Not sure how it goes now, but I used to follow the guidelines/principles in the electric cables handbook by mcallister.
We would build up a thermal resistance circuit including soil sheaths etc and heat inputs etc.
We did take the thermal resistance of armours and sheath to be zero (actually a heat sink rather than resistance), but would include heat capacity for them.
The insulation we would include the thermal insulation , but not thermal capacity
Conductors we would include the heat input (i^2r) and the resistance change on temperature change.
I think to follow it properly you need to consult a reference book as the thermal resistance we used wasn't specific (rho), but calculated based on the geometry
So say its a single core round conductor to a concentric sheath, it would be something like (rho/2 pi) x Ln (1+2r/R) rather than just rho x insulation thickness.
It also used to be detailed quite well in iec 287, but I haven't touched any of this in a while.
Since the semiconductor materials would act more like a thermal insulator than a thermal conductor I would treat it as an insulator (thermal resistance, and no thermal capacity).
Seperate heat sources are just that, in the same way as you would include the 3 heat inputs for 3 phases , include the other heat sources the same.
Thank u for your reply
Not sure how it goes now, but I used to follow the guidelines/principles in the electric cables handbook by mcallister.
We would build up a thermal resistance circuit including soil sheaths etc and heat inputs etc.
We did take the thermal resistance of armours and sheath to be zero (actually a heat sink rather than resistance), but would include heat capacity for them.
The insulation we would include the thermal insulation , but not thermal capacity
Conductors we would include the heat input (i^2r) and the resistance change on temperature change.
I think to follow it properly you need to consult a reference book as the thermal resistance we used wasn't specific (rho), but calculated based on the geometry
So say its a single core round conductor to a concentric sheath, it would be something like (rho/2 pi) x Ln (1+2r/R) rather than just rho x insulation thickness.
It also used to be detailed quite well in iec 287, but I haven't touched any of this in a while.
Since the semiconductor materials would act more like a thermal insulator than a thermal conductor I would treat it as an insulator (thermal resistance, and no thermal capacity).
Seperate heat sources are just that, in the same way as you would include the 3 heat inputs for 3 phases , include the other heat sources the same.
Hello
thank u for your reply
let me give more details, please refer to the attached image for the formula
my case is that I need to calculate the short circuit current in combined screens of a cable (Cu wires + lead sheath), forget the conductor
CU/SC/XLPE/
SC/CU WIRES/LEAD/PE
I did the calculation with adiabatic heating effect as per same standard, where we consider that the heat will not dissipate from screens during the short circuit time, the short circuit current will be divided between the two screens based on their resistance, and the one that will arrive to its max temperature first, will be the max short circuit current..... it is not hard to do it.
but the problem is with non-adiabatic heating effect (Refer to the formula attached) where I need to calculate this factor and multiply it by the short circuit current calculated in adiabatic heating effect (the factor will be >1) where we consider that the heat will dissipate which is the real case
If my case was only one screen, for example SC/
CU WIRES/PE, it will be clear, based on the formula I will consider the volumetric specific heat and thermal resistivity of materials either side of the screen which in this case are: SC and PE.
Based on the same standard it specified the values, the thermal resistivity of SC and PE will be 2.5 and 3.5 K.m/W respectively and volumetric specific heat will be 2.4 x 10^6 J/K*m³ for SC and PE
Other required formula factors are clear...
But the problem is the case mentioned above (
SC/CU WIRES/LEAD/PE) where the materials either side of each screens are non-metallic and metallic, this case is not mentioned by BS standard which is weird because that case is very common (two screens)
The standard didn't mention the thermal resistivity of metallic materials which is logic because it is approx. 0 (or 0.0025 for copper for example) and if we put it in the formula, it will gives wrong values
What to do in this case? can I consider that for Cu wires the materials either side are PE and SC and same for Lead ? means do the calculation using the formula for the copper wires and consider that lead doesn't exist and do the calculation for lead and consider that CU wires doesn't exist ?
The short circuit current will be divided between the two as mentioned above, the heat of copper and lead will affects each other ? because the heat will change the resistivity of each metallic screen which means the current will be divided between them in different way
It is an urgent case that I need to find a solution for it
is there a software that calculate this case ?
thank u so much
