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 ~. }

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