8.33 (a) Estimate the activation energy for (i.e., Qc in Equation 8.20) for the S-590 alloy having the creep behavior shown in Figure 8.31. Use data taken at a stress level of 300 MPa (43,500 psi) and temperatures of 650°C and 730°C . Assume that the stress exponent n is independent of temperature. (b) Estimate at 600°C (873 K) and 300 MPa. Solution (a) We are to estimate the activation energy for creep for the S-590 alloy having the steady-state creep behavior shown in Figure 8.31, using data taken at = 300 MPa and temperatures of 650C and 730C. Since is a   constant, Equation 8.20 takes the form where is now a constant. (Note: the exponent has about the same value at these two temperatures per Problem n 8.32.) Taking natural logarithms of the above expression For the case in which we have creep data at two temperatures (denoted as and and their corresponding T1 T2) creep rates (and ), it is possible to set up two simultaneous equations of the form as above, with two unknowns, namely and Solving for yields Qc. Qc Let us choose as 650C (923 K) and as 730C (1003 K); then from Figure 8.31, at = 300 MPa, = 8.9 10-5 T1 T2  h-1 and = 1.3 10-2 h-1. Substitution of these values into the above equation leads to = 480,000 J/mol (b) We are now asked to estimate at 600C (873 K) and 300 MPa. It is first necessary to determine the value of , which is accomplished using the first expression above, the value of and one value each of and (say Qc, T and Thus, T1).
8.33 (a) Estimate the activation energy for (i.e., Qc in Equation 8.20) for the S-590 alloy having the creep behavior shown in Figure 8.31. Use data taken at a stress level of 300 MPa (43,500 psi) and temperatures of 650°C and 730°C . Assume that the stress exponent n is independent of temperature. (b) Estimate at 600°C (873 K) and 300 MPa. Solution (a) We are to estimate the activation energy for creep for the S-590 alloy having the steady-state creep behavior shown in Figure 8.31, using data taken at  = 300 MPa and temperatures of 650C and 730C. Since  is a constant, Equation 8.20 takes the form where is now a constant. (Note: the exponent n has about the same value at these two temperatures per Problem 8.32.) Taking natural logarithms of the above expression For the case in which we have creep data at two temperatures (denoted as T1 and T2) and their corresponding creep rates (and ), it is possible to set up two simultaneous equations of the form as above, with two unknowns, namely and Qc. Solving for Qc yields Let us choose T1 as 650C (923 K) and T2 as 730C (1003 K); then from Figure 8.31, at  = 300 MPa, = 8.9 10-5 h-1 and = 1.3 10-2 h-1. Substitution of these values into the above equation leads to = 480,000 J/mol (b) We are now asked to estimate at 600C (873 K) and 300 MPa. It is first necessary to determine the value of , which is accomplished using the first expression above, the value of Qc, and one value each of and T (say and T1). Thus,