K. Nasr,c:\thermo \module3.doc
Module 3. Gas Turbine Power Cycles – Brayton Cycle
•
Utilize gas as the working fluid. During combustion, mixture of (air + fuel) → combustion products.
They are lighter and more compact than the vapor power plants examined earlier.
•
In the power industry, this all-gas cycle is named combustion turbine (CT). It usually comes as a
complete package ready to be put to use and generate power.
•
Let’s begin this module with a display of some of the web sites which deal with the gas turbine power
cycle. Try the following sites:
http://ares.ame.arizona.edu/publications/ices94-paper.pdf – A research report detailing the design and use
of a reverse Brayton Cycle heat pump.
http://starfire.ne.uiuc.edu/ne201/course/topics/energy_cycles/simple_brayton.html – Covering open and
closed Brayton cycles.
http://www.geocities.com/dthomas599/Project.htm – A paper detailing the design of an isentropic
turbojet.
http://www.mech.utah.edu/~issacson/tutorial/BraytonCycle.html – An overview of Constant and Variable
Specific Heat Models as well as Java applet calculators.
http://www.freebyte.com/cad/utility.htm and http://er-online.co.uk/s-analys.htm – Free engineering
demonstrators for much more than just Thermodynamic applications.
http://members.aol.com/engware/calc3.htm - A page put together by Engineering Software featuring
Constant Specific Heat Model calculators for the Carnot, Brayton, Otto, and Diesel cycles.
The gas turbine power plant can be described by considering air flowing into the compressor, getting
heated in the combustor, and producing power by interacting with the turbine blades. Air is considered to
be the working fluid where it is compressed in the compressor, receives heat from an external source in
the combustor, and expands in the turbine.
Simple gas turbine: (a) Open to the atmosphere (b) Closed [2]
Copyright © 2003, K. Nasr. All Rights Reserved
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K. Nasr,c:\thermo \module3.doc
Problem: Air enters the compressor of a gas turbine power cycle at 100 kPa, 300 K, with a volumetric
flow rate of 5 m3/s. The compressor pressure ratio is 10. The turbine inlet temperature was measured to
be 1400K. Determine:
a). η- thermal efficiency
.
WC
b). . = back work ratio
WT
c). The net power developed, in kW.
Modeling: To solve this problem, we model what is happening with an air-standard Brayton cycle. The
Brayton cycle is the ideal cycle for gas turbines. Assumptions for Brayton cycle analysis: (1) There are
four internally reversible processes, (2) The working fluid is air, (3) Heat is added to the air somehow
(simulating the process in the burner), and (4) the cycle is complete by having a heat exchanger between
the turbine exhaust and the compressor intake.
⊕
The Brayton Cycle: The air- standard Brayton cycle is the ideal cycle for gas turbines.
All 4 processes are internally reversible:
(1) → (2): Isentropic Compression
(2) → (3): Constant Pressure Heat Addition
(3) → (4): Isentropic Expansion
(4) → (1): Constant Pressure Heat Rejection.
Air-standard gas turbine cycle [2]
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K. Nasr,c:\thermo \module3.doc
⊕
The Brayton Cycle (continued):
The thermal efficiency of the Brayton cycle is given by:
.
W net
ηth =
=
.
QH
.
.
W net W net
η =
=
=
th
.
.
Q in
QH
.
.
QH − QL
= 1−
.
QH
.
.
WT − WC
.
QH
.
QL
=
.
QH
(h 3 - h 4 ) − h 1 − h 2
h3 − h2
(I)
Thus, the Variable Specific Heat Model dictates finding all four enthalpy values. In order to analyze the
processes, we use relationships such as:
p 3 p2
= ; and
p4 p1
(1) → (2) is isentropic: p2 / p1 = pr2 / pr1
(3) → (4) is isentropic:
p3 / p4 = pr3 / pr4
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K. Nasr,c:\thermo \module3.doc
⊕
The Brayton Cycle (continued):
• Special Case: Cold-Air Standard Analysis assuming the specific heats to be constant and
evaluated at a base temperature of 300 K or 520 °R.
When the specific heats are assumed to be constant: ∆h = cp ∆T
In order to analyze the processes, we use relationships such as:
p 3 p2
= ; and
p4 p1
p2
T 2 kk−1
(1) → (2) is isentropic: p 1 = ( T1 ) ;
p3
T 3 kk−1
=
(
) ;
(3) → (4) is isentropic: p 4
T4
T1
1
=1−
(I) becomes ηth = 1 −
(II)
k −1
T2
p2 k
( )
p1
i.e. ηth is a function of pressure ratio if constant specific heats are assumed.
Note: Actual cycles differ from ideal cycles by having irreversibilities and pressure drops.
•
Each device (turbine, compressor, heat exchangers) is an open system. In this analysis, we normally
assume:
♦ ∆KE = ∆PE = 0
♦ No heat loss between devices
♦ SSSF through devices
♦ One-inlet, one-exit and 1-Dimensional flow
The first law of thermodynamics for an open system is a statement of the Energy Balance:
.
.
.
dEC .V . .
V2
V2
= QC.V . − WC .V . + ∑ mi ( hi + i + gzi ) − ∑ me ( he + e + gze )
dt
2
2
i
e
Under SSSF, Negligible ∆KE and ∆PE, Single-Inlet, Single-Outlet conditions:
.
Combustor (HX1):
.
Q In = m (h 3 − h 2 )
.
HX2:
Turbine:
Q Out = m (h 1 − h 4 )
& =m
& (h 3 − h 4 )
W
T
.
Compressor:
or
qIn = h3 - h2
∴
p2 = p3;
Heat In
or
qOut = h1 - h4
∴
p1 = p4;
Heat Out
or
wT = h3 - h4
∴
s4 = s3;
Work Out
or
wC = h1 - h2
∴
s1 = s2;
Work In
.
.
W C = m (h 1 − h 2 )
.
.
WT − WC
.
.
(h 3 - h 4 ) − h 1 − h 2
W net W net
η =
=
=
=
th
.
.
.
h3 − h2
Q in
QH
QH
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K. Nasr,c:\thermo \module3.doc
•
We need the enthalpy values at each state.
A −22
T1 Table

→ h1 = 300.19 kJ
, pr1 = 1.386
kg
P2
p
A −22
= r2 ⇒ p r 2 = p r1 *10 = 13.86 Table

→ h2 = 579.87 kJ
, T2 = 574.1K
kg
P1
pr1
A− 22
T3 = 1400 K Table

→ h3 = 1515.42 kJ
, p r 3 = 450.5
kg
P4
p
p
A −22
= r4 ⇒ p r 4 = r3 = 45.05 Table

→ h4 = 808.5 kJ
kg , T4 = 787.7 K
P3
p r3
10
•
&
& and the Q
Now we can calculate the W‘s
‘s.
W& T
W& C
= h3 − h4 = 706.92 kJ
;
= h1 − h2 = −279.68 kJ
kg
kg
m&
m&
Q& In
Q& Out
= h3 − h2 = 935.55 kJ
;
= h1 − h4 = −508.31 kJ
kg
kg
m&
m&
•
Checking for 1st Law satisfaction:
W& Net
kJ
kJ
= 706.92 kJ
kg − 279.68 kg = 427.24 kg
&
m
&
Q Net
= 935.55 kJ
− 508.31 kJ
= 427.24 kJ
kg
kg
kg
m&
h1 −h 2
W& C
=
The Back Work Ratio = & =
WT
h3 − h4
•
•

 W&
Q& Net

Net
=

v
&
m
 m&

279.68
= 39.6%
706.92
Determine the Net Power developed for this cycle:
 W&
W& 
W& Net = m&  T − C 
 m&
m& 

& =
m& = ρ∀
(
)
&
0.287 kgkJ*K (300 K )
∀
RT1
3
1
, v1 =
=
= 0.861 mkg
v1
P1
100 kN 

2

m 
3
m& =
5 ms
0.861
m3
kg
(
)
= 5.81 kgs ⇒ W& Net = 5.81 kgs * 427.24 kJ
= 2481 kJs = 2.48MW
kg
The following table summarizes the solution:
T1
K
300
WC/m
T2
K
574.1
WT/m
T3
K
1400
QIn/m
kJ/kg
kJ/kg
kJ/kg
-279.68
706.92
935.55
T4
K
787.7
bwr
%
39.6
Copyright © 2003, K. Nasr. All Rights Reserved
P1
kPa
100
Wnet/m
kJ/kg
427.24
P2
kPa
1000
η
%
45.67
P3
kPa
1000
Net Power
MW
2.48
P4
kPa
100
5
K. Nasr,c:\thermo \module3.doc
•
Note that compression of a gas requires quite of bit of energy as the back work ratio for this example
is 39.6% and is typically between 40% and 80%. This is a significant back work ratio especially when
compared to a steam power plant having a representative bwr value of say 1%.
Simplified Analysis: Cold Air Standard Analysis - Let’s solve the same problem using the Constant
Specific Heats Model (CSHM), that is using a cold-air-standard analysis with cp evaluated at 300 K. (cp
= 1.005 kJ/kg K, and k = 1.4).
1
1
Using (II), ηth = 1 −
=1−
= 48 .2% as compared to 45.67% using the Variable Specific
k −1
0.4
 P2

 P1
•



(10)1. 4
k
Heat Model.
Let’s validate it by using (I), replacing ∆h = cp ∆T, and finding the temperatures.
T1 = 300 K (Given)
T3 = 1400 K (Given)
P 
T2 = T1  2 
 P1 
k −1
k
= 300 K (10 ) 1. 4 = 579.2 K
0 .4
k −1
0 .4
P  k
T4 = T3  4  = 1400 K (.1) 1.4 = 725.1K
 P3 
(725.1 − 300 ) = 48.2% It checks with the calculated value above.
ηth = 1 −
(1400 − 579.2 )
h − h1 T2 − T1
bwr = 2
=
= 41.4% (using constant specifc heats)
h3 − h4 T3 − T4
W& T
= h3 − h4 = c P (T3 − T4 ) = 1.005 kgkJ*K (1400 − 725.1)K = 678.27 kJ
kg
m&
W& C
= h1 − h2 = cP (T1 − T2 ) = 1.005 kgkJ*K (300 − 579.2)K = −280.6 kJ
kg
m&
Q& In
= h3 − h2 = cP (T3 − T2 ) = 1.005 kgkJ*K (1400 − 579.2)K = 824.9 kJ
kg
m&
Q& Out
= h1 − h4 = cP (T1 − T4 ) = 1.005 kgkJ*K (300 − 725.1)K = −427.23 kJ
kg
m&
W& Net W& T W& C
=
−
= 397.7 kJ
kg
&
m
m&
m&
T1
K
300
WC/m
T2
K
579.2
WT/m
T3
K
1400
QIn/m
kJ/kg
kJ/kg
kJ/kg
-280.6
678.27
824.9
(
)
(
)
(
)
(
)
T4
K
725.1
bwr
%
41.4
Copyright © 2003, K. Nasr. All Rights Reserved
P1
kPa
100
Wnet/m
kJ/kg
397.7
P2
kPa
1000
η
%
48.2
P3
kPa
1000
Net Power
MW
2.308
P4
kPa
100
6
K. Nasr,c:\thermo \module3.doc
•
Irreversibilities Effect in the Compressor and Turbine:
Of course, the gas flowing through the compressor and the turbine does not undergo isentropic
processes. Typical isentropic efficiencies for the turbine are on the order of 90%, while that for an axial
compressor between 70 and 85%. Because of irreversibilities, the cycle’s capacity (Net Power) and
thermal efficiency will be reduced.
•
Let’s redo the previous example allowing the turbine to have an isentropic efficiency of 0.85 and the
compressor to have an isentropic efficiency of 0.75.
•
The 1st Law analysis has not changed and therefore we need the enthalpy values at each state.
A − 22
T1 Table

→ h1 = 300. 19 kJ
, p r 1 = 1.386 (Unchanged )
kg
P2 p r 2
A − 22
=
⇒ pr 2 = pr 1 *10 = 13. 86 Table

→ h2 s = 579.87 kJ
kg
P1
p r1
W&
h − h 2s
 h − h 2s 
 = 673.1 kJ
ηC = s = 1
⇒ h2 a = h1 −  1
, T 2 = 662.46K
kg
&
W a h1 − h2 a
 ηC 
A − 22
T 3 = 1400K Table

→ h3 = 1515. 42 kJ
, p r 3 = 450.5 (Unchanged)
kg
P4 p r 4
A − 22
=
⇒ pr 4 = pr 3 / 10 = 45. 05 Table

→ h4 s = 808. 5 kJ
kg
P3 p r 3
W&
h − h 4a
ηT = a = 3
⇒ h 4a = h3 + η T (h4 s − h3 ) = 914.54 kJ
, T 4 = 883.56 K
kg
h3 − h4 s
W&
s
•
&
& and the Q
Now we can calculate the W
‘s
‘s.
W& T
W& C
= h3 − h 4 = 600.88 kJ
;
= h1 − h 2 = −372.91 kJ
kg
kg
&
&
m
m
Q& In
Q& Out
= h3 − h2 = 842.32 kJ
;
= h1 − h4 = −614.35 kJ
kg
kg
m&
m&
•
Checking for 1st Law satisfaction:
W& Net
kJ
kJ
= 600.88 kJ
kg − 372.91 kg = 227.97 kg
m&
Q& Net
= 842.32 kJ
− 614.35 kJ
= 227.97 kJ
kg
kg
kg
m&
h1 −h 2
W& C
=
The Back Work Ratio = & =
WT
h3 − h4
•

 W&
Q&

Net
= Net v

&
m&
 m

372.91
= 62.0%
600.88
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K. Nasr,c:\thermo \module3.doc
Now, we determine the Net Power developed in the system.
 W&
W& 
W& Net = m&  T − C 
m& 
 m&
(
)
&
0.887 kgkJ*K (300 K )
∀
RT1
3
1
&
m& = ρ∀ =
, v1 =
=
= 0.861 mkg
v1
P1
100 kN 



m2 
3
m& =
5 ms
0.861
m3
(
)
= 5.81 kgs ⇒ W& Net = 5.81 kgs * 227.97 kJ
= 1325 kJs = 1.32 MW
kg
kg
The following table summarizes the solution:
T1
K
300
WC/m
T2
K
662.46
WT/m
T3
K
1400
QIn/m
kJ/kg
kJ/kg
kJ/kg
-372.91
600.88
842.32
•
T4
K
883.56
bwr
%
62.0
P1
kPa
100
Wnet/m
kJ/kg
227.97
P2
kPa
1000
η
%
27.1
P3
kPa
1000
Net Power
MW
1.32
P4
kPa
100
Note that because of irreversibilities in the compressor and in the turbine:
v The efficiency has dropped substantially from 45.7% to 27.1 %
v The cycle net capacity has dropped from 427 kJ/kg to 228 kJ/kg.
v The back work ratio has increased from 39.6 % to 62% while
v The net power developed has dropped from 2.48 MW to 1.32 MW, a decrease of 47 %.
Copyright © 2003, K. Nasr. All Rights Reserved
8