SME
Heat Transfer
Transient conduction
Use this for “Soil at 10 °C (α = 0.15 × 10⁻⁶ m²/s) has its surface held at −15 °C. How cold is it 0.5 m down after 30 days?”
Chart
T(0.5) ≈ −4.268
Level Tᵢ = 10 doesn't meet the curve in the window shown
This is an example. Type your own values to start a new problem.
z
=
x (m)
2
α (m²/s)
×
t (days)
T (°C)
=
Tₛ (°C)
+
(
Tᵢ (°C)
−
T
s
)
×
erf(z)
Formulas
z
=
x
2 ×
√
α
×
t
0.4009
=
0.5
2 ×
√
1.5 ×
10
-7
× 2,592,000
T
=
Tₛ
+
(Tᵢ
−
Tₛ)
×
erf(
z
)
−4.2677 = −15 + (10 − (−15)) × erf(0.4009)
Formulas are worked in s. The step-by-step shows the conversions.
Assumptions
•
The solid is deep next to √(αt), so its far side never feels the change (semi-infinite).
•
The surface jumps to Tₛ at t = 0 and stays there; α is the same everywhere.
•
erf is the error function, from a table or a calculator; erf(0) = 0 and it nears 1 past z = 2.
Step-by-step
WE KNOW
x = 0.5 m, t = 30 days, α = 1.5 × 10⁻⁷ m²/s, Tᵢ = 10 °C, Tₛ = −15 °C
FIND
similarity variable (z), temperature at depth (T)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to s).
t = 30 days = 2,592,000 s (1 day = 86400 s)
Step 2 · Find similarity variable (z)
Use z = x ÷ (2 × √(α × t))
Measure
the
depth
in
units
of
2
√
αt
,
how
far
the
change
has
soaked
in
by
time
t
.
z
=
0.5
2 ×
√
1.5 ×
10
-7
× 2,592,000
z
=
0.5
2 ×
√
0.3888
z
=
0.5
2 × 0.6235
z
=
0.5
1.247
z
=
0.400938
Step 3 · Find temperature at depth (T)
Use T = Tₛ + (Tᵢ − Tₛ) × erf(z)
At
depth
x
the
share
erf(
z
)
of
the
starting
gap
Tᵢ
−
Tₛ
is
still
left;
add
it
to
Tₛ.
T
=
−15
+
(10
−
(−15))
×
erf(0.400938)
T
=
−15
+
25
×
erf(0.4009)
T
=
−4.2677
°C
Check
Put the numbers back in. Both sides should match.
0.400938
=
0.5
2 ×
√
1.5 ×
10
-7
× 2,592,000
✓
−4.2677 = −15 + (10 − (−15)) × erf(0.400938) ✓
← Previous lesson
The same oil cooler in parallel flow
Next lesson →
Centre of a plane wall (one-term solution)
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.