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| author | priyanka | 2015-06-24 15:03:17 +0530 |
|---|---|---|
| committer | priyanka | 2015-06-24 15:03:17 +0530 |
| commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
| tree | ab291cffc65280e58ac82470ba63fbcca7805165 /1055/CH24 | |
| download | Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.tar.gz Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.tar.bz2 Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.zip | |
initial commit / add all books
Diffstat (limited to '1055/CH24')
| -rwxr-xr-x | 1055/CH24/EX24.3/ch24_3.sce | 31 | ||||
| -rwxr-xr-x | 1055/CH24/EX24.4/ch24_4.sce | 37 |
2 files changed, 68 insertions, 0 deletions
diff --git a/1055/CH24/EX24.3/ch24_3.sce b/1055/CH24/EX24.3/ch24_3.sce new file mode 100755 index 000000000..92661eec2 --- /dev/null +++ b/1055/CH24/EX24.3/ch24_3.sce @@ -0,0 +1,31 @@ +// Priority List Method
+clear
+clc;
+Fc1=1.1;//Fuel cost(1)=Rs 1.1/MBtu
+Fc2=1;//Fuel cost(2)=1/MBtu
+Fc3=1.2;//Fuel cost(3)=1.2/MBtu
+P1max=600;
+P1=P1max;
+F1=600+7.1*P1+0.00141*(P1^2);//For P1= Pm1ax
+Favg1=F1*Fc1/600;//Full load average production cost
+P2max=450;
+P2=P2max;
+F2=350+7.8*P2+0.00195*(P2^2);//For P2= P2max
+Favg2=F2*Fc2/450;//Full load average production cost
+P3max=250;
+P3=P3max;
+F3=80+8*P3+0.0049*(P3^2);//For P3= P3max
+Favg3=F3*Fc3/250;//Full load average production cost
+mprintf("Priority List is as follows\n");
+mprintf("Unit Rs/MWhr MinMW Max MW\n")
+mprintf(" 2 %.3f 100 %.0f \n",Favg2,P2max)
+mprintf(" 1 %.4f 60 %.0f \n",Favg1,P1max)
+mprintf(" 3 %.2f 50 %.0f \n\n",Favg3,P3max)
+Fmax1=P1max+P2max+P3max;
+Fmax2=P2max+P1max
+Fmax3=P2max
+mprintf("Unit Commitment Scheme is follows\n")
+mprintf("Combination Min.MW from Combination Max.MW from Combination\n");
+mprintf("2+1+3 310 %.0f \n",Fmax1);
+mprintf("2+1 260 %.0f \n",Fmax2);
+mprintf("2 100 %.0f ",Fmax3);
diff --git a/1055/CH24/EX24.4/ch24_4.sce b/1055/CH24/EX24.4/ch24_4.sce new file mode 100755 index 000000000..438721133 --- /dev/null +++ b/1055/CH24/EX24.4/ch24_4.sce @@ -0,0 +1,37 @@ +// illustrate the dynamic programming for preparing an optimal unit commitment.
+
+clear
+clc;
+function[F1]=F1(P1)
+ F1=7.1*P1+.00141*(P1^2)
+ mprintf("F1(%.0f)=%.1f\n",P1,F1);
+endfunction
+function[f2]=f2(P2)
+ f2=7.8*P2+.00195*(P2^2)
+ mprintf("f2(%.0f)=%.0f\n",P2,f2);
+endfunction
+function[F]=F(P1,P2)
+ F1=7.1*P1+.00141*(P1^2)
+ F2=7.8*P2+.00195*(P2^2)
+ F=F1+F2
+ mprintf("F1(%.0f)+f2(%.0f)=%.0f\n",P1,P2,F);
+ endfunction
+P1max=600;
+P2max=450;
+mprintf("Unit Commitment using Load 500MW\n")
+F1(500);
+mprintf("Since min. Power of second unit is 100MW , we find\n");
+F(400,100);
+F(380,120);
+F(360,140);
+mprintf("Therefore for load 500 MW , the load commitment on unit 1 is 400 MW and that on 2 is 100 MW which gives min. cost\n");
+mprintf("Next we increase the load by 50 MW and loading unit 1 we get, \n");
+F1(550);
+mprintf("Also if we distribute a part of load to unit 2 we get ,\n")
+F(450,100);
+F(400,150);
+F(350,200);
+mprintf("Therefore for load 550 MW , the load commitment on unit 1 is 400 MW and that on 2 is 150 MW which gives min. cost\n");
+
+
+
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