chapter 3 gas turbine cycles for aircraft propulsion

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


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.


FFig. ig. 3.8 Typical Turbojet Cycle Performance 3.8 Typical Turbojet Cycle Performance


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



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



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


FFigig.. 3.93.9. . Performance and Design Considerations for Performance and Design Considerations for Aircraft Gas TurbinesAircraft Gas Turbines



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.


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 ↑↑


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


FFigig.. 3.10.3.10.1 1 Variation of Variation of sfcsfc with Flight Speedwith Flight Speed for for aa TypicalTypical Turbojet EngineTurbojet Engine


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.


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).


FIG.3.11.Twin -FIG.3.11.Twin - S Spool Turbofan Enginepool Turbofan Engine

Vjc

VjhVa



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



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.



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.



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



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.



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..



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


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.


FFigig.. 3.11.3.11. Optimization of a Optimization of a Turbofan Turbofan EEnginenginePerformancePerformance



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.



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..



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 . .



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


Mixing in a Constant Area DuctMixing in a Constant Area Duct


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


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.


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.


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


Cycle of Turbojet with AfterburningCycle of Turbojet with Afterburning


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


The The GGasas G Generatorenerator

Air intake Compressor Combustion Compressor Chamber

Turbine

0 1 2 3 4


TurbojetTurbojet TurbopropTurboprop

6

4 5

6

5


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 %


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


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.


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


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



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



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



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



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)



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



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



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



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



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



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



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



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



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



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



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



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



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



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



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



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



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



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



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



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



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

chapter 3 gas turbine cycles for aircraft propulsion

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