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+//Example 5_3
+clc;clear;funcprot(0);
+// Given values
+m=5000;// Mass flow rate of water in kg/s
+W_eout=1862;//The electric power generated is measured in kW
+rho=1000;// The density of water in kg/m^3
+h=50;// The depth of the water in m
+g=9.81;// m/s^2
+e_min=g*h;// kJ/kg
+e_mout=0;// kJ/kg
+n_gen=0.95;// The generator efficiency
+
+// Calculation 
+//(a)
+dE_mech=(m*(e_min-e_mout))/1000;//kW
+n_o=(W_eout/dE_mech);// The over all efficiency
+printf('(a)The over all efficiency,n_o=%0.2f\n',n_o);
+//(b)
+n_t=n_o/n_gen;// )The mechanical efficiency of the turbine
+printf('(b)The mechanical efficiency of the turbine,n_t=%0.2f\n',n_t);
+//(c)
+W_sout=n_t*dE_mech;// kW
+printf('(c)The shaft power output,W_shaft,out=%0.0f kW\n',W_sout);
+//The answer is a bit different due to rounding off error in textbook