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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
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