Solving 2-D Crystal Process Avrami Equation: Find k

In summary: Is it 1/time^2?Yes, that is correct. So the dimension of k is 1/time^2 and the numerical value is 1/8, making the final answer:In summary, the dimension of k is 1/time^2 and the numerical value is 1/8.
  • #1
Stat313
13
0

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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  • #2
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.
 
  • #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?
 
  • #4
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?
 
  • #5
Can you elaborate more please?
 
  • #6
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).
 
  • #7
Is't k just a constant?
 
  • #8
Stat313 said:
Is't k just a constant?

Sure, but even constants have units if equations using them are to balance.
 
  • #9
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...
 

Related to Solving 2-D Crystal Process Avrami Equation: Find k

1. What is the Avrami equation and why is it important in 2-D crystal process?

The Avrami equation is a mathematical model that describes the kinetics of a phase transformation process, such as the growth of crystals. In 2-D crystal processes, this equation is used to determine the rate constant, k, which is a measure of the growth rate of the crystals. This is important because it allows scientists to understand and control the crystal growth process.

2. How is the Avrami equation derived and what are its assumptions?

The Avrami equation is derived from the Avrami-Erofeev model, which assumes that the crystal growth occurs through a nucleation and growth process. This model also assumes that the nucleation rate is constant and that the growth of the crystals is linear.

3. How is the rate constant, k, determined using the Avrami equation?

The rate constant, k, can be determined by plotting the natural logarithm of the transformed fraction of crystals (ln(1-α)) against time and fitting the data to the Avrami equation. The slope of the resulting line will give the value of k.

4. Are there any limitations to using the Avrami equation to determine k?

Yes, there are a few limitations to using the Avrami equation. It assumes a constant nucleation rate and linear crystal growth, which may not always be the case. It also assumes that the crystals are all of the same size and shape, which may not always be true in real systems.

5. Can the Avrami equation be applied to other processes besides crystal growth?

Yes, the Avrami equation can be applied to other processes besides crystal growth, such as phase transformations, nucleation, and grain growth. However, the equation may need to be modified for each specific process depending on the assumptions and limitations of the model.

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