MODULE-III --- GAS POWER CYCLE APPLIED THERMODYYNAMICS

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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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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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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http://www-old.me.gatech.edu/energy/brett/four.htm

http://sounak4u.weebly.com/gas-power-cycle.html

http://www.freestudy.co.uk/thermodynamics/t3201.pdf

http://en.wikipedia.org/wiki/Thermodynamic_cycle

https://wiki.ucl.ac.uk/display/MechEngThermodyn/Gas+Power+Cycles

http://www.youtube.com/watch?v=DiTiLfR5kDg

http://www.youtube.com/watch?v=WhooEaJhMyQ

http://www.youtube.com/watch?v=aAfBSJObd6Y

http://www.youtube.com/watch?v=kJlmRT4E6R0

http://www.youtube.com/watch?v=TBdUcGYo7XA

http://www.youtube.com/watch?v=WTWdzpwvs7k



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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http://www.youtube.com/watch?v=r9q80sSHxKM

http://www.youtube.com/watch?v=JJVCdTL0zGI

http://www.youtube.com/watch?v=dH7CrxaF_XY

http://www.youtube.com/watch?v=zNc6R0FV4Ww

http://www.youtube.com/watch?v=MrArmbBIe5Q

http://www.youtube.com/watch?v=-wtgJkhYPqQ

http://www.youtube.com/watch?v=10q0akdZ7QQ

http://www.youtube.com/watch?v=jSgwIMVI9ew



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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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


Qr=341.07 Kj/Kg


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


= (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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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


T2 = 870.75K

Process 3-4:

Cut off ratio =

=1.2


=

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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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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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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= 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,


= 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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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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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


= 25%.

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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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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,


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


Qr=580.77 Kj/Kg

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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


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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=

T4=763.03 K


Qr=341.07 Kj/Kg


Wnet= 543.28 Kj/Kg

Efficiency =

Efficiency = 61.43%

MEP =

Apply gas law: P1V1= mRT 1


= (R T 1 / T 1) = 0.826 m

3/Kg

P2V2= mRT 2


= (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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T2 = 870.75K

Process 3-4:

Cut off ratio =

=1.2


=

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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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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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,


= 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,


= 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


= 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


= 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


= 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


= 25%.

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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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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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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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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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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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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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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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Documents

MODULE-III --- GAS POWER CYCLE APPLIED THERMODYYNAMICS