What a jet keeps, and what it collects

A jet is the one viscous flow on this site with nothing to hold on to. There is no plate, no pipe, no film and no corner — just fluid moving through fluid at rest — and viscosity has no wall to enforce its no-slip condition on. What it does instead is hand momentum sideways.

That single sentence generates the entire solution, because it says what does not happen: nothing outside the jet pushes it. The surroundings are at rest and at uniform pressure, so no force acts, so the jet’s momentum flux is a constant of the motion.

The one thing it conserves #

Take a control volume around the jet, from the nozzle to a station downstream, with its sides far out in the quiescent fluid. Fluid crosses the sides — that is the entrainment — but it crosses carrying no streamwise momentum, because it had none. The pressure is uniform, so the sides contribute nothing. What is left is

$$\int u^2 \, dy = \text{constant} = K,$$

the kinematic momentum flux, and it is the only thing about the nozzle that survives downstream. Nozzle shape, exit profile, whether it was round or slotted or ragged: all of it is forgotten within a few diameters, and one number is remembered. This is the same argument the actuator disc uses to get Betz’s limit, applied to a volume with nothing inside it.

The exponents come out of that, and out of one more statement #

A jet has no length of its own once the nozzle is forgotten, so the solution must be self-similar: the width $b$ grows as $x^m$, the centreline speed $u_c$ falls as $x^{-s}$, and the profile is one fixed shape stretched by those two.

Momentum gives one equation — for a plane jet $u_c^2 b$ is constant, so $m = 2s$. The other comes from the boundary-layer balance, inertia against diffusion: $u_c^2/x \sim \nu u_c / b^2$. With molecular viscosity, a constant, that pair solves to

$$m = \tfrac{2}{3}, \qquad s = \tfrac{1}{3},$$

and substituting into the boundary-layer equations gives the profile itself. It is $\operatorname{sech}^2$, exactly:

$$u = AB\,x^{-1/3}\operatorname{sech}^2\!\left(B y x^{-2/3}\right), \qquad A = 6\nu B, \qquad B = \left(\frac{K}{48\nu^2}\right)^{1/3}.$$

The two constants are not free. The first comes from the equations — every term collapses onto $\operatorname{sech}^4\eta - 2\operatorname{sech}^2\eta\tanh^2\eta$ and the balance fixes $A = 6\nu B$ — and the second from the momentum flux, $\int u^2dy = \tfrac{4}{3}A^2B = K$.

What it does not conserve #

The mass flux is $2Ax^{1/3}$ and it grows for ever. There is no contradiction anywhere: mass is conserved globally, and the jet is simply gathering it in from the sides, at a transverse velocity that the solution computes — 5.7 millimetres a second inward at the station drawn, on a centreline speed of 146.

Seen from far away, then, a jet looks like a line sink. That has a consequence worth carrying away: two parallel jets pull towards each other, because each sits in the other’s inflow. It is why a pair of nearby exhausts merge, and it is the same mechanism as a jet on a wall bending towards it — the wall blocks the supply from one side and the jet is pulled towards the blockage.

The round jet’s arithmetic does something startling #

The axisymmetric version, solved by Schlichting four years before Bickley did the plane one, has

$$u = \frac{3K}{8\pi\nu x}\left(1 + \frac{3K}{64\pi\nu^2}\frac{r^2}{x^2}\right)^{-2},$$

and its volume flux is

$$Q = 8\pi\nu x.$$

There is no momentum in that expression. Squirt the jet a hundred times harder and it entrains exactly the same amount of fluid at every station.

The compensation is worth spelling out because it is the kind of result that could easily be a coincidence of arithmetic. The centreline speed goes as $K$ and the width goes as $K^{-1/2}$, so the volume flux — speed times area — goes as $K \times K^{-1} = K^0$. The two exponents that cancel come from different equations: the speed from the momentum flux and the width from the momentum flux again, entering with opposite powers because area is two-dimensional. It is a theorem, not an accident, and it is checked here to a part in a million on jets four orders of magnitude apart.

The plane jet does not do this: its mass flux goes as $K^{1/3}$. The difference is dimensional and nothing more, which is a good reminder that “a jet entrains” is not one statement.

Which means the round jet’s Reynolds number never changes #

A local Reynolds number built on the width and the centreline speed is the natural measure of whether a jet is still in the regime it was solved in. For the plane jet it grows as $x^{1/3}$: 29 at 80 millimetres, 45 at 300, 72 at 1.2 metres. For the round jet it is exactly constant — 199 at every station drawn — because $u_c \propto x^{-1}$ and $b \propto x$ cancel.

The same table has the turbulent cases in it, and they are obtained by changing one thing. If the mixing is done by turbulence rather than by molecules, the effective viscosity is not a constant: the standard closure makes it proportional to the local width times the local velocity, and that carries its own dependence on $x$. Redo the two lines of algebra and the jet spreads linearly whatever its shape, with the plane jet’s centreline going as $x^{-1/2}$ and the round jet’s as $x^{-1}$.

Those are the exponents a real jet has. What the argument cannot supply is the constant in front — the measured spreading rate of about 0.11 for a plane turbulent jet — and that number is drawn on this site in the colour kept for a borrowed claim, because it is a measurement of turbulence rather than a consequence of anything computed here.

The window this solution needs is shut #

Now the honest part, and it is the most important thing about the Bickley jet.

Two conditions have to hold for it to describe anything. The boundary-layer equations require the jet to be slender, $b \ll x$. And a laminar solution is only seen if it is stable.

These are not independent. Multiply them:

$$\mathrm{Re}\times\frac{b}{x} = \left(\operatorname{arccosh}\sqrt{2}\right)^2\frac{A}{\nu B} = 0.8814^2\times 6 = 4.66,$$

with no distance and no momentum flux in it at all. The product is a pure number belonging to the solution, and the solver checks it on three jets sharing no property.

So a jet slender enough for the boundary-layer approximation has a Reynolds number of at least tens. And this profile is one of the least stable in the subject: $\operatorname{sech}^2$ has inflection points on both flanks, so Rayleigh’s criterion says it is unstable even with no viscosity at all, and the measured critical Reynolds number is a handful.

The Bickley jet is therefore an exact solution that almost never describes a jet. It is not useless — it is the base flow every stability calculation of a jet perturbs, it fixes the exponents that the turbulent case inherits, and it supplies the entrainment argument — but a laboratory jet matching this profile over any length is a delicate thing and a laboratory jet is more usually turbulent by the second diameter.

That is worth setting beside the pipe, where the exact laminar solution survives to Reynolds 2,300, and beside the boundary layer, which is stable to around half a million on the length-based number. The three exact solutions have wildly different tenures, and the property that decides it is not how exact they are.

The jet that needs no approximation at all #

The window closes because of the boundary-layer approximation, not because of the jet. Both solutions above drop the streamwise diffusion term and assume the flow is slender, and it is that pair of assumptions — not conservation of momentum, not similarity — that requires a Reynolds number large enough for the profile to be unstable. It is worth knowing that the round jet has a version with neither assumption in it.

Take the idealisation the similarity solution was already using — a point source of momentum in an unbounded fluid at rest, with no mass injected — and solve the full steady Navier–Stokes equations for it. There is an exact answer. Landau found it in 1944 and Squire independently in 1951, and it is one of the very small number of non-trivial exact solutions the equations possess.

Its structure is worth describing because it is unlike anything else in this collection. The flow is conical: the stream function depends on the spherical radius linearly and on the polar angle through one function, so the velocity falls as $1/r$ in every direction and the entire field is a set of self-similar cones about the axis. There is no slenderness anywhere in the construction, no downstream direction singled out, and one parameter, fixed by the momentum flux.

Three things follow that the boundary-layer version cannot say.

It is valid at any Reynolds number. A weak point source produces a broad, gentle, entirely respectable flow that the Bickley-style argument has no business describing, and the exact solution covers it. The window this essay closed is a window on the approximation.

And it contains Schlichting’s answer. Push the momentum up and the exact solution concentrates about the axis, its polar dependence narrows, and near the axis it becomes the inverse-square profile quoted above. So the boundary-layer solution is not a competitor but a limit — which is the reassuring result, and the one that would have been a serious problem had it failed.

It also entrains, and by the same accounting. The volume flux across any surface enclosing the origin grows with distance for the exact solution just as it does for the approximate one, so the “jet as a sink” reading survives without the slender geometry that produced it.

The plane jet has no such companion, which is a genuine asymmetry rather than an omission — the conical construction is available because three-dimensional space has a radius that a point source can be self-similar in, and the plane does not offer the same freedom. So the essay’s honest position differs between its two cases: the round jet is an approximation to something exact, and the plane one is an approximation to nothing.

Where the nozzle goes #

The similarity solution has a singularity at the origin: zero width, infinite speed, and all the momentum concentrated at a point. A real jet has a slot of finite width at a finite speed, so the two are matched by treating the real nozzle as sitting some distance downstream of a virtual origin, and that distance is usually fitted to a measurement. It can also be computed, and the arithmetic is a useful check on the whole picture.

Take a slot 4 millimetres wide discharging at half a metre a second, which is the momentum flux used in every figure here. The solution’s mass flux, $2Ax^{1/3}$, equals the slot’s own $U_0 d$ at $x = 2.2$ millimetres — a little over half a slot width — and at that station the solution’s half-width is 1.2 millimetres against the slot’s half-width of 2. The two bookkeeping quantities agree to within a factor of two at the same place.

That is as much as a solution with a singular origin can be asked for, and it says the virtual origin sits about a slot width upstream of the real one. It also gives the honest answer to “how far downstream does this apply”: the width is only a tenth of the distance by 80 millimetres, twenty slot widths out, and everything nearer than that is the nozzle’s business rather than the similarity solution’s.

The entrainment arithmetic is worth having in the same units. The jet has doubled its mass flux four and a half slot widths downstream and multiplied it by five at seventy. A jet is mostly made of fluid that was not in the nozzle, and it becomes so within a few centimetres.

What a jet is for, and why entrainment is the point #

Almost every industrial use of a jet is a use of its entrainment rather than of its momentum, and the solution above says why that is the sensible way round.

A jet’s momentum is fixed at the nozzle and cannot be added to; its mass flux grows without limit. So a device that wants to move fluid should be built round the second, and the arithmetic gives the design rule directly: the entrained flow grows as $x^{1/3}$ for a plane jet and as $x$ for a round one, so length is what buys entrainment.

That is the whole principle of an ejector, of a fume cupboard’s induced-draught hood, of a Venturi pump, and of the entrainment air in a gas burner. In each the jet is a small, fast stream whose job is to drag a large, slow one along with it, and the ratio of the two is set by how far the jet is allowed to run before the mixture is collected.

The ejector essay computes the momentum balance for exactly that machine, and this essay’s result is the same statement seen from the free-jet end: the momentum is conserved, the mass is acquired, and a machine is what puts a duct round the acquisition.

What the model does not contain #

No nozzle. The similarity solution has a singularity at the origin: as $x \to 0$ the width goes to zero and the speed to infinity, and the source of momentum is a point. It describes the flow only several nozzle widths downstream, and nothing here computes how far “several” is.

No pressure variation. The boundary-layer form assumes uniform pressure across and along the jet. For a free jet in a large room that is very nearly true; for a jet in a duct it is not, because the entrained flow has to come from somewhere and a confined supply builds a gradient. That is the whole mechanism of an ejector, and it is absent here.

Nothing about stability beyond the observation above. No growth rate is computed, no critical Reynolds number, no eigenvalue. The claim in this essay is only that the profile has inflection points and therefore fails Rayleigh’s necessary condition.

Two dimensions or perfect axisymmetry. Real jets are three-dimensional immediately after transition and their structure is what a turbulence essay would have to describe.

Constant density and no buoyancy. A hot jet, a plume or a buoyant jet has a momentum flux that is not constant — buoyancy is a body force acting all the way along — and the exponents change.

The test that would have caught the wrong answer #

There is a plausible-looking wrong jet worth constructing, because it shows what the assertions are for. Take a profile decaying as $x^{-1/2}$ instead of $x^{-1/3}$, and choose its width to keep the momentum flux constant. It is a perfectly respectable curve: correct shape, conserved momentum, entrainment, spreading, everything a picture would show.

It is not a solution of anything. Fed to the residual of the boundary-layer equations, the true solution returns 7 × 10⁻⁷ — the finite-difference noise floor — and the impostor returns 0.37, five and a half orders of magnitude larger. flowcheck requires that separation, on the rule that an assertion that has never rejected anything proves nothing.

Conserving the right quantity is not the same as solving the equations, and this is the cleanest example of that distinction on the site: the wrong jet conserves momentum exactly.

Who found it, and when #

Schlichting solved the round jet in 1933 and Bickley the plane one in 1937, both by the similarity method Blasius had used on the flat plate in 1908 — which is why all three solutions look alike on the page and are about entirely different things.

The entrainment result has an odd history. Its most consequential use came much later and in somebody else’s subject: Morton, Taylor and Turner’s 1956 model of a buoyant plume assumes that the inflow velocity is proportional to the local centreline speed — the entrainment hypothesis — and that assumption, with one constant fitted, is the basis of most plume and jet modelling since. What the laminar solution supplies is the demonstration that entrainment is not an extra physical process to be added to a jet, but a consequence of it.

Where the ladder goes next #

Both of these are steady solutions in which viscosity spreads momentum sideways. Ask the same question about vorticity instead of momentum and the answer is an exact solution too — a spinning core that spreads as the square root of time while the circulation around it never changes at all.