chapter 3 gas turbine cycles for aircraft propulsion
TRANSCRIPT
CHAPTERCHAPTER 33
Gas Turbine CyclesGas Turbine Cycles
ffor Aircraft Propulsionor Aircraft Propulsion
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 22
Simple Simple Turbojet CycleTurbojet Cycle
TT
ss
p03
Δpb
01
02
03
04
5
pa
p01
p02
p04p5
(ΔT0)comp
Va² / 2cp
(ΔT0) turb
V5² / 2cp
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 33
Simple Turbojet CycleSimple Turbojet Cycle
3.3.1 Optimisation of a Turbojet Cycle3.3.1 Optimisation of a Turbojet Cycle When considering the design of a turbojet, the basic When considering the design of a turbojet, the basic
thermodynamic parameters at the disposal of the designer thermodynamic parameters at the disposal of the designer are the are the Turbine Inlet TemperatureTurbine Inlet Temperature and the and the compressor compressor pressure ratio (tpressure ratio (t , r, rcc))
It is common practice to carry out a series of design point It is common practice to carry out a series of design point calculations covering a suitable range of these two calculations covering a suitable range of these two variables (tvariables (t , r, rcc) using ) using fixed polytropic efficienciesfixed polytropic efficiencies for the for the
compressor and the turbine and plot compressor and the turbine and plot sfc vs F sfc vs Fss with " with " TITTIT““
((TT0303) and ") and " r rcc " as parameters. " as parameters.
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 44
FFig. ig. 3.8 Typical Turbojet Cycle Performance 3.8 Typical Turbojet Cycle Performance
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 55
Optimisation of a Turbojet CycleOptimisation of a Turbojet Cycle
FFss = f (= f (TT0303) ) strong function strong function
high high TT0303 is desirable for a given is desirable for a given FFss
a small enginea small engine means means small small rrcc or small or small ṁṁ
At At rrcc = const. = const. TT0303↑↑ sfcsfc↑↑ ! F! Fss ↑↑((i.e.i.e. fuel increase ), fuel increase ),
( ( opposite in shaft power opposite in shaft power wwss↑↑ sfc sfc ↓↓
).).
This is because as This is because as TT0303 ↑↑ VVjetjet ↑↑ , ,
ηηpp ↓↓↓↓ (( F Fss ↑↑ )), , ηηee ↑↑ ηηoo ↓↓ andand sfc sfc ↑↑ bubut Ft Fss ↑↑..
Gain inGain in sfc sfc is more important since smaller engine size is is more important since smaller engine size is more desirablemore desirable
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 66
Optimisation of a Turbojet CycleOptimisation of a Turbojet Cycle
rrcc ↑↑ sfc sfc ↓↓ ; at a fixed ; at a fixed TT0303
FFss first first ↑↑ then then ↓↓ ( Optim( Optimumum rrcc ↑↑ for best for best FFss ) as T) as T03 03 ↑↑
At the same altitude At the same altitude Z Z , but , but higherhigher Crusing Speed Crusing Speed VVaa : : i.e i.e VVaa ↑↑ ; for given ; for given rrcc and and TT0303 sfcsfc ↑↑, , FFss ↓↓
becausebecause Momentum DragMomentum Drag ↑↑ , ( , (wwcompcomp ↑↑,, since since TT0101 ↑↑ ) )
At different altitudesAt different altitudes ZZ ↑↑ FFss ↑↑ , , sfc sfc ↓↓
since since TT0101 ↓↓ and and wwss ↓↓..
As As VVaa ↑↑ rrcoptcopt ↓↓ due todue to rrRAMRAM ↑↑ at the intake at the intake
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 77
Optimisation of a Turbojet CycleOptimisation of a Turbojet Cycle
Thermodynamic optimization of the turbojet cycle Thermodynamic optimization of the turbojet cycle can not be isolated from mechanical design can not be isolated from mechanical design considerations and the choice of cycle parameters considerations and the choice of cycle parameters depend very much on the depend very much on the TYPE TYPE of the aircraft.of the aircraft.
01
TIT for high Vj
TIT since T increase
essential for the economic operation of a supersonic aircaft
highhigh
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 88
FFigig.. 3.93.9. . Performance and Design Considerations for Performance and Design Considerations for Aircraft Gas TurbinesAircraft Gas Turbines
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 99
Optimisation of a Turbojet CycleOptimisation of a Turbojet Cycle
high TIThigh TIT thermodynamically desirablethermodynamically desirable causes complexity in mechanical design,causes complexity in mechanical design, such as expensive alloys & cooled such as expensive alloys & cooled
blades.blades.
high rhigh rcc increased weightincreased weight
large number of compressor-turbine large number of compressor-turbine stages stages
i.e i.e multi spool engines.multi spool engines.
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 1010
3.3.2 Variation of Thrust & sfc with Flight Conditions3.3.2 Variation of Thrust & sfc with Flight Conditions
The previous figures represent The previous figures represent design point design point ccalculationsalculations. . At different flight conditions, At different flight conditions,
both both thrust & sfc thrust & sfc vary due to the change in vary due to the change in mmaa with with aa
and variation of and variation of Momentum DragMomentum Drag with forward speed with forward speed VVaa. .
As altitude As altitude ZZ ↑↑ , , FFNetNet ↓↓ due to due to aa decrease as decrease as PPaa ↓↓
Although Although F Fss ↑↑ since since TT0101 ↓↓ , , sfcsfc ↓↓ a little a little
At a fixed altitude At a fixed altitude ZZ,, as as MM ↑↑ FFNN ↓↓ at first due to increased momentum drag, at first due to increased momentum drag,
then then FFNetNet due to benefical effects of due to benefical effects of RamRam pressure pressure ratio. ratio.
For For M >1M >1 increase in increase in FFNetNet is substantial for is substantial for MM ↑↑
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 1111
FFigig.. 3.10.3.10.1 1 Variation of Thrust Variation of Thrust with Flight Speedwith Flight Speed for a for a TypicalTypical Turbojet EngineTurbojet Engine
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 1212
FFigig.. 3.10.3.10.1 1 Variation of Variation of sfcsfc with Flight Speedwith Flight Speed for for aa TypicalTypical Turbojet EngineTurbojet Engine
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 1313
3.4 THE TURBOFAN ENGINE3.4 THE TURBOFAN ENGINE
The The Turbofan engineTurbofan engine was originally conceived as a was originally conceived as a method of impmethod of imprroving the oving the propulsive efficiencypropulsive efficiency of the jet of the jet engine by engine by reducingreducing the the Mean Jet VelocityMean Jet Velocity particularly for particularly for operation at high subsonic speeds.operation at high subsonic speeds.
It was soon realized that reducing jet velocity had a It was soon realized that reducing jet velocity had a considerable effect on considerable effect on Jet NoiseJet Noise , a matter tha , a matter thatt became became critical when large numbers of jet critical when large numbers of jet ppropelled aircraft ropelled aircraft entered commercial service.entered commercial service.
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 1414
The Turbofan EngineThe Turbofan Engine
In In Turbofan enginesTurbofan engines ; ;
a portion of the total flow a portion of the total flow by-passesby-passes part of the compressor, part of the compressor, combustion chamber, turbine and nozzle,combustion chamber, turbine and nozzle, before being ejected through a seperate nozzle.before being ejected through a seperate nozzle.
Turbofan EnginesTurbofan Engines are usually decribed in terms of are usually decribed in terms of
""by-pass ratio"by-pass ratio" defined as defined as : :
the ratio of thethe ratio of the flow through the by-pass duct (cold flow through the by-pass duct (cold stream) to thstream) to that through the at through the high pressure compressor high pressure compressor (HPC) (hot stream).(HPC) (hot stream).
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 1515
FIG.3.11.Twin -FIG.3.11.Twin - S Spool Turbofan Enginepool Turbofan Engine
Vjc
VjhVa
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 1616
The Turbofan EngineThe Turbofan Engine
By pass ratio is given byBy pass ratio is given by ;;
ThenThen ;;
andand ṁṁ = = ṁṁ cc + + ṁṁ hh
If If PPjcjc = P = Pjhjh = P = Paa , , (no pressure thrust)(no pressure thrust)
thenthen ; ; F = (F = (ṁṁ ccVVjcjc + + ṁṁ hhVVjhjh ) - ) - ṁṁ V Vaa for a by-pass enginefor a by-pass engine
h
c
m
mB
1B
mmh1
B
mBmc
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 1717
The Turbofan EngineThe Turbofan Engine
The The design point calculationsdesign point calculations for the turbofan are for the turbofan are similar to those for the turbojet.similar to those for the turbojet.
In view of this only the differences in calculations will be In view of this only the differences in calculations will be outlined. outlined.
a)a) Overall pressure ratio ( rOverall pressure ratio ( rcc )) and and
turbine inlet temperature ( TIT)turbine inlet temperature ( TIT) are specified are specified as beforeas before ; ; but it is but it is alsoalso necessary to specify the necessary to specify the
bypass ratio Bbypass ratio B and the and the fan pressure ratio FPR.fan pressure ratio FPR.
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 1818
The Turbofan EngineThe Turbofan Engine
b)b) From the From the inlet conditionsinlet conditions and and FPRFPR ;; the pressure and the pressure and temperature of the flow leaving the fan and temperature of the flow leaving the fan and entering the entering the by-pass duct can be calculated.by-pass duct can be calculated.
The mass flow down the by-pass duct The mass flow down the by-pass duct ṁṁcc can be can be
established from the total mass flow rateestablished from the total mass flow rate ṁṁ and and BB. . The The cold stream thrustcold stream thrust can then be calculated as for the can then be calculated as for the
jet engine noting that the working jet engine noting that the working fluid isfluid is airair.. It is necessary to It is necessary to checkcheck whether the whether the fan nozzle is fan nozzle is
choked or uncchoked or unchhoked.oked. If chokedIf choked the the pressure thrustpressure thrust must be calculated. must be calculated.
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 1919
The Turbofan EngineThe Turbofan Engine
c)c) In the In the 2-spool 2-spool configurationsconfigurations
the FAN is driven by LP turbinethe FAN is driven by LP turbine
CCalculations for the HP compressor and the turbine are alculations for the HP compressor and the turbine are quite standard,quite standard,
thenthen inlet conditio inlet conditionns to the LP turbine can then be found. s to the LP turbine can then be found.
Considering the Considering the work requirement of the LP rotorwork requirement of the LP rotor ; ;
012 056 056 012m
056 012
1
1 ( 1) B = 0.3 8.0
papa m h pg
h pg
pg
pg m
CmmC T m C T T T
m C
CT B T
C
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 2020
The Turbofan EngineThe Turbofan Engine
The value of The value of BB has a major effect on the temperature has a major effect on the temperature drop and the pressure ratio required from the LP turbinedrop and the pressure ratio required from the LP turbine
Knowing Knowing TT0505, , tt and and TT056056 , LP turbine , LP turbine pressure ratiopressure ratio can can
be found, and conditions at the entry to the be found, and conditions at the entry to the hot stream hot stream nozzlenozzle can be established. can be established.
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 2121
The Turbofan EngineThe Turbofan Engine
d)d) If the two streams are mixed it is necessary to If the two streams are mixed it is necessary to find find the conditions after mixing by means of an enthalpy and the conditions after mixing by means of an enthalpy and momentum balance. momentum balance.
Mixing is essential for a Mixing is essential for a reheated turbofanreheated turbofan..
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 2222
The Turbofan EngineThe Turbofan Engine
3.4.1 Optimization of the Turbofan Cycle3.4.1 Optimization of the Turbofan Cycle
There are 4 thermodynamic parametersThere are 4 thermodynamic parameters the designer can play with.the designer can play with.
i)i) Overall pressure ratio Overall pressure ratio rrpp
ii)ii) Turbine inlet temperature Turbine inlet temperature TITTIT
iii)iii) By-pass Ratio By-pass Ratio BB
iv)iv) Fan pressure ratio Fan pressure ratio FPRFPR
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 2323
Optimization of the Turbofan CycleOptimization of the Turbofan Cycle
At first fix; At first fix; a)a) the overall pressure ratio, the overall pressure ratio, rrpp
b)b) By pass ratio, By pass ratio, BB.. Note that optimum values for each Note that optimum values for each TITTIT
( minimum ( minimum sfcsfc && maxmax FFss ) coincide) coincide because of thebecause of the fixed energy input. fixed energy input. Taking the values of Taking the values of sfc sfc and and FFss
for each of these for each of these FPRFPR values in values in turn,turn,a curve of a curve of sfcsfc vs. vs. F Fss can be plotted. can be plotted.
Note that each point on this curve isNote that each point on this curve is the result of a the result of a previous optimizationprevious optimization
and it is and it is aassociated with a particular value of ssociated with a particular value of FPRFPR and and TIT.TIT.
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 2424
FFigig.. 3.11.3.11. Optimization of a Optimization of a Turbofan Turbofan EEnginenginePerformancePerformance
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 2525
Optimization of the Turbofan CycleOptimization of the Turbofan Cycle
Note that optimum values for each Note that optimum values for each TITTIT
( minimum ( minimum sfcsfc && maxmax FFss ) coincide) coincide
because of the fixed energy input. because of the fixed energy input. Taking the values of Taking the values of sfc sfc and and FFss ,,
for each of these for each of these FPRFPR values in values in turn,turn,
a curve of a curve of sfcsfc vs. vs. F Fss can be plotted. can be plotted.
Note that each point on this curve isNote that each point on this curve is the result of a the result of a previous previous optimizationoptimization and it is and it is aassociated with a ssociated with a particular value of particular value of FPRFPR and and TIT.TIT.
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 2626
Optimization of the Turbofan CycleOptimization of the Turbofan Cycle
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 2727
Optimization of the Turbofan CycleOptimization of the Turbofan Cycle
The foregoing calculations may be repeated for a series The foregoing calculations may be repeated for a series of of BB, still at the same , still at the same rrpp to give a family of curves. to give a family of curves.
This plot yields the optimum variation of This plot yields the optimum variation of sfcsfc with with FFss for for
the selected the selected rrpp as shown by the envelope curve. as shown by the envelope curve.
The procedure can be repeated for a range of The procedure can be repeated for a range of rrpp..
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 2828
Optimization of the Turbofan CycleOptimization of the Turbofan Cycle
The quantitative results are summarized asThe quantitative results are summarized as : :
a)a) BB improve improvess sfcsfc at the expense of at the expense of
significant reduction in significant reduction in FFss, ,
b) b) Optimum Optimum FPRFPR with with TITTIT , ,
c)c) Optimum Optimum FPRFPR with with BB . .
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 2929
The Turbofan EngineThe Turbofan Engine
Long range subsonic transport,Long range subsonic transport, sfcsfc is importantis important B = 4-6 ; B = 4-6 ; high rhigh rpp high TIT. high TIT.
Military AircraftMilitary Aircraft; ; with supersonic dash capability & good subsonic with supersonic dash capability & good subsonic sfc sfc
B = 0.5 -B = 0.5 - 11 to keep the frontal area down,to keep the frontal area down, optional reheat.optional reheat.
Short Haul Commercial AircraftShort Haul Commercial Aircraft,,sfc is not as critical sfc is not as critical B B = 2-3= 2-3
Thrust of engines of high Thrust of engines of high BB is very sensitive to forward is very sensitive to forward speed due to large intake speed due to large intake ṁṁ and momentum drag and momentum drag
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 3030
Mixing in a Constant Area DuctMixing in a Constant Area Duct
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 3131
3.5 AFT3.5 AFT - - FAN CONFIGURATIONFAN CONFIGURATION Some early turbofans were directly developed from existing Some early turbofans were directly developed from existing
turbojets, turbojets, A combined turbine-fan was mounted downstream of the Gas A combined turbine-fan was mounted downstream of the Gas
Generator turbine.Generator turbine.
Vjh
Vjc
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 3232
3.6 TURBO PROP ENGINE3.6 TURBO PROP ENGINE
The turboprop engine differs from the shaft power unit in The turboprop engine differs from the shaft power unit in that some of the useful output appears as jet thrust.that some of the useful output appears as jet thrust.
Power must eventually be delivered to the aircraft in the Power must eventually be delivered to the aircraft in the form of form of thrust power (TP)thrust power (TP) . .
This can be expressed in terms of This can be expressed in terms of equivalentequivalent shaft shaft Power (SP), propeller efficiency Power (SP), propeller efficiency pp, , and and jet thrust F jet thrust F by by
TP = (SP)TP = (SP)prpr + FV + FVaa
The The turboshaftturboshaft engine is of greater importance and is engine is of greater importance and is almost universally used in helicopters because of its low almost universally used in helicopters because of its low weight.weight.
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 3333
3.7 3.7 Thrust AugmentationThrust Augmentation
If the thrust of an engine has to be increased above the If the thrust of an engine has to be increased above the original design value, several alternatives are available.original design value, several alternatives are available.
i)i) IncreaseIncrease of turbine inlet temperature of turbine inlet temperature , , TITTIT
ii)ii) IncreaseIncrease of of mass flow ratemass flow rate through the through the engineengine
Both of these methods imply Both of these methods imply thethe rere--design design of the engine, of the engine, and either oand either off them or both may be used to update the them or both may be used to update the existing engine.existing engine.
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 3434
Thrust AugmentationThrust Augmentation
Frequently there will be a requirement for a Frequently there will be a requirement for a temporary increase in thrust.temporary increase in thrust.
e. g. for take off, for an acceleration from subsonic to e. g. for take off, for an acceleration from subsonic to supersonic speeds or during combat manosupersonic speeds or during combat manoeueuvres.vres.
The problem then becomes one of The problem then becomes one of thrust augmentation.thrust augmentation.
Two methods most widely used are:Two methods most widely used are:i)i) Liquid injection Liquid injection (water+methanol)(water+methanol)ii)ii) Reheat (after burner)Reheat (after burner)
Spraying water to the compressor inlet results in a drop in Spraying water to the compressor inlet results in a drop in inlet temperature in net thrustinlet temperature in net thrust
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 3535
Cycle of Turbojet with AfterburningCycle of Turbojet with Afterburning
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 3636
DESIGN POINT PERFORMANCE CALCULATION DESIGN POINT PERFORMANCE CALCULATION FOR TURBOJET & TURBOPROP ENGINES.FOR TURBOJET & TURBOPROP ENGINES.
A A Turbojet & TurbopropTurbojet & Turboprop unit may be considered as unit may be considered as consisting of 2 parts: consisting of 2 parts:
Thus:-Thus:-
ii / / GAS GENERATORGAS GENERATOR
iiii / / POWER UNIT POWER UNIT a a) Turbojet ) Turbojet Jet Pipe & Final Nozzle Jet Pipe & Final Nozzle
bb) Turboprop ) Turboprop Power Turbine Power Turbine
Jet Jet PPipe & Final Nozzleipe & Final Nozzle
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 3737
The The GGasas G Generatorenerator
Air intake Compressor Combustion Compressor Chamber
Turbine
0 1 2 3 4
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 3838
TurbojetTurbojet TurbopropTurboprop
6
4 5
6
5
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 3939
Problem : Problem : Turbojet & Turboprop EnginesTurbojet & Turboprop Engines
DATA:DATA: AltitudeAltitude Z Z == 0 0 ISA (101.325 kPa; 288.0 K)ISA (101.325 kPa; 288.0 K) True Airspeed (True Airspeed (VVaa)) = 0 = 0 StaticStatic Power Power Output turbojetOutput turbojet = 90 kN Thrust= 90 kN Thrust Power Power OutputOutput turbopropturboprop = 4.5 MW Shaft Power= 4.5 MW Shaft Power CompCompressorressor Press Pressureure Ratio Ratio (P(P0202 // PP0101)) == 10 10 TIT (total) TIT (total) TT0303 = 1500K= 1500K Jet Velocity Jet Velocity VV66 = 220 m/s (turboprop)= 220 m/s (turboprop) CompCompressorressor Isent Isentropicropic eff efficiencyiciency 1212 = = 88%88% Turbine IsentTurbine Isentropicropic eff efficiencyiciency 3434 = = 90%90%
4545 = = 90 %90 %
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 4040
Problem :Problem :Turbojet & Turboprop EnginesTurbojet & Turboprop Engines ; ; DataData
Jet pipe Nozzle IsJet pipe Nozzle Isentropic entropic effefficiencyiciency 5656 = = 100%100%
Combustion effCombustion efficiencyiciency 2323 = = 100%100%
MechMechanical anical effefficiency iciency of Turbo compresorof Turbo compresor drive drive
MM = = 100%100%
Reduction Gear effReduction Gear efficiency iciency GG = = 97%97%
Intake PressIntake Pressureure Reco Recoveryvery PP0101// PP0000 == 0.980.98
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 4141
Problem :Problem :Turbojet & Turboprop EnginesTurbojet & Turboprop Engines ;; DataData
CCombustion Chamber ombustion Chamber tottotal pal pressressureure loss loss : :
ΔΔPP023023 = = 7% of comp7% of compressor ressor outlet totoutlet totalal press pressure ure (P(P0202))
Jet PipeJet Pipe-N-Nozzle ozzle ppressressureure loss loss ::
ΔΔPP056056 = = 3% of turb3% of turbine ine outoutletlet total press total pressureure (P (P0404 or P or P0505))
Nozzle discharge CoeffNozzle discharge Coefficient icient CCdd= = 0.980.98
Cooling air bleed Cooling air bleed rr = = 5% 5% of of CompCompressorressor mass mass flow.flow.
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 4242
Problem : Problem : Turbojet & Turboprop EnginesTurbojet & Turboprop Engines; ; DataData
CCppaa = 1.005 kJ/kg = 1.005 kJ/kg--K for airK for air
CCppgg = 1.150 K= 1.150 KJJ/kg/kg--K for gK for gaass
aa = 1.40 for air = 1.40 for air
gg = 1.33 for gasses = 1.33 for gasses
Calorific value of fuel Calorific value of fuel ΔΔHH = 43.124 M = 43.124 MJJ/kg/kg
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 4343
CalculationsCalculations
a) Aa) Airir
Ram TempRam Temperatureerature Rise Rise ΔΔTT00RRamam= V= Vaa22/2C/2Cpp = 0 K = 0 K
TToaoa = (T = (Taa++ΔΔTT00RRamam) = 288 + 0 = 288K) = 288 + 0 = 288K
PP0101 = P = Poaoa ** P P0101 // PPoaoa = 101.3 * 0.98 = 99.3 kPa = 101.3 * 0.98 = 99.3 kPa
No work is done on or by air No work is done on or by air at theat the Intake Intake TT0101 = T = Toaoa = 288K = 288K
Problem :Problem :Turbojet & Turboprop EnginesTurbojet & Turboprop Engines ; ; CalculationsCalculations
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 4444
Problem :Problem :Turbojet & Turboprop EnginesTurbojet & Turboprop Engines ; ; CalculationsCalculations
b) Compressorb) Compressor
TT0202 = T = T0101 + + ΔΔTT012012 = 288. + 304.6 = 592.6 K= 288. + 304.6 = 592.6 K
PP0202 = = PP0101 * * (P(P0202/P/P0101) = 99.3 ) = 99.3 ** 10 = 993.0 kPa 10 = 993.0 kPa
0.4. 1.4
02 0112
02 01
1
01 02012
12 01
012
'
2881 10 1
0.88
304.6
T T
T T
T PT
P
T K
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 4545
Problem :Problem :Turbojet & Turboprop EnginesTurbojet & Turboprop Engines ; ; CalculationsCalculations
C) Combustion ChamberC) Combustion Chamber
ΔΔPP023023 = = ΔΔPP023023* * PP0202 = 0.07 = 0.07 * 993.0 = 69.5 kPa* 993.0 = 69.5 kPa
PP0303 = P = P0202 - - ΔΔPP023023 = 993.0 - 69.5 = 923.5 = 993.0 - 69.5 = 923.5 kkPaPa
By Heat Balance By Heat Balance
2323 m mff ΔΔH = CH = Cp23p23 (m (maa+m+mff) (T) (T0303-T-T0202))
defining defining : : f f ≡≡ m mff // mmaa ; ; ΔΔTT023023 = T = T0303-T-T0202
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 4646
Problem :Problem :Turbojet & Turboprop EnginesTurbojet & Turboprop Engines ; ; CalculationsCalculations
Using the Combustion Curves Using the Combustion Curves
Ideal TempIdeal Temperatureerature Rise ( Rise (ΔΔ T T2323) vs f ) vs f
(with T(with T0202 as a parameter) as a parameter)
ΔΔTT023023' = ' = ΔΔTT023023 / / 2323 = 907.4 K ; = 907.4 K ; ((2323 =100%) =100%)
TT0202 = 592.6K = 592.6K ff’ ’ = 0.0262= 0.0262
This takes account of the variation of CThis takes account of the variation of Cp23p23
with f with f and temperature and temperature
f = 0.0262 / f = 0.0262 / 2323 = 0.0262 / = 0.0262 / = 0.0262= 0.0262
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 4747
Problem :Problem :Turbojet & Turboprop EnginesTurbojet & Turboprop Engines ; ; CalculationsCalculations
dd) Compressor Turbine) Compressor Turbine
Compressor Turbine Output *MechCompressor Turbine Output *Mechanicalanical eff efficiencyiciency of drive = of drive = = = Compressor inputCompressor input
ṁṁ 11 C Cp12p12 ΔΔTT012012 = = mm ṁṁ 33 C Cp34p34 ΔΔTT034034
ṁṁ 11 = = CCompressor mass flow rateompressor mass flow rate
ṁṁ 33 = C = Compressorompressor turbine mass flow rate turbine mass flow rate
r r = = Cooling air Cooling air bbleed = 0.05 leed = 0.05
ṁṁ 11 = = ṁṁ 22 // ( (11 -- rr)) ṁṁ 33 = = ṁṁ 22 (1 (1 ++ f)f)
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 4848
Problem :Problem :Turbojet & Turboprop EnginesTurbojet & Turboprop Engines ; ; CalculationsCalculations
ṁṁ 11 / / ṁṁ 33 = = 1 1 //((((1-r1-r)*)*(1+f)(1+f)))
∴∴
KT
T
frc
cTT
p
p
m
1.273
)0262.1(*)95.0(
1*
150.1
005.1*
00.1
6.304
)1(*)1(
1**
034
034
34
12012034
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 4949
Problem :Problem :Turbojet & Turboprop EnginesTurbojet & Turboprop Engines ; ; CalculationsCalculations
47.2
1500*90.01.273
1
1
1
1
1
03
04
33.033.11
0334
03403
04
1
03
0403
034'
0403
040334
P
P
TTP
P
PP
T
T
TT
TT
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 5050
Problem :Problem :Turbojet & Turboprop EnginesTurbojet & Turboprop Engines ; ; CalculationsCalculations
PP0303 / P / P04 04 = 2.47= 2.47
TT0404 = T = T0303 - - T T034034 = 1500 -273.1 =1226.9 K = 1500 -273.1 =1226.9 K
PP04 04 = P= P03 03 / (P/ (P0303 / P / P0404) = 923.47 / 2.47) = 923.47 / 2.47
PP04 04 = 373.9 kPa= 373.9 kPa
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 5151
Problem :Problem :Turbojet & Turboprop EnginesTurbojet & Turboprop Engines ; ; CalculationsCalculations
Power SectionPower Section i) Ti) Turbojeturbojet ΔΔPP004646 = = ((ΔΔPP004646/ P/ P0044)) * P * P0404 = 0.03 x 373.93 = 11.22 kPa = 0.03 x 373.93 = 11.22 kPa PP0606 = P = P0404 - - ΔΔPP046046 = 373.93 - 11.22 = 362.71 kPa. = 373.93 - 11.22 = 362.71 kPa. As As 5656 = = 100%100% If PIf P0606// PPaa across the final nozzle exceeds across the final nozzle exceeds PP0606/P/Pcc
PP0606/P/Pcc = 1.85 = 1.85 for for = 1.33 = 1.33 Then the nozzle will be choked thus MThen the nozzle will be choked thus M throatthroat = = 11 Here PHere P0606// PPaa =362.71 =362.71 // 101.33 = 3.58 101.33 = 3.58 the the nozzle is chokednozzle is choked
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 5252
Problem :Problem :Turbojet & Turboprop EnginesTurbojet & Turboprop Engines ; ; CalculationsCalculations
TT66 = T = T0606 – 0.143*T – 0.143*T06 06 = 0.857*T= 0.857*T06 06 =0.857*1226.9 =0.857*1226.9
TT6 6 = 1051.6 K= 1051.6 K
167.12
1
2
11 2
66
06
MT
T
0606
26
0606
606606
26
*143.01
1
2
1
211
2
TTC
V
TT
TTTT
C
V
p
p
since since MM66 =1 =1
we havewe haveSince TSince T0606 = T = T0404
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 5353
Problem :Problem :Turbojet & Turboprop EnginesTurbojet & Turboprop Engines ; ; CalculationsCalculations
6 06 6
6 066 06 06
06 06
36
66
2 * ( ) 2 *1150 *175.3 635 /
362.7195.78
1.863
195.78*100.649 /
287 *1051.6
pg
cr
V c T T m s
P PP P P kPa
P P
P
Pkg s
RT
Flowrate at the throat Flowrate at the throat mm66 = = 66AA66 V V66
where A6 is the Effective Nozzle Throat Areawhere A6 is the Effective Nozzle Throat Area
AA66 / / ṁṁ 66 = 1 / ( = 1 / ( 66*V*V66 ) = 1 / ( 0.649 * 635 ) = 0.00243 m ) = 1 / ( 0.649 * 635 ) = 0.00243 m22s/kgs/kg
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 5454
Problem :Problem :Turbojet & Turboprop EnginesTurbojet & Turboprop Engines ; ; CalculationsCalculations
since the nozzle is choked,since the nozzle is choked, the net thrust has 2 componentsthe net thrust has 2 components i) Momentum Thrust ii) Pressure Thrusti) Momentum Thrust ii) Pressure Thrust
FFNN = = ṁṁ 66 V V66 - - ṁṁ aa V Vaa +(P +(P66-P-Paa) A) A66
)1)(1(
1
6 frm
ma
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 5555
Problem :Problem :Turbojet & Turboprop EnginesTurbojet & Turboprop Engines ; ; CalculationsCalculations
6
66 6
6 6
3
6
1
(1 )(1 )
( ) *(1 )(1 )
635 0 ()195.78 101.32) *10 *0.00243
864.31 /
a
N as a
s
Ns
m
m r f
F V AF V P P
m r f m
F
FF Ns kg
m
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 5656
Problem :Problem :Turbojet & Turboprop EnginesTurbojet & Turboprop Engines ; ; CalculationsCalculations
hNkgsNkgsfc
f
f
mFF
msfc
NN
f
/3600*52.29/52.29
0262.1
0262.0*
31.864
1
1*
1
6
Since FN = 90 kN (required value)
skgr
mm
skgf
mm
skgmF
Fm
N
N
/81.10695.0
47.101
1
/47.1010262.1
13.104
1
/13.10431.864
90000
21
62
66
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 5757
Problem :Problem :Turbojet & Turboprop EnginesTurbojet & Turboprop Engines ; ; CalculationsCalculations
ṁṁ ff = f * = f * ṁṁ 2 2 = 0.262 * 101.47 = 2.66.kg/s= 0.262 * 101.47 = 2.66.kg/s
Effective Nozzle AreaEffective Nozzle Area AA6eff 6eff = = ṁṁ 66*( A*( A6eff6eff / / ṁṁ 66) = 104.13 *0.00243 = 0.253 m) = 104.13 *0.00243 = 0.253 m22
AA6-geometrical6-geometrical = A = A6-effective6-effective/ C/ CDD = 0.253 / 0.98 = 0.253 / 0.98
AA6-geometrical6-geometrical = 0.258 m = 0.258 m22
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 5858
Problem :Problem :Turbojet & Turboprop EnginesTurbojet & Turboprop Engines ; ; CalculationsCalculations
iii) i) TTurboprop urboprop Here the expansion takes place mainly in the power Here the expansion takes place mainly in the power
tturbine, leaving only sufficient pressure ratio across the urbine, leaving only sufficient pressure ratio across the nozzle to produce the specified jet velocity.nozzle to produce the specified jet velocity.
The required The required division of pressure dropdivision of pressure drop through the through the turbineturbine & the & the nozzlenozzle is found by is found by trial and errortrial and error::
As a As a first trialfirst trial,, assume that the power turbine temperature assume that the power turbine temperature drop is drop is ΔΔTT045045 = 295K = 295K
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 5959
Problem :Problem :Turbojet & Turboprop EnginesTurbojet & Turboprop Engines ; ; CalculationsCalculations
and: Tand: T0505 =T =T0606 = T = T0404 - - ΔΔTT045045 = 1222 = 122277 - - 295 = 931.9K295 = 931.9K
PP0505 = =PP0404 ** (P (P0505/P/P0404) = 373.9) = 373.9 // 3.47 = 107.85 kPa.3.47 = 107.85 kPa.
47.3
9.1226*9.0
29511
04
05
33.033.11
0445
045
04
05
P
P
T
T
P
P
ThenThen
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 6060
Problem :Problem :Turbojet & Turboprop EnginesTurbojet & Turboprop Engines ; ; CalculationsCalculations
Also Also ΔΔPP056056 = ( = (ΔΔPP056056 // PP0505)* P)* P0505 = =
== 0.03 * 107.85 = 3.24 kPa0.03 * 107.85 = 3.24 kPa
PP0606 = P = P0505 – – ΔΔPP056056 = 107.85 -3.24 =104.62 kPa = 107.85 -3.24 =104.62 kPa
Since PSince P0606/P/Paa = 104.6 = 104.6 // 101.33 =101.33 = 1.033 < 1.85 1.033 < 1.85
far lessfar less then the critial valuethen the critial value ! !
TThus thus the Nozzle is he Nozzle is unchokedunchoked; ; so so PP66 = = PPaa
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 6161
Problem :Problem :Turbojet & Turboprop EnginesTurbojet & Turboprop Engines ; ; CalculationsCalculations
HenceHence
and and VV66 = = √√ (2*1150*7.4) = 130.4 m/s (2*1150*7.4) = 130.4 m/s
K
P
PT
c
V a
p
4.762.104
33.10119.931*00.1
12
41
1
060656
26
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 6262
Problem :Problem :Turbojet & Turboprop EnginesTurbojet & Turboprop Engines ; ; CalculationsCalculations
But the given value VBut the given value V66= 220 m/s= 220 m/s
SSince thince the founde found value is too low, we now try value is too low, we now try a a somewhat lower value of somewhat lower value of ΔΔTT045045 = 283.2 = 283.2KK
PProceeding as above roceeding as above ;; PP0404 // PP0505 = 3.27 = 3.27
TT0505 = T = T0606 =943.7K =943.7K
PP0505 = 114.28 kPa = 114.28 kPa ΔΔPP056056 = 3.43 kPa = 3.43 kPa
PP0606 =110.85 kPa =110.85 kPa
VV6622 // 2C2Cpp = = 21K21K VV66 = 219.8 m/s = 219.8 m/s ≈≈ 220m/s220m/s
thisthis is close enough is close enough, , with the with the Nozzle UnckokedNozzle Unckoked
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 6363
Problem :Problem :Turbojet & Turboprop EnginesTurbojet & Turboprop Engines ; ; CalculationsCalculations
The shaft powerThe shaft power WWshsh = = GG * * ṁṁ 44 * c * cp45p45 * * TT045 045
WWshsh / / ṁṁ44 = 0.97 * 1.15 * 283.2 = 315.9 kJ/kg = 0.97 * 1.15 * 283.2 = 315.9 kJ/kg
Since the Nozzle is Since the Nozzle is uunchoked, there is only nchoked, there is only mmomentum omentum tthrusthrust
FFNN = = ṁṁ 66* V* V66 – – ṁṁaa* V* Va a
66
219.8 /(1 )(1 )
N as
F VF V N s kg
m r f
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 6464
Problem :Problem :Turbojet & Turboprop EnginesTurbojet & Turboprop Engines ; ; CalculationsCalculations
For the static case For the static case it is it is givengiven that that ;; 1N of jet Thrust is 1N of jet Thrust is equivalent toequivalent to 65 W of propeller 65 W of propeller
shaft Power.shaft Power. Shaft power equivalent of jet thrustShaft power equivalent of jet thrust
per unit mass flowper unit mass flow = ( = (wwjj/ / ṁṁ 66) = (F) = (FNN/ / ṁṁ 66)* 65 / 1000)* 65 / 1000
wwjj/ / ṁṁ 6 6 = 14.3 kJ/kg= 14.3 kJ/kg
Then the equivalent shaft power per unit mass flowThen the equivalent shaft power per unit mass flow
wwjj/ / ṁṁ 6 6 = (= (wwss + + wwjj ) / ) / ṁṁ 6 6 = 315.9 + 14.3 = 315.9 + 14.3
wwjj/ / ṁṁ 6 6 = 330.2 kJ/kg= 330.2 kJ/kg
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 6565
Problem :Problem :Turbojet & Turboprop EnginesTurbojet & Turboprop Engines ; ; CalculationsCalculations
Nozzle Exit :Nozzle Exit :
TT66 = T = T0606 – v – v6622 / 2c / 2cPP = 943.7 - 21 = 922.7 K = 943.7 - 21 = 922.7 K
PP66 = P = Paa =101.33 kPa =101.33 kPa
66 = P = P66 / (RT / (RT66) = 101.33 * 1000 / (287*922.7)) = 101.33 * 1000 / (287*922.7)
66 = 0.383 kg/m = 0.383 kg/m33
26
6 6 6
1 10.0119 /
0.383* 219.8
Am s kg
m V
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 6666
Problem :Problem :Turbojet & Turboprop EnginesTurbojet & Turboprop Engines ; ; CalculationsCalculations
The sfc based on shaft power is ;The sfc based on shaft power is ;
Jkgsfc
mwf
f
w
msfc
sh
shsh
fsh
/10*76.80)(
9.315
1*
0262.1
0262.01*
1)(
9
6
The sfc based on Effective shaft power isThe sfc based on Effective shaft power is ; ;
6
9
1 0.0262 1( ) * *
1 1.0262 330.2
( ) 77.26*10 / 77 /
fef
ef ef
sh
m fsfc
w f w m
sfc kg J g MJ
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 6767
Problem :Problem :Turbojet & Turboprop EnginesTurbojet & Turboprop Engines ; ; CalculationsCalculations
Since the shaft power is specified to be Since the shaft power is specified to be
WWshsh = 4.5 MW = 4.5 MW
6
6 36
62
21
4.5*1014.24 /
315.91*10
14.2413.88 /
1 1.0262
13.8814.61 /
1 0.95
sh
sh
wm kg s
w m
mm kg s
f
mm kg s
r
Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 6868
Problem :Problem :Turbojet & Turboprop EnginesTurbojet & Turboprop Engines ; ; CalculationsCalculations
FFNN = = ṁṁ 66* ( F* ( FNN / / ṁṁ 66 ) = 14.24* 219.8 = 3.13 kg/s ) = 14.24* 219.8 = 3.13 kg/s
Effective Nozzle Area Effective Nozzle Area
AA6-eff6-eff = = ṁṁ 66* ( A* ( A66 / / ṁṁ 66 ) = 14.24 * 0.119 = 0.169 m ) = 14.24 * 0.119 = 0.169 m22
AA6-geometrical6-geometrical = A = A6-effective 6-effective / C/ CDD = 0.169 / 0.98 = 0.169 / 0.98
AA6-geometrical6-geometrical = 0.172 m = 0.172 m22
ṁṁ ff = 0.0262 * 13.88 =0.363 kg/s = 0.0262 * 13.88 =0.363 kg/s