)(234. (624.0)( VVmF p&
preparation. If you are a student using this Manual, you are using it without permission.
9-98
9-132 A turbofan engine operating on an ideal cycle produces 50,000 N of thrust. The air temperature at the fan outlet needed to produce this thrust is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with constant specific heats at room temperature. 4 The turbine work output is equal to the compressor work input.
Properties The properties of air at room temperature are R = 0.287 kPa⋅m3/kg⋅K, cp = 1.005 kJ/kg⋅K and k = 1.4 (Table A-2a).
Analysis The total mass flow rate is
kg/s 1.676
/kgm 452.1m/s 200
4m) 5.2(
4
/kgm 452.1kPa 50
K) 253()mkPa 287.0(
3
2
1
12
1
1
33
1
====
=⋅
==
ππvv
v
VDAVm
PRT
&
s
T
1
2 4
3qin
5
qoutNow,
kg/s 51.8488e
kg/s 1.676===
mm&
&
he mass flow rate through the fan is
order to roduce the specified thrust force, the velocity at the fan exit will be
T
kg/s 6.59151.841.676 =−=−= ef mmm &&&
In p
m/s 284.5N 1m/skg 1
kg/s 591.6N 50,000m/s) (200
)( inletexit −= f VVmF &
2
inletexit =⎟⎟⎠
⎞⎜⎜⎝
⎛ ⋅+=+=
fmFVV&
balance on the stream passing through the fan gives
An energy
K 232.6=
⎟⎠
⎞⎜⎝
⎛⋅
−−=
−−=
−=−
22
22
2inlet
2exit
45
2inlet
2exit
54
/sm 1000kJ/kg 1
)KkJ/kg 005.1(2)m/s 200()m/s 5.284(K 253
2
2)(
p
p
cVV
TT
VVTTc
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
9-133 A pure jet engine operating on an ideal cycle is considered. The velocity at the nozzle exit and the thrust produced are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with constant specific heats at room temperature. 4 The turbine work output is equal to the compressor work input. Properties The properties of air at room temperature are R = 0.287 kPa⋅m3/kg⋅K, cp = 1.005 kJ/kg⋅K and k = 1.4 (Table A-2a). Analysis (a) We assume the aircraft is stationary and the air is moving towards the aircraft at a velocity of V 1 = 240 m/s. Ideally, the air will leave the diffuser with a negligible velocity (V 2 ≅ 0).
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9-100
9-134 A turbojet aircraft flying at an altitude of 9150 m is operating on the ideal jet propulsion cycle. The velocity of exhaust gases, the propulsive power developed, and the rate of fuel consumption are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with constant specific heats at room temperature. 4 Kinetic and potential energies are negligible, except at the diffuser inlet and the nozzle exit. 5 The turbine work output is equal to the compressor work input. Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K and k = 1.4 (Table A-2a). Analysis (a) We assume the aircraft is stationary and the air is moving towards the aircraft at a velocity of V 1 = 320 m/s. Ideally, the air will leave the diffuser with a negligible velocity (V 2 ≅ 0).
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9-101
9-135 A turbojet aircraft is flying at an altitude of 9150 m. The velocity of exhaust gases, the propulsive power developed, and the rate of fuel consumption are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with constant specific heats at room temperature. 4 Kinetic and potential energies are negligible, except at the diffuser inlet and the nozzle exit.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K and k = 1.4 (Table A-2a).
Analysis (a) For convenience, we assume the aircraft is stationary and the air is moving towards the aircraft at a velocity of V 1 = 320 m/s. Ideally, the air will leave the diffuser with a negligible velocity (V 2 ≅ 0).
9-136 A turbojet aircraft that has a pressure rate of 9 is stationary on the ground. The force that must be applied on the brakes to hold the plane stationary is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with variable specific heats. 4 Kinetic and potential energies are negligible, except at the nozzle exit.
Properties The properties of air are given in Table A-17.
Analysis (a) Using variable specific heats for air,
2inletexit m/skg 1N 1m/s0968.6kg/s 20VVm& Brake force = Thrust =
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9-104
9-137 Problem 9-136 is reconsidered. The effect of compressor inlet temperature on the force that must be applied to the brakes to hold the plane stationary is to be investigated.
Analysis Using EES, the problem is solved as follows:
P_ratio =9 T_1 = 7 [C] T[1] = T_1+273 "[K]" P[1]= 95 [kPa] P[5]=P[1] Vel[1]=0 [m/s] V_dot[1] = 18.1 [m^3/s] HV_fuel = 42700 [kJ/kg] m_dot_fuel = 0.5 [kg/s] Eta_c = 1.0 Eta_t = 1.0 Eta_N = 1.0 "Inlet conditions" h[1]=ENTHALPY(Air,T=T[1]) s[1]=ENTROPY(Air,T=T[1],P=P[1]) v[1]=volume(Air,T=T[1],P=P[1]) m_dot = V_dot[1]/v[1] "Compressor anaysis" s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor" P_ratio=P[2]/P[1]"Definition of pressure ratio - to find P[2]" T_s[2]=TEMPERATURE(Air,s=s_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exit" h_s[2]=ENTHALPY(Air,T=T_s[2]) Eta_c =(h_s[2]-h[1])/(h[2]-h[1]) "Compressor adiabatic efficiency; Eta_c = W_dot_c_ideal/W_dot_c_actual. " m_dot*h[1] +W_dot_c=m_dot*h[2] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" "External heat exchanger analysis" P[3]=P[2]"process 2-3 is SSSF constant pressure" h[3]=ENTHALPY(Air,T=T[3]) Q_dot_in = m_dot_fuel*HV_fuel m_dot*h[2] + Q_dot_in= m_dot*h[3]"SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0" "Turbine analysis" s[3]=ENTROPY(Air,T=T[3],P=P[3]) s_s[4]=s[3] "For the ideal case the entropies are constant across the turbine" {P_ratio= P[3] /P[4]} T_s[4]=TEMPERATURE(Air,h=h_s[4]) "Ts[4] is the isentropic value of T[4] at turbine exit" {h_s[4]=ENTHALPY(Air,T=T_s[4])} "Eta_t = W_dot_t /Wts_dot turbine adiabatic efficiency, Wts_dot > W_dot_t" Eta_t=(h[3]-h[4])/(h[3]-h_s[4]) m_dot*h[3] = W_dot_t + m_dot*h[4] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" T[4]=TEMPERATURE(Air,h=h[4]) P[4]=pressure(Air,s=s_s[4],h=h_s[4]) "Cycle analysis" W_dot_net=W_dot_t-W_dot_c"Definition of the net cycle work, kW" W_dot_net = 0 [kW] "Exit nozzle analysis:" s[4]=entropy('air',T=T[4],P=P[4]) s_s[5]=s[4] "For the ideal case the entropies are constant across the nozzle" T_s[5]=TEMPERATURE(Air,s=s_s[5], P=P[5]) "T_s[5] is the isentropic value of T[5] at nozzle exit" h_s[5]=ENTHALPY(Air,T=T_s[5]) Eta_N=(h[4]-h[5])/(h[4]-h_s[5]) m_dot*h[4] = m_dot*(h_s[5] + Vel_s[5]^2/2*convert(m^2/s^2,kJ/kg))
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
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])
9-138 Air enters a turbojet engine. The thrust produced by this turbojet engine is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with variable specific heats. 4 Kinetic and potential energies are negligible, except at the diffuser inlet and the nozzle exit.
Properties The properties of air are given in Table A-17.
Analysis We assume the aircraft is stationary and the air is moving towards the aircraft at a velocity of V 1 = 300 m/s. Taking the entire engine as our control volume and writing the steady-flow energy balance yield
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9-107
Second-Law Analysis of Gas Power Cycles
9-139 The process with the highest exergy destruction for an ideal Otto cycle described in Prob. 9-33 is to be determined.
Analysis From Prob. 9-33, qin = 582.5 kJ/kg, qout = 253.6 kJ/kg, T1 = 288 K, T2 = 661.7 K, T3 = 1473 K, and T4 = 641.2 K. The exergy destruction during a process of the cycle is
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9-109
9-141 The exergy loss of each process for an air-standard Stirling cycle described in Prob. 9-74 is to be determined.
Analysis From Prob. 9-74, qin = 1275 kJ/kg, qout = 212.5 kJ/kg, T1 = T2 = 1788 K, T3 = T4 = 298 K. The exergy destruction during a process of the cycle is
Application of this equation for each process of the cycle gives
KkJ/kg 7132.0)12ln()KkJ/kg 287.0(0
lnln1
2
1
212
⋅=⋅+=
+=−v
vv R
TT
css
0kJ/kg 0.034 ≈=⎟⎠
⎞⎜⎝
⎛ −⋅=
⎟⎟⎠
⎞⎜⎜⎝
⎛−−=
K 1788kJ/kg 1275
KkJ/kg 0.7132K) 298(
source
in1202-1 dest, T
qssTx
s
T
3
2 qin
qout
4
1
KkJ/kg 7132.012 ⎠⎝
1ln)KkJ/kg 287.0(0
lnln3
4
3
434
⋅−=⎟⎞
⎜⎛⋅+=
+=−v
vv R
TT
css
0kJ/kg 0.034 ≈−=⎟⎠
⎞⎜⎝
⎛ +⋅−=
⎟⎟⎠
⎞⎛ outq⎜⎜⎝
+−=
K 298kJ/kg 212.5
KkJ/kg 0.7132K) 298(
sink1202-1 dest, T
ssTx
These results are not surprising since Stirling cycle is totally reversible. Exergy destructions are not calculated for processes 2-3 and 4-1 because there is no interaction with the surroundings during these processes to alter the exergy destruction.
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9-110
9-142 The exergy destruction associated with each of the processes of the Brayton cycle described in Prob. 9-82 is to be determined.
Analysis From Prob. 9-82, qin = 698.3 kJ/kg, qout = 487.9 kJ/kg, and
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9-111
9-143 Exergy analysis is to be used to answer the question in Prob. 9-87.
Analysis From Prob. 9-87, T1 = 288 K, T2s = 585.8 K, T2 = 618.9 K, T3 = 873 K, T4s = 429.2 K, T4 = 473.6 K, rp = 12. The exergy change of a flow stream between an inlet and exit state is given by
)(0 ieie ssThh −−−=∆ψ
4s
s
T
1
2s
4
3 qin
qout
873 K
288 K
2
This is also the expression for reversible work. Application of this equation for isentropic and actual compression processes gives
kJ/kg 16.1=−=−=∆ −− 8.4299.44543 rev,43 rev,Trev, www s
Hence, it is clear that the compressor is a little more sensitive to the irreversibilities than the turbine.
preparation. If you are a student using this Manual, you are using it without permission.
9-112
9-144 The total exergy destruction associated with the Brayton cycle described in Prob. 9-108 and the exergy at the exhaust gases at the turbine exit are to be determined.
Properties The gas constant of air is R = 0.287 kJ/kg·K (Table A-1).
Analysis From Prob. 9-108, qin = 480.82, qout = 372.73 kJ/kg, and
s
T
1
2s 4s
3qin1150 K
310 K
5
6
42
KkJ/kg 08156.2kJ/kg 738.43
KkJ/kg 69407.2kJ/kg 14.803
KkJ/kg 12900.3K 1150
KkJ/kg 42763.2kJ/kg 26.618
KkJ/kg 1.73498K 310
55
44
33
22
11
⋅=⎯→⎯=
⋅=⎯→⎯=
⋅=⎯→⎯=
⋅=⎯→⎯=
⋅=⎯→⎯=
o
o
o
o
o
sh
sh
sT
sh
sT
and, from an energy balance on the heat exchanger,
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9-113
9-145 Prob. 9-144 is reconsidered. The effect of the cycle pressure on the total irreversibility for the cycle and the exergy of the exhaust gas leaving the regenerator is to be investigated.
Analysis Using EES, the problem is solved as follows:
9-146 The exergy loss of each process for a regenerative Brayton cycle with three stages of reheating and intercooling described in Prob. 9-126 is to be determined.
Application of this equation for each process of the cycle gives
0kJ/kg 0.03 ≈=⎥⎦⎤
⎢⎣⎡ −=
⎟⎟⎠
⎞⎜⎜⎝
⎛−===
)4ln()287.0(290
430.9(1.005)ln)290(
lnln1
2
1
206-5 dest,4-3 dest,2-1 dest, P
PR
TT
cTxxx p
kJ/kg 32.1=⎥⎦⎤
⎢⎣⎡ −−=⎟⎟
⎠
⎞⎜⎜⎝
⎛−−=
4.8703000
520.7819.2(1.005)ln)290(lnln
source
8-in,7
7
8
7
808-7 dest, T
qPP
RTT
cTx p
kJ/kg 26.2=⎥⎦⎤
⎢⎣⎡ −−=⎟⎟
⎠
⎞⎜⎜⎝
⎛−−=
4.8703000
551.3849.8(1.005)ln)290(lnln
source
10-in,9
7
8
9
10010-9 dest, T
qPP
RTT
cTx p
kJ/kg 22.5=⎥⎦⎤
⎢⎣⎡ −−=⎟⎟
⎠
⎞⎜⎜⎝
⎛−−=
4.8703000
571.9870.4(1.005)ln)290(lnln
source
12-in,11
11
12
11
12012-11 dest, T
qPP
RTT
cTx p
0kJ/kg 0.05 ≈−=⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛−=⎟⎟
⎠
⎞⎜⎜⎝
⎛−=
41ln)287.0(
819.2551.3(1.005)ln)290(lnln
8
9
8
9098- dest, P
PR
TT
cTx p
0kJ/kg 0.04 ≈−=⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛−=⎟⎟
⎠
⎞⎜⎜⎝
⎛−=
41ln)287.0(
849.8571.9(1.005)ln)290(lnln
10
11
10
11011-10 dest, P
PR
TT
cTx p
0kJ/kg 0.08 ≈−=⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛−=⎟⎟
⎠
⎞⎜⎜⎝
⎛−=
41ln)287.0(
870.4585.7(1.005)ln)290(lnln
12
13
12
13013-12 dest, P
PR
TT
cTx p
kJ/kg 50.6=⎥⎦⎤
⎢⎣⎡ +−=⎟⎟
⎠
⎞⎜⎜⎝
⎛+−=
2909.2060
495.9290(1.005)ln)290(lnln
sink
1-out,14
14
1
14
101-14 dest, T
qPP
RTT
cTx p
kJ/kg 26.2=⎥⎦⎤
⎢⎣⎡ +−=⎟⎟
⎠
⎞⎜⎜⎝
⎛+−==
2906.1410
430.9290(1.005)ln)290(lnln
sink
3-out,2
2
3
2
305-4 dest,3-2 dest, T
qPP
RTT
cTxx p
kJ/kg 6.66=⎥⎦⎤
⎢⎣⎡ +=
⎟⎟⎠
⎞⎜⎜⎝
⎛+=∆+∆= −−
585.7495.9(1.005)ln
430.9520.7(1.005)ln)290(
lnln)(13
14
6
701413760regendest, T
Tc
TT
cTssTx pp
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9-116
9-147 A gas-turbine plant uses diesel fuel and operates on simple Brayton cycle. The isentropic efficiency of the compressor, the net power output, the back work ratio, the thermal efficiency, and the second-law efficiency are to be determined.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at 500ºC = 773 K are cp = 1.093 kJ/kg·K, cv = 0.806 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.357 (Table A-2b). Analysis (a) The isentropic efficiency of the compressor may be determined if we first calculate the exit temperature for the isentropic case
( ) K 6.505kPa 100kPa 700K 303
1)/1.357-(1.357/)1(
1
212 =⎟
⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
− kk
s PP
TT
1
Combustion chamber
Turbine
23
4
Compress.
100 kPa30°C
Diesel fuel
700 kPa260°C
0.881=−
=−
=K)303533(12 TTCη −− K)3036.505(12 TT s
(b) The total mass flowing through the turbine and the rate of heat input are
kg/s 81.12kg/s 21.0kg/s 6.1260AF
kJ/kg)(0.9 00kg/s)(42,0 21.0(== qmQ η&&
kg/s 6.12kg/s 6.12 =+=+=+=+= aafa
mmmmm
&&&&&
HVin =cf
perature at the exit of combustion chamber is
3323 =⎯→⎯ Tp
The temperature at the turbine exit is determined using isentropic efficiency relation
t
kW 85557)
The tem
)K533kJ/kg.K)( 3kg/s)(1.09 81.12(kJ/s 8555)( −=⎯→⎯−= TTTcmQ && K 1144in
kW 5455)K4.754144kJ/kg.K)(1 3kg/s)(1.09 81.12()( 43outT, p
&&ne& kW 2287=−=−= 31685455inC,outT,
0.581===kW 5455kW 3168inC,
bw WW
r &
&
outT,
(c) The thermal efficiency is
0.267===kW 8555kW 2287
in
netth Q
W&
&η
The second-law efficieny of the cycle is defined as the ratio of actual thermal efficiency to the maximum possible thermal efficiency (Carnot efficiency). The maximum temperature for the cycle can be taken to be the turbine inlet temperature. That is,
735.0K 1144K 30311
3
1max =−=−=
TT
η
0.364===735.0267.0
max
thII η
ηη and
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9-117
9-148 A modern compression ignition engine operates on the ideal dual cycle. The maximum temperature in the cycle, the net work output, the thermal efficiency, the mean effective pressure, the net power output, the second-law efficiency of the cycle, and the rate of exergy of the exhaust gases are to be determined.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats.
Properties The properties of air at 1000 K are cp = 1.142 kJ/kg·K, cv = 0.855 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.336 (Table A-2b).
Analysis (a) The clearance volume and the total volume of the engine at the beginning of compression process (state 1) are
xcc
c
c
dcr VVVV
V
V
VV===⎯→⎯
+=⎯→⎯
+= 2
33
m 00012.0m 0018.0
16
43
1 m 00192.00018.000012.0 VVVV ==+=+= dc
1
Qin2
3
4
P
V
Qout
xProcess 1-2: Isentropic compression
( )( )
( )( ) kPa 385916kPa 95
K 7.87016K 343
1.336
2
112
1-1.3361
2
112
==⎟⎟⎠
⎞⎜⎜⎝
⎛=
==⎟⎟⎠
⎞⎜⎜⎝
⎛=
−
k
k
PP
TT
v
v
v
v
Process 2-x and x-3: Constant-volume and constant pressure heat addition processes:
K 1692kPa 3859kPa 7500K) 7.870(
22 ===
PP
TT xx
kJ/kg 6.702K)7.8701692(kJ/kg.K) (0.855)( 2-2 =−=−= TTcq xx v
The net work output and the thermal efficiency are
kJ/kg 835.8=−=−= 3.5691405outinoutnet, qqw
59.5%==== 5948.0kJ/kg 1405kJ/kg 835.8
in
outnet,th q
wη
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9-118
(c) The mean effective pressure is determined to be
( )( )kg 0.001853
K 343K/kgmkPa 0.287)m 92kPa)(0.001 95(
3
3
1
11 =⋅⋅
==RTP
mV
kPa 860.4=⎟⎟⎠
⎞⎜⎜⎝
⎛ ⋅
−=
−=
0001.000192.0(MEP
21 VV kJmkPa
m)2kJ/kg)kg)(835.8 (0.001853 3
3outnet,mw
d) The power for engine speed of 3500 rpm is (
kW 28.39=⎟⎠⎞
⎜⎝
==60rev/cycle) 2(
kJ/kg) kg)(835.8 001853.0(2netnet mwW& ⎛
s min 1(rev/min) 2200n&
(e) The second-law efficieny of the cycle is defined as the ratio of actual thermal efficiency to the maximum possible thermal ef ency (Carnot efficiency). We take the dead state temperature and pressure to be 25ºC and 100 kPa.
Note that there are two revolutions in one cycle in four-stroke engines.
fici
8709.0K 2308
K )27325(113
0max =
+−=−=
TT
η
and
68.3%==== 683.08709.05948.0η
max
th
ηη
The rate of exergy of the exhaust gases is determined as follows
9-149 An Otto cycle with a compression ratio of 7 is considered. The thermal efficiency is to be determined using constant and variable specific heats.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible.
Properties The properties of air at room temperature are R = 0.287 kPa·m3/kg·K, cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-2a).
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9-120
9-150 A simple ideal Brayton cycle with air as the working fluid operates between the specified temperature limits. The net work is to be determined using constant and variable specific heats.
Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2a).
Analysis (a) Constant specific heats:
K 2.549K)(12) 270( 0.4/1.4/)1(12 === − kk
prTT
s
T
1
2
4
3 qin
qout
800 K
270 K
K 3.393121K) 800(1 0.4/1.4/)1(
34 =⎟⎠⎞
⎜⎝⎛=⎟
⎟⎠
⎞⎜⎜⎝
⎛=
− kk
prTT
)(
)()(
2143
1243
compturbnet
TTTTc
TTcTTc
www
p
pp
(b) Variable specific heats: (using air properties from Table A-17)
9-151 A turbocharged four-stroke V-16 diesel engine produces 3500 hp at 1200 rpm. The amount of power produced per cylinder per mechanical and per thermodynamic cycle is to be determined.
Analysis Noting that there are 16 cylinders and each thermodynamic cycle corresponds to 2 mechanical cycles (revolutions), we have
(a)
cycle)mech kJ/cyl 8.16(= hp 1Btu/min 42.41
rev/min) (1200cylinders) (16hp 3500
cycles) mechanical of (No.cylinders) of (No.producedpower Total
cycles) amic thermodynof (No.cylinders) of (No.producedpower Total
micthermodyna
⋅⋅=
⎟⎟⎠
⎞⎜⎜⎝
⎛=
=w
ified temperature limits is considered. The pressure ratio for
2 The air-standard assumptions are applicable. 3 Kinetic and potential nergy cha ges are negligible. 4 Air is an ideal gas with constant specific heats.
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 ompressor and turbine exit can be expressed as
9-152 A simple ideal Brayton cycle operating between the specwhich the compressor and the turbine exit temperature of air are equal is to be determined.
Assumptions 1 Steady operating conditions exist.e n
one per u"The first part of the solution is d1-2 is isentropic compr"Process
s[1]=entropy(air,T=T[1],P=P[1]) s[2]=s[1]
T=T[2], v=v[s[2]=entropy(air, P[2]*v[2]/T[2]=P[1]*P[1]*v[1]=R*T[1] R=0.287 [kJ/kg-K] "Conservation of energy for process 1 to 2: no heat trans w_in = DELTAu_12
)-intenergy(air,DELTAu_12=intenergy(air,T=T[2]e h"Process 2-3 is constant volum
T=T[3], P=P[s[3]=entropy(air, /T[3]=P[{P[3]*v[3]
P[3]*v[3]=R*T[3] v[3]=v[2] "Conservation of energy for process 2 to 3: the work is zeq_in = DELTAu_23
])-inteDELTAu_23=intenergy(air,T=T[33-4 is isentropic expans"Process
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 trans - w_out = DELTAu_34
34=intenergy(air,T=T[4])-intenergy(air,DELTAu_"Process 4-1 is constant volume heat rejection" v[4]=v[1] "Conservation of energy for process 2 to 3: the work is ze- q_out = DELTAu_41 DELTAu_41=intenergy(air,T=T[1])-intenergy(air,T=T[4]) w_net = w_out - w_in
/q_in*Convert(, %) "Thermal efficiency, in percent" Eta_th=w_net"The mass contained in each cylinV_cyl=m*v[1] "The net work done per cycle is:" W_dot_n
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
9-155 An ideal gas Carnot cycle with helium as the working fluid is considered. The pressure ratio, compression ratio, and minimum temperature of the energy source are to be determined.
Assumptions 1 Kinetic and potential energy changes are negligible. 2 Helium is an ideal gas with constant specific heats.
Properties The specific heat ratio of helium is k = 1.667 (Table A-2a).
Analysis From the definition of the thermal efficiency of a Carnot heat engine,
s
T
3
2qin
qout4
1TH
288 K
K 576=−+
=−
=⎯→⎯−=0.501
K 273)(151
1Carnotth,
Carnotth, ηη L
HH
L TT
TT
An isentropic process for an ideal gas is one in which Pvk remains constant. Then, the pressure ratio is
ased on the process equation, the compression ratio is
2
B
2.83==⎟⎟⎠
⎞⎜⎜⎝
⎛= 667.1/1
/1
1
2
2
1 )65.5(k
PP
v
v
9-156 The compression ratio required for an ideal Otto cycle to produce certain amount of work when consuming a given amount of fuel is to be determined.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air isan ideal gas with constant specific heats. 4 The combustion efficiency is 100 percent.
cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-
Analysis The heat input to the cycle for 0.043 grams of fuel consumption is 3
HVfuelin =× −
he thermal efficiency is then
Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K,2).
kJ 505.1kJ/kg) kg)(43,000 10035.0(== qmQ
T
6645.0kJ 1.505
kJ 1netth ===
QW
η in
rom the definition of thermal efficiency, we obtain the required compression ratio to be v
P
4
1
3
2 qout
qin
F
15.3=−
=−
=⎯→⎯−=−−− )14.1/(1)1/(1
th1th
)6645.01(1
)1(111
kkr
r ηη
preparation. If you are a student using this Manual, you are using it without permission.
9-126
9-157 An ideal Otto cycle with air as the working fluid with a compression ratio of 9.2 is considered. The amount of heat transferred to the air, the net work output, the thermal efficiency, and the mean effective pressure are to be determined.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats.
Properties The gas constant of air is R = 0.287 kJ/kg·K (Table A-1). The properties of air are given in Table A-17.
preparation. If you are a student using this Manual, you are using it without permission.
9-127
9-158 An ideal Otto cycle with air as the working fluid with a compression ratio of 9.2 is considered. The amount of heat transferred to the air, the net work output, the thermal efficiency, and the mean effective pressure are to be determined.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats.
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:
( )( )
( ) ( ) kPa 2190kPa 98K 300K 728.8
9.2
K 728.89.2K 300
11
2
2
12
1
11
2
22
0.41
2
112
=⎟⎟⎠
⎞⎜⎜⎝
⎛==⎯→⎯=
==⎟⎟⎠
⎞⎜⎜⎝
⎛=
−
PTT
PT
PT
P
TTk
v
vvv
v
v
v
P
4
1
3
2
qin
qout
Process 2-3: v = constant heat addition.
( )( )
( ) ( )( ) kJ/kg523.3 K728.81457.6KkJ/kg 0.718
K 457.618.72822
2323
222
323 PTT
3223
=−⋅=−=−=
====⎯→⎯=
TTcuuq
TTP
TPP
in v
vv
(b) Process 3-4: isentropic expansion.
3
( ) K 600.09.21K 1457.6
0.413 ⎛⎞⎛
−kv
43 =⎟
⎠⎞
⎜⎝
=⎟⎟⎠
⎜⎜⎝
= TTv
ess 4-1: v = constant heat rejection.
4
Proc
( ) ( )( )
kJ/kg307.9 4.2153.523
kJ/kg 215.4K300600KkJ/kg 0.718
outinnet
1414out
=−=−=
=−⋅=−=−=
qqw
TTcuuq v
(c) 58.8%===kJ/kg 523.3kJ/kg 307.9
in
netth q
wη
( )( )
( ) ( )( )kPa393
kJ 1mkPa 1
1/9.21/kgm 0.879kJ/kg 307.9
/11MEP
/kgm 0.879kPa 98
K 300K/kgmkPa 0.287
3
31
net
21
net
max2min
33
1
11max
=⎟⎟⎠
⎞⎜⎜⎝
⎛ ⋅
−=
−=
−=
==
=⋅⋅
===
rww
r
PRT
vvv
vvv
vv (d)
preparation. If you are a student using this Manual, you are using it without permission.
9-128
9-159 An ideal dual cycle with air as the working fluid with a compression ratio of 12 is considered. The thermal efficiency of the cycle is to be determined.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K, cv = 0.718 kJ/kg.K, and k = 1.4 (Table A-2).
preparation. If you are a student using this Manual, you are using it without permission.
9-129
9-160 An ideal Stirling cycle with air as the working fluid is considered. The maximum pressure in the cycle and the net work output are to be determined.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats.
s
T
3
2
qin = 900 kJ/kg
qout
4
1 1800 K
350 K
Properties The properties of air at room temperature are R = 0.287 kJ/kg.K, cp = 1.005 kJ/kg.K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-2).
Analysis (a) The entropy change during process 1-2 is
KkJ/kg 0.5K 1800
kJ/kg 9001212 ⋅===−
HTq
ss
and
( )
( )( ) kPa 5873=⎟⎟⎠
⎞⎜⎜⎝
⎛===⎯→⎯=
=⎯→⎯⋅=⋅⎯→⎯+=− ln1
12 Tcss v
K 350K 800
5.710kPa 200
710.5lnKkJ/kg 0.287KkJ/kg 0.5ln
3
1
1
23
3
1
1
331
1
11
3
33
1
2
1
2
1
20
2
TT
PTT
PPT
PT
P
RT
v
v
v
vvv
v
v
v
v
v
v
(b) The net work output is
1
( ) kJ/kg 725=⎟⎟⎠
⎞⎜⎜⎝
⎛−=⎟⎟
⎠
⎞⎜⎜⎝
⎛−== kJ/kg 900
K 1800K 350
11 ininthnet qTT
qwH
Lη
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
9-161 A simple ideal Brayton cycle with air as the working fluid is considered. The changes in the net work output per unit mass and the thermal efficiency are to be determined.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats.
Properties The properties of air are given in Table A-17.
preparation. If you are a student using this Manual, you are using it without permission.
9-131
9-162 A simple ideal Brayton cycle with air as the working fluid is considered. The changes in the net work output per unit mass and the thermal efficiency are to be determined.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats.
Properties The properties of air at room temperature are R = 0.287 kJ/kg.K, cp = 1.005 kJ/kg.K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-2).
Analysis Processes 1-2 and 3-4 are isentropic. Therefore, For rp = 6,
( )( )( )
( )( )
( )( )( )
( )( )( )
%1.40kJ/kg 803.4kJ/kg 321.9
kJ/kg 321.95.4814.803
kJ/kg 481.5K300779.1KkJ/kg 1.005
kJ/kg 803.4K500.61300KkJ/kg 1.005
K 779.161K 1300
K 500.66K 300
in
netth
outinnet
1414out
2323in
0.4/1.4/1
3
434
0.4/1.4/1
1
212
===
=−=−=
=−⋅=−=−=
=−⋅=−=−=
=⎟⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
==⎟⎟⎠
⎞⎜⎜⎝
⎛=
−
−
qw
qqw
TTchhq
TTchhq
PP
TT
PP
TT
p
p
kk
kk
η
s
T
1
24
3 qin
qout
2
3′
For rp = 12, ( )
( )( )
( )( )
( )( )( )
( )( )( )
%8.50kJ/kg 693.2kJ/kg 352.3
kJ/kg 352.39.3402.693
kJ/kg 340.9K300639.2KkJ/kg 1.005
kJ/kg 693.2K610.21300KkJ/kg 1.005
K 639.2121K 1300
K 610.212K 300
in
netth
outinnet
1414out
2323in
0.4/1.4/1
3
434
0.4/1.4/1
1
212
===
=−=−=
=−⋅=−=−=
=−⋅=−=−=
=⎟⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
==⎟⎟⎠
⎞⎜⎜⎝
⎛=
−
−
qw
qqw
TTchhq
TTchhq
PP
TT
PP
TT
p
p
kk
kk
η
Thus,
(a) ( )increase9.3213.352net kJ/kg 30.4=−=∆w
( )increase%1.40%8.50th 10.7%=−=∆η (b)
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
9-163 A regenerative gas-turbine engine operating with two stages of compression and two stages of expansion is considered. The back work ratio and the thermal efficiency are to be determined.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats.
preparation. If you are a student using this Manual, you are using it without permission.
9-133
9-164 Problem 9-163 is reconsidered. The effect of the isentropic efficiencies for the compressor and turbine and regenerator effectiveness on net work done and the heat supplied to the cycle is to be investigated. Also, the T-s diagram for
nalysis Using EES, the problem is solved as follows:
[K]
cy" sentropic efficiency"
ies are constant across the compressor"
ssor for the isentropic case: dy-flow"
])
T[2], P=P[2])
ideal case the entropies are constant across the HP compressor"
EEta_t =0.86 "Turbien i "LP Compressor:" "Isentropic Compressor anaysis" s[1]=ENTROPY(Air,T=T[1],P=P[1]) s_s[2]=s[1] "For the ideal case the entropP[2] = Pratio*P[1] s_s[2]=ENTROPY(Air,T=T_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exit" Eta_c = w_compisen_LP/w_comp_LP "compressor adiabatic efficiency, W_comp > W_compisen" "Conservation of energy for the LP compree_in - e_out = DELTAe=0 for steah[1] + w_compisen_LP = h_s[2] h[1]=ENTHALPY(Air,T=T[1]) h_s[2]=ENTHALPY(Air,T=T_s[2 "Actual compressor analysis:"h[1] + w_comp_LP = h[2]
[2]=ENTHALPY(Air,T=T[2]) hs[2]=ENTROPY(Air,T= "HP Compressor:" s[3]=ENTROPY(Air,T=T[3],P=P[3]) s_s[4]=s[3] "For theP[4] = Pratio*P[3] P[3] = P[2] s_s[4]=ENTROPY(Air,T=T_s[4],P=P[4]) "T_s[4] is the isentropic value of T[4] at compressor exit" Eta_c = w_compisen_HP/w_comp_HP "compressor adiabatic efficiency, W_comp > W_compisen" "Conservation of energy for the compressoe_in - e_out = DELTAe=0 for steah[3] + w_compisen_HP = h_s[4] h[3]=ENTHALPY(Air,T=T[3]) h_s[4]=ENTHALPY(Air,T=T_s[4 "Actual compressor analysis:"h[3] + w_comp_HP = h[4] h[4]=ENTHALPY(Air,T=T[4]) s[4]=ENTROPY(Air,T=T[4], P=
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
"Ih[2] = q_out_intercool + h[3] "External heat exchanger analysis" "SSSF First Law for the heh[4] + q_in_noreg = h[6] hP[6]=P[4]"process 4-6 i "HP Turbine analysis" s[6]=ENTROPY(Ais_s[7]=s[6] "For the ideal case the entropies are constant across the turbine" P[7] = P[6] /Pratio sEta_t = w_turb_HP /w_turbisen_HP "turbine adiabatic efficiency, w_turbisen > w_turb" "SSSF First Law for the isentrope_in -e_out = DELTAe_cv = 0 for sh[6] = w_turbisen_HP + h_h_s[7]=ENTHALPY(Air,T"Actual Turbine analysis:" h[6] = w_turb_HP + h[7] hs[7]=ENTROPY "Reheat Q_in:" hh[8]=ENTHALPY(Air,T= "HL Turbin P[8]=P[7] s[8]=ENTROPY(Ais_s[9]=s[8] "For the ideal case the entropies are constant across the turbine" P[9] = P[8] /Pratio sEta_t = w_turb_LP /w_turbisen_LP "turbine adiabatic efficiency, w_turbise "SSSF First Law for the isentrope_in -e_out = DELTAe_cv = 0 for sh[8] = w_turbisen_LP + h_sh_s[9]=ENTHALPY(Air,T"Actual Turbine analysis:" h[8] = w_turb_LP + h[9] hs[9]=ENTROPY(A "Cycle analysis" w_net=w_turb_HP+w_turb_LP - w_comp_HP - w_comp_LP q_in_total_noreg=q_in_noreg+q_in_reheat EBwr=(w_comp_HP + w_comp_LP)/(w_turb_HP+w_turb_LP)"Back work r "With the regenerator, the heah[5] + q_in_withreg = h[6] h[5]=ENTHsP[5]=P[4]
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9-165 A regenerative gas-turbine engine operating with two stages of compression and two stages of expansion is considered. The back work ratio and the thermal efficiency are to be determined.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Helium is an ideal gas with constant specific heats.
preparation. If you are a student using this Manual, you are using it without permission.
9-139
9-166 An ideal gas-turbine cycle with one stage of compression and two stages of expansion and regeneration is considered. The thermal efficiency of the cycle as a function of the compressor pressure ratio and the high-pressure turbine to compressor inlet temperature ratio is to be determined, and to be compared with the efficiency of the standard regenerative cycle.
Analysis The T-s diagram of the cycle is as shown in the figure. If the overall pressure ratio of the cycle is rp, which is the pressure ratio across the compressor, then the pressure ratio across each turbine stage in the ideal case becomes √ rp. Using the isentropic relations, the temperatures at the compressor and turbine exit can be expressed as
( )( )( )
( ) ( )( )
( ) ( )( ) ( ) ( ) ( ) kk
pkk
pkk
pkk
p
kk
p
kk
kkp
kk
p
kk
kkp
kk
rTrrTrTr
TPPTT
rTr
TPPTTT
rTPPTTT
2/11
2/1/11
2/12
/1
5
/16
5
2/13
/1
3
/1
3
4347
/11
/1
1
2125
1
1
−−−−
−−
−
−−
−−
===⎟⎟
⎠
⎞
⎜⎜
⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎛
=
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛==
=⎟⎟⎠
⎞⎜⎜⎝
⎛==
Then,
65⎝
( ) ( )( )( ) ( )( )12/1 −=−=−= − kkrTcTTchhq
1
1161
2/137373in −=−=−= −
ppp
kkppp rTcTTchhq
6out
and thus ( )( )
( )( )kkpp rTcq 2/1
3inth
111
−−−=−=η
which simplifies to
kkpp rTcq 2/1
1out 1− −
( ) kkpr
TT 2/1
3
1th 1 −−=η
The thermal efficiency of the single stage ideal regenerative cycle is given as
( ) kkpr
TT /1
3
1th 1 −−=η
Therefore, the regenerative cycle with two stages of expansion has a higher thermal efficiency than the standard regenerative cycle with a single stage of expansion for any given value of the pressure ratio rp.
s
T
1
2
7
qin
5
6
4
3
qout
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
9-167 A gas-turbine plant operates on the regenerative Brayton cycle with reheating and intercooling. The back work ratio, the net work output, the thermal efficiency, the second-law efficiency, and the exergies at the exits of the combustion chamber and the regenerator are to be determined.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats.
Properties The gas constant of air is R = 0.287 kJ/kg·K.
Analysis (a) For this problem, we use the properties from EES software. Remember that for an ideal gas, enthalpy is a function of temperature only whereas entropy is functions of both temperature and pressure.
(c) The second-law efficieny of the cycle is defined as the ratio of actual thermal efficiency to the maximum possible thermal efficiency (Carnot efficiency). The maximum temperature for the cycle can be taken to be the turbine inlet temperature. That is,
786.0K 1400K 30011
6
1max =−=−=
TT
η
and
0.704===786.0553.0
maxηη
η thII
(d) The exergies at the combustion chamber exit and the regenerator exit are
9-168 The thermal efficiency of a two-stage gas turbine with regeneration, reheating and intercooling to that of a three-stage gas turbine is to be compared.
Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg⋅K and k = 1.4 (Table A-2a).
The temperatures at the end of compression and expansion are
T
p 520.216 3/1
K 5.368K)(2.520) 283( 0.4/1.4/)1(min === − kk
pc rTT
K 4.6702.520
1K) 873(1/)1(
=⎟⎞
⎜⎛
=− kk
TT0.4/1.4
max =⎟⎠⎞
⎜⎝⎛
⎟⎠
⎜⎝ pr
he heat input and heat output are
kg =−⋅
s
T
3
4
1
9
8
2
5
10
11
6
7
12
13
14
e
T
kJ/ 3(1.005)(3 maxin =−= ep TTcq kJ/kg 8.610K )4.670K)(873
out kJ/kg 8.257K )283K)(368.5kJ/kg (1.0053)(3 min =−⋅=−= TTcq cp
The thermal efficiency of the cycle is then
0.578=−=−=8.6108.25711
in
outth q
qη
preparation. If you are a student using this Manual, you are using it without permission.
9-143
9-169 A pure jet engine operating on an ideal cycle is considered. The thrust force produced per unit mass flow rate is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with constant specific heats at room temperature. 4 The turbine work output is equal to the compressor work input.
Properties The properties of air at room temperature are R = 0.287 kPa⋅m3/kg⋅K (Table A-1), cp = 1.005 kJ/kg⋅K and k = 1.4 (Table A-2a).
Analysis Working across the two isentropic processes of the cycle yields
K 6.509K)(9) 272( 0.4/1.4/)1(12 === − kk
prTT
s
T
1
2 4
3qin
5
qout
K 2.34391K 643(35 =⎟
⎟⎠
⎜⎜⎝
=pr
TT )1 0.4/1.4/)1(
=⎟⎠⎞
⎜⎝⎛⎞⎛
− kk
pressor is
The work input to the com
K)(509.6kJ/kg 005.1()( 12C ⋅=−= TTcw p kJ/kg 8.238272)K- =
kJ/kg 005.1(C53
=−−=
p
engine exit is determined from
An energy balance gives the excess enthalpy to be
)( −−=∆ wTTch
kJ/kg 50.62kJ/kg 8.238)K2.343K)(643⋅
The velocity of the air at the
2
2inletexit VV
h−
=∆ 2
earranging,
R
( )
m/s 6.504
)m/s 360(kJ/kg 1
/sm 1000kJ/kg) 50.62(2
22/1
222
2/12inletexit
=
⎥⎥⎦
⎤
⎢⎢⎣
⎡+⎟
⎟⎠
⎞⎜⎜⎝
⎛=
+∆= VhV
The specific impulse is then
m/s 145=−=−= 3606.504inletexit VVmF&
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
9-170 The electricity and the process heat requirements of a manufacturing facility are to be met by a cogeneration plant consisting of a gas-turbine and a heat exchanger for steam production. The mass flow rate of the air in the cycle, the back work ratio, the thermal efficiency, the rate at which steam is produced in the heat exchanger, and the utilization efficiency of the cogeneration plant are to be determined.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats.
Analysis (a) For this problem, we use the properties of air from EES software. Remember that for an ideal gas, enthalpy is a function of temperature only whereas entropy is functions of both temperature and pressure.
Process 1-2: Compression
KkJ/kg 682.5kPa 100
C20kJ/kg 5.293C20
11
1
11
kJ/kg 2.567kJ/kg.K 682.512 == ss
2
kJ/kg 8.6115.293
5.2932.56786.0 2212
12C =⎯→⎯
−−
=⎯→⎯−−
= hhhh
hh sη
Process 3-4: Expansion
kJ/kg 5.738C4504 =⎯→⎯°= hT 4
ss hhh 343T −− h
h
4
3 5.738−
We cann ly. However, using the following lines in EES together with the isentropic c relation, we find h3 = 1262 kJ/kg, T3 = 913.2ºC, s3 = 6.507 kJ/kg.K. The solution by hand would require a trial-
h_3=enthalpy(Air, T=T_3)
5 =→h
The inlet water is compressed liquid at 15ºC and at the saturation pressure of steam at 200ºC (1555 kPa). This is not available in the tables but we can obtain it in EES. The alternative is to use saturated liquid enthalpy at the given temperature.
C15
22
2 =⎭⎬⎫
=°=
=⎫°=
ww hx
T
T
12in, =−=−= hhw
hh 43 88.0 =⎯→⎯−
=η
ot find the enthalpy at state 3 directefficien yerror approach.
s_3=entropy(Air, T=T_3, P=P_2)
h_4s=enthalpy(Air, P=P_1, s=s_3)
Also,
C3255 ⎯⎯°=T kJ/kg 4.605
kJ/kg 27921
C200
kJ/kg 47.64kPa 1555 1
1 ⎭⎬= wh
P1w
The net work output is
kJ/kg 2.3185.2938.611C
kJ/kg 4.5235.738126243outT, =−=−= hhw
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9-146
9-171 A turbojet aircraft flying is considered. The pressure of the gases at the turbine exit, the mass flow rate of the air through the compressor, the velocity of the gases at the nozzle exit, the propulsive power, and the propulsive efficiency of the cycle are to be determined.
Assumptions 1 The air-standard assumptions are applicable. 2 Potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats.
Properties The gas constant of air is R = 0.287 kJ/kg·K (Table A-1).
Analysis (a) For this problem, we use the properties from EES software. Remember that for an ideal gas, enthalpy is a function of temperature only whereas entropy is functions of both temperature and pressure.
Diffuser, Process 1-2:
kJ/kg 23.238C35 11 =⎯→⎯°−= hT
kJ/kg 37.269/sm 1000
kJ/kg 12m/s) (15
/sm 1000kJ/kg 1
2m/s) (900/3.6kJ/kg) 23.238(
22
222
2
222
2
22
2
21
1
=⎯→⎯⎟⎠⎞
⎜⎝⎛+=⎟
⎠⎞
⎜⎝⎛+
+=+
hh
VhVh
s
T
1
2
4
3
qin
5
qout
6
kPa 50kJ/kg 37.2692
=Ph
KkJ/kg 7951.522
⋅=⎭⎬⎫=
s
Compressor, Process 2-3:
=⎭⎬⎫
===
shss
P kJ/kg 19.505
kJ/kg.K 7951.5kPa 450
323
3
kJ/kg 50.55337.269
37.26919.50583.0 3323
23C =⎯→⎯
−−
=⎯→⎯−−
= hhhh
hh sη
urbine, P cess 3-4:
1028.130437.26950.553 555423 =⎯→⎯−=−⎯→⎯−=− hhhhhh
here the ass flow rates through the compressor and the turbine are assumed equal.
T ro
kJ/kg 8.1304C950 =⎯→⎯°= hT 44
.0 kJ/kg 6
w m
kJ/kg 45.9628.1304
6.10208.130483.0 5554
54T =⎯ →⎯
−−
=⎯→⎯−−
= sss
hhhh
hhη
4⎭=P
545 KkJ/kg 7725.6
kgP
ss
(b) The mass flow rate of the air through the compressor is
KkJ/kg 7725.6C9504 ⋅=⎬
⎫°=s
T
kPa 4504
=5 kJ/ 45.962h s kPa 147.4=
⎭⎬⎫
⋅==
kg/s 1.760=−
=−
=kJ/kg )37.26950.553(
kJ/s 500
23
C
hhW
m&
&
(c) Nozzle, Process 5-6:
⎫=h KkJ/kg 8336.6
kPa 4.147 55
⋅=⎭⎬=
sP
kJ/kg 6.10205
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9-148
9-172 The three processes of an air standard cycle are described. The cycle is to be shown on the P-v and T-s diagrams, and the expressions for back work ratio and the thermal efficiency are to be obtained.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats.
Analysis (a) The P-v and T-s diagrams for this cycle are as shown.
(b) The work of compression is found by the first law for process 1-2:
These results show that if there is no compression (i.e. r = 1), there can be no expansion and no net work will be done even though heat may be added to the system.
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
9-173 The three processes of an air standard cycle are described. The cycle is to be shown on the P-v and T-s diagrams, and the expressions for back work ratio and the thermal efficiency are to be obtained.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats.
Analysis (a) The P-v and T-s diagrams for this cycle are as shown.
(b) The work of expansion is found by the first law for process 2-3:
(c) Apply first law to the closed system for processes 1-2 and 3-1 to show:
2 1in vq C T T= −
The cycle thermal efficiency is given by
co
( )( )3 1out pq C T T= −
( )( )
( )( )
3 1 1 3 1
2 1 1 2 1
/ 11 1 1
/ 1poutqηth
in v
C T T T T Tk
q C T T T T T− −
= − = − = −− −
Process 1-2 is constant volume; therefore,
2 2
T1 1 2 2 2
2 1 1 1 3
krT T P P
= ⇒ = = =
The efficiency becomes
PV PV T P P
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9-151
111th k
rkr
η −= −
−
(d) Determine the value of the back work ratio and efficiency as r goes to unity.
( )
( ) ( )
( )
1 1exp
11
1exp
11 1 111 1
1
1lim 1 11
compk k
kr
comp
r
w k rrkw r r r
r kr
wk
w k
− −
−→
→
− − −= − =
− −
−⎭ ⎩ ⎭
⎧ ⎫= − =⎨ ⎬−⎩ ⎭
1 1exp
1 1lim 1 lim 1 limcompkr r
w rk kw r→ →
⎧ ⎫ ⎧ ⎫−= − = −⎨ ⎬ ⎨ ⎬−⎩
11 1 1
1
111
1 1lim 1 lim 1 lim1
1lim 1 0
th k
th k kr r r
thr
rkr
rk kr k
kk
η
η
η
−→ → →
→
r
−= −
−⎧ ⎫ ⎧−
= − = −⎨ ⎬ ⎨−⎩ ⎭ ⎩⎧ ⎫= − =⎨ ⎬⎩ ⎭
⎫⎬⎭
These results show that if there is no compression (i.e. r = 1), there can be no expansion and no net work will be done even though heat may be added to the system.
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
9-174 The four processes of an air-standard cycle are described. The cycle is to be shown on the P-v and T-s diagrams; an expression for the cycle thermal efficiency is to be obtained; and the limit of the efficiency as the volume ratio during heat rejection approaches unity is to be evaluated.
Analysis (a) The P-v and T-s diagrams of the cycle are shown in the figures.
(b) Apply first law to the closed system for processes 2-3 and 4-1 to show:
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9-153
9-175 The four processes of an air-standard cycle are described. The back work ratio and its limit as rp goes to unity are to be determined, and the result is to be compared to the expression for the Otto cycle.
Analysis The work of compression for process 1-2 is found by the first law:
The work of expansion for process 3-4 is found by the first law:
3 4 4 3
exp,3 4 3 4 3 4
v
v
q w u
w u C T T
w w C T T
− − −
− −
− −
− = ∆
= −∆ = − −
= − = −
The back work ratio is
( )( )
1− −
1 2 0( )q isentropic process− =
The work of compression for process 4-1 is found by
( ) (1
,4 1compw w− = −
( )( )
3 4 3 4 3 4
3 4 0( )q isentropic process− =
3 4
( ) ( )( )
( ) ( )
( )4 1 2 1
4 1 2 1 1
exp 3 4 3 4 3
/ 1 / 1
1 /comp v v
v
R T T T Tw R T T C T T CTw C T T T T T
− + −− + −
= =− −
Using data from the previous problem and Cv = R/(k-1)
( ) ( )11
exp 31 1
( 1) 1 1
11
kpcomp
k kpr r⎝ ⎠
k r rw Tw T
−
− −
− − + −=
⎛ ⎞−⎜ ⎟⎜ ⎟
( ) ( ) ( )( ) ( )1 11 1
1 1exp 3 3
11 1
11
1exp 3
1
( 1) 1 1 1 0 1lim lim 11 11
1lim 11
p p
p
k kpcomp
r r
kk kp
kcomp
r
k
k r r k rw T Tw T T
rr r
w T rw T
r
− −
→ →
−− −
−
→
−
⎧ ⎫⎧ ⎫⎪ ⎪⎪ ⎪⎪ ⎪− − + − − + −⎪ ⎪ ⎪= =⎨ ⎬ ⎨
⎛ ⎞⎪ ⎪ ⎪ −−⎜ ⎟⎪ ⎪ ⎪⎜ ⎟ ⎩ ⎭⎝ ⎠⎪ ⎪⎩ ⎭⎧ ⎫⎪ ⎪−
= ⎨ ⎬⎪ ⎪−⎩ ⎭
⎪⎬⎪⎪
This result is the same expression for the back work ratio for the Otto cycle.
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9-154
9-176 The effects of compression ratio on the net work output and the thermal efficiency of the Otto cycle for given
nalysis Using EES, the problem is solved as follows:
]
ession"
P=P[2]) v[1]/T[1]
rocess 1 to 2"
T=T[1]) eat addition"
2 to 3"
nergy(air,T=T[2]) ion"
rocess 3 to 4"
T=T[3]) n"
to 1"
nergy(air,T=T[1])-intenergy(air,T=T[4])
Eta_th=w_net/q_in_total*Convert(, %) "Thermal efficiency, in percent"
operating conditions is to be investigated.
A
"Input Data" T[1]=300 [K]
] P[1]=100 [kPa[3] = 2000 [KT
r_comp = 12
1-2 is isentropic compr"Process s[1]=entropy(air,T=T[1],P=P[1]) s[2]=s[1]
(air, s=s[2],T[2]=temperatureP[2]*v[2]/T[2]=P[1]*P[1]*v[1]=R*T[1] R=0.287 [kJ/kg-K] V[2] = V[1]/ r_comp "Conservation of energy for pq_12 - w_12 = DELTAu_12 q_12 =0"isentropic process"
12=intenergy(air,T=T[2])-intenergy(air,DELTAu_"Process 2-3 is constant volume hv[3]=v[2] s[3]=entropy(air, T=T[3], P=P[3]) P[3]*v[3]=R*T[3] "Conservation of energy for process q_23 - w_23 = DELTAu_23 w_23 =0"constant volume process"
23=intenergy(air,T=T[3])-inteDELTAu_"Process 3-4 is isentropic expanss[4]=s[3] s[4]=entropy(air,T=T[4],P=P[4]) P[4]*v[4]=R*T[4] "Conservation of energy for pq_34 -w_34 = DELTAu_34 q_34 =0"isentropic process"
=intenergy(air,T=T[4])-intenergy(air,DELTAu_34"Process 4-1 is constant volume heat rejectioV[4] = V[1] "Conservation of energy for process 4q_41 - w_41 = DELTAu_41
9-177 The effects of pressure ratio on the net work output and the thermal efficiency of a simple Brayton cycle 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])
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
9-178 The effects of pressure ratio on the net work output and the thermal efficiency of a simple Brayton 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 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 = 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]) s[4]=entropy(air,T=T[4],P=P[4])
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
9-179 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 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:
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" V[4] = V[1] "Conservation of energy for process 4 to 1" q_41 - w_41 = DELTAu_41
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
9-180 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
.
nalysis Using EES, the problem is solved as follows:
m diagram window"
P[1])
for the actual compressor, assuming: adiabatic, ke=pe=0"
ot_t + m_dot*h[4] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0"
ycle work, kW"
determined only to produce a T-s plot"
s[4]=entropy(air,T=T[4],P=P[4])
to be investigated. Variable specific heats are to be used
ta_c = 75/100 EEta_t = 82/100} "Inlet conditions" [1]=ENTHALPY(Air,T=T[1]) h
s[1]=ENTROPY(Air,T=T[1],P= "Compressor anaysis" s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor"
sure ratio - to find P[2]" P_ratio=P[2]/P[1]"Definition of presT_s[2]=TEMPERATURE(Air,s=s_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exit" h_s[2]=ENTHALPY(Air,T=T_s[2])
mpressor adiabatic efficiency; Eta_c = W_dot_c_ideal/W_dot_c_actual. " Eta_c =(h_s[2]-h[1])/(h[2]-h[1]) "Com_dot*h[1] +W_dot_c=m_dot*h[2] "SSSF First Law
lysis" "External heat exchanger anaP[3]=P[2]"process 2-3 is SSSF constant pressure"
ir,T=T[3]) h[3]=ENTHALPY(Am_dot*h[2] + Q_dot_in= m_dot*h[3]"SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0""Turbine analysis"
,T=T[3],P=P[3]) s[3]=ENTROPY(Airs_s[4]=s[3] "For the ideal case the entropies are constant across the turbine" P_ratio= P[3] /P[4]
=s_s[4],P=P[4]) "Ts[4] is the isentropic value of T[4] at turbine exit" T_s[4]=TEMPERATURE(Air,sh_s[4]=ENTHALPY(Air,T=T_s[4]) "Eta_t = W_dot_t /Wts_dot turbine adiabatic efficiency, Wts_dot > W_dot_t"
f the net cW_dot_net=W_dot_t-W_dot_c"Definition oEta=W_dot_net/Q_dot_in"Cycle thermal efficiency"
work ratio" Bwr=W_dot_c/W_dot_t "Back "The following state points are T[2]=temperature('air',h=h[2]) T[4]=temperature('air',h=h[4]) s[2]=entropy(air,T=T[2],P=P[2])
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
9-181 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) endif END Procedure 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" 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])
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
ccur such that T_3 = T_1 and the compressors have"Since intercooling is assumed to oratio, the work input to each comprew_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"
t exchanger + the reheat be"The heat added in the external hea C_P*(T_6 - T_5) +(Nstaq_in_total =
"Reheat is assumedT_8 = T_6 "Turbine analysis" P_7 = P_6 /R_p "T_7s is the isentropic value of T_7 at turbine exit"
mperatures, T_6 and T_1 for this problem." ta_th_Ericsson = (1 - T_1/T_6)*Convert(, %) "[%]"
ηth,Ericksson ηth,Regenerative Nstages
T_4 = T_2 "The regeneEta_reg = (T_5 - T_4)/(T_9 - T_4) T_9 = T_7 "Energy balance on regenerator gives T_10T_4 + T_9=T_5 + T_10 "CEta_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 teE
ccur such that T_3 = T_1 and the compressors have"Since intercooling is assumed to oratio, the work input to each comprew_comp_total = Nstages*w_comp "External heat exchanger analysis" SSSF First Law for the heat exchanger, assum"
e_in - e_out =DELTAe_cv =0 for steady flow"
t exchanger + the reheat be"The heat added in the external hea C_P*(T_6 - T_5) +(Nstaq_in_total =
"Reheat is assumedT_8 = T_6 "Turbine analysis" P_7 = P_6 /R_p "T_7s is the isentropic value of T_7 at turbine exit"
mperatures, T_6 and T_1 for this problem." ta_th_Ericsson = (1 - T_1/T_6)*Convert(, %) "[%]"
ηth,Ericksson ηth,Regenerative Nstages
Bwr=w_co P_4=P_2 P_5=P_4 P_6=P_5 T_4 = T_2 "The regeneEta_reg = (T_5 - T_4)/(T_9 - T_4) T_9 = T_7 "ET_4 + T_9=T_5 + T_10 "CEta_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 teE
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9-170
Fundamentals of Engineering (FE) Exam Problems
9-184 An Otto cycle with air as the working fluid has a compression ratio of 10.4. Under cold air standard conditions, the thermal efficiency of this cycle is
(a) 10% (b) 39% (c) 61% (d) 79% (e) 82%
Answer (c) 61%
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).
r=10.4 k=1.4 Eta_Otto=1-1/r^(k-1) "Some Wrong Solutions with Common Mistakes:" W1_Eta = 1/r "Taking efficiency to be 1/r" W2_Eta = 1/r^(k-1) "Using incorrect relation" W3_Eta = 1-1/r^(k1-1); k1=1.667 "Using wrong k value"
9-185 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
9-186 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
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).
TL=300 "K" TH=2000 "K" Wnet=600 "kJ/s" Wnet= (TH-TL)*DS "Some Wrong Solutions with Common Mistakes:" W1_DS = Wnet/TH "Using TH instead of TH-TL" W2_DS = Wnet/TL "Using TL instead of TH-TL" W3_DS = Wnet/(TH+TL) "Using TH+TL instead of TH-TL"
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9-187 Air in an ideal Diesel cycle is compressed from 2 L to 0.13 L, and then it expands during the constant pressure heat addition process to 0.30 L. Under cold air standard conditions, the thermal efficiency of this cycle is
(a) 41% (b) 59% (c) 66% (d) 70% (e) 78%
Answer (b) 59%
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).
V1=2 "L" V2= 0.13 "L" V3= 0.30 "L" r=V1/V2 rc=V3/V2 k=1.4 Eta_Diesel=1-(1/r^(k-1))*(rc^k-1)/k/(rc-1) "Some Wrong Solutions with Common Mistakes:" 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"
9-188 Helium gas in an ideal Otto cycle is compressed from 20°C and 2.5 L to 0.25 L, and its temperature increases by an additional 700°C during the heat addition process. The temperature of helium before the expansion process is
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).
k=1.667 V1=2.5 V2=0.25 r=V1/V2 T1=20+273 "K" T2=T1*r^(k-1) T3=T2+700-273 "C" "Some Wrong Solutions with Common Mistakes:" W1_T3 =T22+700-273; T22=T1*r^(k1-1); k1=1.4 "Using wrong k value" W2_T3 = T3+273 "Using K instead of C" W3_T3 = T1+700-273 "Disregarding temp rise during compression" W4_T3 = T222+700-273; T222=(T1-273)*r^(k-1) "Using C for T1 instead of K"
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9-189 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
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).
rho1=1.20 "kg/m^3" k=1.4 V1=2.2 V2=0.26 m=rho1*V1/1000 "kg" w_net=440 "kJ/kg" Wtotal=m*w_net MEP=Wtotal/((V1-V2)/1000) "Some Wrong Solutions with Common Mistakes:" W1_MEP = w_net/((V1-V2)/1000) "Disregarding mass" W2_MEP = Wtotal/(V1/1000) "Using V1 instead of V1-V2" W3_MEP = (rho1*V2/1000)*w_net/((V1-V2)/1000); "Finding mass using V2 instead of V1" W4_MEP = Wtotal/((V1+V2)/1000) "Adding V1 and V2 instead of subtracting"
9-190 In an ideal Brayton cycle, air is compressed from 95 kPa and 25°C to 1100 kPa. Under cold air standard conditions, the thermal efficiency of this cycle is
(a) 45% (b) 50% (c) 62% (d) 73% (e) 86%
Answer (b) 50%
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).
P1=95 "kPa" P2=1100 "kPa" T1=25+273 "K" rp=P2/P1 k=1.4 Eta_Brayton=1-1/rp^((k-1)/k) "Some Wrong Solutions with Common Mistakes:" W1_Eta = 1/rp "Taking efficiency to be 1/rp" W2_Eta = 1/rp^((k-1)/k) "Using incorrect relation" W3_Eta = 1-1/rp^((k1-1)/k1); k1=1.667 "Using wrong k value"
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9-191 Consider an ideal Brayton cycle executed between the pressure limits of 1200 kPa and 100 kPa and temperature limits of 20°C and 1000°C with argon as the working fluid. The net work output of the cycle is
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).
P1=100 "kPa" P2=1200 "kPa" T1=20+273 "K" T3=1000+273 "K" rp=P2/P1 k=1.667 Cp=0.5203 "kJ/kg.K" Cv=0.3122 "kJ/kg.K" T2=T1*rp^((k-1)/k) q_in=Cp*(T3-T2) Eta_Brayton=1-1/rp^((k-1)/k) w_net=Eta_Brayton*q_in "Some Wrong Solutions with Common Mistakes:" W1_wnet = (1-1/rp^((k-1)/k))*qin1; qin1=Cv*(T3-T2) "Using Cv instead of Cp" W2_wnet = (1-1/rp^((k-1)/k))*qin2; qin2=1.005*(T3-T2) "Using Cp of air instead of argon" W3_wnet = (1-1/rp^((k1-1)/k1))*Cp*(T3-T22); T22=T1*rp^((k1-1)/k1); k1=1.4 "Using k of air instead of argon" W4_wnet = (1-1/rp^((k-1)/k))*Cp*(T3-T222); T222=(T1-273)*rp^((k-1)/k) "Using C for T1 instead of K"
9-192 An ideal Brayton cycle has a net work output of 150 kJ/kg and a backwork ratio of 0.4. If both the turbine and the compressor had an isentropic efficiency of 85%, the net work output of the cycle would be
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).
wcomp/wturb=0.4 wturb-wcomp=150 "kJ/kg" Eff=0.85 w_net=Eff*wturb-wcomp/Eff "Some Wrong Solutions with Common Mistakes:" W1_wnet = Eff*wturb-wcomp*Eff "Making a mistake in Wnet relation" W2_wnet = (wturb-wcomp)/Eff "Using a wrong relation" W3_wnet = wturb/eff-wcomp*Eff "Using a wrong relation"
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9-193 In an ideal Brayton cycle, air is compressed from 100 kPa and 25°C to 1 MPa, and then heated to 927°C before entering the turbine. Under cold air standard conditions, the air temperature at the turbine exit is
(a) 349°C (b) 426°C (c) 622°C (d) 733°C (e) 825°C
Answer (a) 349°C
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).
P1=100 "kPa" P2=1000 "kPa" T1=25+273 "K" T3=900+273 "K" rp=P2/P1 k=1.4 T4=T3*(1/rp)^((k-1)/k)-273 "Some Wrong Solutions with Common Mistakes:" W1_T4 = T3/rp "Using wrong relation" W2_T4 = (T3-273)/rp "Using wrong relation" W3_T4 = T4+273 "Using K instead of C" W4_T4 = T1+800-273 "Disregarding temp rise during compression"
9-194 In an ideal Brayton cycle with regeneration, argon gas is compressed from 100 kPa and 25°C to 400 kPa, and then heated to 1200°C before entering the turbine. The highest temperature that argon can be heated in the regenerator is
(a) 246°C (b) 846°C (c) 689°C (d) 368°C (e) 573°C
Answer (e) 573°C
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). k=1.667 Cp=0.5203 "kJ/kg.K" P1=100 "kPa" P2=400 "kPa" T1=25+273 "K" T3=1200+273 "K" "The highest temperature that argon can be heated in the regenerator is the turbine exit temperature," rp=P2/P1 T2=T1*rp^((k-1)/k) T4=T3/rp^((k-1)/k)-273 "Some Wrong Solutions with Common Mistakes:" W1_T4 = T3/rp "Using wrong relation" W2_T4 = (T3-273)/rp^((k-1)/k) "Using C instead of K for T3" W3_T4 = T4+273 "Using K instead of C" W4_T4 = T2-273 "Taking compressor exit temp as the answer"
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
9-195 In an ideal Brayton cycle with regeneration, air is compressed from 80 kPa and 10°C to 400 kPa and 175°C, is heated to 450°C in the regenerator, and then further heated to 1000°C before entering the turbine. Under cold air standard conditions, the effectiveness of the regenerator is
(a) 33% (b) 44% (c) 62% (d) 77% (e) 89%
Answer (d) 77%
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).
k=1.4 Cp=1.005 "kJ/kg.K" P1=80 "kPa" P2=400 "kPa" T1=10+273 "K" T2=175+273 "K" T3=1000+273 "K" T5=450+273 "K" "The highest temperature that the gas can be heated in the regenerator is the turbine exit temperature," rp=P2/P1 T2check=T1*rp^((k-1)/k) "Checking the given value of T2. It checks." T4=T3/rp^((k-1)/k) Effective=(T5-T2)/(T4-T2) "Some Wrong Solutions with Common Mistakes:" W1_eff = (T5-T2)/(T3-T2) "Using wrong relation" W2_eff = (T5-T2)/(T44-T2); T44=(T3-273)/rp^((k-1)/k) "Using C instead of K for T3" W3_eff = (T5-T2)/(T444-T2); T444=T3/rp "Using wrong relation for T4"
9-196 Consider a gas turbine that has a pressure ratio of 6 and operates on the Brayton cycle with regeneration between the temperature limits of 20°C and 900°C. If the specific heat ratio of the working fluid is 1.3, the highest thermal efficiency this gas turbine can have is
(a) 38% (b) 46% (c) 62% (d) 58% (e) 97%
Answer (c) 62%
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).
"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)"
Eta_regen=1-(T1/T3)*rp^((k-1)/k)
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9-197 An ideal gas turbine cycle with many stages of compression and expansion and a regenerator of 100 percent effectiveness has an overall pressure ratio of 10. Air enters every stage of compressor at 290 K, and every stage of turbine at 1200 K. The thermal efficiency of this gas-turbine cycle is
(a) 36% (b) 40% (c) 52% (d) 64% (e) 76%
Answer (e) 76%
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).
9-198 Air enters a turbojet engine at 320 m/s at a rate of 30 kg/s, and exits at 650 m/s relative to the aircraft. The thrust developed by the engine is
(a) 5 kN (b) 10 kN (c) 15 kN (d) 20 kN (e) 26 kN
Answer (b) 10 kN
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).
Vel1=320 "m/s" Vel2=650 "m/s" Thrust=m*(Vel2-Vel1)/1000 "kN" m= 30 "kg/s" "Some Wrong Solutions with Common Mistakes:" W1_thrust = (Vel2-Vel1)/1000 "Disregarding mass flow rate" W2_thrust = m*Vel2/1000 "Using incorrect relation"
9-199 ··· 9-207 Design and Essay Problems.
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