When a-iron is subjected to an atmosphere of nitrogen gas, the concentration of nitrogen in the iron, CN (in weight percent), is a function of hydrogen pressure, pN2 (in MPa), and absolute temperature (T) according to CN = 4.90 * 10-31pN2 expa - 37,600 J>mol RT b (5.14) Furthermore, the values of D0 and Qd for this diffu- sion system are 5.0 107 m2 /s and 77,000 J/mol, re- spectively. Consider a thin iron membrane 1.5-mm thick at 300C. 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 on the other side is 5.0 MPa (49.3 atm).

Diffusion Flux = 8.81 x 10^(-13) Kg/m².s

Explanation:

We know that

P_N2 = 5 MPa

J = D•(dc/dx)

Thus, J ≈ D•(Δc/Δx)

Now, D is expressed as;

D = D_o(e^(-Qd/RT))

Put it for D in J equation to get;

J = D_o(e^(-Qd/RT))•(Δc/Δx)

We are also given that;

C_N = 4.9 x 10^(-3) √(P_N2) • (e^(-37,600/RT))

Now, R is gas constant with a value of R = 8.314 J/mol·K

T = 300°C = 300 + 273K = 573 °K

So, % wt of N_2 at face 1 is;

%wt = 4.9 x 10^(-3) √(5) • (e^(-37,600/(8.314 x 573)) %

%wt = 4.092 x 10^(-6) %

Now, let's do the same for face 2 where P_N2 = 0.1 MPa;

% wt of N_2 at phase 2 is;

% wt = 4.9 x 10^(-3) √(0.1) • (e^(-37,600/(8.314 x 573)) %

%wt = 5.79 x 10^(-7) %

Concentration of N_2 at face 1 will be = %wt • ρ_fe

Where, ρ_fe is density of iron and it has a value of 7873 kg/m³

Thus,

Concentration of N_2 at face 1 (C1) = 4.092 x 10^(-6) x 7873 = 0.0322

C1 = 0.0322 kg/m³

Let's do the same for face 2;

Concentration of N_2 at face 2; (C2) = 5.79 x 10^(-7) x 7873 = 0.00456

C2 = 0.00456 kg/m³

So,

Δc = C1 - C2 = 0.0322 - 0.00456 = 0.02764 Kg/m³

Substitute this value for Δc in J equation. Thus;

J = D_o(e^(-Qd/RT))•(Δc/Δx)

J = 5 x 10^(-7)[(e^(-77,000/(8.314x573))] • (0.02764/(1.5 x 10^(-3))) = 8.81 x 10^(-13) Kg/m².s

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