# Reynold's number

## Recommended Posts

Can anybody give me an simple explaination about the significance of the Reynold's number in the context of helicopter rotor design?

best regards.

##### Share on other sites

I took several aerodynamics courses in college, and although I can't remember it's exact definition, a professor explained it to me similarly to this...

It is a number that has to do with efficiency of scale in aerodynamics - As objects get larger, they may be more efficiently pushed through the air. Paraphrased, he said, in the most simplest of terms think of the molecules in the air as grains of sand. Molecules are always the same size and there's nothing we can do to change that fact. As the object you're trying to push through the air gets smaller, the molecules become larger by comparison, and it is more as if you're trying to push that object through large rocks... Then he said, "...it's easier to shovel sand than the large rocks, isn't it?".

Edited by nbit
##### Share on other sites

The reynold's number comes about when making the Naiver-Stokes equations (the governing equations for fluid motion) non-dimensional. Basically, it assumed that all lengths, pressures, densities and speeds have a constant "scale" value and the the variables in the equation represent deviations away from that scale length. These constant values can be moved around in the equation in such a way that they end up in one term of the equation (the diffusion term).

From memory, Re = L*U*density/viscosity, where L is the scale length and U is a scale velocity.

Physically, it represents the ratio between the inertial effect in the fluid and the viscous parts. When Re us small, that is when the viscosity is large, dynamic variations (such as turbulence) are quickly diffused away and the flow remains smooth. At high Reynolds number, viscosity is weak relative to the dynamic effects, the primary result of which is that turbulence is a predominant feature of the flow. A jar of honey is a low Re fluid (if try to stir it up, all motions stop with the stirring), the atmosphere is generally considered to be high Re (although there are some applications, such as the boundary layer around a wing, where length scales are small enough to make it low Re).

##### Share on other sites

... interesting answer. Can you give me some examples in the case of rotor aerodynamics.

##### Share on other sites

... interesting answer. Can you give me some examples in the case of rotor aerodynamics.

Do you have a particular application in mind? Simply the flow around the airfoil, for example? I would probably take the chord length of your airfoil to be the characteristic length - somewhere about 20cm for the R22 I fly in training, or 0.2m.

The normal velocity scale would be the freesteam velocity, which is kind of a funky concept for a rotor application. Normally this would be something akin to the speed of the relative wind, but that changes not only with position along the wing, but also with position of rotation (is it the advancing or retreating blade) and the speed of forward flight. My instinct would be that since we'd probably be that the effects of advancing/retreating blades would be not particularly relevant (since we're probably looking for characteristics of the airflow that are seen all the way around a rotation) and also that the most interesting dynamics will be near the tip. So, the relevant "freestream" velocity would be the tip-speed of the blade: (RPM*pi/30)*R, where R is the radius of the rotor (from hub to tip): if I remember correctly, the R22 has a radius of about 4m. I'm having trouble remembering the RPM of the R22 main rotor in normal operations. I'll take a guess at 500RPM - if I'm wrong about that, correct my math accordingly. So (500*pi/30)*4 is approximately 210m/s.

The standard atmospheric density at sea level (used for convenience - you'd want to use a corrected value based on temperature and altitude) is 1.225 kg/m^3

The standard viscosity of air is 18.67 * 10^-6 Pa-s.

So, rho*U*L is 1.225 kg/m^3 * 210m/s * 0.2m = 51.45 Pa-s

(A note on units 1 Pa = 1 N/m^2; 1 N = 1 kg-m/s^2; 1 N/m^2 = 1 kg/m-s^2; 1 Pa-s = 1 kg/m-s, which is what you get after multiplying through above)

So, Re = 51.45 Pa-s/18.67*10^-6 Pa-s = (51.45/18.67) * 10^6 is about 2.8 * 10^6, or 2.8 million. The transition between laminar (smooth) flow and turbulent flow takes place around a "critical" Reynolds number which depends on application (pipes vs plates vs cylinders vs flat plates vs airflows, etc) but which normally has a value of several thousand. Our Reynolds number is several million - many orders of magnitude above any expected critical number, so we will expect turbulent airflow around the tip of the rotor.

Did that make any sense or did I go through that way too fast?

##### Share on other sites

• 14 years later...

Hi there, If I wonted the lift & drag plots verses AofA for a rotor blade to feed into a spread sheet, would the tip velocity Reynolds number be the one to use? thanks.

##### Share on other sites

On 2/20/2022 at 9:42 AM, Harold said:

Hi there, If I wonted the lift & drag plots verses AofA for a rotor blade to feed into a spread sheet, would the tip velocity Reynolds number be the one to use? thanks.

Blade design comparisons of one airfoil with another concerning drag characteristics should use the same Reynolds number (RN). Suppose you’re doing a comparison test you want to use the same RN on each test. Base your test on given criteria.

Plot your x, y coordinates: x=AOA, y=L/D ratio. Normally the blade is stationary inside a wind tunnel where airflow is the same across the span. In your case, mid span, 3/4, or tip speed would probably yield about the same characteristic curve. Edited by iChris

## Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.
Note: Your post will require moderator approval before it will be visible. ×   Pasted as rich text.   Paste as plain text instead

Only 75 emoji are allowed.