Solving 2-D Crystal Process Avrami Equation: Find k

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The discussion revolves around solving the Avrami equation phi(t) = 1 - exp(-k*t^2) for a two-dimensional crystal process, specifically finding the value and dimensions of k given that the inflection point occurs at t = 2 hours. Participants highlight the need to compute the second derivative to identify the inflection point and discuss the implications of k's units to ensure the exponent in the equation remains dimensionless. It is concluded that k should have units of hr^-2 to balance the dimensions in the equation correctly. The conversation emphasizes the importance of dimensional analysis in mathematical modeling.
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Homework Statement


A two dimensional crystal process is modeled 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.


Homework Equations



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

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.
 
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Stat313 said:

Homework Statement


A two dimensional crystal process is modeled 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.


Homework Equations



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

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.

Inflection points are usually found using derivatives of the curve.
 
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?
 
Stat313 said:
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?

Your method is fine. If the magnitude of k is 1/8, what units will k have?
 
Can you elaborate more please?
 
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).
 
Is't k just a constant?
 
Stat313 said:
Is't k just a constant?

Sure, but even constants have units if equations using them are to balance.
 
When I compute phi(2), it gives me 0.39 which is a real number.
 
  • #10
Any hints?
 
  • #11
Stat313 said:
Any hints?

Hint: The argument x of the exponential function ##e^x## must be a pure number without units (all units in the exponent must cancel).
 
  • #12
Is the dimension of k is (1/time(t)), in this case we have t in hours?
 
  • #13
Stat313 said:
Is the dimension of k is (1/time(t)), in this case we have t in hours?

Test it. If k = 1/(8*hr), does the exponent yield a unitless number if you plug in t = 2 hr?
 
  • #14
No, it is not going to 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.
 
  • #15
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?
 
  • #16
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}##.
 
  • #17
haruspex said:
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?

t^2 has (time)^2 and k has to have (1/(time)^2) ?
 
  • #18
Stat313 said:
t^2 has (time)^2 and k has to have (1/(time)^2) ?
Yes. So the units would be?
 
  • #19
The units will cancel, (time)^2/(time)^2=1

Thanks for the help,
 
  • #20
Stat313 said:
The units will cancel, (time)^2/(time)^2=1
No, I meant the units of k.
 
  • #21
haruspex said:
No, I meant the units of k.

Is it (1/t)^2?
 
  • #22
No, that's a dimension, like distance-squared, 1/time, mass... Units are things like sq metres, Herz, kgs...
 
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