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+//Fluid system - By - Shiv Kumar
+//Chapter 4 - Pelton Turbine (Impulse Turbine)
+//Example 4.17
+
+ clc
+ clear
+
+//Given Data:-
+ H_G=510; //Gross Head, m
+ h_f=(1/3)*H_G; //Head lost in friction in penstock, m
+ d=170; //Diameter of Jet, mm
+ AoD=165; //Angle of Deflection of Jet, degrees
+ Ku=0.45; //Speed ratio
+ Cv=0.98; //Co-efficient of Velocity
+
+//Data Used:-
+ rho=1000; //Density of water, kg/m^3
+ g=9.81; //Acceleration due to gravity, m/s^2
+
+//Computations:-
+ H=H_G-h_f; //Effective Head, m
+ Vi=Cv*sqrt(2*g*H); //m/s
+ Vwi=Vi;
+ u=Ku*sqrt(2*g*H); //m/s
+ ui=u;
+ uo=u;
+ Vri=Vi-u; //m/s
+ Vro=Vri;
+ beta_o=180-AoD; //degrees
+ Vrwo=Vro*cosd(beta_o); //m/s
+ Vwo=Vrwo-uo; //m/s
+ Q=(%pi/4)*(d/1000)^2*Vi; //Discharge, m^3/s
+ P=rho*Q*(Vwi+Vwo)*u/1000; //Power developed by runner, kW
+ eta_H=2*(Vwi+Vwo)*u/Vi^2*100; //Hydraulic efficiency, In percentage
+
+//Results:-
+ printf("(a)Power developed by the runner=%.3f kW \n",P) //The answer provided in the Textbook is wrong
+ printf("(b)Hydraulic efficiency, eta_H=%.2f percent", eta_H) //The answer vary due to round off error
+
+
+