I have a site that is due to increase its loading due to various site upgrades.
The site currently has a 100kVA transformer. I have to calculate the maximum demand per phase by calculation.
The existing supply goes to a main db which has 3 phase and 1 phase MCB's.
eg, If i have a maximum demand of 100A for the most used phase.... how do I then work out what size transformer I will need?
I have searched high and low on the net to no avail.
Should the rated tx size KVA size be converted to Amps and then be divided by three and this number should be higher than the max demand????

Any good refernces for doing these calcs would really help.

Thanks.

Vl*Il = VA

so on a 400V system with a max line current of 100A
400*100 = 40 Kva

so if demand per phase is 100a - red, 110a - white, 120a - blue, then what size transformer do I need and can you please tell me the formula.

sorry, I missed out the root 3.

Root 3 times the line voltage times the line current equals the VA rating.

use the largest line current

Hello.

You take the highest loaded phase current, so the size is:-
square root of 3 x 400V x 120A = 83kVA. The preferred transformer rating would therefore be 100kVA.

A bit off subject but :- I have sized a transformer at 2MVA and using text book calculation have arrived at a nominal fault current of 20kA (13kV/480v, 60h/z, 6% impedance).
Having run the cable calcs using AMTECH the transformer report gives me a Pscc of 16.06kA

Hello electratech.

It looks like you've based the calculation on a 1MVA transformer to get a 20kA fault level.

Does AMTECH introduce a notional source or cable etc. impedance to reduce the fault level (again looks like a 1MVA Tx)?

Hi!
Sorry for the curt (curtailed)posting but I am having pc probs.
I am using the Amtech domestic package and am assuming that notional values will be inserted but I haven't got that far into it.
My longhand calcs., came out at just under 40kA so I assumed I'd boobed somewhere so I took the Amtech values as being correct but that threw everything else out - hence the posting.
Subsequently I am suspecting every other calc., I've done.
My secondary current is 2000A; I have assumed the supply in the states (Utah) is 13kV and transformer impedance 6pc; I know the secondary is 480V, 60Hz - my end result was 27MVA. or 45kA fault current.

I would appreciate if you could show me where I'm going wrong.

Many thanks,

Joe

Hello Joe.

I have no experience with Amtech and so cannot comment about what is going on there.

Hand calcs I am okay with.

3 phase 2MVA transformer with an impedance voltage of 6% and a secondary voltage of 480V.

Full load current = 2MVA divided by (square root of 3 x 480V) to give you 2405 amps.

Fault MVA = 2MVA divided by 0.06 to give you 33.333MVA.

Fault current = 33.333MVA divided by (square root of 3 x 480V) to give you 40,093 amps.

Best regards.

Thanks for that; same result that I got originally - which is a professional relief - but I was completely thrown by the Amtech result which led me to doubt my grasp of the subject or my arithmetic.

A greatly relieved respondent.

Thank you.

P.S. Sorry for delayed response.

What you encountered has been always a challenge for Distribution company , supplying unbalanced load. Although overall kva of three phase is within kva rating of three phase transformer, the transformer fails due to overload in one of the phases. We need to have alarm for unbalanced load condition persisting for a long time.

I would agree with that. As long as you remember that each of the secondary phases is a third of the total transformer rating, you shouldn't go far wrong, even with an unbalanced load. For your 100kVA transformer, each phase will be a maximum of 33kVA.

Regards,

Alan.

You have to consider the power factor, PF not always 1, sizing must be based on max peak demand.

Check what primary fault level you have entered, it has a big affect on the PScc amtech calcs. I think the default is 250MVA which is pretty high, but if you entered something smaller it will limit the secondary fault level.
If you just work out the TX fault from the TX size and impedance, you will get the infinite bus fault level, in reality there is primary impedance that reduces it a bit.

If you want a pretty basic calc tool that does the job, Eaton has one:

http://www.eaton.com/Eaton/Pro...Coordination/index.htm