Discuss Adiabatic equation infinite loop in the Electrical Wiring, Theories and Regulations area at ElectriciansForums.net

Zdb

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So I'm looking into the thermal constraints of every cable in my 2396 project.

I've used the adiabatic equation in a spread sheet and 99% of them are fine with the CPC I have selected.

However, there are a few really short circuits with very low earth fault loop impedance. Because of this the prospective fault current is really high and the adiabatic equation says I need a larger CPC.

Not a problem, I enter the adjusted values for the larger CPC into the spread sheet which automatically does the equation again....

Now because I've used a larger CPC, the loop impedance is even lower and the prospective fault current is even higher. It now says I need a larger CPC (again).

This seems to be an infinite loop.

Please can someone explain how I get around this?

Thanks
 
Quick thought - have you included a practical value for Ze in your calculator? The Ze will not be zero and you aren't making it any smaller, so it should converge ... he says hopefully :)
 
If your using 0.1 seconds as T for say mcbs then there’s your problem if your disconnection time is below 0.1 seconds.
Get the let through energy from the manufacturer.
You’ll need to make sure k2S2 is equal to or greater than let through energy of the OCPD and using t=k2S2/i2 you’ll probably see that if your cable sizes comply, t being the maximum time before damage to the conductors and surrounding insulation starts to occur then your circuits will Probably comply.
 
Can you give an example that won't work for you ?

So the Zs is 0.2315764 ohms (3 metre circuit using 1mm/1mm singles).

Cmin ×nominal voltage = 218.5.

218.5 / 0.2315764 = 943.5331061 amps.

Looking at the curve for 6A type C 60898 on page 326 in the BYB it has to be 0.1s disconnection time.

Therefore the square route of 943.5331061 squared timed 0.1 = 298.3713663.

298.3713663 divided by my k value of 115 = 2.59mm minimum CPC required.

Like I said if I go up a size then use the equation again I will have go another size and so on and so on.

I guess I could have the wiring go round the room a couple of times to increase the Zs value therefore lowering the fault current :confused:
 
Yes as stated you need to find out the let through of the protective devices(the units aren't energy it's more like energy per ohm)
You would have an mcb with a bs fuse upstream so for higher pfc the fuse would blow first. Although with a normal sized lv transformer you would only get a maximum of 10kA or so? even if you were connected directly to the transformer due to the volt drop inside the transformer.
Also bear in mind the cpc size of the final circuit wouldn't affect the fault current because you're considering the case where the fault occurs directly after the mcb (ie worst case) so you would use zdb not zs.
For all other calculations eg disconnection time and volt drop the worst case is the far end of the circuit, but not for adiabatic.
Hope that helps!
 
So the Zs is 0.2315764 ohms (3 metre circuit using 1mm/1mm singles).

Cmin ×nominal voltage = 218.5.

218.5 / 0.2315764 = 943.5331061 amps.

Looking at the curve for 6A type C 60898 on page 326 in the BYB it has to be 0.1s disconnection time.

Therefore the square route of 943.5331061 squared timed 0.1 = 298.3713663.

298.3713663 divided by my k value of 115 = 2.59mm minimum CPC required.

Like I said if I go up a size then use the equation again I will have go another size and so on and so on.

I guess I could have the wiring go round the room a couple of times to increase the Zs value therefore lowering the fault current :confused:
Yes but your disconnection time will be less than 0.1 seconds ,you need the let through energy of the device from the manufacturer.
Also use the thermal constraints calculation I’ve already stated above.
Do not use 0.1 seconds.
Also your mcbs will be current limiting class 3 devices, you shouldn’t have an issue
 
Hi - in practice would you have a 1mm circuit that is only 3m long? Anyway, if pssc was 1000A and we use 100ms disconnect time then I2T is 100k. Looking at the Wylex B6 graph, it shows about 2k let through energy for 1kA which will give a very different result :) .

IMG_1104.jpg
 
Also bear in mind the cpc size of the final circuit wouldn't affect the fault current because you're considering the case where the fault occurs directly after the mcb (ie worst case) so you would use zdb not zs
Just realised from wilkos post even that's wrong, it should not be calculated from the zdb, it should be the PFC which would be the maximum of the PSSC and the PEFC.
But still the final circuit would have nothing to do with the let through (energy per ohm)
 
What does the equation say when using 20mm conduit as the CPC?
 
I'm not quite sure how to work that out TBH.
All you need is the resistance of the conduit per meter, you can then work out your R1+R2 etc, change the K value on the adiabatic.

IIRC the minimum CSA of 20mm galv is around the 53mm area so you have plenty to play with.

Let me see if I can find the information I had.
 
How many people on here think the zdb+r1+r2 ie the Zs of the circuit in question is used in the calculation of the cpc size for adiabatic?
I feel like I'm the only one picked up the fundamental error (and wrong side error) the op described, and I'm not even a sparkly.
 
Hi - not sure I've understood correctly, but it may not be an issue in real life. Using the Wylex data from post #15 a B6 disconnects with I2T of about 10,000 on 10kA short circuit current (worst case, I should think). Plug that into adiabatic and 1mm2 is still ok. So not really something we have to think about every day ?
 
Th
How many people on here think the zdb+r1+r2 ie the Zs of the circuit in question is used in the calculation of the cpc size for adiabatic?
I feel like I'm the only one picked up the fundamental error (and wrong side error) the op described, and I'm not even a sparkly.
the zs of the circuit and resulting fault current is what is used as I in the equation
 
How many people on here think the zdb+r1+r2 ie the Zs of the circuit in question is used in the calculation of the cpc size for adiabatic?
I feel like I'm the only one picked up the fundamental error (and wrong side error) the op described, and I'm not even a sparkly.

That doesn't really make any sense.

The Zs would be even lower due to parallel paths.

All you need is the resistance of the conduit per meter, you can then work out your R1+R2 etc, change the K value on the adiabatic.

IIRC the minimum CSA of 20mm galv is around the 53mm area so you have plenty to play with.

Let me see if I can find the information I had.

I'll look into it. If you do have the info that would be great.
 
How many people on here think the zdb+r1+r2 ie the Zs of the circuit in question is used in the calculation of the cpc size for adiabatic?
I feel like I'm the only one picked up the fundamental error (and wrong side error) the op described, and I'm not even a sparkly.

Zs is used because a cpc which is suitable for conditions at the far end of the circuit can be assumed to be suitable at the supply end of the circuit.

Applying the adiabatic equation at the supply end as you suggest and at the load end results in a larger cpc from the load end calculation.
 
the zs of the circuit and resulting fault current is what is used as I in the equation
I'm saying that's wrong, as the worse case ie highest fault current will be when the fault happens close to the origin.
This is exemplified by the suggestions to increase the length of the circuit arbitrarily in order to reduce the necessary size of the cpc. If the fault happen 1m away from the cpc, the remaining 2m or 20m of the circuit are irrelevant.
That doesn't really make any sense.

The Zs would be even lower due to parallel paths.
Maybe it doesn't make sense at the moment, but by rating it Dumb, I'm not sure if that shows a good attitude to understanding whether you are making a serious and potentially dangerous mistake.i also notice it appears you are using cmin to calculate rather than the actual p(e)fc which is also incorrect.
Zs is used because a cpc which is suitable for conditions at the far end of the circuit can be assumed to be suitable at
I don't think that can be assumed, for example with a mechanical mcb the graph posted above clearly shows that not to be the case.
Applying the adiabatic equation at the supply end as you suggest and at the load end results in a larger cpc from the load end calculation.
Did not know that. Interesting, i will look at that. I would expect it to be either head end or the same.
 
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If we sized the CPC for a fault at the beginning of the circuit then the CPC size would be excessively large for what is required.

I can see where you're coming from but feasibly you could end up with a CPC of 6mm for a 1mm lighting circuit if you calculate for the fault at the origin of the circuit..
 
If we sized the CPC for a fault at the beginning of the circuit then the CPC size would be excessively large for what is required.
I contend that it would be exactly what is required to protect for a fault at that point. The remainder of the circuit is not relevant to the fault current. If the cable insulation are damaged on exit from the DB, they need to be not thermally damaged by the fault current. As with any other fault before the furthest point. Hence the need to find the worst case
I can see where you're coming from but feasibly you could end up with a CPC of 6mm for a 1mm lighting circuit if you calculate for the fault at the origin of the circuit..
If that's what is needed then there are either missing protective devices or they are not properly matched. Or someone has calculated i2t from 0.1ms instead of using the manufacturers chart.
If you check the above chart, you should find that even for 1mm cpc on a 6A class 3 mcb would be fine at 16kA as long as it's backed up by a 100A cutout fuse.
 
I'm saying that's wrong, as the worse case ie highest fault current will be when the fault happens close to the origin.
This is exemplified by the suggestions to increase the length of the circuit arbitrarily in order to reduce the necessary size of the cpc. If the fault happen 1m away from the cpc, the remaining 2m or 20m of the circuit are irrelevant.

Maybe it doesn't make sense at the moment, but by rating it Dumb, I'm not sure if that shows a good attitude to understanding whether you are making a serious and potentially dangerous mistake.i also notice it appears you are using cmin to calculate rather than the actual p(e)fc which is also incorrect.

I don't think that can be assumed, for example with a mechanical mcb the graph posted above clearly shows that not to be the case.

Did not know that. Interesting, i will look at that. I would expect it to be either head end or the same.
It’s not wrong you need the fault current at the most onerous point in the circuit
 
the zs of the circuit and resulting fault current is what is used as I in the equation
I'm saying that's wrong, as the worse case ie highest fault current will be when the fault happens close to the origin.
It’s not wrong you need the fault current at the most onerous point in the circuit
Indeed, so for the purposes of the adiabatic equation, would you consider higher or lower current more onerous?

A higher I2t is certainly more onerous (as csa required is proportional to the square root of I2t).
At disconnection times less than 0.1s, it's clear from the I2t chart posted previously that onerous case is close to the origin.

At disconnection times between 0.1s and 5s we need to calculate I2t from the manufacturer's charts for disconnection time. That is the case I mentioned in my 6:36 post today I needed to look into (but didn't!)

For >5s we would indeed end up needing huge CPCs but that's because adiabatic is not relevant due to the adiabatic (no heat disapated) assumption not holding.
 
Zs is used because a cpc which is suitable for conditions at the far end of the circuit can be assumed to be suitable at the supply end of the circuit.

Applying the adiabatic equation at the supply end as you suggest and at the load end results in a larger cpc from the load end calculation.
So as promised I eventually looked into this, it seems like we were both right!
I found an interesting thread on IET to explain IET Forums - Understanding the adiabatic equation - https://www.------.org/forums/forum/messageview.cfm?catid=205&threadid=65487&enterthread=y
The gist of it is for fuses the worst case is indeed as you state, where the I is lower the I2t will be higher, so you have to check the distant end.
However for MCBs the situation reverses when the magnetic trip comes into play - when the I is higher the I2t will be higher, so you have to check the head end instead.
Cheers, I learnt something today!:)

However, after all that, the OPs situation, as far as I can tell, he still needs to use the actual PEFC rather than the Cmin, and he still needs to use the head end and the tabulated I2t values due to the magnetic trip being relevant.
And also not forgetting that the other conductors are also subject to the same process - which may apply if PEFC and PSSC are not the same.
 
So as promised I eventually looked into this, it seems like we were both right!
I found an interesting thread on IET to explain IET Forums - Understanding the adiabatic equation - https://www.------.org/forums/forum/messageview.cfm?catid=205&threadid=65487&enterthread=y
The gist of it is for fuses the worst case is indeed as you state, where the I is lower the I2t will be higher, so you have to check the distant end.
However for MCBs the situation reverses when the magnetic trip comes into play - when the I is higher the I2t will be higher, so you have to check the head end instead.
Cheers, I learnt something today!:)

However, after all that, the OPs situation, as far as I can tell, he still needs to use the actual PEFC rather than the Cmin, and he still needs to use the head end and the tabulated I2t values due to the magnetic trip being relevant.
And also not forgetting that the other conductors are also subject to the same process - which may apply if PEFC and PSSC are not the same.
In the case of a disconnection time lower than 0.1 seconds like in the op case, yes he needs the let through energy of device, this can be used to make sure K2S2 is e equal to or greater than the let through value.
 
I overcame the problem by removing a PIR and having a light switch instead. This increased the circuit length from 3 metres to 12+ metres which increased the Zs.

I've taken all comments on board and in real life would use the manufacturers data for the protective device disconnection times but for the purpose of this project I wanted to stick with the information found in BS7671.

Thanks for all of your replies :thumbsup:
 
I overcame the problem by removing a PIR and having a light switch instead. This increased the circuit length from 3 metres to 12+ metres which increased the Zs
See the list directly above yours which is spot on.
Based on that please also do the calculation assuming the fault occurs within 1m of the DB. Then make sure the cables still comply with adiabatic.
Bs7671 doesn't contain all the data about class 3 protective device let through, but that information is necessary.
 

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