Reference ID: MET-C707 | Process Engineering Reference Sheets Calculation Guide
Introduction & Context
The calculation predicts the transient temperature response at the centre of an infinite slab that is suddenly exposed to a different surface temperature. In process engineering this analysis is essential for:
Designing heat‑treatment cycles for metal plates, sheets, and foils.
Estimating thermal stresses during rapid heating or cooling of reactor walls.
Evaluating the effectiveness of convective cooling or heating in continuous‑flow furnaces.
The slab is assumed to be thermally symmetric about its mid‑plane, so heat conduction occurs only in the thickness direction. The surrounding fluid imposes a constant convective heat‑transfer coefficient, and the surface temperature is changed instantaneously to a new value.
Methodology & Formulas
The analysis follows the one‑dimensional transient conduction solution for a plane wall with symmetric boundary conditions. The key steps are:
With initial temperature \(T_{i}\) and imposed surface temperature \(T_{s}\), the centre temperature is
\[
T_{c}= T_{s} + \theta\,(T_{i}-T_{s})
\]
Regime Classification Based on Biot Number
Biot Number Range
Physical Interpretation
Solution Approach
\(\text{Bi} \le 0.1\)
Surface resistance dominates; slab behaves as a lumped system.
Use lumped‑capacitance model or first‑term eigenvalue \(\lambda_{1}\approx\pi/2\).
\(0.1 < \text{Bi} < 10\)
Both surface convection and internal conduction are comparable.
Retain multiple terms of the eigenfunction series; solve transcendental equation for \(\lambda_{n}\).
\(\text{Bi} \ge 10\)
Internal conduction resistance dominates; surface is effectively isothermal.
Apply the semi‑infinite solid solution or use higher‑order eigenvalues.
By substituting the appropriate material data, geometry, and heating time into the expressions above, the centre temperature \(T_{c}\) can be obtained without iterative numerical methods, providing a quick engineering estimate for process design and safety assessments.
The term “infinite” only means that edge effects are ignored; heat flows perpendicular to the faces.
If the Biot number based on half‑thickness L is Bi = hL/k < 0.1, centre‑plane temperature history will be within 1 % of the 1‑D solution regardless of width.
For Bi > 0.1, check the Fourier number Fo = αt/L². When Fo > 0.2, temperature penetration has not yet reached the edges, so the infinite‑slab solution is still valid for the first few minutes of heating.
Once Fo > 0.4 and the thermal boundary layer becomes comparable to half‑width, switch to a 2‑D or 3‑D model or use shape‑factor corrections.
For Fo > 0.2 the first term of the Fourier series is enough; the error is <0.3 %.
Evaluate λ₁ from the transcendental equation λ₁ tan λ₁ = Bi.
Centre temperature: Tc −T∞ / T0 −T∞ = A₁ exp(−λ₁² Fo) with A₁ = 2 sin λ₁ / (λ₁ + sin λ₁ cos λ₁).
Any off‑centre position x/L uses the same λ₁ but multiply by cos(λ₁ x/L).
For Fo < 0.2 use the semi‑infinite solid solution or at least five terms of the series.
Yes, but redefine the length scale.
Use the full thickness L instead of half‑thickness; the insulated surface is equivalent to a symmetry plane.
Replace Bi with hL/k (not hL/2k) and use the same charts.
The “centre” plane in the charts now corresponds to the insulated surface, while x/L = 1 is the convective face.
Record the centre temperature vs. time and convert to dimensionless temperature θc = (Tc −T∞)/(T0 −T∞).
Estimate Fo at each time step using the known α and L.
Use the first‑term approximation θc ≈ A₁ exp(−λ₁² Fo) and fit λ₁ by non‑linear regression.
Convert λ₁ to Bi via λ₁ tan λ₁ = Bi, then h = Bi k/L.
Uncertainty is typically ±8 % if Fo > 0.2 and temperature error <0.5 °C.
Worked Example – Unsteady‑State Heating of an Infinite Slab
Scenario: A thin aluminum slab (thermal conductivity 167 W m⁻¹ K⁻¹, density 2700 kg m⁻³, specific heat 900 J kg⁻¹ K⁻¹) initially at 25 °C is placed in a furnace that maintains the surface temperature at 150 °C. The slab thickness is 20 mm (half‑thickness = 10 mm). Determine the temperature at the slab centre after 60 s, assuming one‑dimensional heat transfer and convection at the surfaces with a heat‑transfer coefficient of 30 W m⁻² K⁻¹.
