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+//Fluid Systems - By - Shiv Kumar
+//Chapter 4 - Pelton Turbine (Impulse Turbine)
+//Example 4.18
+  
+      clc
+       clear
+
+//Given Data:-
+       Ns=15;         //Specific Speed
+      P=1200;        //Shaft Power, kW
+      Ht=500;     //Total Head at reservoir, m
+      Loss_per=5;       //Percentage of Head loss in Pipe friction
+      Cv=0.98;        //Co-efficient of Velocity
+      Ku=0.45;        //Speed Ratio
+     eta_o=85/100;       //Overall Efficiency
+      n=2;       //Number of Jets
+
+//Data Used:-
+        rho=1000;          //Density of water, kg/m^3
+         g=9.81;            //Acceleration due to gravity, m/s^2
+
+//Computations:-
+          H=Ht-Loss_per/100*Ht;          //Effective Head, m
+         
+      //(a)Speed of Runner, N
+              N=Ns*H^(5/4)/sqrt(P/n);        //rpm
+
+     //(b)Diameter od each Jet, d
+           Q=P*1000/(rho*g*H*eta_o);          //Net Discharge, m^3/s
+            q=Q/n;             //Net Discharge per Jet, m^3/s
+            Vi=Cv*sqrt(2*g*H);        //m/s
+           d=sqrt(q/((%pi/4)*Vi));         //m
+    
+      //(c)Mean Diameter of Bucket Circle, D
+          D=Ku*60*sqrt(2*g*H)/(%pi*N);         //m
+
+     //(d)Number of Buckets in the Runner, Z
+            Z=round(0.5*D/d+15);         
+
+//Results:-
+         printf(" (a)Speed of the Runner, N=%.f rpm\n",N)
+         printf(" (b)Diameter od each Jet, d =%.3f m\n",d)
+         printf(" (c)Mean Diameter of Bucket Circle, D =%.3f m\n",D)
+         printf(" (d)Number of Buckets on the Runner, Z =%.f \n",Z)
+