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Circuit Protective Conductor CPC calculation and Design

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Edelma_1

Edelma_1

Electrical
Apr 16, 2024
31
I came across a formula for circuit protective conductors in bS76761 18th edition page 196 which states S=sqrt(12*t)/k

where S is the csa (mm2) of the conductor
I is the fault current
t is the time in seconds for disconnection
k is the factor taking account of coefficient, temperature and the likes

what I don`t understand is which fault current is being talked about here, is it the earth fault?
 
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It's whatever current that is flowing through the conductor.
 
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I have with me only BS7671 17th edition. Here, on page 160 Regulation 543.1.3 is this formula exposed. The formula may be used for short-circuit in any type of conductor -line or protective-earthing, indeed. However, in this chapter k factor it is only for protective conductor.
For line conductor protection against fault current see Regulation 434.
In this chapter this formula-modified accordingly-is used in order to state the minimum clearing time of fault current.

 
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I only need the K factor for Circuit Protective Conductor
 
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k factor is in Table 54.2 or 54.3 [from 17th edition].
 
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I found out the fault current is calculated by first converting all parameters (fault MVA, Transformer(s), cable(s)) into their various rectangular impedances and then finding total impedance Zs and subsequent fault current for the neutral cable and CPC cable.
 
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As far as I know, the neutral is not, always, a CPC conductor also.
The standard IEC 60364 [ all the parts] it is only for up to 1000 V a.c. and 1500 V d.c.
In chapter 11 of the part 1 of this standard is written:
11.2 It covers:
a) circuits supplied at nominal voltages up to and including 1 000 V a.c. or 1 500 V d.c.
For medium voltage cables see IEC 60502-2
For high voltage cables see IEC 60840.
For earthing details see BS 7430 and BS-EN-50522
 
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