Question 1 of 7: Leaching caffeine from coffee beans — stagnant vs. stirred solvent

Question 1: Leaching caffeine from coffee beans — stagnant vs. stirred solvent (Part A — 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. A caffeine-laden spherical bean is immersed in a large excess of solvent, so the bulk solvent stays essentially caffeine-free. The extraction rate is set by two resistances in series — diffusion through the bean and the external liquid film — whose relative importance is fixed by the partition coefficient $\alpha$.

Find. (a) extraction time with a stagnant solvent; (b) order-of-magnitude time with a well-stirred solvent; (c) ways to speed both up.

Approach. Compare two limiting regimes: internal-diffusion control (well-stirred, one-term sphere series) and external-film control (stagnant, lumped bean). The mass-transfer Biot number $Bi_m=\alpha=0.1\ll1$ tells us which limit each case falls in.

1. (b) Well-stirred solvent — internal-diffusion control. A very high external coefficient drives the surface concentration to zero, so the bean empties by transient internal diffusion. For $\bar c/c_0<0.15$ the sphere average-concentration series keeps only its first term: $$\frac{\bar c}{c_0}=\frac{6}{\pi^2}\,e^{-\pi^2\mathrm{Fo}},\qquad \mathrm{Fo}=\frac{D_{AB}t}{R^2}.$$ Solving for $\mathrm{Fo}$ at $\bar c/c_0=0.03$: $$\mathrm{Fo}=-\frac{1}{\pi^2}\ln\!\left(\frac{0.03\,\pi^2}{6}\right)=0.305.$$
2. (b) Time, stirred case. $$t_b=\frac{\mathrm{Fo}\,R^2}{D_{AB}}=\frac{(0.305)(0.30)^2}{1.8\times10^{-6}}=1.52\times10^{4}\ \text{s}=\boxed{4.2\ \text{h}}.$$
3. (a) Stagnant solvent — external-film control. With no forced convection the film coefficient is the stagnant-sphere limit $Sh=2$, i.e. $k_c=2D_{AB}/d=D_{AB}/R$. The mass-transfer Biot number $Bi_m=k_c\alpha R/D_{AB}=\alpha=0.1\ll1$, so the bean is nearly uniform (lumped) and the external film controls. A whole-bean mass balance $V\,dc/dt=-k_c A\,\alpha c$ with $V=\tfrac43\pi R^3$, $A=4\pi R^2$ gives $$\frac{c}{c_0}=\exp\!\left(-\frac{3\alpha k_c}{R}\,t\right)=\exp\!\left(-\frac{3\alpha D_{AB}}{R^2}\,t\right)=\exp\!\left(-\frac{0.3\,D_{AB}}{R^2}\,t\right).$$
4. (a) Time, stagnant case. Setting $c/c_0=0.03$: $$t_a=\frac{-\ln(0.03)\,R^2}{0.3\,D_{AB}}=\frac{(3.507)(0.30)^2}{0.3\,(1.8\times10^{-6})}=5.84\times10^{5}\ \text{s}=\boxed{6.8\ \text{days}}.$$ The stagnant film makes extraction about 38× slower than the stirred bean. (Check: the exact finite-Biot sphere series at $Bi_m=0.1$, first eigenvalue $\lambda_1=0.542$, gives 6.9 days, so the lumped estimate is within 2%.)
5. (c) How to speed it up. Time scales as $R^2/D_{AB}$ in both limits, so the strongest lever is particle size: grinding the beans to a smaller radius cuts the time with the square. Raising the temperature increases $D_{AB}$ (roughly Arrhenius) and helps both cases. For the stagnant case specifically, agitating or circulating the solvent removes the controlling external film (moving it toward the 4.2-h internal-diffusion limit), and continuously renewing/replacing the solvent keeps the bulk concentration near zero to preserve the driving force.