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+//Fluid Systems - By Shiv Kumar
+//Chapter 16- Hydraulic Power and Its Transmissions
+//Example 16.9
+//To Deternmine the Diameter of the ram.
+ clc
+ clear
+
+//Given Data:-
+ d=125; //Diameter of Pipe, mm
+ l=2; //Lenght of Pipe, km
+ P=35; //Power Transmitted, kW
+ W=1250; //Load on ram, kN
+ loss_per=3; //Percentage of Power Loss due to friction
+ f_dash=0.04; //Pipe Friction Factor
+
+//Data Used:-
+ rho=1000; //Density of Water, kg/m^3
+ g=9.81; //Acceleration due to gravity, m/s^2
+
+//Computations:-
+ Delta_P=loss_per/100*P*1000; //Power Loss due to friction , W
+ //By Darcy's Formula,
+ hf_by_V2=f_dash*(l*1000)/(2*g*d/1000); //hf/V^2
+
+ QbyV=(%pi/4)*(d/1000)^2; //Q/V
+ V=( Delta_P/(rho*g*QbyV*hf_by_V2))^(1/3); //m/s
+ Q=QbyV*V; //m^3/s
+ p=P*1000/Q; //N/m^2
+ D=sqrt(W*1000/(p*%pi/4))*1000; //mm
+
+//Result:-
+ printf("The Diameter of ram, D=%.2f mm",D) //The answer vary due to round off error
+
+