 # 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 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):

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.

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 $C\ (\frac{kg\,m^2}{s^2 K})$ 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 $h\ (\frac{kg}{s^3 K})$, 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.