2.1: The Diffusion Equation
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2: Diffusion and Darcy's Law
Thermal Conduction
1. Heat Flow Equation: 1-D Rod
2. Diffusion Equation: 3-D Volume
3. Full Form of the Diffusion Equation
4. Length-scale of Diffusion
5. Time-scale of Diffusion
There are many, many applications and uses of the diffusion equation in geosciences, from diffusion of an element within a solid at the lattice-scale, to diffusion of heat at a local to regional scale, to diffusion of topography, and diffusion of a chemical species in the crust. Before we dive head first into the diffusion equation, let's look at four real world examples of how it can be applied.
1. Heat (\(T(\overrightarrow{x},t\))): A localized temperature perturbation will slowly cool off or heat up the surrounding region. The temperature profile (T(x)) at any time, t, depends on the initial temperature distribution and how quickly or how far heat diffuses into the surrounding material.
a. Dike emplacement in the crust: How long does it take for the heat to diffuse into the surrounding region? FIGURE dike curves
b. Cooling/aging of the crust/lithosphere: As the crust and lithosphere cool by diffusion, they become denser and subside. Subsidence of the crust beneath large sediment filled basins with organic material are of particular interest to oil/natural gas exploration.
2c. Erosion of a fault scarp (\(H(\overrightarrow{x},t\))): The sharp topography formed at the surface when a fault ruptures with vertical displacement will slowly erode away with time. At first, the cross section will look like a step function. After some amount of time, the sharp corners will fill in or erode away, until slowly the sharp offset becomes a smoothly dipping slope. The shape of this eroding offset with time is described by the diffusion equation. FIGURE fault scarp curves
3d. Chemical (\(C(\overrightarrow{x},t\))): Chemical diffusion of elements within a rock lattice structure is also described by a diffusion equation for the concentration of a chemical species. Chemical diffusion is important for understanding radiogenic dating of rocks (when does loss by diffusion stop?), diffusion of toxic chemicals from dump sites and diffusion of "chemical plumes" associated with nuclear waste leaks. FIGURE Chemical diffusion
The diffusion equation models transport of quantities. It can model far distances by small (local) interactions. An example of this is heat.
Diffusion of heat through a crystal lattice occurs through two processes.
First, heat can be carried by electrons in the form of kinetic energy in metallic rocks. If a region of the rock is heated at one end, the thermal energy excites the electron, giving them extra energy to move through the rock from regions of high heat to regions of low heat. As the electron moves, it loses some of this extra energy as heat, thus heating up the region it has moved to.
FIGURE Diffusion lecture 1 lattice
Second, heating a lattice structure adds kinetic energy to the lattice itself, by increasing the vibrational bonds between atoms (think of atoms connected to each other by little springs). The movement of this vibrational energy through the lattice structure is termed a phonon. Both modes of passing energy through the lattice structure act to conduct heat from regions of higher temperature to regions of lower temperature, changing the distribution (and perhaps the total amount) of heat within the rock.
The rate of transport depends on the gradient, a larger gradient ⇒ faster transport, and a smaller gradient ⇒ slower transport.
Start with a cylindrical rod, of length \(L\), and cross-sectional area \(A\), aligned with the z axis. Fix the temperature at \(z=0\) to \(T_1\) and the temperature at \(z=L\) to \(T_2\), where \(T_2>T_1\). The heat, \(Q\), will move from the high temperature region to the low temperature region, in the -z direction. The change in heat within the rod, \(\Delta Q\), is given by:
\[\Delta Q=kA(\frac{T_2-T_1}{L})\Delta t\]
where k is the thermal conductivity with units of \(\frac{W}{mK}\)(watts per meter Kelvin). Therefore, \(\Delta Q\) has units of joules (note 1W=1\(\frac{J}{s}\)). Rearrange the equation above as,
\[\frac{\Delta Q}{A\Delta t}=k\frac{\Delta T}{L}\]
FIGURE variable explanation pipe
The left hand side is the change in heat per unit area per unit time, which is a heat flux, given the symbol, q, and can be written in terms of derivatives as:
\[q=\frac{1}{A} \frac{dQ}{dt}\]
The right hand side includes the thermal gradient across the rod, \(\frac{\Delta T}{L}\), which can be rewritten in terms of derivatives at \(\frac{dT}{dz}\). Substituting,
\[q=-k\frac{dT}{dz}\]
This is the Heat Flow Equation
where the minus sign shows that the heat moves in the -z direction.
More generally, we should consider a 3-D volume element (of a rock) and examine how the heat content changes within this volume as a function of time. Consider a cubic volume element with sides of length dx, dy, and dz and volume, \(dV=dxdydz\). If the heat flows along the z axis, the heat that enters the volume at z+dz is \(Q_z\) and the heat that leaves the volume at z is \(Q_z -dQ_z\), then the change in heat within the volume is:
\(\Delta Q=Q_z-(Q_z-dQz)\)
FIGURE cube with xyz axes
rewrite as:
\[\Delta Q=Q_z-(Q_z-(\frac{dQz}{dz})(-dz))\]
\[\Delta Q=-(\frac{dQz}{dz})dz\]
From the heat flow equation we also know that
\[dQ=qAdt\]
\[q=-k\frac{dT}{dz}.\]
Substituting,
\[\Delta Q=-\frac{d}{dz}(qAdt)dz=-\frac{d}{dz}(-k\frac{dT}{dz})dVdt\]
where \(dV=Adz=dxdydz\). If k is independent of position (a constant throughout the rock), then this can be written as:
\[\Delta Q=k(\frac{d^2T}{dz^2})dV dt\]
where both negative signs have canceled out so that \(\Delta Q\) is positive. The total heat within the rock can also be considered from the perspective of the total heat capacity of the rock, which depends on the specific heat, c, and the density, \(\rho\), of the rock:
\[\Delta Q=c\rho dV\Delta T\]
Setting the two expressions for \(\Delta Q\) equal:
\[c\rho dV\Delta T= k(\frac{d^2T}{dz^2})dV dt\]
which can be rearranged as:
\[\frac{dT}{dt}=\frac{k}{c\rho}\frac{d^2T}{dz^2}\]
where dT has been substituted for \(\Delta T\). The combination of physical constants, \(\frac{k}{c\rho}\) is the thermal diffusivity and has units of \((\frac{m^2}{s})\) and is given the symbol \(\kappa\) (kappa). Substituting in \(\kappa\) we get the expression for the 1-D diffusion equation:
\[\frac{dT}{dt}=\kappa\frac{d^2T}{dz^2}\]
This is the Diffusion Equation
Note, this equation relates the change in temperature per unit time to the change in temperature over an area perpendicular to the direction of heat flow.
In general, heat flow can come from any direction, so the temperature will depend on x, y, z, and t. Because \(T=T(x, y, z, t)\) and is not just dependent on one variable, it is necessary to rewrite the derivatives in the diffusion equation as partial derivatives:
\[\frac{\partial T}{\partial t}=\kappa(\frac{\partial^2 T}{\partial x^2}+\frac{\partial^2 T}{\partial y^2}+\frac{\partial^2 T}{\partial z^2})\]
In this case for example, for the derivative with respect to time, each of the variables (x, y, z) are treated as constants, and so on for each other variable. Don't let the partial derivatives confuse you, they are just derivatives. In general, you still want to think of this equation as a bunch of small changes in temperature, \(\Delta T\), occurring over small intervals of time, \(\Delta t\), and over small areas (\(\Delta z^2\)):
\[\frac{\Delta T}{\Delta t}=\kappa\frac{\Delta T}{(\Delta z)^2}\]
This is the "Simplified Diffusion Equation"
The above simplified version of the diffusion equation can be solved for \(\Delta z\) to give an indication of how far heat can move in a given amount of time:
\[\Delta z\simeq\sqrt{\kappa\Delta t}\]
This is the Diffusion Length-scale
For the earth, a good average value for the thermal diffusivity is \(\kappa=1x10^{-6}\frac{m^2}{s}\). How far does heat diffuse through a rock in 1 million years?
\[\Delta z=\sqrt{(1x10^{-6})(1x10^6)(3.15x10^7)}\]
\(\Delta z\simeq 5.6 km\) where there are about \(3.15x10^7\) sec per year
What does this length-scale tell us? It does not tell us the temperature at any particular distance, but instead tells us that a noticeable amount of heat will travel this distance in the time specified (we will look more at this in the homework).
Another way of building your intuition for diffusion is to solve the simplified diffusion equation for \(\Delta t\):
\[\Delta t\simeq\frac{\Delta z^2}{\kappa}\]
This is the Diffusion Timescale
How long does it take to cool off a dike intrusion that is 100 meters wide?
\[\Delta t=\frac{(100^2)}{(1x10^{-6})(3.15x10^7)}\]
\[\Delta t=318\) years.
What does this timescale tell us? It does not tell us the temperature at the distance specified, but instead tells us that a noticeable amount of heat will have reached the distance specified in this time (we will look more at this in the homework). A rule of thumb is that cooling is often exponential.
FIGURE exponential heat loss
Most heat is lost or moved within \(t=3\cdot\Delta t\)
2.2: Geological Applications of the Diffusion Equation
Diffusion Equation