Discussion Forum

Don't define your metal (or metal surface) to be a perfect conductor. Set it to be a material (e.g., copper). For the surface, set the boundary condition on it as an impedance boundary condition. For the microwave part of the computation, you don't have to compute fields within the volume of the metal. After the computation, you will find quantities related to power loss in the surface(s), available for selection and plotting in the Results section.

Don't define your metal (or metal surface) to be a perfect conductor. Set it to be a material (e.g., copper). For the surface, set the boundary condition on it as an impedance boundary condition. For the microwave part of the computation, you don't have to compute fields within the volume of the metal. After the computation, you will find quantities related to power loss in the surface(s), available for selection and plotting in the Results section.

Thank you for your answer, I was expecting the result of microwave heating of the metal to produce a very high temperature rise in a very short time. For impedance boundary conditions I usually just set the microwave cavity boundary as impedance boundary condition and I will try to set the metal to be heated as impedance. One more question, after the calculation is done, can I see the current on the surface of the metal medium? Will there be a large temperature rise in the metal medium? Thanks sincerely!

I only addressed the RF computation part of the problem (but including the loss term) with my answer. To explore thermal evolution (temperature distribution vs. time), you need to properly configure this as a multiphysics problem. It is important that you set the relationships/linkages between these different physics calculations correctly. I encourage you to look at the examples in the Comsol-supplied Applications Library. In particular, see the ones listed under RF Module -- Microwave Heating.

I am wondering why copper should heat up strongly in a microwave field. It is often used for coaxial cables and waveguides with low loss. So unless you apply extremely strong fields copper shouldn't heat up that much. Iron and other less conductive metals are different animals though.