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//Fluid Systems- By Shiv Kumar
//Chapter 7- Performance of Water Turbine
//Example 7.12
// To Find (a)The Speed ,Discharge and Power required for the Actual Pump (b) The Discharge of the model.
clc
clear
//Given:-
Lr=5; //Scale Ratio
DpbyDm=Lr;
DmbyDp=1/DpbyDm;
//For Model
Pm=22; //Power Required, kW
Hm=7; //Head, m
Nm=410; //Speed, rpm
eta_m=1; //Assumption that efficiency of the model is 100%
//For Prototype
Hp=35; //Head, m
//Data Required:-
rho=1000; //Density of Water, Kg/m^3
g=9.81; //Acceleration due to gravity, m/s^2
//Computations:-
Np=Nm*DmbyDp*(Hp/Hm)^(1/2); //rpm
Pp=Pm*(Np/Nm)^3*DpbyDm^5; //KW
Qm=Pm*1000*eta_m/(rho*g*Hm); //m^3/s
Qp=Qm*(Np/Nm)^2*DpbyDm^2; //m^3/s
//Results:-
printf("(a)For the Actual Pump(Prototype):\n Speed, Np=%.2f rpm , \n Discharge, Qp=%.3f m^3/s and \n Power,Pp=%.2fKW\n",Np,Qp,Pp) //The Answer vary due to Round off Error(For Qp), The Answer Provided in the Textbook is Wrong (For Np and Pp).
printf("(b)The Discharge of the Model, Qm=%.4f m^3/s",Qm) //The Answer vary due to Round off Error
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