Mains electrical Q: transformer sizing?

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
Reply to
Jon Elson
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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.
Reply to
James T.
Not sure where you came up with the sqrt(3) in your equations. Take a look at
formatting link
(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:
Reply to
JeffB
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.
Reply to
James T.
You were right the first time. The sqrt(3) accounts for the fact that this is a 3-phase system.
Reply to
Ned Simmons
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
Reply to
co_farmer
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.
Reply to
James T.
Looks right to me.
No, a purely resistive load is best case.
Reply to
Ned Simmons
| 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.
Reply to
phil-news-nospam
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
Reply to
Tom M
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
Reply to
Lew Hodgett
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?
Reply to
James T.
My notes (taken from the name plate) say "150C".
Reply to
James T.
Sorry, the power factor can never exceed 1 because it is the cosine of the phase angle between the current and the voltage.
Jim
Reply to
Jim
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
Reply to
Mark Rand
----------------------------
------------------------------ 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.
Reply to
Don Kelly
That's a misconception. If you'd turned the oven ON before powering the transformer, the fuse would'nt blow. The 'load' that is important is the magnetic field in the transformer, which depends on input current MINUS scaled output current. With no output current, and possibly some remnant field, a newly-connected transformer takes its field to maximum, and that causes saturation and the fuse blows.
If the core of a transformer staturates, the electrical circuit becomes just a copper resistor. Better to blow a $50 fuse than melt that copper resistor.
Reply to
whit3rd
I am sure you are correct, in theory. However, the oven is computer controlled and can't control until it gets power. Darn computers!
Paul
Reply to
co_farmer
I see several things in your contention that I have problems with.
Assuming the applied voltage is the same or nearly so, the inrush magnetising current will be the same as before. Adding load won't change this. For AC, unlike DC, it is the applied voltage that determines the peak flux (E=4.44*f*N*phimax or phimax=E/(4.44*f*N) for sinusoidal excitation) and this flux must be there independent of the core. The magnetising current is that needed to provide the flux for the particular core. It will be high with a lousy core or under saturation but the flux will be determined by the voltage. The total input current will be increased by turning the load on before energisation because you will have load current in parallel with the magnetising current. However, given that, whenever switching occurs, there is a transient which can produce a transient DC offset and this can lead to saturation with a resulting current that can be excessive. The transformer will appear as a non-linear reactor, rather than a resistor. Adding a load would not change this as during this transient condition, the load voltage may be quite small as would the load current because at high inrush currents. Rather than say that the magnetising current is the difference between the load current equivalent and the primary current -which is true- consider that the primary current consists of the load current equivalent + the magnetising current which depends on the flux which depends in turn upon the voltage in the AC case (Faraday's law, which applies, relates (rate of change of) flux and voltage, not flux and current.
Reply to
Don Kelly
The turnon transient isn't similar to normal AC operation; a half- cycle of positive potential in normal AC conditions always occurs right after the core achieves maximum negative flux. If the core flux starts at zero (or even slightly positive due to remnant magnetism) and THEN is powered at the beginning of a half-cycle of positive potential, the end of that half-cycle of excitation will occur with 150% of the normal flux in the core.
If there is a substantial load attached, the flux will be less than 150%, and (of course) if there is a short circuit attached to the output, there will be zero flux (but despite the non-saturation of the core, that won't help the blown-fuse situation much).
Reply to
whit3rd

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