Question

Nitrogen from a gaseous phase is to be diffused into pure iron at 700°C. If the surface concentration is maintained at 0.26 wt% N, what will be the concentration (in weight percent) 2.8 mm from the surface after 6.3 h? The diffusion coefficient for nitrogen in iron at 700°C is 2.2 × 10-10 m2/s.

Answer 1

Answer:

$Cx = .376 \times 0.26 = 0.0978 wt% N$

Explanation:

Given details

concentration of surface is Cs 0.26 wt%

distance from surface is x 2.8 mm

time t = 3.6 h

original concentration is Co 0

we know that

$\frac{Cx - Co}{Cs - Co} = 1 - erf(\frac{x}{2\sqrt{Dt}}$

$\frac{Cx - 0}{0.26 - Co} = 1 - erf(\frac{2\times 10^{-3}}{2\sqrt{2.2\times 10^{-11} 5 hr (3600 /1 hr)}}$

$\frac{Cx}{0.26} = 1 - erf(0.626)$

FROM TABLE OF ERROR FUNCTION WE HAVE

Z           erf(z)

0.60        0.6039

0.65         0.6420

so by interpolation technique we have

for z = 0.626 we have value of erf(0.626) =   0.623

$\frac{Cx}{0.26} = 1 - 0.623$

$Cx = .376 \times 0.26 = 0.0978 wt% N$

Answer 2

The concentration in weight percent from the surface is 0.0962%.

Given the following data:

• Temperature = 700°C.

• Surface concentration = 0.26 wt% N.

• Distance = 2.8 mm.

• Time = 6.3 hours.

• Diffusion coefficient for nitrogen = $2.2 \times 10^{-10}\;m^2/s$

Note: The original concentration of iron (Co) is 0.

To determine the concentration in weight percent:

How to calculate the concentration.

In this exercise, we would apply Fick's second law and error function:

$\frac{C_x - C_o}{C_s - C_o} =1-erf\frac{x}{2\sqrt{Dt} }$

Substituting the given parameters into the formula, we have;

$\frac{C_x - 0}{0.26 - 0} =1-erf(\frac{2.8 \times 10^{-3}}{2\sqrt{2.2 \times 10^{-10} \times 22680} })\\\\\frac{C_x - 0}{0.26 - 0} =1-erf(\frac{2.8 \times 10^{-3}}{2\sqrt{4.99 \times 10^{-6}} })\\\\\frac{C_x }{0.26 } =1-erf(\frac{2.8 \times 10^{-3}}{2\times 0.0022 })\\\\\frac{C_x }{0.26} =1-erf(0.6364)\\\\\frac{C_x}{0.26 } =1-0.63\\\\\frac{C_x}{0.26 } =0.37\\\\C_x = 0.26 \times 0.37$

Cx = 0.0962%.

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