The continuity equation in fluid mechanics states that

$$

\frac{\partial\rho}{\partial t} + \nabla\cdot(ρ\mathbf u)=0

$$

Can you explain to me what is the physical meaning of each term of the equation and what are the direct outcomes?

Keep in mind that i do not have much knowledge about differential calculus.

Thank you!

EDIT:

After receiving some answers(for those who want to find out the answer,check the comments),i have only one question left.If the density gets bigger and bigger as time passes,then the first term of the equation is positive.What happens with the second term?What changes in that control volume?I mean,the density in the CV gets bigger,so what happens to velocity(intuitively)?I think that the velocity must get bigger in order to the fluid to get out of the CV.But that means that the second term is also positive and the equation is not right.So what is wrong with my thinking?

## Best Answer

Consider a control volume $\Sigma(t)$ of a fluid with density $\rho(\mathbf x,t)$. The mass inside $\Sigma(t)$ is clearly given by $$M(t):=\int_{\Sigma(t)}\rho(\mathbf x,t)\text d^3\mathbf x.$$ The way $\Sigma(t)$ is defined is that its mass content doesn't change with time, that is, a control volume is representing the time evolution of a certain amount of mass. Therefore the time derivative of the above integral should be zero. For a small time lapse $\epsilon$, the difference between $M(t)$ and $M(t+\epsilon)$ can be estimated by $$\int_{\Sigma(t)}\frac{\partial\rho}{\partial t}(\mathbf x,t)\epsilon + \int_{\partial\Sigma(t)}\epsilon\rho(\mathbf x,t)\mathbf v\cdot\hat{\mathbf n}\text dS,$$ where the first term is the integral of the difference of density at different times on the common domain, i.e. intersection, between $\Sigma(t+\epsilon)$ and $\Sigma(t)$, while the second term counts contributions from the bits that are in $\Sigma(t+\epsilon)$ and not in $\Sigma(t)$ and viceversa (the symmetric difference $\Sigma(t+\epsilon)\triangle\Sigma(t)$), whose volume is estimated by $\epsilon\mathbf v\cdot\hat{\mathbf n}\text dS$, $\hat{\mathbf n}$ being the normal to $\partial\Sigma(t)$.

Divide by $\epsilon$ and take the limit $\epsilon\to0$ together with Gauss theorem on the boundary term to conclude that $$\int_{\Sigma(t)}\left[\frac{\partial\rho}{\partial t}+\nabla\cdot(\rho\mathbf v)\right]\text dV=0$$ for any control volume $\Sigma(t)$, whence the continuity equation.

Physically, the integral form suggests that the continuity equation should be interpreted as relating the density at each point to the flow of matter through a

fixedsurface. In other terms, consider a fixed (not a control) volume $V$ of fluid. The motion of the fluid will bring some matter in and take some matter out of this volume, thus one can expect the density inside it, at a certain point $\mathbf x$ to change over time. The continuity equation is then saying that $$\frac{\partial\rho}{\partial t} = - \nabla\cdot(\rho\mathbf v)$$ since the RHS can be interpreted as the flow of the current $\rho\mathbf v$ across the boundary of $V$, we see that the variation of density in time inside $V$ is precisely related to the net flow of matter across the boundary of $V$ itself.ExampleIf the net flow is outward directed, the flow across the boundary will be positive (by definition of flow, since the velocity vector and the normal to the boundary form an acute angle) and as expected this lead to a drop in density, since the volume $V$ is losing matter.