Answer / pankaj_tikariha

P=no. Of pole,f=frequncy ,n=speed Ns=p/2*N/60

Answer / arup manna

n=120f/p


f=n*p/120

Answer / k.prakashchandra

In other words , 2 poles make one cycle ,P/2 cycles for one
revolution . Rpm ( N) /60 = rps ( rev per second )

At site, we have 30Kw motor (3 phase, 400V,50Hz) and FLC 54A as per the nameplate. At the time of starting it draws 840A instead of 390A (7.2%). As per the data sheet also its 390A starting current. it thru a dol starter from the MCC. can anyone tell what is the problem in the motor

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