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//Chapter 5:Dc Motor Drives
//Example 21
clc;
//Variable Initialization
//Ratings of the series motor
N=600 //speed in rpm
Vs=220 //source voltage in V
Ra_Rf=0.12 //combine armature resistance field resistance
//Magnetisation curve at N
If=[10, 20,30, 40, 50, 60, 70, 80] //field current in A
E =[64,118,150,170,184,194,202,210] //terminal voltage in V
//Solution
//(i)When the duty ratio is 0.6 and motor current is 60 A
delta=0.6 //duty ratio
Ia1=60 //motor current in A
Va1=delta*Vs //terminal voltage for the given duty ratio
E1=Va1-Ia1*Ra_Rf //back emf for the given duty ratio
//For Ia1=60 A the terminal voltage is 194 V as given in the magnetization curve
N1=E1/E(6)*N //motor speed for the given duty ratio
//(ii)When the speed is 400rpm and the duty ratio is 0.65
delta=0.65 //duty ratio
N2=400 //speed in rpm
Va1=delta*Vs //terminal voltage for the given duty ratio
//From the magnetization characteristic for the speed of 400rpm the current Ia=70 A
E1=Va1-If(7)*Ra_Rf //back emf for the given duty ratio
T=(E1*If(7))/N2/(2*%pi/60) //required torque
//Results
mprintf("(i)Hence the motor speed is :%.1f rpm",N1)
mprintf("\n(ii)Hence the required torque is :%.1f N-m",T)
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