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Calculating 3-phase fault from earth fault

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AusLee

Electrical
Sep 22, 2004
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I want to calculate the short circuit at my main switchboard.

I have the transformer's data (2MVA, 6%) and I have the network upstream fault so I use:

1 / MVA system = 1 / MVA utility + 1 / MVA transformer

I presume that's correct, right?

Only one more thing: for the utility they gave me the utility's earth fault kA not the 3 phase fault kA.

Is there a way to work out the utility's 3-phase fault kA from the single phase to earth fault kA they gave me and how?
 
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If you're sizing equipment for your main switchboard, then MVA / (System Voltage * Impedance) will give you your worst case, assuming that everything is fed from your utility connection.
In your case this would give you about 46kA, assuming all is at LV at your location. This is adequate for sizing equipment, but not as much use for a coordination study, particularly if the upstream network is restricted.

Inclusion of any upstream impedance will reduce the expected fault current, so the answer to your question depends on what you're trying to achieve.

There would be ways of evaluating 3 phase fault levels based on a single phase value, but that would depend on what the system impedances are. I'm assuming they gave you the HV EF data, likely for the purposes of evaluating step and touch potential.
 

From the query I can make out that you want a go by.

In a utility fed LV power distribution system, as a thumb rule (exceptions prevail) the three phase fault current and the earth fault (sometimes called as single phase fault) are considered to be the same. So reverse your query as, estimating the earth fault current from the three phase fault current. This would give you an answer.
 
@FreddyNurk

You are correct, I am not designing for fault withstand capacity I am designing for grading. I know, the inclusion of the upstream impedance will reduce the fault level to its actual value not the theoretical maximum based on infinite upsteram bus.

Yes I have your same assumption, they only gave me the EF because that is what they have worked out for earthing purposes.

I need the 3-phase fault for my grading study. I do not want to guess it, I was told that there is an equation that work it out.



 
If anyone else is going to post, I appreciate that it is actually the formula how to calculate the 3 phase fault in HV from the earth fault thank you.
 
The 3ø pu fault is 1/Z1, where Z1 is the pos-sequence pu source impedance. The 1ø pu fault is 3/(Z1+Z2+Z0), where Z2 is the neg-seq pu source impedance, which you can assume = Z1. Z0 is the zero-seq source impedance. If all you know is the 1ø fault current, you can't calculate the 3ø fault current.
 
There is NO formula for calculating I''k(3Ph) from the value of I''k(1Ph).

Your options are: request the value of I''k (3Ph) or accept considering I''k (3Ph) = I''k (1Ph)

JBC
 
I got these formulas but wonderinig how to use them in HV:
Z = impedance per phase
ULN = phase to neutral voltage
ULL = sqrt(3) x ULN

3-Phase fault: Ik = ULN / Z

Earth Fault: Ik = ULN / (Z+Z0)

and just for the record 2-phase fault: Ik = ULL / (2 x Z) which I think is easiest to understand: voltage is phase to phase (ULL = sqrt(3)) and the impedance of the circuit is made of twice the copper path

In the earth fault formula, it is also easy to understand that the voltage applied prior to the fault is ULN and the impedance of the circuit is one leg the copper path and the second leg is the earth resistance. Since they cacluated the step potential, they can give me Z0. I have ULN, Ik and Z0 so I can work out Z and apply it in the 3-phase fault formula.

My problem is this: these formulas make perfect sense for me in low voltage but what do they mean in high voltage or how do you apply them?

The upstream network is 11kV and has no neutral, so how do I apply these formulas in this case, say earth fault was 2 kA:

2 kA = [11 / sqrt(3)] / (Z+Z0)?

And then which votlage to use in 3-Phase
 
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