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December 4, 2012 7:52am UTC in response to Sean Wang

Re: Please help me find how to achieve this function.

Hi

you cannot set a constant T BC and a convective cooling BC on the same boundary, these two conflict, its one or the other.

The easiest is to define the two BC, then use two time series solver sequences in series, the first one (0<t<15 sec) you disable the Convecting cooling and enable the constant temperature, then for the second (t>15sec) you enable the convective cooling and disable the constant temperature BC. This you can do in the solver Study step1 node under Physics and variable selection in v4.3a

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Good luck
Ivar

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December 4, 2012 7:46pm UTC in response to Sean Wang

Re: Please help me find how to achieve this function.

Hi

first some tricks ;)

When you have such nice geometries, you should rather use the layer, its much quicker to build your model like that, the layer have thickness that are cumulative (you define each layer per say) in this way you need only 1 rectangle

Your T232 cannot be defined in a Global parameter as it depends on "z" a local coordinate so it cannot be evaluated and Parameters are all evaluated once at study launch
But you can define a "model definitions variables" T232 valid on some Domain(s) or Boundaries, or all

Then you should move your Convective cooling up one level, before the Temperature BC definition. THis will make the convective cooling operate, but be disabled on the burn region while you are heating this one up with the "lower" Temperature BC case. And for your first solver case leave all BC's active.

Then enable your 2nd time study step, make it disable the Temperature BC (as you have done), but then turn ON the Initial values solved for, and point it to Study 1 Time dependent, you want the temperatures from study 1 too, not only all the other variables

When you now make a default solver, you get both steps, but indeed COMSOL does not store the first one ;)
So try to put it in yourself: right click study 1 add "Other Store solution", and move it up just after the node Time dependent solver 1 and before the Compile equations Time dependent 2, then SOLVE, you get now 2 data set, one per time solver

You have chosen a BDF free time stepping, I would suggest to use a "Intermediate" time stepping, or to use a time stepping as a power series (I often use a 2^{range(-2,1,8} to step between 0.25 and 256 sec in steps of power of 2, this gives a reasonable stepping change for diffusion problems like thermal decays)

One thing, your mesh is still rather coarse, try having at least 3 (or more) elements in the thickness, per layer, if you really want to resolve the gradient of T reasonably

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Good luck
Ivar

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