Gas turbine cycles for aircraft propulsion • In shaft power cycles, power is in form of generated power.
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Transcript Gas turbine cycles for aircraft propulsion • In shaft power cycles, power is in form of generated power.
Gas turbine cycles for aircraft propulsion
• In shaft power cycles, power is in form of generated power. In
air craft cycles, whole power is in the form of thrust.
• Propulsion units include turbojets, turbofans and turboprops
• In turbojets and turbofans, the whole thrust is generated in
propelling nozzles. In turboprops, most of the thrust is
produced by a propeller with only a small contribution from
exhaust nozzle.
Gas turbine cycles for aircraft propulsion
Gas turbine cycles for aircraft propulsion
• Turbojet
The turbine is designed to produce just enough power to drive the
compressor. The gas leaving the turbine at high pressure and
temperature is expanded to atmospheric pressure in a propelling
nozzle to produce high velocity jet. The propelling nozzle refers to
the component in which the working fluid is expanded to give a high
velocity jet.
Gas turbine cycles for aircraft propulsion
• Turbojet
Gas turbine parts
Gas turbine parts
Compressor and turbine of a Gas turbine
Gas turbine cycles for aircraft propulsion
• Turbojet
Turbojet
Turbojet Operation
Temperature and pressure
distributions
Thrust
Turbofan
• Turbofan
Part of the air delivered by an LP compressor or fan
bypasses the core of the engine (HP compressor,
combustion and turbines) to form an annular propulsive
jet or cooler air surrounding the hot jet. This results in a
jet of lower mean velocity resulting in better propulsive
efficiency and reduced noise.
Turbofan
Flow in a turbofan
Turbofan Thrust
Turboprop
• Turboprop
For lower speed, a combination of propeller and exhaust
jet provides the best propulsive efficiency. It has two
stage compressor and ‘can-type’ combustion chamber.
Turboprops are also designed with a free turbine driving
the propeller or propeller plus LP compressor (called
free-turbine turboprop).
Turboprop
Flow in a turboprop
Turboprop
Comparison
Performance Criteria
•
The net momentum thrust is due
to the rate of change of
momentum
.
F m(C j Ca )
•
•
•
Ca is the velocity of air at inlet
relative to engine
Cj Velocity of air at exit relative
to engine.
The net pressure thrust is
Aj (Pj Pa )
• Thus, the total thrust is
.
F m(C j Ca ) Aj ( Pj Pa )
The propulsion efficiency
useful propulsiveenergy (or thrustpower),FCa
.
• Propulsive efficiency p
.
FCa unused K.E.of the jet, m(C j Ca ) 2 / 2
is a measure of the
.
effectiveness with
m Ca (C j Ca )
.
which the propulsive
2
m
[(
C
(
C
C
)
(
C
C
)
/ 2]
a
j
a
j
a
dust is being used for
2
propelling the aircraft
1 (C j / Ca )
but it is not the
efficiency of energy
Thrust power
p
conversion.
Changein K.E.
FCa
.
m[(C 2j Ca2 ) / 2]
The propulsion efficiency
• Energy conversion
efficiency
useful K.E.for propulsion
e
Rat e of enrgy supplied
.
m(C 2j Ca2 ) / 2
.
m f Qnet
• Overall efficiency
useful work
o
p e
energy supplied
.
FCa
.
m f Qnet
m C a (C j C a ) / 2
.
m f Qnet
The propulsion efficiency
• Specific fuel combustion:
fuel consumption per unit
thrust, i.e. kg/h N = 0.12
Ca 1
o
sfc Qnet
• Specific thrust, Fs
Thrust
Fs
Mass flow rat eof air
mf
m f / ma
f
F
F / ma
Fs
Thermodynamics of air craft engines
• Diffuser: Velocity
decreases in diffuser
while pressure increases
• Nozzle: Velocity
increases in nozzle while
pressure decreases
To1 Toa Ta C a2 / 2c p , but c p
To1 Toa Ta [1 C a2 / 2(
Ta [1
1
2
R
Ta )]
1
C / RTa )] Ta [1
2
a
R
1
1
2
M 2]
Thermodynamics of air craft engines
• Isentropic efficiency of a diffuser
To1 Ta
'
i
To1 Ta
To ' 1
1
Pa Ta
Po1
(Ta To1 Ta ) / Ta
'
1
1 (To1 Ta ) / Ta
'
1
1 i (To1 Ta ) / Ta ) 1
'
1
Ca
2
1 1
M
i
[1 i
]
a
2c p Ta
2
Thermodynamics of air craft engines
The rest of the components ( compressor, turbine combustion
chamber) are treated before.
r
The ram efficiency is
Po1 Pa
Poa Pa
Propelling nozzle
Propelling nozzle is the component in which the working fluid is
expanded to give a high velocity jet.
Nozzle Efficiency
j
for adiabatic flow
To4 T5
To4 T5
'
To5 To4
Thermodynamics of air craft engines
But P Po5 Po4 due tofrictionlosses.
To4 T5 j (To4 )(1
j To4 [1
1
'
To4 / T5 )
for unchoked nozzle (Mj<1); P5=Pa
For choked nozzle ( Max. rate is reached)
M=1, P5=Pc
To check if it is choked or not
To5 To4
1
( Po4 / P5 )
1
Thermodynamics of air craft engines
cs
1 2
1
1
Ms
T5
T5
2c pTs
2
2
To5
To4
for choked condition M=1
To4
Tc
1
j
1
2
(1)
2
To4 Tc
To4 Tc
1
But isentropic efficiency is
2
or Tc To 4
'
'
'
Tc
Tc
1 j (1
) Tc
T
T
'
1
j
(To4 Tc )
Thermodynamics of air craft engines
Pc is calculated as
'
Pc Tc 1
Tc 1
1 j (1
)
Po4 To4
To4
substituting for
1
Tc
To4
Pc
1
2 1
1 (1
Po4 j
1)
1
1 1
1
j 1
Thermodynamics of air craft engines
if Pa Pc Ps Pa (unchoked)
Pa Pc Ps Pc (choked)
To calculate A5 of nozzle
.
m 5 C5 As As m/ 5 C5
.
For choked nozzle, As m/ c Cc where Cc RTc
Thermodynamics of air craft engines
Example
Simple turbojet cycle
Ta 255.7 K , c 0.87, i 0.93; b 0.98
r 8; To3 1200K , t 0.90; j 0.95
m 0.99; Pb 4% of com pressor P
C a 270 m/s
Required sfc,
To1 Ta (1
M Ca /
1
2
M 2)
RTa 0.84
Thermodynamics of air craft engines
2
Ca
To1 Ta
292K
2c p
2
2
Po1
C
0
.
93
*
270
a
p 1 2c T 1 2 *1.005*1000* 255.7
p a
a
(1.132)3.5 1.54
Po 2 rPo1 6.67bar
To2 To1 [1
1
c
1
c
(r
1] 564.5K
To3 1200 K ( given)
Thermodynamics of air craft engines
Pb
Po3 Po2 (1
) 6.4bar
PD2
m wT wc ; To To C p (To 2 To1 ) / m Cpg
4
3
a
To4 ` 95 9K
po 4
Po4
1
/ p O 3 1
(1 To4 / To3
t
2.32 7ba r
Po 4
Po 4
g
g 1
)
, 1.33
1 1 1
/ Pc 1 / 1
1.19 4
j 1
/ Pa Po 4 / Pc ch okin g no zzle; Pc Pa
Thermodynamics of air craft engines
State 5; Pc Pa , choking; M 5 1, Ps Pc Pa
To 5
Cs
1 2
1
1
Ms
Tc
2c p Ts
2
2
2
Ts Tc
(To5 To4 ), no heat loss & m ech. work
1
2
Tc To4 (
) 822.01K
1
Po4
Po4
P5 Pc
1.215bar,
1.914
( Po4 / Pc )
Pc
Pc
s c
0.515 kg / m 3 , R 287
RTc
C 5 C c M c RTc 560.8m / s, M 1.0, M 0.84
Thermodynamics of air craft engines
Notes : Cs Ca (560 270)
.
m As Cs A5 / m 1 / 5C5 0.00346m 2 s / kg
m As C s A5 / m 1 / 5 c5 0.00346 m 2 s / kg
As
sp. thrust; Fs (C s C a )
( p s p a ) 525.2
m
To2 564.5, To3 To2 635.5
chart :
f 0.0174 f fth / b 0.0178
Thermodynamics of air craft engines
3600f
sfc
0.122kg / N
Fs
FCa
Ca 1
270
1
0.185kg / sn
0.122 43000*1000
m f net sfc net
(
)
3600
Thermodynamics of air craft engines
Example:2: Turbofan Analysis
Overall pressure ratio given
Po3
Po1
19, sf st sc 0.9
Pb Po4 Po3 1.25
sea level Pa =1 bar
Ta=288 K
m c
B3
m h
n 0.95
m 0.99
m a 115kg / s
C a 0.
Thermodynamics of air craft engines
State 1 is sea level since Ca=0.0
Required: sfc, Fs
Po2
S 2 : Po2
Po
1
Po 1.65bar
1
To2 / To1 ( Po2 / Po1 )
n 1
n
1
To 337.7 K , 1.4,
2
S 3 : Po3 / Po1 ) p o1 19bar
n 1
To3 / To2 ( Po3 / Po2 ) (
) To3 734K
n
S 4 : To4 1300K , Po4 p o3 Pb 17.75
Thermodynamics of air craft engines
n 1
1
S5 :
t
, 1.333
n
c HPT m
m m h C p (To To ) m h (C P )(To To )
g
4
5
a
3
To5 949.7 K
To5
Po5 / Po4
To
4
n
n 1
, Po5 4.415bar
2
Thermodynamics of air craft engines
S 6 : f m LPT
m a C PA (TO2 To1 ) m m h C Pg (TO5 TO6 )
TO6 To5 C Pa (1 B)(To2 To1 ) / m C Pg 773.7
To6
Po5 To5
Po6
n 1
n
Po6 1.78bar
check for choking of both nozzles ( hot and cold)
Thermodynamics of air craft engines
Pa Pc choking
Pa Pc unchoked
1
1 1
S 7: :
1
1.914; Po6 / / Pa 1.78
Pc n 1
com pare; Po6 / / p a p o6 / p c Pa Pc , unchoked
Po6
Pa 1
To6 T7 nTo6 1 ( )
98.5, 1.333
Po6
T7 To6 98.5 675.2 K
C 7 2c P (To7 T7 ); c P 1147, To 7 To 6 ,
2
adiabatic and no mech. work
Thermodynamics of air craft engines
C7= 476 m/s
Notes: a7 RT7 508.2 m / s M 7 1
for cold nozzle ( do same)
1
Po2
1 1
1
1.965, 1.4; but
Pc N
Pa
Po2 Po2
orPa Pc , unchoked; P8 Pa 1bar
Pa
Pc
Po2
1.65
note: Nozzles are independent of each other regarding
choking.
Thermodynamics of air
craft
engines
1
P
To2 T8 N To2 1 a
Po2
T8 294.9 K
C8 2c Pa (To2 T8 ), c Pa 1007; C8 293m / s
2
Notes: a8=344.2; M8<1
m a
Bm a
m h
28.75kg / s; m c
86.25kg / s
1 B
1 B
Fh m h C7 13700N ; Fc m c C8 25300N
Ftotal 39000N ; Fs 39000/ 115 339.13N / kg / s
f (To3 / o 4 ) 566K , To3 734K ) Fth 0.016
Thermodynamics of air craft engines
.
fact f th / b ( 1.0 assum ed); m f 3600fm h 1656kg fuel / h
sfc
m f
Ftotal
0.0425kg / h.N