7. Virtual mass¶

Referred from

Summary

Subject: (Fluid) particles feels their masses heavier than the actual.
Main conclusion: Particle mass is virtually \(\rho_{P} + C_{VM} \rho\), where \(\rho\) is the density of surrounding fluid, and \(C_{VM} = 1/2\) for sphere.
Key idea Particles must accelerate surroundings, thereby the accelerating motion gives an inertial force.
Reference: Landau and Lifshitz [LL87] (Chap. 1)

In this section, a solid sphere moving in an incompressible inviscid fluid is considered. Therefore, let us introduce the velocity potential to express the fluid velocity \(\mathbf{v}\):

(7.1)¶\[ \mathbf{v} = \nabla \phi\]

Substituting this expression into the continuity equation

(7.2)¶\[ \nabla \cdot \mathbf{v} = 0\]

yields the Laplace equation of \(\phi\):

(7.3)¶\[ \nabla^{2} \phi = 0\]

_images/cvm.png — Fig. 7.1 Spherical body in infinite fluid¶

The center of the sphere is set at the origin of the spherical coordinates. At positions far from the sphere, we assume that \(\phi\) is a function of \(r\). \(\phi_{0} = 1/r\) obviously satisfies the Laplace equation:

(7.4)¶\[ \nabla^{2} \phi_{0} \rightarrow \frac{1}{r^{2}} \frac{d}{dr} r^{2} \frac{d \phi_{0}}{dr} = \frac{1}{r^{2}} \frac{d}{dr} r^{2} \left( - \frac{1}{r^{2}} \right) = 0\]

Derivatives of \(\phi_{0}\), can also be solution of the Laplace equation since

(7.5)¶\[ \nabla^{2} \frac{\partial^{n} \phi_{0}}{\partial x_{i_{1}} \partial x_{i_{2}} \cdots \partial x_{i_{n}}} = \frac{\partial^{n} }{\partial x_{i_{1}} \partial x_{i_{2}} \cdots \partial x_{i_{n}}} \nabla^{2} \phi_{0} = 0\]

where \(n \ge 1\). The velocity component \(\mathbf{v}^{(1)}\) for \(- a \phi_{0}\) is given by

(7.6)¶\[ \mathbf{v}^{(1)} = \nabla \left( - \frac{\alpha}{r} \right) = \frac{\partial r}{\partial x_{i}} \frac{d}{dr} \left( - \frac{\alpha}{r} \right) = \alpha \frac{x_{i}}{r} \frac{1}{r^{2}} = \frac{\alpha x_{i}}{r^{3}}\]

where \(a\) is a constant. The mass flow rate through the surface of a large sphere \(S_{\infty}\) set at the origin is

(7.7)¶\[\begin{split} \iint_{S_{\infty}} \rho \mathbf{v}^{(1)} \cdot \mathbf{n} dS &= \iint_{S_{\infty}} \rho \mathbf{v}^{(1)} \cdot \mathbf{e}_{r} dS \rightarrow \rho \alpha \iint_{S_{\infty}} \frac{x_{i} x_{i}}{r^{4}} dS = \rho \alpha \iint_{S_{\infty}} \frac{1}{r^{2}} dS \\ &= 4 \pi \rho \alpha\end{split}\]

where \(\mathbf{n}\) is the outward unit normal to \(S_{\infty}\). However, this must vanish because of the incompressibility; therefore, \(\alpha = 0\). The next candidate is \(\nabla \phi_{0}\). Let us therefore set \(\mathbf{A} \cdot \nabla \phi_{0}\), where \(\mathbf{A}\) is a constant vector. The vector component for this velocity potential is given by

(7.8)¶\[\begin{split}\begin{split} \mathbf{v}^{(2)} &= \nabla ( \mathbf{A} \cdot \nabla \phi_{0} ) \rightarrow \frac{\partial }{\partial x_{i}} \left( A_{j} \frac{\partial }{\partial x_{j}} \frac{1}{r} \right) = \frac{\partial }{\partial x_{i}} \left( A_{j} \frac{\partial r}{\partial x_{j}} \frac{d}{dr} \frac{1}{r} \right) = A_{j} \frac{\partial }{\partial x_{i}} \left( \frac{x_{j}}{r} \frac{d}{dr} \frac{1}{r} \right) = - A_{j} \frac{\partial }{\partial x_{i}} \left( \frac{x_{j}}{r^{3}} \right) \\ &= - A_{j} \left( \frac{\delta_{ij}}{r^{3}} - x_{j} \frac{3 x_{i}}{r^{5}} \right) = \frac{3 A_{j} x_{j} x_{i}}{r^{5}} - \frac{A_{i}}{r^{3}} \rightarrow \frac{3 ( \mathbf{A} \cdot \mathbf{e}_{r} ) \mathbf{e}_{r} - \mathbf{A}}{r^{3}} \end{split}\end{split}\]

In the far field, velocity components given by the velocity potentials of higher order negligibly small compared with \(\mathbf{v}^{(2)}\). Therefore, in the following, we may write \(\mathbf{v}\) instead of \(\mathbf{v}^{(2)}\).

The velocity component normal to the solid sphere \(v_{r}~(= \mathbf{v} \cdot \mathbf{e}_{r})\) must be equal to that of the sphere \(u_{r}~(= \mathbf{u} \cdot \mathbf{e}_{r})\), i.e. the boundary condition \(( \mathbf{v} - \mathbf{u} ) \cdot \mathbf{e}_{r} = 0\).

(7.9)¶\[\begin{split}\begin{split} &\left. \left( \frac{3 A_{j} x_{j} x_{i}}{r^{5}} - \frac{A_{i}}{r^{3}} - u_{i} \right) \frac{x_{i}}{r} \right|_{r=a} = 0 \rightarrow \left. \frac{2 A_{i} x_{i} }{r^{4}} - u_{i} \frac{x_{i}}{r} \right|_{r=a} = 0 \rightarrow \left. \left( \frac{2 A_{i}}{r^{3}} - u_{i} \right ) \frac{x_{i}}{r} \right|_{r=a} = 0 \\ &\rightarrow A_{i} = \frac{a^{3} u_{i}}{2} \end{split}\end{split}\]

The total kinetic energy \(E\) of the fluid is given by

(7.10)¶\[ E = \iiint_{V_{\infty}-V} \frac{\rho v^{2}}{2} dV\]

where \(V_{\infty}\) is the fluid volume of a sphere of large size, while its radius will be taken as infinity later to cover the whole system. \(V\) is the volume of the sphere, \(V = 4 \pi a^{3} / 3\). \(( \mathbf{v} - \mathbf{u} ) \cdot ( \mathbf{v} + \mathbf{u} ) = v^{2} - u^{2}\). Therefore, \(v^{2} = u^{2} + ( \mathbf{v} - \mathbf{u} ) \cdot ( \mathbf{v} + \mathbf{u} )\). The integral of the \(v\) square is therefore rewritten as

(7.11)¶\[ \iiint_{V_{\infty}-V} \left\{ u^{2} + ( \mathbf{v} - \mathbf{u} ) \cdot ( \mathbf{v} + \mathbf{u} ) \right\} dV = \iiint_{V_{\infty}-V} u^{2} dV + \iiint_{V_{\infty}-V} ( \mathbf{v} - \mathbf{u} ) \cdot ( \mathbf{v} + \mathbf{u} ) dV\]

The first term can be immediately integrated as

(7.12)¶\[ \iiint_{V_{\infty}-V} u^{2} dV = u^{2} \iiint_{V_{\infty}-V} dV = u^{2} \left( V_{\infty} - V \right)\]

For the second term, by writing \(\mathbf{v} = \nabla \phi\) and \(\mathbf{u} = \nabla ( \mathbf{u} \cdot \mathbf{r} )\), we have

(7.13)¶\[ \mathbf{v} + \mathbf{u} = \nabla \phi + \nabla ( \mathbf{u} \cdot \mathbf{r} ) = \nabla ( \phi + \mathbf{u} \cdot \mathbf{r} )\]

Then,

(7.14)¶\[\begin{split}\begin{split} ( \mathbf{v} - \mathbf{u} ) \cdot ( \mathbf{v} + \mathbf{u} ) &= ( \mathbf{v} - \mathbf{u} ) \cdot \nabla ( \phi + \mathbf{u} \cdot \mathbf{r} ) = \nabla \cdot ( \mathbf{v} - \mathbf{u} ) ( \phi + \mathbf{u} \cdot \mathbf{r} ) - ( \phi + \mathbf{u} \cdot \mathbf{r} ) \nabla \cdot ( \mathbf{v} - \mathbf{u} ) \\ &= \nabla \cdot ( \mathbf{v} - \mathbf{u} ) ( \phi + \mathbf{u} \cdot \mathbf{r} ) \end{split}\end{split}\]

The second term in the third term vanishes since \(\nabla \cdot \mathbf{v} = \nabla \cdot \mathbf{u} = 0\). Applying the divergence theorem yields

(7.15)¶\[\begin{split}\begin{split} &\iiint_{V_{\infty}-V} u^{2} dV + \iiint_{V_{\infty}-V} ( \mathbf{v} - \mathbf{u} ) \cdot ( \mathbf{v} + \mathbf{u} ) dV \\ &= u^{2} \left( V_{\infty} - V \right) + \iiint_{V_{\infty}-V} \nabla \cdot ( \mathbf{v} - \mathbf{u} ) ( \phi + \mathbf{u} \cdot \mathbf{r} ) dV \\ &= u^{2} \left( V_{\infty} - V \right) + \iint_{S_{\infty}} ( \mathbf{v} - \mathbf{u} ) ( \phi + \mathbf{u} \cdot \mathbf{r} ) \cdot \mathbf{e}_{r} dS - \iint_{S} ( \mathbf{v} - \mathbf{u} ) ( \phi + \mathbf{u} \cdot \mathbf{r} ) \cdot \mathbf{e}_{r} dS \end{split}\end{split}\]

where \(S_{\infty}\) is the surface surrounding \(V_{\infty}\). However, \(( \mathbf{v} - \mathbf{u} ) \cdot \mathbf{e}_{r} = 0\) on \(S\).

\[\begin{equation*} \begin{split} u^{2} \left( V_{\infty} - V \right) + \iint_{S_{\infty}} ( \mathbf{v} - \mathbf{u} ) ( \phi + \mathbf{u} \cdot \mathbf{r} ) \cdot \mathbf{e}_{r} dS \end{split} \end{equation*}\]

Expanding the second term gives

(7.16)¶\[\begin{split}\begin{split} \iint_{S_{\infty}} ( \mathbf{v} - \mathbf{u} ) ( \phi + \mathbf{u} \cdot \mathbf{r} ) \cdot \mathbf{e}_{r} dS &= \iint_{S_{\infty}} \left( \frac{3 ( \mathbf{A} \cdot \mathbf{e}_{r} ) \mathbf{e}_{r} - \mathbf{A}}{r^{3}} - \mathbf{u} \right) \left( \mathbf{A} \cdot \nabla \phi_{0} + \mathbf{u} \cdot \mathbf{r} \right) \cdot \mathbf{e}_{r} dS \\ &\rightarrow \iint_{S_{\infty}} \left( \frac{3 A_{j} x_{j} x_{i} - A_{i} r^{2}}{r^{5}} - u_{i} \right) \left( A_{k} \frac{\partial}{\partial x_{k}} \frac{1}{r} + u_{k} x_{k} \right) \frac{x_{i}}{r} dS \\ &= \iint_{S_{\infty}} \left( \frac{2 A_{j} x_{j}}{r^{3}} - u_{j} x_{j} \right) \left( - \frac{A_{k} x_{k}}{r^{3}} + u_{k} x_{k} \right) \frac{ dS }{r} \\ &= \iint_{S_{\infty}} \left( - \frac{2 ( A_{j} x_{j} )^{2}}{r^{3}} + 3 A_{j} x_{j} u_{k} x_{k} - r^{3} ( u_{j} x_{j} )^{2} \right) \frac{dS}{r^{4}} \end{split}\end{split}\]

Since the radius \(a_{\infty}\) of the sphere \(S_{\infty}\) is large, the first term of the integrand is negligible in comparison with the other two terms, so that,

(7.17)¶\[\begin{split}\begin{split} \iint_{S_{\infty}} \left( - \frac{2 ( A_{j} x_{j} )^{2}}{r^{3}} + 3 A_{j} x_{j} u_{k} x_{k} - r^{3} ( u_{j} x_{j} )^{2} \right) \frac{dS}{r^{4}} &\rightarrow \iint_{S_{\infty}} \left( 3 A_{j} x_{j} u_{k} x_{k} - r^{3} ( u_{j} x_{j} )^{2} \right) \frac{dS}{r^{4}} \\ &= \frac{3 A_{j} u_{k} - a_{\infty}^{3} u_{j} u_{k}}{a_{\infty}^{2}} \iint_{S_{\infty}} \frac{x_{j} x_{k}}{r^{2}} dS \\ \end{split}\end{split}\]

The integrand of the surface integral is the dyadic \(\mathbf{e}_{r} \mathbf{e}_{r}\).

(7.18)¶\[\begin{split} \iint_{S_{\infty}} \frac{x_{j} x_{k}}{r^{2}} dS = \iint_{S_{\infty}} \mathbf{e}_{r} \mathbf{e}_{r} dS \end{split}\]

For the \(xx\) component,

(7.19)¶\[\begin{split} \int_{0}^{2 \pi} \int_{0}^{\pi} ( \sin \theta \cos \varphi )^{2} a_{\infty}^{2} \sin \theta d\theta d\varphi = a_{\infty}^{2} \int_{0}^{\pi} \sin^{3} \theta d\theta \int_{0}^{2 \pi} \cos^{2} \varphi d\varphi = \cdot \frac{4 \pi a_{\infty}^{2}}{3} \end{split}\]

For the \(xy\) component,

(7.20)¶\[\begin{split} \int_{0}^{2 \pi} \int_{0}^{\pi} ( \sin \theta \cos \varphi ) ( \sin \theta \sin \varphi ) a_{\infty}^{2} \sin \theta d\theta d\varphi = a_{\infty}^{2} \int_{0}^{\pi} \sin^{3} \theta d\theta \int_{0}^{2 \pi} \cos \varphi \sin \varphi d\varphi = 0 \end{split}\]

In summary,

(7.21)¶\[\begin{split} \iint_{S_{\infty}} \mathbf{e}_{r} \mathbf{e}_{r} dS \rightarrow \frac{4 \pi a_{\infty}^{2}}{3} \delta_{jk} \end{split}\]

Therefore,

(7.22)¶\[\begin{split} \frac{3 A_{j} u_{k} - a_{\infty}^{3} u_{j} u_{k}}{a_{\infty}^{2}} \iint_{S_{\infty}} \frac{x_{j} x_{k}}{r^{2}} dS = 4 \pi A_{j} u_{j} - \frac{ 4 \pi a_{\infty}^{3} }{3} u_{j} u_{j} = 4 \pi A_{j} u_{j} - V_{\infty} u_{j} u_{j} \end{split}\]

Thus,

(7.23)¶\[\begin{split} E = \iiint_{V_{\infty}-V} \frac{\rho v^{2}}{2} dV = \frac{\rho}{2} \left( 4 \pi A_{j} u_{j} - V u^{2} \right) = \frac{\rho}{2} ( 4 \pi A_{j} - V u_{j} ) u_{j} \end{split}\]

As we have confirmed (Eq. (7.9)), \(\mathbf{A}\) is proportional to \(\mathbf{u}\). Therefore, we may write

(7.24)¶\[ P_{j} = m_{ij} u_{i} = 4 \pi \rho A_{j} - \rho V u_{j}\]

where \(\mathbf{P}\) is the total momentum of the fluid, and the symmetric tensor \(m_{ij}\) is referred to as the induced-mass tensor. Thus, the total kinetic energy of the fluid is expressed as

(7.25)¶\[ E = \frac{1}{2} m_{ij} u_{i} u_{j}\]

By substituting Eq. (7.9) into Eq. (7.24) we obtain

(7.26)¶\[ m_{ij} u_{i} = ( 2 \pi \rho a^{3} - \rho V ) u_{j}\]

Hence,

(7.27)¶\[ m_{ij}= \rho \frac{2 \pi a^{3}}{3} \delta_{ij} = \frac{1}{2} \rho V \delta_{ij}\]

For the sphere, the induced mass is the half of the fluid mass removed by the sphere.

Consider a spherical particle oscillating in a fluid under the action of an external force \(\mathbf{f}\), which is the source of the change in the total momentum in the system, i.e., the momentum of the fluid and that of the particle. Therefore,

(7.28)¶\[ \frac{d ( M \mathbf{u} + \mathbf{P} )}{dt} = \mathbf{f}\]

where \(M\) is the mass of the particle. Substituting Eq. (7.24) yields

(7.29)¶\[ \left( M \delta_{ij} + m_{ij} \right) \frac{d u_{j}}{dt} = f_{i}\]

We then consider a spherical particle in a fluid under oscillation. If this particle was the fluid, the momentum of this volume is \(\rho V \mathbf{v}\) and the force acting on the fluid particle of volume \(V\) is given by

(7.30)¶\[ \rho V \frac{d v_{i}}{dt}\]

where the particle is assumed to be much smaller than the length scale for the spatial change in \(\mathbf{v}\). This volume is actually the particle and may have velocity \(\mathbf{u}\) different from the fluid velocity \(\mathbf{v}\). We therefore need to account for the force due to the relative motion \(\mathbf{u} - \mathbf{v}\):

(7.31)¶\[ - m_{ij} \frac{d (u_{j} - v_{j})}{dt} \]

where the negative sign is added since this is a force acting on the particle as the reaction.

When the relative velocity \(\mathbf{u} - \mathbf{v}\) increases, the force acts on the sphere so as to retard the relative velocity.

The equation of motion of the particle is therefore given by

(7.32)¶\[ M \frac{d u_{i}}{dt} = \rho V \frac{d v_{i}}{dt} - m_{ij} \frac{d (u_{j} - v_{j})}{dt} \]

The second term on the R.H.S. is the so-called virtual (added) mass force. By making use of Eq. (7.27), we have

(7.33)¶\[ M \frac{d u_{i}}{dt} = \rho V \frac{d v_{i}}{dt} - \frac{1}{2} \rho V \frac{d (u_{i} - v_{i})}{dt} \]

Denoting the particle density as \(\rho_{P}~(=M/V)\), one can write

(7.34)¶\[ \left( \rho_{P} + \rho C_{VM} \right) \frac{d u_{i}}{dt} = \rho \left( 1 + C_{VM} \right) \frac{d v_{i}}{dt} \]

where \(C_{VM} = 1/2\) is the virtual mass coefficient of the sphere.