### Dual and Brayton Cycle

```EGR 334 Thermodynamics
Chapter 9: Sections 5-6
Lecture 35:
Gas Turbine modeling with the
Brayton Cycle
Quiz Today?
Today’s main concepts:
• Be able to recognize Dual and Brayton Cycles
• Understand what system may be modeled using Brayton
Cycle.
• Be able to perform a 1st Law analysis of the Brayton Cycle
and determine its thermal efficiency.
• Be able to explain how regeneration may be applied to a
Brayton Cycle model.
Homework Assignment:
Problems from Chap 9: 42, 47, 55
3
OK….Quick Matching Quiz
a) Carnot
b) Rankine
C
c) Otto
d) Diesel
B
p
.
.
4
3
1
.
1’
.
2
2’
v
D
A
4
repertoire.
Dual Cycle
Used as a hybrid cycle which
includes elements of both the
Otto and Diesel cycles. Used
to model internal combustion
engines
Brayton cycle.
Used as a model for gas
turbines (such as jet engines).
Sec 9.4 : Air-Standard Duel Cycle
5
Neither the Otto or Diesel cycle describe the actual P-v diagrams of an engine
Heat addition occurs in two steps
• 2 – 3 : Constant volume heat addition
• 3 – 4 : Constant pressure heat addition (first part of power stroke)
Process 1 – 2 : Isentropic compression
Process 2 – 3 : Constant
volume heat transfer
Process 3 – 4 : Constant
pressure heat transfer
Process 4 – 5 : Isentropic
expansion
Process 5 – 1 : Constant
volume heat rejection
To set state 3: Use ideal gas
law with V3 = V2.
T3  T 2
P3
P2
and P2  P1
Pr 2
Pr 1
Sec 9.4 : Air-Standard Duel Cycle
6
Dual Cycle analysis
process 1-2: s1 = s2
W 12   U 12  m u 1  u 2 
process 2-3: v2 = v3
Q 23   U 23  m u 3  u 2 
process 3-4: p3 = p4
W 34  p  v 4  v3 
Q 34  m  h3  h 2 
process 4-5: s4 = s5
W 45   U 45  m u 4  u 5 
process 5-1: v5 = v1
Q 51   U 51  m u 5  u 1 
 
W cycle

W 34  W 45  W12
Q in
 1
Q 51
Q 23  Q 34
Q 23  Q 34
 u 5  u1 
 1
 u 3  u 2    h 4  h3 
7
Example (9.38): The pressure and temperature at the beginning of
compression in an air-standard dual cycle are 14 psi, 520°R. The
compression ratio is 15 and the heat addition per unit mass is 800 Btu/lbm.
At the end of the constant volume heat addition process the pressure is
1200 psi. Determine,
(a) Wcycle, in BTU/lb.
(b) Qout, in BTU/lb.
(c) The thermal efficiency.
(d) The cut off ratio
State
1
T (R)
520
p (psi)
14
u (Btu/lb)
h (Btu/lb)
vr
pr
P
2
3
4
1200
1200
5\
8
Example (9.38):
Given Information:
compression ratio, r = 15
Qin= Q23 + Q34 = 800 Btu
Qout = - Q51
State
State
T
(R)
T (R)
1
1
520
520
p
p (psi)
(psi)
u
u (Btu/lb)
(Btu/lb)
14
14
88.62
h
h (Btu/lb)
(Btu/lb)
vvrr
158.58
P
Prr
1.2147
2
2
3
3
4
4
1200
1200
1200
1200
Identify State Properties
State
State
State
State
State
1: p1 = 14 psi, T1 = 520 R
2: s2 = s1 v2 = v1/r
3: v3 = v2 and p3 = 1200 psi
4: p4 = p3 = 1200 psi
5: s5 =s4 and v5 = v1
Use Table A22E to fill in many of the
other properties.
5
5\\
9
Example (9.38):
State 1: given T = 520 R
look up u, h, vr, and pr
State 2: use r to find v2
and since 1-2 is isentropic
find vr2
vr 2 
v r1

158 . 58
State
1
2
T (R)
520
1468.8
p (psi)
14
594.26
u (Btu/lb)
88.62
260.26
h (Btu/lb)
124.27
361.53
vr
158.58
10.572
pr
1.2147
51.561
 10 . 572
r
15
then use Table A22E to look up T2, pr2, u2, and h2:
Pressure p2, can then be calculated using
p 2  p1
pr 2
p r1
 5 1 .5 6 1 
 1 4 p si
 1 .2 1 4 7 
 5 9 4 .2 6 p si
3
4
1200
1200
5\
10
State
1
T (R)
520
1468.8 2966
p (psi)
14
594.26
1200
u (Btu/lb)
88.62
260.26
577.4
h (Btu/lb)
124.27
361.53
780.7
p3
vr
158.58
10.572
p2
pr
1.2147
51.561
Example (9.38):
State 3: given v3 = v2 and
p3 = 1200 psi, use ideal
gas law: p v  R T
T3  T 2
 1200 
 1 4 6 8 .8  R 

5
9
4
.2
6


 2 9 6 5 .9 7  R
then use Table A22E to look up u3 and h3:
2
3
4
1200
5\
11
State
State
11
TT (R)
(R)
520
520
1468.8
1468.8 2966
2966
4577.6
pp (psi)
(psi)
14
14
594.26
594.26 1200
1200
1200
1200
uu (Btu/lb)
(Btu/lb)
88.62
88.62
260.26
260.26 577.4
577.4
949.7
hh (Btu/lb)
(Btu/lb)
124.27
124.27 361.53
361.53 780.7
780.7 1263.6
vvrr
158.58
158.58 10.572
10.572
0.2848
m ( u 4  u 2 )  Q 2 4  W 2 4 pprr
1.2147
1.2147 51.561
51.561
5961.6
Example (9.38):
State 4: Knowing p4=p3
and the heat in:
Qin= 800 Btu/lb
use the 1st Law:
 u3
 u2    u4  u3  
 u3
 u2    u4  u3  
Q in
m
Q 23
m
Q in
m

Q 34
m

W 23
m
22
O

33
44
55\\
W 34
m
 p (v 4  v3 )
  u 3  u 2    p ( v 4  v 3 )   u 4  u 3    u 3  u 2    h 4  h3 
h4 
Q in
m
  u 3  u 2   h3
 800   577.4  260.26   780.7  1263.56 B tu / lb m
Use Table A-22E
to find T4 ,u4, pr4,
and v4r
12
Example (9.38):
State 5:
process 4-5 is also
isentropic
V5
V4
V5

V5 V 2

V 5
V2 V4
 r
V4
T3
T4
vr 5
 15 
State
1
2
4
5\
T (R)
520
1468.8
2966
4577.6
2299
p (psi)
14
594.26
1200
1200
61.44
u (Btu/lb)
88.62
260.26
577.4
949.7
431.0
h (Btu/lb)
124.27
361.53
780.7
1263.6
601.48
vr
158.58
10.572
0.2848
2.768
pr
1.2147
51.561
5961.6
305.24
 V1  V 2  V 3 
V2
 2966 
 4577
.6 
V4
 9 . 7187

V1 V 3
V2 V4

3
V1 T 3
V 2 T4
 r
T3
T4
Replace V’s using ideal gas.
 V5 

 v r 4   9 .7 1 8 7   0 .2 8 4 8   2 .7 6 8
 V4 
Use Table A-22E to look up T5, u5, h5, and pr5 and then find p5:
 p 
 3 0 5 .2 4 
p5  p 4  r 5   1 2 0 0 
  6 1 .4 4 p si
 5 9 6 1 .6 
 pr 4 
13
Example (9.38):
(a) Wcycle, in Btu/lb.
(b) Qout, in Btu/lb.
(c) The thermal eff.
(d) The cut off ratio
W cycle

Q in
m
Q out
m
W cycle
m

m
 
Q 51
m
Q out
m

State
1
2
4
5\
T (R)
520
1468.8
2966
4577.6
2299
p (psi)
14
594.26
1200
1200
61.44
u (Btu/lb)
88.62
260.26
577.4
949.7
431.0
h (Btu/lb)
124.27
361.53
780.7
1263.6
601.48
vr
158.58
10.572
0.2848
2.768
pr
1.2147
51.561
5961.6
305.24
Q 23  Q 34
m

3
Q 51
m
 u 5  u 1  4 3 1 .0  8 8 .6 2  3 4 2 .4 B tu / lb m
 8 0 0  3 4 2 .4  4 5 7 .6 B tu / lb m
Example (9.38):
(a) Wcycle, in Btu/lb.
(b) Qout, in Btu/lb.
(c) Thermal efficiency
(d) The cut off ratio
 
W cycle
Q in
 1
Q in

4577.6
2299
1200
1200
61.44
260.26
577.4
949.7
431.0
124.27
361.53
780.7
1263.6
601.48
vr
158.58
10.572
0.2848
2.768
pr
1.2147
51.561
5961.6
305.24
State
1
2
3
T (R)
520
1468.8
2966
p (psi)
14
594.26
u (Btu/lb)
88.62
h (Btu/lb)
 u 5  u1 
 1
 u 3  u 2    h 4  h3 
Q out
W cycle
4
14
5\

Q in
4 5 7 .6
 0 .5 7 2
800
Cut off ratio: from ideal gas equation at constant
pressure: p V  m R T
V3
T3

mR
p

V4
T4
rc 
V4
V3

4 5 7 7 .6
2 9 6 5 .9
 1 .5 4 3
Sec 9.5 : Modeling Gas Turbine Power Plants
Air-Standard analysis of Gas Turbine Power plants.
Gas power plants are lighter
and more compact than vapor
power plants.
Used in aircraft propulsion
& marine power plants.
15
Sec 9.5 : Modeling Gas Turbine Power Plants
Air-Standard analysis:
Working fluid is air
Heat transfer from an external source (assumes there is no reaction)
Jet engine:
Suck (intake)
Squeeze (compressor)
Bang/Burn (combustion)
Blow (turbine/exhaust)
Heat Ex
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
16
Sec 9.5 : Modeling Gas Turbine Power Plants
17
Gas Turbine Analysis
process 1-2: s1 = s2
W12  W C   H 12  m  h 2  h1 
process 2-3: p2 = p3
Q 23  Q in   H 23  m  h3  h 2 
process 3-4: s3 = s4
W 34  W T   H 34  m  h3  h 4 
process 4-1: p4 = p1
Q 41  Q out   H 4 `  m  h 4  h1 
 
W cycle
Q in
h3  h 4   h 2  h1 
W 34  W12



 h3  h 2 
Q 23
For a gas turbine, the back work ratio is
much larger than that in a steam cycle since
vair>>vliquid
h 2  h1 
W C
W 34  W12
bwr 




 h3  h 4 
WT
Q 23
bwr for a gas turbine power cycle is typically
40-80% vs. 1-2% for a steam power cycle.
Sec 9.3 : Air-Standard Diesel Cycle
18
Gas Turbine Analysis
Given T1 & T3  use table to find h1 & h3 .
Find state 2.
p r 2  p r1
Find state 4.
pr 4  pr3
Compressor
pressure ratio:
p2
p1
p4
p3
p2
p1
For Cold-Air Standard analysis:
For state 2.
 k 1 k
 T2   p 2 



 T1   p1 
For state 4.
 T4   p 4 



 T3   p 3 
 k 1
k
 p1 


 p2 
 k 1
k
Sec 9.3 : Air-Standard Diesel Cycle
Gas Turbine Analysis
Effect of Compressor pressure on efficiency.
h3  h 4   h 2  h1 
 

 h3  h 2 
Q in
c P T 3  T 4   c P T 2  T1 

c P T 3  T 2 
T 4  T1 
T1 T 4 T1  1 
T1
 1
 1
 1
T3  T 2 
T 2 T 3 T 2  1 
T2
W cycle
with T 4

T1
  1
T3
T2
1

p 2 p1 
 k 1
k
Max T3 is approximately 1700 K
19
20
Example : Air enters the compressor of an ideal cold air-standard Brayton
cycle at 500°R with an energy input of 3.4x106 Btu/hr. The compression
ratio is 14 and the max T is 3000°R. For k=1.4 calculate
(a) The thermal efficiency
State
1
2
3
4
(b) The back work ratio.
T (R)
500
3000
(c) The net power developed.
h (BTU/lb)
21
Example : Air enters the compressor of an ideal cold air-standard Brayton
cycle at 500°R with an energy input of 3.4x106 BTU/hr. The compression
ratio is 14 and the max T is 3000°R. For k=1.4 calculate
(a) The thermal efficiency
State
1
2
3
4
(b) The back work ratio.
T (R)
500
3000
(c) The net power developed.
Since we are given k=1.4, use a cold-air standard analysis.
Temperatures for states 1 and 3 are given.
 k 1
For state 2.
 p2 
T 2  T1 

 p1 
For state 4.
 p1 
T 4  T3 

 p2 
 k 1
k
k
  5 0 0  1 4 
1 .4  1 
 1 
  3000  

1
4


1 .4
1 .4  1 
 1 0 6 2 .7 6  R
1 .4
 1 4 1 1 .4 2  R
22
Example : Air enters the compressor of an ideal cold air-standard Brayton
cycle at 500°R with an energy input of 3.4x106 BTU/hr. The compression
ratio is 14 and the max T is 3000°R. For k=1.4 calculate
(a) The thermal efficiency
State
1
2
3
4
(b) The back work ratio.
T (R)
500
1063
3000
1411
(c) The net power developed.
  1
T1
T2
W C
bwr 
W
T

m T 2  T1 
m T 3  T 4 
 1
500
 0 . 529
1063

c P T 2  T1 
c P T 3  T 4 

T 2  T1 
T3  T 4 

1063
3000

 0 . 354
 1411 
 500
23
Example : Air enters the compressor of an ideal cold air-standard Brayton
cycle at 500°R with an energy input of 3.4x106 BTU/hr. The compression
ratio is 14 and the max T is 3000°R. For k=1.4 calculate
(a) The thermal efficiency
State
1
2
3
4
(b) The back work ratio.
T (R)
500
1063
3000
1411
(c) The net power developed.
W Cycle  W T  W C  m  h 3  h 4    h 2  h1   m c P T 3  T 4   T 2  T1 
But need the mass flow rate.
m 
Q in
c P  T3  T 2 
Q in  m  h3  h 2   m c P T3  T 2 
(3.4  10 B tu / hr )
6

(0.248 B tu / lbm  R )(3000  1063) R
 7078 lb m / hr
W C ycle   7 0 7 8 lb m / h r   0 .2 4 8 B tu / lb m  R    3 0 0 0  1 4 1 1   1 0 6 3  5 0 0   R
W C ycle  1 .8 0  1 0 B tu / h r
6
24
Example (9.43): The rate of heat addition to an air-standard Brayton cycle is
3.4x109 BTU/hr. The pressure ratio for the cycle is 14 and the minimum and
maximum temperatures are 520°R and 3000°R, respectively. Determine
(a) The thermal efficiency
(b) The net power developed.
State
1
T (R)
520
Pr
h (Btu/lb)
2
3
3000
4
25
Example (9.43): The rate of heat addition to an air-standard Brayton cycle is
3.4x109 BTU/hr. The pressure ratio for the cycle is 14 and the minimum and
maximum temperatures are 520°R and 3000°R, respectively. Determine
(a) The thermal efficiency
(b) The net power developed.
State
11
22
3
44
T (R)
520
520
1092
3000
1573
pr
1.2147 17.01
1.2147
941.4
67.24
h (Btu/lb)
124.27 264.12 790.68 388.63
124.27
Temperatures for states 1 and 3 are given. Relative pressure
and enthalpy values from Table A-22E
Find state 2.
p2
p r 2  p r1
 1 . 2147
p1
14   17 . 0058
Find state 4.
p4
 1 
pr 4  pr3
 941 . 4 
  67 . 24
p3
 14 
26
Example (9.43): The rate of heat addition to an air-standard Brayton cycle is
3.4x109 BTU/hr. The pressure ratio for the cycle is 14 and the minimum and
maximum temperatures are 520°R and 3000°R, respectively. Determine
(a) The thermal efficiency
State
(b) The net power developed. T (R)
 
W cycle
1
2
3
4
520
1092
3000
1573
pr
1.2147
17.01
941.4
67.24
h (Btu/lb)
124.27
264.12
790.68 388.63

h3  h 4   h 2  h1 
 h3  h 2 

790 . 68  388 . 63   264 . 12  124 . 27 
790 . 68  264 . 12 
Q in
 0 . 498
27
Example (9.43): The rate of heat addition to an air-standard Brayton cycle is
3.4x109 BTU/hr. The pressure ratio for the cycle is 14 and the minimum and
maximum temperatures are 520°R and 3000°R, respectively. Determine
(a) The thermal efficiency
State
(b) The net power developed. T (R)
1
2
3
4
520
1092
3000
1573
pr
1.2147
17.01
941.4
67.24
h (Btu/lb)
124.27
264.12
790.68 388.63
W Cycle  W T  W C  m  h 3  h 4    h1  h 2 
But need the mass flow rate. Q in  m  h3  h 2 
m 
3 .4  1 0 B tu / h r
9
Q in
 h3  h 2 

 7 9 0 .6 8  2 6 4 .1 2  B tu / lb m
 6 .4 6  1 0 lb m / h r
6
W C ycle   1 .8 0  1 0 lb m / h r    7 9 0 .6 8  3 8 8 .6 3    2 6 4 .1 2  1 2 4 .2 7   B tu / lb m
6
W C ycle  1 .6 9  1 0 B tu / h r
9
28
End of Slides for Lecture 35
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