Shadow wall in the heat transfer problem

AbubakarAbubakar Member Posts: 46

What is the significance of the shadow wall in the heat transfer problem? I understood that when we have two-sided walls the fluent will automatically create the shadow wall of the same thickness as the original wall but my question is why we need that? we will be having the same flux for the shadow wall and the original wall. Please help. Thanks in Advance.


Thank you,

Abubakar khan

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  • RobRob UKPosts: 11,770Forum Coordinator

    It's the "other side" of the wall. It's mainly to give us a unique label for each side of the wall. It may be smooth on one side and rough on the other (turbulence/wall shear) or white on one side and black on the other (radiation). If the thin wall has a finite thickness you will also find it useful when plotting the temperature on each side.

  • AbubakarAbubakar Posts: 140Member

    Thanks, @Rob for your response still it's unclear to me I am not getting the significance I mean when I am going to get the same results on the original wall and the shadow wall, then why do I need that shadow wall? TIA

  • RobRob UKPosts: 11,770Forum Coordinator

    Find a piece of sand paper. One side is rough, the other smooth: one is the wall, the other the shadow. It's purely a labelling system that also allows us to specify different surface properties on either side. In most cases we don't do anything with either surface. They're linked in the way they are because heat going in one side comes out the other.

  • AbubakarAbubakar Posts: 140Member

    Understood @Rob sir. Just one follow-up question What if I delete that shadow wall? Will that affect my results? TIA

  • RobRob UKPosts: 11,770Forum Coordinator

    Yes and no. In the unlikely event that you're successful the solver will fall over as you'll have corrupted the mesh. Internal walls MUST have a wall and wall:shadow pair; if you change the boundary type to interior etc the two faces are merged to the new boundary. Turning an interior surface to a wall forms the wall and wall:shadow pair.

  • ibmb2020ibmb2020 Posts: 4Member

    Hello Rob,

    I want to just pop up this thread, as I understand and as you explained in previous posts the heat flux through the coupled wall must be equal if there is no source on wall.

    I have question now, I simulate conjugate heat transfer (coupled wall) with radiation on opaque wall (DO method), I do post process radiation heat transfer rate on both side of wall, I get followings;


    Radiation Heat Transfer Rate (w)

    ------------------------------------------------------------

    wall_adjacent_to_fluid -23000

    same_wall_adjacent_to_solid -0

    Net radiation heat source 100000

    --------------------------------------------------------------

    Net 77000


    Why do I have different numbers on both sides? Is my radiating flux comes from fluid, really transfer to the solid? If yes why solid wall heat transfer is zero? My internal emissivity is 1 on that wall. In solid zones "Participates in Radiation" is not selected since I don't know what it means, and user guide says very few things about that.

    If you can clarify, I will be appreciated. Have a nice day.

  • RobRob UKPosts: 11,770Forum Coordinator

    Radiation passes to the wall and becomes heat which may conduct into the solid, or be re-emitted into the fluid. No radiation passes to the solid zone as it's not participating.

  • DrAmineDrAmine GermanyPosts: 7,893Forum Coordinator

    Double check the total heat 🔥 rate and not only a portion of it

  • ibmb2020ibmb2020 Posts: 4Member

    Thank you all for fast repsonpe,

    I actually looking for the heat flux passing to my solid zone via radiation and compare with my convection portion of heat flux. My solid zone is not transmitting, so I believe radiation heat flux conducted to solid zone is absorbed one minus emitted one. If I wrote as in textbooks;

    qrad'' = -(Internal_Emissivity*Surface_Incident_Radiation) + n^2*Internal_Emissivity*Boltzman_Constant*Tw^4

    Fluent gives "Surface Incident Radiation", " Absorbed Radiation Heat Flux", "Reflected Radiation Heat Flux" and "Radiation Heat Flux". But I cant pass into radiation heat flux from the equation above, as i calculated and what fluent gives are a little different. I am calculating emissive power less, and fluent doesn't show it unfortunately.

    Secondly when I integrate "Radiation Heat Flux" from fluent over surface, it gives exactly same with post processed value in "Flux Report> Radiation Heat Transfer Rate>Results". But that screen also shows "Net Radiation Source" and "Net Results", this is where I am confused. I just wonder what portion of heat flux conducted to solid comes from convection and radiation. Am I suppose to just compare "Radiation Heat Flux" and "Total Surface Heat Flux" appear in "Surface Integrals>Wall Fluxes" and do not bother with Flux Reports screen?

    Lastly, as I understood no radiation heat flux in shadow wall (neighbour to solid) is ok, since fluent conducts the radiation heat flux (comes into wall neighbour to fluid) to the solid zone anyway?

  • DrAmineDrAmine GermanyPosts: 7,893Forum Coordinator

    The Net of Radiation Heat Flux might be different than zero as it represent the amount of energy being absorbed by the material.

    For an opaque wall, net radiative heat flux can be regarded as Absorbed heat flux-emitted heat flux.

    Or Incoming heat flux - outgoing heat flux= Surface incident radiation- reflected radiative heat flux - emitted radiative heat flux

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