Transformer upgrade

[Tony S]

Hi Everyone,

Long time reader of this forum but first time posting.

I have a question regarding a transformer installation. Currently there's a 1500KVA transformer installed feeding a 2000A main disconnect. A proposal was written to upgrade to a 3000KVA transformer in order to have room for future expansion. I am reviewing the proposal and everything seems fine with the cable sizing and load side fused protection.

My question is the AIC rating for the 2000A main disconnect. The main disconnect is rated for 100KA or short circuit current rating, which isn't the same as the AIC rating. According to the calculation the max short circuit fault current should be around 62.2KA for the 3000KVA transformer (based on 480V secondary with a 5.8% impedance). If the short circuit current rating is sufficient to carry the fault current, does it matter what the AIC rating is if we size the primary protection of the transformer accordingly? Or do we have to make sure that the AIC for the main disconnect is rated at least 65KA?

Thanks in advance!

Your fault current for phase to earth calculation agrees with my spreadsheet, have you calculated the phase to phase value? I get 107kA which will have a bearing on the rest of the system if you go for increasing the transformer size.

 

[Jason T]

Your fault current for phase to earth calculation agrees with my spreadsheet, have you calculated the phase to phase value? I get 107kA which will have a bearing on the rest of the system if you go for increasing the transformer size.

I guess I haven't thought about the phase to phase fault value. I would assume the SCCR would have to be above that value also in case a phase to phase fault happens?

 

[Tony S]

I guess I haven't thought about the phase to phase fault value. I would assume the SCCR would have to be above that value also in case a phase to phase fault happens?

It does but the calculation is a worse case scenario based on the infinite source method where the supply impedance is ignored. It will never reach that level unless you’re sat on the doorstep of a primary substation.

 

[Jason T]

It does but the calculation is a worse case scenario based on the infinite source method where the supply impedance is ignored. It will never reach that level unless you’re sat on the doorstep of a primary substation.

What if I put the transformer about 150 feet away from the panel and use two 500 kcmil conductors per phase to power the switchboard. I think that will reduce the SCCR dramatically right?

 

[Phil Corso]

Jason T.

Your proposal is doable! But, it's best if you break up the load into multiple circuits!

However, regardless of the number of circuits, the total impedance between the Xfmr bus, and any load will require the insertion of an impedance (for 150ft) of about 0.08 Ohm (using NEC Table 9 for cu cable)!

An earlier post presented the formula, I'd give you the ID tag, but I'm unable to determine the "Thread" ID!

Phil

 

[Toppcatt22]

My Mistake

My Mistake

Yeah. So, I looked back at it, and I was not considering the root(3) from the 3-phase current. Sorry. You are good.

No, he correctly used 1.732 as part of the equation that gives us full load current based on KVA and voltage. Full load current equals KVA divided by voltage (line-to-line, or 480) and divided again by the square root of 3 (approximately 1.732).
Consider it verified.

 

[Tony S]

What if I put the transformer about 150 feet away from the panel and use two 500 kcmil conductors per phase to power the switchboard. I think that will reduce the SCCR dramatically right?

Feasible but if the transformer ever gets near full load you’re looking at sizable heat losses.

I will say, it goes against anything I’ve been taught regarding transformer tails.

 

[Phil Corso]

Jason T.
I found the example using the approximate formula, based on impedance magnitudes, not vectors:

Given:
o Existing System, Isc(e) = SC-Current Available (300kVA Xfmr) = 42,000A.
o Proposed Systm, Isc(p) = SC-Current Available (500kVA Xfmr) = 60,000A.

Find:
o Zi = Impedance to limit SC duty, inserted between new Xfmr and existing layout!

Equation(s):
o Eq 1) : Ze = Vs/Isc(e) = Vs / [Sqrt(3) x Isc(e)] = 2.86 mOhm.
o Eq 2) : Zp = Vs/Isc(p) = Vs / [Sqrt(3) x Isc(p)] = 1.91 mOhm.

Solution:
o Zi = Ze-Zp = 0.91 mOhm.

Regards, Phil Corso