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Fault current for earth grid

Fault current for earth grid

Fault current for earth grid

(OP)
When we designing an earth system we need to find the portion of  fault current that goes through earth grid and portion of fault current that goes back to the source through earth wire or cable screen?
 How can we calculate the portion of fault current that returns back through the screen of incoming cables? any reference or document would be appreciated

 

RE: Fault current for earth grid

Hello,
You may take a look at IEEE and try to find some papers in which this is explained.
As jghirst mentioned, looking ate some software developers may be usefull to understand wht they do.

RE: Fault current for earth grid

(OP)
to jghrist: Thank you for your advice.I have access to  split but can not find  know to use it for this purpose.
Could you please where I should put Number , length and impedance of incoming cable and where it gives the fault current?
can you suggest any tutorial or detailed manual ?

to lume7006: IEEE only talks about fult current returning from overhead earth wire. it does not tell you how calculate it when you have cable

 

RE: Fault current for earth grid

Steve,
Actually, we don't have SPLITS, but use the less rigorous FCDIST (Fault Current Distribution Analysis) program to determine the fault current distribution.  I can't help you with SPLITS, but FCDIST is relatively straightforward.  You can't define an underground cable, but you can define one just above the ground with multiple grounds.  Or you can specify the neutral impedance and mutual impedance to the phase conductor.

SES technical support is pretty good.  Give them a call.

RE: Fault current for earth grid

According to DIN-VDE 0141 the split factor [they call this "reducing factor"] for a cable of 150 sqr.mm AL copper tape shielded 20 KV will be 0.5.
For cable shield split factor calculation see:
http://www.emo.org.tr/ekler/8895f0da0edf4da_ek.pdf
 

RE: Fault current for earth grid

IEEE 80 Section 15.9 has quite a detailed discussion on computing the current division factor. It can depend on a number of factors, e.g. impedance of the earth grid (with respect to remote earth), fault location, alternative return paths (earth wires, buried pipes, etc), and so on.

Ultimately, the factor that you choose will have some degree of "engineering judgement", but you should be able to back it up with some good arguments. Worse case, you assume that most (or all) of the fault current goes back to the source via earth.

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