SME
Heat Transfer
Mass transfer by diffusion
Use this for “Hydrogen (D = 2 × 10⁻⁹ m²/s) diffuses through a 2 mm membrane from 50 to 5 mol/m³. Find the flux and the rate through 0.01 m².”
Diagram
In
steady
diffusion
the
line
is
straight:
the
flux
is
the
same
at
every
depth,
NA
=
D(cA1 − cA2)
L
=
4.5 ×
10
-5
mol/(
m
2
·s).
This is an example. Type your own values to start a new problem.
J (mol/(m²·s))
=
D (m²/s)
×
C₁ (mol/m³)
−
C₂ (mol/m³)
L (mm)
Ṅ (mol/s)
=
J
×
A (m²)
Formulas
J
=
D
×
C₁
−
C₂
L
4.5 ×
10
-5
=
2 ×
10
-9
×
50 − 5
0.002
Ṅ
=
J
×
A
4.5 ×
10
-7
=
4.5 ×
10
-5
×
0.01
Formulas are worked in m. The step-by-step shows the conversions.
Assumptions
•
Steady and one-dimensional: the concentration falls in a straight line across the wall.
•
The diffusivity D is the same all through the wall.
•
C₁ and C₂ are the concentrations just inside each face (solubility already applied).
Step-by-step
WE KNOW
D = 2 × 10⁻⁹ m²/s, C₁ = 50 mol/m³, C₂ = 5 mol/m³, L = 2 mm, A = 0.01 m²
FIND
molar flux (J), molar rate (Ṅ)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to m).
L = 2 mm = 0.002 m (1 m = 1000 mm)
Step 2 · Find molar flux (J)
Use J = D × (C₁ − C₂) ÷ L
Fick’s
law:
the
flux
is
D
times
the
drop
in
concentration
over
the
thickness.
J
=
2 ×
10
-9
×
50 − 5
0.002
J
=
(2
×
10
-9
)
×
45
0.002
J
=
9 ×
10
-8
0.002
J
=
4.5 ×
10
-5
mol/(
m
2
·s)
Step 3 · Find molar rate (Ṅ)
Use Ṅ = J × A
The flux is per square metre each second; times the area gives the whole rate.
Ṅ
=
4.5 ×
10
-5
×
0.01
Ṅ
=
4.5 ×
10
-7
mol/s
Check
Put the numbers back in. Both sides should match.
4.5 ×
10
-5
=
2 ×
10
-9
×
50 − 5
0.002
✓
4.5 ×
10
-7
=
4.5 ×
10
-5
×
0.01
✓
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Evaporation by the heat–mass transfer analogy
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