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End of Fault current for Parallel lines in Ligting

Nick Parker

How to find the end of line fault current for the second case,

If this is a straight forward like below. I calculate all the resistance and reactance of cables in series and arrive at the end of fault current.

However if there are two parallel lines from the JB like in the below image, how do I find the end of fault current?

How do I calculate the combined resistance and combined reactance value in this case?

Should I take the maximum resistance which comes out of the three parallel lines for calculating the end of fault current? Also, In general the resistance and the reactance value decreases when connected parallelly

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    If I've understood your setup correctly then the you don't have anything in parallel - they're just three branches of the same circuit. A single fault can only occur in one place - if it's at the end of one of the branches then the fault current won't be flowing in the other branches, so they can be ignored. So yes, just pick the worst case - i.e. the one with the highest impedance - usually the longest.

    Cables in parallel are connected together at both ends - so current is shared between the distinct cables (sometimes in strange ways if there's a fault part way along just one of the cables) - but that's a different kettle of fish.

    If the cable is of the same type throughout, the calculation is usually somewhat simpler - just use the overall length (to the furthest point) and the per metre characteristics of the cable - usually no need to split it down into each individual section.

       - Andy.

  • 0 in reply to AJJewsbury

    Thank you! your understanding is correct. I just need to split it down with individual section. The question arise because I'm working on the spreadsheet created by some one. What is baffling me is, that some one averaged the resistance and reactance in the equation. I could not paste the excel equation here as it runs long. In simple terms, they summed up the resistances based on the individual section and divided it by the total length (summing up all the lengths from individual sections). Similarly done to reactance values. Why would someone do it?

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  • 0 in reply to AJJewsbury

    Thank you! your understanding is correct. I just need to split it down with individual section. The question arise because I'm working on the spreadsheet created by some one. What is baffling me is, that some one averaged the resistance and reactance in the equation. I could not paste the excel equation here as it runs long. In simple terms, they summed up the resistances based on the individual section and divided it by the total length (summing up all the lengths from individual sections). Similarly done to reactance values. Why would someone do it?

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