Thermodynamic Cycles

Air-standard analysis is a simplification of the real cyclethat includes the following assumptions:

1) Working fluid consists of fixed amount of air (ideal gas)

2) Combustion process represented by heat transfer into and out of the cylinder from an external source

3) Differences between intake and exhaust processes not considered (i.e. no pumping work)

4) Engine friction and heat losses not considered

SI Engine Cycle vs Air Standard Otto Cycle

A I R

CombustionProducts

Ignition

IntakeStroke

FUEL

Fuel/AirMixture

AirTC

BC

CompressionStroke

PowerStroke

ExhaustStroke

Qin Qout

CompressionProcess

Const volume heat addition

Process

ExpansionProcess

Const volume heat rejection

Process

ActualCycle

OttoCycle

Process 1 2 Isentropic compressionProcess 2 3 Constant volume heat additionProcess 3 4 Isentropic expansionProcess 4 1 Constant volume heat rejection

v2TC

TCv1BC

BC

Qout

Qin

Air-Standard Otto cycle

3

4

2

1

v

v

v

vr

Compression ratio:

First Law Analysis of Otto Cycle

12 Isentropic Compression

)()( 12 m

W

m

Quu in

rv

v

v

v

r

r 1

1

2

1

2 2

1

1

2

1

2

1

11

2

22

v

v

T

T

P

P

T

vP

T

vPR

AIR

)( 12 uum

Win

First Law Analysis of Otto Cycle

23 Constant Volume Heat Addition

m

W

m

Quu in )()( 23

)( 23 uum

Qin

2

3

2

3

3

3

2

2

T

T

P

P

RT

P

RT

Pv

AIRQin

TC

3 4 Isentropic Expansion

AIR)()( 34 m

W

m

Quu out

)( 43 uum

Wout

rv

v

v

v

r

r 3

4

3

4

4

3

3

4

3

4

3

33

4

44

v

v

T

T

P

P

T

vP

T

vP

First Law Analysis of Otto Cycle

4 1 Constant Volume Heat Removal

AIRQoutm

W

m

Quu out )()( 41

)( 14 uum

Qout

1

1

4

4

1

1

4

4

T

P

T

P

RT

P

RT

Pv

BC

First Law Analysis of Otto Cycle

23

1243

uu

uuuu

Q

WW

Q

W

in

inout

in

cycleth

23

14

23

1423 1uu

uu

uu

uuuuth

Cycle thermal efficiency:

Net cycle work:

1243 uumuumWWW inoutcycle

First Law Analysis

• For a cold air-standard analysis the specific heats are assumed to be constant evaluated at ambient temperature values (k = cp/cv = 1.4).

• For the two isentropic processes in the cycle, assuming ideal gas with constant specific heat using yields:

Cold Air-Standard Analysis

1 2 : 11

2

1

1

2

k

k

rv

v

T

T k

k

P

P

T

T1

1

2

1

2

3 4 : 11

4

3

3

4 1

kk

rv

v

T

T k

k

P

P

T

T1

3

4

3

4

12

1

23

14 111

)(

)(1

kv

v

const cth rT

T

TTc

TTc

V

RTPvconstPvk .

Effect of Specific Heat Ratio

1

11 k

const cth r

V

Specific heat ratio (k)

Cylinder temperatures vary between 20K and 2000K where 1.2 < k < 1.4