1. Gas Turbine Cycles for Aircraft Propulsion
CHAPTER 3
Gas Turbine Cycles
for Aircraft Propulsion

2. Chapter2 Shaft Power Cycles
Simple Turbojet Cycle T 03
p03
(ΔT0)turb
V5² / 2cp 04
Δpb
p04 5 p5 02
(ΔT0)comp
p02
Va² / 2cp 01
p01 pa s
Chapter Shaft Power Cycles

3. Chapter2 Shaft Power Cycles
Simple Turbojet Cycle
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.
Chapter Shaft Power Cycles

4. Fig. 3.8 Typical Turbojet Cycle Performance
Chapter Shaft Power Cycles

5. Optimisation of a Turbojet Cycle
Fs = f (T03) 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 ),
( opposite in shaft power  ws↑ sfc ↓ ).
This is because as T03 ↑  Vjet ↑ ,
ηp ↓↓ ( Fs ↑ ), ηe ↑  ηo ↓ and sfc ↑ but Fs ↑.
Gain in sfc is more important since smaller engine size is more desirable
Chapter Shaft Power Cycles

6. 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
Chapter Shaft Power Cycles

7. Optimisation of a Turbojet Cycle
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.
Chapter Shaft Power Cycles

8. Chapter2 Shaft Power Cycles
Fig Performance and Design Considerations for Aircraft Gas Turbines
Chapter Shaft Power Cycles

9. 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.
Chapter Shaft Power Cycles

10. 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 ↑
Chapter Shaft Power Cycles

11. Chapter2 Shaft Power Cycles
Fig Variation of Thrust with Flight Speed for a Typical Turbojet Engine
Chapter Shaft Power Cycles

12. Chapter2 Shaft Power Cycles
Fig Variation of sfc with Flight Speed for a Typical Turbojet Engine
Chapter Shaft Power Cycles

13. Chapter2 Shaft Power Cycles
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.
Chapter Shaft Power Cycles

14. Chapter2 Shaft Power Cycles
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).
Chapter Shaft Power Cycles

15. FIG.3.11.Twin - Spool Turbofan Engine
Vjc Va
Vjh
FIG.3.11.Twin - Spool Turbofan Engine
Chapter Shaft Power Cycles

16. Chapter2 Shaft Power Cycles
The Turbofan Engine
By pass ratio is given by ;
Then ;
and ṁ = ṁ c + ṁ h
If Pjc = Pjh = Pa , (no pressure thrust)
then ;
F = (ṁ cVjc + ṁ hVjh ) - ṁ Va for a by-pass engine
Chapter Shaft Power Cycles

17. Chapter2 Shaft Power Cycles
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.
Chapter Shaft Power Cycles

18. Chapter2 Shaft Power Cycles
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.
Chapter Shaft Power Cycles

19. Chapter2 Shaft Power Cycles
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 ;
Chapter Shaft Power Cycles

20. Chapter2 Shaft Power Cycles
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, ht and T056 , LP turbine pressure ratio can be found, and conditions at the entry to the hot stream nozzle can be established.
Chapter Shaft Power Cycles

21. Chapter2 Shaft Power Cycles
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.
Chapter Shaft Power Cycles

22. Chapter2 Shaft Power Cycles
The Turbofan Engine
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
Chapter Shaft Power Cycles

23. 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.
Chapter Shaft Power Cycles

24. Fig. 3.11. Optimization of a Turbofan EnginePerformance
Chapter Shaft Power Cycles

25. 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.
Chapter Shaft Power Cycles

26. Optimization of the Turbofan Cycle
Chapter Shaft Power Cycles

27. 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.
Chapter Shaft Power Cycles

28. 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 .
Chapter Shaft Power Cycles

29. Chapter2 Shaft Power Cycles
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 = 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
Chapter Shaft Power Cycles

30. Mixing in a Constant Area Duct
Chapter Shaft Power Cycles

31. Chapter2 Shaft Power Cycles
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
Chapter Shaft Power Cycles

32. Chapter2 Shaft Power Cycles
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 hp, 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.
Chapter Shaft Power Cycles

33. Chapter2 Shaft Power Cycles
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.
Chapter Shaft Power Cycles

34. Chapter2 Shaft Power Cycles
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
Chapter Shaft Power Cycles

35. Cycle of Turbojet with Afterburning
Chapter Shaft Power Cycles

36. 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
Chapter Shaft Power Cycles

37. Chapter2 Shaft Power Cycles
The Gas Generator
Air intake Compressor Combustion Compressor Chamber
Turbine
Chapter Shaft Power Cycles

38. Chapter2 Shaft Power Cycles
Turbojet Turboprop 5
4 5 6 6
Chapter Shaft Power Cycles

39. Problem : Turbojet & Turboprop Engines
DATA:
Altitude Z = 0 ISA ( kPa; K)
True Airspeed (Va) = 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 h12 = 88%
Turbine Isentropic efficiency h34 = 90%
h45 = 90 %
Chapter Shaft Power Cycles

40. Problem :Turbojet & Turboprop Engines ; Data
Jet pipe Nozzle Isentropic efficiency h56 = 100%
Combustion efficiency h23 = 100%
Mechanical efficiency of Turbo compresor drive
hM = 100%
Reduction Gear efficiency hG = 97%
Intake Pressure Recovery P01/ P00 = 0.98
Chapter Shaft Power Cycles

41. 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.
Chapter Shaft Power Cycles

42. Problem : Turbojet & Turboprop Engines ; Data
Cpa = kJ/kg-K for air
Cpg = KJ/kg-K for gas
ga = 1.40 for air
gg = 1.33 for gasses
Calorific value of fuel
ΔH = MJ/kg
Chapter Shaft Power Cycles

43. Chapter2 Shaft Power Cycles
Problem :Turbojet & Turboprop Engines ; Calculations
Calculations
a) Air
Ram Temperature Rise ΔT0Ram= Va2/2Cp = 0 K
Toa = (Ta+ΔT0Ram) = = 288K
P01 = Poa * P01 / Poa = * 0.98 = 99.3 kPa
No work is done on or by air at the Intake
T01 = Toa = 288K
Chapter Shaft Power Cycles

44. Problem :Turbojet & Turboprop Engines ; Calculations
b) Compressor
T02 = T01 + ΔT = = K
P02 = P01 * (P02/P01) = 99.3 * 10 = kPa
Chapter Shaft Power Cycles

45. Problem :Turbojet & Turboprop Engines ; Calculations
C) Combustion Chamber
ΔP023 = ΔP023* P02 = 0.07 * = 69.5 kPa
P03 = P02 - ΔP = = kPa
By Heat Balance
h 23 mf ΔH = Cp23 (ma+mf) (T03-T02)
defining : f ≡ mf / ma ; ΔT023 = T03-T02
Chapter Shaft Power Cycles

46. Problem :Turbojet & Turboprop Engines ; Calculations
Using the Combustion Curves
Ideal Temperature Rise (Δ T23) vs f
(with T02 as a parameter)
ΔT023' = ΔT023 / h23 = K ; (h23 =100%)
T02 = 592.6K  f’ =
This takes account of the variation of Cp23
with f and temperature
f = / h23 = / 1.00 =
Chapter Shaft Power Cycles

47. Problem :Turbojet & Turboprop Engines ; Calculations
d) Compressor Turbine
Compressor Turbine Output *Mechanical efficiency of drive = = Compressor input
ṁ 1 Cp12 ΔT012 = hm 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)
Chapter Shaft Power Cycles

48. Problem :Turbojet & Turboprop Engines ; Calculations
ṁ 1 / ṁ 3 = 1 /((1-r)*(1+f)) ∴
Chapter Shaft Power Cycles

49. Problem :Turbojet & Turboprop Engines ; Calculations
Chapter Shaft Power Cycles

50. Problem :Turbojet & Turboprop Engines ; Calculations
P03 / P04 = 2.47
T04 = T03 - D T034 = = K
P04 = P03 / (P03 / P04) = / 2.47
P04 = kPa
Chapter Shaft Power Cycles

51. Problem :Turbojet & Turboprop Engines ; Calculations
Power Section
i) Turbojet
ΔP046 = (ΔP046/ P04) * P04 = 0.03 x = kPa
P06 = P04 - ΔP046 = = kPa.
As h56 = 100%
If P06/ Pa across the final nozzle exceeds P06/Pc
 P06/Pc = for g = 1.33
Then the nozzle will be choked thus Mthroat = 1
Here P06/ Pa = / = 
 the nozzle is choked
Chapter Shaft Power Cycles

52. Problem :Turbojet & Turboprop Engines ; Calculations
since M6 =1
we have
Since T06 = T04 
T6 = T06 – 0.143*T06 = 0.857*T06 =0.857*1226.9
T6 = K
Chapter Shaft Power Cycles

53. Problem :Turbojet & Turboprop Engines ; Calculations
Flowrate at the throat m6 = r6A6 V6
where A6 is the Effective Nozzle Throat Area
A6 / ṁ 6 = 1 / ( r6*V6 ) = 1 / ( * 635 ) = m2s/kg
Chapter Shaft Power Cycles

54. 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
Chapter Shaft Power Cycles

55. Problem :Turbojet & Turboprop Engines ; Calculations
Chapter Shaft Power Cycles

56. Problem :Turbojet & Turboprop Engines ; Calculations
Since FN = 90 kN (required value)
Chapter Shaft Power Cycles

57. Problem :Turbojet & Turboprop Engines ; Calculations
ṁ f = f * ṁ 2 = * = 2.66.kg/s
Effective Nozzle Area
A6eff = ṁ 6*( A6eff / ṁ 6) = * = m2
A6-geometrical = A6-effective/ CD = / 0.98
A6-geometrical = m2
Chapter Shaft Power Cycles

58. 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
Chapter Shaft Power Cycles

59. Problem :Turbojet & Turboprop Engines ; Calculations
Then
and: T05 =T06 = T04 - ΔT045 = = 931.9K
P05 =P04 * (P05/P04) = / 3.47 = kPa.
Chapter Shaft Power Cycles

60. Problem :Turbojet & Turboprop Engines ; Calculations
Also ΔP056 = (ΔP056 / P05)* P05 =
= * = 3.24 kPa
P06 = P05 – ΔP056 = = kPa
Since P06/Pa = / = < 1.85
far less then the critial value !
Thus the Nozzle is unchoked; so P6 = Pa
Chapter Shaft Power Cycles

61. Problem :Turbojet & Turboprop Engines ; Calculations
Hence
and V6 = √ (2*1150*7.4) = m/s
Chapter Shaft Power Cycles

62. 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 = kPa ΔP056 = 3.43 kPa
P06 = kPa
V62 / 2Cp = 21K V6 = m/s ≈ 220m/s
this is close enough, with the Nozzle Unckoked
Chapter Shaft Power Cycles

63. Problem :Turbojet & Turboprop Engines ; Calculations
The shaft power
Wsh = hG * ṁ 4 * cp45 * DT045
Wsh / ṁ4 = 0.97 * 1.15 * = kJ/kg
Since the Nozzle is unchoked, there is only momentum thrust
FN = ṁ 6* V6 – ṁa* Va
Chapter Shaft Power Cycles

64. 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 =
wj/ ṁ 6 = kJ/kg
Chapter Shaft Power Cycles

65. Problem :Turbojet & Turboprop Engines ; Calculations
Nozzle Exit :
T6 = T06 – v62 / 2cP = = K
P6 = Pa = kPa
r6 = P6 / (RT6) = * 1000 / (287*922.7)
r6 = kg/m3
Chapter Shaft Power Cycles

66. Problem :Turbojet & Turboprop Engines ; Calculations
The sfc based on shaft power is ;
The sfc based on Effective shaft power is ;
Chapter Shaft Power Cycles

67. Problem :Turbojet & Turboprop Engines ; Calculations
Since the shaft power is specified to be
Wsh = 4.5 MW
Chapter Shaft Power Cycles

68. Problem :Turbojet & Turboprop Engines ; Calculations
FN = ṁ 6* ( FN / ṁ 6 ) = 14.24* = 3.13 kg/s
Effective Nozzle Area
A6-eff = ṁ 6* ( A6 / ṁ 6 ) = * = m2
A6-geometrical = A6-effective / CD = / 0.98
A6-geometrical = m2
ṁ f = * =0.363 kg/s
Chapter Shaft Power Cycles