Thermal conductivity tensor

In summary, the conversation discusses the relationship between temperature gradient and heat flow in materials. It is mentioned that anisotropic materials are more likely to have heat flowing in a direction perpendicular to the temperature gradient. Graphene is given as an example of an anisotropic material with anisotropic thermal conductivity, and the discussion delves into the mathematical explanation of this property. It is also mentioned that in some cases, heat flow may not be exactly orthogonal to the temperature gradient, but it will have a component that is perpendicular to it.
  • #1
dRic2
Gold Member
883
225
Hi, yesterday a professor of mine told me that if you have a temperature gradient along the x-axsis you could have heat flowing in the y direction. Mathematically it is pretty straightforward to find the thermal conductivity tensor required, but in real life can you name some materials that share this property ?
 
Science news on Phys.org
  • #2
Essentially most anisotropic materials.

In more technical terms, if the decomposition of the symmetric traceless representation of rotations includes any singlets of the symmetry group of the material, you would have a harder time explaining why that material should not have an anisotropic thermal conductivity than why it should be isotropic. An example that I would expect to have anisotropic thermal conductivity is graphene, but I am no expert on thermal conductivity properties of materials - this is a pure symmetry argument. @Chestermiller ?
 
  • #3
Consider even a uniform material.

If I have a long rod with both tips at 0C heat baths the rod will reach 0C. If I replace the ice at one tip with a blowtorch at the center of the tip's surface, a temperature gradient will develop along the rod. The surface of the rod along the long direction will also heat up. That means heat is being transferred radially, not just longitudinally.

Is that an example? If not, why not?
 
  • #4
Vanadium 50 said:
That means heat is being transferred radially, not just longitudinally.

Is that an example? If not, why not?
Heat in this case is being transferred radially because there will be a radial component of the temperature gradient.
 
  • Like
Likes Chestermiller and dRic2
  • #5
Vanadium 50 said:
Is that an example? If not, why not?
No because the surface of the rod is initially at 0C and so there is also a radial component of the temperature gradient as the rod warms up that drives the heat flow.

Orodruin said:
An example that I would expect to have anisotropic thermal conductivity is graphene
Consider a very very thin slab of graphene. Heat will flow poorly through the layers of graphene while it will flow without problems along the layers. If you apply a temperature gradient across the faces of the slab I think you will only see an high resistance to the flow but I don't think you will measure heat flowing in the perpendicular direction. IMHO

PS: As I interpreted the statement of my professor the heat must flow only in the perpendicular direction, there is no heat flow along the direction of the temperature gradient
 
  • #6
dRic2 said:
Consider a very very thin slab of graphene. Heat will flow poorly through the layers of graphene while it will flow without problems along the layers. If you apply a temperature gradient across the faces of the slab I think you will only see an high resistance to the flow but I don't think you will measure heat flowing in the perpendicular direction. IMHO
You will never get a flow that is exactly orthogonal to the gradient. However, you will get a flow that is not parallel to the gradient (i.e., it has a component that is orthogonal to the gradient) if you apply a temperature gradient that is not along one of the principal directions. In the case of graphene, apply a gradient that is not along the sheets and not perpendicular to the sheets, i.e., taking ##\vec n## to be the direction perpendicular to the sheets and ##\vec t## to be tangent to the sheets, you apply a gradient ##\nabla T = T^t \vec t + T^n \vec n##. Conductivity is a linear map from a temperature gradient to a heat flux density and therefore
$$
\vec J = -K(\nabla T) = - K(T^t \vec t + T^n \vec n) = - T^t K(\vec t) - T^n K(\vec n).
$$
The directions ##\vec t## and ##\vec n## are in the principal directions of the material so ##K(\vec t) = K^t \vec t## and ##K(\vec n) = K^n\vec n##, where ##K^t## and ##K^n## are material constants. It follows that
##
\vec J = - T^t K^t \vec t - T^n K^n \vec n,
##
which is not parallel to ##\vec J## unless ##K^t = K^n## or ##\nabla T## is along one of the principal directions. In your example, you took ##\nabla T## to be in a principal direction.

Edit: So, the underlying reason that the heat flow is not parallel to the temperature gradient is that it is much easier for the temperature gradient to drive a flow in a direction along the sheets as compared to driving a flow between the sheets.
 
  • Like
Likes anorlunda
  • #7
Orodruin said:
You will never get a flow that is exactly orthogonal to the gradient. However, you will get a flow that is not parallel to the gradient (i.e., it has a component that is orthogonal to the gradient) if you apply a temperature gradient that is not along one of the principal directions. In the case of graphene, apply a gradient that is not along the sheets and not perpendicular to the sheets, i.e., taking ##\vec n## to be the direction perpendicular to the sheets and ##\vec t## to be tangent to the sheets, you apply a gradient ##\nabla T = T^t \vec t + T^n \vec n##. Conductivity is a linear map from a temperature gradient to a heat flux density and therefore
$$
\vec J = -K(\nabla T) = - K(T^t \vec t + T^n \vec n) = - T^t K(\vec t) - T^n K(\vec n).
$$
The directions ##\vec t## and ##\vec n## are in the principal directions of the material so ##K(\vec t) = K^t \vec t## and ##K(\vec n) = K^n\vec n##, where ##K^t## and ##K^n## are material constants. It follows that
##
\vec J = - T^t K^t \vec t - T^n K^n \vec n,
##
which is not parallel to ##\vec J## unless ##K^t = K^n## or ##\nabla T## is along one of the principal directions. In your example, you took ##\nabla T## to be in a principal direction.

Edit: So, the underlying reason that the heat flow is not parallel to the temperature gradient is that it is much easier for the temperature gradient to drive a flow in a direction along the sheets as compared to driving a flow between the sheets.

That is obvious: it is just the vector nature of the flow. I was asking the other thing but you said it was impossible.

For example can a conductivity tensor be like this:
$$ \mathbf k = \begin {bmatrix}
0 & k_{xy} \\
k_{yx} & 0 \\
\end {bmatrix}
$$ ?

In this way if you take a temperature gradient only on the x direction

$$\mathbf Q = \mathbf k \cdot ∇ T =
\begin {bmatrix}
0 & k_{xy} \\
k_{yx} & 0 \\
\end {bmatrix} \cdot \begin{bmatrix} \partial_x T \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ k_{yx} \partial_x T \end{bmatrix}
$$
 
  • #8
That would imply a negative eigenvalue and therefore heat could flow from lower to higher temperature, which breaks the laws of thermodynamics. The conductivity tensor must be positive definite.
 
  • #9
dRic2 said:
Hi, yesterday a professor of mine told me that if you have a temperature gradient along the x-axsis you could have heat flowing in the y direction.
dRic2 said:
That is obvious: it is just the vector nature of the flow.
If you think it is obvious I don't understand your confusion. The statement, as you have recounted it, does not say that the flux is exclusively in the y-direction.
 
  • #10
If you have a composite material, the thermal conductivity can typically be anisotropic. For example, if you have rebar in cement, the preferred direction of heat conduction will be in the long direction of the rebar. This is analogous to the anisotropic mechanical properties of a composite.
 
  • #11
dRic2 said:
Hi, yesterday a professor of mine told me that if you have a temperature gradient along the x-axsis you could have heat flowing in the y direction. Mathematically it is pretty straightforward to find the thermal conductivity tensor required, but in real life can you name some materials that share this property ?

Anisotropic crystalline materials:

https://arxiv.org/ftp/arxiv/papers/1809/1809.04762.pdf
https://journals.aps.org/prb/abstract/10.1103/PhysRevB.66.064525
https://dspace.mit.edu/bitstream/handle/1721.1/4445/rle-tr-377-04740983.pdf?sequence=1
 
  • #12
Orodruin said:
If you think it is obvious I don't understand your confusion. The statement, as you have recounted it, does not say that the flux is exclusively in the y-direction.
I think I solved it. Probably I misunderstood what the professor said: tomorrow I'll talk to him.

Can you tell me more about the problem of negative eigenvalues because I didn't get it very much.

Thanks to everyone for the replies. :D
 
  • #13
dRic2 said:
Can you tell me more about the problem of negative eigenvalues because I didn't get it very much.
In essence, if you have a conductivity on the form you suggested, then it would have negative eigenvalues. This would mean that there is a direction in which you can apply a temperature gradient and get heat to flow in the same direction as the gradient, i.e., from low to high temperature. If you can do this you break the laws of thermodynamics.
 
  • #14
Orodruin said:
In essence, if you have a conductivity on the form you suggested, then it would have negative eigenvalues. This would mean that there is a direction in which you can apply a temperature gradient and get heat to flow in the same direction as the gradient, i.e., from low to high temperature. If you can do this you break the laws of thermodynamics.

I got it now. I have one more question : Does the tensor need to be symmetric ? I'm a bit confused over this one... Is this property shared by anisotropic materials linked to the symmetry of the tensor ?
 
  • #15
dRic2 said:
I got it now. I have one more question : Does the tensor need to be symmetric ? I'm a bit confused over this one... Is this property shared by anisotropic materials linked to the symmetry of the tensor ?
Even if the material is anisotropic, the thermal conductivity will typically be symmetric, with 3 real principal- directions and magnitudes.
 
  • Like
Likes dRic2

1. What is a thermal conductivity tensor?

A thermal conductivity tensor is a mathematical representation of the thermal conductivity of a material. It describes the directional dependence of heat transfer in a material, taking into account both the magnitude and direction of heat flow.

2. How is the thermal conductivity tensor measured?

The thermal conductivity tensor can be measured using various experimental techniques, such as the transient hot-wire method or the laser flash method. These methods involve measuring the temperature gradient and heat flux in different directions in a material, and using this data to calculate the thermal conductivity tensor.

3. What factors affect the thermal conductivity tensor?

The thermal conductivity tensor is affected by a variety of factors, including the material's composition, structure, and temperature. Anisotropic materials, such as crystals, may also have different thermal conductivity tensors in different crystallographic directions.

4. How does the thermal conductivity tensor relate to thermal conductivity?

The thermal conductivity tensor is a more comprehensive measure of a material's thermal conductivity compared to a single value. It takes into account the directional dependence of heat transfer, whereas thermal conductivity only gives an overall average value. The thermal conductivity tensor can be used to calculate the thermal conductivity of a material in any direction.

5. Why is the thermal conductivity tensor important in materials science?

The thermal conductivity tensor is important in materials science because it helps engineers and scientists understand and predict how heat will transfer within a material. This information is crucial for designing and optimizing materials for various applications, such as in thermoelectric devices, heat exchangers, and electronic devices.

Similar threads

Replies
2
Views
1K
  • Thermodynamics
Replies
2
Views
997
Replies
16
Views
12K
Replies
5
Views
980
Replies
13
Views
3K
  • Introductory Physics Homework Help
Replies
2
Views
236
  • Thermodynamics
Replies
5
Views
945
Replies
26
Views
3K
Replies
1
Views
447
  • Thermodynamics
Replies
3
Views
34K
Back
Top