Determining transformer resistance and reactance.

Pac1Man

Hi guys, I need some help regarding a fictional 200 kVA 3-phase transformer. I am trying to find the resistance and reactance values for this transformer as part of an overall short circuit calculation, in the form of: a + j a where 'a' is the resistance value in ohms and 'j a' is the reactance value in ohms (and j signifies a complex number).

The information I have is as follows:

Primary/Secondary = 10000V/400V
% Impedance = 4%
Load Losses = 2850 W
No-Load Losses = 550 W

I can't seem to get a definitive solution online so I thought I may ask here. I'm guessing it could have something to do with the power formula P =(I^2)(R), however that is just a stab in the dark.

Any help would be great!

Bruthal

Pac1Man wrote: »

Hi guys, I need some help regarding a fictional 200 kVA 3-phase transformer. I am trying to find the resistance and reactance values for this transformer as part of an overall short circuit calculation, in the form of: a + j a where 'a' is the resistance value in ohms and 'j a' is the reactance value in ohms (and j signifies a complex number).

The information I have is as follows:

Primary/Secondary = 10000V/400V
% Impedance = 4%
Load Losses = 2850 W
No-Load Losses = 550 W

I can't seem to get a definitive solution online so I thought I may ask here. I'm guessing it could have something to do with the power formula P =(I^2)(R), however that is just a stab in the dark.

Any help would be great!

200 KVA IS 200,000 VA.

Full load amps is 200,000/(400 x root 3) = 288 amps per phase.

Impedence 4% probably means if the secondary is short circuited when the secondary voltage is 4% of 400 (16v), full load current will flow.

So with the primary voltage raised to a level to give secondary voltage of 16v, and the secondary being direct shorted, full load current will flow, assuming the primary supply level to give such an output, can supply the demand.

So impedance = V/I = 16/288 = 0.0555 ohms.

Short circuit current is 400/0.05714 = 7200 amps.

Another way to get the short circuit current is to get the full load current, which = 4% of the short circuit current, due to 4% of the secondary voltage giving full load current when the secondary side is shorted. So the 288 amps full load = 4% of the short circuit current. 280 x 100/4 = 7200 amps.

For resistance, the full load copper losses are I squared x R.

I^2 x R = 2850watts.
280 x 280 x R = 2850, so R = 2850/(280 x 280) = 0.0364 ohms.

Of course, someone might come along and fix all the mistakes. But its something along those lines.

Pac1Man

Thanks a million Bruthal, much appreciated. I tried to research it online but the more I read, the more confused I got.

Bruthal

No bother, just updated it a little. I was a bit rough with getting the full load as 280 which should have been 288 amps.