Problem-solution (Tutorial 1)

8/10/2019 Problem-solution (Tutorial 1)

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In this lecture ...

Solve problems

Ideal cycle analysis of air breathingengines

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

Lect-34


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Problem # 1

The following data apply to a turbojetflying at an altitude where the ambientconditions 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 m

2

Heat of reaction of the fuel: 43 MJ/kg

Find the thrust and TSFC assuming cp as1.005 kJ/kgK and as 1.4


Lect-34


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Ideal cycle for jet engines


Lect-34

a 1 2 3 4 5 6 7

CompressorDiffuserCombustion chamber/burner

Turbine

Afterburner

Nozzle

Schematic of a turbojet engine andstation numbering scheme


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Ideal cycle for jet engines


Lect-34

s

T

2

4

5

3

a

7

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


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Solution: Problem # 1

Speed of the aircraft =805x1000/3600=223.6 m/s

Mach number = 223.6/(RT)

= 223.6/ (1.4x287x248)= 0.708

Intake:


Lect-34

bar639.0)248/86.272(458.0

K86.272708.02

14.1

12482

1

1

)14.1/(4.1

)1/(

0202

22

02

==

=

=

+=

+=

a

a

a

T

TPP

MTT


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Lect-34

Compressor:

Combustion chamber: From energy balance,

( ) K63.405)4(86.272

bar556.2639.04

4.1/)14.1(/)1(

0203

0203

===

===

c

c

TT

PP

017.063.405/1100)63.4051005/1043(

163.405/1100

//

1/,

6

030403

0304

0304

=

=

=

+=

TTTcQ

TTfor

fQhh

pR

R


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Lect-34

Turbine: Since the turbine produces work todrive the compressor, Wturbine= Wcompressor

bar642.1

)1100/45.969(556.2Hence,

K45.969)017.01/()86.27263.405(1100

)1/()(

)()(

)14.1/(4.1

)1/(

04

050405

02030405

02030504

=

=

=

=+=

+=

=

T

TPP

fTTTT

TTcmTTcm papt


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Lect-34

kg/s92.19is,rateflowmassThe

m/s75.56992.8072874.1uTherefore,

kg/m374.0)92.807287/(10867.0/

867.0893.1642.1

/1

92.8075.96914.1

2

1

2

77

7e

35

777

*

04

05

*

7

05

*

7

=====

===

==

==

=+

=

+==

euAm

RT

RTP

PPPPP

KTTT


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Lect-34

[ ][ ]

hkg/N111.0kg/Ns101.3/TSFCTherefore,

kg/s3387.092.19017.0rate,flowFuel

kN912.10

10)458.0867.0(0935.0

6.22375.569)017.01(92.19

)(f)u(1misdevelopedthrustThe

5

5

*

7e

===

===

=+

+=

++=

f

af

a

m

mfm

PPAu


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Since we are required to find the staticthrust, the Mach number is zero.

Intake:

Fan: Fan pressure ratio is known:


Lect-34

bar1

K2882

11

)1/(

'02'02

2

'02

=

=

=

+=

a

a

a

T

TPP

MTT

'02'03/ PPf =

( ) K35.332)65.1(288

bar65.1

4.1/)14.1(/)1(

'02'03

'02'03

===

==

f

f

TT

PP


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Lect-34

Compressor:

Combustion chamber: From energy balance,

( ) K53.668)515.11(35.332

bar0.1965.15151.11

515.1165.1/19ratio/1.65pressureOverall

4.1/)14.1(/)1(0203

0203

===

===

===

c

c

c

TT

PP

01522.053.668/1300)53.6681005/1043(

153.668/1300

//

1/

6

030403

0304

=

=

=

TTTcQ

TT

fpR


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Lect-34

High pressure turbine:

bar79.61300

04.96919Hence,

K04.969)01522.01/()53.33253.668(1300

)1/()(exit.HPTat theretemperatutheisHere,

)()(

)14.1/(4.1)1/(

04

'0504'05

020304'05

'05

0203'0504

=

=

=

=+=

+=

=

T

TPP

fTTTTT

TTcmTTcm paHpt


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Lect-34

Low pressure turbine:

bar08.404.969

98.83779.6And,

K98.837)01522.01/()28835.332(304.969

where,),1/()(

inlet.exit/LPTHPTat theretemperatutheisHere,

)()(

)14.1/(4.1)1/(

'05

05'0505

'02'03'0505

'05

'02'0305'05

=

=

=

=+=

=+=

=

T

TPP

m

mfTTTT

T

TTcmTTcm

aH

aC

paCpt


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Lect-34

Primary nozzle: we first check for choking ofthe nozzle.

The nozzle pressure ratio isP05/Pa=4.08/1=4.08 bar

The critical pressure ratio is

Therefore the nozzle is choking.

The nozzle exit conditions will be determinedby the critical properties.

893.12

14.1

2

1 )14.1/(4.1)1/(

*

05 =

+=

+=

P

P


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Lect-34

m/s7.52932.6982874.1uTherefore,

bar155.2893.1

08.4

/

1

K32.69898.83714.1

2

1

2

7e

*

05

05

*

7

05

*

7

===

==

==

=+

=

+==

RT

PPPPP

TTT


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Lect-34

Secondary nozzle:

The nozzle pressure ratio isP03/Pa=1.65/1=1.65 bar

The critical pressure ratio is

Therefore the nozzle is not choking.

893.12

14.1

2

1 )14.1/(4.1)1/(

*

05 =

+=

+=

P

P

( )[ ]

[ ] 52.298)65.1/1(135.33210052

/12

4.1/)14.1(

/)1(

'03'03

==

=

PPTcu apef


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Lect-34

Thrust,

[ ]

kN74.40

)052.298(75.283]07.529)01522.01[(75.28

kg/s75.284/115

kg/s115,0.3/

.negligiblebeto)(assuming

)()1(

=

+ +=

==

=+=

++=

aH

aCaHaHaC

eae

efaHeaH

m

mmmm

APP

uumuufm


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Lect-34

Exercise: calculate the thrust by factoring thepressure thrust term as well. Hint: you cancalculate the exit area from mass flow, densityand exhaust velocity.

TSFC,

hkg/N0388.0kg/Ns10075.1/TSFCTherefore,

kg/s4376.075.2801522.0rate,flowFuel

5

===

===

f

af

m

mfm


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Problem # 3

A helicopter using a turboshaft engine isflying at 300 km/h at an altitude where theambient temperature is 5oC. Determine thespecific power output and thermalefficiency. The specifications of the engineare: compressor pressure ratio=9.0,turbine inlet temperature = 800oC.


Lect-34


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Problem # 3

For a turboshaft engine, there is no nozzle thrust. u=300x1000/3600= 83.33 m/s

Ta=278 K

Therefore, Mach numberM=83.33/(1.4x287x278) =0.25

Intake:


Lect-34

bar835.0278

48.2818.0

K48.28125.0

2

1-1.41278

2

11

)14.1/(4.1)1/(

0202

22

02

=

=

=

=

+=

+=

a

a

a

T

TPP

MTT


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Problem # 3


Lect-34

Compressor:

Combustor:

( )

kJ/kg42.247)48.28167.527(005.1)(

,compressorthedrivetorequiredworkSpecific

K67.527)0.9(48.281

bar52.7835.00.9

0203

4.1/)14.1(/)1(

0203

0203

===

===

===

TTcW

TT

PP

pc

c

c

013.067.527/1073)67.5271005/1043(

167.527/1073

//

1/

6

030403

0304

=

=

= TTTcQ

TTf

pR


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Problem # 3


Lect-34

Turbine:

kJ/kg54.516

)63.5651073(005.1)013.01()()1(turbine,by thedoneWork

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

=

+=+=

=

==

=

====

TTcfW

KT

P

P

T

T

P

P

P

P

P

P

P

P

pt

aa


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Problem # 3


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%


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Exercise Problem # 1

A turbojet engine inducts 51 kg of air persecond and propels an aircraft with auniform flight speed of 912 km/h. Theenthalpy change for the nozzle is 200kJ/kg. The fuel-air ratio is 0.0119 and theheating value of the fuel is 42 MJ/kg.Determine the thermal efficiency, TSFC,

propulsive power. Ans: 0.34, 0.1034 kg/Nh, 8012 kW.


Lect-34


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Exercise Problem # 3

An aircraft using a turboprop engine isflying at 800 km/h at an altitude where theambient conditions are 0.567 bar and -20oC. Compressor pressure ratio is 8.0 andthe turbine inlet temperature is 1100 K.Assuming that the turboprop does notgenerate any nozzle thrust, determine the

specific power output and the thermalefficiency.

Ans: 311 kJ/kg, 0.44

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Lect-34

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Problem-solution (Tutorial 1)