SME
Materials Science & Material Properties · Problem type
Steady diffusion flux (Fick’s first law)
Use this for “Carbon diffuses through a 5 mm iron plate from 1.2 to 0.8 kg/m³ with D = 3 × 10⁻¹¹ 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
=
2.4 ×
10
-9
mol/(
m
2
·s).
This is an example. Type your own values to start a new problem.
J (kg/(m²·s))
=
D (m²/s)
×
C₁ (kg/m³)
−
C₂ (kg/m³)
Δx (m)
Formulas
J
=
D
×
C₁
−
C₂
Δx
2.4 ×
10
-9
=
3 ×
10
-11
×
1.2 − 0.8
0.005
Assumptions
•
Steady state: each face is held at its concentration, so the profile is a straight line.
•
J is the mass crossing each square meter each second, from the richer face to the leaner one.
•
D belongs to this species in this host at this temperature (see the Arrhenius page).
Step-by-step
WE KNOW
D = 3 × 10⁻¹¹ m²/s, C₁ = 1.2 kg/m³, C₂ = 0.8 kg/m³, Δx = 0.005 m
FIND
diffusion flux (J)
Step 1 · Find diffusion flux (J)
Use J = D × (C₁ − C₂) ÷ Δx
Fick’s
first
law:
the
flux
is
D
times
the
drop
in
concentration
per
meter,
from
high
to
low.
J
=
3 ×
10
-11
×
1.2 − 0.8
0.005
J
=
(3
×
10
-11
)
×
0.4
0.005
J
=
1.2 ×
10
-11
0.005
J
=
2.4 ×
10
-9
kg/(
m
2
·s)
Check
Put the numbers back in. Both sides should match.
2.4 ×
10
-9
=
3 ×
10
-11
×
1.2 − 0.8
0.005
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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