Chapter 9, Problem 9.4P

To determine

The activation enthalpy between $\text{973 K}$ and $\text{1053 K}$ .

Answer to Problem 9.4P

The activation enthalpy between $\text{973 K}$ and $\text{1053 K}$ for zirconium is $2.953\text{ eV/atom}$ .

Explanation of Solution

Given:

The logarithm of stress is $7.10$ .

Concept used:

Write the expression for tensile stage II creep rate.

  $\frac{d{\epsilon }_{II}}{dt}=A{\sigma }^n\exp -\left(\frac{\Delta H_p}{kT}\right)$ …… (1)

Here, $\frac{d{\epsilon }_{II}}{dt}$ is the tensile stage II creep rate, $A$ is the constant and $\sigma$ is the stress, $\Delta H_p$ is the activation enthalpy, $k$ is the Boltzmann constant and $T$ is the temperature.

Calculation:

Refer to Figure 9.5 “The log of the stage II creep strain rate of zirconium as function of the log of stress, at three temperatures” to obtain the value of strain rate corresponding to logarithm of stress as $7.10$ at $973\text{ K}$ and $1053\text{ K}$ as $1.0\times {10}^{-5}{\text{ s}}^{-1}$ and $1.55\times {10}^{-4}{\text{ s}}^{-1}$ respectively.

Substitute $1.0\times {10}^{-5}{\text{ s}}^{-1}$ for $\frac{d{\epsilon }_{II}}{dt}$ , $8.62\times {10}^{-5}\text{ eV/atom}\cdot \text{K}$ for $k$ and $973\text{ K}$ for $T$ in equation (1).

  $1.0\times {10}^{-5}{\text{ s}}^{-1}=A{\left(\sigma \right)}^n\exp -\left(\frac{\Delta H_p}{\left(8.62\times {10}^{-5}\text{ eV/atom}\cdot \text{K}\right)\left(973\text{ K}\right)}\right)$ …… (2)

Substitute $1.55\times {10}^{-4}{\text{ s}}^{-1}$ for $\frac{d{\epsilon }_{II}}{dt}$ , $8.62\times {10}^{-5}\text{ eV/atom}\cdot \text{K}$ for $k$ and $1053\text{ K}$ for $T$ in equation (1).

  $1.55\times {10}^{-4}{\text{ s}}^{-1}=A{\left(\sigma \right)}^n\exp -\left(\frac{\Delta H_p}{\left(8.62\times {10}^{-5}\text{ eV/atom}\cdot \text{K}\right)\left(1053\text{ K}\right)}\right)$ …… (3)

Divide equation (2) by equation (3).

  $\begin{array}{c}\frac{1.0\times {10}^{-5}{\text{ s}}^{-1}}{1.55\times {10}^{-4}{\text{ s}}^{-1}}=\frac{A{\left(\sigma \right)}^n\exp -\left(\frac{\Delta H_p}{\left(8.62\times {10}^{-5}\text{ eV/atom}\cdot \text{K}\right)\left(973\text{ K}\right)}\right)}{A{\left(\sigma \right)}^n\exp -\left(\frac{\Delta H_p}{\left(8.62\times {10}^{-5}\text{ eV/atom}\cdot \text{K}\right)\left(1053\text{ K}\right)}\right)}\\ 0.0645=\exp \left(\frac{-\Delta H_p}{\left(8.62\times {10}^{-5}\text{ eV/atom}\cdot \text{K}\right)}\left[\frac{1}{973\text{ K}}-\frac{1}{1073\text{ K}}\right]\right)\\ \Delta H_p=-\frac{\ln \left(0.0645\right)\left(8.62\times {10}^{-5}\text{ eV/atom}\cdot \text{K}\right)}{0.8\times {10}^{-4}{\text{ K}}^{-1}}\\ \Delta H_p=2.953\text{ eV/atom}\end{array}$

Conclusion:

Thus, the activation enthalpy between $\text{973 K}$ and $\text{1053 K}$ for zirconium is $2.953\text{ eV/atom}$ .