A jet along a wall keeps the flux of its momentum flux

What a jet keeps found the round jet’s one conserved quantity, its momentum flux, and showed that everything else — its spreading angle, its slowing, the fluid it gathers — follows from that number and the viscosity. A point force makes a jet only when it is strong carried the same idea to the exact conical solution, Landau and Squire’s, and ended on the case where the fluid has a wall: a jet issuing through a hole in a plane, which cannot draw fluid from behind its source.

That case resists. In Landau and Squire’s form the conical flow obeys a Riccati equation with three constants of integration, and a jet regular along its axis needs two relations between them, while a wall that admits no flow and allows no slip needs a third that leaves only the trivial flow. Whether some weaker condition at the wall rescues a solution is a question this essay leaves where it found it. There is, however, a wall jet with a clean exact solution — the jet blown along a wall rather than out of one — and it answers the same underlying question, what a wall does to what a jet conserves, with a surprise in it.

Why the hole in the wall resists #

The obstruction is worth a paragraph, because it says something about which exact solutions exist. Landau and Squire’s conical flows have a stream function $\psi = \nu r f(\mu)$, with $\mu$ the cosine of the angle from the axis, and the momentum equations integrate twice to $f^2 - 4\mu f - 2(1-\mu^2)f' = 2(c_1\mu^2 + c_2\mu + c_3)$. On the axis the flow must be regular: no line source there means $f(1) = 0$, which asks $c_1 + c_2 + c_3 = 0$, and a finite speed along the axis asks $2c_1 + c_2 = 0$ as well. On a wall through the origin at right angles to the axis, no flow through it is $f(0) = 0$ and no slip along it is $f'(0) = 0$, and together they ask $c_3 = 0$. Three linear conditions on three constants leave $c_1 = c_2 = c_3 = 0$, which is Landau’s free jet — and Landau’s jet has $f(0) = 2/A$, flow through the wall. A conical jet from a hole in a no-slip wall would have to give up one of these, and the plane analogue, Jeffery and Hamel’s flows in a wedge, shows what giving up symmetry can buy; which condition the real flow relaxes near its hole is the question this one leaves open.

A free plane jet, for reference #

A jet from a long slot into open fluid is a plane jet, and its boundary-layer solution is Bickley’s of 1939: $u = (3K^2/32\nu x)^{1/3}\,\mathrm{sech}^2\xi$ with $\xi = (K/48\nu^2x^2)^{1/3}y$, where $K = \int u^2\,dy$ is the momentum flux per unit width. Like the round jet, it conserves $K$ exactly, because nothing outside it exerts a force. Its thickness grows as $x^{2/3}$, its speed falls as $x^{-1/3}$, and its volume flux, $(36K\nu x)^{1/3}$, grows as $x^{1/3}$: it gathers fluid from either side, more slowly the further it goes.

What the wall takes #

Lay the same slot flush with a wall and blow along it. The jet now has a no-slip surface on one side, and that surface pulls back on it with a shear stress at every station. Its momentum flux is no longer conserved: whatever the wall’s shear takes per unit length is lost from the jet’s momentum per unit length. And its volume flux is not conserved either, since it still entrains from the open side. A free jet’s conserved quantity is gone and nothing obvious replaces it.

Something does. Integrate the boundary-layer momentum equation from a height $y$ up to infinity: the momentum flux carried above $y$ changes along the jet because fluid crosses the line at $y$, carrying its streamwise velocity with it, and because the shear stress acts across that line. Weight that balance by the speed at $y$ and integrate over the whole thickness, and the two contributions cancel, leaving

$$\frac{d}{dx}\int_0^\infty u\left(\int_y^\infty u^2\,dy'\right)dy = 0.$$

The wall’s own shear drops out because the weight, the speed at the wall, is zero there. Glauert found this in 1956: a wall jet conserves the flux, through each height, of the momentum flux carried above that height — a quantity with the dimensions of a speed cubed times a length squared, which has no name in any other flow.

The invariant, and what it fixes #

As the free jet’s momentum fixes its growth, the wall jet’s invariant $F$ fixes its. With $\psi = (40F\nu x)^{1/4}f(\eta)$ and $\eta = (5F/32\nu^3x^3)^{1/4}y$ the equations reduce to one,

$$f''' + f f'' + 2f'^2 = 0, \qquad f(0) = f'(0) = 0, \quad f'(\infty) = 0,$$

and every growth law can be read off the scalings: thickness as $x^{3/4}$, peak speed as $x^{-1/2}$, volume flux as $x^{1/4}$, momentum flux as $x^{-1/4}$. The wall jet spreads faster than the free jet, $x^{3/4}$ against $x^{2/3}$, and gathers fluid more slowly, $x^{1/4}$ against $x^{1/3}$ — one open side instead of two, and a no-slip side that drags the slowest fluid to a stop.

The exponents need nothing but dimensions once the invariant is known. $F$ is a speed cubed times a length squared, the viscosity a length squared per time, and the only thickness that can be built from them and the distance is $(\nu^3x^3/F)^{1/4}$; the only speed is $(F/\nu x)^{1/2}$. Their product is the volume flux, $(F\nu x)^{1/4}$, and the speed squared times the thickness is the momentum flux, $(F^3/\nu x)^{1/4}$. The free plane jet does the same with $K$, a speed squared times a length, and gets its thirds. What the equations add is only the profile and the constants — the 40 and the 5/32 — and the proof that the invariant exists.

The figure plots the four fluxes from one slot. The free jet’s momentum is flat and its volume flux rises with slope one-third. The wall jet’s momentum falls with slope minus a quarter and its volume flux rises with slope a quarter, while its invariant, flat, is the only line that does not move.

The profile has a boundary layer under a free jet #

Shot from the wall — integrate the third-order equation with $f'' (0)$ set to one, find where the speed returns to zero, and rescale using the equation’s own symmetry so that $f(\infty) = 1$ — the profile has a wall curvature $f''(0)$ of 0.222222, Glauert’s 2/9, and its stream function agrees with his closed form, $\eta = \ln[(1 + \sqrt f + f)^{1/2}/(1 - \sqrt f)] + \sqrt3\arctan[\sqrt{3f}/(2+\sqrt f)]$, to six parts in a hundred million.

Scaled on its peak speed and on its half-width above the peak, the wall jet is two flows joined at the peak. Above it the profile follows the free jet’s closely, falling to half speed at one half-width and to nothing by three. Below it a no-slip layer brings the fluid to rest within 0.55 of the half-width: the peak sits at a little more than half the distance to the outer half-speed point, at a similarity variable of 2.03. The lower part is a boundary layer under a favourable push from the fluid above it; how thick is thin is the same no-slip arithmetic on a plate in a uniform stream, and here the stream it lies under is the jet itself.

Where the invariant lives #

The invariant’s integrand is a product of two things that peak in different places. The speed peaks at 0.55 of the half-width. The momentum flux carried above a height is largest at the wall and falls to nothing far out. Their product is zero at the wall, because the speed is, and zero far out, because nothing is above; it peaks at 0.38 of the half-width, inside the boundary-layer half of the jet, below the speed’s peak. The conserved quantity weights fast fluid that has a great deal of momentum still above it, which is the fluid under the peak — the region that feeds the wall’s shear and is fed by the jet above, and the region where the two halves of the balance cancel.

The momentum goes into the wall, and only there #

The momentum flux falls by 60 per cent between five centimetres and two metres from the slot. Add to it the drag the wall has exerted over the same stretch — the wall shear stress integrated along the wall — and the sum is constant to five parts in ten thousand, the error of the trapezoid that integrates the drag; the derivative of the momentum flux equals minus the wall shear to a part in ten million at a single station. The loss is all drag. None of it goes into the entrained fluid, which arrives from above with no streamwise momentum and leaves with the jet’s, exactly as it does in the free jet; there the jet’s momentum is shared over more fluid and not lost, and here it is shared and also paid out to the wall.

The same slot, two jets #

To compare the two from one slot each needs its own invariant, taken from a top-hat jet of speed $U_0$ across a slot of width $b$: $K = U_0^2b$ for the free jet and $F = U_0^3b^2/2$ for the wall jet. From a two-millimetre slot at five centimetres a second in water — a slot Reynolds number of a hundred — both jets thicken by tens of times over a metre, the wall jet faster. A metre downstream the free jet’s centre moves at 1.33 centimetres a second and the wall jet’s peak at 0.79; the free jet carries 5.65 square centimetres a second per unit width and the wall jet 3.16. The wall jet is slower, thicker and has gathered less. Both are similarity solutions measured from a virtual origin at the slot, which is right far downstream and only roughly right within a few hundred slot widths of it.

When a wall jet is slender enough to be one #

Every number here is a boundary-layer solution, and that needs the jet to be thin compared with the distance it has travelled. Its thickness over its distance falls as $x^{-1/4}$, so a wall jet becomes more slender as it goes, but slowly and from a start set by the fluid. From the water slot, thirty centimetres downstream, the wall jet’s peak is ten millimetres up and moves at 1.4 centimetres a second: a local Reynolds number, peak speed times peak height over the viscosity, of 150, and a thickness a thirtieth of the distance. The same slot blowing air at half a metre a second gives a peak fourteen millimetres up and a Reynolds number of 110. Glycerine blown at the same speed from a five-millimetre slot, a slot Reynolds number of 2.5, would put its peak twenty centimetres up at thirty centimetres downstream: a jet as thick as it is long, which is not a jet at all but a creeping flow, and for which the solution above is meaningless.

So the laminar wall jet lives in a window. Too viscous and it is not slender; not viscous enough and it is not laminar. The window is a range of jet Reynolds numbers from roughly ten to roughly a few hundred, the same window as the free plane jet’s, and the reason the thin layer and the full equations agree inside it and nowhere else. Von Kármán’s spinning disc is the rare exact solution with no such window, because its similarity holds at every Reynolds number; the jets’ hold only asymptotically.

Why the Coandă jet stays thin #

The wall jet is the flow every account of the Coandă effect is about, and the effect that is real and where it stops found that a jet on a convex wall makes its whole pressure deficit inside its own thickness. The growth laws here say how thick that is and how it changes: on a flat wall a laminar wall jet’s thickness grows as the three-quarter power of distance, so a jet that has travelled four times as far is 2.8 times as thick and half as fast. On a curved wall the centrifugal pressure the jet’s own speed sustains falls with that speed squared, while the curvature it must follow does not change, which is why a jet clings over a short arc and then lets go. The flat-wall invariant is the reference against which that letting-go is measured.

The other connection is to the free jets’ family. A round jet conserves its momentum and its volume flux grows linearly, $8\pi\nu x$, as exactly similar, and one number short computed; a plane jet conserves its momentum and gathers as $x^{1/3}$; a wall jet loses momentum and gathers as $x^{1/4}$. In each case one integral of the profile is held and the exponents fall out of dimensions. The wall jet is the one where the conserved integral had to be found rather than written down, because the obvious candidate, momentum, is being drained.

How the jets were checked #

The wall jet was shot from the wall with a fourth-order integrator, so the closed form is an independent check rather than the source: the two agree to $6\times10^{-8}$ at six heights through the jet, and the wall curvature to $10^{-13}$ of 2/9. The invariant computed back from the profile by its definition — the speed times the momentum flux above each height, integrated over the thickness — is the value the scaling was built on to a part in ten million. The momentum balance against the wall shear holds to the same precision. Bickley’s jet was integrated across its width by quadrature and returns its momentum flux and its closed-form volume flux to $10^{-14}$.

The convention: per unit width, from a virtual origin #

All fluxes are per unit width of slot. The free jet’s $K$ is its kinematic momentum flux $\int u^2 dy$ across its whole width, and the wall jet’s $F$ is $\int_0^\infty u(\int_y^\infty u^2dy')dy$ from the wall outwards. Distances are from a virtual origin taken at the slot. The half-width of the wall jet is measured on its outer side, where the speed has fallen to half its peak; the free jet’s half-width is to half its centre speed.

What the similarity solution cannot show #

Stability. Laminar plane jets and wall jets are unstable at Reynolds numbers in the tens to low hundreds, and a jet from a slot at a Reynolds number of a hundred is near that edge; downstream, as the jet’s own Reynolds number based on its thickness and speed grows as $x^{1/4}$ for the wall jet, it eventually becomes turbulent, and the cost of going turbulent then applies to its wall shear. The slot. The similarity profile forgets the slot’s shape, which is right only many slot widths downstream. The far field. The entrained fluid arrives by an outer flow the boundary-layer solution does not compute, and its pressure acts back on the jet at higher order. Curvature and gravity, which change the invariant’s balance, are absent: this is a flat wall in a fluid of one density.

Glauert’s invariant #

Bickley solved the free plane jet in 1939, after Schlichting’s round jet of 1933. M. B. Glauert’s 1956 paper on the wall jet found the exterior momentum flux invariant, solved the laminar profile in closed form, and gave the radial wall jet — a round jet impinging on a plate and spreading over it — by the same method; Akatnov had found the plane laminar solution independently in 1953. The wall jet has since become the standard model for film cooling, Coandă devices and the outflow of a jet striking a surface, almost always in its turbulent form, where an analogue of the invariant holds only approximately because the shear stress no longer vanishes in the same way.

Still open: a jet that strikes a wall #

The wall jet here begins at a slot flush with the wall. A jet that arrives at the wall from outside — a plane jet impinging normally on a plate — splits at a stagnation point and becomes two wall jets, and the question is what invariant each inherits from the free jet’s momentum. The next calculation joins Hiemenz’s stagnation-point flow at the plate to Bickley’s jet arriving and Glauert’s jets leaving, and asks what fraction of the arriving momentum appears as the two wall jets’ invariants, how far from the stagnation point their similarity sets in, and whether the round version — a jet striking a plate and spreading radially — leaves the plate’s inner region with the same invariant as Glauert’s radial wall jet or a different one.