# Aerospace Flow Separation of Airfoil in terms of Reynolds Number

1. Jun 1, 2012

### Red_CCF

Hi

I know that for an airfoil, it's typical to describe stall in terms of angle of attack.

I'm wondering, how does Re play a role in flow separation of an airfoil (and for objects in general such as a plate, sphere, or cylinder)? One Wikipedia (http://en.wikipedia.org/wiki/Flow_separation) it states that flow separation location is independent of Re for laminar flow, why is this and what about turbulent flow?

Thanks

2. Jun 2, 2012

The Reynolds number is the ratio of inertial forces to viscous forces in the flow, and flow separation does not depend on this ratio. It depends on the pressure gradient over the surface in question. The phenomenon isn't strongly tied to the Reynolds number in laminar or turbulent flow except in the sense that turbulent flows are more resistant to separation on a given surface.

3. Jun 3, 2012

### RandomGuy88

The Reynolds number is VERY important for determining when and how an airfoil will stall!

It is not uncommon for the maximum lift coefficient of an airfoil to change by 20-30% or more when the Reynolds number is changed and the stalling angle of attack can change by several degrees. It certainly depends on the airfoil and the magnitude of the Reynolds number increase but the stalling characteristics (separation) of the airfoil are strongly influenced by the Reynolds number.

For example at low Reynolds number there may be a laminar separation bubble near the leading-edge and as the angle of attack increases this bubble can pop causing leading-edge type stall. However as the Reynolds number increases this bubble will no longer form and then the airfoil may undergo trailing-edge type stall at a higher lift coefficient and angle of attack.

4. Jun 3, 2012

### Aero51

For lower Reynolds numbers you will get laminar separation. In this case the flow separations from the surface of the object as if it is traveling on another surface. This is known as a separation bubble in aerodynamics. In the case of a sphere, under highly laminar flow, separation will occur suddenly and abruptly at the top of the curve. It will look as if someone drew a tangent line off the top of the sphere.

In the case of turbulent separation, the air will continue to rotate and stick to the object, separating later. While this will delay stall considerably you will also see an overall rise in drag. In the case of the sphere, the flow will wrap around the the ball, separating maybe 1/6 of the way down the diameter on each side.

A very good paper published from Iowa State goes over the effects of turbulence and flow separation:
http://www.public.iastate.edu/~huhui/paper/journal/2008-JA-Corrugation.pdf

What it comes down to is that flow separation is more dependent on the turbulence levelsof the freestream flow as opposed to the actual Reynolds Number. It is erroneously noted that stall is dependent on RE because turbulence levels tend to increase with increasing RE.

Remember, Reynolds number (or any dimensionless parameter) is just the ratio of two formulas and does not take precedence over the physical context of the problem.

Lastly, for an airfoil the general pattern is that maximum lift coefficient increases with increasing RE.

Last edited: Jun 3, 2012
5. Jun 4, 2012

RandomGuy88, the phenomena you describe are determined more by the state of the boundary layer (laminar or turbulent) than the Reynolds number. True, the transition between states is Reynolds number dependent, but there isn't a direct physical link that I am aware of between Reynolds number and separation. Of course you are more familiar with separation phenomena than I, so if I'm missing something I'm all ears.

Aero51, the turbulence levels of the free-stream have very little to do with separation, especially when it comes to geometries like a sphere that are stable to nearly all instabilities until well after the flow typically separates anyway. Turbulence in a boundary layer is not quite the same as turbulence in the free stream.

6. Jun 5, 2012

### RandomGuy88

I see where you are coming from Boneh3ad but I would argue that since separation is a boundary layer phenomena then something that directly effects the boundary layer directly effects separation.

Separation is not only dependent of whether the boundary layer is laminar or turbulent. For example laminar separation bubbles responsible for leading edge stall may disappear at higher Reynolds number even though the flow near the leading edge is still laminar. Similarly if a turbulent boundary layer is separating near the trailing edge the aoa at which it begins to separate can be delayed as the reynolds number is increased. A boundary layer is more resistant to separation as the Reynolds number is increased regardless of whether it is laminar or turbulent.

In addition, the size of laminar separation bubbles and wether or not they burst will depend on the Reynolds number. In many cases it is not actually the bubble that bursts but rather the shear layer reattaches as a turbulent boundary layer and then quickly reseparates and this process can depend on the Reynolds number.

If an increase in Reynolds number causes the transition point to move forward this can lead to adverse scale effects where the performance actually decreases with an increase in Reynolds number.

The overall magnitude of these effects and whether or not they even happen depend on the geometry and the Reynolds number range being considered as well as surface quality and various other factors.

Turbulence in the free stream can effect where the boundary layer transitions which can influence when and how the boundary layer will separate.

7. Jun 5, 2012

True to a point. Leading edge separation would not see the effects of this, for example. That region is subcritical to Tollmien-Schlichting waves and modern airfoils are typically designed such that the crossflow instability doesn't cause transition until a bit farther downstream than a typical separation bubble would occur. The same thing goes for a sphere or a cylinder, where the separation occurs before the boundary layer would typically naturally transition.

Of course you could conceivably have a free stream that was so turbulent on the right scales that the boundary layer transitions almost immediately. However, even in a conventional wind tunnel this isn't the case and it certainly isn't in flight where the air is essentially non-turbulent.

That is also an incredibly complicated problem. There is currently no real way to connect free-stream disturbances to the equivalent initial disturbance amplitudes in the boundary layer since the receptivity process is effectively a black box as understand it currently (or perhaps a slightly translucent black box). Then it is even murkier since there is no real way to predict transition in general. Of course here I go off on one of my stability tangents. I'll stop, haha.

8. Jun 6, 2012

### Aero51

Sorry, what I meant to say was the turbulence levels of the flow near the body (not necessarily exclusively the boundary layer). However, I feel that turbulence in the freestream does affect the flow field and probably where separation will occur. In wind tunnels (at least the one's I've worked in) for example it is crucial that you have a uniform velocity distribution (which translates to a turbulent flow inside the tunnel) to ensure your experiment yeilds reliable results.

In thinking about this question, I think its also worth asking how do you define turbulence in a flow that is still before an object passes though it. That is to say, is it valid to assume that air passing a stationary object is physically identical to a body passing through a stationary body of air?

9. Jun 6, 2012

How do you propose that turbulence outside of the boundary layer affect separation, which is a function of pressure gradient, the boundary layer profile and the state of the boundary layer? Except at the leading edge, they can't. Near the leading edge, the free-stream turbulence can interact with surface roughness and produce the initial conditions for the disturbances that eventually grow and break down into turbulence (a process called receptivity), but this process generally becomes unimportant as you move downstream.

For that reason (and a few others), the atmosphere effectively has zero turbulence in the free stream. The motion it does have in the form of wind can be turbulent, but the spatial scales over which these eddies occur are much too large to enter into the boundary layer on an aircraft. Instead, they are more of the scale that tends to make the plane bounce up and down. This is part the reason why wind tunnel results are not always directly applicable to flight. In a wind tunnel, the object is often to reduce the turbulence intensity of the free stream, defined as:

$$\mathrm{Tu} = \frac{u^{\prime}}{U}$$
where
$$u^{\prime} = \sqrt{\frac{1}{3}\left(u_x^{\prime 2}+u_y^{\prime 2}+u_z^{\prime 2}\right)}$$

Of course it is crucial to have a uniform velocity distribution, but this hardly requires turbulence. In fact, there are a whole class of tunnels, quiet tunnels, which strive to create an environment as free of turbulence as possible. One wind tunnel with which I am intimately familiar have a turbulence intensity of $\mathrm{Tu} \approx 0.025\%$. That is hardly turbulent.

10. Jun 6, 2012

### Aero51

Well what about sudden changes in angle of attack due to gusts? In fact, if you have an unsteady freestream flow and your effective angle of attack is changing with time then separation must depend on the state of the freestream flow. In addition, correct me if I'm wrong, but should an unsteady freestream flow also create an unsteady boundary layer?

11. Jun 6, 2012

### Aero51

I'm also thinking that if you have a flow passing over an airfoil and it is highly turbulent, wouldn't the air have a tendency to wrap around the body more readily? You may get separation but it will be sporadic and perhaps of a smaller magnitude as opposed to the case when you have a very "clean" freestream. I would think this is because the air is already rotating and diffusing, so if a body passes through a turbulent flow it does not change the freestream as readily as say an airfoil passing through a laminar flow.

12. Jun 18, 2012

### AIR&SPACE

13. Jun 18, 2012

### Red_CCF

Hi

Thanks very much for your responses.

So I'm wondering, is the separation location independent of Reynolds number only for practical ranges or for all ranges (0 to infinity). I found sources that support that separation location is independent of Re but only for a finite range the experiment was conducted.

I looked at some videos where streamlined objects were put through glycerine and thus at very low Re (possibly under 10) and there was no noticeable separation at all (although I'm sure that regardless of Re there is a small wake present). However, putting the same object through air intuitively I would think that (for both to be laminar flow) that the wake would be much bigger and separation would occur much earlier. I think that was the reason why streamlined objects have a higher drag coefficients than a sphere at low Re. I always thought that Re indicates how far the fluid can travel in an adverse pressure gradient at the boundary layer, Is this correct?

Thanks very much

14. Jun 19, 2012

### meganlove19

It was very interesting to read this post and inspired me to brush up on my boundary layer and aero theory.

Reynolds number has no effect on the laminar separation point. Increasing Reynolds number can cause a transition to turbulent boundary layer ahead of separation and effectively move the separation point downstream. This holds for all Reynolds numbers theoretically speaking.

"I think that was the reason why streamlined objects have a higher drag coefficients than a sphere at low Re"

Drag is affected by more factors than just Reynolds number. Streamlined refers simply to the shape of the object but neglects surface effects that can also effect the flow. The drag on a sphere is dominated by pressure drag and the drag on a streamlined object by friction drag. Stoke's Law I believe outlines drag on spherical objects at small Reynolds numbers. Also it should be noted that experimental data (from Schlichting Boundary Layer Theory) shows drag on a sphere to be on the order of 200 for Reynolds numbers of ~0.1.

"I always thought that Re indicates how far the fluid can travel in an adverse pressure gradient at the boundary layer, Is this correct?"

This is very interesting I have never heard Reynolds number termed this way (I’m still a bit of a newbie though). The definition of Reynolds number as previously stated is the ratio of inertia forces to viscous forces. Breaking that down we see

inertia forces → ρ u L → density, velocity, and length
viscous forces → μ → dynamic viscosity = a measure of the resistance to flow of a fluid under an applied force

ρ = m/v → (m/v)*u*L → (m*u)/v *L = p/v * L where p is momentum

If we start examining the terms and units we see that there is a momentum term (mass*velocity) and an inverse unit of area (length^ -2)

On the bottom we have resistance to movement essentially in terms of (force *time/(length^2))

Canceling out the area terms we're left with (mass*length / time) / (force * time). Without canceling out all the terms seeing as Reynolds is dimensionless we can evaluate this verses your statement of distance traveled per pressure gradient (Pascal/m)

A pressure gradient has units of pascals/m or force/ length^3 . In order to have distance per pressure gradient we would need units resembling force/length^2 which I don’t see a way to get from Reynolds number. In fact I would say from a dimensional analysis stand point that this statement is not correct. Just a thought.
Out of curiosity where or how did you come to this understanding of Re?

15. Jun 21, 2012

### Red_CCF

Hi

Thanks for the response. I am trying to brush up on my fluids as well so any insight is appreciated. I'm recalling a few things from memory so it's very possible that they may be wrong (which is why I started this thread).

Can you go into more detail from a flow perspective on why separation location is independent of Re for laminar flow as this is a bit counter-intuitive for me.

The point I was trying to make with the comparison of drag is that a blunt object has a lower drag than a streamlined object at low Re (from Fluid Dynamics of Drag series by Ascher Shapiro), which means that there is likely a very small wake (such that only friction drag is significant) for both blunt and streamlined object yet for higher Re this isn't the case, and the only explanation I can think of is that pressure drag must have increase significantly (bigger wake) and thus the separation point must have moved for the blunt object, which is intuitively why I thought that there's a Re dependence on separation location.

I forgot where I heard this from. I was just thinking that intuitively, higher Re means higher average velocity in the boundary layer as boundary layer thickness decrease with increasing Re (ignoring laminar to turbulent transition) and free stream velocity increases (keeping viscosity constant), meaning more momentum in the boundary layer and thus it can travel a greater distance against an adverse pressure gradient. I'm not sure if this is correct though.

I actually had another question about flow separation, not really related to the above. I was wondering, once separation occurs, the wake of the object has a lower pressure than the "atmosphere" above the separated flow so how come the pressure above the flow does not "collapse" the flow back down?

Thanks

Last edited: Jun 21, 2012
16. Jun 22, 2012

### Aero51

If you really want to get a good grasp on flow separation your best bet may be to visit the AIAA website and search the archives for technical papers. Most of the search results will say that you have to pay for the articles, dont be fooled! Simply type the name of the paper in google and you will probably find a free copy. Here is the URL to the AIAA Technical Paper search engine:

https://www.aiaa.org/IframeOneColumn.aspx?id=4745

17. Jun 27, 2012

### Red_CCF

Hi

Thanks for the response. I tried to look at some papers but they seem to be a bit too complex for me as I don't think I have enough basic understanding for them. Also, it seems that my question does have an answer but I just cannot find a source that explains the physics behind it; they more or less just state that flow separation is independent of Re and leave it as is, which I find confusing since the same object in different fluids appear to have different sized wakes.

18. Jun 27, 2012

### Aero51

The different sized wakes in different fluids may be attributed to differences in viscosity, though I cannot be certain unless you provide the name/link to the paper. In general, a higher viscosity in the flow will correspond to higher diffusion rates and a quicker dissipation of the wake. Again, this goes back to the debate on how turbulence levels affect the flow/separation. If there is a specific concept or question that you have regarding the papers you have read feel free to ask.

19. Jun 27, 2012

Might I just point out that different viscosity means a different Reynolds number if all else is held constant?

As long as the Reynolds numbers are the same, the fluid shouldn't matter for the size of the wake so long as it is incompressible regardless of the viscosity. If the viscosity is higher for one, just increase the speed or size of the model a little bit.

20. Jun 27, 2012

### Aero51

Actually you made me realize that if the viscosities are the same and the reynolds numbers are constant then differences in mach number may affect the wake. However neither of us have the paper to make a firm statement.

21. Jun 27, 2012

### Red_CCF

Hi

I was actually not referring to a paper, but this video:

http://vimeo.com/7507698 [Broken]

An experiment was done for two objects (one streamlined, one spherical) traveling in air(around 15:00) and glycerin (around 19:00). The streamlined object traveled slower than the spherical in glycerin for the same drag force applied (also I could not see the wake for either object) which wasn't the case when air was used. Speed was decreased and viscosity increased in the glycerin, so the Re must be extremely low.

From pictures of separation in air for a sphere, I can see that there's usually quite a large wake. I'm not sure if separation was actually occurring at the same location but the glycerin simply dissipated the effects or something else altogether.

Thanks very much

Last edited by a moderator: May 6, 2017
22. Jun 27, 2012

Given the speeds and Reynolds numbers involved, that teardrop shape is certainly not separated at all and I would actually contend that even the sphere likely is not separated, as it appears that the experiment satisfies the requirements for Stokes flow. It has no wake and therefore effectively has only viscous drag. As a result, you have one shape with small surface area and one with large surface area, both dominated by viscous drag. Of course the one with large surface area (the teardrop) will move more slowly. In the air, the Reynolds number is going to be much, much larger and while the teardrop will still be dominated by viscous drag, the sphere will be dominated by pressure drag. In these situations, pressure drag is dominant and so the sphere had higher drag in the air flow.

At the kind of low Reynolds numbers ($\mathrm{Re} \ll 1$) seen in the glycerine experiment, there certainly is an effect due to Reynolds number. You have to remember what the Reynolds number represents. Recall the definition
$$\mathrm{Re} = \frac{\rho U_{\infty} D}{\mu}$$
The number represents a ratio of the inertial forces due to the fluid motion to the viscous forces. For the extremely low Reynolds number, the viscous forces are dominant. For most practical flows such as on cars, planes, etc., the Reynolds number typically falls more within the range $10^3 \leq \mathrm{Re} \leq 10^7$. Anywhere in that range, the inertial forces are many orders of magnitude more important than viscous forces.

The behavior when $\mathrm{Re} \ll 1$ and $10^3 \leq \mathrm{Re} \leq 10^7$ are fundamentally different in essentially all regards. However, within one range or the other, the fluid behaves fundamentally the same regardless of where you fall in that range.

23. Jun 28, 2012

### Aero51

Last night I happened to be looking through my aircraft design book and came across an excerpt that is very pertinent to the discussion of this thread. The Book is "Aircraft Design: A Conceptual Approach 4th Edition" from page 306, the paragraphs (truncated) read:

And one more thing I wanted to ask/add with regards to the notion that a fluid behaves essentially the same when E3≤Re≤E7:

I was always under the impression that, for a generic wing, the flow is assumed to be fully (or close to fully) turbulent when the Reynolds number is Re≥E6. The flow shares a laminar and turbulent boundary layer when (~E5)≤Re≤E6 and when Re≤E4 this is in the aircraft and micro-air vehicle range which is considered a fully laminar flow regime.

24. Jun 28, 2012

This is a very misleading statement. I understand what the author is trying to say, which is that a turbulent boundary layer diffuses momentum (and thus kinetic energy) much more efficiently than a laminar boundary layer so it tends to have higher momentum lower in the boundary layer, but saying it has more energy can lead to some pretty wild interpretations. That's just an interesting way to put it.

If it is in the range of 104, then the flow is almost certainly laminar. There are probably some ways you could still trip that boundary layer (transient growth, for example), but at that low Reynolds number the eigenmodes that arise in the linear stability problem are pretty much all stable and you would need an extraordinarily long airfoil or surface to see transition from crossflow or Görtler vortices. However, fundamentally, the flows are the same. They still are dominated by inertial forces and are subject to instabilities, even if those instabilities are stable in parts of the regime.

Otherwise, there is no generic Reynolds number where the flow is assumed laminar or turbulent. The problem is extraordinarily more complicated than that. We don't even have a general model for the transition location on a zero-pressure-gradient flat plate at zero angle of attack as a function of Reynolds number, let alone an airfoil.

25. Jul 2, 2012

### Aero51

Sorry for the late response, it was a busy weekend. Anyway, when I was speaking about the Reynolds number ranges I was coming from a historical perspective. Many papers cite those ranges typical to the aircraft mentioned above.

I cant say I know much about Görtler vortices or a lot of details about turbulence/turbulence modeling as I have not taken any classes on the material yet. However, in lieu of these facts I will proceed to read some more papers on the subject.