Micromechanics of composites/Conservation 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,

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

Micromechanics of composites

Balance of linear momentum →