Example 10-8 reproduced from Ref. 1
Example 10-8 reproduced from Ref. 1
Consider the cogeneration plant shown in Fig. 10–25. Steam enters the turbine at 7 MPa and 500°C. Some steam is extracted from the turbine at 500 kPa for process heating. The remaining steam continues to expand to 5 kPa. Steam is then condensed at constant pressure and pumped to the boiler pressure of 7 MPa. At times of high demand for process heat, some steam leaving the boiler is throttled to 500 kPa and is routed to the process heater. The extraction fractions are adjusted so that steam leaves the process heater as a saturated liquid at 500 kPa. It is subsequently pumped to 7 MPa. The mass flow rate of steam through the boiler is 15 kg/s. Disregarding any pressure drops and heat losses in the piping and assuming the turbine and the pump to be isentropic, determine (a) the maximum rate at which process heat can be supplied, (b) the power produced and the utilization factor when no process heat is supplied, and (c) the rate of process heat supply when 10 percent of the steam is extracted before it enters the turbine and 70 percent of the steam is extracted from the turbine at 500 kPa for process heating.
H2O saturation properties - pressure increment Temperature (C) Pressure (bar) Enthalpy (kJ/kg) Entropy (J/g*K) Enthalpy (kJ/kg) Entropy (J/g*K)
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,InterpolationOrder1];fsl=Interpolation[tblsl,InterpolationOrder1];fvl=Interpolation[tblvl,InterpolationOrder1];
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 = 70 bar
Temperature (C) Pressure (bar) Enthalpy (kJ/kg) Entropy (J/g*K) Phase
Temperature (C) Pressure (bar) Enthalpy (kJ/kg) Entropy (J/g*K) Phase
dataisobaric=;
In[]:=
pos=Position[dataisobaric,285.83]
Out[]=
{{101,1},{102,1}}
In[]:=
tblhsv=Table[{dataisobaric[[i]][[1]],dataisobaric[[i]][[3]]},{i,102,Length[dataisobaric]}];tblssv=Table[{dataisobaric[[i]][[1]],dataisobaric[[i]][[4]]},{i,102,Length[dataisobaric]}];tblshsv=Table[{dataisobaric[[i]][[4]],dataisobaric[[i]][[3]]},{i,102,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];
In[]:=
FindRoot[ftp[x]==70,{x,290}]
Out[]=
{x285.828}
In[]:=
FindRoot[ftp[x]==5,{x,100}]
Out[]=
{x151.831}
In[]:=
FindRoot[ftp[x]==0.05,{x,1}]
Out[]=
{x32.8741}
In[]:=
v8=fvl[0.05]
Out[]=
0.00100534
In[]:=
v7=fvl[5]
Out[]=
0.00109258
In[]:=
P8=5;P9=7000;
In[]:=
Wpump1=v8(P9-P8)(*inkJ/kg*)
Out[]=
7.03238
In[]:=
P7=500;P10=7000;
In[]:=
Wpump2=v7(P10-P7)(*inkJ/kg*)
Out[]=
7.10179
In[]:=
H2=H1;H3=H1;H4=H1;
In[]:=
H1=fhsv[500]
Out[]=
3411.37
In[]:=
S1=fssv[500]
Out[]=
6.79993
In[]:=
S5=S1;S6=S1;
In[]:=
S5l=fsl[5]
Out[]=
1.86038
In[]:=
S5v=fsv[5]
Out[]=
6.82069
In[]:=
x5=(S1-S5l)/(S5v-S5l)
Out[]=
0.995814
In[]:=
H5l=fhl[5]
Out[]=
640.08
In[]:=
H5v=fhv[5]
Out[]=
2748.09
In[]:=
H5=x5(H5v-H5l)+H5l
Out[]=
2739.26
In[]:=
S6l=fsl[0.05]
Out[]=
0.476153
In[]:=
S6v=fsv[0.05]
Out[]=
8.39381
In[]:=
x6=(S1-S6l)/(S6v-S6l)
Out[]=
0.798693
In[]:=
H6l=fhl[0.05]
Out[]=
137.74
In[]:=
H6v=fhv[0.05]
Out[]=
2560.76
In[]:=
H6=x6(H6v-H6l)+H6l
Out[]=
2072.99
In[]:=
H7=fhl[5]
Out[]=
640.08
In[]:=
H8=fhl[0.05]
Out[]=
137.74
In[]:=
H9=H8+Wpump1
Out[]=
144.772
In[]:=
H10=H7+Wpump2
Out[]=
647.182
The maximum rate of process heat is achieved when all the steam leaving the boiler is throttled and sent to the process heater and none is sent to the turbine (that is, mdot4 = mdot7 = mdot1 = 15 kg/s and mdot3 = mdot5 = mdot6 = 0).
In[]:=
QDOTProcessHeat=15(H4-H7)
Out[]=
41569.4
When no process heat is supplied, all the steam leaving the boiler passes through the turbine and expands to the condenser pressure of 5 kPa (that is, mdot3 = mdot6 = mdot1 = 15 kg/s and mdot2 = mdot5 = 0).
In[]:=
WDOTTurbine=15(H3-H6)
Out[]=
20075.8
In[]:=
WDOTPUMP1=15Wpump1
Out[]=
105.486
In[]:=
WDOTNET=WDOTTurbine-WDOTPUMP1
Out[]=
19970.3
Let us find the rate of process heat supply when 10 percent of the steam is extracted before it enters the turbine and 70 percent of the steam is extracted from the turbine at 500 kPa for process heating.
26.2 MW of the heat transferred will be utilized in the process heater. Let us also show that 11.0 MW of power is produced in this case, and the rate of heat input in the boiler is 43.0 MW.
Thus the utilization factor is 86.5 percent.
Thus the utilization factor is 86.5 percent.
References
References
[1] Çengel, Yunus A., Boles, Michael A., Kanoğlu, Mehmet, Thermodynamics: An Engineering Approach, Ninth Edition, NY: McGraw-Hill, 2019.
CITE THIS NOTEBOOK
CITE THIS NOTEBOOK
An ideal cogeneration plant
by Housam Binous and Ahmed Bellagi
Wolfram Community, STAFF PICKS, February 27, 2025
https://community.wolfram.com/groups/-/m/t/3401420
by Housam Binous and Ahmed Bellagi
Wolfram Community, STAFF PICKS, February 27, 2025
https://community.wolfram.com/groups/-/m/t/3401420