What a jet cannot push sideways

A jet of water hits a flat plate at an angle and divides. Most of it goes one way along the surface and the rest goes the other, and the question of how much goes each way looks like the sort of thing that would need the flow solved — a stagnation region, two turning sheets, a dividing streamline somewhere in the middle.

It needs none of that. The split follows from two sentences, one of which is mass conservation and the other of which is a statement about what a fluid with no viscosity cannot do.

The two sentences #

The sheets leave at the jet’s own speed. Both sheets have free surfaces at ambient pressure, and so does the incoming jet; the streamline along each surface therefore runs between two points at the same pressure, and Bernoulli’s equation — which holds here, along a streamline in a steady inviscid flow — allows only one conclusion. The speed is unchanged. Not approximately: the fluid may be doing anything it likes in the stagnation region, but where it emerges it emerges at V.

The plate exerts no force along itself. In a fluid with no viscosity the only stress on a surface is the pressure, and pressure acts normal to a surface by definition. There is no tangential component available. This is the same statement that makes the exact theory predict no drag at all, arriving in a place where it is a tool rather than an embarrassment.

Everything else is arithmetic. Momentum along the plate, with nothing to supply any:

$$\dot{m}_1 - \dot{m}_2 = \dot{m}\cos\beta$$

Mass, which has nowhere else to go:

$$\dot{m}_1 + \dot{m}_2 = \dot{m}$$

Two equations, two unknowns, and the split is

$$\frac{\dot{m}_1}{\dot{m}} = \frac{1 + \cos\beta}{2}$$

with the force normal to the plate ṁV sin β. At ninety degrees the split is even and the force is the whole of the jet’s momentum flux; at a grazing angle almost everything carries on forwards and the plate feels almost nothing.

The zero that does the work #

It is worth being clear about which of these is the assumption and which is the result, because the site’s own gate checks the tangential force and a reader could reasonably ask what that proves.

The tangential force is not a prediction that a real plate can test. It is the input: the split was computed by setting it to zero, so finding it to be zero afterwards would be circular if it were offered as evidence. What the check does is different and worth having — it verifies that the two sheets returned by the solver are consistent with the statement they were derived from. A sign error in one sheet, a factor of two in the other, an angle measured from the wrong axis: each of those produces a plausible split with a non-zero tangential residual, and each would draw a figure that looked entirely reasonable.

The physics of a real plate is a separate question with a known answer. A real fluid does drag the plate along, by skin friction in the two sheets, and the size of that force is set by the thin layer at the surface: it scales like the square root of the inverse Reynolds number, which for a garden hose against a paving slab is a per cent or so of the normal force. The model is not claiming that force does not exist. It is claiming that the split is decided before that force gets a say, and it is right about that for exactly the same reason that a wing’s lift is decided by the inviscid flow while its drag is not.

The factor of two, which is where the money is #

The force a jet exerts is the change in its momentum, so what matters is not that it arrives but where it goes afterwards. A flat plate takes away the axial momentum and leaves the water travelling sideways: force ṁV. A cup that turns the jet right round takes away twice as much, because the water leaves going backwards: force 2ṁV.

Same jet, same speed, same flow rate, twice the force — and nothing about the plate has changed except the shape of its edges.

The general expression is ṁV(1 − cos θ) for a jet deflected through θ, and it is worth noticing what it does at the ends. At θ = 0 nothing has happened and the force is zero. At θ = π the cosine is −1 and the factor is two. Between them the curve is flat near the top, which is why a bucket that turns the flow through 165° rather than 180° gives away so little: the missing fifteen degrees cost (1 − cos 165°)/2 = 0.983 of the ideal, and the reason to give them away is that a bucket which reverses the jet completely sends the water straight back into the oncoming jet and into the bucket behind it.

That is a real engineering constraint arriving out of a momentum balance rather than out of a handbook, and it is the same shape of finding as the sailing polar’s optimum heading: the physics gives a maximum, the geometry of the machine forbids reaching it, and the cost of the compromise is computable.

Two routes to one force #

The normal force can be written down directly — the incoming momentum’s normal component, with nothing leaving normal to the plate, is ṁV sin β — or it can be assembled from the split by summing the vector momentum fluxes of the two sheets and the incoming jet and taking the component perpendicular to the surface.

The two share no algebra. The first never mentions the split; the second is built entirely out of it. They agree to 10⁻¹⁶ at every angle, which is the site’s usual standard for a check of this kind: compute the same number twice along routes that share no arithmetic, and where the two are supposed to differ, the size of the difference is the physics.

Here they are supposed to agree exactly, and a discrepancy would mean the split was wrong.

Where the argument fails, and what fails with it #

The model has one assumption doing all the work and it is worth saying where that assumption breaks.

A jet that does not stay a sheet. The derivation assumes the water leaves as two coherent sheets along the surface. A jet striking a small plate splashes: it leaves as a spray in every direction, the momentum flux out is no longer confined to the plate’s plane, and the force falls below ṁV because some of the departing water is still travelling forwards. The criterion is a ratio of the plate’s size to the jet’s diameter, and this site computes neither — the assumption is stated and the number that would test it is not available here.

A moving plate that is not moving steadily. Everything above is a steady-flow balance in the frame of the plate. That frame is only inertial if the plate’s speed is constant, which is fine for a bucket at design speed and false during a start-up, and nothing in this essay applies to the second case.

Compressibility, which arrives sooner than expected. The speed is unchanged around the plate because the pressure at the free surface is, and that reasoning needs the density to be constant. For a gas jet at any appreciable Mach number the sheets do not leave at the jet speed and the split is different — the relevant machinery is in the compressible field and not here.

What this has to do with the wing #

The refutation index of this site carries an essay about air being pushed down, whose complaint is that the popular momentum account of lift names the right mechanism and does the accounting badly — it counts the momentum flux through some surface and forgets the pressure acting on the rest of the control volume.

This essay is the case where that accounting is easy, and putting the two side by side is the point of having both. Here the control volume can be closed exactly, because the boundaries of the flow are all either free surfaces at a known pressure or the plate itself, and the momentum sum has nothing hidden in it. There the boundaries are at infinity in every direction, the pressure on them decays slowly, and an account that stops at the momentum flux through a plane below the wing has thrown away a term of the same size as the one it kept.

A momentum argument is only as good as the surface it is drawn on. That is the lesson this field keeps making in different clothes, and a jet on a plate is the cleanest possible statement of it.

The same balance, at a grazing angle #

The lopsided end of the sweep is worth a picture of its own, because it is where the result stops matching anybody’s intuition about what a jet does.

Two things follow from that arithmetic, and both are used elsewhere on this site.

A deflector is cheap and a stopper is expensive. Turning a stream aside costs a force proportional to sin β; stopping it costs the whole momentum flux. That is why a splitter plate in a duct can be thin and a blast wall cannot, and it is the same trade a yacht’s sails make against its keel — a small force applied athwart a large momentum flux, rather than a large force applied against it.

The split is where the plate’s pressure distribution is decided. The dividing streamline lands at the stagnation point, and everything upstream of it goes one way; the pressure peak sits there and falls to ambient at both ends of the wetted region. This essay does not compute that distribution — Rayleigh’s free-streamline solution does — but the integral of it is fixed by the momentum balance above, which is why the force can be known exactly while the pressure profile is not.

What happens to the sheets afterwards #

The control volume stops at the edge of the plate, and what the water does past that point is worth a paragraph because it is where every real application lives.

Both sheets leave the plate travelling along it at the jet’s own speed. In the absence of gravity, surface tension and air they would continue forever as flat films of constant thickness. Real sheets do none of those things: they thin, they develop a rim held together by surface tension, and they break into drops at a distance set by the competition between inertia and tension — a Weber number, which is not a quantity this site computes.

None of that changes anything above. The momentum balance is taken over a control volume that already contains the whole of the wetted surface, so what happens beyond it cannot reach back into the force on the plate. That is a general and slightly surprising property of momentum arguments: the force is decided by what crosses the faces, and the faces can be drawn before the interesting physics starts.

The one exception is when the departing water comes back. A jet in a confined space, a bucket in a rotating wheel, a spray in a chamber: in each of those the sheets can strike another surface, and the control volume drawn round one plate is no longer closed. That is exactly why a Pelton bucket is split down the middle and turned through 165° rather than 180°, and it is the subject of the rung after next.

Who found it, and when #

The result is Rayleigh’s, in the free-streamline tradition his 1876 paper extended from Kirchhoff’s slot. The full inviscid problem of a jet striking an inclined plate has a closed-form solution by the same hodograph method that gives the contraction coefficient, and it yields the shape of both sheets, the position of the dividing streamline and the point on the plate where the pressure peaks.

The mass split does not need any of it. That is what makes it worth teaching: the hodograph solution is beautiful and hard, and the one number a designer actually wants — how much water goes each way — is available from a control volume in two lines, because the tangential force is zero and there is nothing else the momentum can do.

How far away the plate is, which turns out not to matter #

Nothing in the arithmetic says how far the plate stands from the nozzle, and the omission is a result rather than a gap.

A free jet issuing into still fluid conserves its momentum flux — that is the one thing a jet keeps, and it holds because the surrounding fluid is at rest and at uniform pressure, so nothing pushes on the jet at all. Everything else about the jet changes: it slows, it spreads, and it collects fluid it did not start with, so that a few diameters downstream most of what is moving was never in the nozzle. The product $\rho\!\int\! u^2\,\mathrm dA$ does not move.

So a plate placed at ten nozzle diameters and one at a hundred feel the same force, provided each catches the whole jet. That is a strange sentence and it is exact. At the far station the water arrives much slower and there is a great deal more of it, and the two changes compensate precisely, because the quantity that was conserved is the one the force is made of.

The proviso is the only thing that varies with distance, and it varies fast. The jet spreads roughly linearly, so a plate that catches all of it at ten diameters is small and one at a hundred must be ten times wider. Below that size the water passing outside the plate is still travelling forwards, and the force falls — which is the splashing failure listed above, arriving as a statement about standoff rather than about plate size.

The same observation read from the other end is a familiar one. The reaction on the nozzle is the same momentum flux, so the thrust a jet produces and the force it can deliver to a target are the same number, at any separation. A fire monitor pushes back on its own mounting exactly as hard as it pushes on the wall it is aimed at, and neither force knows how far apart they are.

Why the plate’s own shape does not appear #

One absence in the arithmetic deserves comment, because it is the kind of thing that looks like an oversight and is a result.

Nowhere above does the plate’s size, thickness, curvature or finish appear. The force is ṁV sin β and the split is (1 + cos β)/2 for a plate of any extent, provided only that both sheets leave along it. That is a strong claim and it is exactly what a control volume is entitled to say: the box was drawn round the wetted region and everything crossing its faces was written down, so anything the plate does inside the box — a curved surface, a rough one, a stagnation region of one shape or another — cannot change the sum.

Where the claim fails is where the premise does. A plate too small to turn the whole jet loses some of the water past its edges still travelling forwards, and the force falls. A plate curved enough to turn the sheets back on themselves is a bucket rather than a plate, which is the case the next rung starts from. And a plate with a hole in it is not a plate at all.

Reading the assumptions as the specification rather than as fine print is the useful habit here: the model does not require a large plate because large plates are convenient, but because “both sheets leave along the surface” is the sentence the whole computation rests on.

Where the ladder goes next #

Let the plate run away from the jet. The force falls, because the relative speed falls, but the plate is now moving and the product of the two is a power — and the speed at which that product is largest is exactly half the jet speed, for every bucket shape and every flow rate. That machine is a Pelton wheel, and at its optimum the water leaves it stationary.