Question

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.

Answer 1

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

Step-by-step 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^(^-^(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^(^-^(Q_d)/(R*T_1)^)*t_1 = D_o * e^(^-^(Q_d)/(R*T_2)^)*t_2`

- Simplify:

`e^(^-^(Q_d)/(R*T_1)^)*t_1 = e^(^-^(Q_d)/(R*T_2)^)*t_2\\\\(t_2)/(t_1) = e^(^+^(Q_d)/(R*T_2)^ - ^(Q_d)/(R*T_1)^)`

- Take Natural Logs:

`Ln ( (t_2)/(t_1) ) = (Q_d)/(R*T_2) - (Q_d)/(R*T_1)\\\\(R*Ln ( (t_2)/(t_1) ))/(Q_d) = (1)/(T_2) - (1)/(T_1)\\\\(R*Ln ( (t_2)/(t_1) ))/(Q_d) + (1)/(T_1) = (1)/(T_2)`

- Plug in values:

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

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