Chapter 12, Problem 12.4P

Interpretation Introduction

Interpretation:

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

Concept Introduction:

At constant temperature, the determinationof 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 12.4P

The value of $\text{c = 6}{\text{.18×10}}^{\text{-4}}$ and $\text{n = 3}\text{.19}$.

Explanation of Solution

Given:

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

Temperature = $\text{T=135}°\text{C}$

Assume -

Elapsed time − $\text{t}$

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

The effect of temperature on crystallization is:

Fig. $2$ The effect of temperature on crystallization

To calculate the constants 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 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 given equations of curves, the values are:

$\text{f}$	$\text{t}$ (min)	$\text{ln}\left(\text{t}\right)$	$\ln \left[\text{-ln}\left(\text{1-f}\right)\right]$
$0.1$	$5$	$1.61$	$-2.25$
$0.2$	$6.6$	$1.89$	$-1.50$
$0.3$	$7.7$	$2.04$	$-1.03$
$0.4$	$8.5$	$2.14$	$-0.67$
$0.5$	$9$	$2.20$	$-0.37$
$0.6$	$10$	$2.30$	$-0.09$
$0.7$	$10.5$	$2.35$	$0.19$
$0.8$	$11.5$	$2.35$	$0.48$
$0.9$	$13.7$	$2.62$	$0.83$

Table $1$

From above table, plot the solpe of the graph:

Fig. $2$ slope of calculated values

Equation of a straight line −

  $\text{y = mx+c}$

Then to find the value of n,

  $\text{n = }\frac{0.83-\left(-1.50\right)}{2.62-1.89}$

  $\text{n = 3}\text{.19}$

Then to find value of $\text{c}$,

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

  $\begin{array}{l}\text{ln}\left[\text{-ln}\left(0.5\right)\right]\text{= ln}\left(\text{c}\right)\text{+2}\text{.81×ln}\left(9\right)\\ \text{ln}\left(\text{c}\right)\text{= -7}\text{.39}\\ \text{c}=6.18\times {10}^{-4}\end{array}$

Conclusion

The value of c and n are $6.18\times {10}^{-4}$ and $3.19$ respectively and are obtained by using the slope.