A pump with no engine

A hydraulic ram is a lump of iron with two valves in it. Water from a stream runs down a pipe into it; the ram clanks about once a second; and a thin stream of water comes out of a second pipe, going uphill, sometimes a hundred metres above the stream it came from. There is no engine, no electricity, no fuel and nothing that needs adjusting. Some installed in the nineteenth century are still running.

The machine invites the wrong question, which is where the energy comes from, and the wrong answer, which is that it must be getting something for nothing. The right question is what limits it, and the answer is an audit with one line in it.

The audit #

Everything that enters the ram has fallen through h, the height of the supply above the machine. Everything it delivers is lifted through H, the height of the delivery above the machine. Nothing else crosses the boundary except the waste water, which leaves at the ram’s own level and therefore carries no potential energy away.

$$\rho g H q \le \rho g h Q \quad\Longrightarrow\quad \frac{q}{Q}\le\frac{h}{H}$$

That is the whole constraint. It is conservation of energy with every loss set to zero, so it is a bound rather than a performance, and nothing about the machine appears in it: not the valve timing, not the beat rate, not the size of the air vessel, not the length or diameter of the drive pipe, not the material.

The site’s solver refuses an operating point above the ceiling rather than reporting one. That is the same discipline the rest of the field uses — a figure drawn from an impossible operating point would look entirely normal, with a plausible curve and a sensible axis, and the only clue would be an efficiency above one that nobody was reading.

Two efficiencies, and the generous one is in the catalogue #

There are two efficiencies in circulation for rams and they are both honest. They answer different questions, and the difference is the datum.

D’Aubuisson’s measures from the ram: of the power arriving at the machine, how much leaves usefully?

$$\eta_A = \frac{qH}{Qh}$$

Rankine’s measures from the supply’s own surface: of the energy the waste water gives up, how much is delivered to water lifted above where it started?

$$\eta_R = \frac{q(H-h)}{(Q-q)h}$$

Neither is wrong. They are answers to two different questions, and which one a reader wants depends on whether the supply level or the ram level is the thing they could have chosen.

The site computes the gap and asserts its closed form, which is worth doing because the identity is the sort of thing that is easy to state and easy to state wrongly. What comes out is an ordering: η < 1 implies η_A > η_R, always, with equality only for a perfect machine. So a manufacturer quoting D’Aubuisson’s number is quoting the higher of two true figures — which is what manufacturers do, and is worth knowing when comparing two catalogues.

How high, and why the answer is surprising #

The audit says what fraction can be delivered. It does not say how high, and the height limit comes from somewhere else entirely: the pressure the machine can generate, which is the pressure of its own water hammer.

The drive column accelerates under the supply head until it is running at a metre or two per second, the waste valve slams shut, and Joukowsky’s ρaΔV appears at once. That pressure opens the delivery valve and pushes water into the delivery pipe for as long as it lasts.

$$H_{\max}\approx \frac{aV}{g}$$

A hundred and twenty metres from a stream falling two. That is the fact that makes the machine seem impossible, and the audit is what makes it possible: the ram can lift a little water very high precisely because it wastes most of it. The head ratio and the flow ratio multiply to something less than one, and the machine chooses which of them to spend.

The cycle, and why the picture is a list of boxes #

There is no flow field in this essay and there could not honestly be one. The interior of a ram during the slam is an unsteady, cavitating, valve-dominated flow that nothing on this site can compute, and a drawing of streamlines through it would be an invention. What can be computed exactly is what crosses the boundary, which is the whole of the machine’s usefulness.

That is the applied field’s method stated once more, and this is perhaps its cleanest instance: a control volume does not need to know what is inside it, and the parts of a ram nobody can model are precisely the parts the audit does not ask about.

What a working installation looks like in numbers #

The audit is abstract until it is put against a real duty, and the arithmetic is small enough to do in the margin.

Two things about that number are worth drawing out.

The first is that the waste is the point rather than a defect. Ninety-six per cent of the water goes straight back to the stream, a few metres below where it was taken, having lost nothing but its height. Nothing is consumed and nothing is polluted; the machine borrows a head from a large flow and sells it back as a larger head on a small one.

The second is that the arrangement is a transformer, and the analogy is exact rather than decorative: a quantity conserved (energy) is traded between two factors (flow and head) whose product is fixed, with a ratio set by the machine and a loss that shows up as a fraction. The site’s branching essay makes the same kind of statement about a cost function, and Betz’s disc about an extraction limit — the recurring shape of this field is that the interesting number is a ratio and the interesting constraint is a product.

What the model leaves out #

The beat rate is not in it. A ram runs at some rate between twenty and a hundred and twenty beats a minute, set by the drive pipe’s length, the supply head and the waste valve’s weight. None of those appears anywhere above, because the audit is a statement about the ratio of two flows rather than about either of them. Tuning the beat rate changes the flows and the efficiency; it cannot change the ceiling.

The air vessel is not in it either. Without one, the delivery pipe would have to be accelerated from rest at every beat and the machine would tear itself apart; the vessel turns a series of impacts into a nearly steady delivery. It is essential and it is a smoothing device: it changes when the water is delivered, not how much.

The waste jet’s own energy is thrown away and could not easily be kept. It leaves at the speed the drive column had reached, which is a metre or two per second, so it carries about a tenth of a metre of velocity head — small compared with h, and recovering it would need a second machine of the kind an ejector provides and would cost more than it returned.

No loss is computed anywhere. The efficiencies above are the ratio of two ideal quantities, and a real machine reaches 60 to 80 per cent of the D’Aubuisson ceiling. Where the rest goes — friction in the drive pipe, the energy carried away by the waste jet, valve leakage, the air vessel’s own losses — is not calculated here and would need a model of the machine rather than a box round it.

Cavitation is a real failure mode and is not in the audit. The pressure at the ram during the recoil phase falls sharply, and if it reaches the vapour pressure the column separates and slams back — which is the same collapse a cavitation bubble makes and does the same kind of damage to the valve seat. A ram’s snifting valve, which admits a small bubble of air each beat, exists partly to cushion that and partly to keep the air vessel charged.

The drive pipe has to be rigid and it has to be straight. Everything in the height ceiling rests on the wave speed, and that is a property of the pipe rather than of the water. A compliant, kinked or air-entrained drive pipe carries a slower wave and a smaller slam, which is the commonest reason an installed ram underperforms.

The class of machine this belongs to #

Three machines on this site turn a large flow at low grade into a small flow at high grade, and it is worth putting them together because the arithmetic is the same shape in each.

A hydraulic ram takes Q at head h and delivers q at head H, bounded by qH ≤ Qh.

An ejector, which is the next rung’s subject, takes a small fast stream and a large slow one and delivers everything at an intermediate pressure, bounded by the momentum balance across a mixing tube.

An actuator disc takes a wide slow stream and extracts work, bounded by Betz’s 16/27.

A fourth belongs on the list and is the reason this field exists: a weir or a hydraulic jump, where the bound is on how much energy a free surface can destroy rather than on how much it can deliver.

In every case the bound comes from a control volume with every loss set to zero, in every case the bound is reachable only by a machine that does nothing useful at the margin, and in every case the real machine’s efficiency is a fraction of a number that was never attainable. What the list does not contain is a rotating machine, and the reason is that a rotor’s duty is chosen by a dimensionless number rather than bounded by a conservation law — the constraint there is which shape of runner a flow and a head allow, not how much of either can be had. That is the shape of this whole field, and the ram is the one where the arithmetic is simplest and the machine is strangest.

The claim about free energy, answered properly #

Rams attract perpetual-motion enthusiasm, and it is worth answering the claim in the form it is usually made rather than dismissing it.

The claim is that a ram needs no external power, so the energy must be coming from somewhere unaccounted, and a better design could therefore deliver more. The first clause is true and the inference is false, and the audit says exactly where the energy comes from: the waste water. Ninety per cent or more of the supply falls through h and leaves at the ram’s own level, giving up its potential energy; the machine’s whole function is to transfer some of that to the small fraction it sends uphill.

An engine is unnecessary because the stream is the engine. A ram is a device for collecting a head that is already there, and if the stream stops falling the ram stops working within one beat.

This is the same shape of argument as the sailing rung’s answer to the claim that a boat cannot outrun the wind driving it: nothing is being created, two reservoirs at different grades are being connected, and the surprising number is a ratio rather than a total. It is also the reason Betz’s limit is a limit rather than an efficiency — the constraint is on what a machine may take from a stream, not on how well it takes it.

Who found it, and when #

Joseph Michel Montgolfier — of the balloon — built the first working ram in 1796, and patented it with his brother; the name bélier hydraulique is his. Whitehurst had made a hand-operated version in 1772. D’Aubuisson’s efficiency dates from an 1840 treatise on hydraulics and Rankine’s from his Applied Mechanics of 1858, and the two definitions have coexisted, unreconciled, in the literature ever since — which is why an essay about a machine with one moving part needs a section on which of its two efficiencies is being quoted.

The ram’s continued use is not nostalgia. In a place with a stream, a hill and no electricity it is still the correct answer, and its two failure modes after a century of operation are a perished valve rubber and a silted drive pipe.

The two ceilings are not independent #

The essay has two limits in it and treats them separately. They interact, and the interaction is what a practical installation runs into first.

The energy ceiling, $q/Q \le h/H$, is a statement about fractions and permits any delivery height at all — send a small enough trickle and the audit is satisfied to a kilometre. The height ceiling $aV/g$ is a statement about pressure, and it says the machine simply cannot open its delivery valve above a certain head whatever fraction is asked for.

And the two are linked through $V$. The drive column’s speed at closure is what the supply head has been able to build up in the time available, so a larger $h$ raises the slam as well as the energy allowance. A ram is not free to trade the two: the same quantity is buying both.

The sharper point is that the machine delivers nothing at its maximum head. At $H = aV/g$ the pressure in the ram only just equals the pressure in the delivery pipe, so the delivery valve opens for an instant and passes nothing. Reduce the delivery height and the valve stays open longer each beat and more water goes up. So the ram has a characteristic — a falling curve of delivery rate against delivery head, meeting the axis at $aV/g$ — exactly like any other pump, and its operating point is where that curve meets whatever the delivery pipe demands.

Which explains a failure that puzzles people who have done the energy arithmetic. A ram installed to a height the audit permits comfortably can deliver nothing at all, because the head is near the top of its characteristic rather than near the top of its energy budget. The practical rule that keeps $H/h$ below about twenty is a statement about the slam rather than about conservation, and it is the binding constraint on nearly every real installation.

What tuning a ram actually does #

The audit says nothing about the beat, and a practical installation is largely a matter of tuning it, so it is worth saying what the tuning is for.

Each cycle has to accelerate the drive column from rest to the speed at which the waste valve shuts. That takes time — the column’s inertia against the supply head — and the time is proportional to the drive pipe’s length. A long drive pipe therefore gives a slow beat and a large slam; a short one gives a fast beat and a gentle one. The standard rule is a drive pipe five to ten times the supply head in length, which is a statement about that balance and not about friction.

The waste valve’s weight sets the speed at which it closes, and therefore ΔV. A heavier valve stays open longer, the column reaches a higher speed, and the slam is bigger — more delivery per beat, at fewer beats. A lighter valve gives more beats of less each.

What none of that can do is move the ceiling. Every combination of pipe length, valve weight and beat rate produces some operating point on the plot, and every operating point lies under q/Q = h/H. Tuning chooses where on the curve the machine sits; it cannot lift the curve.

That distinction — between the envelope a conservation law draws and the point a design chooses inside it — is the whole reason this field computes envelopes. A designer who knows the envelope knows when to stop tuning.

Where the ladder goes next #

Water hammer is finished: feared in one rung and sold in the other. The next machine has no moving parts at all — not one — and its pumping mechanism is a loss: the same Borda–Carnot expression that makes an orifice plate expensive, running in a direction that makes it useful.