1

34

In this lecture ...

• Solve problems – Ideal cycle analysis of air breathing

engines

Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay2

Lect-34

Problem # 1• The following data apply to a turbojet

flying at an altitude where the ambient conditions are 0.458 bar and 248 K.

• Speed of the aircraft: 805 km/h• Compressor pressure ratio: 4:1• Turbine inlet temperature: 1100 K• Nozzle outlet area 0.0935 m2

• Heat of reaction of the fuel: 43 MJ/kgFind the thrust and TSFC assuming cp as

1.005 kJ/kgK and γ as 1.4Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay

3

Lect-34

Ideal cycle for jet engines

Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay4

Lect-34

a 1 2 3 4 5 6 7

CompressorDiffuser

Combustion chamber/burner

Turbine

Afterburner

Nozzle

Schematic of a turbojet engine and station numbering scheme

Ideal cycle for jet engines

Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay5

Lect-34

s

T

2

4

5

3

a

7

Ideal turbojet cycle (without afterburning) on a T-s diagram

Solution: Problem # 1• Speed of the aircraft =

805x1000/3600=223.6 m/s• Mach number = 223.6/√(γRT)

= 223.6/ √(1.4x287x248)= 0.708

• Intake:

Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay6

Lect-34

bar639.0)248/86.272(458.0

K86.272708.02

14.112482

11

)14.1/(4.1)1/(

0202

2202

==

=

=

−+=

−+=

−

−γγ

γ

aa

a

TTPP

MTT

Solution: Problem # 1

Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay7

Lect-34

• Compressor:

• Combustion chamber: From energy balance,

( ) K63.405)4(86.272

bar556.2639.044.1/)14.1(/)1(

0203

0203

===

=×==−− γγπ

π

c

c

TT

PP

017.063.405/1100)63.4051005/1043(

163.405/1100

//1/,

6

030403

0304

0304

=−××−

=

−−

=

+=

TTTcQTTfor

fQhh

pR

R

Solution: Problem # 1

Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay8

Lect-34

• Turbine: Since the turbine produces work to drive the compressor, Wturbine = Wcompressor

bar642.1

)1100/45.969(556.2 Hence,

K45.969)017.01/()86.27263.405(1100)1/()(

)()(

)14.1/(4.1)1/(

04

050405

02030405

02030504

=

=

=

=+−−=+−−=

−=−

−

−γγ

TTPP

fTTTTTTcmTTcm papt

Solution: Problem # 1

Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay9

Lect-34

• Nozzle: we first check for choking of the nozzle.

• The nozzle pressure ratio is P05/Pa=1.642/0.458=3.58

• The critical pressure ratio is

• Therefore the nozzle is choking.• The nozzle exit conditions will be determined

by the critical properties.

893.12

14.12

1 )14.1/(4.1)1/(

*05 =

+

=

+

=−−γγγ

PP

Solution: Problem # 1

Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay10

Lect-34

kg/s 92.19 is, rate flow mass Them/s 75.56992.8072874.1u Therefore,

kg/m374.0)92.807287/(10867.0/

867.0893.1642.1

/1

92.8075.96914.1

21

2

77

7e

35777

*04

05*

7

05*

7

==

=××==

=××==

==

==

=+

=

+

==

euAmRT

RTP

PPPPP

KTTT

ργ

ρ

γ

Solution: Problem # 1

Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay11

Lect-34

[ ][ ]

h kg/N 111.0 kg/Ns101.3/TSFC Therefore,

kg/s 3387.092.19017.0 rate, flow FuelkN 912.10

10)458.0867.0(0935.06.22375.569)017.01(92.19

)(f)u(1m is developed thrust The

5

5

*7e

=×=ℑ=

=×=×==

×−+

−+=−+−+=ℑ

−f

af

a

m

mfm

PPAu

Problem # 2• The following data apply to a twin spool turbofan

engine, with the fan driven by the LP turbine and the compressor by the HP turbine. Separate hot and cold nozzles are used.

• Overall pressure ratio: 19.0• Fan pressure ratio: 1.65• Bypass ratio: 3.0• Turbine inlet temperature: 1300 K• Air mass flow: 115 kg/s• Find the sea level static thrust and TSFC if the

ambient pressure and temperature are 1 bar and 288 K. Heat of reaction of the fuel: 43 MJ/kg

Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay12

Lect-34

Ideal turbofan engine

Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay13

Lect-34

Schematic of an unmixed turbofan engine and station numbering scheme

a 1 2 3 4 5 6 7

3’ 7’

Compressor

Diffuser

Combustion chamber/burnerTurbine

Primary nozzle

Fan

Secondarynozzle

2’

Solution: Problem # 2• Since we are required to find the static

thrust, the Mach number is zero.• Intake:

• Fan: Fan pressure ratio is known:

Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay14

Lect-34

bar 1

K2882

11

)1/(

'02'02

2'02

=

=

=

−+=

−γγ

γ

aa

a

TTPP

MTT

'02'03 / PPf =π

( ) K 35.332)65.1(288

bar65.14.1/)14.1(/)1(

'02'03

'02'03

===

==−− γγπ

π

f

f

TT

PP

Solution: Problem # 2

Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay15

Lect-34

• Compressor:

• Combustion chamber: From energy balance,

( ) K 53.668)515.11(35.332

bar 0.1965.15151.11515.1165.1/19ratio/1.65 pressure Overall

4.1/)14.1(/)1(0203

0203

=×==

=×=====

−− γγπ

ππ

c

c

c

TT

PP

01522.053.668/1300)53.6681005/1043(

153.668/1300

//1/

6

030403

0304

=−××−

=

−−

=TTTcQ

TTfpR

Solution: Problem # 2

Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay16

Lect-34

• High pressure turbine:

bar 79.61300

04.96919 Hence,

K 04.969)01522.01/()53.33253.668(1300)1/()(

exit. HPT at the re temperatu theis Here,

)()(

)14.1/(4.1)1/(

04

'0504'05

020304'05

'05

0203'0504

=

=

=

=+−−=+−−=∴

−=−

−−γγ

TTPP

fTTTTT

TTcmTTcm paHpt

Solution: Problem # 2

Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay17

Lect-34

• Low pressure turbine:

bar 08.404.96998.83779.6 And,

K 98.837)01522.01/()28835.332(304.969

where,),1/()(

inlet. exit/LPT HPT at the re temperatu theis Here,

)()(

)14.1/(4.1)1/(

'05

05'0505

'02'03'0505

'05

'02'0305'05

=

=

=

=+−×−=

=Β+−Β−=∴

−=−

−−γγ

TTPP

mmfTTTT

TTTcmTTcm

aH

aC

paCpt

Solution: Problem # 2

Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay18

Lect-34

• Primary nozzle: we first check for choking of the nozzle.

• The nozzle pressure ratio is P05/Pa=4.08/1=4.08 bar

• The critical pressure ratio is

• Therefore the nozzle is choking.• The nozzle exit conditions will be determined

by the critical properties.

893.12

14.12

1 )14.1/(4.1)1/(

*05 =

+

=

+

=−−γγγ

PP

Solution: Problem # 2

Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay19

Lect-34

m/s 7.52932.6982874.1u Therefore,

bar 155.2893.108.4

/1

K32.69898.83714.1

21

2

7e

*05

05*

7

05*

7

=××==

==

==

=+

=

+

==

RT

PPPPP

TTT

γ

γ

Solution: Problem # 2

Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay20

Lect-34

• Secondary nozzle:• The nozzle pressure ratio is

P03’/Pa=1.65/1=1.65 bar• The critical pressure ratio is

• Therefore the nozzle is not choking.

893.12

14.12

1 )14.1/(4.1)1/(

*05 =

+

=

+

=−−γγγ

PP

( )[ ][ ] 52.298)65.1/1(135.33210052

/124.1/)14.1(

/)1('03'03

=−××=

−=∴

−

− γγPPTcu apef

Solution: Problem # 2

Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay21

Lect-34

• Thrust, [ ]

kN 74.40)052.298(75.283

]07.529)01522.01[(75.28kg/s75.284/115

kg/s 115,0.3/.negligible be to)( assuming

)()1(

=−×+

−×+=ℑ==∴

=+=−

−Β+−+=ℑ

aH

aCaHaHaC

eae

efaHeaH

mmmmm

APPuumuufm

Solution: Problem # 2

Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay22

Lect-34

• Exercise: calculate the thrust by factoring the pressure thrust term as well. Hint: you can calculate the exit area from mass flow, density and exhaust velocity.

• TSFC,

h kg/N 0388.0 kg/Ns10075.1/TSFC Therefore,

kg/s 4376.075.2801522.0 rate, flow Fuel5 =×=ℑ=

=×=×=−

f

af

m

mfm

Problem # 3• A helicopter using a turboshaft engine is

flying at 300 km/h at an altitude where the ambient temperature is 5oC. Determine the specific power output and thermal efficiency. The specifications of the engine are: compressor pressure ratio=9.0, turbine inlet temperature = 800oC.

Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay23

Lect-34

Problem # 3• For a turboshaft engine, there is no nozzle thrust.• u=300x1000/3600= 83.33 m/s• Ta=278 K• Therefore, Mach number

M=83.33/√(1.4x287x278) =0.25• Intake:

Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay24

Lect-34

bar835.0278

48.2818.0

K48.28125.02

1-1.412782

11

)14.1/(4.1)1/(

0202

2202

=

=

=

=

+=

−+=

−−γγ

γ

aa

a

TTPP

MTT

Problem # 3

Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay25

Lect-34

• Compressor:

• Combustor:

( )

kJ/kg42.247)48.28167.527(005.1)(,compressor thedrive torequired work Specific

K 67.527)0.9(48.281

bar 52.7835.00.9

0203

4.1/)14.1(/)1(0203

0203

=−=−=

=×==

=×==−−

TTcW

TT

PP

pc

c

cγγπ

π

013.067.527/1073)67.5271005/1043(

167.527/1073

//1/

6

030403

0304

=−××−

=

−−

=TTTcQ

TTfpR

Problem # 3

Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay26

Lect-34

• Turbine:

kJ/kg 54.516)63.5651073(005.1)013.01(

)()1( turbine,by the doneWork 63.565

897.1394.9

394.98.0

835.09

0504

05

4.1/)14.1(/)1(

05

04

05

04

02

02

0303

05

04

=−××+=

−+==

==

=

=×===

−

−

TTcfWKT

PP

TT

PP

PP

PP

PP

pt

aaγγ

Problem # 3

Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay27

Lect-34

• Specific work output, Wnet =Wt – Wc

=516.54-247.42=269.12 kJ/kg

• Thermal efficiency: Wnet/Qin

• Qin=cp(T04-T03)= 1.005(1073-527.67) =548.05 kJ/kg

• Therefore, thermal efficiency =269.12/548.05=0.49 or 49%

Exercise Problem # 1• A turbojet engine inducts 51 kg of air per

second and propels an aircraft with a uniform flight speed of 912 km/h. The enthalpy change for the nozzle is 200 kJ/kg. The fuel-air ratio is 0.0119 and the heating value of the fuel is 42 MJ/kg. Determine the thermal efficiency, TSFC, propulsive power.

• Ans: 0.34, 0.1034 kg/Nh, 8012 kW.

Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay28

Lect-34

Exercise Problem # 2• A twin spool mixed turbofan engine

operates with an overall pressure ratio of 18. The fan operates with a pressure ratio is 1.5 and the bypass ratio is 5.0. The turbine inlet temperature is 1200 K. If the engine is operating at a Mach number of 0.75 at an altitude where the ambient temperature and pressure are 240 K and 0.5 bar.

• Determine the thrust and the SFC.• Ans: 74 kN, 0.027 kg/N h

Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay29

Lect-34

Exercise Problem # 3• An aircraft using a turboprop engine is

flying at 800 km/h at an altitude where the ambient conditions are 0.567 bar and -20oC. Compressor pressure ratio is 8.0 and the turbine inlet temperature is 1100 K. Assuming that the turboprop does not generate any nozzle thrust, determine the specific power output and the thermal efficiency.

• Ans: 311 kJ/kg, 0.44

Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay30

Lect-34