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23-Chem-B1 Transport Phenomena · May 2017

Question 6 of 6: C2 — O&sub2; diffusion with zeroth-order consumption in a sphere

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

Notes on this paper

Paper format. EGBC 16-CHEM-B1 Transport Phenomena, May 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) — laminar and turbulent boundary layers, flat-plate drag, and film transfer coefficients; W. M. Deen, Analysis of Transport Phenomena (Oxford) — the physiological transport problems (stenosis flow, reaction–diffusion of O&sub2; in tissue); F. P. Incropera & D. P. DeWitt, Fundamentals of Heat and Mass Transfer (Wiley) — convection correlations and the mixed-convection $Gr/Re^2$ criterion; R. H. Perry & D. W. Green, Perry’s Chemical Engineers’ Handbook — transport properties.

Question 6: C2 — O&sub2; diffusion with zeroth-order consumption in a 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. Steady diffusion of O&sub2; (diffusivity $D$) into a sphere of radius $r_0$ with a constant (zeroth-order) volumetric consumption rate $R_0$ (mol m$^{-3}$s$^{-1}$); the surface concentration is held at $C_0$; spherical symmetry, no convection.

Find. (a) the radial concentration profile $C(r)$; (b) the criterion for an oxygen-free core and an expression locating its radius $r_c$.

O₂-free core r < r_c r₀ r_c C = C₀ at surface; C falls inward, reaching 0 at r_c
Fig. C2: Oxygen diffuses inward and is consumed at a constant rate. If the tissue is large or hungry enough, the concentration hits zero at $r_c$, leaving an anoxic core in which the zeroth-order model no longer holds.

Approach. Reduce the spherical species-continuity equation (Table A.5) to a steady balance of diffusion against the constant sink, integrate for the no-core profile, then split off a reacting outer region when the centre would go anoxic and apply zero-concentration / zero-flux conditions at $r_c$.

  1. Governing equation. With no convection, steady state, constant $D$ and a constant sink $R_0$, the spherical species balance becomes $$D\,\frac{1}{r^2}\frac{d}{dr}\!\Big(r^2\frac{dC}{dr}\Big)=R_0.$$
  2. (a) Integrate for the whole sphere. Two integrations give $C=\dfrac{R_0 r^2}{6D}-\dfrac{A}{r}+B$. Finiteness at the centre forces $A=0$; the surface condition $C(r_0)=C_0$ fixes $B$, giving $$\boxed{C(r)=C_0-\frac{R_0}{6D}\big(r_0^{2}-r^{2}\big).}$$ The profile is convex with its minimum at the centre.
  3. (a→b) Centre concentration and the onset criterion. Setting $r=0$, $C(0)=C_0-\dfrac{R_0 r_0^{2}}{6D}$. A physically impossible negative centre concentration signals an anoxic core; it first appears when $C(0)=0$, i.e. when $$\boxed{\Phi\equiv\frac{R_0\,r_0^{2}}{6D\,C_0}\ge1.}$$ ($\Phi$ is a Thiele-type modulus comparing reaction demand with diffusive supply.)
  4. (b) Reacting region with a dead core. When $\Phi>1$, solve the balance only over $r_cand zero flux at the core edge ($C(r_c)=0$, $dC/dr|_{r_c}=0$, since no O&sub2; crosses into the dead core). These give $$C(r)=\frac{R_0}{6D}\Big(r^{2}+\frac{2r_c^{3}}{r}-3r_c^{2}\Big).$$
  5. (b) Locate $r_c$ from the surface condition. Imposing $C(r_0)=C_0$ yields $\dfrac{R_0}{6D}\big(r_0^{2}+2r_c^{3}/r_0-3r_c^{2}\big)=C_0$, which factors neatly. Writing $\xi=r_c/r_0$, $$\boxed{(1-\xi)^{2}(1+2\xi)=\frac{1}{\Phi}=\frac{6D\,C_0}{R_0\,r_0^{2}}.}$$ This implicit relation gives $r_c=\xi r_0$. It behaves correctly at the limits: $\Phi=1\Rightarrow\xi=0$ (the core just appears), and $\Phi\to\infty\Rightarrow\xi\to1$ (almost the whole tissue is anoxic). For example, $\Phi=2$ gives $\xi=\tfrac12$, i.e. $r_c=r_0/2$.
ResultExpression / Value
(a) Concentration profile (no core)$C(r)=C_0-\dfrac{R_0}{6D}(r_0^{2}-r^{2})$
(b) Anoxic-core criterion$\Phi=R_0 r_0^{2}/(6DC_0)\ge1$
(b) Core radius (implicit)$(1-\xi)^2(1+2\xi)=1/\Phi$, $\ \xi=r_c/r_0$
Illustration ($\Phi=2$)$r_c=r_0/2$
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