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+//Fluid Systems- By Shiv Kumar
+//Chapter 5- Francis Turbine
+//Example 5.15
+ clc
+ clear
+
+//Given Data:-
+ Vo=4.5; //Velocity of water at Tube Entrance, m/s
+ D1=0.4; //Diameter of Tube at Upper End, m
+ D2=0.65; //Diameter of Tube at Lower End, m
+ l=4.8; //Length of Tube, m
+ h_f=0.14; //Head Losses due to Friction, m
+ h=1; //Length of Tube immersed in Tail-race, m
+
+//Data Required:-
+ pa=1.013e5; //Air(Atmospheric) Pressure, Pa
+ rho=1000; //Density of Water, Kg/m^3
+ g=9.81; //Acceleration due to gravity, m/s^2
+
+//Calculations:-
+ A1=(%pi/4)*D1^2; //Cross-sectional Area at Upper End, m^2
+ A2=(%pi/4)*D2^2; //Cross-sectional Area at Lower End, m^2
+ //Using Continuity Equation,
+ Vd=A1*Vo/A2; //Velocity of Water at Outlet, m/s
+
+ //(a) Using Bernoulli's Equation,
+ p1_by_rho_g=pa/(rho*g)+h+(Vd^2-Vo^2)/(2*g)-l+h_f; //Absolute Pressure Head at Inlet, m
+
+ //For Vaccum Pressure Head,
+ pa_by_rho_g=0;
+ p1_by_rho_g_v=pa_by_rho_g+h+(Vd^2-Vo^2)/(2*g)-l+h_f; //Vaccum Pressure Head at Inlet, m
+
+ //(b)Efficiency of Draft Tube:
+ eta_d=(Vo^2-Vd^2-2*h_f*g)*100/Vo^2; //In Percentage
+
+//Results:-
+ printf("(a)The Pressure Head at Tube Entrance is\n\t")
+ printf(" p1/(rho*g)=%.3f m (Absolute)\n\t p1/(rho*g)=%.3f m (Vaccum)\n ",p1_by_rho_g, p1_by_rho_g_v) //The Answer Vary due to Round off Error
+ printf("(b)Efficiency of Draft Tube, eta_d=%.2f Percent\n",eta_d) //The Answer Vary due to Round off Error
+
+
+