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)

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 850 °C  = 1123 K

Diffusion time = 10 min

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

Preexponential = 1.1 × 10⁻⁶ m²/s

Activation Energy = 87400 J/mol

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

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

  ------ equation (1)

where;

concentration of the diffusing solute atom before diffusion

= Constant surface concentration

= Concentration at depth x after time t

= Gaussian error function

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

i.e

Then ;   = constant

= constant

Dt = constant

Thus;

Therefore, the time at 650°C( = 923 K) required to produce the same diffusion on result as at 850°C ( = 1123 K) for = 10 min is

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

At = 1123 K , Diffusion coefficient is calculated by the equation       (equation from temperature dependence of the diffusion coefficient)

it means there are 3 atoms of oxygen

the awnser is group 12