A jet striking a plate leaves a stronger wall jet the further it travelled

A jet along a wall keeps the flux of its momentum flux set two jets from one slot side by side. Blown into open fluid, a plane jet keeps its momentum flux $K$; blown along a wall, it cannot, because the wall’s shear takes momentum out of it all the way down, and what it keeps instead is Glauert’s invariant $F = \int_0^\infty u\left(\int_y^\infty u^2\,\mathrm{d}y'\right)\mathrm{d}y$, the flux across each height of the momentum flux carried above it. That wall jet started at a slot in the wall. The essay ended on the other way a wall jet is made, the one in every jet-cooled turbine blade and every air knife: a jet that arrives from outside, strikes a plate head on, and splits into two wall jets running away from the stagnation point. What does each inherit?

Turned without loss #

At a high enough Reynolds number the turning at the stagnation point is quick and inviscid. The jet is thin against the distance it has travelled, the region where it bends is a few of its own widths across, and the viscous layer that the wall imposes there is thin against the jet. Along every streamline the total head is then kept, and once the turned flow runs parallel to the wall again the pressure is back to ambient. So each streamline leaves at the speed it arrived with. Written as a function of the stream function, the wall jet’s speed $u(\psi)$ is the arriving jet’s, with one change: the jet’s axis, the dividing streamline that ran into the stagnation point, now lies along the wall.

It is worth being precise about why that holds in a jet that is full of vorticity, since the usual statement of Bernoulli’s equation is for irrotational flow. The statement needed here is the narrower one, that in steady inviscid flow the total head is constant along each streamline, whether or not it changes from one streamline to the next — the form where Bernoulli’s equation applies is careful to separate from the irrotational one. A jet’s streamlines each carry their own head, the centreline’s the largest; the stagnation pressure at the plate is the centreline’s head; and the turned streamlines, once the pressure has relaxed to ambient, recover their own speeds one by one. What the turning cannot do is mix streamlines, so it hands the wall jet the free jet’s profile in the stream function rather than in the distance from the wall.

For Bickley’s jet that is exact and simple. Its speed is $U\,\mathrm{sech}^2(ay)$ and its stream function $(U/a)\tanh(ay)$, so $u = U(1 - \psi^2/\Psi^2)$ with $\Psi = U/a$ the volume flux of each half. And since the distance from the wall is $\int\mathrm{d}\psi/u$, the turned wall jet is literally the arriving jet’s half-profile, laid on its side.

That profile slides on the wall at its full speed, which no real wall allows. Glauert’s similarity profile, which the wall jet must become, is held at rest at the wall and peaks about half a half-width out. The turned jet starts as the one and has to become the other, and what carries over between them is the invariant.

What each wall jet inherits #

In terms of the stream function Glauert’s invariant takes a simple form. The momentum flux carried above a height is $\int_y^\infty u^2\,\mathrm{d}y' = \int_\psi^\Psi u\,\mathrm{d}\psi'$, and $u\,\mathrm{d}y = \mathrm{d}\psi$, so

$$F = \int_0^\Psi\!\!\int_\psi^\Psi u\,\mathrm{d}\psi'\,\mathrm{d}\psi = \int_0^\Psi \psi\,u(\psi)\,\mathrm{d}\psi = \frac{U\Psi^2}{4} = \frac{3KQ}{32},$$

the last step using the free jet’s momentum flux $K = \tfrac43 U\Psi$ and its volume flux $Q = 2\Psi$. The same number comes out of Glauert’s definition evaluated directly in $y$ on the laid-down half-profile, by a separate quadrature, to seven decimal places.

Each wall jet inherits the momentum the jet brought times the fluid the jet had gathered. And the free jet gathers fluid as it travels: its volume flux grows as $(36K\nu h)^{1/3}$ with the distance $h$ from the nozzle. So the invariant each wall jet starts with grows as the cube root of the nozzle’s distance from the plate.

Further from the plate, a stronger wall jet #

The figure puts the three together for the slot of the earlier essay, two millimetres wide and blowing at five centimetres a second in water, a laminar jet whose Reynolds number on the slot is a hundred. The jet that arrives at the plate is slower the further it has come, as $h^{-1/3}$: at 2 centimetres it is 1.71 times as fast as at 10, at 2 metres 0.37. But the wall jet it leaves is stronger, as $h^{1/3}$ in its invariant and $h^{1/6}$ in its speed far along the plate, because Glauert’s peak speed goes as the square root of $F$. A metre along the plate the wall jet from a nozzle 2 metres away is 1.65 times as fast as the one from a nozzle 10 centimetres away, and the one from 2 centimetres 0.77 times.

That is the surprising connection, and it runs through what a jet keeps. A free jet is a machine for acquiring fluid it did not start with, and in open fluid the acquisition costs it speed and changes nothing it conserves. Strike a wall, and the fluid it has acquired becomes part of what the wall jet conserves. The invariant is a product of the two things the free jet carried, one conserved and one growing, and the wall converts the growing one into a permanent property of everything downstream.

The march that checks the inheritance #

The argument has two steps, the inviscid turning and the conservation of $F$ downstream, and the second can be checked. The wall jet is marched from the turned profile by the boundary-layer equations, written in Glauert’s own variables — $\psi = (40F\nu x)^{1/4}f$, $\eta = (5F/32\nu^3x^3)^{1/4}y$ and $\xi = \ln x$ — in which they become

$$f''' + ff'' + 2f'^2 = 4\left(f'\frac{\partial f'}{\partial\xi} - f''\frac{\partial f}{\partial\xi}\right),$$

Glauert’s similarity equation with the departure from similarity on the right. Each step along the wall is a two-point problem across it, solved by the same box scheme and Newton that solved the natural-convection plates, with second-order steps in $\xi$. The march starts a two-hundredth of a half-width along the wall with the no-slip layer already grown that far by Rayleigh’s error function — a slip at the wall is a discontinuity Newton cannot start from, since a jet at rest also solves every step — and that pre-grown layer costs between a fifth of a per cent and two per cent of $F$, more at the lower Reynolds numbers, where it is thicker.

From there $F$ is held to three parts in ten thousand all the way to thirty thousand half-widths, as the equations say it must be — a first-order step had let it drift by fifteen per cent, which is the reason the steps are second order. And the profile becomes Glauert’s: the ratio of its peak speed to Glauert’s, for the invariant it carries, climbs from nearly nothing — the slipping profile is far wider than Glauert’s in his variables — through a broad approach and settles to within a per cent.

How far until it forgets it was free #

The approach is slow, and its length is set by viscosity. The peak speed comes within five per cent of Glauert’s after 74 arriving half-widths at a Reynolds number of 100 on the arriving jet’s speed and half-width, 814 at 1,000 and 8,970 at 10,000: in proportion to the Reynolds number, between three-quarters and nine-tenths of a half-width for every unit of it. That is the distance a no-slip layer, growing as $\sqrt{\nu x/U}$, needs to grow through a jet of the arriving width. Until then the wall jet carries Glauert’s invariant but not his profile — a slipping core over a growing wall layer — and its peak speed is not the one his formula gives for its $F$.

The proportionality has a direct reading. The wall layer reaches the slipping core’s full thickness when $\sqrt{\nu x/U}$ is comparable with the half-width $1/a$, which is at $x \sim U/(\nu a^2)$, a distance of the Reynolds number in half-widths. The march finds the coefficient between 0.74 and 0.90 for a five per cent criterion across the three Reynolds numbers, rising slowly as the starting wall layer’s share shrinks, so the estimate is most of the physics: the wall jet becomes Glauert’s once viscosity has had time to reach across it, and not before.

The practical reading is that the similarity solution is a far-field statement for laminar impinging jets of any useful Reynolds number. A jet at a Reynolds number of a thousand needs most of a thousand of its own widths of plate before Glauert’s formula applies, and over the first few hundred the wall jet’s peak speed is higher than his formula predicts for its invariant, because the slipping core has not yet been slowed.

What the wall takes along the way #

The invariant survives because of what it measures, and the march makes the bookkeeping visible. The wall jet’s momentum flux does not survive: the wall’s shear takes it out at a rate equal to the shear itself, so the momentum flux of each half, $K/2$ at the start, falls as $x^{-1/4}$ once the jet is Glauert’s. What the shear cannot touch is the weighting $F$ applies — each layer’s speed times the momentum carried above it — because the shear acts at the bottom of the jet and the weighting gives the bottom nothing to lose. The fast core of the turned profile, sitting at the wall where the shear is largest, loses its momentum first; the slower outer layers, which carry the most weight in $F$, are hardly touched while the wall layer grows into them. A layer that is an integral of everything upstream is the general form of the same fact: a boundary layer’s state at a station is a weighted memory of what happened before it, and here the weighting happens to be exact.

Where impinging jets are used #

The arrangement is everywhere jets meet surfaces on purpose: an air knife stripping liquid from a moving sheet, a row of jets cooling the inside of a turbine blade’s leading edge, a jet drying a printed web. In most of them the region of interest is the stagnation point, where the heat transfer is highest, and the wall jets downstream are what carries the spent fluid away. The inheritance says something useful about the second region. A designer who moves the nozzle closer to the plate to raise the stagnation heating also weakens the wall jets that clear the surface downstream, by the cube root of the distance in their invariant; one who moves it away for the sake of uniform coverage gets stronger wall jets in exchange for a slower, wider jet at the plate. For a laminar jet the trade is exact; for the turbulent jets of most practice the same two quantities, momentum delivered and fluid gathered, still set the wall jet, with an entrainment rate that turbulence makes much larger.

The same slot, two ways #

The earlier essay blew its 2 mm slot along the wall and gave the wall jet $F = U_0^3b^2/2$. Point the same slot at the plate and each of the two wall jets carries 0.39 of that at 5 centimetres’ standoff and 0.84 at 50, reaching it at 84 centimetres. But an impinging jet makes two wall jets, and together their invariants pass the one blown jet’s at 10.5 centimetres. Which arrangement puts more of the slot’s flow into wall jet is a question of how far the jet travels first, and the answer turns on the fluid it gathers.

A wall jet on a curved target #

Many impinging jets strike curved surfaces — a cylinder in a cross-jet, the concave inside of a blade’s leading edge — and their wall jets then run round a curve. A wall jet on a convex surface is the flow the Coandă effect is usually invoked for, and it stays attached for as long as its pressure gradient across the curve can turn it, which depends on how fast and how thin it is. The inheritance says what the jet brings to that contest — an invariant set by the standoff, and a profile that slides on the surface for the first several hundred widths — but how far round a curve that carries it is the curvature’s question rather than the flat plate’s, and the wall layer’s slow growth means the answer near the stagnation point is not Glauert’s at all.

The convention: per unit width, from the stagnation point #

All quantities are per unit width of a plane jet. The jet’s half-width is $1/a$, the distance at which its speed has fallen to $\mathrm{sech}^2 1 = 0.42$ of its centreline value — not the half-velocity width, which is 0.88 of it — and the Reynolds number is the arriving centreline speed times that half-width over the viscosity. Distances along the wall are from the stagnation point, with no virtual origin fitted; Glauert’s formula is used with the invariant the march actually carries. The nozzle’s distance is measured to the plate, and the arriving jet is Bickley’s at that distance, which assumes the jet has travelled many slot widths.

What the inviscid turning leaves out #

The stagnation region’s own viscous layer. Hiemenz’s layer at the stagnation point is $\sqrt{\nu/k}$ thick for a strain rate $k$, a fraction of the jet width at a Reynolds number of a hundred; it removes a part of $F$ of order the inverse square root of the Reynolds number, which the march’s starting layer stands in for. A jet that is not yet Bickley’s. A nozzle a few slot widths from the plate delivers its exit profile, not the similarity profile, and the inheritance is then the exit profile’s $\int\psi u\,\mathrm{d}\psi$. Instability. A laminar plane jet goes unstable at modest Reynolds numbers, and impinging jets are prone to self-sustained oscillation when the nozzle is close to the plate; the jet’s sinuous mode, whose growth every unstable wave inside one circle bounds for Bickley’s profile, and the plate’s feedback on it are outside a steady march. Turbulence, in which a wall jet still has a flux of momentum flux but no exact invariant. And the round jet, whose radial wall jet inherits a different combination, the question left at the end.

Heat and mass transfer, too, are outside it, though they are why impinging jets are used at all: to cool turbine blades and electronics, to dry paper and to clean surfaces. The transfer coefficient beneath a wall jet follows the jet’s speed and thickness, so a stronger wall jet from a longer approach should carry some of its advantage into the transfer rate; but the thermal layer has an invariant and an inheritance of its own from the stagnation region, and this calculation stops at the velocity field.

Glauert’s invariant, put to a new use #

Hiemenz described the stagnation-point flow in 1911 and Bickley the plane jet in 1939. Glauert’s paper on the wall jet, in 1956, found the invariant and the closed-form profile, and noted that the jet arising from an impinging stream would carry such an invariant from its source. That the inheritance is a single integral, the arriving jet’s $\int\psi u\,\mathrm{d}\psi$, follows from the inviscid turning, which the theory of the jet that halves its flow at a hole used in a different form: a free surface or a turned stream keeps its speed along each streamline, so what a stream carries in its stream-function profile is what it delivers.

Still open: the round jet’s radial wall jet #

A round jet striking a plate spreads radially, and its wall jet thins as it spreads round a growing circumference. Glauert found an invariant for that radial wall jet too, weighted by the radius, and a round jet carries a different profile and a volume flux that grows in proportion to the distance it travels rather than as a cube root. The next calculation turns Schlichting’s round jet at a plate, finds the radial invariant it hands on in terms of its momentum flux and the standoff, and asks whether the round version’s dependence on the nozzle’s distance — linear in the gathered fluid where the plane jet’s was a cube root — makes the far radial wall jet even more sensitive to how far the jet travelled before it struck.