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 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 $\langle v \rangle = \frac{G_1 R^2}{8 \mu }$.

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:

What the hell is going on? Well, as the existence failure for the pressure had been warning us, it appears as if there’s another fundamental problem we didn’t spot when we began to set up pipe flow; non-uniqueness! As it turns out, there might not be 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.

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, precisely determine a theoretical criteria for this transition, or even determine if this non-uniqueness hypothesis is true for certain—and so we are left with experiments and numerical simulations to fill this knowledge gap for now.

But what experimentalists could conclude 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.