When alpha -iron is subjected to an atmosphere of nitrogen gas, the concentratio

Question

When alpha -iron is subjected to an atmosphere of nitrogen gas, the concentration of nitrogen in the iron, C_N(in weight percent), is a function of hydrogen pressure, p_N_2 (in MPa). and absolute temperature according to C_N = 4.90 times 10^-3 Squareroot PN_2 exp (- 37,600 J/mol / RT) Furthermore, the values of D_0 and Q_d for this diffusion system are 5.0 times 10 m^2/s and 77,000 J/mol, respectively. Consider a thin iron membrane 1.5-mm thick that is at 300 degree C. Compute the diffusion flux through this membrane if the nitrogen pressure on one side of the membrane is 0.10 MPa (0.99 atm) and that on the other side is 5.0 MPa (49.3 atm).

Explanation / Answer

Given that,

pN2on side A = 5MPa

pN2on side B = 0.1MPa

temperature 3000C = 573.15K

First, we need to determine the concentration of nitrogen at each face using the given equation.

CN = 4.9 *10-7 pN2 exp (-37600/RT J/mol)

At the low pressure (or B) side

CN(B) = 4.9 *10-7 0.1MPa exp (-37600/8.314 * 573.15 J/mol)

CN(B) = 1.22*10-6 wt%

Whereas, for the high pressure (or A) side

CN(A) = 4.9 *10-7 5MPa exp (-37600/8.314 * 573.15 J/mol)

CN(A) = 8.64*10-6 wt%

We now have to convert concentrations in weight percent to mass of nitrogen per unit volume of solid. At face B there are 1.22*10-6 g (or 1.22*10-6 kg) of nitrogen in 100 g of Fe, which is virtually pure iron. From the density of iron (7.87 g/cm3), the volume iron in 100 g (VB) is just

VB = 100/7.87 = 12.7 cc = 1.27*10-5 m3

Therefore, the concentration of nitrogen at the B face in kilograms of N per cubic meter of alloy [CN(B)'] is just

CN(B)' = CN(B) / VB

CN(B)' = 1.22*10-9 / 1.27*10-5 =9.6*10-4 kg/m3

At the A face the volume of iron in 100 g (VA) will also be 1.27*10-5 m3, and

CN(A)' = CN(A) / VB

CN(A)' = 8.64*10-9 / 1.27*10-5 =6.8*10-3 kg/m3

Thus, the concentration gradient is just the difference between these concentrations of nitrogen divided by the thickness of the iron membrane; that is

dC/dx = CN(B)' - CN(A)' / xB - xA

dC/dx = (9.6*10-4 - 6.8*10-3) / (1.5*10-3)

dC/dx = -3.89 kg/m4

and we calculate the value of the diffusion coefficient at 300°C using equation,

D = D0 exp (-Qd/RT)

D = 5*10-7 exp (77000/8.314*573.15)

D = 4.79*10-14 m2/s

And, finally, the diffusion flux is computed using below fick's law as

J = -D dC/dx

J = -4.79*10-14 (-3.89)

J = 1.86*10-13 kg/m2s

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