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23-Chem-B1 Transport Phenomena · December 2017

Question 5 of 6: C1 — Convective mass-transfer coefficient: flat plate and sphere

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

Notes on this paper

Paper format. EGBC 16-CHEM-B1 Transport Phenomena, December 2017, 3 hours, open-book (one textbook permitted, no loose notes). Six problems in three sections (A Fluid Mechanics, B Heat Transfer, C Mass Transfer), each worth 25 marks; candidates attempt one problem from each section plus a fourth from any section, and only the first four in the answer book are marked. All six problems are solved below as a complete study resource. Appendix A of the paper supplies the equations of change (continuity, Navier–Stokes, energy and species-continuity tables) in rectangular, cylindrical and spherical coordinates; these are quoted rather than re-derived.

Reference texts: R. B. Bird, W. E. Stewart & E. N. Lightfoot, Transport Phenomena (2nd ed., Wiley) — the equations of change and the differential momentum / energy / species balances; J. R. Welty, C. E. Wicks, R. E. Wilson & G. L. Rorrer, Fundamentals of Momentum, Heat and Mass Transfer (Wiley) — pipe friction, the transfer analogies, boundary-layer and film coefficients; F. P. Incropera & D. P. DeWitt, Fundamentals of Heat and Mass Transfer (Wiley) — internal-flow convection and the sphere/flat-plate correlations; F. M. White, Fluid Mechanics (McGraw-Hill) — the three-reservoir branching-pipe problem; R. H. Perry & D. W. Green, Perry’s Chemical Engineers’ Handbook — transport properties.

Question 5: C1 — Convective mass-transfer coefficient: flat plate and sphere (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.

QuantitySymbolValue
Free-stream velocity$U$$210\ \text{ft/s}$
Plate length / sphere diameter$L,\ D_s$$2\ \text{ft};\ 3\ \text{in}=0.25\ \text{ft}$
Air viscosity (printed unit “lbf/ft.s” is read as lbm/ft·s, the value for air at 75 °F)$\mu$$1.24\times10^{-5}\ \text{lb}/(\text{ft}\cdot\text{s})$
Air density$\rho$$7.44\times10^{-2}\ \text{lb}/\text{ft}^3$
Diffusivity (H&sub2;O–air)$D_{AB}$$2.37\times10^{-4}\ \text{ft}^2/\text{s}$

Find. (a) the length-average convective mass-transfer coefficient $k_c$ for the turbulent flat plate; (b) $k_c$ for the 3-inch sphere under identical conditions.

(a) flat plate, L = 2 ft U = 210 ft/s concentration boundary layer (b) sphere, D = 3 in R Sh = 2 + 0.6 Re^0.5 Sc^{1/3}
Fig. C1: The same air stream over (a) a flat plate — turbulent, so $\overline{Sh}_L=0.037\,Re_L^{0.8}Sc^{1/3}$ — and (b) a sphere, where the Frössling correlation adds a conduction/diffusion floor of $Sh=2$.

Approach. Both parts use the heat–mass transfer analogy: compute the kinematic viscosity and Schmidt number, form the appropriate Reynolds number, apply the flat-plate (turbulent) or sphere (Frössling) Sherwood correlation, then convert $Sh$ to $k_c$ with the relevant length scale.

  1. Fluid properties. $\nu=\mu/\rho=1.24\times10^{-5}/7.44\times10^{-2}=1.667\times10^{-4}\ \text{ft}^2/\text{s}$, and the Schmidt number $$Sc=\frac{\nu}{D_{AB}}=\frac{1.667\times10^{-4}}{2.37\times10^{-4}}=0.703.$$
  2. (a) Plate Reynolds number. $$Re_L=\frac{UL}{\nu}=\frac{(210)(2)}{1.667\times10^{-4}}=2.52\times10^{6}\gg5\times10^{5},$$ so the boundary layer is turbulent, as stated.
  3. (a) Length-average Sherwood number. For a turbulent flat plate (mass analogue of $\overline{Nu}=0.037\,Re^{0.8}Pr^{1/3}$), $$\overline{Sh}_L=0.037\,Re_L^{0.8}Sc^{1/3}=0.037(2.52\times10^{6})^{0.8}(0.703)^{1/3}\ \Longrightarrow\ \boxed{\overline{Sh}_L\approx4.35\times10^{3}.}$$
  4. (a) Mass-transfer coefficient. $$k_c=\frac{\overline{Sh}_L\,D_{AB}}{L}=\frac{(4349)(2.37\times10^{-4})}{2}\ \Longrightarrow\ \boxed{k_c\approx0.515\ \text{ft/s}.}$$
  5. (b) Sphere Reynolds number. With the diameter as the length scale, $Re_D=UD_s/\nu=(210)(0.25)/1.667\times10^{-4}=3.15\times10^{5}$.
  6. (b) Frössling (Ranz–Marshall) correlation. $$Sh=2+0.6\,Re_D^{1/2}Sc^{1/3}=2+0.6(561)(0.889)=301.5,$$ so $$k_c=\frac{Sh\,D_{AB}}{D_s}=\frac{(301.5)(2.37\times10^{-4})}{0.25}\ \Longrightarrow\ \boxed{k_c\approx0.286\ \text{ft/s}.}$$
CaseSherwood numberMass-transfer coefficient $k_c$
(a) Turbulent flat plate ($L=2$ ft)$\overline{Sh}_L\approx4.35\times10^{3}$$\approx0.515\ \text{ft/s}$
(b) Sphere ($D=3$ in)$Sh\approx302$$\approx0.286\ \text{ft/s}$
Common properties$\nu=1.667\times10^{-4}\ \text{ft}^2/\text{s}$, $Sc=0.703$
Check — correlation ranges

The Frössling/Ranz–Marshall sphere correlation is nominally validated to $Re_D\lesssim7\times10^{4}$; here $Re_D=3.15\times10^{5}$ lies above that, so the sphere $k_c$ is an extrapolation — a high-$Re$ form (e.g. Whitaker) would give a modestly larger coefficient. The flat-plate result assumes the layer is turbulent from the leading edge; retaining the laminar leading run (mixed correlation, $-871$ term) would lower $\overline{Sh}_L$ by a few percent. The dilute assumption $P_{bm}/P\approx1$ lets $k_c$ be used directly without the log-mean drift correction.