Section 3.2 Conservation of mass

Our task in this section is to prove the following equation for the conservation of mass of a fluid:

Theorem 3.2.1 Continuity equation.

The differential form of the law of conservation of mass, otherwise known as the continuity equation, is

\begin{equation} \pd{\rho}{t} + \nabla \cdot (\rho \bu) = 0,\tag{3.2.1} \end{equation}

where $\rho = \rho(\bx,t)$ is the density of the fluid and $\bu = \bu(\bx, t)$ is the velocity of the fluid.

The above form is equivalent, on expanding $\nabla \cdot (\rho \bu)$ and using the definition of the material derivative, to

\begin{equation} \DD{\rho}{t} + \rho(\nabla \cdot \bu) = 0.\tag{3.2.2} \end{equation}

In fact, as it turns out, the proof of this result is trivial if we use Reynolds’ Transport Theorem and Lagrangian formulation following Theorem 3.1.3.

Proof.

Consider the mass contained in an arbitrary material volume $V(t)\text{,}$

\begin{equation*} \iiint_{V(t)} \rho \, \de{V}. \end{equation*}

Since $V(t)$ always consists of the same fluid particles, conservation of mass requires that this mass does not change in time, i.e. its time derivative is zero.

By Reynolds’ Transport Theorem (Theorem 3.1.3), we can pass the time derivative through the integral. This gives

\begin{equation*} 0 = \dd{}{t}\iiint_{V(t)} \rho \, \de{V} = \iiint_{V(t)} \left[ \pd{\rho}{t} + \nabla \cdot (\rho \bu)\right] \, \de{V}. \end{equation*}

Since this holds for every material volume $V(t) \subseteq D\text{,}$ the bump lemma (Lemma 3.2.2) implies that the integrand must be zero. This gives (3.2.1).

The proof of (3.2.2) follows immediately from application of the definition of the material derivative (2.1.1).

Within the above proof, we use an idea used throughout this chapter, which is that if an integral of a quantity (the integrand) is zero for all possible domains of integration, then the integrand, itself, is zero. This is sometimes referred to as the "Bump lemma".

Lemma 3.2.2 Bump lemma.

Let $f(\bx)$ be a sufficiently smooth function defined on $\Omega \subseteq \mathbb{R}^n\text{.}$ Suppose it is the case that

\begin{equation*} \int_V f(\bx) \, \de{V} = 0, \end{equation*}

for all $V \subseteq \Omega\text{.}$ Then

\begin{equation*} f(\bx) \equiv 0 \end{equation*}

in $\Omega\text{.}$

In Exercise 3.6.1, you will prove the above lemma.

Subsection 3.2.1 Derivation of mass conservation using Eulerian methods

The derivation we have just shown for Theorem 3.2.1, using the Lagrangian viewpoint and Reynolds’ Transport Theorem, is deceptively simple, and it is instructive to see how the result is proved purely from the perspective of Eulerian coordinates.

For this, let us consider $V$ to be a fixed and closed subregion of the overall fluid, $D\text{,}$ that does not change with time. An illustration of this is shown in Figure 3.3.

Fluid volume with outward normal and velocity on a boundary patch — Figure 3.3. The fixed fluid volume, $V$ (blue), with boundary $\partial V\text{.}$ On a small patch of the boundary, the velocity $\bu$ has normal component $\bu \cdot \bn\text{,}$ where $\bn$ is the outward unit normal; this gives the outward flux.

We want to prove the following result, which essentially equates the change in mass, due to density changes, to the flow of mass in or out of the volume.

Theorem 3.2.3 Integral form of the law of conservation of mass.

Given a fixed volume $V$ with boundary $\partial V\text{,}$ the integral form of the law of mass conservation is

\begin{equation} \dd{}{t} \iiint_V \rho \, \de{V} = -\iint_{\partial V} \rho \bu \cdot \bn \, \de{S}.\tag{3.2.3} \end{equation}

Proof.

The rate of change of mass in $V$ is

\begin{equation*} \dd{}{t} \iiint_V \rho(\bx, t) \, \de V = \iiint_V \pd{\rho}{t} \, \de V, \end{equation*}

and note the derivative passes through the integral since the volume, $V\text{,}$ does not change with time.

Let the boundary of $V$ be given by $\partial V\text{,}$ and let $\bn$ denote the outward unit normal defined along the boundary $\partial V\text{.}$ At each point on the boundary, the volume flow rate per unit area (known as the flux) across the boundary is given by $\bu \cdot \bn\text{,}$ and therefore the mass flow rate per unit area is $\rho \bu \cdot \bn\text{.}$

We now sum the total mass flow across the entire boundary. This is given by the surface integral

\begin{equation*} \text{rate of mass flow outward across $\partial V$} = \iint_{\partial V} \rho \bu \cdot \bn \, \de{S}. \end{equation*}

The flux out of the boundary is also sketched in Figure 3.3.

Finally, we apply mass conservation: the rate of change of mass in $V$ must equal the rate at which mass flows inward across $\partial V\text{.}$ Since the surface integral above measures the outward flow, this gives the minus sign in (3.2.3).

We now want to transform the integral form in (3.2.3) into the form of a partial differential equation. To do this, apply the Divergence theorem to the right-hand side of the above integral, converting the surface integral to a volume integral. Moving all quantities to the left-hand side now yields

\begin{equation*} \iiint_V \left[ \pd{\rho}{t} + \nabla \cdot (\rho \bu)\right] \, \de{V} = 0. \end{equation*}

Since the above integral equation holds for all possible $V\text{,}$ it must be equivalent to the integrand equated to zero (Lemma 3.2.2). This recovers the continuity equation of Theorem 3.2.1.

Subsection 3.2.2 Corollary of the Transport Theorem

There is a useful corollary to Reynolds’ Transport Theorem (Theorem 3.1.3) in the case where the integrand is proportional to the density, i.e. $f = \rho h\text{.}$

Corollary 3.2.4.

Let $V(t)$ be a material volume, let $\rho$ satisfy the continuity equation (3.2.1), and let $h(\bx, t)$ be any continuously differentiable function. Then

\begin{equation} \dd{}{t} \iiint_{V(t)} \rho h \, \de{V} = \iiint_{V(t)} \rho \DD{h}{t} \, \de{V}.\tag{3.2.4} \end{equation}

Proof.

This proof just follows from expansion. Pass the derivative through the integral and use the formula of Reynolds’ Transport Theorem. Then expand the quantity:

\begin{equation*} \nabla \cdot (f\bu) = \nabla f \cdot \bu + f \nabla \cdot \bu, \end{equation*}

where $f = \rho h\text{.}$ Finally use the continuity equation on $\rho\text{.}$

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