# CHAPTER 46 : ELECTRONIC CONTROL OF A.C MOTORS¶

## EXAMPLE 46.1 , PAGE NO :- 1827¶

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'''The wound-rotor induction motor of Fig.43.5 is rated at 30-kW,975 rpm,440-V,50 Hz.The open-circuit line voltage is 400V and the load
resistance is 0.5 ohm.If chopper frequency is 200 Hz,calculate Ton so that the motor develops a gross torque of 200 N-m at 750 rpm.Also,
calculate the magnitude of the current pulses drawn from the capacitor.'''

import math as m
Ns = 1000.0   #rpm       (Synchronus speed)
N1 = 750.0    #rpm       (rotating speed)
E2 = 400.0    #V         (OC line voltage)
Tg = 200.0    #N-m       (gross torque)
R0 = 0.5      #ohm       (load resistance)
f = 200.0     #Hz        (chopper frequency)
s = (Ns-N1)/Ns   # slip
Vrl = s*E2       #V       (rotor line voltage)
Vdc = 135.0      #V       (DC voltage of 3-phase bridge rectifier)
#Now, Tg = P^2/(2*3.14*Ns).Therefore,
P2 = Tg*2*3.14*Ns/60           #W
#Power dissipated as heat
sP2 = s*P2                      #W
#Power is actually dissipated in R and is equal to rectifier output Vdc*Idc.Therefore,
Idc =  sP2/Vdc         #A
#The apparent resistance at the input of chopper is
Ra = Vdc/Idc                     #ohm
#Now, Ra = Ro/(f*Ton)^2 .Therefore,
Ton = m.sqrt(R0/(f*f*Ra))*1000           #ms
#Current in R0 can be found from relation,I0^2*R0 = sP2
I0 = m.sqrt(sP2/R0)               #A
print "Ton = ",round(Ton,2),"ms."
print "magnitude of the current pulse I0 =",round(I0,2),"A."

Ton =  1.89 ms.
magnitude of the current pulse I0 = 102.31 A.

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