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Avrami Equation

  1. Jan 20, 2013 #1
    1. The problem statement, all variables and given/known data
    A two dimensional crystal process is modelled by an Avrami equation phi(t)= 1-exp(-k*t^2), where k>0. Experimental observation suggests that the inflection point of phi occurs at t=2 hours. Find the dimension and numerical value of k.


    2. Relevant equations

    I know how to take the log-log of this equation in order to linearize it.

    3. The attempt at a solution

    I tried to have a system of two equations with 2 unknowns, but failed because I don't know the value of phi at any time t. Any help would be greatly appreciated.
     
  2. jcsd
  3. Jan 20, 2013 #2

    gneill

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    Inflection points are usually found using derivatives of the curve.
     
  4. Jan 20, 2013 #3
    Thanks for the reply,
    I computed the second derivative with respect to t, I got
    phi^2(t)= 2*k*exp(-k*t^2)-4*k^2*t^2*exp(-k*t^2)

    Since we know that an inflection point occurs at t = 2, I solved for phi^2(2) = 0, which gives me k=0,1/8. so we choose k>0. Is my answer correct?
     
  5. Jan 20, 2013 #4

    gneill

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    Your method is fine. If the magnitude of k is 1/8, what units will k have?
     
  6. Jan 20, 2013 #5
    Can you elaborate more please?
     
  7. Jan 20, 2013 #6

    gneill

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    The equation contains the exponential term
    $$e^{-k\;t^2}$$
    What units should be associated with k in order to make this term correct (mathematically, so that when evaluated numerically will yield a real number).
     
  8. Jan 20, 2013 #7
    Is't k just a constant?
     
  9. Jan 20, 2013 #8

    gneill

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    Sure, but even constants have units if equations using them are to balance.
     
  10. Jan 20, 2013 #9
    When I compute phi(2), it gives me 0.39 which is a real number.
     
  11. Jan 20, 2013 #10
    Any hints?
     
  12. Jan 20, 2013 #11

    gneill

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    Hint: The argument x of the exponential function ##e^x## must be a pure number without units (all units in the exponent must cancel).
     
  13. Jan 20, 2013 #12
    Is the dimension of k is (1/time(t)), in this case we have t in hours?
     
  14. Jan 20, 2013 #13

    gneill

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    Test it. If k = 1/(8*hr), does the exponent yield a unitless number if you plug in t = 2 hr?
     
  15. Jan 20, 2013 #14
    No, it is not gonna cancel, but if I have k 1/time in (half hours), it does cancel? By the way, I am not a physics major, so bare with me please.
     
  16. Jan 20, 2013 #15

    haruspex

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    t has dimension time, so what dimension does t2 have? So what dimension do you need k to have in order to produce a dimensionless kt2?
     
  17. Jan 20, 2013 #16

    gneill

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    In the exponent you have the expression ##-k t^2##. That means the time is squared, so if t is in hours, the units of ##t^2## are ##hr^2##. To make the units of k cancel this, it should have units of ##1/(hr^2)##, or ##hr^{-2}##.
     
  18. Jan 20, 2013 #17
    t^2 has (time)^2 and k has to have (1/(time)^2) ?
     
  19. Jan 20, 2013 #18

    haruspex

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    Yes. So the units would be?
     
  20. Jan 20, 2013 #19
    The units will cancel, (time)^2/(time)^2=1

    Thanks for the help,
     
  21. Jan 20, 2013 #20

    haruspex

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    No, I meant the units of k.
     
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