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// Example 5.6
// Determine (a) New operating speed if a system disturbance causes a 10% drop
// in voltage and 6% drop in frequency (b) New shaft horsepower.
// Page No. 190
clc;
clear;
close;
// Given data
etaV=0.90; // Efficiency related to voltage
V=230; // Voltage
etaF=0.94; // Efficiency related to voltage
f=60; // Frequency
N=6; // Number of poles
nr1=1175; // Speed of motor
P=20; // Horsepower of motor
// (a) New operating speed if a system disturbance causes a 10% drop in
// voltage and 6% drop in frequency
V2=etaV*V; // New voltage after 10% drop
f2=etaF*f; // New frequency after 6% drop
ns1=120*f/N;
ns2=120*0.94*f/N;
s1=(ns1-nr1)/ns1; // Speed difference
s2=s1*((V/V2)^2)*(f2/f);
nr2=ns2*(1-s2); // New speed
// (b) New shaft horsepower
P2=P*(nr2/nr1); // With a constant torque load T2=T1
// Display result on command window
printf("\n New operating speed in case of voltage and frequency drop = %0.0f r/min ",nr2);
printf("\n New shaft horsepower = %0.1f hp ",P2);
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