We have 33KV in 4MVA and 33/6kv Trans and 6/0.4kv Trans.
How We Can calculate Short Circuit Impedance ?

Hire a qualified PE to do this. You would be a fool to get someone unlicensed, and anyone unlicensed would be a fool to accommodate you.

Bill -- Ferme le Bush

We have an Industrial plant with primery voltage 33KV and secondry 6KV and
th-ry voltage 400V .
I want to calculate short circuit current.
It means that i should to know equal Impedance of all circuits.
all of consumers loads totaly is 4MVA.
i Want an Algorythm or program or sample sheet calculation.
thnks.

Please would be nice.

Jaymack

eoots2000 wrote:

insuficient data.... you can not get there from here.

the smart ass answer is zero current because the breakers will have tripped.

insuficient data.... you can not get there from here.

the smart ass answer is zero current because the breakers will have tripped.

You need to know more about the transformer. They typically include synchronous reactance, as well as transient and sub-transient reactances. And the line feeding the transformer primary.

This really is an involved calculation, it is best to hire a pro rather than get advice on USENET for this.

daestrom

and

than

Transient and subtransient for a transformer? Really?

Ah, you got me. I slipped up. I was thinking he had an industrial generator feeding a transformer (I work mostly with power plants, and that is the common configuration we work with for such calcs). Then the generator is feeding the fault through the transformer.

But as I look again, I guess he's talking about a primary line feeding a substation into his 'industrial plant'. The other factor he would need then is the line impedance and its short-circuit capability.

daestrom

daestrom wrote:

If the plant has any good sized rotating machinery, that can contribute to the fault current as well. Like others have said, the OP needs a PE to look at the problem.

If the plant has any good sized rotating machinery, that can contribute to the fault current as well. Like others have said, the OP needs a PE to look at the problem.

--

Paul Hovnanian mailto: snipped-for-privacy@Hovnanian.com

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Paul Hovnanian mailto: snipped-for-privacy@Hovnanian.com

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Short out the power plant, measure the current, and you can calculate
the impedance, probably by grunting at the nurse that follows your
limbless body around.

Easy! The impedance of a short circuit is zero!

wrote:

Is it?

Ben Miller

Is it?

Ben Miller

--

Benjamin D. Miller, PE

B. MILLER ENGINEERING

Benjamin D. Miller, PE

B. MILLER ENGINEERING

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Well, technically, you have to use the special, non-inductive short circuits.

The impedance of the short circuit itself is zero. But this would mean
you have an infinitely high short circuit current, which is of course
impossible.

You still have the impedance of the power lines, the transformer,...

Take these into account, and then you can calculate the short circuit current by dividing the voltage by the total impedance.

Kristof

Doofus wrote:

You still have the impedance of the power lines, the transformer,...

Take these into account, and then you can calculate the short circuit current by dividing the voltage by the total impedance.

Kristof

Doofus wrote:

Only in text books. In the real world, there is always contact resistance, arc impedance, etc. at the fault. Add to that all of the other impedances that you mentioned.

Ben Miller

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Benjamin D. Miller, PE

B. MILLER ENGINEERING

Benjamin D. Miller, PE

B. MILLER ENGINEERING

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True, but in fault analysis and when determining interrupting rating of protective devices, generally only the source impedance and intervening lines are used, the fault is assumed 'bolted' with zero impedance. Slightly conservative, but there you are.

daestrom

True. We use bolted faults because we can't know what will actiually occur. I was just taking exception to the statements that the short-circuit impedance is zero. A more accurate statement is "we assume for calculations that it is zero".

Ben Miller

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Benjamin D. Miller, PE

B. MILLER ENGINEERING

Benjamin D. Miller, PE

B. MILLER ENGINEERING

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occur.

calculations

In customer fault level calculations here we are generally required to assume that the Utility impedance is zero as well. Plus we are forced to use low impedance assumptions for utility transformers. I wonder if they will lighten up for calculating arc flash hazards when that becomes mandatory. I don't think I'll hold my breath.

j

'arc flash'

Oooooh, now

daestrom

of

intervening

to

they

I recall a distributor mentioning a new product they have, high speed relaying or something solid state, that they use for tough arc flash applications. He used 600V MCCs as the example. They knocked 100cal/cm^2 down to 8cal/cm^2 in one instance.

So far I haven't had to do any arc flash calculations. What are the dominant factors in a calculation? I? I^2t? SCMVA? V? Got any advice to share for situations that look like a high cal/cm^2 value is likely?

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