10-1 Chapter 10 VAPOR AND COMBINED POWER CYCLES Carnot Vapor Cycle 10-1C Because excessive moisture in steam causes erosion on the turbine blades. The highest moisture content allowed is about 10%. 10-2C The Carnot cycle is not a realistic model for steam power plants because (1) limiting the heat transfer processes to two-phase systems to maintain isothermal conditions severely limits the maximum temperature that can be used in the cycle, (2) the turbine will have to handle steam with a high moisture content which causes erosion, and (3) it is not practical to design a compressor that will handle two phases. 10-3E A steady-flow Carnot engine with water as the working fluid operates at specified conditions. The thermal efficiency, the quality at the end of the heat rejection process, and the net work output are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) We note that and 19.3% R 1 . 833 R 0 . 672 1 1 R 0 . 672 F 0 . 212 R 1 . 833 F 1 . 373 C th, psia 14.7 @ sat psia 180 @ sat H L L H T T T T T T T 14.7 psia 180 psia q in 1 4 2 3 (b) Noting that s 4 = s 1 = s f @ 180 psia = 0.53274 Btu/lbm·R, 0.153 44441 . 1 31215 . 0 53274 . 0 4 4 fg f s s s x s (c) The enthalpies before and after the heat addition process are Btu/lbm 2 . 1112 16 . 851 90 . 0 14 . 346 Btu/lbm 14 . 346 2 2 psia 180 @ 1 fg f f h x h h h h Thus, and, Btu/lbm 148.1 Btu/lbm 0 . 766 1934 . 0 Btu/lbm 0 . 766 14 . 346 2 . 1112 in th net 1 2 in q w h h q
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
10-1
Chapter 10 VAPOR AND COMBINED POWER CYCLES
Carnot Vapor Cycle
10-1C Because excessive moisture in steam causes erosion on the turbine blades. The highest moisturecontent allowed is about 10%.
10-2C The Carnot cycle is not a realistic model for steam power plants because (1) limiting the heattransfer processes to two-phase systems to maintain isothermal conditions severely limits the maximumtemperature that can be used in the cycle, (2) the turbine will have to handle steam with a high moisturecontent which causes erosion, and (3) it is not practical to design a compressor that will handle two phases.
10-3E A steady-flow Carnot engine with water as the working fluid operates at specified conditions. Thethermal efficiency, the quality at the end of the heat rejection process, and the net work output are to bedetermined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) We note that
and
19.3%R1.833R0.67211
R0.672F0.212R1.833F1.373
Cth,
psia14.7@sat
psia180@sat
H
L
L
H
TT
TTTT
T
14.7 psia
180 psiaqin
1
4
2
3(b) Noting that s4 = s1 = sf @ 180 psia = 0.53274 Btu/lbm·R,
0.15344441.1
31215.053274.044
fg
f
sss
x s
(c) The enthalpies before and after the heat addition process are
Btu/lbm2.111216.85190.014.346Btu/lbm14.346
22
psia180@1
fgf
f
hxhhhh
Thus,
and,Btu/lbm148.1Btu/lbm0.7661934.0
Btu/lbm0.76614.3462.1112
inthnet
12in
qw
hhq
10-2
10-4 A steady-flow Carnot engine with water as the working fluid operates at specified conditions. Thethermal efficiency, the amount of heat rejected, and the net work output are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) Noting that TH = 250°C = 523 K and TL = Tsat @ 20 kPa = 60.06°C = 333.1 K, the thermalefficiency becomes
36.3%3632.0K523K333.1
11Cth,H
L
TT
T
qout
qin
1
4
2
320 kPa
(b) The heat supplied during this cycle is simply the enthalpy of vaporization ,
Thus,
kJ/kg1092.3kJ/kg3.1715K523K333.1
kJ/kg3.1715
inout
250@in
qTT
qq
hq
H
LL
Cfg
250 C
s(c) The net work output of this cycle is
kJ/kg623.0kJ/kg3.17153632.0inthnet qw
10-5 A steady-flow Carnot engine with water as the working fluid operates at specified conditions. Thethermal efficiency, the amount of heat rejected, and the net work output are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) Noting that TH = 250°C = 523 K and TL = Tsat @ 10 kPa = 45.81°C = 318.8 K, the thermalefficiency becomes
39.04%K523K318.811Cth,
H
L
TT
T
qout
qin
1
4
2
310 kPa
(b) The heat supplied during this cycle is simply the enthalpy of vaporization ,
Thus,
kJ/kg1045.6kJ/kg3.1715K523K318.8
kJ/kg3.1715
inout
C250@in
qTT
qq
hq
H
LL
fg
250 C
s(c) The net work output of this cycle is
kJ/kg669.7kJ/kg3.17153904.0inthnet qw
10-3
10-6 A steady-flow Carnot engine with water as the working fluid operates at specified conditions. Thethermal efficiency, the pressure at the turbine inlet, and the net work output are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) The thermal efficiency is determined from
(c) The net work can be determined by calculating the enclosed area on the T-s diagram,
Thus,kJ/kg16235390.11368.760350Area
KkJ/kg5390.10769.71.08313.0
43net
44
ssTTw
sxss
LH
fgf
The Simple Rankine Cycle
10-7C The four processes that make up the simple ideal cycle are (1) Isentropic compression in a pump,(2) P = constant heat addition in a boiler, (3) Isentropic expansion in a turbine, and (4) P = constant heatrejection in a condenser.
10-10C The pump work remains the same, the moisture content decreases, everything else increases.
10-11C The actual vapor power cycles differ from the idealized ones in that the actual cycles involvefriction and pressure drops in various components and the piping, and heat loss to the surrounding mediumfrom these components and piping.
10-12C The boiler exit pressure will be (a) lower than the boiler inlet pressure in actual cycles, and (b) thesame as the boiler inlet pressure in ideal cycles.
10-13C We would reject this proposal because wturb = h1 - h2 - qout, and any heat loss from the steam willadversely affect the turbine work output.
10-14C Yes, because the saturation temperature of steam at 10 kPa is 45.81°C, which is much higher thanthe temperature of the cooling water.
10-4
10-15 A steam power plant operates on a simple ideal Rankine cycle between the specified pressure limits. The thermal efficiency of the cycle and the net power output of the plant are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),
10-16 A steam power plant that operates on a simple ideal Rankine cycle is considered. The quality of thesteam at the turbine exit, the thermal efficiency of the cycle, and the mass flow rate of the steam are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),
10-17 A steam power plant that operates on a simple nonideal Rankine cycle is considered. The quality ofthe steam at the turbine exit, the thermal efficiency of the cycle, and the mass flow rate of the steam are tobe determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),
10-18E A steam power plant that operates on a simple ideal Rankine cycle between the specified pressurelimits is considered. The minimum turbine inlet temperature, the rate of heat input in the boiler, and thethermal efficiency of the cycle are to be determined.Assumptions 1 Steady operating conditions exist.2 Kinetic and potential energy changes are negligible.
T
4x4 = 0.9
2
3
1 Qout
2 psia ·
Qin
1250 psia·
Analysis (a) From the steam tables (Tables A-4E,A-5E, and A-6E),
Btu/lbm77.9775.302.94Btu/lbm75.3
ftpsia5.4039Btu1psia21250/lbmft0.01623
/lbmft01623.0
Btu/lbm02.94
in,12
33
121in,
3psia2@1
psia2@1
p
p
f
f
whh
PPw
hh
v
vv
s
F13373
3
43
3
44
44
Btu/lbm4.1693psia1250
RBtu/lbm7450.174444.19.017499.0
Btu/lbm6.10137.10219.002.94
Th
ssP
sxss
hxhh
fgf
fgf
(b) Btu/s119,67297.771693.4lbm/s7523in hhmQ
(c)
42.4%Btu/s119,672Btu/s68,96711
Btu/s68,96794.021013.6lbm/s75
in
out
14out
QQ
hhmQ
th
10-8
10-19E A steam power plant operates on a simple nonideal Rankine cycle between the specified pressurelimits. The minimum turbine inlet temperature, the rate of heat input in the boiler, and the thermalefficiency of the cycle are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kineticand potential energy changes are negligible. T
Qout·
2s
s
1250 psia
Qin·
4s 4x4 = 0.9
2
3
12 psia
Analysis (a) From the steam tables (Tables A-4E, A-5E,and A-6E),
The turbine inlet temperature is determined by trial and error ,
Try 1:
8171.04.9180.14396.10130.1439
Btu/lbm4.9187.10218069.002.94
8069.074444.1
17499.05826.1
Btu/lbm.R5826.1Btu/lbm0.1439
F900psia1250
43
43
44
344
3
3
3
3
sT
fgsfs
fg
f
fg
fss
hhhh
hxhh
sss
sss
x
sh
TP
Try 2:
8734.03.9436.14986.10136.1498
Btu/lbm3.9437.10218312.002.94
8312.074444.1
17499.06249.1
Btu/lbm.R6249.1Btu/lbm6.1498
F1000psia1250
43
43
44
344
3
3
3
3
sT
fgsfs
fg
f
fg
fss
hhhh
hxhh
sss
sss
x
sh
TP
By linear interpolation, at T = 0.85 we obtain T3 = 958.4 F. This is approximate. We can determine state 3exactly using EES software with these results: T3 = 955.7 F, h3 = 1472.5 Btu/lbm.
(b) Btu/s103,05598.431472.5lbm/s7523in hhmQ
(c)
33.1%Btu/s103,055Btu/s68,96911
Btu/s969,6894.021013.6lbm/s75
in
outth
14out
QQ
hhmQ
10-9
10-20 A 300-MW coal-fired steam power plant operates on a simple ideal Rankine cycle between thespecified pressure limits. The overall plant efficiency and the required rate of the coal supply are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),
T
kJ/kg.0327707.596.271kJ/kg5.07
mkPa1kJ1
kPa255000/kgm00102.0
/kgm000102.0
kJ/kg96.271
in,12
33
121in,
3kPa25@1
kPa25@1
p
p
f
f
whh
PPw
hh
v
vv
kJ/kg2.22765.23458545.096.271
8545.09370.6
8932.08210.6kPa25
KkJ/kg8210.6kJ/kg2.3317
C450MPa5
44
44
34
4
3
3
3
3
fgf
fg
f
hxhh
sss
xss
P
sh
TP
4
2
3
1 Q·25 kPa
out
Qin
5 MPa ·
s
The thermal efficiency is determined from
kJ/kg2.200496.2712.2276kJ/kg2.304003.2772.3317
14out
23inhhqhhq
and
Thus,24.5%96.075.03407.0
3407.02.30402.200411
gencombthoverall
in
outth q
q
(b) Then the required rate of coal supply becomes
and
tons/h150.3tons/s04174.0kg1000
ton1kJ/kg29,300
kJ/s1,222,992
kJ/s992,222,12453.0
kJ/s300,000
coal
incoal
overall
netin
CQ
m
WQ
10-10
10-21 A solar-pond power plant that operates on a simple ideal Rankine cycle with refrigerant-134a as theworking fluid is considered. The thermal efficiency of the cycle and the power output of the plant are to bedetermined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the refrigerant tables (Tables A-11, A-12, and A-13),
10-22 A steam power plant operates on a simple ideal Rankine cycle between the specified pressure limits. The thermal efficiency of the cycle, the mass flow rate of the steam, and the temperature rise of the coolingwater are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),
(c) The rate of heat rejection to the cooling water and its temperature rise are
C8.4CkJ/kg4.18kg/s2000
kJ/s70,586)(
kJ/s,58670kJ/kg1961.8kg/s35.98
watercooling
outwatercooling
outout
cmQ
T
qmQ
10-12
10-23 A steam power plant operates on a simple nonideal Rankine cycle between the specified pressurelimits. The thermal efficiency of the cycle, the mass flow rate of the steam, and the temperature rise of the cooling water are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),
(c) The rate of heat rejection to the cooling water and its temperature rise are
C10.5CkJ/kg4.18kg/s2000
kJ/s88,051)(
kJ/s,05188kJ/kg2125.3kg/s41.43
watercooling
outwatercooling
outout
cmQ
T
qmQ
10-13
10-24 The net work outputs and the thermal efficiencies for a Carnot cycle and a simple ideal Rankinecycle are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) Rankine cycle analysis: From the steam tables (Tables A-4, A-5, and A-6),
10-25 A binary geothermal power operates on the simple Rankine cycle with isobutane as the workingfluid. The isentropic efficiency of the turbine, the net power output, and the thermal efficiency of the cycleare to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Properties The specific heat of geothermal water is taken to be 4.18 kJ/kg.ºC.Analysis (a) We need properties of isobutane,which are not available in the book. However,we can obtain the properties from EES. Turbine:
10-26 A single-flash geothermal power plant uses hot geothermal water at 230ºC as the heat source. Themass flow rate of steam through the turbine, the isentropic efficiency of the turbine, the power output fromthe turbine, and the thermal efficiency of the plant are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) We use properties of water forgeothermal water (Tables A-4 through A-6)
1661.02108
09.64014.990kJ/kg14.990
kPa500
kJ/kg14.9900
C230
22
12
2
11
1
fg
f
hhh
xhh
P
hxT
The mass flow rate of steam through the turbine is kg/s38.20kg/s)230)(1661.0(123 mxm
10-27 A double-flash geothermal power plant uses hot geothermal water at 230ºC as the heat source. Thetemperature of the steam at the exit of the second flash chamber, the power produced from the secondturbine, and the thermal efficiency of the plant are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) We use properties of water for geothermal water (Tables A-4 through A-6)
1661.0kJ/kg14.990
kPa500
kJ/kg14.9900
C230
212
2
11
1
xhh
P
hxT
kg/s80.1911661.0230kg/s38.20kg/s)230)(1661.0(
316
123
mmmmxm
Flashchamber
separator
9
8
7
Flashchamber
6
5
4
3
2condenser
1
steamturbine
separator
reinjectionwell
productionwell
kJ/kg7.234490.0kPa10
kJ/kg1.27481
kPa500
44
4
33
3
hxP
hxP
kJ/kg1.26931
kPa150
0777.0kPa150
kJ/kg09.6400
kPa500
88
8
7
7
67
7
66
6
hxP
xT
hhP
hxP
C111.35
(b) The mass flow rate at the lower stage of the turbine is kg/s.9014kg/s)80.191)(0777.0(678 mxm
The power outputs from the high and low pressure stages of the turbine are
10-28 A combined flash-binary geothermal power plant uses hot geothermal water at 230ºC as the heatsource. The mass flow rate of isobutane in the binary cycle, the net power outputs from the steam turbineand the binary cycle, and the thermal efficiencies for the binary cycle and the combined plant are to bedetermined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) We use properties of water for geothermal water (Tables A-4 through A-6)
10-29C The pump work remains the same, the moisture content decreases, everything else increases.
10-30C The T-s diagram of the ideal Rankinecycle with 3 stages of reheat is shown on theside. The cycle efficiency will increase as thenumber of reheating stages increases.
10-31C The thermal efficiency of the simple ideal Rankine cycle will probably be higher since the averagetemperature at which heat is added will be higher in this case.
1
I
2
10 s
T
3
4
5
6
7
8
II III9
10-19
10-32 [Also solved by EES on enclosed CD] A steam power plant that operates on the ideal reheat Rankinecycle is considered. The turbine work output and the thermal efficiency of the cycle are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis From the steam tables (Tables A-4, A-5, and A-6),
T
kJ/kg.5425912.842.251
kJ/kg8.12mkPa1
kJ1kPa208000/kgm001017.0
/kgm700101.0
kJ/kg42.251
in,12
33
121in,
3kPa20@1
kPa20@1
p
p
f
f
whh
PPw
hh
v
vv
kJ/kg2.23855.23579051.042.251
9051.00752.7
8320.02359.7kPa20
KkJ/kg2359.7kJ/kg2.3457
C500MPa3
kJ/kg1.3105MPa3
KkJ/kg7266.6kJ/kg5.3399
C500MPa8
66
66
56
6
5
5
5
5
434
4
3
3
3
3
fgf
fg
f
hxhh
sss
x
ssP
sh
TP
hss
P
sh
TP
20 kPa
8 MPa 4
3
6
2
5
1s
The turbine work output and the thermal efficiency are determined from
andkJ/kg0.34921.31052.345754.2595.3399
2.23852.34571.31055.3399
4523in
6543outT,
hhhhq
hhhhw kJ/kg1366.4
Thus,
38.9%kJ/kg3492.5kJ/kg1358.3
kJ/kg3.135812.84.1366
in
netth
in,,net
qw
www poutT
10-20
10-33 EES Problem 10-32 is reconsidered. The problem is to be solved by the diagram window data entryfeature of EES by including the effects of the turbine and pump efficiencies and reheat on the steam quality at the low-pressure turbine exit Also, the T-s diagram is to be plotted.Analysis The problem is solved using EES, and the solution is given below.
"Pump analysis"function x6$(x6) "this function returns a string to indicate the state of steam at point 6" x6$=''
if (x6>1) then x6$='(superheated)'if (x6<0) then x6$='(subcooled)'
end
Fluid$='Steam_IAPWS'
P[1] = P[6] P[2]=P[3]x[1]=0 "Sat'd liquid"h[1]=enthalpy(Fluid$,P=P[1],x=x[1])v[1]=volume(Fluid$,P=P[1],x=x[1])s[1]=entropy(Fluid$,P=P[1],x=x[1])T[1]=temperature(Fluid$,P=P[1],x=x[1])W_p_s=v[1]*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume"W_p=W_p_s/Eta_ph[2]=h[1]+W_p "SSSF First Law for the pump"v[2]=volume(Fluid$,P=P[2],h=h[2])s[2]=entropy(Fluid$,P=P[2],h=h[2])T[2]=temperature(Fluid$,P=P[2],h=h[2])"High Pressure Turbine analysis"h[3]=enthalpy(Fluid$,T=T[3],P=P[3])s[3]=entropy(Fluid$,T=T[3],P=P[3])v[3]=volume(Fluid$,T=T[3],P=P[3])s_s[4]=s[3]hs[4]=enthalpy(Fluid$,s=s_s[4],P=P[4])Ts[4]=temperature(Fluid$,s=s_s[4],P=P[4])Eta_t=(h[3]-h[4])/(h[3]-hs[4])"Definition of turbine efficiency"T[4]=temperature(Fluid$,P=P[4],h=h[4])s[4]=entropy(Fluid$,T=T[4],P=P[4])v[4]=volume(Fluid$,s=s[4],P=P[4])h[3] =W_t_hp+h[4]"SSSF First Law for the high pressure turbine""Low Pressure Turbine analysis"P[5]=P[4]s[5]=entropy(Fluid$,T=T[5],P=P[5])h[5]=enthalpy(Fluid$,T=T[5],P=P[5])s_s[6]=s[5]hs[6]=enthalpy(Fluid$,s=s_s[6],P=P[6])Ts[6]=temperature(Fluid$,s=s_s[6],P=P[6])
10-21
vs[6]=volume(Fluid$,s=s_s[6],P=P[6])Eta_t=(h[5]-h[6])/(h[5]-hs[6])"Definition of turbine efficiency"h[5]=W_t_lp+h[6]"SSSF First Law for the low pressure turbine"x[6]=QUALITY(Fluid$,h=h[6],P=P[6])"Boiler analysis"Q_in + h[2]+h[4]=h[3]+h[5]"SSSF First Law for the Boiler""Condenser analysis"h[6]=Q_out+h[1]"SSSF First Law for the Condenser"T[6]=temperature(Fluid$,h=h[6],P=P[6])s[6]=entropy(Fluid$,h=h[6],P=P[6])x6s$=x6$(x[6])
10-34 A steam power plant that operates on a reheat Rankine cycle is considered. The quality (ortemperature, if superheated) of the steam at the turbine exit, the thermal efficiency of the cycle, and themass flow rate of the steam are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),
kJ/kg43.20262.1081.191
kJ/kg10.62
0.95/mkPa1
kJ1kPa1010,000/kgm0.00101
/
/kgm001010.0
kJ/kg81.191
in,12
33
121in,
3kPa10@1
kPa10@1
p
pp
f
f
whh
PPw
hh
v
vvT
KkJ/kg7642.7kJ/kg1.3479
C500MPa1
kJ/kg0.29027.27831.337580.01.3375
kJ/kg8.2783MPa1
KkJ/kg5995.6kJ/kg75.133
C500MPa10
5
5
5
5
433443
43
434
4
3
3
3
3
sh
TP
hhhhhhhh
hss
P
sh
TP
sTs
T
ss
s
2
610 kPa
10 MPa 4s4
3
6
2
5
1s
vapordsuperheatekJ/kg8.26642.24611.347980.01.3479
kJ/kg2.24611.23929487.081.191
exitturbineat9487.04996.7
6492.07642.7kPa10
655665
65
66
66
56
6
g
sTs
T
fgsfs
fg
fss
s
s
h
hhhhhhhh
hxhh
sss
xss
P
From steam tables at 10 kPa we read T6 = 88.1 C.(b)
kJ/kg8.127662.104.1287
kJ/kg8.37490.29021.347943.2021.3375
kJ/kg4.12878.26641.34790.29021.3375
inp,outT,net
4523in
6543outT,
www
hhhhq
hhhhw
Thus the thermal efficiency is
34.1%kJ/kg3749.8kJ/kg1276.8
in
netth q
w
(c) The mass flow rate of the steam is
kg/s62.7kJ/kg1276.9kJ/s80,000
net
net
wW
m
10-23
10-35 A steam power plant that operates on the ideal reheat Rankine cycle is considered. The quality (or temperature, if superheated) of the steam at the turbine exit, the thermal efficiency of the cycle, and themass flow rate of the steam are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),
kJ/kg90.20109.1081.191
kJ/kg10.09mkPa1
kJ1kPa01000,10/kgm00101.0
/kgm00101.0
kJ/kg81.191
in,12
33
121in,
3kPa10@1
kPa10@1
p
p
f
f
whh
PPw
hh
v
vvT
kJ/kg2.24611.23929487.081.191
exitturbineat4996.7
6492.07642.7kPa10
KkJ/kg7642.7kJ/kg1.3479
C500MPa1
kJ/kg8.2783MPa1
KkJ/kg5995.6kJ/kg1.3375
C500MPa10
66
66
56
6
5
5
5
5
434
4
3
3
3
3
fgf
fg
f
hxhh
sss
x
ssP
sh
TP
hss
P
sh
TP
0.9487
10 kPa
10 MPa 4
3
6
2
5
1s
(b)
kJ/kg3.159909.104.1609
kJ/kg5.38687.27831.347990.2011.3375
kJ/kg3.16092.24611.34797.27831.3375
in,outT,net
4523in
6543outT,
pwww
hhhhq
hhhhw
Thus the thermal efficiency is
41.3%kJ/kg3868.5kJ/kg1599.3
in
netth q
w
(c) The mass flow rate of the steam is
skg050 /.kJ/kg1599.3kJ/s80,000
net
net
wW
m
10-24
10-36E A steam power plant that operates on the ideal reheat Rankine cycle is considered. The pressure at which reheating takes place, the net power output, the thermal efficiency, and the minimum mass flow rateof the cooling water required are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the steam tables (Tables A-4E, A-5E, and A-6E),
Btu/lbm11.7239.272.69
Btu/lbm2.39ftpsia5.4039
Btu1psia1800/lbmft01614.0
F69.101
/lbmft01614.0
Btu/lbm72.69
in,12
33
121in,
psia1@sat1
3psia1@sat1
psia1@sat1
p
p
whh
PPw
TT
hh
v
vvT
1 psia
800 psia4
3
6
2
5
1s
Btu/lbm0.10617.10359572.072.69
9572.084495.1
13262.08985.1psia1
RBtu/lbm8985.1Btu/lbm4.1431
F800psia23.62
pressure)reheat(the
Btu/lbm5.1178
vaporsat.
RBtu/lbm6413.1Btu/lbm0.1456
F900psia800
66
66
56
6
5
5
5
5
@sat4
@434
3
3
3
3
4
4
fgf
fg
f
ss
ssg
hxhh
sss
xss
P
sh
TP
PP
hhss
sh
TP
g
g
psia62.23
(b)
Btu/lbm3.99172.690.1061
Btu/lbm8.16365.11784.143111.720.1456
16out
4523in
hhq
hhhhq
Thus,
39.4%Btu/lbm1636.8Btu/lbm991.3
11in
outth q
q
(c) The mass flow rate of the cooling water will be minimum when it is heated to the temperature of thesteam in the condenser, which is 101.7 F,
lbm/s641.0F45101.69FBtu/lbm1.0
Btu/s103.634
Btu/s10634.3Btu/s1063943.0114
outcool
44inthnetinout
TcQ
m
QWQQ
10-25
10-37 A steam power plant that operates on an ideal reheat Rankine cycle between the specified pressurelimits is considered. The pressure at which reheating takes place, the total rate of heat input in the boiler,and the thermal efficiency of the cycle are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),
10-38 A steam power plant that operates on a reheat Rankine cycle is considered. The condenser pressure,the net power output, and the thermal efficiency are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),
kJ/kg3.30271.29482.335885.02.3358
?
95.0?
KkJ/kg2815.7kJ/kg2.3358
C450MPa2
kJ/kg3.30271.29485.347685.05.3476
kJ/kg1.2948MPa2
KkJ/kg6317.6kJ/kg5.3476
C550MPa5.12
655665
65
656
6
66
6
5
5
5
5
4334
43
43
434
4
3
3
3
3
sTs
T
s
sT
sT
ss
hhhhhhhh
hss
P
hxP
sh
TP
hhhh
hhhh
hss
P
sh
TP
2
4
6P = ?
12.5 MPa4s
3
T
4
5
PumpCondenser
BoilerTurbine
21
6
3
s6s
2s
5
1
The pressure at state 6 may be determined by a trial-error approach from the steam tables or by using EESfrom the above equations:
P6 = 9.73 kPa, h6 = 2463.3 kJ/kg,(b) Then,
kJ/kg59.20302.1457.189
kJ/kg14.020.90/
mkPa1kJ1kPa73.912,500/kgm0.00101
/
/kgm001010.0
kJ/kg57.189
in,12
33
121in,
3kPa10@1
kPa73.9@1
p
pp
f
f
whh
PPw
hh
v
vv
Cycle analysis:
kW10,242kg2273.7)kJ/-.8kg/s)(36037.7()(
kJ/kg7.227357.1893.3027
kJ/kg8.36033.24632.33583.30275.3476
outinnet
16out
4523in
qqmW
hhq
hhhhq
(c) The thermal efficiency is
36.9%369.0kJ/kg3603.8kJ/kg2273.7
11in
outth q
q
10-27
Regenerative Rankine Cycle
10-39C Moisture content remains the same, everything else decreases.
10-40C This is a smart idea because we waste little work potential but we save a lot from the heat input. The extracted steam has little work potential left, and most of its energy would be part of the heat rejectedanyway. Therefore, by regeneration, we utilize a considerable amount of heat by sacrificing little workoutput.
10-41C In open feedwater heaters, the two fluids actually mix, but in closed feedwater heaters there is no mixing.
10-42C Both cycles would have the same efficiency.
10-43C To have the same thermal efficiency as the Carnot cycle, the cycle must receive and reject heat isothermally.Thus the liquid should be brought to the saturated liquidstate at the boiler pressure isothermally, and the steam mustbe a saturated vapor at the turbine inlet. This will requirean infinite number of heat exchangers (feedwater heaters),as shown on the T-s diagram.
Boilerexit
s
TBoilerinlet
qin
qout
10-28
10-44 A steam power plant that operates on an ideal regenerative Rankine cycle with an open feedwaterheater is considered. The net work output per kg of steam and the thermal efficiency of the cycle are to bedetermined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),
kJ/kg81.25139.042.251
kJ/kg0.39mkPa1
kJ1kPa20400/kgm001017.0
/kgm001017.0
kJ/kg42.251
in,12
33
121in,
3kPa20@1
kPa20@1
pI
pI
f
f
whh
PPw
hh
v
vv
T
kJ/kg0.22145.23578325.042.251
8325.00752.7
8320.07219.6kPa20
kJ/kg7.26654.21339661.066.604
9661.01191.5
7765.17219.6MPa4.0
KkJ/kg7219.6kJ/kg9.3302
C450MPa6
kJ/kg73.61007.666.604
kJ/kg6.07mkPa1
kJ1kPa4006000/kgm0.001084
/kgm001084.0
kJ/kg66.604
liquidsat.MPa4.0
77
77
57
7
66
66
56
6
5
5
5
5
in,34
33
343in,
3MPa4.0@3
MPa4.0@33
fgf
fg
f
fgf
fg
f
pII
pII
f
f
hxhh
sss
xss
P
hxhh
sss
xss
P
sh
TP
whh
PPw
hhP
v
vvqout
3 y
4
1-y
0.4 MPa
20 kPa
6MPa
6
5
7
2
qin
1s
The fraction of steam extracted is determined from the steady-flow energy balance equation applied to thefeedwater heater. Noting that Q W ,ke pe 0
326332266
(steady)0
11
0
hhyyhhmhmhmhmhm
EE
EEE
eeii
outin
systemoutin
where y is the fraction of steam extracted from the turbine ( /m m6 3 ). Solving for y,
And kJ/kg1016.57.16752.2692outinnet qqw(b) The thermal efficiency is determined from
37.8%kJ/kg2692.2kJ/kg1675.7
11in
outth q
q
10-29
10-45 A steam power plant that operates on an ideal regenerative Rankine cycle with a closed feedwaterheater is considered. The net work output per kg of steam and the thermal efficiency of the cycle are to bedetermined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),
kJ/kg50.25708.642.251
kJ/kg6.08mkPa1
kJ1kPa206000/kgm0.001017
/kgm001017.0
kJ/kg42.251
in,12
33
121in,
3kPa20@1
kPa20@1
pI
pI
f
f
whh
PPw
hh
v
vv
T
9
qou
3y
4
1-y
0.4 MPa
20 kPa
6MPa
6
5
7
28
qin
1
/kgm400108.0
kJ/kg66.604
liquidsat.MPa4.0
3MPa4.0@3
MPa4.0@33
f
fhhPvv
s
kJ/kg73.610
kJ/kg73.61007.666.604
kJ/kg07.6mkPa1
kJ1kPa4006000/kgm001084.0
938338
in,39
33
393in,
hPPhh
whh
PPw
pII
pII
v
v
Also, h4 = h9 = h8¸ = 610.73 kJ/kg since the two fluid streams which are being mixed have the sameenthalpy.
kJ/kg0.22145.23578325.042.251
8325.00752.7
8320.07219.6kPa20
kJ/kg7.26654.21339661.066.604
9661.01191.5
7765.17219.6MPa4.0
KkJ/kg7219.6kJ/kg9.3302
C450MPa6
77
77
57
7
66
66
56
6
5
5
5
5
fgf
fg
f
fgf
fg
f
hxhh
sss
xss
P
hxhh
sss
xss
P
sh
TP
The fraction of steam extracted is determined from the steady-flow energy balance equation applied to thefeedwater heater. Noting that Q ,0pekeW
3628366282
outin
(steady)0outin
1
0
hhyhhyhhmhhmhmhm
EE
EEE
eeii
system
where y is the fraction of steam extracted from the turbine ( /m m6 5 ). Solving for y,
1463.050.25773.61066.6047.2665
50.25773.610
2836
28
hhhhhh
y
Then,kJ/kg4.167542.2510.22141463.011
kJ/kg2.269273.6109.3302
17out
45in
hhyqhhq
And kJ/kg1016.84.16752.2692outinnet qqw(b) The thermal efficiency is determined from
37.8%kJ/kg2692.2kJ/kg1675.411
in
outth q
q
10-30
10-46 A steam power plant operates on an ideal regenerative Rankine cycle with two open feedwaterheaters. The net power output of the power plant and the thermal efficiency of the cycle are to bedetermined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis
P III P II P I
fwh fwh I Condenser
BoilerTurbine
6
5
43
21
10
9
8
7 T
10 MPa
1 - y
85
6
3y
4
0.2 MPa
5 kPa
0.6 MPa
91 - y - z
7
10
2
1s
(a) From the steam tables (Tables A-4, A-5, and A-6),
kJ/kg95.13720.075.137
kJ/kg02.0mkPa1
kJ1kPa5200/kgm0.001005
/kgm001005.0
kJ/kg75.137
in,12
33
121in,
3kPa5@1
kPa5@1
pI
pI
f
f
whh
PPw
hh
v
vv
kJ/kg10.35mkPa1
kJ1kPa60010,000/kgm0.001101
/kgm001101.0
kJ/kg38.670
liquidsat.MPa6.0
kJ/kg13.50542.071.504kJ/kg0.42
mkPa1kJ1
kPa200600/kgm0.001061
/kgm001061.0
kJ/kg71.504
liquidsat.MPa2.0
33
565in,
3MPa6.0@5
MPa6.0@55
in,34
33
343in,
3MPa2.0@3
MPa2.0@33
PPw
hhP
whh
PPw
hhP
pIII
f
f
pII
pII
f
f
v
vv
v
vv
kJ/kg7.2618
6.22019602.071.504
9602.05968.5
5302.19045.6
MPa2.0
kJ/kg8.2821MPa6.0
KkJ/kg9045.6kJ/kg8.3625
C600MPa10
kJ/kg73.68035.1038.670
99
99
79
9
878
8
7
7
7
7
in,56
fgf
fg
f
pIII
hxhh
sss
x
ssP
hss
P
sh
TP
whh
10-31
kJ/kg0.21050.24238119.075.137
8119.09176.7
4762.09045.6kPa5
1010
1010
710
10
fgf
fg
f
hxhh
sss
xss
P
The fraction of steam extracted is determined from the steady-flow energy balance equation applied to thefeedwater heaters. Noting that Q ,0pekeW
FWH-2:
548554488
outin
(steady)0outin
11
0
hhyyhhmhmhmhmhm
EE
EEE
eeii
system
where y is the fraction of steam extracted from the turbine ( /m m8 5 ). Solving for y,
07133.013.5058.282113.50538.670
48
45
hhhhy
FWH-1: 329332299 11 hyhzyzhhmhmhmhmhm eeii
where z is the fraction of steam extracted from the turbine ( /m m9 5 ) at the second stage. Solving for z,
10-47 [Also solved by EES on enclosed CD] A steam power plant operates on an ideal regenerativeRankine cycle with two feedwater heaters, one closed and one open. The mass flow rate of steam throughthe boiler for a net power output of 250 MW and the thermal efficiency of the cycle are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),
kJ/kg10.19229.081.191kJ/kg29.0
mkPa1kJ1
kPa10300/kgm0.00101
/kgm00101.0
kJ/kg81.191
in,12
33
121in,
3kPa10@1
kPa10@1
pI
pI
f
f
whh
PPw
hh
v
vv
kJ/kg0.27555.20479935.087.720
9935.06160.4
0457.26317.6MPa8.0
KkJ/kg6317.6kJ/kg5.3476
C550MPa5.12
kJ/kg727.83MPa12.5,
C4.170
/kgm001115.0
kJ/kg87.720
liquidsat.MPa8.0
kJ/kg52.57409.1343.561kJ/kg13.09
mkPa1kJ1
kPa30012,500/kgm0.001073
/kgm001073.0
kJ/kg43.561
liquidsat.MPa3.0
99
99
89
9
8
8
8
8
5556
MPa0.8@sat6
3MP8.0@6
MPa8.0@766
in,34
33
343in,
3MPa3.0@3
MPa3.0@33
fgf
fg
f
af
f
pII
pII
f
f
hxhh
sss
xss
P
sh
TP
hPTT
TT
hhhP
whh
PPw
hhP
vv
v
vv
1-y-zz
y
3Closed
fwh P II
P I
Openfwh
Condenser
BoilerTurbine
5
6
4
72
1
11
10
9
8
T
7
95
6
3
40.3 MPa
z10 kPa
0.8 MPa
12.5MPa
y
101 - y - z
8
11
2
1s
kJ/kg0.21001.23927977.081.191
7977.04996.7
6492.06317.kPa10
kJ/kg5.25785.21639323.043.561
9323.03200.5
6717.16317.6MPa3.0
1111
1111
811
11
1010
1010
810
10
fgf
fg
f
fgf
fg
f
hxhh
sss
x
ssP
hxhh
sss
x
ssP
The fraction of steam extracted is determined from the steady-flow energy balance equation applied to thefeedwater heaters. Noting that Q ,0pekeW
4569455699
outin
(steady)0systemoutin 0
hhhhyhhmhhmhmhm
EE
EEE
eeii
10-33
where y is the fraction of steam extracted from the turbine ( 510 / mm ). Solving for y,
0753.087.7200.275552.57483.727
69
45
hhhhy
For the open FWH,
310273310102277
outin
(steady)0systemoutin
11
0
hzhhzyyhhmhmhmhmhmhm
EE
EEE
eeii
where z is the fraction of steam extracted from the turbine ( /m m9 5 ) at the second stage. Solving for z,
10-48 EES Problem 10-47 is reconsidered. The effects of turbine and pump efficiencies on the mass flow rate and thermal efficiency are to be investigated. Also, the T-s diagram is to be plotted.Analysis The problem is solved using EES, and the solution is given below.
Fluid$='Steam_IAPWS'P[1] = P[11] P[2]=P[10]h[1]=enthalpy(Fluid$,P=P[1],x=0) {Sat'd liquid}v1=volume(Fluid$,P=P[1],x=0)s[1]=entropy(Fluid$,P=P[1],x=0)T[1]=temperature(Fluid$,P=P[1],x=0)w_pump1_s=v1*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume"w_pump1=w_pump1_s/Eta_pump "Definition of pump efficiency"h[1]+w_pump1= h[2] "Steady-flow conservation of energy"s[2]=entropy(Fluid$,P=P[2],h=h[2])T[2]=temperature(Fluid$,P=P[2],h=h[2])
"Open Feedwater Heater analysis"z*h[10] + y*h[7] + (1-y-z)*h[2] = 1*h[3] "Steady-flow conservation of energy"h[3]=enthalpy(Fluid$,P=P[3],x=0)T[3]=temperature(Fluid$,P=P[3],x=0) "Condensate leaves heater as sat. liquid at P[3]"s[3]=entropy(Fluid$,P=P[3],x=0)
"Boiler condensate pump or Pump 2 analysis"P[5]=P[8]P[4] = P[5] P[3]=P[10]v3=volume(Fluid$,P=P[3],x=0)w_pump2_s=v3*(P[4]-P[3])"SSSF isentropic pump work assuming constant specific volume"w_pump2=w_pump2_s/Eta_pump "Definition of pump efficiency"h[3]+w_pump2= h[4] "Steady-flow conservation of energy"s[4]=entropy(Fluid$,P=P[4],h=h[4])T[4]=temperature(Fluid$,P=P[4],h=h[4])
"Closed Feedwater Heater analysis"P[6]=P[9] y*h[9] + 1*h[4] = 1*h[5] + y*h[6] "Steady-flow conservation of energy"h[5]=enthalpy(Fluid$,P=P[6],x=0) "h[5] = h(T[5], P[5]) where T[5]=Tsat at P[9]"
10-35
T[5]=temperature(Fluid$,P=P[5],h=h[5]) "Condensate leaves heater as sat. liquid at P[6]"s[5]=entropy(Fluid$,P=P[6],h=h[5])h[6]=enthalpy(Fluid$,P=P[6],x=0)T[6]=temperature(Fluid$,P=P[6],x=0) "Condensate leaves heater as sat. liquid at P[6]"s[6]=entropy(Fluid$,P=P[6],x=0)
"Trap analysis"P[7] = P[10] y*h[6] = y*h[7] "Steady-flow conservation of energy for the trap operating as a throttle"T[7]=temperature(Fluid$,P=P[7],h=h[7])s[7]=entropy(Fluid$,P=P[7],h=h[7])
"Boiler analysis"q_in + h[5]=h[8]"SSSF conservation of energy for the Boiler"h[8]=enthalpy(Fluid$, T=T[8], P=P[8]) s[8]=entropy(Fluid$, T=T[8], P=P[8])
"Turbine analysis"ss[9]=s[8]hs[9]=enthalpy(Fluid$,s=ss[9],P=P[9])Ts[9]=temperature(Fluid$,s=ss[9],P=P[9])h[9]=h[8]-Eta_turb_hp*(h[8]-hs[9])"Definition of turbine efficiency for high pressure stages"T[9]=temperature(Fluid$,P=P[9],h=h[9])s[9]=entropy(Fluid$,P=P[9],h=h[9])ss[10]=s[8]hs[10]=enthalpy(Fluid$,s=ss[10],P=P[10])Ts[10]=temperature(Fluid$,s=ss[10],P=P[10])h[10]=h[9]-Eta_turb_ip*(h[9]-hs[10])"Definition of turbine efficiency for Intermediate pressure stages"T[10]=temperature(Fluid$,P=P[10],h=h[10])s[10]=entropy(Fluid$,P=P[10],h=h[10])ss[11]=s[8]hs[11]=enthalpy(Fluid$,s=ss[11],P=P[11])Ts[11]=temperature(Fluid$,s=ss[11],P=P[11])h[11]=h[10]-Eta_turb_lp*(h[10]-hs[11])"Definition of turbine efficiency for low pressure stages"T[11]=temperature(Fluid$,P=P[11],h=h[11])s[11]=entropy(Fluid$,P=P[11],h=h[11])h[8] =y*h[9] + z*h[10] + (1-y-z)*h[11] + w_turb "SSSF conservation of energy for turbine"
"Condenser analysis"(1-y-z)*h[11]=q_out+(1-y-z)*h[1]"SSSF First Law for the Condenser"
10-49 A steam power plant operates on an ideal reheat-regenerative Rankine cycle with an open feedwaterheater. The mass flow rate of steam through the boiler and the thermal efficiency of the cycle are to bedetermined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),
kJ/kg61.19280.081.191kJ/kg0.80
mkPa1kJ1
kPa10800/kgm0.00101
/kgm00101.0
kJ/kg81.191
in,12
33
121in,
3kPa10@1
kPa10@1
pI
pI
f
f
whh
PPw
hh
v
vv
T
kJ/kg7.24941.23929627.081.191
9627.04996.7
6492.08692.7kPa10
KkJ/kg8692.7kJ/kg3.3481
C500MPa8.0
kJ/kg1.2812MPa8.0
KkJ/kg7585.6kJ/kg0.3502
C550MPa10
kJ/kg12.73126.1087.720kJ/kg10.26
mkPa1kJ1
kPa80010,000/kgm0.001115
/kgm001115.0
kJ/kg87.720
liquidsat.MPa8.0
88
88
78
8
7
7
7
7
656
6
5
5
5
5
in,34
33
343in,
3MPa8.0@3
MPa8.0@33
fgf
fg
f
pII
pII
f
f
hxhh
sss
xss
P
sh
TP
hss
P
sh
TP
whh
PPw
hhP
v
vv
10 MPa
1 - y
63
4
y
10 kPa
0.8 MPa 7
5
8
2
1s
6
1-y7
6
P II P I
Openfwh
Condenser
BoilerTurbine
4
32
1
8
5
y
The fraction of steam extracted is determined from the steady-flow energy balance equation applied to thefeedwater heaters. Noting that Q ,0pekeW
326332266
outin(steady)0
systemoutin
11
0
hhyyhhmhmhmhmhm
EEEEE
eeii
where y is the fraction of steam extracted from the turbine ( /m m6 3 ). Solving for y,
10-50 A steam power plant operates on an ideal reheat-regenerative Rankine cycle with a closed feedwater heater. The mass flow rate of steam through the boiler and the thermal efficiency of the cycle are to bedetermined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis
9
10
Mixingchamber
1-y
7y
P II P I
Closedfwh
Condenser
BoilerTurbine
4
32
1
8
5
6
T
10 MPa
1 - y
6
3
410
9y
10 kPa
0.8 MPa 7
5
8
2
1s
(a) From the steam tables (Tables A-4, A-5, and A-6),
kJ/kg13.73126.1087.720kJ/kg10.26
mkPa1kJ1kPa80010,000/kgm0.001115
/kgm001115.0
kJ/kg87.720
liquidsat.MPa8.0
kJ/kg90.20109.1081.191kJ/kg10.09
mkPa1kJ1kPa1010,000/kgm0.00101
/kgm00101.0
kJ/kg81.191
in,34
33
343in,
3MP8.0@3
MPa8.0@33
in,12
33
121in,
3kPa10@1
kPa10@1
pII
pII
af
f
pI
pI
f
f
whh
PPw
hhP
whh
PPw
hh
v
vv
v
vv
Also, h4 = h9 = h10 = 731.12 kJ/kg since the two fluid streams that are being mixed have the same enthalpy.
kJ/kg7.24941.23929627.081.191
9627.04996.7
6492.08692.7kPa10
KkJ/kg8692.7kJ/kg3.3481
C500MPa8.0
kJ/kg7.2812MPa8.0
KkJ/kg7585.6kJ/kg0.3502
C550MPa10
88
88
78
8
7
7
7
7
656
6
5
5
5
5
fgf
fg
f
hxhh
sss
x
ssP
sh
TP
hss
P
sh
TP
10-39
The fraction of steam extracted is determined from the steady-flow energy balance equation applied to thefeedwater heaters. Noting that Q ,0pekeW
3629363292
outin
(steady)0systemoutin
1
0
hhyhhyhhmhhmhmhm
EE
EEE
eeii
where y is the fraction of steam extracted from the turbine ( /m m3 4 ). Solving for y,
10-51E A steam power plant operates on an ideal reheat-regenerative Rankine cycle with one reheater andtwo open feedwater heaters. The mass flow rate of steam through the boiler, the net power output of theplant, and the thermal efficiency of the cycle are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis
High-PTurbine
5P III
1-y-z1211
z
10
4
Openfwh II
1-y
y
P II P I
Openfwh I
Condenser
BoilerLow-PTurbine
6
32
1
8
7
9
T
9
10z
1500 81 - y5
6
3
y4 250 psia
1 psia
40 psia 140 psia
1 - y - z
11
7
12
2
1s
(a) From the steam tables (Tables A-4E, A-5E, and A-6E),
Btu/lbm84.6912.072.69Btu/lbm0.12
ftpsia5.4039Btu1
psia140/lbmft0.01614
/lbmft01614.0
Btu/lbm72.69
in,12
33
121in,
3psia1@1
psia1@1
pI
pI
f
f
whh
PPw
hh
v
vv
Btu/lbm41.38031.409.376Btu/lbm4.31
ftpsia5.4039Btu1
psia2501500/lbmft0.01865
/lbmft01865.0Btu/lbm09.376
liquidsat.psia250
Btu/lbm81.23667.014.236
Btu/lbm0.67ftpsia5.4039
Btu1psia40250/lbmft0.01715
/lbmft01715.0Btu/lbm14.236
liquidsat.psia40
in,56
33
565in,
3psia250@5
psia250@55
i,34
33
343in,
3psia40@3
psia40@33
pIII
pIII
f
f
npII
pII
f
f
whh
PPw
hhP
whh
PPw
hhP
v
vv
v
vv
Btu/lbm5.1308psia250
RBtu/lbm6402.1Btu/lbm5.1550
F1100psia1500
878
8
7
7
7
7
hss
P
sh
TP
10-41
Btu/lbm4.1052
7.10359488.072.69
9488.084495.1
13262.08832.1
psia1
Btu/lbm0.1356psia40
RBtu/lbm8832.1Btu/lbm3.1531
F1000psia140
Btu/lbm8.1248psia140
1212
1212
1012
12
111011
11
10
10
10
10
979
9
fgf
fg
f
hxhh
sss
x
ssP
hss
P
sh
TP
hss
P
The fraction of steam extracted is determined from the steady-flow energy balance equation applied to thefeedwater heaters. Noting that Q ,0pekeW
FWH-2:
548554488
outin
(steady)0systemoutin
11
0
hhyyhhmhmhmhmhm
EE
EEE
eeii
where y is the fraction of steam extracted from the turbine ( /m m8 5 ). Solving for y,
1300.081.2365.130881.23609.376
48
45
hhhh
y
FWH-1
321133221111
outin
(steady)0systemoutin
11
0
hyhzyzhhmhmhmhmhm
EE
EEE
eeii
where z is the fraction of steam extracted from the turbine ( /m m9 5 ) at the second stage. Solving for z,
10-52 A steam power plant that operates on an ideal regenerative Rankine cycle with a closed feedwater heater is considered. The temperature of the steam at the inlet of the closed feedwater heater, the mass flow rate of the steam extracted from the turbine for the closed feedwater heater, the net power output, and thethermal efficiency are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),
kJ/kg85.26543.1442.251
kJ/kg.431488.0/)kPa2012,500)(/kgm0.001017(
/
/kgm001017.0
kJ/kg42.251
in,12
3121in,
3kPa20@1
kPa20@1
pI
ppI
f
f
whh
PPw
hh
v
vvLow-Pturbine
910
11
MixingCham.
1-y7
y
PII PI
Closedfwh Cond.
Boiler
High-Pturbine
4
32
1
8
5
6
/kgm001127.0
kJ/kg51.762
liquidsat.MPa1
3MPa1@3
MPa1@33
f
fhhPvv
kJ/kg25.77773.1451.762kJ/kg73.14
88.0/)kPa001012,500)(/kgm001127.0(
/
in,311
3
3113in,
pII
ppII
whh
PPw v
Also, h4 = h10 = h11 = 777.25 kJ/kg since the two fluid streams which are being mixed have the sameenthalpy.
kJ/kg5.32206.31855.347688.05.3476
kJ/kg6.3185MPa5
KkJ/kg6317.6kJ/kg5.3476
C550MPa5.12
655665
65
656
6
5
5
5
5
sTs
T
s
hhhhhhhh
hss
P
sh
TP
C32888
8
877887
87
878
8
7
7
7
7
kJ/kg1.3111MPa1
kJ/kg1.31111.30519.355088.09.3550
kJ/kg1.3051MPa1
KkJ/kg1238.7kJ/kg9.3550
C550MPa5
ThP
hhhhhhhh
hss
P
sh
TP
sTs
T
s
10-43
kJ/kg2.24929.23479.355088.09.3550
kJ/kg9.2347kPa20
977997
97
979
9
sTs
T
s
hhhhhhhh
hss
P
The fraction of steam extracted from the low pressure turbine for closed feedwater heater is determinedfrom the steady-flow energy balance equation applied to the feedwater heater. Noting that
,Q W ke pe 0
1788.0)51.7621.3111()85.26525.777)(1(
1 38210
yyy
hhyhhy
The corresponding mass flow rate iskg/s4.29kg/s)24)(1788.0(58 mym
10-53C In the simple ideal Rankine cycle, irreversibilities occur during heat addition and heat rejectionprocesses in the boiler and the condenser, respectively, and both are due to temperature difference.Therefore, the irreversibilities can be decreased and thus the 2nd law efficiency can be increased byminimizing the temperature differences during heat transfer in the boiler and the condenser. One way ofdoing that is regeneration.
10-54 The exergy destructions associated with each of the processes of the Rankine cycle described in Prob. 10-15 are to be determined for the specified source and sink temperatures.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis From Problem 10-15,
kJ/kg8.1931kJ/kg72.2650
KkJ/kg5412.6
KkJ/kg0912.1
out
43
kPa50@21
qq
ss
sss
in
f
Processes 1-2 and 3-4 are isentropic. Thus, i12 = 0 and i34 = 0. Also,
kJ/kg351.3
kJ/kg1068
K290kJ/kg1931.85412.60912.1K290
K1500kJ/kg2650.80912.15412.6K290
41,41041destroyed,
23,23023destroyed,
R
R
R
R
Tq
ssTx
Tq
ssTx
10-55 The exergy destructions associated with each of the processes of the Rankine cycle described in Prob. 10-16 are to be determined for the specified source and sink temperatures.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis From Problem 10-16,
kJ/kg9.1897kJ/kg2.3173
KkJ/kg5995.6
KkJ/kg6492.0
out
in
43
kPa10@21
qq
ss
sss f
Processes 1-2 and 3-4 are isentropic. Thus, i12 = 0 and i34 = 0. Also,
kJ/kg172.3
kJ/kg1112.1
K290kJ/kg1897.9
5995.66492.0K290
K1500kJ/kg3173.2
6492.05995.6K290
41,41041destroyed,
23,23023destroyed,
R
R
R
R
Tq
ssTx
Tq
ssTx
10-45
10-56 The exergy destruction associated with the heat rejection process in Prob. 10-22 is to be determinedfor the specified source and sink temperatures. The exergy of the steam at the boiler exit is also to bedetermined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis From Problem 10-22,
kJ/kg8.1961kJ/kg4.3411
KkJ/kg8000.6
KkJ/kg6492.0
out
3
43
kPa10@21
qh
ss
sss f
The exergy destruction associated with the heat rejection process is
kJ/kg178.0K290kJ/kg1961.8
8000.66492.0K29041,41041destroyed,
R
R
Tq
ssTx
The exergy of the steam at the boiler exit is simply the flow exergy,
10-57 The exergy destructions associated with each of the processes of the reheat Rankine cycle described in Prob. 10-32 are to be determined for the specified source and sink temperatures.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis From Problem 10-32,
kJ/kg8.213342.2512.2385
kJ/kg1.3521.31052.3457
kJ/kg0.314054.2595.3399KkJ/kg2359.7KkJ/kg7266.6
KkJ/kg8320.0
16out
in,45
in,23
65
43
kPa20@21
hhq
q
qssss
sss f
Processes 1-2, 3-4, and 5-6 are isentropic. Thus, i12 = i34 = i56 = 0. Also,
kJ/kg212.6
kJ/kg94.1
kJ/kg1245.0
K300kJ/kg2133.82359.78320.0K300
K1800kJ/kg352.57266.62359.7K300
K1800kJ/kg3140.08320.07266.6K300
61,61061destroyed,
45,45045destroyed,
23,23023destroyed,
R
R
R
R
R
R
Tq
ssTx
Tq
ssTx
Tq
ssTx
10-46
10-58 EES Problem 10-57 is reconsidered. The problem is to be solved by the diagram window data entryfeature of EES by including the effects of the turbine and pump efficiencies. Also, the T-s diagram is to beplotted.Analysis The problem is solved using EES, and the solution is given below.
function x6$(x6) "this function returns a string to indicate the state of steam at point 6" x6$=''
if (x6>1) then x6$='(superheated)'if (x6<0) then x6$='(subcooled)'
end"Input Data - from diagram window"{P[6] = 20 [kPa] P[3] = 8000 [kPa] T[3] = 500 [C] P[4] = 3000 [kPa] T[5] = 500 [C] Eta_t = 100/100 "Turbine isentropic efficiency"Eta_p = 100/100 "Pump isentropic efficiency"}"Data for the irreversibility calculations:"T_o = 300 [K] T_R_L = 300 [K] T_R_H = 1800 [K] "Pump analysis"Fluid$='Steam_IAPWS'P[1] = P[6] P[2]=P[3]x[1]=0 "Sat'd liquid"h[1]=enthalpy(Fluid$,P=P[1],x=x[1])v[1]=volume(Fluid$,P=P[1],x=x[1])s[1]=entropy(Fluid$,P=P[1],x=x[1])T[1]=temperature(Fluid$,P=P[1],x=x[1])W_p_s=v[1]*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume"W_p=W_p_s/Eta_ph[2]=h[1]+W_p "SSSF First Law for the pump"v[2]=volume(Fluid$,P=P[2],h=h[2])s[2]=entropy(Fluid$,P=P[2],h=h[2])T[2]=temperature(Fluid$,P=P[2],h=h[2])"High Pressure Turbine analysis"h[3]=enthalpy(Fluid$,T=T[3],P=P[3])s[3]=entropy(Fluid$,T=T[3],P=P[3])v[3]=volume(Fluid$,T=T[3],P=P[3])s_s[4]=s[3]hs[4]=enthalpy(Fluid$,s=s_s[4],P=P[4])Ts[4]=temperature(Fluid$,s=s_s[4],P=P[4])Eta_t=(h[3]-h[4])/(h[3]-hs[4])"Definition of turbine efficiency"T[4]=temperature(Fluid$,P=P[4],h=h[4])s[4]=entropy(Fluid$,T=T[4],P=P[4])v[4]=volume(Fluid$,s=s[4],P=P[4])h[3] =W_t_hp+h[4]"SSSF First Law for the high pressure turbine""Low Pressure Turbine analysis"P[5]=P[4]s[5]=entropy(Fluid$,T=T[5],P=P[5])h[5]=enthalpy(Fluid$,T=T[5],P=P[5])s_s[6]=s[5]hs[6]=enthalpy(Fluid$,s=s_s[6],P=P[6])Ts[6]=temperature(Fluid$,s=s_s[6],P=P[6])vs[6]=volume(Fluid$,s=s_s[6],P=P[6])Eta_t=(h[5]-h[6])/(h[5]-hs[6])"Definition of turbine efficiency"h[5]=W_t_lp+h[6]"SSSF First Law for the low pressure turbine"x[6]=QUALITY(Fluid$,h=h[6],P=P[6])
10-47
"Boiler analysis"Q_in + h[2]+h[4]=h[3]+h[5]"SSSF First Law for the Boiler""Condenser analysis"h[6]=Q_out+h[1]"SSSF First Law for the Condenser"T[6]=temperature(Fluid$,h=h[6],P=P[6])s[6]=entropy(Fluid$,h=h[6],P=P[6])x6s$=x6$(x[6])"Cycle Statistics"W_net=W_t_hp+W_t_lp-W_pEff=W_net/Q_in"The irreversibilities (or exergy destruction) for each of the processes are:"q_R_23 = - (h[3] - h[2]) "Heat transfer for the high temperature reservoir to process 2-3"i_23 = T_o*(s[3] -s[2] + q_R_23/T_R_H)q_R_45 = - (h[5] - h[4]) "Heat transfer for the high temperature reservoir to process 4-5"i_45 = T_o*(s[5] -s[4] + q_R_45/T_R_H)q_R_61 = (h[6] - h[1]) "Heat transfer to the low temperature reservoir in process 6-1"i_61 = T_o*(s[1] -s[6] + q_R_61/T_R_L)i_34 = T_o*(s[4] -s[3])i_56 = T_o*(s[6] -s[5])i_12 = T_o*(s[2] -s[1])
10-59 The exergy destruction associated with the heat addition process and the expansion process in Prob. 10-34 are to be determined for the specified source and sink temperatures. The exergy of the steam at the boiler exit is also to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis From Problem 10-34,
kJ/kg8.3749kJ/kg1.3375
kJ/kg8.2664,kPa10KkJ/kg3870.8KkJ/kg7642.7
kJ/kg0.2902,MPa1KkJ/kg8464.6KkJ/kg5995.6
KkJ/kg6492.0
in
3
666
5
444
3
kPa10@21
qh
hPss
hPss
sss f
The exergy destruction associated with the combined pumping and the heat addition processes is
kJ/kg5.1289K1600kJ/kg3749.8
8464.67642.76492.05995.6K285
15,45130destroyed
R
R
Tq
ssssTx
The exergy destruction associated with the pumping process is
Thus,kJ/kg12895.05.1289
kJ/kg53.009.1062.10
12destroyed,destroyedheatingdestroyed,
,,,12destroyed,
xxx
Pvwwwx apspap
The exergy destruction associated with the expansion process is
kJ/kg247.9KkJ/kg7642.73870.85995.68464.6K285
036,
5634034destroyed,R
R
Tq
ssssTx
The exergy of the steam at the boiler exit is determined from
10-60 The exergy destruction associated with the regenerative cycle described in Prob. 10-44 is to bedetermined for the specified source and sink temperatures.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis From Problem 10-44, qin = 2692.2 kJ/kg and qout = 1675.7 kJ/kg. Then the exergy destructionassociated with this regenerative cycle is
kJ/kg1155K1500
kJ/kg2692.2K290kJ/kg1675.7
K290inout0destroyed,
HLcycle T
qTq
Tx
10-61 The exergy destruction associated with the reheating and regeneration processes described in Prob. 10-49 are to be determined for the specified source and sink temperatures.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis From Problem 10-49 and the steam tables,
kJ/kg6.6687.28123.3481
KkJ/kg6492.0KkJ/kg8692.7
KkJ/kg7585.6
KkJ/kg0457.22016.0
67reheat
kPa10@21
7
65
MPa8.0@3
hhq
ssss
ss
ssy
f
f
Then the exergy destruction associated with reheat and regeneration processes are
kJ/kg47.8
kJ/kg214.3
6492.02016.017585.62016.00457.2K290
1
K1800kJ/kg668.6
7585.68692.7K290
2630
0
0
surr0gen0regendestroyed,
67,670reheatdestroyed,
syyssTT
qsmsmTsTx
Tq
ssTx
iiee
R
R
10-50
10-62 A single-flash geothermal power plant uses hot geothermal water at 230ºC as the heat source. Thepower output from the turbine, the thermal efficiency of the plant, the exergy of the geothermal liquid atthe exit of the flash chamber, and the exergy destructions and exergy efficiencies for the flash chamber, theturbine, and the entire plant are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) We use properties of water for geothermal water (Tables A-4, A-5, and A-6)
10-63C The utilization factor of a cogeneration plant is the ratio of the energy utilized for a useful purposeto the total energy supplied. It could be unity for a plant that does not produce any power.
10-64C No. A cogeneration plant may involve throttling, friction, and heat transfer through a finite temperature difference, and still have a utilization factor of unity.
10-65C Yes, if the cycle involves no irreversibilities such as throttling, friction, and heat transfer through a finite temperature difference.
10-66C Cogeneration is the production of more than one useful form of energy from the same energysource. Regeneration is the transfer of heat from the working fluid at some stage to the working fluid atsome other stage.
10-52
10-67 A cogeneration plant is to generate power and process heat. Part of the steam extracted from theturbine at a relatively high pressure is used for process heating. The net power produced and the utilizationfactor of the plant are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis From the steam tables (Tables A-4, A-5, and A-6),
10-68E A large food-processing plant requires steam at a relatively high pressure, which is extracted fromthe turbine of a cogeneration plant. The rate of heat transfer to the boiler and the power output of thecogeneration plant are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis(a) From the steam tables (Tables A-4E, A-5E, and A-6E),
10-69 A cogeneration plant has two modes of operation. In the first mode, all the steam leaving the turbineat a relatively high pressure is routed to the process heater. In the second mode, 60 percent of the steam isrouted to the process heater and remaining is expanded to the condenser pressure. The power produced andthe rate at which process heat is supplied in the first mode, and the power produced and the rate of processheat supplied in the second mode are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),
kJ/kg47.65038.1009.640kJ/kg10.38
mkPa1kJ1
kPa50010,000/kgm0.001093
/kgm001093.0kJ/kg09.640
kJ/kg57.26115.1042.251kJ/kg10.15
mkPa1kJ1
kPa0210,000/kgm0.001017
/kgm001017.0kJ/kg42.251
inpII,34
33
343inpII,
3MPa5.0@3
MPa5.0@3
inpI,12
33
121inpI,
3kPa20@1
kPa20@1
whh
PPw
hh
whh
PPw
hh
f
f
f
f
v
vv
v
vv
7
3
2
4
P IIP I
Processheater Condens.
BoilerTurbine
5
1
8
6
T
Mixing chamber:E E E E E
m h m h m h m h m hi i e e
in out system (steady)
in out0
5 5 2 2 4 4
0
or, kJ/kg91.4945
47.650357.2612
5
44225 m
hmhmh
34
5
7
6
8
2
1s
KkJ/kg4219.6kJ/kg4.3242
C450MPa10
6
6
6
6sh
TP
kJ/kg0.21145.23577901.042.251
7901.00752.7
8320.04219.6kPa20
kJ/kg6.25780.21089196.009.640
9196.09603.4
8604.14219.6MPa5.0
88
88
68
8
77
77
67
7
fgf
fg
f
fgf
fg
f
hxhhs
ssx
ssP
hxhhs
ssx
ssP
When the entire steam is routed to the process heater,
kW9693
kW3319
kJ/kg09.6406.2578kg/s5
kJ/kg6.25784.3242kg/s5
377process
766outT,
hhmQ
hhmW
(b) When only 60% of the steam is routed to the process heater,
kW5816
kW4248
kJ/kg09.6406.2578kg/s3
kJ/kg0.21146.2578kg/s2kJ/kg6.25784.3242kg/s5
377process
878766outT,
hhmQ
hhmhhmW
10-55
10-70 A cogeneration plant modified with regeneration is to generate power and process heat. The massflow rate of steam through the boiler for a net power output of 15 MW is to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.AnalysisFrom the steam tables (Tables A-4, A-5, and A-6),
kJ/kg73.61007.666.604kJ/kg07.6
mkPa1kJ1kPa4006000/kgm0.001084
/kgm001084.0
kJ/kg66.604
kJ/kg20.19239.081.191kJ/kg0.39
mkPa1kJ1kPa10400/kgm0.00101
/kgm00101.0kJ/kg81.191
inpII,45
33
454inpII,
3MPa4.0@4
MPa4.0@943
inpI,12
33
121inpI,
3kPa10@1
kPa10@1
whh
PPw
hhhh
whh
PPw
hh
f
f
f
f
v
vv
v
vv
fwh4
7
3
2
9
P II P I
Processheater Condenser
BoilerTurbine
5
1
8
6
T
kJ/kg7.21281.23928097.081.191
8097.04996.7
6492.07219.6kPa10
kJ/kg7.26654.21339661.066.604
9661.01191.5
7765.17219.6MPa4.0
KkJ/kg7219.6kJ/kg9.3302
C450MPa6
88
88
68
8
77
77
67
7
6
6
6
6
fgf
fg
f
fgf
fg
f
hxhhs
ssx
ssP
hxhhs
ssx
ssP
sh
TP
5
0.4 MPa
10 kPa
6 MPa
7
6
8
23,4,9
1s
Then, per kg of steam flowing through the boiler, we have
kJ/kg852.0kJ/kg7.21287.26654.0kJ/kg7.26659.3302
4.0 8776outT, hhhhw
kJ/kg8.84523.60.852
kJ/kg6.23kJ/kg6.07kJ/kg0.394.0
4.0
inp,outT,net
inpII,inpI,inp,
www
www
Thus,
kg/s17.73kJ/kg845.8
kJ/s15,000
net
net
wWm
10-56
10-71 EES Problem 10-70 is reconsidered. The effect of the extraction pressure for removing steam fromthe turbine to be used for the process heater and open feedwater heater on the required mass flow rate is tobe investigated.Analysis The problem is solved using EES, and the solution is given below.
"Input Data"y = 0.6 "fraction of steam extracted from turbine for feedwater heater and process heater"P[6] = 6000 [kPa] T[6] = 450 [C] P_extract=400 [kPa] P[7] = P_extractP_cond=10 [kPa]P[8] = P_condW_dot_net=15 [MW]*Convert(MW, kW)Eta_turb= 100/100 "Turbine isentropic efficiency"Eta_pump = 100/100 "Pump isentropic efficiency"P[1] = P[8] P[2]=P[7]P[3]=P[7]P[4] = P[7] P[5]=P[6]P[9] = P[7]
"Condenser exit pump or Pump 1 analysis"Fluid$='Steam_IAPWS'
h[1]=enthalpy(Fluid$,P=P[1],x=0) {Sat'd liquid}v1=volume(Fluid$,P=P[1],x=0)s[1]=entropy(Fluid$,P=P[1],x=0)T[1]=temperature(Fluid$,P=P[1],x=0)w_pump1_s=v1*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume"w_pump1=w_pump1_s/Eta_pump "Definition of pump efficiency"h[1]+w_pump1= h[2] "Steady-flow conservation of energy"s[2]=entropy(Fluid$,P=P[2],h=h[2])T[2]=temperature(Fluid$,P=P[2],h=h[2])
"Open Feedwater Heater analysis:"z*h[7] + (1- y)*h[2] = (1- y + z)*h[3] "Steady-flow conservation of energy"h[3]=enthalpy(Fluid$,P=P[3],x=0)T[3]=temperature(Fluid$,P=P[3],x=0) "Condensate leaves heater as sat. liquid at P[3]"s[3]=entropy(Fluid$,P=P[3],x=0)
"Process heater analysis:"(y - z)*h[7] = q_process + (y - z)*h[9] "Steady-flow conservation of energy"Q_dot_process = m_dot*(y - z)*q_process"[kW]"h[9]=enthalpy(Fluid$,P=P[9],x=0)T[9]=temperature(Fluid$,P=P[9],x=0) "Condensate leaves heater as sat. liquid at P[3]"s[9]=entropy(Fluid$,P=P[9],x=0)
"Mixing chamber at 3, 4, and 9:"(y-z)*h[9] + (1-y+z)*h[3] = 1*h[4] "Steady-flow conservation of energy"T[4]=temperature(Fluid$,P=P[4],h=h[4]) "Condensate leaves heater as sat. liquid at P[3]"
10-57
s[4]=entropy(Fluid$,P=P[4],h=h[4])
"Boiler condensate pump or Pump 2 analysis"v4=volume(Fluid$,P=P[4],x=0)w_pump2_s=v4*(P[5]-P[4])"SSSF isentropic pump work assuming constant specific volume"w_pump2=w_pump2_s/Eta_pump "Definition of pump efficiency"h[4]+w_pump2= h[5] "Steady-flow conservation of energy"s[5]=entropy(Fluid$,P=P[5],h=h[5])T[5]=temperature(Fluid$,P=P[5],h=h[5])
"Boiler analysis"q_in + h[5]=h[6]"SSSF conservation of energy for the Boiler"h[6]=enthalpy(Fluid$, T=T[6], P=P[6]) s[6]=entropy(Fluid$, T=T[6], P=P[6])
"Turbine analysis"ss[7]=s[6]hs[7]=enthalpy(Fluid$,s=ss[7],P=P[7])Ts[7]=temperature(Fluid$,s=ss[7],P=P[7])h[7]=h[6]-Eta_turb*(h[6]-hs[7])"Definition of turbine efficiency for high pressure stages"T[7]=temperature(Fluid$,P=P[7],h=h[7])s[7]=entropy(Fluid$,P=P[7],h=h[7])ss[8]=s[7]hs[8]=enthalpy(Fluid$,s=ss[8],P=P[8])Ts[8]=temperature(Fluid$,s=ss[8],P=P[8])h[8]=h[7]-Eta_turb*(h[7]-hs[8])"Definition of turbine efficiency for low pressure stages"T[8]=temperature(Fluid$,P=P[8],h=h[8])s[8]=entropy(Fluid$,P=P[8],h=h[8])h[6] =y*h[7] + (1- y)*h[8] + w_turb "SSSF conservation of energy for turbine"
"Condenser analysis"(1- y)*h[8]=q_out+(1- y)*h[1]"SSSF First Law for the Condenser"
10-72E A cogeneration plant is to generate power while meeting the process steam requirements for a certain industrial application. The net power produced, the rate of process heat supply, and the utilizationfactor of this plant are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis(a) From the steam tables (Tables A-4E, A-5E, and A-6E),
Btu/lbm5.1229psia120
RBtu/lbm6348.1Btu/lbm0.1408
F800psia600
Btu/lbm49.208
737
7
6543
753
3
3
3
12
F240@1
hss
P
hhhh
sssh
TP
hhhh f
7
3 54
P
Processheater
Boiler Turbine
2
1
6
kW2260Btu/s2142Btu/lbm5.12290.1408lbm/s12
755net hhmW
(b)
Btu/s19,45049.208185.1229120.14086
117766
process
hmhmhm
hmhmQ eeii T
1
3,4,52
120 psia
600 psia
76
Btu/s19,4505.1229120.1408649.20818
776611process hmhmhmhmhmQ iiee
s(c) u = 1 since all the energy is utilized.
10-60
10-73 A cogeneration plant is to generate power and process heat. Part of the steam extracted from theturbine at a relatively high pressure is used for process heating. The mass flow rate of steam that must besupplied by the boiler, the net power produced, and the utilization factor of the plant are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
7
3
24P II P I
Processheater Condenser
BoilerTurbine
5
1
8
6 T
Qout·
3
5
4
10 kPa
7 MPa
0.6 MPa Qproces·
Qin·
7
6
8
2
1s
Analysis From the steam tables (Tables A-4, A-5, and A-6),
10-74C The energy source of the steam is the waste energy of the exhausted combustion gases.
10-75C Because the combined gas-steam cycle takes advantage of the desirable characteristics of the gascycle at high temperature, and those of steam cycle at low temperature, and combines them. The result is a cycle that is more efficient than either cycle executed operated alone.
10-62
10-76 A combined gas-steam power cycle is considered. The topping cycle is a gas-turbine cycle and thebottoming cycle is a simple ideal Rankine cycle. The mass flow rate of the steam, the net power output, andthe thermal efficiency of the combined cycle are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3Air is an ideal gas with constant specific heats.Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2).Analysis (a) The analysis of gas cycle yields
kW6447100,5547,11
kW11,547K3.6791500KkJ/kg1.005kg/s14
K3.679161K1500
kW5100K3005.662KkJ/kg1.005kg/s14
kW784,11K5.6621500KkJ/kg1.005kg/s14
K5.66216K300
gas,gas,gasnet,
87air87airgas,
4.1/4.0/1
7
878
56air56airgas,
67air67airin
4.1/4.0/1
5
656
CT
pT
kk
pC
p
kk
WWW
TTcmhhmW
PP
TT
TTcmhhmW
TTcmhhmQ
PP
TT
420 K
1500 K
STEAMCYCLE
GASCYCLE
7
8
9
5
Qin·
Qout·
6
15 kPa
10 MPa
3400 C
T
42
1
300 K
s
From the steam tables (Tables A-4, A-5, and A-6),
kJ/kg06.23613.1094.225
kJ/kg10.12mkPa1
kJ1kPa1510,000/kgm0.001014
/kgm001014.0kJ/kg94.225
inpI,12
33
121inpI,
3kPa15@1
kPa15@1
whh
PPw
hh
f
f
v
vv
kJ/kg8.20113.23727528.094.225
7528.02522.7
7549.02141.6kPa15
KkJ/kg2141.6kJ/kg0.3097
C400MPa10
44
44
34
4
3
3
3
3
fgf
fg
f
hxhhs
ssx
ssP
sh
TP
Noting that Q for the heat exchanger, the steady-flow energy balance equation yields0pekeW
kg/s1.275kg/s14kJ/kg06.2360.3097
K4203.679KkJ/kg1.005
0
air23
98air
23
98
98air23
outin(steady)0
systemoutin
mhh
TTcm
hhhh
m
hhmhhmhmhm
EEEEE
ps
seeii
(b)
kW13719.121384
kW9.12kJ/kg10.12kg/s1.275
kW1384kJ/kg5.20110.3097kg/s1.275
steamp,steamT,steamnet,
steamp,
43steamT,
WWW
wmW
hhmW
ps
s
and kW781964481371gasnet,steamnet,net WWW
(c) 66.4%kW11,784
kW7819
in
netth Q
W
10-63
10-77 [Also solved by EES on enclosed CD] A 450-MW combined gas-steam power plant is considered.The topping cycle is a gas-turbine cycle and the bottoming cycle is an ideal Rankine cycle with an openfeedwater heater. The mass flow rate of air to steam, the required rate of heat input in the combustionchamber, and the thermal efficiency of the combined cycle are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3Air is an ideal gas with variable specific heats.Analysis (a) The analysis of gas cycle yields (Table A-17)
kJ/kg02462 K460
kJ/kg873518325450141
5450kJ/kg421515K1400
kJ/kg56354019386114
3861kJ/kg19300K300
1212
1110
11
1010
98
9
88
1011
10
89
8
.hT
.h..PPPP
.P.hT
.h..PPP
P
.P.hT
rr
r
rr
r
From the steam tables (Tables A-4, A-5, A-6),
kJ/kg01.25259.042.251kJ/kg0.59
mkPa1kJ1
kPa20600/kgm0.001017
/kgm001017.0kJ/kg42.251
inpI,12
33
121inpI,
3kPa20@1
kPa20@1
whh
PPw
hh
f
f
v
vv
1400 K
GASCYCLE
10
11
8
Qin·
3
9
4
20 kPa Qout·
0.6 MPa 460 K
STEAMCYCLE
12
8 MPa
6
5400 C
T
7
2
1
300 K
s
kJ/kg53.67815.838.670kJ/kg8.15
mkPa1kJ1kPa6008,000/kgm0.001101
/kgm001101.0kJ/kg38.670
inpI,34
33
343inpII,
3MPa6.0@3
MPa6.0@3
whh
PPw
hh
f
f
v
vv
kJ/kg2.20955.23577821.042.251
7821.00752.7
8320.03658.6kPa20
kJ/kg1.25868.20859185.038.670
9185.08285.4
9308.13658.6MPa6.0
KkJ/kg3658.6kJ/kg4.3139
C400MPa8
77
77
57
7
66
66
56
6
5
5
5
5
fgf
fg
f
fgf
fg
f
hxhhs
ssx
ssP
hxhhs
ssx
ssP
sh
TP
Noting that Q for the heat exchanger, the steady-flow energy balance equation yields0pekeW
10-64
steamkgair /kg8.9902.46280.73553.6784.3139
0
1211
45air
1211air45
outin
(steady)0systemoutin
hhhh
mm
hhmhhmhmhm
EE
EEE
s
seeii
(b) Noting that for the open FWH, the steady-flow energy balance equation yieldsQ W ke pe 0
10-78 EES Problem 10-77 is reconsidered. The effect of the gas cycle pressure ratio on the ratio of gasflow rate to steam flow rate and cycle thermal efficiency is to be investigated.Analysis The problem is solved using EES, and the solution is given below.
"Input data"T[8] = 300 [K] "Gas compressor inlet"P[8] = 14.7 [kPa] "Assumed air inlet pressure""Pratio = 14" "Pressure ratio for gas compressor"T[10] = 1400 [K] "Gas turbine inlet"T[12] = 460 [K] "Gas exit temperature from Gas-to-steam heat exchanger "P[12] = P[8] "Assumed air exit pressure"W_dot_net=450 [MW] Eta_comp = 1.0Eta_gas_turb = 1.0 Eta_pump = 1.0Eta_steam_turb = 1.0P[5] = 8000 [kPa] "Steam turbine inlet"T[5] =(400+273) "[K]" "Steam turbine inlet"P[6] = 600 [kPa] "Extraction pressure for steam open feedwater heater"P[7] = 20 [kPa] "Steam condenser pressure"
"GAS POWER CYCLE ANALYSIS"
"Gas Compressor anaysis"s[8]=ENTROPY(Air,T=T[8],P=P[8])ss9=s[8] "For the ideal case the entropies are constant across the compressor"P[9] = Pratio*P[8] Ts9=temperature(Air,s=ss9,P=P[9])"Ts9 is the isentropic value of T[9] at compressor exit"Eta_comp = w_gas_comp_isen/w_gas_comp "compressor adiabatic efficiency, w_comp >w_comp_isen"h[8] + w_gas_comp_isen =hs9"SSSF conservation of energy for the isentropic compressor,assuming: adiabatic, ke=pe=0 per unit gas mass flow rate in kg/s"h[8]=ENTHALPY(Air,T=T[8])hs9=ENTHALPY(Air,T=Ts9)h[8] + w_gas_comp = h[9]"SSSF conservation of energy for the actual compressor, assuming:adiabatic, ke=pe=0"T[9]=temperature(Air,h=h[9])s[9]=ENTROPY(Air,T=T[9],P=P[9])
"Gas Cycle External heat exchanger analysis"h[9] + q_in = h[10]"SSSF conservation of energy for the external heat exchanger, assuming W=0, ke=pe=0"h[10]=ENTHALPY(Air,T=T[10])P[10]=P[9] "Assume process 9-10 is SSSF constant pressure"Q_dot_in"MW"*1000"kW/MW"=m_dot_gas*q_in
"Gas Turbine analysis"s[10]=ENTROPY(Air,T=T[10],P=P[10])ss11=s[10] "For the ideal case the entropies are constant across the turbine"P[11] = P[10] /PratioTs11=temperature(Air,s=ss11,P=P[11])"Ts11 is the isentropic value of T[11] at gas turbine exit"Eta_gas_turb = w_gas_turb /w_gas_turb_isen "gas turbine adiabatic efficiency, w_gas_turb_isen> w_gas_turb"h[10] = w_gas_turb_isen + hs11"SSSF conservation of energy for the isentropic gas turbine, assuming: adiabatic, ke=pe=0"
10-66
hs11=ENTHALPY(Air,T=Ts11)h[10] = w_gas_turb + h[11]"SSSF conservation of energy for the actual gas turbine, assuming:adiabatic, ke=pe=0"T[11]=temperature(Air,h=h[11])s[11]=ENTROPY(Air,T=T[11],P=P[11])
"Gas-to-Steam Heat Exchanger""SSSF conservation of energy for the gas-to-steam heat exchanger, assuming: adiabatic,W=0, ke=pe=0"m_dot_gas*h[11] + m_dot_steam*h[4] = m_dot_gas*h[12] + m_dot_steam*h[5]h[12]=ENTHALPY(Air, T=T[12]) s[12]=ENTROPY(Air,T=T[12],P=P[12])
"STEAM CYCLE ANALYSIS""Steam Condenser exit pump or Pump 1 analysis"Fluid$='Steam_IAPWS'P[1] = P[7] P[2]=P[6]h[1]=enthalpy(Fluid$,P=P[1],x=0) {Saturated liquid}v1=volume(Fluid$,P=P[1],x=0)s[1]=entropy(Fluid$,P=P[1],x=0)T[1]=temperature(Fluid$,P=P[1],x=0)w_pump1_s=v1*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume"w_pump1=w_pump1_s/Eta_pump "Definition of pump efficiency"h[1]+w_pump1= h[2] "Steady-flow conservation of energy"s[2]=entropy(Fluid$,P=P[2],h=h[2])T[2]=temperature(Fluid$,P=P[2],h=h[2])"Open Feedwater Heater analysis"y*h[6] + (1-y)*h[2] = 1*h[3] "Steady-flow conservation of energy"P[3]=P[6]h[3]=enthalpy(Fluid$,P=P[3],x=0) "Condensate leaves heater as sat. liquid at P[3]"T[3]=temperature(Fluid$,P=P[3],x=0)s[3]=entropy(Fluid$,P=P[3],x=0)"Boiler condensate pump or Pump 2 analysis"P[4] = P[5] v3=volume(Fluid$,P=P[3],x=0)w_pump2_s=v3*(P[4]-P[3])"SSSF isentropic pump work assuming constant specific volume"w_pump2=w_pump2_s/Eta_pump "Definition of pump efficiency"h[3]+w_pump2= h[4] "Steady-flow conservation of energy"s[4]=entropy(Fluid$,P=P[4],h=h[4])T[4]=temperature(Fluid$,P=P[4],h=h[4])w_steam_pumps = (1-y)*w_pump1+ w_pump2 "Total steam pump work input/ mass steam""Steam Turbine analysis"h[5]=enthalpy(Fluid$,T=T[5],P=P[5])s[5]=entropy(Fluid$,P=P[5],T=T[5])ss6=s[5]hs6=enthalpy(Fluid$,s=ss6,P=P[6])Ts6=temperature(Fluid$,s=ss6,P=P[6])h[6]=h[5]-Eta_steam_turb*(h[5]-hs6)"Definition of steam turbine efficiency"T[6]=temperature(Fluid$,P=P[6],h=h[6])s[6]=entropy(Fluid$,P=P[6],h=h[6])ss7=s[5]hs7=enthalpy(Fluid$,s=ss7,P=P[7])Ts7=temperature(Fluid$,s=ss7,P=P[7])h[7]=h[5]-Eta_steam_turb*(h[5]-hs7)"Definition of steam turbine efficiency"T[7]=temperature(Fluid$,P=P[7],h=h[7])
10-67
s[7]=entropy(Fluid$,P=P[7],h=h[7])"SSSF conservation of energy for the steam turbine: adiabatic, neglect ke and pe"h[5] = w_steam_turb + y*h[6] +(1-y)*h[7]"Steam Condenser analysis"(1-y)*h[7]=q_out+(1-y)*h[1]"SSSF conservation of energy for the Condenser per unit mass"Q_dot_out*Convert(MW, kW)=m_dot_steam*q_out"Cycle Statistics"MassRatio_gastosteam =m_dot_gas/m_dot_steamW_dot_net*Convert(MW, kW)=m_dot_gas*(w_gas_turb-w_gas_comp)+m_dot_steam*(w_steam_turb - w_steam_pumps)"definition of the net cycle work"Eta_th=W_dot_net/Q_dot_in*Convert(, %) "Cycle thermal efficiency, in percent"Bwr=(m_dot_gas*w_gas_comp + m_dot_steam*w_steam_pumps)/(m_dot_gas*w_gas_turb +m_dot_steam*w_steam_turb) "Back work ratio"W_dot_net_steam = m_dot_steam*(w_steam_turb - w_steam_pumps)W_dot_net_gas = m_dot_gas*(w_gas_turb - w_gas_comp)NetWorkRatio_gastosteam = W_dot_net_gas/W_dot_net_steam
Ratio of Gas Flow Rate to Steam Flow Rate vs Gas Pressure Ratio
10-69
10-79 A 450-MW combined gas-steam power plant is considered. The topping cycle is a gas-turbine cycleand the bottoming cycle is a nonideal Rankine cycle with an open feedwater heater. The mass flow rate ofair to steam, the required rate of heat input in the combustion chamber, and the thermal efficiency of thecombined cycle are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3Air is an ideal gas with variable specific heats.Analysis (a) Using the properties of air from Table A-17, the analysis of gas cycle yields
kJ/kg02462 K460
kJ/kg844.958735421515860421515
kJ/kg873518325450141
5450kJ/kg421515 K1400
kJ/kg709.182019300563519300
kJ/kg56354019386114
3861kJ/kg19300 K300
1212
111010111110
1110
1110
11
1010
898989
89
98
9
88
1011
10
89
8
.hT
....hhhh
hhhh
.h..PPPP
.P.hT
./.../hhhh
hhhh
.h..PPP
P
.P.hT
sTs
T
srr
r
Css
C
srr
r T
10
Qin·
GASCYCLE
511
11s9
9s
124 STEAMCYCLE 63 6s2
8Qout· 7s 71
s
From the steam tables (Tables A-4, A-5, and A-6),
kJ/kg01.25259.042.251kJ/kg0.59
mkPa1kJ1
kPa20600/kgm0.001017
/kgm001017.0kJ/kg42.251
inpI,12
33
121inpI,
3kPa20@1
kPa20@1
whh
PPw
hh
f
f
v
vv
kJ/kg52.67815.838.670kJ/kg8.15
mkPa1kJ1
kPa6008,000/kgm0.001101
/kgm001101.0kJ/kg38.670
inpI,34
33
343inpII,
3MPa6.0@3
MPa6.0@3
whh
PPw
hh
f
f
v
vv
KkJ/kg3658.6kJ/kg4.3139
C400MPa8
5
5
5
5sh
TP
10-70
kJ/kg3.22411.20954.313986.04.3139
kJ/kg1.20955.23577820.042.251
7820.00752.7
8320.03658.6kPa20
kJ/kg3.26639.25854.313986.04.3139
kJ/kg9.25858.20859184.038.670
9184.08285.4
9308.13658.6MPa6.0
755775
75
77
77
57
7
655665
65
66
66
56
6
sTs
T
fgfs
fg
fs
sTs
T
fgsfs
fg
fss
s
hhhhhhhh
hxhhs
ssx
ssP
hhhhhhhh
hxhhs
ssx
ssP
Noting that Q for the heat exchanger, the steady-flow energy balance equation yields0pekeW
steamkgair /kg6.42502.46295.84452.6784.3139
0
1211
45air
1211air45
outin
(steady)0systemoutin
hhhh
mm
hhmhhmhmhm
EE
EEE
s
seeii
(b) Noting that for the open FWH, the steady-flow energy balance equation yields0pekeWQ
and kW933,850kJ/kg1.70942.1515kg/s1158.2910airin hhmQ
(c) 48.2%kW933,850kW450,000
in
net
QW
th
10-71
10-80 EES Problem 10-79 is reconsidered. The effect of the gas cycle pressure ratio on the ratio of gasflow rate to steam flow rate and cycle thermal efficiency is to be investigated.Analysis The problem is solved using EES, and the solution is given below.
"Input data"T[8] = 300 [K] "Gas compressor inlet"P[8] = 14.7 [kPa] "Assumed air inlet pressure""Pratio = 14" "Pressure ratio for gas compressor"T[10] = 1400 [K] "Gas turbine inlet"T[12] = 460 [K] "Gas exit temperature from Gas-to-steam heat exchanger "P[12] = P[8] "Assumed air exit pressure"W_dot_net=450 [MW] Eta_comp = 0.82Eta_gas_turb = 0.86 Eta_pump = 1.0Eta_steam_turb = 0.86P[5] = 8000 [kPa] "Steam turbine inlet"T[5] =(400+273) "K" "Steam turbine inlet"P[6] = 600 [kPa] "Extraction pressure for steam open feedwater heater"P[7] = 20 [kPa] "Steam condenser pressure"
"GAS POWER CYCLE ANALYSIS"
"Gas Compressor anaysis"s[8]=ENTROPY(Air,T=T[8],P=P[8])ss9=s[8] "For the ideal case the entropies are constant across the compressor"P[9] = Pratio*P[8] Ts9=temperature(Air,s=ss9,P=P[9])"Ts9 is the isentropic value of T[9] at compressor exit"Eta_comp = w_gas_comp_isen/w_gas_comp "compressor adiabatic efficiency, w_comp >w_comp_isen"h[8] + w_gas_comp_isen =hs9"SSSF conservation of energy for the isentropic compressor,assuming: adiabatic, ke=pe=0 per unit gas mass flow rate in kg/s"h[8]=ENTHALPY(Air,T=T[8])hs9=ENTHALPY(Air,T=Ts9)h[8] + w_gas_comp = h[9]"SSSF conservation of energy for the actual compressor, assuming:adiabatic, ke=pe=0"T[9]=temperature(Air,h=h[9])s[9]=ENTROPY(Air,T=T[9],P=P[9])
"Gas Cycle External heat exchanger analysis"h[9] + q_in = h[10]"SSSF conservation of energy for the external heat exchanger, assuming W=0, ke=pe=0"h[10]=ENTHALPY(Air,T=T[10])P[10]=P[9] "Assume process 9-10 is SSSF constant pressure"Q_dot_in"MW"*1000"kW/MW"=m_dot_gas*q_in
"Gas Turbine analysis"s[10]=ENTROPY(Air,T=T[10],P=P[10])ss11=s[10] "For the ideal case the entropies are constant across the turbine"P[11] = P[10] /PratioTs11=temperature(Air,s=ss11,P=P[11])"Ts11 is the isentropic value of T[11] at gas turbine exit"Eta_gas_turb = w_gas_turb /w_gas_turb_isen "gas turbine adiabatic efficiency, w_gas_turb_isen> w_gas_turb"h[10] = w_gas_turb_isen + hs11"SSSF conservation of energy for the isentropic gas turbine, assuming: adiabatic, ke=pe=0"
10-72
hs11=ENTHALPY(Air,T=Ts11)h[10] = w_gas_turb + h[11]"SSSF conservation of energy for the actual gas turbine, assuming:adiabatic, ke=pe=0"T[11]=temperature(Air,h=h[11])s[11]=ENTROPY(Air,T=T[11],P=P[11])
"Gas-to-Steam Heat Exchanger""SSSF conservation of energy for the gas-to-steam heat exchanger, assuming: adiabatic,W=0, ke=pe=0"m_dot_gas*h[11] + m_dot_steam*h[4] = m_dot_gas*h[12] + m_dot_steam*h[5]h[12]=ENTHALPY(Air, T=T[12]) s[12]=ENTROPY(Air,T=T[12],P=P[12])
"STEAM CYCLE ANALYSIS""Steam Condenser exit pump or Pump 1 analysis"Fluid$='Steam_IAPWS'P[1] = P[7] P[2]=P[6]h[1]=enthalpy(Fluid$,P=P[1],x=0) {Saturated liquid}v1=volume(Fluid$,P=P[1],x=0)s[1]=entropy(Fluid$,P=P[1],x=0)T[1]=temperature(Fluid$,P=P[1],x=0)w_pump1_s=v1*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume"w_pump1=w_pump1_s/Eta_pump "Definition of pump efficiency"h[1]+w_pump1= h[2] "Steady-flow conservation of energy"s[2]=entropy(Fluid$,P=P[2],h=h[2])T[2]=temperature(Fluid$,P=P[2],h=h[2])"Open Feedwater Heater analysis"y*h[6] + (1-y)*h[2] = 1*h[3] "Steady-flow conservation of energy"P[3]=P[6]h[3]=enthalpy(Fluid$,P=P[3],x=0) "Condensate leaves heater as sat. liquid at P[3]"T[3]=temperature(Fluid$,P=P[3],x=0)s[3]=entropy(Fluid$,P=P[3],x=0)"Boiler condensate pump or Pump 2 analysis"P[4] = P[5] v3=volume(Fluid$,P=P[3],x=0)w_pump2_s=v3*(P[4]-P[3])"SSSF isentropic pump work assuming constant specific volume"w_pump2=w_pump2_s/Eta_pump "Definition of pump efficiency"h[3]+w_pump2= h[4] "Steady-flow conservation of energy"s[4]=entropy(Fluid$,P=P[4],h=h[4])T[4]=temperature(Fluid$,P=P[4],h=h[4])w_steam_pumps = (1-y)*w_pump1+ w_pump2 "Total steam pump work input/ mass steam""Steam Turbine analysis"h[5]=enthalpy(Fluid$,T=T[5],P=P[5])s[5]=entropy(Fluid$,P=P[5],T=T[5])ss6=s[5]hs6=enthalpy(Fluid$,s=ss6,P=P[6])Ts6=temperature(Fluid$,s=ss6,P=P[6])h[6]=h[5]-Eta_steam_turb*(h[5]-hs6)"Definition of steam turbine efficiency"T[6]=temperature(Fluid$,P=P[6],h=h[6])s[6]=entropy(Fluid$,P=P[6],h=h[6])ss7=s[5]hs7=enthalpy(Fluid$,s=ss7,P=P[7])Ts7=temperature(Fluid$,s=ss7,P=P[7])h[7]=h[5]-Eta_steam_turb*(h[5]-hs7)"Definition of steam turbine efficiency"T[7]=temperature(Fluid$,P=P[7],h=h[7])
10-73
s[7]=entropy(Fluid$,P=P[7],h=h[7])"SSSF conservation of energy for the steam turbine: adiabatic, neglect ke and pe"h[5] = w_steam_turb + y*h[6] +(1-y)*h[7]"Steam Condenser analysis"(1-y)*h[7]=q_out+(1-y)*h[1]"SSSF conservation of energy for the Condenser per unit mass"Q_dot_out*Convert(MW, kW)=m_dot_steam*q_out"Cycle Statistics"MassRatio_gastosteam =m_dot_gas/m_dot_steamW_dot_net*Convert(MW, kW)=m_dot_gas*(w_gas_turb-w_gas_comp)+m_dot_steam*(w_steam_turb - w_steam_pumps)"definition of the net cycle work"Eta_th=W_dot_net/Q_dot_in*Convert(, %) "Cycle thermal efficiency, in percent"Bwr=(m_dot_gas*w_gas_comp + m_dot_steam*w_steam_pumps)/(m_dot_gas*w_gas_turb +m_dot_steam*w_steam_turb) "Back work ratio"W_dot_net_steam = m_dot_steam*(w_steam_turb - w_steam_pumps)W_dot_net_gas = m_dot_gas*(w_gas_turb - w_gas_comp)NetWorkRatio_gastosteam = W_dot_net_gas/W_dot_net_steam
Cycle Thermal Efficiency vs Gas Cycle Pressure Ratio
5 9 12 16 19 231.4
1.5
1.6
1.7
1.8
1.9
2.0
2.1
2.2
2.3
Pratio
Net
Wor
kRat
ioga
stos
team
W dot,gas / W dot,steam vs Gas Pressure Ratio
5 9 12 16 19 234.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
8.0
8.5
Pratio
Mas
sRat
ioga
stos
team
Ratio of Gas Flow Rate to Steam Flow Rate vs Gas Pressure Ratio
10-75
10-81 A combined gas-steam power plant is considered. The topping cycle is a gas-turbine cycle and thebottoming cycle is a nonideal reheat Rankine cycle. The moisture percentage at the exit of the low-pressureturbine, the steam temperature at the inlet of the high-pressure turbine, and the thermal efficiency of thecombined cycle are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3Air is an ideal gas with variable specific heats.Analysis (a) We obtain the air properties fromEES. The analysis of gas cycle is as follows
kJ/kg62.475C200
kJ/kg98.87179.7638.130480.08.1304
kJ/kg79.763kPa100
kJ/kg6456.6kPa700C950
kJ/kg8.1304C950
kJ/kg21.5570/16.29047.50316.290
/
kJ/kg47.503kPa700
kJ/kg6648.5kPa100
C15kJ/kg50.288C15
1111
109910109
109
10910
10
99
9
99
787878
78
878
8
77
7
77
hT
hhhhhhhh
hss
P
sPT
hT
hhhhhhhh
hss
P
sPT
hT
sTs
T
s
Css
C
s
80.
Combustionchamber
89
Compressor Gasturbine
710
Heatexchanger11
3Steamturbine
4
6
5 Condenserpump2
1
T
From the steam tables (Tables A-4, A-5,and A-6 or from EES),
kJ/kg37.19965.781.191kJ/kg.567
80.0/mkPa1
kJ1kPa106000/kgm0.00101
/
/kgm00101.0kJ/kg81.191
inpI,12
33
121inpI,
3kPa10@1
kPa10@1
whh
PPw
hh
p
f
f
v
vv
kJ/kg4.23661.23929091.081.191
9091.04996.7
6492.04670.7kPa10
KkJ/kg4670.7kJ/kg5.3264
C400MPa1
66
66
56
6
5
5
5
5
fgsfs
fg
fss
s hxhhs
ssx
ssP
sh
TP
6
4s
810
1 MPa
4
5
GASCYCLE
9
10s
11
7
Qin·
8s
10 kPa
STEAMCYCLE
Qout·
6 MPa3
6s2
1
950 C
15 C
s
10-76
1.6%0158.09842.011PercentageMoisture
9842.0kJ/kg5.2546kPa10
kJ/kg0.25464.23665.326480.05.3264
6
66
6
655665
65
x
xhP
hhhhhhhh
sTs
T
(b) Noting that Q for the heat exchanger, the steady-flow energy balance equationyields
The temperature at the inlet of the high-pressure turbine may be obtained by a trial-error approach or usingEES from the above relations. The answer is T3 = 468.0ºC. Then, the enthalpy at state 3 becomes: h3 = 3346.5 kJ/kg
10-82C Binary power cycle is a cycle which is actually a combination of two cycles; one in the hightemperature region, and the other in the low temperature region. Its purpose is to increase thermalefficiency.
10-83C Consider the heat exchanger of a binary power cycle. The working fluid of the topping cycle(cycle A) enters the heat exchanger at state 1 and leaves at state 2. The working fluid of the bottomingcycle (cycle B) enters at state 3 and leaves at state 4. Neglecting any changes in kinetic and potentialenergies, and assuming the heat exchanger is well-insulated, the steady-flow energy balance relation yields
43123142
outin
(steady)0systemoutin 0
hhmhhmorhmhmhmhm
hmhm
EE
EEE
BABABA
iiee
Thus,
mm
h hh h
A
B
3 4
2 1
10-84C Steam is not an ideal fluid for vapor power cycles because its critical temperature is low, its saturation dome resembles an inverted V, and its condenser pressure is too low.
10-85C Because mercury has a high critical temperature, relatively low critical pressure, but a very low condenser pressure. It is also toxic, expensive, and has a low enthalpy of vaporization.
10-86C In binary vapor power cycles, both cycles are vapor cycles. In the combined gas-steam power cycle, one of the cycles is a gas cycle.
10-78
Review Problems
10-87 It is to be demonstrated that the thermal efficiency of a combined gas-steam power plant cc can be expressed as where d are the thermal efficiencies of the gas and steam cycles, respectively, and the efficiency of a combined cycle is to be obtained.
cc g s g s g g iW Q/ n an s s g,outW Q/
Analysis The thermal efficiencies of gas, steam, and combined cycles can be expressed as
cctotal
in
out
in
gg
in
g,out
in
ss
g,out
out
g,out
WQ
QQ
WQ
QQ
WQ
QQ
1
1
1
where Qin is the heat supplied to the gas cycle, where Qout is the heat rejected by the steam cycle, and where Qg,out is the heat rejected from the gas cycle and supplied to the steam cycle.
Using the relations above, the expression can be expressed asg s g s
cc
QQ
QQ
QQ
QQ
QQ
QQ
QQ
QQ
QQ
QQ
in
out
in
out
outg,
out
in
outg,
outg,
out
in
outg,
outg,
out
in
outg,
outg,
out
in
outg,sgsg
1
111
1111
Therefore, the proof is complete. Using the relation above, the thermal efficiency of the given combinedcycle is determined to be
cc g s g s 0 4 0 30 0 40 0 30. . . . 0.58
10-88 The thermal efficiency of a combined gas-steam power plant cc can be expressed in terms of thethermal efficiencies of the gas and the steam turbine cycles as . It is to be shown that
the value ofcc g s g s
cc is greater than either of .g s or
Analysis By factoring out terms, the relation can be expressed as cc g s g s
cc g s g s g s g
Positive since <1
g
g
( )1
or cc g s g s s g s
Positive since <1
s
s
( )1
Thus we conclude that the combined cycle is more efficient than either of the gas turbine or steam turbinecycles alone.
10-79
10-89 A steam power plant operating on the ideal Rankine cycle with reheating is considered. The reheat pressures of the cycle are to be determined for the cases of single and double reheat.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) Single Reheat: From the steam tables (Tables A-4, A-5, and A-6),
Any pressure Px selected between the limits of 25 MPa and 2.78 MPa will satisfy the requirements, and canbe used for the double reheat pressure.
10-80
10-90E A geothermal power plant operating on the simple Rankine cycle using an organic fluid as theworking fluid is considered. The exit temperature of the geothermal water from the vaporizer, the rate of heat rejection from the working fluid in the condenser, the mass flow rate of geothermal water at thepreheater, and the thermal efficiency of the Level I cycle of this plant are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) The exit temperature of geothermal water from the vaporizer is determined from the steady-flow energy balance on the geothermal water (brine),
F267.42
2
12brinebrine
F325FBtu/lbm1.03lbm/h384,286Btu/h000,790,22T
T
TTcmQ p
(b) The rate of heat rejection from the working fluid to the air in the condenser is determined from thesteady-flow energy balance on air,
MBtu/h29.7F555.84FBtu/lbm0.24lbm/h4,195,100
89airair TTcmQ p
(c) The mass flow rate of geothermal water at the preheater is determined from the steady-flow energy balance on the geothermal water,
lbm/h187,120geo
geo
inoutgeogeo
F8.2110.154FBtu/lbm1.03Btu/h000,140,11
m
m
TTcmQ p
(d) The rate of heat input is
and
, , , ,
, ,
Q Q Q
W
in vaporizer reheater
net
Btu / h
kW
22 790 000 11140 000
33 930 000
1271 200 1071
Then,
10.8%kWh1
Btu3412.14Btu/h33,930,000
kW1071
in
netth Q
W
10-81
10-91 A steam power plant operates on the simple ideal Rankine cycle. The turbine inlet temperature, thenet power output, the thermal efficiency, and the minimum mass flow rate of the cooling water required areto be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),
(c) The mass flow rate of the cooling water will be minimum when it is heated to the temperature of thesteam in the condenser, which is 40.29 C,
kg/s194.6C1540.29CkJ/kg4.18
kJ/s20,572
kJ/s572,20428,19000,40
outcool
netinout
TcQm
WQQ
10-82
10-92 A steam power plant operating on an ideal Rankine cycle with two stages of reheat is considered. The thermal efficiency of the cycle and the mass flow rate of the steam are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),
kJ/kg82.15207.1575.137kJ/kg15.07
mkPa1kJ1kPa515,000/kgm0.001005
/kgm001005.0kJ/kg75.137
inp,12
33
121inp,
3kPa5@1
kPa5@1
whh
PPw
hh
f
f
v
vv T
kJ/kg9.23670.24239204.075.137
9204.09176.7
4762.07642.7kPa5
KkJ/kg7642.7kJ/kg1.3479
C500MPa1
kJ/kg3.2971MPa1
KkJ/kg9781.6kJ/kg7.3434
C500MPa5
kJ/kg4.3007MPa5
KkJ/kg3480.6kJ/kg8.3310
C500MPa15
88
88
78
8
7
7
7
7
656
6
5
5
5
5
434
4
3
3
3
3
fgf
fg
f
hxhhs
ssx
ssP
sh
TP
hssP
sh
TP
hssP
sh
TP
1 MPa 5 MPa
6
7
5 kPa
15 MPa 4
3
8
2
5
1s
Then,
kJ/kg9.18622.22301.4093kJ/kg2.223075.1379.2367
kJ/kg1.40933.29711.34794.30077.343482.1528.3310
outinnet
18out
674523in
qqwhhq
hhhhhhq
Thus,
45.5%kJ/kg4093.1kJ/kg1862.9
in
netth q
w
(b) kg/s64.4kJ/kg1862.9
kJ/s120,000
net
net
wWm
10-83
10-93 An 150-MW steam power plant operating on a regenerative Rankine cycle with an open feedwaterheater is considered. The mass flow rate of steam through the boiler, the thermal efficiency of the cycle,and the irreversibility associated with the regeneration process are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis
1-y6
P II P I
Openfwh
Condenser
BoilerTurbine
4
32
1
7
5
y
T
6
7qout
3 y
4
1-y0.5 MPa
10 kPa
10 MPa
6s
5
7s
2
qin
1s
(a) From the steam tables (Tables A-4, A-5, and A-6),
/kgm001093.0kJ/kg09.640
liquidsat.MPa5.0
kJ/kg33.19252.081.191kJ/kg0.52
95.0/mkPa1
kJ1kPa10500/kgm0.00101
/
/kgm00101.0kJ/kg81.191
3MPa5.0@3
MPa5.0@33
inpI,12
33
121inpI,
3kPa10@1
kPa10@1
f
f
p
f
f
hhP
whh
PPw
hh
vv
v
vv
kJ/kg02.65193.1009.640kJ/kg10.93
95.0/mkPa1
kJ1kPa50010,000/kgm0.001093
/
inpII,34
33
343inpII,
whh
PPw pv
kJ/kg1.26540.21089554.009.640
9554.09603.4
8604.15995.6
MPa5.0
KkJ/kg5995.6kJ/kg1.3375
C500MPa10
66
66
56
6
5
5
5
5
fgsfs
fg
fss
s
s hxhhs
ssx
ssP
sh
TP
kJ/kg3.27981.26541.337580.01.3375
655665
65sT
sT hhhh
hhhh
10-84
kJ/kg7.20891.23927934.081.191
7934.04996.7
6492.05995.6
kPa1077
77
57
7fgsfs
fg
fss
s
s hxhhs
ssx
ssP
kJ/kg8.23467.20891.337580.01.3375
755775
75sT
sT hhhh
hhhh
The fraction of steam extracted is determined from the steady-flow energy balance equation applied to thefeedwater heaters. Noting that ,Q W ke pe 0
326332266
outin
(steady)0systemoutin
11
0
hhyyhhmhmhmhmhm
EE
EEE
eeii
where y is the fraction of steam extracted from the turbine ( /m m6 3 ). Solving for y,
Then the irreversibility (or exergy destruction) associated with this regeneration process is
kJ/kg39.256492.01718.019453.61718.08604.1K303
1 2630
0surr
0gen0regen syyssTT
qsmsmTsTiL
iiee
10-85
10-94 An 150-MW steam power plant operating on an ideal regenerative Rankine cycle with an openfeedwater heater is considered. The mass flow rate of steam through the boiler, the thermal efficiency of the cycle, and the irreversibility associated with the regeneration process are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis
1-y6
P II P I
Openfwh
Condenser
BoilerTurbine
4
32
1
7
5
y
T
qout
3 y
4
1-y
0.5 MPa
10 kPa
10 MPa
6
5
7
2
qin
1s
(a) From the steam tables (Tables A-4, A-5, and A-6),
/kgm001093.0kJ/kg09.640
liquidsat.MPa5.0
kJ/kg30.19250.081.191
kJ/kg0.50mkPa1
kJ1kPa10500/kgm0.00101
/kgm00101.0kJ/k81.191
3MPa5.0@3
MPa5.0@33
inpI,12
33
121inpI,
3kPa10@1
kPa10@1
f
f
f
f
hhP
whh
PPw
ghh
vv
v
vv
kJ/kg7.20891.23927934.081.191
7934.04996.7
6492.05995.6kPa10
kJ/kg1.26540.21089554.009.640
9554.09603.4
8604.15995.6MPa5.0
KkJ/kg5995.6kJ/kg1.3375
C500MPa10
kJ/kg47.65038.1009.640
kJ/kg.3810mkPa1
kJ1kPa50010,000/kgm0.001093
77
77
57
7
66
66
56
6
5
5
5
5
inpII,34
33
343inpII,
fgf
fg
f
fgf
fg
f
hxhhs
ssx
ssP
hxhhs
ssx
ssP
sh
TP
whh
PPw v
10-86
The fraction of steam extracted is determined from the steady-flow energy equation applied to thefeedwater heaters. Noting that Q ,0pekeW
326332266
outin(steady)0
systemoutin
11
0
hhyyhhmhmhmhmhm
EEEEE
eeii
where y is the fraction of steam extracted from the turbine ( /m m6 3 ). Solving for y,
Then the irreversibility (or exergy destruction) associated with this regeneration process is
kJ/kg39.06492.01819.015995.61819.08604.1K303
1 2630
0surr
0gen0regen syyssTT
qsmsmTsTiL
iiee
10-87
10-95 An ideal reheat-regenerative Rankine cycle with one open feedwater heater is considered. Thefraction of steam extracted for regeneration and the thermal efficiency of the cycle are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),
/kgm001101.0kJ/k38.670
liquidsat.MPa6.0
kJ/kg53.22659.094.225kJ/kg0.59
mkPa1kJ1
kPa15600/kgm0.001014
/kgm001014.0kJ/kg94.225
3MPa6.0@3
MPa6.0@33
inpI,12
33
121inpI,
3kPa15@1
kPa15@1
f
f
f
f
ghhP
whh
PPw
hh
vv
v
vv
kJ/kg8.25183.23729665.094.225
9665.02522.7
7549.07642.7kPa15
kJ/kg2.3310MPa6.0
KkJ/kg7642.7kJ/kg1.3479
C500MPa0.1
kJ/kg8.2783MPa0.1
KkJ/kg5995.6kJ/kg1.3375
C500MPa10
kJ/kg73.68035.1038.670kJ/kg10.35
mkPa1kJ1kPa60010,000/kgm0.001101
99
99
79
9
878
8
7
7
7
7
656
6
5
5
5
5
inpII,34
33
343inpII,
fgf
fg
f
hxhhs
ssx
ssP
hssP
sh
TP
hssP
sh
TP
whh
PPw v
7
6
1-y8
P II P I
Openfwh
Condens.
BoilerTurbine
4
32
1
9
5
y
3
48
1 MPa
0.6 MPa
15 kPa
10 MPa6
5
9
2
7
1
T
s
The fraction of steam extracted is determined from the steady-flow energy balance equation applied to thefeedwater heaters. Noting that Q ,0pekeW
328332288
outin(steady)0
systemoutin
11
0
hhyyhhmhmhmhmhm
EEEEE
eeii
where y is the fraction of steam extracted from the turbine ( /m m8 3 ). Solving for y,
10-96 A nonideal reheat-regenerative Rankine cycle with one open feedwater heater is considered. The fraction of steam extracted for regeneration and the thermal efficiency of the cycle are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis
7
6
1-y8
P II P I
Openfwh
Condenser
BoilerTurbine
4
32
1
9
5
y
T
1-y3
4 866s 8sy
5
99s
2
7
1s
(a) From the steam tables (Tables A-4, A-5, and A-6),
/kgm001101.0kJ/kg38.670
liquidsat.MPa6.0
kJ/kg54.22659.094.225kJ/kg0.59
mkPa1kJ1
kPa15600/kgm0.001014
/kgm001014.0kJ/kg94.225
3MPa6.0@3
MPa6.0@33
in,12
33
121in,
3kPa15@1
kPa15@1
f
f
pI
pI
f
f
hhP
whh
PPw
hh
vv
v
vv
kJ/kg8.2783MPa0.1
KkJ/kg5995.6kJ/kg1.3375
C500MPa10
kJ/kg73.68035.1038.670kJ/kg10.35
mkPa1kJ1kPa60010,000/kgm0.001101
656
6
5
5
5
5
inpII,34
33
343inpII,
ss
s hssP
sh
TP
whh
PPw v
kJ/kg4.28788.27831.337584.01.3375
655665
65sT
sT hhhh
hhhh
10-89
kJ/kg2.33372.33101.347984.01.3479
kJ/kg2.3310MPa6.0
KkJ/kg7642.7kJ/kg1.3479
C500MPa0.1
877887
87
878
8
7
7
7
7
sTs
T
ss
s
hhhhhhhh
hssP
shP
T
kJ/kg5.26728.25181.347984.01.3479
kJ/kg8.25183.23729665.094.225
9665.02522.7
7549.07642.7kPa15
977997
97
99
99
79
9
sTs
T
fgsfs
fg
fss
s
s
hhhhhhhh
hxhhs
ssx
ssP
The fraction of steam extracted is determined from the steady-flow energy balance equation applied to thefeedwater heaters. Noting that Q ,0pekeW
328332288
outin
(steady)0systemoutin
11
0
hhyyhhmhmhmhmhm
EE
EEE
eeii
where y is the fraction of steam extracted from the turbine ( /m m8 3 ). Solving for y,
10-97 A steam power plant operates on an ideal reheat-regenerative Rankine cycle with one reheater and two feedwater heaters, one open and one closed. The fraction of steam extracted from the turbine for theopen feedwater heater, the thermal efficiency of the cycle, and the net power output are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis
(a) From the steam tables (Tables A-4, A-5, and A-6),
10-98 A cogeneration power plant is modified with reheat and that produces 3 MW of power and supplies7 MW of process heat. The rate of heat input in the boiler and the fraction of steam extracted for process heating are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the steamtables (Tables A-4, A-5, and A-6),
kJ/kg8.25183.23729665.094.225
9665.02522.7
7549.07642.7
kPa15
KkJ/kg7642.7kJ/kg1.3479
C500MPa1
kJ/kg7.2843MPa1
KkJ/kg7266.6kJ/kg5.3399
C500MPa8
kJ/kg81.503
kJ/kg94.225
99
99
89
9
8
8
8
8
767
7
6
6
6
6
C120@3
12
kPa15@1
fgf
fg
f
f
f
hxhhs
ssx
ssP
shP
hssP
sh
TP
hhhhhh
T4
7
8
P II P I
Processheater
Condenser
BoilerTurbine
5
3
2
1
9
6
T
6 88 MPa
31 MPa5 7
42The mass flow rate through the process heater is
(b) The fraction of steam extracted for process heating is
77.3%kg/s3.873kg/s2.992
total
3
mm
y
10-93
10-99 A combined gas-steam power plant is considered. The topping cycle is an ideal gas-turbine cycle and the bottoming cycle is an ideal reheat Rankine cycle. The mass flow rate of air in the gas-turbine cycle, the rate of totalheat input, and the thermal efficiency of the combined cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats.Analysis (a) The analysis of gas cycle yields
kJ/kg63523 K520
kJ/kg35860356545081
5450kJ/kg421515 K1400
kJ/kg125268499231118
23111kJ/kg16290 K290
1111
109
10
99
87
8
77
910
9
78
7
.hT
.h..PPPP
.P.hT
.h..PPP
P
.P.hT
rr
r
rr
r
From the steam tables (Tables A-4, A-5, and A-6),
gwhh
PPw
hh
f
f
kJ/k95.20614.1581.191
kJ/kg15.14mkPa1
kJ1kPa1015,000/kgm0.00101
/kgm00101.0
kJ/kg81.191
inpI,12
33
121inpI,
3kPa10@1
kPa10@1
v
vv
3 MPa
4
5
1400 K
GASCYCLE
9
10
11
7
Qin·
8
10 kPa
STEAMCYCLE
Qout·
15 MPa
3
T
62
1
290 K
s
kJ/kg8.22921.23928783.081.191
8783.04996.7
6492.02355.7kPa10
KkJ/kg2359.7kJ/kg2.3457
C500MPa3
kJ/kg7.27819.17949880.03.1008
9880.05402.3
6454.21434.6MPa3
KkJ/kg1434.6kJ/kg9.3157
C450MPa15
66
66
56
6
5
5
5
5
44
44
34
4
3
3
3
3
fgf
fg
f
fgf
fg
f
hxhhs
ssx
ssP
sh
TP
hxhhs
ssx
ssP
sh
TP
Noting that for the heat exchanger, the steady-flow energy balance equation yields0pekeWQ
10-100 A combined gas-steam power plant is considered. The topping cycle is a gas-turbine cycle and thebottoming cycle is a nonideal reheat Rankine cycle. The mass flow rate of air in the gas-turbine cycle, therate of total heat input, and the thermal efficiency of the combined cycle are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3Air is an ideal gas with variable specific heats.Analysis (a) The analysis of gas cycle yields (Table A-17)
kJ/kg63.523K520
kJ/kg4.95835.86042.151585.042.1515
kJ/kg35.8603.565.45081
5.450kJ/kg42.1515K1400
kJ/kg1.58580.0/16.29012.52616.290
/
kJ/kg12.526849.92311.18
2311.1kJ/kg16.290K290
1111
109910109
109
109
10
99
787878
78
87
8
77
910
9
78
7
hT
hhhhhhhh
hPP
PP
PhT
hhhhhhhh
hPPPP
PhT
sTs
T
srs
r
r
Css
C
srs
r
r
s
s
T91400 K
Qin·
GASCYCLE
1010s 38
8s 3 MPa15 MPa 5
11
10 kPa
STEAMCYCLE
Qout·
44s
290 K 276s 61
s
From the steam tables (Tables A-4, A-5, and A-6),
kJ/kg95.20614.1581.191kJ/kg15.14
mkPa1kJ1kPa1015,000/kgm0.00101
/kgm00101.0kJ/kg81.191
inpI,12
33
121inpI,
3kPa10@1
kPa10@1
whh
PPw
hh
f
f
v
vv
kJ/kg1.28387.27819.315785.09.3157
kJ/kg7.27819.17949879.03.1008
9880.05402.3
6454.21434.6MPa3
KkJ/kg1428.6kJ/kg9.3157
C450MPa15
433443
43
44
44
34
4
3
3
3
3
sTs
T
fgsfs
fg
fss
s
hhhhhhhh
hxhhs
ssx
ssP
sh
TP
10-95
kJ/kg5.24678.22922.345785.02.3457
kJ/kg8.22921.23928782.081.191
8783.04996.7
6492.02359.7kPa10
KkJ/kg2359.7kJ/kg2.3457
C500MPa3
655665
65
66
66
56
6
5
5
5
5
sTs
T
fgsfs
fg
fss
s
hhhhhhhh
hxhhs
ssx
ssP
sh
TP
Noting that Q for the heat exchanger, the steady-flow energy balance equation yields0pekeW
10-101 It is to be shown tha the exergy destruction associated with a simple ideal Rankine cycle can be expressed as
tthinqx Carnotth,destroyed , where th is efficiency of the Rankine cycle and th, Carnot is
the efficiency of the Carnot cycle operating between the same temperature limits.Analysis The exergy destruction associated with a cycle is given on a unit mass basis as
R
R
Tq
Tx 0destroyed
where the direction of qin is determined with respect to the reservoir (positive if to the reservoir andnegative if from the reservoir). For a cycle that involves heat transfer only with a source at TH and a sink atT0, the irreversibility becomes
thCthCthth
HHH
qqTT
qqqq
TTq
Tq
TqTx
,in,in
0
in
outinin
0out
in
0
out0destroyed
11
10-96
10-102 A cogeneration plant is to produce power and process heat. There are two turbines in the cycle: a high-pressure turbine and a low-pressure turbine. The temperature, pressure, and mass flow rate of steam atthe inlet of high-pressure turbine are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis From the steam tables (Tables A-4, A-5, and A-6),
kJ/kg1.23446.20479.278860.09.2788
kJ/kg6.20471.23927758.081.191
7758.04996.7
6492.04675.6
kPa10
KkJ/kg4675.6kJ/kg9.2788
vaporsat.MPa4.1
544554
54
55
45
45
5
MPa4.1@4
MPa4.1@44
sTs
T
fgsfs
fg
fss
s
g
g
hhhhhhhh
hxhhs
ssx
ssP
sshhP
T
3
44
55s
2
1s
and
kg/min9.107kg/s799.1kJ/kg444.8kJ/s800
kJ/kg8.4441.23449.2788
lowturb,
IIturb,turblow
54lowturb,
wW
m
hhw
Therefore ,
kJ/kg0.28439.278815.54
kJ/kg54.15kg/s18.47kJ/s1000
=kg/min11081081000
4highturb,3
43turbhigh,
,turbhighturb,
total
hwh
hhmW
w
m
I
kg/s18.47
KkJ/kg4289.61840.49908.02835.2
9908.09.1958
96.8298.2770MPa4.1
kJ/kg8.277075.0/9.27880.28430.2843
/
44
44
34
4
433443
43
fgsfs
fg
fss
s
s
Tss
T
sxssh
hhx
ssP
hhhhhhhh
Then from the tables or the software, the turbine inlet temperature and pressure becomes
C227.5MPa2
3
3
3
3KkJ/kg4289.6
kJ/kg0.2843TP
sh
10-97
10-103 A cogeneration plant is to generate power and process heat. Part of the steam extracted from theturbine at a relatively high pressure is used for process heating. The rate of process heat, the net power produced, and the utilization factor of the plant are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis From the steam tables (Tables A-4, A-5, and A-6),
10-104 EES The effect of the condenser pressure on the performance a simple ideal Rankine cycle is to beinvestigated.Analysis The problem is solved using EES, and the solution is given below.
function x4$(x4) "this function returns a string to indicate the state of steam at point 4" x4$=''
if (x4>1) then x4$='(superheated)' if (x4<0) then x4$='(compressed)'
"Pump analysis"Fluid$='Steam_IAPWS'P[1] = P[4] P[2]=P[3]x[1]=0 "Sat'd liquid"h[1]=enthalpy(Fluid$,P=P[1],x=x[1])v[1]=volume(Fluid$,P=P[1],x=x[1])s[1]=entropy(Fluid$,P=P[1],x=x[1])T[1]=temperature(Fluid$,P=P[1],x=x[1])W_p_s=v[1]*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume"W_p=W_p_s/Eta_ph[2]=h[1]+W_p "SSSF First Law for the pump"s[2]=entropy(Fluid$,P=P[2],h=h[2])T[2]=temperature(Fluid$,P=P[2],h=h[2])
"Turbine analysis"h[3]=enthalpy(Fluid$,T=T[3],P=P[3])s[3]=entropy(Fluid$,T=T[3],P=P[3])s_s[4]=s[3]hs[4]=enthalpy(Fluid$,s=s_s[4],P=P[4])Ts[4]=temperature(Fluid$,s=s_s[4],P=P[4])Eta_t=(h[3]-h[4])/(h[3]-hs[4])"Definition of turbine efficiency"T[4]=temperature(Fluid$,P=P[4],h=h[4])s[4]=entropy(Fluid$,h=h[4],P=P[4])x[4]=quality(Fluid$,h=h[4],P=P[4])h[3] =W_t+h[4]"SSSF First Law for the turbine"x4s$=x4$(x[4])
"Boiler analysis"Q_in + h[2]=h[3]"SSSF First Law for the Boiler"
"Condenser analysis"h[4]=Q_out+h[1]"SSSF First Law for the Condenser"
10-105 EES The effect of the boiler pressure on the performance a simple ideal Rankine cycle is to be investigated.Analysis The problem is solved using EES, and the solution is given below.
function x4$(x4) "this function returns a string to indicate the state of steam at point 4" x4$=''
if (x4>1) then x4$='(superheated)' if (x4<0) then x4$='(compressed)'
"Pump analysis"Fluid$='Steam_IAPWS'P[1] = P[4] P[2]=P[3]x[1]=0 "Sat'd liquid"h[1]=enthalpy(Fluid$,P=P[1],x=x[1])v[1]=volume(Fluid$,P=P[1],x=x[1])s[1]=entropy(Fluid$,P=P[1],x=x[1])T[1]=temperature(Fluid$,P=P[1],x=x[1])W_p_s=v[1]*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume"W_p=W_p_s/Eta_ph[2]=h[1]+W_p "SSSF First Law for the pump"s[2]=entropy(Fluid$,P=P[2],h=h[2])T[2]=temperature(Fluid$,P=P[2],h=h[2])
"Turbine analysis"h[3]=enthalpy(Fluid$,T=T[3],P=P[3])s[3]=entropy(Fluid$,T=T[3],P=P[3])s_s[4]=s[3]hs[4]=enthalpy(Fluid$,s=s_s[4],P=P[4])Ts[4]=temperature(Fluid$,s=s_s[4],P=P[4])Eta_t=(h[3]-h[4])/(h[3]-hs[4])"Definition of turbine efficiency"T[4]=temperature(Fluid$,P=P[4],h=h[4])s[4]=entropy(Fluid$,h=h[4],P=P[4])x[4]=quality(Fluid$,h=h[4],P=P[4])h[3] =W_t+h[4] "SSSF First Law for the turbine"x4s$=x4$(x[4])
"Boiler analysis"Q_in + h[2]=h[3] "SSSF First Law for the Boiler"
"Condenser analysis"h[4]=Q_out+h[1] "SSSF First Law for the Condenser"
10-106 EES The effect of superheating the steam on the performance a simple ideal Rankine cycle is to be investigated.Analysis The problem is solved using EES, and the solution is given below.
function x4$(x4) "this function returns a string to indicate the state of steam at point 4" x4$=''
if (x4>1) then x4$='(superheated)' if (x4<0) then x4$='(compressed)'
"Pump analysis"Fluid$='Steam_IAPWS'P[1] = P[4] P[2]=P[3]x[1]=0 "Sat'd liquid"h[1]=enthalpy(Fluid$,P=P[1],x=x[1])v[1]=volume(Fluid$,P=P[1],x=x[1])s[1]=entropy(Fluid$,P=P[1],x=x[1])T[1]=temperature(Fluid$,P=P[1],x=x[1])W_p_s=v[1]*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume"W_p=W_p_s/Eta_ph[2]=h[1]+W_p "SSSF First Law for the pump"s[2]=entropy(Fluid$,P=P[2],h=h[2])T[2]=temperature(Fluid$,P=P[2],h=h[2])
"Turbine analysis"h[3]=enthalpy(Fluid$,T=T[3],P=P[3])s[3]=entropy(Fluid$,T=T[3],P=P[3])s_s[4]=s[3]hs[4]=enthalpy(Fluid$,s=s_s[4],P=P[4])Ts[4]=temperature(Fluid$,s=s_s[4],P=P[4])Eta_t=(h[3]-h[4])/(h[3]-hs[4])"Definition of turbine efficiency"T[4]=temperature(Fluid$,P=P[4],h=h[4])s[4]=entropy(Fluid$,h=h[4],P=P[4])x[4]=quality(Fluid$,h=h[4],P=P[4])h[3] =W_t+h[4]"SSSF First Law for the turbine"x4s$=x4$(x[4])
"Boiler analysis"Q_in + h[2]=h[3]"SSSF First Law for the Boiler"
"Condenser analysis"h[4]=Q_out+h[1]"SSSF First Law for the Condenser"
"Cycle Statistics"W_net=W_t-W_pEta_th=W_net/Q_in
10-105
0 2 4 6 8 10 120
100
200
300
400
500
600
700
s [kJ/kg-K]
T [°
C]
3000 kPa
10 kPa
Steam
1
2
3
4
T3
[C]th Wnet
[kJ/kg]x4
250 0.3241
862.8 0.752
344.4 0.3338
970.6 0.81
438.9 0.3466
1083 0.8536
533.3 0.3614
1206 0.8909
627.8 0.3774
1340 0.9244
722.2 0.3939
1485 0.955
816.7 0.4106
1639 0.9835
911.1 0.4272
1803 100
1006 0.4424
1970 100
1100 0.456 2139 100
200 300 400 500 600 700 800 900 1000 11000.32
0.34
0.36
0.38
0.4
0.42
0.44
0.46
T[3] [C]
th
10-106
200 300 400 500 600 700 800 900 1000 1100750
1050
1350
1650
1950
2250
T[3] [C]
Wne
t [k
J/kg
]
10-107
10-107 EES The effect of reheat pressure on the performance an ideal Rankine cycle is to be investigated.Analysis The problem is solved using EES, and the solution is given below.
function x6$(x6) "this function returns a string to indicate the state of steam at point 6" x6$=''
if (x6>1) then x6$='(superheated)' if (x6<0) then x6$='(subcooled)'
"Pump analysis"Fluid$='Steam_IAPWS'P[1] = P[6] P[2]=P[3]x[1]=0 "Sat'd liquid"h[1]=enthalpy(Fluid$,P=P[1],x=x[1])v[1]=volume(Fluid$,P=P[1],x=x[1])s[1]=entropy(Fluid$,P=P[1],x=x[1])T[1]=temperature(Fluid$,P=P[1],x=x[1])W_p_s=v[1]*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume"W_p=W_p_s/Eta_ph[2]=h[1]+W_p "SSSF First Law for the pump"v[2]=volume(Fluid$,P=P[2],h=h[2])s[2]=entropy(Fluid$,P=P[2],h=h[2])T[2]=temperature(Fluid$,P=P[2],h=h[2])
"High Pressure Turbine analysis"h[3]=enthalpy(Fluid$,T=T[3],P=P[3])s[3]=entropy(Fluid$,T=T[3],P=P[3])v[3]=volume(Fluid$,T=T[3],P=P[3])s_s[4]=s[3]hs[4]=enthalpy(Fluid$,s=s_s[4],P=P[4])Ts[4]=temperature(Fluid$,s=s_s[4],P=P[4])Eta_t=(h[3]-h[4])/(h[3]-hs[4])"Definition of turbine efficiency"T[4]=temperature(Fluid$,P=P[4],h=h[4])s[4]=entropy(Fluid$,h=h[4],P=P[4])v[4]=volume(Fluid$,s=s[4],P=P[4])h[3] =W_t_hp+h[4]"SSSF First Law for the high pressure turbine"
"Low Pressure Turbine analysis"P[5]=P[4]s[5]=entropy(Fluid$,T=T[5],P=P[5])h[5]=enthalpy(Fluid$,T=T[5],P=P[5])s_s[6]=s[5]hs[6]=enthalpy(Fluid$,s=s_s[6],P=P[6])Ts[6]=temperature(Fluid$,s=s_s[6],P=P[6])vs[6]=volume(Fluid$,s=s_s[6],P=P[6])Eta_t=(h[5]-h[6])/(h[5]-hs[6])"Definition of turbine efficiency"
10-108
h[5]=W_t_lp+h[6]"SSSF First Law for the low pressure turbine"x[6]=quality(Fluid$,h=h[6],P=P[6])
"Boiler analysis"Q_in + h[2]+h[4]=h[3]+h[5]"SSSF First Law for the Boiler"
"Condenser analysis"h[6]=Q_out+h[1]"SSSF First Law for the Condenser"T[6]=temperature(Fluid$,h=h[6],P=P[6])s[6]=entropy(Fluid$,h=h[6],P=P[6])x6s$=x6$(x[6])
10-108 EES The effect of number of reheat stages on the performance an ideal Rankine cycle is to be investigated.Analysis The problem is solved using EES, and the solution is given below.
function x6$(x6) "this function returns a string to indicate the state of steam at point 6" x6$=''
if (x6>1) then x6$='(superheated)' if (x6<0) then x6$='(subcooled)'
Fluid$='Steam_IAPWS's5=entropy(Fluid$,T=T5,P=P5)h5=enthalpy(Fluid$,T=T5,P=P5)s_s6=s5hs6=enthalpy(Fluid$,s=s_s6,P=P6)Ts6=temperature(Fluid$,s=s_s6,P=P6)vs6=volume(Fluid$,s=s_s6,P=P6)"Eta_t=(h5-h6)/(h5-hs6)""Definition of turbine efficiency"h6=h5-Eta_t*(h5-hs6)W_t_lp=W_t_lp+h5-h6"SSSF First Law for the low pressure turbine"x6=QUALITY(Fluid$,h=h6,P=P6)Q_in_reheat =Q_in_reheat + (h5 - h4) P3=P4
"Pump analysis"P[1] = P[6] P[2]=P[3]x[1]=0 "Sat'd liquid"h[1]=enthalpy(Fluid$,P=P[1],x=x[1])v[1]=volume(Fluid$,P=P[1],x=x[1])s[1]=entropy(Fluid$,P=P[1],x=x[1])T[1]=temperature(Fluid$,P=P[1],x=x[1])W_p_s=v[1]*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume"W_p=W_p_s/Eta_ph[2]=h[1]+W_p "SSSF First Law for the pump"v[2]=volume(Fluid$,P=P[2],h=h[2])s[2]=entropy(Fluid$,P=P[2],h=h[2])T[2]=temperature(Fluid$,P=P[2],h=h[2])
"High Pressure Turbine analysis"h[3]=enthalpy(Fluid$,T=T[3],P=P[3])s[3]=entropy(Fluid$,T=T[3],P=P[3])v[3]=volume(Fluid$,T=T[3],P=P[3])s_s[4]=s[3]hs[4]=enthalpy(Fluid$,s=s_s[4],P=P[4])Ts[4]=temperature(Fluid$,s=s_s[4],P=P[4])Eta_t=(h[3]-h[4])/(h[3]-hs[4])"Definition of turbine efficiency"T[4]=temperature(Fluid$,P=P[4],h=h[4])s[4]=entropy(Fluid$,h=h[4],P=P[4])v[4]=volume(Fluid$,s=s[4],P=P[4])h[3] =W_t_hp+h[4]"SSSF First Law for the high pressure turbine"
{P[5]=P[4]s[5]=entropy(Fluid$,T=T[5],P=P[5])h[5]=enthalpy(Fluid$,T=T[5],P=P[5])s_s[6]=s[5]hs[6]=enthalpy(Fluid$,s=s_s[6],P=P[6])Ts[6]=temperature(Fluid$,s=s_s[6],P=P[6])vs[6]=volume(Fluid$,s=s_s[6],P=P[6])Eta_t=(h[5]-h[6])/(h[5]-hs[6])"Definition of turbine efficiency"h[5]=W_t_lp+h[6]"SSSF First Law for the low pressure turbine" x[6]=QUALITY(Fluid$,h=h[6],P=P[6])W_t_lp_total = NoRHStages*W_t_lpQ_in_reheat = NoRHStages*(h[5] - h[4])}
"Boiler analysis"Q_in_boiler + h[2]=h[3]"SSSF First Law for the Boiler"Q_in = Q_in_boiler+Q_in_reheat
"Condenser analysis"h[6]=Q_out+h[1]"SSSF First Law for the Condenser"T[6]=temperature(Fluid$,h=h[6],P=P[6])s[6]=entropy(Fluid$,h=h[6],P=P[6])
10-109 EES The effect of extraction pressure on the performance an ideal regenerative Rankine cycle with one open feedwater heater is to be investigated.Analysis The problem is solved using EES, and the solution is given below.
"Condenser exit pump or Pump 1 analysis"Fluid$='Steam_IAPWS'h[1]=enthalpy(Fluid$,P=P[1],x=0) {Sat'd liquid}v1=volume(Fluid$,P=P[1],x=0)s[1]=entropy(Fluid$,P=P[1],x=0)T[1]=temperature(Fluid$,P=P[1],x=0)w_pump1_s=v1*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume"w_pump1=w_pump1_s/Eta_pump "Definition of pump efficiency"h[1]+w_pump1= h[2] "Steady-flow conservation of energy"s[2]=entropy(Fluid$,P=P[2],h=h[2])T[2]=temperature(Fluid$,P=P[2],h=h[2])
"Open Feedwater Heater analysis:"y*h[6] + (1- y)*h[2] = 1*h[3] "Steady-flow conservation of energy"h[3]=enthalpy(Fluid$,P=P[3],x=0)T[3]=temperature(Fluid$,P=P[3],x=0) "Condensate leaves heater as sat. liquid at P[3]"s[3]=entropy(Fluid$,P=P[3],x=0)
"Boiler condensate pump or Pump 2 analysis"v3=volume(Fluid$,P=P[3],x=0)w_pump2_s=v3*(P[4]-P[3])"SSSF isentropic pump work assuming constant specific volume"w_pump2=w_pump2_s/Eta_pump "Definition of pump efficiency"h[3]+w_pump2= h[4] "Steady-flow conservation of energy"s[4]=entropy(Fluid$,P=P[4],h=h[4])T[4]=temperature(Fluid$,P=P[4],h=h[4])
"Boiler analysis"q_in + h[4]=h[5]"SSSF conservation of energy for the Boiler"h[5]=enthalpy(Fluid$, T=T[5], P=P[5]) s[5]=entropy(Fluid$, T=T[5], P=P[5])
"Turbine analysis"ss[6]=s[5]hs[6]=enthalpy(Fluid$,s=ss[6],P=P[6])Ts[6]=temperature(Fluid$,s=ss[6],P=P[6])h[6]=h[5]-Eta_turb*(h[5]-hs[6])"Definition of turbine efficiency for high pressure stages"T[6]=temperature(Fluid$,P=P[6],h=h[6])s[6]=entropy(Fluid$,P=P[6],h=h[6])
10-115
ss[7]=s[6]hs[7]=enthalpy(Fluid$,s=ss[7],P=P[7])Ts[7]=temperature(Fluid$,s=ss[7],P=P[7])h[7]=h[6]-Eta_turb*(h[6]-hs[7])"Definition of turbine efficiency for low pressure stages"T[7]=temperature(Fluid$,P=P[7],h=h[7])s[7]=entropy(Fluid$,P=P[7],h=h[7])h[5] =y*h[6] + (1- y)*h[7] + w_turb "SSSF conservation of energy for turbine"
"Condenser analysis"(1- y)*h[7]=q_out+(1- y)*h[1]"SSSF First Law for the Condenser"
10-110 EES The effect of number of regeneration stages on the performance an ideal regenerative Rankine cycle with one open feedwater heater is to be investigated.Analysis The problem is solved using EES, and the solution is given below.
i=0REPEATi=i+1"The following maintains the same temperature difference between any two regeneration stages."T_FWH[i] = (NoFWH +1 - i)*DELTAT_cond_boiler/(NoFWH + 1)+Tcond"[C]"P_extract[i] = pressure(Fluid$,T=T_FWH[i],x=0)"[kPa]"P3[i]=P_extract[i]P6[i]=P_extract[i]If i > 1 then P4[i] = P6[i - 1]
"Boiler condensate pump or the Pumps 2 between feedwater heaters analysis"h3[i]=enthalpy(Fluid$,P=P3[i],x=0)v3[i]=volume(Fluid$,P=P3[i],x=0)w_pump2_s=v3[i]*(P4[i]-P3[i])"SSSF isentropic pump work assuming constant specific volume"w_pump2[i]=w_pump2_s/Eta_pump "Definition of pump efficiency"h4[i]= w_pump2[i] +h3[i] "Steady-flow conservation of energy"s4[i]=entropy(Fluid$,P=P4[i],h=h4[i])T4[i]=temperature(Fluid$,P=P4[i],h=h4[i])Until i = NoFWH
10-118
i=0REPEATi=i+1"Open Feedwater Heater analysis:"{h2[i] = h6[i]}s5[i] = s[5]ss6[i]=s5[i]hs6[i]=enthalpy(Fluid$,s=ss6[i],P=P6[i])Ts6[i]=temperature(Fluid$,s=ss6[i],P=P6[i])h6[i]=h[5]-Eta_turb*(h[5]-hs6[i])"Definition of turbine efficiency for high pressure stages"If i=1 then y[1]=(h3[1] - h4[2])/(h6[1] - h4[2]) "Steady-flow conservation of energy for the FWH"If i > 1 then js = i -1 j = 0 sumyj = 0 REPEAT j = j+1 sumyj = sumyj + y[ j ] UNTIL j = jsy[i] =(1- sumyj)*(h3[i] - h4[i+1])/(h6[i] - h4[i+1])
ENDIFT3[i]=temperature(Fluid$,P=P3[i],x=0) "Condensate leaves heater as sat. liquid at P[3]"s3[i]=entropy(Fluid$,P=P3[i],x=0)
"Turbine analysis"T6[i]=temperature(Fluid$,P=P6[i],h=h6[i])s6[i]=entropy(Fluid$,P=P6[i],h=h6[i])yh6[i] = y[i]*h6[i]UNTIL i=NoFWHss[7]=s6[i]hs[7]=enthalpy(Fluid$,s=ss[7],P=P[7])Ts[7]=temperature(Fluid$,s=ss[7],P=P[7])h[7]=h6[i]-Eta_turb*(h6[i]-hs[7])"Definition of turbine efficiency for low pressure stages"T[7]=temperature(Fluid$,P=P[7],h=h[7])s[7]=entropy(Fluid$,P=P[7],h=h[7])
"Condenser Pump---Pump_1 Analysis:"P[2] = P6 [ NoFWH] P[1] = P_cond h[1]=enthalpy(Fluid$,P=P[1],x=0) {Sat'd liquid}v1=volume(Fluid$,P=P[1],x=0)s[1]=entropy(Fluid$,P=P[1],x=0)T[1]=temperature(Fluid$,P=P[1],x=0)w_pump1_s=v1*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume"w_pump1=w_pump1_s/Eta_pump "Definition of pump efficiency"
10-119
h[2]=w_pump1+ h[1] "Steady-flow conservation of energy"s[2]=entropy(Fluid$,P=P[2],h=h[2])T[2]=temperature(Fluid$,P=P[2],h=h[2])
"Boiler analysis"q_in = h[5] - h4[1]"SSSF conservation of energy for the Boiler"w_turb = h[5] - sumyh6i - (1- sumyi)*h[7] "SSSF conservation of energy for turbine"
"Condenser analysis"q_out=(1- sumyi)*(h[7] - h[1])"SSSF First Law for the Condenser"
10-111 Consider a steady-flow Carnot cycle with water as the working fluid executed under the saturationdome between the pressure limits of 8 MPa and 20 kPa. Water changes from saturated liquid to saturatedvapor during the heat addition process. The net work output of this cycle is(a) 494 kJ/kg (b) 975 kJ/kg (c) 596 kJ/kg (d) 845 kJ/kg (e) 1148 kJ/kg
Answer (c) 596 kJ/kg
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).
"Some Wrong Solutions with Common Mistakes:"W1_work = Eta1*q_in; Eta1=T2/T1 "Taking Carnot efficiency to be T2/T1"W2_work = Eta2*q_in; Eta2=1-(T2-273)/(T1-273) "Using C instead of K"W3_work = Eta_Carnot*ENTHALPY(Steam_IAPWS,x=1,P=P1) "Using h_g instead of h_fg"W4_work = Eta_Carnot*q2; q2=ENTHALPY(Steam_IAPWS,x=1,P=P2)-ENTHALPY(Steam_IAPWS,x=0,P=P2) "Using h_fg at P2"
10-112 A simple ideal Rankine cycle operates between the pressure limits of 10 kPa and 3 MPa, with a turbine inlet temperature of 600 C. Disregarding the pump work, the cycle efficiency is(a) 24% (b) 37% (c) 52% (d) 63% (e) 71%
Answer (b) 37%
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).
10-113 A simple ideal Rankine cycle operates between the pressure limits of 10 kPa and 5 MPa, with a turbine inlet temperature of 600 C. The mass fraction of steam that condenses at the turbine exit is(a) 6% (b) 9% (c) 12% (d) 15% (e) 18%
Answer (c) 12%
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).
"Some Wrong Solutions with Common Mistakes:"W1_moisture = x4 "Taking quality as moisture"W2_moisture = 0 "Assuming superheated vapor"
10-114 A steam power plant operates on the simple ideal Rankine cycle between the pressure limits of 10kPa and 10 MPa, with a turbine inlet temperature of 600 C. The rate of heat transfer in the boiler is 800 kJ/s. Disregarding the pump work, the power output of this plant is(a) 243 kW (b) 284 kW (c) 508 kW (d) 335 kW (e) 800 kW
Answer (d) 335 kW
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).
P1=10 "kPa"P2=10000 "kPa"
10-123
P3=P2P4=P1T3=600 "C"s4=s3Q_rate=800 "kJ/s"m=Q_rate/q_inh1=ENTHALPY(Steam_IAPWS,x=0,P=P1)h2=h1 "pump work is neglected""v1=VOLUME(Steam_IAPWS,x=0,P=P1)w_pump=v1*(P2-P1)h2=h1+w_pump"h3=ENTHALPY(Steam_IAPWS,T=T3,P=P3)s3=ENTROPY(Steam_IAPWS,T=T3,P=P3)h4=ENTHALPY(Steam_IAPWS,s=s4,P=P4)q_in=h3-h2W_turb=m*(h3-h4)"Some Wrong Solutions with Common Mistakes:"W1_power = Q_rate "Assuming all heat is converted to power"W3_power = Q_rate*Carnot; Carnot = 1-(T1+273)/(T3+273); T1=TEMPERATURE(Steam_IAPWS,x=0,P=P1) "Using Carnot efficiency"W4_power = m*(h3-h44); h44 = ENTHALPY(Steam_IAPWS,x=1,P=P4) "Taking h4=h_g"
10-115 Consider a combined gas-steam power plant. Water for the steam cycle is heated in a well-insulatedheat exchanger by the exhaust gases that enter at 800 K at a rate of 60 kg/s and leave at 400 K. Waterenters the heat exchanger at 200 C and 8 MPa and leaves at 350 C and 8 MPa. If the exhaust gases are treated as air with constant specific heats at room temperature, the mass flow rate of water through the heatexchanger becomes(a) 11 kg/s (b) 24 kg/s (c) 46 kg/s (d) 53 kg/s (e) 60 kg/s
Answer (a) 11 kg/s
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).
"Some Wrong Solutions with Common Mistakes:"m_gas*Cp*(T3 -T4)=W1_msteam*4.18*(T2-T1) "Assuming no evaporation of liquid water"m_gas*Cv*(T3 -T4)=W2_msteam*(h2-h1); Cv=0.718 "Using Cv for air instead of Cp"W3_msteam = m_gas "Taking the mass flow rates of two fluids to be equal"m_gas*Cp*(T3 -T4)=W4_msteam*(h2-h11); h11=ENTHALPY(Steam_IAPWS,x=0,P=P1) "Takingh1=hf@P1"
10-124
10-116 An ideal reheat Rankine cycle operates between the pressure limits of 10 kPa and 8 MPa, withreheat occurring at 4 MPa. The temperature of steam at the inlets of both turbines is 500 C, and theenthalpy of steam is 3185 kJ/kg at the exit of the high-pressure turbine, and 2247 kJ/kg at the exit of thelow-pressure turbine. Disregarding the pump work, the cycle efficiency is(a) 29% (b) 32% (c) 36% (d) 41% (e) 49%
Answer (d) 41%
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).
P1=10 "kPa"P2=8000 "kPa"P3=P2P4=4000 "kPa"P5=P4P6=P1T3=500 "C"T5=500 "C"s4=s3s6=s5h1=ENTHALPY(Steam_IAPWS,x=0,P=P1)h2=h1h44=3185 "kJ/kg - for checking given data"h66=2247 "kJ/kg - for checking given data"h3=ENTHALPY(Steam_IAPWS,T=T3,P=P3)s3=ENTROPY(Steam_IAPWS,T=T3,P=P3)h4=ENTHALPY(Steam_IAPWS,s=s4,P=P4)h5=ENTHALPY(Steam_IAPWS,T=T5,P=P5)s5=ENTROPY(Steam_IAPWS,T=T5,P=P5)h6=ENTHALPY(Steam_IAPWS,s=s6,P=P6)q_in=(h3-h2)+(h5-h4)q_out=h6-h1Eta_th=1-q_out/q_in
"Some Wrong Solutions with Common Mistakes:"W1_Eff = q_out/q_in "Using wrong relation"W2_Eff = 1-q_out/(h3-h2) "Disregarding heat input during reheat"W3_Eff = 1-(T1+273)/(T3+273); T1=TEMPERATURE(Steam_IAPWS,x=0,P=P1) "Using Carnot efficiency"W4_Eff = 1-q_out/(h5-h2) "Using wrong relation for q_in"
10-125
10-117 Pressurized feedwater in a steam power plant is to be heated in an ideal open feedwater heater that operates at a pressure of 0.5 MPa with steam extracted from the turbine. If the enthalpy of feedwater is 252 kJ/kg and the enthalpy of extracted steam is 2665 kJ/kg, the mass fraction of steam extracted from theturbine is (a) 4% (b) 10% (c) 16% (d) 27% (e) 12%
Answer (c) 16%
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).
h_feed=252 "kJ/kg"h_extracted=2665 "kJ/kg"P3=500 "kPa"h3=ENTHALPY(Steam_IAPWS,x=0,P=P3)"Energy balance on the FWH"h3=x_ext*h_extracted+(1-x_ext)*h_feed
10-118 Consider a steam power plant that operates on the regenerative Rankine cycle with one openfeedwater heater. The enthalpy of the steam is 3374 kJ/kg at the turbine inlet, 2797 kJ/kg at the location ofbleeding, and 2346 kJ/kg at the turbine exit. The net power output of the plant is 120 MW, and the fractionof steam bled off the turbine for regeneration is 0.172. If the pump work is negligible, the mass flow rateof steam at the turbine inlet is(a) 117 kg/s (b) 126 kg/s (c) 219 kg/s (d) 288 kg/s (e) 679 kg/s
Answer (b) 126 kg/s
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).
h_in=3374 "kJ/kg"h_out=2346 "kJ/kg"h_extracted=2797 "kJ/kg"Wnet_out=120000 "kW"x_bleed=0.172w_turb=(h_in-h_extracted)+(1-x_bleed)*(h_extracted-h_out)m=Wnet_out/w_turb"Some Wrong Solutions with Common Mistakes:"W1_mass = Wnet_out/(h_in-h_out) "Disregarding extraction of steam"W2_mass = Wnet_out/(x_bleed*(h_in-h_out)) "Assuming steam is extracted at trubine inlet"W3_mass = Wnet_out/(h_in-h_out-x_bleed*h_extracted) "Using wrong relation"
10-126
10-119 Consider a simple ideal Rankine cycle. If the condenser pressure is lowered while keeping turbineinlet state the same, (select the correct statement)
(a) the turbine work output will decrease.(b) the amount of heat rejected will decrease. (c) the cycle efficiency will decrease.(d) the moisture content at turbine exit will decrease.(e) the pump work input will decrease.
Answer (b) the amount of heat rejected will decrease.
10-120 Consider a simple ideal Rankine cycle with fixed boiler and condenser pressures. If the steam issuperheated to a higher temperature, (select the correct statement)
(a) the turbine work output will decrease.(b) the amount of heat rejected will decrease. (c) the cycle efficiency will decrease.(d) the moisture content at turbine exit will decrease.(e) the amount of heat input will decrease.
Answer (d) the moisture content at turbine exit will decrease.
10-121 Consider a simple ideal Rankine cycle with fixed boiler and condenser pressures . If the cycle ismodified with reheating, (select the correct statement)
(a) the turbine work output will decrease.(b) the amount of heat rejected will decrease. (c) the pump work input will decrease. (d) the moisture content at turbine exit will decrease.(e) the amount of heat input will decrease.
Answer (d) the moisture content at turbine exit will decrease.
10-122 Consider a simple ideal Rankine cycle with fixed boiler and condenser pressures . If the cycle ismodified with regeneration that involves one open feed water heater, (select the correct statement per unitmass of steam flowing through the boiler)
(a) the turbine work output will decrease.(b) the amount of heat rejected will increase. (c) the cycle thermal efficiency will decrease.(d) the quality of steam at turbine exit will decrease.(e) the amount of heat input will increase.
Answer (a) the turbine work output will decrease.
10-127
10-123 Consider a cogeneration power plant modified with regeneration. Steam enters the turbine at 6 MPa and 450 C at a rate of 20 kg/s and expands to a pressure of 0.4 MPa. At this pressure, 60% of the steam isextracted from the turbine, and the remainder expands to a pressure of 10 kPa. Part of the extracted steamis used to heat feedwater in an open feedwater heater. The rest of the extracted steam is used for process heating and leaves the process heater as a saturated liquid at 0.4 MPa. It is subsequently mixed with thefeedwater leaving the feedwater heater, and the mixture is pumped to the boiler pressure. The steam in thecondenser is cooled and condensed by the cooling water from a nearby river, which enters the adiabaticcondenser at a rate of 463 kg/s.
1. The total power output of the turbine is(a) 17.0 MW (b) 8.4 MW (c) 12.2 MW (d) 20.0 MW (e) 3.4 MW
Answer (a) 17.0 MW
2. The temperature rise of the cooling water from the river in the condenser is(a) 8.0 C (b) 5.2 C (c) 9.6 C (d) 12.9 C (e) 16.2 C
Answer (a) 8.0 C
3. The mass flow rate of steam through the process heater is(a) 1.6 kg/s (b) 3.8 kg/s (c) 5.2 kg/s (d) 7.6 kg/s (e) 10.4 kg/s
Answer (e) 10.4 kg/s
4. The rate of heat supply from the process heater per unit mass of steam passing through it is(a) 246 kJ/kg (b) 893 kJ/kg (c) 1344 kJ/kg (d) 1891 kJ/kg (e) 2060 kJ/kg
Answer (e) 2060 kJ/kg
10-128
5. The rate of heat transfer to the steam in the boiler is (a) 26.0 MJ/s (b) 53.8 MJ/s (c) 39.5 MJ/s (d) 62.8 MJ/s (e) 125.4 MJ/s
Answer (b) 53.8 MJ/s
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).
Note: The solution given below also evaluates all enthalpies given on the figure.