# 6.2: Mass conservation

In the introduction to this chapter we listed three assumptions that our theory of flow will be based on. We now have the tools to convert those assumptions into Eulerian form.

Our first assumption is that mass is neither created nor destroyed, which appears to be true absent nuclear reactions. In that case, the time derivative of the mass of a fluid parcel, taken on the parcel’s trajectory, is

$\frac{D}{D t} \int_{V_{m}(t)} \rho(\vec{x}, t) d V=0,\label{eqn:1}$

where $$V_m$$ is a material volume. This statement is in Lagrangian form. It is physically clear but difficult to handle mathematically. Invoking (6.1.6), we can write this equation as a volume integral:

$\int_{V_{m}(t)}\left(\frac{\partial \rho}{\partial t}+\vec{\nabla} \cdot \rho \vec{u}\right) d V=0 . \quad \forall V_{m}\label{eqn:2}$

Now here is a critical point that we will encounter repeatedly: the material volume $$V_m$$ is chosen arbitrarily; we could just as easily choose some different volume and the integral would still be zero. So, in what case may an integral be zero over every possible domain of integration? This can happen only if the integrand itself is zero everywhere. Therefore, mass conservation requires that

$\frac{\partial \rho}{\partial t}+\vec{\nabla} \cdot \rho \vec{u}=0\label{eqn:3}$

at every point. This is commonly known as the continuity equation. Note that the product $$\rho\vec{u}$$ in Equation $$\ref{eqn:3}$$ is the mass flux. The equation therefore tells us that the density at each point increases or decreases depending on whether the mass flux converges or diverges. We have now succeeded in converting the Lagrangian statement of mass conservation Equation $$\ref{eqn:1}$$ into the Eulerian form Equation $$\ref{eqn:3}$$. In other words, we have a partial differential equation which can, in principle at least, be solved.

A second form of Equation $$\ref{eqn:3}$$ is equally useful. We use the product rule to split the second term into two,

$\frac{\partial \rho}{\partial t}+\vec{u} \cdot \vec{\nabla} \rho+\rho \vec{\nabla} \cdot \vec{u}=0,$

and note that the first two terms now form the material derivative of $$\rho$$. Thus,

$\frac{D \rho}{D t}+\rho \vec{\nabla} \cdot \vec{u}=0.\label{eqn:4}$

An important special case is that of an incompressible fluid, for which $$\vec{\nabla}\cdot\vec{u}=0$$. In that case, the mass of any material volume remains constant:

$\frac{D \rho}{D t}=0.\label{eqn:5}$

Density can still vary in space and time, but if you follow a fluid particle along its trajectory, its density will not change.