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//Fluid Systems - By Shiv Kumar
//Chapter 16- Hydraulic Power and Its Transmissions
//Example 16.10
//To Find (i)The Weight of Loaded Cylinder and energy stored by the Cylinder (ii)Ther Power supplied by the Accumulator (iii)The Diameter of ram of an ordinary Accumulator.
clc
clear
//Given Data:-
D=180; // mm
d=150; //mm
L=1.25; //Stroke length, m
p=100; //Pressure of Water, bar
//Computations:-
D=D/1000; //m
d=d/1000; //m
p=p*10^5; //N/m^2
A=(%pi/4)*(D^2-d^2); //Annular area of Ram, m^2
//(i)
W=p*A/1000; //Weight of Loaded Cylinder, kN
Energy=W*L; //Energy stored in the Accumulator, kNm
//(ii)
t=90; //Time taken by ram to complete the stroke, seconds
P=W*L/t; //kW
//(iii)
D=(W*1000/(p*%pi/4))^(1/2)*1000; //mm
//Results:-
printf("(i)Weight of Loaded Cylinder, W=%.2f kN\n",W) //The answer vary due to round off error
printf(" Energy stored in the Accumulator=%.3f kNm\n",Energy) //The answer vary due to round off error
printf("(ii)Power Supplied by the Accumulator=%.3f kW\n",P) //The answer vary due to round off error
printf("(iii)Ram Diameter (In case of Ordinary Accumulator) = %.2f mm\n",D) //The answer vary due to round off error
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