Assumptions of Reynolds transport theorem

Question

I have a question about Reynolds transport theorem https://en.wikipedia.org/wiki/Reynolds_transport_theorem

$$\frac{d}{dt}\int_{\Omega(t)}fdV = \int_{\Omega(t)} \frac{\partial f}{\partial t} dV + \int_{\partial\Omega(t)}(v^b . n)fdA $$

where $f = f(x,t)$ , $\Omega(t)$ is time-dependent region that has boundary $ \partial \Omega(t)$ , $ v^b = v^b(x,t)$ is the velocity of the area element.

In particular, I wonder what is the main assumption that makes this equality true? is it the time dependence of the control volume $\Omega(t)$? but I've seen cases where $\Omega(t)$ is replaced by $\Omega$, is this be possible? If yes, under what assumptions on $f$ or $v^b$?

Also if anyone could explain the intuition behind the theorem in brief ... like what does each integral on the r.h.s represent? below is what was written on the Wikipedia page but I'm not sure I understand the issue and how the theorem addresses it

If we wish to move the derivative within the integral, there are two issues: the time dependence of f, and the introduction of and removal of space from Ω due to its dynamic boundary

Thank you :)

Answer 1

The Reynolds theorem is used to describe the basic conservation laws of continuum quantities, i.e., that was something changed inside the focused volume plus/minus what goes in/out of the surface is equal to the total change.

In a simple explanation, I would use the $f(\vec{x},t)$ for the physical quantity function and $\delta V$ as the infinitesimal volume element. The product rule (or Leibniz rule) can well be used to express this: \begin{align} D\left[f(\vec{x},t)\delta V\right]=\delta V{\partial}f+f{\partial}(\delta V)\;\; (1) \end{align} The first differential term on the right is further divided by $\delta t$ and integral over the volume \begin{align} \int_V\delta V\frac{\partial f}{\partial t}=\int_V\frac{\partial f}{\partial t}dV \end{align} and the last term in (1) is the differentiation of the volume over time which is nothing else than the volume change with the normal velocity $U_n$ times the surface element $dS$, therefore, it reads \begin{align} \int_{dV}f\frac{\partial \delta V}{\partial t}=\int_SfU_ndS \end{align} The normal velocity $U_n$ has the positive outward of the volume. The volume considered can be, for instance, a cuboid which has one side moving with the velocity $v$ in the direction normal to its surface say area $a$, like a double-acting piston bellows. The time-rate of volume change of the cuboid is $dV/dt=av=aU_n$. Likely, the same idea can be used for 2D and 1D as well.

The full version can be readily obtained \begin{align} \frac{D}{Dt}\int_VfdV=\int_V\frac{\partial f}{\partial t}dV+\int_SfU_ndS \end{align} Note the surface integral in the last term states the change due to purely the control surface's movements not the quantity's.