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authorprashantsinalkar2017-10-10 12:27:19 +0530
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+//Chapter 6 Example1//
+clc
+clear
+//catchment area of reservoir=a,average rainfall=ar,percent of rainfall utilized=pu//
+//average available water for electricity production=we//
+a=50;// in km^2//
+ar=150;// in cm/year //
+pu=75;// in percent//
+we=(a*10^6)*(ar/100)*(pu/100);// to convert in terms of 10^6//
+printf("\n Total available water for electricity production = %.3f m^3\n",we);
+//quantity available=qa//
+qa=we/(365*24*60*60);
+//power generated=p,efficiency of turbine=te,efficiency of generator=ge,load factor=lf,mean head=mh//
+te=88;// in percent//
+ge=93;// in percent//
+lf=75;// in percent//
+mh=40;
+p=0.736*qa*1000*mh*(ge/100)*(te/100)/75;
+printf("\n Total Power generated in kW = %.2f kW\n",p);
+// installed capacity=ic//
+ic=p/(lf/100);
+printf("\n Installed capacity of the generators = %.2f kW\n",ic);
+// the values given in the book are approximated to the nearest decimal// \ No newline at end of file