# Mass Continuity Equation for Fluid - Running Into a Problem

I'm running into a problem when trying to show the mass continuity equation for a fluid, which says

$$\frac{\partial \rho}{\partial t} + \left(\nabla \cdot \rho \textbf{u}\right) = 0$$

Where $\rho=\rho(x,y,z,t)$ is the density of the fluid and $\textbf{u} = \textbf{u}(x,y,z,t)$ is the velocity vector of the infinitesimal unit of volume [or mass].

I start by noting that $m=\iiint_V\rho dV$. Differentiating that (with chain rule) gives us

\begin{align*} \frac{\partial m}{\partial t} &= \frac{\partial}{\partial t}\iiint_V \rho dV\\ &= \iiint_V \frac{\partial \rho}{\partial t} dV\\ &= \iiint_V \left(\frac{\partial \rho}{\partial x}x'(t)+\frac{\partial \rho}{\partial y}y'(t)+\frac{\partial \rho}{\partial z}z'(t)+\rho_t\right)dV\\ &=\iiint_V(\nabla \rho \cdot \textbf{u} + \rho_t)dV\\ \end{align*}

And as you can see, that doesn't really match up with the originally stated continuity equation. I have $(\nabla \rho \cdot \textbf{u})$ instead of $(\nabla \cdot \rho \textbf{u})$. Where have I gone wrong? What needs to be corrected?

• You should have terms like $\frac{\partial (\rho x'(t))}{\partial x}$ in your second to last line. The result follows from that in vector form. Commented Nov 3, 2014 at 22:55
• Sorry I cannot write an explicit answer at the moment since I'm on my phone but you are setting up your mass balance incorrectly. The total rate of change inside the volume (your line 1) is set equal to the flux over the surface of the volume. And then an application of Gauss's theorem is used to recover the desired form. See: nptel.ac.in/courses/Webcourse-contents/IIT-KANPUR/… Commented Nov 3, 2014 at 23:04
• Ah ok @LuisCosta, I got it. Thanks! Commented Nov 3, 2014 at 23:15

What we want is $\displaystyle \frac{dm}{dt} = \iiint_V \frac{\partial \rho}{\partial t} dV$. But the integrand here is not equal to the expression in the next line as you've written it; that would be the case if were writing out the complete derivative$\displaystyle \frac{d\rho}{dt}$.

Instead, the instantaneous change in the mass moving into the volume is equal to flow into the closed volume $V$ over its boundary $\partial V$. If $\vec{u}$ is the velocity vector of the fluid, then

$$\frac{dm}{dt} = -\iint_{\partial V} \rho\vec{u}.d\vec{A}$$
There is a negative sign here because we want mass moving in have a positive effect, not a negative effect and the convention with a closed surface is the to have the normal vectors $d\vec{A}$ pointing out.

Applying now Stokes' theorem, this last integral equal to $$\frac{dm}{dt} = \iiint_V -\nabla(\rho\vec{u}) \ dV$$

Equating now the two expressions for $dm/dt$ we have the desired equation:

$$\frac{\partial \rho}{\partial t} + \nabla(\rho\vec{u}) = 0$$

I've never seen the derivation written down this way, but now that I have, I like it as much as the usual approach using infinitesimal cubes, of size $dx \ dy \ dz$, which in effect rederive Stokes' theorem for this case.