From charlesreid1

Revision as of 19:44, 31 October 2010 by Admin (talk | contribs)

Reynolds Transport Theorem is always a safe starting point.

In a way, it is the most fundamental starting point.

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, of arbitrary shape, moving at an arbitrary velocity.

A derivation of Reynolds Transport Theorem is given here: Reynolds Transport Theorem Derivation

Given an extensive property $ F $ and a corresponding intensive property $ \mathcal{F} = \frac{\partial F}{\partial t} $, $ F $ can be defined in terms of $ \mathcal{F} $:

$ F = \iiint_{V_{\mathcal{F}}(t)} \rho \mathcal{F} dV $

where $ \rho $ is the fluid density and $ V_{\mathcal{F}} $ is the material volume that corresponds to the property $ F $ (which moves at a velocity $ u_{\mathcal{F}} $.

The partial derivative with respect to volume for the intensive property $ \mathcal{F} $ is,

$ \frac{ \partial \mathcal{F} }{ \partial V } = 0 $

and for the extensive property $ F $,

$ \frac{ \partial F }{ \partial V } = \frac{ \partial F }{ \partial m } \frac{ \partial m }{ \partial V } = \frac{1}{\rho} \frac{ \partial F}{\partial m} $

Then Reynolds Transport Theorem is given by:

$ \frac{d}{dt} \iiint_{V_{\mathcal{F}}(t)} = \iiint_{V_{\mathcal{F}}(t)} \rho \frac{ \partial \mathcal{F} }{\partial t} dV + \iint_{S_{\mathcal{F}}(t)} \rho \mathcal{F} \boldsymbol{v}_{\mathcal{F}} \cdot \boldsymbol{n} dS $

where $ \boldsymbol{v}_{\mathcal{F}} $ is the velocity vector of the property $ \mathcal{F} $.