8.3. Drag acting on a spherical bubble at high \(Re\)

Summary

• Subject: Levich drag [Lev62]

• Main conclusion: Drag \(F_{D} = 12 \pi \mu u a\) for \(Re \rightarrow \infty\).

• Key idea Drag balances with viscous dissipation, which concentrates at bubble surface due to high \(Re\).

• Reference: Derivation presented here is given by G. K. Batchelor [Bat00] (Chap. 5).

Consider a spherical bubble steadily rising through stagnant incompressible fluid at a high Reynolds number. The bubble is rising along the \(z\) axis toward the \(+z\) side (\(\theta = 0\)) at a uniform velocity \(\mathbf{u}~(= u \mathbf{e}_{z})\). The drag force acting on the bubble is \(\mathbf{F}~(= - F_{D} \mathbf{e}_{z})\), which directs \(-z\), so that the reaction on the fluid is \(-\mathbf{F}\). The rise motion of the bubble pushing the fluid in the direction parallel to \(\mathbf{u}\). Therefore, the work done by the reaction on the fluid is given by \(- \mathbf{F} \cdot \mathbf{u} = F_{D} u\).

If we ride on the bubble we may see \(\mathbf{u} = - u \mathbf{e}_{z}\). The drag force \(\mathbf{F}~(= - F_{D} \mathbf{e}_{z})\) directing toward \(-z\), and therefore, the reaction on the fluid is \(- \mathbf{F}\). The fluid is traveling downward (\(-z\)) while experiencing the reaction, so that the work done on the fluid is \(- \mathbf{F} \cdot \mathbf{u} = F_{D} u\).

Since the bubble velocity is constant the work produced by the rise motion is dissipated by the viscous dissipation in the fluid, that is,

(8.103)\[ F_{D} u \delta t = \delta t \iiint_{V} 2 \mu e_{ij} e_{ij} dV\]

where \(\delta t\) is an infinitesimal time variation, \(e_{ij}\) is the rate of strain tensor defined by

(8.104)\[ e_{ij} = \frac{1}{2} \left( \frac{\partial v_{i}}{\partial x_{j}} + \frac{\partial v_{j}}{\partial x_{i}} \right)\]

A brief description of the viscous dissipation is given in Appendix Viscous dissipation. In the limiting case of \(Re \rightarrow \infty\) the vorticity produced at the bubble surface is limited only in a thin layer on the surface, so that the velocity field in the bulk fluid can be assumed to be irrotational. Therefore, there exists a velocity potential deriving the velocity field:

(8.105)\[ \mathbf{v} = \nabla \phi\]

By the potential flow theory (Appendix Potential flow about sphere), \(\phi\) for a spherical body moving along the \(z\) axis (\(\theta = 0\)) at the constant speed \(u\) is given by

(8.106)\[ \phi = - \frac{u a^{3}}{2 r^{2}} \cos \theta\]

Note that the bubble center at this instant locates the origin of the coordinate. The velocity components for this potential are

(8.107)\[\begin{split}\begin{split} &v_{r} = \frac{\partial \phi}{\partial r} = \frac{u a^{3}}{r^{3}} \cos \theta \\ &v_{\theta} = \frac{1}{r} \frac{\partial \phi}{\partial \theta} = \frac{u a^{3}}{2 r^{3}} \sin \theta \end{split}\end{split}\]

The rate of strain tensor is rewritten as

(8.108)\[ e_{ij} = \frac{\partial^{2} \phi}{\partial x_{i} \partial x_{j}}\]

Therefore,

(8.109)\[ e_{ij} e_{ij} = \frac{\partial^{2} \phi}{\partial x_{i} \partial x_{j}} \frac{\partial^{2} \phi}{\partial x_{i} \partial x_{j}}\]

The R.H.S. can be rewritten as

(8.110)\[ \frac{\partial^{2} \phi}{\partial x_{i} \partial x_{j}} \frac{\partial^{2} \phi}{\partial x_{i} \partial x_{j}} = \left( \frac{\partial}{\partial x_{i}} \frac{\partial \phi}{\partial x_{j}} \right) \frac{\partial^{2} \phi}{\partial x_{i} \partial x_{j}} = \frac{\partial}{\partial x_{i}} \left( \frac{\partial \phi}{\partial x_{j}} \frac{\partial^{2} \phi}{\partial x_{i} \partial x_{j}} \right) - \frac{\partial \phi}{\partial x_{j}} \frac{\partial^{3} \phi}{\partial x_{i} \partial x_{i} \partial x_{j}}\]

However, the last term vanishes since \(\nabla^{2} \phi = 0\) by the continuity equation, \(\nabla \cdot \mathbf{v} = 0\). Then, the remaining term becomes

(8.111)\[\begin{split}\begin{split} \frac{\partial}{\partial x_{i}} \left( \frac{\partial \phi}{\partial x_{j}} \frac{\partial^{2} \phi}{\partial x_{i} \partial x_{j}} \right) &= \frac{\partial}{\partial x_{i}} \frac{\partial}{\partial x_{i}} \left( \frac{\partial \phi}{\partial x_{j}} \frac{\partial \phi}{\partial x_{j}} \right) - \frac{\partial}{\partial x_{i}} \left( \frac{\partial \phi}{\partial x_{j}} \frac{\partial^{2} \phi}{\partial x_{i} \partial x_{j}} \right) \\ &= \frac{\partial}{\partial x_{i}} \frac{\partial}{\partial x_{i}} \left( \frac{\partial \phi}{\partial x_{j}} \frac{\partial \phi}{\partial x_{j}} \right) - \frac{\partial^{2} \phi}{\partial x_{i} \partial x_{j}} \frac{\partial^{2} \phi}{\partial x_{i} \partial x_{j}} - \frac{\partial \phi}{\partial x_{j}} \frac{\partial^{3} \phi}{\partial x_{i} \partial x_{i} \partial x_{j}} \end{split}\end{split}\]

The third term vanishes and the second term is the same as the L.H.S. pf Eq. (8.110), so that,

(8.112)\[ \frac{\partial^{2} \phi}{\partial x_{i} \partial x_{j}} \frac{\partial^{2} \phi}{\partial x_{i} \partial x_{j}} = \frac{1}{2} \frac{\partial}{\partial x_{i}} \frac{\partial}{\partial x_{i}} \left( \frac{\partial \phi}{\partial x_{j}} \frac{\partial \phi}{\partial x_{j}} \right)\]

Therefore,

(8.113)\[ \iiint_{V} 2 \mu e_{ij} e_{ij} dV = \mu \iiint_{V} \frac{\partial}{\partial x_{i}} \frac{\partial}{\partial x_{i}} \left( \frac{\partial \phi}{\partial x_{j}} \frac{\partial \phi}{\partial x_{j}} \right) dV\]

Applying the divergence theorem yields

(8.114)\[ \mu \iiint_{V} \frac{\partial}{\partial x_{i}} \frac{\partial}{\partial x_{i}} \left( \frac{\partial \phi}{\partial x_{j}} \frac{\partial \phi}{\partial x_{j}} \right) dV = \mu \iint_{S_{\infty}} \frac{\partial}{\partial x_{i}} \left( \frac{\partial \phi}{\partial x_{j}} \frac{\partial \phi}{\partial x_{j}} \right) n_{i} dS - \mu \iint_{S} \mathbf{e}_{r} \cdot \nabla \left( \frac{\partial \phi}{\partial x_{j}} \frac{\partial \phi}{\partial x_{j}} \right) dS\]

where \(n_{i}\) is the unit outward normal to \(V\), \(S_{\infty}\) is the surface area surrounding \(V\), and \(S\) is the surface area of the bubble. The fluid is at rest in the far field; hence, the surface integral for \(S_{\infty}\) vanishes:

(8.115)\[ \mu \iiint_{V} \frac{\partial}{\partial x_{i}} \frac{\partial}{\partial x_{i}} \left( \frac{\partial \phi}{\partial x_{j}} \frac{\partial \phi}{\partial x_{j}} \right) dV = - \mu \iint_{S} \mathbf{e}_{r} \cdot \nabla \left( \frac{\partial \phi}{\partial x_{j}} \frac{\partial \phi}{\partial x_{j}} \right) dS\]

The operator \(\mathbf{e}_{r} \cdot \nabla\) is the gradient in the \(r\) direction and \(\nabla \phi \cdot \nabla \phi\) is the square, \(v^{2}\), of the velocity magnitude. We therefore have

(8.116)\[ F_{D} u = - \mu \int_{0}^{2 \pi} \int_{0}^{\pi} \frac{\partial v^{2}}{\partial r} a^{2} \sin \theta d\theta d\varphi\]

The velocity square is

(8.117)\[ v^{2} = v_{r}^{2} + v_{\theta}^{2} = \frac{u^{2} a^{6}}{r^{6}} \left( \cos^{2} \theta + \frac{\sin^{2} \theta}{4}\right)\]

The derivative \(\partial v^{2} / \partial r\) is then given by

(8.118)\[ \frac{\partial v^{2}}{\partial r} = - \frac{6 u^{2} a^{6}}{r^{7}} \left( \cos^{2} \theta + \frac{\sin^{2} \theta}{4} \right)\]

At the bubble surface,

(8.119)\[ \left. \frac{\partial v^{2}}{\partial r} \right|_{r=a} = - \frac{6 u^{2}}{a} \left( \cos^{2} \theta + \frac{\sin^{2} \theta}{4} \right)\]

By substituting this into Eq. (8.116) we obtain

(8.120)\[\begin{split}\begin{split} F_{D} u &= 12 \pi \mu u^{2} a \int_{0}^{\pi} \left( \cos^{2} \theta + \frac{\sin^{2} \theta}{4} \right) \sin \theta d\theta \\ &= 12 \pi \mu u^{2} a \left( \frac{2}{3} + \frac{1}{3} \right) = 12 \pi \mu u^{2} a \end{split}\end{split}\]

and

(8.121)\[ F_{D} = 12 \pi \mu u a\]

The drag coefficient is given by

(8.122)\[ C_{D} = \frac{48}{Re}\]

This result was obtained by Levich. In an air-water system bubbles from 0.5 to 1 mm in diameter behave to obey this theory.