Department of Mechanical EngineeringME 322 – Mechanical Engineering

Thermodynamics

Lecture 28

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

TurbopropSmall commuter planes

Turbojethigh speeds

3

TurbofanLarger passenger airliners

4

The TurbojetIdeal 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)

5

The Turbojet with an Afterburner

The turbine exhaust is already hot. The afterburner reheats this exhaust to a higher temperature which provides a higher nozzle exit velocity

6

Turbojet Irreversibilities

Isentropic efficiencies - Diffuser - Compressor - Turbine - Nozzle

Fluid Friction effects - Combustion chamber

7

The Turbojet Model

2 21

1 12 2a

a ac c

V Vh h V Vg g

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

3 2inQ m h h

2254

4 5 4 52 2c c

VVh h V Vg 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

1 2a

ac

Vh hg

0net t cW W W

5 4 52 cV g h h

8

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

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,

9

Turbojet PerformancePropulsive Force (Thrust)

c cexit in

mV mVFg g

Propulsive Power

The power developed from thethrust 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 ac

mF V Vg

5p a a

c

mW V V Vg

10

Turbojet Performance – Efficiencies Overall Efficiency

5

ppropulsive

air a

Wm ke ke

overall thermal propulsive

Thermal Efficiency

5

HVair a

thermalfuel fuel

m ke kem

Propulsive Efficiency

Kinetic energy production rate

Thermal power available from

the fuel

Kinetic energy production rate

Propulsive power

5

2 25

/

2

air c a a

aira

c

m g V V Vm V Vg

5

5 5

2 a a

a a

V V VV V V V

5 5

2 2/ 1

apropulsive

a a

VV 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 20 1

0 12 2c c

V Vm h m hg g

Diffuser

20

0 12 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

sc

h hh h

Combustion Chamber

Turbine

3 4

3 4t 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 Velocity22

544 52 2c c

VVm h m hg g

25

4 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 ac

mF V Vg

0p aircraftW FV FV

5 0

ppropulsive

air

Wm 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):

19

Turbojet Example – AnalysisHow 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 25 0 11,548 kW 47.8%

2netmm ke V V

4,213 kW 36.5%pW

7,335 kW 63.5%excessm ke

excess thermal energy transfer

kinetic energy production rate