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

• Jbari
In summary, the conversation discusses finding a solution to the 3D heat equation for pulsed surface heating of a semi-infinite solid with an insulated surface. The method of reflection and a solution for a point source in an infinite solid are mentioned, but the question is about finding a solution for a surface heat source. It is explained that in this case, the solution for a point source in an infinite solid still applies due to the symmetry of heat flow. The effect of an insulated boundary on heat flow is also discussed.
Jbari
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:

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

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

Jbari said:
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:

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

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).

Chestermiller said:
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?

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.

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?

Jbari said:
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.

Ok, thanks a lot! I didn't realize I had the correct solution lying around all this time :).

AlephZero said:
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

Chestermiller said:
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?

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 interested in the heat propagation on the boundary plane.
Hope this isn't too much to ask?

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

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 each other, 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?

Jbari said:
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 each other, 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?

Chestermiller said:
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..

Jbari said:
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.

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?

rollingstein said:
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 until now).

rollingstein said:
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 $kt\sqrt{\kappa t})$. This would give the correct units for temperature.

Jbari said:
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 until now).

Try "Conduction of Heat in Solids" [H. S. Carslaw, J. C. Jaeger] . Another possibility is to solve it yourself. Of course, there won't be a nice similarity solution like the one you have given.

Chestermiller said:
Try "Conduction of Heat in Solids" [H. S. Carslaw, J. C. Jaeger] . Another possibility is to solve it yourself. Of course, there won't be a nice similarity solution like the one you have given.

+1 for that book. If it is documented anywhere it is in there.

I'll add a possibility of trying to take help from one of the Symbolic solvers like Maxima / Mathematica etc. after you frame the problem.

## 1. How do you solve the heat equation for a heat-pulse in a point on the surface?

The heat equation can be solved using various numerical methods, such as the finite difference method or the finite element method. These methods involve discretizing the heat equation into smaller, solvable equations and then using algorithms to solve for the temperature at each point on the surface.

## 2. What are the boundary conditions for solving the heat equation for a heat-pulse in a point on the surface?

The boundary conditions for solving the heat equation typically include the initial temperature distribution on the surface, the thermal properties of the material, and any external heat sources or sinks. These conditions are necessary to determine the temperature at each point on the surface as the heat pulse propagates through the material.

## 3. How does the thermal diffusivity affect the solution of the heat equation for a heat-pulse in a point on the surface?

The thermal diffusivity, which is a measure of how quickly heat can diffuse through a material, plays a crucial role in the solution of the heat equation. A higher thermal diffusivity will result in a faster propagation of the heat pulse, while a lower thermal diffusivity will result in a slower propagation.

## 4. Can the heat equation be solved analytically for a heat-pulse in a point on the surface?

In most cases, the heat equation cannot be solved analytically for a heat-pulse in a point on the surface. This is because the heat equation is a partial differential equation, and the solution requires knowledge of the temperature at all points on the surface at all times. However, for simpler geometries and boundary conditions, analytical solutions may be possible.

## 5. How is the solution of the heat equation for a heat-pulse in a point on the surface used in real-world applications?

The solution of the heat equation for a heat-pulse in a point on the surface is used in a wide range of real-world applications, such as predicting the temperature distribution in materials during manufacturing processes, designing thermal insulation systems, and understanding the behavior of Earth's subsurface temperatures. It is also an important tool in studying heat transfer in various engineering and scientific fields.

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