As mentioned, the adiabatic equation assumes no heat is dissipated from the cable during a fault.  While putting the calculation on the safe side, in some situations, particularly for longer fault duration there is the potential to be able to get away with a smaller cross section.  In these instances, it is possible to do a more accurate calculation.

Considering non-adiabatic effects is more complex.  Unless there is some driver, using the adiabatic equations is just easier.  Software is available to consider non adiabatic effects, however, there is a cost, time and complexity associated with this.

The IEC also publish a standard which deals with non-adiabatic equations:

• IEC 60949 "Calculation of thermally permissible short-circuit current, taking into account non-adiabatic heating effects".

The method adopted by IEC 60949 is to use the adiabatic equation and apply a factor to cater for the non-adiabatic effects:

$I=ε I AD$

where I - permissible short circuit current, A (or kA)
ε - factor to allow for heat dissipation from cable

The bulk of the IEC 60949 standard is concerned with the calculation of ε.

Shadi Abd Al Azim Feb 06, 2021 8:05 PM
Regarding copper wire screen, which factor i have to use 0.5 or 0.7?
which is better for short circuit rating copper wire screen and followed by tapes or covered by PVC?
what do IEC mean in sub clause 5.3.1 "the screen wires are fully surrounded by non—metallic materials."?
Steven McFadyen Feb 07, 2021 1:03 PM
Good point. The IEC clause is not overly clear.

I would interpret 5.3.1 as each screen wire individually surrounded by insulation.  Whereas in 5.3.2, the between-wire insolation is achieved by an air-gap.

Choose 0.7 or 0.5 depending on the situation you have in your cable.
Jhon M Jan 15, 2022 9:29 PM
Hello
Concerning short circuit current with combined copper wires screen and lead sheath (they are  in parallel, electrically connected ) and taking into consideration the non-adiabatic method

As per paragraph 6, to calculate the non adiabatic coefficient for screen/sheath, we need the thermal resistivity and volumetric heat capacity of materials that are under and over the screens

In this case, copper wire screen has Semi-conductor under it but lead sheath above it, and lead sheath has copper screen under it and PE outer sheath above it
How to do it in this case? Which thermal resistivity and volumetric heat capacity we need to consider for both copper screen and lead sheath which share the total short circuit current ?

We need to consider that the lead and copper are metals and have approx 0 thermal resistivity and in this case we take for copper screen the properties of SC layer and PE layer and for the lead the same? Is it correct?

to be more clear, if a screen has non metallic material under it but a metallic material above it like: XLPE/SC/CU WIRES/LEAD/PE
this case was not mentioned in IEC 60949
For CU wires, we need the  thermal resistivity and volumetric heat capacity of SC and PE or for SC and Lead for the formula ? and same in case of lead for combined screen ?

Is there anything else we need to take into consideration as we have here 2 sources of heat ? Thank u