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authorpriyanka2015-06-24 15:03:17 +0530
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+
+
+//example 3.21
+//calculate reservior capacity
+clc;
+//given
+W=0.4; //amount of water available from precipitation
+Cl=0.15; //Channel loss
+RL=0.1; //reservior loss
+B=[120 320 120 200 100]; //Base period
+D=[1800 800 900 1400 1200];//Duty at field
+A=[500 600 300 1200 500]; //Area under crop
+
+for i=1:5
+ delta(i)=8.64*B(i)/D(i);
+end
+
+for i=1:5
+ V(i)=delta(i)*A(i);
+
+end
+s=0;
+for i=1:5
+ s=s+V(i);
+end
+C=s*(1-W)/((1-Cl)*(1-RL));
+
+mprintf("Reservior capacity=%i ha-m.",C);
+
+