Air at a valve softens the hammer only in quantity

Twice the margin, on top of the hammer shut a valve on a water main fast enough for the pressure behind it to fall to the vapour pressure, let a cavity open there, and followed it exactly. The result was a staircase. While the cavity lives, the valve can do only one thing to an arriving wave — hold the head at the vapour value — so each round trip of the pressure wave adds the same step to the velocity of the water rushing back, and when the cavity finally closes, the first pulse can stand as much as twice the margin to vapour pressure above the Joukowsky rise. The tallest of those pulses were the briefest, and later pulses in the frictionless line climbed without limit, because nothing in the model could round a sharp front.

Its last section named what a real line puts in the cavity besides vapour. Water carries dissolved air, and some of it comes out wherever the pressure falls, the same fall that tears the water round a fast body; a line also traps free air at high points and valves. A pocket of gas does not hold the head at one value. Its pressure rises as it is squeezed and falls as it expands, so the valve’s rule stops being a fixed head and becomes a spring. This essay puts a pocket of free gas at the closed valve and asks what the spring does to the staircase. The expected answer is that gas cushions the hammer, and in the end it does; on the way there it does something else first.

A pocket is a spring with a time of its own #

The line is the one the column-separation essay used: 600 m of 100 mm steel pipe, a wave speed of 1,200 m/s, a reservoir at one end and a valve at the other, a flow of half a metre per second shut off instantly, and no friction. Heads are measured above the steady head in Joukowsky rises, 61.2 m here; time in round trips of the wave, one second; and the margin $\varepsilon$ is the distance from the steady head down to the vapour head, also in rises — 0.574 for this main.

At the valve sits a pocket of gas. Its pressure is the absolute pressure there less the vapour pressure, it follows Boyle’s law — isothermal, since a small pocket in water exchanges heat with it fast — and at the steady head it holds a volume $v_0$. The pocket’s volume changes at the rate the water arrives at the valve, and the head at the valve is whatever the gas’s pressure is. With the pressure waves travelling up and down the pipe unchanged and the reservoir sending back what left the valve one round trip earlier, the whole problem becomes one equation for the pocket’s volume, a delay equation in which the present depends on the head at the valve exactly one round trip ago.

Two limits check it. When the pocket is vanishingly small, its pressure is enormous whenever it is squeezed and near the vapour value whenever it has grown, which is the vapour cavity’s rule — and the computed first pulse matches the exact vapour solution to two parts in a hundred thousand. When the margin is large, the pocket never grows much and behaves as a linear spring whose stiffness is the margin over the volume; its spring time is then $v_0/\varepsilon$ round trips, the time it takes the arriving water to squeeze it appreciably. That time is the one number the essay turns on. A pocket responds to anything slower than its spring time and cannot respond to anything faster — the same distinction between a threshold and a duration that a cavitation pressure turned out to need.

A useful unit for the pocket is its share of the pipe’s volume. The main holds 4.7 cubic metres; a hundred-thousandth of that is 47 cubic centimetres, a bubble 4.5 cm across, and its spring time at this main’s margin is 21 milliseconds.

Gas that cannot keep up #

The first figure lowers the main’s steady head to 19.9 m, a margin of 0.49 rises. That places the vapour cavity’s collapse at the worst moment in the staircase: the cavity lives for two round trips, closes just after the third begins, and the pulse that follows reaches 1.94 rises — 118 m above the steady head — for 28 milliseconds. The vapour history is the faint square wave: the Joukowsky rise, the drop to the vapour head, two round trips held there, the collapse and a step to 0.96 that lasts most of a round trip, and then the tall narrow pulse.

A millionth of gas, with a spring time of 2.5 ms, follows the square wave almost exactly and takes the top off the pulse, to 1.83. A hundred-thousandth, with a spring time of 25 ms — about the pulse’s own length — takes it to 1.62, a third of the excess gone. A ten-thousandth, 245 ms, rounds every corner of the history, slows the collapse and lets the pulse arrive a round trip later at 1.23. Each pocket is invisible during the long flat parts of the history, where nothing is changing faster than it can follow; it shows only at the fronts.

The long pulse keeps itself #

The second figure is the main at its ordinary steady head. Its vapour pulse is lower and far longer: 1.30 rises, lasting 0.59 of a round trip. A millionth of gas changes it by one per cent. A hundred-thousandth lowers it to 1.18, which is a little over a third of the excess, much as for the brief pulse. The ten-thousandth does something the first figure did not suggest: the pulse arrives later and is taller than the vapour pulse, 1.34 rises.

The reason is timing. A vapour cavity closes at a moment fixed by the staircase: the round trip in which the returning water fills the volume the cavity has grown. A pocket of gas has its own volume to fill besides, and it keeps pushing back as it is squeezed, so the collapse is delayed and falls against a different part of the arriving wave. The column-separation essay’s sawtooth came precisely from where in the round trip the collapse fell, so moving it moves the pulse up as well as down.

Length decides what survives #

The third figure gathers the comparison from nine margins and five gas fractions, from a hundred-millionth of the pipe to three hundred-thousandths. Each point is one pair: across, how many of the pocket’s spring times the vapour pulse lasts; up, what share of its excess over the Joukowsky rise is left once the pocket is there.

The order is clear even though the points do not lie on one curve. A pulse thousands of spring times long keeps all of its excess. At a few hundred, a few per cent goes. At ten spring times a quarter to a half goes, and at one or two about half. The scatter is the timing effect again: the pocket shifts when the cavity closes by an amount that depends on the margin as well as the gas, and that shift moves the pulse in either direction. What the length sets is how much of the pulse the pocket can remove; what the timing sets is where the remainder lands.

That is the first half of the answer, and it is the half the usual advice has in mind. A little gas does soften the hammer, but only its tallest and briefest pulses — and the tallest pulses of the staircase are exactly the briefest, so those are the ones it takes.

The teeth move as well as shrink #

The fourth figure is the column-separation essay’s sawtooth, with gas. The vapour pulse against the margin is exactly $2\varepsilon\lceil 1/\varepsilon\rceil - 1$: it rises linearly within each band of margins and drops sharply wherever $1/\varepsilon$ passes a whole number, because the cavity then closes in one round trip fewer. The tallest teeth sit just short of each drop.

A millionth of gas trims the tips of the tallest teeth. A hundred-thousandth takes about a third off each tall tooth and moves the teeth to larger margins — the delayed collapse means that a margin just past a drop, where the vapour pulse is at its lowest, 1.04 rises, now gives a tall pulse, 1.37 at a margin of 0.34 and 1.58 at 0.51. A ten-thousandth leaves no teeth: every margin between a quarter and one gives a pulse between 1.05 and 1.45 rises.

So the claim that gas lowers the pulse is true margin by margin only for the smallest amounts. For a hundred-thousandth it is true for the margins that had the tallest vapour pulses and false for the ones just beside them. A pipeline designer who has sized a line to sit in a trough of the vapour sawtooth — which is easy to do by accident, since the troughs are where the staircase is gentlest — finds the gas has moved a tooth onto it.

The same spring at the scale of a bubble #

The pocket’s two behaviours have a small-scale twin. The bubble that hammers collapses under the liquid’s pressure with nothing inside it to resist, and its wall arrives at the centre at a speed without bound. Put gas inside the same bubble and the collapse stops short: the gas is squeezed until its pressure turns the wall round, and the bubble rebounds instead of closing. How hard the rebound is depends on how much gas there is, in the same way as here. A trace of gas cushions the very last instant of the collapse, which is the briefest and most violent part; more gas stops it sooner and sends it back out, so that a gas bubble oscillates where a vapour bubble would have hammered. The water column in a pipe is the bubble’s wall drawn out to 600 metres, and the gas at the valve is what is left in the bubble.

A notch that comes back as a spike #

The fifth figure removes the cavity altogether. With a margin of 1.1 rises the head never falls to the vapour head, and a rigid valve gives the textbook square wave: plus one rise, minus one, plus one, for ever. Put a ten-millionth of the pipe’s volume in gas at the valve — a bubble under a centimetre across — and at three round trips the head spikes to 2.3 rises.

It is not a numerical artefact, and it is not cavitation. When a step in pressure arrives at a pocket, the pocket first yields — for the first fraction of its spring time the valve behaves as an open end and reflects the step inverted — and then stiffens and reflects it as a closed end. The reflected wave therefore carries a brief notch at its front. The notch travels to the reservoir, where the fixed head turns it over, and returns to the valve one round trip later as a spike riding on the next front. With a larger pocket the notch is longer and the spike broader: a ten-thousandth of gas on this line gives a pulse of 1.56 rises lasting more than a third of a round trip.

A trapped air pocket that raises pressures above the Joukowsky value is a hazard pipeline engineers know from measurement: small pockets at dead ends and closed valves are recorded producing peaks well above what the same line without air produces. The calculation here shows one way it happens, with nothing but a lossless line and a spring.

How much gas it takes to remove the hammer #

The sixth figure sweeps the gas across six decades for the main at its own steady head. Up to a millionth of the pipe’s volume, nothing changes: the event is the vapour cavity’s. At a hundred-thousandth the largest head falls to 1.18 rises. At a ten-thousandth it rises again, to 1.34, above the vapour value. Only past about half a thousandth of the pipe’s volume does the largest head fall below the Joukowsky rise — 0.79 at a thousandth — and there the event has changed character: the pocket is soft enough that the whole column rocks on it like a mass on a spring, taking several round trips over each swing, and the lowest head stays well clear of the vapour head — the head never has to fall to where a liquid does pull.

A thousandth of this main is 4.7 litres of gas at the steady head. That is not a trace of dissolved air: water at room temperature, saturated at atmospheric pressure, holds about a fiftieth of its volume in air, and releasing a twentieth of all of it within a second is more than the pressure drops in a hammer event can drive. It is instead the kind of gas volume put there on purpose. An air vessel — a closed tank with a cushion of air, fitted beside a pump or a valve — is exactly this pocket made large deliberately, and it works for the reason the right-hand end of the figure shows: the column swings slowly against a soft spring instead of being stopped against a hard one. It is the gas-filled cousin of the open surge tank that turns a hammer into a swing — whose sizing, the better tunnel needs the bigger tank showed, is set by the same slow swing — and the same air chamber sits on the hydraulic ram to smooth its delivery.

What was checked #

The seventh figure is the ledger. The delay equation is integrated by the trapezoidal rule with sub-steps sized to the pocket’s stiffness, and it passes four checks. With a pocket of a millionth of a round trip’s delivery, the first pulse matches the exact vapour-cavity pulse at three margins to two parts in a hundred thousand. The line and the pocket together conserve energy — the wave energy in the pipe plus the work stored in the gas — to one part in ten thousand over ten round trips. The first round trip, during which the arriving wave is still the steady one, has an exact solution for the pocket’s compression, and the integration matches it to a millionth of a round trip. And halving the step moves the first pulse by a part in a million.

One trap is worth recording. The trapezoidal rule is stable for any step but not damping: with a step much longer than the spring time it rings about the right answer instead of settling on it, and the ringing conserves energy, so the energy check cannot see it. The exact first-round-trip solution can, and the sub-steps are what it required.

What the lossless line exaggerates #

Sharp features. Every notch and spike here is as sharp as the pocket’s spring time and persists for ever, because the line has no friction, no unsteady friction and no wall damping. Measured traces round such features within a few round trips, and a spike a millisecond long in the model is lower and wider in a real pipe.

One pocket, at the valve. Gas released along the pipe, wherever the pressure falls, lowers the wave speed there and spreads the fronts everywhere, which is the other thing measured traces show and which a single pocket cannot.

Isothermal gas. A large pocket squeezed in milliseconds heats and behaves more nearly adiabatically, which stiffens it; small pockets are closer to isothermal. The exponent changes the spring time by at most the ratio of specific heats.

Gas that stays. The pocket’s mass is fixed. The gas coming out of solution as the pressure falls, and going back as it rises, is not modelled, and it is the part a real line adds most of.

The convention: gas measured against the pipe #

Heads are above the steady head, in Joukowsky rises $aV_0/g$; time is in round trips $2L/a$. The margin $\varepsilon$ is the steady head less the vapour head, in rises. The gas fraction is the pocket’s volume at the steady head divided by the pipe’s volume; in the equation it appears as $v_0 = \alpha a/2V_0$, the pocket’s volume in units of the water delivered in one round trip, and the spring time is $v_0/\varepsilon$ round trips.

Who worked it out #

Column separation with gas was taken up in the 1970s and 80s, once computed vapour-cavity models were seen to produce spikes that measured traces did not. Martin’s studies of entrapped air in pipelines in the 1970s documented that small air pockets can raise transient pressures rather than lower them. Wylie in 1984 introduced the discrete gas cavity model, which places a small free-gas volume at each computational node and is still the standard way of adding gas to the method of characteristics; Bergant, Simpson and Tijsseling’s review of 2006 traces the whole subject from Joukowsky onwards.

Still open: the gas the pressure drop releases #

The pocket here has a fixed amount of gas, and the question the calculation leaves is where that amount comes from. Dissolved air leaves solution at a rate set by the supersaturation and by how much bubble surface there is to leave through, and it redissolves more slowly than it comes out. The next calculation gives the pocket a release rate — proportional to how far the pressure has fallen below the saturation pressure of the dissolved air, with a slower rate of return — and asks how much gas a single cavity phase lasting two round trips actually liberates, whether it is enough to reach the hundred-thousandth that trims the tall pulses, and whether the gas left after the first collapse makes the second pulse lower or, by the timing effect, taller.