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//Fluid Systems - By Shiv Kumar
//Chapter 12- Reciprocating Pumps
//Example 12.1
clc
clear
//Given Data:-
Hs_th=4.8; //Suction Head (Theoretical), m
Hd_th=12; //Delivery Head (Theoretical), m
N=90; //Speed of Pump, rpm
D=100; //Piston Diameter, mm
L=150; //Length of Stroke, mm
Q=102; //Actual Discharge, lit./min
eta_s=60/100; //Efficiency of Suction Stroke
eta_d=75/100; //Efficiency of Delivery Stroke
//Data Used:-
rho=1000; //Density of Water, kg/m^3
g=9.81; //Accelerationdue to gravity, m/s^2
//Computations:-
Vs=(%pi/4)*(D/1000)^2*(L/1000); //Swept volume in one revolution, m^3
Vth=Vs*N/60; //Theoritical Volume of Water pumped per second, m^3
m=Vth*rho; //Theoritical Mass Flow rate, kg/s
m_act=Q*1000/(60*1000); //Actual mas flow rate, kg/s
Slip=(m-m_act)*100/m; //Slip in Percentage
Cd=m_act/m*100; //Co-efficient of Discharge in Percentage
Hs=Hs_th/eta_s; //Suction Head taking suction efficiency in account, m
Hd=Hd_th/eta_d; //Delivery Head taking delivery efficiency in account, m
H=Hs+Hd; //Total Head, m
Pth=m*g*H; //Theoritical power required to Drive the Pump, W
A=(%pi/4)*(D/1000)^2; //Cross section Area of piston, m^2
Fs=rho*g*Hs*A; //Average Force during Suction, N
Fd=rho*g*Hd*A; //Average Force during Delivery, N
P=(Fs+Fd)*L*N/60; //Power required by Pump (Same as Pth), W
//Results:-
printf(" 1. Slip=%.2f Percent \n",Slip) //The answer vary due to round off error
printf(" 2. The Co-efficient of Discharge =%.2f Percent \n",Cd) //The answer vary due to round off error
printf(" 3. Theoretical Power Required to Drive the Pump =%.2f W \n",Pth) //The answer vary due to round off error
printf(" 4. Force Required to Work the Piston during Suction Stroke =%.2f N \n",Fs)
printf(" 5. Force Required to Work the Piston during Delivery Stroke =%.2f N \n",Fd)
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