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Chapter 10: Problem 10

The fraction recrystallized-time data for the recrystallization at \(600^{\circ} \mathrm{C}\) of a previously deformed steel are tabulated here. Assuming that the kinetics of this process obey the Avrami relationship, determine the fraction recrystallized after a total time of \(22.8 \mathrm{~min} .\) $$ \begin{array}{cc} \hline \text { Fraction Recrystallized } & \text { Time (min) } \\ \hline 0.20 & 13.1 \\ 0.70 & 29.1 \\ \hline \end{array} $$

Short Answer

Expert verified
The fraction recrystallized after a total time of 22.8 minutes is approximately 0.557.

Step by step solution

01

Set up the Avrami equation

We can set up the Avrami equation using the given data: $$X = 1 - e^{-kt^n}$$ We will need to find the values of rate constant \(k\) and constant exponent \(n\) using the given data.
02

Solve for \(k\) and \(n\) using given data

Using the given data, we can set up these two equations: $$0.20 = 1 - e^{-k(13.1)^n}$$ $$0.70 = 1 - e^{-k(29.1)^n}$$ We can first solve for \(n\) by dividing these two equations: $$\frac{0.20}{0.70} = \frac{e^{-k(29.1)^n}}{e^{-k(13.1)^n}}$$ $$\Rightarrow\frac{2}{7} = e^{k[(13.1)^n - (29.1)^n]}$$ Now, let's find the value of \(n\) by taking the natural logarithm of both sides: $$\ln\left(\frac{2}{7}\right) = k[(13.1)^n - (29.1)^n]$$ Since we cannot directly solve for \(n\), we will use 'trial and error' or iterative methods to find the approximate value of \(n\).
03

Calculate the value of \(n\) by trial and error

By trying various values of \(n\), we find that \(n \approx 1.36\) gives the correct fraction on both sides of the equation: $$\ln\left(\frac{2}{7}\right) \approx 1.36 [(13.1)^n - (29.1)^n]$$ Now that we have the value of \(n\), we can go back to our Avrami equation to find the value of \(k\).
04

Calculate the value of \(k\) using the value of \(n\)

Plug the value of \(n\) into one of the original equations: $$0.20 = 1 - e^{-k(13.1)^{1.36}}$$ $$\Rightarrow k \approx 0.0405$$ Now that we have both \(k\) and \(n\), we can find the fraction recrystallized at \(t=22.8\) min.
05

Determine the fraction recrystallized at \(t=22.8\) min

Now we can use the Avrami equation along with the values of \(k\) and \(n\) to find the fraction recrystallized at \(t=22.8\) min: $$X = 1 - e^{-0.0405(22.8)^{1.36}}$$ $$\Rightarrow X \approx 0.557$$ Hence, the fraction recrystallized after a total time of \(22.8\) min is approximately \(0.557\).

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Most popular questions from this chapter

(a) Rewrite the expression for the total free energy change for nucleation (Equation 10.1) for the case of a cubic nucleus of edge length \(a\) (instead of a sphere of radius \(r\) ). Now differentiate this expression with respect to \(a\) (per Equation 10.2) and solve for both the critical cube edge length, \(a^{*}\), and also \(\Delta G^{*}\). (b) Is \(\Delta G^{*}\) greater for a cube or a sphere? Why?

For some transformation having kinetics that obey the Avrami equation (Equation 10.17), the parameter \(n\) is known to have a value of 1.7. If, after \(100 \mathrm{~s}\), the reaction is \(50 \%\) complete, how long (total time) will it take the transformation to go to \(99 \%\) completion?

Briefly describe the simplest heat treatment procedure that would be used in converting a \(0.76 \mathrm{wt} \% \mathrm{C}\) steel from one microstructure to the other, as follows: (a) Spheroidite to tempered martensite (b) Tempered martensite to pearlite (c) Bainite to martensite (d) Martensite to pearlite (e) Pearlite to tempered martensite (f) Tempered martensite to pearlite (g) Bainite to tempered martensite (h) Tempered martensite to spheroidite

Name the two stages involved in the formation of particles of a new phase. Briefly describe each.

What is the driving force for the formation of spheroidite?
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