Example 10-5 reproduced from Ref. 1
Consider a steam power plant operating on the ideal regenerative Rankine cycle with one open feedwater heater. Steam enters the turbine at 15 MPa and 600°C and is condensed in the condenser at a pressure of 10 kPa. Some steam leaves the turbine at a pressure of 1.2 MPa and enters the open feedwater heater. Determine the fraction of steam extracted from the turbine and 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 operating on the ideal regenerative Rankine cycle with one open feedwater heater. Steam enters the turbine at 15 MPa and 600°C and is condensed in the condenser at a pressure of 10 kPa. Some steam leaves the turbine at a pressure of 1.2 MPa and enters the open feedwater heater. Determine the fraction of steam extracted from the turbine and 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)
In[]:=
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
In[]:=
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 = 1.2 MPaTemperature (C) Pressure (bar) Enthalpy (kJ/kg) Entropy (J/g*K) State
In[]:=
In[]:=
pos=Position[dataisobaric,187.96]
Out[]=
{{67,1},{68,1}}
In[]:=
tblhsv=Table[{dataisobaric[[i]][[1]],dataisobaric[[i]][[3]]},{i,68,Length[dataisobaric]}];tblssv=Table[{dataisobaric[[i]][[1]],dataisobaric[[i]][[4]]},{i,68,Length[dataisobaric]}];tblshsv=Table[{dataisobaric[[i]][[4]],dataisobaric[[i]][[3]]},{i,68,Length[dataisobaric]}];
In[]:=
fshsv2=Interpolation[tblshsv];
In[]:=
fhsv2=Interpolation[tblhsv];
In[]:=
fssv2=Interpolation[tblssv];
In[]:=
FindRoot[ftp[x]==15,{x,300}]
Out[]=
{x342.154}
In[]:=
FindRoot[ftp[x]==1.2,{x,200}]
Out[]=
{x187.957}
In[]:=
FindRoot[ftp[x]==0.01,{x,1}]
Out[]=
{x45.8057}
In[]:=
H1=fhl[0.01]
Out[]=
191.806
In[]:=
S1=fsl[0.01]
Out[]=
0.649185
In[]:=
v1=fvl[0.01]
Out[]=
0.0010103
In[]:=
P1=0.01;P2=1.2;
In[]:=
Wpump1=v1(P2-P1)1000(*expressedinkJ/kg*)
Out[]=
1.20226
In[]:=
H2=H1+Wpump1
Out[]=
193.008
In[]:=
P3=1.2;P4=15;
In[]:=
v3=fvl[1.2]
Out[]=
0.00113852
In[]:=
Wpump2=v3(P4-P3)1000(*expressedinkJ/kg*)
Out[]=
15.7116
In[]:=
H3=fhl[1.2]
Out[]=
798.335
In[]:=
H4=H3+Wpump2
Out[]=
814.047
In[]:=
H5=fhsv[600]
Out[]=
3583.11
In[]:=
S5=fssv[600]
Out[]=
6.67953
In[]:=
S6=S5
Out[]=
6.67953
In[]:=
H6=fshsv2[S6]
Out[]=
2858.77
In[]:=
T6=x/.FindRoot[fssv2[x]==S6,{x,270}]
Out[]=
217.119
In[]:=
S7=S5
Out[]=
6.67953
In[]:=
S7l=fsl[0.01]
Out[]=
0.649185
In[]:=
S7v=fsv[0.01]
Out[]=
8.14887
In[]:=
x7=(S7-S7l)/(S7v-S7l)
Out[]=
0.80408