Transcript Document

CHAPTER 3
Gas Turbine Cycles
for Aircraft Propulsion
Simple Turbojet Cycle
T
p03
s
Chapter2
Shaft Power Cycles
2
Simple Turbojet Cycle
 3.3.1 Optimisation of a Turbojet Cycle
 When considering the design of a turbojet, the basic
thermodynamic parameters at the disposal of the designer
are the Turbine Inlet Temperature and the compressor
pressure ratio (t , rc)
 It is common practice to carry out a series of design point
calculations covering a suitable range of these two
variables (t , rc) using fixed polytropic efficiencies for the
compressor and the turbine and plot sfc vs Fs with " TIT“
(T03) and " rc " as parameters.
Chapter2
Shaft Power Cycles
3
Fig. 3.8 Typical Turbojet Cycle Performance
Chapter2
Shaft Power Cycles
4
Optimisation of a Turbojet Cycle



strong function
high T03 is desirable for a given Fs
a small engine means small rc or small ṁ

At rc = const. T03↑  sfc↑ ! Fs ↑(i.e. fuel increase ),
Fs = f (T03)
( opposite in shaft power  ws↑ sfc ↓ ).
 This is because as

ηp ↓↓ ( Fs ↑ ),
T03 ↑  Vjet ↑ ,
ηe ↑  ηo ↓ and sfc ↑ but Fs ↑.
 Gain in sfc is more important since smaller engine size is
more desirable
Chapter2
Shaft Power Cycles
5
Optimisation of a Turbojet Cycle
 rc ↑  sfc ↓ ; at a fixed T03
 Fs first ↑ then ↓ ( Optimum rc ↑ for best Fs ) as T03 ↑
 At the same altitude Z , but higher Crusing Speed Va :
 i.e Va ↑ ; for given rc and T03  sfc ↑, Fs ↓
because Momentum Drag ↑ , (wcomp ↑, since T01 ↑ )
 At different altitudes Z ↑  Fs ↑ , sfc ↓
since T01 ↓ and ws ↓.
 As Va ↑  rcopt ↓
due to rRAM ↑ at the intake
Chapter2
Shaft Power Cycles
6
Optimisation of a Turbojet Cycle
high TIT for high Vj
high TIT since T01 increase






 essential for the economic operation of a supersonic aircaft
 Thermodynamic optimization of the turbojet cycle
can not be isolated from mechanical design
considerations and the choice of cycle parameters
depend very much on the TYPE of the aircraft.
Chapter2
Shaft Power Cycles
7
Fig. 3.9. Performance and Design Considerations for
Aircraft Gas Turbines
Chapter2
Shaft Power Cycles
8
Optimisation of a Turbojet Cycle
 high TIT


thermodynamically desirable
causes complexity in mechanical design,
such as expensive alloys & cooled blades.
 high rc


increased weight
large number of compressor-turbine stages
i.e multi spool engines.
Chapter2
Shaft Power Cycles
9
3.3.2 Variation of Thrust & sfc with Flight Conditions
 The previous figures represent design point calculations.
 At different flight conditions,

both thrust & sfc vary due to the change in ma with ra

and variation of Momentum Drag with forward speed Va.
 As altitude Z ↑ , FNet ↓ due to ra decrease as Pa ↓
 Although Fs ↑ since T01 ↓ , sfc ↓ a little
 At a fixed altitude Z,
 as M ↑  FN ↓ at first due to increased momentum drag,
then FNet due to benefical effects of Ram pressure ratio.
 For M >1 increase in FNet is substantial for M ↑
Chapter2
Shaft Power Cycles
10
Fig. 3.10.1 Variation of Thrust with Flight Speed for a
Typical Turbojet Engine
Chapter2
Shaft Power Cycles
11
Fig. 3.10.1 Variation of sfc with Flight Speed for
a Typical Turbojet Engine
Chapter2
Shaft Power Cycles
12
3.4 THE TURBOFAN ENGINE
 The Turbofan engine was originally conceived as a
method of improving the propulsive efficiency of the jet
engine by reducing the Mean Jet Velocity particularly for
operation at high subsonic speeds.
 It was soon realized that reducing jet velocity had a
considerable effect on Jet Noise , a matter that became
critical when large numbers of jet propelled aircraft
entered commercial service.
Chapter2
Shaft Power Cycles
13
The Turbofan Engine
 In Turbofan engines ;
a portion of the total flow by-passes part of the compressor,
combustion chamber, turbine and nozzle,
before being ejected through a seperate nozzle.
 Turbofan Engines are usually decribed in terms of
"by-pass ratio" defined as :
the ratio of the flow through the by-pass duct (cold
stream) to that through the high pressure compressor
(HPC) (hot stream).
Chapter2
Shaft Power Cycles
14
Vjc
Va
Vjh
FIG.3.11.Twin - Spool Turbofan Engine
Chapter2
Shaft Power Cycles
15
The Turbofan Engine
 By pass ratio is given by ;
 Then ;
and
mB
mc 
B 1
mc
B
mh
m
mh 
B 1
ṁ=ṁc+ṁh
 If
Pjc = Pjh = Pa ,
(no pressure thrust)
then ;
 F = (ṁ cVjc + ṁ hVjh ) - ṁ Va
for a by-pass engine
Chapter2
Shaft Power Cycles
16
The Turbofan Engine
 The design point calculations for the turbofan are
similar to those for the turbojet.
 In view of this only the differences in calculations will be
outlined.
 a) Overall pressure ratio ( rc ) and
turbine inlet temperature ( TIT) are specified
as before ; but it is also necessary to specify the
bypass ratio B and the fan pressure ratio FPR.
Chapter2
Shaft Power Cycles
17
The Turbofan Engine
 b) From the inlet conditions and FPR ; the pressure and
temperature of the flow leaving the fan and entering the
by-pass duct can be calculated.
 The mass flow down the by-pass duct ṁc can be
established from the total mass flow rate ṁ and B.
 The cold stream thrust can then be calculated as for the
jet engine noting that the working fluid is air.
 It is necessary to check whether the fan nozzle is
choked or unchoked.
 If choked the pressure thrust must be calculated.
Chapter2
Shaft Power Cycles
18
The Turbofan Engine
 c) In the 2-spool configurations
the FAN is driven by LP turbine
Calculations for the HP compressor and the turbine are
quite standard,
then inlet conditions to the LP turbine can then be found.
Considering the work requirement of the LP rotor ;
mC pa T012  m mh C pg T056
 T056
 T056
C pg 1
 ( B  1)
T012
C pg m
Chapter2

Shaft Power Cycles
m C pa 1

T012
mh C pg m
B = 0.3  8.0
19
The Turbofan Engine
 The value of B has a major effect on the temperature
drop and the pressure ratio required from the LP turbine
 Knowing T05, t and T056 , LP turbine pressure ratio can
be found, and conditions at the entry to the hot stream
nozzle can be established.
Chapter2
Shaft Power Cycles
20
The Turbofan Engine
 d) If the two streams are mixed it is necessary to find the
conditions after mixing by means of an enthalpy and
momentum balance.
 Mixing is essential for a reheated turbofan.
Chapter2
Shaft Power Cycles
21
The Turbofan Engine
3.4.1 Optimization of the Turbofan Cycle
 There are 4 thermodynamic parameters
the designer can play with.

i) Overall pressure ratio rp

ii) Turbine inlet temperature TIT

iii) By-pass Ratio B

iv) Fan pressure ratio FPR
Chapter2
Shaft Power Cycles
22
Optimization of the Turbofan Cycle
 At first fix;

a) the overall pressure ratio, rp

b) By pass ratio, B.
 Note that optimum values for each TIT
( minimum sfc & max Fs ) coincide
because of the fixed energy input.
 Taking the values of sfc and Fs
for each of these FPR values in turn,
a curve of sfc vs. Fs can be plotted.
 Note that each point on this curve is the result of a
previous optimization
and it is associated with a particular value of FPR and TIT.
Chapter2
Shaft Power Cycles
23
Fig. 3.11. Optimization of a Turbofan EnginePerformance
Chapter2
Shaft Power Cycles
24
Optimization of the Turbofan Cycle
 Note that optimum values for each TIT
( minimum sfc & max Fs ) coincide
because of the fixed energy input.
 Taking the values of sfc and Fs ,
for each of these FPR values in turn,
a curve of sfc vs. Fs can be plotted.
 Note that each point on this curve is the result of a
previous optimization and it is associated with a
particular value of FPR and TIT.
Chapter2
Shaft Power Cycles
25
Optimization of the Turbofan Cycle
Chapter2
Shaft Power Cycles
26
Optimization of the Turbofan Cycle
 The foregoing calculations may be repeated for a series
of B, still at the same rp to give a family of curves.
 This plot yields the optimum variation of sfc with Fs for
the selected rp as shown by the envelope curve.
 The procedure can be repeated for a range of rp.
Chapter2
Shaft Power Cycles
27
Optimization of the Turbofan Cycle
 The quantitative results are summarized as :
a) B improves sfc at the expense of
significant reduction in Fs,
b) Optimum FPR with TIT ,
c) Optimum FPR with B .
Chapter2
Shaft Power Cycles
28
The Turbofan Engine
 Long range subsonic transport,  sfc is important

B = 4-6 ; high rp high TIT.
 Military Aircraft;
with supersonic dash capability & good subsonic sfc
B = 0.5 - 1 to keep the frontal area down,
optional reheat.
 Short Haul Commercial Aircraft,
sfc is not as critical
B = 2-3
 Thrust of engines of high B is very sensitive to forward
speed due to large intake ṁ and momentum drag
Chapter2
Shaft Power Cycles
29
Mixing in a Constant Area Duct
Chapter2
Shaft Power Cycles
30
3.5 AFT - FAN CONFIGURATION
 Some early turbofans were directly developed from existing
turbojets,
 A combined turbine-fan was mounted downstream of the Gas
Generator turbine.
Vjc
Vjh
Chapter2
Shaft Power Cycles
31
3.6 TURBO PROP ENGINE
 The turboprop engine differs from the shaft power unit in
that some of the useful output appears as jet thrust.
 Power must eventually be delivered to the aircraft in the
form of thrust power (TP) .
 This can be expressed in terms of equivalent shaft
Power (SP), propeller efficiency p, and jet thrust F by
TP = (SP)pr + FVa
 The turboshaft engine is of greater importance and is
almost universally used in helicopters because of its low
weight.
Chapter2
Shaft Power Cycles
32
3.7 Thrust Augmentation
 If the thrust of an engine has to be increased above the
original design value, several alternatives are available.
i) Increase of turbine inlet temperature , TIT
ii) Increase of mass flow rate through the engine
 Both of these methods imply the re-design of the engine,
and either of them or both may be used to update the
existing engine.
Chapter2
Shaft Power Cycles
33
Thrust Augmentation
 Frequently there will be a requirement for a
temporary increase in thrust.
e. g. for take off, for an acceleration from subsonic to
supersonic speeds or during combat manoeuvres.
 The problem then becomes one of thrust augmentation.
 Two methods most widely used are:
i) Liquid injection (water+methanol)
ii) Reheat (after burner)
 Spraying water to the compressor inlet results in a drop in
inlet temperature in net thrust
Chapter2
Shaft Power Cycles
34
Cycle of Turbojet with Afterburning
Chapter2
Shaft Power Cycles
35
DESIGN POINT PERFORMANCE CALCULATION
FOR TURBOJET & TURBOPROP ENGINES.
A Turbojet & Turboprop unit may be considered as
consisting of 2 parts:
 Thus:i / GAS GENERATOR
ii / POWER UNIT a) Turbojet  Jet Pipe & Final Nozzle
b) Turboprop  Power Turbine
Jet Pipe & Final Nozzle
Chapter2
Shaft Power Cycles
36
The Gas Generator
Air intake
Compressor
Compressor
Combustion
Chamber
Turbine
0
1
2
Chapter2
Shaft Power Cycles
3 4
37
Turbojet
Turboprop
5
4 5
6
6
Chapter2
Shaft Power Cycles
38
Problem : Turbojet & Turboprop Engines










DATA:
Altitude Z = 0 ISA (101.325 kPa; 288.0 K)
True Airspeed (Va) = 0 Static
Power Output turbojet = 90 kN Thrust
Power Output turboprop = 4.5 MW Shaft Power
Compressor Pressure Ratio (P02 / P01) = 10
TIT (total)
T03 = 1500K
Jet Velocity
V6 = 220 m/s (turboprop)
Compressor Isentropic efficiency
12 = 88%
Turbine Isentropic efficiency
34 = 90%
45 = 90 %
Chapter2
Shaft Power Cycles
39
Problem :Turbojet & Turboprop Engines ;
Data
 Jet pipe Nozzle Isentropic efficiency 56 = 100%
 Combustion efficiency
23 = 100%
 Mechanical efficiency of Turbo compresor drive
M = 100%
 Reduction Gear efficiency
G = 97%
 Intake Pressure Recovery
P01/ P00 = 0.98
Chapter2
Shaft Power Cycles
40
Problem :Turbojet & Turboprop Engines ;
Data
 Combustion Chamber total pressure loss :
ΔP023 = 7% of compressor outlet total pressure (P02)
 Jet Pipe-Nozzle pressure loss :
ΔP056 = 3% of turbine outlet total pressure (P04 or P05)
 Nozzle discharge Coefficient
Cd= 0.98
 Cooling air bleed
r = 5% of Compressor mass flow.
Chapter2
Shaft Power Cycles
41
Problem : Turbojet & Turboprop Engines ;





Data
Cpa = 1.005 kJ/kg-K for air
Cpg = 1.150 KJ/kg-K for gas
ga = 1.40 for air
gg = 1.33 for gasses
Calorific value of fuel
ΔH = 43.124 MJ/kg
Chapter2
Shaft Power Cycles
42
Problem :Turbojet & Turboprop Engines ; Calculations
Calculations
a) Air
Ram Temperature Rise ΔT0Ram= Va2/2Cp = 0 K

Toa = (Ta+ΔT0Ram) = 288 + 0 = 288K

P01 = Poa * P01 / Poa = 101.3 * 0.98 = 99.3 kPa

No work is done on or by air at the Intake

T01 = Toa = 288K
Chapter2
Shaft Power Cycles
43
Problem :Turbojet & Turboprop Engines ; Calculations
 b) Compressor
T02 ' T01
12 
T02  T01
T012
T012
g 1


g
.0.4


T01  P02
288
1.4


1 
10
1


 0.88
12  P01 


 304.6 K


 T02 = T01 + ΔT012
= 288. + 304.6 = 592.6 K
 P02 = P01 * (P02/P01) = 99.3 * 10 = 993.0 kPa
Chapter2
Shaft Power Cycles
44
Problem :Turbojet & Turboprop Engines ; Calculations
 C) Combustion Chamber
 ΔP023 = ΔP023* P02 = 0.07 * 993.0 = 69.5 kPa
 P03 = P02 - ΔP023
= 993.0 - 69.5 = 923.5 kPa
 By Heat Balance
 23 mf ΔH = Cp23 (ma+mf) (T03-T02)
 defining :
f ≡ mf / ma ;
Chapter2
ΔT023 = T03-T02
Shaft Power Cycles
45
Problem :Turbojet & Turboprop Engines ; Calculations
 Using the Combustion Curves
Ideal Temperature Rise (Δ T23) vs f
 (with T02 as a parameter)

ΔT023' = ΔT023 / 23 = 907.4 K ;
T02 = 592.6K

(23 =100%)
f’ = 0.0262
 This takes account of the variation of Cp23
with f and temperature
f = 0.0262 / 23 = 0.0262 / 1.00 = 0.0262
Chapter2
Shaft Power Cycles
46
Problem :Turbojet & Turboprop Engines ; Calculations
 d) Compressor Turbine
 Compressor Turbine Output *Mechanical efficiency of drive =
= Compressor input
 ṁ 1 Cp12 ΔT012 = m ṁ 3 Cp34 ΔT034
 ṁ 1 = Compressor mass flow rate
 ṁ 3 = Compressor turbine mass flow rate
 r = Cooling air bleed = 0.05
 ṁ 1 = ṁ 2 / (1 - r)
ṁ 3 = ṁ 2 (1 + f)
Chapter2
Shaft Power Cycles
47
Problem :Turbojet & Turboprop Engines ; Calculations
 ṁ 1 / ṁ 3 = 1 /((1-r)*(1+f))
 ∴
T034 
T034
T034
T012
m
c p12
1
*
*
c p 34 (1  r ) * (1  f )
304.6 1.005
1

*
*
1.00 1.150 (0.95) * (1.0262)
 273.1K
Chapter2
Shaft Power Cycles
48
Problem :Turbojet & Turboprop Engines ; Calculations
 34
P04
P03
T03  T04


'
T03  T04


1

1  T034
  34T03
T034

 P04

T03 1  
  P03






g
g 1



g 1
g








1

273.1 
1 

0.90 *1500

1.33
0.33
P04
 2.47
P03
Chapter2
Shaft Power Cycles
49
Problem :Turbojet & Turboprop Engines ; Calculations
 P03 / P04 = 2.47
 T04 = T03 -  T034 = 1500 -273.1 =1226.9 K
 P04 = P03 / (P03 / P04) = 923.47 / 2.47
 P04 = 373.9 kPa
Chapter2
Shaft Power Cycles
50
Problem :Turbojet & Turboprop Engines ; Calculations
 Power Section
 i) Turbojet
 ΔP046 = (ΔP046/ P04) * P04 = 0.03 x 373.93 = 11.22 kPa
 P06 = P04 - ΔP046 = 373.93 - 11.22 = 362.71 kPa.
 As 56 = 100%
 If P06/ Pa across the final nozzle exceeds P06/Pc

P06/Pc = 1.85
for g = 1.33
 Then the nozzle will be choked thus Mthroat = 1
 Here P06/ Pa =362.71 / 101.33 = 3.58

  the nozzle is choked
Chapter2
Shaft Power Cycles
51
Problem :Turbojet & Turboprop Engines ; Calculations
since M6 =1
we have
Since T06 = T04

T06
g 1 2 g  1
 1
M6 
 1.167
T6
2
2
 T6 

V62
2 
  T06 1 
 
 T06  T6  T06 1 
2C p
 g 1
 T06 
 g 1 
V62
  0.143* T06
 T06 
2C p
 g 1
T6 = T06 – 0.143*T06 = 0.857*T06 =0.857*1226.9
T6 = 1051.6 K
Chapter2
Shaft Power Cycles
52
Problem :Turbojet & Turboprop Engines ; Calculations
V6 
2 * c pg (T06  T6 ) 

P6  P06 

P6
r6 
RT6
2 *1150 *175.3  635m / s




P6 
P06  362.7


 195.78kPa
  P06 

P06 
1.863
 P06 


 P 
  cr  
195.78 *103

 0.649kg / s
287 *1051.6
Flowrate at the throat
m6 = r6A6 V6
where A6 is the Effective Nozzle Throat Area
A6 / ṁ 6 = 1 / ( r6*V6 ) = 1 / ( 0.649 * 635 ) = 0.00243 m2s/kg
Chapter2
Shaft Power Cycles
53
Problem :Turbojet & Turboprop Engines ; Calculations
 since the nozzle is choked,
the net thrust has 2 components
 i) Momentum Thrust ii) Pressure Thrust
FN = ṁ 6 V6 - ṁ a Va +(P6-Pa) A6
ma
1

m6 (1  r )(1  f )
Chapter2
Shaft Power Cycles
54
Problem :Turbojet & Turboprop Engines ; Calculations
ma
1

m6 (1  r )(1  f )
FN
Va
A6
Fs 
 V6 
 ( P6  Pa ) *
m6
(1  r )(1  f )
m6
Fs  635  0  ()195.78  101.32) *10 * 0.00243
3
FN
Fs 
 864.31Ns / kg
m6
Chapter2
Shaft Power Cycles
55
Problem :Turbojet & Turboprop Engines ; Calculations
mf
1
f
1
0.0262
sfc 

*

*
FN FN m6 f  1 864.31 1.0262
sfc  29.52kg / N  s  29.52* 3600kg / N  h
Since FN = 90 kN (required value)
FN
90000
m6 

 104.13kg / s
FN m6 864.31
m6
104.13
m2 

 101.47kg / s
1  f 1.0262
m2 101.47
m1 

 106.81kg / s
1 r
0.95
Chapter2
Shaft Power Cycles
56
Problem :Turbojet & Turboprop Engines ; Calculations
 ṁ f = f * ṁ 2 = 0.262 * 101.47 = 2.66.kg/s
 Effective Nozzle Area
 A6eff = ṁ 6*( A6eff / ṁ 6) = 104.13 *0.00243 = 0.253 m2
 A6-geometrical = A6-effective/ CD = 0.253 / 0.98
 A6-geometrical = 0.258 m2
Chapter2
Shaft Power Cycles
57
Problem :Turbojet & Turboprop Engines ; Calculations
ii) Turboprop
 Here the expansion takes place mainly in the power
turbine, leaving only sufficient pressure ratio across the
nozzle to produce the specified jet velocity.
 The required division of pressure drop through the
turbine & the nozzle is found by trial and error:
 As a first trial, assume that the power turbine temperature
drop is ΔT045 = 295K
Chapter2
Shaft Power Cycles
58
Problem :Turbojet & Turboprop Engines ; Calculations
Then
P05  T045 

 1 
P04   45T04 
P05
 3.47
P04
g
g 1
1.33
295


 1 

 0.9 *1226.9 
0.33
 and: T05 =T06 = T04 - ΔT045 = 12227 - 295 = 931.9K

P05 =P04 * (P05/P04) = 373.9 / 3.47 = 107.85 kPa.
Chapter2
Shaft Power Cycles
59
Problem :Turbojet & Turboprop Engines ; Calculations
 Also
ΔP056 = (ΔP056 / P05)* P05 =
= 0.03 * 107.85 = 3.24 kPa
 P06 = P05 – ΔP056 = 107.85 -3.24 =104.62 kPa
 Since P06/Pa = 104.6 / 101.33 = 1.033 < 1.85
far less then the critial value !
 Thus the Nozzle is unchoked;
Chapter2
so
Shaft Power Cycles
P6 = Pa
60
Problem :Turbojet & Turboprop Engines ; Calculations
 Hence
g 1

2
g 
 Pa 
V6



 56T06 1  
  P06 

2c p


1
  101.33  4 
 1.00* 931.91  
   7.4 K
  104.62  
 and V6 = √ (2*1150*7.4) = 130.4 m/s
Chapter2
Shaft Power Cycles
61
Problem :Turbojet & Turboprop Engines ; Calculations
 But the given value V6= 220 m/s
 Since the found value is too low, we now try a
somewhat lower value of ΔT045 = 283.2K
 Proceeding as above ;
P04 / P05 = 3.27
T05 = T06 =943.7K

P05 = 114.28 kPa
ΔP056 = 3.43 kPa

P06 =110.85 kPa

V62 / 2Cp = 21K
V6 = 219.8 m/s ≈ 220m/s

this is close enough, with the Nozzle Unckoked
Chapter2
Shaft Power Cycles
62
Problem :Turbojet & Turboprop Engines ; Calculations
 The shaft power

Wsh = G * ṁ 4 * cp45 * T045

Wsh / ṁ4 = 0.97 * 1.15 * 283.2 = 315.9 kJ/kg
 Since the Nozzle is unchoked, there is only momentum thrust
 FN = ṁ 6* V6 – ṁa* Va
FN
Va
Fs 
 V6 
 219.8N  s / kg
m6
(1  r )(1  f )
Chapter2
Shaft Power Cycles
63
Problem :Turbojet & Turboprop Engines ; Calculations
 For the static case it is given that ;
 1N of jet Thrust is equivalent to 65 W of propeller
shaft Power.
 Shaft power equivalent of jet thrust
per unit mass flow = (wj/ ṁ 6) = (FN/ ṁ 6)* 65 / 1000
wj/ ṁ 6 = 14.3 kJ/kg
 Then the equivalent shaft power per unit mass flow
wj/ ṁ 6 = (ws + wj ) / ṁ 6 = 315.9 + 14.3
wj/ ṁ 6 = 330.2 kJ/kg
Chapter2
Shaft Power Cycles
64
Problem :Turbojet & Turboprop Engines ; Calculations
 Nozzle Exit :
 T6 = T06 – v62 / 2cP = 943.7 - 21 = 922.7 K
 P6 = Pa =101.33 kPa
 r6 = P6 / (RT6) = 101.33 * 1000 / (287*922.7)
 r6 = 0.383 kg/m3
A6
1
1
2


 0.0119m s / kg
m6 r6V6 0.383* 219.8
Chapter2
Shaft Power Cycles
65
Problem :Turbojet & Turboprop Engines ; Calculations
 The sfc based on shaft power is ;
( sfc ) sh 
mf
wsh

f
1
0.0262
1
*

*
1  f wsh m6 1.0262 315.9
( sfc ) sh  80.76*109 kg / J
 The sfc based on Effective shaft power is
;
mf
f
1
0.0262
1
( sfc)ef 

*

*
wef 1  f wef m6 1.0262 330.2
( sfc) sh  77.26 *109 kg / J  77 g / MJ
Chapter2
Shaft Power Cycles
66
Problem :Turbojet & Turboprop Engines ; Calculations
 Since the shaft power is specified to be
Wsh = 4.5 MW
wsh
4.5*106
m6 

 14.24kg / s
3
wsh m6 315.91*10
m6
14.24
m2 

 13.88kg / s
1  f 1.0262
m2
13.88
m1 

 14.61kg / s
1  r 0.95
Chapter2
Shaft Power Cycles
67
Problem :Turbojet & Turboprop Engines ; Calculations
 FN = ṁ 6* ( FN / ṁ 6 ) = 14.24* 219.8 = 3.13 kg/s
 Effective Nozzle Area
 A6-eff = ṁ 6* ( A6 / ṁ 6 ) = 14.24 * 0.119 = 0.169 m2
 A6-geometrical = A6-effective / CD = 0.169 / 0.98
 A6-geometrical = 0.172 m2
 ṁ f = 0.0262 * 13.88 =0.363 kg/s
Chapter2
Shaft Power Cycles
68