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23-Chem-B1 Transport Phenomena · Undated paper

Question 5 of 6: C1 — Metal oxide dissolving into a flowing melt over a flat plate

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper. National Examination (Engineers Canada) — 16-Chem-B1 Transport Phenomena, May 2019. Open book, 3 hours. Six 25-point problems in three sections — A Fluid Mechanics (A1, A2), B Heat Transfer (B1, B2), C Mass Transfer (C1, C2); one problem from each section must be attempted, plus a fourth from any section. All six problems are solved and fully worked below.

Reference texts: R. B. Bird, W. E. Stewart & E. N. Lightfoot, Transport Phenomena (2nd ed., Wiley) — the equations of change, the power-law slit-flow momentum balance and the film/annular diffusion balances (A2, B2, C2); J. R. Welty, C. E. Wicks, R. E. Wilson & G. L. Rorrer, Fundamentals of Momentum, Heat and Mass Transfer (Wiley) — pipe friction, the Moody chart, and the external-flow convection/mass-transfer correlations (A1, B1, C1); F. P. Incropera & D. P. DeWitt, Fundamentals of Heat and Mass Transfer (8th ed., Wiley) — Churchill–Chu free convection, horizontal-plate correlations, slug-flow internal convection and air properties (B1, B2); C. J. Geankoplis, Transport Processes and Separation Process Principles (Prentice Hall) — boundary-layer mass transfer and the film model (C1); R. H. Perry & D. W. Green, Perry’s Chemical Engineers’ Handbook — transport properties.

Question 5: C1 — Metal oxide dissolving into a flowing melt over a flat plate (25 marks: 15 + 10)

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.

QuantitySymbolValue
Equilibrium constant$K=C_MC_O$$1\times10^{-6}\ (\text{mol/cm}^3)^2$
Plate length, bulk velocity$L,\ U$$3\ \text{m},\ 3\ \text{m/s}$
Melt density, viscosity$\rho,\ \mu$$8000\ \text{kg/m}^3,\ 1.24\times10^{-3}\ \text{Pa}\cdot\text{s}$
Diffusivity (M and O)$D_{AB}$$5\times10^{-9}\ \text{m}^2/\text{s}$

Find. (a) the length-averaged convective mass-transfer coefficient $\bar k_c$ for the dissolving metal; (b) the average molar flux of $MO$ dissolving from the plate.

dissolving MO plate (L = 3 m, surface C = √K = 1000 mol/m³) melt, U = 3 m/s, C = 0 concentration boundary layer δᶜ x = 0 x = L
Fig. C1: Melt sweeps a dissolving $MO$ plate. At the surface, local equilibrium fixes $C_M=C_O=\sqrt{K}$; far away $C_M=0$. A concentration boundary layer grows from the leading edge, and the length-averaged Sherwood number sets $\bar k_c$ and hence the dissolution flux.

Approach. Get the surface concentration of metal from the equilibrium condition $C_M=C_O$, form $Re_L$ and $Sc$, use the flat-plate average Sherwood correlation (the flow is turbulent), extract $\bar k_c$, then compute the flux $\bar N=\bar k_c\,\Delta C$.

  1. Surface concentration. At the plate, $C_M=C_O$ and $C_MC_O=K$, so $C_{M,s}=\sqrt K=\sqrt{10^{-6}}=1\times10^{-3}\ \text{mol/cm}^3=1000\ \text{mol/m}^3$; ahead of the plate and in the bulk, $C_{M,\infty}=0$.
  2. Reynolds and Schmidt numbers. With $\nu=\mu/\rho=1.24\times10^{-3}/8000=1.55\times10^{-7}\ \text{m}^2/\text{s}$, $$Re_L=\frac{UL}{\nu}=\frac{3(3)}{1.55\times10^{-7}}=5.81\times10^{7},\qquad Sc=\frac{\nu}{D_{AB}}=\frac{1.55\times10^{-7}}{5\times10^{-9}}=31.0.$$ $Re_L$ far exceeds the transition value ($\sim5\times10^5$), so the plate is largely turbulent.
  3. Average Sherwood number. Using the mixed laminar–turbulent flat-plate average ($Re_{x,c}=5\times10^5$): $$\overline{Sh}_L=\bigl(0.037\,Re_L^{0.8}-871\bigr)Sc^{1/3}=\bigl(0.037(5.81\times10^7)^{0.8}-871\bigr)(31.0)^{1/3}=1.86\times10^{5}.$$
  4. (a) Average mass-transfer coefficient. $$\boxed{\;\bar k_c=\frac{\overline{Sh}_L\,D_{AB}}{L}=\frac{1.86\times10^5(5\times10^{-9})}{3}\approx3.10\times10^{-4}\ \text{m/s}.\;}$$
  5. (b) Average dissolution flux. The metal flux is $\bar N_M=\bar k_c(C_{M,s}-C_{M,\infty})=3.10\times10^{-4}(1000-0)$. Because each mole of $MO$ that dissolves releases exactly one mole of $M$, the $MO$ dissolution flux equals the metal flux: $$\boxed{\;\bar N_{MO}=\bar N_M=\bar k_c\,C_{M,s}\approx0.31\ \text{mol/(m}^2\text{s}).\;}$$
QuantityValue
Surface metal concentration$C_{M,s}=1000\ \text{mol/m}^3$
$Re_L$ / $Sc$$5.81\times10^7$ / $31.0$
Average Sherwood number$\overline{Sh}_L=1.86\times10^5$
(a) Average mass-transfer coefficient$\bar k_c\approx3.10\times10^{-4}\ \text{m/s}$
(b) Average $MO$ dissolution flux$\approx0.31\ \text{mol/(m}^2\text{s})$
Check — turbulent plate; thermal data are distractors

$Re_L=5.8\times10^7$ places nearly the whole plate in the turbulent regime, so the mixed-boundary-layer average $\overline{Sh}=(0.037Re_L^{0.8}-871)Sc^{1/3}$ is used; a purely-laminar $0.664Re_L^{1/2}Sc^{1/3}$ would under-predict $\bar k_c$ by more than an order of magnitude ($2.6\times10^{-5}$ m/s) and is not appropriate here. The supplied thermal conductivity and heat capacity are not needed for the mass-transfer calculation (they would only matter for a heat–mass analogy). Dilute-solute behaviour is assumed, so $\bar N=\bar k_c\,\Delta C$ with no drift correction; the surface stays at local equilibrium, which is the “fast-dissolution” limit.