Chapter 9, Problem 9.3P

To determine

The value of power $n$ at the logarithms of stress of $7$ and logarithms of stress of $7.3$.of stresses.

Answer to Problem 9.3P

The value of power $n$ at the logarithms of stress of $7$ and logarithms of stress of $7.3$.of stresses are $11$ and $7.11$ respectively.

Explanation of Solution

Given:

The temperature is $973\text{ K}$ .

Power of stress changes between logarithms of stress of $7$ and logarithms of stress of $7.3$.

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 stress and strain for various points.

At $\log {\sigma }_1=6.93$ the strain rate is $\frac{d{\epsilon }_1}{dt}=2.4\times {10}^{-7}$ and stress ${\sigma }_1$ is $8.51\times {10}^6\text{}\text{Pa}$ .

At $\log {\sigma }_2=7.02$ the strain rate is $\frac{d{\epsilon }_2}{dt}=2.4\times {10}^{-6}$ and stress ${\sigma }_2$ is $1.05\times {10}^7\text{}\text{Pa}$ .

At $\log {\sigma }_3=7.23$ the strain rate is $\frac{d{\epsilon }_3}{dt}=6.5\times {10}^{-5}$ and stress ${\sigma }_3$ is $1.7\times {10}^7\text{}\text{Pa}$ .

At $\log {\sigma }_4=7.39$ the strain rate is $\frac{d{\epsilon }_4}{dt}=8.7\times {10}^{-4}$ and stress ${\sigma }_4$ is $2.45\times {10}^7\text{}\text{Pa}$ .

Substitute $8.51\times {10}^6\text{}\text{Pa}$ for $\sigma$ , $2.4\times {10}^{-7}$ for $\frac{d{\epsilon }_{II}}{dt}$ and $973\text{ K}$ for $T$ in equation (1).

  $2.4\times {10}^{-7}=A{\left(8.51\times {10}^6\text{}\text{Pa}\right)}^n\exp -\left(\frac{\Delta H_p}{k\left(973\text{ K}\right)}\right)$ …… (2)

Substitute $1.05\times {10}^7\text{}\text{Pa}$ for $\sigma$ , $2.4\times {10}^{-6}$ for $\frac{d{\epsilon }_{II}}{dt}$ and $973\text{ K}$ for $T$ in equation (1).

  $2.4\times {10}^{-6}=A{\left(1.05\times {10}^7\text{}\text{Pa}\right)}^n\exp -\left(\frac{\Delta H_p}{k\left(973\text{ K}\right)}\right)$ …… (3)

Divide equation (2) by equation (3).

  $\begin{array}{c}\frac{2.4\times {10}^{-7}}{2.4\times {10}^{-6}}=\frac{A{\left(8.51\times {10}^6\text{}\text{Pa}\right)}^n\exp -\left(\frac{\Delta H_p}{k\left(973\text{ K}\right)}\right)}{A{\left(1.05\times {10}^7\text{}\text{Pa}\right)}^n\exp -\left(\frac{\Delta H_p}{k\left(973\text{ K}\right)}\right)}\\ 0.1={\left(\frac{8.51\times {10}^6\text{}\text{Pa}}{1.05\times {10}^7\text{}\text{Pa}}\right)}^n\end{array}$

Take log on both sides.

  $\begin{array}{c}\log \left(0.1\right)=n\log \left(\frac{8.51\times {10}^6\text{}\text{Pa}}{1.05\times {10}^7\text{}\text{Pa}}\right)\\ n=\frac{-1}{-0.0912}\\ n=11\end{array}$

Substitute $1.7\times {10}^7\text{}\text{Pa}$ for $\sigma$ , $6.5\times {10}^{-5}$ for $\frac{d{\epsilon }_{II}}{dt}$ and $973\text{ K}$ for $T$ in equation (1).

  $6.5\times {10}^{-5}=A{\left(1.7\times {10}^7\text{}\text{Pa}\right)}^n\exp -\left(\frac{\Delta H_p}{k\left(973\text{ K}\right)}\right)$ …… (4)

Substitute $2.45\times {10}^7\text{}\text{Pa}$ for $\sigma$ , $8.7\times {10}^{-4}$ for $\frac{d{\epsilon }_{II}}{dt}$ and $973\text{ K}$ for $T$ in equation (1).

  $8.7\times {10}^{-4}=A{\left(2.45\times {10}^7\text{}\text{Pa}\text{}\text{Pa}\right)}^n\exp -\left(\frac{\Delta H_p}{k\left(973\text{ K}\right)}\right)$ …… (5)

Divide equation (4) by equation (5).

  $\begin{array}{c}\frac{6.5\times {10}^{-5}}{8.7\times {10}^{-4}}=\frac{A{\left(1.7\times {10}^7\text{}\text{Pa}\right)}^n\exp -\left(\frac{\Delta H_p}{k\left(973\text{ K}\right)}\right)}{A{\left(2.45\times {10}^7\text{}\text{Pa}\right)}^n\exp -\left(\frac{\Delta H_p}{k\left(973\text{ K}\right)}\right)}\\ 0.075={\left(\frac{1.7\times {10}^7\text{}\text{Pa}}{2.45\times {10}^7\text{}\text{Pa}}\right)}^n\end{array}$

Take log on both sides.

  $\begin{array}{c}\log \left(0.075\right)=n\log \left(\frac{1.7\times {10}^7\text{}\text{Pa}}{2.45\times {10}^7\text{}\text{Pa}}\right)\\ n=\frac{-1.13}{-0.1587}\\ n=7.11\end{array}$

Conclusion:

Thus, the value of power $n$ at the logarithms of stress of $7$ and logarithms of stress of $7.3$.of stresses are $11$ and $7.11$ respectively.