Currently I am studying the reynolds transport theorem. From my research I could find two depictions of the theorem. My problem lies withing tying these two formulations together.
- "Mathematical/Wikipedia" formulation(https://en.wikipedia.org/wiki/Reynolds_transport_theorem):
$\frac{d}{dt}\int_{\Omega(t)} \mathbf{f}\,dV = \int_{\Omega(t)} \frac{\partial \mathbf{f}}{\partial t}\,dV + \int_{\partial \Omega(t)} \left(\mathbf{v}_b\cdot\mathbf{n}\right)\mathbf{f}\,dA$
- "Textbook" formulation:
$\frac{D}{Dt}B_{sys} = \frac{d}{dt}\int_{CV} \beta\rho \,dV + \int_{CS} \beta\rho \left(\mathbf{v}\cdot\mathbf{n}\right)\,dA$
CV = control volume, CS = control surface
For the first term on the right hand side I just can't understand how the time derivative gets moved in front of the integral for the "textbook" formulation.
Are these formulations even equivalent?
A counter example: Let's assume we want to describe the mass in our system. That means for the "Wikipedia" formulation $\mathbf{f} = \rho$ (where $\rho$ is the density) and for the "Textbook" formulation $\beta = 1$. Lets further assume that we have a incrompressible fluid. This means that the density $\rho$ is a constant (-> time invariant). In this case the first term on the right hand side of each formulation would evaluate to different results. "Wikipedia" formulation: $\frac{\partial \rho}{\partial t} = 0$ so $\int_{\Omega(t)} 0\,dV = 0$. However the "Textbook" formulation would evaluate to $\frac{d}{dt}\int_{CV} 1\rho \,dV = \frac{d}{dt} m(t)$ which is unequal to zero.
Are these two formulations really describing the reynolds transport theorem? Where are the differences in what they describe?