What does the Reynolds Number of a flow represent physically?

I am having trouble understanding the meaning and the utility of the Reynolds number for a certain flow, could someone please tell me how this type of dimensionless factor is significant and what it tells us about a problem?


From the Wikipedia article for Reynolds number:

In fluid mechanics, the Reynolds number (Re) is a dimensionless number that gives a measure of the ratio of inertial forces to viscous forces and consequently quantifies the relative importance of these two types of forces for given flow conditions.

In addition to measuring the ratio of inertial to viscous forces in a flow, the incompressible Navier-Stokes equations can be written in non-dimensional form such that the only parameter is the Reynolds number (ignoring body forces). This is very nice because it is the basis for the validity of wind tunnel testing.

Suppose we would like to measure the aerodynamics of the flow around a Boeing 747. Two (at least) options exist:

  1. Build your very own full size 747, instrument it, and fly it. (extremely expensive)
  2. Build a small scale model of a 747, instrument it, test inside a wind tunnel (much less expensive)

But how do we know that the flow we measure in the wind tunnel is what really happens in flight? We match the Reynolds numbers and the exact same equations model both situations--therefore the aerodynamics must be the same. (Ignoring compressibility effects.)

  • $\begingroup$ And what dimensionless numbers rule compressibility effects if I'd like to take them into account? Mach number? $\endgroup$ – firtree May 11 '13 at 7:08
  • $\begingroup$ @firtree The Mach number shows up in the Energy equation with the $Dp/Dt$ and dissipation terms. For low Mach numbers, these terms are negligible. For high Mach numbers these terms increase the spatial temperature gradients and couple the momentum and energy equations. This coupling makes it hard (perhaps impossible) to match Reynolds number and Mach number for compressible experiments. At this point it is up to the experience of the experimentalists to decide the parameters of the tests so that reality is best approximated for scale models. $\endgroup$ – OSE May 12 '13 at 20:39
  • $\begingroup$ So for high Mach numbers scaled experiments are either impossible or (extremely?) inaccurate? Or I got you wrong? $\endgroup$ – firtree May 13 '13 at 5:39
  • $\begingroup$ @firtree Someone please correct me if I'm wrong: I believe for scaled experiments at high Mach numbers, it is most important to match Mach number so the shock structure is the same. If the Mach number is matched the scaled model's Reynolds number will be less than that of the full size model. To approximate higher Reynolds numbers, trip strips are placed where transition from laminar to turbulent flow is expected. If the transition location is unknown, an array of various trip strip locations could be tested. $\endgroup$ – OSE May 13 '13 at 14:34
  • $\begingroup$ Also incompressible measurements can be extended to compressible flow using the Prandtl-Glauert Transformation. $\endgroup$ – OSE May 13 '13 at 14:34

Reynold's number is defined to be: $$ \text{Re} = \frac{ v D }{ \nu } $$ where $v$ is the characteristic velocity for the flow, $D$ is a characteristic size and $\nu$ is the kinematic viscosity.

Now, why should we care? Why is Reynold's number important? Well, the first thing to realize is that the Reynolds number is a dimensionless number. This means that is has a certain power that dimensionful numbers don't. It is a pure number and doesn't depend on any way on your particular choice of units. This means it has some kind of intrinsic or universal meaning outside of any human constructs.

In particular, Reynold's number can be thought to measure relative velocity of the flow. We would expect that the physics of fluids should be different for slow and fast flows, but this question in and of itself is not well defined. Slow or fast compared to what? This is what Reynold's number tells us. It tells us whether flow is slow or fast by forming a natural nondimensional measure of flow speed. Since its a pure number, we expect qualitatively different behaviors if $\text{Re} \ll 1$ and $\text{Re} \gg 1$.

And this is precisely what we observe. The low Reynold's number limit corresponds to things like marbles falling in corn syrup, or cloud droplets in air, or bacteria in water. This is slow viscous flows where drag forces are proportional to velocity.

On the other hand, in the high flow limit, we have turbulent flow, where eddies are created behind our object or around edges in pipes, this is the general limit most things in air on human scales corresponds to, so you are familiar with turbulent flow intuitively. In this limit, drag is proportional to $v^2$. Large things like people will be in this turbulent regime in air even at speeds as small as 0.1 m/s or so. This is the limit in which viscosity becomes unimportant, and for the most part we can imagine flow in a fluid as corresponding to just sweeping up the fluid in front of our bodies of interest.

For instance, look at the drag force felt by a sphere as a function of Reynold's number (from wikipedia )

Drag Coefficient as a function of Reynold's number

In the low reynold's number limit, drag coefficient scales as $\text{Re}^{-1}$ whereas in the high limit it is roughly constant.

Momentum Flow

Consider it another way. Kinematic viscosity is the diffusion constant for momentum in a fluid. It is how fast the momentum spreads out due to collisions between the different molecules in a fluid. Let's take a look at a couple relevant times for fluid flow.

First, we note that $\nu/D$ has the dimensions of a velocity, so $D^2/\nu$ has the dimensions of time. (Here $D$ is a characteristic size of the object and $\nu$ is the kinematic viscosity). What does this time represent? Since kinematic viscosity is a diffusion constant for momentum, the ratio $D^2/\nu$ tells us the time scale for the momentum to move a characteristic distance $D$. Since $D$ is the size of our object, this should correspond, roughly, to the time it takes for presence of the object to be transferred through the fluid from one end of the object to the other. It is the time it takes for the fluid to "flow" around the object. (More accurately, it is the time it takes for the momentum disturbances in the fluid to flow around the object).

But there is another characteristic time: $D/v$. This second time corresponds to the time it takes an object to move a distance equal to its size. $v$ is the velocity it is travelling (relative to the fluid) and $D$ is its size, so it will move a distance $D$ in time $D/v$.

Reynolds number is the ratio of these two times $$ \text{Re} = \frac{ D^2 / \nu}{ D/v} = \frac{ v D }{ \nu} $$ So it measures the ratio of the time it would take the fluid to flow around an object, over the time it takes the object to move a distance equal to its size. Clearly here if that ratio is large, we expect the fluid doesn't move out of the way at all and is just swept up, while if it is low, we expect appreciable flow around the material.

Small Aside

In fact, using this idea you can "derive" the normal air drag equation for force. We can just assume in the simplest case that a ball travelling through air just bumps into all of the air molecules in front of it. Each of these molecules impart a momentum change of $mv$ to the object (where $m$ is the mass of an air molecule). How many molecules do we hit? Well, if we move for a time $\Delta t$, if our object has a cross sectional area of $A$, it sweeps out a volume of $A v \Delta t$, so the mass of air in that volume is $\rho A v \Delta t$, and so the number of air molecules is $\rho A v \Delta t / m$. Do the total change in our momentum is $$ \Delta p = ( \rho A v dt / m ) ( m v ) = 2 \rho A v^2 \Delta t $$ and we know force is the rate of change of momentum $$ F = \frac{ \Delta p }{ \Delta t } = \rho A v^2 $$ which is correct save a factor of 2 and a drag coefficient which on dimensional grounds should depend only on the characteristic of our body (shape, surface) and the Reynold's number.

Navier Stokes

We can also see the importance of Reynold's number directly in the Navier Stokes equation. If you start with the Navier Stokes equation for incompressible flow: $$ \frac{\partial \vec v}{\partial t} + ( \vec v \cdot \nabla )\vec v = -\frac{1}{\rho} \nabla p + \nu \nabla^2 \vec v, \qquad \nabla \cdot \vec v = 0 $$ and nondimensionalize them by choosing a characteristic size $D$ and velocity $V$, you obtain: $$ \frac{\partial \vec v}{\partial t} + ( \vec v \cdot \nabla) \vec v = - \nabla p + \frac{1}{\text{Re}} \nabla^2 \vec v, \qquad \nabla \cdot \vec v = 0$$ Where here it becomes clear the the Reynold's number is just the importance of the $\nabla^2 v$ term in the equation. That is whether you need to consider the laplacian of the velocity field. That is, how much the fluid tries to make its velocities in nearby regions consistent. If Reynold's number is high, this term drops out, so we can have very stark local changes in the velocity field, i.e. turbulent flow.

  • $\begingroup$ This is a good answer. Very detailed and shows conceptually what the meaning of the Reynolds number is. Would you be able to explain in more detail your first paragraph in "Momentum Flow" Section? Namely, what is R (I presume D is diameter and we are talking about the diameter of the object) and how did you arrive at the times that take a fluid parcel to move around the object and the time it takes for the object to move through the flow an equal distance to its size. $\endgroup$ – Isopycnal Oscillation Aug 27 '14 at 19:52
  • $\begingroup$ @IsopycnalOscillation I had goofed, $R$ should be $D$. I've also tried to reword the section to make it more clear. Is that better? Anything more I can do to clarify? $\endgroup$ – alemi Aug 27 '14 at 20:51
  • $\begingroup$ I think one thing that was confusing me is that I was assuming v to be the characteristic velocity of the fluid, but I see that in your definition v is the velocity of the object relative to the fluid. $\endgroup$ – Isopycnal Oscillation Aug 27 '14 at 21:00
  • $\begingroup$ @IsopycnalOscillation ah. I will point out though, that if the object is standing still, $v$ is the velocity of the fluid in that case. I suppose I kept switching frames of reference when I was describing things, but you can either consider the object at rest and the fluid moving, or the fluid at rest and the object moving, or anywhere in between. $\endgroup$ – alemi Aug 27 '14 at 21:03
  • $\begingroup$ thats true, so then wouldn't $D/v$ also be the time it takes the fluid to flow around the object, just like $D^2/\nu$? $\endgroup$ – Isopycnal Oscillation Aug 27 '14 at 21:15

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