SME
Numerical Methods for Engineers
Finite differences for PDEs
Use this for “A rod with α = 1 × 10⁻⁴ m²/s has nodes 0.05 m apart at 80, 50 and 40 °C. With Δt = 10 s, find the middle node one step later.”
Chart
This is an example. Type your own values to start a new problem.
r
=
α (m²/s)
×
Δt (s)
(
Δx (m)
)
2
Tᵢⁿ⁺¹ (°C)
=
Tᵢ (°C)
+
r
×
(
Tᵢ₋₁ (°C)
−
2
×
Tᵢ (°C)
+
Tᵢ₊₁ (°C)
)
The bars are the rod’s nodes: Δx is the distance between neighboring nodes and Δt the time from one step to the next.
Formulas
r
=
α
×
Δt
Δx²
0.4
=
1 ×
10
-4
×
10
0.05
2
Tᵢⁿ⁺¹
=
Tᵢ
+
r
×
(Tᵢ₋₁
−
2
×
Tᵢ
+
Tᵢ₊₁)
58 = 50 + 0.4 × (80 − 2 × 50 + 40)
Assumptions
•
One-dimensional conduction, ∂T ÷ ∂t = α ∂²T ÷ ∂x², with α the same everywhere.
•
Forward in time, centred in space (FTCS): the new Tᵢ uses only the old node values.
•
Stable only for r ≤ 0.5; past 0.5 the explicit scheme blows up, so shorten Δt.
Step-by-step
WE KNOW
Tᵢ = 50 °C, Tᵢ₋₁ = 80 °C, Tᵢ₊₁ = 40 °C, α = 1 × 10⁻⁴ m²/s, Δt = 10 s, Δx = 0.05 m
FIND
mesh ratio (r), new middle node (Tᵢⁿ⁺¹)
Step 1 · Find mesh ratio (r)
Use r = α × Δt ÷ Δx²
The mesh ratio compares the time step with the time heat takes to cross one spacing.
r
=
10
-4
×
10
0.05
2
r
=
10
-4
×
10
0.0025
r
=
0.001
0.0025
r
=
0.4
Step 2 · Find new middle node (Tᵢⁿ⁺¹)
Use Tᵢⁿ⁺¹ = Tᵢ + r × (Tᵢ₋₁ − 2 × Tᵢ + Tᵢ₊₁)
The
bracket
is
the
bend
of
the
profile
at
the
node;
the
node
moves
r
times
that.
Tᵢⁿ⁺¹ = 50 + 0.4 × (80 − 2 × 50 + 40)
Tᵢⁿ⁺¹ = 50 + 0.4 × (80 − 100 + 40)
Tᵢⁿ⁺¹ = 50 + 0.4 × 20
Tᵢⁿ⁺¹ = 50 + 8
Tᵢⁿ⁺¹ = 58 °C
Check
Put the numbers back in. Both sides should match.
0.4
=
10
-4
×
10
0.05
2
✓
58 = 50 + 0.4 × (80 − 2 × 50 + 40) ✓
Topic check
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