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According to Wikipedia the rate of heat flow is:$$\frac{\Delta Q}{\Delta t}=-kA\frac{\Delta T}{\Delta x}$$

  • Where to evaluate?

    I can't understand where we must evaluate this rate. In the hottest side of the object or in the coldest side or somewhere else?

  • Thickness dependance

    When we have 1 material (no conductor between the hottest and coldest side of the material) then why the rate depends on the thickness of the material?

  • How $A$ is defined?

    In the case of 1 material what is "the surface area of the surface emitting heat" $A$? It is an imaginary area?

  • What $k$ to choose

    What if we have two different materials in contact with different temperature with $k_1$, $k_2$. Will $k=k_1$ or $k=k_2$?

The only way that this rate makes sense to me is if we consider that the temperature doesn't change over time in the hottest and the coldest side. Any ideas?

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1 Answer 1

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The equation cited is Fourier's Law of Conduction, in this case, for steady heat heat flow. It is generally written as

$$\dot Q=-kA\frac{dT}{dx}$$

In words, it says "the time rate of heat transfer through a material is proportional to the negative gradient in the temperature and the area"

Where to evaluate?

I can't understand where we must evaluate this rate. In the hottest side of the object or in the coldest side or somewhere else?

Typically, it is used to evaluate steady heat flow through a flat wall. See the diagram below. For steady heat flow through a wall of thickness L with the temperatures at each surface of the wall held constant, the equation can be written.

$$\dot Q=\frac{-kA(T_{2}-T_{1})}{L}$$

An example is the exterior wall of a building. The interior wall surface would essentially be the same temperature of the air inside in contact with the wall. The exterior wall temperature the same as the temperature of the air outside in contact with the wall.

You can see that if $T_{1}>T_{2}$ the direction of the heat flow is positive as shown in the diagram, that is from the hotter wall surface to the cooler wall surface.

Thickness dependance

When we have 1 material (no conductor between the hottest and coldest side of the material) then why the rate depends on the thickness of the material?

I don't know if you are familiar with Ohms law

$$I=\frac{V}{R}$$

Roughly speaking, current $I$ is analogous to heat flow $\dot Q$, Voltage $V$ is analogous to temperature difference, $\Delta T$.

Electrical resistance

$$R_{e}=\frac {ρL}{A}$$

Is analogous to thermal resistance

$$R_{T}=\frac{L}{kA}$$

Where ρ is the electrical resistivity of the material of a resistor, and is analogous to the inverse of the thermal conductivity $k$ of the wall material.

The greater the length $L$ (thickness) of the wall the greater its thermal resistance and the lower the heat transfer rate, all other things being equal.

  • How $A$ is defined?

    In the case of 1 material what is "the surface area of the surface emitting heat" $A$? It is an imaginary area?

$A$ is the area of the surface perpendicular to the direction of heat flow. In the figure below, it would be the area of the wall going into the page. It is not an area "emitting" heat. It is the area through which heat transfers across the higher temperature wall surface and out the lower temperature surface. The assumption is there are no active heat sources in the wall.

  • What $k$ to choose

    What if we have two different materials in contact with different temperature with $k_1$, $k_2$. Will $k=k_1$ or $k=k_2$?

This would be the equivalent of a composite wall in the diagram below. Each material has its own thermal conductivity $k$ and its own thickness $L$. The temperature gradient will then be different for each slab of the composite wall. Meaning there will be a different "temperature drop" across each component of the composite wall.

The electrical analogy are resistors connected in series. The voltage drop across each resistor equals the current through the resistor times its electrical resistance. The thermal analogy is there will be a temperature drop across each thermal resistance (component of the wall) equal to the heat transfer through the wall component times its thermal resistance.

The only way that this rate makes sense to me is if we consider that the temperature doesn't change over time in the hottest and the coldest side.

That is exactly correct, provided we are talking about steady state heat flow where the heat flow into one surface equals that out of the other and temperatures within the material are not changing in time. The more general form of the Fourier heat conduction equation allows for the temperatures within the wall to be not only a function of $x$, but also time $t$. It's a second order differential equation. In that case you need to know more than $k$ for the material. You also need to know the specific heat and density of the material. Combined in a certain way they become the thermal diffusivity of the material. The thermal diffusivity of a material is a measure of its ability to conduct thermal energy (a function of $k$) relative to its ability to store thermal energy (a function of density and specific heat).

Hope this helps.

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  • $\begingroup$ Thank you so much for the answer. So this form is only valid as you said for steady state flow. In the 1 material case can we pick a surface different than the surface between the boundaries of the hottest and coldest side (as no bulk solid goes between them)? In that case I can't see why thickness matters. It is because the heat that goes in must equal the heat goes out? $\endgroup$ Commented Apr 12, 2020 at 11:31
  • $\begingroup$ At steady flow yes, the rate heat that goes in always equals the rate heat out regardless of the thickness. But the magnitude of the rate of heat that goes through the wall does depend on the thickness. Think about the exterior wall of a heated building in winter. The thinner the wall is, the greater the rate of heat loss from the building. $\endgroup$
    – Bob D
    Commented Apr 12, 2020 at 11:55
  • $\begingroup$ Thanks for the reply. I think I got it. $\endgroup$ Commented Apr 12, 2020 at 12:12
  • $\begingroup$ Great. Glad I could help $\endgroup$
    – Bob D
    Commented Apr 12, 2020 at 13:04
  • $\begingroup$ Wiki says that in the limiting case that the rate becomes: $$\frac{dq}{dt}=-kA\frac{dT}{dx}$$ or in 3 dimensions $$\overrightarrow{q}=-k\nabla T$$ I can't understand why the left side must be a vector? Is it multiplied by a unit vector in the direction tangent to the gradient vector? $\endgroup$ Commented Apr 12, 2020 at 20:14

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