## 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.

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 V: Hydraulics

The concept of control volume analysis we developed to analyze the way pressure, velocity, and density interact in fluids is surprisingly powerful. And in fact, such an analysis—coupled with a little bit of experimental data—forms the backbone of nearly all engineering applications of fluid mechanics. The simplest example of this, and certainly the most ubiquitous, is in hydraulics.

We had briefly seen hydraulics at the end of Part II, where things like hydraulic jacks exploited the relationship of external forces and pressure. However, the study of hydraulics is far broader than that; in a nutshell, hydraulics is the field of engineering that exploits or manipulates enclosed fluids. The critical distinction between then and now is that we were exclusively looking at fluids that weren’t moving, where the only thing we could manipulate was the pressure. Now that we have the machinery of control volume analysis, we can explore the full gamut of hydraulic phenomena that involve both pressure and velocity.

Let’s consider a section of pipe again, this time filled with a moving incompressible fluid (so the density is constant everywhere). If we perform a control volume analysis for the mass of fluid inside the pipe section, and assume that the pipe flow has been running sufficiently long such that the total fluid mass in the pipe doesn’t change, we find as we did in Part IV that $\displaystyle \int_{\text{inlet}}v_1\ dA\ = \int_{\text{outlet}} v_2\ dA$. And although we don’t know how the velocity changes at each point of the inlet/outlet—we don’t have the mathematical machinery for that yet—we can define some average speeds at the inlet/outlet of the pipe $\langle v_i \rangle$ such that those integrals turn into simple multiplications, with the assumption that all the flow at the inlet/outlet is perpendicular to the pipe cross-section. With that, we find the following equation:

$\displaystyle \boxed{\langle v_1 \rangle A_1 = \langle v_2 \rangle A_2}$

Although this is just a simplified and specific version of the conservation of mass equation we derived in Part IV, it is also what I like to call the first fundamental equation of hydraulics. It is fundamental because it is giving us a very simple and straightforward statement about the behavior of fluids that is very useful for engineering applications: the change of fluid speed between two ends of a steady hydraulic system involving incompressible fluids depends only on the geometry of that system.

This is relatively shocking, since one would certainly expect that pressure or gravity or external forces would play some kind of role in speeding up or slowing down the fluid. The fact of the matter is that they do, but only while the system is in the process of becoming steady; once flow is constant, the geometry determines flow speeds. The imperative thing to do now is to determine whether or not such a steady flow is attainable in a given system, and for that we need to figure out what’s happening with the pressure.

To that end, if we perform a control volume analysis for the momentum of the fluid in that same pipe, we find in a general sense:

$\displaystyle \frac{\partial \vec{P}}{\partial t} = \int_V \vec{b}\ dV + \int_A \vec{t}\ dA - \int_A \left(\vec{j}^{\left[\rho \vec{v}\right]}_{\text{diff}} \cdot \hat{n}\right) dA - \int_A \rho \vec{v} \left(\vec{v} \cdot \hat{n}\right)\ dA$

We can be far more specific now than we were in Part IV. However, we are going to consider an extra effect we didn’t consider before; a momentum loss term $\vec{H}$. This momentum loss term is going to be extremely general, and is essentially accounting for all the phenomena that we don’t have the tools to understand yet; friction, turbulence, recirculation zones, etcetra. With that stated, we make the following assumptions:

• Tractions on the inlet/outlet are pressure-induced.
• The only body force on the pipe section is gravity.
• The system has been in motion for a long time, so everything in the pipe section is in equilibrium ($\frac{\partial \vec{P}}{\partial t} = 0$).
• The pipe section is relatively straight, such that defining a flow direction $\hat{x}$ through the pipe makes sense.
• All forces perpendicular to the pipe inlet/outlet, and all tractions acting on pipe surfaces that aren’t the pipe inlet/outlet, are balanced out by forces from the pipe itself (i.e. the solid structure surrounding the water).
• Diffusive momentum fluxes at the inlet/outlet are meager in comparison to the advective fluxes.
• The momentum loss term is always opposite to the flow direction, and always represents a loss ($\vec{H}\cdot\hat{x} < 0$)

This makes our expression a bit cleaner when we take the dot product with the flow direction $\hat{x}$:

$\displaystyle 0 = -|\vec{H}|+ \int_V \rho \left(\vec{g} \cdot \hat{x}\right) dV - \int_A p \hat{n} \cdot \hat{x}\ dA - \int_A \rho \vec{v} \left(\vec{v} \cdot \hat{n} \right) \cdot \hat{x}\ dA$

Defining average quantities just like we did before, we can swap out those integrals with products:

$\displaystyle 0 = -|\vec{H}| + \rho \vec{g} \cdot \hat{x} V + \langle p_{1} \rangle A_1 - \langle p_{2} \rangle A_2 + \rho \langle v_{1} \rangle^2 A_1 - \rho \langle v_{2} \rangle^2 A_2$

To simplify the gravitational term, the volume $V$ of the pipe section can be reasonably approximated as the average cross-sectional area of the inlet/outlet times the pipe length. Combining that with the dot-product term and doing some behind-the-scenes algebra, we can equivalently approximate the gravitational term as:

$\rho \left(\vec{g} \cdot \hat{x}\right) V \approx \rho |\vec{g}| \frac{A_1 + A_2}{2} L \cos{\theta} = \rho |\vec{g}| \frac{A_1 + A_2}{2} \left(z_{1} - z_{2}\right)$

where each $z$ represents the height of the center of the inlet/outlet relative to some fixed point, like the ground.

Inserting back, and moving the momentum loss term to the left-hand side, we find the following equation:

$\displaystyle \boxed{|\vec{H}| = \rho|\vec{g}|\frac{A_1 + A_2}{2} \left(z_{1} - z_{2}\right) + \langle p_{1} \rangle A_1 - \langle p_{2} \rangle A_2 + \rho \langle v_{1} \rangle^2 A_1 - \rho \langle v_{2} \rangle^2 A_2}$

I usually refer to this equation, which came from a control volume analysis for the momentum of a pipe section of fluid in the direction of flow, as the second fundamental equation of hydraulics. This is because it is explicitly giving us the remaining pieces of the puzzle of understanding how the pressure changes in a pipe section; we know the changes in the velocity can be found entirely from geometric arguments thanks to the first fundamental equation, and everything coming from the gravitational term is geometric as well since it only depends on pipe section heights and cross-sectional areas. The only other thing we’d need to figure out the average pressure at the outlet, given some average pressure and velocity at the inlet, is that momentum loss term.

The momentum loss term $|\vec{H}|$ is, as we stated before, extremely general; it is essentially accounting for every physical phenomenon associated with momentum loss in the fluid that we don’t know how to describe, and is consequently a function of the pressures, velocities, and geometries of the pipe section in question. Luckily, decades upon decades of experimental work have empirically determined accurate values for $|\vec{H}|$ associated with specific phenomena and pipe geometries, such that hydraulicists only need to look up the values of $|\vec{H}|$ for some given geometry, average flow pressure, and average flow speed. This momentum loss $|\vec{H}|$ is almost universally approximated as proportional to the cross-sectional area, in which case hydraulicists define a momentum loss per area $h =\frac{|\vec{H}|}{\langle A \rangle}$, commonly referred to as hydraulic loss. This hydraulic loss $h$ is usually further split into two confusingly named components; major hydraulic loss $h_M$, and minor hydraulic loss $h_m$. Major hydraulic loss should actually be called frictional loss or length-proportional hydraulic loss, as it is defined to be proportional to the pipe length and stems largely from friction, and minor hydraulic loss should be called geometric or length-independent hydraulic loss, as it accounts for geometrically-induced flow oddities like recirculation zones in pipe bends that are not proportional to pipe length. We can then rewrite the second fundamental equation of hydraulics as:

$\displaystyle \frac{A_1 + A_2}{2}\left(h_M + h_m\right) = \rho|\vec{g}|\frac{A_1 + A_2}{2} \left(z_{1} - z_{2}\right) + \langle p_{1} \rangle A_1 - \langle p_{2} \rangle A_2 + \rho \langle v_{1} \rangle^2 A_1 - \rho \langle v_{2} \rangle^2 A_2$

We can do a couple of rapid-fire qualitative observations from these fundamental equations given some simple assumptions:

• If losses are negligible, and the pipe inlet/outlet are at the same height but the outlet is smaller than the inlet, the pressure at the outlet has to drop relative to the inlet. If the outlet is bigger, the pressure has to increase. This is called the Venturi effect.
• If there is no flow, the second fundamental equation becomes an approximation of the integrated form of the hydrostatic equation.
• If losses are negligible, and the inlet/outlet areas are all the same, one can divide by the area to obtain something called Bernoulli’s equation. I don’t like that equation very much nor do I think it’s useful, but it’s important to mention it so that you know what other people mean when they reference it.

So, how do engineers apply these equations to an aqueduct or a network of pipes? Although there are many ways to do this, the common scenario is to consider some sort of pump or reservoir at a high and fixed pressure $p_{in}$ and some outlet at a fixed lower pressure $p_{out}$, most of the time at atmospheric pressure. Then you guess a “reasonable” value for the velocity at that reservoir $\langle v_{in} \rangle$, use the fundamental equations of hydraulics to determine the inlet pressure and velocity of the next pipe section, and so on until you reach the outlet. Then, you see whether or not the outlet pressure you wind up obtaining with that trial guess of the reservoir velocity $\langle v_{in} \rangle$ matches the correct outlet pressure; if it matches, great, if it doesn’t, you keep trying with different reservoir velocities. If you can’t find a match at all, then steady flow in the system is likely impossible, and you need to tweak your hydraulic system. This sort of thing is called hydraulic circuit analysis or pipe network analysis, depending on the engineer’s background.

Using control volume analysis for the mass and momentum of fluid in a section of pipe at steady-state, along with some experimental data about momentum losses in pipes, we were able to get everything we needed to design and understand basic hydraulic systems. If we want to get a fundamental understanding of what those losses are and how they occur, though—if we want to get a fundamental understanding of the laws of fluid mechanics—we need to give our theoretical machinery of control volume analysis a little upgrade. That upgrade is called transport theory.

1. Why do you think this kind of approach would be tricky when we’re thinking about flow past something (like say, a plane)?

2. What details about the flow that we haven’t found out yet would help us make this analysis even more specific and precise?

3. What processes do you think contribute to momentum loss in a curved pipe? What do you think they depend on?

4. Let’s say you are trying to pump water into the 6th story of a building using just a pump, a curved piece of pipe, and a piece of pipe going straight up. How would you design the pump and the pipe geometry?

5. What do you think momentum loss due to friction depends on?

## Part IV: Control Volume Analysis

As we showed previously, the idea of adding up small contributions on a surface is a powerful concept in fluid mechanics. As a matter of fact, studying the way that moving fluids apply forces to objects also relies on this principle, and we can promptly use the techniques we saw utilized for buoyancy to this purpose.

To do so, consider an imaginary surface enclosing some amount of unmoving fluid, through which fluid could potentially move in and out of—very similar conceptually to the “net” we discussed in Part I. This surface, and the volume it encloses, is commonly referred to as a control volume. Just like a solid object submerged in a fluid, the fluid inside of the surface feels a net force as a result of the contributions from the pressure-induced tractions on the imaginary surface bounding it. In addition, the fluid can also feel net forces due to force densities distributed inside of it, like gravity. Stating this mathematically,

$\displaystyle \vec{F} = \int_V \vec{b}\ dV + \int_A \vec{t}\ dA$

where $b$ represents the force densities within the fluid.

Let’s take a look at a quick example of using this equation by analyzing static fluid in a pipe floating in space. Let’s say that the pressure at one end of our control volume in the pipe is $p_1$, and the pressure at the other end is $p_2$. Since there isn’t any gravity to worry about (thanks space!), the only forces acting on the fluid are the pressure-induced tractions acting on the surface of the control volume. If these pressures are different, a net force acts on the fluid inside the control volume—causing it to flow. As you might have expected, we’ve proved something that is usually intuitive for most; differences in pressure within a fluid, in the absence of countering forces, generate fluid flow.

But forces aren’t the only thing we can analyze with this trick. In fact, we can use this technique to analyze the change of essentially any net property within a control volume!

For example, take into consideration another control volume surrounding a pipe, again in space, filled with a moving fluid containing Chemical X. How would you determine if the amount of Chemical X within the control volume is accumulating (or decreasing) over time, and by how much? Well, there are only two possible ways through which the chemical can enter/exit the control volume; if it was somehow being spontaneously created within the pipe, and by moving into or out of the pipe section. Writing this down mathematically, you’d get the following equation for the rate of change of the mass of chemical X in the control volume over time:

$\displaystyle \frac{\partial m_X}{\partial t} = \dot{m}_{X\text{generated}}\ +\ \dot{m}_{X\text{moving in}}\ -\ \dot{m}_{X\text{moving out}}$

where the dots indicate a rate of change over time. This equation is fairly useless in its current form, but coming up with specific forms for each term will lead to some useful expressions.

Because Chemical X might not be generated in a uniform way within the control volume, the first term in the equation above should actually be the sum of a bunch of tiny contributions within the control volume, each generating (or destroying) some small amount of Chemical X at each fluid point. In other words, the net amount of Chemical X generated or destroyed in the control volume over time comes from adding up changes in the density of Chemical X at each point inside of the fluid volume. Such a contribution is usually called a generation, reaction, or more generally a source, in allusion to the fact that they represent sources (or sinks) of production of a quantity.

Regarding the other two terms, which determine the rate of movement of Chemical X into the control volume, we again find that they come from adding up small contributions of Chemical X, this time moving into or out of points on the surface of the control volume. These small contributions are usually referred to as fluxes (or occasionally flux densities), and are usually mathematically represented with the symbol $\vec{j}$.

To calculate these, we first need to determine the amount of moving Chemical X at a given point, which occurs as a result of two distinct physical processes. The first is advective flux, which is due to Chemical X flowing into (or out of) a point along with the fluid its immersed in; this is just the density $\rho_X$ of Chemical X at the point, multiplied by the velocity of the fluid at that point $\vec{v}$. The second is diffusive flux, which is a type of motion we haven’t discussed previously, which is when molecules of Chemical X spread out from random motion to eliminate gradients in the density of Chemical X, independently from the fluid’s flow. This flux is calculated by multiplying the negative gradient of Chemical X at a point, $-\nabla \rho_X$, with a constant $D_X$ commonly referred to as a diffusivity.

But to determine how much of that movement is going in or out of the control volume, we need to add some kind of extra quantifier that makes the flux contribution zero if the movement is parallel to the surface, positive if the movement is going into the volume, and negative if the movement is going out of the volume. Mathematically, we do this by taking the velocity vector at a point on the surface and dot-producting it with a unit normal vector, which always has a magnitude of 1 (hence unit) and is defined to be always pointing out of the surface of the control volume (hence normal). Doing this has the nifty benefit that we don’t have to worry about distinguishing which terms represent movement into the control volume and which represent flow out; the dot product takes care of that for us.

Finally, we note that the total mass of Chemical X inside the control volume can be calculated by summing up all the density contributions within the control volume. With this in mind, and combining all of these expressions, we get the following equation:

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

$\displaystyle \frac{\partial}{\partial t}\int_V \rho_X\ dV = \int_V \frac{d \rho_X}{d t}\ dV - \int_A \rho_X \left(\vec{v}_X \cdot \hat{n}\right)\ dA + \int_A D_X \left(\nabla \rho_X \cdot \hat{n}\right)\ dA$

This first equation is commonly referred to as the Reynolds transport theorem, and it is incredibly general—so let’s put this equation’s usefulness to the test! Consider a pipe in space again, but this time filled with moving water, and with inlets and outlets of different cross-sectional surface. Let’s say we’ve been running water through this pipe for a relatively long time, so nothing within the pipe is changing. Let’s also assume that water is incompressible, so the density of water is the same everywhere. Can we make any statements about the speed of water coming out of the pipe relative to the speed of water coming in?

We can try to do so by performing a control volume analysis around the pipe. Doing so, we find the following:

1. The mass of water inside the pipe isn’t changing, since we’ve been running it for a long time: $\frac{\partial}{\partial t}\int_V \rho_X\ dV = 0$
2. Water isn’t being generated or destroyed in the pipe through a chemical reaction or anything like that: $\int_V \frac{d \rho_X}{d t}\ dV = 0$
3. The velocity of the water is always pointing either directly into or out of the pipe: $\vec{v}_{\text{inlet}} \cdot \hat{n} = -v_1, \vec{v}_{\text{outlet}} \cdot \hat{n} = v_2$
4. The water is incompressible, so the water’s density at the inlet and outlet are the same and there aren’t any density gradients anywhere: $\rho_{\text{inlet}} = \rho_{\text{outlet}} = \rho, \nabla \rho = 0$

Plugging all of this in, we find:

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

$\displaystyle 0 = - \int_A \rho \left(\vec{v} \cdot \hat{n}\right)\ dA$

$\displaystyle 0 = -\int_{\text{inlet}}\rho_{\text{inlet}} \left(\vec{v}_{\text{inlet}} \cdot \hat{n}\right)\ dA\ -\int_{\text{outlet}}\rho_{\text{outlet}} \left(\vec{v}_{\text{outlet}} \cdot \hat{n}\right)\ dA$

$\displaystyle 0 = \rho\left(\int_{\text{inlet}}v_1\ dA\ -\int_{\text{outlet}} v_2\ dA\right)$

$\displaystyle \boxed{\int_{\text{inlet}}v_1\ dA\ = \int_{\text{outlet}} v_2\ dA}$

Take a look at that; it turns out that the integral of the water’s speed over the inlet has to match the speed integral over the outlet! Roughly speaking, that means that when you have a pipe with steadily flowing water whose inlet is larger than its outlet, the average water speed at the outlet will be bigger than at the inlet. This is why putting your thumb over your sink’s faucet will cause the water to shoot out quickly! A device that exploits this speed-changing phenomenon in engineering is called a nozzle, and you can find one in just about anything that involves fluids flowing inside of something.

Let’s conclude by considering what happens if we use the concepts above to construct an equation for the rate of change of the momentum $\displaystyle \vec{P}$ of the fluid inside a control volume:

• The net momentum inside the control volume $\vec{P}$ is the sum of contributions of momentum density $\rho \vec{v}$ within it, and so the rate of change of momentum in the control volume is equal to the rate of change of those contributions: $\frac{\partial \vec{P}}{\partial t} = \int_V \frac{\partial \left(\rho \vec{v}\right)}{\partial t}\ dV$.
• Momentum is generated or destroyed inside of the control volume only as a result of body forces within the control volume, per Newton’s second law: $\displaystyle \int_V \frac{d\left(\rho \vec{v}\right)}{d t}\ dV = \int_V \vec{b}\ dV$
• Momentum moves into the control volume through momentum density fluxes on points on the surface, either through directly imposed tractions on the surface, or momentum density flowing & diffusing in: $\displaystyle -\int_A \rho \vec{v} \left(\vec{v} \cdot \hat{n}\right)\ dA - \int_A (\vec{j}^{\left[\rho \vec{v}\right]}_{\text{diff}} \cdot \hat{n}) dA + \int_A \vec{t}\ dA$.
I’m keeping the explicit mathematical form of the diffusive momentum density flux term, $\left(\vec{j}^{\left[\rho \vec{v}\right]}_{\text{diff}} \cdot \hat{n}\right)$, hidden for now. All you need to know now is what the term represents.

Putting everything together, we find an equation that tells us how the momentum of a control volume changes under the effects of external forces and fluid flow:

$\displaystyle \frac{\partial \vec{P}}{\partial t} = \int_V \vec{b}\ dV + \int_A \vec{t}\ dA - \int_A \left(\vec{j}^{\left[\rho \vec{v}\right]}_{\text{diff}} \cdot \hat{n}\right) dA - \int_A \rho \vec{v} \left(\vec{v} \cdot \hat{n}\right)\ dA$

That last term is indicating to us that fluid flow can induce forces! It’s also telling us that fluid flow can induce changes in the momentum of the control volume that contains it—to see this in action, just consider some kind of device (in space again, to ignore gravity) that is shooting out incompressible fluid at some constant velocity $\vec{v}$ directly behind it. Making a control volume analysis around the device, we find that the change of the momentum of the object has to be $\displaystyle \frac{\partial \vec{P}}{\partial t} = -\int_A \rho \vec{v} |\vec{v}|\ dA$, which pushes the device in the opposite direction of the fluid’s speed. Using a fluid to induce movement in this way is commonly referred to as jet propulsion, and the device itself can be referred to as a jet (or more arguably, a rocket).

Hearkening back to our conclusion in Part I, the control volume analysis we’ve performed here is letting us understand the different ways pressure, velocity, and density affect each other in a variety of different specific contexts. And to verify its utility, we should spend some time applying this analysis towards the principal engineering application of fluid mechanics; hydraulics.

1. What happens when you include gravity in the examples? Does anything change?

2. Why is the diffusive flux proportional to the negative of the gradient of the diffusing quantity?

3. Say you wanted to design a jet/rocket, and were trying to maximize the amount of force your jet/rocket experiences. Does it benefit you to use a nozzle? If so, what kind? Try to figure it out with control volume analysis.

4. Could you use that last equation to get an estimate for the kind of motion a pressure difference across a pipe induces? What would you need to consider?

5. Can you try to use the techniques above to derive an equation for the net torque on a fluid in a control volume? What would each term represent?

6. Let’s say you shoot a stream of water horizontally onto a wall, and the wall deflects the water straight up and down in two identical streams. How much force is the wall experiencing?

## Part III: Buoyancy

With a decent understanding of hydrostatics and the way pressure and forces interact under our belt, we can then begin to ask some important questions about how objects react to the presence of fluids. For example, how do I know if a boat will sink or float? Is thinking about sinking and floating the only thing I need to thinking about when I’m designing a boat?

To be able to answer those questions, we need to recall the example we saw before of the fluid point at the bottom of the cup. There, we saw that a surface force $\vec{f}_s$ of the same magnitude as the pressure within the fluid point ensures that the fluid point remained stationary. This fact turns out to generalize for any interface between two continuum objects, be they solid or fluid; at every point of the interface between two continua, there is a traction with magnitude equal to the pressure at that point, pointing perpendicular to the interface in the direction of the continua of interest. This is what we saw in the cup example— in that scenario, our continuum of interest was actually the water, and the cup was exerting a traction on the fluid point proportional to its local pressure $|\vec{t}_s| = p$.

This concept allows us to understand & calculate the hydrostatic forces acting on an object embedded in a fluid through the following analysis:

1. Determine the pressure distribution of the fluid.
2. Determine the interface between the continuum of interest and the fluid.
3. Determine the pressure of the fluid at the interface.
4. Calculate the pressure-induced traction at each point on the interface.
5. “Add up” (integrate) those tractions to get the total hydrostatic force on the object.

We can express this last step of the process succinctly using mathematical language:

$\displaystyle \vec{F}_b = \int_{A} \vec{t}_s\ dA$

This procedure involves applying tools from vector calculus, which beginning practitioners of fluid mechanics might find daunting. Luckily, we can obtain the total buoyancy force on any object embedded in a static liquid on Earth without having to do so. In short, the buoyancy force on an object in a fluid is always equal to the weight of the fluid it displaces, and always points up. This is mathematically represented as Archimedes’ law:

$\vec{F}_{\text{b}} = -\rho V_{\text{displaced}} \vec{g}$

For the sake of ideological consistency, we can show how Archimedes’ simple law is derived from the more complicated process of adding up tractions described above. This relies chiefly on the fact that the pressure distribution in a given static fluid on Earth is essentially always the same, stemming from the solution to $\nabla p = \rho \vec{g}$. Since the gravitational force points strictly downwards, we can simply integrate in the “depth” direction $z$ to find that:

$p = p_{0} + \rho|\vec{g}|z$

where $p_0$ represents the pressure at the surface of the fluid and $z$ represents the depth of the fluid at which the fluid point is located. Clearly, the pressure of a fluid point here only depends on its depth within the fluid. As a result, horizontal pressure-induced tractions that act on the surface of an object in such a fluid must cancel out, since a pressure-induced tractions on the “left” side of the object will inevitably be canceled by tractions on the “right” side with the same net magnitude.

Now consider a cube of side length $a$ within the fluid. Its interface is determined by six distinct surfaces; the top, bottom, and four sides. Because the traction on the side surfaces must necessarily cancel by the argument above, the only contributions to the buoyancy force are going to come from the top and bottom, in which the tractions are uniform but distinct.

Therefore, the total traction on the cube is just the difference in tractions between the top and bottom of the cube, multiplied by the area over which they act:

$\vec{F}_{\text{b}} = \int_{A} \vec{t}_s\ dA = a^2 \vec{t}_{\text{top}} + a^2 \vec{t}_{\text{bottom}}$

$\vec{F}_{\text{b}} \cdot \hat{y} = a^2(p_{0} + \rho|\vec{g}|z_{\text{bottom}}) - a^2(p_{0} + \rho|\vec{g}|z_{\text{top}})$

$\vec{F}_{\text{b}} \cdot \hat{y} = a^2 \rho|\vec{g}|(z_{\text{bottom}} - z_{\text{top}})$

$\vec{F}_{\text{b}} \cdot \hat{y} = a^3 \rho|\vec{g}|=V_{\text{cube}} \rho|\vec{g}|$

Because this force is additive in the volume, and because every solid body can be approximated to arbitrary accuracy as a combination of sufficiently small cubes, we have by consequence derived Archimedes’ law for arbitrarily shaped solid objects. Voilà!

This begs the question; why bother thinking about this in a way that requires vector calculus if we don’t need it to calculate the buoyancy force? Luckily, the answer is simple—rotation!

Think of a cylinder wrapped in string. If you pull the string on both sides with equal and opposite force, the center of mass will surely remain stationary by virtue of Newton’s law, but the cylinder as a whole will spin about its axis. Clearly, just the total force on an object doesn’t paint the whole picture of how the cylinder moves; the location of those forces on the object also matters! And since we don’t want our boats to spontaneously capsize as we sail on the ocean, understanding this phenomenon is imperative for any practical application of buoyancy.

In essence, hydrostatic pressure-induced tractions don’t only induce a net buoyancy force $\vec{F}_b$ on an immersed object, but a buoyancy torque $\vec{\tau}_b$ as well. And unlike the buoyancy force, we don’t really have a neat torque version of Archimedes’ law that lets us calculate this buoyancy torque without using vector calculus. As a result, we find the process of calculating it nearly identical to the original process of determining the net buoyancy force:

1. Determine the pressure distribution of the fluid.
2. Determine the interface between the continuum of interest and the fluid.
3. Determine the pressure of the fluid at the interface.
4. Calculate the pressure-induced traction torques ($\vec{r}\times \vec{t}_s$) at each point on the interface.
5. “Add up” (integrate) those traction torques to get the total hydrostatic torque on the object.

We can again list out this last step succinctly using mathematical language:

$\displaystyle \vec{\tau}_b = \int_{A} \vec{r}_{cm} \times \vec{t}_s\ dA$

where $\vec{r}_{cm}$ represents the position of the point being analyzed relative to the object’s center of mass.

As it turns out, this process of summing up small contributions through integration is a critical tool in all areas of fluid mechanics, which we shall soon observe.

1. How would you determine if an object will sink or float? How would you determine if an object will spin when it either sinks or floats, and in which direction?

2. A static, floating object is always partially submerged in the liquid below it. Can you determine the volume of the submerged part of the object using the techniques above? (Ignore air, then ask yourself why you could do that.)

3. Think of the pressure-induced traction distribution on a triangle (◣). Convince yourself that the horizontal tractions cancel out. Which way would the triangle spin if it floats? Is it the same direction if it sinks?

4. Use the result above to justify why boat hulls look the way they do. Try to think of reasons for the design of boat hulls in general (shape, length, width, depth, etcetra).

5. If an object possesses a uniform identical density to water, it will neither sink nor float—its center of mass simply remaining stationary. In this scenario, could it rotate due to hydrostatic forces?

## Part II: Pressure & Hydrostatics

If our objective as fluid mechanicians is to understand how pressure, density, and velocity are related, we should then begin to look at simple examples of fluid mechanical phenomena to get a sense of how these properties interact with each other and with external forces—and in particular, how pressure and external forces interact.

Phenomenologically, it’s important to note that the amount of molecular energy present in some sample of fluid molecules is humongous here on Earth—even in fluids which are completely at rest at the continuum scale. As a result, fluid molecules tend to be pretty good at distributing themselves locally in a way that minimizes that molecular energy. And thanks to all the pushing and pulling between molecules, they tend to rapidly space themselves out so consistently that we can usually always define the number of particles in a given fluid point—and quantities connected to it, like density—as an exclusive function of the molecular energy density/pressure of that point. In short, $\rho = \rho(p)$.

This may lead you to believe that applying a force to a fluid will lead to a change in its density. However, the amount of energy you are giving to a fluid molecule by applying a continuum-scale force to it is positively meager in comparison to the molecular energy the molecule already possesses through pressure. As a result, external forces and other continuum-scale effects almost never alter the density of a fluid—they usually just trigger continuum-scale fluid motion precisely so that the fluid can preserve its density. This assumption or property of most fluids is usually referred to as incompressiblity, and it causes the pressure to be uniquely defined solely by the requirement that the density be the same everywhere.

An everyday example of this can be found in a glass of water. Every point of water in the cup is feeling an identical downwards force density $\vec{f}$, proportional to the density of the point and the gravitational constant ($\vec{f} = \rho \vec{g}$), which we assume is the same everywhere. For future reference, a force distributed through a fluid like this is commonly referred to as a body force. So if Newtonian mechanics holds, then how are those points of water not moving? Why doesn’t all the water simply accumulate into a highly dense thin film at the bottom of the cup?

Common sense indicates that another force, namely a molecular force, needs to be countering gravity in order for the water in the cup to retain its density: $\rho \vec{g} + \vec{f}_{molecular} = 0$. This molecular force is coming from pressure—the molecular energy density of the water is redistributing itself to ensure the density stays the same. This redistribution causes gradients in pressure that manifest as continuum force densities, leading us to obtain the force balance equation for fluids not in motion, or the hydrostatic equation:

$\displaystyle \vec{f} + \nabla p = 0$ (in general)

$\displaystyle \rho \vec{g} + \nabla p = 0$ (when the external force is gravity)

Heuristically, this means that pressure increases in the direction of forces for static fluids. Noting gravity points “downwards”, this is why deep-sea divers can’t go too far down into the ocean without a submersible (they’d get crushed by the increased pressure/molecular energy of the water) and why astronauts wear full-body suits (our bodies’ molecular energy would get dumped out into the very low-pressure environment of space).

But body forces aren’t the only way forces can influence a fluid; we should also consider the influence from external forces that, rather than being distributed through a fluid like gravity, are concentrated on solid surfaces in contact with the fluid (which is how a cup holds water). These surface forces are a little trickier to interpret, but easy to describe with the right conceptual machinery.

Consider a point of fluid located right at the bottom of a cup, in contact with a tiny patch of cup. The point itself isn’t moving, but the molecules within the point are—and something needs to compensate for the lack of fluid points below the one of interest to ensure the fluid point doesn’t move. That compensation is coming from an increase of molecular energy inside that patch of cup to “match” the point’s surroundings. That increase, and its effect in ensuring the fluid point remains stationary, can be mathematically represented as a local force per area pointing into the fluid with the same magnitude as the local pressure $\left(|\vec{t}_s| = p\right)$.

We can call that local force per area a surface traction, contact pressure, or simply a traction; there isn’t really a standard clear name for the concept, and many sources incorrectly refer to it as pressure without making the distinction between the local force per unit area and its molecular source.

To humanity’s benefit, this principle of induced surface forces due to pressure works both ways: if we impose a surface force on a fluid, a static fluid will locally increase its pressure by the magnitude of that force per unit area to compensate. This bidirectional principle lets us manipulate forces acting on objects in useful ways, all of which are variations of the following sequence of phenomena:

1. An object imposes a net force on a fluid, in the form of a force per unit area distributed over a surface.
2. The surface force generates a pressure increase throughout the fluid.
3. The extra pressure is transmitted as a force per unit area onto another object with a different surface area, leading to a different net force acting on that second object.

The field of engineering that utilizes this force-multiplying principle to solve problems (among many other fluid-mechanical tools) is called hydraulics, and has many applications throughout varying fields of science & technology. Notable examples include hydraulic jacks, hydraulic suspensions, hydraulic presses, etcetra.

The arguments we’ve made here are quite general, but don’t really account for what happens when the traction acting on a surface isn’t the same everywhere. To understand what happens in that situation, we’ll need to look at one of the first great scientific discoveries of fluid mechanics; buoyancy.

1. Think of some simple examples of force fields and calculate/infer what the pressure distribution they induce in a still fluid might look like. Can you think of any force field that leads to nonsense pressure results? What would that mean?

2. Can you think of a way to determine when the incompressibility assumption is correct? What properties would you need to know? Can you think of a “metric” to determine how correct the assumption is?

3. If you heat up a liquid in a kettle or frying pan, you would usually reduce its density quite noticeably. How does this not contradict the statement that external continuum forces don’t affect fluid densities?

4. Solids can store energy in molecular bonds when pushed or pulled, behaving in a way similar to a tiny bunch of connected springs. How is this behavior different from the behavior of a fluid? How is it similar?

5. Why isn’t the density of a fluid a function of the fluid velocity—especially if we stated well-defined densities occurs as a result of molecular motion?

6. Can you think of a way to mathematically derive that $|\vec{t}_s| = p$? What would you need to consider?

7. What are the units of pressure, and it is a scalar or a vector? What are the units of traction, and is it a scalar or a vector? Can you think of reasons why people might confuse the two?

8. Can you think of some applications for the concept of hydraulics described above? Are there any you can identify that have already been made?

## Part I: Fluid Concepts

Before delving into the study of fluid mechanics⁠—or the study of anything, really⁠—it’s imperative to look at the underlying assumptions we make when we try to analyze the phenomena we wish to learn about. Namely, what is a fluid, in a conceptual/philosophical sense? And how does that description of a fluid create other related concepts that we can observe and measure?

For starters, most of the matter we interact with on a daily basis tends to come in incomprehensibly large clumps of smaller buildings blocks called atoms or molecules. These clumps can be either mostly stiff and with a specific shape, in which case we would usually describe the clump as solid, or flowing and with uncertain, changing forms, which we would describe as fluid.

In either case, trying to understand the behavior of a single clump really entails understanding the collective behavior of a hundred sextillion of these building blocks, each performing their own complicated dance through time and space. And if you think that trying to understand what each of these molecules is doing is effectively impossible, I’d agree with you! Even though it would be possible to describe the physics of each of these constituents fairly straightforwardly using models of molecular physics, not even the most powerful supercomputer would be able to easily and faithfully obtain the motion of these molecules from them due to their sheer numbers.

However, this incomprehensible complexity runs contrary to most of our daily experiences. I don’t expect my morning coffee to spontaneously crawl up the side of my cup and spill itself—and if I were to tilt my cup of coffee, I would reliably see its liquid surface stay parallel to the ground. Water from the faucet usually comes out in a steady stream, ketchup bottles doesn’t spontaneously explode or dissolve, and the ocean mostly stays put where it’s usually been. In short, the minuscule collective randomness we would expect to see from these massive assemblies of molecules averages out into behavior that is fairly uniform and easy to understand at the scales that we can see and feel.

As a result, the theory of fluids (and solids) is a theory that only cares about those scales that we can directly experience, and considers those tiny building blocks only when it really needs to. This means that the way we mathematically describe large clumps of fluid matter is as precisely that—continuous clumps. Formally, we refer to the theory of fluid mechanics as a continuum theory.

In a continuum theory, we usually focus on some abstract blob of “something” (in this case, matter) distributed over space, which has some properties that also vary over the space the blob occupies. For example, the atmosphere can be thought of as a moving continuum “blob” of air (and other chemicals) whose speed and density changes as you move around and above the Earth. But this brings up another question; what properties do we generally care about in a fluid?

Since the distinguishing characteristic of fluids is motion, surely you’ll agree that velocity is a property we care about. In the theory of fluid mechanics, we associate each point $(x,y,z)$ in a fluid at a time $t$ with a velocity $\vec{v}(x,y,z,t)$. This fluid velocity has three distinct components, $v_x(x,y,z,t)$, $v_y(x,y,z,t)$, and $v_z(x,y,z,t)$, each representing the speed of the fluid point in the $x$ / $y$ / $z$ direction. Its interpretation is straightforward; we expect a “point” of fluid located at $(x,y,z)$ to move with velocity $\vec{v}(x,y,z,t)$ at time $t$. But what does a “point” of fluid even mean?

Recall that, at the scales we experience, what we think is a “point” of fluid is actually a humongous mess of molecules randomly bouncing around at a molecular scale. Therefore, in order to make any sense of this at the continuum scale, we need to somehow “smooth” out all the molecular unpredictability into something predictable and measurable at the continuum scale. Luckily, we can do this without losing too much accuracy by selecting some specific volume size that is very small at the continuum scale, and treating that volume like a continuum “point”. We can then define the properties of that fluid “point” as an average of the properties of the molecules inside it. This is commonly referred to as the continuum approximation, and the properties we obtain from these averages are called continuum or fluid properties.

To be more precise, the properties of fluids we obtain using the continuum approximation are limits of ratios; ratios of volumes and the sum of the properties of molecules inside them. To demonstrate, consider a spherical molecular “net” in a fluid through which molecules pass in and out. At every instant, the net contains some well-defined amount of fluid molecules, and therefore some well-defined total molecular mass. If this net is small at the molecular scale, random perturbations to the total fluid mass within the net as a result of molecules moving in and out of it screw up our ability to define a consistent value of total fluid mass in the net over time. However, as you make the net bigger and bigger at the molecular scale—while keeping it point-like in the continuum scale—the randomness smooths itself out due to the sheer number of molecules, and one observes a total fluid mass within the net that is only a function of the net volume. The ratio of that total fluid mass to the net volume is called the density $\rho(x,y,z,t)$ of the fluid.

Likewise, we could do the same thought experiment but now counting up the energy of the fluid molecules in the net, specifically considering the energy coming from the unpredictable back-and-forth motions of molecules at the nanoscale. As the net becomes bigger, the ratio of the total energy divided by the net volume approximates a kind of molecular energy density usually referred to as pressure* $p(x,y,z,t)$. Note that this is unrelated to the energy from motion of fluid “points” at the continuum scale; pressure is entirely molecular.

*As I’ll talk about in Part II, most people get this definition wrong; even Wikipedia! It doesn’t affect much, but the Wikipedia definition obfuscates the molecular nature of pressure as an energy density, which is an important concept.

Finally, as you might expect, the velocity of a fluid point is simply the average velocity of the molecules within a molecular net of the appropriate size. Interestingly, although each individual molecule possesses a decent instantaneous velocity from random molecular motion, it turns out that the average velocity (i.e. the fluid velocity) of a group of molecules is very often quite low in comparison. As a result, almost all molecular motion is irrelevant to fluid motion. This doesn’t mean that what happens at the molecular scale is completely disconnected from what happens at the continuum scale, though; it just means that we should be able to describe most of what we see in fluid mechanics using only the averaged, continuum properties we described above.

This trinity of properties associated with points of fluid—fluid velocity, density, and pressure—are usually all we need in elementary fluid mechanics to be able to describe the motion of fluids. As a result, we expect the chief enterprise of any fluid mechanician to be describing how each of these properties is related to each other, and how things external to a fluid (forces, etc.) change them.