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calculating voltage drop in distribution transformer

calculating voltage drop in distribution transformer

calculating voltage drop in distribution transformer

(OP)
hi,
I would like to know how could calculate the voltage drop in a distribution transformer based on the data mentioned on the name plate and the drawn current.
Best regards.

RE: calculating voltage drop in distribution transformer

You would need the OEM test certificate for the 'Regulation' of the trafo which is expressed as % drop from no-load to full load for a specific power factor. It is different from the % impedance which is stamped in the name-plate.

RE: calculating voltage drop in distribution transformer

The voltage drop at full load current will never exceed the percent impedance but to get that much drop you need a rather low power factor load. Load power factor would have to match the power factor of the transformer impedance which is APPROXIMATELY what happens when motors start.

You do not need an exact calculation of transformer voltage drop anyways because your upstream and downstream wiring needs to have reasonable voltage drop corresponding to reasonable efficiency.

RE: calculating voltage drop in distribution transformer

edison123 has the correct answer...

RE: calculating voltage drop in distribution transformer

why do you need the voltage drop of the transformer? What are you trying to calculate.

RE: calculating voltage drop in distribution transformer

Assume a low side voltage of 1 pu (100%) at an angle of 0°.  The high side voltage in per unit (pu) will be the vector sum of 1 pu and I·Z, where I is the load current in pu and Z is the transformer impedance in pu (%Z/100).  

I and Z must include the angles.  The magnitude of I in pu is the load kVA in pu divided by the rated kVA.  The angle of I is -acos(pf) (assuming a lagging pf, angle is + if the pf is leading).  The angle of Z is atan(X/R).  

The voltage drop in pu is the magnitude of the high side voltage in pu minus 1.

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