# I Perpendicular loading exerted by a moving fluid to a solid

Tags:
1. Dec 17, 2016

### vco

Picture water flowing in a duct. In these kinds of scenarios, the loading exerted by the moving fluid to the solid in the perpendicular direction is often taken as being equal to the pressure of the fluid at the wall.

I agree that the above assumption is true for a stationary fluid. For a moving fluid, however, should not the perpendicular loading be equal to the corresponding normal stress component of the fluid at the wall?

Since pressure is defined as the negative average of the normal stresses, surely the pressure and the aforementioned normal stress component are most often not equal for a moving fluid? Or is the difference just very small so that the assumption is justified?

Last edited: Dec 17, 2016
2. Dec 17, 2016

Assuming you mean perpidicular to the wall, then years, it is the normal stress component of the fluid at the wall. Of course, the normal stress component of a moving fluid is the pressure for a Newtonian fluid.

3. Dec 17, 2016

### vco

Yes, by perpendicular I mean perpendicular to the wall. Could you elaborate how the normal stress component of a moving fluid is the pressure for a Newtonian fluid? I thought this only holds for a fluid at rest.

4. Dec 18, 2016

How familiar are you with index notation? That's the easiest way to show this. Generally speaking, for a Newtonian fluid, the constitutive equation describing the stress is
$$\tau_{ij} = -p\delta_{ij} + \mu\left( \dfrac{\partial u_i}{\partial x_j} + \dfrac{\partial u_j}{\partial x_i} \right) + \delta_{ij}\lambda\nabla \cdot \vec{v}.$$
If you take all of the normal stresses and sum them and divide by 3 (in other words, if you average all of the normal stresses) then you can define what is sometimes called the mechanical pressure
$$\bar{p} = -\dfrac{1}{3}\left( \tau_{xx} + \tau_{yy} + \tau_{zz} \right) = -\dfrac{1}{3}\mathrm{tr}(\tau_{ij}).$$
This makes sense to do because $\mathrm{tr}(\tau_{ij})$ is a tensor invariant and so is independent of rotations (the choice of coordinate systems). The end result is that you get
$$\bar{p} = p - \left(\lambda + \dfrac{2}{3}\mu\right)\nabla\cdot\vec{v}.$$
So, in the most general sense, no, the mechanical pressure (the actual isotropic normal stress in a moving fluid) is not equal to the thermodynamic pressure, $p$. Further, the second coefficient of viscosity ($\lambda$) is a very nebulous property that is not all that well-understood, even after over a century of viscous flow research. The primary reason for that is that it just isn't generally very important in practice. Consider the following two situations:
1. Stokes' hypothesis: $\lambda + \frac{2}{3}\mu = 0$
Stokes made this assumption some 170 or so years ago and it is still commonly made today, and in most cases, it works well. It hasn't really been proven or even supported all that strongly in any experiment, which is why it is still called a hypothesis, but practical calculations made using that assumption seem to match reality an overwhelming majority of the time. The only major hole in it that I know of is attenuated while propagating through a fluid. In that case, this hypothesis does not work, but most other cases it is sufficient.
2. Incompressible flow: $\nabla\cdot\vec{v} = 0$
Obviously in this case the term containing $\lambda$ is zero and $\bar{p} = p$. This is effectively true in any liquid and any incompressible gas flow. Nothing is every perfectly incompressible, so it's still just an approximation, but decades of experimental results have by and large borne this out that $\bar{p} = p$. In flows that are compressible, $\nabla\cdot\vec{v} \neq 0$ and the bulk viscosity can be important sometimes, but in the overwhelming majority of cases, the viscous normal stresses are effectively zero and $\bar{p}=p$ still holds. The primary exception is in shocks, in which case $\bar{p}\neq p$.

Anyway, the short answer, in case all of that was a bit beyond your familiarity level with fluid mechanics, is that the $\nabla\cdot\vec{v}$ is almost always negligibly small, whether due to incompressiblity or luck that Stoke's hypothesis holds. In the rare cases where it does have considerable magnitude, then you are correct that the actual mechanical pressure, $\bar{p}$, cannot be treated as being equal to $p$.

There is a bit more discussion in https://www.amazon.com/dp/0072402318 and some more in both https://www.amazon.com/dp/0750627670/ and in https://www.amazon.com/dp/1118013433/.

Last edited by a moderator: May 8, 2017
5. Dec 18, 2016

### vco

Thanks for the long answer and reminding me of the related fact of the mechanical pressure generally not being equal to the thermodynamic pressure. I am quite familiar with the index notation. However, I still do not understand why the pressure (be it mechanical or thermodynamic) should be treated as equal to the normal stress perpendicular to the wall.

Let's assume that the flow is incompressible. Now the thermodynamic pressure equals the mechanical pressure and also the constitutive relations are simple. Let's choose that direction 1 is the main flow direction and direction 2 is perpendicular to the wall. In this case, the normal stress perpendicular to the wall is
$$\tau_{22} = -p\ + 2\mu\dfrac{\partial u_2}{\partial x_2}.$$
Clearly, the normal stress is not generally equal to the pressure since the last term can be non-zero? I have a feeling that I am missing something very basic here.

Edit: Clarifications.

Last edited: Dec 18, 2016
6. Dec 18, 2016

### vco

Actually, the basic thing that I am missing here might be that the velocity is constant (zero) along the wall. Therefore, the spatial derivative of the main flow velocity along its direction is zero at the wall: $$\dfrac{\partial u_1}{\partial x_1}=0.$$ And, assuming that direction 3 is parallel to the wall (like direction 1), the corresponding spatial derivative of that velocity component is also zero at the wall: $$\dfrac{\partial u_3}{\partial x_3}=0.$$ Since it was assumed that the flow is incompressible ($\nabla\cdot\vec{v} = 0$), the spatial derivative of the velocity component perpendicular to the wall must also be zero in its direction at the wall: $$\dfrac{\partial u_2}{\partial x_2}=0.$$ Because $\dfrac{\partial u_2}{\partial x_2}=0$, the perpendicular normal stress at the wall simplifies to $$\tau_{22} = -p.$$ So, at least for an incompressible flow, the claim that the perpendicular normal stress at the wall equals the pressure appears indeed to be true. To me, it appears to be a lucky coincidence since pressure is something we can measure from a flow.

However, the claim that "the normal stress component of a moving fluid is the pressure for a Newtonian fluid" does not appear to be generally true. It seems to be true at the wall for an incompressible flow, though.

Last edited: Dec 18, 2016
7. Dec 18, 2016

### vco

To sum up my above post:
It appears to me that at the wall for an incompressible flow, the normal stresses perpendicular and parallel to the wall have a magnitude equal to the pressure. This is the reason why we can treat the pressure as the loading that is exerted to the solid perpendicular to it.

8. Dec 19, 2016

### Staff: Mentor

This sounds correct. Still, since for an incompressible fluid, the pressure is determined only up to an arbitrary constant (which must be determined in practice from the boundary conditions for the flow), the quantity that we call the pressure p should not be regarded as the thermodynamic pressure.