Understanding Free Expansion of Gas: Velocity Distribution Function Evolution

[thepolishman]

TL;DR Summary

Evolution of velocity distribution of a gas undergoing free expansion

Hey all. I'm trying to understand the evolution of the velocity distribution function of a gas undergoing free expansion. I know that at t=0, the velocity distribution function is given by $u(t=0, x)=Ax^{1/2}$, where $A$ is a known constant.

From the momentum equation, I have:

$du/dt+u du/dx=0$

Using separation of variables, I get:

$u=F(x)G(t)$

$FdG/dt=-FG^2dF/dx$
$1/G^2dG/dt=-dF/dx$

$1/G^2dG/dt=M$
$-dF/dx=M$

$F=-Mx+f(t)$
$G=-1/(Mt+g(x))$

$u=(Mx+f(t))/(Mt+g(x))$

Using the initial condition that $u(t=0, x)=Ax^{1/2}$, I get:

$u(t=0, x)=(Mx+f(0))/g(x)$
$Ax^{1/2}=(Mx+f(0))/g(x)$

From which I get that:

$f(0)=0, g(x)=x^{1/2}, M=A$

Thus, my velocity distribution becomes:

$u=(Ax+f(t))/(At+x^{1/2})$

Assuming I did everything correctly until this point, my issue is with figuring out what f(t) is. Anyone know how I can proceed?

 

[erobz]

You have that:

$$G(t) = -\frac{1}{Mt + G(x) }$$

But don't you instead need:

$$G(t) = -\frac{1}{Mt + G(t=0) }$$

$G$ is supposed to just be a pure function of time? You have it as a function of both time and position?

Same with $F$. You have it as a function of time, when it should be just a function of $x$?

I know very little about PDE's though, so perhaps I'm not catching what you are doing (or I don't understand the technique).

Either way, it would help others ( who definitely know what they are doing) if you could convert your write up to LaTeX. That much I know.

 

[thepolishman]

I'm in the same boat, not great at PDEs. I also initially took the approach that $g(x)$ and $f(t)$ were simply constants, but then there was no way to match the result with the initial condition. I may just be approaching this problem incorrectly.

 

[erobz]

I'm in the same boat, not great at PDEs. I also initially took the approach that g(x) and f(t) were simply constants, but then there was no way to match the result with the initial conditions. I may just be approaching this problem incorrectly.

I’m just thinking that you are looking for a solution where $F$ and $G$ are purely functions of $x$ and $t$ respectively. So I think the idea here is just solving two regular ODE’s with initial condition $F(x=0)$ and $G(t=0)$. But I could be mistaken.

 

[thepolishman]

I’m just thinking that you are looking for a solution where $F$ and $G$ are purely functions of $x$ and $t$ respectively. So I think the idea here is just solving two regular ODE’s with initial condition $F(x=0)$ and $G(t=0)$. But I could be mistaken.

Right, but doing that gives:

$F=-Mx+C$

But since the initial condition is:

$u=Ax^{1/2}$

There doesn't seem to be any way for me to match the initial condition and the function $F$ since one has an $x$ with an exponent of 1/2 and the other an exponent of 1.

 

[kuruman]

The problem is that you start assuming that you can separate functions of x and t only, but once you do that you mix the two back in. When you say $u = F(x)G(t),$ you assert that $F$ is a function of $x$ only and $G$ is a function of $t$ only. Then you separate variables and come up with $$\frac{1}{G^2}\frac{\partial G}{\partial t}=-\frac{\partial F}{\partial x}$$So far so good. On the left you have a function of $t$ only and on the right a function of $x$ only. This can only be if each side is equal to a constant which you call $M$. Great!

But - big but here - now you solve $$\frac{d F}{d x}=-M$$ and you come up with $$F=-Mx+f(t)$$But, - big but here - you have already stipulated when you separated variables the $F$ is a function of $x$ only. So what does $f(t)$ doing there? At best, it must be a constant, call it $C$ and proceed.

 

[erobz]

I get:

$$u(x,t) = \frac{Mx - C_f}{Mt + C_g}$$

Applying the initial condition, you have that

$$ u(x,0) = A x^{1/2} = \frac{Mx - C_f}{ C_g}$$

So, it does seem like you are an initial condition shy of a completing the solution?

 

[Frabjous]

Physically, separation of variables does not make any sense. Assume that you know F(x). G(t) is is essentially a “time varying magnitude“ and consequently the ”shape” of u does not change.

Also, using separation of variables implies F(x) equals the initial condition which does not satisfy the ODE.

I haven‘t tried to solve it, but I would expect wave-like solutions and would also need an EOS. There should be a pressure gradient term in your PDE.

 

[Chestermiller]

What makes you think that the initial velocity distribution for a gas experiencing free expansion is as you have specified, and what makes you think that, in a gas experiencing free expansion, viscous stresses are zero?

 

[thepolishman]

What makes you think that the initial velocity distribution for a gas experiencing free expansion is as you have specified, and what makes you think that, in a gas experiencing free expansion, viscous stresses are zero?

The initial velocity distribution is mostly just assumed to be a known distribution function. The distribution function I specified was just given as an example.

As for the viscous stress and pressure terms, I assumed them to be zero since the gas density is very low (<1E17 cm^-3 initially, and decreasing during expansion) and the background pressure is vacuum. Essentially, I assumed the gas to be non-interacting and all particles to be freely expanding away from a point source with a known initial velocity distribution. I tried approaching the problem initially from a 1D perspective to make things easier on myself.

That said, my approach may be incorrect.

 

[Chestermiller]

The initial velocity distribution is mostly just assumed to be a known distribution function. The distribution function I specified was just given as an example.

As for the viscous stress and pressure terms, I assumed them to be zero since the gas density is very low (<1E17 cm^-3 initially, and decreasing during expansion) and the background pressure is vacuum. Essentially, I assumed the gas to be non-interacting and all particles to be freely expanding away from a point source with a known initial velocity distribution. I tried approaching the problem initially from a 1D perspective to make things easier on myself.

That said, my approach may be incorrect.

Free expansion of a gas is not an inviscid deformation. At 1 bar, where air obeys ideal behavior almost perfectly, the mean free path is tiny. What is your estimate of the mean free path for molecules of air at 1 bar? Clearly, collisions play a large role in air free expansion from 1 bar.

 

[Sailor Al]

Hey all. I'm trying to understand the evolution of the velocity distribution function of a gas undergoing free expansion.

I wonder if you have thought about what you mean by velocity of a gas.
I know that this is heresy to the aerodynamicists, but I don't think it is a useful concept, or even if it exists at all.
Clouds have a speed: with a bit of trigonometry and the knowledge of their height and distance, we can use the speed of clouds to deduce the velocity of the wind.
We can measure the speed of an atmospheric low, by analysing the records of the various locations over which it has passed.
We are familiar with the meteorological forecasts of tomorrow's wind speed and direction.
But, take a closer look at that cloud. Its boundary is constantly changing: swirling and moving. If you look even more closely you will see that the cloud's boundary is not sharp at all, it is a gentle change from clear air, through mistiness to thick cloud (which, you recall from your flights, is only really a thick fog after all).
If you look closely, the front is moving at a different speed to the back, the top from the bottom and from side to side.
So whilst you may be able to determine its approximate velocity, even with the most precise measurements, the lack of defined, fixed boundaries make it impossible to make an accurate measurement of the wind in which it is moving.
So it is with the consideration of any parcel of air. Except on a still day or in a closed room, where the air is stationary and the velocity is zero, the concept of velocity of a gas is at best challenging and, I suggest, that at a scale of anything below tens of metres, is undefinable. (Even in a closed room, a light fog and a laser beam will reveal the the swirling motion of thermal convection.)
Certainly the idea of the velocity of a small, or infinitesimally small parcel of air that is the foundation of fluid dynamics is doubtful.
We know that a gas has density, pressure and temperature. But only when it is in a container does it have a volume (that of the container) and thus a mass. Until it is contained it doesn't have momentum, or velocity.
I don't think the concept of it having a velocity when undergoing "free expansion" has any meaning.

 

[russ_watters]

But, take a closer look at that cloud. Its boundary is constantly changing: swirling and moving. If you look even more closely you will see that the cloud's boundary is not sharp at all, it is a gentle change from clear air, through mistiness to thick cloud (which, you recall from your flights, is only really a thick fog after all).
If you look closely, the front is moving at a different speed to the back, the top from the bottom and from side to side.
So whilst you may be able to determine its approximate velocity, even with the most precise measurements, the lack of defined, fixed boundaries make it impossible to make an accurate measurement of the wind in which it is moving.
So it is with the consideration of any parcel of air. Except on a still day or in a closed room, where the air is stationary and the velocity is zero, the concept of velocity of a gas is at best challenging and, I suggest, that at a scale of anything below tens of metres, is undefinable. (Even in a closed room, a light fog and a laser beam will reveal the the swirling motion of thermal convection.)
Certainly the idea of the velocity of a small, or infinitesimally small parcel of air that is the foundation of fluid dynamics is doubtful.
We know that a gas has density, pressure and temperature. But only when it is in a container does it have a volume (that of the container) and thus a mass. Until it is contained it doesn't have momentum, or velocity.
I don't think the concept of it having a velocity when undergoing "free expansion" has any meaning.

Your comparison to a cloud is a good one, but you've interpreted it backwards. Clouds are many meters to several kilometers in size (length/width). Yes, assigning a velocity to a large flow field becomes problematic if you try to assign one velocity to the entire field, then zoom in on a small part of it. That's a pretty obvious flaw in the analysis of it, not a flaw in fluid/aero-dynamics. Heck, that flaw applies to analysis of basically anything with scale. The universe is mostly hydrogen but Earth has very little. Flaw in the Standard Model? No, it's a flaw in the analysis/logic.

[Computational] Fluid dynamics works. Aerodynamics works. I don't get how/why you can simply not believe in them.

 

[Chestermiller]

We know that a gas has density, pressure and temperature. But only when it is in a container does it have a volume (that of the container) and thus a mass. Until it is contained it doesn't have momentum, or velocity.

This is totally incorrect. A small volume of gas has mass, and thus specific volume whether a container is present of not. And temperature and pressure are measurable parameters of a gas, irrespective of the presence of a container. Moreover, what is being discussed in this thread is velocity distribution of gas molecules which move much faster than average velocities of air in a small scale parcel.

I don't think the concept of it having a velocity when undergoing "free expansion" has any meaning.

Also wrong. My advice is for you to take a course in fluid dynamics or gas dynamics. See Transport Phenomena by Bird et al.

 

[russ_watters]

I'm going to lock this as it is a bit of a hijack and necropost, and is also headed in a bad direction. @Sailor Al you may start a new thread asking for help learning about this subject, but it needs to be a sincere attempt at learning the known science, not questioning/challenging it.