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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