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Unformatted text preview: Taking the natural logarithm of Equation (8.20) yields 196 ln Ý ε s = ln K 2 + n ln σ 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 Ý ε s at 83 MPa and 1300 K using Equation (8.20) as Ý ε s = K 2 σ 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...
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This note was uploaded on 08/30/2011 for the course EGN 3365 taught by Professor Staff during the Spring '11 term at University of Florida.
 Spring '11
 Staff
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