Fluid Mechanics: Friction of a moving surface

Well what you said is only true for what whose depth is greater than the boundary layer thickness. If that is the case then the shear at the interface in the water is zero because there is no velocity gradient.

 

" flowing horizontally in laminar steady state motion, where its surface is also moving. Above it is air."
Without any driving gradient, either from pressure or gravity?

The flow won't be steady at all, energy will be continually dissipated, until the fluid is at rest..

Thus, you can get such a steady case where you either have some plate on top of the fluid enforcing a velocity there, or on an inclined plane.

 

The assumption that it is both horizontal and steady state in the first place implies there is a pressure gradient. He never implied otherwise.

 

It might be in that case that a strictly horizontal surface is an unstable feature, and ripples will form at the surface.
I'll think about it..

 

You seem to be tying your self in mathematical knots here. Forget about fluids and go back to Calc I.

If you have a function ##f## defined on ##0 \le x \le 1## by ##f(x) = x^2## (and not defined if ##x < 0## or ##x > 1##), then the derivative is ##f'(x) = 2x## when ##0 \le x \le 1##. Nothing special happens at the end points.

The limit $$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$ only makes sense when both ##0 \le x \le 1## and ##0 \le (x+h) \le 1##, but that isn't a problem. ##h## can be negative.

You can call it a "one-sided derivative" at the end points it you like, but it behaves just the same as a "two sided derivative".

 

Well you said you were ignoring the effect of the air, so in this hypothetical situation the flowing high-viscosity fluid would simply not feel the effect of there being air. That is a non-issue. I think the problem is you are formulating the thought experiment based on ignoring the effect of air, and then running into trouble when you try and figure out afterward why the air is not affecting your flow. In this case, the shear stress in the oil or water or whatever your more viscous fluid happens to be will not care whether the velocity gradient in the air is defined since you are ignoring the presence of the air entirely and the shear stress will depend on the velocity gradient in the oil or water.

Now, if you didn't ignore the air, you would have a small bit of effect from the air on the water. The phenomenon you describe is called a free surface, and at that surface, one of the boundary conditions is that the stress be continuous across the surface. In situation, the shear stress is not necessarily nonzero (though it can be). In this case, you can still get the shear stress in the flowing oil or water solely from the velocity gradient in that fluid (assuming you know it), it will simply likely be nonzero.

You may be thinking of (or interested in reading about) the Kelvin-Helmholtz instability, which would absolutely apply in the case of the situation proposed by the OP, and in this situation the shear stress would certainly be finite. You can't make the assumption that the OP does that you ignore the effects of the air, however.

 

I think we've all been there before. Hopefully you've got it all straightened out now.

 

You were right up until this point. While this is the case in many situations (e.g. water flowing over a stationary surface), it is not necessarily true. In particular, it is not necessarily true in the case the OP mentioned where the bottom surface is moving. That surface could very well be moving faster than the water at the interface is moving, in which case it would be a minimum.

And the typo here. I think you mean it varies linearly with [itex]\partial u/\partial y[/itex]. Just clarifying for the OP.