I'm trying to understand an 'approximate derivation' of the flow rate of gas down a pipe at low pressure from my book (an example problem). It uses the result that the flux of a gas is given by $\displaystyle{F=\frac{n\langle v \rangle}{4}}$ where $n$ is the number density of the gas and $\langle v \rangle$ is the mean speed of the gas molecules. The idea is that because the pressure is low, we can treat this using effusion (mean free path is much less than the diameter).

I'll write down the first section, which I can't understand the logic behind.

Problem: Estimate the mass flow rate of gas down a long pipe of length $L$ and diameter $D$ at very low pressures in terms of the difference in pressures ($P_1-P_2$) between the two ends of the pipe.

Solution: This type if flow is known as Knudsen flow. At very low pressures, molecules collide with the walls of the tube much more often than they do with each other. Let us define a coordinate $x$, which measures the distance along the pipe. The net flux $F(x)$ of molecules flowing down the pipe at position $x$ can be estimated by subtracting the molecules effusing down the pipe since their last collision (roughly a distance $D$ upstream) from the molecules effusing up the pipe since their last collision (roughly a distance $D$ downstream). Thus

$$\displaystyle{F(x)=\langle v \rangle\frac{[n(x-D)-n(x+D)]}{4}}$$ where $n(x)$ is the number density of molecules at position $x$.

So I don't really understand why we may estimate the flux at $x$ via subtracting the flux at $x+D$ from the flux at $x-D$.

Later it states that in the steady state, $F(x$) is the same all along the tube, which I understand, but then how does this fit in with the above, where if this was the case, we could get $F(x)=0$ because it is the flux at $x+D$ minus the flux at $x-D$ which should be equal.

Thanks for any help in advance...


1 Answer 1


The density of the gas changes along the length of the pipe - at higher pressures we think of this as the pressure drop along the pipe, and if temperature is constant this means that a drop in pressure is equivalent to a drop in density. That is why $F(x)$ is not zero. (Note - in your last paragraph you stated that the flux is the same everywhere; that's true, but it does not mean that the number density needs to be the same everywhere. It's OK to have a gradient in density.)

At any point, particles can hit the wall, and when they do, they will scatter in a random direction - some forward, some backward (equal amounts of each since they don't notice the pressure differential at the time of collision - remember this is low pressure, so long mean free path). Looking ahead in the pipe (dowstream of the observer), we can see particles "scattering back upstream"; from behind us (upstream of the observer), particles are scattering downstream. The net flux is the difference between particles flowing downstream and upstream. This is expressed by the equation given:

$$F(x) = <v> (n(x-D) - n(x+D))$$

Note that the use of the distance $x-D$ is not explained very rigorously - simply "roughly how far away the last interaction occurred". You really need to do an integral to convince yourself this is the right distance - but I don't believe that's what you were asking for, right?

  • $\begingroup$ My problem is that if we have say a gas in a container, then the flux through a small hole is simply n< v >/4 where n is the number density in the container. So here, why can we not just say the flux down the pipe is F(x)=n(x)< v >/4 (you forgot to divide by 4 on your F(x) by the way)? Also, if only half move upstream, and half downstream, why do you not need to divide each number density by 2? $\endgroup$
    – Watw
    Commented Dec 27, 2014 at 12:35
  • $\begingroup$ Also since number density changes, shouldn't $\left< v \right>$ increase downstream of the pipe to maintain constant flux? $\endgroup$
    – Deep
    Commented Aug 13, 2016 at 7:24
  • $\begingroup$ @Zero yes, but given the other approximations made in the "derivation" (which is explicitly called "approximate")I think that is probably a second order effect. $\endgroup$
    – Floris
    Commented Aug 13, 2016 at 13:21

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