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+// Example 2_12
+clc;funcprot(0);
+//Given data
+H=30;// m
+A=250;// sq.km
+Ar=125;// Annual rainfall in cm
+Tr=70/100;// Total rainfall
+F_l=50/100;// Load factor
+h_l=8/100;// Head loss
+n_m=90/100;// Mechanical efficiency of the turbine
+n_g=95/100;// Generator efficiency
+rho_w=1000;// kg/m^3
+g=9.81;// m/s^2
+
+//Calculation
+V=A*10^6*(Ar/100)*Tr;//Water available during the year in m^3
+Q=(V)/(8760*3600);// Water flow per second in m^3/sec
+Q=Q*1000;// kg/sec
+n_h=(1-h_l);// Hydraulic efficiency
+n_o=n_h*n_m*n_g;//The over all efficiency
+P=(Q*9.81*H*n_o)/(1000);// kW
+//With 50% load factor
+Gc=P/F_l;// Generator capacity in kW
+printf('\nThe power=%0.0f kW \nGenerator capacity=%0.1f kW',P,Gc);
+// The answer provided in the textbook is wrong
+