Ir. TRI TJAHJONO, MT/INTERNAL COMBUSTION ENGINE

OTTO CYCLE

The Otto cycle is the theoretical cycle for the spark-ignition (SI) engine

0-1 suction stroke 1-2 compressed isentropically 2-3 heat is added at constant volume

3-4 isentropic expansion 4-1 the heat is rejected

Consider 1 kg of air.

Heat added during process 2-3 = cv (T3 - T2)

Heat rejected during process 4-1 = cv(T4 – T1)

∴Work done = heat added – heat rejected

= cv (T4 – T2) – cv(T4 – T1)

Thermal efficiency, suppliedheat

donework η =

Or( )

)TT(c

)TT(cTTc

23v

14v23v

−−−−

=η

)TT(

)TT(1

23

14

−−

−=

Now compression ratio 2

1

v

v= Expansion ratio r

v

v

3

4 =

For ideal gas RTpv = and .ttanconspv =γ

These equations yield

1

4

3

1

3

4

1

2

1

1

2 rT

T

v

v

v

v

T

T −−−

==

=

= γ

γγ

21

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1γ12

1γ43 rTT and rTT −− ==∴

Hence, substituting

( ) ( ) 1114

141

11

4

14

r

11

rTT

TT1

rTrT

TT1 −−−− −=

−−−=

−−−= γγγγη

The efficiency can also be expressed in terms of temperatures T3 and T4.

Now, since 4

3

2

1

v

v

v

v= , it follows that

4

1

3

2

4

3

1

2

T

T

T

Tor

T

T

T

T==

4

1

3

2

T

T

T

T =

and hence,

4

14

3

23

4

1

3

2

T

TT

T

TTor

T

T1

T

T1

−=

−−=−

Therefore,

3

4

23

14

T

T

TT

TT=

−−

Hence, substituting, in (a) we get

2

1

3

4

T

T1

T

T1 −=−=η

Otto cycle efficiency at different compression ratios and γ

Helium (γ = 1.66) monoatomic gas

Argon (γ = 1.97) monoatomic gas

22

-1

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Mean effective pressure (mep) of Otto cycle

Let clearance volume be unity, i.e, V2 = V3 = 1, and V1 – V4 = r

and α=2

3

p

p

Now 4

3

1

2

p

pr

p

p == γ

α==∴1

4

2

3

p

p

p

p

Work done = area of the p-v diagram

1

vpvp

1

vpvp 11224433

−−

−−−

=γγ

−−

−

−= 1

rp

prp1

rp

prp

1

1

1

21

4

34γ

( ) ( )[ ]1rp1rp1

r 11

14 −−−

−= −− γγ

γ

( )( )141 pp1r

1

r −−−

= −γ

γ

( ) ( )1

pp1rp 141

−−−

=γ

α

Length of the diagram 1r −=

23

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diagram theoflength

diagram theof areamep =∴

( ) ( )( ) ( )1r1

1r1rp 11

−−−−

=−

γα γ

1. Two Stroke

24

Exhaust blow down Scavenging

Crankcase

Inlet valve

Spark plug

Transfer port

Deflector

P3

VV

CV

S

15

2

4

Ideal diagram

Piston travel in uncovering port

P

V

Actual diagram

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2. The Diesel Cycle

Work done in compression 1-2 = area 1256 Heat supplied in 2-3 = area 2365

Work done in expansion 3-4 = area 23456 heat rejected in 4-1 = area 4156

Net work done = area 1234

The thermal efficiency of the ideal Diesel cycle is given by

addedheat

rejectedheat addedheat η

−=

( ) ( )( )

( )( )

( )( )23

14

v

p23

14

p

v

23p

14v23p

TT

TT

c

c1

1TT

TT

c

c1

TTc

TTcTTc

−−−=

−−−=

−−−−

=

−−−=

23

14

TT

TT11

γ

−

−−=

1T

T

1T

T

T γ

T1

2

3

1

4

2

1

For isentropic compression and expansion process

1

4

3

3

4

1γ

1

2

2

1

v

v

T

T and

v

v

T

T−−

=

=

γ

For constant pressure heat addition 2-3, 2

3

2

3

v

v

T

T=

Also v4 = v1

25

T

s

2

1

T

T:

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Thus γγγ

=

=

=

−−

2

3

1

2

3

2

3

1

12

43

2

3

1

4

v

v

v

v

v

v

v/v

v/v

T

T

T

T

Substituting these values in Eq.(the thermal efficiency), we get

( )

−

−−= −

1T

T1v

v/v

11

2

3

21

21γγ

η

( )

−−−= − 1

1

r

11

1 ργρη

γ

γ

Note that the efficiency of the Diesel cycle differs from that of Otto cycle only by the

bracketed term, which is always greater than unity (except when ρ = 1 and there is no heat

addition).Thus the Diesel cycle always has a lower efficiency than the Otto cycle of the same

compression ratio.

Fig ..Efficiency of the Diesel cycle for various cut-off ratios and compression ratios.

Mean Effective Pressure (mep) of Diesel cycle.

Work done = area of the p-v diagram

( )1

vpvp

1

vpvpvvp 11224433

232 −−

−−−

+−=γγ

( ) ( )1

rpprpp1p 1243

2 −−−−

+−=γ

ρρ

( ) ( ) ( ) ( ) ( ) ( )[ ]1r11

p

1

r1prp11p 121

21

22 −−−−

−−

−−−+−−= −

−−γγ

γγγ

ρργγγ

ρργρ

diagram indicator theoflength

diagram indicator theof areamep =

( ) ( )[ ]( ) ( )1r1

1r1p 12

−−−−−

=−

γρργ γγ

26

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( ) ( )[ ]( ) ( )1r1

1r1rp 11

−−−−−

=−

γρργ γγγ

27