Introduction & Context

The calculation implements Fourier’s second law for transient (unsteady‑state) heat conduction in a plane slab that is cooled by convection on both faces. It is a fundamental tool in process engineering for predicting temperature evolution of solid components during quenching, heat‑treatment, or rapid cooling operations. Accurate temperature predictions are essential for controlling material properties, avoiding thermal stresses, and ensuring product quality in industries such as metalworking, glass manufacturing, and semiconductor processing.

Methodology & Formulas

The slab of total thickness L_tot is symmetric, so the analysis is performed on a half‑thickness L_c = L_tot/2. The governing equation for one‑dimensional conduction with convective boundary conditions is

\[ \frac{\partial T}{\partial t}= \alpha \,\frac{\partial^{2} T}{\partial z^{2}} \] where the thermal diffusivity \(\alpha\) is defined as \[ \alpha = \frac{k}{\rho\,c_{p}} \] with k the thermal conductivity, \(\rho\) the density, and \(c_{p}\) the specific heat.

Two dimensionless groups are introduced:

\[ \text{Biot number:}\quad \mathrm{Bi}= \frac{h\,L_{c}}{k} \] \[ \text{Fourier number:}\quad \mathrm{Fo}= \frac{\alpha\,t}{L_{c}^{2}} \] where h is the convection coefficient and t the elapsed time.

The solution of the transient problem is expressed as an infinite series of eigenfunctions. The eigenvalues \(\lambda_{n}\) satisfy the transcendental equation

\[ \lambda_{n}\,\tan(\lambda_{n}) = \mathrm{Bi} \] The first positive root \(\lambda_{1}\) dominates the response for moderate times and is obtained iteratively (e.g., Newton‑Raphson).

The corresponding Fourier coefficient for the first term is

\[ A_{1}= \frac{2\,\sin(\lambda_{1})}{2\,\lambda_{1}+ \sin(2\lambda_{1})} \]

Introducing the dimensionless spatial coordinate \(\eta = z/L_{c}\) (with \(\eta = 1\) at the centre plane) the dimensionless temperature ratio \(\theta\) at the centre after time t is

\[ \theta = A_{1}\,\exp\!\big(-\lambda_{1}^{2}\,\mathrm{Fo}\big)\,\cos(\lambda_{1}\,\eta) \]

The physical temperature at the centre is recovered by

\[ T_{\text{center}} = T_{\infty} + (T_{i}-T_{\infty})\,\theta \] where \(T_{i}\) is the initial temperature and \(T_{\infty}\) the ambient temperature.

The maximum possible heat that can be removed per unit surface area (if the slab were cooled to the ambient temperature) follows from an energy balance:

\[ Q_{\max}= \rho\,L_{\text{tot}}\,c_{p}\,(T_{i}-T_{\infty}) \]

Regime Classification (Biot Number)

Regime	Condition (Biot number)	Implication
Lumped‑capacitance approximation	\(\mathrm{Bi} \ll 1	Temperature within the solid is spatially uniform; simple exponential decay applies.
One‑dimensional transient conduction (series solution)	\(\mathrm{Bi} \gtrsim 1	Internal temperature gradients are significant; eigenfunction expansion required.

Solution Procedure (Algorithmic Summary)

Compute thermal diffusivity \(\alpha = k/(\rho c_{p})\).
Form the Biot and Fourier numbers using the definitions above.
Solve \(\lambda\,\tan\lambda = \mathrm{Bi}\) for the first positive root \(\lambda_{1}\) (Newton‑Raphson or other root‑finding method).
Evaluate the coefficient \(A_{1}\) with the obtained \(\lambda_{1}\).
Set \(\eta = 1\) for the centre plane and compute the dimensionless temperature \(\theta\).
Obtain the centre temperature \(T_{\text{center}}\) from the dimensional relation.
Optionally calculate the maximum extractable heat per unit area \(Q_{\max}\) for energy‑budget assessments.

Fourier's Second Law describes unsteady‑state (transient) heat conduction. It relates the rate of change of temperature within a material to the spatial temperature gradient and thermal diffusivity. In contrast, Fourier's First Law deals with steady‑state conduction where temperature fields do not vary with time.

First Law: q = –k ∇T (heat flux proportional to temperature gradient, no time term).
Second Law: ∂T/∂t = α ∇²T (temperature change over time equals thermal diffusivity times the Laplacian of temperature).
Implication: The second law introduces a time‑dependent term, requiring initial conditions and often solution by analytical or numerical methods.

To model transient heat flow in a pipe, follow these steps:

Identify geometry: radius r, length L (assume L≫r for one‑dimensional radial analysis).
Write the governing equation in cylindrical coordinates: ∂T/∂t = α (1/r) ∂/∂r(r ∂T/∂r).
Specify initial temperature distribution T(r,0) = T₀.
Apply appropriate boundary conditions, e.g.:
- Inner surface (r = r_i): prescribed temperature or convective heat transfer.
- Outer surface (r = r_o): convective heat transfer to the surrounding fluid.
Choose a solution method:
- Analytical (Separation of Variables, Bessel functions) for simple BCs.
- Numerical (finite difference or finite element) for complex BCs or variable properties.
Validate the model against experimental data or published correlations.

Common boundary conditions include:

Dirichlet (Prescribed Temperature): T = T_s(t) on the surface.
Neumann (Prescribed Heat Flux): –k ∂T/∂n = q''(t) on the surface.
Robin (Convective): –k ∂T/∂n = h [T – T_∞(t)], where h is the convection coefficient and T_∞ is the ambient temperature.
Symmetry (Zero Flux): ∂T/∂n = 0 at a plane of symmetry or at the centerline of a cylinder.

These conditions are combined with the initial temperature field to fully define a transient heat‑transfer problem.

The characteristic time constant (τ) indicates how quickly a slab approaches thermal equilibrium. Compute it as follows:

Identify slab thickness 2L (half‑thickness L) and material thermal diffusivity α = k/(ρ c_p).
Use the first‑term approximation of the series solution for a slab with convective boundaries: τ ≈ L²/π² α.
If both surfaces are insulated, the eigenvalue changes and τ ≈ L²/α.
Validate the estimate by comparing the predicted temperature response with a detailed numerical simulation or experimental data.

Worked Example – Transient Cooling of a Thin AISI 1010 Steel Strip after Annealing

A 100 mm-thick steel plate leaves an annealing furnace at 650 °C and is suddenly exposed to 15 °C air on both faces. Plant engineers need to know the centre-plane temperature after 3 min to decide when the strip can be handled safely. Use Fourier’s second law for an infinite slab with convection on both surfaces.

Knowns

Total thickness, L_tot = 0.100 m
Characteristic length, L_c = 0.050 m (half-thickness)
Initial temperature, T_i = 650 °C
Air temperature, T_∞ = 15 °C
Convection coefficient, h = 220 W m⁻² K⁻¹
Thermal conductivity, k = 110 W m⁻¹ K⁻¹
Density, ρ = 8530 kg m⁻³
Specific heat, c_p = 380 J kg⁻¹ K⁻¹
Elapsed time, t = 180 s

Step-by-step calculation

Compute thermal diffusivity α: \[ \alpha = \frac{k}{\rho\,c_p} = \frac{110}{8530 \times 380} = 3.394 \times 10^{-5}\ \text{m}^2\ \text{s}^{-1} \]
Evaluate Biot number: \[ Bi = \frac{h\,L_c}{k} = \frac{220 \times 0.050}{110} = 0.100 \]
Compute Fourier number: \[ Fo = \frac{\alpha\,t}{L_c^2} = \frac{3.394 \times 10^{-5} \times 180}{0.050^2} = 2.443 \]
Find the first eigenvalue λ₁ from the transcendental equation: \[ \lambda_1 \tan \lambda_1 = Bi \quad \Rightarrow \quad \lambda_1 = 0.311 \]
Determine the coefficient A₁: \[ A_1 = \frac{2 \sin \lambda_1}{\lambda_1 + \sin \lambda_1 \cos \lambda_1} = 0.508 \]
Centre-plane dimensionless temperature: \[ \theta_0 = A_1 \exp(-\lambda_1^2 Fo) = 0.508 \exp(-0.311^2 \times 2.443) = 0.382 \]
Convert to actual temperature: \[ T_0 = T_\infty + \theta_0 (T_i - T_\infty) = 15 + 0.382 (650 - 15) = 257.5\ °C \]

Final Answer

After 3 min the centre of the steel plate has cooled to 257 °C, low enough for safe handling with insulated gloves.

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