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diff --git a/3751/CH11/EX11.26/Ex11_26.sce b/3751/CH11/EX11.26/Ex11_26.sce new file mode 100644 index 000000000..742e16626 --- /dev/null +++ b/3751/CH11/EX11.26/Ex11_26.sce @@ -0,0 +1,37 @@ +//Fluid Systems - By - Shiv Kumar +//Chapter 11 - Centrifugal Pumps +//Example 11.26 +//To Calculate the Specific Speed of Pump and the Power Input and Find the Head, Discharge and Power required at 900 rpm. + + clc + clear + +//Given Data:- + + N=1500; //Speed, rpm + Q=0.2; //Discharge, m^3/s + H=15; //Head, m + eta_o=0.68; //Overall Efficiency + N2=900; //Speed, rpm + + //Data Used:- + rho=1000; //Density of water, kg/m^3 + g=9.81; //Acceleratio due to gravity, m/s^2 + + +//Computations:- + Ns=N*Q^(1/2)/(H^(3/4)); //Specific Speed of Pump, SI Units + P=rho*g*Q*H /eta_o; //Power Input, W + + Q1=Q; H1=H; N1=N; P1=P; + Q2=Q1*(N2/N1); // m^3/s + H2=H1*(N2/N1)^2; //m + P2=P1*(N2/N1)^3; //W + +//Results:- + printf("Specific Speed of Pump, Ns=%.2f (SI Units)\n",Ns) + printf(" Power Input, P=%.2f W\n",P) + printf(" At 900 rpm (Condition 2)\n\t ") + printf(" Head, H2=%.1f m \n\t Discharge, Q2=%.2f m^3/s,\n\t Power required, P2=%.2f W",H2,Q2,P2) + + |