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Rate of Tungsten Evaporation

  1. Mar 3, 2015 #1
    1. The problem statement, all variables and given/known data
    Ina light bulb, billions of electrons flow through the tungsten wire bumping around, causing the filament to glow hot. Hot tungsten evaporates slowly, so its initial thickness profile r(x, t=0) can change over time. Consider a volume V of tungsten, with N bonds total, each requiring energy E to break. We will (rather incorrectly) assume that a constant fraction f of electric power is spent on breaking these bonds, and the remaining (1-f) is spent to heat and illuminate. We will also assume that the temperature remains constant.

    a. How much energy must be spent to evaporate a volume dV?
    b. As the wire evaporates over time, does the current increase or decrease? Is there anything in this system that remains constant?
    c. Derive an equation governing the thickness r(x, t) of the tungsten wire as a function of position and time. Don't assume that the thickness is uniform.
    d. Assume now that the thickness r(x, t) has no x dependence (i.e. we start with a perfectly cylindrical wire). Solver for r(t).
    e. plot, and discuss your results

    notes:
    Since the evaporation happens from the surface, you might find it helpful to think in terms of n = number of bonds per area, instead of number of bonds per volume. It's perfectly ok if you write your answer in terms of n, instead of number of bonds per volume.

    2. Relevant equations
    Not that I know of

    3. The attempt at a solution
    Don't even know how to start with this problem
     
  2. jcsd
  3. Mar 3, 2015 #2

    haruspex

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    Take it one step at a time. a. is quite easy. If a volume V has N bonds, how many does a volume dV have? How much energy to break them all?
     
  4. Mar 3, 2015 #3
    dN? so I need to create a function which relates N to dt? Okay so would that look something like fPt? or do I need to relate it to surface area because it evaporates from the outside in?
     
  5. Mar 3, 2015 #4

    haruspex

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    No, the answer to the first of my two questions only involves the three variables provided, V, N, and dV.
     
  6. Mar 3, 2015 #5
    In a volume dV there are (N/V)dV bonds. So it takes (N/V)dV*E energy to break them
     
    Last edited: Mar 3, 2015
  7. Mar 4, 2015 #6
    What about the thickness equation though? I think that's the hardest part.
     
  8. Mar 4, 2015 #7

    haruspex

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    Sure, but you need an answer to b first.
     
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