Continuum mechanics/Balance of mass

Statement of the balance of mass

The balance of mass can be expressed as:

$\dot{\rho} + \rho~\boldsymbol{\nabla} \bullet \mathbf{v} = 0$

where $\rho(\mathbf{x},t)$ is the current mass density, $\dot{\rho}$ is the material time derivative of $\rho$, and $\mathbf{v}(\mathbf{x},t)$ is the velocity of physical particles in the body $\Omega$ bounded by the surface $\partial{\Omega}$.

Proof

We can show how this relation is derived by recalling that the general equation for the balance of a physical quantity $f(\mathbf{x},t)$ is given by

$\cfrac{d}{dt}\left[\int_{\Omega} f(\mathbf{x},t)~\text{dV}\right] = \int_{\partial{\Omega}} f(\mathbf{x},t)[u_n(\mathbf{x},t) - \mathbf{v}(\mathbf{x},t)\cdot\mathbf{n}(\mathbf{x},t)]~\text{dA} + \int_{\partial{\Omega}} g(\mathbf{x},t)~\text{dA} + \int_{\Omega} h(\mathbf{x},t)~\text{dV} ~.$

To derive the equation for the balance of mass, we assume that the physical quantity of interest is the mass density $\rho(\mathbf{x},t)$. Since mass is neither created or destroyed, the surface and interior sources are zero, i.e., $g(\mathbf{x}, t) = h(\mathbf{x},t) = 0$. Therefore, we have

$\cfrac{d}{dt}\left[\int_{\Omega} \rho(\mathbf{x},t)~\text{dV}\right] = \int_{\partial{\Omega}} \rho(\mathbf{x},t)[u_n(\mathbf{x},t) - \mathbf{v}(\mathbf{x},t)\cdot\mathbf{n}(\mathbf{x},t)]~\text{dA}~.$

Let us assume that the volume $\Omega$ is a control volume (i.e., it does not change with time). Then the surface $\partial{\Omega}$ has a zero velocity ($u_n = 0$) and we get

$\int_{\Omega} \frac{\partial \rho}{\partial t}~\text{dV} = - \int_{\partial{\Omega}} \rho~(\mathbf{v}\cdot\mathbf{n})~\text{dA}~.$

Using the divergence theorem

$\int_{\Omega} \boldsymbol{\nabla} \bullet \mathbf{v}~\text{dV} = \int_{\partial{\Omega}} \mathbf{v}\cdot\mathbf{n}~\text{dA}$

we get

$\int_{\Omega} \frac{\partial \rho}{\partial t}~\text{dV} = - \int_{\Omega} \boldsymbol{\nabla} \bullet (\rho~\mathbf{v})~\text{dV}.$

or,

$\int_{\Omega} \left[\frac{\partial \rho}{\partial t} + \boldsymbol{\nabla} \bullet (\rho~\mathbf{v})\right]~\text{dV} = 0 ~.$

Since $\Omega$ is arbitrary, we must have

$\frac{\partial \rho}{\partial t} + \boldsymbol{\nabla} \bullet (\rho~\mathbf{v}) = 0 ~.$

Using the identity

$\boldsymbol{\nabla} \bullet (\varphi~\mathbf{v}) = \varphi~\boldsymbol{\nabla} \bullet \mathbf{v} + \boldsymbol{\nabla} \varphi\cdot\mathbf{v}$

we have

$\frac{\partial \rho}{\partial t} + \rho~\boldsymbol{\nabla} \bullet \mathbf{v} + \boldsymbol{\nabla} \rho\cdot\mathbf{v} = 0 ~.$

Now, the material time derivative of $\rho$ is defined as

$\dot{\rho} = \frac{\partial \rho}{\partial t} + \boldsymbol{\nabla} \rho\cdot\mathbf{v} ~.$

Therefore,

${ \dot{\rho} + \rho~\boldsymbol{\nabla} \bullet \mathbf{v}= 0 ~. }$