Non-adiabatic short circuit current for cable screen/sheath BS 7454

[Jhon M]

Hello all,

I have a question please concerning the calculation of the non-adiabatic factor for screen/sheath with a case that is not mentioned clearly in BS 7454

If we have a cable with below construction:

CU/SC/XLPE/SC/CU WIRES/LEAD/PE outer sheath

As per BS 7454 paragraph 6, the formula requires the thermal resistivity and volumetric specific heat of materials either side of the screen/sheath

In this case, the Cu wires has non-metallic material under it: SC (semi-conductor) but a metallic material above it: the lead sheath, and same case for lead sheath

how to do it in this case? Do we need to consider that the thermal resistivity of lead and copper is approx 0 and take into consideration for the formula the thermal resistivity and volumetric specific heat of the SC and PE materials for CU wires short circuit calculation ? is that correct ? and same case for lead sheath?

because I have a case of combined screen/sheath: CU wires + lead sheath to handle together the total short circuit current in non-adiabatic heating effects

and any comment concerning two sources of heat working together ?


thank u so much

 

[Julie.]

Hello all,

I have a question please concerning the calculation of the non-adiabatic factor for screen/sheath with a case that is not mentioned clearly in BS 7454

If we have a cable with below construction:

CU/SC/XLPE/SC/CU WIRES/LEAD/PE outer sheath

As per BS 7454 paragraph 6, the formula requires the thermal resistivity and volumetric specific heat of materials either side of the screen/sheath

In this case, the Cu wires has non-metallic material under it: SC (semi-conductor) but a metallic material above it: the lead sheath, and same case for lead sheath

how to do it in this case? Do we need to consider that the thermal resistivity of lead and copper is approx 0 and take into consideration for the formula the thermal resistivity and volumetric specific heat of the SC and PE materials for CU wires short circuit calculation ? is that correct ? and same case for lead sheath?

because I have a case of combined screen/sheath: CU wires + lead sheath to handle together the total short circuit current in non-adiabatic heating effects

and any comment concerning two sources of heat working together ?


thank u so much

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.

 

[pc1966]

For the non-adiabatic values (i.e. long term faults like open PEN, etc) then you base it on the resistance factor. So the Cu equivalent of SWA steel armour is about 1/8 of the armour cross-section.

For the adiabatic case you use the ratio of k factors as in the BBB and that takes in to account both the higher R (so more heating) as well as the greater thermal inertia from more CSA, so it is of the order of sqrt(resistivity ratios)

EDIT: Julie has just given far more detail about other aspects!

 

[Jhon M]

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