SME
Materials Science & Material Properties
Defects and diffusion
Use this for “Steel at 0.20 wt% C is carburized with its surface held at 1.00 wt% for 10 h (D = 1.6 × 10⁻¹¹ m²/s). What is the carbon 0.5 mm down?”
Chart
C(0.5) ≈ 0.7131
C = 1 at the surface, falling toward C₀ = 0.2 deep inside
This is an example. Type your own values to start a new problem.
Cₓ (wt%)
=
Cₛ (wt%)
−
(
Cₛ (wt%)
−
C₀ (wt%)
)
erf(
z
)
z
=
x (mm)
2
D (m²/s)
×
t (h)
Formulas
z
=
x
×
0.001
2 ×
√
D
×
t
× 3,600
0.3294
=
0.5
×
0.001
2 ×
√
1.6 ×
10
-11
× 10 × 3,600
Cₓ
=
Cₛ
−
(Cₛ
−
C₀
)
×
erf(
z
)
0.7131 = 1 − (1 − 0.2) × erf(0.3294)
Assumptions
•
The surface holds Cₛ from the start, and the part is thick next to 2√(Dt) (a semi-infinite solid).
•
D does not change with the carbon content: one temperature, one D.
•
erf is the error function, read from a table or a calculator; erf(0) = 0 and it nears 1 past z = 2.
Step-by-step
WE KNOW
x = 0.5 mm, C₀ = 0.2 wt%, Cₛ = 1 wt%, D = 1.6 × 10⁻¹¹ m²/s, t = 10 h
FIND
depth over 2√(Dt) (z), carbon at the depth (Cₓ)
Step 1 · Find depth over 2√(Dt): z
Use z = x × 0.001 ÷ (2 × √(D × t × 3,600))
Turn
the
depth
into
metres
and
the
hours
into
seconds,
then
measure
the
depth
in
units
of
2
√
Dt
.
z
=
0.5
×
0.001
2 ×
√
1.6 ×
10
-11
× 10 × 3,600
z
=
0.5
×
0.001
÷
(2
×
√((1.6
×
10
-10
)
×
3,600))
z
=
0.5
×
0.001
2 ×
√
5.76 ×
10
-7
z
=
0.5
×
0.001
2 × 0.0007589
z
=
0.5
×
0.001
0.001518
z
=
0.0005
0.001518
z
=
0.3294
Step 2 · Find carbon at the depth (Cₓ)
Use Cₓ = Cₛ − (Cₛ − C₀) × erf(z)
Fick’s
second
law
with
a
fixed
surface:
Cₓ −
C₀
Cₛ −
C₀
=
1
−
erf(
z
).
Cₓ = 1 − (1 − 0.2) × erf(0.3294)
Cₓ = 1 − 0.8 × erf(0.3294)
Cₓ = 1 − 0.8 × 0.3587
Cₓ = 1 − 0.287
Cₓ = 0.7131 wt%
Check
Put the numbers back in. Both sides should match.
0.3294
=
0.5
×
0.001
2 ×
√
1.6 ×
10
-11
× 10 × 3,600
✓
0.7131 = 1 − (1 − 0.2) × erf(0.3294) ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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