Chapter 9, Problem 9.5P

(a)

To determine

The total stage II creep strain theory for the alloy.

Answer to Problem 9.5P

The total stage II creep strain theory for the alloy at $913\text{K}$ is $6.3$ .

Explanation of Solution

Given:

The tensile stress of super alloy is $200\text{Mpa}$

The creep-rate equation is $n=5$

Formula Used:

Write the expression for the total stage II creep strain as:

  ${\epsilon }_{\text{u}}=\stackrel{.}{\epsilon }T$ …… (I)

Here, ${\epsilon }_{\text{u}}$ is the total stage II creep strain for the alloy, $T$ is the total time and $\stackrel{.}{\epsilon }$ is the strain rate of alloy.

Calculation:

Substitute $365\text{days}$ for $T$ and $2\times {10}^{-7}{\text{s}}^{-1}$ for $\stackrel{.}{\epsilon }$ in equation (I)

  $\begin{array}{c}{\epsilon }_u=2\times {10}^{-7}{\text{s}}^{-1}\times \left(1\text{year}\times \frac{365\text{days }}{1\text{year}}\times \frac{24\text{hr}}{1\text{day}}\times \frac{3600\text{s}}{1\text{hr}}\right)\\ =\left(2\times {10}^{-7}{\text{s}}^{-1}\right)\left(31536000\text{s}\right)\\ =6.3\end{array}$

Conclusion:

Thus, the total stage II creep strain theory for the alloy at $913\text{K}$ is $6.3$ .

(b)

To determine

The stage II creep rate of the alloy at $891\text{K}$ and at a tensile stress.

Answer to Problem 9.5P

The stage II creep rate of the alloy at $891\text{K}$ and at a tensile stress of $300\text{Mpa}$ is $7.57\times {10}^{-7}{\text{s}}^{-1}$ .

Explanation of Solution

Given:

The tensile stress of super alloy is $300\text{Mpa}$

The creep-rate equation is $5$

Formula Used:

Write the expression for the tensile stage II creep strain.

  ${\epsilon }_{\text{u}}=A{\sigma }^n\exp -\left(\frac{\Delta H_{\text{p}}}{kT}\right)$ …… (II)

Here, ${\epsilon }_{\text{u}}$ is the tensile stage II creep rate, $A$ is the pre-exponential constant, $\sigma$ is the applied stress, $n$ is the log of applied stress, $\Delta H_{\text{p}}$ is the enthalpy of activation, $k$ is the Boltzmann’s constant and $T$ is the temperature at which the stress applied.

Write the expression for the ratio of strain rates of $200\text{Mpa}$ and $300\text{Mpa}$ at temperature $891\text{K}$ .

  $\begin{array}{c}\frac{{\epsilon }_{\text{u}}{}_{\left(200\text{Mpa}\right)}}{{\epsilon }_{\text{u}}{}_{\left(300\text{Mpa}\right)}}=\frac{A{\sigma }_1^n\exp -\left(-\frac{\Delta H_p}{kT}\right)}{A{\sigma }_2^1\exp \left(-\frac{\Delta H_p}{kT}\right)}\\ \frac{{\epsilon }_{\text{u}}{}_{\left(200\text{Mpa}\right)}}{{\epsilon }_{\text{u}}{}_{\left(300\text{Mpa}\right)}}={\left(\frac{{\sigma }_1}{{\sigma }_2}\right)}^n\end{array}$ …… (III)

Here, $A$ is the pre-exponential constant, $\sigma$ is the applied stress, $n$ is the log of applied stress , $\Delta H_p$ is the enthalpy of activation , $k$ is the Boltzmann’s constant and $T$ is the temperature at which the stress applied.

Calculation:

Substitute $1.0\times {10}^{-7}{\text{s}}^{-1}$ for ${\stackrel{.}{{\epsilon }_u}}_{\left(200\text{Mpa}\right)}$ , $200\text{Mpa}$ for ${\sigma }_1$ , $300\text{Mpa}$ for ${\sigma }_2$ and $5$ for $n$ in equation (III).

  $\begin{array}{c}\frac{1.0\times {10}^{-7}{\text{s}}^{-1}}{{\epsilon }_{\text{u}}{}_{\left(300\text{Mpa}\right)}}={\left(\frac{200\text{Mpa}}{300\text{Mpa}}\right)}^5\\ \frac{1.0\times {10}^{-7}{\text{s}}^{-1}}{{\epsilon }_{\text{u}}{}_{\left(300\text{Mpa}\right)}}=0.132\\ {\epsilon }_{\text{u}}{}_{\left(300\text{Mpa}\right)}=\frac{1.0\times {10}^{-7}{\text{s}}^{-1}}{0.132}\\ =7.57\times {10}^{-7}{\text{s}}^{-1}\end{array}$

Conclusion:

Thus, the stage II creep rate of the alloy at $891\text{K}$ and at a tensile stress of $300\text{Mpa}$ is $7.57\times {10}^{-7}{\text{s}}^{-1}$ .

(c)

To determine

The activation enthalpy for creep in $\text{eV}/\text{atom}$ for the alloy.

Answer to Problem 9.5P

The activation enthalpy for creep in $\text{eV}/\text{atom}$ for the alloy is $3.0\text{eV}/\text{atom}$ .

Explanation of Solution

Formula Used:

Write the expression for the ratio of strain rates of $200\text{Mpa}$ at temperature $891\text{K}$ and $913\text{K}$ .

  $\frac{{\epsilon }_{\text{u}}{}_{\left(891\text{K}\right)}}{{\epsilon }_{\text{u}}{}_{\left(913\text{K}\right)}}=\frac{A{\sigma }^n\exp -\left(-\frac{\Delta H_p}{kT}\right)}{A{\sigma }^n\exp \left(-\frac{\Delta H_p}{kT}\right)}$ …… (IV)

Here, $A$ is the pre-exponential constant , $\sigma$ is the applied stress, $n$ is the log of applied stress , $\Delta H_p$ is the enthalpy of activation , $k$ is the Boltzmann’s constant and $T$ is the temperature at which the stress applied.

Calculation:

Substitute $1.0\times {10}^{-7}{\text{s}}^{-1}$ for ${\epsilon }_{\text{u}}{}_{\left(891\text{K}\right)}$ , $2\times {10}^{-7}{\text{s}}^{-1}$ for ${\epsilon }_{\text{u}}{}_{\left(913\text{K}\right)}$ , $891\text{K}$ for $T_1$ and $913\text{K}$ for $T_2$ in equation (IV).

  $\begin{array}{c}\frac{1.0\times {10}^{-7}{\text{s}}^{-1}}{2\times {10}^{-7}{\text{s}}^{-1}}=\left(-\frac{\Delta H_p}{k}\left(\frac{1}{891\text{K}}-\frac{1}{913\text{K}}\right)\right)\\ 0.5=\exp \left(-\frac{\Delta H_p}{k}\left(1.12\times {10}^{-3}-1.10\times {10}^{-3}\right){\text{K}}^{-1}\right)\end{array}$

Take log both side,

  $\begin{array}{c}\ln \left(0.5\right)=\ln \left[\exp \left(-\frac{\Delta H_p}{k}\left(0.02\times {10}^{-3}\right){\text{K}}^{-1}\right)\right]\\ -0.693=\frac{-\Delta H_p}{8.62\times {10}^{-5}\frac{\text{eV}}{\text{atom}\cdot \text{K}}}\left(0.02\times {10}^{-3}{\text{K}}^{-1}\right)\\ \Delta H_p=3.0\text{eV}/\text{atom}\end{array}$

Conclusion:

Thus, the activation enthalpy for creep in $\text{eV}/\text{atom}$ for the alloy is $3.0\text{eV}/\text{atom}$ .