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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
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