Monday, June 28, 2021

Continuity Equation

We had introduced the Euler flow equations, and saw it gave us (in $n$ dimensions) $n$ equations in $n+2$ unknowns. This requires introducing another 2 equations for us to solve the equations of motion. Today we will do half the required job, we will introduce one additional equation: the conservation of mass, or continuity equation.

Euler's Derivation

A relatively clever derivation begins by examining a fluid parcel of "small volume" $V$ and density $\rho$, so small such that $\int_{V}\rho\,\D^{n}x\approx\rho V$. Since the particles in the fluid parcel remain in the parcel over time, we have its total mass be constant: \begin{equation} \frac{\mathrm{D}(\rho V)}{\mathrm{D} t} = 0. \end{equation} Since this fluid parcel is nonempty, $\rho V\gt0$, so taking the logarithm of the mass and then taking its time derivative gives us: \begin{equation} \frac{1}{\rho}\frac{\mathrm{D}\rho}{\mathrm{D} t} + \frac{1}{V}\frac{\mathrm{D} V}{\mathrm{D} t} = 0. \end{equation} Supposing we have the fluid parcel be described as a rectangle with edges $\delta x$, $\delta y$, $\delta z$, then we see \begin{equation} V = \delta x\,\delta y\,\delta z \end{equation} hence Taylor expanding the volume as a function of time gives us \begin{equation} V + \frac{\mathrm{D} V}{\mathrm{D} t}\delta t = \left[1 + \left(\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} + \frac{\partial w}{\partial z}\right)\delta t\right]\delta x\,\delta y\,\delta z. \end{equation} Collecting these together give us the equation \begin{equation} \frac{\mathrm{D}\rho}{\mathrm{D} t} + \rho\left(\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} + \frac{\partial w}{\partial z}\right) = 0. \end{equation} This is the continuity equation.

Puzzle 2. This derivation seems to use the Lagrangian description; in general, however, we would need to include a factor of the Jacobian to describe the volume of the fluid parcel. Work out this derivation. Cheaters may consult Childress.

Modern Textbook Derivation

Consider now a fixed cube in the fluid, whose side lengths are again $\delta x$, $\delta y$, $\delta z$. What amount of mass flows into the cube across the $\delta y\delta z$ face?

We find, assuming the flow moves away from the origin, \begin{equation} \begin{pmatrix}\mbox{mass which}\\ \mbox{flows in}\end{pmatrix} = \left(\rho - \frac{1}{2}\frac{\partial(\rho u)}{\partial x}\delta x\right)\delta y\,\delta z. \end{equation} Similarly, the amount flowing out \begin{equation} \begin{pmatrix}\mbox{mass which}\\ \mbox{flows out}\end{pmatrix} = \left(\rho + \frac{1}{2}\frac{\partial(\rho u)}{\partial x}\delta x\right)\delta y\,\delta z. \end{equation} We then have the net mass gained be $-\delta x\,\delta y\,\delta z\partial_{x}(\rho u)$. Doing this for the other faces, we get \begin{equation} \begin{pmatrix}\mbox{net mass}\\ \mbox{flows}\end{pmatrix} = \nabla\cdot(\rho\vec{u})\,\delta x\,\delta y\,\delta z. \end{equation} On the other hand, the mass flux varies over time as \begin{equation} \begin{pmatrix}\mbox{mass}\\ \mbox{flux}\end{pmatrix} = \frac{\partial}{\partial t}(\rho\,\delta x\,\delta y\,\delta z). \end{equation} Setting equals to equals, and simplifying, we find \begin{equation} \frac{\partial\rho}{\partial t} + \nabla\cdot(\rho\vec{u})=0 \end{equation} which is another, equivalent, form of the continuity equation.

Exercise 3. Prove Eq (10) is equivalent to Eq (5).

References

I have profited most from Lamb's presentation, though Childress takes greater care when working through the derivation in the Lagrangian description.

  1. Leonhard Euler, "Principes généraux du mouvement des fluides". Mémoires de l'académie des sciences de Berlin 11 (1757) 274–315; English translation arXiv:0802.2383
  2. Horace Lamb, Hydrodynamics. Dover, sixth ed., 1932; §7 (archive)
  3. Stephen Childress, An Introduction to Theoretical Fluid Mechanics. AMS Press, 2009.

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