//Fluid Systems - By Shiv Kumar //Chapter 16- Hydraulic Power and Its Transmissions //Example 16.3 //To Determine the Minimum Number of Pipes. clc clear //Given Data:- l=7500; //Length of each Pipe, m d=125; //Diameter of each Pipe, mm Pr=6000; //Pressure at Discharge End, kPa eta=85/100; //Efficiency P=156; //Power Delivered, kW f=0.006; //Data Required:- rho=1000; //Density of Water, Kg/m^3 g=9.81; //Acceleration due to gravity, m/s^2 //Computations:- H_minus_hf=Pr*10^3/(rho*g); //H-hf, m H=H_minus_hf/eta; //m hf=H-H_minus_hf; //m Q=P*1000/(rho*g*(H-hf)); //m^3/s q=sqrt((hf*2*g*%pi^2*(d/1000)^5)/(64*f*l)); //Discharge in each Pipe, m^3/s n=Q/q; //Number of Pipes //Results:- printf("The Minimum Number of Pipes Required=%.f\n",n)