Department of Mechanical EngineeringME 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
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