Reynolds transport theorem

Introduction

Reynolds transport theorem [1] is a fundamental theorem used in formulating the basic laws of fluid mechanics. We will enunciate and demonstrate in this entry the referred theorem. For our purpose, let us consider a fluid flow, characterized by its streamlines, in the Euclidean vector space $({ℝ}^3,\parallel \cdot \parallel )$ and embedded on it we consider, a continuum body $ℬ$ occupying a volume $𝒱$ whose particles are fixed by their material (Lagrangian) coordinates $𝐗$, and a region $\mathrm{\Re }$ where a control volume $𝔳$ is defined whose points are fixed by it spatial (Eulerian) coordinates $𝐱$ and bounded by the control surface $\partial 𝔳$. An arbitrary tensor field of any rank is defined over the fluid flow according to the following definition.

Definition 1.

We call an extensive tensor property to the expression

$\mathrm{\Psi }(𝐱,t):={\displaystyle {\int }_{𝔳}}\psi (𝐱,t)\rho (𝐱,t)𝑑v,$(1)

where $\psi \mathit{}\mathrm{(}\mathrm{x}\mathrm{,}t\mathrm{)}$ is the respective intensive tensor property.

Theorem’s hypothesis

The kinematics of the continuum can be described by a diffeomorphism $\chi$ which, at any given instant $t\in [0,\mathrm{\infty })\subset ℝ$, gives the spatial coordinates $𝐱$ of the material particle $𝐗$,

$𝒱\times [0,\mathrm{\infty })\to 𝔳\times [0,\mathrm{\infty }),t\mapsto t,𝐗\mapsto 𝐱=\chi (𝐗,t).$

Indeed the above sentence corresponds to a change of coordinates which must verify

$J=\left|{\displaystyle \frac{\partial x_i}{\partial X_j}}\right|\equiv \left|F_{ij}\right|\ne 0,F_{ij}:={\displaystyle \frac{\partial x_i}{\partial X_j}},$

$J$ being the Jacobian of transformation and $F_{ij}$ the Cartesian components of the so-called strain gradient tensor $𝐅$.

Reynolds transport theorem 1.

The material rate of an extensive tensor property associate to a continuum body $\mathrm{B}$ is equal to the local rate of such property in a control volume $\mathrm{v}$ plus the efflux of the respective intensive property across its control surface $\mathrm{\partial }\mathit{}\mathrm{v}$.

Proof.

By taking on Eq.(1) the material time derivative,

${\displaystyle \frac{D\mathrm{\Psi }}{Dt}}=\dot{\mathrm{\Psi }}=\dot{\overline{{\displaystyle {\int }_{𝔳}}\psi \rho 𝑑v}}=\dot{\overline{{\displaystyle {\int }_{𝒱}}\psi \rho J𝑑V}}={\displaystyle {\int }_{𝒱}}\dot{\overline{\psi \rho J}}𝑑V={\displaystyle {\int }_{𝒱}}(\dot{\overline{\psi \rho }}J+\psi \rho \dot{J})𝑑V=$

${\displaystyle {\int }_{𝒱}}\left\{J\left[{\displaystyle \frac{\partial }{\partial t}}(\psi \rho )+𝐯\cdot {\nabla }_x(\psi \rho )\right]+\psi \rho (J{\nabla }_x\cdot 𝐯)\right\}𝑑V={\displaystyle {\int }_{𝒱}}\left\{\left[{\displaystyle \frac{\partial }{\partial t}}(\psi \rho )\right]+\left[𝐯\cdot {\nabla }_x(\psi \rho )+(\psi \rho ){\nabla }_x\cdot 𝐯\right]\right\}(JdV)$

$={\displaystyle {\int }_{𝔳}}{\displaystyle \frac{\partial }{\partial t}}(\psi \rho )𝑑v+{\displaystyle {\int }_{𝔳}}{\nabla }_x\cdot (\psi \rho 𝐯)𝑑v={\displaystyle \frac{\partial }{\partial t}}{\displaystyle {\int }_{𝔳}}\psi \rho 𝑑v+{\displaystyle {\int }_{\partial 𝔳}}\psi \rho 𝐯\cdot 𝐧𝑑a,$

since ${\partial }_t(dv)=0$ ($𝐱$ fixed) on the first integral and by applying the Gauss-Green divergence theorem on the second integral at the left-hand side. Finally, by substituting Eq.(1) on the first integral at the right-hand side, we obtain

$\dot{\mathrm{\Psi }}={\displaystyle \frac{\partial \mathrm{\Psi }}{\partial t}}+{\displaystyle {\int }_{\partial 𝔳}}\psi \rho 𝐯\cdot 𝐧𝑑a,$(2)

endorsing the theorem statement.∎