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//Fluid System By Shiv Kumar
//Chapter 6 - Kaplan and Propeller Turbines
//Example 6.7
//To Find (a)Diameter of Runner (b)Speed (c)Specific Speed
clc
clear
//Given:
H=32; //Head, m
P=16000; //Shaft Power, KW
D_per=190; //Percentage by which Diameter of Runner(D)is Larger than diameter of Boss(d)
eta_0=91/100; //Overall Efficiency
Ku=2; //Speed Ratio
Kf=0.64; //Flow Ratio
//Data Required:
rho=1000; //Density of Water, Kg/m^3
g=9.81; //Acceleration due to gravity, m/s^2
//Computations
Vfi=Kf*sqrt(2*g*H); //Velocity of Flow at Inlet, m/s
ui= Ku*sqrt(2*g*H); //Velocity of Runner at Inlet, m/s
Q=P*10^3/(rho*g*H*eta_0); //Discharge, m^3/s
d=sqrt(Q/((%pi/4)*Kf*sqrt(2*g*H)*((D_per/100+1)^2-1))); // Diameter of Hub ,m
//(a) Diameter of Runner ,D
D=d+(D_per/100)*d; //m
//(b) Speed,N
N=ui*60/(%pi*D); // rpm
//(iii) Specific Speed of Turbine, Ns
Ns=N*P^(1/2)/(H^(5/4)); // SI Units
//Results
printf("(a)Diameter of Runner , D=%.3f m\n",D)
printf(" (b)Speed, N =%.2f rpm\n",N) //The answer vary due to round off error
printf(" (c)Specific Speed, Ns =%.2f (SI Units)\n",Ns) //The answer provided in the textbook is wrong.
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