I am not sure about what has gone wrong, but I think you have not passed the power loss by the electric currents into the heating problem.

The thermal problem requires a heating power input. This is calculated in the electric problem, and must be given to the heat problem as its input.

Dear Malcolm White :

I think I understand what you are saying.

I already could solve successfully the Electrostatics analysis, as well as the Electric Currents analysis.

Cannot I write the heating power input using the expression (V^2)/R ?

In my problem V is a fixed given value, and also is R .

I am trying to use one "Heat Source" Domain Condition. And I am selecting the option of "Heat Rate" in the "Heat Source" section. This option ask you to write the value (or expression using the parameters table) in Watts , so is what I am doing.

I'm writing in this field: (V0^2)/R_elec

where I have defined in my parameters table:

V0 as 35 000 V

L as 1[m]

r as 0.002032[m]

Can you help me further ?

Thank You

V0^2/R is for lumped elements, like a whole cable or a resistor, which has a resistance and a voltage across it. It gives you watts, which is a total power

You need to calculate E^2/rho, electric field squared over resistivity, or J^2rho, which is current density times resistivity. These give watts per cubic metre at the place which has that resistivity and current density or electric field. I don't think you need the electrostatic solver.

The electric currents analysis should give you current density in all places, perhaps also in the regions surrounding the cable. These, together with the resistivity or conductivity will give a heat power distribution. I don't think you need an electrostatic solution, but I'm not sure. The heat power distribution is the input to the thermal problem.

The thermal problem needs to know where the power is input, not just how much there is.

In the simplest solution, you could just use the mean current density and the conductivity of the cable to work out by hand the average heating of the cable, then tell the thermal problem that this much heat was input equally spread out over the conductor cross-section.

Dear Malcolm White :

Thank you, I'm trying this approach now,

1) I disabled the Electrostatic, but I get this :

(first attached image)

Maybe I do need this module to be activated, since I selected "Electromagnetic Heating (emh)" since the begining, which is a coupled module ?

then,

2) I therefore switched the Heat Source to the option of: "General source" Qo, to imput this in W/m3 as you said.

But If I try to solve it selecting "Volumetric loss density, electromagnetic /ecs)" (Attached image 2) I can run the solution, but I don't get any Temperature Distribution (see please attached image no.3), The temperature is the same across all the model.

3) If I come back to the definition of the "Heat Source" and I select "User defined" (see please attached image no.4)

I don't know what expression I have to write down on the box for "User defined".

Thank you Malcom, hope you can help me

Regards ! :)

Maybe you do need the electrostatics.

What boundary conditions have you given the thermal problem?

Have you set the boundaries to a fixed temperature? Like 280K? It looks like you might have insulating boundaries, so the heat just builds up with nowhere to go.

Are the units in the two problems the same? Is this a 2D problem so the heating is per metre? Are you using m for one and mm for the other? Are you using the right units and values for thermal capacity and conductivity?

Dear Malcolm White :

Thank you for replying

"What boundary conditions have you given the thermal problem?

The following:

for the Electrostatics (es) :

- 1 Terminal (Domain), to set up the Value of Voltage = 35kV at the centered conductor.

- 1 Ground (Boundary), to set up the V=0 at the outer edges of the Soil.

for the Electric Currents (ecs) :

For this I only set up the same Terminal (Domain):

- 1 Terminal (Domain), to set up the Value of Voltage = 35kV at the centered conductor.

(defined exactly in the same way as in the (es) solution)

An for the Heat Transfer in Solids (ht)

I am trying to set it up with the Domain condition "Heat Source"

Have you set the boundaries to a fixed temperature? Like 280K? It looks like you might have insulating boundaries, so the heat just builds up with nowhere to go.

You are right, I didn't set the boundaries to a fixed temperature, I assigned the Thermal Insulation condition to the Boundaries (which are the outer edges of the Soil, I will try to assign a fixed value of temperature next.

Are the units in the two problems the same? I didn't changed anything about the units from one physics to the other. I have "SI" for "Uni System" in the root settings, and "m" for "length unit" in the main setting window for "Geometry", didn't change anything furthur regarding the units.

Is this a 2D problem Yes it is, this is a 2D model so the heating is per metre? But I don't know where to consider this unit, in the "Heat Source" condition, under the "Heat Transfer in Solid physics, the unit for "Heat Source" is for defalut as "W/m3"

Are you using m for one and mm for the other?, Nope, m for both Are you using the right units and values for thermal capacity and conductivity?" Yes, all my units are consistent in all the "Materials" section

Do you think I can send you my model so you can see where the mistake is ?

I'm sure this is caused by an evident mistake :(

Thank You

Malcolm White :

These are the most "Zoom Out" view of my model, and the "Zoom In" view of it,

So you can see the whole domain,

It is a burried cable, and I am trying to consider a whole section of Soil (1m X 1m)

This is a 2D model.

Dear Malcolm White :

I did change the Thermal Insulation BC for a fixed Temperature BC at the boundaries. And I been able to run the solution now.

Now what I need is to specify correctly the value of the "Heat Source"

You said to me that I need to fix this as ; either E^2/rho , or J^2*rho

How can I do this ? What type of "Heat Source" I need to select ?: "Linear Source" ? or "General Source"

And then, where I can find the functions "E" , "J" and "rho" ?

I'll appreciate the help,

Best Regards !

The electromagnetic solution may do this itself. It should I think it is the volumetric loss density from the electromagnetic solver that you need as input to the thermal solver. That is certainly what I would expect.