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+//Fluid Systems by Shiv Kumar
+//Chapter 7 - Performance of water turbine
+//Example 7.5
+//To Find (a) Number of Units to be installed (b) Diameter of Wheel  (c) Diameter of Jet
+    clc
+     clear
+
+//Given:
+
+     H_G= 310;  //Gross Head,m
+     l=2.5;     // Length, km
+     h_f=25;    // Friction Loses, J/N=m
+     TO=20;   //Total Output Power, MW
+     N=660;   // Speed, rpm
+     ubyVi=0.46    //Ratio of bucket to jet speed
+     eta_o=88/100;   //Overall Efficiency
+     Ns=28;     //Specific Speed, SI Units
+     Cv=0.97;   
+     Cd=0.94;   
+     
+     
+//Data Required:
+     rho=1000;      //Density of water, kg/m^3
+     g=9.81;       //Acceleration due to gravity, m/s^2
+      
+
+//Computations:
+   H=H_G-h_f;   //Effective Head, m
+   P=(Ns*H^(5/4)/N)^2;   //Power Output of each Unit, KW
+  //(a) The no. of units to be lnstalled,n
+   n=round(TO*1000/P);   
+  //(b)Diameter of Wheel,D
+    Vi=Cv*sqrt(2*g*H);    //m/s
+    D=ubyVi*Vi*60/(%pi*N);  //m
+
+//(c) Diameter of Jet,  d
+    Q=TO*10^6/(rho*g*H*eta_o);   //Net Discharge, m^3/s
+    q=Q/n;  // Discharge through one unit, m^/s
+    d=sqrt(q/((%pi/4)*Cd*sqrt(2*g*H)))*1000;  //mm
+
+
+//Results
+   printf("(a) The no. of units to be Installed=%.f Units\n",n)
+   printf("(b) Diameter of Wheel, D=%.3f m\n",D)
+   printf("(c) Diameter of Jet, d=%.1f mm\n",d)      //The Answer Vary due to round off Error
+
+