Inputs
Formula Used
This tool solves one-dimensional transient conduction in a plane wall with convection on both surfaces. It assumes uniform initial temperature, constant properties, and negligible internal heat generation.
- alpha = k/(rho·cp)
- Bi = h·L/k and Fo = alpha·t/L²
- Eigenvalues lambda_n satisfy lambda_n·tan(lambda_n) = Bi
- Local ratio: theta(x,t)/theta_i = Σ Cn·cos(lambda_n·x/L)·exp(-lambda_n²·Fo)
- Average ratio: thetā(t)/theta_i = Σ Cn·(sin(lambda_n)/lambda_n)·exp(-lambda_n²·Fo)
- Surface heat flux magnitude: |q"| = |k·(Ti-T∞)/L · Σ Cn·lambda_n·sin(lambda_n)·exp(-lambda_n²·Fo)|
If Bi is very small, the slab behaves nearly lumped; if Bi is large, surface cooling is strong and gradients are steeper.
How to Use This Calculator
- Enter slab thickness and material properties (k, ρ, cₚ).
- Enter convection coefficient h and ambient temperature T∞.
- Enter initial temperature Tᵢ and the elapsed time t.
- Choose position x from the center (0 is center, L is surface).
- Pick the number of series terms. Use 10–20 for most cases.
- Press Compute. Results appear above the form.
- Use the export buttons to save CSV or PDF.
Example Data Table
Example values for a brick-like material (illustrative).
| Thickness (m) | k (W/m·K) | ρ (kg/m³) | cₚ (J/kg·K) | h (W/m²·K) | Tᵢ | T∞ | t (s) | x (m) | Bi | Fo | T(x,t) |
|---|---|---|---|---|---|---|---|---|---|---|---|
| 0.04 | 1.4 | 1800 | 840 | 15 | 200 | 25 | 60 | 0 | 0.214 | 0.003 | 195.8 |
| 0.04 | 1.4 | 1800 | 840 | 15 | 200 | 25 | 600 | 0 | 0.214 | 0.033 | 184.0 |
| 0.04 | 1.4 | 1800 | 840 | 15 | 200 | 25 | 1800 | 0.02 | 0.214 | 0.098 | 113.6 |
The example temperatures depend on the chosen series terms.