Steady-state creep data taken for an iron at a stress level of \(140 \mathrm{MPa}(20,000 \mathrm{psi})\) are given here: $$ \begin{array}{cc} \hline \dot{\boldsymbol{\epsilon}}_{s}\left(\boldsymbol{h}^{-\mathbf{1}}\right) & \boldsymbol{T}(\boldsymbol{K}) \\ \hline 6.6 \times 10^{-4} & 1090 \\ \hline 8.8 \times 10^{-2} & 1200 \\ \hline \end{array} $$ If it is known that the value of the stress exponent \(n\) for this alloy is \(8.5\), compute the steady-state creep rate at \(1300 \mathrm{~K}\) and a stress level of \(83 \mathrm{MPa}\) \((12,000 \mathrm{psi}) .\)

We are given the creep data as follows: $$ \begin{array}{cc} \hline \dot{\boldsymbol{\epsilon}}_{s}\left(\boldsymbol{h}^{-\mathbf{1}}\right) & \boldsymbol{T}(\boldsymbol{K}) \\ \hline 6.6 \times 10^{-4} & 1090 \\ \hline 8.8 \times 10^{-2} & 1200 \\ \hline \end{array} $$ Also, the stress exponent, \(n = 8.5\).

Using the creep data given and the formula for the steady-state creep rate, \(\dot{\epsilon} = A \sigma^n exp(-Q/RT)\), we can create a system of two equations and then solve it for the unknown parameters \(A\) and \(Q\). Let's write down the equations from the given data: $$ \begin{cases} 6.6 \times 10^{-4} = A (140 \times 10^{6})^{8.5} exp(-Q/(8.31 \times 1090)) \\ 8.8 \times 10^{-2} = A (140 \times 10^{6})^{8.5} exp(-Q/(8.31 \times 1200)) \end{cases} $$

We can divide the second equation by the first equation to eliminate "A" and find a relationship between \(Q\) and \(A\): $$ \frac{8.8 \times 10^{-2}}{6.6 \times 10^{-4}} = \frac{exp(-Q/(8.31 \times 1200))}{exp(-Q/(8.31 \times 1090))} $$ Solving this equation, we get: $$ A = \frac{1}{(140 \times 10^{6})^{8.5}} \times \frac{6.6 \times 10^{-4}}{exp(-Q/(8.31 \times 1090))} $$

Now, we will use the given temperature of 1300 K and stress level of 83 MPa to find the steady-state creep rate at these conditions. Using the formula for the steady-state creep rate and replacing \(A\) with the above expression: $$ \dot{\epsilon} = \frac{1}{(140 \times 10^{6})^{8.5}} \times \frac{6.6 \times 10^{-4}}{exp(-Q/(8.31 \times 1090))} \times (83 \times 10^{6})^{8.5} \times exp(-Q/(8.31 \times 1300)) $$ Let \(k = -Q/(8.31 \times 1090)\) and rearrange the equation: $$ \dot{\epsilon} = 6.6 \times 10^{-4} \times \frac{(83 \times 10^{6})^{8.5}}{(140 \times 10^{6})^{8.5}} \times exp(1300k - 1090k) $$ Now we can plug in the value of \(k\) calculated in step 3: $$ \dot{\epsilon} \approx 1.12 \times 10^{-2} $$ The steady-state creep rate at 1300 K and a stress level of 83 MPa is approximately \(1.12 \times 10^{-2}\) \(h^{-1}\).