Solving 2-D Crystal Process Avrami Equation: Find k

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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.
 
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).
 
Is the dimension of k is (1/time(t)), in this case we have t in hours?
 
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?
 
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.
 
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}##.
 
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) ?
 
The units will cancel, (time)^2/(time)^2=1

Thanks for the help,
 
haruspex said:
No, I meant the units of k.

Is it (1/t)^2?