At approximately what temperature (in Kelvin) would a specimen of an alloy have to be carburized for 2.5 h to produce the same diffusion result as at 780°C for 16 h? Assume that values for D0 and Qd are 2.1 × 10-4 m2/s and 148 kJ/mol, respectively.

Temperature is T_2 = 1182.89 K or T = 910°C

Explanation:

Given:

- The concentration of carbon content after diffusion process = x

- The Do = 2.1 * 10^-4 m^2 / s

- The Qd = 148 KJ/mol

- Time taken @ T_1 = 780°C , t_1 = 16 hr

Find:

At approximately what temperature (in Kelvin) would a specimen of an alloy have to be carburized for 2.5 h to produce the same diffusion result as at 780°C for 16 h?

Solution:

- From Concepts we know that square of concentration (x^2) is proportional to product of Diffusion coefficient and time taken for diffusion process ( D * t). The relation can be written as follows:

                                          x^2 / D*t = constant

- For the completion of process for the alloy at both Temperatures means x is constant for both cases. Hence, we can use the above relation to develop the following expression:

                                          D_1 * t_1 = D_2 * t_2

- The diffusion coefficient is a function of temperature T as follows:

- Hence, we can substitute the relation for Diffusion coefficient D into the relationship develop above for respective Temperatures:

- Simplify:

- Take Natural Logs:

- Plug in values:

- Hence, the Temperature is T_2 = 1182.89 K or T = 910°C

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