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

Answer:

1.25 mm

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 = $D_0\exp\frac{-Q_d}{RT}$

here,

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

T is the temperature

Thus

D₁₀₅₀ = $2.4\times10^{-4}\exp\frac{-154\times10^3}{8.314\times1323}$

or

D₁₀₅₀ = 1.98 × 10⁻¹⁰

and,

Diffusion coefficient at 711°C , D₇₁₁ = $D_0\exp\frac{-Q_d}{RT}$

D₇₁₁ = $2.4\times10^{-4}\exp\frac{-154\times10^3}{8.314\times(711+273)}$

or

D₇₁₁ = 1.58 × 10⁻¹²

also,

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

here, x is the position

thus,

$\frac{x_{1050}^2}{x_{711}^2} = \frac{D_{1050}}{D_{711}}$

or

$\frac{14^2}{x_{711}^2} = \frac{1.98\times10^{-10}}{1.58\times10^{-12}}$

or

x₇₁₁ = 1.25 mm