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My lecture notes contain the idea below, but I'm finding it difficult to understand one aspect.

Consider a finite volume $V$, bounded by a surface $S$ and outward pointing normal $\hat{\bf{n}}$.

A fluid occupies the space of which $V$ is a subset.

The fluid has velocity $u(\boldsymbol{r},t)$ and denisty $\rho(\boldsymbol{r},t)$.

By the principle of conservation of mass, we have that $$ \frac{d}{dt}\int_V \rho(\boldsymbol{r},t) \, dV=-\int_S \rho\boldsymbol{u}\cdot\hat{\boldsymbol{n}}\, dS $$

I understand that the left-hand side represents the rate of change of mass in $V$, and the right-hand side represents the rate at which mass flows into $V$ through $S$.

I don't understand why these two quantities are equal though.

Surely the rate of change of mass (i.e., the LHS) should be equal to the rate in minus the rate out? Rather than just the rate out.

Perhaps I have a fundamental misunderstanding of something here?

MHW
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    On the part of the boundary where $\boldsymbol{u}\cdot\hat{\boldsymbol{n}} < 0$ we have mass flow in and where $\boldsymbol{u}\cdot\hat{\boldsymbol{n}} > 0$ we have mass flow out. Note that the outward pointing normal is used. – RRL Jul 25 '22 at 21:53
  • @RRL Thanks, I think that makes sense. So am I correct in thinking that the RHS represents the net mass flow? Moreover, why is the $-$ necessary on the RHS? – MHW Jul 25 '22 at 22:05
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    Yes -- it is the net mass flow rate. The minus sign is necessary because if the dot product of the velocity and the outward normal at the surface is positive at a point then the fluid is leaving the region in a neighborhood of that point and we want the contribution to net flow to be negative. – RRL Jul 25 '22 at 22:16

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