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+//Fluid Systems- By Shiv Kumar
+//Chapter 7- Performance of Water Turbine
+//Example 7.12
+// To Find    (a)The Speed ,Discharge and Power required for the Actual Pump    (b) The Discharge of the model.
+   clc
+   clear
+
+//Given:-
+      Lr=5;        //Scale Ratio
+      DpbyDm=Lr;
+       DmbyDp=1/DpbyDm;
+        //For Model
+             Pm=22;         //Power Required, kW
+             Hm=7;        //Head, m
+             Nm=410;     //Speed, rpm
+             eta_m=1;     //Assumption that  efficiency of the model is 100%
+      //For Prototype
+        Hp=35;         //Head, m
+  //Data Required:-
+       rho=1000;     //Density of Water, Kg/m^3
+       g=9.81;       //Acceleration due to gravity, m/s^2
+
+//Computations:-
+       Np=Nm*DmbyDp*(Hp/Hm)^(1/2);     //rpm
+       Pp=Pm*(Np/Nm)^3*DpbyDm^5;    //KW
+       Qm=Pm*1000*eta_m/(rho*g*Hm);    //m^3/s
+       Qp=Qm*(Np/Nm)^2*DpbyDm^2;       //m^3/s
+
+//Results:-
+       printf("(a)For the Actual Pump(Prototype):\n         Speed, Np=%.2f rpm  , \n          Discharge, Qp=%.3f m^3/s  and \n         Power,Pp=%.2fKW\n",Np,Qp,Pp)         //The Answer vary due to Round off Error(For Qp),  The Answer Provided in the Textbook is Wrong (For Np and Pp).
+
+       printf("(b)The Discharge of the Model, Qm=%.4f m^3/s",Qm)        //The Answer vary due to Round off Error
+
+