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authorpriyanka2015-06-24 15:03:17 +0530
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+
+
+//example 3.17
+//calculate resvior capacity
+clc;
+//Given
+CL=0.2; //Canal loss
+RL=0.12; //Reservior loss
+A=[4800 5600 2400 3200 1400]; //given values of area under crop
+D=[1800 800 1400 900 700]; //given values of duty at field
+B=[120 360 200 120 120]; //given values of base period
+
+//(a) Wheat
+d=A(1)/D(1);
+V1=d*B(1);
+//(b) Sugarcane
+d=A(2)/D(2);
+V2=d*B(2);
+//(c) Cotton
+d=A(3)/D(3);
+V3=round(d*B(3));
+//(d) Rice
+d=A(4)/D(4);
+V4=round(d*B(4));
+//(e) vegetables
+d=A(5)/D(5);
+V5=d*B(5);
+
+Vd=(V1+V2+V3+V4+V5)*8.64;
+SC=Vd/((1-CL)*(1-RL));
+mprintf("Reservior capacity=%f hectare-metres.",SC);
+
+//Alternative method
+
+for i=1:5
+ delta(i)=8.64*B(i)/D(i);
+end
+
+for j=1:5
+ V(j)=A(j)*delta(j);
+end
+s=0;
+for k=1:5
+ s=s+V(k);
+end
+SC=s/((1-CL)*(1-RL));
+SC=round(SC);
+mprintf("\n By Alternative method.\nStorage capacity=%f hectare-metres.",SC);
+
+
+
+
+
+