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Topic Title: Adiabatic equation
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Created On: 25 January 2019 06:55 PM
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 25 January 2019 06:55 PM
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Bewlec

Posts: 34
Joined: 11 June 2017

This will be a quick one for most of you, but trying to work out the minimum acceptable size of the main earthing conductor.
TNS supply protected by 60Amp fuse ( Not Verified as sealed ), measured pfc of 1.80kA.
At present it is connected onto the lead sheath via what looks to be an approx 2.5mm bare conductor.
I calculate this at 3.98mm, but not sure if my sums are correct, any advise is appreciated. Thanks

Edited: 25 January 2019 at 07:42 PM by Bewlec
 25 January 2019 08:37 PM
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Zoomup

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Is the earthing conductor tinned copper and stranded and soldered onto the lead sheath?

Z.
 25 January 2019 08:46 PM
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Bewlec

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Definitely soldered onto sheath, but not 100% sure if it's tinned copper or not
 25 January 2019 08:57 PM
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Zoomup

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Taking t as 0.01s, and k as 159 (table 54.6) I make S at just over 1.1mm2, which seems far too small. I will be interested in finding the correct answer though. The final temperature of the copper conductor will be 200 degrees though.

Z.

Edit bad arithmetic.

Edited: 25 January 2019 at 09:04 PM by Zoomup
 25 January 2019 09:06 PM
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Zoomup

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I am basing t on 0.01s with a B.S. 1361 60 Amp fuse.

Z.
 25 January 2019 09:09 PM
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Zoomup

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Of course it may well be a TN-C-S supply in the road and just presenting as a TN-S supply in the premises. In which case table 54.7 might best be used.

Z.
 25 January 2019 09:25 PM
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Bewlec

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I have used table 54.2. Bare conductor in contact with cable covering as soldered onto sheath but cable has a wrap of some sort around it. Also I used 0.1 seconds as this is the lowest available on the curve characteristics for 1361 60amp. Like mentioned I'm not sure these are the correct entries to use.
 25 January 2019 10:15 PM
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UKPN

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This is a DNO situation.16mmsq reqd.

Regards, UKPN.

Edited: 25 January 2019 at 10:42 PM by UKPN
 25 January 2019 10:22 PM
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chrispearson

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Any chance of a photo?
 26 January 2019 12:10 AM
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mapj1

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The modern way is a flat braid and a constant force spring, as hot works are now considered dangerous. If it needs replacing its really a call to the DNO.
But the Adiabatic eqn., well lets assume we start at 30c and the lead tin solder is soft solder and melts at 200c
(235 is a good no to remember, both for uranium (atomic mass) and for most solders being soft, including the higher lead fraction ones, and the modern lead free stuff. we had a tech who used to set all the soldering tools for what he called a "uranium run" when we switched to lead free in the mid 2000s.)
where the '2' denotes a square

I2t<=K2S2

S is the minimum protective conductor cross-sectional area (mm2)
I is the fault current (A)
t is the opening time of the protective device (s) now that is tricky - the curves go to 0.1 sec normally, and then it is "see maker's data"
page 26 here suggests 900A and 0.01 seconds for a 60A fuse.
if it got no faster, so 0.01s* 1800A*1800A
k is a factor depending on the conductor material and insulation, and the initial and maximum insulation temperatures. take k= 160
for 30 to 200C

page 26 here suggests 900A and 0.01 seconds for a 60A fuse.
if it got no faster, so 0.01s* 1800A*1800A, so I2t is 32400 amps squared per second.
So S= root of that divided by k 2

32400 /(160*160) = 1.22 mm2, so S= 1.125mm.

But really you need more info, if anyone on the substation has PME, as there may be unfused and therefore very long duration currents between plumbing and the CPC, that are not coming via your company fuse over which you have little control.These diverted neutral currents can overheat inadequate earthing arrangements.

-------------------------
regards Mike
 26 January 2019 08:41 AM
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Zoomup

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UKPN is correct. Please see above.

Z.
 26 January 2019 08:53 AM
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UKPN

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Thank you Zoomup, have a good weekend.

Regards, UKPN.
 26 January 2019 03:14 PM
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Timeserved

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So S= root of that divided by k 2 

543.1.3 shows that it's divided k and not k squared.

But I would use table 54.7 in this case.

I normally only use the adiabatic equation if I'm calculating submain CPC size.

Either way I would welcome someone to confirm if I'm correct or if it's a typo in the big blue book.

Cheers
Ts
 26 January 2019 03:41 PM
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Zoomup

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543.1.3

Multiply I squared by t, then find the square root of that number. Then divide that answer by k. That will give you S.

Z.

Edit. Corrected for bad eyes' error.

(Don't get confused with I2t<=k2S2)

Edited: 26 January 2019 at 03:55 PM by Zoomup
 27 January 2019 01:29 AM
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Timeserved

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Thanks Z, I'm happy I'm doing this correctly. ??
 27 January 2019 08:11 AM
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lyledunn

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It would be inappropriate to have the earthing conductor selected using a method which depends on another authority. The situation is one where the requirements of the DNO must be met.

-------------------------
Regards,

Lyle Dunn
 27 January 2019 03:18 PM
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alanblaby

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Originally posted by: lyledunn

It would be inappropriate to have the earthing conductor selected using a method which depends on another authority. The situation is one where the requirements of the DNO must be met.


Yes, but in the case of the OP, he thinks the DNO soldered connection to the sheath is too small. In that case he cannot do anything without DNO involvement, who will probably not do anything about it. (Using the equation, shows it may well be fine anyway)

However, I think our responsibility starts (our earthing conductor) at the connection joint to this soldered conductor from the sheath, and from that point we need our 10/16/25mm earth conductor cable to our installation.
 27 January 2019 04:13 PM
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Bewlec

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The reason I am asking if the equation is correct is because this forms part of a domestic condition report and therefore I need to confirm that the earthing conductor is of an acceptable size and code it accordingly.
I know that if undersized, it is a matter for the DNO and I would not attempt to up grade the conductor myself.
Thanks.
 27 January 2019 08:49 PM
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UKPN

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This is a SNE earthing facility which is available for the consumer if he chooses to use it. The DNO have no responsibility that is the consumers. Probably a 60/70s sheath, with the old 7.029 cable, around 20 amp. One second rating, 275 amp. Best to sleeve the bare cable, fit an earth block if not already, give us a ring and we will deal with it. We may PME the service, so bonding must be to the ESQCR regs. Hope this helps.

Regards, UKPN.
 28 January 2019 09:55 AM
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chrispearson

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Originally posted by: Bewlec

I calculate this at 3.98mm, but not sure if my sums are correct, any advise is appreciated. Thanks


Is there only one strand? If a piece of copper wire has been wrapped around the sheath, there ought to be at least two ends.
IET » Wiring and the regulations » Adiabatic equation

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