Engineering, 25.02.2020 03:37 jnaifuson

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.

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Answer from: sarinawhitaker

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:

$D = D_o * e^(^-^\frac{Q_d}{R*T}^)$

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

$D_o * e^(^-^\frac{Q_d}{R*T_1}^)*t_1 = D_o * e^(^-^\frac{Q_d}{R*T_2}^)*t_2$

- Simplify:

$e^(^-^\frac{Q_d}{R*T_1}^)*t_1 = e^(^-^\frac{Q_d}{R*T_2}^)*t_2\\\\\frac{t_2}{t_1} = e^(^+^\frac{Q_d}{R*T_2}^ - ^\frac{Q_d}{R*T_1}^)$

- Take Natural Logs:

$Ln ( \frac{t_2}{t_1} ) = \frac{Q_d}{R*T_2} - \frac{Q_d}{R*T_1}\\\\\frac{R*Ln ( \frac{t_2}{t_1} )}{Q_d} = \frac{1}{T_2} - \frac{1}{T_1}\\\\\frac{R*Ln ( \frac{t_2}{t_1} )}{Q_d} + \frac{1}{T_1} = \frac{1}{T_2}$

- Plug in values:

$\frac{8.3145*Ln ( \frac{2.5}{16} )}{148000} + \frac{1}{(780+273)} = \frac{1}{T_2}\\\\-1.04285*10^-^4 + 9.496676*10^-^4 = \frac{1}{T_2}\\\\ \frac{1}{T_2} = 8.453888225*10^-^4\\\\T_2 = \frac{1}{8.453888225*10^-^4} = 1182.89 K$

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

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