I want to connect a 3 phase 480V motor to a 600V supply.
Lets say for example the load of the motor is 100A at 480V.
600V - 480V = 120V is the voltage difference.
Typically the rating of the transformer would be expected to be
1.73 x 120V x 100A = approx 20 KVA, and I would add a safety margin of
125%
but this is a very small transformer with small lugs and a small
enclosure. Am I right in assuming that this transformer would be
acceptable to start and run this motor load? It seems that there are
other issues that might come into play like line impedence etc.
I look at the size of the wire I need to feed the motor and the size
of the wires and lugs inside the transformer and there is a huge
difference.
Here is another hypothetical situation just to show the direction
things go. Lets say I want to change a 200A load from 600V to 575V, a
25V difference.
1.73 x 25V x 200A = 8.6KVA.
You would never get a 250 Kcmil cable into this box.
How would I size that transformer? Just upsize it until the case and
lugs are big enough?
Is this an acceptable means of doing this calculation?
Thanks
Remove dashes "-----------" to e-mail.
You size a buck-boost transformer for the power it is providing. If it
is supplying single phase 25V at 200A to boost 208V to 233V it would
need to be at least 25x200 = 5KVA. That is the power it is supplying to
the load for boost or back to the line for buck.
bud--
FYI, if you are going 600 to 480, there is a trick you can do where you
reverse the polarity of a 120V transformer and connect it in series
with each leg of the 600, since 600 minus 120 = 480. Is that what you
are trying to do? If so, you can reduce the kVA of the transformer,
since you need rated motor current through the 120V winding. In other
words, you would need three transformers
120V * 100A = 12kVA each.
Get somebody who knows this stuff to do you a drawing before you blow
something up.
Asuming this is covered by the NEC:
If this is an "industrial occupancy where conditions and supervision
ensure only qualified persons service the installation" you could use 2
buck wired as autotransformers (NEC210.9). Assuming "industrial":
B'
/
/
B
/ \
/ \
/ \
/ \
A -------- C----C'
AB'C' is the 600V source, ABC is the derived 480V.
ABB' is an buck/boost transformer connected as an autotransformer. AB'
is a 600V winding, BB' is a 120V winding. They are connected so the
voltages subtract to 480V at AB.
ACC' is the same.
If the current to the ABC motor is 100A the current at A B' and C'
should be 100 x (480/600) = 80A. The rating of EACH transformer should
be 80A x 120V x 1.732 approx equals 17KVA minimum. If continuous that
should be increased by 20%. The (IIRC) 1.732 is because the current in
the autotransformers is out of phase (low power factor). Would that
supply starting current? I would guess yes but don't know.
This could also be done with 480/120 transformers with slightly more
involved calculation. The transformer would have to be rated for a
higher voltage to ground than a conventional 480V transformer.
I havn't done these calculations for a long time and they may look
diferent in a few hours. And some of your questions are not inspiring.
Someone who has significant experience should design this. 600V and 480V
can do funny (or not so funny) things.
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If not "industrial" and if 600V is Y connected (I never got to play with
600V) this could be done with 3 neutral to phase autotransformers
connected for the odd voltages of 345V to 277V.
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Not obvious why you have 250Kcmil wire with 200A load.
bud--