3 Phase Short Circuit Current

[big john]

When calculating SCC on a 3 phase transformer all the equations seem to be:

(kVA * 1000) / (V * 1.732)

This results in less than the current on a single phase 480V fault.

What is happening that is necessary to calculate the square-root of 3 into this fault current?

 

[Ingenieur]

That will give rated output i
not fault
you need to divide that by the pu Z
that will give you a L-L fault i

 

[big john]

That will give rated output i
not fault
you need to divide that by the pu Z
that will give you a L-L fault i

I know, I'm just asking about the 1.732 part of calculating FLA.

 

[Ingenieur]

I know, I'm just asking about the 1.732 part of calculating FLA.

because it is 3 ph, correct?

S va = sqrt3 x v (l-l) x i (line)

a 3 ph 1000 kva xfmr is 3 x 333.3 kva xfmrs
so the 'single ph' i is 333300/480
3 ph is 3 x 333300 / (1.732 x 480)

 

[big john]

But if the fault is L-L, then it more than one winding would be contributing power.

That's why I don't follow the point of the single winding calculation.

 

[Ingenieur]

But if the fault is L-L, then it more than one winding would be contributing power.

That's why I don't follow the point of the single winding calculation.

you can't compare a 1000 3 ph with a 1000 1 ph

you are correct 2 x 333.3 kva would contribute
but they are not in ph so the Z/coil does not simply sum

it's times like this I wish I had video and a chalk board lol

3 ph 277/480 wye 1 MVA
let's assume each coil has Z = 0.01152 Ohm (pu 5%)
line-line involves 2 coils
but not 2 x 0.0115 = 0.02304
but 2 x sin120 x 0.01152 = 0.01995

so fault = 480/0.01995 = 24,060 A
same as 100000/(1.732 480 0.05) = 24.060 A
identical

1 ph 480 666667 va ( same capacity as the 3 ph)
same pu 5% so Z = 0.01728 for the single coil

fault = 480/0.01728 = 27,778 A
a bit more

 

[Ingenieur]

One more step
assume rated current is the same for both the 1 ph and 3 ph xfmrs
3 ph 1 MVA 277/480 wye 1203 A
1 ph kva = 480 x 1203 = 577,350
pu for both is 5%

the 3 ph fault remains 24,060 A

For the new 1 ph
Z = 0.05 x 480^2/577350 = 0.020
fault = 480/0.020 = 24,060 A
same as 3 ph

 

[big john]

So if I'm understanding this right, the short answer is that even though power goes up by 1.732, the series winding impedance increases by the same value, effectively making it like only 1 winding is contributing fault current?

 

[Ingenieur]

So if I'm understanding this right, the short answer is that even though power goes up by 1.732, the series winding impedance increases by the same value, effectively making it like only 1 winding is contributing fault current?

Assuming the current rating (not power) is the same you could have a 1 ph with low pu Z and a 3 ph with high Z

so the the phase fault on the 1 ph could be larger

you can't draw a fixed relationship between 1 ph and 3 ph xfmr faults
you could I guess, but it would an algebraic expression based on (for a common v)
pu Z's
kva

 

[big john]

You can't draw a fixed relationship between 1 ph and 3 ph xfmr faults

But isn't that what the original equation is doing, that's why we multiply by 1.732?

If I have a 1kVA 277V transformer with a 5% individual impedance, then I've got a SCC of 72A. (1000VA/277V/0.05Z)

But if I gang three of those then I've got a 3kVA 480V with a total 5% impedance and it gives me a SCC of 72A (3000VA/480V/[0.05Z*1.732])

Does that work?

 

[Ingenieur]

But isn't that what the original equation is doing, that's why we multiply by 1.732?

If I have a 1kVA 277V transformer with a 5% individual impedance, then I've got a SCC of 72A. (1000VA/277V/0.05Z)

But if I gang three of those then I've got a 3kVA 480V with a total 5% impedance and it gives me a SCC of 72A (3000VA/480V/[0.05Z*1.732])

Does that work?

We mult by sqrt3 because we have 3 signals offset by 120 deg

Yes that works for the i to be equal
as long as
kva 1 ph = 1/3 kva 3 ph
v 1 ph = vl-l 3 ph / sqrt3

 

[Julius Right]

The short-circuit current [supertransient, transient and steady state] is as following:
single phase Isc1=sqrt(3)*VLL/(X1 +X2+Xo)
phase-to-phase Isc2=VLL/(X1+X2)
three phase : Isc3=VLL/sqrt(3)/(X1)
If Xo~0 and X1=X2 [far from generator] then Isc1/Isc2=1.5
If Xo~X1 and X1=X2 [far from generator] then Isc1/Isc2=1
If Xo>X1 Isc1/Isc3<1