Exactly similar, and one number short

What a jet keeps, and what it collects is the collection’s account of the two things a free jet does: it conserves its momentum flux exactly and it gathers mass without limit. This essay is about the solution those two facts admit, which is exact, closed-form and remarkable — and about the constant in it that nothing determines.

The solution, and the three exact things about it #

Schlichting’s round jet, in boundary-layer form, is

$$u(x, r) = \frac{3}{8\pi}\frac{K}{\rho\nu}\frac{1}{x}\frac{1}{\left(1 + \eta^2/4\right)^2}, \qquad \eta = \frac{1}{4}\sqrt{\frac{3K}{\pi\rho}}\,\frac{r}{\nu x}.$$

It is an exact solution — Landau’s and Squire’s full version solves the Navier–Stokes equations without the boundary-layer approximation, and this is its slender limit. Three things about it are exact and none is quoted here without being checked.

The momentum flux is conserved. A free jet has no boundary anywhere, so nothing can push on it, so the integral of $\rho u^2$ across it is the same at every station. Computed at five distances spanning a factor of sixteen, it returns the value it started with to two parts in ten million — and the residual is the quadrature, not the solution.

The spreading is exactly linear and the decay exactly hyperbolic. The half-width grows in exact proportion to $x$ and the centreline speed falls in exact inverse proportion to it, so their product is constant — which is what conserving a momentum flux while keeping a fixed profile shape means. The half-width in similarity variables is $\eta_{1/2} = 2\sqrt{\sqrt{2}-1} = 1.2871885$, the root of $(1+\eta^2/4)^{-2} = \tfrac12$.

And the entrainment is $8\pi\mu x$ — with no $K$ in it at all. That is the strangest exact result in the essay. Three jets fired with momentum fluxes differing by a factor of ten thousand entrain identically: the same mass, at the same distance, to the last digit computed. How hard a jet was fired decides how fast it goes, and not at all how much fluid it collects.

The reason is a cancellation. A stronger jet is faster and narrower in exact compensation — the velocity goes as $K$ and the area over which it acts as $1/K$ — so the product that gives the mass flux is independent of it. That kind of cancellation is what a similarity solution is for, and it is also why it is worth checking rather than trusting: the closed form $8\pi\mu x$ and the numerical integral agree to nine parts in ten million.

How a similarity solution is found, and what it assumes #

The route to the solution is worth having, because it shows precisely where the free constant enters and why nothing in the method can fix it.

Start from the boundary-layer equations for an axisymmetric jet and ask whether there is a solution of the form $u = f(x)\,F(\eta)$ with $\eta = r/g(x)$ — that is, a solution whose profile shape is the same everywhere and only its amplitude and width change. Substituting that ansatz turns two partial differential equations into an ordinary one for $F$, provided $f$ and $g$ take particular powers of $x$; any other powers leave $x$ in the reduced equation and the ansatz fails.

Two conditions then fix those powers. Conservation of momentum flux requires $f^2 g^2$ to be constant. The balance between convection and diffusion in the reduced equation requires $f g^2/x$ to be constant. Together they give $f \propto 1/x$ and $g \propto x$, which is the hyperbolic decay and the linear spreading, before any equation has been solved.

Then the ordinary differential equation for $F$ is integrated — and it happens to integrate in closed form, which is the piece of luck that makes this jet famous. Nowhere in that chain is the viscosity determined. It appears only in the constant relating $\eta$ to $r$, which is to say in the scale, which is to say in exactly the place the turbulent model puts its fitted number.

Self-similarity is therefore an assumption about the existence of a shape, and the method that finds the shape cannot also find the scale. That is not a defect of this problem; it is what a similarity solution is, and the same structure appears wherever the collection meets one — most sharply in an exponent dimensions cannot give, where the missing number is an exponent rather than a coefficient and has to be found by solving an eigenvalue problem instead.

And the shape is not the theory #

Now the difficulty. A turbulent jet is routinely modelled by taking the same solution and replacing $\nu$ with a constant eddy viscosity $\nu_t$, fitted so that the predicted spreading rate matches the measured one. That substitution is the entirety of the model.

Scaled that way the two are identical to seven parts in $10^{16}$, which is round-off. They have to be: the scaling removes the only place the viscosity appears.

Unscaled they differ by a factor of forty in spreading rate and by a factor of forty in centreline speed at a given station. So everything the similarity solution fixes is shared and everything dimensional is not — and the shared half is the half a profile measurement tests.

A collapse of measured profiles onto the similarity shape therefore confirms self-similarity and nothing else. It is a real result: self-similarity is an assumption and could be false, and jets that are confined, buoyant or swirling do fail it. But it is the cheap half. The rate is the half that carries the physics, and the rate is the quantity the eddy viscosity was chosen to reproduce. There is no independent prediction in the turbulent version at all — which is what the ladder that never closes is about, seen in the simplest possible flow.

There is in fact a second constant. The solution is singular at $x = 0$ and a real jet has a nozzle of finite size, so the distance is measured from a virtual origin fitted to the data. Four choices of it, all describing the same far field, give centreline decays that differ by tens of per cent within ten diameters. So the model that is fitted to a spreading rate is fitted to two numbers, and a two-parameter fit to a decaying curve is not a demanding test.

What the constant is worth, numerically #

It is worth putting a number on how much the fitted constant is doing, because “one free parameter” sounds modest.

A round turbulent jet spreads at about 0.094 half-widths per unit distance, and a laminar one at $1.29\,\nu/\sqrt{K/\rho}$ — a ratio that at any ordinary Reynolds number is enormous. Matching the observed spreading rate therefore requires an eddy viscosity of order $0.015\,u_c\,r_{1/2}$, which for a laboratory jet is a few hundred to a few thousand times the molecular value.

So the fitted constant is not a correction to the physics; it is the physics, in the sense that everything dimensional about the predicted jet is proportional to it. The molecular viscosity has dropped out of the answer entirely — which is a proper statement about turbulence and is the same statement the limit that is not the value makes about dissipation — but what has replaced it is not derived from anything.

The one real content in the model is that a single constant suffices: the same $\nu_t$, uniform across the jet and unchanged with distance, reproduces the whole profile at every station. That is a non-trivial claim and it is approximately true, and it is the reason the model is used. It is also the reason it fails where it fails, since a uniform eddy viscosity is exactly wrong at the jet’s edge, where there is intermittently no turbulence at all.

Where the exact shape is exactly wrong #

The shape function decays as $\eta^{-4}$ — algebraically. Measured jets decay faster, close to a Gaussian, and the two are not close outside the core.

Matched at the half-width the two agree to 0.2 per cent inside it. At three half-widths the algebraic profile is 23 times the Gaussian; at six, $2.7\times10^8$ times. So the exact solution is exactly right where the jet is and increasingly wrong where it is not — which is exactly the region a measurement has the least signal in, and exactly the region an entrainment calculation depends on.

That is not a small caveat. The entrainment integral $\int u\,r\,dr$ converges for an $\eta^{-4}$ tail and converges faster for a Gaussian one, so the exact $8\pi\mu x$ above is a property of a tail nobody observes. It is right for the laminar jet, where the solution is exact; for the turbulent model it inherits a shape that is wrong precisely where the integral collects.

The tail is a probability, not a velocity #

The disagreement about tails deserves a mechanism, because a measured jet’s edge is not a smoothly decaying profile and never was.

A turbulent jet at any instant has a sharp, deeply convoluted boundary — a thin interface across which the vorticity drops to nothing — and outside it the fluid is irrotational and, in the mean, being drawn quietly inward. That boundary is not at a fixed radius. It writhes, so a probe held near the edge is inside the jet for part of the record and outside for the rest, and the fraction of time it spends inside falls from one on the axis to zero well outside.

So the mean profile near the edge is not a velocity anything has. It is a time average over two different states, weighted by how often each occurred, and its shape out there is largely the statistics of where the interface was rather than a decay law. A quantity that is essentially the distribution of a wandering boundary will look roughly Gaussian for reasons that have nothing to do with viscosity, molecular or eddy.

Two things follow. The eddy-viscosity model’s failure at the edge is structural rather than a matter of the wrong constant — a single $\nu_t$ cannot describe a region that is turbulent only part of the time. And the entrainment happens at that interface, by nibbling and engulfment across it, in a place the mean profile has averaged out of existence.

Why the entrainment result is worth more than it looks #

Of the three exact statements, the one about entrainment is the one that changes how a reader thinks, and it is worth turning over once more.

The obvious expectation is that a harder-fired jet gathers more fluid: it is moving faster, it is more energetic, it makes more noise. The solution says it gathers exactly as much, and the reason is that the two things which decide the gathering pull in opposite directions and cancel exactly rather than approximately.

A jet with ten thousand times the momentum flux is a hundred times faster on its axis and a hundred times narrower at a given station. The mass it drags in per unit length goes as the speed times the width — so the factor of a hundred and the factor of a hundredth meet, and the answer contains only the viscosity and the distance.

Read forward that is a design statement: the entrainment of a laminar jet is set by the fluid and the geometry and not by the pump. Read backwards it is a warning about scaling arguments, since almost any dimensional reasoning would predict that the entrainment grows with the jet’s strength, and it does not.

The plane jet, for contrast #

Running the same argument for a two-dimensional jet is instructive, because one of the exact results survives and one does not.

Momentum flux is still conserved — that follows from there being no boundary, and it does not care about the dimension. But the similarity exponents change: a plane jet spreads as $x^{2/3}$ and decays as $x^{-1/3}$, and its entrained mass grows as $x^{1/3}$ rather than linearly. So the entrainment depends on the momentum flux in two dimensions and not in three, and the clean cancellation above is a property of the round geometry rather than of jets.

That is a useful check on how much to read into an exact result. The conservation law is general; the elegance is not. It is the same lesson as three dimensions are kinder in the ideal-flow chapter — that the dimension is a hypothesis, and the results that look like general truths often turn out to be statements about how a disturbance decays in a particular number of directions.

What to take from it #

The laminar solution is a genuine exact solution and deserves the word. It solves the equations, it conserves what it must, and every number in it is a consequence rather than a fit. There are perhaps a dozen such solutions in the whole subject and this is one.

The turbulent version is a shape borrowed from it. That is a legitimate and useful thing to do — the shape is right in the core, and having the right shape with one fitted scale is much better than having neither. What it is not is a derivation, and describing a measured collapse as confirming it mistakes a shared shape for a shared mechanism.

The exact results are not equally exposed. The momentum conservation is a consequence of there being no boundary and survives every model of what is happening inside the jet — it is true of a turbulent jet, a laminar one, and a jet nobody can compute. The linear spreading and hyperbolic decay follow from momentum conservation plus self-similarity, so they survive as long as the flow is self-similar. The entrainment law needs the profile shape as well, and the shape is the part that is borrowed. Sorting an exact result by how many hypotheses it needs is most of what reading one carefully consists of, and it is what what a jet keeps does for the first of these three.

And the distinction is testable, in principle. The two differ in their tails, and a measurement with enough dynamic range at three or more half-widths distinguishes an algebraic decay from a Gaussian one. Such measurements exist and they favour the Gaussian, which is the honest reason for saying that the eddy-viscosity jet is a model rather than a solution.

One sentence on why this jet is famous #

Of the handful of exact solutions the Navier–Stokes equations possess, most are trivial in the sense that the nonlinear term vanishes identically — Couette flow, Poiseuille flow, the Stokes layer, the flow between rotating cylinders. In all of them the convective acceleration is exactly zero and what is being solved is a linear equation wearing a nonlinear equation’s clothes.

The round jet is not like that. Its convective term is not zero anywhere, the equation being solved is genuinely nonlinear, and it still integrates in closed form. That is why it is in every textbook and why it is worth checking rather than quoting: an exact solution of a nonlinear equation is rare enough that the temptation to lean on it beyond its hypotheses is strong, and this essay is largely about where that leaning goes wrong.

What is not claimed #

The boundary-layer form, not Landau’s full solution. The slender approximation is used throughout; Landau and Squire’s exact Navier–Stokes solution has a slightly different profile and the same qualitative content, and the difference is of order the square of the spreading angle.

No buoyancy, no swirl, no confinement. A heated jet becomes a plume with different exponents; a swirling one has a second conserved quantity and can break down; a jet in a duct entrains until it runs out of room and then behaves completely differently, which is the mechanism mixing is a pump is about.

The entrainment is the solution’s, not a measurement’s. $8\pi\mu x$ for the laminar case is exact; the turbulent entrainment rate is measured, is about 0.32 times the local centreline speed times the width, and is not derived from anything here.

No account is taken of the near field. Everything here is the far field, tens of diameters downstream, where a jet has forgotten its nozzle. The first few diameters contain a potential core, a developing shear layer and — if the jet is loud — an instability with its own preferred frequency, which is the frequency a wake chooses in a different geometry and is entirely absent from a similarity solution.

And the Gaussian comparison is a stand-in. Real jet profiles are not exactly Gaussian either. What the figure demonstrates is that two shapes agreeing to a fraction of a per cent in the core can differ by orders of magnitude outside it, not that the Gaussian is the truth.