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Topic Title: Use of adiabatic
Topic Summary: What figures to use
Created On: 16 April 2014 07:02 PM
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 16 April 2014 07:02 PM
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Thripster

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This enquiry comes about because of a discussion being had on Diynot.com entitled 'Relying on loads not overloading' (or similar). It is an interesting discussion and well argued by JohnW and EFLI but I'm not sure we are reaching a conclusion so thought to ask here. If an MCB is providing short cct protection only what calculation would you undertake to give the minimum Ze at the MCB to ensure adiabatic protection of the CPC?

Coates and Jenkins advocate 0.1 Uo/kS, assuming a minimum disconnection time of 0.01 seconds. By the way, I am not seeking corroboration for my views - would just like to know one way or another. Some seem to say use 0.1s as the disconnection time - is there a definitive/correct way to approach it?
 16 April 2014 07:41 PM
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spinlondon

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Is this for design, or verification?
 16 April 2014 07:59 PM
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Thripster

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Hi Spin,

Design

Thanks
 16 April 2014 08:38 PM
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spinlondon

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I would always use the values for the maximum disconnection time for a particular circuit as per the requirements of BS7671.
Designing to the worst case scenario would to my mind be a safer option.
If however you have measured values to use in your design, I would always prefer to use the disconnection times that those values would provide.
Determining the disconnection times based on the instantaneous disconnection times for an MCB won't be much use if the PEFC is lower than that required to cause instantaneous disconnection.
 16 April 2014 09:14 PM
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Parsley

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Is the question actually what's the maximum fault current/lowest earth fault loop of the circuit that the CPC can withstand?
i.e. if the disconnection time is less than 0.01 secs (fault close to the origin of the circuit) will the energy let through of the protective device be less than the K2S2 of the cpc?.

Or have I read the post wrong?

Regards
 16 April 2014 09:50 PM
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Thripster

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Thank you folks.

My question is that Coates and Jenkins say that minimum Ze at MCB to protect CPC is given by 0.1 Uo/kS derived from standard adiabatic, substituting V/R for I and understanding that the square root of 0.01 (minimum disconnection time of MCB) is 0.1 seconds. Others are saying that standard formula with disconnection time of 0.1 second be used. Wondered what others do. Might be worth reading the thread in the other place to get the gist.

Regards
 16 April 2014 10:33 PM
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spinlondon

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If you use the disconnection time of 0.01 secs as opposed to 0.1 secs, you will arrive at a smaller CSA.
In most cases, it won't make much odds, as both methods will produce a smaller CSA than the Regulations allow.
Other problems with using the instantaneous disconnection time for an MCB, is that it is not transferable to fuses, or where the supply frequency differs.
 16 April 2014 10:50 PM
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Thripster

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Sorry - I do not follow. If using an MCB for fault protection only, then it is necessary to understand whether the CPC is protected by using the adiabatic equation. That is, for both minimum and maximum Zs values. Given that we are talking domestic installations (at the moment), if not by use of the adiabatic, how do you check that CPC is adequately sized?

Regards
 17 April 2014 09:03 AM
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John Peckham

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Thripster

The regulation that needs to be satisfied is 434.5.2. That says a fault occurring in any point in a circuit shall not exceed the limiting temperature of the conductor or cable.

That said most people and software, and the City and Guilds in the 2396, calculate the circuit impedance at the far end of the cable and use this to calculate the fault current. The using the re-arranged adiabatic equation ensure that the I squared t is less than k squared S squared. This takes no account of the energy let through for the particular manufacture's device which will limit the fault current. So if you have the manufacture's I squared t data you will end up with a smaller calculated CSA.

The adiabatic equation only works for disconnection times between 0.1s and 5s. Below 0.1s the fault current is not symmetrical, ie for the first 5 cycles, above 5s the cable will be radiating heat.

Coates and Jenkins advise after calculating the thermal withstand at the far end of the cable to check the near end withstand by determining the Zs at the near end does not exceed (0.1 x Uo) /kS.

I think most specifiers will put a requirement in their spec. to say the CSA of the CPC shall be the same CSA as the live conductors thus avoiding the need to verify the thermal withstand for the CPC.

-------------------------
John Peckham

http://www.astutetechnicalservices.co.uk/
 17 April 2014 12:24 PM
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Thripster

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Thank you John. Can you point me to the place where it is stated that the Adiabatic equation only works for disconnection times above 0.1 seconds ' thank you. Coates and Jenkins are using a re-arranged version of the Adiabatic equation to work out the minimum impedance at the MCB assuming a minimum MCB disconnection time of 0.01s - so not sure why the adiabatic is not applicable below 0.1s?

Coates and Jenkins provide worked examples - the order is:-

(a) Work out Zs
(b) Work out minimum acceptable loop impedance at the MCB
(c) Work out k²S² and I²t
(d) Work out fault current close to MCB and disconnection time at this point to ensure cable is protected along its length.

(Example 4.8, Fourth edition)

Edited to add more detail

Regards

Edited: 17 April 2014 at 12:35 PM by Thripster
 17 April 2014 01:22 PM
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John Peckham

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The less than 0.1s I quoted from memory. There is a reference to it in GN8 9.2.1. It says there "for short operating times of less than 0.1s, where the current asymmetry is significant e.g for a protective device immediately downstream of the source".If I have to prove what I said further I might have to dig through my text books or summon up Geoff Blackwell for the theory.

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John Peckham

http://www.astutetechnicalservices.co.uk/
 17 April 2014 01:50 PM
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GeoffBlackwell

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I can answer your question about the limits of the adiabatic equation but I have no time to get involved with the rest of it .

I am assuming you know what the adiabatic equation is and why it is used.

Lower limit
This is stated in 434.5.2 as 0.1 seconds

0.1 seconds = 100 ms = 5 cycles of the supply sine wave @ 50 hz.

Work on short circuits shows that, in general, the fault current will be asymmetric during the first five cycles. The waveform climbs up on to the zero reference and then decays back. Its peak value can be around twice its steady state value. This happens because you have just bashed the supply network with a short circuit and its getting angry

Now all this distortion in the sinusoidal waveform during the first five cycles makes the calculation of let through energy impossible. So as 434.5.2 requires we must ask the manufacturer for the data they obtain when testing devices for real. After five cycles calculation becomes reasonably accurate.

The upper limit is not quite so clear cut - it is as JP pointed out due to the fact that the cable is radiating heat. The rate of heat loss is not constant and this makes the maths a bit tough . The heat loss is reducing and ultimately halting the temperature rise of the cable.

The adiabatic equation takes no account of this and so, as time passes, it becomes increasingly inaccurate.

The 5 second figure is from the misty mountains. It was specifically stated in the 15th Edition but it was removed in the 16th . I sat on the committee wot done it! The reason was that, whilst disconnection within 5 seconds was, and still is a good idea, it has no scientific basis.

The 5 second disconnection time for shock protection comes from the same mountain

So the upper limit of 5 seconds is arbitrary but as good a limit as any.

Regards

Geoff Blackwell
 17 April 2014 04:19 PM
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Thripster

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Thank you both. Then what I need to do is get in touch with Coates and Jenkins (although I think I read somewhere that, sadly, one of them had passed away) and find out why they say what they say.

Regards
 17 April 2014 04:37 PM
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John Peckham

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Yes sadly Brina Jenkins is deceased. Perhaps Mark Coates reads this forum?

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John Peckham

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 17 April 2014 05:44 PM
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Thripster

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Have attempted contact and so will advise of any developments.

Regards
 17 April 2014 06:17 PM
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OMS

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Thank you John. Can you point me to the place where it is stated that the Adiabatic equation only works for disconnection times above 0.1 seconds


If you look at 543.1.3 and then at the definition of "I" then you will find reference to due account being taken of the current limiting effect of the circuit impedances and the limiting capacity (I2t) of that protective device"

As Geoff explained above, the assymetry of the fault isn't predictable below about 5 cycles or 100mS or 0.1 seconds.

MCB's as mechanical devices have moving parts that cannot operate faster than (approx) 0.01 seconds ie 10mS as defined in the product standards (such as BS EN 60898)

If you accept that (from the MCB's perspective) that there is no difference between a short circuit or an earth fault then looking back to 434.5.2 will show in para 2 that "for a fault of very short duration (less than 0.1 seconds)............................

Brian Jenkins "says what he says" becaue it's basically true and derives from a knowledge of the device characteristics and the application of BS 7671 in full to the case in point - plus, if memory serves me, the text is full of "assumed" and "approximately" when discussing close up faults as he is using an adiabatic expression technically in the zone between definite minimum time of 0.1 seconds and the 0.1 second boundary of asymmetric and symmetric fault current where it is not accurate - ie the assumption is based on the full prospective fault current rather than the attenuated fault current

Does that help clarify ?

Regards

OMS

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Failure is always an option
 17 April 2014 09:58 PM
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Thripster

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Thanks OMS - it does help to a large degree. A couple of points though:-

(a) in your answer above can I assume you mean 0.01s 'definite minimum time'?
(b) Brian does use 'assumption' in his text but (1) 'if basically true' (2) he shows worked examples as part of a design process (3) the answers produced as a result of using his formula 0.1 Uo/kS result in a far more 'usable' range of values between Zs min and Zs max

- is it safe me to use this formula?
(c) As he recommends energy let through calculations in compliance with BS7671 then I cannot think what harm can come from using the formula.

Thank you for your time.
 18 April 2014 07:50 PM
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spinlondon

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The minimum time is based on the time it takes for one half of a cycle.
Our electricity cycles 50 x a second, so it reaches maximum voltage 100 x a second.
Divide 1 second by 100, and you get the tripping time 0.01 seconds.

You still have the problem that by using the minimum time, you would be under sizing the CPC.
To my mind, unless you use real values, then you might as well not bother with the adiabatic equation.
Just use the rule of thumb table provided in BS7671.

Edited: 18 April 2014 at 07:58 PM by spinlondon
 19 April 2014 05:23 AM
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GeoffBlackwell

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If you expect that the CPD will operate in under 0.1 seconds you should comply with 434.5.2 paragraph 2. This tells you to either use quoted data or manufacturers data.

For BS 60898 mcbs you can use the information given in the standard in Annex ZA (normative) -
Classification of circuit-breakers into energy limiting classes

Typical examples being:

Table ZA .2 16A to 32A

Energy Limiting Class 3
...............................Type B....................Type C
6kA........................45000......................55000.....A2s
10kA......................90000....................110000....A2s

Specific manufacturers data is typically much less than that quoted in BS 60898.

The energy limiting class is the figure in the square box on the mcb.

Regards

Geoff Blackwell

Edited: 19 April 2014 at 05:44 AM by GeoffBlackwell
 19 April 2014 09:20 AM
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Thripster

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Thank you both.

Regards
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