Question

Problem 5.34 (GO Tutorial) 

At approximately what temperature (in Kelvin) would a specimen of an alloy have to be carburized for 1.2 h to produce the same diffusion result as at 890°C for 13 h? Assume that values for D0 and Qd are 2.9 x 10-6 m2/s and 157 kJ/mol, respectively. T=                      K 

Answer 1

Ficks second law of diffusion is used to calculate the temperature

Ti=890°c - 890+273-15 = 1163.15k t = 13hrs = 1383600 = 46800s te= 1.2 his = 112x3600 = 432os. To = ? Do = 2.9x100 ml, ad= 157kJ/mol = 1574103 /mol Temperature dependence of diffusion coefficient, D. Da Do e RT from Ficks second law of diffusion Cot-ert ( ) since, both procedures produce From equation 2) Dit = Date same diffusion result substituting, equation.) in the equation ③

tel laking natural log on both sides. (.)- GF-1) substitute values: 8.314 4320 157x103 46800 1163.15 in T₂ = 1363.21k OY T = 1090.66°c