I'm currently scaling down a model 64 times for a wind tunnel analysis, where I'm measuring drag force (for drag coefficient) and lift force (for lift coefficient). However, I read that scaling down 64 times means that wind velocity must be scaled up by 64 times to keep the Reynolds number constant. I read that using trip strips are a viable solution. How do I calculate the location, size, and number and strips of the slips required to increase the Reynolds number by 64 times so I can get accurate results?
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$\begingroup$ Read about Dr. Max Munk and the variable density wind tunnel at the Langley research center, developed in the early 1920s. $\endgroup$– Robert DiGiovanniOct 12, 2019 at 7:24
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2 Answers
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Trip strips will approximate a higher Reynolds number in the sense that you get turbulent flow at a lower Reynolds number, but they will not increase the Reynolds number themselves. You will have different pressure and density gradients in the boundary layer, as though you were operating at a higher Reynolds number, but the number itself has no way of accounting for a trip.
The equation for the Reynolds number is $$\frac{\rho v d}{\mu}.$$ These are the only parameters you have to work with if you're going to be using the Reynolds number as your similarity parameter. If you absolutely have to use a wind tunnel, then increasing the model size or increasing the velocity are your only options (theoretically you could pressurize the wind tunnel to increase density if it were sealed, or you could change the temperature of the flow within the tunnel, but these have not been realistic options for any wind tunnel I've worked with). If you don't have to use a wind tunnel, a water tunnel is a viable alternative.
See An Assessment of the Usefulness of Water Tunnels for Aerodynamic Investigations for more details on water tunnels.
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3$\begingroup$ Cryogenic wind tunnels are a thing too: sciencedirect.com/science/article/abs/pii/0376042192900086 $\endgroup$ Oct 11, 2019 at 11:03
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$\begingroup$ In the blow-down tunnels I've worked with, we were able to control the Mach number and increase Reynolds number at the same time. I assume that's done by increasing the density? $\endgroup$– JZYLOct 11, 2019 at 11:15
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$\begingroup$ Hi, is it feasible to test at different velocities(and as a result, different Reynold numbers) and extrapolate the results? I know this will create error, but I hope the error will be smaller than just ignoring the Reynold number. $\endgroup$ Oct 13, 2019 at 2:11
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$\begingroup$ @itisyeetimetoday yes, you can do that, although you'll probably end up creating more error than you expect. The results will likely be good enough for a school report, but not for anything that would actually fly. $\endgroup$– zaenOct 15, 2019 at 21:16
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Reynolds Number is a relatively easy calculation to do: Velocity x Chord Length/ Kinematic viscosity. So, changing the fluid from air to water may be helpful. I would also consider developing data points from varying degrees of scale, for example 1/128th, 1/64th, 1/32th for comparative purposes, as well as changing fluid velocity.
Also, a review of relevant performance data of anything similar to what you are testing, with or without "trip strips", may save you many hours in the laboratory researching what has already been done.
Finally, run your tests with a "control" of known performance to see if your methodology is proper for the "experimental" you are trying to test.
answered Oct 11, 2019 at 16:19
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$\begingroup$ Thank you for your response! Do you have any link to research with trip strips and airfoils? $\endgroup$ Oct 13, 2019 at 2:02
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$\begingroup$ @itisyeetimetoday also, because adding test strips, roughness, dimples etc is changing the airfoil, it is very important to do this first on the well known control (Clarke Y, Davis airfoil, NACA 0010, etc) and see if your polars match the higher Reynolds "real life" scale. At that point, you move on to your experimentals. Try to find a control that is as close as possible, and avoid the temptation of making anticipated results fit the experiment. Sometimes, for better or for worse, they don't. $\endgroup$ Oct 13, 2019 at 16:44