Statement of the balance of mass

The balance of mass can be expressed as:

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

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

Proof

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

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

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

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

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

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

Using the divergence theorem

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

we get

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

or,

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

Since ${\displaystyle \Omega }$ is arbitrary, we must have

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

Using the identity

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

we have

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

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

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

Therefore,

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