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//Fluid Systems - By Shiv Kumar
//Chapter 16- Hydraulic Power and Its Transmissions
//Example 16.6
//To Calculate the Rise in Pressure due to Valve Closure in (i)10 seconds, (ii)2.5 seconds.
clc
clear
//Given Data:-
l=2500; //Lenfth of Pipe, m
V=1.2 ; //Velocity of Flow, m/s
K=20*10^8; //Bulk Modulus of Water, N/m^2
//Data Used:-
rho=1000; //Density of Water, Kg/m^3
//Computations:-
a=sqrt(K/rho); //Velocity of Pressure Wave, m/s
t_c=2*l /a; //Critical time, s
// (i)
t=10; // s
//t>t_c. so, This is a case of Gradual valve closure.
p=rho*l*V/(t*1000); //Pressure Rise, kPa
//Result (i)
printf("(i)Pressure Rise, p=%.f kPa\n",p)
//(ii)
t=2.5; // s
// t<t_c. This is a case of Instantaneous Valve Closure.
p=rho*V*a/1000; // Pressure Rise, kPa
//Result (ii)
printf("(ii)Pressure Rise, p=%.2f kPa\n",p) //The answer vary due to round off error
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