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What is the use for plasma pressure?

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  1. Feb 13, 2014 #1
    Hi!

    I just wonder what's the actual use for

    [tex]p=nkT≈nE_k[J/m^3][/tex]
    when it comes to plasma physics?

    What does p help us understand/enable?

    While Ek is related to particle velocity and thus temperature, this "new" concept of pressure tells me nothing other than particle density will have to be increased (at the same T) for higher pressure.

    Or you may see it as an increasement of v (and thus T) for higher p (at the same n).

    Or a combination.

    But what's the use?

    Best regards, Roger
     
  2. jcsd
  3. Feb 13, 2014 #2

    Simon Bridge

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    Consider analogy: what use is "pressure" in general?
    i.e. think about plasma confinement.
     
  4. Feb 14, 2014 #3
    To me pressure in general is force distributed over an area

    [tex]p=F/A[/tex]
    but even here it sounds unneccesary to invent this concept because we all f.i know that if something has mass and is resting on something

    [tex]F=mg[/tex]
    which is simple to comprehend because of F being a vector thus having both direction and magnitude. Calculating a resultant force is also simple.

    But when it comes to pressure things suddenly becomes omnidirectional.

    A useful consequence of pressure is however a drawing-pin's ability to penetrate rather hard materials due to it's pointed edge. But here we have pressure that actually has direction.

    Pressure is therefore not only omnidirectional. It may have direction also while at the same time being defined as a scalar.

    But I probably mix gas pressure and force pressure up.

    Anyway, let's say force is omnidirectional in a magnetical confinement and begin with

    [tex]F=q(E+vXB)[/tex]
    Then a high F affecting the limited area of the Tokamak plasma surface would yield a high plasma pressure.

    While v is huge due to extreme temperatures, F is very high too.

    So we have a situation where F is very high and acting upon a relitively small outer plasma surface.

    This should mean that plasma pressure is very high indeed.

    The little I've heard about Tokamaks however tells me that they often state "we need to get high pressure". But to me this is the same as saying that "we need high temperatures".

    I simply do not get why they "circumvent" kinetic energy for pressure suddenly because the above reasoning seems equally true for temperature (or Ek).

    Finally, would you mind providing me with a link where I can read more about this (I have read the Wikipedia article on pressure but will read it again)?

    Or even better, talk to me and correct my wrongfulness.

    Best regards, Roger
    PS
    But of course we need to have a volume density of particles also otherwise we would not have a plasma at all. And the formulas for Ek only holds v & T.
     
    Last edited: Feb 14, 2014
  5. Feb 14, 2014 #4

    Simon Bridge

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    For plasma pressure think: fluid pressure.
    A high plasma pressure requires more effort in building confinement.
    In fusion research the confinement is electromagnetic - so the bigger the plasma pressure the bigger the magnets you need to confine it just like a high pressure gas needs a thicker walled cylinder.

    Pressure and temperature in a plasma are related of course - just as they are related in a gas.
    But increased pressure is not always the same as increased temperature ... depends on the context.
    Same with gas pressure.

    Plasma pressure is useful for the same reason gas pressure is useful.

    I think you will benefit from looking up college "introduction to plasma physics" courses.
    If you do not have at least a beginning college background i math and physics you will want to back-fill the knowledge as you go. i.e.
    http://silas.psfc.mit.edu/introplasma/
     
  6. Feb 15, 2014 #5
    Ok, plasma pressure equals fluid (which I recently have understood is the name for both gases and liquids) pressure.
    I very much like your simple analogy.

    Simplifying my equation above

    [tex]F=q(vXB)[/tex]
    Where B is the strength of the confinement magnetic flux density.

    I kind of get this. By creating the pressure concept we have two variables (T&n) which may be changed "optionally".

    And as I said in my PS I get that n needs to be there because if n=0 we actually have no plasma/gas at all.

    So pressure is a useful concept all things considered.

    But what I don't get is the actual use for pressure in this case. What does pressure enable? Other than as you have said, making it harder to confine the plasma.

    Is there some limit to pressure where actual ignition of the plasma may occur?

    In other words, how much pressure is needed for DT-fuel to ignite?

    And why is 100 million degrees at the same time "determined"?

    Which also would mean that we need a certain volume density for it all to ignite.

    This I don't get because high density just means that the Hydrogen isotopes are closer to eachother.

    This I obviously have to study some more.

    Thank you very much for that link as well as your nice answer.

    With regard to my stupid questions you may not believe this but I actually hold a Mater's degree in Electrical Engineering :rofl:

    The only excuse I have for my stupidity is that I graduated 18 years ago...

    You should also be aware that I back then even took a couple of courses in Plasma Physics.

    Since I have forgotten so much (or perhaps never understood) I am eager to refresh my knowledges.

    Hope you still want to educate me!

    Best regards, Roger
     
  7. Feb 15, 2014 #6

    Simon Bridge

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    The concept is used in the fluid model for plasma physics - the references I gave you go into detail. TLDR: it has the same basic use as in fluids.

    It enables an understanding of fluids to be used, in a modified form, to help you work with plasmas.
    Just like understanding pressure in fluids helps you doing pretty much anything with fluids.

    Since you have to manipulate a plasma using magnetic fields, you gotta know about the pressures that will be pushing against the fields. It helps you understand the interaction in-bulk so you can tell where the plasma will go and how it will behave.

    Aside:
    Elect.Eng. is a different skill set to plasma physics and 18 years is a long time so I feel for you - things change. Your experience at post-grad level should allow you to bring yourself up to speed quite quickly using the lecture notes I gave you end of post #4. That part is like falling off a bicycle ... you'll get back into the rhythm easily.

    Enjoy.
     
  8. Feb 16, 2014 #7
    I will read the "Introduction to Plasma Physics" link you so nicely have provided and not disturb you again until I'm finished.

    Yet I think my problem is basic understanding of the common gas law. So I think I will begin by reading my shoolar books about the pressure business. I seem to remember something about adiabatic processes but I only remember that

    [tex](pV)^\gamma=constant[/tex]
    or something the like.

    Yet it doesn't really tell me anything. Just a formula :smile:

    Finally, thank you for your encouragement!

    Take care!

    Best regards, Roger
    PS
    I hope I don't need to learn rot and div again :smile:

    Appendix:

    [tex]F=m\frac{dv}{dt}=q(vXB)[/tex]
    [tex]v==v_0e^{jwt}[/tex]
    [tex]jwm=qB[/tex]
    using this imaginary part
    [tex]w_c=\frac{qB}{m}[/tex]
    [tex]v==w_cr_L[/tex]
    [tex]r_L=\frac{mv}{qB}[/tex]

    and the area is thus

    [tex]A=2\pi r_L*C_{iter}[/tex]
    where C is some 20m(?) which would give

    [tex]A=40\pi r_L[/tex]
    And the magnitude of F above is

    [tex]F=qvB[/tex]
    Playing with protons only with a B of 5T and a kT of 10keV (ref: the_wolfman)

    [tex]E_k=\frac{m_pv^2}{2}≈kT[/tex]

    we have
    [tex]v=\sqrt{2kT/m_p}[/tex]

    and thus
    [tex]r_L=m_pv/eB=m_p\sqrt{2kT/m_p}/eB=\sqrt{2kTm_p}/eB[/tex]

    Now,
    [tex]p=F/A=evB/A=e\sqrt{2kT/m_p}B/(2\pi r_L C_{iter})=e\sqrt{2kT/m_p}B/(2\pi C_{iter}\sqrt{2kTm_p}/eB )[/tex]

    getting rediculous...

    [tex]p=e^2B^2\sqrt{2kT/m_p}/(2\pi C_{iter}\sqrt{2kTm_p})=e^2B^2/(2\pi C_{iter}m_p)=(1,6e^{-19}*5)^2/(2\pi*20*1,67e^{-27})≈3pPa[/tex]

    which also is independent of T :rofl:
     
    Last edited: Feb 16, 2014
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