Question 6 of 6: Mean Mass-Transfer Coefficient over a Flat Plate
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2013 — 04-Chem-A3 Mass Transfer Operations. Three-hour, open-book exam; any non-communicating calculator permitted. Format: six questions in two parts — answer two of Q1–Q3 (Part A) and two of Q4–Q6 (Part B); four questions of equal value constitute a complete paper. All six are solved below for completeness. Property data (diffusivities, vapour pressures, psychrometric enthalpies) are stated in each Given block; the psychrometric appendix supplied with the exam furnishes the saturated-air enthalpies used in Q5.
Reference texts: Geankoplis, Transport Processes and Separation Process Principles (4th ed., Prentice Hall) — molecular diffusion, unsteady-state diffusion, absorption and humidification; Treybal, Mass-Transfer Operations (3rd ed., McGraw-Hill) — Stefan diffusion, gas absorption, cooling-tower design; Welty, Wicks, Wilson & Rorrer, Fundamentals of Momentum, Heat and Mass Transfer (boundary-layer mass transfer); supporting property/psychrometric data from Perry's Chemical Engineers' Handbook (9th ed.).
Question 6: Mean Mass-Transfer Coefficient over a Flat Plate (equal value)
Given. Local Sherwood-number correlations for the laminar and turbulent regions of a flat plate, with transition at $x=L_t$. This is a derivation, so the answer is prose followed by the two short integrations — no numerical data are needed.
Find. The length-averaged coefficient $\bar{k_c}=\frac1L\int_0^L k_{c,x}\,dx$, and confirmation that it equals the stated expression.
Figure 6 — Boundary layer growing along the plate: laminar up to the transition $L_t$, turbulent beyond it. The mean coefficient averages the local value over both regions.
The mean coefficient is the length average of the local coefficient. Because the plate carries a laminar layer up to $L_t$ and a turbulent layer beyond it, the average splits into two integrals, each using its own correlation. We first convert each local Sherwood number into a local coefficient, then integrate.
Local coefficient from each correlation. By definition $Sh_x=k_{c,x}x/D_{AB}$, so $k_{c,x}=Sh_x D_{AB}/x$. Substituting $Re_x=ux/\nu$: $$k_{c,x}^{lam}=0.332\,D_{AB}\Big(\tfrac{u}{\nu}\Big)^{1/2}Sc^{1/3}\,x^{-1/2},\qquad k_{c,x}^{turb}=0.0292\,D_{AB}\Big(\tfrac{u}{\nu}\Big)^{4/5}Sc^{1/3}\,x^{-1/5}.$$ Each is a constant times a simple power of $x$.
Average over the two regions. Splitting the length average at the transition, $$\bar{k_c}=\frac1L\Big[\int_0^{L_t}k_{c,x}^{lam}\,dx+\int_{L_t}^{L}k_{c,x}^{turb}\,dx\Big].$$
Laminar integral. $\displaystyle\int_0^{L_t}x^{-1/2}dx=2L_t^{1/2}$, so the first term becomes $0.332\times2\,D_{AB}(u/\nu)^{1/2}Sc^{1/3}L_t^{1/2}=0.664\,D_{AB}(u/\nu)^{1/2}Sc^{1/3}L_t^{1/2}.$
Turbulent integral. $\displaystyle\int_{L_t}^{L}x^{-1/5}dx=\tfrac{5}{4}\big[L^{4/5}-L_t^{4/5}\big]$, so the second term becomes $0.0292\times\tfrac54\,D_{AB}(u/\nu)^{4/5}Sc^{1/3}[L^{4/5}-L_t^{4/5}] =0.0365\,D_{AB}(u/\nu)^{4/5}Sc^{1/3}[L^{4/5}-L_t^{4/5}].$
Assemble. Dividing the sum by $L$, $$\boxed{\bar{k_c}=\frac{0.664\,D_{AB}\big(\tfrac{u}{\nu}\big)^{1/2}Sc^{1/3}L_t^{1/2}+0.0365\,D_{AB}\big(\tfrac{u}{\nu}\big)^{4/5}Sc^{1/3}\big[L^{4/5}-L_t^{4/5}\big]}{L}},$$ which is the required result. The two prefactors 0.664 and 0.0365 are exactly $2\times0.332$ and $\tfrac54\times0.0292$ — the integration constants of the two powers.
