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What's the meaning of the Reynolds number

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Old   September 20, 2012, 10:13
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Nick
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Originally Posted by agd View Post
Just a quick question for my own edification - if the velocity is not changing with distance, then aren't the viscous forces also zero? Wouldn't the whole flow system become trivial, rendering the Reynolds number moot?
Viscous forces are not zero because there s a velocity gradient in the normal to wall direction but they are balanced by pressure difference forces .the.velocity for fully developed flow doesn't change in the parallel to wall direction .that means fluid particles don't accelerate because of the balance of forces
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Old   September 20, 2012, 10:52
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How do you define inertial force? You.previously said that it is the convective term in the navier stokes equations..that term is zero for.the.described channel flow.

Your observation about the null lagrangian acceleration u * grad u is correct provided that you have a fully developed steady flow, this is only an academic framework. In the case you are citing it is the equilibrium of the isotropic to deviatoric part of the stress tensor that balance each other, since the isotropic part (the pressure) is of order rho*u^2 that leads to some misleading in the Re.
Again, I don't see the problem in the definition of Re as ratio between convective and diffusive fluxes (or convective to diffusive time) ... both do not vanish...
But the real flow in channel is far to be the Poiseulle solution, it is turbulent, with unsteady behaviour and inertial term always present, therefore I am used to define the Re in channel as rho*u_tau*H/mu wherein u_tau is the stress velocity that is defined by means of the pressure gradient
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Old   September 20, 2012, 22:28
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I think we can agree on the definition of Re based on the momentum comvective flux thru one.cross section divided by viscous flux ... I'm not convinced of the presence.of any inertial forces ..the flow.could be laminar too .anyways thanks for.your.input and.time
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