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I'm a HS student so please dumb it down. I'm looking into the Reynolds number of a sphere sinking in a fluid, here's what I got.

Inertial drag force = 0.5 * 0.47(CoefficientOfDragSphere) * rho(DensityOfLiquid) * pi r^2(CrossSectionalAreaSphere) * velocity^2

Viscous drag force = 6pir(ConstantKForSpheres) * mu * velocity

Reynolds Number is the ratio between inertial and viscous drag forces so after simplifying it should be = (0.47(Cd) * rho(DensityOfLiquid) * velocity * r) / (12 * mu)

So then how did the equation of Reynolds number = (rho(DensityOfLiquid) * velocity * 2r) / (mu) come to be?

What am I missing, the equations look similar but not quite, is there some sort of "super math" that I'm missing, or are my equations misused? And is my equation of Re correct? Could it be represented that way?

Thanks, A confused HS student trying to write a physics essay

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  • $\begingroup$ What you call the inertial drag force is really the total force from inertia and viscous drag. And your equation for viscous force is the total force from inertia and drag only in the limit of low Reynolds numbers; this equation transitions to the other equation as the Reynolds number increases. $\endgroup$ Jul 7, 2023 at 22:48

1 Answer 1

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The Reynolds Number is a dimensionless quantity used in order to correlate the behaviour of a flow with a pure number.

For example, for flow in pipes if $Re_D = \frac{\rho v D}{\mu} < 2300$ the flow is considered laminar, for $Re > 2900$ it is considered turbulent, and anything in between is intermediate flow. Those numbers are defined empirically and as you can see, engineers are fond of multiples of 100, since they're easier to remember than 2891 for example.

Even though you are right in your claim that it represents the ratio between inertial forces and viscous forces, the Reynolds Number is typically defined only in terms of the quantities with dimensions (which will cancel out in order to make a dimensionless number), so all those constants like 0.47 and 12 are ignored.

Let's say for your system,

$$ Re = \frac{0.47\rho v D}{12\mu} $$

Then you make some experiments and you find out a constant number C, which is the boundary for turbulent flow. But why not redefine your Reynolds number so that turbulent flow starts at $ C^\prime = \frac{12 C}{0.47} $ instead?

This way you get an easy to remember formula for the Reynolds number, which is more consistent with the formulas for other geometries:

$$ Re = \frac{\rho v D}{\mu} $$

Imagine having to look up two tables, one for the right formula for Re in each possible geometry (pipe, sphere, stirred tank, etc.) and another one for the boundaries between laminar, intermediate and turbulent flow. That would not be pratical.

BTW, I have no idea what HS means, could you tell me in the comments? ;)

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  • $\begingroup$ For example, suppose that you were using the definition $Re = \frac{0.47\rho v D}{12\mu}$ and you found experimentally that for $Re > 100$, the flow becomes turbulent in the system you are studying. But you could redefine the number of Reynolds getting rid of the constant numbers by making $ Re = \frac{\rho v D}{\mu} $ and adopting the convention that turbulent flow starts for $Re > 100 \times \frac{12}{0.47} \approx 2400 $ instead. $\endgroup$
    – GFonseca
    Jun 13, 2023 at 8:16
  • $\begingroup$ Thanks so much, HS as in High School student. I understand your explanation but I kinda got lost here: "Then you make some experiments and you find out a constant number C ... 𝐶′=12𝐶0.47 instead?". I don't really understand what you mean by C and C'. Are you trying to say that for my specific scenario, my equation for Re is correct, but the typical equation doesn't only reflect my scenario and is more broad and general? So in my case, C is just a constant that I would multiply by the Re, and in my case, it's ((0.47*2)/12)? Thank you so much again! $\endgroup$
    – axelbendl
    Jun 13, 2023 at 8:17
  • $\begingroup$ I see, so if I were to do an experiment where I drop a ball through a liquid, and I wanted to determine if the flow was Creeping Re<<1 or not. Then I'd use the standard equation and not the one I derived? If so, in my essay what would I say as the reasoning to why I removed the constant, because re is typically only defined in terms of quantities with dimensions? Also, my equation I got wasn't Re = 0.47*rhov*2r/12*mu (where D=2r), it was 0.47*rhovr/12*mu, does that change anything? If I ignore all the constants then I won't get that standard equation, I'd get, rhov*(D/2)/mu. TY btw $\endgroup$
    – axelbendl
    Jun 13, 2023 at 9:14
  • $\begingroup$ Nevermind, I totally get it now, thank you so much for taking time out of your day to respond to me :pray:, very much appreciated! C: $\endgroup$
    – axelbendl
    Jun 13, 2023 at 13:47

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