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Topic Title: Understanding the adiabatic equation
Topic Summary: Calculating minimum CPC size
Created On: 12 May 2016 06:06 PM
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 14 May 2016 05:04 PM
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rubberlegs15

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Thanks Andy, but how do I use the adiabatic equation in this instance? Apparently this is what the examiners are expecting!
 14 May 2016 10:30 PM
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AJJewsbury

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I'd calculate k²S² and compare it with the I²t data you have for the device for the conditions you have. (Although I can't vouch for your examiners! JP or others will know better than me about them.)
- Andy.
 16 May 2016 06:00 PM
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John Peckham

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If this is for the 2396 exam you won't have manufacturers data with you in the exam room as you are only allowed BS7671, GN3 and the OSG to do your calculations.

The typical exam question is to do a cable calc. showing all workings from the load to selecting the actual cable. The last step being to check the thermal with stand for the cable. I teach my students to check the thermal withstand by ensuring the calculated I squared t is less than K squared S squared.

That is for the exam, meanwhile in the real world for below 0.1s disconnection times you would obtain the manufactures energy let through data for the particular device. See Regulation 434.5.2 that explains this.

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

http://www.astutetechnicalservices.co.uk/
 16 May 2016 06:24 PM
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OMS

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Do you base it on 0.1 seconds John

Or on a shorter time not less than minimum time to operate as we know (in this case) that the fault current is significantly greater than the device instantaneous operating threshold

Just curious really as to what the current thinking (or teaching) is

Regards

OMS

-------------------------
Let the wind blow you, across a big floor.
 16 May 2016 06:55 PM
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John Peckham

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OMS

For the exam to use 0.1 to 5s and the exam question will fal within this range and the CPC will be to small requiring an increase which will change the fault current.

For the project the student is expected to select their chosen circuit protection and justify their selection. They have to produce their own spreadsheet for the cable calcs. They are required to include manufacturers information in their project file.

The thermal withstand concept and calculation is always the hardest part to teach and get the students to grasp and do. I use the filling the bath analogy with plug in with the lip of the bath being the 5 second limit for the cable.

I can get the students through the exam but getting them to do their projects is like trying to nail jelly to the ceiling!

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

http://www.astutetechnicalservices.co.uk/
 16 May 2016 07:49 PM
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OMS

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OK - makes sense for the exam conditions

Personally, I always do the CPC check as part of estimating Zs - as (as you say)if it fails then your Zs value is out as well if you change the CPC size

Thanks

OMS

-------------------------
Let the wind blow you, across a big floor.
 16 May 2016 09:26 PM
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rubberlegs15

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Thanks Andy. The information in the question does not specify any particular brand of 60898 device! The papers have been submitted anyhow but it was more to quell my curiosity. The crux of the question being to 'prove' a particular size of CPC is adequate by using the adiabatic equation. It presents no problem where the prospective fault current is below the 'instantaneous' disconnection time but of course checking against the manufacturers data is also impossible with the information given. I'm starting to find there are several such issues with BS7671 where information is contradictory or at least difficult to make sense of. Sigh. . .
 16 May 2016 09:58 PM
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AJJewsbury

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I'm starting to find there are several such issues with BS7671 where information is contradictory or at least difficult to make sense of. Sigh

Indeed - and that's mainly down to history - if you were to look back a few editions of the wiring regs you'd find that there was a lot less calculation involved (no computerised test meters or pocket calculators in those days, let alone spread sheets) and the rules were a lot simpler - e.g. for fault protection the rule was simply that the protective device rating couldn't be more than 3 times the cable rating. And it provded perfectly adequate in practice for many years and many things became "accepted practice" as a result (like the reduced c.p.c. sizes in T&E cables). Now we're a bit more modern and want to calculate things properly (or at least less improperly) things don't necessarily quite meet up. Add to that the fact that most of the MCB manufacturers now international corporations, the UK is a relatively small part of their market and so traditional UK practice isn't necessarily top of their list when it comes to agreeing standards for protective device performance. For example hardly anyone else in the world would use anything less than a 4mm2 c.p.c. on a 32A MCB (by virtue of only having radials and not permitting reduced c.p.c.s on small circuits) - but by using rings and T&E we hope a 1.5mm2 is sufficient.

You'll find that there are many little things that the regs don't quite cover quite as thoroughly as you might have imagined, but there's usually a good reason.

- Andy.
 22 May 2016 09:49 AM
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rubberlegs15

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Thanks John, Andy and OMS.

In the example we were asked to prove the CPC was adequate by calculation (using the adiabatic equation) yet the PFC is greater than that which would cause the BS60898 device to trip 'immediately' so by definition (refer to the manufacturers let through data) the adiabatic equation cannot be used in this case yet apparently this is what the examiner expects us to use! I don't think the question can be answered in the expected way as a result so still don't know how we should have approached this. The project is complete now so changing the answer s no longer possible, its merely something I would like to understand. How do you approach the answer (using the adiabatic equation) or is this simply not possible (an errant question)?

We did a couple of examples using PFC's that were within the disconnection time caused by the thermal part of the device and these work fine at proving the CPC size. The disconnection time in these calculations is > 5s in some examples which raised another interesting question - if a disconnection time of <5s is required for the circuit, how can the device be considered suitable for the circuit? In other words, if the PFC is low enough, a disconnection time of over 5s would occur which would not satisfy the requirement for the device to trip in under 5s (>32A) or 0.4s (<32A). I'm guessing there is something here I haven't quite grasped but I don't know what?!!

In a nutshell, it seems to me that a PFC high enough to cause a disconnection time of under 0.4s (in our example where the device is <32A) removes the ability to prove the CPC size using the adiabatic equation but to add further confusion (where Zs is high enough) it seems that the device would not trip within the required time of 0.4s! Does this mean that in such instances the device is inadequate? Confusion levels rising on this one for sure. It still troubles me that on a TNCS systems with a maximum earth loop impedance of 0.35 ohms a fault could happen to any cable leaving the board (i.e at a point where the Zs is no greater than 0.35 ohms - especially where the Ze is found to be lower) and in such cases that PFC is always greater than that required to 'instantly' cause disconnection of a BS60898 device. Does this not mean that the manufacturers let through data is almost always going to be required to demonstrate that a BS60898 is suitable? Anybody for example living at such a property could easily stick a nail through a cable leaving the back of a cable leaving a CCU while hanging a picture on the wall for example!
 22 May 2016 11:08 AM
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John Peckham

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The adiabatic equation only applies between 0.1s and 5s.

Below 0.1s you need to use the energy let through data from the manufacturer for the actual device in use or for generic circuit breaker the energy let through information for the class of device from the British Standard. The energy let through class is the number sometimes written in the small box below the box containing the fault current rating on some circuit breakers. The circuit breaker or fuse will limit the energy let through and if you look at manufacturers time/ current curves below 0.1st he graph is not a straight line.

Above 5s the cable will start to radiate heat (it will do below 5s but that gets discounted) so is not considered to be adiabatic as the cable is not absorbing all the heat.

The fault current will determine the speed of disconnection, the higher the fault current the faster the operation of the device the limiting factor will be the speed of the mechanical movement of circuit breaker contacts. In addition the arc between the contacts has to be extinguished as energy will continue to flow as the arc is a short circuit.

The fault current in a circuit is limited by the circuit impedance Zs. The current causing operation of the device in a given disconnection time is Ia. The maximum Zs for a given device can be found from Zs = Uo/Ia x Cmin. Cmin is given in BS7671 as 0.95 this is to allow for fluctuations in the supply voltage.

So under fault conditions current will flow in the circuit conductors until the device operates and the arc is extinguished. That excess current flowing for a period may damage the cable if it exceeds the thermal withstand of the cable.

The cable needs protection throughout it's length but usually a fault occurring at the far end of the cable will be the most onerous as the circuit impedance will attenuate the fault current and the cable will be subject to a longer period of heating.

If Isquaredt is equal to or less than Ksquared Ssquared the thermal withstand of the cable is satisfied.

In your example of the fault current not being high enough to operate the device in 0.4s due to a high value of Zs you can use the time current curves in BS7671 to determine the actual disconnection time. Knowing this time you can re-arrange the adiabatic equation to determine the actual time that would cause damage to the cable from t = Ksquared x S squared / I squared. Then compare your calculated time from the time current curves to the calculated time from the re-arranged adiabatic.

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

http://www.astutetechnicalservices.co.uk/
 22 May 2016 12:30 PM
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tillie

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Hi Rubberlegs , have you seen my post " Disconnection in less than 0.1 secs ".

I think we are asking basically the same question.

This is also something that has always troubled me but as OMS has pointed out it is really quite simple.

Compare the K2S2 with the data provided by the manufacturer or use the worst case values provided by the BS 60898 standard.

Use the figure of 0.1 first and if this fails then use the K2S2 > I2t.

I hope this post comes through ok , my computer downloaded Windows 10 yesterday and it is now really playing up.

Regards
 14 April 2018 10:11 PM
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dantheman22002

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Hi everyone,

I have been reading this thread and others plus the responses in the thread i created.

But still having problems with energy let through from manufacturers.

I have selected a schneider 6A Type B mcb ic60h. The fault current at the extremity of the circuit is 700A and at the DB (i havent got the calculation in front of me) but i think was 2000A ish.

Can someone please explain to me how i get the energy let through from the manufacturers data sheet.

Thank you in advance.

Dan
 15 April 2018 10:14 AM
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Jaymack

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Originally posted by: John Peckham
The adiabatic equation only applies between 0.1s and 5s.

To be pedantic: it's not really adiabatic in the sense, that it is one that occurs without transfer of heat or matter, between a thermodynamic system and its surroundings. (From my steam days).

Regards
 15 April 2018 10:45 AM
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AJJewsbury

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Can someone please explain to me how i get the energy let through from the manufacturers data sheet.

I don't have the Schneider data to hand, so I'll use the values from BS EN 60898 itself (the manufacturer's own data can't be worse and the process should be the same anyway).

For a device rated <= 16A, of energy limiting class 3, B-type and a fault current(*) <= 3,000A (which covers your 2000A) the energy let-through value is stated as 15,000 A²s.

So we're looking for k²S² >= 15,000 A²s

You haven't told me what side c.p.c. you have in mind, so I'll work backwards to calculate the minimum. If you had, you could directly calculate k²S² and see if it's at least 15,000A²s

So I'll take k²S² >= 15,000 A²s, presume a value for k and then solve for S...

or S² >= 15,000/k²

or S >= SQRT ( 15,000/ k² )

if k = 115 then we've got

S >= SQRT ( 15,000 / 13,225 )

S >= SQRT ( 1.134 )

S >= 1.06mm²

NOTE (*) - BS EN 60898 tabulates against the device's rated short-circuit capacity rather than the fault current - but I've followed the IET's example by treating that as equivalent to the fault current.

- Andy.
 15 April 2018 11:50 AM
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mapj1

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I think Schneider don't publish a curve li the way that wylex do, rather they say 'meets or exceeds' the EN standard.
That K factor assumes the cable is PVC and already at 70C when the fault comes on, and the insulation nearest it is OK to flash up to no more than 160 degrees for the duration of the disconnection - the whole 'adiabatic' thing assumes that the time it takes is so short that no cooling of the wire occurs. After the power is off, it will cool in some seconds/minutes depending on the arrangement.

If either of the assumptions is not true - for example if it is a CPC it will not start at 70C before the fault comes on, so a lower starting temp may be in order, or if its not PVC, so the maximum temperature is higher (we may allow the middle of a bare bus bar to reach a few hundred if there is nothing other than more metal in contact with it for example) then the method is the same, but the answer is different.
If you assumed a 50 degree starting temp, instead of 70, then instead of a 90 degree rise being the most it could stand, then 110, and then you may decide that a 1mmsq cpc would be OK on that breaker, even for a dead short fault.

So actually it is just fine on a 6A 1mm lighting circuit as that runs cold anyway, and up to 16A breaker on 1.5mm T and E with 1mm core cpc is also fine unless its very stressed.
If the result was anything else we'd be worried

Note that the only common case for k being less than 115 is if it is high temp. PVC, so it starts at say 85 degrees, and still should not exceed 160 during the 10msec of the fault.

For that hot start case, assume 104. (or call it 100 if like me you sometimes like to do quick sanity check mental sums as you walk round, and only bother to go back and do the proper sums for the ones that look a bit close to the limit.)

-------------------------
regards Mike
 15 April 2018 03:56 PM
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dantheman22002

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 15 April 2018 04:47 PM
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mapj1

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Hi Dan, that is the same wylex data that I was alluding to in a previous posr, that as far as I can tell Schneider don't produce in that form.

page 64 top right hand side graph, is the one of I²t going vertically, and fault current horizontally.

There for your 2kA fault you could run along the X axis to 2000A, which is almost graph central, and then look up until it crosses the lne for the breaker you have, example, you see the 6a MCB is about 7000A²s, so quite a bit faster than the minimum standard example that Andy calculated.

The horizontal dotted lines are fuses ; I²t is constant regardless of fault magnitude. The curves are various ratings of breaker they do.
the diagonal line HS is a line that would be a truly constant opening time of 0.01 seconds. All the real breakers are considerably faster than this, but note how the higher rated breakers are rather slower than the low current ones, and the lighter rated ones are curve away more from the constant time slope at large fault - this implies they are still getting faster at the higher currents, where as the heaver models are closer to a constant speed.
.
As it happens the published curves for the wylex 10kA breakers look identical to the 6kA ones, and no distinction for C type or B type.

-------------------------
regards Mike


Edited: 15 April 2018 at 05:15 PM by mapj1
 15 April 2018 08:37 PM
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dantheman22002

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Thanks mike.

Schneider have a similar graph. Its along way down in the data sheets

but I think the reason i am struggling with this is.

You calculate the cpc using the adiabatic equation with a fault current at the extremity of the circuit if the disconnection time is greater than 0.1 sec.

But if the disconnection time is less than 0.1 second
you use the adiabatic equation with let throught energy from the manufacturers charts for this you have to use the fault current at the origin of the circuit ZDB to give you your I2t

Is that correct.

Dan
 16 April 2018 12:06 AM
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mapj1

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sort of - of course in reality we need to protect against dead short faults if they occur at the near end or far end of the circuit, and if we are not sure we could calculate both and use the one that breaks one regs limit or another to determine the circuit's limit.

For a fuse, that is normally the far end, as the blow time is longest, and the let through energy is the same regardless of higher or lower fault impedance. It is also the far end where we need to think about voltage drop - though if you have problems clearing ADS, you may have volt drop problems too.

For a circuit breaker, the far end is still the place for voltage drop sums, and we need to be sure that we can break the ADS in a respectable time but actually it is any fault nearer the origin end that stresses the cable most, as the let though energy rises with higher fault current.

Part of the problem is that a lot of the text books were written in the fuse era, or at least by folk who grew up in it, and the non-constant I²t is not well explained.

Oh, and note 0.1 second is very slow, for a near end fault breaking time, and will almost always lead to calculations that oversize the cable.

-------------------------
regards Mike
IET » Wiring and the regulations » Understanding the adiabatic equation

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