Viper57

OK all you engineering minded (wannabe) phenoms (like me)... many years ago I learned how to calculate the maximum available (infinite bus) fault current on the secondary of a 3-phase transformer by:

1 - Calculate secondary current kVA=(V*A*1.732)/1000 or A=(kVA)*1000/v*1.732

2 - Divide the secondary current by the % impedance*100

A 1500kVA, 5.75%Z, 12,470-480V transformer will have 1804A secondary current and 31,378A maximum fault current.

Here's the problem... when 3-single phase, 500kVA, 2.3%Z, 480V transformers are connected (WYE-WYE) to form a 3-phase transformer, what is the resulting impedance or maximum secondary current. Without involving an engineer, this is a quick and easy way to ensure the OCPD on the secondary will not be over duty.

Thanks!

ampman

install 200,000 aic rated fuses

william1978

I wonder how much a 200k AIC 600V fuse cost?

ampman

a sh!t load

william1978

I can only imagine. I tried Google to see if i could find some.

Magnettica

The OP's question is above my pay grade.

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william1978

It is probably above all of our pay rate.

Benaround

Viper, This is just a guess, but, I feel you would treat it like it is one

transformer.

1st. example: 1500000/831 (480x1.732)=1805, 1805/5.75%=31392 sca

2nd. example: 1500000/831=1805, 1805/2.3%=78478 sca

__________________
Frank, Arizona,USA

It's what you learn after you know it all that counts

micromind

The primary voltage or type of connection doesn't matter at all if we're interested in secondary current.

In a WYE connection, each transformer is a 'stand alone', meaning that the maximum available short circuit current of one transformer is the same as the entire system.

Beenarounds examples are correct.

Are you sure that the transformer secondaries are 480 volt? If so, and they're connected WYE, the phase to phase voltage will be 796. And the short circuit current will be 45,260. (This system does actually exist, it's called a 460/796. I've only seen it used on oilfield pumpjacks.)

Here's how the impedance percentage is arrived at. The transformer secondary is solidly short circuited. Voltage is applied to the primary, and raised until full-load current flows across the shorted secondary. The impedance is the percentage of full primary voltage that resulted in full load secondary current.

If the full load current is divided by this figure (expressed as a decimal), the result will be the current across a bolted short circuit on the secondary if full primary voltage is applied.

This formula works on the primary as well. Using the 1500KVA transformer at 5.75% as an example, the secondary short circuit current will be 31,378 amps assuming full primary voltage. The primary current will be 1207 amps.

In real life, if this transformer was close to a good-sized substation, and the secondary was solidly shorted, the primary voltage would be very close to 12,470, and the secondary current would be close to 31,000 amps. If it were 20 miles away, the primary voltage would fall quite a bit, and so would the secondary current. Quite likely around half.

Around here, the POCO will figure the maximum available short circuit current at the terminals of a switchboard. This can be very useful for specifying the AIC rating of circuit breakers installed in said board. The higher the AIC, the higher the $$$

Rob

P.S. A fuse can interrupt a FAR greater current than a circuit breaker can. 200K AIC is actually fairly common for fuses.

Viper57

Thank you... I knew there was a lot of knowledge here! The system is actually 12,470 - 480/277... sorry for the confusion. I too, am coming up with 78k sca. But when I think of a phase-to-phase fault with the wye secondary, I'm inclined to believe the impedance of 2 transformers would be series and the sca would be reduced. A SLG fault would be 78k sca. This transformer feeds a 100ka braced swbd with KRP-L fuses (200kair). The swbd feeder breakers are SqD MA breakers (30kair). So you can see my concern about adding more MA breakers to this swbd. I think it is time for someone to do an actual short circuit study to calculate the actual sca available on this swbd. And you are correct about the 200kair fuses being common... all RK-1, RK-5 and Class L fuses are standard 200kair... it's those Class H fuses (10kair) you have to watch for!

Zog

They are very common in power systems, depending on the type they can be $10 or $1,000+. The ones I needed yesterday were $787/ea.

Viper57

Yeah, that's about what a Cooper KRPC-l-1200 goes for.

Zog

The impedance is in per unit. So a single transformer is 2.3% impedance on a 500kVA base. The new arrangement of transformers would be 1500kVA with 2.3% on the new base of 1500.

Does seem simplistic but that is the beauty of the per unit system.

Now your problem is you have over 78kA of fault current available. Which I assume exceeds the interuption rating of your breakers, which is why you are asking about this right?

Keep in mind the method you are using assumes an infinite source on the promary side of the transformer.

william1978

$787.

Zog

Are the big eyes because that seem like a lot or a little to you?

I replace $2-3,000 fuses (Per phase) all the time. A double barrell 150E 15kV fuse can be around $3,000 per phase. The $787 fuse I mentioned was simply a 1600A 480V fuse on a breaker, some dolt closed the breaker back into a fault without testing the load and blew 2 of the current limiting fuses.

william1978

That is alot of money to me because I'm a RAT.