Part VI: Transport Theory

The control volume analysis technique we saw and used previously is both astoundingly general and eminently pragmatic. But it has a shortcoming; namely, that we have to define a specific volume before we can make statements about the fluid inside of it. This means that all of the statements we can make about fluids using control volume analysis depend on the specific system we’re analyzing. If we want to make universal statements about the way fluids behave, i.e. describe the physical laws of fluid mechanics, we need to overcome this obstacle. As it turns out, this is easier than it seems.

Let’s briefly summarize what control volume analysis dictates about some quantity in a fluid volume; the rate of change of the total amount of that quantity inside a volume is equal to the total amount of that quantity being generated or destroyed within it, plus the amount of that quantity which is moving in/out of the volume through its boundary surface. If we refer to the quantity we are interested in as $\Xi$ , and its associated density as $\xi$ , we can generate the following equation for the rate of change of $\Xi$ :

$\displaystyle \frac{\partial }{\partial t}\int_V \xi\ dV = \int_V \frac{d \xi}{d t}\ dV - \int_A \left(\vec{j}_\xi \cdot \hat{n}\right)\ dA$

There’s a neat mathematical trick we can do to turn that last surface integral into a volume integral called the divergence theorem. The details don’t matter too much as long as you trust me; the only important thing is that the term still represents what it always did, flow in/out of the control volume. Applying this trick, we find:

$\displaystyle \frac{\partial}{\partial t}\int_V \xi\ dV = \int_V \frac{d\xi}{d t}\ dV - \int_V \nabla \cdot \vec{j}_{\xi}\ dV$

Take a look at what we’ve managed to get; our equation, which represents the change in the total amount of some arbitrary fluid quantity, is exclusively in terms of adding up tiny contributions of different things within the control volume. So if we shrink the control volume down to a point, we don’t have to bother with the integrations—we can look at an equation that relates those contributions directly! Remember, the concept of the equation is the same, we’re just now using it to look at the way quantities accumulate within points rather than volumes.

To save ourselves considerable confusion from mathematical notation, I’m going to do what nearly every scientist does and relabel the $\frac{d \xi}{d t}$ term to something simpler, like $r$ ; it still represents the same thing, which is the generation/destruction of $\Xi$ through some physical process going on in that point. Performing this relabeling, and shrinking things down to a point, we obtain:

$\displaystyle \frac{\partial\xi}{\partial t}\ = r - \nabla \cdot \vec{j}_{\xi}$

This type of equation is called a transport equation, and the process of using this type of equation to understand something is called a transport theory. Sometimes physicists call this kind of equation a continuity equation, which is in my opinion silly and confusing. Moving the flux term to the other side of the equation, we get its final version:

$\displaystyle \boxed{\frac{\partial \xi}{\partial t}\ + \nabla \cdot \vec{j}_{\xi} = r}$

This equation is the most important equation in science. I am biased, sure, but you can model nearly everything with some version of it: fluid and solid mechanics, chemical kinetics, heat transfer, most of electromagnetism, general relativity, and a decent slice of quantum mechanics. Even statistical mechanics, if you play your cards right! This equation lets you mathematically understand the movement of anything through points in space, as long as you’re able to understand the ways that your quantity of interest can move through space (its fluxes) and the ways your quantity is generated or destroyed at certain points (its sources and sinks).

Here’s a simple and very useful example. If we wanted to look at setting up an equation for mass, you would first identify the corresponding quantity representing mass per unit volume (density $\rho$ ) and write a transport equation for it down:

$\displaystyle \frac{\partial \rho}{\partial t}\ + \nabla \cdot \left(\rho \vec{v}\right) = r_{\rho}$

Remember that all fluid mass movement is described by fluid velocity $\vec{v}$ , so no need to include a diffusion term. In addition, in the absence of something goofy like nuclear reactions or relativistic phenomena, we know mass can’t be destroyed or created. That means that the $r_{\rho}$ term representing the generation and destruction of mass has to be zero! Therefore, we get:

$\displaystyle \frac{\partial \rho}{\partial t}\ + \nabla \cdot \left(\rho \vec{v}\right) = 0$

This equation describes the conservation of mass, and is the second-most important equation in fluid mechanics. If we additionally assumed that the liquid was incompressible, something interesting happens; the density can’t change with respect to time or space, so any term proportional to that needs to vanish in the above equation. Expanding it out and getting rid of the density change terms, we get:

$\displaystyle \frac{\partial \rho}{\partial t}\ + \nabla \rho \cdot \vec{v} + \rho \left(\nabla \cdot \vec{v}\right) = 0$

$\displaystyle \nabla \cdot \vec{v} = 0$

This equation refers to conservation of mass when a liquid is incompressible, and is sometimes called the incompressibility equation.

Even though mass in general is conserved, there might be some chemical reactions going on in our system of interest that change mass from one type of chemical to another. In that situation, you would set up transport equations for every chemical of interest in your system, and wind up with a system of equations for the chemical densities:

$\displaystyle \frac{\partial \rho_i}{\partial t}\ + \nabla \cdot \left(\rho_i \vec{v}\right) - D_i \nabla \rho_i = r_{i}$

where the subscripts indicate a specific chemical in the system, and each source/sink term representing chemical reactions can depend on every other chemical density in the system. Such a system of equations is usually referred to as a reaction network. In chemical systems where there’s no fluid flow, only diffusion and reactions occur, which are studied under the (apt) name of reaction-diffusion systems. Most chemical engineers ignore both the movement of fluid within the system and gradients in the densities of the chemicals, leaving only the reactions to drive the physical changes; this approximation is commonly called the CSTR approximation.

We can construct transport equations for energy as well; one very useful type of energy transport equation involves looking at the movement of thermal energy through a solid. The generic version of the equation should look something like this, according to transport theory:

$\displaystyle \frac{\partial e}{\partial t}\ + \nabla \cdot \vec{j}_{e} = r_e$

As it turns out, the heat energy density of a point is equal to the temperature $T$ of the point, times the mass density $\rho$ times the heat capacity $c$ . In addition, because the solid doesn’t flow, the only way energy moves through the solid is diffusively. As a result, we update the equation above to find:

$\displaystyle \frac{\partial e}{\partial t}\ + \nabla \cdot \left[-D_e \nabla e\right] = r_e$

$\displaystyle \frac{\partial \left(c \rho T\right)}{\partial t}\ + \nabla \cdot \left[-D_e \nabla \left(c \rho T \right)\right] = r_e$

Most thermal engineers like to assume the material properties of a solid that’s transferring heat stay constant and are the same everywhere, so we can pull those properties (the heat capacity, density, and heat diffusivity) out of the gradients and simplify:

$\displaystyle c \rho \frac{\partial T}{\partial t}\ - c \rho D_e \nabla \cdot \nabla T = r_e$

$\displaystyle \frac{\partial T}{\partial t}\ - \frac{D_e}{c \rho}\left(\nabla \cdot \nabla T\right) = \frac{r_e}{c \rho}$

Cleaning up the vector calculus term, expressing $\frac{D_e}{c \rho}$ as a single term $\alpha$ representing the thermal diffusivity, and expressing $\frac{r_e}{c \rho}$ as a single thermal generation term $h$ , we find:

$\displaystyle \frac{\partial T}{\partial t}\ - \alpha \nabla^2 T = h$

Congrats! You just derived the heat equation using transport theory. Notice how many assumptions we had to make! And funnily enough, that heat diffusivity term $D_e$ we saw above is almost always never called that; it’s confusingly referred to as the thermal conductivity.

Hopefully this has convinced you of the generality and usefulness of transport theory. And with transport theory, we can now construct the most important equation in fluid mechanics; the transport equation for momentum.

If transport theory is to be believed, all I have to do is identify the momentum density, which is $\rho \vec{v}$ , and write a transport equation for it that looks like this:

$\displaystyle \frac{\partial\left(\rho \vec{v}\right)}{\partial t}\ + \nabla \cdot \vec{j}_{\rho \vec{v}} = r_{\rho \vec{v}}$

So far, so good. But if I go and write the advective flux term explicitly, what we find seems a bit confusing:

$\displaystyle \frac{\partial (\rho \vec{v})}{\partial t}\ + \nabla \cdot \rho \textrm{?`}\vec{v} \vec{v}\,\textrm{?} + \nabla \cdot \vec{j}^{\left[\rho \vec{v}\right]}_{\text{diff}} = r_{\rho \vec{v}}$

How should we multiply these two vectors? It shouldn’t be a dot product, because then we’d get a scalar, and we can’t take the divergence of that. As it turns out, the right way to multiply these two vectors is with something called an outer product, which looks like this: $\otimes$ . And the outer product of two vectors is a weird mathematical object called a tensor, which we will see a lot of in the next section.

Equipped with this knowledge, we rework the momentum transport equation to:

$\displaystyle \frac{\partial (\rho \vec{v})}{\partial t}\ + \nabla \cdot \left(\rho \vec{v} \otimes\vec{v}\right) + \nabla \cdot \vec{j}^{\left[\rho \vec{v}\right]}_{\text{diff}} = r_{\rho \vec{v}}$

It might make you uncomfortable to have that tensor expression up there; most students don’t learn how to do calculus with tensors until after fluid mechanics (if they ever do). Luckily, we can break the term up into terms that can be handled entirely with vector calculus, leading us to the Lamb form of the momentum transport equation:

$\displaystyle \frac{\partial (\rho \vec{v})}{\partial t}\ + \frac{1}{2} \nabla \left(\rho \vec{v} \cdot \vec{v}\right) + \left(\nabla \times \rho \vec{v}\right)\times \vec{v} + \nabla \cdot \vec{j}^{\left[\rho \vec{v}\right]}_{\text{diff}} = r_{\rho \vec{v}}$

This is helpful to understand what the tensor term in the equation means, but I personally don’t find it helpful for keeping track of what the equation means; which is that it’s a transport equation for momentum in a fluid. We’ll revisit the Lamb form later, but right now I’d like to stick with the $\nabla \cdot \left(\rho \vec{v} \otimes\vec{v}\right)$ notation for conceptual clarity.

At this point, it seems tempting to simply say that the only source or sink of momentum in a fluid point comes from body forces that are external to the fluid. But we know that’s not true! We know pressure gradients cause changes in momentum within a fluid, and the momentum for that is coming entirely from the “storage” of molecular energy within the fluid. In addition, we’re missing another huge factor; friction. With friction, the opposite happens—energy in the movement of a macroscale fluid gets dissipated into microscale molecular energy of random motion. The common factor between pressure-induced forces and friction is that both of them are causing macroscale, fluid effects as a result of microscale molecular phenomena.

Noting that momentum density diffusion is also a molecular effect, I find it helpful to lump in all the molecular effects together into a simple term $\vec{r}_{\textrm{molecular}}$ and rewrite the momentum transport equation in the following way:

$\displaystyle \boxed{\frac{\partial(\rho \vec{v})}{\partial t}\ + \nabla \cdot \left(\rho \vec{v} \otimes\vec{v}\right) = \vec{f} + \vec{r}_{\textrm{molecular}}}$

Although it may not look like it yet, this is the most general form of the mathematical basis for essentially all of fluid mechanics: the Cauchy momentum equation.

Things to Think About

1. Let’s say you’re a nuclear engineer and have to deal with the unpleasant situation of handling a radioactive fluid that is routinely converting its mass into radioactive energy emissions. How would you write the mass transport equation for it?

2. How would you describe heat transport in a fluid? Is it more efficient in a solid, or less? Does this help you understand some common design features in thermal engineering?

3. Knowing what you know about the momentum transport equation, what do you think is the form of a transport equation for an arbitrary vector?

4. I never wrote out the momentum diffusive flux term. Knowing what you know about the generic mathematical form of diffusive fluxes, can you figure out why I chose to do that?

Part VI: Transport Theory

Things to Think About

Published by Arnaldo Rodriguez-Gonzalez

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Things to Think About

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