4.5 Dispersion of supension in a steady shear flow

Transcription

1 Notes on.63 Advanced Environmental Fluid Mechanics Instructor: C. C. Mei, November, 4-5disp-sh.tex. Abbreviated version for steady flow only. Refs:. Fischer, List, Koh. Imberger and Brooks : Mixing in Inland and Coastal Waters.. Brenner & Edwards : Macrotransport Processes. Butterworth-Heinemann, 993. Macro 4.5 Dispersion of supension in a steady shear flow Similar to heat, a passive solvent is also transported by convection and diffusion. When the concentration of the solvent is low, the flow affects the transport but the solvent does not affect the flow. We explain a systematic theory of enhanced diffusion by shear- a phenomenon first analysized by G.I. Taylor, hence is called the Taylor dispersion The convective diffusion equation for a passive suspension The spreading of a passive solute in a flow is governed by the convective diffusion equation C t + (qc) =D C (4.5.) where C is the volume concentration of the solute, and D the diffusivity. The derivation is as follows. Consider a material volume V. Conservation of the solute mass requires: D CdV = f nds = fdv (4.5.) Dt V S V The left hand side can be written as V C t + (qc) dv (4.5.3) Assuming Fick s law f = D C (4.5.4) we get C dv V t + (qc) = V dv D C (4.5.5) Since V is arbitrary, (4.5.) follows.

2 4.5. Heuristic picture of shear-enhanced diffusion and scales estimates If a streak of die is put initially straight aross a channel of fluidinsteadyuniformflow, after some time the die streak will become curved due to the transverse variation of velocity, and also becomes thicker due to diffusion in all directions. See figure After sufficiently long time lateral diffusion will be complete and the die distribution is nearly uniform across the channel. Convective diffusion will continue as a one-dimensional process. We wish to predict the effecitive dissfusivity in the last phase. Consider dispersion in a one-dimensional shear Figure 4.5.: Initial stage of convective diffusion of a die streak. From Fischer et al. flow in a long circular pipe resulting from a pressure gradient which is constant in space but has steady and oscillatory parts. The velocity profile is known, The concentration of a solvent is governed by u = U s (r), <r<a. (4.5.6) C t + (uc) = D x C x + r r C r r (4.5.7) There are three time scales : diffusion time across the pipe radius a, convection time across L, anddiffusion time across L. Their ratios are : a D : L U o : L D =: ² : ² (4.5.8)

3 The first time scale for lateral diffusion is very small. Let us focus attention to processes in the longitudinal direction long after lateral diffusion is complete. Hence we choose the following normalization, x = Lx,r= ar,u= U o u,t= L U o t, (4.5.9) 3 where U o is the scale of U. Equation (4.5.7) is normalized to U o a D a C L + (u C ) t x = a C L x + r C r r r (4.5.) Let the Péclét number = U o a/d be of order unity and a/l = ². Equation (4.5.) becomes the boundary conditons C ²P e + (u C ) t x For brevity we op the primes from now on. = ² C x + r Multiple scale analysis-homogenization r C r r (4.5.) r =, r =, (4.5.) u = U s (4.5.3) Ignoring the early stage of lateral difrusion, there are two time scales : convection time across L, anddiffusion time across L. Their ratios are : L U o : L D =: ² (4.5.4) Therfore we introduce the multiple time coordinates t, t = ²t, (4.5.5) and consider C(x, r, t, t )todependont, and t as if they were independent. In particular, Introduce multiple scale expansions. C t C t + C t t t = C t + ²C t (4.5.6) C = C + ²C + ² C +... (4.5.7)

4 4 C i = C i (x, r, t, t ), for all i =,, 3. The perturbation problems are O(² ): = r C (4.5.8) r r r the boundary conditions: =, r =,. (4.5.9) r O(²): C t + (uc ) = r C (4.5.) x r r r : we get O(² ): The solution at O(² )is C r =, r =,. (4.5.) C + C t t + (uc ) = C x x + r C r r r C r At O(²), let the known velocity be C t + U s (4.5.) =, r =,. (4.5.3) C = C (x, t ), (4.5.4) u = U s (r) (4.5.5) = x r r C r r (4.5.6) C =, r =, (4.5.7) r Let us now integrate(or average ) (4.5.6)across the pipe, and get where angle brackets denote averaging over the cross section. t + hu si x = (4.5.8) hhi = π Z πrh

5 5 Now subtract (4.5.8) from (4.5.6) Ũs x = r C r r r (4.5.9) where Ũ = U s (r) hu s i (4.5.3) is the velocity nonuniformity Now C is governed by a linear equation, we can assume the solution to be proportional to the forcing, i.e., C = x B s(r) (4.5.3) then r d r db s = Ũ(r) (4.5.3) the boundary conditions db s =, r =,. (4.5.33) and After solving for B s we go to O(² ), i.e., (4.5.). Note that, in view of (4.5.3), hence, uc x = (hu s i + Ũs)B s (r) C x C + C + (hu s i + t t Ũs)B s (r) C x = C x + r C r r r (4.5.34) which is a linear PDE for C. From (4.5.3) and (4.5.8) we find It follows that C t = C x hu sib s (r) (4.5.35) + C Ũ s B s t x = C x + r C r r r (4.5.36) C = r =, (4.5.37) r Averaging (4.5.36) across the pipe, we get t = E C x (4.5.38)

6 6 E = hũsb s i (4.5.39) which is the effective diffusion coefficient. The first part is of molecular origin; the second part (the dispersion coefficient) is due to fluid shear. Finally we add ² (4.5.8) and (4.5.38) to get: t + ² C + hu s i t x = ²E C (4.5.4) x This describes the convective diffusion of the area averaged concentration, which is certainly of practical value. After the perturbation analysis is complete, there is no need to use multiple scales; we may now write t + hu si x = ²E C (4.5.4) x In dimensionless form, it reads, t + hu si x = DE C (4.5.4) x This equation governs the convective diffusion of the cross-esctional average, after the initial transient is smoothed out The dispersion coefficient for steady flow We demonstrate the details for a pure steady flow. The dimensional velocity profile is U s (r) =U o r a U o begin the maximum at the centerline. It is easy to show that hui = U o We shall use U o as the velocity scale for normalization, then in dimensionless form u(r) = r, hui = (4.5.43) (4.5.44) (4.5.45) and the nonuniform part is The equation for B s is r d ũ = r (4.5.46) r db s = r (4.5.47)

7 which can be easily integrated to give B s = 8 r r4 + B (4.5.48) To detemine the integration constant we impose the condition that Z hb s i = rb s = (4.5.49) This implies =B + 4 (4.5.5) hence B s = 4 + r r4 8 4 (4.5.5) Using (4.5.46) and (4.5.5), we get Z µ " hũb s i = r r 4 + # r r4 = (4.5.5) 7 Thus In physical variables + hu s i C t x = ² + C (4.5.53) 9 x t + hu si C " x = D + 9 µ Uo a # C (4.5.54) D x Denote by K the shear-enhanced dispersion coifficient, K = (U o a) 9 D (4.5.55) If the clet number is not small, the enhanced diffusivity (dispersivity) can be greater than the molecular diffusivity. Since the dispersivity is inversely proportional to D, its importance becomes greater for smaller molecular diffusisivity. Example: Forsaltinwater,D = 5 cm /s. If U o =cm/s, and a =.cm, then K =cm /s, far exceeding the molecular diffusivity. Homework: Show that the shear-enhanced dispersion coefficient E for the uniform flow in a wide and inclined channel of depth h is (8/945)U o h /D. Homework: Find the dispersion coefficient E in the oscillatory flow in a circular pipe. Homework: Find the dispersion coefficient E in the oscillatory flow in a blood vessel elastic wall. Homework: Find the dispersion coefficient for sediments in an open channel flow. The sediments are uniform fall velocity W o.