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+//Fluid Systems- By Shiv Kumar
+//Chapter 7- Performance of Water Turbine
+//Example 7.9
+// To Determine the Performance of the Turbine Under a Head of 20 m
+   clc
+   clear
+
+//Given:-
+       //Condition 1:
+            H1=25;     //Head, m
+            N1=200;    //Speed, rpm
+            Q1=9;        //Discharge, m^3/s
+            eta_o=90/100;     //Overall Efficiency
+
+        //Condition 2:
+           H2=20;     //Head, m
+
+//Data Required:-
+    rho=1000;        //Density of Water, Kg/m^3
+    g=9.81;          //Acceleration due to Gravity, m/s^2
+
+
+//Calculations:-
+         P1=rho*Q1*g*H1*eta_o/1000;           //KW
+         N2=N1*sqrt(H2/H1);          //rpm
+         Q2=Q1*sqrt(H2/H1);          //m^3/s
+         P2=P1*(H2/H1)^(3/2);       //KW
+
+
+//Results:-
+   printf("At Condition 2 (Under a Head of 20 m):\n")  
+   printf("\tSpeed, N2=%.2f rpm\n     Discharge, Q2=%.2f m^3/s\n      Power Developed, P2=%.2f kW",N2,Q2,P2)          //The Answer vary due to Round off Error
+
+