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+// Example 2_7
+clc;funcprot(0);
+// Given data
+A=1200;// The catchment area in km^2
+AR=160;// The annual rainfall in cm
+H=360;// The head available in m
+n_o=75;// Over all efficiency of the plant in %
+F_l=0.5;// Load factor
+PEL=25;// Percoalation and evaporation losses in %
+g=9.81;// m/s^2
+
+// Calculation
+V=A*10^6*(AR/100)*(1-((PEL/100)));// The quantity of water available for power generation per year in cu.m
+Q=V/(365*24*3600);// Average flow per second in m^3/sec
+m=Q*1000;// Discharge in kg/sec
+P_avg=((m*g*H)/1000)*(n_o/100);// Average power developed in kW
+P_avg=P_avg/1000;// MW
+MD=(P_avg/F_l);// Maximum demand in MW
+printf('\nThe average power developed=%0.2f MW \nMaximum demand=%0.1f MW',P_avg,MD);
+// The answer vary due to round off error