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Underground Cable Joule Heating
Posted Apr 14, 2017, 6:30 a.m. EDT Low-Frequency Electromagnetics, Heat Transfer & Phase Change Version 5.2 8 Replies
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Hello,
I am new to Comsol and am doing an underground sea cable simulation. First with a bipolar cable system (two transmission lines, +-320 kV).
I have the geometry ready and activated the "electric currents" as well as the "heat transfer in solids" mode.
My problem is that I do not understand how to get the current through the cables in 2D and how to do the grounding of the cables. Also I need an infinite boundary for the surrounding sediment or make it an infinite domain, that is another point ( I think the open boundary that I chose is not the rifght solution).
The goal is to model the heat dissipation by the cable because of heat losses (joule heating) in a transient simulation (load profile) and the heat transfer through the sediment.
I tried to attach my file, but only got 'file extension error' message there. :-(.
I can try again later to upload it.
I would be delighted to hear some tips of how to do this. Thanks very much and happy easter!
Madlen
I am new to Comsol and am doing an underground sea cable simulation. First with a bipolar cable system (two transmission lines, +-320 kV).
I have the geometry ready and activated the "electric currents" as well as the "heat transfer in solids" mode.
My problem is that I do not understand how to get the current through the cables in 2D and how to do the grounding of the cables. Also I need an infinite boundary for the surrounding sediment or make it an infinite domain, that is another point ( I think the open boundary that I chose is not the rifght solution).
The goal is to model the heat dissipation by the cable because of heat losses (joule heating) in a transient simulation (load profile) and the heat transfer through the sediment.
I tried to attach my file, but only got 'file extension error' message there. :-(.
I can try again later to upload it.
I would be delighted to hear some tips of how to do this. Thanks very much and happy easter!
Madlen
8 Replies Last Post Apr 24, 2017, 11:44 a.m. EDT
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Posted:
8 years ago
This blog on how to model heating in cables may be useful to you:
www.comsol.com/blogs/how-to-model-the-electromagnetic-heating-of-underground-cables/
Best,
Jeff
www.comsol.com/blogs/how-to-model-the-electromagnetic-heating-of-underground-cables/
Best,
Jeff
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Posted:
8 years ago
Unfortunately this is about AC modeling, where it is all about induction, eddy currents, magnetic fields and complicated stuff I don't understand yet. ;-)
Trying to do a simpler DC cable first.
Maybe a shorter question now:
I have found that when using magnetics interface and there an External current densitiy, I can enter a value for the z-direction.
Why is this not possible for the electric currents interface, which is part of the Joule Heating? When I try it there is only x- and y-direction.
I would love a current normal to my 2D model.
Greets
Trying to do a simpler DC cable first.
Maybe a shorter question now:
I have found that when using magnetics interface and there an External current densitiy, I can enter a value for the z-direction.
Why is this not possible for the electric currents interface, which is part of the Joule Heating? When I try it there is only x- and y-direction.
I would love a current normal to my 2D model.
Greets
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Posted:
8 years ago
Hi Lea,
If you know a priori that the current flows perpendicular to the plane, there is nothing to be computed as far as the electric problem goes: by definition there is no current in the plane. So it's purely a thermal analysis, with a heat source in the cable to account for the Joule heating. You can compute its magnitude based on the electric conductivity of the cable material and voltage drop per unit length of the cable ( Q = sigma* (deltaV/deltaL)^2 )).
The magnetics interface accepts different inputs because it solves a structurally different equation ( We have the cross product to blame for the existence of those out of plane effects!).
Best,
Jeff
If you know a priori that the current flows perpendicular to the plane, there is nothing to be computed as far as the electric problem goes: by definition there is no current in the plane. So it's purely a thermal analysis, with a heat source in the cable to account for the Joule heating. You can compute its magnitude based on the electric conductivity of the cable material and voltage drop per unit length of the cable ( Q = sigma* (deltaV/deltaL)^2 )).
The magnetics interface accepts different inputs because it solves a structurally different equation ( We have the cross product to blame for the existence of those out of plane effects!).
Best,
Jeff
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Posted:
8 years ago
Hi Jeff,
thanks a lot. At the moment I don't know dV/dL. But I know the two DC cables operate at +/- 320 kV and I know the current through the cables, which changes over time --> transient study.
Can I use the copper cables as a heat source (general, user defined) in W/m³ calculating the values for the heat dissipation with
= P*time = I²*R (T) and knowing the current and the diameter getting the current density in [A/m²] * the resistance in [ohm*m] would be [W/m³] :-)
Just wondering why m³ works in 2D simulation, but it would be nice if that works.
What does the heat source - general source- total power dissipation density do? Could that be something for me?
Thank you, have a nice week end!
BR
thanks a lot. At the moment I don't know dV/dL. But I know the two DC cables operate at +/- 320 kV and I know the current through the cables, which changes over time --> transient study.
Can I use the copper cables as a heat source (general, user defined) in W/m³ calculating the values for the heat dissipation with
= P*time = I²*R (T) and knowing the current and the diameter getting the current density in [A/m²] * the resistance in [ohm*m] would be [W/m³] :-)
Just wondering why m³ works in 2D simulation, but it would be nice if that works.
What does the heat source - general source- total power dissipation density do? Could that be something for me?
Thank you, have a nice week end!
BR
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Posted:
8 years ago
Sorry I got that my calculation was wrong, the units don't fit.
Maybe you have more ideas? Can you give me advice how to find out the voltage drop per length for the cable? Is it depending on the current also?
Thx
Maybe you have more ideas? Can you give me advice how to find out the voltage drop per length for the cable? Is it depending on the current also?
Thx
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Posted:
8 years ago
Hi Lea,
Yes, you can express the heat source per unit volume in terms of the current density (rho*J^2 where rho is the resistivity and J=I/A is the current density) instead of in terms of the voltage drop. It's equivalent.
Jeff
Yes, you can express the heat source per unit volume in terms of the current density (rho*J^2 where rho is the resistivity and J=I/A is the current density) instead of in terms of the voltage drop. It's equivalent.
Jeff
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Posted:
8 years ago
Ok that is great than I will dot this!
Thanks a million Jeff!
(I might come back to you if I have to do an AC Simulation as well, but maybe I can get around having to simulate the actual losses by induced currents etc. and can use estimations for the total AC losses, as this is just a master thesis :-P)
Best Regards, Lea Madlen
Thanks a million Jeff!
(I might come back to you if I have to do an AC Simulation as well, but maybe I can get around having to simulate the actual losses by induced currents etc. and can use estimations for the total AC losses, as this is just a master thesis :-P)
Best Regards, Lea Madlen
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Posted:
8 years ago
If you can neglect those effects, then the only difference I can think of is that in the AC case your heat source would be 1/2*rho*J^2 with J the peak current density. The factor 1/2 comes from integrating cos^2.
Jeff
Jeff