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//Fluid system - By - Shiv Kumar
//Chapter 4 - Pelton Turbine (Impulse Turbine)
//Example 4.6
clc
clear
//Given Data:-
N=300; //Speed of runner, rpm
H=510; //Head, m
d=200; //Diameter of the Jet, mm
AoD=165; //Angle of Jet(Deflection inside bucket), degrees
Vel_per=15; //Percentage by which velocity is reduced due to friction
Loss_per=3; //Percentage of mechanical Losses (of power Supplied)
//Data Used:-
rho=1000; //Density of water, kg/m^3
g=9.81; //Acceleration due to gravity, m/s^2
Cv=0.98; //Co-efficient of Velocity
Ku=0.46; //Speed ratio
//Computations:-
d=d/1000; //m
beta_O=180-AoD; //degrees
Vro_by_Vri=1-Vel_per/100; //Vro/Vri
Vi=Cv*sqrt(2*g*H); //m/s
Vwi=Vi;
u=Ku*sqrt(2*g*H); //m/s
uo=u;
ui=u;
Vri=Vi-ui; //m/s
Vro=Vri*Vro_by_Vri; //m/s
Vrwo=Vro*cosd(beta_O); //m/s
Vwo=uo-Vrwo; //m/s
Q=(%pi/4)*d^2*Vi; //Discharge, m^3/s
//(i) Resultant Force on bucket, F
F=rho*Q*(Vwi-Vwo)/1000; //kN
//Result (i):-
printf("(i) Resultant Force on bucket, F=%.3f kN \n", F) //The answer vary due to round off error
//(ii) Shaft Power, P
Pr=F*u; //power developed by runner, kW
P=Pr-(Loss_per/100)*Pr; //kW
//Result (ii)
printf("(ii)Shaft Power, P=%.3f kW \n", P) //The answer given in the textbook is wrong (due to round off error in F)
//OR
eta_m=1-Loss_per/100; //Mechanical Efficiency
P=eta_m*Pr; //kW
//(iii) Overall Efficiency, eta_O
eta_O=P*1000/(rho*Q*g*H)*100; //In percentage
//Result (iii)
printf("(iii)Overall efficiency, eta_O=%.2f percent", eta_O)
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