(a) Estimate the activation energy for creep (i.e., \(Q_{c}\) in Equation \(\left.8.25\right)\) for the S-590 alloy having the steady-state creep behavior shown in Figure 8.32. Use data taken at a stress level of 300 MPa (43,500 psi) and temperatures of \(650^{\circ} \mathrm{C}\) and \(730^{\circ} \mathrm{C}\). Assume that the stress exponent \(n\) is independent of temperature. (b) Estimate \(\dot{\epsilon}_{s}\) at \(600^{\circ} \mathrm{C}(873 \mathrm{~K})\) and \(300 \mathrm{MPa}\).

Short Answer

Expert verified
b) What is the estimated steady-state creep rate for S-590 alloy at a temperature of 600°C and a stress level of 300 MPa using the calculated activation energy for creep (Qc)?

Step by step solution

01

$$\dot{\epsilon}_{s} = A \sigma^{n} \exp \left( -\frac{Q_{c}}{RT} \right)$$ In this equation, - \(\dot{\epsilon}_{s}\) is the steady-state creep rate, - \(A\) is a material constant, - \(\sigma\) is the applied stress, - \(n\) is the stress exponent, - \(Q_{c}\) is the activation energy for creep, - \(R\) is the gas constant, - \(T\) is the absolute temperature. #Step 2: Rearrange the equation to find the activation energy for creep# We can rearrange Equation 8.25 to find \(Q_{c}\) as follows:

$$Q_{c}=-R T\ln{\left(\frac{\dot{\epsilon}_{s}}{A \sigma^n}\right)}$$ #Step 3: Use the information provided in the data points# The given data points are at a stress level of 300 MPa and temperatures of \(650^{\circ} C\) and \(730^{\circ} C\). Convert these temperatures to the absolute (Kelvin) scale: - \(T_1 = 650^{\circ} C + 273.15 K = 923.15 K\) - \(T_2 = 730^{\circ} C + 273.15 K = 1003.15 K\) From Figure 8.32, find the corresponding steady-state creep rates (\(\dot{\epsilon}_{s1}\) and \(\dot{\epsilon}_{s2}\)) at the given stress level and temperatures. Let's assume the stress exponent, \(n\), is the same at both temperatures. #Step 4: Apply the equation to find the activation energy for creep# Subtract the equations at two different temperatures:
02

$$\Delta Q_{c} = Q_{c2} - Q_{c1} = -RT_1 \ln{\left(\frac{\dot{\epsilon}_{s1}}{A\sigma^n}\right)} + RT_2 \ln{\left(\frac{\dot{\epsilon}_{s2}}{A\sigma^n}\right)}$$ Solve for \(Q_{c}\):

$$Q_{c} = \frac{\Delta Q_{c}}{R (\frac{1}{T_1}-\frac{1}{T_2})}= \frac{T_1T_2R(\ln{\left(\frac{\dot{\epsilon}_{s2}}{\dot{\epsilon}_{s1}}\right)})}{T_2-T_1}$$ Plug in the known values and calculate the expected activation energy for creep, \(Q_{c}\). #Part (b) - Estimate the steady-state creep rate at 600°C and 300 MPa# #Step 1: Translate the given temperature into Kelvin# The given temperature is \(600^{\circ} C\). Convert it into Kelvin: - \(T = 600^{\circ} C + 273.15 K = 873.15 K\) #Step 2: Rearrange Equation 8.25 to find the steady-state creep rate# Rearrange the equation by keeping the steady-state creep rate the subject of the formula:
03

$$\dot{\epsilon}_{s} = A \sigma^{n} \exp{\left(-\frac{Q_{c}}{RT}\right)}$$ #Step 3: Use the known values and the calculated activation energy for creep (Qc) to find the steady-state creep rate# Plug in the known values for a stress level of 300 MPa, temperature of 873.15 K, and previously calculated \(Q_{c}\):

$$\dot{\epsilon}_{s} = A(300 \ \text{MPa})^{n} \exp{\left(-\frac{Q_{c}}{R * (873.15 K)}\right)}$$ By using Figure 8.32, estimate the material constant \(A\) at the given stress and temperature. Calculate the steady-state creep rate, \(\dot{\epsilon}_{s}\).

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Most popular questions from this chapter

Estimate the theoretical fracture strength of a brittle material if it is known that fracture occurs by the propagation of an elliptically shaped surface crack of length \(0.5 \mathrm{~mm}\) (0.02 in.) and a tip radius of curvature of \(5 \times 10^{-3} \mathrm{~mm}\left(2 \times 10^{-4}\right.\) in.), when a stress of \(1035 \mathrm{MPa}(150,000 \mathrm{psi})\) is applied.

(a) Using Figure 8.31, compute the rupture lifetime for an \(S-590\) alloy that is exposed to a tensile stress of \(400 \mathrm{MPa}\) at \(815^{\circ} \mathrm{C}\). (b) Compare this value to the one determined from the Larson-Miller plot of Figure \(8.33\), which is for this same S-590 alloy.

A cylindrical component constructed from an S-590 alloy (Figure 8.31) has a diameter of \(14.5 \mathrm{~mm}\) (0.57 in.). Determine the maximum load that may be applied for it to survive \(10 \mathrm{~h}\) at \(925^{\circ} \mathrm{C}\left(1700^{\circ} \mathrm{F}\right)\).

A specimen of a 4340 steel alloy with a plane strain fracture toughness of \(54.8 \mathrm{MPa} \sqrt{\mathrm{m}}(50 \mathrm{ksi} \sqrt{\mathrm{in}}\).) is exposed to a stress of \(1030 \mathrm{MPa}\) (150,000 psi). Will this specimen experience fracture if the largest surface crack is \(0.5 \mathrm{~mm}\) (0.02 in.) long? Why or why not? Assume that the parameter \(Y\) has a value of \(1.0 .\)

A cylindrical component constructed from an S-590 alloy (Figure 8.31) is to be exposed to a tensile load of \(20,000 \mathrm{~N}\). What minimum diameter is required for it to have a rupture lifetime of at least \(100 \mathrm{~h}\) at \(925^{\circ} \mathrm{C} ?\)

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