Twice the margin, on top of the hammer

Stopping water costs more than moving it computed the square wave a valve makes when it shuts on a moving column: a rise of $aV_0/g$ at once, a fall of the same size one round trip later, and so on for as long as nothing takes the energy away. It drew the vapour pressure on that wave from outside, as a line the computation ran straight through, and said in words what the water does instead: the column separates, a cavity opens, and what happens when it closes is worse than the hammer that started it. The march through a slow closure left the same event outside its model for the same reason.

That sentence is true and it was not computed. This essay computes it, and the computation says three things the sentence does not. The pulse after the collapse can pass the closure’s own rise by up to twice the margin the line had to vapour pressure. Whether it does depends on how many round trips of the wave the cavity lasts, which makes it a sawtooth in the flow velocity rather than a curve. And the tallest of those pulses are also the briefest, which is why they are both dangerous and hard to measure.

The line is ordinary: 600 metres of 100-millimetre steel pipe, a wave speed of 1200 metres a second, so a round trip of exactly one second, and a reservoir holding 25 metres of head at the upstream end. The water at the valve boils at 10.1 metres of head below the atmosphere.

A closed valve can do only two things #

The head at the valve in that figure is a step function, and it is a step function exactly, not approximately. The reason is the one the characteristics essay gives for a gas: in a frictionless pipe two combinations of head and velocity are carried unchanged along the two families of waves. Measure the head from its steady value in units of the Joukowsky rise $J = aV_0/g$, the velocity in units of $V_0$ and the time in round trips $2L/a$. Then

$$I^{+} = \eta + u \quad\text{arrives at the valve},\qquad I^{-} = \eta - u \quad\text{leaves it},$$

and neither changes on the way. The reservoir holds $\eta = 0$, so whatever leaves the valve comes back one round trip later with its sign reversed: $I^{+}(t) = -I^{-}(t-1)$.

At the valve there are exactly two rules. Shut and full of liquid, the velocity is zero, so the valve sends back what arrives and shows it as the head: $\eta = I^{+}$. Once the head would fall below the vapour head, the liquid boils instead of stretching, the head is held at the vapour value, and the velocity beside the valve is whatever that head leaves over. Writing the vapour head as $-\varepsilon$,

$$\varepsilon = \frac{H_0 - H_v}{aV_0/g},$$

the line’s margin to vapour pressure measured in Joukowsky rises, the second rule is $\eta = -\varepsilon$, $u = I^{+} + \varepsilon$, and the valve sends out $I^{-} = -2\varepsilon - I^{+}$. The cavity’s volume grows while $u$ is negative, as the water beside the valve moves away from it, and shrinks while $u$ is positive.

That is the whole model, and it is small enough to solve without a grid. Every quantity is constant between events — a wave arriving, a cavity opening, a cavity closing — so the calculation follows the events and never takes a time step. For the line above, $J = 44.05$ metres, the margin is $H_0 - H_v = 35.1$ metres, and $\varepsilon = 0.797$: less than one rise, so the closure’s reflection reaches the vapour head.

A staircase, one step per round trip #

The cavity changes what the valve sends back, and the reservoir returns it inverted. The consequence is easiest to see at a larger velocity, where the cavity lasts several round trips.

At one round trip the closure’s reflection arrives at $I^{+} = -1$, the valve cannot show it, and the cavity opens. The valve now sends out $I^{-} = -2\varepsilon - (-1) = 1 - 2\varepsilon$, which the reservoir returns a round trip later as $-1 + 2\varepsilon$. The cavity is still open, so the valve sends out $1 - 4\varepsilon$, which comes back as $-1 + 4\varepsilon$, and so on. Each round trip adds $2\varepsilon$ to the wave arriving at the valve, because a cavity holding the head fixed turns every arrival into a departure shifted by twice the margin, and the reservoir hands that shift straight back.

Nothing about the pipe enters the size of the step. Its length sets how often a step arrives; its wave speed and the velocity set the rise the steps are counted in. The steps themselves are the margin, twice, and the margin is a fact about the pressure the line runs at, not about its hydraulics.

The same climbing sequence has a direct physical reading. The velocity beside the valve is $u = I^{+} + \varepsilon$, so it runs $-(1-\varepsilon)$, $-(1-3\varepsilon)$, $-(1-5\varepsilon)$: the water that was rushing back towards the reservoir is being slowed, round trip by round trip, by a head difference of exactly the margin, and at some round trip it turns and comes back towards the valve.

The round trip in which the cavity closes #

A cavity that grows while the water moves away and shrinks while it returns closes when the two volumes balance, and at a velocity that climbs by $2\varepsilon$ each round trip, that balance has a closed form.

In units of $V_0 A \cdot 2L/a$, the volume after $n$ round trips of cavity is the sum of $1-(2j+1)\varepsilon$ over $j$ from $0$ to $n-1$, which is $n(1 - n\varepsilon)$. It is positive until $n$ reaches $1/\varepsilon$ and zero there. So the cavity closes during the round trip in which $n$ passes $1/\varepsilon$, which is the $\lceil 1/\varepsilon\rceil$-th, and how far into that round trip follows from the rate within it.

For 1 m/s the margin is 0.287 rises, $1/\varepsilon = 3.49$, and the cavity closes during its fourth round trip: at 4.415 round trips after the closure, 3.415 after it opened. The largest volume, 6.70 litres in a pipe that holds 4.7 cubic metres, is reached as the velocity beside the valve passes through zero.

That cavity is small, and it is worth noticing how small. A column 600 metres long, stopped and reversed, has opened a gap at its end roughly the size of a bucket. The pressure it produces when it closes is not set by the size of the gap at all.

The collapse is not the peak #

The obvious reading of a cavity’s collapse is the one the earlier essay gave: the column crashes back into the valve, and the crash is a second closure. The steps say something more specific, and in one respect the opposite.

When the cavity closes the valve returns to the first rule, $u = 0$, and shows the arriving wave as its head. In the round trip the cavity closes in, that wave is $-1 + 2k\varepsilon$, with $k = \lceil 1/\varepsilon \rceil - 1$. For 0.36 m/s that is 0.594 rises: a head of 51.2 metres, which is the step at 2.146 seconds in the first figure, and it is below the Joukowsky head. Measured from the vapour head the collapse is a violent event — the water arrives at 0.50 metres a second, 1.39 times the flow that was shut off — but measured from the steady head it is smaller than the closure was.

The pulse comes one step later. The cavity’s last round trip sent a wave towards the reservoir that returns at the end of that round trip as $-1 + 2(k+1)\varepsilon$, and the valve, now shut and full, shows it:

$$\eta_{\text{pulse}} = 2\varepsilon\,\lceil 1/\varepsilon\rceil - 1.$$

Writing $1/\varepsilon = k + \delta$ with $0 < \delta \le 1$, the excess over the closure’s rise is $2\varepsilon(1-\delta)$ rises — in metres, $2(H_0 - H_v)(1 - \delta)$. The pulse passes the Joukowsky head by up to twice the margin to vapour pressure, and it lasts only as long as the cavity took to close in its last round trip, which works out to $k\delta/(k+1-\delta)$ of a round trip. At 0.36 m/s the excess is 52.3 metres and the pulse lasts 146 milliseconds.

So the damage is done not by the crash but by superposition: the collapse restores a reflecting end to the line at a moment when a step the cavity built is still on its way back. Remove the staircase and there is no pulse above the closure’s rise; remove the collapse and the step never reaches a reflecting valve.

A sawtooth in the flow velocity #

Because the pulse depends on $\lceil 1/\varepsilon\rceil$, it is not a smooth function of anything.

The margin in rises is $\varepsilon = g(H_0 - H_v)/aV_0$, so $1/\varepsilon$ is proportional to the velocity, and a whole number of round trips fits into the cavity’s life at every multiple of $g(H_0-H_v)/a$ — 0.287 metres a second for this line. Just above each multiple, $\delta$ is small, the cavity closes at the very start of a round trip, and the pulse is nearly twice the margin above the Joukowsky head. As the velocity rises towards the next multiple, $\delta$ grows, the excess falls linearly, and at the multiple itself it is exactly zero. Then one more round trip fits, and the tooth begins again.

Two consequences follow, and neither is what a designer expects. A faster flow can make a lower peak: 0.55 metres a second peaks at 98.1 metres here and 0.36 at 121.3. And the teeth never go away: the excess can always reach twice the margin, at every velocity above the first tooth, although it becomes a smaller fraction of a Joukowsky rise that keeps growing.

The dots are the check. They come from an ordinary characteristics grid of 120 reaches, with the vapour head applied at the valve node by clamping and the cavity volume integrated there — the discrete vapour cavity model used in practice — and it was told nothing about the closed form. At ten velocities spread across four teeth it agrees with $2\varepsilon\lceil 1/\varepsilon\rceil - 1$ to better than a millimetre of head, including the collapse times. The same check with a cavity allowed at every node rather than only at the valve gives the same peaks at the three velocities it was run at: in a frictionless line the wave train never takes an interior point below the vapour head, which the invariants also say, since every head in the pipe is an average of an arriving and a departing value that differ by at most one step.

Tall and brief #

A pulse that passes the Joukowsky head by fifty metres sounds like a design case, and whether it is depends on how long it lasts.

The two curves are the same parameter read two ways. The height share is $1-\delta$ and the length share is $k\delta/(k+1-\delta)$, and as $\delta \to 0$ the first tends to one while the second tends to zero. The tallest pulses are the shortest, and the reason is in the volume figure: a pulse is tall when the cavity closes right at the start of a round trip, and then there is almost nothing of that round trip left for the returning step to fill.

That has two practical faces. A pipe wall or a fitting does not respond to a pressure instantly: it has its own natural periods, and a pulse much shorter than them loads it more like an impulse than like a pressure, so its effect scales with height times length rather than with height alone. The product here is largest in the middle of each tooth, not at its start.

And a pulse lasting a few milliseconds has to be seen by the instrument recording it. A pressure transducer sampled every ten milliseconds, or one with a slow mechanical response, reports a pulse shorter than that as something much lower or not at all. This is the instrument being part of the answer in a hydraulic setting, and it is one reason laboratory measurements of these pulses were contested for years: the tallest ones are exactly the ones a slow gauge cannot see.

More steady pressure, less hammer #

The margin to vapour pressure is set by the pressure the line runs at, which is something an operator can change. The sawtooth in velocity has a counterpart in that pressure, and it contains the most counter-intuitive number in the essay.

The threshold is where the closure’s reflection just reaches the vapour head, $H_0 = H_v + aV_0/g$, and the margin there is exactly one rise. Just below it $\lceil 1/\varepsilon\rceil = 2$, the cavity lives for a single round trip and a sliver, and the pulse is $4\varepsilon - 1$ rises: nearly three rises above the steady head, or the vapour head plus four. Just above it the cavity never opens and the peak is the ordinary one rise.

So raising the steady head by a fifth of a metre, from 50.98 to 51.18 metres, lowers the worst transient from 234.1 metres to 112.4 — and lowering it by the same fifth of a metre from above the threshold does the reverse. Two metres either side, at 50.08 and 52.08, the peaks are still 229.6 and 113.3. The same design choice that looks conservative in steady operation, running a line at a lower pressure to reduce the load on its fittings, is the one that brings the separation and the pulse that follows it.

The surge tank that turns a hammer into a slow swing is the large-scale version of the same remedy, holding the head near its steady value rather than letting a wave take it anywhere. This is also the precise form of a warning the earlier essay gave in general terms, that low-pressure systems suffer most because a surge is an absolute quantity and not a percentage of the working pressure. It is also why the remedies for column separation are mostly about the margin rather than about the valve: surge vessels and air chambers that stop the head falling, and pressure maintained at high points where the margin is smallest.

What friction does, and does not do #

Every figure so far has been frictionless. A real line has friction, and the obvious question is whether friction removes the pulse.

It does not, at least not as friction is ordinarily modelled. A friction factor of 0.06 is a rough old pipe, it costs several metres of head in steady flow, and it moves the first pulse by a few per cent of its rise — lower at two of the three velocities and higher at the third. The direction is not even consistent, because friction does two things at once: it drains energy from the waves, and it shifts their timing, so the collapse can move closer to or further from the start of a round trip and the sawtooth’s phase moves with it.

The reason the effect is small is the time scale. A friction factor is measured in steady flow, and applied at each instant to the instantaneous velocity it acts on the pulse for a few hundredths of a second. The pulse is over long before a quasi-steady friction term has taken much from it.

Measured pulses are lower and rounder than these, and the difference has to come from something the model leaves out. A wall whose shear depends on the history of the flow is the first candidate: in a rapidly reversing flow the velocity profile is nothing like the steady one, and unsteady friction damps sharp fronts far more than the steady factor does.

A model that sharpens what it should smooth #

The first pulse is where the calculation can be trusted. After it the line has a second cavity, a third, and a wave train cut into more and more pieces, and a frictionless model does something with those pieces that is worth seeing because it is plainly wrong.

Every head the exact calculation produces lies on a lattice: $-\varepsilon$, or $\pm 1 + 2m\varepsilon$ for a whole number $m$. That is forced by the two rules — the valve can only send back what arrives or shift it by twice the margin — and it is checked on every segment of forty round trips at seventy margins between 0.05 and 0.995, 7,536 segments, none further from the lattice than four parts in $10^{15}$. What the lattice does not limit is $m$. Each new cavity cuts the train at a new place, the pieces recombine a step higher, and at this margin the peak reaches $-1 + 20\varepsilon$ within forty round trips, on pieces as short as three quarters of a millisecond.

The grid solver climbs with it, which shows the climb belongs to the model and not to the arithmetic. Nothing in a frictionless, dispersionless line can round off a step, so steps pile up without limit. In a real pipe many things round them off — friction, the wall’s own viscoelasticity, the gas that comes out of solution in every cavity, and the spreading of a front over the pipe’s cross-section — and with only a quasi-steady friction factor of 0.02 the ratchet is already gone. The first pulse is a prediction; the later ones are a description of what the model lacks.

What a real line adds #

The water does not boil the instant it could. The cavity opens exactly at the vapour head here. Water with few nuclei in it can be taken below that pressure for a moment — how far, and for how long, is a question of duration as much as of pressure — and a line whose water holds a brief tension separates later and at a lower head than the rule assumes, which changes the margin every step is made of. The threshold itself is the vapour-pressure argument a propeller blade obeys.

The valve takes time to close. Every case here is instantaneous. A closure shorter than a round trip is the same event as far as the first rise is concerned, but it changes the shape of the first step, and a slower one lowers the rise and moves the whole sawtooth.

Gas comes out of solution. Water holds dissolved air, and a cavity at a few kilopascals releases some of it. A bubble with gas in it does not collapse to nothing, so it cushions the collapse, and a small volume fraction of free gas slows the wave far below either the water’s or the air’s own sound speed, which lengthens every round trip and flattens every step.

The cavity is not always at the valve. In a frictionless horizontal line it is, and the check with a cavity allowed everywhere agrees. A line with high points, or with enough friction to tilt the head along its length, can separate in the middle, and two column halves then meet at a place no valve sets.

The collapse is not instantaneous. A vapour cavity at the scale of litres closes as a bubble does, with its wall accelerating inwards; the bubble that hammers computes that collapse and the point at which its own model fails. The pipe calculation treats the cavity as a volume and nothing else.

Who worked it out #

Column separation was seen in water mains as soon as water hammer was measured, in the Moscow tests of the 1890s, and the pressure after a collapse exceeding the closure’s own was reported long before it was explained. The discrete vapour cavity model — a characteristics grid with the head clamped at vapour pressure and a volume carried at a node — was set out by Wylie and Streeter in the 1970s and became the standard engineering calculation.

The explanation drawn here, that the largest pulses are short and come from a collapse superposed on a returning wave rather than from the collapse alone, was made quantitative by Simpson and Wylie in 1991, and the laboratory work that separated the short pulses from instrument artefacts is Bergant and Simpson’s, through the 1990s. The closed form in this essay is a restatement of that understanding for the simplest line in which it can be solved without a grid at all.

Still open: what the dissolved air does to the staircase #

Every step in the staircase is exactly twice the margin, because a vapour cavity holds the head at one value. A cavity with gas in it does not: its pressure rises as it shrinks, so the valve’s second rule becomes a spring rather than a fixed head, and the steps are no longer equal. That turns the sawtooth into something smoother and lower, and whether a small amount of released gas removes most of the pulse or only some of it is a calculation with a real answer — one that needs a rate at which air comes out of solution and goes back in, which is where the model stops being pure wave mechanics.

Beside it is the case most lines are actually destroyed by, which begins with a down-surge rather than a closure: a pump that trips. There the first wave is the low one, the cavity forms at the pump end or at a high point before any valve is involved, and the non-return valve that closes on the returning column is the reflecting end the collapse lands against. The same invariants apply, and the order of events is different enough that the sawtooth here is not the answer there.