1
$\begingroup$

Say that the only pressure differences in the atmosphere was due to the gravitational pull of the earth. That is, say pressure only changed as a function of height.

So we're assuming there's no wind, no differences in humidity, equal composition of air everywhere, the atmosphere is at the same temperature everywhere...etc.

I'm told that in any fluid, the particles are always in random motion, smashing into one another and smashing into whatever surface we insert into the fluid, and that's what causes pressure.

But, how "much" motion" is this "random motion" really?

In this hypothetical scenario, would an air particle from the top of Mount Everest ever make its way down to sea level?

Would air's random motion be mostly horizontally, or vertically? (I think vertically...but I'm not sure...)


Just for context, I'm asking this question because I'm struggling to understand pressure, and especially its relation to gravity.

Thanks!!!

$\endgroup$
  • $\begingroup$ If the motion of the air is mostly downwards, how much air has to go the other way? $\endgroup$ – user207455 Jul 4 '19 at 17:39
  • 1
    $\begingroup$ I think the concepts you are looking for are “random walk” (en.wikipedia.org/wiki/Random_walk) and “mean free path”. (en.wikipedia.org/wiki/Mean_free_path). $\endgroup$ – G. Smith Jul 4 '19 at 17:42
  • $\begingroup$ @SolarMike what do you mean? $\endgroup$ – joshuaronis Jul 4 '19 at 17:43
  • $\begingroup$ @G.Smith Thanks, but I'm not asking about in between collisions, I mean after a LOONG time, how far will an air particle have moved? $\endgroup$ – joshuaronis Jul 4 '19 at 17:44
  • 1
    $\begingroup$ The distance will be proportional to the square root of the time. This is a characteristic of random walks and diffusion. $\endgroup$ – G. Smith Jul 4 '19 at 17:46
1
$\begingroup$

When trying to understand the relative dominance of diffusion vs. advection in a hydrodynamic system, we usually compare the characteristic length scales associated with each process and define their ratio as a dimensionless group often referred to as a Peclét number. This involves the following steps:

  1. Determine what is the system you want to analyze. Are you interested in the motion of a single particle or of a chemical species in a fluid? Are you interested in motion happening in one dimension or in three? Are you looking at the displacements occurring over a fixed interval of time or the displacements that happened within a fixed length/volume?
  2. Estimate the diffusive/advective displacements. Diffusion is almost always a result of Brownian motion, which means that the root-mean-square displacement occurring through diffusion is proportional to the square root of the elapsed time $t$ multiplied by a diffusive constant $D$ (with some numerical factor depending on the dimension of the problem). Advective displacement is almost always the product of the bulk fluid velocity $u$ and the elapsed time $t$; this needs to be altered if the speed is not uniform or changing with time.
  3. Take the ratio and simplify. Using the expressions above, divide the advective displacement by the diffusive displacement and simplify the expression⁠—chances are some terms will simplify or cancel out. This ratio (the Peclét number) now indicates to you which parameters in your physical system influence the balance of advective/diffusive displacement and which don't.

As an example, consider a molecule that is part of a fluid flowing unidirectionally and steadily in 3-D⁠—such a molecule undergoes both bulk advective motion and random Brownian motion. If we wanted to compare the total diffusive and advective displacement taken by that molecule over some specific time interval, we would find the RMS diffusive displacement in 3D to be $d_{diff} = \sqrt{6Dt}$, where $D$ is the diffusivity constant associated with the molecule. The advective displacement would simply be $d_{adv} = ut$ where $u$ is the bulk fluid velocity.

Hence, when taking the ratio of each, we find

$$Pe = \frac{d_{adv}}{d_{diff}}=\frac{ut}{\sqrt{6Dt}}=u\sqrt{\frac{t}{6D}}$$

The most common definition of the Peclét number is for 1-D motion over a fixed length interval, in which case the $6$ above becomes a $2$ and the $t$ is usually replaced with $\frac{L}{u}$, where $L$ is the length of the interval.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you! +1 !!! I bet your answer is amazing, but it's going right over my head! I've been learning fluids for like a day. Could you maybe summarize a little in an edit? $\endgroup$ – joshuaronis Jul 5 '19 at 22:32
  • $\begingroup$ @G.Smith you were interested in other's answers - do you understand this? $\endgroup$ – joshuaronis Jul 5 '19 at 22:33
  • $\begingroup$ Certainly @JoshuaRonis! $\endgroup$ – aghostinthefigures Jul 5 '19 at 22:34
-1
$\begingroup$

But, how "much" motion" is this "random motion" really?

So basically the average velocity of a molecule is the speed of sound. That should not be surprising. But, of course, it keeps bumping into things. So that's a random walk.

If one assumes a random walk, then the distance travelled over time is d=(6Dt)^1/2, where D is the diffusion coefficient, which for oxygen at STP is ~0.18cm^2/s.

So, solving, that's about 4 seconds per cm.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ There isn't a linear relationship between $d$ and $t$, so "4 seconds per cm" doesn't make sense. $\endgroup$ – G. Smith Jul 4 '19 at 18:55
  • $\begingroup$ true, but that's one solution. $\endgroup$ – Maury Markowitz Jul 4 '19 at 18:59
  • $\begingroup$ No, it isn’t a solution, and it is extremely misleading. Calculate $t$ for a $d$ of 8000 meters, and then compare that with the time to go 8000 meters at 1/4 cm/s. $\endgroup$ – G. Smith Jul 4 '19 at 19:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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