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8/14/2019 Power Plant 1/2547
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6 5 20 1. 5 MW
compressor 97 k Pa, 30 0C compressor 2 k Pa compressor 5.5 : 1 isentropic compressor 84 % (transmission efficiency) 98 % turbine 900 0C (C12 H26) 43,100 kJ/kg
97 % 3 % isentropic turbine 88 % turbine 100 k Pa
specific fuel consumption thermal efficiency
(CPa) = 1.005 kJ/kg K , (CPg) = 1.147 kJ/kg K , isentropic index (a) = 1.4 , (g) = 1.33 gas constant ( R ) = 0.287 kJ/kg K
(20 )
1
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c = 84 %
m = 98 %T03 = 900
oC = 1,173 KHV = 43,100 kJ/kg
b = 97 %
compressor
adiabatic process stagnationh0a = h01 T0a = T01
T0a ~ Ta = T01 = 30 + 273 = 303 K
P1 ~ P 01 P01 = 97 2 = 95 k Pa
P02 = 5.5 P01 = 5.5 x 95 = 522.5 kPa P03 = P02 0.03 P02 = 0.97 x 522.5
= 506.8 k Pa
2
Wnet = 5 MW
Pa = 97 kPaTa = 30
oC
Pi = P0a - P01 = 2 k Pa
01
02
= 5.5
Pb = P03 - P02 = 3 % of P02t = 88 %
P04
= 100 k Pa
a = 1.4
g = 1.33R = 0.287 kJ/kg K Cpa = 1.005 kJ/kg KCpg = 1.147 kJ/kg K
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compressor
01
20
T
T
= a1a
01
02
P
P
= 4.1
14.1
)5.5(
= (5.5)0.2857 = 1.6275
=20T 1.6275 x 303 = 493.1 K
c = 0.84 =0102
0120
TT
TT
T02 - T01 = 0.84303493.1
= 226.4 K
T02 = 529.4 K
Turbine
40
03
T
T
= g
g
1
)(04
03
= 1.33
11.33
)100
506.8(
= (5.068)0.248
= 1.4955
'04T =1.49551,173
= 784.3 K
t
= 0.88 =
4003
0403
TT
TT
T03- T04 = 342 K
T04 = 1,173 342 = 830.97 K
W = 5 MW = m a )(m
cW
tW
=
)T(T
0.98
1.005-)T1.147(Tm
01020403a
=
(226.4)
0.98
1.005-1.147(342)m a
3
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= 160.1 am
am = 31.2 kg / s
bfHVm = m a Cpg ( T03 T02)
m f = 0.97x43,100)529.41,173(1.147x31.2
= 0.763 kg / s
f = af
m
m
= 31.20.763
= 0.024 kgf / kg
Specific fuel consumption (sfc) =Wf
= 600,3X160.1
0.024
= 0.55 kgf / kWh
Thermal efficieney () =HVfW
=43,100x0.024
160.1
= 15.5 %
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2. Single stage gas turbine nozzle rotor (designed conditions) ( ) nozzle 700 stagnation pressure temperature 311 k Pa 850 0C static pressure 100 kPa total (stagnation) to static efficiency 87 % mean blade speed =500 m/s
stage 1. specific work done 2. Mach Number nozzle3. 4. total to total efficiency5. degree of reaction stage
(CP) isentropic ( ) 1.148 kJ/kgK 1.33
(20 )
Ca = 900 km/h = 250 m/s
F = A)P(P]CCf)[(1m aeaja ++am = = 100 kg/s
f = = 0.018 kgf /kgCj =
Ca = = 250 m/sPe = (nozzle)Pa = = 69 kPaA =
5
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T04 = 970 K
T04 = T05 = T5 +gp
25
2C
C1
Check choking
2.49
69
172
P
P
a
04 ==
+
=
1g
g
)1
1
(1-1
1
cP
P
g
g
j
04
= 4
)]2.330.33(
0.951[1
1
Pc j
c
04
P
P= 1.914
choked
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T5 = Tc =
+12
g
T04 = x 970 = 832.6 K
P5 = Pc = 1.914P04
= 1.914172 = 89.86 k Pa 1
25C = (970 832.6) x 2 x 1.147 = 315.2 x 10 3
C5 = Cj = 561.4 m/s
5 =5
5
TR
P=
832.6x0.287
89.86 = 0.375 kg/m3
5m = 5 A5 C5 = 0.375 A5 x 561.4 = am ( 1 + f )
= 100 x 1.018 = 101.8 kg/s A5 = 0.4836 m
2
F = 100 [ (1.018) 561.4 250 ] + (89.86 69) 0.4836
. = 42.24 kN
. (propulsive efficiency) , p
p =aC
C1
2
j+ =250
561.41
2
+ = 61.62 %
. (energy conversion efficiency) , e
e =(HV)m
/2)C(Cam
f
2a
2j
= 322
10x43,930x1.8
/2](250)[(561.4)100
. (overall efficiency) , o
o = 0.6162 x 0.1597 = 9.84 %
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22.33
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( 3- 4)
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6. ( Combined Cycle Power Plant ) (Gass Turbine ) ( Steam Turbine) Heat Recovery Steam Generator, HRSG q q q Percentage
(20 )
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