Reynolds Transport Theorem

Reynolds Transport Theorem is a good starting point for deriving governing equations of quantities in a fluid.

Fundamental balance equation:

$ \mbox{in} - \mbox{out} + \mbox{generation} - \mbox{consumption} = \mbox{accumulation} $

Reynolds Transport Theorem is a formal way of performing this balance over a fluid control volume, or fluid material volume, of arbitrary shape, moving at an arbitrary velocity.

Definitions and derivations of some important terms and identities used to construct Reynolds Transport Theorem are given here: Reynolds Transport Theorem Derivation

Definition

For any extensive fluid property $ B $, there is a corresponding intensive property $ b = \frac{\partial B}{\partial m} $.

A material volume corresponding to a mass of fluid with the property $ B $ can be written as $ V_{B}^{M}(t) $

$ B(t) = \iiint_{V_{B}^{M}} \rho b(\boldsymbol{x}, t) dV $

where $ \rho $ is the fluid density.

The quantity of interest for the material volume is material derivative of the extensive property, $ \frac{dB}{dt} $ - that is, the rate of change of $ B $ in the material volume $ V_{B}^{M} $ (which is a volume whose boundaries move such that the flux of $ B $ across the boundaries is 0, see Reynolds Transport Theorem Derivation page for details).

This quantity can be written as:

$ \frac{dB}{dt} = \frac{d}{dt} \iiint_{V_{B}^{M}(t)} \rho b(\boldsymbol{x}, t) dV $

Because the integral is with time, the integral $ \iiint_{V_{B}^{M}(t)} and the derivative <math>\frac{d}{dt} $ do not commute.

However, if the integration were over a fixed volume that did not change in time (for example, $ V_{B,0} $, the integral and the derivative would commute.

The Jacobian $ J = \frac{ d V_{B}^{M} }{ d V_{0,B} } $ allows us to do this.

The volume integral can be converted as follows:

$ \begin{array}{rcl} \frac{ dB }{ dt } &=& \frac{d}{dt} \iiint_{V_{B}^{M}(t)} \rho b(\boldsymbol{x},t) dV \\ &=& \frac{d}{dt} \iiint_{V_{B,0}} \rho b(\boldsymbol{x},t) J dV \end{array} $

Next, the temporal derivative can be taken inside the integral (since they now commute):

$ \begin{array}{rcl} \frac{d}{dt} \iiint_{V_{B,0}} \rho b(\boldsymbol{x},t) J dV &=& \iiint_{V_{B,0}} \left[ \frac{ d \rho b }{ dt } J + \rho b \frac{ dJ }{ dt } \right] dV \\ &=& \iiint_{V_{B,0}} \left[ \frac{ d \rho b }{ dt } J + \rho b \frac{ 1 }{ J } \frac{dJ}{dt} \right] J \end{array} $

Next, the Euler expansion formula can be used to simplify the second term in the integral to yield:

$ \frac{dB}{dt} = \iiint_{V_{B,0}} \left[ \frac{ d \rho b }{ dt } + \rho b \left( \nabla \cdot \boldsymbol{v}_{B} \right] J dV $

and this can now be transformed back to the integral over the material volume $ V_{B}^{M} $ to yield Reynolds' Transport Theorem:

$ \frac{dB}{dt} = \frac{d}{dt} \iiint_{V_{B}^{M}(t)} \rho b dV = \iiint_{V_{B}^{M}(t)} \left[ \frac{d \rho b}{ dt } + \rho b \left( \nabla \cdot \boldsymbol{v}_{B} \right) \right] dV $

Now, the definition of the substantial derivative can be used to put this into a different (more familiar) form:

$ \frac{dB}{dt} = \iiint_{V_{B}^{M}(t)} \left[ \frac{ \partial \rho b }{\partial t} + \nabla \cdot \left( \rho b \boldsymbol{v}_{B} \right) \right] dV $

Green's Theorem can now be applied to this last divergence term, so that

$ \frac{dB}{dt} = \iiint_{V_{B}^{M}(t)} \frac{ \partial \rho b }{\partial t} dV + \iint_{S_{B}^{M}(t)} \rho b \boldsymbol{v}_{B} \cdot \boldsymbol{n} dS $

where $ S_{B}^{M} $ is the surface of the material volume $ V_{B}^{M} $, $ \boldsymbol{n} $ is the vector normal to this surface, and $ dS $ is a differential area on the surface.