The siphon that does not break

The first rung of this ladder establishes a coupling that is not obvious and is exact: lowering the outlet costs hump height, one metre for one. Running a siphon faster makes it break sooner, because the velocity head at the crown comes out of the same atmosphere that is holding the column up.

It also excludes, by name, the case every real siphon is in:

Nothing unsteady. A siphon whose source level is falling is unsteady, its flow rate declines as the drop shrinks, and the quasi-steady treatment used here is an approximation whose error nobody has computed on this site.

A falling source does two things at once. It shrinks the drop — the same variable whose increase was just shown to eat the crown’s margin — and it grows the rise, which eats the margin directly. Both push the same way. The expectation is a siphon that gets closer to breaking as it empties, and a designer who has read the previous rung has every reason to hold it.

The cancellation #

The arithmetic is four lines and it has no solver in it.

The pressure at the crown of a running siphon, with the source surface at height $z$ and the outlet fixed at $z_{\text{exit}}$, is atmospheric less the hydrostatic rise less the velocity head:

$$p_{\text{crown}} = p_{\text{atm}} - \rho g\,(z_{\text{crown}} - z) - \tfrac{1}{2}\rho V^2.$$

The velocity is Torricelli’s, from the current drop:

$$V^2 = 2g\,(z - z_{\text{exit}}).$$

Substitute:

$$p_{\text{crown}} = p_{\text{atm}} - \rho g\,z_{\text{crown}} + \rho g\,z - \rho g\,z + \rho g\,z_{\text{exit}},$$

and the source level cancels identically. What is left is

$$p_{\text{crown}} = p_{\text{atm}} - \rho g\,\left(z_{\text{crown}} - z_{\text{exit}}\right),$$

a constant containing the crown and the outlet and nothing else.

Reading what just happened #

Two effects of equal size in opposite directions is a coincidence until it is explained, and the explanation is one sentence per term.

Losing a metre of source level grows the rise by a metre, which lowers the crown pressure by $\rho g$ per metre — one atmosphere per 10.3 metres of water.

Losing the same metre shrinks the drop by a metre, which reduces $V^2$ by $2g$ per metre, which reduces the velocity head $\tfrac{1}{2}\rho V^2$ by exactly $\rho g$ per metre — the same coefficient, because the half and the two cancel.

They are the same number because both are $\rho g$ times a height, and the height being lost is the same height in both cases. That is why the cancellation is exact rather than approximate: it does not depend on the geometry, the bore, the liquid or the flow rate, and it would survive any change to any of them that left the tube of one section throughout.

Why the first rung’s coupling is not contradicted #

There is an apparent conflict here and resolving it is the useful part.

The first rung says lowering the outlet costs a metre of hump per metre, and this rung says lowering the source costs nothing at all. Both are read off the same expression, which contains $z_{\text{crown}} - z_{\text{exit}}$ and no $z$.

Lowering the outlet increases $z_{\text{crown}} - z_{\text{exit}}$, and the crown pressure falls. Lowering the source changes neither of the two heights in the expression: it moves the reference the crown is measured above and it moves the drop, and those are the two effects that cancel.

So the quantity that matters is the crown’s height above the outlet, not above the source. That is a genuinely different statement from the one every account of a siphon makes, including the first rung’s own figures, which draw the rise from the source. The rise from the source decides the crown pressure only when the flow rate is also known — and once Torricelli is substituted, it disappears.

The three ways a siphon stops #

With breaking removed from the list for any siphon that started, what remains is geometry, and there are exactly three endings.

The level reaches the outlet’s height. There is no drop left, the flow rate is zero, and the siphon stops with the tube still full. Restarting it needs only that the source be topped up, since the prime was never lost.

The level reaches the inlet. The tube swallows air, the column parts, and the siphon stops with liquid still above the outlet’s level in the tank. This is the ending that wastes liquid, and how much is left is the difference between the inlet’s height and the outlet’s.

The column breaks at the crown. This requires $z_{\text{crown}} - z_{\text{exit}}$ to exceed the barometric height — and since that quantity does not change while the tank drains, a siphon that breaks does so at the instant it starts or not at all.

That third statement is worth pausing on because it is a design rule and it is a stronger one than the usual advice. It is not that a siphon should be set up with margin in case it drifts towards its limit; it is that the margin does not drift. A siphon that runs for a minute will run until it runs out of liquid, and any failure after that first minute is a bubble, a leak or a blockage rather than the column reaching its limit.

Where the quasi-steady approximation actually costs something #

The first rung’s exclusion said the error had not been computed, so it is worth saying what the approximation is and where it bites, since the answer is not where a reader would guess.

The treatment above solves each instant as though it were steady: the flow rate is Torricelli’s at the current drop, with no term for the column’s own acceleration. The neglected term is $L\,dV/dt$ over the tube’s length, and it matters when the flow is changing fast compared with the time a pressure signal takes to run the tube.

Over the run it is negligible and at the ends it is not. The tank in the figures empties over an hour — a draining vessel being a wait rather than an event and the velocity changes over that timescale, so the acceleration term is smaller than the terms kept by many orders. At the very start — the moment the outlet is opened — the velocity goes from zero to ten metres a second in the time it takes the column to accelerate, and nothing here describes that at all — an acceleration through a duct being a pressure that depends on the past rather than on the present state. At the very end, as the drop approaches zero, Torricelli’s expression predicts a rate approaching zero and the real column coasts on its own momentum, overshooting slightly.

And the cancellation survives both, which is the point worth carrying. The unsteady term appears in the momentum balance and not in the two heights, so a siphon accelerating or decelerating has a crown pressure differing from the constant above by the acceleration term — which is a transient of a few seconds at the start and nothing thereafter.

The same cancellation, checked a different way #

An algebraic cancellation is worth confirming by a route that does not use the algebra, because the easiest way to produce one is to make an error in the substitution.

Take the energy statement instead. A parcel of liquid at the source surface has pressure energy $p_{\text{atm}}/\rho$ and potential energy $gz$, and it arrives at the crown with pressure energy $p_{\text{crown}}/\rho$, potential $gz_{\text{crown}}$, and kinetic energy $V^2/2$. The sum is conserved, so

$$\frac{p_{\text{atm}}}{\rho} + gz = \frac{p_{\text{crown}}}{\rho} + gz_{\text{crown}} + \frac{V^2}{2}.$$

Now note that $V^2/2 = g(z - z_{\text{exit}})$, which says the kinetic energy at any point in the tube is exactly the potential energy the parcel will lose between the source and the outlet. Substituting gives the same constant, and the reason is now visible in the accounting rather than in the cancellation: the parcel’s kinetic energy at the crown is borrowed against a fall it has not yet taken, and that borrowing is measured from the outlet. The source’s height enters twice, once as where the parcel started and once as where the borrowing began, with opposite signs.

That is a better way to hold it than the algebra. The crown pressure is what is left of the atmosphere after paying for the height above the outlet, because the parcel is on its way there and has already committed the whole fall.

And it says immediately what would break the cancellation. Anything that spends energy between the source and the crown, rather than merely storing it — friction, most obviously — is a payment that is not recovered on the way down, and it enters the crown’s balance without a matching term. That is why the final section of this essay names friction as the thing that breaks this result, and it is why the frictionless case is the one where the result is exact rather than approximate.

What a designer should measure instead #

The practical upshot is a change in what is worth checking, and it is short.

Measure the crown above the outlet, not above the source. That difference is the whole of the column’s margin and it is fixed by the installation. Where the reservoir’s surface happens to be does not enter, so a siphon specified at high water and a siphon specified at low water have the same margin and a specification written against the upstream level is measuring the wrong height.

Then check it once, against the barometric height for the liquid at its working temperature. If the crown-to-outlet difference is under that, the siphon will run to whichever of the two geometric endings comes first. If it is over, the siphon will not start.

And place the inlet as low as the liquid is wanted removed. How much is left in the tank is the inlet’s height above the outlet’s, and nothing else in the arithmetic touches it. That is the one design choice in a siphon that is genuinely free — it costs nothing, it is decided entirely by where the tube’s end is put, and it is the difference between draining a tank and half-draining one.

What the flow rate does instead #

The crown pressure holds still and something else does not, and it is worth drawing the contrast because the two together are what a person watching a siphon actually sees.

The flow rate is $\sqrt{2g\,\Delta z}$ with $\Delta z$ the drop from the current surface to the outlet, so it falls as the square root of the remaining drop — which means a tank empties with a decelerating level and a characteristic long tail. In the run drawn above, the exit speed starts at 10.8 metres a second and finishes at 0.17: a factor of sixty, over the same six metres in which the crown pressure did not move by a nanopascal.

That contrast is the essay in one sentence. The thing that is obviously changing has nothing to do with the failure mode, and the thing that decides the failure mode is not changing at all.

The emptying time follows and has a closed form for a tank of constant plan area. Integrating $A\,dz/dt = -a\sqrt{2g(z - z_{\text{exit}})}$ from the initial level down to the outlet gives a time proportional to the square root of the initial drop and to the ratio of the tank’s area to the tube’s — which is the same integral a tank draining through a hole in its side obeys, with the siphon’s geometry supplying a different drop and nothing else changed.

So a siphon and a hole in the bottom of a tank empty it on the same law, and what a siphon buys is that the hole may be somewhere the tank does not have one — over a wall, out of a sealed vessel, or below a level the tank’s own floor is above.

What the picture cannot show #

No friction. The tube is frictionless throughout, so the flow rates are upper bounds and the emptying times lower ones. On a real hose the friction term dominates — the first rung computes it at twelve times the unity it sits beside — and the emptying time is several times longer. Friction also reduces the velocity at the crown, which raises the margin, and it costs pressure along the ascending leg, which lowers it; the two have opposite signs and neither is here.

No bubble. The margin against the vapour pressure is exactly the cavitation margin the applied field prices, and what decides whether a bubble actually forms is whether the liquid has anything to nucleate on. The failure mode the previous rung identifies as the real enemy of a long siphon — gas coming out of solution at the crown — is not in this model, and it is the thing that actually stops siphons that have run for hours. This essay’s finding is about what the heights do, and it is silent about what the liquid does.

The tank has vertical walls. A tank of varying plan area drains at a rate this integration would get wrong, and the shape enters only through the area.

Only one bore. The cancellation used the fact that the velocity at the crown is the velocity at the outlet, which needs a tube of one section throughout. A siphon with a constriction somewhere has a different velocity at the crown, the two terms no longer carry the same coefficient, and the source level returns to the expression. That is a real device — a siphon with a valve part-closed is one — and this result does not cover it.

And the endings are treated as instantaneous. An inlet being uncovered is not a moment; it is a period of gulping in which air and liquid enter together, and what happens in it decides whether the siphon re-establishes or dies.

The assertion behind the figures is the one that could reject and is the essay’s whole claim: the crown pressure across the entire drain must be constant to machine precision, while the exit velocity must demonstrably fall. Checking only the first would pass a model in which nothing happened at all.

Who noticed, and when #

This particular cancellation does not appear to have a name or a discoverer, which is unsurprising: it falls out of two expressions that have both been standard since Torricelli and Bernoulli, as soon as one is substituted into the other.

What it does have is a history of being got wrong in practice. Siphon spillways and inverted siphons in irrigation systems are routinely specified with the crown height measured above the upstream water level, which is the quantity that varies with the reservoir and which the expression above says is irrelevant. The quantity that matters is the crown above the downstream outlet, and specifying against the wrong one produces designs that appear to have margin at high water and appear to lose it at low water, when in fact nothing has changed.

That is a small error with a large signature: an operator watching a siphon at falling reservoir levels is watching a device whose margin is constant and whose flow rate is falling, and the falling flow rate looks like the onset of trouble.

Where the ladder goes next #

The heights are now settled, and every remaining question about a siphon is about the liquid rather than about the geometry.

The rung above is the bubble. Gas comes out of solution at the crown because the crown is where the pressure is lowest, it accumulates over hours, and it eventually breaks a column that the arithmetic above says can never break. That is a mass-transfer problem — Henry’s law, a diffusion rate, a concentration gradient — with the fluid mechanics supplying only the pressure, and it is the one failure mode of a working siphon that is genuinely unavoidable.

The one beside it is the friction this essay refused. Putting the pipe loss into the quasi-steady integration turns the emptying time from an analytic result into a numerical one, changes the shape of the level-against-time curve, and — the interesting part — breaks the cancellation, because the friction term depends on the velocity and therefore on the drop, and it enters the crown pressure with a coefficient that is not $\rho g$. How much it breaks it, and in which direction, is a rung with a computation in it.