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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
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