Example 10-6 reproduced from Ref. 1
Consider a steam power plant that operates on an ideal reheat–regenerative Rankine cycle with one open feedwater heater, one closed feedwater heater, and one reheater. Steam enters the turbine at 15 MPa and 600°C and is condensed in the condenser at a pressure of 10 kPa. Some steam is extracted from the turbine at 4 MPa for the closed feedwater heater, and the remaining steam is reheated at the same pressure to 600°C. The extracted steam is completely condensed in the heater and is pumped to 15 MPa before it mixes with the feedwater at the same pressure. Steam for the open feedwater heater is extracted from the low-pressure turbine at a pressure of 0.5 MPa. Determine the fractions of steam extracted from the turbine as well as the thermal efficiency of the cycle.
Reference:
[1] Çengel, Yunus A., Boles, Michael A., Kanoğlu, Mehmet, Thermodynamics: An Engineering Approach, Ninth Edition, NY: McGraw-Hill, 2019.
Consider a steam power plant that operates on an ideal reheat–regenerative Rankine cycle with one open feedwater heater, one closed feedwater heater, and one reheater. Steam enters the turbine at 15 MPa and 600°C and is condensed in the condenser at a pressure of 10 kPa. Some steam is extracted from the turbine at 4 MPa for the closed feedwater heater, and the remaining steam is reheated at the same pressure to 600°C. The extracted steam is completely condensed in the heater and is pumped to 15 MPa before it mixes with the feedwater at the same pressure. Steam for the open feedwater heater is extracted from the low-pressure turbine at a pressure of 0.5 MPa. Determine the fractions of steam extracted from the turbine as well as the thermal efficiency of the cycle.
Reference:
[1] Çengel, Yunus A., Boles, Michael A., Kanoğlu, Mehmet, Thermodynamics: An Engineering Approach, Ninth Edition, NY: McGraw-Hill, 2019.
H2O saturation properties - pressure increment
Temperature (C), Pressure (MPa), Enthalpy (kJ/kg), Entropy (J/g*K), Enthalpy (kJ/kg), Entropy (J/g*K), Volume (m3/kg), Volume (m3/kg)
Temperature (C), Pressure (MPa), Enthalpy (kJ/kg), Entropy (J/g*K), Enthalpy (kJ/kg), Entropy (J/g*K), Volume (m3/kg), Volume (m3/kg)
datasat=;
In[]:=
tblhl=Table[{datasat[[i]][[2]],datasat[[i]][[3]]},{i,1,Length[datasat]}];tblsl=Table[{datasat[[i]][[2]],datasat[[i]][[4]]},{i,1,Length[datasat]}];tblvl=Table[{datasat[[i]][[2]],datasat[[i]][[7]]},{i,1,Length[datasat]}];
In[]:=
fhl=Interpolation[tblhl];fsl=Interpolation[tblsl];fvl=Interpolation[tblvl];
In[]:=
tblhv=Table[{datasat[[i]][[2]],datasat[[i]][[5]]},{i,1,Length[datasat]}];tblsv=Table[{datasat[[i]][[2]],datasat[[i]][[6]]},{i,1,Length[datasat]}];
In[]:=
fhv=Interpolation[tblhv];fsv=Interpolation[tblsv];
In[]:=
tbltp=Table[{datasat[[i]][[1]],datasat[[i]][[2]]},{i,1,Length[datasat]}];
In[]:=
ftp=Interpolation[tbltp];
In[]:=
fhl[0.01]
Out[]=
191.806
H2O isobaric properties - pressure = 15 MPaTemperature (C) Pressure (bar) Enthalpy (kJ/kg) Entropy (J/g*K) State
dataisobaric=;
In[]:=
pos=Position[dataisobaric,342.16]
Out[]=
{{120,1},{121,1}}
In[]:=
tblhsv=Table[{dataisobaric[[i]][[1]],dataisobaric[[i]][[3]]},{i,121,Length[dataisobaric]}];tblssv=Table[{dataisobaric[[i]][[1]],dataisobaric[[i]][[4]]},{i,121,Length[dataisobaric]}];tblshsv=Table[{dataisobaric[[i]][[4]],dataisobaric[[i]][[3]]},{i,121,Length[dataisobaric]}];
In[]:=
fshsv=Interpolation[tblshsv];
In[]:=
tblhl1=Table[{datasat[[i]][[3]],datasat[[i]][[2]]},{i,1,Length[datasat]}];tblsl1=Table[{datasat[[i]][[4]],datasat[[i]][[1]]},{i,1,Length[datasat]}];
In[]:=
tblhv1=Table[{datasat[[i]][[5]],datasat[[i]][[2]]},{i,1,Length[datasat]}];tblsv1=Table[{datasat[[i]][[6]],datasat[[i]][[1]]},{i,1,Length[datasat]}];
In[]:=
fhsv=Interpolation[tblhsv];
In[]:=
fssv=Interpolation[tblssv];
H2O isobaric properties - pressure = 4 MPaTemperature (C) Pressure (bar) Enthalpy (kJ/kg) Entropy (J/g*K) State
dataisobaric=;
In[]:=
pos=Position[dataisobaric,250.35]
Out[]=
{{88,1},{89,1}}
In[]:=
tblhsv=Table[{dataisobaric[[i]][[1]],dataisobaric[[i]][[3]]},{i,89,Length[dataisobaric]}];tblssv=Table[{dataisobaric[[i]][[1]],dataisobaric[[i]][[4]]},{i,89,Length[dataisobaric]}];tblshsv=Table[{dataisobaric[[i]][[4]],dataisobaric[[i]][[3]]},{i,89,Length[dataisobaric]}];
In[]:=
fshsv2=Interpolation[tblshsv];
In[]:=
fhsv2=Interpolation[tblhsv];
In[]:=
fssv2=Interpolation[tblssv];
H2O isobaric properties - pressure = 0.5 MPaTemperature (C) Pressure (bar) Enthalpy (kJ/kg) Entropy (J/g*K) State
dataisobaric=;
In[]:=
pos=Position[dataisobaric,151.83]
Out[]=
{{54,1},{55,1}}
In[]:=
tblhsv=Table[{dataisobaric[[i]][[1]],dataisobaric[[i]][[3]]},{i,55,Length[dataisobaric]}];tblssv=Table[{dataisobaric[[i]][[1]],dataisobaric[[i]][[4]]},{i,55,Length[dataisobaric]}];tblshsv=Table[{dataisobaric[[i]][[4]],dataisobaric[[i]][[3]]},{i,55,Length[dataisobaric]}];
In[]:=
fshsv3=Interpolation[tblshsv];
In[]:=
fhsv3=Interpolation[tblhsv];
In[]:=
fssv3=Interpolation[tblssv];
In[]:=
H9=fhsv[600]
Out[]=
3583.11
In[]:=
S9=fssv[600]
Out[]=
6.67953
In[]:=
H1=fhl[0.01]
Out[]=
191.806
In[]:=
v1=fvl[0.01]
Out[]=
0.0010103
In[]:=
P2=500;P1=10;
In[]:=
wpump1=v1(P2-P1)(*INkJ/kg*)
Out[]=
0.495046
In[]:=
H2=H1+wpump1
Out[]=
192.301
In[]:=
H3=fhl[0.5]
Out[]=
640.087
In[]:=
v3=fvl[0.5]
Out[]=
0.00109259
In[]:=
P4=4000;P3=500;
In[]:=
wpump2=v3(P4-P3)(*INkJ/kg*)
Out[]=
3.82405
In[]:=
H4=H3+wpump2
Out[]=
643.911
In[]:=
S10=S9
Out[]=
6.67953
In[]:=
H10=fshsv2[S10]
Out[]=
3153.87
In[]:=
T10=x/.FindRoot[fhsv2[x]H10,{x,300}]
Out[]=
374.647
In[]:=
H11=fhsv2[600]
Out[]=
3674.89
In[]:=
S11=x/.FindRoot[fshsv2[x]==H11,{x,8}]
Out[]=
7.37051
In[]:=
S12=S11
Out[]=
7.37051
The fractions of steam extracted are determined from the mass and energy balances of the feedwater heaters