Convection-Diffusion in Microchannels

33443 Microfluidics Theory and Simulation Convection-Diffusion in Microchannels Catharina Hartmann (s97176) Luigi Sasso (s06094) Institute of Micro and Nanotechnology, DTU, 800 Kongens Lyngby, DK June, 007 1

Abstract This report is motivated by an article recently published, regarding the effect of cross-sectional width on Taylor dispersion in shallow microchannels [1]. The mentioned article reveals an expression for the effective dispersion coefficient as a function of the first and second moments of the concentration. The first part of the report deals with a theoretical background explanation. The second part tries to calculated the moments utilizing a mathematical tool known as the method of moments, as described by Aris in 1956 []. The solutions for the 0 th and 1 st moments for the case of a rectangular channel are outlined, but unfortunately not solved completely because of the complexity of the problem. Lyngby, 1 st June 007,........................... Catharina Hartmann s97176........................... Luigi Sasso s06094

1 Introductory Theory 1.1 The Equations of Motion The continuity equation is an expression of mass flowing through an arbitrarily shaped region. In this case the mass will be given by ρ multiplied by the volume. Here, ρ is the density of the fluid. If we imagine a fixed region Ω in space and time, it is possible to calculate the mass flowing through this region by M(Ω, t) = ρ(r, t)dr (1) Since mass can not appear or disappear spontaneously, the mass can only vary by flowing into or out of the surface of the fixed region. So we can define an expression for mass flow per oriented unit area per unit time. This is given by Ω J(r, t) = ρ(r, t)v(r; t) () where v is an expression for the Eulerian velocity field. Since we have a fixed time-derivative of mass, the mass can be calculated by either equation (1) as a volume integral or by a surface integral using equation (). The calculations will not be shown here but the results give us the continuity equation which is true for any choice of region Ω. But this is only true if the integration yields zero as shown below: dr [ t ρ(r, t) + (ρ(r, t)v(r, t))] = 0 (3) Ω From this expression we arrive at the continuity equation written as: t ρ + (ρv) = 0 or t ρ + J = 0 (4) where J is the mass current density and is defined as the mass flow per oriented unit area per unit time. Next we move on to the basic of the Navier-Stokes equation which is a expression for the motion in the Eulerian velocity field. It is very difficult to solve the Navier-Stokes system analytically since it is a non-linear differential equation. First we will have a look on how the actual equation is found and further more look at how this equation can be included into a mechanical equilibrium for fluids. To get the most correct expression for this motion we must take under consideration the time-derivative of the velocity field. This is because the fluid system of the Eulerian velocity field is not for any particular valocity field, so we have to introduce a more physically correct equation of motion. We therefore introduce the material time-derivative D t. The material time-derivative is obtained by the fact that the total differential of the Eulerian velocity field is given by dv = dt t v + (dr )v. If we secondly insist on following a particular 3

particle and the changes in flow due to this particular particle, we find that dr = vdt. By combining these expressions to we get that The calculation is shown in the following: D t = t + (v ) (5) Our starting point is newtons second law given by m dv dt = j F j ρd t v = j f j ρ( t v + (v )v) = j f j, where f j is an expression for the force density. 1. Poiseuille Flow The Poiseuille flow consists of fluids driven through a long straight channel while imposing a pressure difference between the two ends of the linear channel. To give an example on how a flow of this kind can be solved analytically we look at the flow through a arbitrary cross sectional shape in steady state, and we later on elaborate on this topic. The constant cross-section in the xy-plane is denoted C on the figure shown with the boundary C. As seen there is a constant pressure difference P over the segment L of the channel. Because the gravitational force is balanced by the hydrostatic pressure gradient in the vertical direction, the two forces can be left out of the equation. In other words, the velocity field will be independent of x, due to the translation invariants of the channel in the x-direction and the vanishing of the forces in the zy-plane. Therefore, only the x-component can be non-zero v(r) = v x (y, z)e x so that (v )v = 0 and the steady state Navier-Stokes equation becomes 0 = η [v x (y, z)e x ] P (6) The pressure field is now only dependent on x, since the velocity field in both z and y-direction are zero. So the Navier-Stokes equation becomes η[ y + z]v x (y, z) = x P (x) (7) 4

Both sides of the equation need to be equal to the same constant, so this problem can be solved by seeing that the constant pressure gradient x P (x) implies that the pressure must be a linear function of x and by using the boundary conditions for the pressure it is seen that [3] P (r) = P L (L x) + P 0 (8) Finally using this in equation(7) it is seen that a second order partial differential equation can be solved for a velocity v x (y, z) in the domain C, using the no-slip boundary conditions at the walls. The calculation is seen below [ y + z]v x (y, z) = P, for (y, z) in C. (9) ηl v x (y, z) = 0, for (y, z) in C (10) For further details look at the Henrik Bruus notes [3]. 1.3 Diffusion Until now our main flows have only been convectional steady state. Here we introduce another motion, diffusion. Diffusion is a motion where a solute in a solvent moves from high concentrations to low. For further reading we again refer to Henrik Bruss notes [3] for the detailed calculation for the equation of the diffusion. In this paper we present the actual equation for a system in steady state which is given by t c = D c (11) here, D is the diffusion constant of the solute in the solvent and is given by D = L 0 T 0 (1) with L 0 and T 0 being the characteristic length and time scales over which the concentration c(r, t) varies. In the next sections we see examples of two different systems where the topics we briefly have been introducing, convection and diffusion, is put to use. 5

1.4 The H-Filter This is the first actual microsystem used to separate two solutes from each other. The method combines laminar flow and fast diffusion in a very small area. The H-filter has two inlets channels to the left, which are kept at a constant pressure P + P 0 and two outlet channels on the right hand side kept at the constant pressure P 0. In the crossbar of the H-filter the actual diffusion take place. In our example we have two different solutes, the light gray is the solvent of small ions(black dots) in water, where the darker gray represent solvent of the 30-base-pair DNA molecules (white spots ) in water. All channels have the same width and hight h << w and the central channel has the length L. In behavior of a solute in the H-filter, two timescales becomes relevant for this system, the τ convection, which is the time referring to the down-stream of the inlet, and secondly the τ diffusion which is the time it takes to diffuse across the half width of the cross bar channel. This is given by τ convection = L ( 1 0 τ diffusion = w) ( 1 v 0 D = w) D = w 4D (13) Some ground rules can be said as if a solute with τ convection << τ diffusion does not have enough time to act, and will remain in the original water stream. If, on the other hand, τ convection τ diffusion the solute has time to diffuse across the central channel and the concentration of the solute will be the same in both streams. So in total we have that if we are operating a H-filter with two solutes it is important that the solute fulfil τ convection,1 << τ diffusion,1 τ convection, τ diffusion,. This way, it is possible to separate the two different solutes from each other. Below we see an example of how the system works in real life given the odds already mentioned, furthermore the length L = 1 mm, v 0 = 1 mm 9 m D 1 = 10 s, D = 4 10fracm s, height h = 1µ m and witdh ω = 100µ m. By calculation we find that τ convection,1 = 1 10 3 1 10 3 = 1 sec. s, 6

τ diffusion,1 = (10 10 6 ) 10 9 = 5 sec. τ convection, = 1 10 3 1 10 3 = 1 sec. τ diffusion, = (10 10 6 ) 4 10 11 = 50 sec. The optimal situation would be given by τ diffusion, << τ convection << τ diffusion,1. So, to obtain this we have to change either the length or the average speed of the system. The speed of the system depends on the pressure chosen for the system which we will calculate next. Again by looking at the figure it is seen there are three places for pressure and therefore three equations are obtained for the system: P 0 + P P a = 1 R 1Q P a P b = R Q P b P 0 = 1 R 1Q where R i refers to the hydraulic resistance for fluid i and Q is the flowrate. By solving these equations we arrive at The flowrate can be calculated from the equation, P = (R 1 + R ) Q (14) Q = whv ave = ( 10 10 6) ( 10 1 10 6) 1.0 10 3 1 m3 = 1.0 10 sec The hydraulic resistances for two plates then become, (15) R T OT = 1ηL 1 ( h 3 w = 1 10 3 1.0 10 3 The difference in pressure is then, 1 10 1 10 6 ) 3 10 10 6 14 P a sec = 1. 10 m 3 (16) P = R T OT Q = 1. 10 14 1.0 10 1 = 10P a (17) The hydrostatic pressure equation P = ρg h allows us to calculate the height for a pressure difference of 10 Pa. This yields h = 1. cm. In conclusion, we found that the system needs a longer time to diffuse small ions since we arrived at a diffusion time much larger than the convection time (about 5 times). The problem can be solved by either optimizing the speed or the length of the channel. 7

1.5 Taylor Dispersion Taylor dispersion is an example of combined convection and diffusion. This phenomenon is the behavioral result of a concentration being present in a solution flowing at a certain velocity. An obvious example of the preceding statement is a small pluck of finite concentration in a microchannel with a present Poiseuille flow. As time increases, the concentration profile of the pluck disperses axially as result of convection and it diffuses across the width and height of the channel because of concentration gradients. At the back of the paraboloid pluck whose shape is reflective of the Poiseuille flow velocity gradients, the concentration gradients encourage diffusion towards the middle of the channel. At its front-end, instead, the pluck diffuses towards the channel walls. The result is an evenly shaped plug moving downstream with a speed equal to the average Poiseuille flow velocity [3]. Motivation An article was published in 006 regarding the effect of cross-sectional shape on Taylor dispersion in microchannels [1]. The article illustrates how hydrodynamic dispersion for shallow rectangular microchannels is mostly affected by the width of the channel and not by the height, which is usually much smaller than the width. The writers of the article described the flow and transport of molecules in the channel by a time dependent effective dispersion coefficient, which they were able to obtain by calculating the first and second moments of the concentration distribution. The article claims to utilize a mathematical procedure, called the method of moments, described by Aris in 1956 []. This report tries to work out the calculations of the moments of the concentration, hoping to find a common ground with the recent article discussed above [1]. 3 Aris s Method of Moments In 1956, R. Aris described the distribution of a solute in terms of its moments in the direction of flow. He described a sequence of mathematical steps used to find the p th moment, c p. The effective diffusion coefficient was then shown to be proportional to the rate of growth of the second moment (p = ), which is closely related to the variance of the concentration. 3.1 Equations and Conditions In a cartesian coordinate system, a tube of arbitrary cross sectional shape is positioned with its axis along the x-axis and with his cross sectional area along the y z plane, with the z-axis being vertical. The velocity relative to the mean is given by, χ(y, z) = u(y, z) U 8 1 (18)

where u(y, z) is the velocity field, assumed for uniform flow to have only an x component, and U is the mean velocity. At time t, the concentration is denoted as C(x, y, z, t) and the diffusion coefficient as Dψ(y, z). A new dimensionless coordinate system, relative to a characteristic length parameter a, is introduced so that (x Ut) ξ =, η = y a a, ζ = z a, τ = Dt a, The concentration equation, in these new units is then µ = Ua D. (19) τ C = ψ ξ C µχ ξ C + η (ψ η C) + ζ (ψ ζ C) (0) with i denoting i. The conditions are the initial condition C (ξ, η, ζ, 0) = C 0 (ξ, η, ζ) (1), and the continuity of boundary, which involves the vanishing of the first derivative of the concentration along the normal to the curve describing the perimeter of the cross section of the tube. The p th moment of the distribution of solute in the tube through η, ζ at time t can be found by c p (η, ζ, τ) = + ξ p C(ξ, η, ζ, τ)dξ (). Applying the governing equation (0) to () yields an equation for the evolution of the p th moment of the concentration τ c p = η (ψ η c p ) + ζ (ψ ζ c p ) + p(p 1)ψc p + pµχc p 1 (3) Equation (3) can then be solved by applying the continuity of boundary condition and the initial condition in addition to an eigenfunction expansion method [] to yield c p (η, ζ, 0) = c p0 (η, ζ) (4) The mathematical tools described above constitute a step-by-step method which can be used to find any of the moments c p. In particular, the second moment c is of concern since it is closely related to the variance of the distribution of solute about the moving origin, σ (τ), by σ (τ) = 1 c (η, ζ, τ)dηdζ. (5) S S where S denotes the domain occupied by the interior of the tube in the plane η-ζ. The effective diffusion coefficient is then simply 9

D eff (τ) = 1 d dt σ (τ) (6) 3. Example: The Tube of Circular Cross-Section Applying the method of moments to a tube of circular cross-section, Aris illustrates results which agreed with previous work done by Taylor []. The new polar coordinate system is then η = ρcosθ, ζ = ρsinθ. (7) The characteristic dimension a is taken to be the radius of the tube, D is defined as the molecular diffusion coefficient, ψ = 1. For a tube of this shape, the velocity field in the axial direction has the form u(ρ) = P ( a 4η ρ ), (8) L where P refers to a difference in pressure across the tube, L is the length of the tube and η the viscosity of the fluid [3]. The mean velocity can then be found by averaging over the cross section of the tube U = 1 a a 0 u(ρ)dρ = 1 a a 0 P 4η L a P ( a 4η ρ ) dρ = P a L 4η L. (9) The velocity relative to the mean is then ( P 4η χ(ρ) = L a ρ ) 1 = 1 ρ. (30) The p th moment can then be found from the evolution equation τ c p = 1 ρ ρ (ρ ρ c p ) + 1 ρ θc p + p(p 1)c p + pµ(1 ρ )c p 1, (31) the boundary condition and the initial condition c p (ρ, θ, 0) = c p0 (ρ, θ), (3) For p = 0, equations (31), (3) and (33) become ρ c p = 0 at ρ = 1. (33) τ c 0 = 1 ρ ρ (ρ ρ c 0 ) + 1 ρ θc 0, (34) c 0 (ρ, θ, 0) = c 00 (ρ, θ), (35) 10

The solution is then of the form ρ c 0 = 0 at ρ = 1. (36) c 0 (ρ, θ, τ) = 1 + m=0 n=1 [ A 0 mn cos(mθ) + Bmnsin(mθ) 0 ] J m (α mn ρ)e α mn τ (37) with α mn being the n nt zero of djm dρ (ρ). The constants A and B can be found by applying the initial condition (36). Multiplying both sides by J k (α kn ρ) and integrating with respect to ρ from 0 to 1 then yields A 0 mn = B 0 mn = π 1 0 0 cos(mθ)ρj m(α mn ρ)c 00 (ρ, θ)dρdθ ( ), π 1 m α [J m (α mn )] mn (38) π 1 0 0 sin(mθ)ρj m(α mn ρ)c 00 (ρ, θ)dρdθ ( ). π 1 m α [J m (α mn )] mn (39) Similarly, for p = 1, equation (31) becomes τ c 1 = 1 ρ ρ (ρ ρ c 1 ) + 1 ρ θc 1 + µ ( 1 ρ ) c 0, (40) whose solution consists of 3 parts: The homogeneous solution of the form [ A 1 mn cos(mθ) + Bmnsin(mθ) 1 ] J m (α mn ρ)e α mn τ ; (41) m n A particular term coming from the constant part of c 0 1 4 µρ ( 1 1 ρ ) ; (4) A particular term coming from the non-constant part of c 0 of the form [ A 0 mn cos(mθ) + Bmnsin(mθ) 0 ] φ mn (ρ)e α mn τ, (43) µ m n with φ mn satisfying equation (40) and therefore 1 ρ ρ (ρ ρ φ nm ) + (α mn m ρ ) φ mn = J m (α mn ρ) (1 ρ ). (44) 11

As can be seen above, the more p is increased, the more complex the solutions for c p become. The method discussed can obviously also be used to find the nd moment, c, which in term allows the calculation of the effective diffusion coefficient. In his article [], Aris shows how the D eff calculated by his method of moments is equivalent to the results found by Taylor [4]. 4 Tube of Rectangular Cross Section The same work done by Aris for both a general case and for a tube of circular cross section [] was performed for the case of a rectangular geometry. 4.1 The Velocity Relative to the Mean The velocity field for the Poiseuille flow in such a channel is of the form ( ) u(η, ζ) = 4h P 1 cosh kπη ( ) h 1 π 3 η L k 3 cosh ( ) kπζ sin, (45) kπw h h k,odd with h being the vertical height of the channel, w the width, L the length, and a coordinate system chosen so that the channel lies on 0 < ζ < h and 1 w < η < 1 w [3]. The mean velocity is then U = 1 w hw w h 0 u(η, ζ)dζdη = 4h P π 3 η L k,odd [ 1 k 3 1 ( )] 1 kπw kπ kπ w tanh, h (46) so that the velocity relative to the mean, from equation (18), is { [ ] kπη 1 cosh( ( ) h ) } χ(η, ζ) = π k,odd k 1 sin kπζ 3 cosh( kπw h ) h { 1 [ k,odd k 1 1 4 kπ w tanh ( )]} 1. (47) kπw h 4. The 0 th Moment For p = 0 and ψ = 1, equation (3) becomes with initial and boundary conditions τ c 0 = ηc 0 + ζ c 0 (48) c 0 (η, ζ, 0) = c 00 (η, ζ), (49) η c 0 = 0, η = ± 1 w, (50) 1

and ζ c 0 = 0, ζ = 0, h. (51) Similarly as for the case of a tube with a circular cross section, the solution for the 0 th moment can be found to be with c 0 (η, ζ, τ) = n=0 m=0 ( ) nπ ( mπ ) γnm 0 cos w η cos h ζ β(τ) (5) and h γnm 0 = 4 0 wh w w c 00 (η, ζ)cos ( nπ w η) cos ( mπ h ζ) dηdζ [ 1 + 1 mπ sin ( )] mπ (53) [ ( n π β(τ) = exp 16h + m π ) ] 4w τ (54) 4.3 The 1 st Moment Increasing to just p = 1 makes the expressions a lot more complicated. In rectangular coordinates, the first moment of the concentration evolves according to the following equation with initial and boundary conditions τ c 1 [ η + ζ ] c1 = µχ(η, ζ)c 0 (η, ζ, τ). (55) c 1 (η, ζ, 0) = c 10 (η, ζ), (56) η c 1 = 0, η = ± 1 w, (57) and ζ c 1 = 0, ζ = 0, h. (58) The solution to the homogeneous problem can be found using a similar method used for c 0 (η, ζ, τ). It is then of the form with n=0 m=0 ( ) nπ ( mπ ) γnm 1 cos w η cos h ζ β(τ) (59) 13

h γnm 1 = 4 0 wh w w The particular solution can be divided into 4 parts: c 10 (η, ζ)cos ( nπ w η) cos ( mπ h ζ) dηdζ [ 1 + 1 mπ sin ( )] mπ (60) The term arising from the constant part of c 0, in other words when n = 0, m = 0. Equation (55) then becomes τ c 1 [ η + ζ ] c1 = µχ(η, ζ) (61) The term independent of ζ, when n = 1, m = 0 satisfying the equation τ c 1 [ η + ζ ] c1 = µχ(η, ζ) n=1 ( ) nπ γn0 0 cos w η β(τ) (6) The term independent of η, when n = 0, m = 1 satisfying the equation τ c 1 [ η + ζ ] c1 = µχ(η, ζ) m=1 The rest of the particular solution, satisfying ( mπ ) γ0m 0 cos h ζ β(τ) (63) τ c 1 [ η + ζ ] c1 = µχ(η, ζ) 5 Discussion n=1 m=1 ( ) nπ ( mπ ) γnm 0 cos w η cos h ζ β(τ) (64) Considering the form of χ(η, ζ) for a rectangular geometry, as can be seen from equation (47), the 4 differential equations obtained are extremely difficult to solve. The final solution for the first moment of the concentration for a rectangular channel is then a combination of equation (59) and the solutions to the differential equations (61), (6), (63) and (64) with the initial condition (56) and boundary conditions (57) and (58). Although a similar procedure could be done in order to obtain solutions to the second moment of the concentration, the report has to stop proceeding because of the complexity of the problem. Theoretically, and perhaps with the possibility of much more time available, a solution to the second moment could be found by making several assumptions and therefore eliminating most of the terms present in the calculations. It would be a possibility of future work establishing exactly the assumptions made in the article mentioned in section in order to then obtain a similar expression for the effective diffusion coefficient. 14

References [1] Ajdari, A.; Bontoux, N.; Stone, H. A. Hydrodynamic Dispersion in Shallow Microchannels: the Effect of Cross-Sectional Shape. Analytical Chemistry 006, 78, 387-39. [] Aris, R. On the dispersion of a solute in a fluid flowing through a tube. Proceedings of the Royal Society of London. Series A, Mathematical and Physical Sciences 1956, 35, 67-77. [3] Bruus, H. Theoretical microfluidics; Lecture notes, 3rd edition, Fall 006. [4] Taylor, Sir G. Proceedings of the Royal Society of London. Series A, Mathematical and Physical Sciences 1953, 19, 186. 15