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//Fluid Systems - By - Shiv Kumar
//Chapter 4 - Pelton Turbine (Impulse Turbine)
//Example 4.20
clc
clear
//Given Data:-
D=1.6; //Mean Diameter of Bucket Circle, m
P=3200; //Power Developed, kW
n=2; //Number of Wheels
H=300; //Effective Head, m
N=410; //Speed, rpm
eta_o=0.89; //Overall Efficiency
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:-
Q=P*1000/(rho*g*H*eta_o); //Discharge, m^3/s
//(a)Diameter of the Nozzle, d
Vi=Cv*sqrt(2*g*H); //m/s
d=sqrt(Q/((%pi/4)*Vi))*1000; //mm
//(b)Speed Ratio, Ku
u=%pi*D*N/60; //m/s
Ku=u/sqrt(2*g*H);
//(c)Specific Speed, Ns
Ns=N*sqrt(P/n)/(H^(5/4)); // In SI Units
//Results:-
printf(" (a)Diameter of the Nozzle, d=%.2f mm\n",d) //The answer vary due to round off error
printf(" (b)Speed Ratio, Ku =%.3f \n",Ku) //The answer vary due to round off error
printf(" (c)Specific Speed, Ns =%.f (SI Units)\n",Ns)
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