Conduction -- Fourier's law


In our discussion of the flow of thermal energy through a medium resulting from a temperature difference, we wrote a "local" equation — an equation that said the flow at a particular point in space was due to how the temperature was changing at that point: the derivative of the temperature. Our equation is the Fourier heat law:

$$J_{conduction} = - \kappa \frac{dT}{dx}$$

But if we are going to use this in real world examples, often we want to have a "global" equation — one that works over finite distances. Basically, we need to change from our derivative form of the law to a "Delta" form of the law more like the HP equation. Remember that the HP equation tells how much flow you have in a pipe if the flow is driven by a pressure difference from one end of the pipe to another. The HP equation looks like this:

$$\Delta P = ZQ \quad \quad Z = \frac{8 \mu L}{\pi R^4}$$

Reading the equations, they say that a fluid flow ($Q$ — volume per second) is created by a driving difference ($ΔP$ — pressure drop), and that they are related by a resistance ($Z$). The second equation says that the resistance depends on the geometry of the pipe ($L/R^4$) times a factor that only depends on the fluid ($8μ/π$ — where $μ$ is the viscosity of the fluid).

We have a similar situation with heat. A change in a scalar field over space (ΔT — temperature drop) drives a flow. Let's see if we can make an analogous equation. (Of course we can!)

As usual, we'll make the simplest possible model to see how things work. We'll assume that for the area A the only thing between the inside and outside is a block of of thickness $L$, area $A$, and thermal conductivity $κ$. We''ll ignore other parts of the world,  leakage through the edges, etc. and we'll assume that while we are watching both the hot and cold regions stay at their original temperature despite the flow of heat.

Starting from Fourier's differential law of heat flow, let's find an equation that gives the amount of heat flowing through the area A per second in terms of the parameters of the block of insulation and the temperature difference ($L$, $A$, and $κ$). Well use the symbol $\Phi$ (capital Greek phi) to represent the rate of heat flow through the block.

To get this result, we have to see how our basic equation — the Fourier heat law — applies in this situation, and how its variables relate to the ones we need to use. The Fourier law at the top of the page gives the flux — the energy per area per second. We want the energy per second. Clearly we have to multiply by the area. So we must have 

$$\Phi = J_{conduction} A = \kappa A \frac{dT}{dx}$$

(We're not going to worry about the sign here, since we know the direction of the flow — from high T to low.) 

If we're in a steady state, the flow of energy through the block will be constant, and we expect the flow to be the same at every place along the block. This means that $dT/dx$ is a constant so we expect $T$ to be proportional to $x$ inside the block. We can replace our "dT" and "dx" with deltas. (To prove this rigorously, integrate both sides of the equation over $x$ from the beginning to the end of the block.)

$$\Phi = \kappa A \frac{\Delta T}{\Delta x}$$

$Δx$ is just the thickness of the block, $L$, so we have

$$\Phi = \frac{\kappa A}{L} \Delta T$$

Solving for $ΔT$we have our desired equation:

$$\Delta T = \bigg(\frac{L}{\kappa A}\bigg)\Phi = Z \Phi$$

where we have grouped the constants having to do with the properties of the thermally conducting block into a "thermal resistance", Z

BOur equation looks just like the HP equation — or any other gradient driven flow equation: a driving difference (the temperature) equals a resistance times a flow (of heat). The resistance depends on the geometric properties (size and shape) of the block through the factor $L/A$. The HP equation is for a cylindrical pipe, so instead of $A$ it talks about $R$. To compare, lets assume a cylindrical block. Then our geometrical factor here would be $L/R^2$ while for the HP equation it is $L/R^4$. This tells us that the flow of a liquid is more sensitive to the radius (or area) of the carrying object than is the flow of heat.

Joe Redish 11/24/14


Article 554
Last Modified: March 4, 2019