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//Fluid Systems - By - Shiv Kumar
//Chapter 12- Reciprocating Pumps
//Example 12.10
//To Find the Power required to overcome the friction of Delivery pipe when (a)No air vessel is fitted on it , (b)A large air vessel is fitted at the centre line of the pump.
clc
clear
//Given Data:-
N=60; //Speed of the Pump, rpm
D=250; //Plunger Diameter, mm
L=450; //Stroke Length, mm
d_d=112.5; //Diameter of Delivery Pipe, mm
l_d=48; //Length of Delivery Pipe, m
f=0.04; //Co-efficient of friction
//Data Used:-
g=9.81; //Acceleration due to gravity, m/s^2
rho=1000; //Density of water, kg/m^3
//Computations:-
d_d=d_d/1000; //m
D=D/1000; //m
L=L/1000; //m
a=(%pi/4)*d_d^2; //m^2
A=(%pi/4)*D^2; //m^2
omega=2*%pi*N/60; //rad/s
r=L/2; //m
//(a)Without Air Vessel
H_fd=f*(l_d/d_d)*(omega*r*A/a)^2/(2*g); //Maximum loss of head due to friction in delivery pipe, m
m=rho*A*L*N/60; //Mass of water lifted, kg/s
Power=(2/3)*H_fd*m; //W
//Result (a)
printf("(a)Without Air Vessel\n\t")
printf("Power Required to Overcome Friction=%.2f W\n\n",Power) //The answer provided in the textbook is wrong
//(b)With Air Vessel
Ud=A*L*N/(a*60); //m/s
H_fd=f*(l_d/d_d)*(Ud^2/(2*g)); //m
Power=m*H_fd; //W
//Result (a)
printf("(a)With Air Vessel\n\t")
printf("Power Required to Overcome Friction=%.2f W\n",Power) //The answer vary due to round off error
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