= Incompressible flow
{title2=$\nabla\cdot\mathbf u=0$}
{wiki}
= Incompressibility
{synonym}
= Incompressibility condition
{synonym}
An <incompressible flow> preserves the <volume> of each moving <fluid element>. Its local condition is zero <velocity divergence>, $\nabla\cdot\mathbf u=0$. Combined with <mass conservation>, it says that each parcel's <mass density> stays unchanged as it moves. This does not require every parcel to have the same <mass density>: an initially nonuniform distribution can be transported without compression. Nor does <incompressibility> require a constant <velocity> or constant <pressure>.
For positive <mass density>, the <continuity equation> makes the equivalence precise:
$$
\frac{\partial\rho}{\partial t}+\nabla\cdot(\rho\mathbf u)=0,
\qquad
\frac{D\rho}{Dt}+\rho\nabla\cdot\mathbf u=0,
\qquad
\frac{D}{Dt}=\frac{\partial}{\partial t}+\mathbf u\cdot\nabla.
$$
The <material derivative> follows a <fluid element>. Thus $D\rho/Dt=0$ if and only if the <velocity field> is a <divergence-free vector field>. The distinction between following a parcel and observing a fixed position prevents confusing <incompressibility> with an unchanging <density> field. https://uclnatsci.github.io/2021/Electromagnetism-Fluids-and-Waves/fluids/conservation.html[UCL, Conservation law derivations, sections 30.1–30.2]{ref}
For a rigid pipe with impermeable walls, no leaks and <steady flow>, <mass conservation> at constant <mass density> implies a constant <volumetric flow rate>:
$$
Q=\int_A\mathbf u\cdot\mathbf n\,dA=A\overline U,
\qquad A_1\overline U_1=A_2\overline U_2.
$$
Here $\mathbf n$ points downstream, $A$ is the section's <area> and $\overline U$ is the mean normal <velocity>, not necessarily the centreline speed. No assumption of <inviscid flow> or a uniform <velocity profile> is needed. If unequal amounts entered and left a fixed segment, its <mass> would accumulate, contradicting <steady flow>.
For a worked example, take horizontal water flow with stipulated <mass density> $1000\,\mathrm{kg\,m^{-3}}$, inlet radius $R_1=0.010\,\mathrm m$, outlet radius $R_2=0.005\,\mathrm m$ and inlet mean <velocity> $\overline U_1=0.20\,\mathrm{m\,s^{-1}}$. The circular <areas> give
$$
Q=\pi R_1^2\overline U_1=2\pi\times10^{-5}\,\mathrm{m^3\,s^{-1}},
\qquad
\overline U_2=\frac{Q}{\pi R_2^2}=0.80\,\mathrm{m\,s^{-1}}.
$$
The radius halves, the <area> falls to one quarter, and the mean <velocity> quadruples. The <mass flow rate> is $\rho Q=0.02\pi\,\mathrm{kg\,s^{-1}}$. The same amount of water passes both sections each second; narrowing changes its speed rather than compressing it. The <Reynolds number> and <Bernoulli's principle> discussions use this same example to examine viscous effects and the additional assumptions needed to calculate <pressure>.
\Image[/fluid-flow-constriction.png]
{title=Continuity fixes the mean speeds; Bernoulli's principle gives a conditional ideal pressure drop}
{description=The pipe outline shows the radius ratio. Arrows represent section means, not a computed velocity field. The bars show the constant $Q$ and the ideal exchange between pressure and kinetic energy density at equal height; $p_2$ is the arbitrary pressure reference.}
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