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
Ideal Turbojet
Pressure drop with
acceleration
Ram effect - pressure rise
with deceleration
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)
4
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
5
Turbojet Irreversibilities
Isentropic efficiencies
- Diffuser
- Compressor
- Turbine
- Nozzle
Fluid Friction effects
- Combustion chamber
6
The Turbojet Model
First Law analysis of the components in the cycle
Q in  m  h3  h 2 
Combustion is replaced with a
heat transfer
Wt  m  h 3  h 4 
Wc  m  h 2  h1 
Wnet  0
 Wt  Wc
Air is the working
fluid throughout the
complete cycle
Va2
V12
ha 
 h1 
2 gc
2 gc
 V1
Va2
 h1  h a 
2 gc
7
Va
V52
V42
h4 
 h5 
2 gc
2 gc
 V4
 V5  2 gc  h4  h5 
V5
Turbojet Performance
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,
• 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
Turbojet Performance
Propulsive Force (Thrust)
in (a)
exit (5)
 mV 
F 

g
 c exit
 mV 


g
 c in
m
F  V5  Va 
gc
In this equation, the velocities are
relative to the aircraft (engine).
For an aircraft traveling in still air,
Vaircraft  Vin  Va
9
Propulsive Power
The power developed from the
thrust of the engine
Wp  FVaircraft
m

Wp   V5  Va  Va
 gc

Turbojet Performance – Efficiencies
Overall Efficiency
 overall  thermal  propulsive
Thermal Efficiency  thermal 
mair  ke5  kea 
m fuel  HVfuel 
Propulsive Efficiency
Kinetic energy
production rate
Thermal power
available from
the fuel
Propulsive power
 propulsive 
Wp
mair  ke5  kea 
 m / g V  V V
 air c 5 a a
Kinetic energy
production rate
  propulsive
10
mair 2
V5  Va2 

2 gc
2Va
2


V5  Va V5 / Va  1
2 V5  Va  Va

V5  Va V5  Va 
Turbojet Example
Given: A turbojet engine operating as shown below
Find:
T3  1400 K
PR  11
P0  26 kPa
T0  230 K
V0  220 m/s
m  25 kg/s
0
11
P5  26 kPa
c  0.85
d  1.0
t  0.90
n  1.0
(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
T3  1400 K
PR  11
P0  26 kPa
T0  230 K
V0  220 m/s
m  25 kg/s
0
P5  26 kPa
c  0.85
d  1.0
t  0.90
n  1.0
Note: An array position of [0] is
allowed in EES!
12
Turbojet Example
Strategy: Build the property table
first. This will require some
thermodynamic analysis. Consider
each component in the cycle.
Diffuser


V02 
V12 
m  h0 
  m  h1 

2
g
2
g
c 
c 


V02
h0 
 h1
2 gc
13
T3  1400 K
PR  11
P0  26 kPa
T0  230 K
V0  220 m/s
m  25 kg/s
0
P5  26 kPa
c  0.85
d  1.0
t  0.90
n  1.0
Turbojet Example
Compressor
c 
h1  h 2 s
T3  1400 K
PR  11
P0  26 kPa
T0  230 K
V0  220 m/s
m  25 kg/s
0
P5  26 kPa
c  0.85
t  0.90
d  1.0
n  1.0
h1  h2
Turbine
t 
Combustion Chamber
14
h3  h 4
h3  h 4 s
wt  wc
Turbojet Example
Nozzle
T3  1400 K
PR  11
P0  26 kPa
T0  230 K
V0  220 m/s
m  25 kg/s
0
P5  26 kPa
c  0.85
t  0.90
d  1.0
n  1.0
At this point, the property table is complete!
3
2
2s
0
15
4
4s
Turbojet Example
Now, we can continue with
the rest of the thermodynamic
analysis.
Combustion Heat Transfer Rate
Q in  m  h3  h 2 
T3  1400 K
PR  11
P0  26 kPa
T0  230 K
V0  220 m/s
m  25 kg/s
0
P5  26 kPa
c  0.85
t  0.90
d  1.0
n  1.0
Nozzle – Exit Velocity


V52 
V42 
m  h4 
  m  h5 

2
g
2
g
c 
c 


V52
h 4  h5 
2 gc
16
Turbojet Example
Now the propulsive
parameters can be calculated,
P0  26 kPa
T0  230 K
V0  220 m/s
m  25 kg/s
0
F
m
V5  Va 
gc
Wp  FVaircraft  FV0
 propulsive 
17
Wp
mair  ke5  ke0 
T3  1400 K
PR  11
P5  26 kPa
c  0.85
d  1.0
t  0.90
n  1.0
Turbojet Example
Solution (Key Variables):
P0  26 kPa
T0  230 K
V0  220 m/s
m  25 kg/s
0
18
T3  1400 K
PR  11
P5  26 kPa
c  0.85
d  1.0
t  0.90
n  1.0
Turbojet Example – Analysis
How is the energy input to this engine distributed?
Qin  24,164 kW
P0  26 kPa
T0  230 K
V0  220 m/s
m  25 kg/s
19
P5  26 kPa
T5  719.5 K
V5  986 m/s
m  25 kg/s
excess thermal energy transfer
Q out  m  h5  h 0   12, 617 kW
 52.2% 
kinetic energy production rate
m
m  kenet   V52  V02   11,548 kW
2
Wp  4, 213 kW
m  keexcess   7,335 kW
 47.8% 
36.5%
 63.5%