Parametric Cycle Analysis Lecture 10 Parametric Cycle Analysis

Parametric CycleAnalysis

IntroductionRecap

Meaning of ParametricCycle Analysis

Turbojet CycleOverview

Components

Performance

Turbofan CycleOverview

Performance

Carpet Plots

Conclusions

10.162

Lecture 10Parametric Cycle AnalysisText:

Motori AeronauticiMar. 26, 2015

Mauro ValoraniUniveristà La Sapienza


IntroductionRecap



Components

Performance


Performance

Carpet Plots

Conclusions

10.163

Agenda

1 IntroductionRecapMeaning of Parametric Cycle Analysis

2 Turbojet CycleOverviewComponentsPerformance

3 Turbofan CycleOverviewPerformance

4 Carpet Plots

5 Conclusions


IntroductionRecap



Components

Performance


Performance

Carpet Plots

Conclusions

10.164

Design Process

MASS FLOW

CONSTRAINT&

MISSIONANALYSIS

PARAMETRICCYCLE

ANALYSIS

Mission Specs Efficiencies (1st attempt)

Thrust Cycle parameters (βc, T4, BPR, …)

Specific Thrust Ia

Componentsizing

Assumed TSFCbehavior with

h, V, δT

OFF-DESIGN

Geometries

Efficiencies (Actual)

Cross-section,blade profiles,combustor, …

Desired TSFCReference flight conditionTech limitation

Figure: Design process, schematic


IntroductionRecap



Components

Performance


Performance

Carpet Plots

Conclusions

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A Roadmap

DESIGN PROCESS⇒ Constraint and Mission AnalysisChoice of (TSL/WTO) and (WTO/S)Estimation of WTO to obtain TSL

ENGINE SELECTION⇒ Parametric Cycle Analysis and Performance

ENGINE COMPONENTS⇒ Components Sizing


IntroductionRecap



Components

Performance


Performance

Carpet Plots

Conclusions

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What we have done so far and what’s next

Done:

RFP⇒ Constraints on TSL/WTO and WTO/S

First attempt choice of TSL/WTO and WTO/S

Mission Analysis⇒WTO ⇒ TSL

At this point, the designer knows:

TSL

Assumed behaviors of T and TSFC with h and M

Next: Parametric Cycle Analysis and Performance estimationWe are looking for the best set of design choices (BPR, βF , βC , etc.) in termsof:

design limitations (T4max , βCmax , etc.)

flight conditions (h, M)

required performance (TSFC)


IntroductionRecap



Components

Performance


Performance

Carpet Plots

Conclusions

10.167

Parametric Cycle Analysis: things to keep in mind

A flight condition has to be fixedImagine to have a knob for each design parameter and to fine tune themuntil satisfactory engine performance are reached.

Cycle Analysis assumes 1-D flow, perfect gas and non-ideal componentsare modeled through realistic efficiencies

We obtain specific terms (ratios) for the performance. The engine sizewill then be chosen assigning a Massflow or a Thrust.


IntroductionRecap



Components

Performance


Performance

Carpet Plots

Conclusions

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On-Design / Reference Point

Parametric Cycle Analysis is often referred to as Design-Point Analysis.

BE CAREFUL !

It doesn’t really make sense to identify a single point that dictates thedesign.

The final design is based on engine performance over the entire aircraftmission profile

The "winner" is chosen because of its balanced behavior over the wholeflight conditions spectrum

⇒ The operating point at which all design quantities are chosen is aReference Point rather than a Design Point.

The reference point is needed to size the engine and explore its off-designbehavior


IntroductionRecap



Components

Performance


Performance

Carpet Plots

Conclusions

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Turbojet Cycle


IntroductionRecap



Components

Performance


Performance

Carpet Plots

Conclusions

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Intake

External (a–1) and internal (1–2) compressions

Energy Eqn: ∆h0 = 0 but ∆h = h2 − h1 6= 0p0 decreases (friction); p incr., u decr.

Efficiencies: 1) P0 ratio εd , 2) Adiabatic Efficiency ηad

a≡1

0a

2

V2

2cp

2’

pa

p2≡p

02

p0a

T

s

T02’

≡T2’

T0a

≡T02

Ta

T2 ≈ T02 = T0a = Ta(1 + δM2

0

)T2 depends on M0 e Ta (altitude)

p2 ≈ p02 depends on intake efficiency


IntroductionRecap



Components

Performance


Performance

Carpet Plots

Conclusions

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Intake I/O


IntroductionRecap



Components

Performance


Performance

Carpet Plots

Conclusions

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Compressor

Energy Eqn. (Q = 0, M � 1):

∆h0 '∆h = Lc

Total Quantities ' Static Quantities

βc := p3/p2 (Pressure Ratio) =⇒ p3 = βc p2

ηc := L′c/Lc (Adiabatic Efficiency)

Ideal compression:

T ′3 = β(γ−1)/γc T2 = τc T2 ; L′c = cp T2 (τc − 1)

Real compression:

T3 =

(1 +

τc − 1ηc

)T2 ; Lc = cp T2 (τc − 1)/ηc


IntroductionRecap



Components

Performance


Performance

Carpet Plots

Conclusions

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Compressor I/O


IntroductionRecap



Components

Performance


Performance

Carpet Plots

Conclusions

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Burner

Energy Eqn. (Ls = 0, M � 1): ∆h0 '∆h = ∆Qηb := Q/(mf Qf ) Combustion efficiency

T4: Turbine Inlet Temperature (TIT)

ηpb := p4/p3 Pneumatic efficiency

f := mf /ma Fuel / Air ratio

mah3 + mf hf + Q = (ma + mf )h4

f � 1⇒ mah3 + Q = mah4 ⇒ macp(T4 − T3) = Q

Q = ηb mf Qf

f =cp (T4 − T3)

ηb Qf=

cpT3

ηb Qf

(T4

T3− 1

)

p4 = ηpb p3

f decreases if βc is increased, with a fixed T4 !!


IntroductionRecap



Components

Performance


Performance

Carpet Plots

Conclusions

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Burner I/O


IntroductionRecap



Components

Performance


Performance

Carpet Plots

Conclusions

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Turbine

Energy Eqn. (Q = 0, M � 1): ∆h0 '∆h = −Lt

Power balance :

−Lt = Lc ⇒ (ma + mf )Lt = ma Lc

Mechanical efficiencies of compressor and turbine: ηmc , ηmt

−ηmt Lt = ηmt(ma + mf )(h4 − h5) =Lc

ηmc=

ma(h3 − h2)

ηmc

ηmtηmc(1 + f)(h4 − h5) = h3 − h2

Turbine Exit Temperature (TET)

T5 = T4 −T3 − T2

ηmtηmc(1 + f )' T4 −

T3 − T2

ηmtηmc

Turbine Exit Pressure

ηt =Lt

L′t=

cp(T4 − T5)

cp(T4 − T5′)→ T5′ = T4 −

T4 − T5

ηt→

p4

p5=

p4

p5′=

(T4

T5′

) γγ−1

⇒ p5 = p4

(1−

1− T5/T4

ηt

) γγ−1


IntroductionRecap



Components

Performance


Performance

Carpet Plots

Conclusions

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Turbine I/O


IntroductionRecap



Components

Performance


Performance

Carpet Plots

Conclusions

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Nozzle

Energy Eqn. (Ls = Q = 0): ∆h0 = 0 , ma: ∆h = h9 − h5 6= 0

h05 = h5 +u2

52 ' h5, h09 = h9 +

u29

2 =⇒ u9(= ue)

Adiabatic efficiency:

ηn :=h5−h9h5−h9′

=T5−T9T5−T9′

, T9′ = T5

(p9p5

) γ−1γ

Exit flow speed:

ue = u9 '√

2 cp (T5 − T9) =√

2 ηn cp (T5 − T ′9) =

=

√√√√√2 ηn cp T5

1−(

p9

p5

) γ−1γ

Adapted nozzle: p9 = pa[

T9 = T5 − ηn(T5 − T9′

)]


IntroductionRecap



Components

Performance


Performance

Carpet Plots

Conclusions

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Nozzle I/O


IntroductionRecap



Components

Performance


Performance

Carpet Plots

Conclusions

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Overall I/O


IntroductionRecap



Components

Performance


Performance

Carpet Plots

Conclusions

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Turbojet Performance (inflight)

Thrust (Adapted nozzle, f � 1):F = ma[(1 + f) ue − V0] ' ma (ue − V0)

Thermal efficiency (f � 1):

ηth =Pj

Pav' ma (u2

e/2− V 20 /2)

mf Qf=

u2e − V 2

02 f Qf

Propulsive efficiency (f � 1):ηp =

PpPj' 2 ν/(1 + ν)

Overall efficiency (f � 1):ηo =

PpPav

= ηth ηp ' (ue − V0) V0f Qf

Specific Thrust (f � 1):Ia = F

ma= (1 + f) ue − V0 ' ue − V0

TSFC (f � 1):

TSFC =mfF = f ma

F = f(1 + f ) ue−V0

'f

ue − V0


IntroductionRecap



Components

Performance


Performance

Carpet Plots

Conclusions

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Turbojet Performance (inflight): Specific Thrust

Higher Ia allows for lighter engines for a given Thrust.

Ia has a maximum for a certain βc (corresponding to the max of ue - or Lu).

βc

I a(m

/s)

10 20 300

200

400

600

T4/T

a

For any fixed βc , Ia increases withT4/TaThe maximum slightly moves to theright for increasing T4/Ta, because theoptimum of Lu moves to higher βc

βc

I a(m

/s)

10 20 300

200

400

600

M

Ia decreases with flight Mach number.That’s because V0 increases with M(also ue, because of the higher intakecompression, but less)


IntroductionRecap



Components

Performance


Performance

Carpet Plots

Conclusions

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Turbojet Performance (inflight): TSFC

TSFC goes like f/ue. f decreases with βc for a fixed T4. ue has a maximumwith βc .

βc

TS

FC

(kg

/h/N

)

10 20 300

0.1

0.2

0.3

0.4

0.5

T4/T

a

For a fixed βc , TSFC decreases withincreasing T4/Ta (because ue isincreasing)

βc

TS

FC

(kg

/h/N

)

10 20 300

0.1

0.2

0.3

0.4

0.5

M

TSFC increases with flight Machnumber. That’s because V0 increaseswith M (also ue, because of the higherintake compression, but less)


IntroductionRecap



Components

Performance


Performance

Carpet Plots

Conclusions

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Turbofan Cycle


IntroductionRecap



Components

Performance


Performance

Carpet Plots

Conclusions

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Why a Turbofan?

ηp =1

1 + F2 ma V0

=1

1 + 12

IaV0

→ for a fixed V0 : ηp ↑ if Ia ↓

Does it makes sense to reduce Ia in a turbojet? NO:larger massflow means bigger and heavier turbomachinesu9 (thus cycle useful work) decreases⇒ lower ηth

Work-around: part of the available work at the turbine exit is used to moveanother turbine that powers a fan. This fan gives a slight pressure ratio to a bigmassflow that does not enter the turbomachines.

F/mV0

ηp

0 2 4 6 8 100.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

.

Figure: Andamento del redimento propulsivo in funzione della spinta specifica Ia.


IntroductionRecap



Components

Performance


Performance

Carpet Plots

Conclusions

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Turbofan: things to remember

BPR =ma2ma1

The fan treats both massflows 1 and 2

HPC, Burner, HPT, LPT treat massflow 1 only (plus fuel from HPT on)

HP spool: power balance HPC-HPT

LP spool: power balance Fan/LPC-LPT

Separate flows: two nozzles

Mixed flows: mixing chamber and single nozzle


IntroductionRecap



Components

Performance


Performance

Carpet Plots

Conclusions

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Overall I/O


IntroductionRecap



Components

Performance


Performance

Carpet Plots

Conclusions

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TURBOFAN A FLUSSI SEPARATIPRESTAZIONI

F = F1+F2 = ma1[(1+f)u9−V0

]+(p9−pa)A9+ma2

(u19−V0

)+(p19−pa)A19

PER UGELLI ADATTATI⇒ F = ma1{[

(1 + f)u9 − V0]+ BPR

(u19 − V0

)}

Ia =

[(1+f )u9−V0

]+BPR

(u19−V0

)1+BPR '

[u9−V0

]+BPR

(u19−V0

)1+BPR

riferita

allaportata

complessiva

TSFC = f[

(1+f )u9−V0]+BPR

(u19−V0

) ' f[u9−V0

]+BPR

(u19−V0

)


IntroductionRecap



Components

Performance


Performance

Carpet Plots

Conclusions

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Turbofan Performance: effect of BPR

ηth

TSFC

Ia

BPR

η

TS

FC

(kg

/h/N

)

I a(m

/s)

0 1 2 3 4 5 60.0

0.2

0.4

0.6

0.8

0.00

0.04

0.08

0

200

400

600

Figure: Rendimento termico, TSFC, e spinta specifica in funzione del BPR per unturbofan a flussi separati con βf = 1.5; βc = 20; Ta = 290 K; T4 = 1400 K; M = 0.0.


IntroductionRecap



Components

Performance


Performance

Carpet Plots

Conclusions

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Turbofan Performance: effect of BPR

BPR = 0 means Turbojet

Specific Thrust decreases with BPR because:

ue goes from the pure u9 to a lower value in between u9 and u19

TSFC (= fF/ma1

) decreases with BPR because:

f is constant with BPR (the cycle does not vary with BPR)

F/ma1 grows with BPR:

Fma1

= (u9 − V0) + BPR(u19 − V0)


IntroductionRecap



Components

Performance


Performance

Carpet Plots

Conclusions

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Choosing the BPR

High-BPR Turbofan: (subsonic) commercial aircrafts. Efficiency and flightendurance are of key importance

Low-BPR Turbofan: (sub/supersonic) military aircrafts. High maximumthrust (possibly w/ post-combustion) is the main goal


IntroductionRecap



Components

Performance


Performance

Carpet Plots

Conclusions

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Turbofan Carpet Plot: effect of βc and T4

Figure: Separate flows Turbofan, BPR = 9, βf = 1.5, M = 0.8, h = 10000 m


IntroductionRecap



Components

Performance


Performance

Carpet Plots

Conclusions

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Turbofan Carpet Plot: effect of βc and BPR

Figure: Separate flows Turbofan, T4 = 1500 K, βf = 1.5, M = 0.8, h = 10000 m


IntroductionRecap



Components

Performance


Performance

Carpet Plots

Conclusions

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Conclusions

Once the carpet plots are available:

Fix the goal TSFC

Choose the best/optimum set of parameters that maximizes the SpecificThrust

Very important remarks:

Note/1: the parameters space can be very high-dimensional. Suitableoptimization algorithms might be needed to find the set of design choices

Note/2: efficiencies (η) values are needed for the cycle calculations.Being this the first step of the design process, we use estimates (firstattempts). Clearly, the component sizing will give us realistic efficienciesthat we can re-use here (iterative design process)

Parametric Cycle Analysis Lecture 10 Parametric Cycle Analysis

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