A choked throat buys time, not silence

The venturi that stops listening downstream found a meter turning into a limiter. Lower the downstream pressure below a critical value and the throat reaches vapour pressure, the flow rate stops rising, and the upstream pipe carries a fixed flow whatever happens below the throat. It ended with a question that the steady calculation could not answer: whether the same throat also blocks a transient.

The question matters because the steady result is exactly the kind that gets generalised. A cavitating venturi is used deliberately as a flow limiter in rocket feed lines, in fire-protection systems and in process plant, and the property people want from it is not only a fixed flow but protection — a guarantee that trouble downstream cannot propagate up the line. The steady statement seems to promise that. It does not, and the reason is in the one thing the steady statement leaves out: the cavity has a volume.

The line, the valve and the node with a volume #

The line is the one the steady essay used, extended in both directions: a reservoir at 5 bar, 20 metres of 25 mm pipe to the venturi, the venturi itself with its 5 mm throat, then 40 metres of pipe to a valve discharging to a receiver at 2 bar. The wave speed is 1,200 metres a second and pipe friction is left out, so every wave arrives at full strength.

With the valve open, the throat is choked at 0.608 litres a second, the upstream pipe sits at 5 bar, the downstream pipe at 3 bar, and a vapour cavity occupies part of the throat and diffuser. At time zero the valve shuts in five milliseconds.

The pipes are computed by the method of characteristics, with the discrete vapour cavity model at every node, so that any low-pressure wave that would take the water below its vapour pressure boils it there instead — the calculation used in twice the margin on top of the hammer. The venturi is a node with a volume. While its cavity exists, the throat is at vapour pressure; the upstream pipe feeds it through the contraction, and the downstream pipe draws from it through the diffuser or, if the water is driven back, returns to it as a jet into a vapour space that recovers nothing. The cavity’s volume changes at the difference between the two flows. When the volume reaches zero the throat is liquid-full, one flow passes through the whole meter, and the upstream pipe and the downstream pipe are connected.

Two numbers in that model are prescribed rather than computed, and both are named here because the result depends on them. The first is the cavity’s starting volume. A one-dimensional balance fixes the flow through a choked throat and says nothing at all about how far the vapour extends into the diffuser, so the volume is a parameter, swept from about a millilitre to a quarter of a litre. The second is how much of the throat’s pressure drop the diffuser recovers while a cavity reaches into it; that is set at 60 per cent, the value that puts the steady downstream pipe at 3 bar, against 85 per cent for a liquid-full diffuser.

What the upstream pipe hears #

The valve sees what any valve shut quickly on a moving column sees: a pressure rise of $\rho a V$, 14.8 bar, arriving at the moment of closure. That rise travels back up the downstream pipe, bringing the water behind it to rest, and it reaches the venturi 33.3 milliseconds after the valve starts to move — the pipe length over the wave speed.

For the next eight milliseconds the upstream pipe hears nothing. Its pressure stays at exactly 5 bar and its flow stays at exactly the choked rate. The steady statement is holding through the transient, and it is holding for the reason it held in steady flow: the throat is at vapour pressure, and a throat at vapour pressure transmits nothing upstream. The surge arriving from downstream is spending itself on the cavity.

Then the cavity is gone, the throat is liquid-full, and the upstream face of the venturi jumps to 18.8 bar absolute. The upstream pipe receives 13.8 bar of the 14.8-bar surge — 93 per cent of it. The missing 7 per cent is the meter’s own loss: the liquid-full contraction and diffuser absorb part of the wave as it passes.

Why the cavity closes so fast #

The flows show the mechanism. On the upstream side nothing changes: the choked flow keeps arriving at 0.608 litres a second, as it must, because the throat is still at vapour pressure. On the downstream side the surge brings the water to rest and then, because the pressure it leaves behind is far above the cavity’s, drives the downstream column backwards into the throat at 0.56 litres a second.

So the cavity is filled at nearly twice the choked flow, from the pipe it was supposed to be isolating and from the pipe whose surge it was supposed to be absorbing. A 20 mL cavity lasts 21 milliseconds after the surge arrives. The protection is measured in milliseconds per millilitre, and the volume that would protect a line for a second is most of a litre — a vapour pocket far larger than any venturi’s diffuser holds.

The steady statement is not wrong in any of this. It is that the steady statement is about a state, and a transient is a question about how long a state lasts. A choked throat stays choked only while the cavity that chokes it exists, and the arriving surge is precisely the thing that removes the cavity.

How long the cavity lasts #

For cavities from about twenty millilitres to a quarter of a litre, the survival time after the surge arrives is between a half and six-tenths of $V/Q$ — 21 milliseconds for 20 mL, 79 for 85 mL, 280 for 256 mL — between the time the choked flow would take to fill the cavity alone and the time it would take at twice that rate. The downstream column’s return is slightly slower than the choked flow — 0.56 against 0.61 litres a second — because its speed is set by the surge’s own strength rather than by the throat. For cavities of a millilitre or two the survival time levels off near five milliseconds, which is the valve’s closure time: the surge’s front is five milliseconds long, and a cavity that small is filled before the front has finished arriving.

When a cavity is big enough to matter #

A cavity that survives longer than eight milliseconds changes nothing about the size of the pulse; the upstream pipe still receives 13.8 bar, only later. The pulse gets smaller only once the cavity survives long enough for something else to happen in the downstream pipe first.

That something is the relief wave. The surge that arrived at the venturi reflects off the cavity — a constant-pressure boundary at vapour pressure — as a wave of low pressure, which runs back to the closed valve and returns. After one round trip of the downstream pipe, $2L/a$, the column driving back into the cavity is being slowed by its own reflection. A cavity that closes before that sees the column at full return speed; one that closes after sees it slowed.

The threshold falls where that argument puts it. For every one of three downstream lengths, a cavity smaller than about 1.7 times the choked flow times the round-trip time passes 93 per cent of the surge, and one larger than about 2.1 times passes 65 per cent. The survival time at that volume is one round trip. The volume a cavitating throat needs to weaken a water hammer is the volume the choked flow delivers in the time the downstream wave takes to go and come back — 40 millilitres for this 40-metre line, and proportionally more for a longer one.

Past the threshold the pulse falls in steps rather than smoothly, because in a line with no friction the downstream column’s speed changes only when a reflection arrives, and each additional round trip the cavity survives is one more reflection. Friction would round the steps off. It would not change the essential point, which is in the figure’s lowest points: a 256 mL cavity, six times the scale, still passes 7.1 bar. There is no cavity size at which the upstream pipe is protected, because in a frictionless line every cavity eventually closes.

The long run makes the delay visible. For over three hundred milliseconds the upstream pipe is perfectly quiet — the steady statement at its most convincing — while the downstream pipe rings between 17.8 bar and vapour pressure every 67 milliseconds, its column reversing each time. Every reversal back towards the venturi fills the cavity further. When the last of it goes, the upstream pipe gets its pulse.

The wave that is worse than the one that caused it #

The first pulse is not the largest the upstream pipe ever sees. In the frictionless line, within the first 0.6 seconds, the upstream face later reaches between 15 and 35 bar above its steady pressure, depending on the cavity — up to two and a half times the original surge. That is the column-separation mechanism of the hammer essay arriving at the venturi: the transmitted pulse reflects from the reservoir as a relief wave, the relief takes the upstream pipe’s water to vapour pressure at the venturi face, a new cavity forms and closes, and each collapse adds the returning columns’ momentum to a pressure already raised.

Those later peaks belong to a line with no friction and are larger than a real line would produce; they are not quoted as predictions. What they show is the character of the thing. A cavitating throat is not a barrier that removes a transient. It is a store that delays one, and when the store empties it releases the transient into a line that now has a cavity-forming node in it.

Why a choked gas throat really does isolate, and a choked liquid throat does not #

The misconception has a respectable source, and it is worth tracing because the source is correct in its own setting.

A gas nozzle whose throat is sonic genuinely does not hear what happens downstream. A small pressure disturbance travels at the speed of sound relative to the gas, the gas at the throat moves downstream at exactly that speed, and so the disturbance’s net speed there is zero: it cannot pass. That isolation is a property of speeds. It does not consume anything, it does not run out, and it holds for every disturbance small enough not to unchoke the throat, for as long as the flow lasts.

A cavitating liquid throat reaches the same steady conclusion — a fixed flow, deaf to the downstream pressure — by a completely different route. It is deaf because its pressure is pinned at the vapour pressure, and its pressure is pinned only because a vapour cavity exists to pin it. The steady essay showed that the vapour-laden mixture there is also, in a sense, supersonic, since its sound speed falls to a few metres a second. But a surge arriving from downstream does not need to propagate through the mixture. It condenses it. The gas throat’s isolation is a property of speeds; the liquid throat’s is a property of a volume, and a surge spends volumes.

That is why the two devices, which share a steady characteristic almost exactly, behave so differently to a transient. A disturbance small enough to leave a gas throat sonic is blocked for ever. A disturbance of any size arriving at a cavitating throat is blocked for a time set by the volume it has to fill, and the time is short.

A store that empties, against one that does not #

The line in a tank that turns a hammer into a swing protects a pipe from its valve with a surge tank: an open standpipe that takes in the column’s water when the valve shuts, lets the level rise, and converts a sharp pressure wave into a slow oscillation of the level. The cavity at a venturi throat is doing the same job — taking in water that has nowhere else to go — and the comparison shows exactly where it falls short.

A surge tank has a free surface open to the atmosphere, so it can take in water indefinitely at a pressure that rises only as its level rises; it never empties in the sense that matters. A vapour cavity takes in water at vapour pressure until its volume is gone, and then it has no capacity at all. The tank’s protection is sized by its cross-section, which sets how slowly the level rises. The cavity’s protection is sized by its volume, which sets how soon it runs out — and the volume that matters, $Q\cdot 2L/a$, is forty millilitres here, or about the size of a small espresso, while the cavity a real venturi carries is perhaps a millilitre.

A ledger of the line #

The checks are chosen to fail if the venturi node were wrong. An open valve moving the line would mean the node’s steady relations disagree with the pipes’ initial state. A valve pressure other than ρaV would mean the downstream characteristics or the valve boundary were mishandled. And a large cavity letting any pressure through before it closes would mean the node was transmitting across a throat at vapour pressure, which is the one thing the model must not do.

What the one-dimensional line cannot show #

Where the cavity is. The throat’s vapour pocket is a single node with a volume. In a real venturi the cavity is attached to the throat wall and sheds into the diffuser, and a surge does not fill it uniformly; it collapses it from the downstream end, probably faster than a lumped volume would suggest, and with the damage concentrated where collapse happens.

The cavity’s starting volume. It is a parameter here because nothing in a one-dimensional calculation fixes it. The scale that matters, $Q\cdot 2L/a$, is 40 millilitres for this line; a real cavitating venturi of this size carries a vapour pocket of perhaps a millilitre or two, which puts it firmly in the regime where 93 per cent of the surge passes.

Pipe friction and wave damping. Without friction the steps in the pulse are sharp and the later peaks are too large. A real line damps each reflection, and the threshold’s position would move a little while its scaling on $Q\cdot 2L/a$ would not.

Dissolved gas. Water that has come out of solution at the throat does not recondense when the surge arrives. A gas pocket cushions the surge rather than collapsing, and a throat with enough released gas would protect the upstream pipe far better than a vapour cavity can.

The wall. Pipe elasticity is folded into the wave speed, and nothing about the venturi body itself responds.

Still open: whether shaping the diffuser can keep the cavity alive #

The cavity’s survival is set by how fast it is filled from both sides, and the downstream side fills it because the diffuser lets the returning column in freely. A diffuser that resisted reverse flow — a check valve in effect, or a shape that loses much more head backwards than forwards — would slow the return, and the cavity would last closer to $V/Q$ than to half of it.

The calculation that follows gives the diffuser a reverse-flow loss of its own and asks how much of the protection it buys: whether a strongly asymmetric diffuser can double a cavity’s survival, and whether that moves the threshold enough to matter for a line of a realistic length. Beside it is the case with dissolved gas, where the cavity does not collapse but compresses, and the question becomes how much gas turns a delay into a real reduction of the pulse.