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//Fluid Systems- By Shiv Kumar
//Chapter 7- Performance of Water Turbine
//Example 7.9
// To Determine the Performance of the Turbine Under a Head of 20 m
clc
clear
//Given:-
//Condition 1:
H1=25; //Head, m
N1=200; //Speed, rpm
Q1=9; //Discharge, m^3/s
eta_o=90/100; //Overall Efficiency
//Condition 2:
H2=20; //Head, m
//Data Required:-
rho=1000; //Density of Water, Kg/m^3
g=9.81; //Acceleration due to Gravity, m/s^2
//Calculations:-
P1=rho*Q1*g*H1*eta_o/1000; //KW
N2=N1*sqrt(H2/H1); //rpm
Q2=Q1*sqrt(H2/H1); //m^3/s
P2=P1*(H2/H1)^(3/2); //KW
//Results:-
printf("At Condition 2 (Under a Head of 20 m):\n")
printf("\tSpeed, N2=%.2f rpm\n Discharge, Q2=%.2f m^3/s\n Power Developed, P2=%.2f kW",N2,Q2,P2) //The Answer vary due to Round off Error
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