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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)