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Ideal Rankine Cycle

(a) Schematic representation of an ideal Rankine

cycle (b) T-s diagram of an ideal Rankine cycle

Application of the First law of thermodynamics

to the control volume (pump, steam generator,

turbine and condenser), gives

Work done on pump, per kg of water, WP= h2-h1

Energy added in steam generator, q1= h3-h2

Work delivered by turbine, WT= h3-h4

Energy rejected in the condenser, q2= h4-h1

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The thermal efficiency of the Rankine cycle is

given by,

23

1243

23

1423

1

21 )()()()(

hh

hhhh

hh

hhhh

q

qq

−−−−=

−−−−=−=η 

η= Net work done

----------------------

Energy absorbed

Practical Rankine cycle

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Pump and Turbine do not operate isentropically

in practice.

The practical Rankine cycle is shown as 1-2’-3-

4’-1.

In the actual turbine, the work delivered is less

than the isentropic turbine. Similarly, the work 

consumed by an actual pump is greater than the

work consumed by an isentropic pump.

That is,

h3-h4’ < h3-h4

h2’-h1 > h2-h1

Thermal efficiency of a practical Rankine cycle,

'

23

1

'

2

'

43 )()(

hh

hhhh

−−−−

=η 

The performance of an actual turbine or pump is

usually expressed in terms of isentropic

efficiency.

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Isentropic efficiency of turbine (ηT) is defined as

the ratio of ‘Work delivered by actual turbine’ to

‘Work delivered by an isentropic turbine’.

43

'

43

hh

hhT  −

−=η 

Isentropic efficiency of pump (ηP) is defined as

the ratio of ‘Work required by isentropic pump’

to ‘Work required by actual pump’

1

'

2

12

hh

hh P  −

−=η 

Methods to increase the efficiency of the

Rankine cycle

Basic idea: Increase the average temperature at

which heat is transferred to the working fluid inthe boiler, or decrease the average temperature

at which heat is rejected from the working fluid

in the condenser.

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1. Lowering the condenser Pressure:-

Lowering the operating pressure of the

condenser lowers the temperature at which heat

is rejected. The overall effect of lowering the

condenser pressure is an increase in the thermal

efficiency of the cycle.

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2. Superheating the steam to high temperatures:-

The average temperature at which heat is added

to the steam can be increased without increasing

the boiler pressure by superheating the steam to

high temperatures.

Superheating the steam to higher temperatures

has another very desirable effect: It decreases

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the moisture content of the steam at the turbine

exit.

3. Increasing the Boiler pressure:-

Increasing the operating pressure of the boiler,

automatically raises the temperature at which

 boiling takes place.

This raises the average temperature at which

heat is added to the steam and thus raises the

thermal efficiency of the cycle..

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Reheat Rankine Cycle

(a) schematic representation of a reheat Rankine

cycle (b) T-s diagram of a reheat Rankine cycle

The energy added ( per unit mass of steam ) in

the steam generator is given by,

)()( 45231 hhhhq −+−=

The energy rejected in the condenser,

162 hhq −=

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The thermal efficiency,

)()(

)()()(

4523

164523

hhhh

hhhhhh

−+−−−−+−=η 

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Regenerative Cycle

(a) schematic diagram (b) T-s diagram

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Consider the feed water heater as the control

volume and apply the first law of  

thermodynamics to obtain,

382 mmm =+  

and 338822 hmhmhm =+

or  8828333

)( hmhmmhm

+−=

or  2

3

88

3

83 1 h

m

mh

m

mh   

 

  

 −+  

 

  

 =

Let, 3

8

m

m

=Y’= the fraction of steam extractedfrom the turbine for preheating

( ) 283 1 hY hY h ′−+′=

Energy added in the boiler per unit mass of the

working fluid,

471 hhq −=

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Energy rejected in the condenser,

( ) )('1 192 hhY q −−=Thermal efficiency,

( )

)(

)('1)(

47

1947

hh

hhY hh

−−−−−

=η 

The work output of the turbines =

( ) )('1)( 9887 hhY hh −−−−

Work spent on the pumps =

( ) )('1)( 1234 hhY hh −−−−

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Air standard Otto Cycle

Air standard Otto cycle on (a) P-v diagram (b)

T-s diagram

 Processes: -

0-1: a fresh mixture of fuel-air is drawn into the

cylinder at constant pressure

1-2: isentropic compression

2-3: energy addition at constant volume

3-4: isentropic expansion

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4-1: combustion products leave the cylinder 

1-0: the piston pushes out the remaining

combustion products at constant pressure

Since the net work done in processes 0-1 and 1-0

is zero, for thermodynamic analysis, we consider 

the 1-2-3-4 only.

The thermal efficiency of the cycle is given by

1

21

1 Q

QQ

Q

W net  −==η 

where Q1 and Q2 denote the energy absorbed andrejected as heat respectively.

For a constant volume process Q=∆U. If ‘m’ is

the mass of the air which is undergoing the

cyclic process,

T mC U  v∆=∆

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Energy is absorbed during the process 2-3

Energy is rejected during the process 4-1

Hence,

( )

23

14

14142

23231

1

)(

T T T T 

T T mC U U Q

T T mC U U Q

v

v

−−−=∴

−=−=−=−=

η 

For an ideal gas undergoing an isentropic

 process (process 1-2 and 3-4),

1−γ  Tv = constant

Hence,

1

1

2

2

1

−

   

 

 

 =

γ  

v

v

T 

T 

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and

1

4

3

3

4

−

   

  

 =

γ  

v

v

T 

T 

But v1=v4 and v2=v3. Hence we get,

3

4

2

1

T 

T 

T 

T = or 

3

2

4

1

T 

T 

T 

T =

3

2

4

1 11T 

T 

T 

T −=− or 

3

23

4

14

T 

T T 

T 

T T  −=

−

2

1

3

4

23

14

T 

T 

T 

T 

T T 

T T 

==−−

Hence,

1

0

1

1

2

2

1 1111

−−

   

  

 −=  

 

  

 −=−=

γ  γ  

η r v

v

T 

T 

Where the compression ratio r 0 is defined as

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2

1

0v

vr  =

Sometimes it is convenient to express the

 performance of an engine in terms of  Mean

effective Pressure, Pm, defined as the ratio of 

“Net work done” to “Displacement volume”

21 vvW  P m −

=  

)( 21 vv P W  m −=

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Thermal efficiency of the ideal Otto cycle as a

function of compression ratio (γ =1.4)

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The thermal efficiency of the Otto cycle

increases with the specific heat ratio, γ  of the

working fluid.

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Air standard Diesel cycle

Diesel cycle on (a) P-v diagram (b) T-s diagram

 Processes: -

0-1: fresh air is drawn into the cylinder 

1-2: isentropic compression

2-3: constant pressure energy addition

3-4: isentropic expansion

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4-1: combustion products leave the cylinder 

1-0: remaining combustion products are

exhausted at constant pressure

Defining cutoff ratio, r c as,

2

3

v

v

r c =

For a constant pressure process (2-3),

Q=∆H.

Hence, the energy addition during process 2-3,

)()( 2323231 T T mC hhm H  H Q  p −=−=−=

where ‘m’ is the mass of gas undergoing the

cyclic change.

The energy rejection during the process 4-1,

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)()( 1414142 T T mC uumU U Q v −=−=−=

The thermal efficiency, η is given by

   

  

 −

  

 

 

 

 −

−=−−−=

−

−−−=

−=

1

1

1)(

1

)(

)()(

2

3

2

1

4

1

23

14

23

1423

1

21

T 

T T 

T 

T T 

T T T T 

T T mC 

T T mC T T mC 

Q

QQ

 p

v p

γ  γ  

η 

η 

Since the process 1-2 is isentropic,

1

0

1

1

2

2

1 1−−

   

  

 =  

 

  

 =

γ  γ  

r v

v

T 

T 

Since the process 4-1 is a constant volume

 process,

   

  

 =  

 

  

 ==

1

2

3

4

1

3

3

4

1

4

1

4

 P 

 P 

 P 

 P 

 P 

 P 

 P 

 P 

 P 

 P 

T 

T 

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since P2=P3

The processes 1-2 and 3-4 are isentropic. Hence,

γ  

   

  

 =

4

3

3

4

v

v

 P 

 P and

γ  

   

  

 =

2

1

1

2

v

v

 P 

 P 

Hence we get,

γ  

γ  γ  γ  

cr v

v

v

v

v

v

T 

T =  

 

  

 =  

 

  

    

  

 =

2

3

2

1

4

3

1

4

For the constant pressure process,

cr v

v

T 

T ==

2

3

2

3

Hence the efficiency becomes,

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−−

−= − 1

111

1

0 c

c

r 

r 

r 

γ  

γ  γ  

η 

The mean effective pressure of an air standard

diesel cycle is given by,

( )[ ]

)1)(1(

1()1

0

001

−−

−−−=

r 

r r r r  P  P  cc

m

γ  

γ  γ  γ  

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Thermal efficiency of the ideal diesel cycle as a

function of compression and cutoff ratios

(γ =1.4)

Air standard Dual cycle

Dual cycle on (a) P-v diagram (b) T-s diagram

Energy addition is in two stages: Part of energy

is added at constant volume and part of the

energy is added at constant pressure

Energy added, q1

)()( 34231 T T C T T C q  pv −+−=

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Energy rejected, q2

)( 152 T T C q v −=

Thermal efficiency, η

)()(

)(1

)()(

)(1

1

3423

15

3423

15

1

2

T T T T 

T T T T C T T C 

T T C 

q

q

 pv

v

−+−−

−=−+−

−−=

−=

γ  η 

η 

η 

The efficiency can be expressed also in terms of,

Compression ratio, r 0 = V1/V2

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Cut-off ratio, r c = V4/V3

Constant volume pressure ratio, r vp= P3/P2