# 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.