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authorprashantsinalkar2017-10-10 12:27:19 +0530
committerprashantsinalkar2017-10-10 12:27:19 +0530
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+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)