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This is taken from Versteeg and Malalasekhara CFD book. δx, δy, and δz are given as the dimensions of the sides of the fluid element under consideration along the three axes respectively.

On the other hand, I've read the same mass flow rate derivation in Anderson's Fundamentals of Aerodynamics and understood it perfectly. enter image description here

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    $\begingroup$ Dont you just need $ \rho \delta x \delta y \delta z = \rho \delta V = \delta m$ for the derivation? $\endgroup$ – John H. K. Oct 13 '15 at 6:56

So $\rho\delta x\delta y \delta z$ is just the mass of a cube of size $\delta x\delta y \delta z$.

Then $\frac{\partial}{\partial t}\rho\delta x\delta y \delta z$ is the change in mass over time of this cube.

Expending out this equation we get: $$\frac{\partial}{\partial t}\rho\delta x\delta y \delta z = \frac{\partial\rho}{\partial t} \delta x\delta y \delta z + \rho \frac{\partial \delta x}{\partial t} \delta y \delta z + ...$$

You can then conclude that $\frac{\partial \delta x}{\partial t} = 0$ as your fluid element is fixed in size. Using the same argument for $y$ and $z$ you get the equation given.

Your next step will be to find the net mass flow through the fluid element, which is $-\nabla \rho \textbf{v} \delta x\delta y \delta z$

Then equating the two terms to get the conservation of mass.


The confusion grounds in a different frame of reference (see Wikipedia on Lagrangian and Eulerian description).

The first equation describes the rate of change of a fluid element.
One might think of it as "sitting on the fluid element".

The second equation describes the rate of change within a system.
Here one is an observer, looking at multiple fluid elements flying through the observed system.


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