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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
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay3
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.
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay4
Lect-11
Problem # 1
s
T
2
4
5
3
a
7
Real turbojet cycle (without afterburning) on a T-s diagram
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay5
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
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay6
Lect-11
Solution: Problem # 1
• 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
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay7
Lect-11
Solution: Problem # 1
• 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
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay8
Lect-11
Solution: Problem # 1
• 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
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay9
Lect-11
Solution: Problem # 1
• 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
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay10
Lect-11
Solution: Problem # 1
• 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
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay11
Lect-11
Solution: Problem # 1
( )
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
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay12
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.
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay13
Lect-11
Solution: Problem # 2
s
T
2
4
5, 63
a
7
Real turbojet cycle (with afterburning) on a T-s diagram
6a
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay14
Lect-11
Solution: Problem # 2
• 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.
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay15
Lect-11
Solution: Problem # 2
• 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
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay16
Lect-11
Solution: Problem # 2
• 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
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay17
Lect-11
Solution: Problem # 2
• 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
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay18
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.
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay19
Lect-11
Solution: Problem # 3
• 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
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay20
Lect-11
Solution: Problem # 3
choking.not is nozzle theTherefore965.1
14.114.1
95.011
1
1111
1is ratio pressure critical The
1.65ratio pressurefan theis ratio pressure nozzle cold The
)14.1/(4.1)1/('02
=
+−
−
=
+−
−
=
=
−−γγ
γγ
ηn
cPP
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay21
Lect-11
Solution: Problem # 3
( )[ ]( )[ ]
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
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay22
Lect-11
Solution: Problem # 3
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
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay23
Lect-11
Solution: Problem # 3
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
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay24
Lect-11
Solution: Problem # 3
( )[ ] 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
1is ratio pressure critical The
/)1(0505
)133.1/(33.1)1/(05
=−=
=
+−
−
=
+−
−
=
−
−−
γγ
γγ
η
γγ
η
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay25
Lect-11
Solution: Problem # 3
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,,
,
=+=
=×=
=+
=
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay26
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?
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay27
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
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay28
Lect-11
Solution: Problem # 4
• 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
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay29
Lect-11
Solution: Problem # 4
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
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay30
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
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay31
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 %
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay32
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
Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay33
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