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Topic Title: Maximum ZS Permitted By BS7671
Topic Summary: What is the correct value
Created On: 07 February 2015 09:00 AM
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 12 February 2015 10:49 AM
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Jaymack

Posts: 5377
Joined: 07 April 2004

Originally posted by: Jaymack
don't know the GS add on.

After thought - it's probably Gold (5%) and Silver (10%) for the tolerances but I also needed to develop a mnemonic for matching the numerals, for whether starting at 0 or 1?: I had a guy in the class, whose surname was Green, and remember allocating the number 5 to this.

Regards
 13 February 2015 12:55 PM
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mapj1

Posts: 9692
Joined: 22 July 2004

Big Ben rings out young Girls Buy Volkswagens (or violets), George Washington.

The one I may know, that I hope no one will print ends 'but virgins go without' and the first words are unsuitable for an open forum, and may just about belong in a rugby club song book.

We are of course discussing the number to colour code,
commonly seen as stripes on resistors.
Black 0
Brown 1
Red 2
Orange 3
yellow 4
Green 5
Blue 6
Violet 7
Grey 8
White 9

Never had a problem with black holes being 'nothing' to remember to count from zero.
Nowadays, bare (20% tol) and silver (10% tol) are actually unusual - even gold (5%) are pretty cheap.
Persojnally I tend to design in 1 % metal film by default these days, unless its a cost reduction job.

-------------------------
regards Mike
 13 February 2015 03:42 PM
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Spark68

Posts: 26
Joined: 20 April 2012

Gold and silver are used as multipliers too though on the lower values.

Eg1 Blue Grey Gold = 6.8 ohms
Eg2 Blue Grey Silver = 0.68 0hms
 13 April 2016 01:22 PM
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SpencerHenry7671

Posts: 7
Joined: 29 June 2015

Hi all. Here is my take on the op. It's based on a test with a similar question I have asked my self when testing.

I like to be methodical so some of this is sucking eggs I know.

I tested a circuit that comprised of a 4mm t&e with a 2.5mm Earth clipped direct. The circuit was supplied by a 32A type C 30mA RCBO. My measured Zs was 1.30 ohms.

As we know table 41.3 requires a Zs of no more than 0.68 (or 0.544 for 80%) and 411.4.9 allows the use of table 41.5 for the RCBO to BS EN 61009 for a max Zs of 1667 ohms. However the device must still provide overload current and fault current protection.

Overload current - as the cable is rated to 37A this is covered completing the calculation in appendix 4 shows compliance with Reg 433

Fault current - this is the bit I usually find the problems. And it's in Reg 434.5.2 K^2 x S^2 must be greater than I^2 x t

Therefore 115^2 x 2.5^2 = 82656.25 And I^2 x t = 110000 so the calculation is false and Reg 411.4.9 has not been satisfied for the use of table 41.5

Now just for fun we can also check this way. We can calculate the fault current

Uo x Cmin / Zs = 230 x 0.95 / 1.30 = 168.07A

From this we can calculate the time it will take for the cable to be destroyed using a variation of the adiabatic

t = K^2 x S^2 / I^2 = 115^2 x 2.5^2 / 168.07A = 2.93 seconds

If we look at appendix 3 Fig 3A5 we can see it will take approximately 4 seconds for the device to operate. Just a little to late.

-------------------------
Spencer Henry

http://www.spencerhenryelectrical.co.uk/
contact@spencerhenryelectrical.co.uk
https://uk.linkedin.com/in/spencerhenry
 13 April 2016 01:36 PM
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mapj1

Posts: 9692
Joined: 22 July 2004

ahahha. True but luckily over a time of a more than a few seconds, it's no longer adiabatic, as the heat is not all staying in the copper core only, some has soaked through the plastic on a cable that thin and is already being sweated off - so the copper cores will likely not reach the final temp you expect , and the cable may in practice just about survive to live another day, though it may pong a bit.
(and the time curves for the breaking time are worst case)


Also in this special case, what sort of fault heats the 2.5mm cpc but does not fire the RCD part of the RCBO near-instantly? An L-N fault will only heat the 4mm cores of course.

-------------------------
regards Mike
 13 April 2016 01:58 PM
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AJJewsbury

Posts: 16102
Joined: 13 August 2003

Uo x Cmin / Zs = 230 x 0.95 / 1.30 = 168.07A

Why Cmin? While it's required for ADS (411.4.5) - and I can see why you might want to be consistent, but regs wise I don't see any mention of it in section 434 when it comes to the protection of cables.

If we look at appendix 3 Fig 3A5 we can see it will take approximately 4 seconds for the device to operate.

OK, I'm not good with log scales, but my copy seems to go instantaneous below 5s. At around 160A (i.e. above the vertical section for 16A MCBs) the 32A curve looks to cross at between 10s and 20s to me - well beyond adiabatic usefulness.

As Mike says, with an RCBO, the disconnection time will likey be 40ms max (table 3A of appendix 3), given the earth fault current well exceeds 150mA.

- Andy.
 13 April 2016 02:24 PM
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SpencerHenry7671

Posts: 7
Joined: 29 June 2015

I could also of used the mesured voltage instead of nominal voltage but I just followed the method I use. There isn't much difference

As for. The bit your quoting on its just another angle. The key point is And it's in Reg 434.5.2 K^2 x S^2 must be greater than I^2 x t

-------------------------
Spencer Henry

http://www.spencerhenryelectrical.co.uk/
contact@spencerhenryelectrical.co.uk
https://uk.linkedin.com/in/spencerhenry
 13 April 2016 03:23 PM
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leckie

Posts: 4439
Joined: 21 November 2008

Blimey, this thread is a blast from the past!

I forgot how much I have missed Phantom!!
 13 April 2016 05:16 PM
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AJJewsbury

Posts: 16102
Joined: 13 August 2003

The key point is And it's in Reg 434.5.2 K^2 x S^2 must be greater than I^2 x t

Which is will be - easily - when you use the correct 40ms disconnection time for an RCBO.
I²t = 1252 A²s.
k²S² = 82 656 A²s.

Although at this point you should notice that the worst case is not at the far end of the circuit (max Zs) but at the supply end - as I will increase but t remains at 40ms. (It's similarly true for most MCBs - energy let-through increases with fault current - unlike most fuses where energy let-though tends to be flat or decreases with fault current).

If your 1.3 Ohms was the result of an extraordinarly long final circuit and Zs at the start of the circuit (Zdb) was say 0.1 Ohms then fault current would be about 2300A and so the energy-let through on the RCD element would be around 211 600 A²s. By then of course the MCB element should have taken over - I'll leave it as an excercise to work out the cross-over point and wherther your 2.5mm² c.p.c. is protected there.

- Andy.
 13 April 2016 06:15 PM
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Legh

Posts: 4064
Joined: 17 December 2004

Well that was a stimulating and entertaining thread

Jolly good chaps keep up the good work !

Legh

-------------------------

http://www.leghrichardson.co.uk

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IET » Wiring and the regulations » Maximum ZS Permitted By BS7671

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