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Solving heat equation for heat-pulse in a point on the surface

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Jbari
#1
Jan18-13, 08:12 AM
P: 8
Hi everybody, I'm trying to find a solution for the 3D heat equation for pulsed surface heating of a semi-infinte solid with insulated surface. I know the method of reflection is required, and that a point source in an infinite solid gives the following solution:

[itex]U(x,y,z,t)= \frac{Q}{8\sqrt{(πκt)^3}}*e^{-\frac{x^2+y^2+z^2}{4κt}}[/itex]

Where κ is thermal conductivity and Q is a measure for the strength of the heat source. However, I have only found a solution for a semi-infinite solid with surface temperature zero and a heat source inside the solid.
In my case however, the heat source is on the surface, (let's say in point (0,0,0)), hence surface temperature cannot be zero, yet to make matters (a little less) complicated, let's assume a perfectly insulated surface with no heat transfer..

Thanks in advance for help, or tips for usefull literature
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Chestermiller
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Jan18-13, 10:09 AM
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Quote Quote by Jbari View Post
Hi everybody, I'm trying to find a solution for the 3D heat equation for pulsed surface heating of a semi-infinte solid with insulated surface. I know the method of reflection is required, and that a point source in an infinite solid gives the following solution:

[itex]U(x,y,z,t)= \frac{Q}{8\sqrt{(πκt)^3}}*e^{-\frac{x^2+y^2+z^2}{4κt}}[/itex]

Where κ is thermal conductivity and Q is a measure for the strength of the heat source. However, I have only found a solution for a semi-infinite solid with surface temperature zero and a heat source inside the solid.
In my case however, the heat source is on the surface, (let's say in point (0,0,0)), hence surface temperature cannot be zero, yet to make matters (a little less) complicated, let's assume a perfectly insulated surface with no heat transfer..

Thanks in advance for help, or tips for usefull literature
Your solution looks OK to me. The partial derivative of U with respect to z is zero, so the surface heat flux (normal to the x-y plane) is zero. Your semi-infinite solid lies above the x-y plane (z > = 0).
Jbari
#3
Jan21-13, 02:35 AM
P: 8
Quote Quote by Chestermiller View Post
Your solution looks OK to me. The partial derivative of U with respect to z is zero, so the surface heat flux (normal to the x-y plane) is zero. Your semi-infinite solid lies above the x-y plane (z > = 0).
Thanks for the quick response, but are you sure this is correct? Because this is the solution for a point source inside an infinite solid, and it looks quite odd to me that a surface heat source in a semi-infinite solid has te same solution. I mean, intuitively, one would expect a different heat propagation in this semi-infinite solid since there is only one way the heat is dissipated (in the part where z>=0) and the heat is 'blocked' in the other direction (where z<0)? Or is this assumption not correct?

Chestermiller
#4
Jan21-13, 08:26 AM
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Solving heat equation for heat-pulse in a point on the surface

Think of it first as an infinite solid. You have a sudden instantaneous spherically symmetric injection of heat at the origin. Then you let the heat diffuse away. Half the previously injected heat goes up, and half the previously injected heat goes down. So there is symmetry of the temperature distribution with respect to the x-y plane (z = 0). No heat crosses this boundary (after t = 0). This is exactly what your solution tells you. The upward heat flux at z = 0 is -kdU/dz, but dU/dz is zero at all times after t = 0.
Jbari
#5
Jan21-13, 08:57 AM
P: 8
Thanks a lot for the extra information, your explanation looks correct indeed, and I understand that in this case my equation covers the problem since I cannot see a theoretical error in your explanation .
However, I still find it odd that there is no effect of the insulated boundary (although I might of course be mistaken): Isn't the heat that normally goes down in the z<0 area (when the solid is infinite), 'trapped' due to thermal insulation of the surface, resulting in the fact that it is dissipated to the other side, leading to a higher heat input in the z>=0 area? And if this is the case, does this only affect the factor Q (heat input) in the equation?
AlephZero
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Jan21-13, 09:07 AM
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Quote Quote by Jbari View Post
However, I still find it odd that there is no effect of the insulated boundary
The point is that, because of the symmetry of the heat flow, there would be no heat flow across the "boundary" plane even if the solid was infinite, not semi-infinite.

If there is no heat flow through a surface, replacing that surface with an insulated boundary doesn't change the heat flow.
Jbari
#7
Jan21-13, 09:16 AM
P: 8
Ok, thanks a lot! I didn't realise I had the correct solution lying around all this time :).
Chestermiller
#8
Jan21-13, 12:55 PM
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Quote Quote by AlephZero View Post
The point is that, because of the symmetry of the heat flow, there would be no heat flow across the "boundary" plane even if the solid was infinite, not semi-infinite.

If there is no heat flow through a surface, replacing that surface with an insulated boundary doesn't change the heat flow.
I like your explanation much better than mine. I was struggling to say just this, but was unable to articulate it as simply and concisely.

Chet
rollingstein
#9
Jan24-13, 11:37 PM
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Quote Quote by Chestermiller View Post
I like your explanation much better than mine. I was struggling to say just this, but was unable to articulate it as simply and concisely.

Chet
Very counter-intuitive. So adding a thin insulating plane to an infinite solid at z=0 would have no influence at all on the T distribution?
Jbari
#10
Feb26-13, 03:20 AM
P: 8
About the insulation: in my question I assumed a perfect insulation with no heat transfer across the boundary, but what if we try to make the situation more realistic and insulation is not perfect.. So instead of a semi-infinite solid, we now actually have 'two semi-infinite solids with their surfaces in thermal contact' (or something like that). Does this change anything for the heat transfer?
Again, intuitively, one would assume it does, but I cannot rely on my intuition :).
Furthermore, small differences are quite important here, because I'm also intrested in the heat propagation on the boundary plane.
Hope this isn't too much to ask?
Chestermiller
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Feb26-13, 06:35 AM
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Let me say this back so that I understand correctly. You now have 2 semi-infinite solids with an imperfect insulating material of thickness ?? between then, and you release the heat at a point on the surface of one of the solids, but not at the mirror image point on the other solid. And the thermal properties of the insulating material is different from that of the two semi-infinite solids. Correct?

Chet
Jbari
#12
Feb27-13, 02:21 AM
P: 8
Ah no, sorry for the confusion, there is 1 semi-infinite solid, and the other semi infinite solid IS the insulating material.. So 2 materials in contact with eachother, with different thermal properties, and with a point heat source on the boundary surface between them. I hope this makes it more clear to you?
Chestermiller
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Feb27-13, 05:52 PM
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Quote Quote by Jbari View Post
Ah no, sorry for the confusion, there is 1 semi-infinite solid, and the other semi infinite solid IS the insulating material.. So 2 materials in contact with eachother, with different thermal properties, and with a point heat source on the boundary surface between them. I hope this makes it more clear to you?
Is the other semi-infinite solid a perfect insulator?
Jbari
#14
Feb28-13, 03:20 AM
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Quote Quote by Chestermiller View Post
Is the other semi-infinite solid a perfect insulator?
No, it is a 'realistic' insulator, so just a material with a lower thermal conductivity than the other semi-infinite solid, so from the mathematical point of view it could be any material..
Chestermiller
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Feb28-13, 06:40 AM
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Quote Quote by Jbari View Post
No, it is a 'realistic' insulator, so just a material with a lower thermal conductivity than the other semi-infinite solid, so from the mathematical point of view it could be any material..
If the thermal conductivities are not equal, you can still have zero heat flux at the interface if the thermal diffusivities are equal. The heat pulse will just initially partition between the two slabs in proportion to the thermal conductivities. After that, no heat flow will occur across the interface.

If the thermal diffusivities are unequal, there will be heat flow across the interface. I'm not sure whether this problem has an analytic solution. Of course, it can always be solved numerically.
rollingstein
#16
Feb28-13, 08:01 AM
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What confuses me is how come the OP's original equation does not have a diffusion term at all? Shouldn't k, Cp and density always occur in combination (i.e. Diffusivity) in the solutions of the heat equation?
Jbari
#17
Feb28-13, 08:18 AM
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Quote Quote by rollingstein View Post
What confuses me is how come the OP's original equation does not have a diffusion term at all? Shouldn't k, Cp and density always occur in combination (i.e. Diffusivity) in the solutions of the heat equation?
You are correct, I ment κ to be thermal diffusivity (= K/(ρ*Cp)) but I defined it incorrectly in my explanation as thermal conductivity..

And to return to the problem: the thermal diffusivities are NOT equal, but I would still like to find an analytical solution (if possible of course). Any suggestions on how/where to find it (maybe in literature, but my search has been fruitless untill now).
Chestermiller
#18
Feb28-13, 08:33 AM
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Quote Quote by rollingstein View Post
What confuses me is how come the OP's original equation does not have a diffusion term at all? Shouldn't k, Cp and density always occur in combination (i.e. Diffusivity) in the solutions of the heat equation?
Two reasons. First of all, in the OP's relationship, κ (kappa) is the thermal diffusivity, not, as he stated, the thermal conductivity k. Secondly, the term involving Q is not quite right. The units don't properly give temperature. I'm too lazy to look up what the correct expression for what this term should be, but, at the very least, there should be a k (thermal conductivity) in the denominator (if U has units of temperature and Q has units of energy). I'm guessing that the denominator should be a constant times [itex]kt\sqrt{\kappa t})[/itex]. This would give the correct units for temperature.


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