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Fault Current Sufficient to Operate a Protective Device.

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I was undertaking some work today in a chalet park. I was testing R.C.D.s. These are supplied in each holiday chalet at the chalet meter position before any consumer unit. The chalet owners are responsible for any internal electrics after the park's R.C.D.


I wondered why R.C.D.s were first installed by the park owners. After all an R.C.D. or R.C.B.O.s could be incorporated in the consumer units. The supplies to the chalets is TT. O.K. so we do need an R.C.D.


But why TT the permanent wooden chalets? This is not a caravan or camp site at all. And the chalets are not transportable. They are not residential park homes according to the B.S. 7671 definition.


Then it dawned on me. At the last chalet, the remotest from the main intake room the P.S.C.C. is only 400 Amps. It is protected by a B50 type M.C.B. So, under fault conditions the fault may take a long time to clear. Evidently a TT system is safer.


So, what effect would a heavily loaded distribution cable have on the operation of the B50 M.C.B. if the park was fully occupied and the supply Voltage dropped? Could it ever become dangerous and prevent a B50 M.C.B. from not opening, or opening too slowly,  at the remotest chalet say due to a short from L to N?


Z.






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    A B-type should trip immediately at 5x In - so 250A for a B50 - which hopefully you should have at the far end (otherwise you're likely to have v.d. issues for normal working) - unless you've got a significantly reduced c.p.c. (SWA perhaps).


    Loads shouldn't make a huge difference - given the (assumed) negligible impedance of the fault itself almost all of the available current will be drawn through the fault at the expense of the normal loads.


       - Andy.
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  • 0
    A B-type should trip immediately at 5x In - so 250A for a B50 - which hopefully you should have at the far end (otherwise you're likely to have v.d. issues for normal working) - unless you've got a significantly reduced c.p.c. (SWA perhaps).


    Loads shouldn't make a huge difference - given the (assumed) negligible impedance of the fault itself almost all of the available current will be drawn through the fault at the expense of the normal loads.


       - Andy.
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