Chapter 12, Problem 12.6P

Interpretation Introduction

Interpretation:

The value of constants n and c for the given equation needs to be calculated.

Concept Introduction:

At constant temperature, the determination of solid phase transformation is done by using the Avarmi equation. Generally, it is used to describe the rate of kinetic crystallization.

This equation is also known as the JMAK equation as it was founded by Johnson-Mehl-Avrami-Kolmology. This equation was first found in the year $1937$ by Kolmogorov. And the first paper was published by Melvin Avrami during the years $1939$ and $1941$ in the Journal of chemical equations.

Answer to Problem 12.6P

The constants value of $\text{c}$ and $\text{n}$ are $0$ and $1.177$ respectively

Explanation of Solution

Given Information:

Eqauation: ${\text{f=1-exp(-ct}}^{\text{n}}\text{)}$

Elapsed time − $\text{t = 10 sec}$

Assuming, elapsed time − $\text{t = 10 sec}$

Temperature constants - $\text{c }$ and $\text{n}$

From the given sigmoidal curve, we can obtain values that are found out in Table 1

To calculate the value of c and n by using the below formula,

  ${\text{f=1-exp(-ct}}^{\text{n}}\text{)}$

  $\begin{array}{l}\text{1-f = exp}\left({\text{-ct}}^{\text{n}}\right)\\ \text{ln}\left(\text{1-f}\right)\text{= -ct}\\ \end{array}$

  

Taking natural logarithm on both the sides,

  $\ln \left[\text{-ln}\left(\text{1-f}\right)\right]\text{= ln}\left(\text{c}\right)\text{+n×ln}\left(\text{t}\right)$

From the above sigmoidal curve, we can find time taken for solidification,

$\text{f}$	$\text{t}$ (min)	$\text{ln}\left(\text{t}\right)$	$\ln \left[\text{-ln}\left(\text{1-f}\right)\right]$
$0.1$	$65$	$4.174$	$-2.25$
$0.2$	$85$	$4.442$	$-1.50$
$0.3$	$90$	$4.5$	$-1.03$
$0.4$	$105$	$4.554$	$-0.67$
$0.5$	$200$	$4.654$	$-0.37$
$0.6$	$250$	$5.3$	$-0.09$
$0.7$	$250$	$5.521$	$0.19$
$0.8$	$350$	$5.858$	$0.48$
$0.9$	$600$	$5.396$	$0.83$

Table $1$

Equation of the straight line −

  $\text{n = }\frac{{\text{y}}_{\text{2}}{\text{-y}}_{\text{1}}}{{\text{x}}_{\text{2}}{\text{-x}}_{\text{1}}}$

Then to find value of $\text{n}$, we can use −

  $\text{n = }\frac{0.83-\left(-1.50\right)}{6.396-4.442}$

  $\text{n = 1}\text{.177}$

Then to find the value of $\text{c}$, we can use −

  $\ln \left[\text{-ln}\left(\text{1-f}\right)\right]\text{= ln}\left(\text{c}\right)\text{+n×ln}\left(\text{t}\right)$

  $\begin{array}{l}\text{ln}\left[\text{-ln}\left(\text{0}\right)\right]\text{= ln}\left(\text{c}\right)\text{+1}\text{.177×ln}\left(\text{10}\right)\\ \text{0 = - c}{\left(\text{10}\right)}^{\text{1}\text{.177}}\\ \text{c = 0}\end{array}$

Conclusion

Thus, the value of are $\text{c}$ and $\text{n}$

  $1.177$ respectively and are obtained by using the slope of the graph drawn above.