Question

A diffusion couple of two metals, A and B, was fashioned as shown in Figure 5.1. After a 26 hour heat treatment at 1050°C (and subsequent cooling to room temperature) the concentration of A is 4.0 wt% at the 14-mm position within metal B. If another heat treatment is conducted on an identical diffusion couple, only at 711°C for 26 hours, at what position (in mm) will the composition be 4.0 wt% A? The preexponential and activation energy values for the diffusion of A in B are 2.4 × 10-4 m2/s and 154 kJ/mol, respectively.

Answer 1

1.25 mm

Step-by-step explanation:

Given:

Temperature = 1050°C = 1050 + 273 = 1323 K

concentration of A = 4.0 wt%

preexponential value for the diffusion, D₀ = 2.4 × 10⁻⁴ m²/s

activation energy value for the diffusion,
`Q_d`= 154 kJ/mol = 154 × 10³ J/mol

Diffusion coefficient at 1050°C , D₁₀₅₀ =
`D_0\exp(-Q_d)/(RT)`

here,

R is the ideal gas constant = 8.314 J/mol·K.

T is the temperature

Thus

D₁₀₅₀ =
`2.4*10^(-4)\exp(-154*10^3)/(8.314*1323)`

or

D₁₀₅₀ = 1.98 × 10⁻¹⁰

and,

Diffusion coefficient at 711°C , D₇₁₁ =
`D_0\exp(-Q_d)/(RT)`

D₇₁₁ =
`2.4*10^(-4)\exp(-154*10^3)/(8.314*(711+273))`

or

D₇₁₁ = 1.58 × 10⁻¹²

also,

`\frac{\textup{x}^2}{\textup{D}}` = constant

here, x is the position

thus,

`(x_(1050)^2)/(x_(711)^2) = (D_(1050))/(D_(711))`

or

`(14^2)/(x_(711)^2) = (1.98*10^(-10))/(1.58*10^(-12))`

or

x₇₁₁ = 1.25 mm