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//Fluid System By Shiv Kumar
//Chapter 6 - Kaplan and Propeller Turbines
//Example 6.6
//To Find (a)Diameter of Runner (b)Speed of Turbine (c)Specific Speed of the turbine
clc
clear
//Given:
P=9100; //Shaft Power, KW
H=5.6; //Net Available Head, m
Ku=2.09; //Speed Ratio
Kf=0.68; //Flow Ratio
eta_0=86/100; //Overall Efficiency
dbyD=1/3; //Ratio of Diameters of Hub and Runner
//Data Required:
rho=1000; //Density of Water, Kg/m^3
g=9.81; //Acceleration due to gravity, m/s^2
//Computations
Q=P*10^3/(rho*g*H*eta_0); //Discharge, m^3/s
d=sqrt(Q/((%pi/4)*Kf*sqrt(2*g*H)*(dbyD^-2-1))); // Diameter of Hub ,m
//(i) Diameter of Runner ,D
D=d/dbyD; //m
//(ii) Speed of Turbine,N
N=Ku*60*sqrt(2*g*H)/(%pi*D); // rpm
//(iii) Specific Speed of Turbine, Ns
Ns=N*(P)^(1/2)/(H^(5/4)); // SI Units
//Results
printf("(i)Diameter of Runner , D=%.2f m\n",D)
printf("(ii)Speed of Turbine, N =%.2f rpm\n",N) //The answer vary due to round off error
printf("(iii) Specific Speed of Turbine, Ns =%.2f (SI Units)\n",Ns) //The answer provided in the textbook is wrong(Due to error in N)
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