Chalmers University of Technology Lecture 5 Shaft power cycles –Cycle selection –Technology...

Chalmers University of Technology

Lecture 5

• Shaft power cycles– Cycle selection– Technology trends

• Aircraft engine performance– Thrust and propulsion efficiency– Intakes and engine installation

• Theory 5.1 and 5.2

• Problem 3.1


Simple ideal cycle – max. efficiency• Maximum thermal efficiency

when compressor exit temp. = max allowed turbine inlet temp.

– T3 = 1500 => η = 81% efficiency attained at rc = 320

• In practice titanium alloy compressor rotors withstand around 870 K and nickel alloys around 990 K– T3 = 990 K => η = 71% efficiency

attained at rc = 75

T2 => Tmax


Simple real cycle – max. efficiency• Setting:

– T3 = 1500 => η∞,c = 87%, η∞,t = 85%, 5% burner pressure drop and 99% mechanical efficiency => no power delivered at rc = 143 (far above allowable t2).

• For real cycle maximum efficiency obtained for (with data above)– rc = 38 at which a cycle efficiency of

η = 43.8% is obtained.

Conservative assumptions (old technology)


• Curve is fairly flat around optimum

• Lower pressure ratio may be taken with small perfor-mance penalty– Lower pressure

ratio is cheaper to manufacture and maintain


• Suitable compromise in this case is:– rc = 20

• Selecting a lower rc also– Reduces the number of

required turbomachinerystages

– Allows more efficientcooling (Tc low)

Between 0.6-0.7 is current

state of the art.

coolantgasgasmetal

coolantgas

metalgas

TTTT

TT

TT


GTX 100

π = 19.2T3 = 1570 K


Technology improvements – permissible T3

• T3 is increasing with 8 K/year


Technology improvements – permissible T3


Cycle Efficiency• Optimal pressure ratio

increase with t3.

• Gain in efficiency becomes marginal as T3 increases


Specific output• Considerable increase

in specific outputwith increasing t3.

• Trend:– Increase T3 to increase

specific output

– Follow with rc to obtain high efficiency


Heat exchange cycle

)1(21

tr

• Recall:– Heat exchanger

useful when:

– Optimum for r > 1in real cycle.

– Optimum rc increasewith t3.

– Gain in efficiency with t3 is greater than for simple cycle

– Power output curves about the same

– Cooling simplified – t2 low


131

• Heat exchanger cycle:– IRA (intercooled recuperated

promises increased fuelefficiency).

– More efficient cooling– Heavy and bulky

Heat exchanger versus simple cycle


Which cycle has the highest ideal efficiency?

3

1

1

3

1

3

1

,

3

1

1

3

1

2,

11

111

11

11

1

T

T

TT

r

TTr

T

T

TT

TT

cc

exchangerheatth

simpleth

Theoretically the same!


What are the fundamentals of flight?

• The performance of the jet engine

• Aircraft aerodynamics– Covered by the Henrik Ekstrand

material

• Characteristics of the atmosphere


drag momentum intake thrustmomentum gross

aj mCmC

ThrustNet

momentumofchangeofRate

Aircraft propulsion – thrust generation


thrustpressure

ajjaj ppACCm

ThrustNet

Jet engine – principles of thrust generation


Efficiency considerationsHow much of the power in the jet is transformed to thrust ?:

22

22

aj mCmCKEenergykineticinChange

ajaa CCmCFCaircrafttoPower

22

efficiency Propulsive22

aj

aja

mCmC

CCmC


Efficiency considerations• ηp is at maximum

when Cj=Ca but then the thrust is zero.

• Make difference as small as possible, still obtaining the necessary thrust =>classes of engines!!!

a

jaj

a

ajaj

aja

aj

ajap

C

CCC

C

CCCC

CCC

mCmC

CCmC

1

2

21

21

22

efficiency Propulsive22


Further efficiency considerations

pnetf

aj

e Qm

mCmC

conversionenergyofEfficiency,

22

2

pnet

a

pnetf

a

QSFC

C

Qm

CFefficiencyOverall

,,o

1

2

2

,,

22

22 opnetf

a

pnetf

aj

aj

aep Qm

FC

Qm

mCmC

mCmC

FC

Note that:


How should we design the engine

• Decrease T3 => Decrease jet velocity

– Poor cycle efficiency – Poor specific output => high engine weight

• What if we could:– Use high T3 and rc cycle and still obtain an average

low Cj, optimized for the aircraft speed Ca !?


Propulsion engines – families

The turbofan:BPR 0+-10

h

c

m

m bpr


The turboprop: BPR typically around 25-30

Propulsion engines – families


RM12 engine powering the Swedish GRIPEN fighter – Military turbofan (low bpr)

m

F sF

High speed flight requires high specific thrust


Variable intake optimize aerodynamicperformance of “shock-compression”system

Very high speed flight requires very high specific thrust


• Approximation to conditions averaged over location and season

• Deviations largest at sea level

• Deviations largest in temperature

The International Standard Atmosphere (ISA)


The International Standard Atmosphere (ISA)

• Temperature drops with 6.5K per 1000 meters

• At 11000 m variation stops and T remains constant up to 20000 meters

• Pressure variation can be computed by simple integration of hydrostatic effects.

gdhRT

PlawgasIdealgdhdP

hTT SL3105.6

h

SL

P

P hTR

gdh

P

dP

SL 03105.6


Hydrostatic integration....

R

g

SL

SL

SL

SL

h

SLPP

T

hT

T

hT

R

g

hTR

gP

SL

3105.63

SL

3

3SL

0 3

3

105.6PP

105.6

ln105.6P

Pln

105.6ln105.6

ln


Intakes• Adiabatic duct used to recover kinetic energy in air

at minimal pressure loss, i.e. we have

0

2200

01020102

21

1

22

2

0102

TTchh

Vh

Vhwq

p

hh

Another example of th

e

first la

w for o

pen systems

with no heat o

r work

exchange (same id

ea

as for th

e nozzle)


Intake efficiencies• The available stagnation temperature is:

a

ai TT

TT

01

01

201

1201

2

11

2

11

aa

aia

MT

T

MP

P

p

aa c

CTT

2

2

01

T´01is the stagnation temperature that would have been necessary to achieve P01 under isentropic conditions, i.e.:

Some algebra gives (as well as definition of Mach number and a relation for cp):


Intakes• Design criteria:

– Minimize inlet compressor inlet distortion

– Distortion may lead to surge => flame out or mechanical damages


Functionality• Static conditions, very low aircraft speeds

– Intake acts as a nozzle

• Cruise – normal forward speeds– Intake performs as diffuser

• Supersonic operation– System of shock waves

followed by a subsonic diffusion section


Supersonic intakes• Pressure recovery factor

is used: a

a

aa P

P

P

P

P

P 0

factorrecovery Pressure

0

0101

120

2

11

a

a

a MP

Pwhere:

A rough rule of thumb published by the Department of Defense is:

5 1for 1075.00.1 35.1

0

01

aa

shocka

MMP

P


SR71 – intake ram pressure ratio


SR71 – intake ram pressure ratioThe ram pressure rise is estimated using the following expression:

a

a

aa P

P

P

P

P

P 0


0

0101

where the second factor in the left hand expression is obtained from:

120

2

11

a

a

a MP

P

shockasubsonicaa P

P

P

P

P

P

0

01

0

01

0

01the first factor is obtained from:

Includes both shock and viscous losses



960.0]part) subson.(for - 1[

21

1

21

1

12

12

0

01

a

a

ai

subsonica

M

M

M

P

P

The subsonic part is calculated from from our “universal” assumption of ηi=0.93, i.e:

The shock pressure recovery factor is estimated by the crude formula stated by the Department of Defence (assuming a cruise Mach number of 3.0):

0.809 1075.00.1 35.1

0

01

a

shocka

MP

P


and thus the pressure recovery factor is: 0.776 809.0960.00

01 aP

P


The pressure ratio over the intake can finally be estimated to:

!!! 5.287.36776.00


0

0101 a

a

aa P

P

P

P

P

P

This is a very crude approximation methodology, but it gives a demonstration of the considerable pressure ratio that a successfully designed inlet may give. It also illustrates why the ram jet engine provides a thermodynamically attractive cycle at very high speeds.


Some examples of engine installation


Engine installation examples

• Wing mounted pod installation (attached to the wing by pylons):

• Third engine buried in the tail fuselage


Theory 5.1 – Stagnation pressure for isentropic compression

We have already introduced the stagnation temperature as:

pc

VTT

2

2

0

and shown that (revision task): Rcc vp

The Mach number is defined as:

RT

V

soundofspeed

V

a

VM

*

The specific heat ratio γ is defined:

v

p

c

c


Thus:

120

2

11

M

P

P

2

11

2

1

2

1

12

202

22

0

MT

TTMT

TMT

R

c

c

RTMTT p

p

100

T

T

P

Pbut we have: which directly gives:

Theory 5.1 – Stagnation pressure for isentropic compression


Theory 5.2 - Continuity in stagnation property form

)1(2

)1(

2

0

02

1

12

0

0

2

0

120

/1

2

11

2

11

21

1

21

1

]properties stagnation use[

][

MRT

APMM

RT

APM

A

M

TR

MPMA

RT

PM

ART

PMRTAM

RT

P

RTMVAVRT

PRTvPAVm


),(2

11

)1(2

)1(

2

0

0 MMMAP

RTm

Thus:

extremely powerful.

Theory 5.2 - Continuity in stagnation property form


Learning goals

• Have an understanding for the propulsive efficiency concept and how it:– relates to the total efficiency– relates to the different jet engine types available

• Have a quantitative understanding of how real cycle effects impact cycle efficiency and choice of design conditions

• Have a basic understanding of how intakes work and know how engines can be integrated in aircraft

Chalmers University of Technology Lecture 5 Shaft power cycles –Cycle selection –Technology...

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