•   Wikis

# Continuum mechanics/Balance of mass: Wikis

Note: Many of our articles have direct quotes from sources you can cite, within the Wikipedia article! This article doesn't yet, but we're working on it! See more info or our list of citable articles.

# Study guide

Up to date as of January 14, 2010

## 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 ρ, and $\mathbf{v}(\mathbf{x},t)$ is the velocity of physical particles in the body Ω 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 Ω is a control volume (i.e., it does not change with time). Then the surface $\partial{\Omega}$ has a zero velocity (un = 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 Ω 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 ρ is defined as

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

Therefore,

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