Understanding Gas Pressure: Exploring the Mechanics and Energy of Gases

[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