Dual and Brayton Cycle
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Transcript 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.
Reading Assignment:
Read Chapter 9, Sections 7-8
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
Today you get to add two more cycles to your cycle
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
W12 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 W12
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 W12
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