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Adiabatic Equation

Horace Charles

I am not an electrician but have a personal technical interest in understanding the Wiring Regulations.

I have become a bit confused by some contributions on the internet which seem to suggest that the adiabatic calculation will determine the size of CPC that will "cope with" a temperature rise to the maximum temperature permitted for a given cable type.

But as I understand it, it is the insulation which has to tolerate the temperature rise. This rise is caused by the heating effect (IsquaredR) of the current flowing in the CPC (and live conductor) under a dead short earth fault condition and which flows for the time it takes for the overcurrent protective device to disconnect the supply. Exceeding the maximum permitted temperature would permanently damage the insulation which I assume would no longer insulate.

Outside the parameters of the adiabatic conditions I assume that the conductors themselves could tolerate a very much higher and longer duration current, and hence temperature rise, before they themselves would become damaged. The only damage I can think of is that an unprotected conductor would melt like a fuse.

But for the adiabatic conditions of a known maximum level of earth fault current and a known OCPD disconnection time, the equation will determine the minimum size of CPC which would generate a temperature rise in the cable to the maximum permissible temperature for the insulation. In the case of a multi core cable like twin and earth, I assume that for an earth fault the heat contribution from the live conductor is not considered because the contribution from the smaller and hence higher resistance CPC will be much greater and more significant.

I will be pleased to hear from expert readers to confirm or otherwise clarify if I have understood the theory of this correctly . Thanks

  • 0 in reply to Horace Charles

    For copper, the square root of the first bit to the left of ln is always 226. So, its only the square root to the right of ln that needs to be determined before multiplying. 

    In your calculation, square root of the right hand side (0.258948) is approx 0.51 and 0.51*226 = approx 115.

    At least that is how, I do it. 

    By the way, the left for aluminium is 228 and steel 202.

  • 0 in reply to mapj1

    Thank you Mike for your reply with the example of a 1 metre length of 2.5mm2 copper cable.

    I'm working from first principles and here's what I understand from the figures you have used.

    The density of copper is approx 9 grams per cubic centimetre so for the 1 metre length of cable whose volume is 2.5 cubic centimetres, its approx. mass is 9 x 2.5 = 22.5 grams.

    The specific heat of copper is approx 0.38 joules per gram per degree C so to raise the temperature of the 1 metre length by 90 degrees will require 0.38 x 22.5 x 90 =approx. 760 joules which is 760 watt seconds.

    The next entry re watts/R having the units I2t has got me a bit confused. Did you mean watt seconds/R, that is, joules/R ?

    I had understood that Power = I2R = joules/second which if rearranged becomes joules/R = I2t

    So 760 joules is generated in the 0.0075 ohm of resistance by an I2t value of 760/0.0075 =approx. 100,000

    I'm sorry but I don't understand the note about the book value being 77,000, where does this figure come from?

    I understand the point about rounding errors and resistance increases due to temperature.

    As an experiment I inserted these values into the rearranged adiabatic formula to see what value of k they would produce.

    K = sq. root I2t/2.5 = sq.root 100,000/2.5 = 126.5

    Quite a way off from 115.

    Thanks again and I hope I have correctly understood the information you have provided

  • 0 in reply to lyledunn
    • Thanks for your reply, that's a useful tip. I can see that for each conductor material, that part of the equation is constant and it is the logarithmic part which will vary dependent on the change in temperature. Thanks 
  • 0 in reply to Horace Charles

    The 'book value' is in a number of publications that seem to self reference , guide to the regs, wiring matters article etc.

    I'm not sure why its that much lower, but the point is that apart from the mangling of units (joules per ohm into A2t)  and having square roots that mean the constant does not immediately have units that relate to an obvious physical quantity, what we are looking at is the same.- I was trying to show the physics under the wrappers as it were.

    Mike. 

  • 0 in reply to mapj1

    Thanks Mike for your expert feedback which is much appreciated and has added to my understanding. I will keep studying this topic. Thanks

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