# Interpretation of Reynolds Transport Theorem?

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:

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

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.

## 1 Answer

The system consists of the material within your control volume at time t. You get an instantaneous rate of change of the property of that system by analyzing the control volume through which it flows.

At a later time, if your control volume is fixed, you get get an instantaneous rate of change of the property of the new system that occupies the control volume at the later time.

• That makes sense. I was trying to think of a substantial change in time when the RTT refers to the instantaneous rate of change. – nwsteg Mar 5 '17 at 1:32