What is the true nature of gas pressure and its energy mechanics?

[Chestermiller]

What adjacent parcels?

The figure above shows a chamber (the solid boundary) containing a gas, with the gas subdivided into smaller volumetric parcels by the hypothetical dotted boundaries. A and B are two adjacent gas volumetric parcels within the chamber. The gas in parcel A exerts a pressure force on the gas in parcel B at the area interface between these parcels (and vice versa). This happens because gas moleclues from A cross the interface from A to B (carrying momentum), but gas molecules from B cross this same interface in the opposite direction (carrying an equal amount of momentum). As far as the gas in parcel A is concerned, this is the same as if the molecules from A had bounded off a solid boundary at the interface.

 

[rogerk8]

Ok

 

[Chestermiller]

Today it however struck me that the minus sign must be due to this reasoning:

The normal unit vector points form the on-side to the by-side so if pressure is from the by-side to the on-side, there has to be a negative sign.

I am very uncertain about the terminology but feel like something like this must be true.

Yes. This is correct.

Now to my homework :)

I find it a hard one, but I happily will try to solve it.

We have
\vec{σ}=-p\vec{i}_x\vec{i}_x-p\vec{i}_y\vec{i}_y-p\vec{i}_z\vec{i}_z
and dotting this with
\vec{i}_x
from the right gives
\vec{σ}=-p\vec{i}_x
while the rest are ortogonal (is this an english word?) and thus zero.
To be a little lazy, dotting with $\vec{i}_y$ or $\vec{i}_z$ will in the same manner get the stress vector in each respective direction.

Yes. This is all correct. Well done.

So you see that, now matter which coordinate direction unit vector you dot the stress tensor with (in our case of hydrostatic equilibrium), you end up with a pressure force in that coordinate direction. This suggests that "pressure is isotropic."

Now for your next homework problem:

Suppose you have a differential element of area dA within a gas that is oriented spatially in such a way that the unit normal to the area is given by:
$$\vec{n}=n_x\vec{i}_x+n_y\vec{i}_y+n_z\vec{i}_z$$

What is the force vector exerted by the gas on the side of dA toward which $\vec{n}$ is pointing on the material on the side of dA from which $\vec{n}$ is pointing?

Chet

 

[rogerk8]

Yes. This is correct.

Yes. This is all correct. Well done.

So you see that, now matter which coordinate direction unit vector you dot the stress tensor with (in our case of hydrostatic equilibrium), you end up with a pressure force in that coordinate direction. This suggests that "pressure is isotropic."

Now for your next homework problem:

Suppose you have a differential element of area dA within a gas that is oriented spatially in such a way that the unit normal to the area is given by:
$$\vec{n}=n_x\vec{i}_x+n_y\vec{i}_y+n_z\vec{i}_z$$

What is the force vector exerted by the gas on the side of dA toward which $\vec{n}$ is pointing on the material on the side of dA from which $\vec{n}$ is pointing?

Chet

Chet,

Thank you for encourageing me!

I am very happy I was right :)

Getting down to my new lesson:

This first bold part is clear to me.

The other part is however a bit unclear.

Yet, I think you are just testing me :)

Quoting your formula:

\vec{dF}=(\vec{σ}\centerdot \vec{n}) dA

and if

\vec{n}=n_x\vec{i}_x+n_y\vec{i}_y+n_z\vec{i}_z

while using

\vec{σ}=-p\vec{i}_x\vec{i}_x-p\vec{i}_y\vec{i}_y-p\vec{i}_z\vec{i}_z

the force vector has to be

\vec{dF}=-pn_x\vec{i}_x-pn_y\vec{i}_y-pn_z\vec{i}_z

I hope I'm right :)

Roger

 

[Chestermiller]

Chet,

Thank you for encourageing me!

I am very happy I was right :)

Getting down to my new lesson:

This first bold part is clear to me.

The other part is however a bit unclear.

I don't know what to do about this. I worded it as best I could. It's probably not important, and you will soon get the idea.

Quoting your formula:

\vec{dF}=(\vec{σ}\centerdot \vec{n}) dA

and if

\vec{n}=n_x\vec{i}_x+n_y\vec{i}_y+n_z\vec{i}_z

while using

\vec{σ}=-p\vec{i}_x\vec{i}_x-p\vec{i}_y\vec{i}_y-p\vec{i}_z\vec{i}_z

the force vector has to be

\vec{dF}=-pn_x\vec{i}_x-pn_y\vec{i}_y-pn_z\vec{i}_z

Almost right. You left out the dA. Also, I would like you to factor out the -p from the three terms and see what you get.

Chet

 

[rogerk8]

Sloppy of me again :)

My idea of the correct expression would then be:

\vec{dF}=-p(n_x\vec{i}_x+n_y\vec{i}_y+n_z\vec{i}_z)dA

Is this correct?

Roger

 

[Chestermiller]

Sloppy of me again :)

My idea of the correct expression would then be:
\vec{dF}=-p(n_x\vec{i}_x+n_y\vec{i}_y+n_z\vec{i}_z)dA
Is this correct?

Yes. Do you recognize the term in parenthesis in this equation? If so, please replace it with the vector that it represents.

Chet

 

[rogerk8]

Yes. Do you recognize the term in parenthesis in this equation? If so, please replace it with the vector that it represents.

Chet

Yes, it's the normal vector $\vec{n}$ which gives

\vec{dF}=-p\vec{n}dA

Right?

Roger

 

[Chestermiller]

Yes, it's the normal vector $\vec{n}$ which gives

\vec{dF}=-p\vec{n}dA

Right?

Roger

Yes. We're close to the end now.

So you can see that the force per unit area on an arbitrarily oriented element of area in a gas or liquid at hydrostatic equilibrium is pointing in the direction of the normal to the area (i.e., perpendicular to the area element). The equation I gave you for the stress tensor in this situation automatically delivers this result.

The important thing to remember here is that tension in a wire and pressure in a gas or liquid both feature a kind of directionality, because both are determined by their own unique special forms of the stress tensor, which itself features a directional nature (via the dyads that represent it).

Chet

 

[rogerk8]

Yes. Do you recognize the term in parenthesis in this equation? If so, please replace it with the vector that it represents.

Chet

Yes, it's the normal vector $\vec{n}$ which gives

\vec{dF}=-p\vec{n}dA

Right?

Roger

Yes. We're close to the end now.

So you can see that the force per unit area on an arbitrarily oriented element of area in a gas or liquid at hydrostatic equilibrium is pointing in the direction of the normal to the area (i.e., perpendicular to the area element). The equation I gave you for the stress tensor in this situation automatically delivers this result.

The important thing to remember here is that tension in a wire and pressure in a gas or liquid both feature a kind of directionality, because both are determined by their own unique special forms of the stress tensor, which itself features a directional nature (via the dyads that represent it).

Chet

So the fun lessons are over?

Anyway, I seam to understand that pressure force (force per unit area) actually does have direction (normal to the area element) while pressure itself is isotropic (from the stress tensor).

Roger

 

[Chestermiller]

Yes, it's the normal vector $\vec{n}$ which gives

\vec{dF}=-p\vec{n}dA

Right?

Roger

So the fun lessons are over?

I guess for now, unless you have any other specific questions.

Anyway, I seam to understand that pressure force (force per unit area) actually does have direction (normal to the area element) while pressure itself is isotropic (from the stress tensor).

Yes, it is isotropic in the sense the pressure force on an area of any arbitrary orientation is perpendicular to that area. This is, it acts the same in all directions.

Chet

 

[rogerk8]

I guess for now, unless you have any other specific questions.

Yes, it is isotropic in the sense the pressure force on an area of any arbitrary orientation is perpendicular to that area. This is, it acts the same in all directions.

Chet

Here comes a specific question.

Just to warm up, I have gotten interested in the nature of pressure because this is one way of looking at a Tokamak Fusion Reactor.

There seams to be several approaches in how to try to understand how a plasma behaves.

One approach is to consider it as a fluid.

Anyway, pressure is a key ingredience in all approaches (p=nkT, for instance).

And up to now I haven't had a clue what pressure really is.

The studies I mentioned I took back in the 90's where both about plasma physics.

I took them because I really love the idea of harvesting our sun's way of producing energy.

After that we can say goodbye to oil (and other inefficient ways of producing energy, like windmills for instance).

Well these are grandios thoughts, but I love them.

I should point out that I was not so good in school.

While I was attending Chalmers University of Technology here in Gothenburg (Sweden) I almost only got 3's (on a scale from 3 to 5).

But I am very proud to say that in Electromagnetic Field Theory i got 4's in both courses.

I am not only proud about this but 25 years later I long to work with Maxwell's equations again.

I love Maxwell's equations!

Which I still don't understand though :D

So I guess my question to you is if you may want to guide me through personal MHD studies which I now will move on to.

Roger

 

[Chestermiller]

Sorry. MHD is not my area. But maybe other members of PF can help you. Try starting some threads and see it your questions are satisfactorily answered.

Chet

 

[rogerk8]

Sorry. MHD is not my area. But maybe other members of PF can help you. Try starting some threads and see it your questions are satisfactorily answered.

Chet

But please remember Chet, you have opened my eyes when it comes to pressure.

Thanks!

Roger

 

[rogerk8]

Hi Chet, my friend!

I really like you and all the effort you have putten down into helping me.

All of this without a single referation to basic links or litterature, thanks!

I have a new problem that you might want to help me with:

Let's say we have:

\vec{E}=E_x\vec{i}_x+E_y\vec{i}_y+E_z\vec{i}_z

and

\vec{B}=B_x\vec{i}_x+B_y\vec{i}_y+B_z\vec{i}_z

and the Lorentz Force

0=q(\vec{E}+\vec{v}X\vec{B})

which due to

\vec{E}X\vec{B}=\vec{B}X(\vec{v}X\vec{B})=vB^2-B(\vec{v}\cdot \vec{B})

and transverse components only, gives

v_{gc}=\frac{\vec{E}X\vec{B}}{B^2}

where Vgc is the guiding center drift of the charged particles in a magnetic field with an electric field.

My question now is how to calculate

\vec{C}=\vec{E}X\vec{B}

I could have chosen pure math for this but I'm tired of theory that is hard to see the practical use of.

So what is C, with my definitions of E & B?

And how do I calculate it (the manitude and resulting direction is easy but I whish to see it in math)?

Roger

 

[Chestermiller]

Hi Roger,

I think you should start a new thread with this question. I kind of get the gist of what you are asking, but I would rather others with more E&M experience than mine weigh in on this one.

Chet