Let's break down the problem step by step. ## Problem Rest Given: Steady-state creep data for an iron at a stress level of 48 MPa and two temperatures (110 K and 120 K). - Creep rate at 110 K, 48 MPa is \(8.8 \times 10^{-10}\) 1/s. - Creep rate at 120 K, 48 MPa is \(6.2 \times 10^{-8}\) 1/s. - Stress exponent \(n\) for both temps = 8.1. - You need to estimate the steady-state creep rate at 120 K at a stress level of 79 MPa. --- ## Step 1: Creep Equation The steady-state creep rate is generally expressed as: \[ \dot{\epsilon} = A \sigma^n e^{-Q/(RT)} \] Where: - \(\dot{\epsilon}\) = steady-state creep rate - \(A\) = material constant - \(\sigma\) = applied stress - \(n\) = stress exponent - \(Q\) = activation energy - \(R\) = gas constant (\(8.314\,\text{J/mol·K}\)) - \(T\) = absolute temperature (K) --- ## Step 2: Find the Ratio of Creep Rates at Different Stresses (Same Temperature) Since you need the creep rate at **120 K** at a higher stress (\(79\,\text{MPa}\)), use the ratio: \[ \frac{\dot{\epsilon}_2}{\dot{\epsilon}_1} = \left(\frac{\sigma_2}{\sigma_1}\right)^n \] Where: - \(\dot{\epsilon}_1\) = known creep rate at 120 K, 48 MPa = \(6.2 \times 10^{-8}\) 1/s - \(\sigma_1 = 48\,\text{MPa}\) - \(\sigma_2 = 79\,\text{MPa}\) - \(n = 8.1\) --- ## Step 3: Plug in Values \[ \frac{\dot{\epsilon}_2}{6.2 \times 10^{-8}} = \left(\frac{79}{48}\right)^{8.1} \] First, calculate \(\frac{79}{48}\): \[ \frac{79}{48} \approx 1.6458 \] Now, raise this to the power of 8.1: \[ (1.6458)^{8.1} \approx e^{8.1 \times \ln(1.6458)} \] \[ \ln(1.6458) \approx .4979 \] \[ 8.1 \times .4979 \approx 4.033 \] \[ e^{4.033} \approx 56.43 \] --- ## Step 4: Solve for New Creep Rate \[ \dot{\epsilon}_2 = 6.2 \times 10^{-8} \times 56.43 \] \[ \dot{\epsilon}_2 \approx 3.5 \times 10^{-6} \text{ 1/s} \] --- ## **Final Answer** \[ \boxed{3.5 \times 10^{-6} \text{ 1/s}} \] This is the estimated steady-state creep rate at 120 K and 79 MPa stress.