Getting the Reynolds Number Right

One of the most common mistakes I see in my Fluid Mechanics lab course is incorrect computation of the Reynolds number. This is a brief treatment of this subject.

First, the Reynolds number is defined as

$R_e = \frac{VD\rho}{\mu}$

$R_e = \frac{VD}{\nu}$

where

$R_e =$ Reynolds number (yes, I know the newer textbooks use a different notation, but I’m a dinosaur)
$V =$ Average or reference velocity
$D =$ “Diameter” or other governing dimension
$\mu =$ Dynamic viscosity
$\rho =$ Fluid Density
$\nu =$ Kinematic viscosity

Let’s go over this variable by variable.

Velocity

The velocity is generally some kind of “average” velocity, which basically makes the fluid velocity through or around the object under analysis uniform. In the case of closed ducts or pipes, such as are described here, it’s an average velocity, which can be computed by dividing the fluid flow per unit time by the cross-sectional area of the pipe or duct through which the fluid flows. For Wind Tunnel Testing (actual and virtual, such as in CFD) it’s the “free stream” velocity, or the velocity through which the airfoil or object moves through the fluid.

Diameter

The “diameter” used to compute Reynolds numbers would seem to be clear-cut, but it’s more a matter of convention than anything else.

With straight pipes of uniform diameter, it’s just the diameter of the pipe. For flow meters such as orifices or venturi meters, it’s customarily the diameter of the incoming pipe, although in the past that wasn’t always the case. Non-circular pipes have their own rules as well, which related to the “circular equivalent” of a cross-section.
For round objects in free-flow condition, such as cylinders or spheres, the diameter is pretty straightfoward.
For airfoils the rules are more complicated and are described in Wind Tunnel Testing. For airfoils, the diameter is generally taken as the chord.
For “odd shaped” objects, it’s a matter of judgement and convention to determine the diameter.

Density

This is simply the density at wherever the velocity is chosen. For incompressible fluids, that’s pretty simple. For compressible fluids it’s more complicated. With Wind Tunnel Testing it’s the free-stream density of the fluid.

Viscosity

This, I think, is where most students get into trouble. If you use dynamic viscosity and density, it’s easy to get into trouble with the units. My advice to students is to use the kinematic viscosity $\nu$ whenever possible, which means that we use the second form of the Reynolds number. The units for this (unless you’re given it in stokes or centistokes, in which cases you’ll need to convert it) are $\frac{length^2}{time}$ , which cancel nicely with the other dimensions. For water and air, the two fluids we mostly test in my course, the properties are in my monograph Variation in Viscosity, along with a discussion of viscosity in general.

Some Examples

Let’s take a couple of examples. The first one comes from the valve loss coefficient example in our discussion on flow metering. We have a 1.48″ diameter pipe with water flowing through it at 15.9 gallons per minute. What is the Reynolds Number?

From that example, we computed that the average velocity in the pipe was 35.38 in/sec. From the monograph Variation in Viscosity, the kinematic viscosity of water is $\nu=10.877\times10^{-6}\frac{ft^{2}}{sec}$ . Using the formula above, by direct substitution (and a quick units conversion) we have

$R_{e}=\frac{VD}{\nu}=\frac{35.58\frac{in}{sec}\times1.48\,in}{10.877\times10^{-6}\frac{ft^{2}}{sec}\times\left(12\frac{in}{ft}\right)^{2}}=33,431$

As another example, consider a USGA standard golf ball, which is 1.68 in. in diameter. If it flies through still air at 50 mph at an altitude of 600′, on a standard day what is the Reynolds Number?

Working with US units, one of the trickiest decisions we have to make in fluid mechanics (and sometimes in solid mechanics too) is whether to work in feet or inches. The previous example did so in inches, converting the kinematic viscosity to $\frac{in^2}{sec}$ . For this one we’ll do it in feet. Which one you pick is up to you; the one you use is the one where error is least likely. (For very small values of distance, the significant digits problem comes into play.)

In this case we convert the diameter 1.68 in = 0.14 ft and 50 mph = 73.3 ft/sec. Using the kinematic viscosity for air (again from Variation in Viscosity) $\nu=161.6\times10^{-6}\frac{ft^{2}}{sec}$ , we substitute and solve