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2D Modeling in COMSOL

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Hi,

I am developing a 2D model in COMSOL Multiphysics using Heat Transfer Module. I have a question about how COMSOL treat 2D model because I saw that it virtually adds 3rd dimension of depth. I am bit confused like how do I handle it because if COMSOl virtually adds 3rd dimension then my heat gets distributed in all the 3 dimensions and not 2 which I want to do and ultimately I see the temperature drop in the material different than what I expect it to be.
Is there anyone who faced similar problem?

Regards,
--
MKSharma

9 Replies Last Post 3 ott 2012, 02:59 GMT-4
Ivar KJELBERG COMSOL Multiphysics(r) fan, retired, former "Senior Expert" at CSEM SA (CH)

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Posted: 1 decade ago 25 set 2012, 10:47 GMT-4
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)

--
Good luck
Ivar
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) -- Good luck Ivar

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Posted: 1 decade ago 25 set 2012, 11:06 GMT-4
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

--
MKSharma
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 -- MKSharma

Ivar KJELBERG COMSOL Multiphysics(r) fan, retired, former "Senior Expert" at CSEM SA (CH)

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Posted: 1 decade ago 26 set 2012, 01:15 GMT-4
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
--
Good luck
Ivar
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 -- Good luck Ivar

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Posted: 1 decade ago 26 set 2012, 05:09 GMT-4
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
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

Ivar KJELBERG COMSOL Multiphysics(r) fan, retired, former "Senior Expert" at CSEM SA (CH)

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Posted: 1 decade ago 26 set 2012, 05:23 GMT-4
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

--
Good luck
Ivar
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 -- Good luck Ivar

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Posted: 1 decade ago 26 set 2012, 05:53 GMT-4
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,
--
MKSharma
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, -- MKSharma

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Posted: 1 decade ago 2 ott 2012, 10:13 GMT-4
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 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

Ivar KJELBERG COMSOL Multiphysics(r) fan, retired, former "Senior Expert" at CSEM SA (CH)

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Posted: 1 decade ago 2 ott 2012, 15:25 GMT-4
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

--
Good luck
Ivar
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 -- Good luck Ivar

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Posted: 1 decade ago 3 ott 2012, 02:59 GMT-4
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
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

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