4 gas turbine cycles for aircraft propulsion

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.


• 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.


• Turbojet

Compressor and turbine of a Gas turbine

Temperature and pressure
distributions

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.

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).

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

A j ( Pj − Pa )

• Thus, the total thrust is
.
F = m(C j − Ca ) + A j ( Pj − Pa )

The propulsion efficiency
useful propulsive energy (or thrust power), FCa
• Propulsive efficiency η p = .
.

is a measure of the FCa + unused K.E. of the jet, m(C j − Ca ) 2 / 2
effectiveness with .
m Ca (C j − Ca )
which the propulsive = .
dust is being used for m[(Ca (C j − Ca ) + (C j − Ca ) 2 / 2]
propelling the aircraft 2
=
but it is not the 1 + (C j / Ca )
efficiency of energy
Thrust power
conversion. ηp =
Change in K.E.
FCa
= .
m[(C 2 −Ca ) / 2]
j
2

• Energy conversion ηe =
useful K.E. for propulsion
efficiency Rate of enrgy supplied
.
m(C 2 − Ca ) / 2
j
2

= .
m f Qnet
• Overall efficiency
useful work
ηo = =ηp ηe
energy supplied
.
FCa m Ca (C j −Ca ) / 2
= .
= .
m f Qnet m f Qnet

• Specific fuel combustion: Ca 1
fuel consumption per unit η =
o
thrust, i.e. kg/h N = 0.12 sfc Qnet

• Specific thrust, Fs

Thrust
Fs =
Mass flow rate of 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

γR
To1 = Toa = Ta + C a / 2c p , but c p =
2

γ −1
γR
To1 = Toa = Ta [1 + C a / 2(
2
Ta )]
γ −1
γ −1 2 γ −1 2
= Ta [1 + C a / γRTa )] = Ta [1 + M ]
2 2

• Isentropic efficiency of a diffuser
'
To1 − Ta
ηi =
To1 − Ta
γ
Po1 To '  γ −1
= 1 
Pa  Ta 
 
[ ]
γ
'
= (Ta + To1 − Ta ) / Ta γ −1

[ ]
γ
'
= 1 + (To1 − Ta ) / Ta γ −1

2 γ
Ca  γ −1 2

[ ]
γ
= [1 + η i ] γ −1 = 1 + η i Ma 
= 1 − η i (To1 − Ta ) / Ta )
' γ −1
 γ 
2c p Ta

The rest of the components ( compressor, turbine combustion
chamber) are treated before.
Po1 − Pa
The ram efficiency is ηr =
Poa − Pa
Propelling nozzle

Propelling nozzle is the component in which the working fluid is
expanded to give a high velocity jet.

Nozzle Efficiency
To4 − T5
ηj = '
To4 − T5
for adiabatic flow
To5 = To4

But P Po5 ≠ Po4 due to friction losses.
1
1 = η j To4 [1 − γ −1
To4 − T5 = η j (To4 )(1 − '
To4 / T5 ) γ
( Po4 / P5 )

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

2
To4 To5 cs γ −1 2
= = 1+ = 1+ Ms
T5 T5 2c p Ts 2

for choked condition M=1
To4 γ −1 2 γ +1
= 1+ (1) = But isentropic efficiency is

Tc 2 2
To4 − Tc 1
ηj =
'
or Tc = To 4 − (To4 − Tc )
To4 − Tc
'
ηj
'
Tc Tc
= 1 − η j (1 −
'
) → Tc
T T


Pc is calculated as
γ γ

Pc  Tc   Tc 
' γ −1 γ −1
=  = 1 − η j (1 − )
Po4  To4 
  
 To4 

To4 γ +1
substituting for =
Tc γ
γ

Pc  1 2  γ −1 γ
= 1 − (1 − 
Po4  η j
 γ + 1) 

 1 γ − 1 γ −1
= 1 − 
 η j γ + 1
 


if Pa > Pc → Ps = Pa (unchoked)

Pa ≤ Pc → Ps = Pc (choked )

To calculate A5 of nozzle

.
m = ρ 5 C 5 As → As = m/ ρ 5 C 5
.
For choked nozzle, As = m/ ρ c C c where C c = γRTc

Example
Simple turbojet cycle
Ta = 255.7 K , η c = 0.87, η i = 0.93; η b = 0.98
r = 8; To3 1200 K , η t = 0.90; η j = 0.95
η m = 0.99; ∆Pb = 4% of compressor ∆P
C a = 270 m/s
Required sfc, η

γ −1 2
To1 = Ta (1 + M )
2
M =C a / γRTa =0.84


2
Ca
To1 = Ta + = 292 K
2c p
 Po   Ca  
2
0.93 * 270 2 
  1 + η  = 1 +
 p =
1

 a  2c p Ta   2 *1.005 *1000 * 255.7 

= (1.132) 3.5 = 1.54

Po 2 = rPo1 = 6.67bar
1 γ −1
To2 = To1 [1 + (rc γ − 1] = 564.5 K
ηc
To 3 = 1200 K ( given)

∆Pb
Po3 = Po2 (1 − ) = 6.4bar
PD2
η m wT = wc ; To4 = To3 − C pa (To 2 − To1 ) / η m Cpg
To4 ` =959 K
γg
 1 γg −1
po 4 / p O 3 = −
1 (1 − o4 / To3 ) 
T , γ = .33
1
 η t 
Po4 =2.327bar
γ
 1 γ − 
1 γ− 1
Po 4 / Pc = /  −
1 1 
γ +   = .194
1

 ηj  1 

Po 4 / Pa >Po 4 / Pc → choking nozzle; Pc >Pa

State 5; Pc > Pa , choking ; M 5 = 1, Ps = Pc ≠ Pa
2
To 5 Cs γ −1 2
= 1+ = 1+ Ms
Tc 2c p Ts 2
2
Ts = Tc = (To5 = To4 ), no heat loss & mech. work
γ +1
2
Tc = To4 ( ) = 822.01K
γ +1
Po4 Po4
P5 = Pc = = 1.215bar , = 1.914
( Po4 / Pc ) Pc
Pc
ρs = ρc = = 0.515 kg / m 3 , R = 287
RTc
C 5 = C c = M c γRTc = 560.8m / s, M = 1.0, M = 0.84


Notes : Cs > Ca (560 > 270)
.
m = ρAs Cs → A5 / m = 1 / ρ 5C5 = 0.00346m 2 s / kg


m = ρAs C s → A5 / m = 1 / ρ 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 / η b = 0.0178

3600 f
sfc = = 0.122kg / N
Fs
FC a Ca 1 270 1
η= = = = 0.185kg / sn
m f ϕ net sfc ϕ net 0.122 43000 *1000
( )
3600

Example:2: Turbofan Analysis
Overall pressure ratio given mc

B=3=
Po3
mh

= 19,ηsf = ηst = ηsc = 0.9 η n = 0.95
Po1
∆Pb = Po4 − Po3 = 1.25 η m = 0.99
ma = 115kg / s

sea level Pa =1 bar C a = 0.
Ta=288 K

State 1 is sea level since Ca=0.0
Required: sfc, Fs
 Po2 
S 2 : Po2 =   Po = 1.65bar
 Po  1
 1 
n −1
∧ γ −1
To2 / To1 = ( Po2 / Po1 ) n
→ To2 = 337.7 K , γ = 1.4,
γ
S 3 : Po3 / Po1 ) p o1 = 19bar
∧ n −1
To3 / To2 = ( Po3 / Po2 ) ( ) → To3 = 734 K
n
S 4 : To4 = 1300 K , Po4 = p o3 − ∆Pb = 17.75

n −1 γ −1
S5 : = η αt , γ = 1.333
n γ
ω c = ω HPTη m
η m mh C p g (To4 − To5 ) = mh (C Pa )(To3 − To2 )
 
To5 = 949.7 K
n
 To5  n −1
Po5 / Po4 =   , Po5 = 4.415bar
 To 
 4 


S 6 : ω f = η mω LPT
ma C PA (TO2 − To1 ) = η m mh C Pg (TO5 − TO6 )
 
TO6 = To5 − C Pa (1 + B )(To2 − To1 ) / η m C Pg = 773.7
n −1
∧ n
Po6  To6 
=  → Po6 = 1.78bar
Po5  To5




check for choking of both nozzles ( hot and cold)

Pa ≤ Pc → choking
Pa < Pc → unchoked
γ
− γ −1
Po6  1 γ − 1
S 7: : = 1 −  = 1.914; Po6 / / Pa = 1.78
Pc  η n γ + 1
compare; Po6 / / p a < p o6 / p c → Pa > Pc , unchoked
 Pa − γ γ−1 
To6 − T7 = η nTo6 1 − ( )  = 98.5, γ = 1.333

 Po6 

∴ T7 = To6 − 98.5 = 675.2 K
2
C 7 = 2c P (To7 − T7 ); c P = 1147, To 7 = To 6 ,
adiabatic and no mech. work

C7= 476 m/s

Notes : a7 = γRT7 = 508.2 m / s → M 7 < 1
for cold nozzle ( do same)
γ
− γ −1
Po2  1 γ − 1  Po2 
= 1 −  = 1.965, γ = 1.4; but 
P
 = 1.65

Pc  η N γ   a 
Po2 Po2
< orPa > Pc , unchoked ; ∴ P8 = Pa = 1bar
Pa Pc
note: Nozzles are independent of each other regarding
choking.

Thermodynamics ofγ −1 craft engines
air
 P  γ
To2 − T8 = η N To2 1 −  a  → T8 = 294.9 K
  2 Po 
 
2
C8 = 2c Pa (To2 − T8 ), c Pa = 1007; C8 = 293m / s
Notes: a8=344.2; M8<1
ma
 Bma

mh =
 = 28.75kg / s; mc =
 = 86.25kg / s
1+ B 1+ B
Fh = mh C 7 = 13700 N ; Fc = mc C8 = 25300 N
 
Ftotal = 39000 N ; Fs = 39000 / 115 = 339.13 N / kg / s
f → (∆To 3 / o 4 ) = 566 K , To3 = 734 K ) → Fth = 0.016

.
fact = f th / η b → (= 1.0 assumed ); m f = 3600 fmh = 1656kg fuel / h

mf

sfc = = 0.0425kg / h.N
Ftotal

4 gas turbine cycles for aircraft propulsion

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