Unsteady-State Leaching from a Porous Slab: 23-Chem-A3 Heat and Mass Transfer, December 2013, Q3

Question 3: Unsteady-State Leaching from a Porous Slab (equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Transient diffusion out of a slab with both faces held at the surface (equilibrium) concentration. Well-agitated, dilute solvent ⇒ surface concentration $C_s\approx0$.

Quantity	Value
Slab thickness / half-thickness	25 mm / $L=0.0125$ m (both faces exposed)
Effective diffusivity $D_{eff}$	$5.0\times10^{-10}$ m²/s
Initial content $C_0$	0.5 kg A/kg solid
Target centre content $C_c$	0.005 kg A/kg solid
Surface content $C_s$	≈ 0 (large, well-agitated bath)

Find. The time for the centre-plane content to fall from 0.5 to 0.005 kg/kg.

Figure 3 — Slab leached from both faces into a well-agitated solvent; the centre plane x = 12.5 mm is the slowest point to deplete.

Approach. Form the dimensionless centre concentration, invert the one-term centre-plane series solution of the transient diffusion equation for the Fourier number, then convert to time. Confirm the one-term truncation is accurate at this Fourier number.

Unaccomplished centre change. With $C_s=0$, $E_c = \dfrac{C_c-C_s}{C_0-C_s} = \dfrac{0.005}{0.5} = 0.010.$
Centre-plane series solution. For a slab of half-thickness $L$ initially uniform, both faces at $C_s$, the centre concentration is $E_c=\dfrac{4}{\pi}\sum_{n=0}^{\infty}\dfrac{(-1)^n}{2n+1}\exp\!\big[-(2n+1)^2\tfrac{\pi^2}{4}Fo\big]$, $Fo=D_{eff}\,t/L^2$. At the small $E_c$ here only the first term matters: $E_c\approx\dfrac{4}{\pi}\exp\!\big(-\tfrac{\pi^2}{4}Fo\big).$
Solve for the Fourier number. $$Fo = -\frac{4}{\pi^2}\ln\!\Big(\frac{\pi}{4}E_c\Big) = -\frac{4}{\pi^2}\ln(0.00785) = \boxed{1.96}.$$ Since $Fo\gg0.2$, dropping the higher terms changes the answer by less than 0.1% (verified against the full series).
Convert to time. $$t = \frac{Fo\,L^2}{D_{eff}} = \frac{1.96\,(0.0125)^2}{5.0\times10^{-10}} = 6.14\times10^{5}\ \text{s} = \boxed{170\ \text{h}\ (7.1\ \text{days})}.$$

Quantity	Result
Centre unaccomplished change $E_c$	0.010
Fourier number $Fo$	1.96
Leaching time	≈ 6.14×10⁵ s (170 h, 7.1 days)

Concept & Theory

Topic. Transient (unsteady-state) diffusion in a slab — centre-plane depletion.

Key relations. $\partial C/\partial t=D_{eff}\partial^2C/\partial x^2$; centre solution $E_c=(4/\pi)\exp(-\pi^2Fo/4)$ for $Fo>0.2$; $Fo=D_{eff}t/L^2$, $L=$ half-thickness.

Why this works. Once the initial sharp gradients have smoothed (large $Fo$), the whole profile decays as the slowest (first) eigenmode, so a single exponential describes the centre. Setting $C_s=0$ is justified because the well-agitated bath keeps the faces at essentially zero solute.

Common pitfall. Using the full slab thickness (25 mm) instead of the half-thickness for $L$ — that quadruples $L^2$ and inflates the time fourfold. With both faces exposed the diffusion length is half the slab.

Source. Geankoplis §7.1 (unsteady-state diffusion, charts & series); Crank, Mathematics of Diffusion §4.3.

Question 3 of 6: Unsteady-State Leaching from a Porous Slab

Question 3: Unsteady-State Leaching from a Porous Slab (equal value)