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
B3
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