1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Reynolds Transport Theorem Notational Issue

  1. Oct 14, 2009 #1
    Perhaps someone can clear up some confusion for me. EQs (1) and (2) are synonymous, but use different notations for the derivative.

    In my textbook, the RTT is written in the following form:

    Given some fluid property 'B,' let [itex]\beta[/itex] be the intensive property [itex]\beta = d\, B/d\, m[/itex]. Then the rate of change of B of the system can be written in terms of a control volume:

    [tex]\frac{d}{dt}(B_{sys}) = \frac{d}{dt}(\int_{cv}\beta\rho\,dV) + \int_{cs}\beta\rho(\mathbf{v}\cdot\mathbf{n})\, dA \,\,\,\,\,\,\,\,\,(1)[/tex]

    But, I have seen it elsewhere using partials on the RHS and material derivative on the LHS:

    [tex]\frac{D}{Dt}(B_{sys}) = \frac{\partial}{\partial{t}}(\int_{cv}\beta\rho\,dV) + \int_{cs}\beta\rho(\mathbf{v}\cdot\mathbf{n})\, dA\,\,\,\,\,\,\,\,(2)[/tex]

    So my questions are :

    i.) Comparing the left-hand-sides of (1) and (2) I presume that [itex]d/dt \equiv D/Dt[/itex] ?

    ii.) Why are we using partials, [itex]\partial/\partial{t}[/itex], in (2) and not in (1) ?

    iii.) With regard to the material derivative, what is it exactly? Am I correct in saying that with respect to fluid motion, EQs (1) and (2) tell us about the change in a fluid property 'B' as we follow the system upstream?
    Last edited: Oct 14, 2009
  2. jcsd
  3. Oct 14, 2009 #2
    I have reformulated my question in attempt to attract some responses. I hope my questions are now clear and concise.

    Any ideas are welcome :smile:
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook