Steady-state creep data taken for an iron at a stress level of \(140 \mathrm{MPa}(20,000 \mathrm{psi})\) are given here $$\begin{array}{cc}\hline \dot{\epsilon}_{s}\left(h^{-1}\right) & T(K) \\\\\hline 6.6 \times 10^{-4} & 1090 \\\8.8 \times 10^{-2} & 1200 \\\\\hline\end{array}.$$ 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 MPa \((12,000 \text { psi })\).

We will use the steady-state creep equation which relates the creep rate, stress, temperature, and the stress exponent. The equation is given by: $$ \dot{\epsilon}_{s}=A\left(\frac{\sigma}{R}\right)^{n}e^{\frac{-Q}{RT}} $$ where \(\dot{\epsilon}_{s}\): steady-state creep rate, \(A\): constant, \(\sigma\): applied stress, \(n\): stress exponent, \(Q\): activation energy, \(R\): gas constant \((8.314\, \mathrm{J \, mol^{-1} \, K^{-1}})\), \(T\): temperature in Kelvin.

First, we need to find the activation energy (Q). Then, we can use it to find the creep rate at the desired conditions. Rearrange the steady-state creep equation to solve for Q: $$ Q = -RT \ln{\left(\frac{\dot{\epsilon}_{s}}{A\left(\frac{\sigma}{R}\right)^n}\right)} $$ Since we have 2 sets of data, we can write two equations with the given values: $$ Q = -RT_1 \ln{\left(\frac{\dot{\epsilon}_{s1}}{A\left(\frac{\sigma_1}{R}\right)^n}\right)} $$ and $$ Q = -RT_2 \ln{\left(\frac{\dot{\epsilon}_{s2}}{A\left(\frac{\sigma_2}{R}\right)^n}\right)} $$

Now, we will plug in the given data into two equations to solve for Q: $$ Q = -1090 \cdot 8.314 \ln{\left(\frac{6.6 \times 10^{-4}}{A\left(\frac{140\times 10^6}{8.314}\right)^{8.5}}\right)} $$ and $$ Q = -1200 \cdot 8.314 \ln{\left(\frac{8.8 \times 10^{-2}}{A\left(\frac{140\times 10^6}{8.314}\right)^{8.5}}\right)} $$ Solving these two equations simultaneously to get the value of Q: $$ Q \approx 428000 \, \mathrm{J \, mol^{-1}} $$

Now that we know the activation energy, we can use the steady-state creep equation to compute the creep rate at \(1300 \mathrm{K}\) and 83 MPa stress level. Let's denote the steady-state creep rate at these conditions as \(\dot{\epsilon}_{s3}\): $$ \dot{\epsilon}_{s3} = A\left(\frac{\sigma_3}{R}\right)^{n} e^{\frac{-Q}{RT_3}} $$ In this equation, we already know the values of \(\sigma_3 = 83 MPa\), temperature \(T_3 = 1300 \mathrm{K}\), activation energy \(Q = 428000 \, \mathrm{J \, mol^{-1}}\), and stress exponent \(n = 8.5\). We can use one of the previous equations to find the constant \(A\). Using the first data set: $$ A = \frac{6.6 \times 10^{-4}}{\left(\frac{140\times 10^6}{8.314}\right)^{8.5}e^{-\frac{428000}{8.314 \cdot 1090}}} $$ Now, substituting A and other values into the equation to find the creep rate at the given conditions: $$ \dot{\epsilon}_{s3} \approx 2.02 \times 10^{-4}\,h^{-1} $$

Thus, the steady-state creep rate of the iron alloy at a temperature of \(1300 \mathrm{K}\) and a stress level of 83 MPa is approximately \(2.02 \times 10^{-4} \,h^{-1}\).