Discussion Forum

Hi

the 3rd dimension Z is constant (invariant) but you would need it to define total power flux, as you lines = boundaries in 2D give you fluxes that must be multiplied by the depth to get total power out, no ?

So COMSOL's default is "always 3D". So for 2D: with 1m depth Z or 1m^2 area YZ (in 1D)

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Good luck
Ivar

Hi Ivar,

Thank you very much for your reply.
So taking the 3rd dimension into consideration I need to increase my heat input as a part of it goes into the extra dimension? Also how do we accurately predict its result?. e.g predicting the melt depth in a 2D model?
Thanks in advance for your help

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MKSharma

Hi

there is no change in Z, the value you read in X & Y is INVARIANT or constant along Z. You only need to define the depth Z for absolute values. If you enter a power flux of 1[W], on a boundary, it's 1W per meter depth implicitely.

Check the units, if you want to get absolute values and integrate over boundaries in 2D you get mostly units/meter so you should multiply by the true "depth" OR set the true Z depth as you can in given phyiscs under the main physics node.

Read the doc, and check the units, as well as the formula COMSOl is using
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Good luck
Ivar

Dear Ivar,

Thank you very much for your valuable time and suggestion.
Let us say I create a 2D geometry of 10X10 microns and I want to apply a volumetric heat source which is a function of space and time with unit [W/m^3]. The doubt I have is, with 2D how is it possible to apply volumetric heat source?. I cannot use heat flux or boundary heat source as the unit of my heat input is W/m^3 whereas heat flux/boundary heat source has a unit of W/m^2.
Thanks again for your help

Manoj

Hi

but in 2D your "domain" is not a volume but a surface so you deposit

P0[W/m^2] = Pvol[W/m^3]*depth_in_Z[m]

You must multiply your volumetric power by the depth to get the average surface power you deposit in the 2D domain = surface

This assumes that the power is constant over the depth

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Good luck
Ivar

Dear Ivar,

Thank you very much for your help.
I shall try that and will get back to you in case of any difficulty.
Thanks alot for your help again.

Regards,
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MKSharma

Hi Ivar,

I got a basic doubt.
I apply gaussian beam as my heat source [W/m^3] for a 2D model. I have question like if I draw a rectangle with upper boundary at 0 and lower boundary at -20. When we simulate where does the laser beam hit the target material (top or bottom). Its a silly doubt but I was just curious and thought you can help me understand it.

Regards,
Manoj

Hi

I would say its an interesting question. As a laser beam, of not hitting a transparent glass wall, will stop on the surface, hence be rather an "out of plane" 2D boundary source of X-Y extent (in 2D) hence W/m^2 power density

But then you should probably define the laser beam penetration via a 2D-axi model, and convert the result to a 2D x-y beam model. I wounder if that was the way proposed n the model library laser beam example

a 2D model is a simplification so it will never be as precise as a full 3D case, particularly for such types of interactions that are not fully 2D and Z invariant

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Good luck
Ivar

Hi Ivar,

Thanks again for your valuable suggestion. You are right, for a 2D model it makes more sense to use a power density [W/m^2], but when we include the parameters like absorption coefficient and reflectivity and Beer Lambert's law of absorption in our heat source the unit becomes [W/m^3] and because the absorption coefficient has a unit of [1/m] the input power becomes very high. I am stuck here with this problem and not getting an idea how to deal with it.
Hope you can suggest me some solution for this.

Kind Regards,
Manoj