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

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.

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:

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.

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

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.

| 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.

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

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

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?

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

----------------------------
------------------------------
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.

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.

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.

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).

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