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Dr A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINEERING ,RAMANAGARAM
MODULE-III --- GAS POWER CYCLE APPLIED THERMODYYNAMICS 2014
VTU-NPTEL-NMEICT
Project Progress Report
The Project on Development of Remaining Three Quadrants to NPTEL
Phase-I under grant in aid NMEICT, MHRD, New Delhi
DEPARTMENT OF MECHANICAL ENGINEERING,
GHOUSIA COLLEGE OF ENGINEERING,
RAMANARAM -562159
Subject Matter Expert Details
SME Name : Dr.A.R.ANWAR KHAN
Prof & H.O.D Dept of Mechanical Engineering
Course Name:
Applied Thermodynamics
Type of the Course
web
Module
3
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Dr A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINEERING ,RAMANAGARAM
MODULE-III --- GAS POWER CYCLE APPLIED THERMODYYNAMICS 2014
CONTENTS
Sl.
No. DISCRETION
1. Quadrant -2
a. Animations.
b. Videos.
c. Illustrations.
2. Quadrant -3
a. Wikis.
b. Open Contents
3. Quadrant -4
a. Problems.
b. Assignments
c. Self Assigned Q & A.
d. Test your Skills.
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Dr A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINEERING ,RAMANAGARAM
MODULE-III --- GAS POWER CYCLE APPLIED THERMODYYNAMICS 2014
Module-3
GAS POWER CYCLE
UADRANT-2
Animations
(Animation links related ,Gas Power Cycle)
1.Thermodynamic Power Cycles:
http://www-old.me.gatech.edu/energy/brett/four.htm
2. Gas Power Cycles:
http://sounak4u.weebly.com/gas-power-cycle.html
3. Gas Turbine Power Cycles:
http://www.freestudy.co.uk/thermodynamics/t3201.pdf
4. Thermodynamic Cycle:
http://en.wikipedia.org/wiki/Thermodynamic_cycle
5. Gas Power Cycles:
https://wiki.ucl.ac.uk/display/MechEngThermodyn/Gas+Power+Cycles
Videos
(Animation links related , Gas Power Cycle)
1.Introduction to Gas power cycles:
http://www.youtube.com/watch?v=DiTiLfR5kDg
2. Carnot Cycle:
http://www.youtube.com/watch?v=WhooEaJhMyQ
http://www.youtube.com/watch?v=aAfBSJObd6Y
3. Carnot Engine:
http://www.youtube.com/watch?v=kJlmRT4E6R0
4. Gas Turbine Animation:
http://www.youtube.com/watch?v=TBdUcGYo7XA
5. Failure of gas Turbine:
http://www.youtube.com/watch?v=WTWdzpwvs7k
6. Gas Turbine Power Plant:
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Dr A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINEERING ,RAMANAGARAM
MODULE-III --- GAS POWER CYCLE APPLIED THERMODYYNAMICS 2014
http://www.youtube.com/watch?v=r9q80sSHxKM
7. Working of Otto Cycle:
http://www.youtube.com/watch?v=JJVCdTL0zGI
8. Working of Diesel Cycle:
http://www.youtube.com/watch?v=dH7CrxaF_XY
9. Working of Sterling cycle:
http://www.youtube.com/watch?v=zNc6R0FV4Ww
http://www.youtube.com/watch?v=MrArmbBIe5Q
10. p-v diagram:
http://www.youtube.com/watch?v=-wtgJkhYPqQ
11. T-s diagram:
http://www.youtube.com/watch?v=10q0akdZ7QQ
12. Regenerative gas turbine cycle:
http://www.youtube.com/watch?v=jSgwIMVI9ew
ILLUSTRATIONS
1) The compression ratio of an ideal Otto cycle engine is 6.2:1. The pressure and temperature at
the beginning of the compression are 1 bar & 280C. Heat added during the constant volume is
1250 Kj/Kg. Determine the peak pressure and temperature, work output per Kg of air and air
standard efficiency. Assume Cv= 0.717 Kj/Kg K and 1.4 = ال.
Data: P1=100 KPa, T1=301 K, Qs=1250 Kj/Kg,
Process 1-2: adiabatic process
P2=1286.31 KPa
T2 = 624.4K
Heat supply Qs =
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Dr A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINEERING ,RAMANAGARAM
MODULE-III --- GAS POWER CYCLE APPLIED THERMODYYNAMICS 2014
T3=2305.01K
Process 2-3:
P3=4748.6 KPa
Process 3-4:
T4=1111.0 K
Heat rejection Qr= =
Qr=580.77 Kj/Kg
Work output, Wnet = (Qs-Qr)
Wnet= 624.23 Kj/Kg
Efficiency =
Efficiency = 51.8%
2) In an air standard diesel cycle the compression ratio is 16. At the beginning of isentropic
compression the temperature is 150C and pressure is 0.1 MPa. Heat is added until the
temperature reaches to 14800C. Calculate i) cut off ratio, ii) heat supply per Kg of air, iii) cycle
efficiency, iv) work done per Kg of air, v) MEP.
Data : T1=288K, P1=100KPa, T3=1753K, compression ratio r =16
Process 1-2: adiabatic
process
P2=4850.29 KPa
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Dr A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINEERING ,RAMANAGARAM
MODULE-III --- GAS POWER CYCLE APPLIED THERMODYYNAMICS 2014
T2 = 873.05K
Heat supply Qs =
Qs = 884.34 Kj/Kg
Process 2-3:
Cut off ratio =
=2
Process 3-4:
=
T4=763.03 K
Heat rejection Qr= =
Qr=341.07 Kj/Kg
Work output, Wnet = (Qs-Qr)
Wnet= 543.28 Kj/Kg
Efficiency =
Efficiency = 61.43%
MEP =
Apply gas law: P1V1= mRT 1
Specific volume at 1, v1 =
= (R T 1 / T 1) = 0.826 m
3/Kg
P2V2= mRT 2
Specific volume at 2, v2 =
= (R T 2 / p 2) = 0.00516 m
3/Kg
specific swept volume Vs= ( v1-v2) = 0.7744 m3/Kg
MEP = 701.54 KPa
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Dr A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINEERING ,RAMANAGARAM
MODULE-III --- GAS POWER CYCLE APPLIED THERMODYYNAMICS 2014
3) An air standard dual cycle uses one Kg of air and has a compression ratio of 14. The pressure
and temperature at the beginning of compression are 1 bar and 300C respectively. The
temperature at the end of constant volume and constant pressure heat addition are 12000C and
15000C respectively. Calculate heat supply, net work done, efficiency. Take Cp=1.01 Kj/Kg K
and Cv= 0.716 Kj/Kg K.
Data :m=1 Kg, P1=100 KPa, T1=303K, T3=1473K, T4= 1773K, = 1.4
Process 1-2: adiabatic process
T2 = 870.75K
Process 3-4:
Cut off ratio =
=1.2
Process 4-5: adiabatic process
=
T5 = 663.63K
Heat supply: Qs= (heat supply in 2-3 ) + ( heat supply in 3-4 process)
Qs= cv (T3-T2) + cp (T4-T3)
Qs=734.21 Kj/Kg
Heat rejection: Qr= cv (T5-T1)
Qr= 258.21 Kj/Kg
Wnet= (Qs-Qr)
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Dr A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINEERING ,RAMANAGARAM
MODULE-III --- GAS POWER CYCLE APPLIED THERMODYYNAMICS 2014
Wnet = 476 Kj/Kg
Efficiency =
Efficiency = 64.84%
4) A striling regenerative engine working between the temperature of 4000C and 150C. Ratio of
isothermal expansion is 3. Calculate) ideal efficiency, ii) when the efficiency of regenerator is
0.8, calculate efficiency of the engine. Take cp= 0.2375 Kj/Kg K and cv=0.1691 Kj/Kg K.
Data: TH=673K, TL=298K, r=
=
= 3, R= cp - cv = 0.0684 Kj/Kg K,
Regenerative efficiency = 0.8
Case (i) ideal efficiency
Heat supply: Qs= R TH (ln(r))
Qs=50.57 Kj/Kg
Heat rejection: Qr= R TL (ln(r))
Qr= 21.64 Kj/Kg
Wnet= (Qs-Qr)
Wnet = 28.92 Kj/Kg
Efficiency =
Efficiency = 75.2%
Case (ii) the efficiency of regenerator is 0.8
Heat supply: Qs= R TH (ln(r))+ (1- )cv(TH-TL)
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Dr A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINEERING ,RAMANAGARAM
MODULE-III --- GAS POWER CYCLE APPLIED THERMODYYNAMICS 2014
Qs=63.25 Kj/Kg
Efficiency =
Efficiency = 45.5%.
5) A gas turbine unit has a pressure ratio of 6:1 and maximum cycle temperature of 6100C. The
isentropic efficiency of the compressor and turbine are 0.8 and 0.82 respectively. Calculate the
power output, when the air enters the compressor at 150C at the rate of 16 Kg/Sec. Take =1.4 &
cp=1.005 Kj/Kg k for compression, =1.333 & cp=1.11 Kj/Kg k for expansion.
Data: T1=288K, T3=883K,
= 6, comp=0.8, turb=0.8, =1.4 & cp=1.005 Kj/Kg k for compression,
=1.333 & cp=1.11 Kj/Kg k for expansion,
Procee 1-2: Adiabatic
= 1.67
compressor work input, Wcomp=Cp (T2’-T1)
T2 = 481 K Wcomp=242.2 Kj/Kg
Also, Turbine work, Wturb= Cp (T3-T4’)
comp=(T2-T1) / (T2’-T1) Wturb= 290.4 Kj/Kg
T2’ = 529 K Net work, Wnet= (Wturb - Wcomp) = 48.2 Kj/Kg
Procee 3-4: Adiabatic Power, P=(mx Wnet) = 771.2 Kw.
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Dr A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINEERING ,RAMANAGARAM
MODULE-III --- GAS POWER CYCLE APPLIED THERMODYYNAMICS 2014
= 1.565
T4 = 564 K
Also,
turb=(T3-T4’) / (T3-T4)
T4’ = 621.4 K
6) In an air standard regenerative gas turbine cycle the pressure ratio is 5. Air enters the compressor at
1 bar and 300 K and leaves at 490 K. the maximum temperature in the cycle is 1000 K. calculate
the cycle efficiency, given that efficiency of regenerator and adiabatic efficiency of turbine are each
80%. Assume for air, = 1.4.
Data: P1=100KPa, T1=300K, T2’=490, T3=1000K,
= 5 =
, turbine=0.8, =0.8,
Procee 3-4: Adiabatic
= 1.5838
T4 = 631.4 K
Also,
turb=(T3-T4’) / (T3-T4)
T4’ = 705 K
Effectiveness of heat exchanger = (T5-T2’) / (T4’-T2’)
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Dr A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINEERING ,RAMANAGARAM
MODULE-III --- GAS POWER CYCLE APPLIED THERMODYYNAMICS 2014
T5 = 662 K
Compressor work input, Wcomp=Cp (T2’-T1) = 190.9 Kj/Kg
Turbine work, Wturb= Cp (T3-T4’) = 296.5 Kj/Kg
Net work, Wnet= (Wturb - Wcomp) = 105.6 Kj/Kg
Heat supply Qs = Cp (T3-T5) = 339.7 Kj/Kg
Cycle efficiency, cycle=
= 31%.
7) The pressure ratio of an open cycle gas turbine power plant is 5.6. Air is taken at 1 bar and 300C.
The compression is carried out in two stages with perfect intercooling in between. The maximum
temperature in the cycle is limited to 7000C. Assuming the isentropic efficiency of each compressor
as 85% and that of turbine is 90%, determine the power developed and cycle efficiency, if air flow
is 1.2 Kg/Sec. Mass of fuel may be neglected and assume that cp=1.02 Kj/Kg K and = 1.41.
Data: T1=303K, P1=100 KPa, T5=973K, (P4/P1)=5.6, turbine=0.9, comp=0.85, ma=1.2 Kg/Sec,
cp=1.02 Kj/Kg K and = 1.41.
Pressure ratio in each stage of
compressor
,
= = 2.366
Process 1-2: Adiabatic
= 1.284
T2 = 389.23 K = T4
Also,
comp=(T2-T1) / (T2’-T1)
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Dr A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINEERING ,RAMANAGARAM
MODULE-III --- GAS POWER CYCLE APPLIED THERMODYYNAMICS 2014
T2’ = 404.44 K = T4’
Work input to 2 stage of compressor, Wcomp= 2 x m x Cp (T2’-T1) = 248.32 Kw
Process 5-6: Adiabatic
= 1.5838
T6 = 589.7 K
Also,
turb=(T5-T6’) / (T5-T6)
T6’ = 628 K
Turbine work, Wturb= m x Cp (T5-T6’) = 422.28 Kw
Net work, Wnet= (Wturb - Wcomp) = 173.96 Kw
Heat supply Qs = m x Cp (T5-T4’) = 695.92 Kw
Cycle efficiency, cycle=
= 25%.
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Dr A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINEERING ,RAMANAGARAM
MODULE-III --- GAS POWER CYCLE APPLIED THERMODYYNAMICS 2014
QUADRANT-3
Wikis:
(This includes wikis related to Gas Power Cycle contains practical application and research trends in
gas power cycle)
1.http://au.search.yahoo.com/r/_ylt=AwrSbmIWGN5SjjoAgqa7HAx.;_ylu=X3oDMTE1bnFoNmUxB
HNlYwNzcgRwb3MDMQRjb2xvA2dxMQR2dGlkA1ZJUElOMDJfNzQ/SIG=131jqsvjt/EXP=13903
15670/**http%3a//highered.mcgrawhill.com/sites/dl/free/0072884959/240292/Chapter09.ppt
2. Otto Cycle:
http://www.powershow.com/view/109acf-
ZDk5O/The_Otto_and_Dual_Engine_Cycles_powerpoint_ppt_presentation
List of Questions (FAQ):
3.Vapor power cycle:
http://www.powershow.com/view/109b32-
ZGI1Y/Vapor_Power_Cycles_powerpoint_ppt_presentation
Open Contents:
(This includes wikis related to introduction to Gas power cycle contains practical application and
research trends)
1. Modern Engineering Thermodynamics By Robert T. Balmer / Chapter -13 / Vapor and Gas power
cycles / Pages-447 to 525
2.Engineering Thermodynamics By P. K. Nag / Chapter-13 / Gas Power Cycles / Pages-482 to
540
3. Thermal Engineering By R.K. Rajput / Chapter-21 / Gas Power Cycles / Pages-932 to 1003
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Dr A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINEERING ,RAMANAGARAM
MODULE-III --- GAS POWER CYCLE APPLIED THERMODYYNAMICS 2014
QUADRANT-4
Problems
1) The compression ratio of an ideal Otto cycle engine is 6.2:1. The pressure and temperature at
the beginning of the compression are 1 bar & 280C. Heat added during the constant volume is 1250
Kj/Kg. Determine the peak pressure and temperature, work output per Kg of air and air standard
efficiency. Assume Cv= 0.717 Kj/Kg K and 1.4 = ال.
Data: P1=100 KPa, T1=301 K, Qs=1250 Kj/Kg,
Process 1-2: adiabatic process
P2=1286.31 KPa
T2 = 624.4K
Heat supply Qs =
T3=2305.01K
Process 2-3:
P3=4748.6 KPa
Process 3-4:
T4=1111.0 K
Heat rejection Qr= =
Qr=580.77 Kj/Kg
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Dr A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINEERING ,RAMANAGARAM
MODULE-III --- GAS POWER CYCLE APPLIED THERMODYYNAMICS 2014
Work output, Wnet = (Qs-Qr)
Wnet= 624.23 Kj/Kg
Efficiency =
Efficiency = 51.8%
2) In an air standard diesel cycle the compression ratio is 16. At the beginning of isentropic
compression the temperature is 150C and pressure is 0.1 MPa. Heat is added until the temperature
reaches to 14800C. Calculate i) cut off ratio, ii) heat supply per Kg of air, iii) cycle efficiency, iv)
work done per Kg of air, v) MEP.
Data : T1=288K, P1=100KPa, T3=1753K, compression ratio r =16
Process 1-2: adiabatic process
P2=4850.29 KPa
T2 = 873.05K
Heat supply Qs =
Qs = 884.34 Kj/Kg
Process 2-3:
Cut off ratio =
=2
Process 3-4:
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Dr A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINEERING ,RAMANAGARAM
MODULE-III --- GAS POWER CYCLE APPLIED THERMODYYNAMICS 2014
=
T4=763.03 K
Heat rejection Qr= =
Qr=341.07 Kj/Kg
Work output, Wnet = (Qs-Qr)
Wnet= 543.28 Kj/Kg
Efficiency =
Efficiency = 61.43%
MEP =
Apply gas law: P1V1= mRT 1
Specific volume at 1, v1 =
= (R T 1 / T 1) = 0.826 m
3/Kg
P2V2= mRT 2
Specific volume at 2, v2 =
= (R T 2 / p 2) = 0.00516 m
3/Kg
specific swept volume Vs= ( v1-v2) = 0.7744 m3/Kg
MEP = 701.54 KPa
3) An air standard dual cycle uses one Kg of air and has a compression ratio of 14. The pressure
and temperature at the beginning of compression are 1 bar and 300C respectively. The temperature
at the end of constant volume and constant pressure heat addition are 12000C and 15000C
respectively. Calculate heat supply, net work done, efficiency. Take Cp=1.01 Kj/Kg K and Cv=
0.716 Kj/Kg K.
Data :m=1 Kg, P1=100 KPa, T1=303K, T3=1473K, T4= 1773K, = 1.4
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Dr A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINEERING ,RAMANAGARAM
MODULE-III --- GAS POWER CYCLE APPLIED THERMODYYNAMICS 2014
Process 1-2: adiabatic process
T2 = 870.75K
Process 3-4:
Cut off ratio =
=1.2
Process 4-5: adiabatic process
=
T5 = 663.63K
Heat supply: Qs= (heat supply in 2-3 ) + ( heat supply in 3-4 process)
Qs= cv (T3-T2) + cp (T4-T3)
Qs=734.21 Kj/Kg
Heat rejection: Qr= cv (T5-T1)
Qr= 258.21 Kj/Kg
Wnet= (Qs-Qr)
Wnet = 476 Kj/Kg
Efficiency =
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Dr A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINEERING ,RAMANAGARAM
MODULE-III --- GAS POWER CYCLE APPLIED THERMODYYNAMICS 2014
Efficiency = 64.84%
4) A striling regenerative engine working between the temperature of 4000C and 150C. Ratio of
isothermal expansion is 3. Calculate) ideal efficiency, ii) when the efficiency of regenerator is 0.8,
calculate efficiency of the engine. Take cp= 0.2375 Kj/Kg K and cv=0.1691 Kj/Kg K.
Data: TH=673K, TL=298K, r=
=
= 3, R= cp - cv = 0.0684 Kj/Kg K,
Regenerative efficiency = 0.8
Case (i) ideal efficiency
Heat supply: Qs= R TH (ln(r))
Qs=50.57 Kj/Kg
Heat rejection: Qr= R TL (ln(r))
Qr= 21.64 Kj/Kg
Wnet= (Qs-Qr)
Wnet = 28.92 Kj/Kg
Efficiency =
Efficiency = 75.2%
Case (ii) the efficiency of regenerator is 0.8
Heat supply: Qs= R TH (ln(r))+ (1- )cv(TH-TL)
Qs=63.25 Kj/Kg
Efficiency =
Efficiency = 45.5%.
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Dr A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINEERING ,RAMANAGARAM
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5) A gas turbine unit has a pressure ratio of 6:1 and maximum cycle temperature of 6100C. The
isentropic efficiency of the compressor and turbine are 0.8 and 0.82 respectively. Calculate the
power output, when the air enters the compressor at 150C at the rate of 16 Kg/Sec. Take =1.4 &
cp=1.005 Kj/Kg k for compression, =1.333 & cp=1.11 Kj/Kg k for expansion.
Data: T1=288K, T3=883K,
= 6, comp=0.8, turb=0.8, =1.4 & cp=1.005 Kj/Kg k for compression,
=1.333 & cp=1.11 Kj/Kg k for expansion,
Procee 1-2: Adiabatic
= 1.67 compressor work input, Wcomp=Cp (T2’-T1)
T2 = 481 K Wcomp=242.2 Kj/Kg
Also, Turbine work, Wturb= Cp (T3-T4’)
comp=(T2-T1) / (T2’-T1) Wturb= 290.4 Kj/Kg
T2’ = 529 K Net work, Wnet= (Wturb - Wcomp) = 48.2 Kj/Kg
Procee 3-4: Adiabatic Power, P=(mx Wnet) = 771.2 Kw.
= 1.565
T4 = 564 K
Also,
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turb=(T3-T4’) / (T3-T4)
T4’ = 621.4 K
6) In an air standard regenerative gas turbine cycle the pressure ratio is 5. Air enters the compressor at
1 bar and 300 K and leaves at 490 K. the maximum temperature in the cycle is 1000 K. calculate
the cycle efficiency, given that efficiency of regenerator and adiabatic efficiency of turbine are each
80%. Assume for air, = 1.4.
Data: P1=100KPa, T1=300K, T2’=490, T3=1000K,
= 5 =
, turbine=0.8, =0.8,
Procee 3-4: Adiabatic
= 1.5838
T4 = 631.4 K
Also,
turb=(T3-T4’) / (T3-T4)
T4’ = 705 K
Effectiveness of heat exchanger = (T5-T2’) / (T4’-T2’)
T5 = 662 K
Compressor work input, Wcomp=Cp (T2’-T1) = 190.9 Kj/Kg
Turbine work, Wturb= Cp (T3-T4’) = 296.5 Kj/Kg
Net work, Wnet= (Wturb - Wcomp) = 105.6 Kj/Kg
Heat supply Qs = Cp (T3-T5) = 339.7 Kj/Kg
Cycle efficiency, cycle=
= 31%.
7) The pressure ratio of an open cycle gas turbine power plant is 5.6. Air is taken at 1 bar and 300C.
The compression is carried out in two stages with perfect intercooling in between. The maximum
temperature in the cycle is limited to 7000C. Assuming the isentropic efficiency of each compressor
as 85% and that of turbine is 90%, determine the power developed and cycle efficiency, if air flow
is 1.2 Kg/Sec. Mass of fuel may be neglected and assume that cp=1.02 Kj/Kg K and = 1.41.
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Data: T1=303K, P1=100 KPa, T5=973K, (P4/P1)=5.6, turbine=0.9, comp=0.85, ma=1.2 Kg/Sec,
cp=1.02 Kj/Kg K and = 1.41.
Pressure ratio in each stage of
compressor,
=
= 2.366
Procee 1-2: Adiabatic
= 1.284
T2 = 389.23 K = T4
Also,
comp=(T2-T1) / (T2’-T1)
T2’ = 404.44 K = T4’
Work input to 2 stage of compressor, Wcomp= 2 x m x Cp (T2’-T1) = 248.32 Kw
Procee 5-6: Adiabatic
= 1.5838
T6 = 589.7 K
Also,
turb=(T5-T6’) / (T5-T6)
T6’ = 628 K
Turbine work, Wturb= m x Cp (T5-T6’) = 422.28 Kw
Net work, Wnet= (Wturb - Wcomp) = 173.96 Kw
Heat supply Qs = m x Cp (T5-T4’) = 695.92 Kw
Cycle efficiency, cycle=
= 25%.
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Dr A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINEERING ,RAMANAGARAM
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Frequently asked Questions.
(1) Derive an expression for thermal efficiency of an air standard Otto cycle.
(2) Derive an expression for thermal efficiency of an air standard Diesel cycle.
(3) Derive an expression for thermal efficiency of an air standard Dual cycle.
(4) What are the drawbacks of Carnot cycle?
(5) Derive an expression for efficiency of Bryton cycle in terms of pressure ratio.
(6) Derive an expression for pressure ratio for maximum specific work output of Bryton cycle.
(7) Comparison of Otto, Diesel and Dual cycles for same compression ratio.
(8) Comparison of Otto, Diesel and Dual cycles for same maximum pressure and temperature.
(9) Explain the methods of improving the efficiency of Bryton cycle.
(10) Two engines are operates on Otto and Diesel cycle with the following date. Maximum temperature
1400K, exhaust temperature 700K. At the beginning of compression air at 0.1MPa, 300K. Estimate the
compression ratio, the maximum pressure, cycle efficiency and rate of work output (for 1 Kg/Sec of
air) of the respective cycle.
(11) The minimum pressure and temperature of an air standard Carnot cycle are 1 bar and 150C respectively.
The pressure after isothermal compression is 3.5 bar and the pressure after isentropic compression is
10.5bar. determine i) efficiency, ii) M.E.P., iii) power developed if engine takes 2 cycles per sec.
(12) A gas turbine plant draws in air at 1.013bar and 100C and has a pressure ratio of 5.5. The maximum
temperature in the cycle is limited to 7500C. Compressor, turbine and heat exchanger isentropic
efficiencies are 82%, 85% and 70% respectively. For an air flow of 40 Kg/sec, find i) overall
efficiency, ii) Turbine output, iii) Air- fuel ratio if the calorific value of fuel is 45.22 MJ/Kg.
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Dr A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINEERING ,RAMANAGARAM
MODULE-III --- GAS POWER CYCLE APPLIED THERMODYYNAMICS 2014
Assignments:
1) A Carnot engine rejects heat to the sink at 320C and has a thermal efficiency of 52.3%. the
work output from the engine is 120 Kj, determine i) maximum temperature of the engine, ii)
heat added, iii) change in entropy during heat supply.
Solution: T l=305K,W net=120 Kj, efficiency=52.3%
For Carnot cycle, Efficiency= [Th/ ( T h-Tl)]
i)Maximum temperature of the engine Th = 639.41 K
Also,efficiency = [Wnet / heat added]
ii)Heat added = 229.44 K
iii)Change in entropy, (dS) [heaty added / Th]
Therefore, dS=0.358 Kj/ Kg K
2) An engine working on Otto cycle has a volume of 0.45 m3, pressure 1 bar, and temperature
300C at the beginning of compression. At the end of compression stroke pressure is 11 bar. 210
Kj / cycle heat is added at constant volume. Determine
i) temperature and pressure at all silent points ii) % clearance, iii) cycle efficiency
iv) net work / cycle, v) M.E.P, vi) ideal power developed if number of working cycle is 210
per minutes.
Answers: i) temperature and pressure at all silent points T1=303K, T2= 601.6K, T3=
1167.35K, T4= 587.9K, P1=100KPa, P2=1100 KPa, P3= 2134.44 KPa, P4=193.49 KPa ii) %
clearance – 21.95%, iii) cycle efficiency – 49.43%,
iv) net work / cycle -104.25 Kj/cycle, v) M.E.P,-282.52 KPa, vi) ideal power developed if
number of working cycle is 210 per minutes-364.89 KW.
3) Derive an expression for thermal efficiency of an air standard Otto cycle.
4) Derive an expression for thermal efficiency of an air standard Diesel cycle.
5) Derive an expression for thermal efficiency of an air standard Dual cycle.
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Dr A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINEERING ,RAMANAGARAM
MODULE-III --- GAS POWER CYCLE APPLIED THERMODYYNAMICS 2014
Self Answered Question & Answer
(1) In an air standard diesel cycle, the compression ratio is 16. At the beginning of isentropic
compression, the temperature is 150C and pressure is 0.1 MPa. Heat is added until the temperature at
the end of the constant pressure process is 14800C. calculate i) cut-off ratio, ii) heat supply per Kg of
air, iii) cycle efficiency and iv) M.E.P.
Answers : cut-off ratio- 2.01, heat supply per Kg of air – 884.4 Kj/Kg, cycle efficiency – 61.2%,
M.E.P.- 698.45 KPa
(2) Two engines are operates on Otto and Diesel cycle with the following date. Maximum temperature
1400K, exhaust temperature 700K. At the beginning of compression air at 0.1MPa, 300K. Estimate the
compression ratio, the maximum pressure, cycle efficiency and rate of work output (for 1 Kg/Sec of
air) of the respective cycle.
Answers : For Otto cycle: Compression ratio-5.656, the maximum pressure – 2.639 bar, cycle
efficiency – 49.9%, rate of work output – 287.22 Kw.
For Diesel cycle: Compression ratio-10.36, the maximum pressure – 2.639 MPa, cut off ratio
- cycle efficiency – 55%, rate of work output – 351.68 Kw
(3) The minimum pressure and temperature of an air standard Carnot cycle are 1 bar and 150C respectively.
The pressure after isothermal compression is 3.5 bar and the pressure after isentropic compression is
10.5bar. determine i) efficiency, ii) M.E.P., iii) power developed if engine takes 2 cycles per sec.
Answers : efficiency – 26.85%, M.E.P. – 0.529 bar, power developed – 76 Kw.
(4) A gas turbine power plant operates on the simple Bryton cycle with air as the working fluid
delivers 32 MW of power. The minimum and maximum temperatures in the cycle are 310 and 900 K,
and the pressure of the air at the compressor exit is 8 times the value at the compressor inlet. Assume
an isentropic efficiency of 80% for the compressor and 86% for turbine; determine the mass flow rate
of air through the cycle.
Answers:
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Dr A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINEERING ,RAMANAGARAM
MODULE-III --- GAS POWER CYCLE APPLIED THERMODYYNAMICS 2014
5) A Stirling regenerative engine working between the temperature of 4000C and 15
0C. Ratio of
isothermal expansion is 3. Calculate 1) ideal efficiency 2) when the efficiency of the regenerative is
0.8, calculate efficiency of the engine. Take Cp= 0.2375 Kj / Kg K and Cv = 0.1691 Kj / Kg K
Answers: Ideal efficiency – 57.2%, when the efficiency of the regenerative is 0.8
efficiency of the engine – 45.5%
6) Air standard limited pressure cycle has a compression ratio of 15 and compression begins at 0.1
MPa and 400C. the maximum pressure is limited to 6 MPa and heat added is 1.675 Mj / Kg. calculate
heat supply at constant volume per Kg of air, heat supply at constant pressure per Kg of air, work
done, cycle efficiency, cut off ratio and M.E.P.
Answers: heat supply at constant volume per Kg of air – 235.03 Kj / Kg, heat supply at constant
pressure per Kg of air – 1440 Kj / Kg, cycle efficiency – 60.56%, cut off ratio – 2.14 and
M.E.P.- 1221.7 KPa.
7) The stroke and bore of C.I. engine cylinder is 250 mm and 150mm respectively. If the clearance
volume is 0.0004 m3 and fuel injection takes place at constant pressure for 5% of the stroke.
Determine the air standard efficiency of the engine.
Answers: the air standard efficiency of the engine – 59.33%
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Dr A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINEERING ,RAMANAGARAM
MODULE-III --- GAS POWER CYCLE APPLIED THERMODYYNAMICS 2014
Test Your Skills:
Fill up the blanks
(1) Compression ratio is the ratio of …………………………………………..
(2) Otto cycle is also called as ……………………………………………….
(3) Working fluid in air standard cycle is ………………………………..
(4) Cut off ratio is the ratio of ………………………………………..
(5) In air standard dual cycle heat supply takes during …………………… and ……………… process
(6) ………………………. Cycle is maximum efficiency air standard cycle.
(7) Bryton cycle efficiency is depends on …………………..
(8) Compression during air standard Stirling cycle ……………………
(9) For given compression ratio …………………… cycle is most efficient.
(10) For same maximum pressure and temperature …………………. Cycle is most efficient.
(11) Air standard dual cycle is also called as ……………………………….
ANSWERS:
(1) Total volume to the clearance volume
(2) Constant volume cycle
(3) Air
(4) Volume after heat supply to the volume before heat supply
(5) Isochoric and Isobaric
(6) Carnot
(7) Pressure ratio
(8) Reversible isothermal process
(9) Otto
(10) Diesel
(11) Limited Pressure or Semi Diesel cycle.
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Dr A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINEERING ,RAMANAGARAM
MODULE-III --- GAS POWER CYCLE APPLIED THERMODYYNAMICS 2014
Match the following
Part A Part B
1. Thermal efficiency of a heat engine a. Work done per cycle to the stroke
Volume.
2. Mean effective pressure of an engine b. Compression ratio
3. Air standard diesel cycle c. Locomotive engine
4. Gas power cycle is not used for d. Ratio of work done to heat supplied.
5. Sterling cycle consists e. Constant volume cycle
6. Air standard duel cycle f. Two isothermal |& isentropic process.
7. Air standard otto cycle g. Heat supplied at constant pressure
8. Effeciency of air standard otto cycle depends on h. Two isothermal & constant volume
process.
9. Cornot cycle i. Semi diesel cycle.
10. Air standard dual cycle is also called j. Heat supplied at constant volume &
constant pressure process.
Answers :
1-d, 2-a, 3-g, 4-c, 5-h, 6-j, 7-e, 8-b, 9-f, 10-I.
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Dr A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINEERING ,RAMANAGARAM
MODULE-III --- GAS POWER CYCLE APPLIED THERMODYYNAMICS 2014
TRUE (T)/ FALSE (F)
1) Carnot cycle is 100% efficiency cycle. ------ (F)
2) Otto cycle is constant volume cycle. ---------- (T)
3) In Diesel cycle heat supply takes at constant volume process. ----- (F)
4) Dual cycle is also called as semi diesel cycle. ----------- (T)
5) In regenerative gas turbine cycle, heat supply is reduced. ---- (T)
6) In dual cycle heat rejection takes at constant volume process. ---- (T)
7) For given compression ratio diesel cycle is more efficient. ---- (F)
8) Compression ratio is ratio of clearance volume to the total volume of the cylinder. ----------- (F)
9) In Intercooling gas turbine cycle, cooling of working fluid takes between two compressors. ---- (T)
10) In Regenerator working fluid is heated by exhaust gas. ----- (T)
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