1
$\begingroup$

Consider the simple problem of a flow between two plates, one at $x_2=0$ and one at $x_2=h$ with the bottom one held stationary and the top plate moving in the $x_1$ direction with velocity $V$. Also, the velocity field is in direction $e_1$: $v(x,y)=v_1(x_2)e_1$ with boundary conditions: $$v_1(0)=0$$ $$v_1(h)=V$$

Assuming no pressure drop in the $x_1$ direction, we obtain from Navier Stokes $$v_1(x_2)=\frac{V}{h}x_2$$

We can now see the stress tensor in matrix form

$$[T]=- \pi \begin{bmatrix}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1 \end{bmatrix} + \frac{\mu V}{h} \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$

So far so good.

Question: I can't understand the following sentence, which I found on Gurtin's book:

The force per unit area exerted by the fluid on the to plate has a tangential shearing component $-\frac{\mu V}{h}$

Indeed my professor drew the following picture (notice the arrow from right to left)

enter image description here

I think this is because by definition the force per unit area exerted by the fluid on the top plate is given by $T (-e_2)= (\frac{-\mu V}{h},\pi,0)$. Now, I see that the first component of this vector is precisely what the book states, i.e $\frac{- \mu V}{h}$. Why is the first component referred to the shearing?

$\endgroup$
6
  • $\begingroup$ What particle?.. $\endgroup$ May 18, 2021 at 21:16
  • $\begingroup$ Sorry, I meant "on the top plate", I've just fixed it @ChetMiller $\endgroup$ May 18, 2021 at 21:29
  • $\begingroup$ The force per unit area component you are referring to is the force per unit area exerted by the fluid on a plane of constant x2 in the x1 direction. That is why it is referred to as a shear force. The second component in the vector is the force per unit area exerted by the fluid on the plane of constant x2 in the x2 direction. It is a normal force. $\endgroup$ May 18, 2021 at 23:09
  • $\begingroup$ @ChetMiller Thanks for your comment. I still have a minor question about this: Why do I have to take $-e_2$ as normal direction when I want to compute the stress, and not $e_2$? I mean, if the fluid is pushing to the plate, I'd say the outer normal is $e_2$, not $-e_2$. $\endgroup$ May 19, 2021 at 7:41
  • $\begingroup$ Are you familiar with the Cauchy Stress Relationship? If so, please state it according to your understanding in your own words. $\endgroup$ May 19, 2021 at 10:24

1 Answer 1

1
$\begingroup$

If you want to find the force per unit area that the fluid exerts on the plate, the outward normal is directed from the plate to the fluid; that would be -e2. More, specifically, at a surface S, the force per unit area exerted by the material on side B of the surface on the material on side A of the surface is obtained by drawing a unit normal n from side A to side B, and matrix multiplying this normal by the stress tensor matrix Tn. The normal is drawn from the side receiving the force to the side exerting the force.

$\endgroup$
12
  • $\begingroup$ Thanks @ChetMiller for your answer. Actually, I'd like to understand why the normal is drawn from the side receiving the force to the side exerting the force. $\endgroup$ May 19, 2021 at 12:22
  • $\begingroup$ For that, you would have to go back to the derivation of the Cauchy Stress Relationship. When you are working with the "tensile stress tensor," this is the way it works out. If you were working with the "compressive stress tensor" (which is opposite in sign to the compressive stress tensor), you with use the normal in the opposite direction. Bird, Stewart, and Lightfoot (Transport Phenomena) uses the compressive stress tensor. $\endgroup$ May 19, 2021 at 12:35
  • $\begingroup$ Okay, so if I set $n = -e_2$, then the positive part is the inside of the channel, while the negative part is the "outside". By definition, $s(n)$ gives the force per unit area exerted by the positive part on the negative part, so the force exerted by the fluid on the wall. Right? $\endgroup$ May 19, 2021 at 13:35
  • $\begingroup$ If I understand you correctly, yes. $\endgroup$ May 19, 2021 at 14:01
  • 1
    $\begingroup$ Probably you read right $-e_2$, but I edited after some seconds with $e_2$. Thanks Chet :-) $\endgroup$ May 19, 2021 at 15:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.