Example 10-4 reproduced from Ref. 1
​
​Consider a steam power plant that operates on the ideal reheat Rankine cycle. The plant maintains the inlet of the high-pressure turbine at 600 psia and 600°F, the inlet of the low-pressure turbine at 200 psia and 600°F, and the condenser at 10 psia. The net power produced by this plant is 5000 kW. Determine the rate of heat addition and rejection 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.
Thermophysical Properties from: https://webbook.nist.gov/
H2O saturation properties - pressure increment ​
{Temperature (F), Pressure (psia), Enthalpy (Btu/lbm), Entropy (Btu/lbm*R), Enthalpy (Btu/lbm), Entropy (Btu/lbm*R), Volume (ft3/lbm), Volume (ft3/lbm)}
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];
H2O isobaric properties - pressure = 200 psia
{Temperature (F), Pressure (psia), Enthalpy (Btu/lbm), Entropy (Btu/lbm*R) State}
In[]:=
In[]:=
pos=Position[dataisobaric,381.80]
Out[]=
{{69,1},{70,1}}
In[]:=
tblhsv=Table[{dataisobaric[[i]][[1]],dataisobaric[[i]][[3]]},{i,70,Length[dataisobaric]}];​​tblssv=Table[{dataisobaric[[i]][[1]],dataisobaric[[i]][[4]]},{i,70,Length[dataisobaric]}];​​tblshsv=Table[{dataisobaric[[i]][[4]],dataisobaric[[i]][[3]]},{i,70,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 = 600 psia
{Temperature (F), Pressure (psia), Enthalpy (Btu/lbm), Entropy (Btu/lbm*R) State}
In[]:=
In[]:=
pos=Position[dataisobaric,486.24]
Out[]=
{{89,1},{90,1}}
In[]:=
tblhsv=Table[{dataisobaric[[i]][[1]],dataisobaric[[i]][[3]]},{i,90,Length[dataisobaric]}];​​tblssv=Table[{dataisobaric[[i]][[1]],dataisobaric[[i]][[4]]},{i,90,Length[dataisobaric]}];​​tblshsv=Table[{dataisobaric[[i]][[4]],dataisobaric[[i]][[3]]},{i,90,Length[dataisobaric]}];
In[]:=
fshsv2=Interpolation[tblshsv];
In[]:=
fhsv2=Interpolation[tblhsv];
In[]:=
fssv2=Interpolation[tblssv];
In[]:=
FindRoot[ftp[x]==600,{x,590}]
Out[]=
{x486.243}
In[]:=
FindRoot[ftp[x]==200,{x,400}]
Out[]=
{x381.799}
In[]:=
FindRoot[ftp[x]==10,{x,200}]
Out[]=
{x193.162}
In[]:=
H3=fhsv2[600]
Out[]=
1290.77
In[]:=
S3=fssv2[600]
Out[]=
1.53354
In[]:=
H5=fhsv[600]
Out[]=
1323.19
In[]:=
S5=fssv[600]
Out[]=
1.67826
In[]:=
S4=S3
Out[]=
1.53354
In[]:=
S6=S5
Out[]=
1.67826
In[]:=
S4l=fsl[200]
Out[]=
0.544169
In[]:=
S4v=fsv[200]
Out[]=
1.54701
In[]:=
x4=(S4-S4l)/(S4v-S4l)
Out[]=
0.986565
In[]:=
H4l=fhl[200]
Out[]=
355.703
In[]:=
H4v=fhv[200]
Out[]=
1199.56
In[]:=
H4=x4(H4v-H4l)+H4l
Out[]=
1188.22
In[]:=
S6l=fsl[10]
Out[]=
0.283807
In[]:=
S6v=fsv[10]
Out[]=
1.78878
In[]:=
x6=(S6-S6l)/(S6v-S6l)
Out[]=
0.926568
In[]:=
H6l=fhl[10]
Out[]=
161.357
In[]:=
H6v=fhv[10]
Out[]=
1143.86
In[]:=
H6=x6(H6v-H6l)+H6l
Out[]=
1071.72
In[]:=
H1=fhl[10]
Out[]=
161.357
In[]:=
v1=fvl[10]
Out[]=
0.0165892
In[]:=
P1=10;P2=600;