Continuity Equation

Figure 2.1: A volume enclosed by a surface $S$. The normal to the surface is ${\bf dS}$.

If the mass is neither created or destroyed in the volume, then the mass inside a fixed surface, $S$, bounding the closed volume $V$ will increase if mass (or density) flows into the volume and decrease if it flows out. This can be expressed mathematically as

• rate of increase of mass inside the volume + rate at which mass is flowing out across the surface = zero.

$${d\over dt}\left \{ \int_{V}\rho dV\right \} + \int_{S} \rho {\bf v} \cdot {\bf dS} = 0.$$ (2.3)

By the divergence theorem (1.18), this becomes

$$\int_{V}{\partial \rho\over \partial t} + \nabla \cdot (\rho {\bf v}) dV = 0.$$

This holds for an arbitrary volume and so must be true at each point in space. Thus,

$${\partial \rho \over \partial t} + \nabla \cdot (\rho {\bf v}) = 0.$$ (2.4)

This is the mass continuity equation.

In 1-D problems this reduces to

$${\partial \rho \over \partial t} + {\partial \over \partial x}(\rho v) = 0,$$ (2.5)

and we can obtain this equation another way by simply considering how a fluid behaves in a small region lying between the positions $x$ and $x + \delta x$. Consider a fluid with a spatially and temporally varying density, $\rho (x, t)$ and velocity $v(x,t)$. The mass within this small region, as shown in Figure 2.2,

Figure 2.2: Consider the mass lying inside the region between $x$ and $x + \delta x$.

The mass in the region at time $t$ is simply

$$\int_{x}^{x+\delta x} \rho (x,t) dx \approx \rho (x,t) \delta x.$$

Similarly, the mass in this region at the later time $t + \delta t$ is

$$\int_{x}^{x+\delta x} \rho (x,t+\delta t) dx \approx \rho (x,t + \delta t) \delta x.$$

So why has the mass changed? It changes because there is a flow across the ends of this region. Consider the mass flowing through the left hand edge in the time interval $\delta t$. This is given by

$$\rho (x,t) v(x,t) \delta t.$$

It is straightforward to check that the units do indeed give density times a length. At the other edge the mass flowing out of the system is

$$\rho (x + \delta x, t) v(x+\delta x, t)\delta t.$$

Thus the change in mass in the time $\delta t$ is

$$\rho (x, t+\delta t) \delta x - \rho (x, t) \delta x.$$

This change is due to the mass flowing out through the right hand edge - the mass flowing into the region through the left hand edge, namely

$$\rho (x + \delta x,t ) v(x + \delta x, t)\delta t - \rho (x,t)v(x,t) \delta t.$$

Equating this two expressions and dividing by $\delta x \delta t$ results in

$${\rho (x, t+\delta t) - \rho (x,t)\over \delta t} = - {(\rh... ... x, t)v(x + \delta x, t) - \rho (x,t) v(x,t) )\over \delta x}.$$

Taking the limit of $\delta x \rightarrow 0$ and $\delta t \rightarrow 0$ we obtain

$${\partial \rho \over \partial t} = - {\partial \over \partial x}(\rho v),$$

which gives (2.5).

For a steady state $\rho$ and ${\bf v}$ do not change in time and so

$${\partial \rho \over \partial t} = 0,$$

and

$$\nabla \cdot (\rho {\bf v}) = 0.$$

Hence, the mass flowing into the region is equal to the mass flowing out of the region.