Micromechanics of composites/Conservation of mass

Statement of the balance of mass

The balance of mass can be expressed as:

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$\dot{\rho }+\rho \mathrm{\nabla }\bullet 𝐯=0$
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where $\rho (𝐱,t)$ is the current mass density, $\dot{\rho }$ is the material time derivative of $\rho$, and $𝐯(𝐱,t)$ is the velocity of physical particles in the body $\mathrm{\Omega }$ bounded by the surface $\partial \mathrm{\Omega }$.

Proof

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

$\frac{d}{dt}[\underset{\mathrm{\Omega }}{\int }f(𝐱,t)\text{dV}]=\underset{\partial \mathrm{\Omega }}{\int }f(𝐱,t)[u_n(𝐱,t)-𝐯(𝐱,t)\cdot 𝐧(𝐱,t)]\text{dA}+\underset{\partial \mathrm{\Omega }}{\int }g(𝐱,t)\text{dA}+\underset{\mathrm{\Omega }}{\int }h(𝐱,t)\text{dV}.$

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

$\frac{d}{dt}[\underset{\mathrm{\Omega }}{\int }\rho (𝐱,t)\text{dV}]=\underset{\partial \mathrm{\Omega }}{\int }\rho (𝐱,t)[u_n(𝐱,t)-𝐯(𝐱,t)\cdot 𝐧(𝐱,t)]\text{dA}.$

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

$\underset{\mathrm{\Omega }}{\int }\frac{\partial \rho }{\partial t}\text{dV}=-\underset{\partial \mathrm{\Omega }}{\int }\rho (𝐯\cdot 𝐧)\text{dA}.$

Using the divergence theorem

$\underset{\mathrm{\Omega }}{\int }\mathrm{\nabla }\bullet 𝐯\text{dV}=\underset{\partial \mathrm{\Omega }}{\int }𝐯\cdot 𝐧\text{dA}$

we get

$\underset{\mathrm{\Omega }}{\int }\frac{\partial \rho }{\partial t}\text{dV}=-\underset{\mathrm{\Omega }}{\int }\mathrm{\nabla }\bullet (\rho 𝐯)\text{dV}.$

or,

$\underset{\mathrm{\Omega }}{\int }[\frac{\partial \rho }{\partial t}+\mathrm{\nabla }\bullet (\rho 𝐯)]\text{dV}=0.$

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

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

Using the identity

$\mathrm{\nabla }\bullet (\phi 𝐯)=\phi \mathrm{\nabla }\bullet 𝐯+\mathrm{\nabla }\phi \cdot 𝐯$

we have

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

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

$\dot{\rho }=\frac{\partial \rho }{\partial t}+\mathrm{\nabla }\rho \cdot 𝐯.$

Therefore,

$\dot{\rho }+\rho \mathrm{\nabla }\bullet 𝐯=0.$

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