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Adiabatic Equation - different starting temperature

I have recetly had to do some calcs with the adiabatic equation, and this got me wondering on the following:


As-written, this assumes an initial conductor temperature of the max operating temp of the condictor for that cable type, so for a 90 degree Thermoplastic cable it's 90 degrees.  This is clearly a worst case scenario for a heavily loaded cable bit in most cases cables are operating at much lower temperatures than that.  In my use-case, the design current was approx 0.6A, so the cable is likely to be barely above ambient, so there would be a good 60 degrees of additional 'headroom' for an energy let-thorugh before hitting the stated final temperature for the cable type.

Has anyone expanded this calcuation to account for this?  If you were just the wrong side of the line and needed some 'help'  then this could maybe do it for you.  It could be a bit complicated though - factors that occur to me are:

 



  • You would need to know what the worst-case operating temp of the conductor is under the design current - and this could be raised at a point where it passed through some insulation

  • Then you would need to work in the specific heat capcity of the conductor material and the heat losses through the insulation and sheathing (although these could probably be neglected for a short duration fault).

  • That would then allow you to calcualte the energy (in Joules) needed to raise the conductor temp from operating temp to max temp

  • Then work that back to give you the I2t figure equivalent.


Anyone seen this done or worked it through?


This is just an interesting thought experiment for me....I didn't need the headroom in my example, and arguably if you are that close then you should probably go up a conductor size anyway....


Jason.
Parents
  • It may help to realise that  12t has the same units as Joules per ohm you are just looking at the heating in joules of a given mass of copper - the 12t *R/heat capacity.

    Clearly a colder start either leads to a colder final temp, or allows a bigger energy dissipation. The problem, as alluded above, is that you do not really know the start temp - or at least the reg writers do not, so they assume worst case. However, if you know better for your specific case then you can make use of the extra margin.

    I suggest if you do you also leave a copy of the sums next to the layout diagrams, or for ever after each inspector who sees it will try and fail it.


Reply
  • It may help to realise that  12t has the same units as Joules per ohm you are just looking at the heating in joules of a given mass of copper - the 12t *R/heat capacity.

    Clearly a colder start either leads to a colder final temp, or allows a bigger energy dissipation. The problem, as alluded above, is that you do not really know the start temp - or at least the reg writers do not, so they assume worst case. However, if you know better for your specific case then you can make use of the extra margin.

    I suggest if you do you also leave a copy of the sums next to the layout diagrams, or for ever after each inspector who sees it will try and fail it.


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