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K2s2 - Adiabatic for parallel cables

While verifying the cable selection for let through energy withstand, How should it be calculated for parallel cables? for example, 120sqmm of 2 runs (XLPE, copper), the K2s2 value is 294465600 (143 x 143 x 120 x 120) - Should I account for the 2 run by multiplying 2 to this value?

  • While verifying the cable selection for let through energy withstand, How should it be calculated for parallel cables? for example, 120sqmm of 2 runs (XLPE, copper), the K2s2 value is 294465600 (143 x 143 x 120 x 120) - Should I account for the 2 run by multiplying 2 to this value?



    It all depends on where your worst case fault could be. If it is thought that a fault could only occur downstream of the point where the parallel cables are connected back together again - so that any fault current would be expected to be shared equally between the two conductors - then multiplying the withstand of one conductor by two would make sense.


    If, as is usually the case, you need to consider that a fault could occur anywhere - then it's likely that a more challenging situation would be when just one of the parallel conductors is involved in the fault (either to earth or to another conductor). If the fault is close to the point where the parallel run starts then it's likely that just one conductor of the paralleled group will take almost all the fault current on its own - so you certainly wouldn't want to assume a doubled  withstand for that case. Faults further along start to get the fault current taking several paths - some direct along the faulted conductor, some via the other conductors in the group to the end of the parallel section and then 'backwards' to the fault location. Some of the diagrams in appendix 10 of BS 7671 probably paint a clearer picture than my words. Figuring out where the worst case then becomes far from trivial - especially when some protective devices (e.g. fuses) tend to have the greatest energy let-though for furthest faults (highest loop impedance, longest disconnection times) while others (e.g. MCBs) tend to have the greatest energy let-though for nearest faults (highest fault current).


       - Andy.

  • then multiplying the withstand of one conductor by two would make sense.



    On second thoughts perhaps that should be the withstand calculated for 2x the c.s.a. (k²(Sx2)²) rather than 2x the withstand (2x k²S²).


       - Andy.