A 10KVArated stepup isolation transformer is used to connect a 480vdelta
resistive heater to 208vY 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 overtemperature 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 http://www.powerstream.com/VAWatts.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:
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.
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.
But since the transformer is (presumably) designed with a worstcase PF (or
at least some PF value greater than 1), can't it be rerated (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
>> 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 worstcase PF (or
 at least some PF value greater than 1), can't it be rerated (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.


 Phil Howard KA9WGN  http://linuxhomepage.com/ http://ham.org/ 
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
 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 LN 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.


 Phil Howard KA9WGN  http://linuxhomepage.com/ http://ham.org/ 
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
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.

Don Kelly snippedforprivacy@shawcross.ca
remove the X to answer
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 newlyconnected 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 am sure you are correct, in theory. However, the oven is computer
 controlled and can't control until it gets power. Darn computers!
Batteries! Capacitors!


 Phil Howard KA9WGN  http://linuxhomepage.com/ http://ham.org/ 
HomeOwnersHub.com is a website for homeowners and building and maintenance pros. It is not affiliated with any of the manufacturers or service providers discussed here.
All logos and trade names are the property of their respective owners.