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Topic Title: Zs for RCD fault protection
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Created On: 03 September 2013 09:58 AM
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 03 September 2013 09:58 AM
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marclambert

Posts: 310
Joined: 23 June 2010

Hi all, I remember seeing a thread which stated that the max Zs if the RCD is used for fault rather than additional protection could be 7666 ohms. This being Uo/Ia where Uo =230v and Ia = 0.03 A.
This makes sense from reg 411.4.5. Great....but...
411.4.9 seems to suggest that table 41.5 may be applied? Which of course gives Ra<= 50v/I delta n = 1667ohms.
Could someone please elucidate.
Is the clue in the word "may" be applied?
Or am I barking up the wrong tree
thanks in advance
Regards
Marc
 03 September 2013 10:20 AM
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AJJewsbury

Posts: 11553
Joined: 13 August 2003

Is the clue in the word "may" be applied?

Yup.

Table 41.5 takes account of both ADS (230V/Ia for required disconnection time, Ia might be higher than Idn - e.g. 2x for 0.2s disconnection) and 50V/Ia for 5s disconnection (i.e. Ia = Idn in almost all circumstances).

They could have produced another table for TN systems that didn't take into account the 50V rule, which would generally yield higher allowable figures - certainly no lower, but since it's highly unlikely that you'd get even within 1/100th of the TT figures on a TN system anyway, it probably wasn't worth the paper. Likewise 41.5 gives Zs, whereas the 50V requirement is based on Ra - but again the difference usually isn't worth the effort to calculate precisely, and the error errs on the side of safety anyway.

It's a bit like having table 54.7 for c.p.c. - you may use that - and that's quick and easy, or you have the option to calculate it and get a more generous answer.

- Andy.
 03 September 2013 11:07 AM
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marclambert

Posts: 310
Joined: 23 June 2010

Thanks Andy.
It is as we suspected, at least an ambiguity, at worst a contradiction
Just to be awkward, can you think of any scenario where the Uo/Ia would actually be used, allowing 7666 ohms for 230v and 0.03 A?
In a TN system surely a Zs this high would be just wrong??
In TT the 50v/Idn would apply so back to table 41.5
Beginning to regret raising this in the office now. Can't wait for the students to start back so I can "explain" this to them! Or I might not
regards
Marc
 03 September 2013 11:27 AM
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Parsley

Posts: 1035
Joined: 04 November 2004

I think this needs careful consideration.
Why would you need to use if the circuits been designed probably?

Longer cable lengths than would normally be used, reduced CPC's? volt drop and 434.5.2 requirements.

A fault occurring at any point in a circuit shall be interrupted within a time such that the fault current does not cause the permitted limiting temperature of any conductor or cable to be exceeded.


Regards
 03 September 2013 11:40 AM
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AJJewsbury

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Just to be awkward, can you think of any scenario where the Uo/Ia would actually be used, allowing 7666 ohms for 230v and 0.03 A?

To legitimately have a Zs higher than 1667 Ohms on a TN system you'd need to have extraordinarily long thin conductors - then to have acceptable voltage drop they would need to supply an extraordinarily small load. Back of an envelope reckoning suggests that using 1.5mm2 L + c.p.c. the cable would have to be around 69,000m long to have a R1+R2 of over 1667 Ohms - and then to keep the v.d. at 230V to below 5% - the load would have to be below around 6.9mA (i.e. about 1.5W). So theoretically, it's possible, but I'm struggling to think of a plausible situation for such a setup (power small electronics at the top of an offshore mountain?) - and suspect that the capacitance of that length of cable would produced enough earth leakage to defeat attempts to use 30mA protection anyway.

- Andy.
 03 September 2013 12:20 PM
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Parsley

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My comments where in relation to using the 1667 ohm figure instead of the maximum Zs values according to the tables in chapter 41.

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 03 September 2013 12:32 PM
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marclambert

Posts: 310
Joined: 23 June 2010

Originally posted by: AJJewsbury

Just to be awkward, can you think of any scenario where the Uo/Ia would actually be used, allowing 7666 ohms for 230v and 0.03 A?


To legitimately have a Zs higher than 1667 Ohms on a TN system you'd need to have extraordinarily long thin conductors - then to have acceptable voltage drop they would need to supply an extraordinarily small load. Back of an envelope reckoning suggests that using 1.5mm2 L + c.p.c. the cable would have to be around 69,000m long to have a R1+R2 of over 1667 Ohms - and then to keep the v.d. at 230V to below 5% - the load would have to be below around 6.9mA (i.e. about 1.5W). So theoretically, it's possible, but I'm struggling to think of a plausible situation for such a setup (power small electronics at the top of an offshore mountain?) - and suspect that the capacitance of that length of cable would produced enough earth leakage to defeat attempts to use 30mA protection anyway.



- Andy.


So that'll be a no then
 03 September 2013 12:42 PM
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geoffsd

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Originally posted by: marclambert

Just to be awkward, can you think of any scenario where the Uo/Ia would actually be used, allowing 7666 ohms for 230v and 0.03 A?


No, just because a theoretical mathematical limit (7666.6666?) results from a calculation using a very small current (30mA) does not mean the situation would ever occur.
You are never going to have to base a decision on whether a reading is 7665 or 7668.


It's comparable to 23k? (230/10mA) to determine if main bonding is required.
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