The Boundary Element Method in Fluid Mechanics: Application to Bubble Growth

Abstract

The origin of the numerical implementation of boundary integral equations can be traced from fifty years earlier, when the electronic computers had become available. The full emergence of the numerical technique known as the boundary element method occurred in the late 1970s. In implementing the method, only the boundary of the solution domain has to be discretized into elements. In the case of a two-dimensional problem, this is really easy to do: put closely packed points on the boundary (a curve) and join up two consecutive neighboring points to form straight line elements. In this chapter we present one of the applications of this method, namely, the growth and detachment of bubbles generated by the continuous injection of gas into a quiescent liquid and the effect of partial confinement on the shape and volume of bubbles generated by injection of a constant flow rate of gas. In the problem of bubble generation, the contours are the surfaces of the bubbles and the solid surfaces of the reservoir, which are all surfaces of revolution. The unknowns involved in the formulation of the boundary element are fluid particle velocities that define surfaces of the bubbles and the stresses on the vessel wall. First, we neglect viscous effects and assume the flow to be irrotational so that a velocity potential exists. In second case we solve the Stokes equations for the liquid and the evolution equation for the surface of a bubble. Experiments with two different liquids show that cylindrical and conical walls and cylinder walls with periodic concentric corrugations with a gas injected through an orifice at the bottom of the liquid may strongly affect the shape and volume of the bubbles, and can be used to control the size of the generated bubbles without changing the flow rate of gas.

Appendix A. The Green’s Functions for Axisymmetric Flow

This Appendix lists Green functions for axisymmetric flow generated by a ring force of unit strength located at $(x_{0}, r_{0})$ and pointing in the direction $\mathbf{e }_{k}$ with $k = r, x$. Defining the following quantities in cylindrical coordinares

$$\begin{aligned} Z&= x - x_{0} \\ L&= \sqrt{Z^{2} + \left( {r + r_{0} } \right) } \\ D&= \sqrt{Z^{2} + \sqrt{Z^{2} + \left( {r + r_{0} } \right) } }\\ S&= \sqrt{Z^{2} + r^{2} + r_{0}^{2} } \\ m&= \frac{{2\left( {rr_{0} } \right) ^{{\frac{1}{2}}} }}{L} \end{aligned}$$

and elliptic integrals

$$\begin{aligned} K\left( m \right)&= \int _{0}^{{\frac{\pi }{2}}} {\frac{{d\theta }}{{\sqrt{1 - m^{2} sen^{2} \theta } }}} \\ E\left( m \right)&= \int _{0}^{{\frac{\pi }{2}}} {\sqrt{1 - m^{2} sen^{2} \theta d\theta } } , \end{aligned}$$

we have

$$\begin{aligned} G_{x}^{x}&= 4\frac{r}{L} \left( K + E\frac{Z^{2}}{D^{2}} \right) \\ G_{r}^{x}&= 2\frac{Z}{L} \left( K - E\frac{S^{2} - r^{2}}{D^{2}}\right) \\ G_{x}^{r}&= 2\frac{{rZ}}{{r_{0} L}}\left( { - K + E\frac{{S^{2} - 2r_{0} ^{2} }}{{D^{2} }}} \right) \\ G_{r}^{r}&= 2\frac{1}{{r_{0} L}}\left[ { - k\left( {S^{2} + Z^{2} } \right) - E\left( {L^{2} + \frac{{Z^{2} S^{2} }}{{D^{2} }}} \right) } \right] \\ {T_{x}^{x}}_{x}&= 8\frac{{rZ^{3} }}{{D^{2} }}\left( {K - E\frac{{4S^{2} }}{{D^{2} }}} \right) \\ {T_{x}^{x}}_{r}&= {T_{r}^{x}}_{x} = - 4\frac{{Z^{2} }}{{D^{2}L }}\left[ {K\frac{{S^{2} - 2r^{2} }}{{L^{2} }} - E\left( {1 + \frac{{8r_{0}^{2} \left( {2r_{0}^{2} - s^{2} } \right) }}{{D^{2} L^{2} }}} \right) } \right] \\ {T_{r}^{x}}_{r}&= - 4\frac{{rZ}}{L}\left[ {K\left( {\frac{1}{{r^{2} }} + \frac{{2Z^{2} }}{{D^{2} L^{2} }}} \right) - \frac{E}{{D^{2} }}\left( {6 - S^{2} \left( {\frac{1}{{r^{2} }} + \frac{{8Z^{2} }}{{D^{2} L^{2} }}} \right) } \right) } \right] \\ {T_{x}^{r}}_{r}&= - 4\frac{{rZ}}{{r_{0} D^{2} L}}\left[ {K\frac{{2r_{0}^{2} - S^{2} }}{{L^{2} }} - E\left( {1 + \frac{{8r_{0}^{2} \left( {2r^{2} - S^{2} } \right) }}{{D^{2} L^{2} }}} \right) } \right] \\ {T_{x}^{r}}_{r}&= {T_{r}^{r}}_{x} = - 4\frac{Z}{{r_{0} }}\left[ {K\left( {\frac{{Z^{2} S^{2} }}{{D^{2} L^{2} }} - 2} \right) + \frac{E}{{D^{2} }}\left( {2S^{2} - Z^{2} - \frac{{16r^{2} r_{0}^{2} Z^{2} }}{{D^{2} L^{2} }}} \right) } \right] \\ \qquad {T_{r}^{r}}_{r}&= - 4\frac{r}{L}\left[ {\frac{K}{{r_{0} }}\left( {\frac{{Z^{2} \left( {S^{2} - 2r_{0}^{2} } \right) }}{{D^{2} L^{2} }} - \frac{{r^{2} - r_{0}^{2} - 2Z^{2} }}{{r^{2} }}} \right) } \right. \\&\quad + \left. {\frac{E}{{D^{2} }}\left( {\frac{{8r_{0} Z^{2} \left( {S^{2} - 2r^{2} } \right) }}{{D^{2} L^{2} }} + \frac{{r^{2} \left( {r^{2} + r_{0}^{2} } \right) - S^{2} \left( {r_{0}^{2} + 2Z^{2} } \right) }}{{r^{2} r_{0} }}} \right) } \right] \end{aligned}$$