blob: 378937223d99e8d61544fda5db51a4dc1cdf6a5b (
plain)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
|
clc
// Variable Initialization
Vm=230//Supply Voltage in Volts
Ra=1//Armature circuit resistance in Ohm
I=12 //Rated current in Amp
Nr=1500 //Motor speed in Rpm
Va=210 //Motor voltage in volts
N1=1200 //Speed in RPM
//Solution
Eb=Va-(I*Ra) //Back emf in volts
N=(Nr*2*%pi)/60 //Speed in Rad/sec
Ka=Eb/N //Constant
//Current at rated torque
//At 1200 Rpm
Eb1=(N1*Eb)/Nr
af=acosd((((I*Ra)+Eb1)*%pi)/(2*Vm*1.414))//Firing angle in Degree
//At Eb=-198 V at speed 1500 RPM
af1=acosd((((I*Ra)-Eb)*%pi)/(2*Vm*1.414))//Firing angle in Degree
//But reversal of field of armature forward regeneration is obtained
Ka1=-Ka
W=-Eb/Ka1 //Angular speed in Rad/sec(Wrongly calculated in book,wrong value of Eb is taken)
N1=(W*60)/(2*%pi) //Speed in Rpm
//Results
printf('\n\n The Firing Angle=%0.1f Degree\n\n',af)
printf('\n\n The Firing Angle=%0.1f Degree\n\n',af1)
printf('\n\n The motor Speed =%0.1f RPM \n\n',N1)
//The answer provided in the textbook is wrong(3rd answer omly)
|
