# How can mass diffuse out of a lagrangian control volume when it always tracks the same mass?

I am trying to understand how mass conservation in fluid flow works and how it relates to lagrangian and eulerian control volumes.

The way I understand eulerian control volumes is that they are stationary and feel the change in time at that constant location and lagrangian control volumes follow a parcel of fluid that contains the same particles. I also understand the basic idea behind the material derivative:

$$\frac{Df}{Dt} = \nabla f \cdot \vec{v} + \frac{\partial f} {\partial t}$$

aka the total change of the quantity in time is the one experienced due to motion through space with a certain velocity plus the change at that location.

What I don't get is how this ties exactly with with conservation and the lagrangian and eulerian framework. Is the material derivative refering to a lagrangian or eulerian control volume or is it relating them to each other?

In my class notes it says that mass is in general not conserved in an eulerian control volume but it always is in a Lagrangian control volume (since it always contains the same particles) and that this means that conservation implies $$\frac{Df}{Dt} = 0$$

Does this statement about the material derivative being equal to zero refer now to an eulerian control volume? How can it refer to a Lagrangian control volume, when it implies mass exchange and hence different particles?

I am further confused, because later on in these notes it is implied that there is mass exchange in lagrangian control volumes. The example of advection diffusion systems with the governing equation $$\frac{\partial f} {\partial t} = \nabla \cdot (D \nabla f) - \nabla \cdot (f \vec{v})$$

is studied and a particle methods algorithm to simulate them is explained. They state that the computational particles in the method correspond to Lagrangian control volumes and store extensive quantities. It is then stated that from the perspective of the particle $$\frac{Df}{Dt} = \nabla \cdot (D \nabla f)$$ $$\frac{dx}{dt} = \vec{v}$$

Does this mean that the change in mass a particle experiences is only due to diffusion, since it flows through the field with velocity $$\vec{v}$$ and does not experience any advective flux? If so, is there any diffusion across the boundary of the Lagrangian control volume? Doesn't that mean not the same mass particles are tracked?

This post is super confused and long since I myself am quite confused. Assume you are explaining to someone without foundational knowledge.

Consider binary diffusion of species A and B, in the Lagrangian formalism you would track an N number of particles but a fraction will be A and the rest will be B. If there are gradients in the concentration of A between lagrangian points then A will diffuse from one point to another. Simultaneously, if there are gradients in A then there are equal but opposite gradients in B, so B will diffuse into the langrangian point, effectively replacing the particles of A which have diffused out; this is the definition of binary diffusion.

In the end, A and B will diffuse but the total number N of particles is constant and conserved. So you in essence have two equations of conservation for A and B or for the total number of particles and either A and B (since they are linearly dependent)

• This is how I understand conservation as well: the total amount of mass remains constant. I am confused because this for me is not equivalent to always tracking the same particles. Why do we say we track the same particles, when what we mean is we track the same sized parcel of mass the whole time? Or is this wrong, can you please clarify? Commented Jul 13, 2021 at 12:08

What we track in the Lagrangian view of fluids is the fluid parcel (Lagrangian control volume). In the fluid dynamics, we assume that the fluid is continuous thus there is a shape boundary, i.e., we can define a control volume and we then keep an eye on it.

The new particles coming in or out will change the density of the fuild parcel also may change the size of the control volume. But it does not matter, because we do not treat a collection of particles as the studying object but the control volume formed by the continuous fluids.

There is a subtle difference between the original Lagrangian view. In the original Lagrangian view, we keep an eye on one particle, but here we keep an eye on a control volume thanks to the assumption of the continuous fluids. And actually, if you treat a collection of the water/air particles as the studying object, you will be totally lost because of the mess and discreterization of these particles.

Further, I would like to say, the continuous assumption is good enough but may fail (leading to high error) in extreme conditions such as the very high atmospheric air with low density.

Appendix: In my opinion, the most clear description about this part is the first 11 pages of the book "A Mathematical Introduction to Fluid Mechanics", see https://link.springer.com/book/10.1007/978-1-4612-0883-9. Note that there is a typo at the bottom equation on page 10. The $$\nabla\cdot u$$ should be $$\frac{\partial u}{\partial \xi} + \frac{\partial u}{\partial \eta} + \frac{\partial u}{\partial \zeta}$$. It can explain why we use Lagrangian control volume and how we apply Newton's laws on the control volume and how we finnaly get the right form of the equations by transport theorem which is also proved concisely.