Gas Turbine Cycles for Aircraft Propulsion

CHAPTERCHAPTER 33
Gas Turbine CyclesGas Turbine Cycles
for Aircraft Propulsionfor Aircraft Propulsion
By: Ebrahem EshtaiweBy: Ebrahem Eshtaiwe

Chapter2 Shaft Power CyclesChapter2 Shaft Power Cycles 22
SimpleSimple Turbojet CycleTurbojet Cycle
TT
ss
p03
Δpb
01
02
03
04
5
pa
p01
p02
p04
p5
(ΔT0
)comp
Va
² / 2cp
(ΔT0
)turb
V5
² / 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 basicWhen considering the design of a turbojet, the basic
thermodynamic parameters at the disposal of the designerthermodynamic parameters at the disposal of the designer
are theare the Turbine Inlet TemperatureTurbine Inlet Temperature and theand the compressorcompressor
pressure ratio (t , rpressure ratio (t , rcc))
 It is common practice to carry out a series of design pointIt is common practice to carry out a series of design point
calculations covering a suitable range of these twocalculations covering a suitable range of these two
variables (t , rvariables (t , rcc) using) using fixed polytropic efficienciesfixed polytropic efficiencies for thefor the
compressor and the turbine and plotcompressor and the turbine and plot sfc vs Fsfc vs Fss with "with " TITTIT““
((TT0303) and ") and " rrcc " as parameters." as parameters.

Fig. 3.8 Typical Turbojet Cycle PerformanceFig. 3.8 Typical Turbojet Cycle Performance

Optimisation of a Turbojet CycleOptimisation of a Turbojet Cycle
 Thermodynamic optimization of the turbojet cycleThermodynamic optimization of the turbojet cycle
can not be isolated from mechanical designcan not be isolated from mechanical design
considerations and the choice of cycle parametersconsiderations and the choice of cycle parameters
depend very much on thedepend very much on the TYPETYPE of the aircraft.of the aircraft.
01
TIT for high Vj
TIT since T increase
essential for the economic operation of a supersonic aircaft
high
high





⇒
⇒

Fig. 3.9. Performance and Design Considerations forFig. 3.9. Performance and Design Considerations for
Aircraft Gas TurbinesAircraft Gas Turbines

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 blades.such as expensive alloys & cooled blades.
 high rhigh rcc increased weightincreased weight
 large number of compressor-turbine stageslarge number of compressor-turbine stages
 i.ei.e multi spool engines.multi spool engines.

Fig. 3.10.1 Variation of Thrust with Flight Speed for aFig. 3.10.1 Variation of Thrust with Flight Speed for a
Typical Turbojet EngineTypical Turbojet Engine

Fig. 3.10.1 Variation of sfc with Flight SpeedFig. 3.10.1 Variation of sfc with Flight Speed
for a Typical Turbojet Enginefor a Typical Turbojet Engine

3.4 THE TURBOFAN ENGINE3.4 THE TURBOFAN ENGINE
 TheThe Turbofan engineTurbofan engine was originally conceived as awas originally conceived as a
method of improving themethod of improving the propulsive efficiencypropulsive efficiency of the jetof the jet
engine byengine by reducingreducing thethe Mean Jet VelocityMean Jet Velocity particularly forparticularly for
operation at high subsonic speeds.operation at high subsonic speeds.
 It was soon realized that reducing jet velocity had aIt was soon realized that reducing jet velocity had a
considerable effect onconsiderable effect on Jet NoiseJet Noise , a matter that became, a matter that became
critical when large numbers of jet propelled aircraftcritical when large numbers of jet propelled aircraft
entered commercial service.entered commercial service.

The Turbofan EngineThe Turbofan Engine
 InIn Turbofan enginesTurbofan engines ;;
a portion of the total flowa 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 ofare usually decribed in terms of
""by-pass ratio"by-pass ratio" defined as :defined as :
the ratio of the flow through the by-pass duct (coldthe ratio of the flow through the by-pass duct (cold
stream) to that through the high pressure compressorstream) to that through the high pressure compressor
(HPC) (hot stream).(HPC) (hot stream).

FIG.3.11.Twin - Spool Turbofan EngineFIG.3.11.Twin - Spool Turbofan Engine
Vjc
Vjh
Va

 By pass ratio is given by ;By pass ratio is given by ;
 Then ;Then ;
andand ṁṁ == ṁṁ cc ++ ṁṁ hh
 IfIf PPjcjc = P= Pjhjh = P= Paa ,, (no pressure thrust)(no pressure thrust)
then ;then ;
 F = (F = (ṁṁ ccVVjcjc ++ ṁṁ hhVVjhjh ) -) - ṁṁ VVaa for a by-pass enginefor a by-pass engine
h
c
m
m
B ≡
1+
=
B
m
mh
1+
=
B
mB
mc

 TheThe design point calculationsdesign point calculations for the turbofan arefor 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 beIn view of this only the differences in calculations will be
outlined.outlined.
 a)a) Overall pressure ratio ( rOverall pressure ratio ( rcc )) andand
turbine inlet temperature ( TIT)turbine inlet temperature ( TIT) are specifiedare specified
as before ;as before ; but it isbut it is alsoalso necessary to specify thenecessary to specify the
bypass ratio Bbypass ratio B and theand the fan pressure ratio FPR.fan pressure ratio FPR.

 b)b) From theFrom the inlet conditionsinlet conditions andand FPRFPR ; the pressure and; the pressure and
temperature of the flow leaving the fan andtemperature of the flow leaving the fan and entering theentering the
by-pass duct can be calculated.by-pass duct can be calculated.
 The mass flow down the by-pass ductThe mass flow down the by-pass duct ṁṁcc can becan be
established from the total mass flow rateestablished from the total mass flow rate ṁṁ andand BB..
 TheThe cold stream thrustcold stream thrust can then be calculated as for thecan then be calculated as for the
jet engine noting that the workingjet engine noting that the working fluid isfluid is airair..
 It is necessary toIt is necessary to checkcheck whether thewhether the fan nozzle isfan nozzle is
choked or unchoked.choked or unchoked.
 If chokedIf choked thethe pressure thrustpressure thrust must be calculated.must be calculated.

 c)c) In theIn the 2-spool2-spool configurationsconfigurations
the FAN is driven by LP turbinethe FAN is driven by LP turbine
Calculations for the HP compressor and the turbine areCalculations for the HP compressor and the turbine are
quite standard,quite standard,
then inlet conditions to the LP turbine can then be found.then inlet conditions to the LP turbine can then be found.
Considering theConsidering the work requirement of the LP rotorwork requirement of the LP rotor ;;
012 056 056 012
m
056 012
1
1
( 1) B = 0.3 8.0
pa
pa m h pg
h pg
pg
pg m
Cm
mC T m C T T T
m C
C
T B T
C
η
η
η
∆ = ∆ ∴ ∆ = ∆
⇒ ∆ = + ∆ ⇒ →

 The value ofThe value of BB has a major effect on the temperaturehas 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
 KnowingKnowing TT0505,, ηηtt andand TT056056 , LP turbine, LP turbine pressure ratiopressure ratio can becan be
found, and conditions at the entry to thefound, and conditions at the entry to the hot streamhot stream
nozzlenozzle can be established.can be established.

 d)d) If the two streams are mixed it is necessary toIf the two streams are mixed it is necessary to find thefind the
conditions after mixing by means of an enthalpy andconditions after mixing by means of an enthalpy and
momentum balance.momentum balance.
 Mixing is essential for aMixing 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 ratioOverall pressure ratio rrpp
 ii)ii) Turbine inlet temperatureTurbine inlet temperature TITTIT
 iii)iii) By-pass RatioBy-pass Ratio BB
 iv)iv) Fan pressure ratioFan 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 eachNote 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 ofTaking the values of sfcsfc andand FFss
for each of thesefor each of these FPRFPR values in turn,values in turn,
a curve ofa curve of sfcsfc vs.vs. FFss can be plotted.can be plotted.
 Note that each point on this curve is the result of aNote that each point on this curve is the result of a
previous optimizationprevious optimization
and it is associated with a particular value ofand it is associated with a particular value of FPRFPR andand TIT.TIT.

Fig. 3.11. Optimization of a Turbofan EnginePerformanceFig. 3.11. Optimization of a Turbofan EnginePerformance

 Note that optimum values for eachNote 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 ofTaking the values of sfcsfc andand FFss ,,
for each of thesefor each of these FPRFPR values in turn,values in turn,
a curve ofa curve of sfcsfc vs.vs. FFss can be plotted.can be plotted.
 Note that each point on this curve is the result of aNote that each point on this curve is the result of a
previousprevious optimizationoptimization and it is associated with aand it is associated with a
particular value ofparticular value of FPRFPR andand TIT.TIT.

 The foregoing calculations may be repeated for a seriesThe foregoing calculations may be repeated for a series
ofof 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 ofThis plot yields the optimum variation of sfcsfc withwith FFss forfor
the selectedthe selected rrpp as shown by the envelope curve.as shown by the envelope curve.
 The procedure can be repeated for a range ofThe procedure can be repeated for a range of rrpp..

 The quantitative results are summarized as :The quantitative results are summarized as :
a)a) BB improvesimproves sfcsfc at the expense ofat the expense of
significant reduction insignificant reduction in FFss,,
b)b) OptimumOptimum FPRFPR withwith TITTIT ,,
c)c) OptimumOptimum FPRFPR withwith 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 subsonicwith supersonic dash capability & good subsonic sfcsfc
B = 0.5 - 1B = 0.5 - 1 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 criticalsfc is not as critical B = 2-3B = 2-3
 Thrust of engines of highThrust of engines of high BB is very sensitive to forwardis very sensitive to forward
speed due to large intakespeed due to large intake ṁṁ and momentum dragand momentum drag

Mixing in a Constant Area DuctMixing in a Constant Area Duct

3.5 AFT - FAN CONFIGURATION3.5 AFT - FAN CONFIGURATION
 Some early turbofans were directly developed from existingSome early turbofans were directly developed from existing
turbojets,turbojets,
 A combined turbine-fan was mounted downstream of the GasA 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 inThe 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 thePower must eventually be delivered to the aircraft in the
form ofform of thrust power (TP)thrust power (TP) ..
 This can be expressed in terms ofThis can be expressed in terms of equivalentequivalent shaftshaft
Power (SP), propeller efficiencyPower (SP), propeller efficiency ηηpp,, andand jet thrust Fjet thrust F byby
TP = (SP)TP = (SP)prpr + FV+ FVaa
 TheThe turboshaftturboshaft engine is of greater importance and isengine is of greater importance and is
almost universally used in helicopters because of its lowalmost universally used in helicopters because of its low
weight.weight.

3.7 Thrust Augmentation3.7 Thrust Augmentation
 If the thrust of an engine has to be increased above theIf 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 ofof mass flow ratemass flow rate through the enginethrough the engine
 Both of these methods imply theBoth of these methods imply the re-designre-design of the engine,of the engine,
and either of them or both may be used to update theand either of them or both may be used to update the
existing engine.existing engine.

Thrust AugmentationThrust Augmentation
 Frequently there will be a requirement for aFrequently 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 toe. g. for take off, for an acceleration from subsonic to
supersonic speeds or during combat manoeuvres.supersonic speeds or during combat manoeuvres.
 The problem then becomes one ofThe problem then becomes one of thrust augmentation.thrust augmentation.
 Two methods most widely used are:Two methods most widely used are:
i)i) Liquid injectionLiquid injection (water+methanol)(water+methanol)
ii)ii) Reheat (after burner)Reheat (after burner)
 Spraying water to the compressor inlet results in a drop inSpraying 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 CALCULATIONDESIGN POINT PERFORMANCE CALCULATION
FOR TURBOJET & TURBOPROP ENGINES.FOR TURBOJET & TURBOPROP ENGINES.
AA Turbojet & TurbopropTurbojet & Turboprop unit may be considered asunit may be considered as
consisting of 2 parts:consisting of 2 parts:
 Thus:-Thus:-
i /i / GAS GENERATORGAS GENERATOR
ii /ii / POWER UNIT a) TurbojetPOWER UNIT a) Turbojet ⇒⇒ Jet Pipe & Final NozzleJet Pipe & Final Nozzle
b) Turbopropb) Turboprop ⇒⇒ Power TurbinePower Turbine
Jet Pipe & Final NozzleJet Pipe & Final Nozzle

The Gas GeneratorThe Gas Generator
Air intake Compressor Combustion
Compressor Chamber
Turbine
0 1 2 3 4

TurbojetTurbojet TurbopropTurboprop
6
4 5
6
5

Problem : Turbojet & Turboprop EnginesProblem : Turbojet & Turboprop Engines
 DATA:DATA:
 AltitudeAltitude ZZ = 0 ISA (101.325 kPa; 288.0 K)= 0 ISA (101.325 kPa; 288.0 K)
 True Airspeed (True Airspeed (VVaa)) = 0 Static= 0 Static
 Power Output turbojetPower Output turbojet = 90 kN Thrust= 90 kN Thrust
 Power OutputPower Output turbopropturboprop = 4.5 MW Shaft Power= 4.5 MW Shaft Power
 Compressor Pressure RatioCompressor Pressure Ratio (P(P0202 / P/ P0101)) = 10= 10
 TIT (total)TIT (total) TT0303 = 1500K= 1500K
 Jet VelocityJet Velocity VV66 = 220 m/s (turboprop)= 220 m/s (turboprop)
 Compressor Isentropic efficiencyCompressor Isentropic efficiency ηη1212 = 88%= 88%
 Turbine Isentropic efficiencyTurbine Isentropic efficiency ηη3434 = 90%= 90%
ηη4545 = 90 %= 90 %

Problem :Turbojet & Turboprop Engines ;Problem :Turbojet & Turboprop Engines ; DataData
 Jet pipe Nozzle Isentropic efficiencyJet pipe Nozzle Isentropic efficiency ηη5656 = 100%= 100%
 Combustion efficiencyCombustion efficiency ηη2323 = 100%= 100%
 Mechanical efficiency of Turbo compresor driveMechanical efficiency of Turbo compresor drive
ηηMM = 100%= 100%
 Reduction Gear efficiencyReduction Gear efficiency ηηGG = 97%= 97%
 Intake Pressure RecoveryIntake Pressure Recovery PP0101/ P/ P0000 ==0.980.98

Problem :Turbojet & Turboprop EnginesProblem :Turbojet & Turboprop Engines ;; DataData
 Combustion Chamber total pressure loss :Combustion Chamber total pressure loss :
ΔΔPP023023 = 7% of compressor outlet total pressure (P= 7% of compressor outlet total pressure (P0202))
 Jet Pipe-Nozzle pressure loss :Jet Pipe-Nozzle pressure loss :
ΔΔPP056056 = 3% of turbine outlet total pressure (P= 3% of turbine outlet total pressure (P0404 or Por P0505))
 Nozzle discharge CoefficientNozzle discharge Coefficient CCdd= 0.98= 0.98
 Cooling air bleedCooling air bleed rr = 5% of Compressor mass flow.= 5% of Compressor mass flow.

Problem : Turbojet & Turboprop EnginesProblem : Turbojet & Turboprop Engines;; DataData
 CCpapa = 1.005 kJ/kg-K for air= 1.005 kJ/kg-K for air
 CCpgpg = 1.150 KJ/kg-K for gas= 1.150 KJ/kg-K for gas
 γγaa = 1.40 for air= 1.40 for air
 γγgg = 1.33 for gasses= 1.33 for gasses
Calorific value of fuelCalorific value of fuel
 ΔΔHH = 43.124 MJ/kg= 43.124 MJ/kg

CalculationsCalculations
a) Aira) Air
Ram Temperature RiseRam Temperature Rise ΔΔTT0Ram0Ram= V= Vaa
22
/2C/2Cpp = 0 K= 0 K
 TToaoa = (T= (Taa++ΔΔTT0Ram0Ram) = 288 + 0 = 288K) = 288 + 0 = 288K
 PP0101 = P= Poaoa * P* P0101 / P/ Poaoa = 101.3 * 0.98 = 99.3 kPa= 101.3 * 0.98 = 99.3 kPa
 No work is done on or by air at the IntakeNo work is done on or by air at the Intake
 TT0101 = T= Toaoa = 288K= 288K
Problem :Turbojet & Turboprop EnginesProblem :Turbojet & Turboprop Engines ; Calculations; Calculations

Problem :Turbojet & Turboprop Engines ; CalculationsProblem :Turbojet & Turboprop Engines ; Calculations
 b) Compressorb) Compressor
 TT0202 = T= T0101 ++ ΔΔTT012012 = 288. + 304.6 = 592.6 K= 288. + 304.6 = 592.6 K
 PP0202 = P= P0101 * (P* (P0202/P/P0101) = 99.3 * 10 = 993.0 kPa) = 99.3 * 10 = 993.0 kPa
( )
0.4.
1.4
02 01
12
02 01
1
01 02
012
12 01
012
'
288
1 10 1
0.88
304.6
T T
T T
T P
T
P
T K
γ
γ
η
η
−
−
=
−
 
  ∆ = − = − ÷    
∆ =

 C) Combustion ChamberC) Combustion Chamber
 ΔΔPP023023 == ΔΔPP023023* P* P0202 = 0.07 * 993.0 = 69.5 kPa= 0.07 * 993.0 = 69.5 kPa
 PP0303 = P= P0202 -- ΔΔPP023023 = 993.0 - 69.5 = 923.5 kPa= 993.0 - 69.5 = 923.5 kPa
 By Heat BalanceBy Heat Balance
ηη 2323 mmff ΔΔH = CH = Cp23p23 (m(maa+m+mff) (T) (T0303-T-T0202))
 defining :defining : ff ≡≡ mmff / m/ maa ;; ΔΔTT023023 = T= T0303-T-T0202

 Using the Combustion CurvesUsing the Combustion Curves
⇒Ideal Temperature Rise (Ideal Temperature Rise (ΔΔ TT2323) vs f) vs f
⇒ (with T(with T0202 as a parameter)as a parameter)
 ΔΔTT023023' =' = ΔΔTT023023 // ηη2323 = 907.4 K ;= 907.4 K ; ((ηη2323 =100%)=100%)
TT0202 = 592.6K= 592.6K ⇒⇒ f’ = 0.0262f’ = 0.0262
 This takes account of the variation of CThis takes account of the variation of Cp23p23
with f and temperaturewith f and temperature
f = 0.0262 /f = 0.0262 / ηη2323 = 0.0262 /= 0.0262 / 1.001.00 = 0.0262= 0.0262

 d) Compressor Turbined) Compressor Turbine
 Compressor Turbine Output *Mechanical efficiency of drive =Compressor Turbine Output *Mechanical efficiency of drive =
= Compressor input= Compressor input
 ṁṁ 11 CCp12p12 ΔΔTT012012 == ηηmm ṁṁ 33 CCp34p34 ΔΔTT034034
 ṁṁ 11 = Compressor mass flow rate= Compressor mass flow rate
 ṁṁ 33 = Compressor turbine mass flow rate= Compressor turbine mass flow rate
 r = Cooling air bleed = 0.05r = Cooling air bleed = 0.05
 ṁṁ 11 == ṁṁ 22 / (1 - r)/ (1 - r) ṁṁ 33 == ṁṁ 22 (1 + f)(1 + f)

 ṁṁ 11 // ṁṁ 33 = 1 /((1-r)*(1+f))= 1 /((1-r)*(1+f))
 ∴∴
KT
T
frc
cT
T
p
p
m
1.273
)0262.1(*)95.0(
1
*
150.1
005.1
*
00.1
6.304
)1(*)1(
1
**
034
034
34
12012
034
=∆
=∆
+−
∆
=∆
η

47.2
1500*90.0
1.273
1
1
1
1
1
03
04
33.0
33.11
0334
03403
04
1
03
04
03
034
'
0403
0403
34
=










−
=












∆
−
=














−
∆
=
−
−
=
−
−
P
P
T
TP
P
P
P
T
T
TT
TT
γ
γ
γ
γ
η
η

 PP0303 / P/ P0404 = 2.47= 2.47
 TT0404 = T= T0303 -- ∆∆ TT034034 = 1500 -273.1 =1226.9 K= 1500 -273.1 =1226.9 K
 PP0404 = P= P0303 / (P/ (P0303 / P/ P0404) = 923.47 / 2.47) = 923.47 / 2.47
 PP0404 = 373.9 kPa= 373.9 kPa

 Power SectionPower Section
 i) Turbojeti) Turbojet
 ΔΔPP046046 = (= (ΔΔPP046046/ P/ P0404) * 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.
 AsAs ηη5656 = 100%= 100%
 If PIf P0606/ P/ Paa across the final nozzle exceeds Pacross the final nozzle exceeds P0606/P/Pcc
⇒⇒ PP0606/P/Pcc = 1.85= 1.85 forfor γγ = 1.33= 1.33
 Then the nozzle will be choked thus MThen the nozzle will be choked thus Mthroatthroat = 1= 1
 Here PHere P0606/ P/ Paa =362.71 / 101.33 = 3.58=362.71 / 101.33 = 3.58 ⇒⇒
 ⇒⇒ thethe nozzle is chokednozzle is choked

TT66 = T= T0606 – 0.143*T– 0.143*T0606 = 0.857*T= 0.857*T0606 =0.857*1226.9=0.857*1226.9
TT66 = 1051.6 K= 1051.6 K
167.1
2
1
2
1
1 2
6
6
06
=
+
=
−
+=
γγ
M
T
T
0606
2
6
06
06
6
06606
2
6
*143.0
1
1
2
1
2
11
2
TT
C
V
T
T
T
TTT
C
V
p
p
=





+
−
=
=





+
−=





−=−=
γ
γ
γ
since Msince M66 =1=1
we havewe have
Since TSince T0606 = T= T0404
⇒⇒

6 06 6
6 06
6 06 06
06 06
3
6
6
6
2 * ( ) 2 *1150 *175.3 635 /
362.7
195.78
1.863
195.78*10
0.649 /
287 *1051.6
pg
cr
V c T T m s
P P
P P P kPa
P P
P
P
kg s
RT
ρ
= − = =
 
 ÷
   ÷= = = = ÷  ÷    ÷ ÷ ÷
  
= = =
Flowrate at the throatFlowrate at the throat mm66 == ρρ66AA66 VV66
where A6 is the Effective Nozzle Throat Areawhere A6 is the Effective Nozzle Throat Area
AA66 // ṁṁ 66 = 1 / (= 1 / ( ρρ66*V*V66 ) = 1 / ( 0.649 * 635 ) = 0.00243 m) = 1 / ( 0.649 * 635 ) = 0.00243 m22
s/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 VV66 -- ṁṁ aa VVaa +(P+(P66-P-Paa) A) A66
)1)(1(
1
6 frm
ma
+−
=

6
6
6 6
6 6
3
6
1
(1 )(1 )
( )*
(1 )(1 )
635 0 ()195.78 101.32)*10 *0.00243
864.31 /
a
N a
s a
s
N
s
m
m r f
F V A
F V P P
m r f m
F
F
F Ns kg
m
=
− +
= = − + −
− +
= − + −
= =

hNkgsNkgsfc
f
f
mFF
m
sfc
NN
f
−=−=
=
+
==
/3600*52.29/52.29
0262.1
0262.0
*
31.864
1
1
*
1
6
Since FN = 90 kN (required value)
skg
r
m
m
skg
f
m
m
skg
mF
F
m
N
N
/81.106
95.0
47.101
1
/47.101
0262.1
13.104
1
/13.104
31.864
90000
2
1
6
2
6
6
==
−
=
==
+
=
===

 ṁṁ ff = f *= f * ṁṁ 22 = 0.262 * 101.47 = 2.66.kg/s= 0.262 * 101.47 = 2.66.kg/s
 Effective Nozzle AreaEffective Nozzle Area
 AA6eff6eff == 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

ii) Turbopropii) Turboprop
 Here the expansion takes place mainly in the powerHere the expansion takes place mainly in the power
turbine, leaving only sufficient pressure ratio across theturbine, leaving only sufficient pressure ratio across the
nozzle to produce the specified jet velocity.nozzle to produce the specified jet velocity.
 The requiredThe required division of pressure dropdivision of pressure drop through thethrough the
turbineturbine & the& the nozzlenozzle is found byis found by trial and errortrial and error::
 As aAs a first trial,first trial, assume that the power turbine temperatureassume that the power turbine temperature
drop isdrop is ΔΔTT045045 = 295K= 295K

 and: Tand: T0505 =T=T0606 = T= T0404 -- ΔΔTT045045 = 12227 - 295 = 931.9K= 12227 - 295 = 931.9K
 PP0505 =P=P0404 * (P* (P0505/P/P0404) = 373.9 / 3.47 = 107.85 kPa.) = 373.9 / 3.47 = 107.85 kPa.
47.3
9.1226*9.0
295
11
04
05
33.0
33.1
1
0445
045
04
05
=






−=




 ∆
−=
−
P
P
T
T
P
P γ
γ
η
ThenThen

 AlsoAlso ΔΔPP056056 = (= (ΔΔPP056056 / P/ P0505)* P)* P0505 ==
= 0.03 * 107.85 = 3.24 kPa= 0.03 * 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 / 101.33 = 1.033 < 1.85= 104.6 / 101.33 = 1.033 < 1.85
far less then the critial value !far less then the critial value !
 Thus the Nozzle isThus the Nozzle is unchoked; so Punchoked; so P66 = P= Paa

 HenceHence
 andand VV66 == √√ (2*1150*7.4) = 130.4 m/s(2*1150*7.4) = 130.4 m/s
K
P
P
T
c
V a
p
4.7
62.104
33.101
19.931*00.1
1
2
4
1
1
06
0656
2
6
=














−=
=














−=
−
γ
γ
η

 But the given value VBut the given value V66= 220 m/s= 220 m/s
 Since the found value is too low, we now try aSince the found value is too low, we now try a
somewhat lower value ofsomewhat lower value of ΔΔTT045045 = 283.2K= 283.2K
 Proceeding as above ; PProceeding as above ; P0404 / P/ P0505 = 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
 VV66
22
/ 2C/ 2Cpp = 21K= 21K VV66 = 219.8 m/s= 219.8 m/s ≈≈ 220m/s220m/s
 this is close enough, with thethis is close enough, with the Nozzle UnckokedNozzle Unckoked

 The shaft powerThe shaft power
 WWshsh == ηηGG ** ṁṁ 44 * c* cp45p45 ** ∆∆TT045045
 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 unchoked, there is only momentum thrustSince the Nozzle is unchoked, there is only momentum thrust
 FFNN == ṁṁ 66* V* V66 –– ṁṁaa* V* Vaa
6
6
219.8 /
(1 )(1 )
N a
s
F V
F V N s kg
m r f
= = − = −
− +

 For the static case it isFor the static case it is givengiven that ;that ;
 1N of jet Thrust is1N of jet Thrust is equivalent toequivalent to 65 W of propeller65 W of propeller
shaft Power.shaft Power.
 Shaft power equivalent of jet thrustShaft power equivalent of jet thrust
per unit mass flow = (per unit mass flow = (wwjj// ṁṁ 66) = (F) = (FNN// ṁṁ 66)* 65 / 1000)* 65 / 1000
wwjj// ṁṁ 66 = 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 66 = (= (wwss ++ wwjj ) /) / ṁṁ 66 = 315.9 + 14.3= 315.9 + 14.3
wwjj// ṁṁ 66 = 330.2 kJ/kg= 330.2 kJ/kg

 Nozzle Exit :Nozzle Exit :
 TT66 = T= T0606 – v– v66
22
/ 2c/ 2cPP = 943.7 - 21 = 922.7 K= 943.7 - 21 = 922.7 K
 PP66 = P= Paa =101.33 kPa=101.33 kPa
 ρρ66 = P= P66 / (RT/ (RT66) = 101.33 * 1000 / (287*922.7)) = 101.33 * 1000 / (287*922.7)
 ρρ66 = 0.383 kg/m= 0.383 kg/m33
26
6 6 6
1 1
0.0119 /
0.383*219.8
A
m s kg
m Vρ
= = =

 The sfc based on shaft power is ;The sfc based on shaft power is ;
Jkgsfc
mwf
f
w
m
sfc
sh
shsh
f
sh
/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 /
f
ef
ef ef
sh
m f
sfc
w f w m
sfc kg J g MJ−
= = =
+
= =

 Since the shaft power is specified to beSince the shaft power is specified to be
WWshsh = 4.5 MW= 4.5 MW
6
6 3
6
6
2
2
1
4.5*10
14.24 /
315.91*10
14.24
13.88 /
1 1.0262
13.88
14.61 /
1 0.95
sh
sh
w
m kg s
w m
m
m kg s
f
m
m 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 AreaEffective 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-effective6-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

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