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If a liquid is freezing, is equation of continuity violated? As the liquid flows, some portion of it is getting frozen. The mass of the fluid thus keeps dropping. Similarly, when a molten fluid flows over a solid, its mass may increase because the solid over which it flows also melts. In either of these flows, can we assume that div u = 0, where u is the velocity vector.

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1 Answer 1

The continuity equation is not violated in either of the situations you describe above. The generic continuity equation for some scalar quantity (such as density) can be written as

$\frac{\partial \psi}{\partial t} + \nabla . \psi\mathbf{u} = \sigma$, (1)

where here $\psi$ is the some conseverd scalar quantity, $\mathbf{u}$ is the velocity field and $\sigma$ is the is the generation of $\psi$ per unit volume per unit time (or source term). This source term is introduced when you are adding or taking away from the substance under consideration (or indeed changing its state).

For a ordinary incompressible flow (with the fluid not changing state), the continuity equation can be written as

$\frac{\partial \rho}{\partial t} + \nabla . \rho\mathbf{u} = 0$. (2)

Taking the density of the fluid to be constant we find (as you point out)

$\nabla . \mathbf{u} = 0$. (3)

However, for the situation describe above, it is not this simplified version that will model the flow correctly as the density is not constant. As long as you model the substance with correct continuity equation, nothing will be violated. In fact, for the cases you speak of, you can still treat the molten metal, or melting ice and as the same continuous substance, but instead of the simplified version of the continuity equation (the divergence free condition you have stated (3)), would need to use equation (2). To model the flow only, where you would treat the solidification as a change in volume/mass you would also need to include the source term (equation (1)).

I hope this helps.

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Thanks, your explanation was indeed helpful. I was actually looking at a flow of a liquid under gravity, for which $\vec{u}=w\vec{e}_z$. If eqn(3) was valid then w would be independent of z and Navier-Stokes equation will be $0=-\nabla{p}+\eta\Delta\vec{u}+\rho\vec{g}$. However, in case of a melting flow, I would have to use the general form of continuity equation (1) and I will no longer conclude that $w$ is a function of $x$ and $y$ alone. Am I thinking correctly? –  Amey Joshi Jul 21 '12 at 11:46
    
$w$ here is a scalar. This is giving magnitude to the unit vector $\mathbf{e}_{z}$. The equation $\mathrm{div}\;\mathbf{u} = 0$ is just saying that the net flux though the surface of the relevent volume is zero. As you have written the defintition of $\mathbf{u}$ above, it has no dependency on x or y coords. If $w = w(x, y)$ so $\mathbf{u} = w(x, y)\mathbf{e}_{z}$ changing the type of flow considered will not nessecaraly change your definition of $w$. Also note, that the Navier-Stokes eqn. is a conservation of momentum eqn., the continuity eqn. for density above are conservation of mass... –  Killercam Jul 21 '12 at 12:19

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