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
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
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
T
1
4
2
3
th, C60 273 K350 273 K
1 1TT
L
H46.5%
(b) Note that350 C
s2 = s3 = sf + x3sfg
= 0.8313 + 0.891 7.0769 = 7.1368 kJ/kg·KThus , 60 C
s(Table A-6)MPa1.402
2
2
KkJ/kg1368.7C350
PsT
(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.
10-8C Heat rejected decreases; everything else increases.
10-9C Heat rejected decreases; everything else increases.
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),
kJ/kg58.34304.354.340kJ/kg04.3
mkPa1kJ1
kPa503000/kgm001030.0
/kgm001030.0
kJ/kg54.340
in,12
33
121in,
3kPa50@1
kPa50@1
p
p
f
f
whh
PPw
hh
v
vvT
kJ/kg3.2272
7.23048382.054.340
8382.05019.6
0912.15412.6kPa50
KkJ/kg5412.6kJ/kg3.2994
C300MPa3
44
44
34
4
3
3
3
3
fgf
fg
f
hxhh
sss
xss
P
sh
TP
3 MPa
4
2
3
150 kPa
qout
qin
s
Thus,
kJ/kg9.7188.19317.2650kJ/kg8.193154.3403.2272kJ/kg7.265058.3433.2994
outinnet
14out
23in
qqwhhqhhq
and
27.1%7.26508.193111
in
outth q
q
(b) MW25.2kJ/kg9.718kg/s35netnet wmW
10-5
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),
kJ/kg90.20109.1081.191kJ/kg09.10
mkPa1kJ1kPa10000,10/kgm00101.0
/kgm00101.0
kJ/kg81.191
in,12
33
121in,
3kPa10@1
kPa10@1
p
p
f
f
whh
PPw
hh
v
vv T
kJ/kg2089.72392.10.7934191.81
4996.70.64926.5995kPa10
KkJ/kg6.5995kJ/kg3375.1
C500MPa10
44
44
34
4
3
3
3
3
fgf
fg
f
hxhh
sss
xss
P
sh
TP
0.7934
10 MPa
4
2
3
110 kPa
qout
qin
s
(b)
kJ/kg4.12759.18972.3173kJ/kg9.189781.1917.2089kJ/kg2.317390.2011.3375
outinnet
14out
23in
qqwhhqhhq
and
40.2%kJ/kg3173.2kJ/kg1275.4
in
netth q
w
(c) skg164.7 /kJ/kg1275.4
kJ/s210,000
net
net
wW
m
10-6
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),
kJ/kg68.20387.1181.191kJ/kg11.87
0.85/mkPa1
kJ1kPa1010,000/kgm0.00101
/
/kgm00101.0
kJ/kg81.191
in,12
33
121in,
3kPa10@1
kPa10@1
p
pp
f
f
whh
PPw
hh
v
vvT
0.874
0.793
123928119152282
kJ/kg5.2282kPa10
kJ/kg5.22827.20891.337585.01.3375
kJ/kg7.20891.23927934.081.191
44996.7
6492.05995.6kPa10
KkJ/kg5995.6kJ/kg1.3375
C500MPa10
44
4
4
433443
43
44
44
34
4
3
3
3
3
...
hhh
xhP
hhhhhhhh
hxhh
sss
xss
P
sh
TP
fg
f
sTs
T
fgfs
fg
fss
s
s
qin
qout
2
4
s
10 MPa
4
2
3
110 kPa
(b)
kJ/kg7.10807.20904.3171kJ/kg7.209081.1915.2282kJ/kg4.317168.2031.3375
outinnet
14out
23in
qqwhhqhhq
and
34.1%kJ/kg3171.5kJ/kg1080.7
in
netth q
w
(c) kg/s194.3kJ/kg1080.7
kJ/s210,000
net
net
wWm
10-7
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
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),
Btu/lbm43.9841.402.94Btu/lbm41.4
85.0/ftpsia5.4039
Btu1psia21250/lbmft0.01623
/
/lbmft01623.0
Btu/lbm02.94
in,12
33
121in,
3psia2@1
psia2@1
p
Pp
f
f
whh
PPw
hh
v
vv
RBtu/lbm7450.174444.19.017499.0Btu/lbm6.10137.10219.002.94
44
44
fgf
fgf
sxsshxhh
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
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),
kJ/kg40.8958.082.88
kJ/kg58.0mkPa1
kJ1kPa7001400/kgm0008331.0
/kgm0008331.0
kJ/kg82.88
in,12
33
121in,
3MPa7.0@1
MPa7.0@1
p
p
f
f
whh
PPw
hh
v
vv T
kJ/kg.2026221.1769839.082.88
9839.058763.0
33230.09105.0MPa7.0
KkJ/kg9105.0
kJ/kg12.276
vaporsat.MPa4.1
44
44
34
4
MPa4.1@3
MPa4.1@33
fgf
fg
f
g
g
hxhh
sss
xss
P
ss
hhP
R-134a
1.4 MPa
4
2
3
10.7 MPa
qout
qin
s
Thus ,
kJ/kg34.1338.17372.186kJ/kg38.17382.8820.262kJ/kg72.18640.8912.276
outinnet
14out
23in
qqwhhqhhq
and
7.1%kJ/kg186.72kJ/kg13.34
in
netth q
w
(b) kW40.02kJ/kg13.34kg/s3netnet wmW
10-11
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),
kJ/kg198.8706.781.191
kJ/kg7.06mkPa1
kJ1kPa107,000/kgm0.00101
/kgm00101.0
kJ/kg.81191
in,12
33
121in,
3kPa10@1
kPa10@1
p
p
f
f
whh
PPw
hh
v
vv T
kJ/kg3.62151.23928201.081.191
8201.04996.7
6492.08000.6kPa10
KkJ/kg8000.6kJ/kg411.43
C500MPa7
44
44
34
4
3
3
3
3
fgf
fg
f
hxhh
sss
xss
P
sh
TP
7 MPa
4
2
3
110 kPa
qout
qin
s
Thus,
kJ/kg7.12508.19615.3212kJ/kg8.196181.1916.2153kJ/kg5.321287.1984.3411
outinnet
14out
23in
qqwhhqhhq
and 38.9%kJ/kg3212.5kJ/kg1250.7
in
netth q
w
(b) skg036 /.kJ/kg1250.7kJ/s45,000
net
net
wWm
(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),
kJ/kg9.921911.881.191kJ/kg8.11
0.87/mkPa1
kJ1kPa107,000/kgm00101.0
/
/kgm00101.0
kJ/kg91.811
in,12
33
121in,
3kPa10@1
kPa10@1
p
pp
f
f
whh
PPw
hh
v
vvT
kJ/kg1.23176.21534.341187.04.3411
kJ/kg6.21531.2392820.081.191
8201.04996.7
6492.08000.6kPa10
KkJ/kg8000.6kJ/kg.43411
C500MPa7
433443
43
44
44
34
4
3
3
3
3
sTs
T
fgfs
fg
f
hhhhhhhh
hxhh
sss
xss
P
sh
TP
qout
qin2
4
s
7 MPa
4
2
3
110 kPa
Thus,
kJ/kg2.10863.21255.3211kJ/kg3.212581.1911.2317kJ/kg5.321192.1994.3411
outinnet
14out
23in
qqwhhqhhq
and 33.8%kJ/kg3211.5kJ/kg1086.2
in
netth q
w
(b) skg4341 /.kJ/kg1086.2kJ/s45,000
net
net
wWm
(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),
kJ/kg57.26115.1042.251kJ/kg15.10
mkPa1kJ1kPa20000,10/kgm001017.0
/kgm001017.0
kJ/kg42.251
in,12
33
121in,
3kPa20@1
kPa20@1
p
p
f
f
whh
PPw
hh
v
vv
kJ/kg3.1845
5.23576761.042.251
6761.00752.7
8320.06159.5kPa20
KkJ/kg6159.5kJ/kg5.2725
1MPa10
44
44
34
4
3
3
3
3
fgf
fg
f
hxhh
sss
xss
P
sh
xP
s
Rankinecycle
4
3
2
1
T
kJ/kg869.90.15949.2463kJ/kg0.159442.2513.1845kJ/kg9.246357.2615.2725
outinnet
14out
23in
qqwhhqhhq
0.3539.24630.159411
in
outth q
q
(b) Carnot Cycle analysis:
s
Carnotcycle
4
2 3
1
T
kJ/kg9.1093)5.2357)(3574.0(42.251
3574.00752.7
8320.03603.3kPa20
KkJ/kg3603.3kJ/kg8.1407
0C0.311
C0.311kJ/kg5.2725
1MPa10
11
11
21
1
2
2
2
32
3
3
3
3
fgf
fg
f
hxhhs
ssx
ssP
sh
xTT
Th
xP
kJ/kg565.43.7527.1317kJ/kg4.7519.10933.1845
kJ/kg7.13178.14075.2725
outinnet
14out
23in
qqwhhqhhq
0.4307.13174.75111
in
outth q
q
10-14
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:
kJ/kg74.689C5.179
kPa410
kJ/kg40.670kPa410
KkJ/kg5457.2kJ/kg54.761
C147kPa3250
44
4
434
4
3
3
3
3
hTP
hss
P
sh
TP
s
4
3 2
1
Geothermalwater out
Geothermalwater in
heat exchanger
pumpturbine
air-cooledcondenser
0.78840.67054.76174.68954.761
43
43
sT hh
hh
T
qin
qout
2s
4
s
3.25 MPa
4s
2
3
1410 kPa
(b) Pump:
kJ/kg82.27881.501.273kJ/kg81.5
90.0/mkPa1
kJ1kPa4103250/kgm001842.0
/
/kgm001842.0
kJ/kg01.273
in,12
33
121in,
3kPa410@1
kPa014@1
p
Pp
f
f
whh
PPw
hh
v
vv
kW941,21kJ/kg)74.68954.761kJ/kg)(305.6()( 43outT, hhmW
kW1777kJ/kg)1kJ/kg)(5.8305.6()( inp,12inP, wmhhmW
kW20,1651777941,21inP,outT,net WWW
Heat Exchanger:
kW656,162C)90160(C)kJ/kg..184(kJ/kg).9555()( outingeogeoin TTcmQ
(c) 12.4%0.124656,162165,20
in
netth Q
W
10-15
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
(b) Turbine:
kJ/kg7.2344)1.2392)(90.0(81.19190.0kPa10
kJ/kg3.2160kPa10
KkJ/kg8207.6kJ/kg1.2748
1kPa500
444
4
434
4
3
3
3
3
fgf
s
hxhhxP
hss
P
sh
xP
productionwell
Flashchamber
65
4
3
2condenser
1
steamturbine
separator
reinjectionwell
0.6863.21601.27487.23441.2748
43
43
sT hh
hh
(c) The power output from the turbine is
kW15,410kJ/kg)7.23448.1kJ/kg)(27438.20()( 433outT, hhmW
(d) We use saturated liquid state at the standard temperature for dead state enthalpy
kJ/kg83.1040
C250
0
0 hxT
kW622,203kJ/kg)83.104.14kJ/kg)(990230()( 011in hhmE
7.6%0.0757622,203
410,15
in
outT,th E
W
10-16
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
kW15,410kJ/kg)7.23448.1kJ/kg)(27438.20()( 433outT1, hhmW
kW5191kJ/kg)7.23443.1kJ/kg)(269.9014()( 488outT2, hhmW
(c) We use saturated liquid state at the standard temperature for the dead state enthalpy
kJ/kg83.1040
C250
0
0 hxT
kW621,203kJ/kg)83.10414kg/s)(990.230()( 011in hhmE
10.1%0.101621,203
5193410,15
in
outT,th E
W
10-17
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)
1661.0kJ/kg14.990
kPa500
kJ/kg14.9900
C230
212
2
11
1
xhh
P
hxT
kg/s80.19120.38230kg/s38.20kg/s)230)(1661.0(
316
123
mmmmxm
kJ/kg7.234490.0kPa10
kJ/kg1.27481
kPa500
44
4
33
3
hxP
hxP
kJ/kg04.3770
C90
kJ/kg09.6400
kPa500
77
7
66
6
hxT
hxP
The isobutene propertiesare obtained from EES:
/kgm001839.0kJ/kg83.270
0kPa400
kJ/kg01.691C80kPa400
kJ/kg05.755C145kPa3250
310
10
10
10
99
9
88
8
v
hxP
hTP
hTP
1
19
8
7
65
4
3
2
1
flashchamber
condenser
air-cooledcondenser
separator
BINARYCYCLE
pump
heat exchanger
isobutaneturbine
reinjectionwell
productionwell
steamturbine
kJ/kg65.27682.583.270kJ/kg.82.5
90.0/mkPa1
kJ1kPa4003250/kgm001819.0
/
in,1011
33
101110in,
p
pp
whh
PPw v
An energy balance on the heat exchanger gives
kg/s105.46isoiso
118iso766
kg276.65)kJ/-(755.05kg377.04)kJ/-09kg/s)(640.81.191(
)()(
mm
hhmhhm
(b) The power outputs from the steam turbine and the binary cycle are
kW15,410kJ/kg)7.23448.1kJ/kg)(27438.19()( 433steamT, hhmW
10-18
kW6139)kJ/kg82.5)(kg/s46.105(6753
kW6753kJ/kg)01.691.05kJ/kg)(755.46105()(
,isoisoT,binarynet,
98isoT,
inp
iso
wmWW
hhmW
(c) The thermal efficiencies of the binary cycle and the combined plant are
kW454,50kJ/kg)65.276.05kJ/kg)(755.46105()( 118isobinaryin, hhmQ
12.2%0.122454,50
6139
binaryin,
binarynet,binaryth, Q
W
kJ/kg83.1040
C250
0
0 hxT
kW622,203kJ/kg)83.104.14kJ/kg)(990230()( 011in hhmE
10.6%0.106622,203
6139410,15
in
binarynet,steamT,plantth, E
WW
The Reheat Rankine Cycle
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.
"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"}
"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])
"Cycle Statistics"W_net=W_t_hp+W_t_lp-W_pEff=W_net/Q_in
0.0 1.1 2.2 3.3 4.4 5.5 6.6 7.7 8.8 9.9 11.00
100
200
300
400
500
600
700
s [kJ/kg-K ]
T [C
]
8000 kPa
3000 kPa
20 kPa
3
4
5
6
Ideal R ankine cycle w ith reheat
1,2
SOLUTIONEff=0.389 Eta_p=1 Eta_t=1Fluid$='Steam_IAPWS' h[1]=251.4 [kJ/kg] h[2]=259.5 [kJ/kg]h[3]=3400 [kJ/kg] h[4]=3105 [kJ/kg] h[5]=3457 [kJ/kg]h[6]=2385 [kJ/kg] hs[4]=3105 [kJ/kg] hs[6]=2385 [kJ/kg]P[1]=20 [kPa] P[2]=8000 [kPa] P[3]=8000 [kPa]P[4]=3000 [kPa] P[5]=3000 [kPa] P[6]=20 [kPa]Q_in=3493 [kJ/kg] Q_out=2134 [kJ/kg] s[1]=0.832 [kJ/kg-K]s[2]=0.8321 [kJ/kg-K] s[3]=6.727 [kJ/kg-K] s[4]=6.727 [kJ/kg-K]s[5]=7.236 [kJ/kg-K] s[6]=7.236 [kJ/kg-K] s_s[4]=6.727 [kJ/kg-K]s_s[6]=7.236 [kJ/kg-K] T[1]=60.06 [C] T[2]=60.4 [C]T[3]=500 [C] T[4]=345.2 [C] T[5]=500 [C]T[6]=60.06 [C] Ts[4]=345.2 [C] Ts[6]=60.06 [C]v[1]=0.001017 [m^3/kg] v[2]=0.001014 [m^3/kg] v[3]=0.04177 [m^3/kg] v[4]=0.08968 [m^3/kg] vs[6]=6.922 [m^3/kg] W_net=1359 [kJ/kg]W_p=8.117 [kJ/kg] W_p_s=8.117 [kJ/kg] W_t_hp=294.8 [kJ/kg]W_t_lp=1072 [kJ/kg] x6s$='' x[1]=0x[6]=0.9051
10-22
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),
kJ/kg95.20614.1581.191
kJ/kg.1415mkPa1
kJ1kPa10000,15/kgm00101.0
/kgm00101.0
kJ/kg81.191
in,12
33
121in,
3kPa10@sat1
kPa10@sat1
p
p
whh
PPw
hh
v
vv
T
kJ/kg2.2817MPa161.2
kJ/kg53.3466pressurereheattheC500
KkJ/kg3988.74996.790.06492.0
kJ/kg7.23441.239290.081.191kPa10
KkJ/kg3480.6kJ/kg.83310
C500MPa15
434
4
5
5
65
5
66
66
56
6
3
3
3
3
hss
P
hP
ssT
sxss
hxhh
ssP
sh
TP
fgf
fgf
kPa2161
10 kPa
15 MPa4
3
6
2
5
1s
(b) The rate of heat supply is
kW45,038kJ/kg2.281753.346695.2068.3310kg/s124523in hhhhmQ
(c) The thermal efficiency is determined from
Thus,
42.6%kJ/s45,039kJ/s25,834
11
kJ/s,83525kJ/kg81.1917.2344kJ/s12
in
outth
16out
hhmQ
10-26
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,
1462.081.2517.266581.25166.604
26
23
hhhh
y
Then,
kJ/kg7.167542.2510.22141462.011kJ/kg2.269273.6109.3302
17out
45in
hhyqhhq
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,
1373.007136.0195.1377.261895.13771.5041
29
23 yhhhh
z
Then,
kJ/kg2.13888.15560.2945kJ/kg1556.8137.752105.01373.007133.011
kJ/kg0.294573.6808.3625
outinnet
110out
67in
qqwhhzyq
hhq
and
MW30.5kW540,30kJ/kg1388.2kg/s22netnet wmW
(b) 47.1%kJ/kg2945.0kJ/kg1556.8
11in
outth q
q
10-32
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,
1381.010.1925.2578
10.19287.7200753.010.19243.561
210
2723
hhhhyhhz
Then,
kJ/kg12491.15001.2749kJ/kg1.150081.1910.21001381.00753.011
kJ/kg1.274936.7275.3476
outinnet
111out
58in
qqwhhzyq
hhq
and
kg/s200.2kJ/kg1249
kJ/s250,000
net
net
wWm
(b) 45.4%kJ/kg2749.1kJ/kg1500.1
11in
outth q
q
10-34
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.
"Input Data"P[8] = 12500 [kPa] T[8] = 550 [C] P[9] = 800 [kPa] "P_cfwh=300 [kPa]"P[10] = P_cfwhP_cond=10 [kPa] P[11] = P_cond W_dot_net=250 [MW]*Convert(MW, kW) Eta_turb= 100/100 "Turbine isentropic efficiency"Eta_turb_hp = Eta_turb "Turbine isentropic efficiency for high pressure stages"Eta_turb_ip = Eta_turb "Turbine isentropic efficiency for intermediate pressure stages"Eta_turb_lp = Eta_turb "Turbine isentropic efficiency for low pressure stages"Eta_pump = 100/100 "Pump isentropic efficiency"
"Condenser exit pump or Pump 1 analysis"
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"
"Cycle Statistics"w_net=w_turb - ((1-y-z)*w_pump1+ w_pump2) Eta_th=w_net/q_inW_dot_net = m_dot * w_net
10-36
turb turb th m [kg/s]0.7 0.7 0.3916 231.6
0.75 0.75 0.4045 224.30.8 0.8 0.4161 218
0.85 0.85 0.4267 212.60.9 0.9 0.4363 207.9
0.95 0.95 0.4452 203.81 1 0.4535 200.1
0.7 0.75 0.8 0.85 0.9 0.95 10.39
0.4
0.41
0.42
0.43
0.44
0.45
0.46
200
205
210
215
220
225
230
235
turb
th
m[kg/s]
turb = pump
0 2 4 6 8 10 120
100
200
300
400
500
600
s [kJ/kg-K]
T[C] 12500 kPa
800 kPa
300 kPa
10 kPa
Steam
1,2
3,4
5,6
7
8
910
11
10-37
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,
2017.061.1921.281261.19287.720
26
23
hhhh
y
Then,
kJ/kg6.14665.18381.3305kJ/kg5.183881.1917.24942017.011
kJ/kg1.33051.28123.34812017.0112.7310.35021
outinnet
18out
6745in
qqwhhyq
hhyhhq
and kg/s54.5kJ/kg1466.1kJ/s80,000
net
net
wW
m
(b) 44.4%kJ/kg3305.1kJ/kg1466.1
in
netth q
w
10-38
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,
2019.090.20113.73187.7207.2812
90.20113.731
2936
29
hhhhhhy
Then,
kJ/kg6.14668.18375.3304kJ/kg9.183781.1917.24942019.011
kJ/kg5.33047.28123.34812019.0113.7310.35021
outinnet
18out
6745in
qqwhhyq
hhyhhq
and
kg/s54.5kJ/kg1467.1kJ/s80,000
net
net
wW
m
(b) 44.4%kJ/kg3304.5kJ/kg1837.8
11in
outth q
q
10-40
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,
1125.01300.0184.690.135684.6914.2361
211
23 yhhhh
z
Then,
Btu/lbm4.6714.7448.1415Btu/lbm744.469.721052.41125.01300.011
Btu/lbm8.14158.12483.15311300.0141.3805.15501
outinnet
112out
91067in
qqwhhzyq
hhyhhq
and
lbm/s282.5Btu/lbm1415.8
Btu/s104 5
in
in
m
(b) MW200.1Btu1
kJ1.055Btu/lbm671.4lbm/s282.5netnet wmW
(c) 47.4%Btu/lbm1415.8Btu/lbm744.4
11in
outth q
q
10-42
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
(c) Then,
kJ/kg1.184042.2512.24921788.011kJ/kg7.30295.32209.355025.7775.3476
19out
6745in
hhyqhhhhq
and
kW28,550kJ/kg)1.18407.3029)(kg/s24()( outinnet qqmW
(b) The thermal efficiency is determined from
39.3%393.0kJ/kg3029.7kJ/kg1840.1
11in
outth q
q
10-44
Second-Law Analysis of Vapor Power Cycles
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
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
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,
03003
03
023
030033 2ssThhqzssThh V
where KkJ/kg2533.0kJ/kg95.71
K290@kPa100,K290@0
K290@kPa100,K290@0
f
f
ssshhh
Thus, kJ/kg1440.9KkJ/kg2532.0800.6K290kJ/kg95.714.34113
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$(x[6])"Cycle Statistics"W_net=W_t_hp+W_t_lp-W_pEff=W_net/Q_in"The irreversibilities (or exergy destruction) for each of the processes are:"q_R_23 = - (h[3] - h[2]) "Heat transfer for the high temperature reservoir to process 2-3"i_23 = T_o*(s[3] -s[2] + q_R_23/T_R_H)q_R_45 = - (h[5] - h[4]) "Heat transfer for the high temperature reservoir to process 4-5"i_45 = T_o*(s[5] -s[4] + q_R_45/T_R_H)q_R_61 = (h[6] - h[1]) "Heat transfer to the low temperature reservoir in process 6-1"i_61 = T_o*(s[1] -s[6] + q_R_61/T_R_L)i_34 = T_o*(s[4] -s[3])i_56 = T_o*(s[6] -s[5])i_12 = T_o*(s[2] -s[1])
0 .0 1 .1 2 .2 3 .3 4 .4 5 .5 6 .6 7 .7 8.8 9 .9 11 .00
100
200
300
400
500
600
700
s [kJ /kg -K ]
T [C
]
8000 kP a
300 0 kP a
20 kP a
3
4
5
6
Id ea l R an k in e cyc le w ith reh eat
1 ,2
SOLUTION
Eff=0.389Eta_p=1Eta_t=1Fluid$='Steam_IAPWS'h[1]=251.4 [kJ/kg] h[2]=259.5 [kJ/kg] h[3]=3400 [kJ/kg] h[4]=3105 [kJ/kg] h[5]=3457 [kJ/kg] h[6]=2385 [kJ/kg] hs[4]=3105 [kJ/kg] hs[6]=2385 [kJ/kg] i_12=0.012 [kJ/kg] i_23=1245.038 [kJ/kg] i_34=-0.000 [kJ/kg] i_45=94.028 [kJ/kg] i_56=0.000 [kJ/kg] i_61=212.659 [kJ/kg] P[1]=20 [kPa] P[2]=8000 [kPa]
P[3]=8000 [kPa] P[4]=3000 [kPa] P[5]=3000 [kPa] P[6]=20 [kPa] Q_in=3493 [kJ/kg] Q_out=2134 [kJ/kg] q_R_23=-3140 [kJ/kg] q_R_45=-352.5 [kJ/kg] q_R_61=2134 [kJ/kg] s[1]=0.832 [kJ/kg-K] s[2]=0.8321 [kJ/kg-K] s[3]=6.727 [kJ/kg-K] s[4]=6.727 [kJ/kg-K] s[5]=7.236 [kJ/kg-K] s[6]=7.236 [kJ/kg-K] s_s[4]=6.727 [kJ/kg-K] s_s[6]=7.236 [kJ/kg-K] T[1]=60.06 [C] T[2]=60.4 [C] T[3]=500 [C]
T[4]=345.2 [C] T[5]=500 [C] T[6]=60.06 [C] Ts[4]=345.2 [C] Ts[6]=60.06 [C] T_o=300 [K] T_R_H=1800 [K] T_R_L=300 [K] v[1]=0.001017 [m^3/kg] v[2]=0.001014 [m^3/kg] v[3]=0.04177 [m^3/kg] v[4]=0.08968 [m^3/kg] vs[6]=6.922 [m^3/kg] W_net=1359 [kJ/kg] W_p=8.117 [kJ/kg] W_p_s=8.117 [kJ/kg] W_t_hp=294.8 [kJ/kg] W_t_lp=1072 [kJ/kg] x6s$=''x[1]=0
10-31
x[6]=0.9051
10-48
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
03003
03
023
030033 2ssThhqz
VssThh
where
KkJ/kg1806.0kJ/kg51.50
K285@kPa100K,285@0
K285@kPa100K,285@0
f
f
ssshhh
Thus,kJ/kg1495KkJ/kg1806.05995.6K285kJ/kg51.501.33753
10-49
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)
kJ/kg.K6841.21661.0
kJ/kg14.990kPa500
kJ/kg.K6100.2kJ/kg14.990
0C230
2
2
12
2
1
1
1
1
sx
hhP
sh
xT
kg/s38.19kg/s)230)(1661.0(123 mxm
KkJ/kg7739.7kJ/kg3.2464
95.0kPa10
KkJ/kg8207.6kJ/kg1.2748
1kPa500
4
4
4
4
3
3
3
3
sh
xP
sh
xP
KkJ/kg8604.1kJ/kg09.640
0kPa500
6
6
6
6
sh
xP
kg/s81.19119.38230316 mmm
2
productionwell
reinjectionwell
separator
steamturbine
1
Flashchamber
65
4
3
condenser
The power output from the turbine is
kW10,842kJ/kg)3.24648.1kJ/kg)(27438.19()( 433T hhmW
We use saturated liquid state at the standard temperature for dead state properties
kJ/kg3672.0kJ/kg83.104
0C25
0
0
0
0
sh
xT
kW622,203kJ/kg)83.104.14kJ/kg)(990230()( 011in hhmE
5.3%0.0532622,203
842,10
in
outT,th E
W
(b) The specific exergies at various states are kJ/kg53.216kJ/kg.K)3672.0K)(2.6100(298kJ/kg)83.104(990.14)( 010011 ssThh
kJ/kg44.194kJ/kg.K)3672.0K)(2.6841(298kJ/kg)83.104(990.14)( 020022 ssThh
kJ/kg10.719kJ/kg.K)3672.0K)(6.8207(298kJ/kg)83.104(2748.1)( 030033 ssThh
kJ/kg05.151kJ/kg.K)3672.0K)(7.7739(298kJ/kg)83.104(2464.3)( 040044 ssThh
kJ/kg97.89kJ/kg.K)3672.0K)(1.8604(298kJ/kg)83.104(640.09)( 060066 ssThh
The exergy of geothermal water at state 6 is
kW17,257kJ/kg)7kg/s)(89.9.81191(666 mX
10-51
(c) Flash chamber:
kW5080kJ/kg)44.19453kg/s)(216.230()( 211FCdest, mX
89.8%0.89853.21644.194
1
2FCII,
(d) Turbine:
kW10,854kW10,842-kJ/kg)05.15110kg/s)(719.19.38()( T433Tdest, WmX
50.0%0.500)kJ/kg05.15110kg/s)(719.19.38(
kW842,10)( 433
TTII, m
W
(e) Plant:
kW802,49kJ/kg)53kg/s)(216.230(11Plantin, mX
kW38,960842,10802,49TPlantin,Plantdest, WXX
21.8%0.2177kW802,49kW842,10
Plantin,
TPlantII, X
W
Cogeneration
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),
kJ/kg38.670
kJ/kg41.19260.081.191kJ/kg0.60
mkPa1kJ1
kPa10600/kgm0.00101
/kgm00101.0kJ/kg81.191
MPa0.6@3
inpI,12
33
121inpI,
3kPa10@1
kPa10@1
f
f
f
hh
whh
PPw
hh
v
vv
7
3
24P II P I
Processheater Condenser
BoilerTurbine
5
1
8
6
Mixing chamber:
332244
outin(steady)0
systemoutin 0
hmhmhmhmhm
EEEEE
eeii
or,
kJ/kg47.31857.690.311kJ/kg6.57
mkPa1kJ1kPa6007000/kgm0.001026
/kgm001026.0
kJ/kg90.31130
38.67050.741.19250.22
inII,45
33
454inII,
3kJ/kg90.311@4
4
33224
p
p
hf
whh
PPw
mhmhmh
f
v
vv
T
KkJ/kg8000.6kJ/kg4.3411
C500MPa7
6
6
6
6sh
TP
kJ/kg6.21531.23928201.081.191
8201.04996.7
6492.08000.6kPa10
kJ/kg6.2774MPa6.0
88
88
68
8
767
7
fgf
fg
f
hxhhs
ssx
ssP
hss
PQout·
3
5
4
10 kPa
7 MPa
0.6 MPa Qproces·
Qin·
7
6
8
2
1s
Then,
kW32,8666.210077,33
kW210.6kJ/kg6.57kg/s30kJ/kg0.60kg/s22.5
kW077,33kJ/kg6.21536.2774kg/s22.5kJ/kg6.27744.3411kg/s30
inp,outT,net
inpII,4inpI,1inp,
878766outT,
WWW
wmwmW
hhmhhmW
Also, kW782,15kJ/kg38.6706.2774kg/s7.5377process hhmQ
and
52.4%788,92
782,15866,32
kW788,9247.3184.3411kg/s30
in
processnet
565in
QQW
hhmQ
u
10-53
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),
Btu/lbm13.282
Btu/lbm29.9427.002.94Btu/lbm0.27
ftpsia5.4039Btu1
psia2)80)(/lbmft0.01623(86.01
/
/lbmft01623.0Btu/lbm02.94
psia80@3
inpI,12
3
3
121inpI,
3psia2@1
psia2@1
f
p
f
f
hh
whh
PPw
hh
v
vv
7
3
24P II P I
Processheater Condenser
BoilerTurbine
5
1
8
6
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 out
0
4 4 2 2 3 3
0 T
or,
Btu/lbm72.17229.343.169Btu/lbm3.29
86.0/ftpsia5.4039
Btu1psia801000/lbmft0.01664
/
/lbmft01664.0
Btu/lbm43.1695
13.282229.943
in,45
33
454inpII,
3Btu/lbm43.169@4
4
33224
pII
p
hf
whh
PPw
mhmhm
h
f
v
vv8s
7Qin·
3
5
480 psia
Qproces·
2 psia
1000psia
7s
6
8
2
1s
RBtu/lbm6535.1Btu/lbm2.1506
F1000psia1000
6
6
6
6sh
TP
Btu/lbm98.9597.10218475.002.94
8475.074444.1
17499.06535.1psia2
Btu/lbm0.1209psia80
88
88
68
8
767
7
fgsfs
fg
fss
s
s
ss
s
hxhhs
ssx
ssP
hss
P
Then, Btu/s6667Btu/lbm72.1722.1506lbm/s5565in hhmQ
(b)
kW2026Btu/s1921Btu/lbm959.981209.0lbm/s3Btu/lbm1209.01506.2lbm/s586.0
878766,outT, sssTsTT hhmhhmWW
10-54
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"
"Cycle Statistics"w_net=w_turb - ((1- y)*w_pump1+ w_pump2)Eta_th=w_net/q_inW_dot_net = m_dot * w_net
Pextract[kPa]
th m[kg/s]
Qprocess[kW]
100 0.3413 15.26 9508200 0.3284 16.36 9696300 0.3203 17.12 9806400 0.3142 17.74 9882500 0.3092 18.26 9939600 0.305 18.72 9984
0 2 4 6 8 10 120
100
200
300
400
500
600
700
s [kJ/kg-K]
T [°
C]
6000 kPa
400 kPa
10 kPa
Steam
1
2 3,4,9
5
6
7
8
10-58
100 200 300 400 500 60015
15.5
16
16.5
17
17.5
18
18.5
19
Pextract [kPa]
m [
kg/s
]
100 200 300 400 500 6009500
9600
9700
9800
9900
10000
Pextract [kPa]
Qpr
oces
s [k
W]
100 200 300 400 500 6000.305
0.31
0.315
0.32
0.325
0.33
0.335
0.34
0.345
Pextract [kPa]
th
10-59
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),
kJ/kg38.670
kJ/kg40.192596.081.191kJ/kg596.0
mkPa1kJ1
kPa01600/kgm0.00101
/kgm00101.0kJ/kg81.191
MPa6.0@3
inpI,12
33
121inpI,
3kPa01@1
kPa10@1
f
f
f
hh
whh
PPw
hh
v
vv
Mixing chamber:
kJ/kg90.311)1()kJ/kg))(192.4075.0(kJ/kg))(670.3825.0( 44
442233
hh
hmhmhm
kJ/kg47.318563.690.311kJ/kg.5636
mkPa1kJ1kPa0067000/kgm0.001026
/kgm001026.0
inII,45
33
454inII,
3kJ/kg90.311@4
p
p
hf
whh
PPw
f
v
vv
KkJ/kg8000.6kJ/kg4.3411
C500MPa7
6
6
6
6sh
TP
kJ/kg9.2773MPa6.0
767
7 hss
P
kJ/kg6.2153kPa10
868
8 hss
P
10-61
kg/s4.088kJ/kg38.6709.2773kJ/s8600
7
7
377process
mm
hhmQ
This is one-fourth of the mass flowing through the boiler. Thus, the mass flow rate of steam that must besupplied by the boiler becomes
kg/s16.35kg/s)4.088(44 76 mm
(b) Cycle analysis:
kW17,919115033,18
kW6.114kJ/kg6.563kg/s16.35kJ/kg0.596kg/s4.088-16.35
kW033,18kJ/kg6.21534.3411kg/s4.088-16.35kJ/kg9.27734.3411kg/s088.4
inp,outT,net
inpII,4inpI,1inp,
868767outT,
WWW
wmwmW
hhmhhmW
(c) Then,
and
52.4%524.0581,50
8600919,17
kW581,5046.3184.3411kg/s16.35
in
processnet
565in
QQW
hhmQ
u
Combined Gas-Vapor Power Cycles
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
326336622
outin
(steady)0systemoutin
11
0
hhyyhhmhmhmhmhm
EE
EEE
eeii
Thus,
kJ/kg23.9562.20951.25861792.011.25864.31391
extractedsteamoffractionthe1792.001.2521.258601.25238.670
7665
26
23
hhyhhw
hhhhy
T
kJ/kg3.44419.3005.6358.73542.1515
kJ/kg56.94815.859.01792.0123.9561
891110in,gasnet,
,,in,steamnet,
hhhhwww
wwywwww
CT
IIpIpTpT
The net work output per unit mass of gas is
kJ/kg8.54956.9483.444 99.81
steamnet,99.81
gasnet,net www
and
kW720,215kJ/kg5.63542.1515kg/s818.5
kg/s7.188kJ/kg549.7
kJ/s450,000
910airin
net
netair
hhmQ
wWm
(c) 62.5%kW720,215kW450,000
in
net
QW
th
10-65
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
Pratio MassRatiogastosteam
Wnetgas[kW]
Wnetsteam[kW]
th[%]
NetWorkRatiogastosteam
10 7.108 342944 107056 59.92 3.20311 7.574 349014 100986 60.65 3.45612 8.043 354353 95647 61.29 3.70513 8.519 359110 90890 61.86 3.95114 9.001 363394 86606 62.37 4.19615 9.492 367285 82715 62.83 4.4416 9.993 370849 79151 63.24 4.68517 10.51 374135 75865 63.62 4.93218 11.03 377182 72818 63.97 5.1819 11.57 380024 69976 64.28 5.43120 12.12 382687 67313 64.57 5.685
0.0 1.1 2.2 3.3 4.4 5.5 6.6 7.7 8.8 9.9 11.0200
300
400
500
600
700
800
900
1000
1100
1200
1300
1400
1500
1600
s [kJ/kg-K]
T [K
]
8000 kPa
600 kPa
20 kPa
Combined Gas and Steam Power Cycle
8
9
10
11
12
1,23,4
5
6
7
Steam CycleSteam Cycle
Gas CycleGas Cycle
10-68
5 9 13 17 21 2555
57.2
59.4
61.6
63.8
66
P ratio
th[%
]
5 9 14 18 232.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
Pratio
Net
Wor
kRat
ioga
stos
team
W dot,gas / Wdot,steam vs Gas Pressure Ratio
5 9 14 18 235.0
6.0
7.0
8.0
9.0
10.0
11.0
12.0
13.0
14.0
Pratio
Mas
sRat
ioga
stos
team
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
326336622
outin(steady)0
systemoutin
11
0
hhyyhhmhmhmhmhm
EEEEE
eeii
Thus,
kJ/kg5.8243.22413.26631735.013.26634.313986.01
extractedsteamoffractionthe1735.001.2523.266301.25238.670
7665
26
23
hhyhhw
hhhh
y
TT
kJ/kg56.26119.3001.70995.84442.1515
kJ/kg9.81515.859.01735.015.8241
891110in,gasnet,
IIp,Ip,inp,steamnet,
hhhhwww
wwywwww
CT
TT
The net work output per unit mass of gas is
kJ/kg55.3889.81556.261 425.61
steamnet,423.61
gasnet,net www
kg/s2.1158kJ/kg388.55
kJ/s450,000
net
netair w
Wm
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
Pratio MassRatiogastosteam
Wnetgas[kW]
Wnetsteam[kW]
th[%]
NetWorkRatiogastosteam
6 4.463 262595 187405 45.29 1.4018 5.024 279178 170822 46.66 1.634
10 5.528 289639 160361 47.42 1.80612 5.994 296760 153240 47.82 1.93714 6.433 301809 148191 47.99 2.03715 6.644 303780 146220 48.01 2.07816 6.851 305457 144543 47.99 2.11318 7.253 308093 141907 47.87 2.17120 7.642 309960 140040 47.64 2.21322 8.021 311216 138784 47.34 2.242
0.0 1.1 2.2 3.3 4.4 5.5 6.6 7.7 8.8 9.9 11.0200
300
400
500
600
700
800
900
1000
1100
1200
1300
1400
1500
1600
s [kJ/kg-K]
T [K
]
8000 kPa
600 kPa
20 kPa
Combined Gas and Steam Power Cycle
8
9
10
11
12
1,23,4
5
6
7
Steam CycleSteam Cycle
Gas CycleGas Cycle
10-74
5 9 12 16 19 2345.0
45.5
46.0
46.5
47.0
47.5
48.0
48.5
Pratio
th [
%]
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
0pekeW
kJ/kg0.2965)62.47598.871)(10()5.3264()37.1995.3346()15.1( 44
1110air4523
outin
hh
hhmhhmhhm
hmhm
EE
ss
eeii
Also,
sTs
T
ss
hhhhhhhh
hssP
sh
TP
433443
43
434
4
3
3
3
3 MPa1?
MPa6
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
(c) kW4328kJ/kg98.8718.1304kg/s10109airgasT, hhmW
kW2687kJ/kg50.28821.557kg/s1078airgasC, hhmW
kW164126874328gasC,gasT,gasnet, WWW
kW1265kJ/kg0.25465.32640.29655.3346kg/s1.156543ssteamT, hhhhmW
kW7.8kJ/kg564.7kg/s1.15ssteamP, pumpwmW
kW12567.81265steamP,steamT,steamnet, WWW
kW289712561641steamnet,gasnet,plantnet, WWW
(d) kW7476kJ/kg21.5578.1304kg/s1089airin hhmQ
38.8%0.388kW7476kW2897
in
plantnet,th Q
W
10-77
Special Topic: Binary Vapor Cycles
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
WQ
WQ
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
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),
kPa2780565
5
66
66
6
6
C600
KkJ/kg5488.74996.792.06492.0kJ/kg5.23921.239292.081.191
92.0kPa10
Pss
T
sxsshxhh
xP
fgf
fgf
T
600 CTSINGLE
10 kPa
25MPa 4
3
s6
2
5
1
6
7DOUBL
10 kPa
25MPa 4
3
8
2
5
1
600 C
(b) Double Reheat :
C600and
KkJ/kg3637.6C600MPa25
5
5
34
4
33
3
TPP
ssPP
sTP
xx
s
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),
kJ/kg2501.8kJ/kg0.2574
kJ/kg79.17404.675.168kJ/kg6.04
mkPa1kJ1
kPa7.56000/kgm0.001008
C29.40/kgm001008.0
kJ/kg75.168
kPa5.7@4
kPa5.7@4
inp,12
33
121inp,
kPa7.5@sat1
3kPa5.7@1
kPa5.7@1
g
g
f
f
sshh
whh
PPw
TT
hh
v
vv T
6 MPa
4
2
3
17.5 kPa
qout
qin
s
C1089.23
3
43
3 kJ/kg2.4852MPa6Th
ssP
(b)
kJ/kg1.22723.24054.4677kJ/kg3.240575.1680.2574kJ/kg4.467779.1742.4852
outinnet
14out
23in
qqwhhqhhq
and
48.6%kJ/kg4677.4kJ/kg2272.1
in
netth q
w
Thus,
kJ/s19,428kJ/s40,0004857.0inthnet QW
(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,
1718.033.1923.279833.19209.640
26
23
hhhh
y
Then,
kJ/kg4.9397.17841.2724kJ/kg7.178481.1918.23461718.011
kJ/kg1.272402.6511.3375
outinnet
17out
45in
qqwhhyq
hhq
and
skg7159 /.kJ/kg939.4
kJ/s150,000
net
net
wW
m
(b) The thermal efficiency is determined from
34.5%kJ/kg2724.1kJ/kg1784.7
11in
outth q
q
Also,
KkJ/kg6492.0KkJ/kg8604.1
KkJ/kg9453.6kJ/kg3.2798
MPa5.0
kPa10@12
MPa5.0@3
66
6
f
f
sssss
shP
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,
1819.031.1921.265431.19209.640
26
23
hhhh
y
Then,
kJ/kg0.11727.15526.2724kJ/kg7.155281.1917.20891819.011
kJ/kg6.272447.6501.3375
outinnet
17out
45in
qqwhhyq
hhq
and skg0128 /.kJ/kg1171.9
kJ/s150,000
net
net
wWm
(b) The thermal efficiency is determined from
%0.43kJ/kg2724.7kJ/kg1552.7
11in
outth q
q
Also,
KkJ/kg6492.0
KkJ/kg8604.1KkJ/kg5995.6
kPa10@12
MPa5.0@3
56
f
f
sss
ssss
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,
0.14453.2262.331053.22638.670
28
23
hhhhy
(b) The thermal efficiency is determined from
kJ/kg7.196294.2258.25181440.011kJ/kg7.33898.27831.347973.6801.3375
19out
6745in
hhyqhhhhq
and 42.1%kJ/kg3389.7kJ/kg1962.7
11in
outth q
q
10-88
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,
0.142753.2263.333553.22638.670
28
23
hhhhy
(b) The thermal efficiency is determined from
kJ/kg2.209794.2255.26721427.011kJ/kg1.32954.28781.347973.6801.3375
19out
6745in
hhyq
hhhhq
and
36.4%kJ/kg3295.1kJ/kg2097.2
11in
outth q
q
10-90
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),
High-PTurbine
111-y-z
13
12 z
10
4
Closedfwh
5y
P II
P I
Openfwh I
Condenser
BoilerLow-PTurbine
63
21
7
8
9
T 1 MPa
10
1115 MPa
z
8
56
3
y
4
1 - y - z5 kPa 7
0.2 MPa
0.6 MPa
12
9
13
2
1s
kJ/kg95.13720.075.137kJ/kg0.20
mkPa1kJ1
kPa5200/kgm0.001005
/km001005.0kJ/kg75.137
inpI,12
33
121inpI,
3kPa5@1
kPa5@1
whh
PPw
ghh
f
f
v
vv
/kgm100110.0kJ/kg38.670
liquidsat.MPa6.0
kJ/kg41.52070.1571.504kJ/kg15.70
mkPa1kJ1kPa20015,000/kgm0.001061
/kgm001061.0kJ/kg71.504
liquidsat.MPa2.0
3MPa6.0@6
MPa6.0@766
inpII,34
33
343inpII,
3MPa2.0@3
MPa2.0@33
f
f
f
f
hhhP
whh
PPw
hhP
vv
v
vv
KkJ/kg7642.7kJ/kg1.3479
C500MPa0.1
kJ/kg8.2820MPa0.1
KkJ/kg6796.6kJ/kg1.3583
C600MPa15
kJ/kg686.23 mkPa1kJ1kPa60015,000/kgm0.00110138.670
10
10
10
10
989
9
8
8
88
33
6566556
sh
TP
hss
P
sh
TP
PPvhhTT
10-91
kJ/kg1.23680.24239205.075.137
9205.0
9176.74762.07642.7
kPa5
kJ/kg9.3000MPa2.0
kJ/kg2.3310MPa6.0
1313
1313
1013
13
121012
12
111011
11
fgf
fg
f
hxhh
sss
x
ssP
hss
P
hss
P
The fraction of steam extracted is determined from the steady-flow energy balance equation applied to thefeedwater heaters. Noting that Q ,0pekeW
4561145561111
outin
(steady)0systemoutin 0
hhhhyhhmhhmhmhm
EE
EEE
eeii
where y is the fraction of steam extracted from the turbine ( /m m11 5 ). Solving for y,
06287.038.6702.331041.52023.686
611
45
hhhhy
For the open FWH,
31227
3312122277
outin
(steady)0systemoutin
11
0
hzhhzyyhhmhmhmhm
hmhm
EE
EEE
eeii
where z is the fraction of steam extracted from the turbine ( /m m12 5 ) at the second stage. Solving for z,
0.116595.1370.3000
95.13738.67006287.095.13771.504
212
2723
hhhhyhh
z
(b)
kJ/kg9.17244.18303.3555kJ/kg4.183075.1370.23681165.006287.011
kJ/kg3.35558.28201.347923.6861.3583
outinnet
113out
91058in
qqwhhzyq
hhhhq
and
48.5%kJ/kg3555.3kJ/kg1830.4
11in
outth q
q
(c) kW72,447kJ/kg1724.9kg/s42netnet wmW
10-92
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
kg/s929.2kJ/kg81.5037.2843
kJ/s7,000
37
process3 hh
Qm 15 kPa
91sAlso,
or,8.25181.3479992.27.28435.3399kJ/s3000
993.2
66
986766989766
mm
hhmhhmhhmhhmWT
It yields kg/s873.36m
and kg/s881.0992.2873.3369 mmmMixing chamber:
E E E
E E
m h m h m h m h m hi i e e
in out system (steady)
in out
0
4 4 2 2 3 3
0
or, kJ/kg60.440873.3
81.503992.294.225881.0
4
332254 m
hmhmhh
Then,
kW12,020kJ/kg2843.73479.1kg/s0.881kJ/kg440.603399.5kg/s3.873
788566in hhmhhmQ
(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
kg/s262.9kg/s3063.52335.86095.2069.3157
1110
23air
1110air23outin
s
seeii
mhhhhm
hhmhhmhmhmEE
(b)
kW102.80 5kW280,352kJ/kg7.27812.3457kg/s30kJ/kg12.52642.1515kg/s262.9
45reheat89airreheatairin hhmhhmQQQ
(c)
55.6%
,
kW280,352kW124,409
11
kW409124kJ/kg81.1918.2292kg/s30kJ/kg16.29063.523kg/s262.9
in
outth
16711airsteamout,airout,out
hhmhhmQQQ s
10-94
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
kg/s203.6kg/s3063.5234.95895.2069.3157
0
1110
23air
1110air23
outin
(steady)0systemoutin
s
seeii
mhhhhm
hhmhhmhmhm
EE
EEE
(b)kW207,986kJ/kg1.28382.3457kg/s30kJ/kg1.58542.1515kg/s203.6
45reheat89airreheatairin hhmhhmQQQ
(c)
44.3%kW207,986kW115,805
11
kW115,805kJ/kg81.1915.2467kg/s30kJ/kg16.29063.523kg/s203.6
in
outth
16711airsteamout,airout,out
hhmhhmQQ sQ
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),
kJ/kg47.908
kJ/kg71.25329.242.251kJ/kg.292
88.0/mkPa1
kJ1kPa022000/kgm0.001017
/
/kgm001017.0kJ/kg42.251
MPa2@3
inpI,12
33
121inpI,
3kPa02@1
kPa20@1
f
p
f
f
hh
whh
PPw
hh
v
vv
Mixing chamber:
kJ/kg81.491kg/s)11()kJ/kg)71kg/s)(253.411(kJ/kg)47kg/s)(908.4( 44
442233
hh
hmhmhm 24P P I
Conden.
Boiler
3
Processheater
5
1
6
7Turbine
8
kJ/kg02.49921.781.491kJ/kg.217
88.0/mkPa1
kJ1kPa00208000/kgm0.001058
/
/kgm001058.0
inII,45
33
454inII,
3kJ/kg81.491@4
p
pp
hf
whh
PPw
f
v
vv
T
Qout·
3
5
4
20 kPa
8 MPa
2 MPa Qproces·
Qin·
7
6
8s 8
2
1
KkJ/kg7266.6kJ/kg5.3399
C500MPa8
6
6
6
6sh
TP
kJ/kg4.3000MPa2
767
7sh
ssP
s
kJ/kg3.30484.30005.339988.05.3399766776
76sT
sT hhhh
hhhh
kJ/kg5.2215kPa20
868
8sh
ssP
kJ/kg6.23575.22155.339988.05.3399866886
86sT
sT hhhh
hhhh
Then, kW8559kJ/kg47.9083.3048kg/s4377process hhmQ(b) Cycle analysis:
kW8603958698
kW95kJ/kg7.21kg/s11kJ/kg2.29kg/s7
kW8698kJ/kg6.23575.3399kg/s7kJ/kg3.30485.3399kg/s4
inp,outT,net
inpII,4inpI,1inp,
868767outT,
WWW
wmwmW
hhmhhmW
(c) Then,
and53.8%538.0
905,3185598603
kW905,3102.4995.3399kg/s11
in
processnet
565in
Q
QW
hhmQ
u
10-98
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$='(compressed)'
end
P[3] = 5000 [kPa] T[3] = 500 [C] "P[4] = 5 [kPa]"Eta_t = 1.0 "Turbine isentropic efficiency"Eta_p = 1.0 "Pump isentropic efficiency"
"Pump analysis"Fluid$='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$(x[4])
"Boiler analysis"Q_in + h[2]=h[3]"SSSF First Law for the Boiler"
"Condenser analysis"h[4]=Q_out+h[1]"SSSF First Law for the Condenser"
"Cycle Statistics"W_net=W_t-W_pEta_th=W_net/Q_in
10-99
th P4
[kPa]Wnet
[kJ/kg]x4 Qin
[kJ/kg]Qout
[kJ/kg]0.3956 5 1302 0.8212 3292 19900.3646 15 1168 0.8581 3204 20360.3484 25 1100 0.8772 3158 20570.3371 35 1054 0.8905 3125 20720.3283 45 1018 0.9009 3100 20820.321 55 988.3 0.9096 3079 2091
0.3147 65 963.2 0.917 3061 20980.3092 75 941.5 0.9235 3045 21040.3042 85 922.1 0.9293 3031 21090.2976 100 896.5 0.9371 3012 2116
0 2 4 6 8 10 120
100
200
300
400
500
600
700
s [kJ/kg-K]
T [°
C]
5000 kPa
5 kPa
Steam
1
3
4
2
0 20 40 60 80 100800
900
1000
1100
1200
1300
1400
P[4] [kPa]
Wne
t [k
J/kg
]
10-100
0 20 40 60 80 1000.28
0.3
0.32
0.34
0.36
0.38
0.4
P[4] [kPa]
th
0 20 40 60 80 1000.82
0.84
0.86
0.88
0.9
0.92
0.94
P[4] [kPa]
x[4]
10-101
10-105 EES The effect of the boiler pressure on the performance a simple ideal Rankine cycle is to be investigated.Analysis The problem is solved using EES, and the solution is given below.
function x4$(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$='(compressed)'
end
{P[3] = 20000 [kPa]}T[3] = 500 [C] P[4] = 10 [kPa] Eta_t = 1.0 "Turbine isentropic efficiency"Eta_p = 1.0 "Pump isentropic efficiency"
"Pump analysis"Fluid$='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$(x[4])
"Boiler analysis"Q_in + h[2]=h[3] "SSSF First Law for the Boiler"
"Condenser analysis"h[4]=Q_out+h[1] "SSSF First Law for the Condenser"
"Cycle Statistics"W_net=W_t-W_pEta_th=W_net/Q_in
10-102
th Wnet
[kJ/kg]x4 P3
[kPa]Qin
[kJ/kg]Qout
[kJ/kg]Wp
[kJ/kg]Wt
[kJ/kg]0.2792 919.1 0.9921 500 3292 2373 0.495 919.60.3512 1147 0.886 2667 3266 2119 2.684 11500.3752 1216 0.8462 4833 3240 2024 4.873 12210.3893 1251 0.8201 7000 3213 1962 7.062 12580.399 1270 0.8001 9167 3184 1914 9.251 12800.406 1281 0.7835 11333 3155 1874 11.44 1292
0.4114 1286 0.769 13500 3125 1840 13.63 12990.4156 1286 0.756 15667 3094 1808 15.82 13020.4188 1283 0.744 17833 3062 1780 18.01 13010.4214 1276 0.7328 20000 3029 1753 20.2 1297
0 2 4 6 8 10 120
100
200
300
400
500
600
700
s [kJ/kg-K]
T [°
C]
20000 kPa
10 kPa
Steam
1
2
3
4
0 4000 8000 12000 16000 20000900
950
1000
1050
1100
1150
1200
1250
1300
P[3] [kPa]
Wne
t [k
J/kg
]
10-103
0 4000 8000 12000 16000 200000.26
0.28
0.3
0.32
0.34
0.36
0.38
0.4
0.42
0.44
P[3] [kPa]
th
0 4000 8000 12000 16000 200000.7
0.75
0.8
0.85
0.9
0.95
1
P[3] [kPa]
x[4]
10-104
10-106 EES The effect of superheating the steam on the performance a simple ideal Rankine cycle is to be investigated.Analysis The problem is solved using EES, and the solution is given below.
function x4$(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$='(compressed)'
end
P[3] = 3000 [kPa] {T[3] = 600 [C]}P[4] = 10 [kPa] Eta_t = 1.0 "Turbine isentropic efficiency"Eta_p = 1.0 "Pump isentropic efficiency"
"Pump analysis"Fluid$='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$='(subcooled)'
end
P[6] = 10 [kPa] P[3] = 15000 [kPa] T[3] = 500 [C] "P[4] = 3000 [kPa]"T[5] = 500 [C] Eta_t = 100/100 "Turbine isentropic efficiency"Eta_p = 100/100 "Pump isentropic efficiency"
"Pump analysis"Fluid$='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$(x[6])
"Cycle Statistics"W_net=W_t_hp+W_t_lp-W_pEta_th=W_net/Q_in
P4
[kPa]th Wnet
[kJ/kg]X6
500 0.4128 1668 0.99211833 0.4253 1611 0.91023167 0.4283 1567 0.87474500 0.4287 1528 0.85115833 0.428 1492 0.83327167 0.4268 1458 0.81848500 0.4252 1426 0.80589833 0.4233 1395 0.7947
11167 0.4212 1366 0.784712500 0.4189 1337 0.7755
0.0 1.1 2.2 3.3 4.4 5.5 6.6 7.7 8.8 9.9 11.00
100
200
300
400
500
600
700
s [kJ/kg-K]
T[C]
8000 kPa
3000 kPa
20 kPa
3
4
5
6
Ideal Rankine cycle with reheat
1,2
10-109
0 2000 4000 6000 8000 10000 12000 140001300
1350
1400
1450
1500
1550
1600
1650
1700
P[4] [kPa]
Wnet
[kJ/kg]
0 2000 4000 6000 8000 10000 12000 140000.412
0.414
0.416
0.418
0.42
0.422
0.424
0.426
0.428
0.43
P[4] [kPa]
th
0 2000 4000 6000 8000 10000 12000 140000.775
0.8
0.825
0.85
0.875
0.9
0.925
0.95
0.975
1
P[4] [kPa]
x[6]
10-110
10-108 EES The effect of number of reheat stages on the performance an ideal Rankine cycle is to be investigated.Analysis The problem is solved using EES, and the solution is given below.
function x6$(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$='(subcooled)'
end
Procedure Reheat(P[3],T[3],T[5],h[4],NoRHStages,Pratio,Eta_t:Q_in_reheat,W_t_lp,h6)P3=P[3]T5=T[5]h4=h[4]Q_in_reheat =0 W_t_lp = 0 R_P=(1/Pratio)^(1/(NoRHStages+1))
imax:=NoRHStages - 1 i:=0
REPEATi:=i+1
P4 = P3*R_P
P5=P4P6=P5*R_P
Fluid$='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
UNTIL (i>imax)
END
"NoRHStages = 2"P[6] = 10"kPa"P[3] = 15000"kPa"P_extract = P[6] "Select a lower limit on the reheat pressure"T[3] = 500"C"T[5] = 500"C"Eta_t = 1.0 "Turbine isentropic efficiency"Eta_p = 1.0 "Pump isentropic efficiency"Pratio = P[3]/P_extract
10-111
P[4] = P[3]*(1/Pratio)^(1/(NoRHStages+1))"kPa"
Fluid$='Steam_IAPWS'
"Pump analysis"P[1] = P[6] P[2]=P[3]x[1]=0 "Sat'd liquid"h[1]=enthalpy(Fluid$,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"
"Low Pressure Turbine analysis"Call Reheat(P[3],T[3],T[5],h[4],NoRHStages,Pratio,Eta_t:Q_in_reheat,W_t_lp,h6)h[6]=h6
{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])
10-112
x[6]=QUALITY(Fluid$,h=h[6],P=P[6])x6s$=x6$(x[6])
"Cycle Statistics"W_net=W_t_hp+W_t_lp - W_p Eta_th=W_net/Q_in
th NoRHStages
Qin
[kJ/kg]Wnet
[kJ/kg]0.409
71 4085 1674
0.4122
2 4628 1908
0.4085
3 5020 2051
0.4018
4 5333 2143
0.3941
5 5600 2207
0.386 6 5838 22530.377
97 6058 2289
0.3699
8 6264 2317
0.3621
9 6461 2340
0.3546
10 6651 2358
1 2 3 4 5 6 7 8 9 101600
1700
1800
1900
2000
2100
2200
2300
2400
NoRHStages
Wne
t[k
J/kg
]
1 2 3 4 5 6 7 8 9 100.35
0.36
0.37
0.38
0.39
0.4
0.41
0.42
NoRHStages
th
10-113
1 2 3 4 5 6 7 8 9 104000
4500
5000
5500
6000
6500
7000
NoRHStages
Qin
[kJ
/kg]
10-114
10-109 EES The effect of extraction pressure on the performance an ideal regenerative Rankine cycle with one open feedwater heater is to be investigated.Analysis The problem is solved using EES, and the solution is given below.
"Input Data"P[5] = 15000 [kPa] T[5] = 600 [C] P_extract=1400 [kPa] P[6] = P_extract P_cond=10 [kPa] P[7] = P_cond Eta_turb= 1.0 "Turbine isentropic efficiency"Eta_pump = 1.0 "Pump isentropic efficiency"P[1] = P[7] P[2]=P[6]P[3]=P[6]P[4] = P[5]
"Condenser exit pump or Pump 1 analysis"Fluid$='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"
"Cycle Statistics"w_net=w_turb - ((1- y)*w_pump1+ w_pump2) Eta_th=w_net/q_in
th Pextract
[kPa]wnet
[kJ/kg]qin [kJ/kg]
0.4456 50 1438 32270.4512 100 1421 31500.4608 500 1349 29270.4629 1000 1298 28050.4626 2000 1230 26590.4562 5000 1102 24160.4511 7000 1040 23050.4434 10000 961.4 21680.4369 12500 903.5 2068
0 2 4 6 8 10 120
100
200
300
400
500
600
700
s [kJ/kg-K]
T [°
C]
6000 kPa
400 kPa
10 kPa
Steam
1
2 3
4
5
6
7
10-116
0 2000 4000 6000 8000 10000 12000 140000.435
0.44
0.445
0.45
0.455
0.46
0.465
Pextract [kPa]
th
0 2000 4000 6000 8000 10000 12000 14000900
1000
1100
1200
1300
1400
1500
Pextract [kPa]
wne
t [k
J/kg
]
0 2000 4000 6000 8000 10000 12000 140002000
2200
2400
2600
2800
3000
3200
3400
Pextract [kPa]
q in
[kJ/
kg]
10-117
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.
Procedure Reheat(NoFwh,T[5],P[5],P_cond,Eta_turb,Eta_pump:q_in,w_net)
Fluid$='Steam_IAPWS'Tcond = temperature(Fluid$,P=P_cond,x=0)Tboiler = temperature(Fluid$,P=P[5],x=0)P[7] = P_cond s[5]=entropy(Fluid$, T=T[5], P=P[5]) h[5]=enthalpy(Fluid$, T=T[5], P=P[5]) h[1]=enthalpy(Fluid$, P=P[7],x=0)
P4[1] = P[5] "NOTICE THIS IS P4[i] WITH i = 1"
DELTAT_cond_boiler = Tboiler - Tcond
If NoFWH = 0 Then
"the following are h7, h2, w_net, and q_in for zero feedwater heaters, NoFWH = 0"h7=enthalpy(Fluid$, s=s[5],P=P[7])
h2=h[1]+volume(Fluid$, P=P[7],x=0)*(P[5] - P[7])/Eta_pump w_net = Eta_turb*(h[5]-h7)-(h2-h[1]) q_in = h[5] - h2
else
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$,T=T_FWH[i],x=0)"[kPa]"P3[i]=P_extract[i]P6[i]=P_extract[i]If i > 1 then P4[i] = P6[i - 1]
UNTIL i=NoFWH
P4[NoFWH+1]=P6[NoFWH]h4[NoFWH+1]=h[1]+volume(Fluid$, P=P[7],x=0)*(P4[NoFWH+1] - P[7])/Eta_pump
i=0REPEATi=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])
sumyi = 0 sumyh6i = 0 wp2i = W_pump2[1] i=0REPEATi=i+1sumyi = sumyi + y[i]sumyh6i = sumyh6i + yh6[i]If NoFWH > 1 then wp2i = wp2i + (1- sumyi)*W_pump2[i]UNTIL i = NoFWH
"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"
"Cycle Statistics"w_net=w_turb - ((1- sumyi)*w_pump1+ wp2i)
endifEND
"Input Data"NoFWH = 2 P[5] = 15000 [kPa] T[5] = 600 [C] P_cond=5 [kPa]
Eta_turb= 1.0 "Turbine isentropic efficiency"Eta_pump = 1.0 "Pump isentropic efficiency"P[1] = P_condP[4] = P[5]
"Condenser exit pump or Pump 1 analysis"Call Reheat(NoFwh,T[5],P[5],P_cond,Eta_turb,Eta_pump:q_in,w_net)Eta_th=w_net/q_in
0 2 4 6 8 10 120
100
200
300
400
500
600
700
s [kJ/kg-K]
T [°
C]
6000 kPa
400 kPa
10 kPa
Steam
1
2 3
4
5
6
7
NoFWH
th wnet
[kJ/kg]qin
[kJ/kg]0 0.4466 1532 34301 0.4806 1332 27712 0.4902 1243 25363 0.4983 1202 24114 0.5036 1175 23335 0.5073 1157 22806 0.5101 1143 22407 0.5123 1132 22108 0.5141 1124 21869 0.5155 1117 2167
10 0.5167 1111 2151
10-120
0 2 4 6 8 100.44
0.45
0.46
0.47
0.48
0.49
0.5
0.51
0.52
NoFwh
th
0 2 4 6 8 101100
1150
1200
1250
1300
1350
1400
1450
1500
1550
NoFwh
wne
t[k
J/kg
]
0 2 4 6 8 102000
2200
2400
2600
2800
3000
3200
3400
3600
NoFwh
q in
[kJ/
kg]
10-121
Fundamentals of Engineering (FE) Exam Problems
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).
P1=8000 "kPa"P2=20 "kPa"h_fg=ENTHALPY(Steam_IAPWS,x=1,P=P1)-ENTHALPY(Steam_IAPWS,x=0,P=P1)T1=TEMPERATURE(Steam_IAPWS,x=0,P=P1)+273T2=TEMPERATURE(Steam_IAPWS,x=0,P=P2)+273q_in=h_fgEta_Carnot=1-T2/T1w_net=Eta_Carnot*q_in
"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).
P1=10 "kPa"P2=3000 "kPa"P3=P2P4=P1T3=600 "C"s4=s3h1=ENTHALPY(Steam_IAPWS,x=0,P=P1)v1=VOLUME(Steam_IAPWS,x=0,P=P1)w_pump=v1*(P2-P1) "kJ/kg"h2=h1+w_pumph3=ENTHALPY(Steam_IAPWS,T=T3,P=P3)s3=ENTROPY(Steam_IAPWS,T=T3,P=P3)
10-122
h4=ENTHALPY(Steam_IAPWS,s=s4,P=P4)q_in=h3-h2q_out=h4-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-(h44-h1)/(h3-h2); h44 = ENTHALPY(Steam_IAPWS,x=1,P=P4) "Using h_g for h4"W3_Eff = 1-(T1+273)/(T3+273); T1=TEMPERATURE(Steam_IAPWS,x=0,P=P1) "Using Carnot efficiency"W4_Eff = (h3-h4)/q_in "Disregarding pump work"
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).
P1=10 "kPa"P2=5000 "kPa"P3=P2P4=P1T3=600 "C"s4=s3h3=ENTHALPY(Steam_IAPWS,T=T3,P=P3)s3=ENTROPY(Steam_IAPWS,T=T3,P=P3)h4=ENTHALPY(Steam_IAPWS,s=s4,P=P4)x4=QUALITY(Steam_IAPWS,s=s4,P=P4)moisture=1-x4
"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).
m_gas=60 "kg/s"Cp=1.005 "kJ/kg.K"T3=800 "K"T4=400 "K"Q_gas=m_gas*Cp*(T3-T4)P1=8000 "kPa"T1=200 "C"P2=8000 "kPa"T2=350 "C"h1=ENTHALPY(Steam_IAPWS,T=T1,P=P1)h2=ENTHALPY(Steam_IAPWS,T=T2,P=P2)Q_steam=m_steam*(h2-h1)Q_gas=Q_steam
"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
"Some Wrong Solutions with Common Mistakes:"W1_ext = h_feed/h_extracted "Using wrong relation"W2_ext = h3/(h_extracted-h_feed) "Using wrong relation"W3_ext = h_feed/(h_extracted-h_feed) "Using wrong relation"
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.
h1 = 191.81 h2 = 192.20 h3 = h4 = h9 = 604.66 h5 = 610.73 h6 = 3302.9 h7 = h8 = h10 = 2665.6 h11 = 2128.8
710
fwh4
8
3
2
9
P II P I
Processheater Condenser
BoilerTurbine
5
1
11
6
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.
P1=10 "kPa"P11=P1P2=400 "kPa"P3=P2; P4=P2; P7=P2; P8=P2; P9=P2; P10=P2 P5=6000 "kPa"P6=P5T6=450 "C"m_total=20 "kg/s"m7=0.6*m_totalm_cond=0.4*m_totalC=4.18 "kJ/kg.K"m_cooling=463 "kg/s"s7=s6s11=s6h1=ENTHALPY(Steam_IAPWS,x=0,P=P1)v1=VOLUME(Steam_IAPWS,x=0,P=P1)w_pump=v1*(P2-P1)h2=h1+w_pumph3=ENTHALPY(Steam_IAPWS,x=0,P=P3)h4=h3; h9=h3 v4=VOLUME(Steam_IAPWS,x=0,P=P4)w_pump2=v4*(P5-P4)h5=h4+w_pump2h6=ENTHALPY(Steam_IAPWS,T=T6,P=P6)s6=ENTROPY(Steam_IAPWS,T=T6,P=P6)h7=ENTHALPY(Steam_IAPWS,s=s7,P=P7)h8=h7; h10=h7 h11=ENTHALPY(Steam_IAPWS,s=s11,P=P11)W_turb=m_total*(h6-h7)+m_cond*(h7-h11)m_cooling*C*T_rise=m_cond*(h11-h1)m_cond*h2+m_feed*h10=(m_cond+m_feed)*h3m_process=m7-m_feedq_process=h8-h9Q_in=m_total*(h6-h5)
10-124 ··· 10-133 Design and Essay Problems
Recommended