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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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 12 May 2016 06:06 PM
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rubberlegs15

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Hi all, hopefully somebody can shed some light on this problem even our tutor hasn't been able to answer?!

Prospective fault current is used in the adiabatic equation to check the selected CPC size for a circuit is adequate (returns a minimum size required) such that:

s = Square root(I x I x t) / k

where s is the size of CPC required in mm squared
I is the prospective fault current
t is the disconnection time and k a value from another table in the regs.

PFC is calculated by the formula PFC=U(0)/Zs

Where U(0) is the nominal voltage and Zs is well, the Zs!

In a calculation not out of the ordinary say for a single light wired from the board on a 6A MCB (BS60898) the Zs is 0.35 ohms and nominal voltage is 230v, the PFC comes in at 657Amps. Dropping this into the equation gives a CPC size of 1.8mm squared (bigger than the phase conductor of 1mm squared!) and using a disconnection time of 0.1 seconds (the time the graphs for a 6A MCB show - i.e the bit where the line goes flat). K is taken as 115 from another table for thermoplastic copper.

Hopefully enough details here to get the gist of my question? We tried the equation with other low Zs values but have the same problem where a CPC size greater than the phase size is returned.

Can anybody shed any light on where this calculation is going wrong? Should t be the required disconnection time (i.e. 0.4 seconds for a circuit <32A)? This would give an even greater calculated size for the CPC! Should the PFC be ignored and the current required to give instantaneous disconnection used (where the line goes flat on the log graph for BS60898 type B 6A MCB's)?

The regs are very unclear on this usage of high PFC's. Any help greatly appreciated!
 12 May 2016 07:05 PM
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alanblaby

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This is where 543.1.1 (ii), referencing 543.1.4 comes into effect, which says (basically) for a lIne conductor of less than 16mm, then a CPC of the same size is acceptable.
 12 May 2016 07:50 PM
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Mick01

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Assuming its a type B Mcb it will only need about 30A to trip in 0.1 sec. As you have 657A PEFC you are way above this so tripping time will be quicker than 0.1sec, because of this you will need to consult the energy let through value (A2s) from the manufacturer of your particular MCB, then check that your cpc is thermally protected by using A2s<K2S2. However as stated by alanblaby if you can use table 54.7 then this is not necessary as you are deemed to comply.

Edit -spelling

Edited: 12 May 2016 at 09:27 PM by Mick01
 12 May 2016 09:00 PM
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alanblaby

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Then we have 1.5 T+E, with a 1.5mm LIne, and 1mm CPC.
How many people check that it is suitable?
 12 May 2016 10:29 PM
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rubberlegs15

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Hi alanblaby, I've seen that reg and the table that shows the different sizes for different conductors. The table is actually a safe rule of thumb guide that is always adequate. The equation should work for all values but somewhere along the lines BS7671 fails to specify these parameters. For a fault current that trips a device on the main curve I'm sure it always works (i.e a high enough Zs). The high Zs (low PFC) means the adiabatic equation returns a smaller size. The use of the equation can prove that a smaller CPC can sometimes be used (if proven adequate by calculation from the adiabatic equation) - reg 514.1.4 provides a 'catch all' situation (more of a safe minimum without need for calculation).

My question centres more around why the equation does not work correctly. I suspect that it is not correct to use the CPC or 0.1 seconds there but I don't have an explanation why this is!

Hi Mick01,

A type B mcb will not trip with a 30mA current unless protected by an RCD (i.e earth leakage) though this is down to the RCD itself tripping via an unbalanced load and return between line and neutral (earth leakage) - the same would apply to type C's and D's. I'm interested in why the equation fails with the figures I am using. I'm fairly confident it's my usage of the numbers concerned that is the problem but not which. K is 115 almost certainly but t is questionable. The graphs dark black line for each MCB stops at 0.1 seconds. I presume this it the fastest physical time (based on the movement of the spring loaded switch) that a typical MCB can remove contact from the circuit (a physical attribute of the MCB) and this is why I've used the figure in the adiabatic equation. Still puzzled as to why this does not calculate an expected minimum CPC size such as 0.5mm squared.

Thanks for the responses
 12 May 2016 10:33 PM
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Mick01

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30A not mA. 0.1 is not the quickest it's just the quickest listed in app 3.
0.01 is the quickest, however once you get below 0.1s you need energy let through data.

Edited: 12 May 2016 at 10:44 PM by Mick01
 12 May 2016 10:34 PM
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rubberlegs15

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"This is where 543.1.1 (ii), referencing 543.1.4 comes into effect, which says (basically) for a lIne conductor of less than 16mm, then a CPC of the same size is acceptable. "

I agree with this point btw. It makes no sense that a CPC should ever be bigger than the line conductor as the problem occurs with the heat generated in the cables. If the CPC were of greater diameter the line conductor (under fault) would heat more (Be the primary cause of the fire over the CPC!). This is why I am questioning the figures in the equation. Cheers
 12 May 2016 10:38 PM
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rubberlegs15

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"30A not mA "

Without digging out the reg book I'm not sure but it's certainly lower than the PFC! Should trip immediately but if you put 0 in as t into the equation then you need 0mm x-section diameter for the CPC! I thin this is the physical nature of the MCB's. The thermal trip built inside wont immediately reach temperature under a specific fault current - perhaps 0.1 seconds (ish) and the reason why the curves stop at that point?
 12 May 2016 10:47 PM
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Mick01

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At that sort of current it's the magnetic part your interested in. Read my edited post above that should help with the time issue.
 12 May 2016 11:45 PM
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mapj1

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It is important to think about the physics - the adiabatic assumption (== no cooling) is only suitable for very fast events, where energy is dumped into the wire (I.V.t or if you prefer I.I.R.t ) so quickly it has not time to escape out to the surface of the plastic - you need to be at the kilo amp end of the sums.
If the supply impedance is too high, then the breaker does not open anywhere near instantly and you have a problem - a 30A fault is far more serious for a 1mm cable, than a 3000A fault, as the latter will be off in about half a cycle, the former may take long enough on a 10a breaker to toast the wire..

with your 650A fault curretn you are getting near the sticky end , but I dont see a big problem - 650 squared is 420,000, times say 0.01 seconds - 4,200

typical datasheet for MCB here
sqrt is 66, and so Smin is ~ 0.21mmsq.

to reach 1mm, you need a really slowMCB - nearly ten cycles!! or a fault curretn of almost 5 squared times higher = 4000 amps perhaps. By then you are racing the company fuse and may not care if the wire gets a little cooked.
Also your fault now has to be surprisingly close to the consumer unit- 1m of 1 core of 1mmsq cable is ~ 16-18 milliohms, so more than a couple of metres away from the CU , a 4000A fault current is simply not possible on 230Veven if you are right next to the substation.

Note that a 10msec disconnect for an MCB is one half cycle of mains- which assumes the worst case of the fault at the start of the half cycle and the arc not going out until the next zero crossing.
For fuses, you need to look at the real curves.

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


Edited: 12 May 2016 at 11:55 PM by mapj1
 13 May 2016 06:17 AM
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rubberlegs15

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Hi all, thanks for the answers - I've not heard of the energy let through data before?

My question arises from a college assignment where we are asked to prove the CPC size is adequate to protect the circuit. It's actually a very poorly written question as we are given no value for the Ze (so actually impossible to make the calculation!). Without the Ze it's impossible to calculate the Zs and its corresponding PFC. The earthing system is quoted as as TNCS so a maximum impedance of 0.35 makes for the lowest PFC although it could be as low as say 0.01 ohms too which would give a high PFC. There are no specific manufacturers MCB's specified so referral to a specific manufacturers data is not possible either.

I'm interested in how you show this by calculation? The circuit in question is specified as being 8m long and we have already ascertained that 1mm copper is adequate to carry the design current for the circuit. The problem comes with doing the calculation. The graphs in BS7671 for 60898 breakers shows the line stops at 0.1 seconds hence the attempt to use this as a value for t. This is where things get tricky! Should the time in the equation be the required disconnection time (i.e 0.4 seconds), the fastest time shown by the curves in BS7671 (0.1 seconds) or some other value? It's clear to see the device would trip before 0.4 seconds with high PFC's so knowing what value for t to use us questionable (should it be the required disconnection time of 0.4 seconds or the lowest time shown of 0.1 seconds). The same problem exists with the I squared part of the equation - should I be taken as the calculated PFC or should some other lower value be used (for example the current required to trip the MCB at the flat line part of the curve (presumably where the magnetic part of the MCB comes into play)).

I've found examples of the adiabatic equation in use for higher values of Zs where it works as expected but not for low Zs values which is where knowing what value to use for t and PFC become blurred. A 0.35 ohm value for Zs on a short lighting circuit of 8m for example seems quite realistic (adding a low Ze to calculated R1+R2 values for a 1mm sq twin and earth cable for example). It's where these values for calculations using the adiabatic equation are poorly explained rather than for lower fault currents with higher readings for Zs.

In a nutshell I'm looking for answers of what value for t should be used and whether the calculated value of PFC or the current required to trip the magnetic part of the MCB should be used in its place. I can find no reference to this in either BS7671 or the OSG! Thanks
 13 May 2016 09:06 AM
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mapj1

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My copy of the regs has the note
For prospective fault currents in excess of those providing instantaneous operation refer to the manufacturer's let-through energy data.

under the graphs you are referring to. That is what BS7671 expects you to do. If your exam question did not lead you this way, then I agree it is badly worded. (and by instantaneous they don't mean that, they mean the 100 milliseconds, which to me in my role as an electronics guy is long enough to power something up, do a job involving a million clock cycles and shut down again.)

Any MCB meeting or exceeding the requirement of BS 60898 will have a vertical line down to ~10msec, but around there in most designs of MCB the switch off time (or the effective switch off time - as the arc impedance is none constant during opening ) flattens off - more fault current does not make it go much faster, it just moves as fast as the physical design permits.




In a sense the let through energy is not constant, unlike a fuse whcih melts faster with more heat, as obviously if time is constant, as fault current rises, so does energy, as the square of it in fact. So a single I2t number for an MCB is misleading - a maximum let through , at 6kA or 10kA or whatever the breaker is rated for, is often quoted, but really energy versus PFC is a curve.
Its a bit unhelpful of BS7671 not to point this out more clearly - as you say, if you assume 0.1seconds, your cables all end up 10 times bigger than you need.

Actually with higher energy things, where the breakers are much bigger, and the currents are large enough to seriously pull the waveform off sinusoidal, the disconnection times are longer (5 seconds anyone), and the mains industrial things tend to be very much designed as energy trips, plenty of older ones still in service looking like a cross between a traditional spinning disk electricity meter and a car speedometer - the modern equivalent does an integration in electronics, but is still tripping based on a kVA .time product.

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


Edited: 13 May 2016 at 09:22 AM by mapj1
 13 May 2016 05:01 PM
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OMS

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OK - by convention, we are in the region of definite minimum time - taken as 0.01 seconds for a BS EN 60898 device

That's the fastest the CB is ever going to open due to all the mechanical junk we need to make move

The worst that it's going to perform at is 0.1 seconds for fault currents exceed the stated threshold (see Mike's opening comment above)

In the region between 0.1 seconds and 0.01 seconds, we can't use graphic data - we need to know the total energy let through ie the Ampere squared seconds or I2t (for comparison with the cable energy acceptance (the K2S2))

For the larger stuff Mike is talking about above, we have the concept of Inverse definite minimum time - so more fault current (via the CT's forces faster operation - but only up to a point - there is a physical speed limit due to moving "junk" and also a practical delay (as we usually want to achieve discrimination with downstream devices

The inverse bit simply tells you that for more current, it gets faster

We confuse things with inverse and extreme inverse settings and a whole load of points in between - right up to what we used to call High Set overcurrent - so at a threshold of say 20 x the breaker design current In we just bang the thing open as fast as possible based on that absolute minimum time to operate - sod everything, we are in crisis territory and may be about to lose the whole system

Regards

OMS

Regards

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 13 May 2016 07:39 PM
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rubberlegs15

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Thanks for the detailed answers OMS and Mike.

Just out of curiosity, does this mean that the adiabatic equation for checking the size of an earth is practically useless in TNCS earthing arrangements? Any cable leaving a board could potentially develop a fault (be struck / damaged etc) very close to the board meaning that the PFC at that point would be sufficiently high enough to be greater than the 'instantaneous' disconnection current for most MCB's. You could for example calculate a 'suitable' CPC size at the end of a circuit with a Zs of say 1.44 ohms but should a fault occur on the same cable closer to the board, the Zs at that point could be much lower and the PFC much higher rendering the calculation useless?!

The same issue could arise in TNS arrangements too with a low enough Ze! Just trying to grasp the point of the adiabatic equation and it seems based on this information that it is of no use on systems with a low Ze?
 13 May 2016 08:09 PM
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AJJewsbury

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Just out of curiosity, does this mean that the adiabatic equation for checking the size of an earth is practically useless in TNCS earthing arrangements?

Not at all - as you say you need to show that the conductor is protected whereever the fault occurs. Most textbooks seem to use examples based on fuses - which have a helpful characteristic that they always get faster as the fault current increases to the extent that the overall energey let-through (I²t) either remains flat or decreases as fault current increases. So for fuses the worst-case is a fault at the far end of the circuit, so you can conveniently use the same values for checking both worst-case disconnection time for shock protection and worst case energy let-through for conductor protection. As you've correctly spotted, MCBs have a different characteristic the the energy let-through typically increases with fault current, so the worst case as far as conductor protection is concerned is right after the protective device.

If you look in GN 3 or the OSG there are tables of c.p.c. sizes for various MCBs and ranges of fault currents (3kA, 4.5kA, 6kA etc) - these are based not on BS 7671 tables (because they're useless as they stop at 0.1s) nor on manufacutrer's data (as that could only be relied on for devices for that particular manufacturer) but on the standard for MCBs (EN 60898) which stipulates the maximum allowable energy let-through for various fault currents and rating and type of device (B-type, C-type etc). E.g. for a B-type MCB rated 16A or less (and energy limiting class 3) the max permitted is 15,000 A²s for a 3kA fault. For devices over 16A but 32A or under, it's 18,000 A²s and so on.

So using those values you can apply the adiabatic (if rearranged into k²S² >= I²t ) to check you cable sizing against worst case fault current.

(You might notice a small hole in this procedure where fault currents are very high - e.g. getting close(*) to the theoretical maximum 16kA in the UK, but that's covered to some extent by a kludge in the UK version of the standard for consumer units which makes a few assumptions to allow us to use consumer units with 6kA or even lower rated CUs on 16kA supply - partly is asserts that faults in the first 3m of cables from the CU are sufficiently unlikely that they can be ignored and partly by relying on backup protection from an upstream HBC fuse (usually the DNO one).

(*) OK sometimes it's not very close at all - fault levels over 3kA start looking dodgy with some UK-traditional practice sized c.p.c.s

- Andy.
 13 May 2016 08:15 PM
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OMS

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In essence, an MCB has two limiting values of Zs

The first one is obvious - will enough current flow to operate the device for safety of persons and to avoid thermal damage to the cable

The second is less well recognised - if we accept definite minimum time is 0.01 seconds, then you can determine the minimum Zs value from:

(0.1 x Uo)/(kS)

So for your 1.0mm2 conductor and Uo = 230V then (for 70C PVC and Table 54-3) a value of 0.2 Ohms

or put another way, if the fault current exceeds 230V/0.2ohms

The value in Ohms will decrease as the conductor CSA increases - so for virtually all cases fed from a public network, given typical values of Ze it is extremely unlikely that the PEFC will be greater than that corresponding to the effective point of the intersection of the time current curve and a superimposed adiabatic line for the conductor

You can draw an adiabatic line for common conductor sizes on the top of the MCB curves if you like

You should also recognise by now that MCB's are current limiting devices - so again for the region below 0.1 seconds you really do have to evaluate:

I2t <= K2S2

As BS 7671 indicates

It's pretty rare to find any normal earthling arrangement stressed when protected by MCB's - but you must take care with reduced CPC sizes in T&E cable

Regards

OMS

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 13 May 2016 11:53 PM
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rubberlegs15

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Once again, thanks for the detailed answers, I bow down to your superior knowledge on the subject! I think a little more reading is required (on my behalf) to get my head around exactly what you are stating here (not that I'm a thicko!) but I'll look into it further using your guidance above - K2S2 for example is new to me - not something I've yet come across but then I haven't spent nearly enough time with my head in the regs to really get a good understanding of the more complex parts (I'm probably not likely to in the long run but who knows!). All the same, many thanks to all of you for your detailed explanations and time spent responding to my questions. You don't get much for free in life but I'm overwhelmed by the responses to this and truly very grateful for your given time and efforts, thank you. Top forum!
 14 May 2016 03:03 AM
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ChengLi

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Hi everyone A newbie here. Thank you for the additional information. Looking forward read more about like these. Godbless!

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 14 May 2016 10:43 AM
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rubberlegs15

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Penny has just dropped, I2t<=K2S2 is the adiabatic equation rearranged! Doh!

Sorry to bother you all once more but could you possibly walk me through an example of its use in proving a CPC is adequate?

Assuming a Zs of 0.25 ohms on a circuit (radial) of 8m in length (single phase) with a conductor size of 1mm squared on a TNCS earthing arrangement, prove by calculation that your selected CPC size is adequate to protect the circuit. The protective device is a BS60898 type B MCB.

I've posed it as a question as this is similar in nature to what we are being asked to prove by calculation (of course the question is actually ridiculous and ill conceived as no value for Ze is provided but it has sparked my curiosity on how you would apply the adiabatic equation where PFC exceeds the 'instantaneous' disconnection of the MCB!).

Regards
 14 May 2016 01:38 PM
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AJJewsbury

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Assuming a Zs of 0.25 ohms on a circuit (radial) of 8m in length (single phase) with a conductor size of 1mm squared on a TNCS earthing arrangement, prove by calculation that your selected CPC size is adequate to protect the circuit. The protective device is a BS60898 type B MCB.


Strictly speaking you need Zdb rather than Zs, but given the figures you have it's probably safe to assume it'll be at least 0.077 Ohms - i.e. <3kA fault current.

The easiest way is just to look up the c.p.c. size for a B type MCB (presumable) <= 16A in the OSG or GN 3 tables and <= 3kA - which will say 1.0mm² (you could work it out from the BS EN 60898 energy let-through (15,000 A²s) and see that there's been a slight kludge there (by 0.065mm2 ) and ponder why the rest of Europe have a 1.5mm² minimum conductor size and full sized c.p.c.s for small circuits)

- Andy.
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