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In this lecture ...
Solve numerical problems
Gas power cycles: Otto, Diesel, dual cycles
Gas power cycles: Brayton cycle, variants
of Brayton cycle
Thermodynamic property relations
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay2
Lect-20
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Problem 1
In an air standard Otto cycle, thecompression ratio is 7 and thecompression begins at 35oC and 0.1
MPa. The maximum temperature of thecycle is 1100oC. Find (a) thetemperature and the pressure at variouspoints in the cycle, (b) the heat supplied
per kg of air, (c) work done per kg of air,(d) the cycle efficiency and (e) the MEPof the cycle.
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay3
Lect-20
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Solution: Problem 1
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay4
Lect-20
v
P
1
3
42
qin
qout
Isentropic
T1=35oC=308 K
P1=0.1 MpaT3=1100
oC=1373 Kr=v1/v2=7
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Solution: Problem 1
Since process, 1-2 is isentropic,
Hence, P2=1524 kPa
Hence, T2=670.8 K
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay5
Lect-20
24.157 4.1
2
1
1
2 ==
=
v
v
P
P
178.27
14.1
1
2
1
1
2
==
=
v
v
T
T
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Solution: Problem 1
For process, 2-3,
P3=3119.34 kPa. Process 3-4 is again isentropic,
Hence, T2=630.39 K
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay6
Lect-20
34.311915248.607
1373, 2
2
33
3
33
2
22 ==== PT
TP
T
vP
T
vP
K39.630178.2
1373
178.27
4
14.1
1
3
4
4
3
==
==
=
T
v
v
T
T
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Solution: Problem 1
Heat input,
Qin=cv(T3-T2)
=0.718(1373670.8)
=504.18 kJ/kg
Heat rejected,
Qout=cv(T4-T1)
=0.718(630.34308)=231.44 kJ/kg
The net work output, Wnet=Qin-Qout
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay7
Lect-20
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Solution: Problem 1
The net work output,
Wnet=Qin-Qout=272.74 kJ/kg
Thermal efficiency, th,otto=Wnet/Qin=0.54
=54 %
Otto cycle thermal efficiency,th,otto =1-1/r
-1 = 1-1/70.4
= 0.54 or 54 %
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay8
Lect-20
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Solution: Problem 1
v1=RT1/P1=0.287x308/100=0.844 m3/kg
MEP = Wnet/(v1v2) = 272.74/v1 (1-1/r)
=272.74/0.844(1-1/7)
=360 kPa
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay9
Lect-20
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Problem 2
In a Diesel cycle, the compression ratiois 15. Compression begins at 0.1 Mpa,40oC. The heat added is 1.675 MJ/kg.
Find (a) the maximum temperature inthe cycle, (b) work done per kg of air(c) the cycle efficiency (d) thetemperature at the end of the isentropic
expansion (e) the cut-off ratio and (f)the MEP of the cycle.
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay10
Lect-20
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Solution: Problem 2
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay11
Lect-20
v
P
1
3
4
2
qin
qout
Isentropic T1=40oC=313 K
P1=0.1 MpaQin=1675 MJ/kgr=v1/v2=15
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Solution: Problem 2
It is given that Qin = 1675 MJ/kg
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay12
Lect-20
/kgm06.015/898.015/
/kgm898.0100
313287.0
3
12
3
1
11
===
=
==
vv
P
RTv
K66.924954.2313
954.215
)(
2
4.0
1
2
1
1
2
23
==
==
=
=
T
v
v
T
T
TTcQ pin
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Solution: Problem 2
Hence, the maximum temperature is 2591.33 K
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay13
Lect-20
max3
3
K33.2591
)66.924(005.11675
TT
TQin
==
==
/kgm168.006.066.924
33.2591kPa4431
31.4415
3
2
2
33
3
33
2
22
2
4.1
2
1
1
2
====
=
==
=
vT
Tv
T
vP
T
vP
P
v
v
P
P
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Solution: Problem 2
The cut-off ratio is 2.8.
=948.12 kJ/kg
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay14
Lect-20
8.206.0
168.0
2
3 ===v
vrc
88.7261675done,Net work
kJ/kg88.726)3134.1325(718.0)(Q
K37.1325
898.0
168.033.2591
14out
4.01
4
334
=====
=
=
=
outinnet
v
QQW
TTc
v
vTT
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Solution: Problem 2
Therefore, thermal efficiency,
th=Wnet/Qin=948.12/1675=0.566 or 56.6%
The cycle efficiency can also be calculated usingthe Diesel cycle efficiency determined earlier.
The mean effective pressure is 1131. 4 Kpa.
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay15
Lect-20
kPa4.113106.0898.0
12.948
21
=
=
=vv
WMEP net
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Problem 3
An air-standard Ericsson cycle has anideal regenerator. Heat is supplied at1000C and heat is rejected at 20C. Ifthe heat added is 600 kJ/kg, find thecompressor work, the turbine work, andthe cycle efficiency.
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay16
Lect-20
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Solution: Problem 3
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay17
Lect-20
s
qin
qout
1
34
2Regeneration
Isobaric
T
T1=T2=1000C=1273.15 KT3=T4=20C=293.15 K
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Solution: Problem 3
Since the regenerator is given as ideal,
-Q2-3 = Q1-4
Also in an Ericsson cycle, the heat isinput during the isothermal expansionprocess, which is the turbine part of thecycle. Hence the turbine work is 600
kJ/kg.
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay18
Lect-20
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Solution: Problem 3
Thermal efficiency of an Ericsson cycle isequal to the Carnot efficiency.
th=th, Carnot=1-TL/TH
=1-293.15/1273.15=0.7697
Therefore the net work output is equal
to:wnet= thQH
= 0.7697600=461.82 kJ/kg
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay19
Lect-20
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Solution: Problem 3
The compressor work is equal to thedifference between the turbine work andthe net work output:
wc
=wt
-wnet
=600-461.82 =138.2 kJ/kg
In the Ericsson cycle the heat is rejected
isothermally during the compressionprocess. Therefore this compressor workis also equal to the heat rejected duringthe cycle.
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay20
Lect-20
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Problem 4
In a Brayton cycle based power plant,the air at the inlet is at 27oC, 0.1 MPa.The pressure ratio is 6.25 and the
maximum temperature is 800oC. Find(a) the compressor work per kg of air(b) the turbine work per kg or air (c)the heat supplied per kg of air, and (d)the cycle efficiency.
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay21
Lect-20
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Solution: Problem 4
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay22
Lect-20
s
T qin
qout
1
3
42
Isobaric
T1 = 27C = 300 K
P1 = 100 kParp = 6.25T3 = 800C = 1073 K
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Solution: Problem 4
Since process, 1-2 is isentropic,
The compressor work per unit kg of airis 207.72 kJ/kg
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay23
Lect-20
kJ/kg72.207
)30069.506(005.1)(
K69.506
689.125.6
12
2
4.1/)14.1(/)1(
1
2
=
==
=
===
TTcW
T
rT
T
pcomp
p
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Solution: Problem 4
Process 3-4 is also isentropic,
The turbine work per unit kg of air is439.89 kJ/kg
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay24
Lect-20
kJ/kg89.439
)29.6351073(005.1)(K29.635
689.125.6
43
4
4.1/)14.1(/)1(
4
3
=
===
===
TTcWT
rT
T
pturb
p
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Solution: Problem 3
Heat input, Qin,
Heat input per kg of air is 569.14 kJ/kg Cycle efficiency,
th=(Wturb-Wcomp)/Qin
=(439.89-207.72)/569.14=0.408 or 40.8%
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay25
Lect-20
kJ/kg14.569
)69.5061073(005.1)( 23
=
== TTcQ pin
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Problem 5
Solve Problem 3 if a regenerator of 75%effectiveness is added to the plant.
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay26
Lect-20
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Solution: Problem 5
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay27
Lect-20
s
Tqin
qout
1
3
4
2
Regeneration55
qregen
qsaved=qregen
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Solution: Problem 5
T4, Wcomp, Wturb remain unchanged
The new heat input, Qin=cp(T3-T5)
=472.2 kJ/kg
Therefore th=(Wturb-Wcomp)/Qin=(439.89-207.72)/472.2
=0.492 or 49.2 %
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay28
Lect-20
KT
Tor
TT
TT
14.603
75.069.50629.635
69.506,
75.0
5
5
24
25
=
=
=
=
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Exercise Problem 1
A gasoline engine receives air at 10oC,100 kPa, having a compression ratio of9:1 by volume. The heat addition bycombustion gives the highesttemperature as 2500 K. use cold airproperties to find the highest cyclepressure, the specific energy added by
combustion, and the mean effectivepressure.
Ans: 7946.3 kPa, 1303.6 kJ/kg, 0.5847,1055 kPa
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay29
Lect-20
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Exercise Problem 2
A diesel engine has a compression ratioof 20:1 with an inlet of 95 kPa, 290 K,with volume 0.5 L. The maximum cycle
temperature is 1800 K. Find themaximum pressure, the net specific workand the thermal efficiency.
Ans: 6298 kPa , 550.5 kJ/kg, 0.653
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay30
Lect-20
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Exercise Problem 3
Consider an ideal Stirling-cycle engine inwhich the state at the beginning of theisothermal compression process is 100
kPa, 25C, the compression ratio is 6,and the maximum temperature in thecycle is 1100C. Calculate the maximumcycle pressure and the thermal efficiency
of the cycle with and withoutregenerators.
Ans: 2763 kPa, 0.374, 0.783
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay31
Lect-20
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Exercise Problem 4
A large stationary Brayton cycle gas-turbinepower plant delivers a power output of 100 MWto an electric generator. The minimumtemperature in the cycle is 300 K, and the
maximum temperature is 1600 K. The minimumpressure in the cycle is 100 kPa, and thecompressor pressure ratio is 14 to 1. Calculatethe power output of the turbine. What fraction of
the turbine output is required to drive thecompressor? What is the thermal efficiency ofthe cycle?
Ans: 166.32 MW, 0.399, 0.530
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay32
Lect-20
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In the next lecture ...
Properties of pure substances Compressed liquid, saturated liquid,
saturated vapour, superheated vapour
Saturation temperature and pressure
Property diagrams of pure substances
Property tables
Composition of a gas mixture
P-v-T behaviour of gas mixtures Ideal gas and real gas mixtures
Properties of gas mixtures
33
Lect-20
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