Department of Mechanical Engineering

ME 322 – Mechanical Engineering

Thermodynamics

Lect 27b

Jet Aircraft Propulsion

Aircraft Propulsion

• Thrust produced by increasing the kinetic energy of the air in the opposite direction of flight

• Slight acceleration of a large mass of air

– Engine driving a propeller

• Large acceleration of a small mass of air

– Turbojet or turbofan engine

• Combination of both

– Turboprop engine

2

Aircraft Gas Turbine Engines

Turboprop Small commuter planes

Turbojet high speeds

3

Turbofan Larger passenger airliners

The Turbojet

4

Ideal Turbojet

Ram effect - pressure rise

with deceleration

Pressure drop with

acceleration

a-1 Isentropic increase in pressure (diffuser)

1-2 Isentropic compression (compressor)

2-3 Isobaric heat addition (combustion chamber)

3-4 Isentropic expansion (turbine)

4-5 Isentropic decrease in pressure with an increase in fluid velocity (nozzle)

The Turbojet with an Afterburner

5

The turbine exhaust is already hot. The afterburner reheats

this exhaust to a higher temperature which provides a higher

nozzle exit velocity

Turbojet Irreversibilities

6

Isentropic efficiencies

- Diffuser

- Compressor

- Turbine

- Nozzle

Fluid Friction effects

- Combustion chamber

The Turbojet Model

7

2 2

11 1

2 2

aa a

c c

V Vh h V V

g g

2 1cW m h h 3 4tW m h h

3 2inQ m h h

22

544 5 4 5

2 2c c

VVh h V V

g g

First Law analysis of the components in the cycle

Air is the working

fluid throughout the

complete cycle

Combustion is replaced with a

heat transfer

2

12

aa

c

Vh h

g

0net t cW W W

5 4 52 cV g h h

Turbojet Performance

• Propulsive Force (Thrust)

– The force resulting from the velocity at the nozzle exit

• Propulsive Power

– The equivalent power developed by the thrust of the

engine

• Propulsive Efficiency

– Relationship between propulsive power and the rate of

kinetic energy production

8

There is no net power output of the turbojet engine.

Therefore, the idea of net power and thermal efficiency are not

meaningful. In turbojet engines, performance is measured by,

Turbojet Performance

9

Propulsive Force (Thrust)

c cexit in

mV mVF

g g

Propulsive Power

The power developed from the

thrust of the engine

p aircraftW FV

in (a) exit (5)

In this equation, the velocities are

relative to the aircraft (engine).

For an aircraft traveling in still air,

aircraft in aV V V

5 a

c

mF V V

g

5p a a

c

mW V V V

g

Turbojet Performance – Efficiencies

10

Overall Efficiency

5

p

propulsive

air a

W

m ke ke

overall thermal propulsive

Thermal Efficiency

5

HV

air a

thermal

fuel fuel

m ke ke

m

Propulsive Efficiency

Kinetic energy

production rate

Thermal power

available from

the fuel

Kinetic energy

production rate

Propulsive power

5

2 2

5

/

2

air c a a

aira

c

m g V V V

mV V

g

5

5 5

2 a a

a a

V V V

V V V V

5 5

2 2

/ 1

apropulsive

a a

V

V V V V

Turbojet Example

11

Given: A turbojet engine operating as shown below

0

0

0

26 kPa230 K220 m/s25 kg/s

PTVm

11PR

0.85c

1.0d 1.0n

0

3 1400 KT

0.90t

5 26 kPaP

Find:

(a) The properties at all the

state points in the cycle

(b) The heat transfer rate

in the combustion

chamber (kW)

(c) The velocity at the

nozzle exit (m/s)

(d) The propulsive force

(lbf)

(e) The propulsive power

developed (kW)

(f) The propulsive

efficiency of the engine

Turbojet Example

12

0

0

0

26 kPa230 K220 m/s25 kg/s

PTVm

11PR

0.85c

1.0d 1.0n

0

3 1400 KT

0.90t

5 26 kPaP

Note: An array position of [0] is

allowed in EES!

Turbojet Example

13

0

0

0

26 kPa230 K220 m/s25 kg/s

PTVm

11PR

0.85c

1.0d 1.0n

0

3 1400 KT

0.90t

5 26 kPaP

Strategy: Build the property table

first. This will require some

thermodynamic analysis. Consider

each component in the cycle.

2 2

0 10 1

2 2c c

V Vm h m h

g g

Diffuser

2

00 1

2 c

Vh h

g

Turbojet Example

14

0

0

0

26 kPa230 K220 m/s25 kg/s

PTVm

11PR

0.85c

1.0d 1.0n

0

3 1400 KT

0.90t

5 26 kPaP

Compressor

1 2

1 2

s

c

h h

h h

Combustion Chamber

Turbine

3 4

3 4

t t c

s

h hw w

h h

Turbojet Example

15

0

0

0

26 kPa230 K220 m/s25 kg/s

PTVm

11PR

0.85c

1.0d 1.0n

0

3 1400 KT

0.90t

5 26 kPaP

Nozzle

At this point, the property table is complete!

0

22s 4s

4

3

Turbojet Example

16

0

0

0

26 kPa230 K220 m/s25 kg/s

PTVm

11PR

0.85c

1.0d 1.0n

0

3 1400 KT

0.90t

5 26 kPaP

Now, we can continue with

the rest of the thermodynamic

analysis.

Nozzle – Exit Velocity

22

544 5

2 2c c

VVm h m h

g g

2

54 5

2 c

Vh h

g

Combustion Heat Transfer Rate

3 2inQ m h h

Turbojet Example

17

0

0

0

26 kPa230 K220 m/s25 kg/s

PTVm

11PR

0.85c

1.0d 1.0n

0

3 1400 KT

0.90t

5 26 kPaP

Now the propulsive

parameters can be calculated,

5 a

c

mF V V

g

0p aircraftW FV FV

5 0

p

propulsive

air

W

m ke ke

Turbojet Example

18

0

0

0

26 kPa230 K220 m/s25 kg/s

PTVm

11PR

0.85c

1.0d 1.0n

0

3 1400 KT

0.90t

5 26 kPaP

Solution (Key Variables):

Turbojet Example – Analysis

19

How is the energy input to this engine distributed?

24,164 kWinQ

0

0

0

26 kPa230 K220 m/s25 kg/s

PTVm

5

5

5

26 kPa719.5 K986 m/s25 kg/s

PTVm

5 0 12,617 kW 52.2%outQ m h h

2 2

5 0 11,548 kW 47.8%2

net

mm ke V V

4,213 kW 36.5%pW

7,335 kW 63.5%excessm ke

excess thermal energy transfer

kinetic energy production rate