Transformer Overcurrent Protection Question Via a PM

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iwire

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Another question in my PM box. :)

Justin P. Le Blanc said:
Hi, This is my first time using this site, so I'm not sure if my question is ok.
But here goes...

If I have a 480v 3-phase 3-wire distribution panel and I'm wanting to feed a 400amp 120/208v main panel, how do I work out the KVA rating of the necessary transformer?

Is the calculation: 400(A) multiplied by 208(V) which would equal 83.2 KVA ???

And how does one work out the breaker size to protect the transformer in the 480v distribution panel?

Thanks.
 
Justin P. Le Blanc said:
If I have a 480v 3-phase 3-wire distribution panel and I'm wanting to feed a 400amp 120/208v main panel, how do I work out the KVA rating of the necessary transformer?

Is the calculation: 400(A) multiplied by 208(V) which would equal 83.2 KVA ???

Not quite.

400A * 208V * 1.73 = 143,936 watts / 1000 = 143 KVA

The next standard size transformer is a 150 KVA.

Justin P. Le Blanc said:
And how does one work out the breaker size to protect the transformer in the 480v distribution panel?

There are many answers to that question, it really depends on what you want out of the installation.

Bare bones, code minimum, least cost or do you want to go the other way and leave some 'extra' in the design.

With a 150 KVA transformer the primary overcurrent protection at 480 volt might be 1 amp to 450 amps, whatever overcurrent protection you choose will determine the size of the primary supply conductors.

With a 150 KVA transformer the secondary overcurrent protection at 208 volt might be 1 amp to 500 amps, whatever overcurrent protection you choose will determine the size of the secondary supply conductors.
 
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Not really sure what you need to feed but thinking about this I have a couple of ideas. :smile:

Supply a 150 KVA transformer with 500 Kcmil copper protected with a 350 amp overcurrent device.

On the secondary side install two barrel lugs on the transformers 'X' connections and supply two 225 amp main breaker 42 circuit panels using 4/0 copper.

You end up with 84 available spaces and a total of 416 amps available current between the two panels.

Just thinking out loud. :smile:
 
Justin P. Le Blanc

Justin P. Le Blanc

Thank you Bob, for your speedy reply.

The scenario I'm thinking of to be a little more specific is the following:

I came across a 480V distribution panel with a main breaker of 225A. There was a 100A breaker protecting a 75 KVA transformer, that is feeding a 225A 120/208V main panel. This installation was completed in 1969 and all looked in good condition.

Does this installation seem ok to you? If so how did they decide 100A at the distribution to protect the 75KVA transformer? Is it determined by the load on the transformer at the secondary load side (225A 208V)?

The main panel was already full and feeding 2 additional 100A sub panels, but there was a need to add an additional 21,000 watts of new load to the 225A main.

If one wanted to change the 225A main panel out for a single 400A main panel would it be possible to keep the existing system in tact, that is, the 75 KVA transformer, the existing primary conductors, and the primary 100A breaker that protects it?

If not, could the existing transformer support a 2nd 225A panel?

From memory the conduit size between the distribution panel and transformer was 1 1/2 inch GRC and between the transformer and the main panel was 3 inch GRC.
 
First of all, I'm not an engineer, just a journeyman, but I work with xfmrs alot.

The original installation is good. A 75KVA xfrm is 90 amps on the 480 side, and 208 amps on the 208 side, so a 100 and a 225 are about right.

To properly design a system like this, you first calculate the load. This determines the xfmr size. Don't undersize it, go at least one size up, it's a safe bet that more load will be added later. The xfmr size determines the breaker or fuse size. See article 450 in the NEC.

Given the choice, I like to stay at the lower end of protection on the primary side, realistically, the secondary side overcurrent is going to protect the xfmr, the primary side is short circuit and ground fault protection. With copper at $3 a pound, and 480V breakers insanely expensive, it makes economic sense to keep it as small as possible without compromising usability.

In your case, the primary of the 75KVA xfmr is most likely a delta, and the secondary a wye. If there were no secondary protection, you could grossly overload one leg of the secondary, and the primary wouldn't even see full load current, let alone an overload. That's why secondary protection needs to be sized to the particular xfmr.

Just because a panel is full doesn't mean a xfmr is loaded, consider for example, a fire alarm panel. A dedicated breaker for a load less than one amp. You'll need to calculate the total load, and go from there.
 
follow up question

follow up question

iwire said:
Not really sure what you need to feed but thinking about this I have a couple of ideas. :smile:

Supply a 150 KVA transformer with 500 Kcmil copper protected with a 350 amp overcurrent device.

On the secondary side install two barrel lugs on the transformers 'X' connections and supply two 225 amp main breaker 42 circuit panels using 4/0 copper.

You end up with 84 available spaces and a total of 416 amps available current between the two panels.

Just thinking out loud. :smile:


bob, my question is

is the transformer designed to balance the total connected load via the 3 windings in 3 phase applications? so for example: the 150kva has 3 windings, with 416 amps divided by 3= 138amps each winding?
is this correct?:roll:
 
steveng said:
bob, my question is

is the transformer designed to balance the total connected load via the 3 windings in 3 phase applications? so for example: the 150kva has 3 windings, with 416 amps divided by 3= 138amps each winding?
is this correct?:roll:

A transformer may be operated at full load continuously. The primary FLA is 90 A. 90 x 1.25 = 112.5. Use conductors capable of carrying 112.5 Amps on the primary and fuse for the next size up, 125A.

The transformer can be thought of as three 25kVA single phase transformers. So, 25,000/120 = 208 Amps full load output on the secondaries.
 
The above post was for a 75kVA transformer. I forgot to mention that.

For a 150kVA, the output current can be calculated as follows:
150,000 / sqrt(3) / 208 = 417
...or...
50,000 / 120 = 417
 
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