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//Fluid Systems by Shiv Kumar
//Chapter 7 - Performance of water turbine
//Example 7.5
//To Find (a) Number of Units to be installed (b) Diameter of Wheel (c) Diameter of Jet
clc
clear
//Given:
H_G= 310; //Gross Head,m
l=2.5; // Length, km
h_f=25; // Friction Loses, J/N=m
TO=20; //Total Output Power, MW
N=660; // Speed, rpm
ubyVi=0.46 //Ratio of bucket to jet speed
eta_o=88/100; //Overall Efficiency
Ns=28; //Specific Speed, SI Units
Cv=0.97;
Cd=0.94;
//Data Required:
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
P=(Ns*H^(5/4)/N)^2; //Power Output of each Unit, KW
//(a) The no. of units to be lnstalled,n
n=round(TO*1000/P);
//(b)Diameter of Wheel,D
Vi=Cv*sqrt(2*g*H); //m/s
D=ubyVi*Vi*60/(%pi*N); //m
//(c) Diameter of Jet, d
Q=TO*10^6/(rho*g*H*eta_o); //Net Discharge, m^3/s
q=Q/n; // Discharge through one unit, m^/s
d=sqrt(q/((%pi/4)*Cd*sqrt(2*g*H)))*1000; //mm
//Results
printf("(a) The no. of units to be Installed=%.f Units\n",n)
printf("(b) Diameter of Wheel, D=%.3f m\n",D)
printf("(c) Diameter of Jet, d=%.1f mm\n",d) //The Answer Vary due to round off Error
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