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Apologies in advance if the question is trivial. I am accustomed to electromagnetics but an amateur on fluid dynamics.

My understanding is that the Navier-Stokes equations are solved to determined the velocity and the pressure drop of a fluid.

Reynolds and Mach's numbers tell whether the flow is laminar and incompressible respectively, therefore they seem to tell whether the Navier-Stokes equations can be simplified.

These numbers depend on velocity, which is an unknown of the problem.

So how do you calculate Reynolds and Mach's numbers before solving the Navier-Stokes equations?

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In general, you know the system you want to solve. For example, I know I have an airplane flying at 150 m/s at an altitude of 3000 meters and I know what the size of the wing is. From this, I can calculate the Reynolds and Mach numbers.

I don't know what the velocity is at every point. And so I don't know what the local Reynolds or local Mach number is at any given point until I solve the equations. But I know the global numbers based on my boundary conditions and geometry, and that's usually good enough to get an estimate of regimes.

This couples with experience to determine what might happen. For instance, I may know that for a global Mach number of 0.2, my geometry might have a really big flow restriction that could accelerate the flow to a large Mach number. So even though my global Mach number says the problem isn't compressible, I use my experience to determine that I might have a part of the flow that could become compressible, and so I should use the compressible equations.

On the other hand, I may not have that experience and so I solve the incompressible equations. And then I see in my solution that the velocity is bigger than I thought it would be and I should probably redo my calculations with the compressible equations.

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  • $\begingroup$ The key point of your second approach is internal consistency, and it is something that is sometimes under-emphasized in early instruction. Anytime you perform a calculation with a simplifying assumption you should check when you are done that the assumption makes sense in light of the results of the calculation (i.e. that the calculation is internally consistent). $\endgroup$ – dmckee Mar 3 at 20:18
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An alternative understanding says that the Reynolds number is not intrinsically meaningful in the way you are hoping. This is visible in the fact that it makes reference to the size of the system: what size? Is it a radius or diameter or circumference of some round object? What if that object is not round but square? I have a propeller on the front of the plane, do you want the thickness of the blades or the length of the blades or the full length of the airplane? These vary by several orders of magnitude.

Rather, the Reynolds number is a convention tied to a certain geometry. It comes from trying to normalize the Navier-Stokes equations in the context of that geometry: that geometry defines a characteristic fluid velocity (maybe the speed at infinity), and a characteristic size scale, and a characteristic air density. However you define those things, they give you a characteristic length, time, and mass: we divide every unit-ful number in our problem by these units to find a unit-less version of that number in our geometry, and so cast the Navier-Stokes equations into unit-less form. One of the dimensionless parameters that is left in the equations is the Reynolds number, which becomes a promise: if you wanted to study this system in a different fluid or in miniature, if you get these dimensionless parameters to all be exactly the same, the two systems are governed ultimately by the same equation and must do the same thing. The geometry and the dimensionless parameters determine the behavior, always.

Now of course it doesn't represent nothing just because it does not have one universal declaration: it represents in its relative sense either the reciprocal of the viscosity of the fluid, or equivalently we can take the viscosity as a given and it represents the scale of the fluid flow speeds at infinity. Twice the Reynolds number means “I want to study the same system in a more syrupy fluid” or “...at a higher driving speed” equivalently. But different geometries have different senses of how syrupy or fast a measurement of “1” is.

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