# Does separation happen easier with less dense air?

Does less dense air affect when separation happens?

So if you have a cylinder traveling in a straight path like the one above, there is separation happening at a given speed. If you were to change only the air density, would this separation be less at the same speed, the same, or more? If it is different, why?

The Reynolds number is

$$Re=\frac{\rho\, u\, L}{\mu}$$

If the density $$\rho$$ were to be less, you would have a lower Reynolds number (all else equal).

In general, that would tend to make the separation happen earlier.

However, it really depends on where you are starting -- is the flow laminar or turbulent. Perhaps you're at extremely low Reynolds number -- creeping flow even.

Whatever happens, it is determined by the Reynolds number.

• I see. So to summarize, the lower density air will have a lower Reynolds’s number (more laminar). Since laminar flow will separate before a turbulent flow, that would mean less dense air will separate easier? Commented May 7 at 16:20
• Yes, I said in general because this is what happens for the range of Re we usually think about for airplanes and such. At extremely low Reynolds number (less than 200), flow takes on a different character. At even lower Reynolds number (much less than 1), flow takes on even more different character. It is still laminar in these regions, but it is extremely viscous (think glycerine or molasses). At these Re, the flow might not separate at all. Commented May 7 at 17:10

The ubiquitous plot of the drag coefficient of a cylinder in respect to the Reynolds number is worth a thousand words (plot source):

As you can see, at higher Reynolds numbers (i.e. at higher density) the drag coefficient becomes smaller.

Does separation happen easier with less dense air?

This is represented by the jump in the coefficient at around Re $$3 \times 10^5$$: here we have the transition of the boundary layer from laminar to turbulent. A turbulent boundary layer remains attached to the surface a bit longer than a laminar one and this results in a lower drag.

This is even better represented in the following plot (source - many thanks to @ROIMaison) which shows at what angle on the cylinder's surface the flow detaches:

Here 0° means the side of the cylinder invested by the flow, 90° its crest and 180° the opposite side. Also here it can be seen that beyond the Re values as given before, the (turbulent) boundary layer remains attached till higher angles (around 120°) than at lower Re (around 80°).

So yes, separation (and the relevant higher drag) actually happens easier with lower dense air (left part of the plot).

Well, this is true at least for all Reynolds numbers of practical interest. Anyway, when Re becomes very small (smaller than one - so called creeping flow), the flow crosses the body following its shape without separation (180°), as can be seen for example in the following picture taken from the legendary "An Album of Fluid Motion" by "Milton Van Dyke" (pdf here):

• It is a good answer, but I think the reasoning becomes clearer when you show separation angle vs Reynolds numbers (such as this one). That way you can answer the question directly, without making the 'detour' to drag. Commented May 7 at 8:23
• @ROIMaison: uh, that's a nice plot, can I update my answer with it? Thanks for the editing 🙂 Commented May 7 at 8:34
• Yes of course, for reference; here is the link leading to the publication that contains the image Commented May 7 at 8:54
• @ROIMaison: thanks, I've update my answer with your plot Commented May 8 at 4:17
• @RobMcDonald: thanks for the link. I let you put your interesting plots about what happens at very low Reynolds numbers in your own answer Commented May 8 at 4:19