Question

Hydrogen is stored in a steel rectangular container having a steel wall thickness of 25 mm.The molar concentration of hydrogen at the inside and outside surface of steel container are5.5 mol/m² and 2.7 mol/m". Determine the molar diffusion flux for hydrogen through thecontainer, if the diffusivity of hydrogen in steel is 0.26x10-12 m²/s.(A) 2.9x10-11 mol/s. m²(B) 2.4x10-1 mol/s: m(C) 3.2x10-11 mol/s. m²(D) 4.2 x 10-11 mol/s: m?​

Answer 1

Option (A) is correct.

The Molar diffusion flux for hydrogen through the container is 2.9 x 10^(-11) mol/s m².

Given:

The thickness of steel wall = 25mm = 0.025m

The molar concentration of hydrogen inside the steel container

= 5.5 mol/m²

The molar concentration of hydrogen outside the steel container

= 2.7 mol/m²

Diffusivity of Hydrogen in steel = 0.26 x $10^{-12}$ m²/s

To find:

Molar diffusion flux for hydrogen through the container.

Solution:

The question can be solved using Fick's law which states that

$N_{h} = \frac{D_{hs} (C_{h1} - C_{h2})}{x_{2} - x_{1} }$

$N_{h}$ = Molar flux diffusion for hydrogen

$D_{hs}$ = Diffusivity of hydrogen in steel

$C_{h1}$ = Molar concentration of hydrogen inside the container

$C_{h2}$ = Molar concentration of hydrogen outside the container

$(x_{2} - x_{1} )$ = Thickness of steel

Replacing the variables with the given data,

$N_{h}$ = (0.26 x $10^{-12}$) x (5.5 - 2.7) / 0.025 mol/s m²

=> $N_{h}$ = 2.912 x $10^{-11}$ mol/s m²

Hence, the Molar diffusion flux for hydrogen through the container is  2.9 x 10^(-11) mol/s m².

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Answer 2

Answer: Molar flux of hydrogen through steel is $\bold {N_{Ax}=2.912\times 10^{-11} \ mol/(s. m^2)}$.

Given:

• Steel wall thickness is 25 mm.
• The molar concentration of hydrogen at the inside and outside surface of steel container are 5.5 mol/m² and 2.7 mol/m².
• Diffusivity of hydrogen in steel is $0.26\times10^{-12 }$ m²/s.

To Find: Molar diffusion flux for hydrogen through the container.

Explanation:

Step 1: Diffusivity $D_{AB} =0.26\times10^{-12} \ m^2/s$

Hydrogen concentration inside the tank (inside wall) $C_{A_i}=5.5\ mol/m^2$

Hydrogen concentration inside the tank (outside wall) $C_{A_o}=2.7\ mol/m^2$

Wall thickness $\triangle x=25\ mm$

Assumptions:

1. Diffusion is only in one direction i.e. x direction across the wall therefore the system can be assumed to act as a slab.
2. Steady state  with no chemical reaction i.e. $\frac{\partial C_A}{\partial t} =0\ and\ R_A=0$

Step 2: Apply the above assumptions to the equation of continuity of species for rectangular coordinates

$\frac{\partial C_A}{\partial t} +\frac{\partial N_A_x}{\partial x} +\frac{\partial N_A_y}{\partial y} +\frac{\partial N_A_z}{\partial z} =R_A$

To obtain: $\frac{\partial N_{A_x}}{x} =0$

Therefore $N_{A_x}$ is independent of x i.e. it is a constant

$N_{Ax}=-D_{AB}\frac{\partial C_A}{\partial x} + Y_A(N_A+N_B)$

Step 3: Since component A concentration in the container wall is relatively low, $Y_A \approx 0$. Then bulk flow will be negligible.

$N_{Ax}=-D_{AB}\frac{\partial C_A}{\partial x}$

and since $\frac{\partial C_A}{\partial x} =constant$

$N_{Ax}=-D_{AB}\frac{\partial C_A}{\partial x} \implies N_{Ax}=-D_{AB}\frac{C_{Ao}-C_{Ai}}{\triangle x}$

Substituting the given data:

$N_{Ax}=-D_{AB}\frac{C_{Ao}-C_{Ai}}{\triangle x}=-[0.26\times 10^{-12}\ m^2/s][\frac{(2.7-5.5)\frac{mol}{m^3} }{0.25}]$

The molar flux of hydrogen through steel for this case is:

$N_{Ax}=2.912\times 10^{-11} \ mol/(s. m^2)$

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