Upload
warren-pope
View
281
Download
10
Tags:
Embed Size (px)
Citation preview
EGR 334 ThermodynamicsChapter 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:
Homework Assignment:
Read Chapter 9, Sections 7-8
Problems from Chap 9: 42, 47, 55
3
OK….Quick Matching Quiza) Carnot b) Rankine c) Otto d) Diesel
p
v
.
.
.
.
4 1
3
1’
2 2’
AD
BC
4
Today you get to add two more cycles to your cycle repertoire.
Dual Cycle Brayton cycle.
Used as a hybrid cycle which includes elements of both the Otto and Diesel cycles. Used to model internal combustion engines
Used as a model for gas turbines (such as jet engines).
5Sec 9.4 : Air-Standard Duel Cycle
Neither the Otto or Diesel cycle describe the actual P-v diagrams of an engineHeat 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 5 – 1 : Constant volume heat rejection
Process 3 – 4 : Constant pressure heat transfer
Process 4 – 5 : Isentropic expansion
2
323 P
PTT
To set state 3: Use ideal gas law with V3 = V2.
1
212
r
r
P
PPP and
6
Dual Cycle analysis
2334 hhmQ
544545 uumUW
211212 uumUW
34 45 12
23 34
cycle
in
W W W W
Q Q Q
Sec 9.4 : Air-Standard Duel Cycle
232323 uumUQ
34 4 3W p v v
155151 uumUQ
5 151
23 34 3 2 4 3
1 1u uQ
Q Q u u h h
process 1-2: s1 = s2
process 2-3: v2 = v3
process 3-4: p3 = p4
process 4-5: s4 = s5
process 5-1: v5 = v1
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 2 3 4 5\
T (R) 520
p (psi) 14 1200
1200
u (Btu/lb)
h (Btu/lb)
vr
pr
State 1 2 3 4 5\
T (R)
p (psi)
u (Btu/lb)
h (Btu/lb)
vr
Pr
8
Example (9.38): State 1 2 3 4 5\
T (R) 520
p (psi) 14 1200
1200
u (Btu/lb) 88.62
h (Btu/lb)
vr 158.58
Pr 1.2147
Identify State Properties
State 1: p1 = 14 psi, T1 = 520 RState 2: s2 = s1 v2 = v1/rState 3: v3 = v2 and p3 = 1200 psiState 4: p4 = p3 = 1200 psiState 5: s5 =s4 and v5 = v1
Use Table A22E to fill in many of the other properties.
compression ratio, r = 15
Qin= Q23 + Q34 = 800 Btu
Given Information:
Qout = - Q51
State 1 2 3 4 5\
T (R) 520
p (psi) 14 1200
1200
u (Btu/lb)
h (Btu/lb)
vr
Pr
9
Example (9.38):
State 2: use r to find v2
and since 1-2 is isentropicfind vr2
572.1015
58.15812
r
vv rr
22 1
1
51.56114 594.26
1.2147r
r
pp p psi psi
p
State 1 2 3 4 5\
T (R) 520 1468.8
p (psi) 14 594.26
1200
1200
u (Btu/lb) 88.62 260.26
h (Btu/lb) 124.27
361.53
vr 158.58
10.572
pr 1.2147
51.561
State 1: given T = 520 Rlook up u, h, vr, and pr
then use Table A22E to look up T2, pr2, u2, and h2:
Pressure p2, can then be calculated using
State 1 2 3 4 5\
T (R) 520
p (psi) 14 1200
1200
u (Btu/lb) 88.62
h (Btu/lb) 124.27
vr 158.58
pr 1.2147
10
Example (9.38):
33 2
2
pT T
p
pv RT
State 1 2 3 4 5\
T (R) 520 1468.8
2966
p (psi) 14 594.26
1200
1200
u (Btu/lb) 88.62 260.26
577.4
h (Btu/lb) 124.27
361.53
780.7
vr 158.58
10.572
pr 1.2147
51.561
12001468.8
594.26R
State 3: given v3 = v2 and
p3 = 1200 psi, use ideal
gas law:
then use Table A22E to look up u3 and h3:
2965.97 R
State 1 2 3 4 5\
T (R) 520 1468.8
p (psi) 14 594.26
1200
1200
u (Btu/lb) 88.62 260.26
h (Btu/lb) 124.27
361.53
vr 158.58
10.572
pr 1.2147
51.561
11
Example (9.38):
State 4: Knowing p4=p3 and the heat in:
Qin= 800 Btu/lb
use the 1st Law:
4 3 2 3inQh u u hm
800 577.4 260.26 780.7 1263.56 / mBtu lb
Use Table A-22E to find T4 ,u4, pr4,
and v4r
State 1 2 3 4 5\
T (R) 520 1468.8
2966 4577.6
p (psi) 14 594.26
1200
1200
u (Btu/lb) 88.62 260.26
577.4
949.7
h (Btu/lb) 124.27
361.53
780.7
1263.6
vr 158.58
10.572
0.2848
pr 1.2147
51.561
5961.6
4 2 24 24( )m u u Q W
23 34 23 343 2 4 3
Q Q W Wu u u u
m m m m
3 2 4 3 4 3( )inQu u u u p v vm
O
3 2 4 3 4 3 3 2 4 3( )inQ u u p v v u u u u h hm
State 1 2 3 4 5\
T (R) 520 1468.8
2966
p (psi) 14 594.26
1200
1200
u (Btu/lb) 88.62 260.26
577.4
h (Btu/lb) 124.27
361.53
780.7
vr 158.58
10.572
pr 1.2147
51.561
12
Example (9.38):
State 5:
Use Table A-22E to look up T5, u5, h5, and pr5 and then find p5:
4
3
4
3
2
1
4
3
2
1
4
32
2
15
4
2
2
5
4
5
T
Tr
T
T
V
V
V
V
V
V
V
VV
V
VV
V
V
V
V
V
V
Replace V’s using ideal gas. 7187.9
6.4577
296615
4
3
4
5 T
Tr
V
V
55 4
4
9.7187 0.2848 2.768r r
Vv v
V
State 1 2 3 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
process 4-5 is also isentropic
55 4
4
305.241200 61.44
5961.6r
r
pp p psi
p
State 1 2 3 4 5\
T (R) 520 1468.8 2966 4577.6
p (psi) 14 594.26 1200 1200
u (Btu/lb) 88.62 260.26 577.4 949.7
h (Btu/lb) 124.27 361.53 780.7 1263.6
vr 158.58 10.572 0.2848
pr 1.2147 51.561 5961.6
13
Example (9.38):
(a) Wcycle, in Btu/lb.
(b) Qout, in Btu/lb.
(c) The thermal eff. (d) The cut off ratio
23 34 51cycle in outW Q Q Q Q Q
m m m m m
515 1 431.0 88.62 342.4 /out
m
Q Qu u Btu lb
m m
800 342.4 457.6 /cyclem
WBtu lb
m
State 1 2 3 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
14
Example (9.38):
(a) Wcycle, in Btu/lb.
(b) Qout, in Btu/lb.
(c) Thermal efficiency(d) The cut off ratio
cycle
in
W
Q
457.60.572
800cycle
in
W
Q
4
3
4577.61.543
2965.9c
Vr
V
State 1 2 3 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
5 1
3 2 4 3
1u u
u u h h
1 out
in
Q
Q
Cut off ratio: from ideal gas equation at constant pressure: pV mRT
3 4
3 4
V VmR
T p T
15Sec 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.
16
Jet engine:Suck (intake)Squeeze (compressor)Bang/Burn (combustion)Blow (turbine/exhaust)
Air-Standard analysis:Working fluid is airHeat 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
Heat Ex
Sec 9.5 : Modeling Gas Turbine Power Plants
17
Gas Turbine Analysis
121212 hhmHWW C
23
1243
23
1234
hh
hhhh
Q
WW
Q
W
in
cycle
232323 hhmHQQ in
Sec 9.5 : Modeling Gas Turbine Power Plants
433434 hhmHWW T
14`441 hhmHQQ out
43
12
23
1234
hh
hh
Q
WW
W
Wbwr
T
C
For a gas turbine, the back work ratio is much larger than that in a steam cycle since vair>>vliquid
bwr for a gas turbine power cycle is typically 40-80% vs. 1-2% for a steam power cycle.
process 2-3: p2 = p3
process 3-4: s3 = s4
process 4-1: p4 = p1
process 1-2: s1 = s2
18
Gas Turbine Analysis
Given T1 & T3 use table to find h1 & h3 .
Find state 2.
Find state 4.
For Cold-Air Standard analysis:
12 2
1 1
k kT p
T p
1 1
4 4 1
3 3 2
k k k kT p p
T p p
For state 2. For state 4.
22 1
1r r
pp p
p
Sec 9.3 : Air-Standard Diesel Cycle
44 3
3r r
pp p
p
Compressor pressure ratio:
2
1
p
p
19
Gas Turbine Analysis
Effect of Compressor pressure on efficiency.
23
1243
hh
hhhh
Q
W
in
cycle
23
1243
TTc
TTcTTc
P
PP
2
1
23
14
2
1
23
14 11
111
T
T
TT
TT
T
T
TT
TT
with
2
3
1
4
T
T
T
T
12 1
11
k kp p
Max T3 is approximately 1700 K
Sec 9.3 : Air-Standard Diesel Cycle
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(b) The back work ratio.(c) The net power developed.
State 1 2 3 4
T (R) 500 3000
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(b) The back work ratio.(c) The net power developed.
State 1 2 3 4
T (R) 500 3000
Since we are given k=1.4, use a cold-air standard analysis.
Temperatures for states 1 and 3 are given.
1
1.4 1 1.422 1
1
500 14 1062.76k k
pT T R
p
1 1.4 1 1.4
14 3
2
13000 1411.42
14
k kp
T T Rp
For state 2.
For state 4.
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(b) The back work ratio.(c) The net power developed.
State 1 2 3 4
T (R) 500 1063 3000 1411
2
11T
T 529.0
1063
5001
T
C
W
Wbwr
43
12
43
12
43
12
TT
TT
TTc
TTc
TTm
TTm
P
P
354.0
14113000
5001063
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(b) The back work ratio.(c) The net power developed.
State 1 2 3 4
T (R) 500 1063 3000 1411
12431243 TTTTcmhhhhmWWW PCTCycle
But need the mass flow rate. 2323 TTcmhhmQ Pin
6
3 2
(3.4 10 / )7078 /
(0.248 / )(3000 1063)in
mP
Q Btu hrm lb hr
c T T Btu lbm R R
7078 / 0.248 / 3000 1411 1063 500Cycle m mW lb hr Btu lb R R
61.80 10 /CycleW Btu hr
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 2 3 4
T (R) 520 3000
Pr
h (Btu/lb)
25
State 1 2 3 4
T (R) 520 1092 3000 1573
pr 1.2147
17.01 941.4 67.24
h (Btu/lb) 124.27
264.12
790.68
388.63
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 2 3 4
T (R) 520 3000
pr 1.2147
941.4
h (Btu/lb) 124.27
790.68Temperatures for states 1 and 3 are given. Relative
pressure and enthalpy values from Table A-22E
Find state 2.
Find state 4.
22 1
1r r
pp p
p
44 3
3r r
pp p
p
0058.17142147.1
24.6714
14.941
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(b) The net power developed.
23
1243
hh
hhhh
Q
W
in
cycle
12.26468.790
27.12412.26463.38868.790
State 1 2 3 4
T (R) 520 1092 3000 1573
pr 1.2147
17.01 941.4 67.24
h (Btu/lb) 124.27
264.12
790.68
388.63
498.0
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(b) The net power developed.
State 1 2 3 4
T (R) 520 1092 3000 1573
pr 1.2147
17.01 941.4 67.24
h (Btu/lb) 124.27
264.12
790.68
388.63 2143 hhhhmWWW CTCycle
But need the mass flow rate. 23 hhmQin
9
6
3 2
3.4 10 /6.46 10 /
790.68 264.12 /in
mm
Q Btu hrm lb hr
h h Btu lb
61.80 10 / 790.68 388.63 264.12 124.27 /Cycle m mW lb hr Btu lb
91.69 10 /CycleW Btu hr
28
End of Slides for Lecture 35