SME
Transport Phenomena
Mass transport
Use this for “A gas diffuses across a 3 cm film from 5 to 1 mol/m³ with DAB = 1.5 × 10⁻⁵ m²/s. What is the flux?”
Diagram
In
steady
diffusion
the
line
is
straight:
the
flux
is
the
same
at
every
depth,
NA
=
D(cA1 − cA2)
L
=
0.0006
mol/(
m
2
·s).
This is an example. Type your own values to start a new problem.
NA (mol/(m²·s))
=
DAB (m²/s)
L (m)
×
(
cA1 (mol/m³)
−
cA2 (mol/m³)
)
Formulas
NA
=
DAB
×
cA1 − cA2
L
6 ×
10
-4
=
2 ×
10
-5
×
2 − 0.5
0.05
Assumptions
•
Steady, with no reaction in the film: the flux is the same at every depth.
•
Equimolar counterdiffusion: B moves back as fast as A moves on, so no bulk flow.
•
The diffusivity DAB is constant, so the concentration falls in a straight line.
Step-by-step
WE KNOW
DAB = 2 × 10⁻⁵ m²/s, cA1 = 2 mol/m³, cA2 = 0.5 mol/m³, L = 0.05 m
FIND
molar flux (NA)
Step 1 · Find molar flux (NA)
Use NA = DAB × (cA1 − cA2) ÷ L
Fick’s law: the flux is DAB times the straight line’s slope, from high to low.
NA
=
2 ×
10
-5
×
2 − 0.5
0.05
NA
=
(2
×
10
-5
)
×
1.5
0.05
NA
=
3 ×
10
-5
0.05
NA
=
6 ×
10
-4
mol/(
m
2
·s)
Check
Put the numbers back in. Both sides should match.
6 ×
10
-4
=
2 ×
10
-5
×
2 − 0.5
0.05
✓
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