Question

The outer surface of a steel gear is to be hardened by increasing its carbon content; the carbon is to be supplied from an external carbon-rich atmosphere, which is maintained at an elevated temperature. A diffusion heat treatment at 850 °C (1123 K) for 10 min increases the carbon concentration to 0.90 wt% at a position 1.0 mm below the surface. Estimate the diffusion time required at 650 °C (923 K) to achieve this same concentration also at a 1.0-mm position. Assume that the surface carbon content is the same for both heat treatments, which is maintained constant. D = 1.1 x 10-6 m2 /s and Qd = 87,400 J/mol C diffusion in a-Fe. (8 points)

Answer 1

Answer:

$\mathbf{t_2 = 75.696 \ min}$

Explanation:

From the question:

The outer surface of a steel gear is to be hardened by increasing its carbon content

Given that :

Diffusion of heat temperature at $T_1$ 850 °C  = 1123 K

Diffusion time $t_1$ = 10 min

diffusion after the carbon concentration at a position $x_1$ ( 1.0 mm) below the surface =  0.90 wt%

Preexponential = 1.1 × 10⁻⁶ m²/s

Activation Energy $Q_d$ = 87400 J/mol

We are to determine the time $t_2$ at 650  °C (923 K) to achieve the same diffusion result as at 850 °C (1123 K) for $t_1$ = 10 min

Considering Fick's second law for the condition of Constant surface concentration; we have:

$\frac{Cx-C_0}{C_s-C_0} = 1-erf(\frac{x}{2\sqrt{Dt} } )$   ------ equation (1)

where;

$C_0 =$ concentration of the diffusing solute atom before diffusion

$C_s$ = Constant surface concentration

$C_x$ = Concentration at depth x after time t

$erf(\frac{x}{2\sqrt{Dt} } )$ = Gaussian error function

At some desired specific  concentration of solute $C_1$ in an alloy ; the left side of the above equation (1) thus becomes constant ;

i.e $\frac{Cx-C_0}{C_s-C_0} = \mathbf{ constant}$

Then ;  $\frac{x}{2\sqrt{Dt} }$ = constant

$\frac{x^2}{Dt}$ = constant

Dt = constant

Thus; $D_1t_1 = D_2t_2$

Therefore, the time $t_2$ at 650°C($T_2$ = 923 K) required to produce the same diffusion on result as at 850°C ($T_1$ = 1123 K) for $t_1$ = 10 min is $t_2 = \frac{D_1t_1}{D_2}$

We need to first determine the Diffusion coefficient at 1123 K and 923 K ( i.e $D_1$  and  $D_2$)

At $T_1$ = 1123 K , Diffusion coefficient $D_1$ is calculated by the equation $D_1 = D_0 exp ( - \frac{Q_d}{RT_1})$       (equation from temperature dependence of the diffusion coefficient)

$D_1 = 1.1 * 10^{-6} \ exp ( - \frac{87,400}{8.314*1123} )$

$D_1 = 9.462*10^{-11} \ m^2/s$

$D_2 = D_0 exp ( - \frac{Q_d}{RT_2})$

$D_2 = 1.1 * 10^{-6} \ exp ( - \frac{87,400}{8.314*923} )$

$D_2 = 1.25*10^{-11} m^2/s$

$t_2 = \frac{ 9.462*10^{-11}*10}{ 1.25*10^{-11} }$

$\mathbf{t_2 = 75.696 \ min}$