The hole that halves the flow

Water leaving a hole in the side of a tank does not fill the hole. A few diameters downstream the jet is visibly narrower than the opening it came out of, and the ratio of the two areas — the contraction coefficient — is what stands between a hole’s dimensions and the flow through it.

Every handbook prints the number. For a sharp-edged orifice it is about 0.61, for a rounded entry it is 1.0, and the tables carry a dozen more with a decimal place each and no derivation anywhere. The impression a reader takes away is that this is a measured quantity, like a friction factor or a drag coefficient: something the flow does, too complicated to predict, and settled by experiment.

It is not. Two of those numbers are exact, and they are exact for completely different reasons.

Why the jet contracts at all #

The reason is geometric and it is worth stating before any arithmetic.

Fluid arrives at the hole from every direction — from straight ahead, from the sides, from behind the plate. A particle approaching the edge is travelling almost parallel to the wall when it gets there, and it cannot turn a sharp corner: turning requires a centripetal acceleration, that acceleration requires a pressure gradient, and an infinitely sharp turn would require an infinite one. So the fluid leaving near the edge is still moving inwards as it crosses the plane of the hole, and the jet goes on narrowing until every particle in it is going the same way.

That is a mechanism, not a number. Getting a number out of it needs the flow solved, and the flow has a feature no other field on this site has had: its boundary is not given.

Every solved field in the previous nine fields had its shape written down in advance. A cylinder is a circle. A Joukowski aerofoil is the image of a circle under a map. A channel has walls where the drawing says it has walls. The surface of a jet is none of these — it is where the pressure is ambient, and where that surface goes is part of what has to be solved for. A problem whose domain is unknown until the solution is known is a different kind of problem, and it is why this coefficient took Helmholtz and Kirchhoff rather than a handbook.

The mouthpiece that dodges the whole difficulty #

Borda’s answer is to choose a geometry in which the flow need not be solved at all.

Take a tube that projects into the vessel rather than sitting flush with its wall. The fluid in the annulus between the tube and the wall is very nearly at rest: it is a dead pocket, far from anything moving. So the pressure over that whole region — over the entire inner wall of the vessel, in fact, except for the opening itself — is the undisturbed hydrostatic ρgH. That is the trick, and it is the applied field’s method in its purest form: the control volume has been drawn so that everything crossing its faces is known without looking inside.

The horizontal force on the vessel is therefore ρgH·A, with A the tube’s bore — the whole wall pushes one way and the missing patch at the hole means the far side pushes back by that much less. The jet carries momentum away at a rate ρ(C_cAV)V, with V = √(2gH) from Bernoulli’s equation applied where it holds, so the momentum flux is 2ρgH·C_cA. Momentum is conserved. Setting the two equal:

$$\rho g H A = 2\rho g H\, C_c A \quad\Longrightarrow\quad C_c = \tfrac{1}{2}$$

Everything cancels. The head cancels, the area cancels, the density cancels, gravity cancels. The answer is a half exactly, at any scale, in any liquid, under any head — and the site’s own solver returns 0.500000000000000 over four decades of each, which is the check that the algebra above has no hidden dependence in it.

There is no flow field anywhere in that argument. No streamline is drawn, no velocity is computed except at the exit, and the enormous difficulty of the free surface has been sidestepped by putting the hole somewhere the pressure is known. That is exactly the move Betz’s limit makes, and the sudden-expansion loss, and Euler’s turbomachinery equation: draw the box somewhere the faces are simple and refuse to look inside.

Kirchhoff’s slot, where the surface is solved for #

The ordinary case — a hole in a flat wall, with nothing projecting — cannot be dodged that way. The fluid beside the opening is moving, its pressure is not hydrostatic, and there is no face of the control volume where the answer is already known.

Helmholtz and Kirchhoff solved it in the 1860s by a change of variable that is still the standard tool. Instead of working in the plane of the flow, where the boundary is unknown, work in the plane of the velocity, where it is entirely known: on the free surface the speed is the jet speed (because the pressure there is ambient and Bernoulli allows nothing else), and on the wall the direction is the wall’s direction. A boundary that is impossible to draw in one plane is two straight lines and a circular arc in the other.

The whole solution for a two-dimensional slot fits on one line:

$$z(\zeta) = -\frac{2b}{\pi}\left[\frac{1}{\zeta} + \log\frac{\zeta-1}{\zeta+1}\right] + 2bi, \qquad \frac{dw}{dz} = U\zeta$$

on the quarter disc |ζ| ≤ 1 with ζ in the first quadrant. The parameter ζ is the velocity, in units of the jet speed: its modulus is the speed ratio and its argument carries the direction. The quarter circle is the free surface, the imaginary radius is the wall, and the real radius is the symmetry line.

The contraction coefficient falls out of the geometry. The wall’s own equation, read off the map, is y = (2b/π)(1/s + 2 arctan s) with s the speed ratio there; at the edge s = 1, so the edge sits at y = b(π+2)/π. The jet’s half-width is b. Therefore

$$C_c = \frac{\pi}{\pi+2} = 0.611015\ldots$$

and that number is a consequence of the conformal map rather than an input to it. The site’s solver finds the edge two ways — from the wall equation and from the free streamline’s own endpoint — and they agree to 2·10⁻¹⁶.

The momentum theorem, applied to a solution that never heard of it #

The map is a statement about complex analysis. Whether the flow it describes conserves momentum is a completely separate question, and asking it is a real test rather than a restatement.

The control volume is the jet itself: in through the plane of the slot, out through a downstream section where the flow is parallel, with the free surfaces as sides. The sides carry no flux and no excess pressure, so

$$\rho(2b)U^2 - \int \rho u^2\,dy = \int (p - p_\infty)\,dy$$

both integrals over the slot. The left-hand side needs the velocity profile across the plane of the hole, which the closed forms do not hand over: that plane is an interior line of the flow, so its image in the parameter disc has to be found by inverting the map numerically, point by point.

That profile is the mechanism made quantitative. Most of what leaves the hole is not yet going the way the jet goes, so the jet has to keep narrowing until it is. And with it in hand the momentum balance can be closed:

Two independent statements come out of that, and they are worth separating. Mass conservation says the profile carries the right total; momentum conservation says it is distributed correctly. A profile could satisfy the first and fail the second, and a solution that failed either would be a beautiful picture of a flow that cannot exist.

The third case, where the coefficient is one #

Rounding the entry removes the contraction entirely. There is no edge for a streamline to turn round, so nothing is still moving inwards at the throat, and the jet fills the opening: C_c = 1.

This is not a small effect being tidied up. Between Borda’s half and a bell-mouth’s one there is a factor of two in the flow rate, at the same head, through the same bore. The shape of the last millimetre of the hole decides how much water comes out, and it decides it more strongly than anything else in the arrangement — not the length of the pipe, not the roughness, not the viscosity.

What the picture cannot show #

Three things, and the first is the one a reader is most likely to assume away.

There is no viscosity in any of this. The contraction is an inviscid effect, produced entirely by the inability of streamlines to turn a corner, and this site’s own habit of asking which regime a result belongs to gives an unusual answer here: any of them. The coefficient is the same in treacle and in air, provided only that the boundary layer on the plate is thin compared with the hole. That last proviso fails for a small hole in a thick plate, where the flow is dominated by the passage rather than by the edge, and the tables’ fourth decimal place is largely a record of that failure.

The free surface in the solution is exact and the free surface in a real jet is not. Surface tension rounds the edge over a length of a millimetre or so in water; gravity bends the jet down; and a few diameters out the surface breaks into drops, which is a different subject entirely. The solved jet is a two-dimensional, weightless, tensionless idealisation, and its virtue is that the contraction is a property of the corner rather than of any of those.

The 0.611 is two-dimensional. A real orifice is round, and the same hodograph argument does not separate for an axisymmetric flow: there is no closed form, and the measured coefficient of a sharp-edged round hole is 0.61 to 0.62. That the two agree to within a per cent is a fact and not a proof, and every figure here that uses π/(π+2) for a round hole says so. It is a better position than a table lookup — the sensitivity of everything downstream to the coefficient is computable, and it is small — but it is not the same as a derivation.

The hole that gets more flow by being made longer #

The proviso above — that the plate be thin — hides a result worth having on its own, because it runs against every instinct about adding pipework to a system.

Fit a short tube to the hole, a couple of diameters long. The jet contracts inside it exactly as before, reaching a vena contracta a fraction of a diameter in; then, with nowhere else to go, it expands again and reattaches to the tube’s wall, so that what leaves the far end fills the bore. The contraction coefficient at the exit is therefore one, not 0.61.

The flow does not rise by the full factor, because the internal re-expansion is a sudden enlargement and costs the Borda–Carnot head — which is what a jet expanding into a larger passage always costs, whether the passage is a pipe or a tube. That loss shows up as a velocity coefficient near 0.8 instead of the thin plate’s 0.98. Multiplying the two,

$$C_d \approx 1.0 \times 0.8 = 0.8$$

against the thin plate’s $0.61 \times 0.98 = 0.60$.

So adding a stub of pipe to a hole increases the discharge by about a third, and it does so by introducing a loss. That is a genuinely strange sentence and both halves of it are right: the tube throws away head at the internal expansion and recovers more than it spends by removing the contraction at the exit.

Two further consequences follow and both are practical.

The pressure inside the tube is below ambient. At the internal vena contracta the flow is fastest, so by Bernoulli its pressure is lowest — measurably below atmospheric, which is why a tube of this kind can draw air or liquid in through a small side hole, and why under a large head it will cavitate there rather than anywhere else.

And it is bistable. The reattachment needs the tube to be long enough and needs the low-pressure pocket to stay sealed. Make the tube too short, or admit air to the annulus, and the jet does not reattach at all: the coefficient falls back towards 0.6 and stays there. A tube run under the same head can therefore deliver two different flow rates, and which one depends on how it was started.

What a contraction coefficient is not #

Two neighbouring quantities are often confused with it and the difference is worth fixing, because they appear together in every meter’s specification.

The velocity coefficient is the ratio of the actual jet speed to Torricelli’s √(2gH). For a sharp-edged hole in a thin plate it is 0.97 to 0.99: the jet leaves very nearly at the ideal speed, because almost nothing has had a chance to rub against anything. It is a viscous quantity and it is close to one.

The contraction coefficient is the ratio of areas and is the subject of this essay. It is a purely geometric quantity, it has nothing to do with viscosity, and it is nowhere near one.

The discharge coefficient is the product of the two, which is what a flow measurement actually needs, and it is close to 0.6 because the contraction dominates and the velocity coefficient is nearly unity.

Keeping them apart matters because they behave completely differently. Polishing a plate, changing the fluid or raising the Reynolds number moves the velocity coefficient a little and the contraction coefficient not at all. Rounding the edge by half a millimetre leaves the velocity coefficient untouched and moves the contraction coefficient from 0.61 to 1.0.

The lever that matters is geometric and the one that feels like craftsmanship is not, which is the opposite of what an intuition about machined parts would suggest.

Where the number goes next #

A contraction coefficient looks like a curiosity until it is noticed that half the instruments in a process plant depend on one. An orifice plate infers a flow rate from a pressure difference, and the inference uses the area of the jet rather than the area of the hole; the discharge coefficient a meter is calibrated with is this number wearing a different hat, and the permanent pressure loss such a meter causes is the Borda–Carnot head of the jet expanding back into the pipe afterwards.

The same coefficient governs a tank draining, a nozzle’s effective throat, the flow through a perforated plate, and — with the sign of everything reversed — how much of a diffuser’s area a separated flow actually uses.

Who found it, and when #

Jean-Charles de Borda described the re-entrant mouthpiece in 1766, in the same decade and the same tradition as the Borda–Carnot loss, and for the same reason: he was looking for arrangements in which a control volume could be closed exactly. Helmholtz introduced free-streamline surfaces in 1868, Kirchhoff turned them into a method in 1869, and the π/(π+2) for a slot in a plane wall is his. Rayleigh generalised the method to a plate at an angle in 1876, which is the ancestor of the jet-splitting result in the next rung of this ladder.

What survives from all of it is a habit rather than a formula. Where the boundary of a flow is a pressure rather than a wall, the shape of that boundary is part of the unknown — and the two ways to proceed are to choose a geometry where the answer does not depend on it, as Borda did, or to change variables until the unknown boundary becomes a known one, as Kirchhoff did. Neither is a measurement, and the number in the handbook is a theorem with its provenance mislaid.

Where the ladder goes next #

The jet exists now, and it is going somewhere. The next rung puts a plate in front of it and asks how the water divides — a question whose answer follows from a single statement about what an inviscid fluid cannot do — and the one after that lets the plate run away from the jet, which is a Pelton wheel and is the only turbine on this site whose efficiency reaches one.