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authorprashantsinalkar2017-10-10 12:27:19 +0530
committerprashantsinalkar2017-10-10 12:27:19 +0530
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+//Fluid Systems by Shiv Kumar
+//Chapter 7 - Performance of water turbine
+//Example 7.4
+//To Find (a) The Diameter of Runner (b) The Diameter of jet
+ clc
+ clear
+//Given:
+ P=3200; //Power Developed, kW
+ H=310; // Effective Head , m
+ eta_o=82/100; //Overall Efficiency
+ Ku=0.46; // Speed Ratio
+ Cv=0.98 // Co-efficient of Velocity
+ Ns=18; //Specific Speed (SI Units)
+//Data required
+ rho=1000; //Density of Water, Kg/m^3
+ g=9.81 // Acceleration due to gravity, m/s^2
+
+//Computations
+ N=Ns*H^(5/4)/sqrt(P); //Speed, rpm
+ D=Ku*sqrt(2*g*H)*60/(%pi*N); //Diameter of runner, m
+ Q=P*1000/(rho*g*H*eta_o); //Discharge, m^3/s
+ d=sqrt(Q/((%pi/4)*Cv*sqrt(2*g*H))); // Diameter of Jet, m
+
+//Results
+ printf("(a) The Diameter of Runner, D =%.2f m\n",D) //The Answer Vary due to Round Off Error
+ printf("(c) The Diameter of Jet, d=%.3f m \n",d)