Selection on LV and MV

The short_circuit capacity of low voltage circuit breaker is limited to 45-50 kA The short-circuit impedance of a low voltage transformer is limited to 6-8% since the voltage drop is limited to +/- 5%[up to a receiver terminal]. That means the maximum transformer power could be 2500-3000 kVA.

The locked rotor motor current is 6.5-8 times the rated.

A 250 kw [4 poles] rated current is 450 A[around] and the locked rotor current 3000 A.

Transformer Full Load Current Itrf=2500/sqrt(3)/0.4=3608 A.

If a transformer is full load and a motor will be in locked rotor situation[DOL start or blocked]then will be more than 10% voltage drop at transformer terminal[only] and that will disturb the other receivers as other motors , controls, lighting and other[under voltage].

No such phenomenon happens on 6.6[or 11] kV, since the rated motor current is low and the voltage drop of tens -or even hundred-volts does not disturb 6.6 kV System

Good and Interesting answer indeed, Would you elaborate more on how you calculate 10% VD at transformer terminals?

Thank you for your reply

Interview questions, or homework?

Where have you looked, so far?

It appears to be questions asked to a fresher at his interview which he could not answer, now he wants to get an answer from us it seems.

1. Total connected load to the system will decide the voltage selection.

2. Answered already in (1) above

3. There is no hard & fast rule as to the provision of an MCCB or an ACB for a given load. Both are circuit breakers and are covered in the same standard (IEC 60947-2). It bis only designer's choice. ACB is maintainable & MCCB is not.

Thank you for your reply...

Anonymous #1

It is very simply. The transformer rated current it is 3608 A [apparent] =3067-j1901 [p.f.=0.85].

Motor starting current =900-j2862 A.[p.f.=0.3]. Total current 3967+j4762 A.

Transformer impedance[8% for 2500 kVA,400 V]ztrf=0.00512 ohm. xtrf/rtrf=7.

xtrf=ztrf/sqrt(1+1/49)=0.0050685. rtrf=0.0007241.

DV=sqrt(3)*(3967*0.0007241+4762*0.0050685)=46.8 V[11.7%].