The basic form of Reynold's Transport Theorem can be written as: $${DB_{sys}\over Dt}={\partial B_{CV} \over \partial t}−\dot B_{in}+\dot B_{out}$$

Now my question is, shouldn't ${\partial B_{CV} \over \partial t}=0$, since there is no way that a property inside the control volume will just randomly disappear or appear?

For example when we deal with the classic three inlet-one outlet problem, we always equate ${\partial B_{CV} \over \partial t}=0$, when we try to find the velocity of fluid at the outlet.

If this is the case, then why do we include the ${\partial B_{CV} \over \partial t}$ term at all?


1 Answer 1


$\partial B_{CV}/\partial t$ is a generation term. For a classic fluid flow problem such as you describe it's equal to zero, but there are many scenarios where it might not be.

Consider a model of the level of CO$_2$ in a room. The ventilation would give a flux in and out of the room, but the occupants are generating CO$_2$ by breathing, so the generation term is positive.

A control volume in which components of the flow are reacting, such as combustion, would have a generation term, either positive or negative, in the equations for energy, oxygen, fuel and combustion products.

  • $\begingroup$ Thank you for the explanation! It cleared the concept for me! $\endgroup$ Commented Jun 7, 2021 at 18:30
  • $\begingroup$ Excellent, glad to be able to help. If you're happy with my answer, could you mark it as accepted please? $\endgroup$
    – Nick
    Commented Jun 7, 2021 at 19:25
  • $\begingroup$ For people that would still be confused, see here for a detailed explanation $\endgroup$
    – rambi
    Commented Jun 25, 2023 at 12:31

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