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I Interpretation of the Reynolds Transport Theorem?

  1. Mar 4, 2017 #1
    Background:
    I am taking an undergraduate fluid mechanics class. I seem to have a misunderstanding with my interpretation of Reynolds Transport theorem (RTT), which I have written below:

    $$\frac{DB_{sys}}{Dt} = \frac{\partial}{\partial t}\int_{CV}\rho bd V +\int_{CS}\rho b \vec{V}\cdot \vec{n}~A,$$

    where B_sys is the extensive property B of a system, CV repesents the control volume, and CS represents the control volume surface. The value b is B/m, an intensive property.

    If the fluid in question is not static, then we can consider the system to be moving. Let the control volume be static. My book says that the RTT is way of relating the time rate of change of an arbitrary extensive parameter, B, of the system (the left hand side) to the change of the property within the control volume and the amount of the property, B, that gets carried out/in across the surface of the control volume.

    Here is my difficulty:
    We have stated that the system is moving, because the flow is not static. So, at some point, the "system" of fluid that we initially considered will have completely left the boundaries of the control volume. How can the equation still make sense at that point? The change of the system is no longer related to the change of the control volume, because it is nowhere near the control volume.
     
  2. jcsd
  3. Mar 4, 2017 #2

    Andy Resnick

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    I'm not entirely sure I understand your difficulty, but there would be flux in and out of the control volume (the second term of your equation). That said, let's say your system is a fixed control volume, through which a bolus of some solute passes through- surely you would be comfortable with D/Dt = 0 after that bolus has completely passed through the control volume.

    But maybe I didn't understand your question....
     
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