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+//Fluid Systems - By Shiv Kumar
+//Chapter 16- Hydraulic Power and Its Transmissions
+//Example 16.1
+//To Find the Maximum Power Available at the Outlet of Pipe.
+      clc
+      clear
+
+//Given Data:-
+    d=300;        //Diameter of the Pipe, mm
+    l=3000;       //Length of the Pipe, m
+    H=400;       //Total Head at Inlet, m
+     f=0.005;
+
+//Data Required:-
+        rho=1000;     //Density of Water, Kg/m^3
+        g=9.81;       //Acceleration due to gravity, m/s^2
+
+//Computations:-
+        //Condition for Maximum Power transmission     
+              hf=H/3;       //m
+        V=sqrt(hf*(2*g*d/1000)/(4*f*l));       //m/s
+        Q=(%pi/4)*(d/1000)^2*V;       //Discharge, m^3/s
+        Pmax=rho*g*Q*(H-hf)/1000;        //Maximum Power Available at Outlet of Pipe, kW
+
+
+//Results:-
+    printf("The Maximum Power Available at Outlet of Pipe=%.3f kW",Pmax)       //The answer vary due to round off error
+