Part X: Dimensional Analysis

We are, at the present, in a bit of a pickle. We have spent a lot of time deriving the laws of fluid mechanics to understand flow through a pipe, but we’ve found that even in something as simple as pipe flow, the inherent pathology of the laws of fluid mechanics causes pipe flow to spontaneously transition from something we can understand (Hagen-Poiseuille flow) to something we can’t (turbulence). We’ve also commented on the fact that experimentalists have noted that the transition from one to the other occurs when the Reynolds number, which we’ve usually taken to represent an approximation of the ratio of convective to viscous effects in a flow, passes a critical value—even though the actual ratio of convective to viscous effects in Hagen-Poiseuille flow is identically zero. How could this be?

To solve this mystery, we’re going to need apply a little bit of what may appear to be abstract nonsense. Let’s again take a look at a horizontal section of pipe of radius R, with an arbitrarily long length, and a fluid for whom we prescribe a uniform pressure p and a purely longitudinal, uniform velocity with speed |\vec{v}| at the inlet. The fluid itself has a constant density \rho and constant viscosity \mu, and is under the effects of gravity with an acceleration magnitude of |\vec{g}|. These quantities should fully define the flow throughout the pipe, whatever that flow may be.

A horizontal section of pipe with a radius R and an arbitrarily long length, containing a fluid with constant density and viscosity and under the effects of gravity. The pressure and velocity are prescribed uniformly at the inlet; as the flow passes through the pipe, it will either evolve into Hagen-Poiseuille flow or turbulent flow, something fluid-mechanical theory can’t tell us.

What was said above is worth repeating; every single characteristic of the flow downstream should be determined by a unique combination of these physical quantities. As a result, any physical quantity or property that we could possibly “read out” from this flow has to be a function of these 6 parameters and of these 6 parameters only.

One such readout could be some kind of turbulence index \zeta, a simple number that is 0 when the flow is Hagen-Poiseuille and 1 if the flow is turbulent. In accordance with the statement above, we can say this turbulence index \zeta is only a function of the above listed parameters:

\zeta = f(\rho, |\vec{v}|, |\vec{g}|, p, \mu, R)

Alright, this helps us out a little but not much. That being said, we’ve actually constrained the problem far more than it appears we have—to demonstrate, take a look at what happens if I claimed the following:

\zeta = \rho

From a mathematical perspective, there’s no problem with this, but physically, this doesn’t make a lick of sense; a number like the turbulence index can’t be equal to a density! This is like saying “A dog is three oranges” or “The sky is looking pretty book today”—it just doesn’t possess meaning as a physical description, because we aren’t relating objects that are the same type. In our case, whether or not objects are of the same type is determined by their physical dimensions. In fact, notice that it’s impossible to have any expression of the form \zeta = f(\rho) make physical sense, because there’s no mathematical function depending on the density that can take a density and return a simple number! And the same goes for every other physical variable that describes the system.

As a result of this, we find that we need to get a number on the right-hand side of our equality, and to get that to happen, we need to have our right hand side to be a function of just numbers as well. Now we face another question; how many independent number quantities can we construct out of those six physical parameters, and what are they? To do this, let’s list out all the quantities in the system and their units (in SI):

Physical quantityDensity \rhoViscosity \muRadius RPressure pSpeed |\vec{v}|Gravitational acceleration |\vec{g}|
Units (SI)\frac{kg}{m^3} \frac{kg}{m\, s} m \frac{kg}{m\, s^2} \frac{m}{s} \frac{m}{s^2}

Notice that the units of each quantity are built from multiplying and dividing three fundamental units: meters representing distance, seconds representing time, and kilograms representing mass. To get numbers with no units, it follows that we have to multiply and divide these quantities by each other to “cancel out” any and all fundamental units they may possess.

Let’s start out by trying to cancel out the gravitational acceleration term |\vec{g}|. As expected, it possess units of acceleration \frac{m}{s^2}, and so we it seems natural to find quantities with units of meters, seconds, or some combination of them to try and cancel its units out. If we decided to divide |\vec{g}| by the speed squared |\vec{v}|, that would get rid of all of our time units; multiplying that ratio by the radius R nets us something that doesn’t have units! Note that we didn’t need to cancel gravity out in any specific way; I just picked out a process to try and make the unit removal as simple as possible. In any case, let’s go ahead and call this number \Pi_1:

\Pi_1 = \frac{|\vec{g}|R }{ |\vec{v}|^2}

Note that we could do basically whatever we wanted to this quantity and still get a dimensionless number; we could square it, cube it, multiply it by a constant, exponentiate it, you name it; it will still be a totally valid number. In fact, most fluid mechanicists like to use the square root of the inverse of this number out of habit, which they call the Froude number Fr:

Fr = \sqrt{\frac{1}{\Pi_1}} = \frac{|\vec{v}|}{\sqrt{|\vec{g}|R}}

Alright, we’ve got one number down—let’s try to build another one. To guarantee independence between our numbers, let’s try to start the process of canceling units by picking some physical quantity that didn’t show up in our numbers before, like the pressure p. It has units of force per unit area \frac{kg}{m\, s^2}, so we can try canceling out the time units the same way we did for \Pi_1, by dividing by the speed squared. We can then get rid of the mass units by dividing by the density, leaving us with a quantity possessing only units of distance squared, which we can readily get rid of by dividing again by the radius squared. This second number we’ve just found, \Pi_2, is also commonly called the Euler number Eu:

Eu = \Pi_2 = \frac{p}{\rho|\vec{v}|^2 R^2}

Neat! With these two dimensionless numbers in hand, we’ve utilized every physical quantity in the system expect for the viscosity; it makes sense to start our search for a third number there. Viscosity has units of \frac{kg}{m\, s}, so we can try to divide by the speed and by the density to get rid of the time and kilogram units respectively. That leaves behind a single distance unit, which we can get rid of by dividing by the radius. This third number might look a little familiar if we explicitly show it:

\Pi_3 = \frac{\mu}{ \rho |\vec{v}| R} = \frac{1}{Re}

This is just a sneaky inverted form of our good friend the Reynolds number.

Now we can ask, are there any other independent dimensionless numbers we can construct out of these quantities? As it turns out, there isn’t—although there’s an infinity of different numbers we can make out of these quantities, they’d all wind up being functions of the three we saw above. This is because of a famous theorem called the Buckingham pi theorem, which states that the number of dimensionless numbers one can build out of a set of physical quantities is equal to the number of physical quantities minus the number of fundamental units they’re built from. (Proving this involves some relatively simple linear algebra, but it isn’t really germane to what we’re doing here.) In our case, we have 6 quantities and 3 units, so we get 3 dimensionless numbers as I stated before.

Given these numbers, we now have much tighter relationship between the turbulence index \zeta and the variables in the system:

\zeta = f(\Pi_1, \Pi_2, \Pi_3) = g(Fr, Eu, Re)

Remember that changing the specific dimensionless numbers we use as arguments doesn’t really matter, whether they be \Pi_1 or Fr—all the change does is alter the structure of the function, which we don’t know anyways.

Nearly all of the time, fluid mechanicians discard the dependence of the turbulence index on the Froude and Euler numbers by making some assumptions on the role of pressure in the flow. For example, in Hagen-Poiseuille flow, the absolute value of the pressure at either the inlet or the outlet didn’t play a physical role, just their difference; if we assume this to be true for turbulent flow as well, then the flow is independent of the pressure, which means it must be independent of the Euler number. Similarly, gravity didn’t play a role in the flow because all it did was generate a pressure gradient perpendicular to the flow, which doesn’t factor into the equation we used to solve for the Hagen-Poiseuille flow. All it does is make the absolute pressure at the inlet non-uniform, which we’ve already considered irrelevant, making the turbulence index be independent on the Froude number as well.

This gives us the desired relationship that experimentalists observe:

\boxed{\zeta = g(Re)}

As a result, the Reynolds number—which we saw before as an approximation to the ratio of convective to viscous effects—is now playing a dual role as a dimensionless parameter solely responsible for determining whether pipe flow is turbulent or not. And given our predetermined form of the turbulence index \zeta, which outputs one if the flow is turbulent and 0 if the flow is Hagen-Poiseuille flow, we know everything about the function g(...) except for the location where the “switch” from 0 to 1 happens. Experimentalists measure the switch to happen at about Re \approx 2300.

This process that we have just described required understanding absolutely nothing about the nature of the turbulent solutions to the Navier-Stokes equation, or the nature of the process by which flow transitions from Hagen-Poiseuille to turbulent, other than some assumptions on the role of pressure. In fact, it required understanding nothing about fluid mechanics at all except for understanding the physical quantities required to uniquely define a fluid flow.* This process is called dimensional analysis, and is far more powerful than it may appear at first glance.

*Granted, the entire reason we had to resort to this was because we weren’t finding a unique flow from this mathematical setup, but we are hypothetically observing only one type of flow for a given set of conditions, which is the flow we care about.

Consider what would happen if I as an experimentalist wanted to understand the critical average flow speed |\vec{v_{\text{crit}}}| of some fluid through a pipe of radius R_0 at which turbulence occurs, but only had access to a pipe of radius R_1. Well, because the turbulence index is only a function of Reynolds number, all I have to do is match them:

Re_{\text{crit}} =  \frac{ \rho |\vec{v_{\text{crit}}}|_1 R_1} {\mu} =  \frac{ \rho |\vec{v_{\text{crit}}}|_0 R_0} {\mu}

By doing some experiments to find the critical speed |\vec{v_{\text{crit}}}|_1 in the pipe I have access to, I can then solve for the critical speed |\vec{v_{\text{crit}}}|_0 I’m interested in:

|\vec{v_{\text{crit}}}|_0  =  \frac{|\vec{v_{\text{crit}}}|_1 R_1}  {R_0 }

I can now determine the critical speed at which turbulence occurs for a given fluid flowing through a pipe of arbitrary radius by doing experiments on just one pipe. Such a set of systems is said to possess similitude, and it is a key design tool for fluid dynamicists; an engineer designing a humongous oil pipeline can rest assured that an appropriately scaled & sped-up model of the pipeline in his lab will show the same non-turbulent/turbulent behavior as his final gargantuan product. One could also change the densities and viscosities too—as long as the Reynolds number is the same, the non-turbulent/turbulent behavior will be the same.

Two pipes with different radii and inlet velocities. Their inlet velocities are calibrated such that they both possess the same Reynolds number, which implies they will show the same non-turbulent/turbulent behavior. Engineers would refer to such systems as similar.

Most engineers would call it a day here, but we’re not done yet. Understanding how to trigger turbulence is one thing, but it doesn’t really give us any information about the direct physical consequence of it; how it affects the behavior of the flow in the pipe. More specifically, we don’t know anything about the pressure drop in a pipe flow when the flow is turbulent. For this, we can use a slightly different flavor of dimensional analysis.

To make straightforward comparisons with the results we found before from Hagen-Poiseuille flow, we can define some variable \left< \frac{\partial p}{\partial z} \right> representing the average pressure drop per length in a pipe flow, Hagen-Poiseuille or turbulent. This quantity has units of pressure per length, or \frac{kg}{m^2 s^2}. If we ignore the roles of absolute pressure and gravity by the arguments we justified above, we’ll inevitably find an expression for the average pressure drop per unit length of the form:

\left< \frac{\partial p}{\partial z} \right> = f\left(\rho, \mu,  |\vec{v}| , R\right)

We know that dimensional analysis restricts this expression more. In fact, we know that whatever is on the right-hand side has to possess units of pressure per length, and so we should set about constructing physical quantities with units of pressure per length out of the four quantities we listed above.

To save you the trouble, there’s only two such independent quantities we could construct: \frac{\rho|\vec{v}|^2}{R} and \frac{\mu|\vec{v}|}{R^2}. Consequently, one could expect that the expression representing \left< \frac{\partial p}{\partial z} \right> could be something of the following form:

\displaystyle \left< \frac{\partial p}{\partial z} \right> = \sum\limits_{n \in \mathbb{R}} \alpha_n \left( \frac{\rho|\vec{v}|^2}{R} \right)^n  \left(\frac{\mu|\vec{v}|}{R^2}\right)^{1-n}

Those \alpha_n are dimensionless, and so must be a function only of the single dimensionless quantity we can construct out of this system—the Reynolds number.

Now, recall what we said about turbulent flows before; their Reynolds tensors are decidedly non-zero. In fact, it’s reasonable to expect that as we crank up the Reynolds number in a turbulent pipe flow, the Reynolds tensor is going to increase as well. Additionally, we could crank it up enough such that the convective effects completely dominate the viscous effects! In that scenario, we’d find that all of the flow characteristics are virtually independent of the viscosity.

If that’s the case, then the expression for the pressure drop per length we found above is tightly constrained by the fact that every term with the viscosity in it has to vanish, since the pressure drop per length can’t depend on it. This leaves us with just the n = 1 term:

\displaystyle \left< \frac{\partial p}{\partial z} \right> =  \alpha_1 \frac{\rho|\vec{v}|^2}{R}

We had originally stated that \alpha_1, like the other coefficients, had to be a function of just the Reynolds number. But alas—the Reynolds number is a function of the viscosity! The only way to reconcile the fact that the viscosity can’t influence the physics with this Reynolds-viscosity dependence is if \alpha_1 wasn’t a function of anything; if it was just some arbitrary constant number, totally independent of the Reynolds number. I like to call this phenomenon, where a reduction in the number of dependent physical quantities makes it impossible to construct a dimensionless number, a dimensional crisis. Sounds cool, doesn’t it?

Anyways, we’ve found the following result at high-Reynolds numbers for pipe flow:

\displaystyle \left< \frac{\partial p}{\partial z} \right>_{\text{high Re}} = \alpha_1 \frac{\rho|\vec{v}|^2}{R}

where \alpha_1 is just some constant number.

As a sanity check, we could try the same procedure when the Reynolds number is low and see if what we get matches the Hagen-Poiseuille result. In that scenario, the Reynolds tensor should be small (it is in fact identically zero), and so the viscosity-proportional term should dominate the behavior of the system. To that end, the average pressure drop per length should be independent of the only physical variable not present in the \frac{ \mu|\vec{v}| }{R^2} term, the density.

That means we kill off every term in the above sum except for the n=0 term, and again trigger a dimensional crisis, leading to the following expression at the low-Reynolds number limit:

\displaystyle \left< \frac{\partial p}{\partial z} \right>_{\text{low Re}} = \alpha_0 \frac{ \mu|\vec{v}| }{R^2}

This precisely matches the expression we got before for Hagen-Poiseuille flow; that expression lets us determine that \alpha_0 = 8. Up to that factor of 8, we could have avoided all the differential equation-solving we did in the previous Part by making the assumptions we did above and performing dimensional analysis; we would have gotten the same result knowing absolutely nothing about differential equations. That’s pretty awesome!

I should mention that fluid mechanicists do a very cute trick when studying this average pressure drop per length; they divide the pressure drop per unit length by one of those constructed quantities with units of pressure per length, almost always the \frac{\rho|\vec{v}|^2}{R} expression—this expression is called the Fanning friction factor f:

\displaystyle f = \frac{ \left< \frac{\partial p}{\partial z} \right>}{ \frac{\rho|\vec{v}|^2}{R}}

The reason they do this is because now you wind up getting a dimensionless number, and we already showed that this number can only be a function of the only other dimensionless number we can construct in this system, if it’s a function of anything at all—the Reynolds number. Hence, plotting the Fanning friction factor against the Reynolds number tells me the pressure drop behavior of every single possible pipe flow system of the type we describe here by virtue of similarity.

Dividing the previous results we got by this \frac{\rho|\vec{v}|^2}{R} expression, we find the following expressions for the friction factor at low and high Reynolds numbers:

\displaystyle f_{\text{low Re}} =  \frac{\alpha_0 \frac{ \mu|\vec{v}| }{R^2}}{ \frac{\rho|\vec{v}|^2}{R} } = \alpha_0 \frac{\mu}{\rho|\vec{v}|R} = \frac{\alpha_0}{Re}

\displaystyle f_{\text{high Re}} =  \frac{ \alpha_1 \frac{\rho|\vec{v}|^2}{R} }{ \frac{\rho|\vec{v}|^2}{R} } = \alpha_1

Having done no calculus, and relying almost exclusively on dimensional analysis, we have established reasonable hypotheses for the behavior of the Fanning friction factor as a function of the Reynolds number; in a logarithmic plot, it will show up as a straight line with slope of -1 at the start, followed by a vertical kink somewhere due to the discontinuous transition from Hagen-Poiseuille to turbulent flow, while eventually flattening out into a straight horizontal line as the Reynolds number gets higher.

Lo and behold, look what the experimentalists got:

A logarithmic plot of the Fanning friction factor versus the Reynolds number in a pipe as determined by experimental data. The friction factor is proportional to the inverse of the Reynolds number for low-Re, and independent of the Reynolds number at high-Re. This diagram is called a Moody diagram, and fluid mechanicists often also include an empirical value called surface roughness in it to justify pipe irregularity effects. All surface roughness values lead to the same qualitative behavior as shown above.

That’s one hell of a magic trick! Using the behavior at the “ends” to come close to describing the behavior in the “middle” is squarely in the ballpark of a field of math called asymptotics, and for my next trick, we’re going to use it to answer the last remaining question of introductory fluid mechanics—how to characterize flow over an immersed object.

Things to Think About

1. Before we described the Reynolds number as being an approximation of the ratio of convective effects to viscous effects. How would you describe the Froude and Euler numbers in a similar way? Is there a single way to do so?

2. Try to find a physical system you don’t understand and use the dimensional analysis techniques I demonstrated above to get some practical answers from them. When is it helpful? When isn’t it?

3. Sometimes dimensional analysis lets us construct differential equations we can then solve. Consider dipping a hot sphere of radius R\  (m) with a specific heat C\ (\frac{kg}{s^3 K}) into a vat of cold liquid with a fixed ambient temperature far from the sphere. If you assume the rate of change of the temperature with time \frac{dT}{dt} is only a function of these two parameters, of the current temperature of the sphere T (K) , and of the heat transfer coefficient h\  (\frac{kg\,m^2}{s^2 K}) , find an expression for \frac{dT}{dt} using dimensional analysis. Compare what you get to what engineers use. Try solving it, if you can! No knowledge of thermal physics is needed.

4. Physicists in the early 20th century noticed that particles at the quantum scale behave like waves. Knowing nothing about quantum mechanics other than that this wave is a function of the particle’s mass m (kg), the particle’s speed u (\frac{m}{s}), and the Planck constant h (\frac{kg\, m^2}{s}), determine the wavelength (m) and frequency (\frac{1}{s}) of this matter wave up to a proportionality factor. Compare it to what quantum physicists theorized.

5. Do you agree with my arguments for why the Euler and Froude numbers don’t matter for turbulent pipe flow? Can you think of other fluid-mechanical systems where you should have to consider them?

6. Think about how you would exploit similarity to analyze the flow of a substance like oil through a pipe versus a substance like water. What would you do?

Part IX: Pipe Flow & Turbulence

With the analytical tool of the Navier-Stokes equation in hand, we can finally begin to make statements about the precise shapes and speeds of flows. Perhaps the best place where we can make those statements without running afoul of the pathology of the Navier-Stokes equation—and trust me, we will run afoul of it—is in studying flow in pipes or channels, which we described using hydraulic equations in Part V.

Let’s again consider steady incompressible flow through a horizontal section of pipe with length L and with a uniform cross-section. This time, we’re going to try and explicitly define what the momentum loss \vec{H} is using the Navier-Stokes equation, and get a formula for the shape and speed of the flow inside of the pipe as a function of the other free parameters in the system. To do this, we’ll need to do a couple of tricks to bring everything down from the integral formulation we described in Part V—the fundamental equations of hydraulics—to a differential one.

Now, the first fundamental equation of hydraulics tells us that, because the fluid is incompressible, conservation of mass requires that the speed at the inlet of the pipe be the same as the speed at the outlet, meaning that all the momentum coming into the pipe from flux at the inlet must leave at the outlet. In addition, all the momentum being added into the pipe due to gravity is acting perpendicular to the momentum loss; this immediately indicates that the momentum loss in steady flow within such a pipe must necessarily be compensated by a drop in the pressure (and only a pressure drop) along the flow direction.

A horizontal pipe of length L with constant cross-sectional area and the associated tractions/force densities acting on it. Conservation of mass indicates that the speeds on either end are equal; this means the loss coming from \vec{H} can only be balanced by a drop in the pressure between the ends due to conservation of momentum.

A nice thing about this result is that it is totally independent of the length of the pipe section! We can make the pipe as long or as short as we want and the above statement still holds. Now I’m going to do a classic math trick; I’m going to pick some horizontal line across the pipe, and represent the pressure through that arbitrary line as a function of the distance from the inlet z with a Taylor series off of the value at the inlet portion of the line:

p = p_1 + G_1 z + G_2 z^2 ...

If I wanted to represent the pressure drop, I could just subtract by the inlet value:

\Delta p = G_1 z + G_2 z^2 ...

If I take the length of the pipe and shrink it down more and more, the squared term is going to get smaller than the linear term as long as both terms are nonzero. This means that, for an infinitesimally short length of pipe, the pressure drop is always strictly linear in the length, and so the pressure drop per length is a constant, G_1.

Now think about this; if I assume the momentum loss is independent of the pressure, I can take the section of pipe and split it up into a bunch of infinitesimally tiny segments one behind the other, which should all be essentially identical to each other; sure, the pressure at each of the inlets might be different, but the pressure drop should be the same, as well as everything else! This implies the pressure drop is linear in all of the tiny segments, each with the same pressure drop per length G_1, so the pressure drop per length in a finite section of such a pipe must be constant (under the given assumptions, of course).

This gives us what we need to analyze pipe flow through a horizontal section of pipe with constant cross-sectional area using Navier-Stokes. In this scenario, based on the argument above, we are going to impose some arbitrary pressure drop per length through the pipe G_1, and see what the resulting flow field looks like. The argument above guarantees (or at least, suggests) that imposing such a pressure drop is physically sensible, and will lead to a sensible solution of the Navier-Stokes equation. Before we do, however, notice a very curious quirk in our logic—if we assume the pressure drop per length is constant, that means that the pressure in the pipe through some arbitrary horizontal line is going to be of the form:

p = p_1 + G_1 z

where p_1 can be the pressure in an arbitrary location of the pipe. No matter how big that arbitrary reference pressure is, there is always some length of pipe in one direction that causes the pressure to go negative, which makes no physical sense! In that sense, we find that any solution we get for such a problem can’t apply to arbitrarily long pipes, as we will invariably run into a region of pipe where the solution we find simply cannot exist—an existence failure, as we saw in Part VII, and a harbinger of things to come.

Alright, let’s get on with the solution already. We want to solve the incompressible Navier-Stokes equation for steady flow inside of a cylindrical pipe with a uniform cross-sectional pressure and a constant pressure drop per length in the “pipe” direction:

\displaystyle \rho\left(\frac{\partial \vec{v}}{\partial   t}\ + \nabla \vec{v} \cdot \vec{v} \right) = \vec{f} - \nabla p\ +\    \mu \nabla^2 \vec{v}

Let’s list out a couple of observations and “common-sense” assumptions that get us to a point where we can handle this equation:

  1. We assumed the flow is steady already, so the time rate of change term is zero: \displaystyle \rho\frac{\partial \vec{v}}{\partial   t} = 0.
  2. There’s no flow in any direction other than in the “pipe direction”/longitudinal direction; this means there’s no flow in the radial direction or the angular direction (swirling flow): \vec{v} = [v_r\ v_\theta\ v_z] = [0\ 0\ v_z]
  3. Everything about the flow is totally independent of the pipe angle, so we can kill off any terms involving changes in the angular direction: \frac{\partial ...}{\partial \theta} = 0
  4. Gravity is the only external force and acts downwards, so it only causes pressure gradients perpendicular to the flow direction.

If we take a look at the incompressible conservation of mass equation and take into consideration these assumptions, we see something that is fairly obvious:

\nabla \cdot \vec{v} = \frac{1}{r}\frac{\partial(r v_r)}{\partial r} +  \frac{1}{r}\frac{\partial(v_\theta)}{\partial \theta} + \frac{\partial v_z}{\partial z} = 0

\frac{\partial  v_z}{\partial z} = 0

The flow doesn’t change as we move along the pipe/longitudinal direction. This is going to do something drastically nice for our problem; it’s going to kill off the convective flux term we spent so much time whining about!

\nabla \vec{v} \cdot \vec{v} = v_z \frac{\partial v_z}{\partial z} = 0

That means that the Reynolds tensor for this problem is exactly 0, and we shouldn’t have to worry about shocks and all of those other horrible things we saw in the previous Part because the Navier-Stokes equation becomes linear. In fact, it becomes something almost too simple to recognize:

\displaystyle 0 = - \nabla p\ +\    \mu  \nabla^2 \vec{v} +\rho \vec{g}

The Navier-Stokes equation is telling us that, as we saw before, the pressure drop down the pipe length needs to be balanced out by a momentum loss or viceversa; but our knowledge of rheology in a fluid described by Navier-Stokes is telling us specifically how that momentum loss is related to the velocity of fluid in the pipe, giving us the connection to the velocity we needed!

Recalling that the equation we’re dealing with relates vectors, we can now begin to look at what this equation tells us for the individual components of these vectors. If I looked at the component of the vectors pointing in the direction of gravity in this equation, which I’ll label as the \hat{y} direction, I’d get a simple relationship that looks very familiar:

0 = -\frac{\partial p}{\partial y} + \rho g

This is just the hydrostatic equation we saw before! That means that the pressure of a fluid in a horizontal pipe increases in the direction of gravity consistent with what you’d observe in a horizontal pipe full of standing fluid. This also has the convenient effect of ensuring that gravity doesn’t affect the flow; it only affects the absolute value of the pressure, which we assumed previously as not affecting the momentum loss.

Now we can take a look at the components of the vector that are perpendicular to gravity, pointing down the length of the pipe. Applying the assumptions we listed above and doing some simplifications, we find that:

\displaystyle 0 =  -\frac{G_1}{\mu} + \frac{1}{r}\frac{\partial(r \frac{\partial v_z}{\partial r})}{\partial r} = \frac{1}{r}\frac{\partial v_z}{\partial r} + \frac{\partial^2 v_z}{\partial r^2}

Not only is this equation an ordinary differential equation—we’re only taking derivatives in the radial direction r—but it’s also linear in the velocity! Doing some calculus tricks to solve it, we find the general solution:

v_z(r)=c_1 \log (r)+c_2+\frac{G_1 r^2}{4 \mu}

We need c_1 = 0 or the velocity will go to negative infinity at the center of the pipe, and we need the velocity at the edge of the pipe to be zero to ensure that the velocity doesn’t discontinuously change at the pipe/fluid interface—giving us a value of c_2 such that we finally have our velocity for pipe flow:

v_z(r)=\frac{G_1 (r^2 - R^2)}{4 \mu }

\boxed{\vec{v} = \left[0 \enspace 0\enspace \frac{G_1 (r^2 - R^2)}{4 \mu }\right]}

This type of flow is called Hagen-Poiseuille flow, and is perhaps the most famous and well-known flow in fluid mechanics. It’s parabolic in shape, is 0 at the pipe edges by definition, peaks at a value of v_{max}=\frac{G_1 R^2}{4 \mu } at the center of the pipe, and has a cross-sectional average of \langle v \rangle = \frac{G_1 R^2}{8 \mu }.

A view of Hagen-Poiseuille flow within a pipe for specific parameter values. The velocity drops to 0 at the pipe edges, denoted in black, while the velocity peaks in the center. We have assumed there is no flow in any direction other than the longitudinal one.

This finally gives us an expression for the momentum loss in a horizontal, constant cross-section pipe as a function of the velocity, by solving for the pressure drop and noting the drop in pressure is equivalent to the momentum loss from the arguments we made at the start:

|\vec{H}| = \frac{8 \mu \langle v \rangle L}{R^2}

In a kinder, more just world, this would be all there is to it. Unfortunately, when researchers started directly looking at pipe flows, they noticed that everything went according to theory—until they cranked the speed up high enough:

A simulation of pipe flow. Notice how the observed flow profile is totally inconsistent with the Hagen-Poiseuille flow solution we obtained above.

What the hell is going on? Well, as the existence failure for the pressure had been warning us, there’s another fundamental problem we didn’t spot when we began to set up pipe flow; non-uniqueness! As it turns out, there isn’t just one possible flow for a given condition on a pipe inlet, but a potentially infinite number of them—and from the experiments, it seems like those other flows possess all the things we had assumed away when we derived Hagen-Poiseuille flow; they’re unsteady, they have nonzero convective momentum flux, they have nonzero flow in all three pipe directions, etcetra! These flows, which undoubtedly have nonzero Reynolds tensors, are called turbulent flows, and are omnipresent in all of fluid mechanics.

Comparison sketches between Hagen-Poiseuille and turbulent pipe flows. Hagen-Poiseuille flow is orderly and symmetric; turbulent pipe flow possesses eddies, whorls, and few if any symmetries. When the parameters of the fluid/flow are changed in specific ways, the flow transitions from one to the other.

One way fluid mechanicists explain what is happening is that when the average speed is low through the pipe, nature “picks” out the Hagen-Poiseuille flow from all the other possible valid pipe flows because the litany of assumptions we made to get to Hagen-Poiseuille flow made sense; but when the average speed is sufficiently large enough, nature changes its mind and picks out one of the extremely complicated chaotic flows instead. In mathematics, this is commonly referred to as a bifurcation. Nobody has ever been able to come up with an analytical expression for any of these alternate chaotic flows, or precisely determine a theoretical criteria for this transition, and so we are left with experiments and numerical simulations to fill this knowledge gap for now.

But what experimentalists found was curious; they observed that once a specific dimensionless number crossed a threshold, the Hagen-Poiseuille flow would switch into turbulent flow and back. That dimensionless number is the Reynolds number, which we saw in the previous Part as an approximation to the ratio of convective to viscous effects. But that interpretation doesn’t make any sense here; the convective effects in Hagen-Poiseuille flow are exactly zero, so saying that the flow transitions from laminar to turbulent because the ratio of convective to viscous effects in pipe flows becomes too large is gobbledygook. Clearly, the Reynolds number here represents something else, and to discover what it represents, we’ll need to take stroll down a very curious field of mathematical physics called dimensional analysis.

Things To Think About

1. List out all the assumptions I made to justify the Hagen-Poiseuille flow. When do these assumptions make sense? When don’t they?

2. In a real pipe system, what happens when the pressure drops “too much”? Do the Navier-Stokes equations apply then, and if not, what would you need to change?

3. How would you try to understand turbulence? Do you agree or disagree with the explanation above?

4. What happens if you try to include gravity? Are you able to get a solution to the Navier-Stokes equations then?

5. What happens if the pipe has a constant cross-section that isn’t circular? Could you get a solution similar to Hagen-Poiseuille, and if you do, what would drive the differences?

6. Based on the results we got for pressure in the downwards and longitudinal direction, can you write a complete expression for the pressure in a pipe section? If you want to practice changing coordinate frames, use cylindrical coordinates.

Part VIII: Navier-Stokes, Existence & Uniqueness

At long last, we’ve finally managed to derive the famous Navier-Stokes equation; the cornerstone of most of fluid mechanics. Notice all the assumptions we had to make to get to it! To refresh your memory, this is what it looks like, along with descriptions of each term in it:

The Navier-Stokes equation, which describes momentum density transport in a simple, Newtonian, isotropic, incompressible fluid. The terms on the left-hand side are mathematical consequences of the concept of transport, while the terms on the right-hand side (save for the external force density) are defined by intermolecular fluid interactions.

There are many different, equivalent ways to write the Navier-Stokes equation, but the most common is this one, which incorporates a couple of simplifications thanks to conservation of mass:

\boxed{\displaystyle \rho\left(\frac{\partial \vec{v}}{\partial  t}\ + \nabla \vec{v} \cdot \vec{v} \right) = \vec{f} - \nabla p\ +\   \mu \nabla^2 \vec{v}}

(Note that some people write the dot product in the flux term backwards based on their notation for the gradient, but it’s all just preference.) Another equivalent form I briefly mentioned when discussing transport theory, which is particularly useful when discussing the curvature of flows, is commonly known as the Lamb form:

\displaystyle \rho\left(\frac{\partial \vec{v}}{\partial  t}\ + \frac{\nabla \left( \vec{v} \cdot \vec{v}\right)}{2} - \vec{v}\times\left(\nabla\times\vec{v}\right) \right) = \vec{f} - \nabla p\ -\   \mu \nabla \times \left(\nabla \times \vec{v}\right)

In whatever form it takes, this equation (along conservation of mass and varying simplifying assumptions) will be the chief mathematical tool we’ll use to determine the shape and speeds of flows. That being said, it turns out that the Navier-Stokes equation is notoriously difficult to deal with except in the simplest of situations, and it’s all thanks to the flux term:

\rho \nabla \vec{v} \cdot \vec{v}

This little mathematical pest has been the bane of fluid mechanicists for a solid hundred years, and it induces such a monumental headache on any mathematician attempting to solve the Navier-Stokes equation that there is actually a million-dollar bounty out for anyone who is even able to prove that the Navier-Stokes equation always has “sensible” solutions for a physical input—this is one of the most important mathematical problems ever conceived!

The reason this flux term is such a problem is because of a property called nonlinearity, which is best shown rather than told. To demonstrate, let’s consider a silly non-physical problem in 1-D, where we consider two imaginary fluids whose velocity is solely determined by the following equations:

Fluid 1Fluid 2
1-D Equation\rho v\frac{dv}{dx} = f \mu \frac{d^2v}{dx^2} = f
3-D Equivalent \rho \nabla \vec{v} \cdot \vec{v} = \vec{f} \mu \nabla^2 \vec{v} = \vec{f}

Applying some calculus tricks, what you’ll wind up finding is that the expressions for the velocities in each of the fluids is markedly different:

Fluid 1Fluid 2
1-D Solutionv(x) =\pm \frac{\sqrt{2} \sqrt{c_1 \rho +f x}}{\sqrt{\rho }} v(x) = c_2 x+c_1+\frac{f x^2}{2 \mu }

Note that for Fluid 2, if I set the constants to 0, the velocity is linearly proportional to the force. This means that if I double the force, the velocity everywhere is going to be doubled, and so on—a property unsurprisingly called linearity. For Fluid 1, however, this isn’t true; setting the constant to 0 and doubling the force instead leads to an increase in the velocity by a factor of \sqrt{2}! As innocuous as it seems, nearly every single tool ever made by mathematicians and scientists to solve partial differential equations like the Navier-Stokes equation relies on linearity, meaning any attempts to solve the Navier-Stokes equation relies on conceptual machinery on the outside of most of mathematical history. But that’s not even the worst part; as it turns out, nonlinearity is often comorbid with a host of horrendous properties, and the velocity of Fluid 1 is no exception.

For example, consider what would happen if I attempt to find the flow velocity for Fluid 1 if I take the velocity at the origin to be zero, v(0) = 0; because of that \pm sign in front of the velocity expression for Fluid 1, I actually have two possible fluid velocities! This is incredibly strange from a physical perspective; one could conjecture a situation where I have two exact copies of this system, where I’m applying the same force under the same conditions to each of them, and somehow get different flows. This goes straight against any rational understanding of physics, and is called non-uniqueness. Even worse, no matter what value of the constant c_1 I choose based on some physical information, or which expression for the velocity of Fluid 1 I pick, there is always some value of x past which my velocity suddenly turns into an imaginary number! This means that in some regions, we don’t even have an expression for the flow velocity, which is even more mind-boggling from a physical perspective and is referred to by mathematicians as existence failure (this is a really cool name). And this is only in one dimensions, folks—just imagine how much worse it gets in 3.

Two solutions for the velocity of Fluid 1, each solving v(x)\frac{dv(x)}{dx} = 1 with zero velocity at the origin, corresponding to a fluid with f, \rho = 1. Neither solution is defined in the region x < 0.

When you include time into the equations, things get even worse; you can start off with a flow that makes sense, and find it evolving into something seemingly nonsensical! Take, for example, what happens if you now take the ODE describing Fluid 1 and add a time rate-of-change term, turning it into a PDE:

\frac{\partial v}{\partial t} + \rho v\frac{\partial v}{\partial x} = f

This equation represents momentum transport in a one-dimensional fluid with no intermolecular interactions, and is commonly called Burgers’ equation. Take a look at what happens if you take a perfectly nice initial flow field for such a fluid and let it evolve in time for a constant unit f and \rho:

The solution to the PDE \frac{\partial v}{\partial t} +  v\frac{\partial v}{\partial x} = 1 for a smooth initial condition. The function describing the velocity eventually develops a discontinuity in space, causing the solution to become ill-defined. The numerical solver fails to handle this phenomenon, and glitches out once it develops.

After time passes, the velocity becomes discontinuous, and jumps immediately from one value to another; a phenomenon commonly referred to as a shock or shock wave. Mathematically, this is a problem not just because \frac{dv}{dx} = \infty at the discontinuity, but because the velocity has effectively two values there, causing it to be ill-defined! What happens here, from a philosophical perspective, is astounding; we’ve gone and defined an object mathematically using the rules of calculus, and the object has then evolved over time into something that violates the very mathematical principles we used to define it. This is exactly the problem (or at least, one of them) people run into when trying to solve the Navier-Stokes equation; the objects one uses to define flows for fluids can very rapidly outgrow the conceptual framework we used to define them.

Consequently, nearly every analytical attempt at using the Navier-Stokes equations to describe a flow necessitates ensuring that the momentum flux term in the equations is either negligible or handled indirectly. One way to verify this negligiblity is by taking the ratio of the magnitudes of each component of the flux term with the magnitudes of the components of the only other term that involves spatial gradients of the velocity—the viscosity term:

\mathbf{Re} = \rho \nabla \vec{v} \cdot \vec{v} \oslash \mu \nabla^2 \vec{v}

The \oslash symbol represents that we are dividing every term of a tensor by every other term of another tensor, effectively acting like the division counterpart of the outer product \otimes. Since each of these quantities is a 3-D vector, this ratio actually contains 9 distinct ratios, and I refer to it as the Reynolds tensor. The Reynolds tensor is useful in the sense that it is telling us whether or not, for a given flow, we should expect to observe the same kind of mind-boggling behavior we saw above in the Burgers’ equation, where it would be located, and it what directions we expect it to manifest. Because calculations involving the Reynolds tensor are pretty unwieldy, fluid mechanicists almost always single out a specific component of the Reynolds tensor, guess “characteristic” values for the density/viscosity/velocity/velocity gradients/lengths that they assume are valid everywhere in the flow, and come up with a single scalar ratio called the Reynolds number which they use instead. Sometimes this leads to spurious assumptions and misdrawn conclusions, but alas.

The components of the Reynolds tensor in Cartesian coordinates, along with a typical Reynolds number-type approximation. Calculating the Reynolds tensor is often unwieldy, so most scientists simply estimate a single number based on quantities in the problem they expect to see and call it the Reynolds number, shown below.

Nearly all of the time, we use the Reynolds tensor/number to do sanity checks on any simplifying assumptions we make to the Navier-Stokes equation—either by using a conjectured Reynolds tensor/number to justify removing some term from the Navier-Stokes equations, or by calculating the Reynolds tensor/number for a solution to a simplified Navier-Stokes equation to validate whether or not the simplification was sensible in the first place. And one setting where we’ll be able to get rid of that flux term and, as a result, easily construct specific conclusions about flow shapes and speeds is a setting we’ve looked at before—pipe flow.

Things to Think About

1. Some people think adding a term representing viscosity to equations like the ones described above help get rid of the shocks & other strange behaviors we saw just now. Try solving the ODE \rho v(x)\frac{dv}{dx} - \frac{d^2 v}{dx^2} = f for a bunch of different \rho/f/\mu using some math computer programs to see whether or not that line of thinking is correct.

2. How would neglecting the flux term in a problem change the physics of the fluid you would observe, and would it make sense? Is there a fluid that can possess a zero flux term?

2. In what situations do you think simply using the Reynolds number instead of the Reynolds tensor would be sufficient? In what situations would it fail?

3. Try calculating the Reynolds tensor for simple flows; maybe a constant flow, or a flow that changes linearly in one direction in space. What components are zero, and which aren’t? What do they correspond to?

4. Shock waves appear to make no sense in a differential equations sense, but seem to make perfect sense physically; is there some mathematical tool we were using before that would be able to make mathematical sense of shock waves?

5. In Part II, we talked about how pressure ensures liquid doesn’t just accumulate into a hyperdense thin film at the bottom of a cup. How is that connected to the shock wave phenomenon we saw above?

6. If you want to get a glimpse of how linearity is useful for solving differential equations, solve \frac{dv}{dt} = f for f=1 and 2 for arbitrary constants. Can the sum of both of these solutions represent a solution for \frac{dv}{dt} = 3?

Part VII: Rheology

As we saw during our discussion on transport theory, the principal equation of fluid dynamics really boils down to a transport equation for the momentum in a fluid, or for the transport of momentum density through fluid points:

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

where the \vec{r}_{mol} term represents forces stemming from molecular effects. It is this molecular term which exclusively “identifies” a fluid, as everything else in the governing equation above was generated through a very general mathematical framework describing the movement of “things” in space rather than through specific physical insight associated with fluids. Consequently, understanding this connection between continuum-scale momentum and molecular effects is a humongously important task for physicists and engineers alike, and is the principal endeavor of rheology.

It’s often extremely difficult to generate refined models for \vec{r}_{mol} from first principles, so often the process of determining an \vec{r}_{mol} for a given fluid consists of making some basic assumptions about the fluid and then testing the fluid experimentally to see if the behavior of the fluid is consistent with those assumptions. Rheology, as a result, is largely an experimental science.

This doesn’t mean we’re simply left to fiddle around with rheometers from now on, however—in fact, we’ll rapidly find that a couple of simple assumptions and some not-so-simple tensor calculus will lead us to an equation that applies to nearly all fluids.

To that end, the first thing that we can assume is that molecular forces from a fluid point affect only the fluid points immediately around that point. This assumption is referred to by physicists as the principle of locality, but is largely taken for granted by fluid mechanicists; this is because it’s almost always true, since the range of molecular interactions tends to be extremely small, and the strength of those interactions is only notable when molecules are really close together. In other words, this assumption implies that molecules don’t warp momentum around in a fluid from one location to another; they transport it. Mathematically, this means that molecular forces always manifest as momentum density fluxes, and as a result we can rewrite the equation above as:

\displaystyle \frac{\partial\left(\rho\vec{v}\right)}{\partial  t}\ + \nabla \cdot \left(\rho \vec{v} \otimes \vec{v}\right) = \vec{f} +  \nabla \cdot \boldsymbol\sigma

That molecular momentum density flux \boldsymbol\sigma, which is a tensor just like the other molecular momentum density flux term \rho \vec{v} \otimes \vec{v}, is commonly called stress.

A diagram of a series of fluid points, each represented as a cube, with yellow arrows representing the possible interactions between them. The locality assumption indicates that interactions between non-neighboring points of fluid are either negligible or non-existent, meaning forces between two non-neighboring points of fluid have to be transported through points in between them. As a result, intermolecular forces can be described as the divergence of a tensor called the stress.

The stress tensor \boldsymbol\sigma is just a second-order tensor that represents, for a given fluid point, the traction/force per unit area \vec{t} acting on that element thanks to its neighboring fluid point in the \hat{n} direction. If you’ve studied solid mechanics, you’ve likely heard the term stress thrown around before; it means the same thing there as it does here. In fact, the only difference between a solid and a fluid is that the stress is a function of different continuum parameters in each type of substance. Notice we never did anything fluid-specific when we derived the momentum transport equation above!

Luckily, we already know one piece of the puzzle of \boldsymbol\sigma; it has to contain the forces coming from pressure gradients, since we spent two Parts discussing how pressure (and the forces it induces) is an entirely molecular effect. We can make our lives easier and pull that pressure gradient term out, making sure we make it negative so that everything is consistent with the hydrostatic equation we derived in Part II:

\displaystyle \frac{\partial\left(\rho\vec{v}\right)}{\partial   t}\ + \nabla \cdot \left(\rho \vec{v} \otimes \vec{v}\right) = \vec{f} - \nabla p +  \nabla \cdot \boldsymbol\tau

That new tensor \boldsymbol\tau is usually called the deviatoric stress tensor, which I think is a silly name.

The contents of the deviatoric stress tensor are in some sense what we’re really looking for, since the notion of pressure and pressure gradients is universal in solids and liquids. And here, we can classify fluids rheologically into two important & distinct branches based on a key characteristic of their behavior when standing still; fluids whose static behavior is fully characterized by pressure gradients (commonly called simple fluids), and fluids for which it is not (commonly called non-Newtonian fluids). For the latter, the molecules in the fluid will usually have some kind of extra molecular force that “causes” them to try to remain in some specific macroscopic shape or configuration—just like a solid. Some examples of this are ketchup (it doesn’t like to flow out unless you smack the bottle or pump it out), mucus, pancake mix, melted rubber, etcetra; there’s too many to count. A simple experimental way to figure out whether or not a fluid is simple or non-Newtonian is to pour it into onto a surface and then tilt the surface very, very slightly; if the fluid flows at even a slight angle, then it’s safe to say the fluid behaves like a simple fluid. (Professionals use rheometers). Modeling non-Newtonian fluids is very tricky, and usually done on a case-by-case basis; we won’t concern ourselves here with understanding how they’re modeled, other than that their additional static molecular forces usually depend on the same thing that they depend on for solids, which is some displacement off of an “equilibrium” configuration like in a spring.

The “tilt test”; a puddle of simple fluid in blue and a puddle of non-Newtonian fluid in yellow are placed on a surface. When the surface is tilted slightly, the simple fluid loses its form and flows while the non-Newtonian fluid deforms but doesn’t flow/preserves form.

Looking back at simple fluids, modeling them is far easier than modeling non-Newtonian fluids since the assumption on the behavior of simple fluids induces quite a big constraint on that deviatoric stress tensor \boldsymbol\tau; if we take the assumption that every molecular effect that induces forces in an unmoving fluid is by definition captured through pressure gradients, then we find that \boldsymbol\tau can’t possibly contain any terms involving the physics of the fluid when it’s standing still. It is entirely a function of the properties of the fluid’s motion, which is why I like to call \boldsymbol\tau the hydrodynamic stress tensor when modeling simple fluids.

With that in hand, let’s continue with some extra assumptions. The next assumption we can make is that \boldsymbol\tau, which we’ve said depends on the motion of the fluid, is independent of the absolute velocity of the fluid. This is pretty obvious when you think about it; if the stress of a fluid directly depended on the velocity, we would be able to observe a fluid deform differently depending on whether or not we were walking forwards or backwards relative to the fluid, since velocity is always relative to the observer. Instead, we can assume that \boldsymbol\tau is only a function of velocity gradients, which is almost always true unless you’re moving extremely fast. Mathematically, this means:

\displaystyle \frac{\partial\left(\rho\vec{v}\right)}{\partial t}\ + \nabla \cdot \left(\rho \vec{v} \otimes \vec{v}\right) = \vec{f} - \nabla p +  \nabla \cdot \left(\boldsymbol\tau\!\left[\nabla \vec{v}\right]\right)

Here the brackets just indicate that \boldsymbol\tau is a function of \nabla \vec{v}. But remember; when talking about velocity gradients, we have to consider the gradients in each direction of the velocity components in each direction. So instead of representing three pieces of independent information like \vec{v} does, velocity gradients represents nine, meaning that it’s—you guessed it—a tensor.

As a matter of fact, the assumption above puts a further restriction on \boldsymbol\tau; if I decided to somehow spin myself around a static fluid I’m observing in the lab, I would observe a nonzero velocity gradient even though the fluid is standing perfectly still. As a result, a simple fluid’s deviatoric stress tensor can’t depend on every part of the velocity gradient tensor, just some parts. Some behind-the-scenes tensor manipulations show that in a simple fluid, the deviatoric stress tensor can only depend on the symmetric part of the velocity gradient tensor. This property is commonly referred to as objectivity, and the symmetric part of the velocity gradient tensor is usually called the strain-rate tensor \mathbf{E}, calculated from the velocity gradients by:

\displaystyle \mathbf{E} = \frac{\nabla\vec{v} + \nabla\vec{v}^\intercal}{2}

This still doesn’t really get us anywhere; the nine components of \boldsymbol\tau are so far completely arbitrary functions of the nine components of \mathbf{E}. But we can section simple fluids out into two other categories; fluids for which \boldsymbol\tau is strictly proportional to the strain-rate tensor, which are commonly called Newtonian fluids, and fluids for which it is not, which are called generalized Newtonian fluids. (For what it’s worth, I think the naming conventions for Newtonian/non-Newtonian/generalized Newtonian fluids is confusing and outright ridiculous, but there’s not much I can do about it.) Examples of generalized Newtonian fluids include blood, nail polish, paint, syrup, etcetra. For generalized Newtonian fluids, the behavior of the fluid is properly “liquid”, but the magnitude of the molecular forces induced by the flow of the fluid are complicated functions of the components of the gradients in the velocity. Models for these functions are more of an art than a science, based largely on experimental results; here are some examples.

For a Newtonian liquid, however, we now find that there is a linear relationship between the second-order strain-rate tensor \mathbf{E}, with nine independent components, and the second-order deviatoric stress tensor \boldsymbol\tau, also with nine independent components. This means that we can represent the linear relationship by a fourth-order tensor with 81 different components (yikes!) which I’ll call the viscosity tensor \mathbf{M}:

\displaystyle \boldsymbol\tau = \mathbf{M}:\mathbf{E}

For better or worse, there’s an extra assumption that nearly always applies to fluids of this type; the molecular interactions are independent of direction, and as a result the stress tensor is independent of the orientation of the velocity gradients. Fluids for which this holds true are called isotropic, and fluids for which it’s not are called anisotropic. I’ve never heard of an anisotropic simple Newtonian fluid, but maybe you’ll be the first to find one!

For a locally-interacting simple Newtonian isotropic fluid, doing some behind-the-scenes tensor calculus reveals that our updated conservation of momentum equation has to take the following form once we all these assumptions:

\displaystyle \frac{\partial\left(\rho\vec{v}\right)}{\partial  t}\ + \nabla \cdot \left(\rho \vec{v} \otimes \vec{v}\right) = \vec{f} -  \nabla p\ +\  \mu \nabla^2 \vec{v}\ +\ \left(\lambda + \mu\right)\nabla\left(\nabla\cdot\vec{v}\right)

This absolute mess of an equation is called the compressible Navier-Stokes equation, and is the equation nearly everyone uses when dealing with flow of compressible fluids. The two symbols \mu and \lambda are undetermined proportionality factors that are the only surviving components of that viscosity tensor \mathbf{M}, and are usually determined through experiment; the first is universally referred to as the dynamic viscosity or just viscosity, and the second is so inconsistently defined in the literature that the only name we can give this thing that rheologists wouldn’t argue over is the profoundly unhelpful moniker of second Lamé parameter.

If we further assume that the fluid is incompressible and that the viscosity is the same everywhere, we can polish the equation up a lot more and pull the density out of a lot of expressions, get rid of that last term that depends on \lambda, and wind up with:

\boxed{\displaystyle  \frac{\partial\left(\rho\vec{v}\right)}{\partial  t}\ + \nabla \cdot \left(\rho \vec{v} \otimes \vec{v}\right)  = \vec{f} -   \nabla p\ +\  \mu \nabla^2 \vec{v}}

This is the incompressible Navier-Stokes equation, also known as just the Navier-Stokes equation, and it is the most famous equation in fluid dynamics; understanding it and analyzing its components will be the sole subject of the next Part. Given how involved this sequence of assumptions is for an introduction to fluid mechanics, I’ve put a table below describing how each assumption we made brought us from the Cauchy momentum equation at the beginning of this Part to the Navier-Stokes equation above.

AssumptionMathematical ConsequenceWhen It’s Wrong
Locality\vec{r}_{mol} = \nabla \cdot \boldsymbol\sigmaLong-range molecular interactions
Simplicity \vec{r}_{mol} = -\nabla p + \nabla \cdot \boldsymbol\tau Material has solid-like properties, pressure defined differently
Objectivity\boldsymbol\tau =  \boldsymbol\tau[\mathbf{E}]Velocities/rotations are extremely fast
Linearity\boldsymbol\tau = \mathbf{M}:\mathbf{E}Molecular forces have a complicated relationship with velocity gradients
Isotropy\boldsymbol\tau = 2 \mu \mathbf{E} + \lambda \text{Tr}(\mathbf{E})\mathbf{I}I don’t know (molecules would likely have to be very asymmetric)
Incompressibility\boldsymbol\tau = \mu \nabla \vec{v}Density is very low, fluid speeds are comparable to molecular speed (see Part II)

Things to Think About

1. What physical effects do you think the viscosity and second Lamé parameter quantify? Can you describe flows where only one of the terms is nonzero?

2. Can you think of other ways to test if a fluid is simple or not? How about to test if a fluid is Newtonian or not?

3. If you’re brave enough to handle some tensors, can you find why the form of the deviatoric stress tensor goes from 2 \mu \mathbf{E} + \lambda Tr(\mathbf{E})\mathbf{I} to \mu \nabla \vec{v} when we take the incompressibility assumption? Remember the definition of the strain-rate tensor!

4. How would the momentum transport equation change if you considered non-local molecular forces? Can you come up with some kind of parameter to describe if the locality assumption is valid? (This is something I think about often.)

5. What would the Navier-Stokes equation look like if the fluid was a locally-interacting simple Newtonian isotropic fluid but with a non-uniform viscosity?

6. Let’s say that you could describe the forces per unit area in the definition of the stress as coming from an energy density gradient. How would you define the stress in that case, and what would be the associated directions?

Intermezzo: Tensors

Before we continue on with our study of fluid mechanics, it’s best to make a quick pit stop and make sense of that strange concept I mentioned previously; that of a tensor, which we saw in the form of that strange multiplication \rho \vec{v} \otimes \vec{v}. The “proper” way of defining a tensor is hotly debated by mathematicians and physicists of different fields, so my idea here is not necessarily to rigorously define them, but to give you a brief sense of what a tensor is and how we use them in fluid mechanics. In my experience, that is best illustrated by revisiting the concept of a vector, like the velocity \vec{v}.

Knowing the velocity of an object tells us two things; the speed of the object, which is just some number like 20 miles per hour, and the direction in which it is traveling. But as anyone that’s said “No, I meant my left” can attest to, direction is relative—whenever we talk about velocity in its most general sense, we usually describe it in terms of three independent components representing the speeds in specific, perpendicular directions that we’ve arbitrarily defined in the 3-dimensional space we live in (usually denoted with x, y, z). Changing those arbitrarily defined reference directions must then necessarily change the expressions for the components, commonly referred to as projections, even though it doesn’t physically change the velocity of the object itself.

A velocity vector, in black, along with its three components (or projections) onto a set of three distinct directions represented by coordinate axes. Each component represents the speed of the object in the direction the component represents. If I spun the axes around, the components would change, but the vector itself wouldn’t.

That being said, velocity doesn’t intrinsically possess three directions—as we stated above, it possesses just one, the direction in which the object is instantly traveling in. With that in mind, a tensor is just a mathematical object that possesses any number of directions. People usually refer to the amount of directions a tensor possesses as its order, and so we can straightforwardly spot that the velocity is a first-order tensor because of its single “natural” direction.

If the velocity is a first-order tensor, then it might seem obvious to state that the mathematical object \rho \vec{v} \otimes \vec{v}, which we said represented the momentum density flux, possesses two directions, and is as such a second-order tensor. In this case, those directions are straightforward to spot; the direction of the momentum density \rho \vec{v}, and the direction of the flux, which is just the velocity \vec{v}. As a matter of principle, you can always spot the directions of any tensor defined through a sequence of outer products of vectors by looking at the direction of each vector. In fact, both of those directions are the same in this specific case, since \rho \vec{v} \otimes \vec{v} and \vec{v} point in the same direction!

So now that we know what that tensor means, we can think a bit more concretely about how to represent it mathematically. Well, just like the velocity can be represented using three components with reference to some arbitrary coordinate system, the momentum density flux is the product of every component of the 3-D vector \rho \vec{v} with every component of the 3-D vector \vec{v}, so that the tensor \rho \vec{v} \otimes \vec{v} is represented by 3\times3 = 9 independent components. To make sure we distinguish which direction is which, mathematicians usually visually represent second-order tensors using a visual “square” of numbers called a matrix.

These are the nine different components of the momentum density flux tensor, associating every component of the momentum density with every component of the flux. They’ve been placed in a geometric array called a matrix that highlights the different directions associated with each part of the momentum density flux tensor.

This isn’t the end of the story on tensors, though; remember that the Cauchy momentum equation we saw before includes the term \nabla \cdot \left(\rho \vec{v} \otimes \vec{v}\right), not just the momentum density flux. Hence, we need to be able to understand the effects of trying to take derivatives, like the divergence and gradient, on tensors. To do that, let’s repeat our formula on testing it out on the velocity—which we know is a first-order tensor—and see how to extrapolate from there.

The divergence of the velocity at a point, from a purely mathematical standpoint, tells you what the rate of change of the velocity is at that point as you move in each of those predefined spatial coordinates, and then adds all of those rates of changes up into a single direction-less number that tells you if the flow is “accumulating” or “diminishing” at that point. Make sure to notice that the divergence of the velocity is truly directionless; if you looked at a flow upside down, the direction of the flow would certainly change, but the amount by which the flow is “accumulating” at a point doesn’t. This is easily extrapolated to a general, tensorial case; the divergence of a tensor takes the spatial rate of change of the tensor in the direction of one of the tensor’s directional elements, usually defined to be the “last” element, and adds them up. This causes a tensor to “lose” a direction when acted upon by the divergence—meaning that the second-order tensor \rho \vec{v} \otimes \vec{v}, when acted on by the divergence, gives you a first-order tensor/vector, just as the momentum transport equation required.

The divergence of the momentum density flux tensor (order 2). Here, we take the spatial rate of change of each of the 9 elements in the flux direction, and add them up to obtain 3 elements representing a vector (order 1). Note the directions of the partial derivatives match the directions of the flux velocities.

Lastly, let’s take a look at the gradient of the velocity \nabla \vec{v}. Here we are still obtaining the spatial rates of change of the velocity, but this time we’re not adding anything up; in fact, we’re obtaining the spatial rate of change of every component of the velocity in each and every one of those arbitrarily defined directions of space. 3 different possible rates of change for each of the three components indicates that the velocity gradient possesses 9 distinct pieces of information, revealing that the gradient of the velocity is a second-order tensor. As such, we can extrapolate to the general case and state that the gradient of a tensor is another tensor of a single higher order.

The velocity gradient tensor. The two “directions” associated with this tensor are connected to the velocity and to the direction of the rate of change.

Explicitly defining the direction that gets added as a result of taking the gradient is trickier to spot than in the outer product case, where the directions of the resulting tensor are just the directions of each component of the tensor; but the direction exists, and the vector representing it can be found by solving the following equation for the vector \vec{m}:

\nabla \vec{v} \cdot \vec{m} = \vec{v}

(The dot product here is just representing standard matrix-column vector multiplication.) This vector doesn’t really have a name, nor a more intuitive description other than the above as far as I can tell; but I call it the gradial vector for book-keeping, and would love to find a nice application for it.

These explanations are by no means rigorous or comprehensive; there are many incredible and beautiful aspects to the study of tensors that I sadly have to exclude for the sake of coherence, but I invite you to read up on them if you’re interested in the subject. My current favorite book on the topic is Henry Block’s Introduction to Tensor Analysis, which is tragically out-of-print but has been preserved electronically by the thoughtfulness of faculty members in the Theoretical & Applied Mechanics field at Cornell. In any case, this is just about everything we need to know about tensors to get a handle on all the mathematical machinery abound in introductory fluid mechanics; and although the principal language of fluid mechanics is vector calculus, it’s good to get a handle on tensors to deal with an often-ignored part of the world of fluid mechanics—the process by which molecules exert forces on each other when a fluid moves, known as the study of rheology.

Things to Think About

1. What is the difference between a second-order tensor and a matrix? Are all matrices second-order tensors? Are all second-order tensors matrices?

2. If a fluid is incompressible, how does the expression for \nabla \cdot \left(\rho \vec{v} \otimes \vec{v}\right) simplify? Remember the incompressibility equation, \nabla \cdot \vec{v} = 0.

3. How would you describe the gradient of a second-order tensor? Extrapolate from the principles shown here for the velocity gradient.

4. Consider the gradient of any simple scalar (order 0 tensor) function. If you changed around the coordinate axes, how would the gradient change? Would it change in a different way than a velocity vector would?

5. What happens if you take the divergence of the gradient of the velocity? How is that different from taking the gradient of the divergence of the velocity? The first expression is called the vector Laplacian, and will feature prominently in the following sections.

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_S \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 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 \vec{F} =  \int_V \vec{b}\ dV + \int_S \vec{t}\ dA -  \int_S  \left(\vec{j}^{\left[\rho \vec{v}\right]}_{\text{diff}} \cdot  \hat{n}\right) dA - \int_S \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 (\sum\vec{F} = 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 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_S p \hat{n} \cdot \hat{x}\ dA - \int_S \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

A control volume analysis over a hydraulic pipe section; each distinct contribution is in a different color. Gravitational body forces are in brown, pressure-induced tractions are in yellow, advective momentum fluxes are in black. The momentum loss, in green, has an unknown distribution within the fluid section but always has a net direction opposite to the flow.

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 second fundamental equation of hydraulics. Although it appears there are many unknowns, the only variables that remain unknown after applying the first fundamental equation are the pressures, one of which is usually already known, and the 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.

A hydraulic circuit or pipe network. Usually, one knows the reservoir/pump pressure and outlet pressure. Given a guess of the reservoir velocity, the fundamental equations of hydraulics are applied over every segment, starting from the reservoir, until the calculated pressure at the outlet matches the known pressure. Momentum losses for each segment have usually been previously determined by experiment for a given pressure/velocity change.

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.

Things to Think About

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_S \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.

Pressure-induced tractions act on a volume of fluid in a pipe, which is described by a control volume. If the pressures are not equal, the fluid within the control volume experiences a net force, causing it to 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.

A control volume is generated in a section of pipe containing fluid with Chemical X. The total amount of Chemical X in the pipe changes only as a result of two things; the creation/destruction of Chemical X within the pipe (sources/sinks), and the movement of Chemical X into/out of the pipe (fluxes).

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_S \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_S \rho_X \left(\vec{v}_X \cdot \hat{n}\right)\ dA + \int_S 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_S \left(\vec{j}  \cdot \hat{n}\right)\ dA

\displaystyle 0 = - \int_S \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.

In order for fluid to not accumulate within a nozzle, the integrals of speed over the area of the inlet/outlet need to match, which causes fluid at the nozzle outlet to be generally faster than fluid at the nozzle inlet.

Let’s conclude by observing how mathematically similar the equation for the net force on a control volume and the equation for the rate of change of Chemical X’s mass on a control volume are. This isn’t coincidental; it turns out that you can derive that equation using the exact same procedure we did above in deriving the equation for Chemical X’s mass rate of change! Consider 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 is the sum of contributions of momentum density \rho \vec{v}, whose rate of change is the force acting on the control volume as stated by Newton’s second law: \displaystyle \frac{\partial}{\partial t}\int_V \left(\rho \vec{v}\right)\ dV = \frac{\partial\vec{P}}{\partial t} = \vec{F}
  • Momentum is generated or destroyed inside of the control volume through body forces acting on points of fluid: \displaystyle \int_V  \frac{\partial \left(\rho \vec{v}\right)}{\partial 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_S \rho \vec{v} \left(\vec{v} \cdot \hat{n}\right)\ dA  - \int_S (\vec{j}^{\left[\rho \vec{v}\right]}_{\text{diff}} \cdot \hat{n}) dA +  \int_S \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; we’ll get back to it soon. All you need to know now is what the term represents.

Putting everything together, we now find a more general version of the force equation on a control volume that we saw above, which also considers the effect of fluid flow:

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

As it turns out, that last term is indicating to us that fluid flow can induce forces! 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 net force on the object has to be \displaystyle \vec{F} = -\int_S \rho \vec{v} |\vec{v}|\ dV, 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).

A jet/rocket shoots out fluid behind it. Because the fluid is carrying momentum away from the control volume, the fluid in the control volume experiences a net force in the opposite direction, pushing the jet/rocket forward.

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.

Things to Think About

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 floating? 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.

A solid sphere submerged in a cup of water experiences a traction at every point on its surface, which is pointing perpendicular to the surface at that point, and with magnitude equal to the fluid pressure at that point.

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_{S} \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.

Frontal view of a submerged cube with side length a and the pressure-induced tractions it feels. From symmetry, the tractions on the sides are equal and opposite, and the net buoyancy force comes from the difference in tractions between the top and bottom of the cube.

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_{S} \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.

A cylinder wrapped in string is pulled on opposite ends by equal and opposite forces. Although the cylinder’s center of mass doesn’t move due to zero net force, the cylinder spins due to a torque equal to the sum of the moments induced by the forces.

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_{S} \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.

Things to Think About

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?