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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