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Modeling Heat Transfer with Comsol

  1. Jun 10, 2009 #1
    Hi Everyone,

    I'm using Comsol v.3.4 to model a 1-meter copper rod experiencing a Temperature change of 294 Kelvin (4K at bottom and 300K at top). I have a slow computer/connection to the server, so I can't see any 3-D images or the Temperature plots. I'm working in the axial symmetry (2-D) space dimension.

    I think I am doing this(Conduction) correctly, and now I need to add the convection portion of the problem. It is being cooled by Liquid Helium Gas. How do I go about doing this? Do I just change the boundary conditions of the outer surface?

    Appreciate your time,
    Mike
     
  2. jcsd
  3. Jun 10, 2009 #2
    You will need to change the boundary conditions of your material since I am assuming they are insulated. You will also need to create the boundary conditions of the liquid helium as well.
     
  4. Jun 10, 2009 #3
    Thank you for the response Topher.

    Where can I create the boundary conditions for the liquid helium? I assume I need to make that before I can change the boundary conditions because nothing happens when I set the outer surface to convective flux.

    Unless I'm supposed to change the outer boundary to "Heat Flux" and give it a negative value?
     
  5. Jun 10, 2009 #4
    If there is no subdomain in contact with the surface of your bar then there can not be any convective flux.

    If you know the amount of heat being removed from the bar by the liquid helium then you can have a "heat flux" boundary, for example 1600 watts/m^2 of heating transported through that surface.

    If you have your metal bar in a flow of helium and want to model the removal of heat by convection then you need to make a boundary for which that helium will be flowing through. For example, you may have a tube in which the helium will flow through and inside that tube you will have your heated metal bar.
     
  6. Jun 11, 2009 #5
    Since the difference in temperature is so great, the thermal resistance is affected by the purity of the copper. The Residual Resistivity Ratio(RRR) affects both Thermal and Electrical Resistance right? Do you know if Comsol already took that into account? My temperature profile came out linear, so I don't think it does.. Have you or anyone else come across this problem?
     
    Last edited: Jun 11, 2009
  7. Jun 11, 2009 #6

    Mech_Engineer

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    Keep in mind that FEA is not a magic bullet- you'll have to do some up-front analysis. Specifically, you're going to have to calculate the convective coefficient for the liquid helium on the outside of the rod.

    I would also do some analytical calcs to make sure your FEA results seem to be in the ballpark. Its very easy to make a simple mistake which will give you an answer that's off by a factor of 2 (or even an order of magnitude).
     
  8. Jun 11, 2009 #7
    I've been trying to do the analytical calculation, but I don't know what I'm looking for. I'm new to heat transfer and this problem is much more intense than other problems in that the temperature change is so drastic.

    Could you point me in the direction of a good example, or some good reading material on the subject? What's a key word I should look out for?
     
  9. Jun 11, 2009 #8
    Theres a good chance you will get an answer thats off by a factor of two even if you do everything right. It's important to know how to experimentally verify your models for accuracy, otherwise your model is worthless.

    It looks to me that you need to put away Comsol and open up a text book. I would suggest going out and finding a good heat transfer text and at least learn the basics. After that, go find a text on numerical modeling which covers the finite element, finite volume, and finite difference methods. After that you might be ready for COMSOL.
     
  10. Jun 11, 2009 #9
    I've already done research on Heat Transfer. It is a difficult problem because this isn't a simple problem like in the books. A few books suggest a linear relation, just as Comsol does, but that's assuming near room temperature experiments or non-drastic temperatures changes.

    Do you have any suggestions for good books on this topic?
     
    Last edited: Jun 11, 2009
  11. Jun 11, 2009 #10
    Here's the equation I'm working with:

    [tex]A * \frac{\partial}{\partial x}* \left[ \kappa(T(x)) * \frac{\partial T(x)}{\partial x}\right] = 0 [/tex]

    where
    x : Position
    A : Cross sectional area of the rod
    [tex] T(x) [/tex] : Temperature as a function of position
    [tex] \kappa ( T(x) ) [/tex] : the coefficient of thermal conductivity as a function of Temperature


    That covers the Heat transfer by conduction. I haven't solved this yet, so after that I will focus on Helium cooling. I'm trying to solve for [tex] T(x)[/tex].
     
  12. Jun 15, 2009 #11
    The Residual resistivity Ratio(RRR) affects both Thermal and Electrical Resistance right? Do you know if Comsol already took that into account?
     
  13. Jun 15, 2009 #12

    Mech_Engineer

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    Comsol won't take special nonlinear considerations (like the RRR) into account unless you specifically add it, so no.
     
  14. Jun 17, 2009 #13
    This may be a dumb question, but: Where would I add that?

    Would it be under physics->subdomain settings->Thermal Conductivity?

    I tried to input an expression in there with a variable T for temperature, but it did not recognize that variable(which I kind of expected). I know where I can declare constants, is there anywhere to declare(or associate) T with Temperature?
     
  15. Jun 17, 2009 #14
    Since you already added conduction physics to your model, a variable associated with temperature should already be defined. check your output variables by
    solve->solver manager-> output vairables.

    You can also see what variables comsol associates with a specific subdomain by
    Physics->equation systems->subdomain settings->vairables

    This is useful because these are some of the variables comsol already knows, so you can use them to define other functions, ie "couple the physics"
     
  16. Jun 23, 2009 #15
    i would suggest re-doing your model with the fluid structure thermal analysis module. it would probably be set up already for all your inputs and save you a lot of time.
     
  17. Jun 24, 2009 #16
    There are a few fluid thermal packages, could you specify which one you are referring to Nick?

    Also, does anyone know how to access/view the differential equations that Comsol is using? My mentor wants me to find the equation that solves the system, but all I can come up with is a Temperature profile.

    Thanks,
    Mike
     
  18. Jun 24, 2009 #17
    check out Physics-->Subdomain Settings....the equaiton is displayed at the top of the dialouge box.

    Vary the physics being used under multiphysics and repeat the above to view the equations corresponding to the various physics involved
     
  19. Jun 29, 2009 #18
    Thanks Effort,

    Now that I got a temperature profile using the Conduction Multiphysics package, I'd like to try to get the same result by inputting my own equation in the PDE mode. What's the difference between the Weak form, General form, and the coefficient form?
     
  20. Jun 30, 2009 #19
    The General and coefficient PDE's are essemtially the same, there are just written in different formats. I believe one can be used for nonlinear PDEs where the other is only for linear PDEs. I'm not familiar with the weak form, but I am almost certain you wont need to use it to input your own heat equation.
     
  21. Jul 7, 2009 #20
    OK, so I have gone over the user guide and I see that they are the same. So, I'm still trying to solve the equation:

    [tex] A \cdot \frac{\partial}{\partial x} \cdot \left[ \kappa(T(x)) \cdot \frac{\partial T(x)}{\partial x}\right] = 0 [/tex]

    And from what I can gather, I would just change the coefficient form from

    [tex] e_{a} \cdot \frac{\partial^{2} u}{\partial t^{2}} + d_{a} \cdot \frac{\partial u}{\partial t} + \nabla \cdot (- c \cdot \nabla u - \alpha u + \gamma) + \beta \cdot \nabla u + a u = f [/tex]

    to

    [tex] A \cdot \frac{\partial}{\partial x} \cdot \left[ \kappa(T(x)) \cdot \frac{\partial T(x)}{\partial x}\right] = 0 [/tex]

    by making

    [itex] e_{a} = 0 [/itex]
    [itex] d_{a} = 0 [/itex]
    [itex] c = -\kappa(T(x)) [/itex]
    [itex] \beta = 0 [/itex]
    [itex] f = 0 [/itex]
    [itex] \alpha = \gamma = 0 [/itex]

    Is this correct? Also, I wanted to check with you to make sure that it is true that u is now the dependent variable instead of T.
     
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