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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
188
For steady-state: )(0 outinCVCV hhm
Wm
Q−+−=
&
&
&
&
1 2 Adiabatic compression )( 12 hhm
in −=&
&W
2 3 Heat addition )( 23 hhm
Qin −=&
&
3 4 Adiabatic expansion )( 43 hhmout −=&
&W
4 1 Heat removal )( 14 hhm
Qout −=&
&
Cycle Thermal Efficiency:
23
1411hhhh
mQmQ
in
out
cycleBrayton −
−−=−=
&&&&
η
Back work ratio:
43
12
hhhh
mWmWbwr
out
in
−−
==&&&&
189
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
Qin
Qout
For the isentropic process 1 2
=
1
212 P
PPP rr
For the isentropic process 3 4
=
3
434 P
PPP rr
190
Ideal Cold Air-standard Brayton Cycle For isentropic processes 1 2 and 3 4
k
k
PP
TT
1
1
2
1
2
−
= and
kk
PP
TT
1
3
4
3
4
−
=
Since 4
3
1
2
PP
PP
= thus 4
3
1
21
4
31
1
2
TT
TT
TT
TT k
kkk
=→
=
−−
Thermal Efficiency
( )( )
( )( )1/
1/111232
141
23
14
23
14
−−
−=−−
−=−−
−=TTTTTT
TTcTTc
hhhh
P
P
constkBraytonη
recall 2
3
1
4
4
3
1
2
TT
TT
TT
T=→=
T
( ) kk
constkBrayton
PPTT
1
122
1 111 −−=−=η
191
Efficiency increases with increased pressure ratio across the compressor
Back work ratio
( ) 43
12
43
12 )(TTTT
TTcTTc
mWmW
mWmWbwr
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 ∫
21 vdP
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
192
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
Ideal (reversible) processes: 1 - 2s and 3 - 4s
Actual (irreversible) processes:
1 - 2 and 3 - 4
These irreversiblities are taken into account by:
s
s
t
t
turb hhhh
mW
mW
43
43
−−
=
=
&&
&&
η 12
12
hhhh
mW
mW
s
c
s
c
comp −−
=
=
&&
&&
η
193
Efficiency versus Power Consider two Brayton cycles A and B with a similar turbine inlet temperatures T3
Since BA P
PPP
>
1
2
1
2 BA ηη >
Since (enclosed area 1-2-3-4)B > (enclosed area 1-2-3-4)A
A
cycle
B
cycle
mW
mW
>
&
&
&
&
Bcycle
Acycle
B
A
WW
mm
,
,&
&
&
&=
In order for cycle A to produce the same amount of net power as cycle B, i.e., BcycleAcycle WW ,,
&& = , need BA m&& >m . Higher mass flow rate requires larger (heavier) equipment which is a concern in transportation applications
194
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
2 3 a b
Compressor
1 4
The turbine is split into two stages and a second combustor is added where additional heat can be added
195
Recall: '4
3
1
2
TT
TT
= so, isobars on T-s diagram diverge
Note:
&2,inQ&
The tot
The tot
W
Since h
Since t
The rehadditio
T
Q 1,inal turbine work out([ hWbasic
&3=
al turbine work outWW tt
reheatwturbine
&&&,1,
/+=
b - h4 > ha - h4’
he compressor work
rewcycleW&
/
eat cycle efficiencynal heat Q is add2,in
&
3
hb - h4 > ha - h4’
21
p−
p2
W
h
a
ut witho) (hha +
ut with ([ h3 −=
reheatwturbine&
/
h2 - h1 i
bc
eatW&>
is not ned betw
b
4
r
s
ay
ee
s
4’
ut reheat is: )]mha &'4−
eheat is: ) ( )]mhhh ba &4−+
basicW&>
unaffected by reheat
siccle
cessarily higher since en states a and b
196
Compression with Intercooling The compressor power can be reduced by compressing in stages with cooling between stages.
Recall:
'4
3
1
2
TT
TT
= so, isobars on T-s diagram diverge
d
2’
h2’ – hc > h2 – hd
2’
197
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 basic
reheatwcomp WW && <
/
Since the turbine work h3 – h4 is unaffected by intercooling
basiccycle
reheatwcycle WW && >
/
198
Different approach: The reversible work per unit mass for a steady flow device is ∫ vdP , so
Without intercooling : mWc
&
&
With intercooling : mW
w
c
&
&
Since area(b-1-c-2’-a) > ar
basi
c
mW
&
&
2’
-a-c-b-
vdPvdPvdPc
c
basic
'21 area
'22
1 1
=
∫∫ ∫ +==
-a-c-d-b-
vdPvdPvdPd
c
/
21area
22
1 1int
=
∫∫ ∫ +==
ea(b-1-c-d-2-a)
int/w
c
c mW
>&
&
199
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:
NozzleDiffuser 3
541
2
a
200
Normally compression through the diffuser (a-1), and expansion through the nozzle (4-5) are taken as isentropic
&
In the ideal jet propulsioto ambient pressure Pa. Instead the gas expands such that the power prodcompressor, no net cyclethus
( 2h −
After the turbine the gaswhich is the same as Pa.
inQ
&
n eng
to an uced pow
)
mWc =&
&
1h =
expan
outQ
ine the gas is not expanded
intermediate pressure P4 is just sufficient to drive the er produced ( 0=cycleW& ),
(
mWt
&
&
)43 hh −
ds to ambient pressure P5
201
Apply the steady-state conservation of energy equation to the Diffuser and Nozzle
+−
++−=
220
22out
outin
inCVCV VhVhm
Wm
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
cVTT
Vhh
VhVh
+=
+=
+=+
Nozzle accelerates the gas leaving the turbine (turbine exit velocity negligible compared to nozzle exit velocity):
Nozzle (4 5) ( )
( ) kconstant for 2
222
545
545
25
5
24
4
TTcV
hh
VhVh
P −=
−=
+=+
V
202
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!
203
Other Propulsion Systems
Turboprop
Subsonic ramje
In turbofan bypass flow produces additake-off. During cruise thrust comes fr In a ramjet engine there is no compresscompression is achieved gasdynamical Ramjet engines produce no thrust whemust be coupled with a turbojet engineground
Turbofan
t
tional thrust for om turbojet
or or turbine, ly.
n stationary thus to get off the
204
Supersonic Ramjet Engine The flow is decelerated to subsonic velocity before the burner via a series of shock waves. Combustion occurs at constant pressure
Supersonic exhaust flow
T
Supersonic free stream flow
choked flow
urbojet-ramjet combination:
205
Supersonic Combustion Ramjet (SCRAMJET) Engine At very high Mach numbers the air temperature gets extremely hot after deceleration through the diffuser
2
2
1P
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
206
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!
207