Reynolds' transport theorem

Let $ F(\mathbf{r},t)$ represent the amount of some physical property of a continuous material medium per unit volume. The total amount of this property present in a finite region $ {\cal V}$ of the material is obtained through the volume integral.

$\displaystyle \int_{\cal V} F(\mathbf{r},t) \;dV
$

If this property is being transported by the action of the flow of the material with a velocity $ \mathbf{u}(\mathbf{r},t)$ , then Reynolds' transport theorem states that the rate of change of the total amount of $ F$ within the material volume is equal to the volume integral of the instantaneous changes of $ F$ occuring within the volume, plus the surface integral of the rate at which $ F$ is being transported through the surface $ {\cal S}$ (bounding $ {\cal V}$ ) to and from the surrounding region.

$\displaystyle \frac{d}{d t} \int_{\cal V} F(\mathbf{r},t) \;dV =
\int_{\cal V} ...
...\partial F}{\partial t} \;dV +
\int_{\cal S} F\mathbf{u} \cdot \mathbf{n} \;dS
$

Here, $ \mathbf{n}$ is a unit vector indicating the normal direction of the surface (oriented to point out of the volume).



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