This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Taking the natural logarithm of Equation (8.20) yields 196 ln s = ln K 2 + n ln Q c RT With the given data there are two unknowns in this equationnamely K 2 and Q c . Using the data provided in the problem we can set up two independent equations as follows: ln 6.6 x 104 (h)1 [ ] = ln K 2 + (8.5) ln(140 MPa)Q c (8.31 J /mol  K)(1090 K) ln 8.8 x 102 (h)1 [ ] = ln K 2 + (8.5) ln(140 MPa)Q c (8.31 J /mol  K)(1200 K) Now, solving simultaneously for K 2 and Q c leads to K 2 = 57.5 (h)1 and Q c = 483,500 J/mol. Thus, it is now possible to solve for s at 83 MPa and 1300 K using Equation (8.20) as s = K 2 n exp Q c RT = 57.5 (h)1 [ ] (83 MPa) 8.5 exp483,500 J /mol (8.31 J/mol  K)(1300 K) 4.31 x 102 (h)1 197...
View Full
Document
 Spring '11
 Staff
 Stress

Click to edit the document details