introduction to thermodynamics, lecture 27-28 prof. g ......gas turbine power plants gas turbine...
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Source URL: http://me.queensu.ca/Courses/230/LectureNotes.html Saylor URL: http://www.saylor.org/courses/ME103/#6.2
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Introduction to Thermodynamics, Lecture 27-28 Prof. G. Ciccarelli (2012)
Gas Turbine Power Plants
Gas Turbine Power Plants are lighter and more compact
than vapor power plants. The favorable power-output-to-
weight ratio for gas turbines make them suitable for
transportation.
Air-standard Brayton Cycle
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For steady-state: )(0 outinCVCV hhm
W
m
Q
1 2 Adiabatic compression )( 12 hhm
Win
2 3 Heat addition )( 23 hhm
Qin
3 4 Adiabatic expansion )( 43 hhm
Wout
4 1 Heat removal )( 14 hhm
Qout
Cycle Thermal Efficiency:
23
1411hh
hh
mQ
mQ
in
out
cycleBrayton
Back work ratio:
43
12
hh
hh
mW
mWbwr
out
in
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Ideal Air-standard Brayton Cycle (processes are
reversible)
1 2 Isentropic compression
2 3 Constant pressure heat addition
3 4 Isentropic expansion
4 1 Constant pressure heat removal
For the isentropic process 1 2
1
212 P
PPP rr
For the isentropic process 3 4
3
434 P
PPP rr
Qi
n
Qout
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Ideal Cold Air-standard Brayton Cycle
For isentropic processes 1 2 and 3 4
k
k
P
P
T
T1
1
2
1
2
and
k
k
P
P
T
T1
3
4
3
4
Since 4
3
1
2
P
P
P
P thus
4
3
1
21
4
31
1
2
T
T
T
T
T
T
T
T k
k
k
k
Thermal Efficiency
1/
1/111
232
141
23
14
23
14
TTT
TTT
TTc
TTc
hh
hh
P
P
constkBrayton
recall 2
3
1
4
4
3
1
2
T
T
T
T
T
T
T
T
k
kconstkBrayton
PPT
T1
122
1 111
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Efficiency increases with increased pressure ratio across
the compressor
Back work ratio
43
12
43
12 )(
TT
TT
TTc
TTc
mW
mW
mW
mWbwr
P
P
turb
comp
out
in
Typical BWR for the Brayton cycle is 40 - 80% compared
to < 5% for the Rankine cycle.
Recall, reversible compressor work is given by 2
1vdP
Since gas has a much larger specific volume than liquid
much more power is required to compress the gas from P1
to P2 in the Brayton cycle compared to the Rankine cycle
for which liquid is compressed.
The turbine inlet temperature is limited by metallurgical
factors, e.g., Tmax = 1700K
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Gas Turbine Irreversibilities
In the ideal Brayton cycle all 4 processes are assumed
reversible, thus processes 2-3 and 4-1 are constant
pressure and processes1-2 and 3-4 are isentropic.
The constant pressure assumption does not normally incur
any great errors but the compressor and turbine processes
are far from isentropic
These irreversiblities are taken into account by:
s
s
t
t
turbhh
hh
mW
mW
43
43
12
12
hh
hh
mW
mW
s
c
s
c
comp
Ideal (reversible) processes: 1 - 2s and 3 - 4s
Actual (irreversible) processes:
1 - 2 and 3 - 4
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Efficiency versus Power
Consider two Brayton cycles A and B with a similar
turbine inlet temperatures T3
Since
BAP
P
P
P
1
2
1
2 BA
Since (enclosed area 1-2-3-4)B > (enclosed area 1-2-3-4)A
A
cycle
B
cycle
m
W
m
W
Bcycle
Acycle
B
A
W
W
m
m
,
,
In order for cycle A to produce the same amount of net
power as cycle B, i.e., BcycleAcycle WW ,, , need BA mm .
Higher mass flow rate requires larger (heavier) equipment
which is a concern in transportation applications
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Increasing Cycle Power
The net cycle power is: ctcycle WWW
The cycle power can be increased by either increasing the
turbine output power or decreasing the compressor input
power.
Gas Turbine with Reheat
The turbine work can be increased by using reheat, as was
shown in the Rankine cycle
The turbine is split into two stages and a second
combustor is added where additional heat can be added
3 2
Compressor
a b
4 1
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Recall: '4
3
1
2
T
T
T
T so, isobars on T-s diagram diverge
Note:
hb - h4 > ha - h4’
The total turbine work output without reheat is:
mhhhhW aabasic '43
The total turbine work output with reheat is:
mhhhhWWW battreheatw
turbine 432,1,
/
Since hb - h4 > ha - h4’ basicreheatw
turbine WW /
Since the compressor work h2 - h1 is unaffected by reheat
basiccycle
reheatwcycle WW
/
The reheat cycle efficiency is not necessarily higher since
additional heat 2,inQ is added between states a and b
T
3
2
1
a
b
4
s
1,inQ
4’
2,inQ
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Compression with Intercooling
The compressor power can be reduced by compressing in
stages with cooling between stages.
Recall: '4
3
1
2
T
T
T
T so, isobars on T-s diagram diverge
2’
2’
d
h2’ – hc > h2 – hd
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The compressor power input without intercooling is:
mhhhhW ccbasic 1'2
The total compressor power input with intercooling is:
mhhhhWWW dcccreheatw
comp 212,1,/
Since h2’ – hc > h2 – hd basicreheatw
comp WW /
Since the turbine work h3 – h4 is unaffected by
intercooling
basiccycle
reheatwcycle WW
/
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Different approach: The reversible work per unit mass for
a steady flow device is vdP , so
Without intercooling :
-a-c-b-
vdPvdPvdPm
W
c
c
basic
c
'21 area
'22
1 1
With intercooling :
-a-c-d-b-
vdPvdPvdPm
W
d
c
w/
c
21 area
22
1 1int
Since area(b-1-c-2’-a) > area(b-1-c-d-2-a)
int/w
c
basic
c
m
W
m
W
2’
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Aircraft Gas Turbines
Gas turbine engines are widely used to power aircraft
because of their high power-to-weight ratio
Turbojet engines used on most large commercial and
military aircraft
Ideal air-standard jet propulsion cycle:
Nozzle
Diffuser
a
2
1 4 5
3
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Normally compression through the diffuser (a-1), and
expansion through the nozzle (4-5) are taken as isentropic
In the ideal jet propulsion engine the gas is not expanded
to ambient pressure Pa.
Instead the gas expands to an intermediate pressure P4
such that the power produced is just sufficient to drive the
compressor, no net cycle power produced ( 0cycleW ),
thus
4312 hhhh
m
W
m
W tc
After the turbine the gas expands to ambient pressure P5
which is the same as Pa.
inQ
outQ
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Apply the steady-state conservation of energy equation to
the Diffuser and Nozzle
220
22out
outin
inCVCV V
hV
hm
W
m
Q
Diffuser slows the flow to a zero velocity relative to the
engine:
Diffuser (a 1)
kconstant for 2
2
22
2
1
2
1
221
1
P
aa
aa
aa
c
VTT
Vhh
Vh
Vh
Nozzle accelerates the gas leaving the turbine (turbine
exit velocity negligible compared to nozzle exit velocity):
Nozzle (4 5)
kconstant for 2
2
22
545
545
25
5
24
4
TTcV
hhV
Vh
Vh
P
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The gas velocity leaving the nozzle is much higher than
the velocity of the gas entering the diffuser, this change in
momentum produces a propulsive force, or thrust Ft
at VVmF 5
Where V is flow velocity relative to engine
For aircraft under cruise conditions the thrust just
overcomes the drag force on the aircraft fly at high
altitude where the air is thinner and thus less drag
To accelerate the aircraft increase thrust by increasing V5
In military aircraft afterburners are used to get very
large thrust for short take-offs on aircraft carriers
An afterburner is simply a reheat device!
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Other Propulsion Systems
In turbofan bypass flow produces additional thrust for
take-off. During cruise thrust comes from turbojet
In a ramjet engine there is no compressor or turbine,
compression is achieved gasdynamically.
Ramjet engines produce no thrust when stationary thus
must be coupled with a turbojet engine to get off the
ground
Turboprop Turbofan
Subsonic ramjet
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Supersonic Ramjet Engine
The flow is decelerated to subsonic velocity before the
burner via a series of shock waves.
Combustion occurs at constant pressure
Turbojet-ramjet combination:
choked flow
Supersonic free stream flow
Supersonic exhaust flow
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Supersonic Combustion Ramjet (SCRAMJET) Engine
At very high Mach numbers the air temperature gets
extremely hot after deceleration through the diffuser
2
2
1
P
aa
c
VTT
For Mach 6 flight speed, the air temperature just before
the burner reaches about 1550K. At this temperature the
air dissociates resulting in a drop in enthalpy
At flight speeds greater than Mach 6 (hypersonic) better
to burn fuel- in supersonic air stream
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US National Aero Space Plane (X-30)
Was to use 5 scramjet engines to achieve a Mach 12 flight
speed
To be used for travel to space and also as an airliner, a
flight between any two points on earth would take less
than 2 hours
Canceled in 1993!
Several countries have similar planes on the drawing
board, Canada is not one of them!
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