### Reheat and Intercooling for Brayton Cycle

```EGR 334 Thermodynamics
Chapter 9: Sections 7-8
Lecture 36:
Reheat and Intercooling of
Gas Turbine Systems
Quiz Today?
Today’s main concepts:
• Be able to explain the concept and purpose of using
reheat in a gas turbine.
• Be able to explain the concept and purpose of using
intercooling in a gas turbine system.
• Be able draw and explain Brayton cycles with reheat and
intercooling.
Homework Assignment:
Problem 9:80
Sec 9.5 : Modeling Gas Turbine Power Plants
Last time:
Introduced to the Gas Turbine Power Plant
Working fluid is air
Heat transfer from an external source (assumes there is no reaction)
Process 1 – 2 : Isentropic compression of air (compressor).
Process 2 – 3 : Constant pressure heat transfer to the air from an
external source (combustion)
Process 3 – 4 : Isentropic expansion (through turbine)
Process 4 – 1 : Completes cycle by a constant volume pressure in which
heat is rejected from the air
3
4
How a Jet Engine Works:
eature=endscreen&v=ON0sVe1yeOk
Jet Engine
Power Station Gas Turbine
Images from:
http://library.thinkquest.org/C006011/english/sites/gasturbine.php3?v=2
5
Possible Enhancements to the Brayton Cycle:
1) Use of Regenerator to preheat combustion air.
2) Reheating air between successive turbine stages.
3) Intercooling of air between successive compressor stages.
6
Brayton Cycle with Regeneration:
The exhaust air out of the turbine contains significant waste heat
that would normally be discarded to the atmosphere. Use of
regenerator can make use of this heat that would be discarded to
preheat the air on its way to the combustor, allowing the burned
fuel to be used more efficiently.
Sec 9.8 : Regenerative Gas Turbines with Reheat and Intercooling
To limit the temperature of the gas entering the turbine, air is
provided in excess to the primary combustion chamber.
The excess air limits the temperature of the combustion process by
providing a heat sink for the heat produced during combustion.
Q Combustion  m rxn c P  T  Q Combustion  m rxn c P  T  m inert c P  T
This has the added advantage of having the second turbine run at
lower pressure. Processes 2-3, a-b, and 4-1 occur isobarically.
7
Sec 9.8 : Regenerative Gas Turbines with Reheat and Intercooling
Improved efficiency may also be
achieved by decreasing the work
required for the compressors.
It is not practice to achieve such
cooling within a compressor.
Consequently, the compressor is spit
with a heat-exchanger in between.
8
Sec 9.8 : Regenerative Gas Turbines with Reheat and Intercooling
Putting all three
enhancements together,
the diagram show a
process combining
regeneration, reheat,
and intercooling.
9
10
Example (9.74): Air enters the compressor of a gas turbine at 100 kPa,
300 K. The air
is compressed
stages to 900
to4
State
1 in two
2 kPa,
3 with intercooling
c
d
a
b
300 K between the stages at a pressure of 300 kPa. The turbine inlet
300the expansion
300occurs in two
1480stages, with
1420reheat
temperatureTis(K)1480 K and
P (kPa) the stages
100 at300
300 of900
900 The300
300
100
to 1420 K between
a pressure
300 kPa.
compressor
and
turbine stage
Prefficiencies are 84 and 82% respectively, The net power
developed is 1.8 W. Determine
h (kJ/kg)
(a) The volumetric flow rate entering the cycle.
(b) The thermal efficiency of the cycle.
(c) The back work ratio.
11
Example (9.74): The compressor and turbine stage efficiencies are 84 and
82% respectively, The net power developed is 1.8 W. Determine
(a)
of
StateThe thermal
1 efficiency
2
3
c
d
4
a
b the cycle.
(b)
work ratio. 300
T (K)The back300
1480
1420
p (kPa)
100
300
300
900
900
pr
1.3860
1.3860
568.8
h (kJ/kg)
300.19
300.19
1611.79
300
300
100
478.0
1539.44
Find state a, Process 1 – a is Isentropic compression
 300 
 1 . 3860 
  4 . 158
p1
 100 
Using the isentropic compressor efficiency:
From the table haS = 411.26 kJ/kg
 h1 
p ra S  p r 1
 
h a  h1 
h aS
 h1 

 h aS
 h a  h1 
 300 . 19 
pa
411 . 26  300 . 19 
0 . 84
 432 . 42
kJ
kg
12
Example (9.74): The compressor and turbine stage efficiencies are 84 and
82% respectively, The net power developed is 1.8 W. Determine
(a)
of bthe cycle.
StateThe thermal
1 efficiency
2
3
c
d
4
a
(b)
work ratio.
T (K)The back300
300
1480
1420
P (kPa)
100
300
pr
1.3860
h (kJ/kg)
300.19
300
432.42
900
900
1.3860
568.8
300.19
1611.79
300
300
100
478.0
1539.44
Find state a, Process b – 2 is Isentropic compression
p2
 900 
p r 2 S  p rb
 1 . 3860 
  4 . 158
pb
 300 
Using the isentropic compressor efficiency:
From the table h2S = 411.26 kJ/kg
 hb 
h 2  hb 
h 2 S
 
 h 2  hb 
 h 2 S  hb 

 300 . 19 
411 . 26  300 . 19 
0 . 84
 432 . 42
kJ
kg
13
Example (9.74): The compressor and turbine stage efficiencies are 84 and
82% respectively, The net power developed is 1.8 W. Determine
(a)
ofbbthe cycle.
State
StateThe thermal
11 efficiency
22
33
c
dd
44
aa
work ratio. 300
T(b)
T(K)
(K)The back
300
300
300
1480
1480
1420
1420
pp(kPa)
(kPa)
100
100
300
300
300
300
900
900
1.3860
1.3860
900
900
568.8
568.8
300
300
300
300
ppr r
1.3860
1.3860
hh(kJ/kg)
(kJ/kg)
300.19
300.19 432.42
432.42 300.19
300.19 432.42
432.42 1611.79
1611.79 1275.4 1539.44
1539.44
100
100
478.0
478.0
Find state a, Process 3 – c is Isentropic expansion
 300 
 568 . 8 
  189 . 60
p3
 900 
Using the isentropic turbine efficiency:
From the table hcS = 1201.5 kJ/kg
 h3  hc 
p rcS  p r 3
 
pc
 h3  h cS 
hc  h3    h3  h cS   1611 . 79  0 . 82 1611 . 79  1201 . 5   1275 . 4
kJ
kg
14
Example (9.74): The compressor and turbine stage efficiencies are 84 and
82% respectively, The net power developed is 1.8 W. Determine
State
11
T (K)
300
300
p (kPa)
100
100
aa
bb
22
300
300
300
300
300
300
900
900
1.3860
1.3860
33
c c
d d
1480
1480
1420
1420
900
900 300
300
300
300
568.8
568.8
44
100
100
prr
1.3860
1.3860
478.0
478.0
h (kJ/kg)
300.19
300.19 432.42
432.42 300.19
300.19 432.42
432.42 1611.79
1611.79 1275.4
1275.4 1539.44
1539.44 1216.77
Find state a, Process d – 4 is Isentropic expansion
 100 
  478 . 0 
  159 . 33
pd
 300 
Using the isentropic turbine efficiency:
From the table h4S = 1145.94 kJ/kg
h4  hd 
p r 4 S  p rd
 
p4
 h 4  h dS 
h 4  h d    h d  h 4 S   1539 . 44  0 . 82 1539 . 44  1145 . 94   1216 . 77
kJ
kg
15
Example (9.74): The compressor and turbine stage efficiencies are 84 and
82% respectively, The net power developed is 1.8 W. Determine
State
1
2
3
c
d
4
(a) The thermal
efficiency
a
of
b the cycle.
(b) The back
T (K)
300work ratio. 300
1480
1420
p (kPa)
100
300
pr
1.3860
h (kJ/kg)
300.19
300
900
900
1.3860
432.42
300.19
300
568.8
432.42
1611.79 1275.4
300
100
478.0
1539.44
1216.77
Determine the mass flow rate
W cycle  W T 1  W T 2  W C 1  W C 2  1 .8 kJ / s
W cycle
m
W cycle
m
 394 . 6

 h3  hc    h d
 h 4    h a  h1    h 2  hb 
kJ
kg
m 
W cycle
3 9 4 .6 kJ / kg

1 .8 kJ / s
3 9 4 .6 kJ / kg
 4 .5 6 2 kg / s
16
Example (9.74): Air enters the compressor of a gas turbine at 100 kPa,
State
1 is compressed
2
d
4to
a
bin two stages
300 K. The air
to 3900 kPa,cwith intercooling
a pressure of 300
kPa. The turbine
T 300
(K) K between
300 the stages at
300
1480
1420inlet
temperature is 1480 K and the expansion occurs in two stages, with reheat
p (kPa)
100
300
300
900
900
300
300
100
to 1420 K between the stages at a pressure of 300 kPa. The compressor and
prturbine stage
1.3860
568.8
478.0
efficiencies 1.3860
are 84 and 82% respectively,
The net
power
h developed
(kJ/kg) 300.19
432.42
300.19 432.42 1611.79 1275.4 1539.44 1216.77
is 1.8 W.
Determine
(a) The volumetric flow rate.
kg
(b) The thermal efficiency of the cycle.
m  4 . 562
s
(c) The back work ratio.
 A V 1
 A V 1
 R T1 
 0 .2 8 7 0 kJ / kg  K   3 0 0 K
 m
   4 .5 6 2 kg / s 
5
2
p
(1
0
/
m
)
 1 
 3 .9 3
m
s
3

17
Example (9.74): Air enters the compressor of a gas turbine at 100 kPa,
State
1 is compressed
2
d
4to
a
bin two stages
300 K. The air
to 3900 kPa,cwith intercooling
a pressure of 300
kPa. The turbine
T 300
(K) K between
300 the stages at
300
1480
1420inlet
temperature is 1480 K and the expansion occurs in two stages, with reheat
p (kPa)
100
300
300
900
900
300
300
100
to 1420 K between the stages at a pressure of 300 kPa. The compressor and
prturbine stage
1.3860
568.8
478.0
efficiencies 1.3860
are 84 and 82% respectively,
The net
power
h developed
(kJ/kg) 300.19
432.42
300.19 432.42 1611.79 1275.4 1539.44 1216.77
is 1.8 W.
Determine
(a) The volumetric flow rate.
W cycle
kg
(b) The thermal efficiency of the cycle.  

m  4 . 562

Q
(c) The back work ratio.
in
s
kJ

Q in  m  h3  h 2    h d  h c   1443 . 41
kg
W cycle
394 . 6
 

 0 . 2734

Q in
1443 . 41
18
Example (9.74): Air enters the compressor of a gas turbine at 100 kPa,
State
1 is compressed
2
d
4to
a
bin two stages
300 K. The air
to 3900 kPa,cwith intercooling
a pressure of 300
kPa. The turbine
T 300
(K) K between
300 the stages at
300
1480
1420inlet
temperature is 1480 K and the expansion occurs in two stages, with reheat
p (kPa)
100
300
300
900
900
300
300
100
to 1420 K between the stages at a pressure of 300 kPa. The compressor and
prturbine stage
1.3860
568.8
478.0
efficiencies 1.3860
are 84 and 82% respectively,
The net
power
h developed
(kJ/kg) 300.19
432.42
300.19 432.42 1611.79 1275.4 1539.44 1216.77
is 1.8 W.
Determine
(a) The volumetric flow rate.
WC
kg
(b) The thermal efficiency of the cycle. b w r 

m  4 . 562
W
T
(c) The back work ratio.
s
h a  h1   h 2  hb 
W C
bwr 


 h3  h c    h d  h 4 
WT
bwr 
264 . 46
659 . 06
 0 . 401
19
end of slides for Lecture 36
```