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20


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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.


Lect-20


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Solution: Problem 1


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


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


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


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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 %


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


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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.


Lect-20


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Solution: Problem 2


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


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


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


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.


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.


Lect-20


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Solution: Problem 3


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.


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


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.


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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.


Lect-20


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Solution: Problem 4


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


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


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


25/33

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%


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.


Lect-20


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Solution: Problem 5


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 %


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


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


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


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


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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