# Mains electrical Q: transformer sizing?

A 10KVA-rated step-up isolation transformer is used to connect a 480v-delta resistive heater to 208v-Y supply.
Measured currents are:
- primary (208v) current ~35A. - secondary (480v) current ~15A
If my math is correct this transformer is undersized by 20 percent, or so: - primary: 35A x sqrt(3) x 208v = 12.6KVA - secondary: 15A x sqrt(3) x 480v = 12.3KVA
All wires and the breaker are sized to handle the measured currents. While the windings seem quite cool, the laminations are uncomfortably warm to the touch.
The load is turned on for periods of up to 15 minutes at a time and disconnected for between 10 minutes to overnight, depending on the demand for the equipment.
There doesn't seem to be evidence of over-temperature in the windings and wires are sized properly. Is there a risk here?
Thanks.
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James T. wrote:

Overcurrent won't cause the laminations to overheat. On some transformers, the laminations do run surprisingly hot. Because of the duty cycle, you can probably get away with this indefinitely. Of course if an OSHA or NFPA inspector starts poling around, you could have a fine.
Jon
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Jon Elson wrote:

It definitely would - required lamination crossection depends on power and frequency. If it is not enough then it would start overheat and then transformation ratio would drop due to insufficient magnetic flux.
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---------------- Not so.
Core losses which directly heat the core, are nearly independent of load current. They are dependent on the effective applied voltage which will decrease somewhat due to primary impedance voltage drops under load. The core cross section depends on voltage, frequency, number of turns and the maximum flux allowable. It doesn't depend on power except that larger conductors will require more space and a longer core. This is a secondary effect. Hence, the core losses will actually be slightly higher at no load than under full load- not enough to normally be significant. Laminations simply reduce the eddy current component of the core losses.
In the case of a fixed applied AC voltage, the flux is fixed- independent of the core material (Faraday or Maxwell- take your pick). Magnetising current follows as needed.
Yes- I am being bitchy!
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Don Kelly snipped-for-privacy@shawcross.ca
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Not sure where you came up with the sqrt(3) in your equations. Take a look at http://www.powerstream.com/VA-Watts.htm (first page of many when Googled). The power factor for resistive heaters will be near 1. The transformer rating is just fine. -- JeffB remove no.spam. to email
James T. wrote:

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This was from a conversation with an engineer at the transformer manufacturer's support department. (To be fair, I didn't mention the load as purely resistive). He gave the formula as;
KVA / V / sqrt(3) = A

So the formula for a resistive load is:
KVA / V = A
meaning that this 10KVA transformer is good up to a maximum of:
- primary: 48.0A - secondary: 20.8A
Is this right?
Thanks.
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wrote:

You were right the first time. The sqrt(3) accounts for the fact that this is a 3-phase system.
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Ned Simmons

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So this 10KVA transformer is rated to handle a maximum current of:
- primary (208v): 27.8A - secondary (480v): 12.0A
Is this right?
Because this is driving a purely resistive load, this doesn't help bring the parameters within the transformer's rating, does it?
Thanks.
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wrote:

Looks right to me.

No, a purely resistive load is best case.
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Ned Simmons

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But since the transformer is (presumably) designed with a worst-case PF (or at least some PF value greater than 1), can't it be re-rated (what's the opposite of "derated"?) if used with a strictly resistive load?
In other words, a transformer rated at 10KVA for a load with PF = x, should be able to handle a greater KVA if driving a PF = 1 load. Shouldn't it?
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No. kVA means what it says. i.e. kW/cos phi. The kVA rating of the transformer is independent of the load PF. Load _power_ must be reduced for low PF loads.
But this transformer is probably reasonably rated if the duty factor is 50% worst case and the on time 15 minutes worst case.
Mark Rand RTFM
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Sorry, the power factor can never exceed 1 because it is the cosine of the phase angle between the current and the voltage.
Jim
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|>> Because this is driving a purely resistive load, this doesn't help bring |>> the parameters within the transformer's rating, does it? | |> No, a purely resistive load is best case. | | But since the transformer is (presumably) designed with a worst-case PF (or | at least some PF value greater than 1), can't it be re-rated (what's the | opposite of "derated"?) if used with a strictly resistive load? | | In other words, a transformer rated at 10KVA for a load with PF = x, should | be able to handle a greater KVA if driving a PF = 1 load. Shouldn't it?
What is relevant for a transformer rating is both amps and voltage in an independent way. There is a maximum amperage each winding can handle. There is a maximum voltage each winding (and the core) can handle. That is why they are rated in terms of VA, kVA, MVA, etc. They are not rated in W, kW, or MW for this reason. If your PF is 1, then you can get the most power through the transformer ... e.g. you can run 10 kW through a 10 kVA transformer. But if your PF is only 0.5, then the most power you can run through a 10 kVA transformer is 5 kW. Low PF forces you to derate the transformer. Unity PF lets you run it at maximum rated power.
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| Phil Howard KA9WGN | http://linuxhomepage.com/ http://ham.org/ |
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Thanks, Phil. With my convoluted terminology, that is what I wanted to know. Your answer was... elegant, to-the-point.
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The sqrt(3) is the same as the tan 120 deg. for each leg of the 3 phase transformer.
You were told right the first time by the engineer.
Most of those transformers run surprisingly hot. Take a look at the label for the temp rise rating.
Tom

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RE: Subject
Read the spec on heat rise for a X'fmr, typically 55C.
NBD for the iron to be warm, even hot to the touch when in service.
They are designed to operate that way.
Lew
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My notes (taken from the name plate) say "150C".
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| Not sure where you came up with the sqrt(3) in your equations. Take a
Actually he should have _divided_ by sqrt(3) ... after or before _multiplying_ by the number of conductors, which is 3. But since 3/sqrt(3) is sqrt(3), the multiplication by sqrt(3) is actually the correct simpler formula.
If there is no neutral current, you can figure the power simply by multiplying the currents by the L-N voltages. In a delta system, there is no neutral (or at least not one supplied to or used by the load), so no neutral current, but it works out the same as if you had one. Rather than typing in the value for sqrt(3), I just do the arithmetic on things like this as: 35*3*120 = 12600 or 15*3*277 = 12465. Either way that is above the transformer rating.
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| Phil Howard KA9WGN | http://linuxhomepage.com/ http://ham.org/ |
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From practical experience let me tell you what we found doing the same thing to power a convection oven built for 480 volts from a 208 volt distribution.
We asked the electrical contractor next door to do the installation. He sized the transformer just like you did, then asked an expert who told him he didn't consider the transformer loss, so he found a transformer that covered it and then some. It was surplus to another contractor and had been paid for by someone who didn't use it. So we got it cheap. We have a 150 amp circuit breaker between the meter box and the transformer. Works great. The transformer does get warm, but not uncomfortably so.
The reason I am writing is to let you know that we initially turned off the transformer on weekends using the 150 amp circuit breaker. This was great for a while. Then one Monday morning the transformer would not power up. We called the electrician and he eventually discovered a 200 amp fuse in the meter box was open. He replaced it, \$50.. and it worked again. A few weeks later, the same thing happened on a Monday morning. Same fuse problem. He said to stop turning off the transformer. The high inrush current was blowing the fast acting fuse, but not effecting the heat operated circuit breaker. The oven was not powered up, so there was very little load on the transformer. At \$50 per fuse, that could quickly add up to more than the electricity used over a weekend.
So, after you have done all your calculations, go see what fuses are in the meter box for that circuit. They will be your limiting factor.
Paul
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----------------------------
wrote:

------------------------------ Actually the input data given by James includes the transformer losses. The difference between the KVA at 208V and that at 480V is 0.3KVA which includes all real and reactive losses in the transformer- this is no big deal (Real losses producing heat are probably below 0.1KW ). Also ratings are based on output not input KVA so oversizing for losses isn't necessary. If he was are switching the load on the secondary side, the transformer should not have a problem. The core heating will be mainly due to core losses which will be there even at no load. These should be normal. Considering a duty cycle of 15 minutes on and 10 off, the 10KVA rating appears adequate.
The advice that your contractor got meant that you had a larger transformer with (most likely) higher core losses and inrush current than a smaller unit sized transformer as well as a bigger initial price tag except in your case.
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Don Kelly snipped-for-privacy@shawcross.ca