AE245 Intro 06 September 2011

8/4/2019 AE245 Intro 06 September 2011

1/12

Mechanics and Thermodynamics of Propulsion (AE 245)Department of Aerospace Engineering, Indian Institute of Science

Ideal cycle analysis of turbojet enginesCycle analysis Turbojet engine

(8)11

21

1

2

(2)Eq.in(5)and(4)Eqs.Substitute

(7)11

1

1111

,Since

compressorthedrivetorequiredpowerturbineby thedevelopedworkofRate

2

9

0

9

2

0

9

4

0

0

2

4

2

4

2

24

2354

2354

tcr

rc

tcrb

cr

t

r

t

t

t

t

ct

t

tcttt

ttttfueli

ttpittpfueli

MT

T

a

u

T

T

T

T

T

T

T

TTT

TTTTmm

TTCmTTCmm


2/12



(10)

11

2

(9)11

2thrustspecific

(8),Eq.Using

00

0

000

0

90

Ma

T

Q

C

Ffsfc

MaMa

uaF

tcr

rc

rc

R

p

s

tcr

rc

s


3/12



For a given Mo, ao, and T, the previous plot

shows that the specific thrust increases with

compressor pressure ratio, reaches a maximum

and then decreases. Specific thrust is maximum at a particular

compressor pressure ratio.

Dashed line in the plot corresponds to the loci of

optimum compressor pressure ratio.

A lower compressor pressure ratio is desired athigh flight Mach number.


4/12



For a given Mo

, ao

, and T, sfc decreases with

compressor pressure ratio. An increase in

compressor ratio means that the gas enters the

burner with higher temperature. This results in a

decrease in the amount of fuel needed to raise

the temperature of flowing gas.


5/12




6/12



(13)111

2

(12)111

2

(11)011

200

ratiopressurecompressoroptimumfindTo.11

2

maximumiswhenmaximumis,andgivenaFor

thrustSpecific

0

2

0max

2

max

2

0

9

2

9

2

0

9

0

9

0

0

90

MaF

a

u

a

uF

a

u

a

uFaM

Ma

uaF

rs

r

r

ctcr

rccocc

s

tcr

rc

soo

s


7/12


Ideal cycle analysis of turbojet enginesThrust augmentation

Requirement of an additional thrust for a shorter period of time is essential during

flight operation. Typical cases are: take-off, acceleration from subsonic to supersonicspeeds, combat operations, etc. Thrust augmentationis adopted in aircraft engines totackle such cases.

Two different methods are commonly used: Liquid injectionand afterburning.

Liquid injection: Spraying water into the compressor inlet causes a reduction incompressor inlet temperature. This effect increases the compressor pressure ratioand thereby increases the thrust.

Afterburning:The temperature of gas exiting from the turbine is increased by burning

additional fuel in a duct (jet pipe). This provide a larger jet exit velocity and thereby

increases the thrust.

Due to the absence any rotary components, the gas temperature can reach closer to

stoichiometric temperature levels, around 2000 K levels.

Develops higher sfc. Therefore used for only shorter duration. An increase of 44%

increase in thrust results in an increase of 164% in sfc.

Size and weight penalties and additional pressure losses in the jet pipe during withoutafterburning conditions.

Develops higher noise levels due to high temperature exhaust.

Concorde uses afterburning to accelerate from subsonic to supersonic conditions.


8/12


Ideal cycle analysis of turbojet enginesCycle analysis Turbojet engine with afterburner

Without afterburning With afterburning


9/12



The total temperature leaving the afterburner (T9) has ahigher limiting value than the total temperature leavingthe turbine. The value of T4 is limited by the presence of

turbine.

Process 5-7 corresponds to the afterburner unit.

Subscript AB is used for indicate the parameters with

afterburner.


10/12



(14)12

122

11nozzleFor the

11

2rafterburneoutojet withFor turb

modified.getstermther,afterburnewithenginejetFor turbo

1

9

99

2

9

9

99

2

92

9

9

9

2

0

9

2

0

9

t

t

p

t

p

t

tcr

rc

P

PT

C

u

T

TT

C

uM

T

T

a

u

a

u


11/12



(17)

16,and14Eqs.From

(16)12

rafterburneut jet withoFor turbo

(15)

expansion.nozzlecompletenoteAlso.,isentropicareprocessesflowtheSince

5

9

2

0

'

9

2

0

9

5

9

2

'

9

9

1

5

'

95

2'

9

9

9

9

'

9

975

t

t

AB

t

t

t

t

p

ABtt

ttt

T

T

a

u

a

u

T

T

u

u

P

PT

C

u

P

P

P

P

PPP


12/12



7AB

0

9 7

7 09 AB

5 t5 4

4 0

2

9 ABAB

0 t

2

9

0

Define

2 2 11 1

1 1

1

We know 1 1

2

t

t t

tt

t t t

t

r c t

c r r c t AB

r cr

t c

AB

T

T

T T

T TT

T T TT T

u

a

u

a

AB 1 (18)

1 1

r c

r c

AE245 Intro 06 September 2011

Documents