Fluid Mechanics: Friction of a moving surface

Hey! So for newtonian fluids, the shear-stress is given by
$$\tau = \mu \frac{du}{dy}$$

Now, let's assume you have water flowing horizontally in laminar steady state motion, where its surface is also moving. Above it is air. So if we disregard the friction from air, the shear-stress will become zero for the surface.

But why? the formula given above is not even defined for the surface layer, and besides you have shear-stress from the water below the surface as well... So what's up?

Gold Member
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.

arildno
Homework Helper
Gold Member
Dearly Missed
" 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.

Gold Member
arildno said:
Without any driving gradient, either from pressure or gravity?

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

Sorry. I was too quick when writing the OP so I forgot that water has too little viscosity to make that example intuitive. I'm just too much of a newb in fluid dynamics. sry.

Anyway, forget the water. I'll be more rigorous this time:

Let's say it's some kind of fluid with a very high viscosity - like oil or something, flowing horizontally (including the surface). Driving its motion is a flat plate under it which is moving with a velocity V. Above the fluid you have air. Assume steady state and laminar flow. Why will the surface-layer not experience shear-stress, if we ignore the air?

I assume the fluid-blocks making up the surface-layer will move with constant speed, carried by the fluid below them? Thus the answer to my question is: Since we ignore the effect of the air, the blocks will not be deformed and thus the shear-stress is defined as zero??

If that is the case then the shear at the interface in the water is zero because there is no velocity gradient.
How can the gradient (I assume you mean the horizontal velocity differentiated with respect to the vertical direction?) of the velocity be zero, when there is no fluid above the interface? I mean, the function is undefined in that area, isn't it?

Last edited:
arildno
Homework Helper
Gold Member
Dearly Missed
It might be in that case that a strictly horizontal surface is an unstable feature, and ripples will form at the surface.

AlephZero
Homework Helper
How can the gradient (I assume you mean the horizontal velocity differentiated with respect to the vertical direction?) of the velocity be zero, when there is no fluid above the interface? I mean, the function is undefined in that area, isn't 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".

Indeed this derivative is valid only for the direction downwards, and not valid for the direction upwards (because x+h does not exist). Yet I believe bonehead is saying the reason there is no shear-stress for the interface, is because the upwards derivative is zero..

This is exactly what makes me confused: How can there be no shear-stress when the derivative in the downwards direction exists and is non-zero?

Gold Member
Nikitin said:
Let's say it's some kind of fluid with a very high viscosity - like oil or something, flowing horizontally (including the surface). Driving its motion is a flat plate under it which is moving with a velocity V. Above the fluid you have air. Assume steady state and laminar flow. Why will the surface-layer not experience shear-stress, if we ignore the air?

Nikitin said:
Indeed this derivative is valid only for the direction downwards, and not valid for the direction upwards (because x+h does not exist). Yet I believe bonehead is saying the reason there is no shear-stress for the interface, is because the upwards derivative is zero..

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.

arildno said:
It might be in that case that a strictly horizontal surface is an unstable feature, and ripples will form at the surface.

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.

okay I think I understand it now. my problem was a very stupid confusion about the math, completely ignoring the physical aspect.

thanks for the help guys. It's a trivial thing - i get that now. It's just that my brain is getting stupider due to overwork.

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

Chestermiller
Mentor
Here's another way of looking at it. The shear stress in the water at the interface has to match the shear stress in the air. Since the viscosity of the air is very low compared to that of the water, you can neglect the frictional drag of the air. Therefore, the shear stress in the water at the interface must be virtually zero. This means the du/dy in the water must be zero at the interface, and, thus, the water velocity must be maximum at the interface. The shear stress in the water varies linearly with y, from a maximum magnitude at y =0 to a value of zero at the interface (say for steady state flow down an inclined plane).

Chet

Gold Member
Chestermiller said:
the water velocity must be maximum at the interface.

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.

Chestermiller said:
The shear stress in the water varies linearly with y

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

Chestermiller
Mentor
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.
Oh. You're right. I was thinking flow down an inclined plane.

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

Yes. Under transient conditions, the shear stress would not vary linearly with y and neither would $\partial u/\partial y$. At final steady state, the shear stress would be zero throughout the water layer, and the water velocity would be uniform at the bottom wall velocity.