= Axisymmetric pure plume
Consider a steady, axisymmetric <pure plume> in a uniform ambient fluid, using the <Boussinesq approximation> and uniform top-hat cross-sectional profiles. If $r(z)$ is its <radius>, $U(z)>0$ its vertical <velocity> and $g'(z)>0$ its <reduced gravity>, define the physical fluxes
$$
Q=\pi r^2U,\qquad M=\pi r^2U^2,\qquad B=\pi r^2Ug'.
$$
They are respectively <volumetric flow rate>, <momentum flux> per reference <mass density>, and <buoyancy flux>. \b[Only $B$ is conserved] in this unstratified, source-free model. <Fluid entrainment> adds ambient volume to $Q$, and <buoyancy> increases $M$.
For constant entrainment coefficient $\alpha>0$, the <Batchelor entrainment hypothesis> gives an inward boundary speed $\alpha U$. Volume, vertical momentum and buoyancy balances therefore give
$$
\frac{dQ}{dz}=2\pi r\alpha U=2\alpha\sqrt{\pi M},\qquad
\frac{dM}{dz}=\pi r^2g'=\frac{BQ}{M},\qquad
\frac{dB}{dz}=0.
$$
To find a pure <similarity solution>, put $Q\propto z^p$ and $M\propto z^q$ relative to the <plume virtual origin>. The first balance gives $p-1=q/2$; the second gives $q-1=p-q$. Thus $p=5/3$, $q=4/3$, and
$$
Q\propto B^{1/3}z^{5/3},\qquad M\propto B^{2/3}z^{4/3}.
$$
In particular $r=Q/\sqrt{\pi M}$ grows linearly while $U=M/Q$ decreases as $z^{-1/3}$. A finite source shifts the virtual-origin coordinate. Dividing all three fluxes by $\pi$ changes their balance coefficients, so pi-normalized and physical-flux formulas must not be mixed.
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