ETH Aerospace Propulsion Notes

Prof. Dr. R.S. Abhari 1

EidgenössischeTechnische Hochschule

Zürich

Ecole Polytechnique Fédérale de ZurichPolitecnico federale svizzero di Zurigo

Swiss Federal Institute of Technology Zurich

Aerospace Propulsion



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•

Recommended text

- Hill & Peterson

„

Mechanics and thermodynamics of propulsion“

Addison, Wesley

- Kerrebrock

„Aircraft Engines and Gas Turbines“

MIT Press

•

6 Exercises

„Design a core of engine for A3XX

“



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Course Outline

I. Introduction:

Wk1

1. History of Propulsion

2. Reactive Force & Thrust

3. Airbreathing vs. non-airbreathing

II. Airbreathing engines

1. Definitions & Brayton cycle

2. Types of Aircraft engines Wk2

a. Turbojet

b. Ramjet & Scramjet

c. Afterburning

d. Turbofan

e. Afterburning Turbofan Wk3

f. Turboshaft

g. Recuperators

3. Review of Boundary Layers & fluid loss

4. Losses & efficiency Wk4

a. Non-ideal Brayton Cycle

b. Component Efficiency

c. Non-ideal Ramjet

c. Non-ideal Turbofan



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5. Turbomachines Wk5

a. Sources of losses & deviation angle

b. Meanline design

b. Streamline curvature analysis

6. Compressor

Wk6

a. Design concepts

b. Compressor map

c. Stall & stability

7. Combustor Wk7

a. Kinetics of combustions

b. Stability concepts

c. Ignition

d. Emissions and environment

Holidays Wk8

8. Turbine Wk9

a. Design concepts

b. Turbine Cooling

c. Materials & Coating

9. Inlet & Exhaust Wk10

a. Subsonic & supersonic inlet design

b. Exhaust system

c. Jet noise



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10. System considerations Wk11

a. Overall cycle tradeoff

b. Component matching

c. Mechanical Design

III. Non-Airbreathing engines (ROCKETS)

1. Introduction to rocket propulsion Wk12

a. Basic rocket equations

- Payload

- Multistaging

b. Overview of rocket types

- Chemical

- Nuclear

- Electric

c. Performance consideration Wk13

- High trust versus low thrust

- Energy limited versus power limited

2. Rocket processes

a. Thermal engines

- Limiting speed

- Frozen flow

b. Electric devices Wk14

- Electrostatic

- Electromagnetic

3. System design issue



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I Introduction

I.1 History of Aerospace propulsion

Humans have always been fascinated by the idea of flight

.

Uncontrolled flights have been around for thousands of years. There have been many ran-dom take off and landing of objects in the history of mankind.

Chinese have been credited for inventing gunpowder and with it fuel for simple rocket engines. The first attempt at an engineered „Rocket“ is believed to be around 1150 AD.

Over the last millennium, there have been many uses of crude rockets, primarily for pur-pose of war.



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Key technologies to flight

Propulsion, structures, aerodynamics (atmospheric flights), control

First controlled flight: December 17, 1903 by: Wilbur & Orville Wright

Duration: 12 seconds ~ 40 meters traveled

Location: Kitty Hawk, North Carolina, USA

Engine was a key technology. Their engine was an internal combustion engine with 4 cyl-inders, 4 stroke inline engine and drove a propeller. The engine produced 9 kW of power and weighed 82 kg.

Engine power/weight ratio:

Wright Brothers = 0.11 kW/kg

„State of the Art“ ~ 20 kW/kg

About a factor of 200 increase in the power to weight ratio of the engine in less than 100 years!

We shall see later that power to weight ratio is a critical parameter for aerospace propul-sion.



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Historical Development of Different Aircrafts



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Aircraft propulsion was primarily based on propeller driven aircraft in the first half of the century.

In the last 30 to 40 years, gas turbine engines have dominated atmospheric flight vehicles.

History of gas turbine engine development for flight is based on a competition betweeen two groups of two pioneers; Sir Frank Whittle of England and Hans Von Ohain of Ger-many.

First successful jet flight was that of Heinkel 178 in Germany:

August 27, 1937

Space Travel

Key technologies to flight: Propulsion, structures aerodynamics (atmospheric flights), con-trol

First controlled flight to space

October 4, 1957

by: Soviet Union

Duration: Orbit around the earth

Location First satellite in space.



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I.2 Reactive Force & Thrust

Q. What is propulsion?

A. Force to move vehicle forward.

Reactive Force is the force to propel vehicles.

Force = Rate of change of momentum

.

Control Volume

• A region of constant shape which is fixed in space relative to observer.

• Apply conservation of mass, momentum and energy.

Example:

Take a static rocket stand



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Conservation of mass

• Conservation of momentum in x-direction:

or

• In principal it is possible to adjust the exit area of nozzle to obtain . This

condition is known as “matched nozzle” or “ideal expansion” mode.

Then

•

Example

•

Conservation of energy

Chemical energy of heat release + heat loss = kinetic energy of exhaust

= 100 kg/s

⇒

F = 250’000 N

= 2500 m/s

me mf mprop+=

PaS F Pa S Ae–( ) meue peAe+ +=+

F Ae Pe Pa–( ) meue+=∴

F m⁄ e ue Ae

Pe Pa–

me------------------

+=

Pe Pa≅

F m⁄ e ue≅

me

ue



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I.3 Airbreathing vs. non-airbreathing engines

Now let’s study a stationary air-breathing engine stand

• Conservation of mass:

• Conservation of momentum in x-direction :

Now . As then

Again for a matched nozzle,

Thrust , same as a rocket

•

For example

= 100 kg/s

= 2 kg

(or ) = 1.36

Mach # exit = 1 (choked nozzle)

T

stagnation

@

e

= 800 ° k

me min mf+=

PaS minuin F Pa S Ae–( ) PeAe meue+ +=+ +

F Ae Pe Pa–( ) meue minuin–+=∴

uin

min

ρa-------- 1

S---= S ∞→ uin 0→

Pe Pa≈

Fme------ ue=

min

mf

γ λ



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-

Thrust

> Now consider engine flying velocity ua.

• Conservation of mass

, define

f = fuel fraction ratio

• Conservation of momentum in x:

, assuming “matched nozzle”

Typically 0.01 < f < 0.03, f << 1.

Te Tstag 1γ 1–

2-----------M

2+

678°k=⁄=

ue Me γRTe 1.36 287.15 678×× 514m s⁄===

F 100 2+( ) 514 52.471 N.=×=

me min mf+=

me 1( f ) min+= fmf

min--------=

F meue minua–=

F min 1 f+( )ue=⁄ ua–∴

F min ue ua–≅⁄∴



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• Example

Say aircraft flying at Mach # 0.8 and Ta = 220 K

Then

(from previous example):

• The flight thrust is much less than ground stationary value.

ua 0.8 1.4 287.1 220×× 238m s⁄==

Me 1.0 Tstag 800 k ue 514m s⁄ min 100kg s⁄=,=,=,=

F 100 514 238–( ) 27 600N,=×≅∴



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II Airbreathing Engine

Rolls Royce Trent900 Engine



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II.1 Definitions & Brayton cycle

Engine performance

ηo = Overall efficiency =

Propulsive Efficiency

Propulsive Power

Mechanical Power

Mechanical Power

ηpropulsive

Propulsive PowerThermal Power of Fuel--------------------------------------------------------------------

ηoPropulsive PowerMechanical Power---------------------------------------------------- Mechanical Power

Thermal fuel energy per unit time------------------------------------------------------------------------------------------------------×=

ηo Propulsive efficiency Thermal efficiency×=

ηo η propulsive η thermal×=

Thrust speed F ua×=×=

min 1 f+( )ue minua–{ } ua=

∆ kinetic energy of fluid( ) per unit time=

min 1 f+( )ue

2

2----- min

ua2

2-----–=

minua 1 f+( )ue ua–{ }

min12--- 1 f+( )ue

2ua

2–{ }

---------------------------------------------------------=



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for f << 1 we have ηpropulsive

then

Max ηpropulsive @

But

@ No good!

Example:

Note:

Αnother way of getting ηoverall:

QR = energy content of fuel/unit mass of fuel

ue = 514 m/s

⇒ η propulsive = 63.3 %

ua = 238 m/s

2ua ue ua–( )

ue2

ua2

–( )------------------------------=

2ua u( e ua)–

ue ua+( ) u( e ua)–--------------------------------------------=

η prop 2 1ue

ua-----+

⁄=

ue

ua----- 1.0=

F mua

ue

ua----- 1–

≅⁄

max η prop F 0→ →

ηo

2η thermal

1 ue ua⁄+------------------------=

ηoThrust Power

Power from fuel-----------------------------------------------

F ua

mf QR--------------==

ηo

min ue ua–( )ua

mf QR---------------------------------------≅

ηo1

f QR-------------- ue ua–( )ua≅



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Plotting ηo vs ua

• Other definitions

• Measure of efficiency of engine

• Thrust specific fuel consumption, TSFC =

• Specific Impulse:

∴ I @

Trends of ηo and F for aircraft propulsion

ηo max is at ua ue 2⁄=

ηo max

2ua ua–( )ua

f QR-------------------------------=

η∴ o max

ua2

f QR--------------

ue2

4 f QR---------------------==

mf

F-------

TSFCmf

min ue ua–( )---------------------------------≅

TSFCf

ue ua–----------------≅

I1

TSFC--------------- F

mf-------==

I1f--- ue ua–( )=

ηo max ue 2ua=,

ηo max ua f⁄≅



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• Refer to handout on piston engine and propulsion trends.

Airbreathing Gas Turbine Engines

> Describe processes in

where :

is the Total (stagnation) Temperature

is the Total (stagnation) Pressure

M = Mach number

γ = Specific heat ratio

γ = CP/CV

Compressor • inlet

• compressor

Heat • combustor

Expansion • turbine

• nozzle

TT Ts 1γ 1–

2-----------M

2+

=

PT Ps 1γ 1–

2-----------M

2+

γγ 1–-----------

=

TT

PT



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Propulsion

Texts

Articles

* Cohen, Rogers and

Saravanamutto

Gas Turbine Theory Longmann

* Kerrebrock Aircraft Engines & Gas Turbines

M.I.T. Press

Harman Gas Turbine Engineering Macmillan (generation)

* Roll Royce The Jet Engine

Mc Mahon Airctaft Propulsion Pitman

* Hill & Peterson Mechanics & Thermody-namics

of Propulsion

Addison, Wesley

Shephard Aerospace Propulsion Elsevier

Lancaster Jet Propulsion Engines O.U.P. (out of date but of interest)

W.H. Bennett Aero engine development for the future

Proc. I MechEng. (1983) Vol. 197

G. Wilde Future Large Civil Turbo-fans and Power plants

Aeronautical Journal 1978 (July)

Rosen, Koff and Hartmann Airbreathing Propulsion Component Technology

Astronautics and Aero-nautics

1980 (June)

Thomas Jr. New Generation Ramjetsd Astronautics and Aero-nautics 1980 (June)



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Jet Engines

Turboprop

Turbojet + Reheat

Turbofan

Ramjet

Hybrid Engines ramrockets

jet rotor

turbo rocket

turbo ramjet

We shall be considering these



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Ideal Analysis : In all jet engines the basic process is the same.

Combustion - const.

Brayton

The heat addition takes place at constant pressure. Thus the ideal cycle is

Considering an ideal gas

Hence lines of constant P in the T - S diagram are separated by a constant ∆S.

The work done from (1)→(6) is given by the steady state energy equation for unit mass-flow

h = CPTT

S2 S1 Cp=–T2

T1------ R–ln

P2

P1------ln⋅ ⋅

Q1 W hT6+ hT1

–=



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In a turboshaft engine for example and the only consequences are external

work and raising of the enthalpy of the working fluid. (Stationary turbo)

In a turbojet, which is stationary, no external work is done and the net result is an increase in the total enthalpy of the working fluid.

When the turbojet is moving, work is done on the aircraft and is also an increase in the total enthalpy of the working fluid. The working fluid is air and we shall consider this to be an ideal gas, γ = 1,4, and neglect the mass addition of the fuel.

Heat rejection in the cycle occurs when the emerging working fluid gives up its energy to the atmosphere away from the engine. From the steady state energy equation it is obvious that the gain in kinetc energy of the working fluid can be considered to be included in the work done, i.e.

Thrust This follows immediately from momentum considerations in engine coordi-nates. Let the flight speed be V and assume exit conditions are mate.

For a mass flow rate through the engine, ,

momentum (1)

with f<<1

From energy considerations the same result is obtained

u1 u2 0≈ ≈

Wext

u62

2-----

u12

2-----–+

W

h6 h1–

specific enthalpy+

m

Fm---- u6 V–=

Wwork done

unit mass through engine------------------------------------------------------------------------=



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(2)

W must be the same in both cases hence

as expected the same as equation 1

Hence we may derive the thrust from the work done in the cycle

(3)

Thus we can now analyse the thermodynamic cycle to find W and then deduce F from equation 3.

Wu6

2

2----- V

2

2------–=

WFVm

--------u6 V–

2---------------

+=

u62

2-------

V2

2------ FV

m--------

u62

2------- V

2

2------ u6V–+ +=–

Fm---- u6 V–=∴

u62

2----- W

V2

2------+=

u6

2 W V2

2⁄+( )=

Fm---- 2W V

2+[ ]

1 2⁄V–=

Engine coord.

Earth coord.



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Cycle

Consider first the case when the kinetic energy of the air is negligible

where we define : ,

The work done in the cycle is Q1 - Q2 = Cp (T4 - T3) - Cp (T6 - T1)

The compression and expansion processes are isentropic hence

Hence

where i.e. the ratio of maximum temperature to inlet

i.e. a function of the compression ratio

and equals the temperature rise in compression.

tT4

T1------= c

P3

P1------

γ 1–

γ-----------

=

CP const≅

T3

T1------

P3

P1------

γ 1–

γ-----------

andT4

T6------

P3

P1------

γ 1–

γ-----------

c==,=

Wm Cp T4 T6–( ) T3 T4–( )–[ ]=

Wm T1CP

T4

T1------ 1

T6

T4------–

T3

T1------ 1–

–=

Wm∴ TCP t 1 1c---–

c 1–( )–=

tT4

T1------=

cP3

P1------

γ 1–

γ-----------

=P3

P1------



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The thermal efficiency of the cycle is

(4a)

Thus η is a function of solely

W has a maximum for a given t at

This last fact is important for engines have a maximum value of t at which it is possible to

operate and it is necessary to optimise to maximise W.

η WQ1------ 1 1

c---–==

P2

P1------

c t=

P2

P1------



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Basic Brayton cycle : definitions

Assume constant Cp and fuel to air ratio f << 1

Work done =

Specific Work =

where

for a given t ,

upstream

Station 1 ----- inlet conditions

Station 2 ----- compressor inflow

Station 3 ----- combustor inflow

Station 4 ----- turbine inflow

Station 5 ----- nozzle inflow

Station 6 ----- exhaust condition

minCp T4 T6–( ) T3 T1–( )–[ ]

minCpTi

T4

T1------

1T6

T4------–

T3

T1------ 1–

–

=

Wm----- CpT1 t 1 1

c---–

c 1–( )–

=

tT4

T1------ and= c

T3

T1------

T4

T6------

P3

P1------

γ 1–

γ-----------

===

ddc----- W

min----------

CpT1t

c2

----- 1– P3

P1------

T4

T1------

γ– 1–2γ

----------------

==

M∴ ax or Min( ) atddc----- W

min----------

0= ie c t=

c2

2

d

d Wmin----------

2CpT1t

c3

-------------------–= 0< ie max

W min⁄( )max⇒ CpT1 t 1–( )2

=



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II.2 Types of Aicraft engines

II.2.a Turbojet engine

Specific thrust : but we will assume f<<1

Now replace by m and by U, we have

Mechanical work :

Divide through by the speed of sound for the ambient atmosphere condition, which is

Then, specific thrust equals , where M = flight Mach number

From previous computation :

Specific workWmin-------- Also η th 1 1

c---–==

Specific thrustF

min--------=

Fmin-------- 1 f+( )U6 Uin–( )≅

min Uin

Wm2---- U6 U

2–( )=

Ue2Wm

-------- U2

+ 1 2⁄

=∴

Fm---- 2W

m-------- U

2+

1 2⁄U–=∴

a0

a0 γRT1=

F

ma02

---------- 2W

ma02

---------- M2

+1 2⁄

M–

W minCpT1 t 1 1c---–

c 1–( )–

=

Wm-----

γRT1

γ 1–------------- t 1 1

c---–

c 1–( )–

=ao

2

γ 1–----------- t 1 1

c---–

c 1–( )–

=



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TSFC = Thrust Specific Fuel Consumption

Replace

@

Wma0

2----------

1γ 1–----------- t 1 1

c---–

c 1–( )–

=

Fma0----------

∴ 2γ 1–----------- t 1 1

c---– c 1–[ ]–

M2

+1 2⁄

M–=

η thermalW

mf QR-------------- mf

Wη

thQR

--------------- W1 1 c⁄–( )QR

-------------------------------==⇒=

TSFCmf

F-------=

TSFCa0QR

CpT1-------------

t 1 1

c---–

c 1–( )–

1 1c---–

2γ 1–----------- t 1 1

c---– c 1–[ ]–

M2

+ 1 2⁄

M– --------------------------------------------------------------------------------------------------------------------------------=

B t c,( ) t 1 1c---–

c 1–( )–=

Fm----

max

B t c,( ) 2 t 1–( )2

γ 1–-------------------------------- c,= t=

TSFCa0QR

CpT1-------------

B t c,( )

1 1c---–

2γ 1–----------- B t c,( )( M

2)

12---

M–+

------------------------------------------------------------------------------------ andF

ma0---------- 2

γ 1–----------- B( t c,( ) M

2)1 2⁄

M–+==



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@

Example γ = 1.4

T4 = 1.800 k

T1 = 200 k t = 9 ⇒ Copt. = 3

@ M = 0 F/ma0 = 4.47 , TSFC = 1.34

@ M = 1 F/ma0 = 3.58 , TSFC = 1.67

@ M = 2 F/ma0 = 2.90 , TSFC = 2.07

@ M = 3 F/ma0 = 2.38 , TSFC = 2.51

@ M = 4 F/ma0 = 2.00 , TSFC = 3.00

Fma0----------

max

Fma0---------- 5 4 M

2+( )

1 2⁄{ } M–=,



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@ optimum specific thrust

is the overall pressure ratio

assuming isentropic inlet

Pressure ratio across compressor :

where , assuming per-

fectly isentropic compressor

when

Then = 1

When then ∴ Do not need a compressor

i.e. above M = 3.0 : - Do not need a compressor

∴ - Do not need a turbine

The engine then becomes an inlet, a combustor and a nozzle

Example T4 = 1.800 k

T1 = 200 k ∴ t = 9

c t= cT3

T1------

T4

T1------=

P3

P1------

γ 1–

γ-----------

= =

T3

T1------

T2

T1------

T3

T2------× or c

T3

T2------ t

γ 1–2

-----------M2

+

t== =

P3

P2------

P3

P2------

T3

T2------

γ

γ 1–-----------

=

P3

P2------

t

1γ 1–

2-----------M

2+

-----------------------------

γγ 1–-----------

=

M2

γ 1–----------- t 1–( )

1 2⁄

=

P3

P2------

M 10= 3.1=P3

P2------ 1.0=



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II.2.b Ramjet and Scramjet

All the compression is due to Ram effect

So that :

Problem with ramjet is how to get it started.

Because as then

Also as then

C 1γ 1–

2-----------M

2+

=

Fma0---------- 2

γ 1–-----------

γ 1–2

-----------M2t

1γ 1–

2-----------M

2+

-----------------------------γ 1–

2-----------M

2

–

M2

+

1 2⁄

M–=∴

2γ 1–-----------γ 1–

2-----------M

2 t

1γ 1–

2-----------M

2+

----------------------------- 1– 1+ 1 2⁄

M–=

Mt

1γ 1–

2-----------M

2+

-----------------------------

1 2⁄

1–

=

Fma0---------- M t 1

γ 1–2

-----------M2

+ 1 2⁄–

1–=

M 0→ Fma0---------- 0→

M2

γ 1–-----------

t 1–( )

1 2⁄

→ Fma0---------- 0→



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or as :

kinetic energy of incoming fuel

For T4 = 2000 k

Ti = 200 k

as

For Ramjet

The actual mass flow into an engine does increases with Mach number. It is more appropriate to consider the flow Mach num-ber after diffusion is constant.

∴ even though decreases after M = 3,

the actual F does increase well above M = 3

t 1γ 1–

2-----------M

2+→

M2

0.4------- 9×

1 2⁄

6.6∼→

Fma0---------- 0→

TSFCa0QR

CpTo---------------

t 1–( ) γ 1–2

-----------M2

+ t1 2⁄

1γ 1–

2-----------M

2+

1 2⁄1–

1–

=

Fmao----------



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Scramjet

Now, instead of the diffusing flow such that M3 << 1, in a scramjet, the Mach number after the diffuser and in the combustor is maintained in a supersonic range.

Then where M3 is Mach # in the combustor

Then

For scramjet when

M3# M

or as M → for Scramjet

For ramjet as M →

T3

T1------

1γ 1–

2-----------M

2+

1γ 1–

2-----------M3

2+

-----------------------------

C==

Fma0---------- M

2M3

2–( ) t 1

γ 1–2

-----------M2

+ 1–

1γ 1–

2-----------M3

2+

1––

M2

+1 2⁄

M–=

Fmao---------- 0→

M2

M32

–( ) t 1γ 1–

2-----------M

2+

1–1

γ 1–2

-----------M32

+ 1 2⁄

–

M2

M2

=+

t 1γ 1–

2-----------M3

2+

1γ 1–

2-----------M

2+=∴

2γ 1–----------- t 1

γ 1–2

-----------M32

+ 1–

1 2⁄

Fma0---------- 0→ 2

γ 1–----------- t 1–( )

1 2⁄



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Example

T4 = 2000 k

T1 = 200 k

For ramjet as when

For scramjet (M3 ~ 3) as

or

By burning fuel at M3 = 3 we can increase the useful thrust producing range of our engine to a Mach number of 6.5 to 11.

• Practical problem of burning fuel @ M > 1 in a reasonable length and with reasonable

efficiency.

Thus for a given flight speed the propul-sion efficiency can be raised by reducing specific thrust

• for a turbojet

γ 1.4= M2

0.4------- 9×

1 2⁄6.6∼→ F

ma0---------- 0→

M2

0.4------- 10 1 1.8+( ) 1–×( )

1 2⁄→

Fmao---------- 0 as M 11.3→→

η0 ηP η th×=

ηP2U

Ue U+-----------------= and F m Ue U–( )=

ηP1

1 F2mU------------+

----------------------=∴

F m⁄( )

Fma0---------- 2

γ 1–-----------

t 1 1c---–

c 1–( )–

M2

+1 2⁄

M–=



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• To reduce we can reduce t

But as is reduced, the engine size has to increase to increase and hence maintain

overall thrust.

• • We can divert a portion of flow mass frome core.

• Accelerate the “bypass” flow by performing work

using a fan and mix with core flow to exhaust

• Advantage is that the bypass air gives most

thrust without affecting core flow efficiency

∴ Breakup engine to a “core engine” and a “bypass area”.

Fm----

Fm---- m

FmU---------

Ue

U------ 1–=



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II.2.c Afterburning (or reheat)

Take the case of a simple turbojet engine. In an afterburning turbojet, a combustor is added downstream of the turbine.

Afterburning turbojet is similar to a turbojet + ramjet.

Ideal Brayton Cycle

Work : W = (W3 + W2) - W1

so that

assuming f << 1 and Cp ≈ const.

Also Turbine Power = Compressor Power

∆kE W mCp T4 T5–( ) T5a T6a–( ) T3 T1–( )–+{ }=,

mCp T4 T5–( ) mCp T3 T2–( )=

T5 T4 T3 T2–( )–=

W mCp T5a T6a–( ) T2 T1–( )–{ }=∴

mCpT1

T5a

T1-------- 1

T6a

T5a--------–

T2

T5a-------- 1–

–

=



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T4 node Turbing rotating parts higher T4 compared to straight Brayton

Note

Now let’s define as Afterburner temperature ratio

Recall

Thus

To find maximum specific work with respect to cycle pressure ratio, set

At optimum specific work,

and

⇒

tT'4T1-------=

T2

T1------ 1

γ 1–2

-----------M2 T3

T1------ c=,,+=,

T5a

T1-------- taf=

P5

P1------

P5

P3------

P3

P1------×=

isentropic

→T5a

T6a--------∴

T5

T4------

T3

T1------× γ const=,=

T6a

T5a-------- 1

c---

T4

T1------

T1

T5------×

tcr-----

T4

T1------

T3

T1------–

T2

T1------+

–

==∴

t c⁄

t c– 1γ 1–

2-----------M

2+ +

--------------------------------------------------=

W mCpT1 taf 1 t c⁄

t c– 1γ 1–

2-----------M

2+ +

----------------------------------------------–γ 1–

2-----------M

2–

=

dWdc-------- 0=

dW

dC------- 1

c2

t c– 1γ 1–

2-----------M

2+ +

---------------------------------------------------------– 1

c t c– 1γ 1–

2-----------M

2+ +

-------------------------------------------------------+

=

0=

Copt12--- t 1

γ 1–2

-----------M2

+ + =

Wopt mCpT1 taf 1 t

Copt2

----------– γ 1–

2-----------M

2–

=



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Afterburners are generally used to assist take off or to achieve high speed (up to M = 3) maneuvers.

At take off

and

@ take off, with afterburner

Specific thrust ,

without afterburner

Example: t = 8, taf = 11

With a design point of M = 2.2, the only commercial turbo-jet+afterburner is the Concorde (Olympus engine)

Above a flight Mach number of 2.0, a pure turbojet engine loses thrust rapidly. With after-burning, can extend beyond M = 4.0, above which an afterburning turbojet looks more like a ramjet.

Amount of fuel burnt,

But Turbine work = Compressor work

MFm---- Ue U–=,→

Coptt 1+

2-----------= Ue

2Wm

-------- 1 2⁄

≅

Fm----

max

2CpT1taf 1 4t

t 1+( )2------------------–

1 2⁄

=

Fma1----------

max

t 1–t 1–----------

2taf

γ 1–-----------

1 2⁄

= a1 γRT1=

Fma1----------

max

2tγ 1–----------- 1 1

t-----–

1 2⁄=

mf QR mCP T4 T3–[ ] T5a T5–[ ]+( )=

mf

m------- f

CP

QR------- T4 T5–( ) T5a T3–( )+{ }==∴

⇒ T4 T5–( ) T3 T2–( )=



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Afterburner turbojet feeding a ramjet, 2 independent engine, so that do not effect each other much.

And

Thrust specific fuel consumption TSFC =

Recall, ,

and

Combining the three equations above:

Example: t = 8, taf = 11

Above M = 2, afterburning turbojet has a better TSFC than pure turbojets, e.g. Concorde!

fCpT1

QR-------------

T5a

T1--------

T2

T1------–

=∴

fCpT1

QR------------- taf 1–

γ 1–2

-----------M2

–=

f F m⁄( )⁄

Fma1---------- 2W

ma12

---------- M2

+1 2⁄

M–= a1 γRT1=

WmCpT1------------------ taf 1 t c⁄

t c– 1γ 1–

2-----------M

2+ +

----------------------------------------------– γ 1–

2-----------M

2–=

TSFCa1

γ 1–( )QR------------------------

=taf 1–

γ 1–2

-----------M2

–

2taf

γ 1–----------- 1 t c⁄

t c– 1γ 1–

2-----------M

2+ +

----------------------------------------------– 1 2⁄

M–

--------------------------------------------------------------------------------------------------



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II.2.d Turbofan

In the turbofan engine, shaft work is done on the fan by the turbine. In the gas generator everything is like a turbojet.

∴ We have increased of gas generator by of by pass and reduced the effective exhaust velocity.

For a matched nozzle, the total thrust of engine is given by

where f<<1

or

Work done,

For a fixed , and W

Differentiate over ubp

To find the maximum thrust

@ maximum thrust

m αm

Fm u6 U–( )

core

αm u( bp U )–

bypass+=

Fm 1 α+( )----------------------

u6 U–( ) α ubp U–( )+

1 α+( )------------------------------------------------------=

W mu6

2

2----- mα

ubp2

2-------- m

1 α+( )U2

2-------------------------–+=

α m

u62 2W

m-------- αubp

2– 1 α+( )U2

+=

2u6

∂u6

∂ubp----------- 2αubp or

∂u6

∂ubp-----------

αubp–

u6---------------=–=

∂F∂ubp----------- m

∂u6

∂ubp----------- αm m α

ubp

u6--------– α+

0==+=



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or

Define

or Specific thrust

Define

(for commercial engine we have 2< <8)

It can be seen that for , the specific thrust is lower than a turbojet.

but

ubp

ub-------- 1 0=– Max thrust when ubp u6=∴

u6 u6 ubp==

F m u6 U–( ) αm u6 U–( )+=

Fm 1 α+( )-------------------------- u6 U and

Wm 1 α+( )--------------------------

u62

2----- U

2

2-------–=–=

Fm 1 α+( )-------------------------- 2W

m 1 α+( )-------------------------- U

2+

1 2⁄

= U–

B t c,( )2

γ 1–----------- t 1 1

c---–

c 1–( )–

=

Faom 1 α+( )-----------------------------------

B t c,( )1 α+( )

------------------ M2

+1 2⁄

M–= α

α 0>

TSFCmf

F-------

Wm ηQR⁄F

-----------------------------CpT1B t c,( )m

γ 1–2

-----------

η thQR 1 α+( )m Fm 1 α+( )--------------------------

˙------------------------------------------------------------------------== =

η th 1 1c---–=

TSFCaoQR

CpT1-------------

γ 1–

2-----------

B t c,( )

1 1c---–

1 α+( )B t c,( )1 α+( )

------------------ M2

+

1 2⁄M–

----------------------------------------------------------------------------------------------------=∴



Zürich



= 20

= 0.443

Specific thrust, %

for a turbojet of same cone size, we have

Example T4 = 1800 k

T1 = 200 k t = 9

c = 3

= 1.4

M = 0.8

= 9 very efficient !

= 100 kg/s

c t=

γ

α

mcone

B t c),( )2

0.4------- 9 1 1

3---–

3 1–( )–

=

ao 1.4 287.1 200××=

283.5m s⁄ and U 226.5m==

TSFCaoQR

CpT1--------------

0.2 20×

1 1 3⁄–( ) 1 9+( ) 4010------ 0.8

2+

1 2⁄0.8–

--------------------------------------------------------------------------------------------------=

thrust F 100 1 9+( ) 283.5×× 2 20×10

--------------- 0.64+1 2⁄

0.8–

= 383.9 kN=

Fm 1 α+( )-------------------------- 384

Nkg------= ηP 54=

TSFCaoQR

CpT1--------------

1.076=

Thrust F 100 283.5 40 0.64+[ ] 1 2⁄0.8–{ }×=

158.0 kN=



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%

The actual thrust has increased but the specific thrust is 4 times smaller than a turbojet.

• Advantage ; greater propulsive efficiency

• Disadvantage ; bigger engine and inlet area

∴ larger nacelle losses

II.2.e Afterburning Turbofan

Consider a turbofan with an after-burner attached

The exhaust has to be opened before starting. If not, there could be a backflow -> engine stalls !

Pressure in the afterburner inlet can not be higher than fan exit, or the fan will stall

It is common practice to allow core and bypass air to mix before enter-ing afterburner.

T-S diagram looks like two turbojet cycles on top of one another

Specific thrustFm---- 1580N kg⁄= ηP 22=



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Assuming the cone flow and the by pass flow mix at a constant pressure.

assume f << 1, Cp = const.

Also Power from Turbine = Power for Compressor + Fan

or

From isentropic relationships:

Combining above

Specific power

Max. work for taf, t, and M = constant when

Note as , i.e. an afterburning turbojet.

Also specific thrust :

∆kE W mCp T4 T5–( ) 1 α+( )m Cp T5a T6a–( )+=,

mCp T3 T2 f–( )– m 1 α+( )Cp T2 f T1–( )–

mCp T4 T5–( ) mCp T3 T2 f–( ) m 1 α+( )Cp T2 f T2–( )+=

T4 T5 T3 T2 f 1 α+( ) T2 f T2–( )+–=–

T3

T2 f---------

T4

T5------

T5a

T6a--------

T2 f

T1---------

T2

T1------ 1

γ 1–2

-----------M2

+=,=,=

WmCp 1 α+( )------------------------------ taf

γ 1–2

-----------M2

–taf αc t+( )

c 1 α+( ) 1γ 1–

2-----------M

2+

t c–+

--------------------------------------------------------------------------------–=

α

dWdc-------- 0 Copt

tα--- 1 α α 1 α+( )

t---------------------- 1

γ 1–2

-----------M2

+ 1–+ +=⇒=

α 0 Copt t 1γ 1–

2-----------M

2+ +

2⁄→,→

Fm 1 α+( )a1--------------------------------

2taf

γ 1–-----------

1 2⁄

1 αc t+

c t c– 1 α+( ) 1γ 1–

2-----------M

2+

+

------------------------------------------------------------------------------–

1 2⁄

M–=



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and also

• An afterburning turbofan engine has good subsonic cruise and high supersonic thrust. Many modern fighter aircraft use an afterburning turbofan (F111, F14, F15, F16,F18,Tornado, EJ 2000 ...) ( )or less.

• Afterburning turbofan has the advantage of a cooler outercasing compared to a pure

turbojet with afterburning easier installation.

• First afterburning turbofan, had control problems (fan stall, pressure wave of after-

burner of TF 30 engine used in F111, 1964).

TSFCγ 1–( )QR

a1------------------------

t c–1 α+------------- α taf 1–

γ 1–2

-----------M2

– +

1 α+( ) F ma1 1 α+( )⁄( )----------------------------------------------------------------------------=

α 1∼

⇒



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II.2.f Turboshaft

Power HPC = Power HPT

Assume gas expanded in power turbine to Po.

Then

Compressor work = Turbine work

so that

f<<1

Cp ~ const.

∆H5 6,=

Wshaft mCp T5 T6–( )=

mCp T4 T6–( ) T4 T5–( )–{ }=

Wshaft

m------------------- CpT1

T4

T1------ 1

T6

T4------–

1T1------ T4 T5–( )–

=

mCP T3 T2–( ) T4 T5–( )=T3

T1------

T4

T6--------- c==

Wshaft

m------------------- CpT1 1 1

c---–

T3

T1------

T2

T1------–

–

=∴

Wshaft

m------------------- CpT1 t 1 1

c---–

T3

T1------

T2

T1------–

–

=∴

CPT1 t 1 1c---–

c 1γ 1–

2-----------M

2+–

–

=



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Modification to improve thermal efficiency

• Add heat exchanger to recover exhaust heat

• Heat exchanger heats up compressor exit flow (before combustor) lowers Qin

• Without heat addition

II.2.g Recuperators

With heat exchanger for a fixed t

At maximum recuperator efficiency heat exchanger efficiency = 1.0

⇒

η th 1 1c---–=

η thW

Q Qrec–---------------------=

Qrec mCp T3′ T3–( )=

T3′ T6=

η thermal recuperator 11c--- c

2

t-----

cP3

P1------

γ 1–

γ-----------

=,–=

∆η th ηth with recup

η th without recup–=

∆η th 1– 1c---+

11c--- c

2

t-----

–+=

1c--- 1 c

2t⁄–( )=

η th v+ e for c2

t<∆



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as pressure ratio goes up

and heat exchanger is no longer useful.

Recuperator

• Very useful for low pressure ratio cycles, or at idle

• If significant time of operation at idle then use a recuperator (occurs in tanks and ships)

• Generally not useful for aircraft.

(Hybrid Ramjet, Turbofan, Turbojet speed range from low-high)

c2

t→

ETH Aerospace Propulsion Notes

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