The dependence of the cross-sectional shape on the hydraulic resistance of microchannels

Size: px

Start display at page:

Download "The dependence of the cross-sectional shape on the hydraulic resistance of microchannels"

Philippa Arnold
6 years ago
Views:

Supervisor: Niels Asger Mortensen and Henrik Bruus MIC Department

1 3-weeks course report, s0973 The dependence of the cross-sectional shape on the hydraulic resistance of microchannels Hatim Azzouz a Supervisor: Niels Asger Mortensen and Henrik Bruus MIC Department of Micro and Nanotechnology Technical University of Denmark 5 July 004

2 Contents 1 Introduction 3 Theory 4.1 Poiseuille Flow in a circular pipe Velocity field Flow rate Poiseuille Flow in an elliptic pipe Velocity field Flow rate Poiseuille Flow in a rectangular pipe - Fourier Transformation Velocity field Flow rate Hydraulic resistance Elliptic cross-section Rectangular cross-section FEMLAB simulations 13 4 Conclusion 16

3 Chapter 1 Introduction For a steady incompressible flow through a channel the hydraulic resistance of the channel depends on the cross-section shape of the channel. The objective of this project is to investigate the dependence between the hydraulic resistance and the cross-section of the channel in detail as well as to investigate the hydraulic resistance for various cross-section geometries. In the following, Chapter, a description of the Poiseuille flow through a channel with a series of cross-section geometries will be given and a definition of the hydraulic resistance will be introduced. The flow through channels will be simulated by use of FEMLAB. The results obtained in the FEMLAB simulation will be compared with those predicted by theory and discussed in Chapter 3. Finally, conclusions are given. 3

4 Chapter Theory.1 Poiseuille Flow in a circular pipe.1.1 Velocity field We consider an incompressible fluid flowing through a pipe (cylindrical tube) of length L in the x direction and a circular cross section with radius a in the yz plane; see Figure.1. The pressure drop over the length L is p, i.e p(0)=p 0 + p and p(l)=p 0. We introduce cylindrical polar coordinates in the yz-plane, y=rcosφ and z=rsinφ, where r is the radial distance from the cylinder axis and φ is the azimuthal angle (the angle around from the z axis). a Figure.1: A cylindrical tube with radius a. We study a flow along the axis of the tube, so that v = v x (r)e x = (v x (r), 0, 0). (.1) This field depends only on r and it is invariant under translations along the cylinder axis as well as rotations. 4

5 .1. POISEUILLE FLOW IN A CIRCULAR PIPE CHAPTER. THEORY In applying the Navier-Stokes equation 1 we assume that there is neither external forces nor instantaneous acceleration affecting the flow i.e f=0 and v t =0, respectively, giving p = η v (.3) where η is the viscosity of the fluid. Since v=e x v x (r) it follows from projecting of the Navier-Stokes equation p=η v on e r and e φ that p r = p = 0. (.4) φ This means that pressure only depends on the longitudinal coordinate p = p(x). Calculating the Laplacian term acting on a function of r in the Navier-Stokes equation we finally obtain ( ) dvx (r) v x (r) = ( v x (r)) = e x dr dp(x) dx = η ( d v x (r) dr + 1 r = d v x (r) dr + 1 r dv x dr (.5) ) dv x. (.6) dr The left hand side of this equation depends only on x whereas the right hand side depends only on r. Since the pressure difference over the length of the pipe is given by p L it follows that dp(x) dx = p (.7) L where the minus sign is due to the assumption that if the fluid is supposed to run in the positive x-direction the pressure must be higher at the beginning (x = 0) than at the end (x = L). Thus we obtain, with a little rewriting, the following Navier-Stokes equation equation ( 1 d r dv ) x(r) = η p r dr dr L. (.8) The solution to this equation with no-slip boundary conditions, i.e v x (0)=v x (a)=0, is ( ) v x (r) = v 0 1 r a (.9) where v 0 is the integration constant. This can easily be shown to be the solution by inserting dv x (r) = v 0r dr a (.10) 1 Navier-Stokes equation is a partial differential equation expressing the local balance of momentum in a fluid around any point of space at any time. It is given by ρ v t = p + η v + f (.) where the terms v is the instantaneous acceleration, p is the dynamical pressure gradient, t η v is the shear viscosity term and f is the external body force. 5

6 .. POISEUILLE FLOW IN AN ELLIPTIC PIPE CHAPTER. THEORY in the finally obtained Navier-Stokes. The integration constant is then determined to be v 0 = 1 p 4η L a. (.11) Thus, the solution to our Navier-Stokes equation is v x (r) = 1 ) p (1 4η L a r a. (.1).1. Flow rate The total flow rate Q (volume pr. time) carried through the pipe is given by Q = ρ dydz v x (r). (.13) circle In cylindrical polar coordinates it can be rewritten as to get Q = ρ a π 0 0 rdrdφ v x (r) (.14) Q = ρ p 8η L πa4. (.15). Poiseuille Flow in an elliptic pipe..1 Velocity field For an elliptic pipe, as shown in Figure., with an elliptic cross section having axes a and b along the y and the z axis, respectively, the same assumptions as for the circular pipe are valid. Instead of using the cylindrical polar coordinates we introduce the cartesian coordinates. The velocity field is again assumed longitudinal of the form v = v x (y, z)e x = (v x (y, z), 0, 0). (.16) The Navier-Stokes equation is given by the same equation as in a circular pipe. Choosing the cartesian coordinates the Navier-Stokes equation can be expressed as η v x (y, z) = p (.17) where the Laplacian term can be calculated by the same formalism of section.1 to give v x (y, z) = Thus we obtain the following Navier-Stokes equation 1 η ( ) v x (y, z) y + v x (y, z) z. (.18) ( p ) L = v x (y, z) y + v x (y, z) z 6 (.19)

7 .. POISEUILLE FLOW IN AN ELLIPTIC PIPE CHAPTER. THEORY a b Figure.: A pipe with an elliptic cross-section having axes a and b with the solution, satisfying the no-slip condition, given by Inserting v x(y,z) integration constant = v dy 0 a ( ) v x (y, z) = v 0 1 y a z b. (.0) and v x(y,z) dz = v 0 b v o = 1 η Thus, the longitudinal velocity profile in an elliptic pipe becomes v x (y, z) = 1 η p L in the Navier-Stokes equation we obtain the p L (ab) a + b (ab) a + b. (.1) ) (1 y a z b. (.) Notice that for a = b the integration constant for the elliptic pipe becomes the one for the circular pipe indicating that circular pipe is a special case of the elliptical... Flow rate The flow rate is given by Q = ρ dydz v x (y, z). (.3) ellipse Introducing the transformation variables Ỹ = 1 b z and Z= 1 ay, and inserting them in the integral for the flow rate we obtain Q = abv 0 ρ ellipse dỹ d Z 1 (1 ỹ z ) = πabv 0 rdr (1 r ) (.4) After the integration we finally obtain the flow rate for the elliptic pipe Q = ρ 4η p L 0 π(ab) 3 a + b (.5) 7

8 .3. POISEUILLE FLOW IN A RECTANGULAR PIPE - FOURIER TRANSFORMATION CHAPTER. THEORY.3 Poiseuille Flow in a rectangular pipe - Fourier Transformation.3.1 Velocity field The solution of the flow in the rectangular pipe is not simple to obtain. The no-slip boundary condition does not in this case allow for a separation of the D problem into two independent 1D problems. Instead we solve the problem by Fourier analysis, where only sine functions are involved since they guarantee fulfilment of the no-slip boundary condition. We Fourier-expand in the z-axis, so that v x (y, 0) = v x (y, h) = 0. The solution to the Navier-Stokes equation is of the form In applying the Navier-stokes equation v x (y, z) = f n (y) sin(nπ z h ) (.6) n=1 v x (y, z) = P ηl (.7) we get v x (y, z) = [ d f n (y) n=1 dy n π h ] f n(y) sin(nπ z h ) (.8) In the Navier-Stokes equation we also have to Fourier transform the constant P/ηL, or simpler just the constant 1 which can be multiplied by any constant. The Fourier transform of the constant P/ηL thus becomes The solution is only possible if P ηl = P ηl 1 = P 4 ηl π n,odd 1 n sin(nπ z ). (.9) h f n (y) = 0 for n even (.30) d f n (y) dy n π h f n(y) = P 4 for n odd (.31) ηl nπ The first step to solve this nonhomogeneous equation is to find a general solution of the homogeneous equation f n (y) h. The characteristic equation of the homogeneous equation is λ (nπ/h) λ = 0. (.3) It has the roots λ 1 = 0 and λ = nπ/h. Hence a real general solution of the homogenous equation is f n (y) h = Ae yλ 1 + Be yλ. (.33) 8

9 .3. POISEUILLE FLOW IN A RECTANGULAR PIPE - FOURIER TRANSFORMATION CHAPTER. THEORY W H Figure.3: A rectangular pipe of height h and width w. Using the definitions of the hyperbolic functions sinh x = ex e x cosh x = ex + e x the general solution of the homogeneous equation can be rewritten as ( f n (y) h = B cosh nπ y ) ( + B sinh nπ y ) h h (.34) (.35) (.36) The second step is to find a particular solution f n (y) p of the nonhomogeneous equation. Since the equation on the right has the derivative zero, we say f n (y) p = C. (.37) Substitution gives (n π h ) C = P 4 P ηl nπ. By comparison, C = ηl solution of the nonhomogeneous equation is 4h π 3 1 n 3. Hence a general f n (y) = f n (y) h + f n (y) p (.38) ( = B cosh nπ y ) ( + B sinh nπ y ) P 4h 1 h h η π 3 n 3. (.39) From the boundary condition f n ( ω/) = f n (ω/) = 0, we get B = P η resulting in [ ( ) ] cosh f n (y) = P η 4 h π 3 n 3 cosh nπ y z ( ) 1 nπ w h 4h π 3 1 n 3 1 cosh(nπ w h ). (.40) Thus the solution of the flow in the rectangular pipe is v x (y, z) = P 4 [ ( ) 1 cosh nπ y ] ( η h z π 3 n 3 ( ) 1 sin nπ y ). (.41) n,odd cosh nπ w z h 9

10 .4. HYDRAULIC RESISTANCE CHAPTER. THEORY.3. Flow rate The flow rate for a rectangular pipe is given by Using the following we get w 0 h 0 w h Q = ρ dy dz v x (y, z). (.4) 0 0 ( dy cosh nπ y ) h ( dz sinh nπ z ) h n,odd Q = P [ 1η ρwh3 1 = h ( nπ sinh nπ y ) (.43) h = h for n odd (.44) nπ 1 n 4 = π4 96. (.45) n,odd 19 π 5 h w 1 (nπ n 5 tanh w ) ]. (.46) h.4 Hydraulic resistance In analogy with Ohm s law for an electric resistor, it is natural to define the hydraulic resistance R hyd of a pipe carrying some liquid as p = R hyd Q (.47) Since we are interested in the relation between the hydraulic resistance and the crosssection geometry of the pipe it is appropriate to reexpress the hydraulic resistance as R = α η ρa (.48) where α is the correction coefficient. We introduce the compactness β which the measure of the square of the perimeter l over the area A of the cross-section geometry Ω, β = l A = ( Ω ) ( 1 dl dr) (.49) Ω.4.1 Elliptic cross-section Expressing the correction coefficient as a function of the compactness we get for the circular cross-section that α(β) = 8π = β. 10

11 .4. HYDRAULIC RESISTANCE CHAPTER. THEORY Dd Figure.4: A domain ω. subdomain is the area A. The red boundary is the perimeter l of the domain and the blue For the elliptic cross-section the compactness is given by ( ) π 0 1 dθ (1 b /a ) sin θ β = (.50) πab = 16 ( π a dθ 1 (1 b π b /a ) sin θ) (.51) 0 16 a, a b. (.5) π b Thus, the correction coefficient for the elliptic cross-section is given by α(β) = 4π a + b ab π β. (.53) 4.4. Rectangular cross-section ( ) Assuming that w > h then tanh nπ w h 1 and we get the flow rate to Q P ( 1η ρwh ) π 5 ζ(5)w. (.54) h The correction coefficient is then given by α 1π 5 γ π 5 γ 186ζ(5), (.55) where γ is the width-to-hight ratio w/h. Setting the compactness β = 8 + 4γ + 4/γ for a rectangle in the expression of α we get α = From the expression of α it can be shown that for 1π 5 (β/4 1/γ) π 5 (β/4 1/γ) 186ζ(5). (.56) α lim β β = 3 (.57) 11

12 .4. HYDRAULIC RESISTANCE CHAPTER. THEORY we have the asymptotic dependence α 3β + c. Since 3ζ(5) lim (α 3β) = 4 + β π 5 (.58) we get the correction coefficient for the rectangular class to α = 3β (.59) 1

13 Chapter 3 FEMLAB simulations FEMLAB is a partial differential equation solver that builds on MATLAB. FEMLAB - which stands for Finite Element Modelling LABoratory- is an advanced software package for modeling and simulation of any physical process that can be described with partial differential equations. The differential equation must be specified in some geometry, the so-called computational domain, together with the boundary conditions along the surface of the domain as well as the initial values at the starting time. The computational domain is divided into a huge number of finite-sized elements forming the basis computation. Having introduced the analytic description of the Poiseuille flow through a pipe with various cross-sections, we are now interested in simulating the flow in FEMLAB for the same cross-section geometries. The numerical calculations are then compared to those predicted by the theory. Before starting the simulation, a differential equation must be specified. Since the equation expressing the velocity field is of the form P = η v, the the simulation differential model was chosen to be a Piosson s equation in the D-space. To draw the geometry in which the partial differential equation must be specified, we click on Draw Mode on the tool bar. In this window the desired geometry can now be designed. Hereafter, the no-slip boundary is specified by clicking on Boundary Settings under Physics. Specifying the subdomain settings is done by clicking on Subdomain Settings under Physics. The geometry is then refined and the PDE solved. The solution plotted, displays the velocity field; see Figure 3.1. The velocity field of the flow has its maximum in the center of the pipe, but begins to decrease when moving radially from the center. When in contact with the cross-section boundary the velocity of the flow is at its minimum, fulfilling the no-slip condition. Since we also are interested in calculating the the hydraulic resistance, the correction coefficient 1 and the compactness is determined by making subdomain A, boundary l and velocity field integration. The results depicting the correction coefficient versus the com- 1 We saw that the hydraulic resistance was defined as P/L = R hyd Q. This can be rewritten as P = η α ρ dydzv L ρ A x (y, z). Since all the values are required dimensionless in FEMLAB we set the factor A P/Lη = 1 giving us the correction coefficient: α = dydzv x(y,z). 13

CHAPTER 3. FEMLAB SIMULATIONS Figure 3.1: A diagram showing the simulation results of the velocity field for a circular, elliptical, rectangular and triangular cross-section.

14 CHAPTER 3. FEMLAB SIMULATIONS Figure 3.1: A diagram showing the simulation results of the velocity field for a circular, elliptical, rectangular and triangular cross-section. pactness for various cross-sections is shown in Figure 3.. It emerges clearly from Figure 3. that the relation between the correction coefficient and the compactness is linear for the three cross-section classes: circular, elliptic and rectangular. The numeric proportionality factor for all cross-section groups is in good agreement with those predicted by theory. For the elliptic it is.55, close to the theoretical value of.47 The deviation is due to the assumption: a b, made in our calculation. For the rectangular class, the value of the proportionality factor is found to be 3. This value agree completely with the theory. Since the analytic calculations are not quite possible for triangle classes, only simulation was made. The results for this cross-section class is also shown in Figure 3.. Also here, the relation between α and β is linear. The proportionality factor is less than the other cross-section classes, namely For cross-section classes more complicated than the elliptical (circular), α increases for small values of β, leading to an increase of the hydraulic resistance. 14

15 CHAPTER 3. FEMLAB SIMULATIONS Correction coefficient vs. Compactness Theory for rectangular class. Theory for ellipse class. Linear fit to triangular class. 100 For circular l /A Figure 3.: Correction coefficient versus compactness for the elliptical, rectangular and triangular class. The result for the circle is also indicated. 15

16 Chapter 4 Conclusion We have demonstrated a close to linear relation between the hydraulic resistance and the compactness for the following cross-section geometries: circular, elliptic, triangular and rectangular. The results obtained were in good agreement with theory. For geometries more complicated than the circular the resistance increases, since a big part of the fluid is in contact with the inner surface of the channel. 16

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,