BE.430 Tutorial: Molecular Transport/Mass Transport 9/17/2004 Chemical engineers usually have a strong background in transport phenomena and may be familiar with and have a better understanding of the concepts taught in the first few weeks of BE.430. We have therefore designed this tutorial for students who are unfamiliar with the concepts of mass transport. Remember, however, that there are significant differences between transport occurring in chemical systems and biological systems. This tutorial summarizes only the chemical transport for background; the lectures teach you the intrigues of phenomena occurring in biological systems. 0. Differential Operations (more in Deen pp. 573-577) a. Rectangular b. Cylindrical ∂fG ∂fG ∂fG ∂fG ∂fG ∂fGff∇=∂x ex+∂y ey+∂z ez ∇=∂r er+ ∂θeθ+ ∂zez G 1 ∂(rv)+1 ∂vθ∂vz∇⋅v G = ∂∂ vxx+ ∂∂ vyy+∂∂ vzz ∇⋅v = rr ∂θ+ r r ∂z∂ ∂2 ∇2 ∂2 f ∂2 f ∂2 f ∇21 ∂⎛ ∂f⎞ 1 ∂2 f f+⎜r + 2 2f=∂x2 +∂y2 +∂z2 f = rr⎝∂r⎠⎟ r∂θ ∂z2∂ c. Spherical ∂fG 1 ∂fG 1 ∂fGf∇= e + eθ+ ∂r r ∂θ rsin θφ eφr ∂ 1 ∂(vsinθ)+ 1 ∂vφG 2∇⋅v = 1 ∂(r v)+ rsinθθ θ2 rr r ∂ rsin θ φ ∂ ∂ ∂2 ∇2 f = 12 ∂∂ ⎜⎛ r2 ∂∂ fr⎠⎟⎞+ 21 ∂ ⎜⎛sinθ∂f⎞ 1 f 2r r ⎝ rsinθ∂θ⎝ ∂θ⎟⎠ + r2 sin θφ ∂ 2 1. Governing Equations a. Species conservation ∂c G =−∇⋅NR+ ∂tb. Constitutive equation (Fick’s law)G N =−D ⋅∇c c. Rate of chemical reaction (R is positive for generation, negative for consumption) kf+ ZZZX C and we want to find the concentration profile of C.Consider a reaction AB YZZZkr Then by mass conservation of C, dcC = kc B (generation) − k c (consumption) dt fA rC Generation will be dependent on concentrations of A and B, where as loss of C will be dependent on the concentration of C The change in cC over time describes the reaction rate; therefore, at any point in the system, R = kc x B (x)− k C ()fA () xr Summary of most commonly encountered reaction rates: 1. 0th-order kinetics: 2. 1st-order kinetics 3. 2nd-order kinetics conc. time]= [ R = k [ 1time]⋅c (x) R = k2[ ]⋅c (x)⋅a (x)Rk0 1 conc time.⋅14. Michaelis-Menten 5. and others… cx Rmax ()R = xK + c ()m 2. Boundary conditions a. Constant, fixed concentrations at boundary cx0()= c0 c0 e.g. well mixed chamber separated by membrane; at membrane boundaries, we assume constant concentrations b. No flux conditions at boundary dcN =−D = 0 xx0 dx= xx0 = xe.g. no flux of species at indiffusible wall oc. Or more generally…Flux matching ! D−⋅ ∇ c + R = 0 (watch out for the signs, always consider n G !)xx0 = Ie.g. reactions at the wall (heterogeneous reaction!) NOTE: homogeneous vs. heterogeneous reactions Homogeneous reaction: reaction occurs throughout the system G∂c=−∇⋅ NR+ ∂t Heterogeneous reaction: reaction only occurs at system boundaries (or only at a partial volume of the system); R = 0, use flux matching to incorporate heterogeneous reaction. G∂c=−∇⋅ NR+ ∂t d. Geometrical symmetry N = 0 r=0 e.g. at center of sphere. 0 r 3. Assumption/Examples for simplifications (always show in homework!) a. steady-state ⎞ time-independence ⎝⎛ ∂∂ t → 0 ⎟⎜ ⎠ quasi-steady-state approximation by comparison of process time scales b. dimensionality 1-dimensional problem; 2-dimensional problem… , e.g. since z >> xy , assume conc. profile only in x and y. z (,, ,,c. constant diffusivity D ≠ f cxyzetc.) simplifies the governing equations in 1-D to x y∂c ∂2c = D + R ∂t ∂x2 d. no chemical reaction or heterogeneous chemical reaction R = 0 e. no charged species, no convection…4. Solving a mass transport problem Example problem: Steady-state diffusion of oxygen in slab of tissue with first-order homogeneous reaction Consider a slab of tissue of length 2l in a tank with a well-mixed solution of chemoattractant at a concentration of c0. In the tissue, chemoattractant is consumed with first-order reaction rate k. Determine the concentration profile in the tissue. a. Draw schematic of problem and label with variables Recognize the symmetry of the problem! c0 2l c0 x b. State assumptions 1. steady-state2. 1-dimensional problem (thin slab with large surface area) 3. constant diffusivity of chemoattractant 4. homogeneous first-order reaction c. Set up governing equations ∂c ∂N ∂cx0==− +R; N=−D ; R=−kc (consumption)x∂t ∂x ∂x ∂2c → D − kc = 0 ∂x2 d. Set up boundary conditions 1. cx l)(==c0 2. cx=−l)= c( 0 e. Solve governing equations k Dx k xD k− kc =0 → c xD∂2c ()= Ae − + Be = Ccosh x ∂x2 D f. Apply boundary conditions ⎛ k ⎞ 1. c0 = C cosh ⎜⎜l⎟ ⎟⎝ D⎠⎛ k ⎞ ⎛ k ⎞2. c0 = C cosh ⎜⎜− l⎟⎟= C cosh ⎜⎜l⎟ ⎟⎝ D⎠ ⎝ D⎠ Cc0→= ⎛ k⎞ cosh ⎜ l⎟ ⎝ D⎠ g. Solution k cosh x cx ()= c0 D k cosh lD h. Plot for conceptual understanding (not shown