C1 — Dissolution of a rotating iron cylinder: 23-Chem-B1 Transport Phenomena, May 2016, Q5

Question 5: C1 — Dissolution of a rotating iron cylinder (25 marks)

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.

Quantity	Symbol	Value
Cylinder diameter	$d$	1.5 cm
Rotational speed	$N$	77 rpm
Diffusivity (Fe in melt)	$D_{AB}$	$9\times10^{-5}\ \text{cm}^2/\text{s}$
Kinematic viscosity	$\nu$	$1.41\times10^{-2}\ \text{cm}^2/\text{s}$
Driving force	$\Delta X_{Fe}$	0.385
Measured recession rate	$-dr/dt$	$5.57\times10^{-3}\ \text{cm/s}$

Find. The predicted recession rate $-dr/dt$ and its ratio to the measured value.

Fig. C1: A rotating cylinder in a melt develops a thin diffusion boundary layer; the Eisenberg–Tobias–Wilke correlation gives the mass-transfer coefficient, and with equal densities the surface recession equals $k_d\,\Delta X_{Fe}$.

Approach. Compute the peripheral speed, then $Re$ and $Sc$, evaluate the correlation for $k_d$, convert to a recession rate through the mass balance $-dr/dt=k_d\,\Delta X_{Fe}$ (equal densities), and compare with experiment.

Peripheral speed. $U_r=\pi d\,N=\pi(1.5)\dfrac{77}{60}=6.05\ \text{cm/s}.$
Reynolds and Schmidt numbers. $Re=\dfrac{dU_r}{\nu}=\dfrac{1.5(6.05)}{1.41\times10^{-2}}\approx643;\qquad Sc=\dfrac{\nu}{D_{AB}}=\dfrac{1.41\times10^{-2}}{9\times10^{-5}}\approx157.$
Mass-transfer coefficient. $\dfrac{k_d}{U_r}=0.079\,(643)^{-0.30}(157)^{-0.644}\approx4.39\times10^{-4},$ so $k_d=4.39\times10^{-4}(6.05)\approx2.65\times10^{-3}\ \text{cm/s}.$
Predicted recession rate. Equal densities make the moving-boundary mass balance $-\dfrac{dr}{dt}=k_d\,\Delta X_{Fe}$: $$-\frac{dr}{dt}=2.65\times10^{-3}(0.385)\;\Longrightarrow\;\boxed{-\frac{dr}{dt}\approx1.02\times10^{-3}\ \text{cm/s}}.$$
Compare with experiment. $\dfrac{(-dr/dt)_{exp}}{(-dr/dt)_{pred}}=\dfrac{5.57\times10^{-3}}{1.02\times10^{-3}}\approx5.5.$ The measured dissolution is about $\boxed{5.5\times}$ the correlation prediction.

Quantity	Result
Peripheral speed $U_r$	6.05 cm/s
$Re$ / $Sc$	643 / 157
Mass-transfer coefficient $k_d$	$2.65\times10^{-3}$ cm/s
Predicted recession $-dr/dt$	$1.02\times10^{-3}$ cm/s
Measured / predicted	$\approx5.5\times$ (experiment exceeds correlation)

Check — why measured > predicted

A dissolving surface roughens and flutes as it recedes, which trips the boundary layer and augments transport; natural convection from the density/composition gradient and the low, transitional $Re\approx640$ further enhance it. The smooth-cylinder correlation is therefore a conservative lower bound, consistent with the observed $\sim5.5\times$ enhancement rather than an error.

Concept & Theory

Topic. Convective mass transfer from a rotating cylinder (Eisenberg–Tobias–Wilke) applied to a dissolving/receding surface.

Key relations. $U_r=\pi dN$; $Re=dU_r/\nu$; $Sc=\nu/D_{AB}$; correlation $k_d/U_r=0.079\,Re^{-0.30}Sc^{-0.644}$; equal-density recession $-dr/dt=k_d\,\Delta X_{Fe}$.

Why this works. The rotating cylinder is a classical mass-transfer standard because it presents a uniform, well-defined boundary layer whose thickness is set by rotation. The written $k_d/U_r$ is a Stanton-number form (dimensionless coefficient), not a true Sherwood number despite the printed “Sh” label; with equal solid/melt densities every mole leaving the surface simply retreats the interface, giving the tidy $-dr/dt=k_d\Delta X$.

Common pitfall. Treating $0.079\,Re^{-0.3}Sc^{-0.644}$ as a Sherwood number and multiplying by $D_{AB}/d$ — it is already $k_d/U_r$, so $k_d$ follows by multiplying by $U_r$. Also, use radius (not diameter) recession in the mass balance.

Source. Eisenberg, Tobias & Wilke, J. Electrochem. Soc. 101 (1954) 306; Welty et al., Ch. 28 (convective mass-transfer correlations).

Question 5 of 6: C1 — Dissolution of a rotating iron cylinder

Question 5: C1 — Dissolution of a rotating iron cylinder (25 marks)