Fluids: Continuity and Conservation of Mass

  • #1
joshmccraney
Gold Member
1,964
104
Hi PF!

In fluids I've noticed many authors use the continuity equation with an integral form of conservation of volume (assume density is constant). Is this double counting? Example: let fluid velocity inside an idle bubble be ##\vec u = \nabla \phi##. Conservation of mass implies ##\nabla u = 0 \implies \nabla^2\phi = 0##. Let the surface be perturbed, so conservation of volume requires ##\int_S \nabla \phi \cdot \hat n \, dS = 0## where ##S## is the bubble surface and ##\hat n## is a unit normal to ##S##.

Why use both?
 

Answers and Replies

  • #2
vanhees71
Science Advisor
Insights Author
Gold Member
16,770
8,008
It's always easier to work with the local laws first and then translate them to the integral laws using the integral theorems by Gauss, Stokes, and Green.

In your case the point is the continuity equation for the mass density and current,
$$\partial_t \rho + \nabla \cdot \vec{j}=0.$$
In terms of the velocity field you have
$$\partial_t \rho + \nabla \cdot (rho \vec{u})=0.$$
I guess what you call "conservation of volume" is in fact incompressibility, i.e.,
$$\vec{\nabla} \cdot \vec{u}=0,$$
which implies that you can write the continuity equation as
$$\partial_t \rho + \vec{u} \cdot \vec{\nabla} \rho=0.$$
In the 2nd part of your posting, you however only use the incompressibility condition for the special case of an irrotational fluid. Using Stokes's theorem you get
$$0=\int_V \mathrm{d}^3 \vec{x} \vec{\nabla} \cdot \vec{u} = \int_{\partial V} \mathrm{d}^2 \vec{f} \cdot \vec{u}=0.$$
I don't see, where you need the continuity equation for mass to get this result or where you need the assumption of a potential flow.
 
  • Like
Likes Orodruin and BvU
  • #3
joshmccraney
Gold Member
1,964
104
I'm sorry, but I'm not following you. I understand how to derive the differential equation ##\nabla \cdot \vec u = 0## assuming ##\rho = const##, and I also understnad the case where ##rho## is not constant. I also understand applying divergence theorem. Since ##\nabla \cdot \vec u## is conservation of mass (or conservation of volume since density is constant), this gives the same information as ##\int_S \vec u \cdot \hat n \, dS = 0##: both conserve volume. Why use both?
 
  • #4
vanhees71
Science Advisor
Insights Author
Gold Member
16,770
8,008
You mix up different things. The conservation of mass is expressed by the continuity equation
$$\partial_t \rho + \vec{\nabla} \cdot (\rho \vec{u})=0.$$
This holds for any (non-relativistic) fluid.

Then there is the special case of incompressible fluids, which is a kinematical constraint, saying that the volume of a material fluid element is conserved. It is expressed by
$$\vec{\nabla} \cdot \vec{u}=0.$$
For the integral you quote you only need this incompressibility constraint. It is independent of the mass-conservation continuity equation. It is an additional constraint, approximately valid for liquids (hydrodynamics) but not for gases (aerodynamics).

Of course, if mass is conserved and for incompressible fluids the volume of any material fluid element, then the density must be conserved as well. Mathmatically this follows from
$$\vec{\nabla} \cdot (\rho \vec{u})=\vec{u} \cdot \vec{\nabla} \rho + \rho \vec{\nabla} \cdot \vec{u} =\stackrel{\text{incompress}}{=} \vec{u} \cdot \vec{\nabla} \rho,$$
i.e., the continuity equation becomes for incompressible fluids [edit: corrected in response to @joshmccraney 's remark in #5]
$$\partial_t \vec{\rho} + \vec{u} \cdot \vec{\nabla} \rho=0,$$
but on the left-hand side this is nothing else than the material time derivative ##\mathrm{D}_t \rho##, and thus this equation precisely proves formally what's obvious from mass conservation plus incompressibility: the density of a material fluid element doesn't change.
 
Last edited:
  • #5
joshmccraney
Gold Member
1,964
104
If ##\rho## is constant and volume does not change, then mass conservation is
$$\partial_t \int_V \rho dV + \int_{\partial V} \rho \vec u \cdot \hat n dS = 0\implies\\
\partial_t \int_V dV+ \int_{\partial V} \vec u \cdot \hat n dS= 0\implies\\
\int_{\partial V} \vec u \cdot \hat n dS= 0\implies\\
\int_V \nabla \cdot \hat u dV = 0$$
Since this is true for arbitrary ##V##, then it must be that ##\nabla \cdot \hat u = 0##. How is it that the surface flux being zero (3rd equation) and this expression are different if one is derived from the other?

I know they are different because I believe you and the authors I've read, but I'm confused, as noted above.

Also, I think the last equation you wrote has a few typos, but I understand what you meant.
 
  • #6
vanhees71
Science Advisor
Insights Author
Gold Member
16,770
8,008
Indeed, I've corrected the typo. Thanks for pointing it out. Of course, your calculation is correct. One should only mention that ##V=V(t)## is a "material volume", i.e., the volume consisting of a fixed set of particles.
 

Related Threads on Fluids: Continuity and Conservation of Mass

Replies
1
Views
2K
  • Last Post
Replies
6
Views
592
  • Last Post
Replies
6
Views
3K
Replies
1
Views
2K
Replies
9
Views
827
Replies
4
Views
721
Replies
2
Views
6K
Replies
3
Views
761
  • Last Post
Replies
6
Views
6K
  • Last Post
Replies
3
Views
4K
Top