Lect-11 · • An aircraft using a simple turbojet engine, flies at Mach 0.8 where the ambient...

1

Lect-11

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

Lect-11

In this lecture...

• Tutorial– Solve problems involving real cycle

analysis


Lect-11

Problem # 1

• An aircraft using a simple turbojet engine, flies at Mach 0.8 where the ambient temperature and pressure are 223.3 K and 0.265 bar, respectively. The compressor pressure ratio is 8.0 and the turbine inlet temperature is 1200 K. The isentropic efficiencies of: compressor=0.87, turbine=0.90, intake=0.93, nozzle=0.95, mechanical=0.99, combustor=0.98. The pressure loss in the combustor=4% of compressor delivery pressure. Determine the thrust and SFC.


Lect-11

Problem # 1

s

T

2

4

5

3

a

7

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


Lect-11

Solution: Problem # 1

• For the given ambient conditions and the Mach number, the flight speed is V=239.6 m/s

• Intake exit stagnation temperature and pressure

barPor

TT

PP

Kc

VTT

ad

a

Pa

393.0482.12650.0,

482.111

9.25110052

6.2393.2232

02

)1/(

0202

22

02

=×=

=

−+=

=×

+=+=

−γγ

η


Lect-11


• The compressor exit conditions are determined as follows:

[ ]

[ ]K

TT

barPP

cC

c

8.486

11887.019.251

111

144.3393.00.8is pressureexit Compressor

4.1/)14.1(

/)1(0203

0203

=

+−=

+−=

=×==

−

− γγπη

π


Lect-11


• Combustion chamber:

bar.PP

TcTcTcQTcTc

for

QfTcTc

Qfhh

b

papgpafb

papg

fbpapg

fb

018.3144.3960 Also,0.0198 f values, theall ngSubstituti

//1/

,

0304

030403

0304

0304

0304

=×===

−

−=

+=

+=

π

η

η

η


Lect-11


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

barTTPP

K

fTTcTcTTTcmTTcmm

t

mpapg

papgfm

284.1)/1(11

Similarly,3.992

)0198.01(99.0/)9.2518.486(100512001147

)1(/)(

)()()(

)1/(

04050405

02030405

02030504

=

−−=

=+−−×=

+−−=

−=−+

−γγ

η

η

η


Lect-11


• Nozzle: We shall first check for nozzle choking.

choking. is nozzle the,// Since914.1

133.1133.1

95.011

1

1111

1is ratio pressure critical The

845.4265.0284.1:is ratio pressure nozzle The

0505

)133.1/(33.1)1/(05

05

ca

n

c

a

PPPP

PP

PP

>=

+−

−

=

+−

−

=

==

−−γγ

γγ

η


Lect-11


• Therefore the nozzle exit conditions are fixed by the critical parameters.

smRTV

mkgRTP

barPP

PPP

KTTT

ex

cc

c

/5.570

/275.0/

671.0/

1

7.8501

2

7

3777

05057

057

==

==

=

==

=

+

==

γ

ρ

γ


Lect-11


( )

sNkgFf

kgNs

PPmAVVfF

kgsmVm

A

n

ace

exn

ex

e

/1032.325.596

0198.0 SFC

/25.596

)()1( is, thrust Specific

/006374.01

5

2

7

−×===

=

−+−+=∴

==

ρ


Lect-11

Problem # 2

• Determine the thrust and SFC in the above problem if the engine operates with an afterburner. The nozzle inlet temperature in this case is limited to 1800 K. All other parameters and operating conditions remain unchanged.


Lect-11


s

T

2

4

5, 63

a

7

Real turbojet cycle (with afterburning) on a T-s diagram

6a


Lect-11


• Since all other operating conditions and parameters remain unchanged, the cycle analysis up to the turbine exit is exactly the same as discussed for the previous problem.

• The nozzle will be choking and the exit conditions will need to be calculated.

• Besides this the fuel flow rate in the afterburner too needs to be determined.

• The total fuel flow rate will be the sum of the fuel in the main combustor and that of the afterburner.


Lect-11


• To calculate fuel flow rate in the afterburner,

04236.002256.00198.00.02256 f values, theall ngSubstituti

//1/

,

21

2

050605

05062

20506

=+=+=∴=

−

−=

+=

fff

TcTcTcQTcTc

for

Qfhh

pgpgpgfb

pgpg

fb

η

η


Lect-11


• At the nozzle exit,

smRTV

mkgRTP

barPP

PPP

KTTT

ex

cc

c

/9.787

/151.0/

671.0/

1

06.15451

2

7

3777

05057

067

==

==

=

==

=

+

==

γ

ρ

γ


Lect-11


• Afterburning therefore leads to substantial thrust augmentation (about 35%). But this is accompanied by about 28% increase in SFC.

( )

sNkgFf

kgNs

PPmAVVfF

kgsmVm

A

n

ace

exn

ex

e

/1064.456.912

04236.0 SFC

/56.912

)()1( is, thrust Specific

/0084.01

5

2

7

−×===

=

−+−+=∴

==

ρ


Lect-11

Problem # 3

• A twin spool un-mixed turbofan engine has the fan driven by the LP turbine and the compressor by the HP turbine. The overall pressure ratio is 25 and the fan pressure ratio is 1.65. The engine has a bypass ratio of 5.0 and a turbine inlet temperature of 1550 K. The fan, turbine and compressor have polytropicefficiencies of 0.90. The nozzle efficiency is 0.95 and the mechanical efficiency for each spool is 0.99. The combustor pressure loss is 1.5 bar and the total air mass flow is 215 kg/s. Find the thrust under sea level static conditions, where ambient pressure and temperature are 1 bar and 288 K.


Lect-11


• Under static conditions, T01=Ta and P01=Pa.

KPPTT

PP

KTT

comppoly

fanpolyf

1.800

15.1565.1

25 25, is ratio pressure overall theSince

6.337)(

,/)1(

,

02

030203

02

03

/)1(01'02

=

=

==

==

−

−

γηγ

γηγπ


Lect-11


choking.not is nozzle theTherefore965.1

14.114.1

95.011

1

1111


1.65ratio pressurefan theis ratio pressure nozzle cold The

)14.1/(4.1)1/('02

=

+−

−

=

+−

−

=

=

−−γγ

γγ

ηn

cPP


Lect-11


( )[ ]( )[ ]

kNVmF

skgB

Bmm

sm

PPTcV

V

fexCn

C

anpex,f

ex,f

532.52is nozzlesecondary by the developed thrust theTherefore

/2.1791

is flow mass cold the5, is ratio bypass theSince/2.293

965.1/116.33795.010052

/12

locity,exhaust ve nozzlesecondary The

,sec,

4.1/)14.1(

/)1('02'02

==

=+

=

=

−×××=

−=

−

−

γγη


Lect-11


KT

TTc

cBTT

KT

TTc

cTT

pgm

pa

pgm

pa

8.877

)()1(

rotor, LP For the

1141)6.3371.800(114799.0

10051550

)(

turbine,HP For the

05

01'0205'05

'05

0203'0504

=∴

−+=−

=−×

−=∴

−=−

η

η


Lect-11


878.1/is ratio pressure nozzlehot The

878.1)/)(/(

5.2350.10.10.25

208.3

902.3

05

05'05'0504

0405

0304

)1/(

05

'05

05

'05

)1/(

'05

04

'05

04

,

,

=

==

=−×=∆−=

=

=

=

=

−

−

a

b

PP

barPPPP

PP

barPPPTT

PP

TT

PP

turbpoly

turbpoly

γγη

γγη


Lect-11


( )[ ] smPPTcV

V

PP

anpex

ex

n

c

/3.528/12

locity,exhaust ve nozzleprimary The choking.not is nozzle theTherefore

914.1

133.1133.1

95.011

1

1111


/)1(0505

)133.1/(33.1)1/(05

=−=

=

+−

−

=

+−

−

=

−

−−

γγ

γγ

η

γγ

η


Lect-11


kNFFF

kNF

skgB

mm

nprimarynn

primaryn

h

5.71 thus,is thrust totalThe

931.183.52883.35

/83.351

is nozzlehot he through trate flow Mass

sec,,

,

=+=

=×=

=+

=


Lect-11

Problem # 4

• An aircraft operating on a turboprop engine flies at 200 m/s while ingesting a primary mass flow of 20 kg/s. The propeller of the engine having an efficiency of 0.8, generates a thrust of 10000 N, while the jet thrust is 2000N. The power turbine and nozzle have efficiencies of 0.88 and 0.92respectively. If we remove the power turbine and the nozzle, what would be the thrust developed by the engine while operating under the same conditions?


Lect-11

Solution: Problem # 4Lect-10

s

h5

6

7

ΔhΔhα

05

06

7

P05

P06

Pa

Enthalpy-entropy diagram for power turbine-exhaust nozzle analysis


Lect-11


• We know that the thrust power developed by the propeller is given by,

smVorVVVmF

kgJh

mhVF

exex

exnozzlen

prprprn

/300,)200(202000)( is thrust nozzle The

/45.1420452088.08.0

20010000

,

,

=−=

−=

=××

×=∆∴

×∆××=×

α

αηη


Lect-11


Nsm

hV

h

kgJhh

hV

nex

nex

18.7855)20076.592(20 Thrust /76.592

49.19095892.022

nozzle. he through toccurs drop entire theremoved,propeller theand inepower turb With the

/49.190958)1(92.02300

)1(22

=−=∴=

××=∆=∴

∆

=∆∴∆−××=

∆−=

η

α

αη


Lect-11

Exercise Problem # 1• A simple turbojet is operating with a compressor

pressure ratio of 8.0, a turbine inlet temperature of 1200 K and a mass flow rate of 15 kg/s, when the aircraft is flying at 260 m/s at and altitude of 7000m. Assuming the following component efficiencies, calculate the nozzle area required, the net thrust and the SFC: polytropic efficiencies of turbine and compressor: 0.87, intake and nozzle efficiency: 0.95, Mechanical efficiency: 0.99, combustion efficiency: 0.97, combustor pressure loss: 6% of compressor delivery pressure.

• Ans: 0.0713 m3, 7896 N, 0.126 kg/h N


Lect-11

Exercise Problem # 2

• The gases at the turbine exit (given in problem #1) are reheated to 2000 K and the combustion pressure loss is 3% of the pressure at the outlet from the turbine. Calculate the percentage increase in the nozzle area required if the mass flow rate is to remain unchanged and also the percentage increase in the net thrust.

• Ans: 48.3 % and 64.5 %


Lect-11

Exercise Problem # 3• 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, By pass ratio: 3.0, Turbine inlet temperature: 1300 K, Combustor pressure loss: 1.25 bar, Total air mass flow: 115 kg/s. It is required to find out the thrust under sea level static conditions where the ambient pressure and temperature are 1.0 bar and 288 K. Assume fan, compressor and turbine efficiencies as 0.90 and that of each of the nozzle as 0.95.

• Ans: 47.6 kN


Lect-11

Exercise Problem # 4• A turboprop is operating under the following

conditions: Flight speed at standard sea level: 0 m/s; Airflow entering the compressor: 1 kg/s; Compressor pressure ratio: 12; Efficiencies: Diffuser: 100 %, Compressor: 87 %, Turbine to drive the compressor: 89 %, Turbine to drive the propeller: 89 %, Nozzle: 100%, Turbine inlet temperature: 1400 K, Stagnation pressure leaving the second turbine: 4.6 bar. Take into account the mass of fuel added. Calculate:

• (a)the horse power delivered to the propeller• (b)the thrust developed by the gases passing through

the engine. • Ans: 632 kW, 875 N

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Lect-11 · • An aircraft using a simple turbojet engine, flies at Mach 0.8 where the ambient...

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