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+//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)
+