Gas Turbines for Aircraft
Propulsion
Gas turbines
The turbojet engine
consists
of three main sections: the diffuser, the
gas generator, and the nozzle.
The diffuser placed before the compressor decelerates the incoming
air relative to the engine. A pressure rise known as the ram effect is
associated with this deceleration.
Gas turbines
The gas generator section consists of a compressor, combustor, and
turbine, with the same functions as the corresponding components of
a stationary gas turbine power plant
.
The gases leave the turbine at a pressure significantly greater than
atmospheric and expand through the nozzle to a high velocity before
being discharged to the surroundings.
Gas turbines
The
working fluid is air modeled as an ideal gas. The diffuser,
compressor, turbine, and nozzle processes are isentropic, and the
combustor operates at constant pressure.
In
an actual engine, there would be increases in specific entropy
across the diffuser, compressor, turbine, and nozzle.
Gas turbines
Process a
–
1 shows the pressure rise that occurs in the diffuser as the
air decelerates
isentropically
through this component.
Process 1
–
2 is an isentropic compression.
Process 2
–
3 is a constant

pressure heat addition.
Process 3
–
4 is an isentropic expansion through the turbine during
which work is developed.
Process 4
–
5 is an isentropic expansion through the nozzle in which
the air accelerates and the pressure decreases.
Gas turbines
In a typical thermodynamic analysis of a turbojet on an air

standard
basis, the following quantities might be known:
the velocity at the
diffuser inlet, the compressor pressure ratio, and the turbine inlet
temperature (at 3)
.
The objective of the analysis would be to
determine the velocity at the nozzle exit. Once the nozzle exit velocity
is determined, the thrust is determined by applying Newton's second
law of motion in a form suitable for a control volume
Example 9.12
Air enters a turbojet engine at 11.8
lbf
/in.
2
,
430
°
R, and an inlet velocity of 620 miles/h (909.3 ft/s). The
pressure ratio across the compressor is 8. The turbine inlet
temperature is 2150
°
R and the pressure at the nozzle exit is
11.8
lbf
/in.
2
The work developed by the turbine equals the compressor
work input. The diffuser, compressor, turbine, and nozzle
processes are isentropic, and there is no pressure drop for
flow through the combustor.
For operation at steady state, determine the velocity at the
nozzle exit and the pressure at each principal state. Neglect
kinetic energy at the exit of all components except the nozzle
and neglect potential energy throughout.
Air enters a turbojet engine at 11.8
lbf
/in.
2
, 430
°
R, and an
inlet velocity of 620 miles/h (909.3 ft/s).
The pressure ratio across the compressor is 8. The turbine
inlet temperature is 2150
°
R and the pressure at the nozzle
exit is 11.8
lbf
/in.
2
Each component is analyzed as a control volume at steady state. The
control volumes are shown on the
sketch
by dashed lines.
The diffuser, compressor, turbine, and nozzle processes are isentropic.
There is no pressure drop for flow through the combustor.
The turbine work output equals the work required to drive the
compressor.
Except at the inlet and exit of the engine, kinetic energy effects can be
ignored. Potential energy effects are negligible throughout.
The
working fluid is air modeled as an ideal gas.
Analysis:
To determine the velocity at the exit to the nozzle, the mass
and energy rate balances for a control volume enclosing this
component reduce at steady state to give
Except at the inlet and exit of the engine, kinetic energy effects can be
ignored. Potential energy effects are negligible throughout.
The
working fluid is air modeled as an ideal gas.
Analysis:
To determine the velocity at the exit to the nozzle,
the
mass
and energy rate balances for a control volume enclosing
the nozzle
are applied.
Energy rate balances for a control volume enclosing
the nozzle
The inlet kinetic energy is dropped by the assumption that
kinetic
energy effects can be
ignored except at
the inlet and exit of the
engine
.
Solving for V
5
With the operating parameters specified, the determination
h
4
and h
5
values
is accomplished by analyzing each component in turn, beginning
with the diffuser.
The
pressure at each principal state can be evaluated as a part of the
analyses required to find the enthalpies
h
4
and
h
5
.
Energy
rate
balance
for a control volume enclosing the diffuser
gives
Table
A

22E
T=420 R, h = 100.32
btu
/lb
T=440 R, h = 105.11
btu
/lb
h
a
(at 430 R) =102.7 Btu/lb
Table
A

22E
h = 114.69
btu
/lb, p
r
= 0.9182
h = 119.48
btu
/lb, p
r
= 1.0590
Interpolation at h = 119.2
btu
/lb, p
r1
= 1.051
Table
A

22E
T=420 R, p
r
= 0.5760
T=440 R, p
r
= 0.6776
Interpolation p
a
(at 430 R) = 0.6268
The
flow through the diffuser is isentropic, so pressure
p
1
is
p
1
= 19.79
lbf
/in.
2
Using
the given compressor pressure ratio, the pressure at state 2 is
p
2
= 8(19.79
lbf
/in.
2
) = 158.3
lbf
/in.
2
The flow through the compressor is also isentropic. Thus
Table
A

22E
p
r
=
7.761
, h = 211.35 Btu/lb
p
r
=
8.411
, h =
216.26
Btu/lb
Interpolation h
2
(at p
r
= 8.408) =
216.2 Btu/lb
At state 3 the temperature is given as
T
3
= 2150
°
R.
Table
A

22E
T= 2150
°
R,
h
= 546.54 Btu/lb
From
Table A

22E,
h
3
= 546.54 Btu/lb. By assumption
of no
pressure
drop for flow through the combustor
,
p
3
=
p
2
. The work developed by
the turbine is just sufficient to drive the
compressor.
That is
Solving for
h
4
h
4
= 449.5 Btu/lb
Table
A

22E
h =
436.12
btu
/lb, p
r
=
101.98
h =
449.71
btu
/lb, p
r
=
114.0
Interpolation at h
4
= 449.5
btu
/lb, p
r4
= 113.8
The expansion through the turbine is isentropic, so
With
p
3
=
p
2
=
158.3
lbf
/in.
2
and
p
r3
data
from
Table A

22E
at h
3
=
546.54
btu
/lb, p
r3
= 233.5
The expansion through the nozzle is isentropic to
p
5
= 11.8
lbf
/in.
2
thus
From Table
A

22E
,
h
5
= 265.8 Btu/lb, which is the remaining specific
enthalpy value required to determine the velocity at the nozzle exit.
Using the values for
h
4
= 449.5 Btu/lb and
h
5
=
265.8 Btu/lb
determined above, the velocity at the nozzle exit is
Combined Gas Turbine
–
Vapor
Power Cycle
A
combined
power
cycle
couples
two
power
cycles
such
that
the
energy
discharged
by
heat
transfer
from
one
cycle
is
used
partly
or
wholly
as
the
input
for
the
other
cycle
.
The stream exiting the turbine of a
gas turbine is at a high temperature.
This high

temperature gas stream
can be used by the combined cycle
shown, involving a gas turbine cycle
and a vapor power cycle.
The two power cycles are coupled so
that the heat transfer to the vapor
cycle is provided by the gas turbine
cycle, which may be called the
topping cycle.
The combined cycle has the gas
turbine's high average temperature
of heat addition and the vapor
cycle's low average temperature of
heat rejection, and thus a thermal
efficiency greater than either cycle
would have individually.
For
many applications combined
cycles are economical, and they are
increasingly being used worldwide
for electric power generation.
The
thermal efficiency of the
combined cycle is
is
the
net
power
developed by the gas turbine
is
the
net
power developed
by the vapor cycle
is the
total
rate of heat
transfer to the combined cycle,
including additional heat transfer, if
any, to superheat the vapor
entering the vapor turbine.
For steady

state operation, negligible
heat transfer with the surroundings,
and no significant changes in kinetic
and potential energy
, energy balance
around the heat exchanger gives
Combined
cycle performance can
be analyzed using mass and energy
balances. To complete the analysis,
however, the second law is
required to assess the impact of
irreversibilities
and the true
magnitudes of losses.
Combined
cycle performance can
be analyzed using mass and energy
balances. To complete the analysis,
however, the second law is
required to assess the impact of
irreversibilities
and the true
magnitudes of losses
.
Among the
irreversibilities
, the
most significant is the
exergy
destroyed by combustion. About
30% of the
exergy
entering the
combustor with the fuel is
destroyed by combustion
irreversibility.
Definition of
exergy
:
Exergy
is
the
maximum
theoretical
work
obtainable
from
an
overall
system
consisting
of
a
system
and
the
environment
as
the
system
comes
into
equilibrium
with
the
environment
(passes
to
the
dead
state)
.
The
exergy
of a system, E, at a specified state is given by the
expression
The
exergy
of a system, E, at a specified state is given by the
expression
U
, KE, PE,
V
, and
S
denote, respectively, internal energy, kinetic energy,
potential energy, volume, and entropy of the system at the specified
state.
U
0
,
V
0
, and
S
0
denote internal energy, volume, and entropy, respectively,
of the system when at the dead state. In this chapter kinetic and
potential energy are evaluated relative to the environment.
Thus, when the system is at the dead state, it is at rest relative the
environment and the values of its kinetic and potential energies are
zero.
Death state:
the system comes into equilibrium with the environment
—
that is, as the system passes to the dead state.
Neglecting changes in kinetic and potential energies, the
net rate of
exergy
increase for an open system is given by
m
g
= mass flow rate through the system
T
o
= surrounding temperature
Example
9.13:
A combined gas turbine
–
vapor power plant has a net
power output of 45 MW. Air enters the compressor of the gas turbine
at 100
kPa
, 300K, and is compressed to 1200
kPa
.
The
isentropic efficiency of the compressor is 84%. The condition at
the inlet to the turbine is 1200
kPa
, 1400 K. Air expands through the
turbine, which has an isentropic efficiency of 88%, to a pressure of 100
kPa
.
The
air then passes through the interconnecting heat exchanger and is
finally discharged at 400 K. Steam enters the turbine of the vapor
power cycle at 8
MPa
, 400
°
C, and expands to the condenser pressure
of 8
kPa
. Water enters the pump as saturated liquid at 8
kPa
. The
turbine and pump of the vapor cycle have isentropic efficiencies of 90
and 80%, respectively.
A combined gas turbine
–
vapor
power plant has a net power
output of 45 MW. Air enters the
compressor of the gas turbine at
100
kPa
, 300K, and is
compressed to 1200
kPa
(
at 2
).
The isentropic efficiency of the
compressor is 84%. The
condition at the inlet to the
turbine is 1200
kPa
, 1400 K. Air
expands through the turbine,
which has an isentropic
efficiency of 88%, to a pressure
of 100
kPa
(at 4)
.
The air then passes through the
interconnecting heat exchanger
and is finally discharged at 400
K.
Steam enters the turbine of the
vapor power cycle at 8
MPa
,
400
°
C, and expands to the
condenser pressure of 8
kPa
(at
8)
.
Water enters the pump as
saturated liquid at 8
kPa
. The
turbine and pump of the vapor
cycle have isentropic efficiencies
of 90 and 80%, respectively.
Let
T
0
= 300 K,
p
0
= 100
kPa
.
From the temperature of relative pressure, the enthalpy and entropy
at 9 states can be determined and listed in the following table:
State 1:
T
1
= 300 K
h
1
= 300.19 kJ/kg, p
r1
= 1.3860
State 2:
p
2
= 1200
kPa
For adiabatic compression from
p
1
to
p
2
:
Therefore
p
r2
: =
p
r1
= 1.3860
= 16.6320
From Table A

22
p
r
= 16.28,
h
= 607.02 kJ/kg
p
r
= 17.30,
h
= 617.53 kJ/kg
From Table A

22
p
r
= 16.28,
h
= 607.02 kJ/kg
p
r
= 17.30,
h
= 617.53 kJ/kg
h
2,isen
= 607.02 + (16.632
16.28)*(617.53
607.02)/(17.30
16.28)
h
2,isen
= 610.647 kJ/kg
h
2
=
h
1
+ (
h
2,isen
h
1
)/0.84 = 300.19 + (610.647
300.19)/0.84
h
2
= 669.7817 kJ/kg
State 3:
T
3
= 1400 K
h
3
= 1515.42 kJ/kg, p
r3
= 450.5
State 4:
p
= 100
kPa
Adiabatic expansion from 1,200
kPa
to 100
kpa
From Table A

22
p
r
= 37.35,
h
= 767.29 kJ/kg
p
r
= 39.27,
h
= 778.18 kJ/kg
h
4,isen
= 767.29 + (37.5417
37.35)*(778.18
767.29)/(39.27
37.35)
h
4,isen
= 768.3773 kJ/kg
h
4
=
h
3
+ 0.88*(
h
4,isen
h
3
) = 1515.42 + 0.88*(768.3773
1515.42)
h
4
= 858.0224 kJ/kg
State 7:
T
7
= 400
o
C,
p
7
= 8
MPa
From steam table or CATT2 program
(Computer Aided Thermodynamic Tables 2)
h
7
= 3138.3 kJ/kg,
s
7
= 6.3633 kJ/
kg
K
Isentropic expansion to state 8,
p
8
= 8
kPa
h
8,isen
= 1989.9 kJ/kg
h
8
=
h
7
+ 0.90*(
h
8,isen
h
7
) = 3138.3 + 0.90*(1989.9
3138.3)
h
8
= 2104.7 kJ/kg
State 9: Saturated liquid at 8
kPa
:
h
9
= 173.86 kJ/kg,
s
9
= 0.59254 kJ/
kg
K
State 9: Saturated liquid at 8
kPa
:
h
9
= 173.86 kJ/kg,
s
9
= 0.59254 kJ/
kg
K
Isentropic compression to state 6: 8
MPa
h
6,isen
= 181.9 kJ/kg
h
6
=
h
9
+ (
h
6,isen
h
9
)/0.8 = 173.86 + (181.9
173.86)/0.8
h
6
= 183.91 kJ/kg
Apply energy balance to the heat exchanger:
m
g
(
h
4
h
5
) =
m
v
(
h
7
h
6
)
The net power developed by the gas turbine is given by
The net power developed by the steam turbine is given by
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