Answer:
5.12 mm
Explanation:
Let's convert the time taken from hours to seconds
t = 5 h
t = [tex]5h * \frac{3600s}{1h}[/tex]
t = 18000 s
The relation expressing the concentration, position and time together is given by the formula:
[tex]\frac{C_x-C_0}{C_s-C_0} = 1 - erf (\frac{x}{2\sqrt{Dt} } )[/tex]
where:
[tex]C_x[/tex] = concentration at depth = 0.206 wt%
[tex]C_0[/tex] = initial concentration = 0.275 wt%
[tex]C_s[/tex] = concentration at the surface position = 0.0 wt%
D = diffusion coefficient = [tex]5.6*10 ^{-10}m^2/s[/tex]
t = time = 18000 s
Replacing the value into the previous formula; we have:
[tex]\frac{0.206-0.275}{0.0-0.275} = 1 - erf (\frac{x}{2\sqrt{5.6*10^{-10}*18000}})[/tex]
0.2509 = 1 - erf (157.85 x)
erf (157.85 x) = 1 - 0.2509
erf (157.85 x) = 0.7491
So, Let's assume the value of z to be 157.485x ; we have:
z = 157.485 x ------------------ Equation (1)
We obtain the value of Z corresponding to erf (Z) = 0.7491 from the Table 5.1 , 'Table of Error Function Values'
[tex]\frac{0.75-x}{0.75-0.70} = \frac{0.7112-0.7491}{0.7112-0.6778}[/tex]
[tex]\frac{0.75-x}{0.05} = \frac{-0.0379}{0.0334}[/tex]
- 0.001895 = (0.75 - z ) 0.0334
- 0.001895 = 0.02505 - 0.0334 z
0.0334 z = 0.02505 + 0.001895
0.0334 z = 0.026945
z = [tex]\frac{0.026945}{0.0334}[/tex]
z = 0.806737
Substituting 0.806737 for z in equation (1)
0.806737 = 157.485 x
x = [tex]\frac{0.806737}{157.485}[/tex]
x = 0.00512 m to mm; we have
[tex]x = 0.00512 m * \frac{1000mm}{1m}[/tex]
x = 5.12 mm
Thus, the position at which the carbon concentration is 0.206 wt% after a 5 h treatment = 5.12 mm
