Chapter 10 VAPOR AND COMBINED POWER CYCLES · PDF file10-1 Chapter 10 VAPOR AND COMBINED POWER CYCLES Carnot Vapor Cycle 10-1C Because excessive moisture in steam causes erosion on
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10-1
Chapter 10 VAPOR AND COMBINED POWER CYCLES
Carnot Vapor Cycle 10-1C Because excessive moisture in steam causes erosion on the turbine blades. The highest moisture content allowed is about 10%. 10-2C The Carnot cycle is not a realistic model for steam power plants because (1) limiting the heat transfer processes to two-phase systems to maintain isothermal conditions severely limits the maximum temperature that can be used in the cycle, (2) the turbine will have to handle steam with a high moisture content which causes erosion, and (3) it is not practical to design a compressor that will handle two phases. 10-3E A steady-flow Carnot engine with water as the working fluid operates at specified conditions. The thermal efficiency, the quality at the end of the heat rejection process, and the net work output are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) We note that
and
19.3%=−=−=η
=°===°==
R 1.833R 0.67211
R 0.672F0.212R 1.833F1.373
Cth,
psia14.7@sat
psia 180@sat
H
L
L
H
TT
TTTT
T
14.7 psia
180 psia qin
1
4
2
3 (b) Noting that s4 = s1 = sf @ 180 psia = 0.53274 Btu/lbm·R,
0.153=−
=−
=44441.1
31215.053274.044
fg
f
sss
x s
(c) The enthalpies before and after the heat addition process are
10-4 A steady-flow Carnot engine with water as the working fluid operates at specified conditions. The thermal 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 thermal efficiency becomes
36.3%==−=−= 3632.0K 523K 333.1
11Cth,H
L
TT
η T
qout
qin 1
4
2
3 20 kPa
(b) The heat supplied during this cycle is simply the enthalpy of vaporization ,
Thus,
( ) kJ/kg 1092.3=
===
==
kJ/kg 3.1715K 523K 333.1
kJ/kg 3.1715
inout
250 @in
qTT
qq
hq
H
LL
Cfg o
250°C
s(c) The net work output of this cycle is ( )( ) kJ/kg623.0 kJ/kg 3.17153632.0inthnet === qw η
10-5 A steady-flow Carnot engine with water as the working fluid operates at specified conditions. The thermal 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 thermal efficiency becomes
39.04%=−=−=K 523K 318.811C th,
H
L
TT
η T
qout
qin 1
4
2
3 10 kPa
(b) The heat supplied during this cycle is simply the enthalpy of vaporization ,
Thus,
( ) kJ/kg 1045.6=
===
== °
kJ/kg 3.1715K 523K 318.8
kJ/kg 3.1715
inout
C250 @in
qTT
qq
hq
H
LL
fg
250°C
s(c) The net work output of this cycle is ( )( ) kJ/kg 669.7=== kJ/kg 3.17153904.0inthnet qw η
10-6 A steady-flow Carnot engine with water as the working fluid operates at specified conditions. The thermal efficiency, the pressure at the turbine inlet, and the net work output are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) The thermal efficiency is determined from
(c) The net work can be determined by calculating the enclosed area on the T-s diagram,
Thus,
( )( )
( )( ) ( )( ) kJ/kg 1623=−−=−−==
⋅=+=+=
5390.11368.760350Area
KkJ/kg 5390.10769.71.08313.0
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 heat rejection 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 involve friction and pressure drops in various components and the piping, and heat loss to the surrounding medium from these components and piping. 10-12C The boiler exit pressure will be (a) lower than the boiler inlet pressure in actual cycles, and (b) the same 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 will adversely 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 than the temperature of the cooling water.
10-15 A steam power plant operates on a simple ideal Rankine cycle between the specified pressure limits. The thermal efficiency of the cycle and the net power output of the plant are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),
10-16 A steam power plant that operates on a simple ideal Rankine cycle is considered. The quality of the steam at the turbine exit, the thermal efficiency of the cycle, and the mass flow rate of the steam are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),
10-17 A steam power plant that operates on a simple nonideal Rankine cycle is considered. The quality of the steam at the turbine exit, the thermal efficiency of the cycle, and the mass flow rate of the steam are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),
10-18E A steam power plant that operates on a simple ideal Rankine cycle between the specified pressure limits is considered. The minimum turbine inlet temperature, the 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.
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),
10-19E A steam power plant operates on a simple nonideal Rankine cycle between the specified pressure limits. The minimum turbine inlet temperature, the 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. T
Qout ·
2s
s
1250 psia
Qin ·
4s 4 x4 = 0.9
2
3
12 psia
Analysis (a) From the steam tables (Tables A-4E, A-5E, and A-6E),
( )( )( )
Btu/lbm 43.9841.402.94Btu/lbm 41.4
85.0/ftpsia 5.4039
Btu 1psia 21250/lbmft 0.01623
/
/lbmft 01623.0
Btu/lbm 02.94
in,12
33
121in,
3psia 2 @1
psia 2 @1
=+=+==
⋅−=
−=
==
==
p
Pp
f
f
whh
PPw
hh
ηv
vv
( )( )( )( ) RBtu/lbm 7450.174444.19.017499.0
Btu/lbm 6.10137.10219.002.94
44
44
⋅=+=+==+=+=
fgf
fgf
sxsshxhh
The turbine inlet temperature is determined by trial and error ,
Try 1:
( )( )
8171.04.9180.14396.10130.1439
Btu/lbm 4.9187.10218069.002.94
8069.074444.1
17499.05826.1
Btu/lbm.R 5826.1Btu/lbm 0.1439
F900psia 1250
43
43
44
344
3
3
3
3
=−−
=−−
=
=+=+=
=−
=−
=−
=
==
°==
sT
fgsfs
fg
f
fg
fss
hhhh
hxhh
sss
sss
x
sh
TP
η
Try 2:
( )( )
8734.03.9436.14986.10136.1498
Btu/lbm 3.9437.10218312.002.94
8312.074444.1
17499.06249.1
Btu/lbm.R 6249.1Btu/lbm 6.1498
F1000psia 1250
43
43
44
344
3
3
3
3
=−−
=−−
=
=+=+=
=−
=−
=−
=
==
°==
sT
fgsfs
fg
f
fg
fss
hhhh
hxhh
sss
sss
x
sh
TP
η
By linear interpolation, at ηT = 0.85 we obtain T3 = 958.4°F. This is approximate. We can determine state 3 exactly using EES software with these results: T3 = 955.7°F, h3 = 1472.5 Btu/lbm.
10-20 A 300-MW coal-fired steam power plant operates on a simple ideal Rankine cycle between the specified 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),
10-21 A solar-pond power plant that operates on a simple ideal Rankine cycle with refrigerant-134a as the working fluid is considered. The thermal efficiency of the cycle and the 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 refrigerant tables (Tables A-11, A-12, and A-13),
10-22 A steam power plant operates on a simple ideal Rankine cycle between the specified pressure limits. The thermal efficiency of the cycle, the mass flow rate of the steam, and the temperature rise of the 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),
10-23 A steam power plant operates on a simple nonideal 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 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),
10-24 The net work outputs and the thermal efficiencies for a Carnot cycle and a simple ideal Rankine cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) Rankine cycle analysis: From the steam tables (Tables A-4, A-5, and A-6),
10-25 A binary geothermal power operates on the simple Rankine cycle with isobutane as the working fluid. The isentropic efficiency of the turbine, the net power 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. Properties The specific heat of geothermal water is taken to be 4.18 kJ/kg.ºC. Analysis (a) We need properties of isobutane, which are not available in the book. However, we can obtain the properties from EES. Turbine:
10-26 A single-flash geothermal power plant uses hot geothermal water at 230ºC as the heat source. The mass flow rate of steam through the turbine, the isentropic efficiency of the turbine, the power output from the 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 for geothermal water (Tables A-4 through A-6)
1661.02108
09.64014.990kJ/kg 14.990
kPa 500
kJ/kg 14.9900
C230
22
12
2
11
1
=−
=−
=
===
=
=°=
fg
f
hhh
xhh
P
hxT
The mass flow rate of steam through the turbine is kg/s 38.20=== kg/s) 230)(1661.0(123 mxm &&
10-27 A double-flash geothermal power plant uses hot geothermal water at 230ºC as the heat source. The temperature of the steam at the exit of the second flash chamber, the power produced from the second 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 for geothermal water (Tables A-4 through A-6)
1661.0kJ/kg 14.990
kPa 500
kJ/kg 14.9900
C230
212
2
11
1
=
===
=
=°=
xhh
P
hxT
kg/s 80.1911661.0230kg/s 38.20kg/s) 230)(1661.0(
316
123
=−=−====
mmmmxm
&&&
&&
Flash chamber
separator
9
8
7
Flash chamber
6
5
4
3
2condenser
1
steam turbine
separator
reinjection well
production well
kJ/kg 7.234490.0kPa 10
kJ/kg 1.27481
kPa 500
44
4
33
3
=
==
=
==
hxP
hxP
kJ/kg 1.26931
kPa 150
0777.0kPa 150
kJ/kg 09.6400
kPa 500
88
8
7
7
67
7
66
6
=
==
=°=
==
=
==
hxP
xT
hhP
hxP
C 111.35
(b) The mass flow rate at the lower stage of the turbine is kg/s .9014kg/s) 80.191)(0777.0(678 === mxm &&
The power outputs from the high and low pressure stages of the turbine are
10-28 A combined flash-binary geothermal power plant uses hot geothermal water at 230ºC as the heat source. The mass flow rate of isobutane in the binary cycle, the net power outputs from the steam turbine and the binary cycle, and the thermal efficiencies for the binary cycle and the combined 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)
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 Rankine cycle with 3 stages of reheat is shown on the side. The cycle efficiency will increase as the number of reheating stages increases. 10-31C The thermal efficiency of the simple ideal Rankine cycle will probably be higher since the average temperature at which heat is added will be higher in this case.
10-32 [Also solved by EES on enclosed CD] A steam power plant that operates on the ideal reheat Rankine cycle 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/kg 8.12mkPa 1
kJ 1kPa 208000/kgm 001017.0
/kgm 700101.0
kJ/kg 42.251
in,12
33
121in,
3kPa 20 @1
kPa 20 @1
=+=+=
=
⋅−=
−=
==
==
p
p
f
f
whh
PPw
hh
v
vv
( )( ) kJ/kg 2.23855.23579051.042.251
9051.00752.7
8320.02359.7kPa 20
KkJ/kg 2359.7kJ/kg 2.3457
C500MPa 3
kJ/kg 1.3105MPa 3
KkJ/kg 7266.6kJ/kg 5.3399
C500MPa 8
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
10-33 EES Problem 10-32 is reconsidered. The problem is to be solved by the diagram window data entry feature 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$,P=P[1],x=x[1]) W_p_s=v[1]*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume" W_p=W_p_s/Eta_p h[2]=h[1]+W_p "SSSF First Law for the pump" v[2]=volume(Fluid$,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]) "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_p Eff=W_net/Q_in
10-34 A steam power plant that operates on a 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 the mass flow rate of the steam are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),
10-35 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 the mass flow rate of the steam are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),
10-36E 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 rate of 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/lbm 11.7239.272.69
Btu/lbm 2.39ftpsia 5.4039
Btu 1psia 1800/lbmft 01614.0
F69.101
/lbmft 01614.0
Btu/lbm 72.69
in,12
33
121in,
psia 1 @sat1
3psia 1 @sat1
psia 1 @sat1
=+=+=
=
⋅−=
−=
°==
==
==
p
p
whh
PPw
TT
hh
v
vv
T
1 psia
800 psia 4
3
6
2
5
1s
( )
( )( ) Btu/lbm 0.10617.10359572.072.69
9572.084495.1
13262.08985.1psia 1
RBtu/lbm 8985.1Btu/lbm 4.1431
F800psia 23.62
pressure)reheat (the
Btu/lbm 5.1178
vaporsat.
RBtu/lbm 6413.1Btu/lbm 0.1456
F900psia 800
66
66
56
6
5
5
5
5
@sat4
@434
3
3
3
3
4
4
=+=+=
=−
=−
=
==
⋅==
°==
==
==
=
⋅==
°==
=
=
fgf
fg
f
ss
ssg
hxhh
sss
x
ssP
sh
TP
PP
hhss
sh
TP
g
g
psia 62.23
(b) ( ) ( )Btu/lbm 3.99172.690.1061
Btu/lbm 8.16365.11784.143111.720.1456
16out
4523in
=−=−=
=−+−=−+−=
hhq
hhhhq
Thus,
39.4%=−=−=Btu/lbm 1636.8Btu/lbm 991.3
11in
outth q
qη
(c) The mass flow rate of the cooling water will be minimum when it is heated to the temperature of the steam in the condenser, which is 101.7°F,
10-37 A steam power plant that operates on an ideal reheat Rankine cycle between the specified pressure limits is considered. The pressure at which reheating takes place, the total rate of heat input in the boiler, and the thermal efficiency of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),
10-38 A steam power plant that operates on a reheat Rankine cycle is considered. The condenser pressure, the net power output, and the thermal efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),
( )( )( )
( )( )( ) kJ/kg 3.30271.29482.335885.02.3358
?
95.0?
KkJ/kg 2815.7kJ/kg 2.3358
C450MPa 2
kJ/kg 3.30271.29485.347685.05.3476
kJ/kg 1.2948MPa 2
KkJ/kg 6317.6kJ/kg 5.3476
C550MPa 5.12
655665
65
656
6
66
6
5
5
5
5
4334
43
43
434
4
3
3
3
3
=−−=−η−=→
−−
=
=
==
=
==
⋅==
°==
=−−=
−η−=→
−−
=η
=
==
⋅==
°==
sTs
T
s
sT
sT
ss
hhhhhhhh
hss
P
hxP
sh
TP
hhhh
hhhh
hss
P
sh
TP
η
2
4
6 P = ?
12.5 MPa4s
3
T
4
5
Pump Condenser
Boiler Turbine
2 1
6
3
s 6s
2s
5
1
The pressure at state 6 may be determined by a trial-error approach from the steam tables or by using EES from the above equations: P6 = 9.73 kPa, h6 = 2463.3 kJ/kg, (b) Then,
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 rejected anyway. Therefore, by regeneration, we utilize a considerable amount of heat by sacrificing little work output. 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 liquid state at the boiler pressure isothermally, and the steam must be a saturated vapor at the turbine inlet. This will require an infinite number of heat exchangers (feedwater heaters), as shown on the T-s diagram.
10-44 A steam power plant that operates on an ideal regenerative Rankine cycle with an open feedwater heater is considered. The net work output per kg of steam 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/kg 81.25139.042.251
kJ/kg 0.39mkPa 1
kJ 1kPa 20400/kgm 001017.0
/kgm 001017.0
kJ/kg 42.251
in,12
33
121in,
3kPa 20 @1
kPa 20 @1
=+=+=
=
⋅−=
−=
==
==
pI
pI
f
f
whh
PPw
hh
v
vv
T
( ) ( )( )
( )( )
( )( ) kJ/kg 0.22145.23578325.042.251
8325.00752.7
8320.07219.6kPa 20
kJ/kg 7.26654.21339661.066.604
9661.01191.5
7765.17219.6MPa 4.0
KkJ/kg 7219.6kJ/kg 9.3302
C450MPa 6
kJ/kg 73.61007.666.604
kJ/kg 6.07mkPa 1
kJ 1kPa 4006000/kgm 0.001084
/kgm 001084.0
kJ/kg 66.604
liquidsat.MPa 4.0
77
77
57
7
66
66
56
6
5
5
5
5
in,34
33
343in,
3MPa 4.0 @3
MPa 4.0 @33
=+=+=
=−
=−
=
==
=+=+=
=−
=−
=
==
⋅==
°==
=+=+=
=
⋅−=−=
==
==
=
fgf
fg
f
fgf
fg
f
pII
pII
f
f
hxhh
sss
x
ssP
hxhh
sss
xss
P
sh
TP
whh
PPw
hhP
v
vv
qout
3 y
4
1-y
0.4 MPa
20 kPa
6 MPa
6
5
7
2
qin
1s
The fraction of steam extracted is determined from the steady-flow energy balance equation applied to the feedwater 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,
10-45 A steam power plant that operates on an ideal regenerative Rankine cycle with a closed feedwater heater is considered. The net work output per kg of steam 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/kg 50.25708.642.251
kJ/kg 6.08mkPa 1
kJ 1kPa 206000/kgm 0.001017
/kgm 001017.0
kJ/kg 42.251
in,12
33
121in,
3kPa 20 @1
kPa 20 @1
=+=+=
=
⋅−=
−=
==
==
pI
pI
f
f
whh
PPw
hh
v
vv
T
9
qou
3y
4
1-y
0.4 MPa
20 kPa
6 MPa
6
5
7
28
qin
1
/kgm 400108.0
kJ/kg 66.604
liquid sat.MPa 4.0
3MPa 4.0 @3
MPa 4.0 @33
==
==
=
f
fhhPvv
s
( ) ( )( )
( ) kJ/kg 73.610
kJ/kg 73.61007.666.604
kJ/kg 07.6mkPa 1
kJ 1kPa 4006000/kgm 001084.0
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 same enthalpy.
( )( )
( )( ) kJ/kg 0.22145.23578325.042.251
8325.00752.7
8320.07219.6kPa 20
kJ/kg 7.26654.21339661.066.604
9661.01191.5
7765.17219.6MPa 4.0
KkJ/kg 7219.6kJ/kg 9.3302
C450MPa 6
77
77
57
7
66
66
56
6
5
5
5
5
=+=+=
=−
=−
=
==
=+=+=
=−
=−
=
==
⋅==
°==
fgf
fg
f
fgf
fg
f
hxhh
sss
x
ssP
hxhh
sss
x
ssP
sh
TP
The fraction of steam extracted is determined from the steady-flow energy balance equation applied to the feedwater heater. Noting that Q , 0∆pe∆ke ≅≅≅≅W&&
( ) ( ) ( )( ) ( 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,