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A 220 V, 1000 rpm, 60 A separately-excited dc motor has an armature resistance of .5 ω. It is fed from single-phase full converter with an ac source voltage of 230 V, 50Hz. Assuming continuous conduction, the firing angle for rated motor torque at (-400) rpm is _________(a) 122.4°(b) 117.6°(c) 130.1°(d) 102.8°The question was asked during an interview.This is a very interesting question from Dynamics topic in chapter Dynamics of Electrical Drives of Electric Drives

Answer»

Right choice is (d) 102.8°

To explain I would say: During rated operating conditions of the MOTOR, Eb = Vt-Ia×Ra = 220-60×.5=190 V. As Eb=Kmwm so Km=190×60÷(2×3.14×1000) = 1.8152 V-s/rad. Back emf at (-400 rpm) is Kmwm = 1.8152×(2×3.14×(-400))÷60 = -76 V. Now Vt = -76+60×.5 = -46 V. Average voltage of single-phase full converter is 2×Vm×cos(α)÷3.14. The OUTPUT of the converter is connected to the input TERMINAL of the motor so α = cos^-1(-46×3.14÷2×230×1.414) = 102.8^o.



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