Adiabatic equation infinite loop

Zdb · Jul 27, 2018

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

The Ghost · Jul 27, 2018

Smaller earth?

Wilko · Jul 27, 2018

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

Zdb · Jul 27, 2018

I've used the Ze given in the specification. 0.1 ohms

Zdb · Jul 27, 2018

Vortigern said:
Smaller earth?

Yes I was thinking that but do they even do a single core CPC smaller than 1mm?

Strima · Jul 27, 2018

Two options.

Reduce your conductor sizes to the bare minimum or do the calculation size for the required CPC and leave it at that.

Just thought of a third option use conduit as the CPC.

davesparks · Jul 27, 2018

Have you re-checked the calculation manually to ensure there isn’t an error in the spreadsheet?
What disconnection time are you using in the calculation?

Have you tried a different OCPD?

Zdb · Jul 27, 2018

I think I will have to change cable type and use the minimum size conductor like Strima said.

Wilko · Jul 27, 2018

Can you give an example that won't work for you ?

Ian1981 · Jul 27, 2018

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.

The Ghost · Jul 27, 2018

Maybe you could move the circuits to make them longer then?

Zdb · Jul 27, 2018

Wilko said:
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

johnduffell · Jul 27, 2018

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!

Ian1981 · Jul 27, 2018

Zdb said:
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

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

Wilko · Jul 27, 2018

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

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[ElectriciansForums.net] Adiabatic equation infinite loop

Adiabatic equation infinite loop

Zdb

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The Ghost

Wilko

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Zdb

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Zdb

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Strima

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davesparks

Zdb

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Wilko

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Ian1981

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The Ghost

Zdb

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johnduffell

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Ian1981

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Wilko

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