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Joule heating question

  1. Aug 15, 2014 #1
    Ok so as fare as I understand the equation for Joule heating in a material with a DC current is I^2*R.
    I am trying to calculate for the heating of carbon fiber in an argon environment. Carbon fiber has a resistivity similar to graphite (roughly 5.0*10^-7) I am trying to calculate for 3.3VDc. However I am getting a massive value of 14291338.58 watts dissipated. To get the value of I I'm using the equation V/R=I. What am I doing wrong? I feel 14.2 megawatts is a bit off.
     
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  3. Aug 15, 2014 #2

    AlephZero

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    You seem to be confusing resistivity, which is a property of a material, with resistance, which depends of the size and shape of a particular piece of the material. Resistivity and resistance are measured in different units.

    That is rather like confusing the density of steel with the mass of a particular piece of steel.

    See http://hyperphysics.phy-astr.gsu.edu/hbase/electric/resis.html
     
  4. Aug 15, 2014 #3
    Oh I need to edit that I factored in the dimensions of the material to convert resistivity to resistance
     
  5. Aug 15, 2014 #4

    billy_joule

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    It doesn't sound off to me. The resistivity of CF is a very small number, the resistance value for a macroscopic CF conductor will also be a very small number.

    In reality the heating will be limited by the current supply capability of your voltage source.

    If you add the dimensions of the CF someone could check your calculation.
     
  6. Aug 15, 2014 #5
    It was a guesstimate at around 2" by 3"
     
  7. Aug 15, 2014 #6

    billy_joule

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    Giving better information will lead to better help.

    I guessed it's a square bar with voltage applied across it's length and using :

    ρ = 1 x10^-8 (Ωm)

    For CF

    (From wiki, as you didn't give units)

    And

    R = ρl/A

    P=V^2/R

    => P = AV^2/ρl

    I got 36 MW so you are in the right ball park.

    Now all you have to do is find 3.3 VDC supply that'll put out 11 MILLION amps ;-)
     
  8. Aug 15, 2014 #7
    Thank you for the help. I actually think I'll be building an adjustable current supply, I'd rather not fuse any atoms.
     
  9. Aug 16, 2014 #8

    sophiecentaur

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    Power circuits with carbon elements often need current regulation in them because of the Negative Temperature Coefficient of the resistivity, which can give you thermal runaway if you aren't careful.
     
  10. Aug 16, 2014 #9
    So would a current regulator paired with an adjustable current source allow me to gradually increase how hot it is?
     
  11. Aug 16, 2014 #10

    billy_joule

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    Yes.

    CF is such a good conductor that unless your CF has a very small CSA most of the power will be dissipated in the connecting cables and the CF will not heat up much.
    ie your wires will get hot and the CF will stay cool.
     
  12. Aug 17, 2014 #11

    sophiecentaur

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    Carbon filaments were commonly used for lighting before tungsten established itself as a better material. They worked at mains voltage (you can still buy them, in fact). They are pretty fragile, though, because of their necessary small cross sectional area.

    Here's a historical link.
     
  13. Aug 17, 2014 #12
    Cool thanks. I'm going to be using large blocks of graphite to act as the cathode and anode and to sandwich the carbon fiber at both ends. Do you think with a maximum of 35Amps I could get the carbon fiber to melt? The melting point of carbon is 3500 degrees C. All in an inert environment of course.
     
  14. Aug 17, 2014 #13

    sophiecentaur

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    Using carbon / carbon contact could possibly be a good idea as it may avoid hot spots on contact but I would just have chosen copper blocks as they would dissipate any locally generated heat due to uneven contact. But under pressure, the copper would (?) mould around the fibres to give a good contact.
     
  15. Aug 17, 2014 #14
    Cool so the 35 Amps should be sufficient
     
  16. Aug 17, 2014 #15
    That's a good idea with the carbon carbon
     
  17. Aug 17, 2014 #16

    sophiecentaur

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    It depends entirely upon the cross sectional area of your carbon.
     
  18. Aug 17, 2014 #17
    6" by 6"
     
  19. Aug 17, 2014 #18

    sophiecentaur

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    OMG
    Your figure of MW in the OP sounds quite a reasonable answer - it's just an unfeasible project if you don't have a lot of money, I think. Source resistance will really prove to be a problem with such a low resistance load.
    Would it be possible to take a step backward and to approach your requirement in a different way? Could you provide the electrical heating in a more conventional way, perhaps? But what is the CSA of your carbon Fibre??? Aren't they very thin and not 6" X 6"?
    14MW would melt your whole lab!!!
     
  20. Aug 17, 2014 #19
    Ya I don't want to get it nearly that hot it would vaporize I want like 3500 degrees Celsius so I won't be using nearly 14mw
     
  21. Aug 17, 2014 #20
    It would be a group of fibers in a cloth. I only need to get it to absorb about 9 jules in the form of heat to get it to 3500 degrees Celsius based off its specific heat
     
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