Axisymmetric deformation of drops through tubes with symmetric and asymmetric constrictions

Abstract

Drop deformation in constricted passages plays a vital role in porous media, microfluidic devices, lab-on-a-chip applications, etc. Depending on the viscosity ratio, Capillary number, drop volume and geometry of the constriction, drops may break-up by the snap-off mechanism. Different from the previous studies that consider symmetric single/periodic constrictions, we model a tube with a three-dimensional asymmetric constriction which is natural in porous media or can be used in microfluidic devices for the control of deformation and/or break-up mechanism. We model the motion of drops in Stokes regime and integrate the drop evolution using the axisymmetric boundary integral equations. Compared with the symmetric constrictions, the asymmetry of the constriction affects the snap-off time and thus the volume of the generated daughter droplet after snap-off. The volume is particularly affected by the smoothness of the upstream rather than the downstream of the constriction: more drop volume moves into to the constriction before the snap-off occurs. We show that in the case of a drop does not snap-off while passing through a particular constriction, an asymmetric design of this passage promotes snap-off. Pressure drop variation with time across the tube distinguishes the stages of snap-off from escape. Contrary to the use of two-dimensional models for flows in large aspect ratio microfluidic channels, we finally show that, because of the lack of out-of-plane curvature, planar models of tubular flows may be insufficient to observe certain physics such as jet formation and snap-off.

Appendices

Appendix 1

The analytical integration for the axisymmetric problem results with the following components of velocity and traction kernels with K(k) and E(k) being the elliptic integrals of the first and second kind [56]; R and r are the radial coordinates of source and field points, respectively [57].

$$\begin{aligned}&u_{rr}=\frac{1}{2 \pi \sqrt{a+b}}\nonumber \\&\quad \left[ -\left( \frac{a+c^2}{b}\right) K(k)+\left( \frac{b^2-a^2-ac^2}{b}\right) \frac{E(k)}{a-b}\right] \end{aligned}$$

(11)

$$\begin{aligned}&u_{rz}=\frac{1}{2 \pi \sqrt{a+b}}\left[ \left( \frac{cr}{b}\right) K(k)+\left( \frac{cR-acr}{b}\right) \frac{E(k)}{a-b}\right] \end{aligned}$$

(12)

$$\begin{aligned}&u_{zr}=\frac{1}{2 \pi \sqrt{a+b}}\left[ -\left( \frac{cR}{b}\right) K(k)+\left( \frac{-cr+acR}{b}\right) \frac{E(k)}{a-b}\right] \end{aligned}$$

(13)

$$\begin{aligned} u_{zz}&=\frac{1}{2 \pi \sqrt{a+b}}\left[ K(k)+\left( c^2 \right) \frac{E(k)}{a-b}\right] \end{aligned}$$

(14)

$$\begin{aligned} t_{rr}&=\frac{1}{\pi \left( a^2-b^2\right) \sqrt{a+b}}\Bigl [\frac{K(k)}{b^2}\left[ db^3+ab^2Rn_r-2abde \right. \Bigr . \nonumber \\&\quad + \left. \left( 4a^2-6b^2 \right) \left( eRn_r-dRr \right) -\left( 8a^3b-9ab^3 \right) Rn_r\right] \nonumber \\&\quad +\frac{E(k)}{2b^2\left( a-b \right) }\left[ -4ab^3d+\left( a^2b-3b^3 \right) \left( 2de-bRn_r \right) \right. \nonumber \\&\quad +\Bigl . \left. 4a\left( 3b^2-a^2 \right) \left( eRn_r-dRr\right) +\left( 8a^4-15a^2b^2+3b^4 \right) Rn_r\right] \Bigr ] \end{aligned}$$

(15)

$$\begin{aligned} t_{rz}&=\frac{1}{\pi \left( a^2-b^2\right) \sqrt{a+b}}\nonumber \\&\quad \Bigl [\frac{K(k)}{2b}\left[ -2Rbdc-2ac\left( R^2n_r-rd \right) -\left( 2a^2-3b^2 \right) cn_r \right] \Bigr . \nonumber \\&\quad +\Bigl .\frac{cE(k)}{b \left( a-b \right) }\nonumber \\&\quad \left[ -4Rabd +\left( a^2+3b^3 \right) \left( R^n_r-rd\right) +2a\left( 3b^2-a^2\right) n_r\right] \Bigr ] \end{aligned}$$

(16)

$$\begin{aligned} t_{zr}&=\frac{1}{\pi \left( a^2-b^2\right) \sqrt{a+b}}\nonumber \\&\quad \Bigl [\frac{cK(k)}{b^2}+\left[ rb^2d-abR\left( d-rn_r \right) +\left( 2a^2-3b^2 \right) R^2n_r \right] \Bigr .\nonumber \\&\quad +\frac{cE(k)}{b^2 \left( a-b \right) }\left[ -4rab^2d +\left( a^2+3b^3 \right) \left( Rd-Rrn_r\right) \right. \nonumber \\&\quad +\Bigl . \left. 2a\left( 3b^2-a^2\right) R^2n_r\right] \Bigr ] \end{aligned}$$

(17)

$$\begin{aligned} t_{zz}&=\frac{1}{\pi \left( a^2-b^2\right) \sqrt{a+b}}\Bigl [\frac{K(k)}{b}+\left[ -bc^2d-aRc^2n_r\right] \Bigr .\nonumber \\&\quad +\Bigl .\frac{E(k)}{b \left( a-b \right) } \left[ 4abdc^2+\left( a^2+3b^3\right) Rc^2n_r\right] \Bigr ] \end{aligned}$$

(18)

with

$$\begin{aligned} a&=r^2+R^2+c^2, \end{aligned}$$

(19)

$$\begin{aligned} b&=2rR, \end{aligned}$$

(20)

$$\begin{aligned} c&=Z-z, \end{aligned}$$

(21)

$$\begin{aligned} d&=cn_z-rn_r, \end{aligned}$$

(22)

$$\begin{aligned} e&=r^2+R^2, \end{aligned}$$

(23)

$$\begin{aligned} k&=\left[ 2b/\left( a+b\right) \right] ^{1/2}. \end{aligned}$$

(24)

The radial and axial components of normal vector and the interface curvature ($\nabla \cdot {\varvec{n}}$), which is the sum of axial ($\kappa _z$) and azimuthal curvatures ($\kappa _r$), are computed by

$$\begin{aligned} n_r&=\frac{\frac{dz}{ds}}{\left[ \left( \frac{dr}{ds}\right) ^2+\left( \frac{dz}{ds}\right) ^2\right] ^{1/2}},\nonumber \\ n_z&=-\frac{\frac{dr}{ds}}{\left[ \left( \frac{dr}{ds}\right) ^2+\left( \frac{dz}{ds}\right) ^2\right] ^{1/2}} \end{aligned}$$

(25)

$$\begin{aligned} \nabla \cdot&{\varvec{n}}=\kappa _z+\kappa _r=\frac{\frac{dr}{ds}\frac{d^2z}{ds^2}-\frac{dz}{ds}\frac{d^2r}{ds^2}}{\left[ \left( \frac{dr}{ds}\right) ^2+\left( \frac{dz}{ds}\right) ^2\right] ^{3/2}}\nonumber \\&\quad +\frac{\frac{dz}{ds}}{r\left[ \left( \frac{dr}{ds}\right) ^2+\left( \frac{dz}{ds}\right) ^2\right] ^{1/2}} \end{aligned}$$

(26)

where s is the arc length along the generating curve. As the source point ${\varvec{x}}$ goes to zero ($R=0$), we use the limiting values of the kernel components:

$$\begin{aligned} u_{rr}&=0 \end{aligned}$$

(27)

$$\begin{aligned} u_{rz}&=0 \end{aligned}$$

(28)

$$\begin{aligned} u_{zr}&=-\frac{1}{4}\frac{cr}{a^{3/2}} \end{aligned}$$

(29)

$$\begin{aligned} u_{zz}&=\frac{1}{4}\frac{a+c^2}{a^{3/2}} \end{aligned}$$

(30)

$$\begin{aligned} t_{rr}&=0 \end{aligned}$$

(31)

$$\begin{aligned} t_{rz}&=0 \end{aligned}$$

(32)

$$\begin{aligned} t_{zr}&=-\frac{3}{2}\frac{rdc}{a^{5/2}} \end{aligned}$$

(33)

$$\begin{aligned} t_{zz}&=\frac{3}{2}\frac{dc^2}{a^{5/2}} \end{aligned}$$

(34)

Appendix 2

The fundamental solutions of planar Stokes flow, nondimensionalized with same scales as axisymmetric case but $l_s=2R_0$, are

$$\begin{aligned} G_{ij}(\varvec{x,y})=-\frac{1}{4\pi }\left( \delta _{ij} \ln r-\frac{{\hat{r}}_{i}{\hat{r}}_{j}}{r^2}\right) , \end{aligned}$$

(35)

$$\begin{aligned} T_{ijk}(\varvec{x,y})=-\frac{1}{\pi }\frac{{\hat{r}}_{i}{\hat{r}}_{j}{\hat{r}}_{k}}{r^4} \end{aligned}$$

(36)

for velocity and stress, respectively. Unlike the axisymmetric problem, we directly place them into the linear system. The system is similar to the axisymmetric case (10) but in cartesian coordinates and given by

$$\begin{aligned} {\varvec{c}}\begin{bmatrix} u_x({\varvec{x}}) \\ u_y({\varvec{x}}) \end{bmatrix}=(1-\lambda ) \int _{\Gamma _d}\begin{bmatrix} T_{xx} &{} T_{xy} \\ T_{yx} &{} T_{yy} \end{bmatrix} \begin{bmatrix} u_x({\varvec{y}}) \\ u_y({\varvec{y}}) \end{bmatrix}d\Gamma \nonumber \\ + \int _{\Gamma }\begin{bmatrix} T_{xx} &{} T_{xy} \\ T_{yx} &{} T_{yy} \end{bmatrix} \begin{bmatrix} u_x({\varvec{y}}) \\ u_y({\varvec{y}}) \end{bmatrix}d\Gamma \nonumber \\ -\int _{\Gamma }\begin{bmatrix} G_{xx} &{} G_{xy} \\ G_{yx} &{} G_{yy} \end{bmatrix} \begin{bmatrix} t_x({\varvec{y}}) \\ t_y({\varvec{y}}) \end{bmatrix}d\Gamma \nonumber \\ -\int _{\Gamma _d}\frac{\nabla \cdot {\varvec{n}}}{Ca}\begin{bmatrix} G_{xx} &{} G_{xy} \\ G_{yx} &{} G_{yy} \end{bmatrix} \begin{bmatrix} n_x({\varvec{y}}) \\ n_y({\varvec{y}}) \end{bmatrix}d\Gamma . \end{aligned}$$

(37)