Understanding the Reynolds Number

motherh
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Homework Statement



I don't fully understand the Reynolds number and it has arisen in a problem. It says:

Fluid with viscosity µ and density ρ fills the gap between two parallel plates at z = 0 and z = h. The upper plate at z = h moves with speed V in the x direction, while the lower plate at z = 0 is stationary. The fluid is also subject to a pressure gradient −G in the x direction.

Homework Equations



I have solved for the velocity u (no idea if this is needed for the Reynolds number bit at all) to find

u(z) = Gz(h-z)/2µ + Vz/h.

The Attempt at a Solution



The question then says:

What is the Reynolds number for this flow? What are the conditions on V and G for it to be small?

Can anybody help at all? It would be much appreciated.

All my notes really say is that Re = ρUL/µ or Re = ρL^2/µT but I don't understand the U, L or the T.
 
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That Wiki helps a little but I'm still confused by this. So the Reynolds number is given by Re = ρUL/µ? As U is the velocity scale it makes to take U as V here. But after this I don't know how G comes into the Reynolds number.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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