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//Fluid Systems by Shiv Kumar
//Chapter 7 - Performance of water turbine
//Example 7.3
//To Find (a) The Discharge required (b) The Diameter of Wheel (c) The Diameter and number of jets required (d)The Specific Speed
clc
clear
//Given:
P=8200; //Power Developed, kW
H=128; // Head , m
N=210; // Speed, rpm
Cv=0.98; // Co-efficient of Velocity
eta_H=89/100; //Hydraulic Efficiency
Ku=0.45; // Speed Ratio
dbyD=1/10; //Ratio of jet diameter to wheel Diameter
eta_m=92/100; //Mechanical Efficiency
//Data required
rho=1000; //Density of Water, Kg/m^3
g=9.81; // Acceleration due to gravity, m/s^2
//Assumptions:
eta_v=100/100; // Volumetric efficiency is 100%
//Computations
D=Ku*sqrt(2*g*H)*60/(%pi*N); //Wheel Diameter, m
d=D*dbyD; // Jet diameter, m
eta_o=eta_H*eta_m*eta_v; //Overall Efficiency
Q=P*1000/(rho*g*H*eta_o); //Net Discharge, m^3/s
q=(%pi/4)*d^2*Cv*sqrt(2*g*H); //Discharge through one jet, m^3/s
n=round(Q/q); //Number of jets
Ns= N*P^(1/2)/(H^(5/4)); //Specific Speed, SI Units
//Results
printf("(a) The Discharge required, Q =%.3f m^3/s\n",Q)
printf("(b) The Diameter of Wheel, D =%.2f m\n",D)
printf("(c) The Diameter, d=%.3f m and\n number of jets required =%.f \n",d,n)
printf("(d)The Specific Speed, Ns=%.2f (SI Units)\n",Ns) //The Answer Vary due to Round off Error
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