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Is it in the direction of the velocity ?

If it is, can anybody describe it somehow that matches my physical feeling?

How the shear stress affects another layer of fluid ?

I faced a question today that fluid is moving alongside a horizontal tube and and if we write momentum equilibrium we must omit gravity force because it is not in the direction of the velocity, but what about shear stress ?

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2 Answers 2

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Since you are talking about layers of liquid, i am going to assume you are talking about incompressible, laminar pipe flow (i.e. $\mathrm{Re}\ll2000$).

If it is, can anybody describe it somehow that matches my physical feeling?

Shear stress is simply friction between layers of fluid. Imagine rubbing your flat hand across each other, it should be relatively easy to understand that the friction force is in the direction of the movement. This is the same with shear stress. It maybe more convincing to understand that shear stress has units of pressure which can be understood as the friction force exerted per $m^2$ of fluid layer.

How the shear stress affects another layer of fluid ?

Physically, shear stress is the diffusive transport of momentum. As is well-known, diffusion occurs from a high to low concentrations of mass/energy/momentum. More accurately the diffusion transport $j_{\alpha}$ occurs in the direction of a negative mass/energy/momentum concentration gradient:

$$j_{\alpha,m}=-D\partial_{\alpha}\rho \quad j_{\alpha,T}=-k\partial_{\alpha}T \quad j_{\alpha,v_{\beta}}=-\mu\partial_{\alpha}v_{\beta}$$

I write mass/energy/momentum because their diffusive transport is physically similar and there is a clear analogy. The quantity $j_{\alpha,v_{\beta}}$ is what we call the stress; it can be shearing ($j_{y,v_x}$ or $j_{x,v_y}$) but also something else, e.g. ($j_{x,v_x}$ or $j_{y,v_y}$). This something else is a normal (as opposed to shear) contribution to the stress (much like a pressure) which may be important in turbulent or compressible flows. Since we are dealing with incompressible flow you can neglect it.

Now, imagine the pipe and that we have an initially uniform flow; all layers are moving at the same uniform velocity so there is no friction between layers; another way to see this is the gradients are zero so there is no diffusion of momentum between layers. If we now zoom in on the wall, assuming a no-slip condition where the velocity is zero at the wall, we see that there is obviously friction between the wall and the next nearest layer. The next nearest layer will be slowed down by this because it has lost some of its momentum and this has caused friction with the layer above it because that is still going at the uniform speed. This continuous until it reaches the layer exactly at the center of the pipe which sees a layer above and below it with exactly the same (slightly lower) velocity (also known as a symmetry condition). At this point the diffusion has reached an equilibrium and the flow has reached a steady-state. From the perspective of diffusion, friction at the wall is a momentum sink (i.e. source of low momentum concentration) and frictionless layers at the center of the pipe are a momentum source (i.e. source of high momentum concentration).

Now let's bring it all together.

Is it in the direction of the velocity ?

If you read the pipe flow story carefully you should realize that, while the flow is in the $x$-direction, the diffusive transport of momentum is from the center of the pipe to the wall, i.e. in the $y$-direction. Therefore, the shear stress that is relevant in your equations is:

$$j_{y,v_x} = -\mu\partial_y{v_x}$$

which exerts a friction force in the $x$-direction so belongs in the $x$-component of the force balance, but varies in the $y$-direction.

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  • $\begingroup$ I like your answer but don't see how it 'matches a physical feeling'. If anything you've introduced concepts that must be more abstract to the OP. And terms like "diffusive transport of momentum" are hardly intuitive. It's bizarre you berate my simpler answer, then go on to formulate one that is more exact and mathematical than mine! $\endgroup$
    – Gert
    Commented Oct 30, 2015 at 16:39
  • $\begingroup$ @Gert - My assumption was that since he's able to write down a force balance for a fluid (according to his question), he would know the difference between convective and diffusive transport. I don't think that is unreasonable to assume in the subject of transport phenomena. Furthermore, I have used natural laws (Fick/Fourier/Newton) to make my point which is like using $F=ma$ to make a point about projectile motion. If my comments sound(ed) berating that is not my intention and i apologise (i don't mean to start a flame war), but personally i think my comments sound quite calm. $\endgroup$
    – nluigi
    Commented Oct 31, 2015 at 11:38
  • $\begingroup$ @nluigi thanks a lot for your answer,now i correctly understand the directions :) $\endgroup$
    – Mo Samani
    Commented Nov 2, 2015 at 22:17
  • $\begingroup$ @nluigi can you explain for me why the force of fluids on walls of the tube or another surface is always positive? i tried to calculate the force of fluid on the walls and i calculated (shear stress quantity at position R) * (surface of the wall which fluid runs to),but i got a negative force,then i checked the solutions and saw it used -shear stress *surface , i checked some other questions and saw whenever the force is negative using +shear stress, the solution is using -shear stress to obtain a positive one, actually why it happens? thank you. $\endgroup$
    – Mo Samani
    Commented Nov 2, 2015 at 22:25
  • $\begingroup$ @MohammadHoseinKhalili why do you say it's always positive? For a parabolic profile, the gradient at one end will be positive, while at the other end it will be negative. The forces are indeed the shear stress multiplied with the surface area which will be positive on one end and negatives in the other. $\endgroup$
    – nluigi
    Commented Nov 2, 2015 at 22:46
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It should help to go back to a definition here. Consider the diagram below:

Shear stress diagram.

A material is enclosed between two parallel plates with surface area $A$ and distance $h$ apart from each other.

Assume that a force $F$ is applied to the top plate which now starts moving at constant speed $v$ (the bottom plate is kept stationary).

The equation of motion for this situation is:

$$F=\mu A\frac{\partial v}{\partial h}.$$

We can also re-write this as:

$$\sigma = \mu \gamma.$$

Where $\sigma$ is the shear stress, $\gamma$ the shear rate and $\mu$ the dynamic viscosity.

I hope this helps.

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  • $\begingroup$ I don't think this intuitively explains why the shear stress is in a particular direction. The equation you show is the result of the fact the shear stress is in a particular direction not the explanation. $\endgroup$
    – nluigi
    Commented Oct 30, 2015 at 12:57
  • $\begingroup$ @nluigi: I'm not sure what you understand by 'intuitively'. Surely it's better to have an exact understanding of something than an 'intuitive' one? At the end of the day physics is an exact and mathematical description of nature, nothing more and nothing less. $\endgroup$
    – Gert
    Commented Oct 30, 2015 at 14:53
  • $\begingroup$ The question ask for a description which matches his 'physical feeling'; i understood he wants an intuitive way to understand shear stress (and specifically the direction in which it acts). I think for 'complex' physical phenomena, an intuitive understanding is important because it often leads to an exact mathematical understanding. Wouldn't you agree that Schrödinger's cat has contributed to a better understanding of Quantum Mechanics? Thought experiments like that are an intuitive view of an otherwise difficult to understand topic. $\endgroup$
    – nluigi
    Commented Oct 30, 2015 at 15:13
  • $\begingroup$ A 'physical feeling' is not something we can deal with on a site that's about understanding of physics. Schrodinger's cat is a thought experiment that deals with a particular interpretational problem of QM. Personally I can't say it contributed much to my understanding of QM. It's not very relevant today either. $\endgroup$
    – Gert
    Commented Oct 30, 2015 at 15:20
  • $\begingroup$ Maybe a bad choice of thought experiment but the point still stands that, even if it didn't contribute to your understanding or isn't relevant today, it was relevant in the past because the understanding of QM wasn't what it is now, therefore it has contributed. If a 'physical feeling' is something we can't deal with then this question is off-topic (opinion-based or something related) but i disagree with that as giving an intuitive interpretation of a problem can be important in communicating scientific ideas which is exactly what we do on this site. I guess we will have to agree to disagree! $\endgroup$
    – nluigi
    Commented Oct 30, 2015 at 15:55

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