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23-Chem-A3 Heat and Mass Transfer · December 2019

Question 5 of 6: Mass-Transfer Coefficient by the Chilton–Colburn Analogy

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

Notes on this paper

National Exams — December 2019 — 16-Chem-A3 Heat and Mass Transfer. Three-hour, open-book exam (one textbook of the candidate’s choice; any non-communicating calculator). Format: two parts — Part A (Q1–Q3) Heat Transfer and Part B (Q1–Q3) Mass Transfer; at least two questions must be attempted from each part and only the first two in each part are marked, so four questions (each 25 points) constitute a complete paper. All six questions are solved below for completeness. Property values not printed on the paper (molar masses, water latent heat, the dimensionless free-convection peak velocity $f'_{max}$, benzene/toluene physical properties) are stated explicitly in each Given block as open-book look-ups.

Reference texts: Coulson & Richardson (Backhurst, Harker & Richardson), Chemical Engineering, Vol. 1 — Fluid Flow, Heat Transfer and Mass Transfer (6th ed., Butterworth-Heinemann) — the source family for the crystalliser, tube-condenser, Stefan-tube and distillation problems; Incropera & DeWitt, Fundamentals of Heat and Mass Transfer (free- and forced-convection correlations, the Ostrach similarity solution); Treybal, Mass-Transfer Operations (3rd ed.) and McCabe, Smith & Harriott, Unit Operations of Chemical Engineering (7th ed.) — Stefan diffusion, Chilton–Colburn analogy, McCabe–Thiele; supporting property data from Perry’s Chemical Engineers’ Handbook (9th ed.) and the NIST Chemistry WebBook.

The obscured verb in Part B Q1 is taken as “calculate”, and the Part B Q2 water viscosity at 27 °C is $8.76\times10^{-4}$ Pa·s. One column of the Part A Q2 air-property table, the buoyancy group $g\beta/\nu^{2}$, is internally inconsistent: its printed values are about 40% below what $\rho$, $\mu$ and $\beta=1/T_f$ give. That column is not used; the Grashof number is built from the $\rho$ and $\mu$ columns and $\beta=1/T_f$.

Question 5: Mass-Transfer Coefficient by the Chilton–Colburn Analogy (Part B — 25 pts)

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 $D=1.5$ cm NaCl cylinder in cross-flowing water, $v=10$ m/s, $T=27$ °C. The heat-transfer Nusselt correlation is converted to mass transfer by the Chilton–Colburn analogy ($Nu\to Sh$, $Pr\to Sc$).

QuantityValue
Cylinder diameter $D$0.015 m
Water velocity $v$10 m/s
$\mu$ at 18 / 27 °C1.073×10⁻³ / 8.76×10⁻⁴ Pa·s
$D_{AB}$ at 18 °C1.26×10⁻⁵ cm²/s
$\rho$ (water)996 kg/m³

Find. The liquid-phase mass-transfer coefficient $k_L$ at 27 °C.

water v = 10 m/s, 27 CNaCld=1.5 cmdissolving solid: k_L ~ 3.8×10⁻⁴ m/s
Figure 5 — Water in cross-flow over a dissolving NaCl cylinder. The same boundary-layer structure that sets the heat-transfer Nusselt number sets the Sherwood number, so the given $Nu$ correlation carries over with $Pr\to Sc$.

Approach. Correct the diffusivity to 27 °C (Stokes–Einstein), form $Re$ and $Sc$ at 27 °C, apply the analogous Sherwood correlation, and recover $k_L=Sh\,D_{AB}/D$.

  1. Diffusivity at 27 °C. Stokes–Einstein gives $D_{AB}\propto T/\mu$:$$D_{AB,27}=1.26\times10^{-5}\frac{300.15}{291.15}\frac{1.073\times10^{-3}}{8.76\times10^{-4}}=1.59\times10^{-5}\ \text{cm}^2/\text{s}=\boxed{1.59\times10^{-9}\ \text{m}^2/\text{s}}.$$
  2. Reynolds and Schmidt numbers (27 °C).$$Re=\frac{\rho v D}{\mu}=\frac{996(10)(0.015)}{8.76\times10^{-4}}=1.71\times10^{5},\qquad Sc=\frac{\mu}{\rho D_{AB}}=\frac{8.76\times10^{-4}}{996(1.59\times10^{-9})}=553.$$
  3. Sherwood via Chilton–Colburn. The analogy $j_D=j_H$ replaces $Nu\to Sh$ and $Pr\to Sc$ in the given correlation:$$Sh=(0.506\,Re^{0.5}+0.00141\,Re)\,Sc^{0.33}=(0.506\times413+0.00141\times1.71\times10^{5})(553)^{0.33}=449\times8.04=\boxed{3.61\times10^{3}}.$$
  4. Mass-transfer coefficient.$$k_L=\frac{Sh\,D_{AB}}{D}=\frac{3610\,(1.59\times10^{-9})}{0.015}=\boxed{3.83\times10^{-4}\ \text{m/s}}.$$

Check (property temperature): the diffusivity is quoted at 18 °C but everything else is at 27 °C, so the Stokes–Einstein correction (a ~26% increase, mostly from the viscosity drop) is essential — skipping it would under-predict $k_L$ by the same factor. $Re$ and $Sc$ use the 27 °C viscosity consistently.

QuantityResult
$D_{AB}$ at 27 °C1.59×10⁻⁹ m²/s
$Re$ / $Sc$1.71×10⁵ / 553
$Sh$3.61×10³
$k_L$3.83×10⁻⁴ m/s