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//Fluid system - By - Shiv Kumar
//Chapter 4 - Pelton Turbine (Impulse Turbine)
//Example 4.11
clc
clear
//Given Data:-
N=300; //Speed of runner, rpm
H=500; //Head, m
d=200; //Diameter of the Jet, mm
AoD=165; //Angle of Deflection, degrees
Vel_per=15; //percentage by which velocity is reduced
Cv=0.98; //Co-efficient of Velocity
Ku=0.46; //Speed ratio
Loss_per=3; //Percentage of Mechanical losses
//Data Used:-
rho=1000; //Density of water, kg/m^3
g=9.81; //Acceleration due to gravity, m/s^2
//Computations:-
d=d/1000; //m
beta_O=180-AoD; //degrees
Vro_by_Vri=1-Vel_per/100; //Vro/Vri
K=Vro_by_Vri;
Vi=Cv*sqrt(2*g*H); //m/s
Vwi=Vi;
ui=Ku*sqrt(2*g*H); //m/s
uo=ui;
u=ui;
Vri=Vi-ui; //m/s
Vro=K*Vri; //m/s
Vrwo=Vro*cosd(beta_O); //m/s
Vwo=uo-Vrwo; //m/s
//(a) Water power, WP
Q=(%pi/4)*d^2*Vi; //m^3.s
WP=rho*Q*g*H/1000; //kW
//(b)The Force on the bucket in the direction of Jet, F
F=rho*Q*(Vwi-Vwo)/1000; //kN
//(c)Shaft Power, SP
Pr=F*u; //Power developed by the Runner, W
SP=Pr-Loss_per/100*Pr; //kW
//(d)Overall Efficiency, eta_o
eta_o=SP/WP*100; //In percentage
//Results:-
printf("(a) Water power, WP=%.2f kW \n",WP) //The answer provided in the Textbook is wrong
printf("(b)The Force on the bucket in the direction of Jet=%.3f kN \n", F) //The answer vary due to round off error
printf("(c)Shaft Power, SP=%.3f kW\n",SP) //The answer provided in the Textbook is wrong
printf("(d)Overall efficiency, eta_o=%.2f percent", eta_o) //The answer vary due to round off error
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