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Diffstat (limited to '3751/CH16/EX16.6')
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diff --git a/3751/CH16/EX16.6/Ex16_6.sce b/3751/CH16/EX16.6/Ex16_6.sce new file mode 100644 index 000000000..6c14d6b67 --- /dev/null +++ b/3751/CH16/EX16.6/Ex16_6.sce @@ -0,0 +1,37 @@ +//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 + + + |