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Adiabatic Calculation

fiftyhertz

Hi All,


Been looking into the Adiabatic equation, as its been bothering me for a while that im not 100% on the subject. But after reading a lot of the old forums posts and the Amtech handbook i have collated the below information. Hopefully it is correct, and can be used by others. If not i would be very appreciative of any feedback.


Of course, you could just look at table 54g, but where’s the fun in that ?. I would be interested in understanding how that table has been derived, Id assume the line conductor is associated from the selection of an sensible MCB size as there seems to be no input of the variables of the adiabatic equation


Adiabatic Calculation



This calculation is a check to ensure that the cross-section of the CPC is sufficient to allow it to withstand the energy let-through of the Circuit Protective Device (CPD), i.e., fuse or circuit-breaker under earth fault conditions.



S = √(I2t)/ k


where:



S = the minimum section of the conductor in mm2



I = the earth fault current in amperes*



t =the CPD disconnection time at the earth fault I*



k = a factor (k) which ‘takes account of the resistivity, temperature coefficient and heat capacity of the conductor material, and the appropriate initial and final temperatures’.



 



*Where the disconnection time is greater than 0.1 s, I2t is the earth fault current squared times the disconnection time. S = √(I2t)/ k



 



*where the CPD disconnection time is less than 0.1 s, I2t is the CPD energy let-through in ampere squared seconds (A2s).  The Calculation must be transposed as I2t<=K2S2 and manufacturers data/ EN 60898 must be consulted as BS 7671 only gives 0.1s minimum from the tables so the I2t figure cannot be ascertained.



 



Fuse Fault Rules



For fuses the worst-case is a fault at the far end of the circuit, so you can use the same values for checking both worst-case disconnection time for shock protection and worst-case energy let-through for conductor protection.



 



MCB Fault Rules



MCBs have a different characteristic, 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. 



 



For a fault above 0.1 seconds Example



Assuming a Zs of 1.5 ohms on a circuit at supply distribution board with a line conductor size of 4mm2 (90 Degrees Thermosetting (LSF) where the protective device is a 32ABS60898 type B MCB. 



 



S = √(I2t)/ k



I = 230/1.5 = 154A



t =From BS7671, 15 seconds



k = is 176 for thermosetting cable from table 54.2



 



S = √(154 x 154 x 15)/ 176



S= 3.38mm2 (4mm2 CPC required)


 

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 forth.  As previously stated the energy let-through typically increases with fault current for MCBs, as more energy is essentially “let through”, so ultimately a larger cpc will be required.



 



For a fault below 0.1 seconds Example:



 



Assuming a Zs of 0.06 ohms on a circuit at supply distribution board with a line conductor size of 16mm2 (90 Degrees Thermosetting (LSF) where the protective device is a 63ABS60898 type B MCB. 



 



I2t<=K2S2



First calculate Earth Fault Current to establish the let through current or energy from the breaker.:



 



230/0.06 = 3883A Earth Fault Current.



 



From there, the I2t figure can be found from EN 60898.



 



It can be seen from a 63A Type B Breaker that 48000 A2s (I2t) units of energy are let through at this fault for this specific breaker and fault level. Therefore, we can summarise that for a cable to be safe it has to withstand 48000  energy units otherwise it will get too hot and there will be a breakdown in insulation.



 



I2t<=K2S2



Try for a 10mm2 CPC:



 



S = 10mm2



k = is 176 for thermosetting cable from table 54.2



48000 A2s <= (10 x 10) x (176 x 176)



48000A2s <= 3097600 A2s


 

So, we now know the cable can with withstand 3097600 energy units before it gets damaged, as the (I²t) of 48000 energy units is less than this there is no problem. 


 

A 6mm2 cable can be tried:



 



It is also worth noting that the Earth Fault Loop impedance calculation is a completely different calculation to ensure that the breaker disconnects fast enough depending on the fault current and specified disconection times.


 

Thanks


  • 0
    You seem to have got the general picture just about spot on!


    Just a few (very minor) details - the adiabatic is only really good for disconnection times of up to about 5s - after that the amount of heat lost from the cable during disconnection becomes more significant and so the adiabatic will over-estimate the required c.s.a. Thus your 15s example is perhaps not particularly realistic in that respect (although the procedure itself is fine). In any event if a c.p.c. is used for ADS 5s you'll require a disconnection time of 5s or below anyway - just from a shock perspective. B-type and C-type MCBs always have a disconnection time of over 5s if not in the "instantaneous" region, so in practice there's a short-cut there. (Arguably you could use the example with a D-type MCB - but another recent thread seems to suggest that BS 7671's D-type curves aren't quite right and so they might not be good for sub 5s disconnection times in the thermal region either.)


    With the k²S² = I²t approach you don't necessarily need to try out different values for S on a trial and error basis - knowing S and I²t you can re-arrange to S=SQRT(I²t)/k and calculate S directly. (You can't do that trick with fuses as changing the c.s.a. of the conductor changes the loop impedance and hence the fault current, but with MCBs when the worst case is at the start of the circuit, the impedance of the conductor you're calculating for doesn't affect things ?)


      - Andy.
  • 0
    Almost the case.


    A good explanation is available in Chatper 8 of the IET's Electrical Installation Design Guide: https://shop.theiet.org/electrical-installation-design-guide-calculations-forelectricians-and-designers-4thedition


    With mcb's you need to use either the let-through energy in place of I2t - either that published in the standard, or manufacturer's published data which may well lead to a smaller acceptable csa.


    In addition, as Andy says, with fuses, you need to be careful of both ends - usually only need to worry about the far end as the current decreases with distance, and something else comes into play before the cable gets too long (volt-drop or Zs).


    With longer disconnection times, > 5 s, there is an alternative approach called "non-adiabatic",  but as the tables for standard circuits in the OSG shows, often you'll have another limit (volt-drop, or loop impedance for circuits not protected by RCD) that will get to the limit before adiabatic is an issue, and therefore this isn't covered in BS 7671 or associated guidance - however, it may be used by some software packages, and by consultants, in certain circumstances.
  • 0
    I still cant get my head around this adiabatic equation,

    1. why for fuse, the worst case consideration is at far end of the cable and for MCB at the start of the cable? 

    2. Why is it that for less than 0.1s, the let through energy by the manufacturer to be used and between 0.1s and 5s, calculate using adiabatic equation after calculating the PFC.

  • 0
    For a fuse, as you increase the fault current, (and that means consider faults nearer the origin) the fuse speeds up, so the dissipation in the cable in the time it takes to blow the fuse remains substantially constant, unless the cable is very long, and then the current is so low the fuse does not blow at all. ( but of course the voltage drop is 100% by then) A circuit that can blow the fuse without cable damage from a dead short at the far end however, will also do so with a dead short anywhere else along the cable.


    For an MCB, the tripping time does initially get shorter as the fault current rises,  but there is a mechanical limit to how fast contacts will move and break an arc, so as the breaking time no longer keeps up with the rising fault current there may be  a risk to the cable from faults at the near end, if the supply PSSC is high enough, even in  cases when the far end sums work out OK, so the fault current for both end cases should be checked.


    Actually there is a case with fuses that should be checked for near end faults too, and that is with fuses like the old hot wire BS3036 where the maximum safe breaking current is quite low. The fuse will break the circuit OK but some of those would burn your hand if you pushed the carrier back in onto a faulty circuit as the hot metal could be expelled from the ends of the carrier for the wire if the fault current was high enough.

    Mike.


    PS

    Why does BS7671 data run out of steam at 0.1 seconds ? Mainly as that is where the maker's published data for breakers diverges, as the standards only require a few fixed test currents to be passed, and fur fuses, in the old days at least, it was considered fast enough that you could assume that I2t was constant from hat fault current upwards.

    So for example,  knowing that a 30A BS3036 would blow in 0.1 second at 400A, you could safely assume  that at 800A it would blow 4 times faster, i.e. in about 0.025 seconds and that the cable heating while that occurrred would be essentially equal in both cases.  Much higher currents and you could not say anything at all on that case, as they only had a 1kA max rating for the reasons above...

    Mike


  • 0
    Why is it that for less than 0.1s, the let through energy by the manufacturer to be used

    There's also the thought that for very short disconnections times - less than 1/2 the a.c. cycle - there won't be a simple single value for disconnection time for a given r.m.s. fault current - but rather it'll vary depending on quite where in the a.c. cycle the fault starts. If the fault happens to start just when the current is passing through zero then there might be quite a delay (percentage wise) until the current builds back up to a level that'll cause the device to open for instance. If you're just protecting a conductor from overheating it's probably not of much significance - since the heating effect would be similarly delayed so overall energy let-though isn't increased - but if you need to verify the disconnection time for other reasons - perhaps shock protection (where in some circumstances you'd be looking for < 0.07s or <0.04s) then it's perhaps better not glossed over as the overall effects might not be so accommodating. It might be clearer if we had a distinct symbol for energy let-though rather presuming it is always the product of time and the heating effect of the r.m.s. fault current.

       - Andy.
  • 0
    Say, if the fault is at the beginning of the cable, it is thus deemed as damage has already been done, and hence replacing the cable is the only option. If this is so, why to verify the thermal stress at the beginning of the cable?
  • 0
    That is not a wise assumption in every case- it may be true, but a cable may be thermally damaged by an event on a side spur that does not affect the main line. Real circuits are not always simple radials,  often more of a a Christmas tree and the cable size is not always the same all the way along (Indeed in large installations it rarely is. For reasons of voltage drop you end up with heavier arterial lines feeding thinner final drops - not dissimilar to the street mains and house service lines being different sizes)

    Mike
  • 0
    Nick Parker:

    Say, if the fault is at the beginning of the cable, it is thus deemed as damage has already been done, and hence replacing the cable is the only option. If this is so, why to verify the thermal stress at the beginning of the cable?


    Consider not just faults right at the start of the cable, but one at the first accessory - perhaps only a short distance along the cable (socket or light next to the DB for instance) - the fault current will be less than immediately after the MCB, but not by much - and normally there wouldn't be any need to replace the cable. You could calculate individually for every such situation, but it's a lot simpler just to take the absolute worst case (at the MCB) and do it once, which also guards against future alterations.


    Similarly damage to a cable might be repaired or a short replacement spliced in to replace just the damaged section - rather than the entire cable replaced - especially where a cable is buried over a considerable distance or embedded in building materials which might make replacement all the way back to the DB difficult or expensive.


       - Andy.


  • 0
    I would not agree that the worst case is always is for a near end fault, but it might be. It could be that with a long cable a far end short circuit fault may be more onerous as the cable will attenuate the fault current and you have the "long slow blow" situation which may damage the cable.
  • 0
    John is quire right.


    In practice, for simple installations (say max demand < 100 A) the limit of voltage drop, or the need to meet a disconnection time, would mean the near-end is usually the one that's the worst-case in a circuit ... but this is not always the case where a 5 s disconnection time is used, or disconnection times can't be met and supplementary local equipotential bonding is applied in accordance with Regulation 419.3, you might well run into the situation that the far-end is the problem.


    You can see this if you plot the limit values of current-for-time for a given cross-sectional area onto the time-current curves in BS 7671 (per the example in Figure 8.2 of the Electrical Installation Design Guide).

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