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//Fluid Systems by Shiv Kumar
//Chapter 7 - Performance of water turbine
//Example 7.4
//To Find (a) The Diameter of Runner (b) The Diameter of jet
clc
clear
//Given:
P=3200; //Power Developed, kW
H=310; // Effective Head , m
eta_o=82/100; //Overall Efficiency
Ku=0.46; // Speed Ratio
Cv=0.98 // Co-efficient of Velocity
Ns=18; //Specific Speed (SI Units)
//Data required
rho=1000; //Density of Water, Kg/m^3
g=9.81 // Acceleration due to gravity, m/s^2
//Computations
N=Ns*H^(5/4)/sqrt(P); //Speed, rpm
D=Ku*sqrt(2*g*H)*60/(%pi*N); //Diameter of runner, m
Q=P*1000/(rho*g*H*eta_o); //Discharge, m^3/s
d=sqrt(Q/((%pi/4)*Cv*sqrt(2*g*H))); // Diameter of Jet, m
//Results
printf("(a) The Diameter of Runner, D =%.2f m\n",D) //The Answer Vary due to Round Off Error
printf("(c) The Diameter of Jet, d=%.3f m \n",d)
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