I saw this problem somewhere and I wanted to give it a try, but the numbers I've got seem to be pretty outlying.

We have a 25kV / 4160V transformer which is out of circuit for now. To protect it from humidity in a freezing cold conditions, we need to keep it warm by short-circuiting its primary coil and connecting the secondary coil to a 208V, 3ph line. How much heat will be generated in this way? The transformer specifications are as follows.


My try:

Since the primary coil is short-circuited, all the loss occurs at the core. And since it is operating at 5% of its nominal voltage, the coil loss will be 5% of its nominal value, i.e. about 650 watt. Though I suspected that since the loss is proportional to V^2 then it would be even lower, but nonetheless, the numbers are way too lower than what I've been told (which was 57% of the full-load loss).

So I would appreciate if anyone guide me on how to approach such problems.

  • \$\begingroup\$ I'm confused; how can the total power loss be 69005 watts when the transformer has a VA rating of 15000? Your understanding of running a transformer at a low voltage to achieve the required core losses is fundamentally flawed. Shorting a winding induces copper losses and not core losses. \$\endgroup\$ – Andy aka Nov 6 '18 at 9:32
  • \$\begingroup\$ @Andyaka You are right about the flaw in my understanding. But now I am even more confused. Will you elaborate a little bit? \$\endgroup\$ – polfosol Nov 6 '18 at 9:34
  • \$\begingroup\$ I am no expert at this but I would think the law of conversion of energy applies: If you pump X watts into a cable with equipment at the end, apart from some cable loss the equipment will consume the X watts. \$\endgroup\$ – Oldfart Nov 6 '18 at 10:02
  • \$\begingroup\$ I'm not sure what you want me to elaborate about. Try googling short and open circuit transformer testing - results should give you the reasons whay a short circuit test is used to find copper loss and leakage whilst an open circuit test provides information on core losses. \$\endgroup\$ – Andy aka Nov 6 '18 at 10:02
  • \$\begingroup\$ It is standard practice to fit a small heater (~10W) inside equipment cabinets for outdoor use. Would it not be possible to fit (larger) heater(s) to the transformer for this purpose; this would remove the need to switch supplies. \$\endgroup\$ – henros Nov 6 '18 at 10:26

4160*0.066 (6.6% impedance) = 274V on the secondary for full secondary current into a shorted primary.

Then the current will be 76% of full load by simple ratio (208/274), so the power losses will be 0.76^2 = 57.6% of full load losses. This obviously makes an assumption that iron losses are linear with flux, which they are not, but it is a reasonable working model.

Note that there is primary current flowing so all the copper losses work in the usual way, it is NOT just core loss.


With an X/R of 17.5, the transformer will look like an inductor with some resistance when viewed as a load on the 208-volt supply applied to the secondary. I don't think there is enough information to completely determine the equivalent circuit, but you should be able to make a reasonable approximation using the loss data and X/R.


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