//Fluid Systems - By - Shiv Kumar //Chapter 11- Centrifugal Pumps //Example 11.6 //To Calculate the Blade angle at Outlet, Power Required and Overall Efficiency of Pump. clc clear //Given Data:- Do=80; //Outer Diameter of the Impeller, cm Q=1; //Discharge, m^3/s H=80; //Head, m N=1000; //Speed, rpm bo=8; //Width at Outlet, cm Delta_Q_per=3; //Percentage of Leakage Loss(of the Discharge) Delta_P=10; //Mechanical Loss, kW eta_H=80/100; //Hydraulic Efficiency //Data Used:- rho=1000; //Density of water, kg/m^3 g=9.81; //Acceleration due to gravity, m/s^2 //Computations:- Do=Do/100; //m bo=bo/100; //m uo=%pi*Do*N/60; //m/s Vfo=Q/(%pi*Do*bo); //m/s Vwo=g*H/(uo*eta_H); //m/s Vrwo=uo-Vwo; //m/s //(a) beta_o=atand(Vfo/Vrwo); //Blade Angle at Outlet, degrees //Result1 printf(" Blade Angle at Outlet, beta_o=%.2f Degrees \n",beta_o) //The answer vary due to round off error //(b)Power Required Pi=rho*(1+Delta_Q_per/100)*Q*Vwo*uo; //Power delivered by the Impeller, W P=Pi/1000+Delta_P; //Power required, kW //Result2 printf(" Power Required, P =%.3f kW \n",P) //The answer vary due to round off error //(c)Overall Efficiency, eta_o eta_V=1/(1+Delta_Q_per/100); //Volumetric Efficiency eta_m=(P-Delta_P)/P; //Mechanical Efficiency eta_o=eta_H*eta_V*eta_m*100; //In Percentage //Result3 printf(" Overall Efficiency, eta_o =%.2f Percent \n",eta_o) //The answer vary due to round off error //Also, Overall Efficiency eta_o=rho*Q*g*H/(P*1000)*100; //In Percentage printf("Also, Overall Efficiency, eta_o=%.2f Percent\n",eta_o)