I am confused about Reynolds number. I am trying to use the formula to see if NASCAR cars have laminar or turbulent flow. But I am not sure how exactly to use the equation in order to do such calculation.
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$\begingroup$ As CuriousOne said in his previous answer, the air flow is going to be turbulent. $\endgroup$– Chet MillerCommented Jan 27, 2016 at 16:26
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2 Answers
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Use the Reynolds number equation:
$Re={vL \over \nu}$
where $\nu\approx 1.5\times10^{-5}m^2/s$ is the kinematic viscosity for air. If you enter this into the equation, you end up with
$Re\approx 67000{v\over{m/s}}{L\over m}$
i.e. for a race car traveling at 40m/s and with a length of 4m it comes out to be around 10 million, which is certainly $Re>>1$, i.e. in a regime far beyond the requirement for the Stokes formula to be a good approximation.
answered Jan 27, 2016 at 16:30
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1$\begingroup$ The key insight, not stated explicitly in your answer, is that the "length scale" $L$ in the formula is a proxy for "the scale of the problem" - which we usually take to be one of the linear dimensions of the object in question. But for an irregularly shaped object it does beg the question "which dimension should I take" and the answer is "any or all of them...". $\endgroup$– FlorisCommented Jan 27, 2016 at 18:27
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$\begingroup$ @Floris: you are correct and I didn't want to get into the discussion deliberately, especially since the "rule of thumb" approach puts this so far into the realm of turbulent flow that we really don't have to care about which dimension on the car we pick. Basically... even the tiniest of features on the car's body is a source of turbulence. $\endgroup$ Commented Jan 27, 2016 at 18:36
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$\begingroup$ Since OP said "I am not exactly sure how to use the equation" I think it's important to explain what value of $L$ to use. $\endgroup$– FlorisCommented Jan 27, 2016 at 18:39
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$\begingroup$ @Floris: In that case I would have had to reply: "I am not sure, either, but if you plug any of the possible numbers in, you get the same answer.". The real trouble with the Reynolds number is, of course, that the turbulence may develop behind the actual object and there is plenty of false intuition to be had about those cases (like about the famous and wrong answers to "Why can planes fly?"). $\endgroup$ Commented Jan 27, 2016 at 18:45
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$\begingroup$ I believe this number but wind tunnel test seem to look pretty smooth $\endgroup$ Commented Jan 27, 2016 at 19:23
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The Reynolds number gives a ratio between forces of inertial origin and those of viscous origin. For a given geometry of the problem, increasing the Reynolds number will lead to turbulent flow from a certain threshold. However, this threshold is strongly dependent on the geometry: this is actually common knowledge, a better design (more "aerodynamic" we say in common language) will lead to lower drag and delayed transition to turbulence.
So calculating the Reynolds number just gives a rough indication whether the flow will be turbulent or not, unless you find $Re<1$ which would mean that inertial forces are negligible and turbulence can never occur (uncommon for human-scale and common fluids).
However, if you do a wind-tunnel experiment with a reduced-size model having the exact geometry of your car (and size $L_R$ say), and determine the Reynolds number $Re^*= V_R^* L_R/\nu$ from which developed turbulence appears, then you can determine the velocity at which the full size car $L_F$ will create turbulence: it will simply be the velocity $V_F=V_F^*$ giving the same Reynolds, $Re = V_F L_F/\nu = Re^*$.
That said, full size cars are never so well designed that you have laminar flow around them, so yes, expect turbulence around a racing car.
