Answer:
Non steady state process
Because D ∝ 1/t
Explanation:
Here we have
[tex]\frac{c_x -c_D}{c_S-c_D} = 1-erf(\frac{x}{2\sqrt{Dt} } )[/tex]
Where:
[tex]c_D[/tex] = 0.0025 % by weight of N
[tex]c_S[/tex] = 0.45 % by weight of N
[tex]c_x[/tex] = 0.12 % by weight of N
Therefore,
[tex]\frac{0.12 -0.0025}{0.45-0.0025} = 1-erf(\frac{x }{2\sqrt{Dt} } )= 0.2626[/tex]
[tex]erf(\frac{x }{2\sqrt{Dt} } )= 1-0.2626 = 0.7374[/tex]
From the inverse erf function we have
[tex]\frac{x }{2\sqrt{Dt} } = 0.7921[/tex]
Since x = 0.45 mm, we have
[tex]\sqrt{Dt} =\frac{4.5\times10^{-4}}{2\times 0.7921}[/tex] and
Dt = 8.069×10⁻⁸ m²
[tex]D_0exp\frac{-Qd}{RT} \times t = 8.07\times10^{-8} m^2[/tex]
For steady state
[tex]\frac{dC}{dt} =0 =D\frac{d^2C}{dx^2}[/tex]
Since Dt = 8.069×10⁻⁸ m² = Constant, then as the time increases the diffusion rate decreases hence the process is not steady state.
D is inversely proportional to t.
