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Chapter 7 - Gas Turbine Power Plant
Power Plant Engineering (University of the Philippines System)
Studocu is not sponsored or endorsed by any college or university
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CHAPTER 7 - GAS TURBINE POWER PLANT
2
2
1. Calculate the work done per kg of gas expanding from 6.33 kg/cm ab to 1.05 kg/cm ab in a gas
turbine of 82% interna l efficiency. Initial temperature, 750 C;
1.34; MW= 29.
Solut ion:
r=
b
w
er
VI
0..
0
VOLUME
1}, = 0.82
T3 = 750 + 273 = 1023K
r= l.34
MW=29
p3 =6.33kg / cm 1 ab
p4 =1.05kg/cm i ab
wT= 102c;. (T -rJ,, ,
3
_.!,_ =
p
P3
(T
J*
T3
=(l )l.34-1
1.34
1.05
6.33
1023
T4 = 648.5K
C;o -C-. =8.3451/ mole·C
C-
_P
c.
= 1.34
C-
c, --P-= 8.3451/mole·C
1.34
C;; =32.8891/ mole
CP
-
C;;
MW
32.889
29
l. 1341J' ·C
,g
WT =102c ;.(T3-('Jr!,
Wr = 102(1.1341)(1023-648.SX0.82) = 35,524kg •m/kg (answer)
r
of 1.35, 556 K, MW 29, are moving in an exhaust pipe at 174 m/sec,
1.12 kg/cm 2 ab static pressure. Find the total pressure and temperature.
Solution:
2. Products of combustion with
1
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CHAPTER 7 - GAS TURBINE POWER PLANT
7
, p = p[ 1 + ( r; )M
r-1
1
2
]
1
,r=r [ +(r;
M
1)M
2
]
V
R=C-p - C- =8.345J I1mole•C
R 8.345
R= -=--=0. 28781/g•C
V
29
MW
174
3.78
.J(1.35)(9.s1Xo.2s18 X556)
M
:-l
13
, p= l. 12[ 1+ ( ·
1
) 3.78}2]~ =l.SSkg/cm 2 ob (answer)
3
2
,T=556 [ 1+ (1· ~-l }3. 78) ]= 1946 K (answer)
3. The exhaust temperatu re of a gas turbine, taken with a good total temperature probe, is recorded
at 635 C. Air-fuel ratio, 0.015; fue l is oil. The gas velocity having been determined to be 650 fps,
calculate the static temperatu re.
Solut ion:
M-
V
V = 650 fps= 198.11 m/ s
1
,r =r[ +(r;
1)M
2
]
T = 635+ 273= 908K
At 635 C, Air-fuel ratio, 0.015; fuel is oil. Figure 7-2.
cp =1.091/g •C
R=
Cp
R
MW
8 345
= '
=0.28781/g•C
29
=-1!!__
r-1
1.o9 = r(o.2s1s)
r-1
r=l.359
M
l9S.ll
3.3564
.J(l .359)(9.81X0.2878X908)
1 35
1
,T= gos[ 1+ ( • :- )3.3564
)2 ] = 2744 K = 2471 C
2
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CHAPTER 7 - GAS TURBINE POWER PLANT
4. Kerosene is the fuel of a gas turbine plant. f = 0.012, T3 = 972 K pressure ratio , 4.5, exhaust to
atmosphere. Find the available energy, kg-m per kg air flow.
Solution:
f =0.012, T3 =972 K pressure ratio, 4.5, exhaust to atmosphere. Kerosene considered to be C12H26•
By a tria l solution it is guessed that t he final temperature will be about 670 K. Fig. 7-2 gives cp = 1.16
at 972 K; 1.09 at 670 K. An arithmetical average is C-;; = 1.125 J/g-C. The molecular weight of air is
usually given as 28.97. Air-fue l ratios used w ith the gas turbine are lean in order to hold the
temperature to that allowed by the turbine design; hence, the product's molecular weight is but
little more than air's. Taking 29.0 as the molecular weight , mol specific heat C1 = 29 x 1.125 =
32.625 J per mole - C. From the kinetic theory of gases, Cp - C, = 8.245 J per mol (independent of
Hence C, = 24.38 J per mol and r = 32.625 / 24.38 = 1.338 and C;; = 1.125 the
temperature).
following calculations lead to the desired available energy.
1
)7;
1
= (4.5) ~:; =4.5 0.2526 = 1.462
= (P
T,
P,
Since T3 =972 K, T4 =972 / 1.462 =664.8 K
Although the value C;, is based on a 670 K estimate of T4, a recalculation is not considered necessary
3
in this solution.
LIT=972-664.8=307.2K
From Eq. 7-1.
Llh =1.125 x 307.2 =345.6 J per g of gas available energy
In kg-m per kg
Llh = 345.6 x 1000 / 9.81 = 35,229 kg-m per kg of gas available energy (answer)
5. A gas with
1.35, temperature 649 C, is expanded in a gas turbine from 3.52 to 1.05 kg/cm 2 ab. 17r
= 0.80. What is the temperature of the exhaust gas? Is this static or tota l temperature? Give the
r=
reason .
Solution:
T1 =649 + 273
r= 1.35
=922 K
1]r= 0.80
=(::J7
us- 1
922 = ( 3.52)035
7;
1.05
T2= 674 K
Tl -r;,
17r = T.l - T.2
0.80 = 922- Tz•
922-674
Tr= 723.6 K = 450.6 C (answer)
This is an actua l static temperature not a total temperature which represents the sum of the static
and t he increments representing velocity.
3
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CHAPTER 7 - GAS TURBINE POWER PLANT
6. Sketch the p-v and T-s cycles of an ideal open gas turbine cycle having: pressure ratio, S; t3, 649 C.
Solution:
P2/Pi = 5
I; = 649+273 = 922K
Assume f = 0.015
At 922 K, Cp = 1.145 J/g-C
Trial T4 580 K, Cp 1.065 J/g-C
Average Cp = 1.105 J/g-C
Cp =29 x 1.105 =32.045 J/mol-C
Cv =32.045 - 8.345 =23.7 J/mol-C
C
32.045 1.352
=
r= •
c.
=
23.7
1
v, =(p ,, Jk =5 L:,l =3.2884
V2 P1
p3 = Pi, P4 = P1
P3/p, =Pz/P, = 5
V./V3 = v,/V2 = 3. 2884
Sketch:
3
w
a::
4
::::,
i
649 C = 922 K
606K
2
w
a..
w
I-
1
4
ENTROPY
VOLUME
7. An open gas turbine cycle is to be operated w ith a maximum temperature of 750 C. Intake air, 1
2
kg/cm ab, 29.4 C; fue l, C12H26; Ot = 43,155 J/g, 1/r = 0.84, 1Jc = 0.80; combustion efficiency , 95%.
Draw t he full load 1!t vs. pressure ratio characteristics through the pressure ratio range of 4 to 10.
Solution:
4
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CHAPTER 7 - GAS TURBINE POWER PLANT
T3 =750 + 273.15 =1023.15 K
Pi = 1 kg/cm 2 ab
T1 =29.4 + 273.15 =302.55 K
Fuel Cu H26, Q, = 43,155 kJ/kg
1Jr=0.84
T/, = 0.80
T/,om= 0.95
b
w
er
:::>
Ill
Ill
w
er
0.
0
VOLUME
Fm. 7-3 Open cycle of the simple gns
turbine plnnt .
Pressure ratio = 4
T1 = 29.4 + 273.15 = 302.55 K
f= 0.0000
At 302.55 K, Cp = 1.005 J/g-C
Trial T1 =450 K, Cp =1.025 J/g-C
Average Cp = 1.015 J/g-C
Cp =28.97 x 1.015 =29.405 J/mol-C
C, =29.405 - 8.345 =21.06 J/mol-C
r= cp
c.
29.405 1.4
21.06
With a pressure ratio of 4. This gives an ideal T1 of
I=!
7
T=T{::J
2
L4-1
T2 =302 .55(4)T.
T2 =449.6K
Real T1 leaving the compressor and entering combustor
Tz =302.55 + (449.6- 302.55) / 0.80 =486.4 K
Assume f = 0.015 and using Fig. 7-2
Cp at 486.4 K = 1.05 J/g-C
Cp at 1023.15 K = 1.18 J/g-C
Average cp = 1.115 J/g-C
5
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CHAPTER 7 - GAS TURBINE POWER PLANT
a1 needed to raise products
to 1023.15 K
= (1023.15 - 486.4)(1.115) = 598.5 J/g
598.5
0.0146
43,155x0 .95
Fuel-air ratio= ----
By means of a preliminary trial it is estimated that the ideal T4 will approximate 720 K. Now Fig. 7-2
is used to obtain the mean Cp to expansion.
At 643 K, Cp = 1.11 J/g-C
At 1023.15 K, Cp = 1.18 J/g-C
Average Cp = 1.145 J/g-C
For f = 0.0146 and temperature range 1023.15 to 720 K.
Cp = 29 x 1.145 = 33.2 J/mole C
and
r
33.2
1.34
33.2-8.345
1023.15
Calculated
1.34-1
719.8 K
(4)°"u4
Wr =102{1+ /)cP(J;-T4 )
=102(1+0.0146X1.14sX1023.1s-119.s)
Wr =35,946kg •m/kg
302.55 K = 1.02 J/g-C
Cp at 449.6 K = 1.04 J/g•C
Average cp = 1.03 J/g-C
Cpat
Wc = 1Q1cp(",; - ½)
Wc =102(1.03X4B64- 302ss)
W, =19,315kg •m/kg
Tl,
Tl,
TMlcWr -w,
10211,a,
(o.s4)(o.soX35,946) - 19,315
102(0.80X598.5)
1/, =0.09 91=9. 91%
Including combustion losses, plant efficiency= 9.91% x 0.95 = 9.41%
Pressure ratio = 6
With a pressure ratio of 6. This gives an ideal T2 of
I::!
2=r1(::)
T
r
6
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CHAPTER 7 - GAS TURBINE POWER PLANT
l..4- 1
T2 =302.55(6) u
J; = 5048K
Real Ti leaving the compressor and entering combustor
Tr= 302.55 + (504.8 - 302.55) / 0.80 = 555.4 K
Assume f = 0.015 and using Fig. 7-2
Cpat 555.4 K = 1.068 J/g-C
Cp at 1023.15 K = 1.18 J/g-C
Average cp = 1.124 J/g-C
Q1 needed to ra ise products to 1023.15 K
(1023.15- 555.4)(1.124) 525.75 J/g
=
=
525.75
Fuel-air ratio= ---0.013
43,155X0.95
By means of a preliminary trial it is estimated that the ideal T4 will approximate 643 K. Now Fig. 7-2
is used to obtain the mean Cp to expansion.
At 643 K, Cp = 1.09 J/g-C
At 1023.15 K, Cp = 1.18 J/g-C
Average cp = 1.135 J/g-C
For f = 0.013 and temperature range 1023.15 to 643 K.
Cp =2.9 x 1.135 =32.9 J/mole C
and
32.9
=1.34
32.9-8.345
1023.15
Calculated
,.34-l
649 K
r
(6) Ll4
wT =102(1+ f)cp(,; - r4)
WT =102{1 +0.013X1.13SX102315-649)
WT=43,879kg •m/kg
302.55 K = 1.02 J/g-C
Cp at 504.8 K = 1.05 J/g-C
Average Cp = 1.035 J/g-C
Cpat
w, = 102cp(7;-7;)
=102(1.03sXso4.8-3o2ss)
W, =21,357kg•m/kg
1/,
1/,
17r1JcWr -W,
10277, a ,
(0.84 )(o.80X43,879)-21,357
102(0.soX525.75)
7
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CHAPTER 7 - GAS TURBINE POWER PLANT
T/, =0. 1895 =18. 95%
Including combustion losses, plant efficiency= 18.95% x 0.95 = 18.00%
Pressure ratio = 8
With a pressure ratio of 8. This gives an ideal T2 of
L:.!:
T=Ti(::)
2
r
1..4 - 1
72 =302.55(8)7.
T,_ =548.lK
Real T2 leaving the compressor and entering combustor
Tr= 302.55 + (548.1- 302.55) / 0.80 = 609.5 K
Assume f = 0.015 and using Fig. 7-2
Cpat 609.5 K = 1.08 J/g-C
Cp at 1023.15 K = 1.18 J/g-C
Average cp = 1.13 J/g-C
Q1 needed to raise products to 1023.15 K
=(1023.15-
609.5)(1.13) =467.4 J/g
467.4
0.0114
Fuel-air ratio = ---43,155x0 .95
By means of a preliminary trial It Is estimated that the ideal T4 will approximate 600 K. Now Fig. 7-2
is used to obtain the mea n Cp to expansion.
At 600 K, Cp = 1.075 J/g-C
At 1023.15 K, Cp = 1.18 J/g-C
Average cP = 1.128 J/g-C
For f = 0.0114 and temperature range 1023 .15 to 600 K.
Cp = 29 x 1.128 = 32.7 J/mole C
and
32.7
1.34
32.1-8.345
1023.15
Calculated
l.3'-'
603. 7 K
r
(8)u.
WT =102(1+
f)cp(~ - T4)
L¼ =102(1 + o.0114X1.128X1021is
- 6011)
WT=48,810kg •m/kg
302.55 K = 1.02 J/g-C
548 .1 K = 1.065 J/g-C
Average Cp = 1.043 J/g-C
Cpat
Cp at
8
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CHAPTER 7 - GAS TURBINE POWER PLANT
Wc =102c/J;
-7J
w, = 102{1043XS4&1-30255)
Wc =26,123kg•m/kg
11,
.,,,
1Jr1JcWr -Wc
10211,a,
(o.84Xo.8oX48,810)- 26,123
102(0.80X467.4)
.,,, =0.1751=17.51%
Including combustion losses, plant efficiency= 17.51% x 0 ..95 = 16.63%
Pressure ratio = 10
With a pressure ratio of 10. This gives an ideal T2 of
7
T=T{::)
2
U -1
T; =302.55(10)u
J; = 584.lK
Real T2 leaving the compressor and ente ring combustor
T:r =302.55 + (584.1- 302.55) / 0.80
=654.5 K
Assume f = 0.015 and using Fig. 7-2
Cp at 654.5 K = 1.09 J/g-C
Cp at 1023.15 K = 1.18 J/g-C
Average Cp = 1.135 J/g-C
OJ needed
to raise products to 1023.15 K
= (1023.15- 654.5)(1.135) = 418.4 J/g
418.4
43,155x0.95
Fuel-air ratio= ----
0.0102
By means of a preliminary trial It is estimated that the ideal T4 will approximate 570 K. Now Fig. 7-2
is used to obtain the mean Cp to expansion.
At 570 K, Cp = 1.07 J/g-C
At 1023.15 K, Cp = 1.18 J/g-C
Average cp = 1.125 J/g-C
For f = 0.0102 and temperature range 1023.15 to 570 K.
Cp = 29 x 1.125 = 32.6 J/mole C
and
32.6
1.34
32.6-8.345
1023.15
Calculated
L)-H
570.4 K
(10)u,
r
9
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CHAPTER 7 - GAS TURBINE POWER PLANT
wr =102(1+ J)cp(~ -T4 )
WT=102{1+ 0.0102X1.125X1023.15 -57a 4)
Wr =52,4B3kg •m/kg
302.55 K= 1.02 J/g-C
584.1 K= 1.07 J/g-C
Average cP = 1.045 J/g-C
Cp at
Cpat
Li¼: =102cp(7; - I"i)
Li¼: =102{1.045X584.1- 30255)
Wc =30,0l0kg•m/ kg
T/,
T/r'TJcWr -Wc
10211, a,
11
'
(o.s4 )(o.80 X52,483)- 30,010
102(0.80X418.4)
11, ::::0.1540::::15.40%
Including combustion losses, plant efficiency= 15.40% x 0.95 = 14.63%
Plot:
10
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CHAPTER 7 - GAS TURBINE POWER PLANT
18
16
14
12
:,!:!
0
10
1;C:
a,
·u 8
ea,
ti,
E
a,
6
.s:::
t-
4
2
0
4
6
8
10
Pressu re Ratio
8. Consider that Fig. 7-11 is characteristi c of all gas tu rbine plants and determine the efficiency of a
plant at half load. Plant has a pressu re ratio of 6; othe r data same as in Prob. 7.
Solution:
b
...,
er
:::>
V)
V)
...,
0
VOLUME
Fl<;. 7-3
Open cycle of the simple
turbine plant.
F::lS
T3 = 750 + 273 .15 = 1023 .15 K
P1= 1 g/c m2 ab
T1 =29/4 + 273/15 =302.55 K
11
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CHAPTER 7 - GAS TURBINE POWER PLANT
Fuel Cu H26, Ot = 43,155 kJ/kg
7]r=0.84
1/c = 0.80
1/a,m= 0./95
Pressure ratio = 6
T1 = 29.4 + 273.15 = 302.55 K
f= 0.0000
At 302.55 K, Cp = 1.005 J/g-C
Trial T2 = 500 K, Cp = 1.035 J/g-C
Average Cp = 1.0175 J/g-C
Cp = 28.97 x 1.0175 = 29.477 J/mol -C
C, = 29.477 - 8.345 = 21.132 J/mol -C
r c.
c,
29.477
21.132
_
14
With a pressure ratio of 6. This gives an ideal T2 of
r- 1
r,
=r{::)r
1.,4.-1
T2 =302.55( 6 )---i::.
J; ;::::504.BK
Real 72 leaving the compressor and entering combustor
T2' = 302.55 + (504.8 - 302.55) / 0.80 = 555.4 K
Assume/= 0.015 and using Fig. 7-2
Cpat 555.4 K = 1.068 J/g-C
Cp at 1023.15 K = 1.18 J/g-C
Average cp = 1.124 J/g-C
Q1 needed to raise products to 1023.15 K
= (1023 .15- 555.4)(1 .124) = 525.75 J/g
525.75
0.013
Fuel-air ratio = ---43,lSSx0.95
By means of a preliminary trial it is estimated that the ideal T4 will approximate 643 K. Now Fig. 7-2
is used to obtain the mean Cp t o expansion .
At 643 K, Cp = 1.09 J/g-C
At 1023.15 K, Cp = 1.18 J/g-C
Average Cp = 1.135 J/g-C
For f = 0.013 and temperature range 1023.15 to 643 K.
Cp = 29 x 1.135 = 32.9 J/mole C
and
r
32.9
32.9 - 8.345
1.34
12
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CHAPTER 7 - GAS TURBINE POWER PLANT
Calculated
1023.15
_
130 1
(6) L34
=649 K
WT = 102(1+ f)c) ¼-"'4)
WT= 102(1+0.013 Xl.135Xl023.15-649)
WT=43,879kg•m / kg
302.55 K = 1.02 J/g-C
Cpat 504.8 K = 1.05 J/g-C
Average Cp = 1.035 J/g-C
Cp at
w, = 102cJ,; - 1J
w, = 102(1.03sXsoc1.s- 3o255)
'111c =21,357kg -m/ kg
1/,
1/,
1/r'f/cWr -W,
102.,,, a ,
(0.84Xo.soX43,879) - 21,357
102(0.80)(525.75)
1/, =0.1895 = 1&95%
Includ ing com bust ion losses, plant efficiency= 18.95% x 0.95 = 18.00%
I
PE~ tNT RATED LOAD
Fla. 7-11 Part load operating efficiencies
af ga.e turbine power plants.
Percent of Fu ll Load Ther mal Efficien cy At 50 percent rated load
95%
90%
75%
Simp le closed cycle
Compounded open cycle (two shaft)
Simp le open cycle
13
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CHAPTER 7 - GAS TURBINE POWER PLANT
I Regenerative open cycle
Thermal efficiencies (answer)
Simple closed cycle
Compounded open cycle (two shaft)
Simple open cycle
Regenerative open cycle
17.1%
16.2%
13.5%
11.7%
9. Find the full -load efficiency, air rate , and exhaust tem perature of an open -cycle gas turbine plant to
be built for optimum pressure ratio with T3 =1028 K, T1 =278 K, 1Jr= 0.82 , 1Jc =0. 79, Oil fuel w ith Ot
= 43,165 J/g. Combustion efficiency, 94%. Standard sea level atmosphere Specific heat data as in Fig.
7-2.
Solution:
2
3
b
LI.I
r:r
::>
V)
V)
LI.I
a'.
Cl.
0
VOLUME
Fm. 7-3
Open cycle of the simple gas
turbine plant.
For optimum pressure rat io:
Eq. 7-10.
2
2
(21Jr1Jc~TJJ;-["fi-(l-1Jc1Jr)T3}1;=1Jr1JcT3 "fi
[2(0.s2xo.79)(102sx21s)]7;- [21s- (1-o.79xo.s2x102s)]r
; =(o.s2xo.79x102so)2(278)
T2 =464.9K
Optimum pressure ratio:
p =
_!,.
Pi
(T....!.)";:i
Tl
l4
P2 ( -464.9)1.•
-1 =.
6 025 use 6
-=
p1
278
1
6 =( 2i8 J:~
T2 =46 3.8K
14
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CHAPTER 7 - GAS TURBINE POWER PLANT
Real T2 leaving the compressor and entering combustor
Tz =278 + (463.8 - 278) / 0.79 =513.2 K
Assume/= 0.015 and using Fig. 7-2
Cpat 513.2 K = 1.06 J/g-C
Cp at 1028 K = 1.18 J/g-C
Average Cp = 1.12 J/g-C
a1 needed to raise products
to 1028 K
= (1028- 513.2)(1.12) = 576.6 J/g
576.6
43,165x0.94
Fuel-air ratio= ----
0.0142
By means of a preliminary trial it is estimated that the ideal
is used to obtain the mean cp to expansion .
At 650 K, Cp = 1.09 J/g-C
At 1028 K, Cp = 1.18 J/g-C
Average Cp = 1.135 J/g-C
r~ will
approximate 650 K. Now Fig. 7-2
For f = 0.0142 and temperature range 1028 to 650 K.
Cp =29 x 1.135 =32.9 J/mole C
and
r
32.9
1.34
32.9 - 8.345
1028
Calculated T4
uH
652.5K
(6)u,
Wr =102(1+ f)c)J; - T4 )
Wr = 102(1+0. 0142Xl.13SX1028-652 5)
Wr =44,089kg •m/kg
278 K = 1.02 J/g-C
Cp at 463.8 K = 1.045 J/g-C
Average cp = 1.033 J/g-C
Cpat
WC=102c/J;-7J
Wc =102(1.033X463.s - 21s)
Wc =19,577kg-m/kg
.,,, =0.1 934=19.34%
Including combustion losses, plant efficiency= 19.34% x 0.94 = 18.18% (answer)
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lOMoARcPSD|33011865
CHAPTER 7 - GAS TURBINE POWER PLANT
Net output per kg air fl ow
=(1.01 42Xl.135X1028-6525X0.82 )x10 3 -(1.033X463.8-278 )x10 3
3
/o.79
3
=(354.44 - 24295)xl0 = 11149xl0 J/kg
2.648x10 6
- - -x- - - - 1.
11 49 103
Air rate , w 0
23.75 kg per hp hr (answer)
Actua l exhaust temperatu re= 1028- (1028 - 652.5)(0.82) = 720.1 K (answer)
10. An open-cycle regen erative gas turbine plant has: p1, 1.02 kg/cm 2 ab; t 1, 32.2 C; p2, 7.03 kg/cm 2 ab;
t3, 816 C; 1/r, 0.85; TJc, 0.84; fuel, Bunker Coil; combustion efficiency, 90%; Ee., 0.60. Find 1/c and w0 •
Solution:
1°i = 32.2+273.15 = 305.35K
= 816 + 273.15 = 1089.15K
z:!
T2 =T1
(f!l.J
P1
tA - 1
7 03
r =305.35 ( · ) u
1.02
=530K
The actua l Ti is higher on account of compressor losses which add to the entha lpy of the
comp ressed air.
Real Ti leaving compresso r and entering combustion chamber = 305.35 + (530 - 305.35)/0.84 =
572.79 K.
Assume/= 0.015 .
@ 1089.15 K, Cp= 1.19 J/g-K
@ 572.79 K, Cp = 1.07 J/g-K
Average Cp =1.13 for f =0.015 and temperature range 1089.15 - 572.79 K
Q1 needed to raise products to 1089.15 K = (1089.15 - 572.79)(1.13) = 583.49 J/g
Fuel-air ratio= 583.49 / (43,155 x 0.90) = 0.015
By means of a preliminary t rial it is estimated that the ideal T4 will approximate 670 K. Now Fig. 7 .2
is used to obtai n the me.an cp for expansion.
@ 1089 .15 K, Cp = 1.19 J/g-K
@ 670 K, Cp = 1.095 J/g-K
Average cP = 1.143 for f = 0.015 and temperature range 1089.15 - 670 K. Then
Cp = 29 x 1.143 = 33.147 J per mol -C
And
r
33.147
1.337
33.147 - 8.345
Calculate idea l 74 =T3
p1
( Pi
J7
=1089.15 ( -102
· - )L~:;l =670K
7.03
WT =102{1+ f,:p(~ - T.)
Wr = 102(1 +o.0 1sX1.143X108s 1s - 610)
\IVr =49,600kg-m/kg
16
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lOMoARcPSD|33011865
CHAPTER 7 - GAS TURBINE POWER PLANT
@ 530 K, Cp = 1.055 J/g-K
1.02 J/g-K
1.038 for/= 0.015 and temperature range 530 - 305.35 K
@ 305.35 K, Cp =
Average
Cp =
li¼: =102c/ J;-7i)
WC= 102(1.038X530-30S.35)
We =23,78Skg •m/kg
1/rTJcW,-We
T/,
102TJca ,
(o.s5Xo.s4X49,600)-23,1s5
102(0.84X583.49)
T/,
T/, =0.2326=23.26%
Including combustion losses, plant efficiency= 23.26% x 0.90 = 20.93% (answer)
Net output per kg air flow
=(1.015X1.143X1089.15-670X0.8S)x10 3 -(1.038XS30-305.3S)x 103 /0.84
=(486.28-277.60)x10
Air rate,
3
=208.68 X103 J/kg
2.648X10 6
208 68 10
w. - - -.- -x- - -3 12.69 kg per hp hr
(answer)
11. To the plant described in Prob. 9 is added a regenerator of 50% effectiveness. Calculate the plant
efficiency.
Solution:
T4• -T4 ,
E
"" r•. -Tr
Continue Prob. 9
Actua l exhaust temperature, T4 , = 720.09 K
Tr=
E
"'
513.2 K
T4 , -T4 ,
T4, -T2•
o.50 = 120.09-r ••
720.09-513.2
T.· =616 .65K
cpo (r2•
-
r2• )= (1 + J)c119 (T•. - r•• )
f = 0.0142 and using Fig. 7-2
T4,= 720.09 K, Cp = 1.105 J/g-C
T4x = 616.65 K, Cp = 1.08 J/g-C
Average Cp = 1.0925 J/g-C
At
T2•=
513.2 K, Cp = 1.03 J/g-C
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lOMoARcPSD|33011865
CHAPTER 7 - GAS TURBINE POWER PLANT
Trial T;,, = 625 K, Cp = 1.06 J/g-C
Average cp = 1.045 J/g-C
(1.045XT2,-513.2)=(1+0.0142)(1.0925X720.09-616.65)
T;,, = 622.9 K
!= 0.0142
622.9 K = 1.075 J/g-C
1028 K = 1.18 J/g-C
Average cP = 1.1275 J/g-C
Cp at
Cp at
a1 needed to raise prod ucts to 1028 K
=(1028- 622.9)(1.1275) =456.75 J/g
T/,
T/,
T/,TJ,W, -W,
1021'/cO,
(o.82Xo.19 X44,089 )-19,577
102(0.19X456.75)
T/i =0.2 441=24.4 1%
Including combustion losses, plant efficiency= 24.41% x 0.94 = 22.95% (answer)
12. An intercooler of 90% effectiveness is added to the plant of Prob. 10 at p =
available at 21 C. Find T/r•
Solution:
..
E
T,_•• -T,b
T;.-. -Tw
Tw =21+273.15=294.lSK
p1 =1.02kg /cm 2 abs
p2 = 7.03 kg/ cm 2 abs
= Pib = .JP1P2
P10 =p1 b =.J,.....(1-.0.....,2)..,...(7-.0--.-3)
=2.68 kg/cm 2 abs
P io
w; = 102c.(r
2• -
rtb)+102c.(r,_ .• - ri)
T,_ =305.35K
z.:!
l.4- 1
2 68 14
T. =T.( p•• ) r =305.35( · )
=40241K
lo
l
P1
1.02
18
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Cooling wate r
lOMoARcPSD|33011865
CHAPTER 7 - GAS TURBINE POWER PLANT
0.84
402.41 - 305.35
T1,0 -305.35
I;_.0 =420 .90K
420.90 -T,b
420. 90 - 294.15
Tlb = 306.83K
ti
0.90
1.4-1
7 03
72 =T;_b(..E!...Jr =306.83 ( ·
P,b
2.68
T/c =
0.84
)""i,
= 404.17 K
72 -Tlb
72. -T,b
404.17-306.83
T,: -306.83
T,: = 422. 71 K
Figure 7 .2, f = 0.000
At T,: =422 .71K , Cp = 1.02 J/g-C
T,b =306.83K,
Cp
= 1.005 J/g-C
Average cp = 1.0125 J/g-C
At T,,0= 420.90 K , Cp = 1.02 J/g-C
I;_= 305.35 K, Cp = 1.005 J/g-C
Average cp = 1.0125 J/g-C
w; = 102(1.0125 X422. 11- 306.83) + 102(1.0125 )(420.90 - 3o5. 25)
w; =23,90lkg-m/kg
Trial/= 0.019
At ¼=1089.lSK,
Cp = 1.2 J/g-C
Ti =422 . 71K, Cp = 1.045 J/g-C
Average cp = 1.1225 J/g-C
a1 needed to raise products to 1089.15 K
= (1089.15- 422.71)(1.122.5) = 748.1 J/g
Fuel-air ratio = 748.1 / (43,155 x 0.90):: 0.019 ok.
T/,
T/,
-W;
1JrW7
1020,
(0.85)(49,600)-23,901
102(748.1)
0.2393=23.93%
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lOMoARcPSD|33011865
CHAPTER 7 - GAS TURBINE POWER PLANT
Including combustion losses , plant efficiency= 23.93% x 0.90 = 21.54% (answer)
13. Graph the p-v cycle of Prob. 10 with scales of 1 cm = 1 kg/cm 2 and 1 cm = 50 cm 3• Quantity= 1 kg ai r
at intake. Conside r that compressor and turbine losses appear as constant pressure reheats
following isentropic processes.
Solut ion:
p1 = 1.02 kg/cm 2
7; =305.35K
II.= mRT1
1
Pi
8 345
R= ·
=0.288kJ'k · C
28.97
I g
v.
(1Xo.28sX30S.3sX100)
cm
879 3
1.Q2X9.81
1
I
l
1
1 02
3
V2 =( Pi ) V1 =( · )"i:4(879)= 221cm
p2
7.03
Tt
=572. 79 K
Vz· r;_.
- =Vz 7;_
vi'
= s1219
221
530
v2• = 239cm 3
p3 = p2 =7.03kg/cm
3
fa = 1089.15 K
8 345
=0.2878kJ/kg ·C
R= ·
29
V m(l+ f)RT3 (1X1+0.01sxo.2s1sx1089.1SX100)
3
p3
7.03X9.81
V3 = 461cm
3
I
1
1
7 03
v. =(!!J..
) v 3 =( ·
p,
1.02
).1.m(461)= 1953cm
3
v., r•.
v. r.
-=-
r., = Ta - 11r (r3 - r.)
T.· = 1089.15 - 0.85(1089.15 - 670)
20
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