A choke that belongs to two streams

The liquid jet pump reaches a limit when the entrained stream, accelerated into the throat, lowers its own pressure to the point where it boils. That limit freezes the flow ratio and turns the right-hand end of the pump curve into a vertical wall.

A gas has nothing to boil. It does have a speed it cannot be pushed through by a pressure difference acting from downstream, and in an ejector driven by a supersonic jet that speed is reached in a place with no wall around it at all. The entrained gas flows in the annulus between the tube and the jet, the jet is expanding as it goes, and the annulus narrows. Somewhere along it the gas can accelerate no further.

This is the machine that holds steam condensers at vacuum, pulls the air out of evaporators and drives steam-jet refrigeration. Its characteristic has a shape unlike any pump curve, and the model that explains the shape explains two further things about it: why its entrainment and its pressure rise cannot both be chosen, and why the stages of a vacuum train are not alike.

The model, with nothing fitted #

The calculation is the one-dimensional balance the mixing tube used, carried into compressible flow and given a supersonic nozzle.

The primary steam comes from its boiler through a converging–diverging nozzle, which is choked, so its mass flow is fixed by the nozzle throat and the boiler’s pressure and temperature. The jet leaves the nozzle and keeps expanding until its static pressure matches the entrained gas beside it. At that section the two streams share a pressure and occupy the tube between them. From there they mix in a tube of constant area, conserving mass, momentum and total enthalpy, and the mixed stream is decelerated to its stagnation pressure in a diffuser.

Every component is ideal. Steam is treated as a perfect gas with a ratio of specific heats of 1.3, the motive steam at 130 °C and the entrained vapour at 10 °C, and the condensation that real expanding steam undergoes is ignored. Pressures are quoted as fractions of the motive pressure: a suction at 0.01 of a 10-bar supply is 100 millibar.

The picture is the answer to the question a vacuum-system engineer asks first, and it has two parts.

The flat part. Up to a critical back pressure the entrainment does not depend on the back pressure at all. A tube 60 throats in area entrains 0.546 kilograms of vapour for every kilogram of motive steam, and it entrains exactly that at a back pressure of 1.5 times the suction or of 3.4 times. This is what makes a steam ejector a good vacuum pump: its capacity does not sag as the vacuum it is holding deepens.

The end of it. Above the critical back pressure the entrained stream is no longer choked and the machine stops entraining over a very small range of pressure — in this lossless model, the whole fall from full capacity to nothing happens within 1.4 per cent of the critical value. Real ejectors fall more gradually, over a few per cent to a few tens of per cent, because the shock train and the separation in a real mixing tube spread the transition out. But the character is the same, and it is why a steam ejector is specified by its critical back pressure: running above it is not an inefficiency, it is a failure.

The two roots are a shock #

The mixing balance has a feature worth stopping on, because it removes what looks like the hardest part of the problem.

Conserving mass, momentum and total enthalpy through a constant-area tube leaves a quadratic in the mixed velocity. It has two roots: one supersonic and one subsonic. A supersonic mixed stream cannot be decelerated to a high back pressure smoothly, so somewhere in the tube or the diffuser there must be a shock, and it looks as though the model has to decide where.

It does not. The two roots conserve the same mass, momentum and energy flux through the same area, and a normal shock is exactly a jump that does that. So the subsonic root is the state behind a normal shock standing on the supersonic one, wherever that shock happens to sit.

The same statement was met in two ways to choke, where the Fanno and Rayleigh lines through one state cross a second time on the normal-shock relation, and in the jump the equations allow, where the shock is derived as precisely this pair of solutions. In the ejector it has a practical consequence: the shock’s position changes the pressure distribution inside the tube and changes nothing about what leaves it, so a one-dimensional account can be exact about the discharge while saying nothing about the shock’s location. The position becomes a question about losses — the boundary layer the shock interacts with — which is where real ejectors differ from this one.

Where the choke is, and why it is not at Mach one #

The textbook form of this model sets the entrained gas to Mach one at the section where its pressure matches the jet’s, and calls that the choke. That is the obvious place for it. It is not where the choke is.

The way to see this is to do what the textbook form does not: hold the geometry fixed, let the entrained gas’s Mach number at that section take any value up to one, and compute the entrained flow each time.

The entrained flow has a maximum below Mach one, and the reason is the stream beside it.

Drawing the entrained gas faster lowers the pressure the two streams share. For a single stream in a fixed duct that would raise the flux right up to Mach one. But this duct is not fixed. The primary jet is supersonic, and a supersonic stream expanding to a lower pressure gets larger — every drop in the shared pressure lets the jet swell and take more of the tube, leaving less annulus for the entrained gas. Near Mach one, the flux through the annulus is barely increasing, because the flux function is flat at its peak, and the annulus is shrinking at a finite rate. So the product turns over first.

That maximum is a known object under another name. A duct carrying several streams side by side at a common pressure is choked when

$$\sum_i \frac{A_i\,(1 - M_i^2)}{\gamma\,M_i^2} = 0,$$

the compound-choking condition — a supersonic stream contributes a negative term and a subsonic one a positive term, and the flow is choked when they cancel. For a single stream it reduces to Mach one. At the maximum found above, the sum is $3 \times 10^{-8}$ of the entrained stream’s own term, across four machines from 12 to 150 throats in area.

The choke belongs to the pair. Neither stream is at the condition that would choke it alone: the jet is at Mach 3.86, the entrained gas at 0.886. What cannot be increased is the flow of the two together through the tube they share. The size of the correction depends on how much of the tube the jet occupies. In a 150-throat machine the jet is a small part of the tube, the entrained gas chokes at Mach 0.954, and the sonic assumption is 0.2 per cent out; in a 30-throat machine the jet fills most of it, the gas chokes at 0.773, and the assumption is 9.6 per cent out.

This is also what “information cannot pass upstream” means in a device like this. In a choked nozzle a pressure wave from downstream cannot pass the sonic throat. Here no single stream is sonic, but a disturbance from downstream travels through both streams at once, and the compound condition is exactly the one at which the combined system can no longer carry it forward. That is why the entrainment on the flat part ignores the back pressure, and why the flat part is flat.

One area ratio, and what it trades #

With the supply, the suction and the nozzle fixed, the one free choice is the area of the mixing tube. Enlarging it gives the entrained gas more annulus to flow through, so the entrainment rises. It also gives the jet’s momentum more gas to push, so the pressure the mixed stream can recover falls.

No choice of tube buys both. At a suction of a hundredth of the motive pressure, a tube just large enough to admit the jet compresses by 10.6 and entrains 0.037; one six times as large compresses by 2.6 and entrains 0.90. The curve between them is monotone in both directions at once, which means the tube area is not a parameter to be optimised but a position on a trade — and the right position depends on whether the machine is short of capacity or short of pressure.

That is the liquid jet pump’s lesson in a compressible form. There one area ratio answered three questions differently, and the head and the flow had no common optimum. Here the same ratio answers two questions, entrainment and compression, and they are opposed along the whole range.

The deeper-suction curve lying above the others needs a word. Drawing from a lower pressure means the jet expands further before meeting the entrained gas, so it arrives faster — at Mach 3.86 from a hundredth of the motive pressure against 2.47 from a tenth — carrying more momentum per unit of steam. At a fixed entrainment it can therefore compress by more. That is what the next figure uses.

Why the stages of a vacuum train are not alike #

A single stage cannot take a condenser from a millibar to the atmosphere: it would need a compression ratio of a thousand, and the trade curves above end near ten. So steam ejectors are built in trains, each stage discharging into the suction of the next. The question for the designer is how many.

The compression ratio available rises as the suction deepens. A stage drawing from 1 millibar on a 10-bar supply compresses by 7.8 at an entrainment of a quarter; the same design rule at 100 millibar gives 5.2, and at a bar 3.2. The reason is the one just given: the lower the stage’s suction, the more the motive steam expands before it does any work, and the faster it arrives.

So a train from 1 millibar to the atmosphere is not four equal stages, and the arithmetic is worth doing one stage at a time. The first stage takes the gas from 1 to 7.8 millibar. The second, now drawing from 7.8, manages 6.8 and delivers at 53. The third manages 5.6 and delivers at 299. The fourth, working nearest the atmosphere, manages only 4.3 — and that is enough, delivering at 1,286 millibar. Four stages, and the last is the one that very nearly was not enough. A design rule that treats the stage count as the logarithm of the overall ratio divided by one typical stage ratio overcounts the deep stages and undercounts the shallow ones.

What an intercondenser is for #

There is a second thing wrong with treating the stages as independent, and it is much larger than the first.

Each stage’s motive steam leaves in its discharge, mixed with the gas it was sent to move, and enters the suction of the next stage. At an entrainment of a quarter, every kilogram of gas a stage moves is accompanied by four kilograms of the stage’s own motive steam, so the next stage has five kilograms to move instead of one — and it too adds four times its own load.

With an intercondenser after each stage, the motive steam condenses on the cooling water and only the original gas passes on. Every stage then moves the same load, needs four units of steam to do it, and the train uses sixteen.

Without intercondensers the loads compound: 1, 5, 25 and 125, and the steam needed goes 4, 20, 100 and 500 — 624 units in all. The intercondensers cut the steam consumption by a factor of 39. That is why multi-stage steam ejectors on condensing duties always have them, and why the stages downstream of the last condenser are sized on the whole of the steam that reaches them. It is also the reason a stage that follows a condenser can be treated as starting again from its original load, which is what makes the stage-by-stage arithmetic above valid at all.

The count of four is a result of this model and not a general rule. A larger design entrainment makes each stage weaker and the train longer; a smaller one makes each stage stronger and the steam per stage larger. Real trains to a millibar commonly run to four or five stages, which is consistent, and the ideal model’s figure is a lower bound since every loss it omits weakens the stages.

The ledger #

Two of those rows are checks rather than results, and they are what make the others worth quoting. The compound residual says the maximum of entrained flow found by searching is the condition derived from the compressible equations for two streams in one duct, to eight figures. The shock row says the model’s choice of the subsonic root is the normal-shock state computed a different way.

What the one-dimensional machine cannot show #

The path between the two ends. The model fixes the critical back pressure, and it fixes the back pressure at which the entrainment has fallen to nothing. It does not fix the characteristic between them. Swept below the choke, the discharge pressure first falls and then rises again to the shut-off value, so one back pressure corresponds to several entrainments, and a real machine chooses among them through its shock train and its separated flow — neither of which a uniform-profile balance contains. The dashed lines in the first figure join two computed points and are not a prediction of the shape.

Losses. Every component here is ideal, and real ones are not: a primary nozzle is perhaps 90 to 95 per cent efficient, the mixing is incomplete and frictional, and a diffuser recovers part of what it is given. Each loss lowers the critical back pressure and most lower the entrainment. Design practice applies efficiencies to each component, borrowed from tests, and the resulting numbers are lower than these by a margin that depends on the machine.

Condensation. Steam expanding to Mach 3.9 cools far below its saturation temperature, and in a real nozzle it condenses, releasing latent heat into the jet. That changes the jet’s pressure and velocity, and a perfect-gas model cannot represent it.

Non-condensable gas. The intercondenser argument assumes the load is gas that passes the condensers and the motive steam is all that condenses. A real load is air and vapour, the vapour partly condenses too, and the intercondensers’ own pressure drops take from the stage ratios.

Where the shock stands. The model’s discharge does not depend on the shock’s position; a real ejector’s does, because a shock standing in the mixing tube interacts with its boundary layer, and one standing in the diffuser throat has a different loss from one in the diffuser’s divergence.

Who found it #

Steam-jet pumps were in use on condensers before any of this was written down, and Maurice Leblanc built steam-jet refrigeration machines around 1910 on the principle that an ejector can hold water at a vacuum low enough for it to boil at a few degrees. The one-dimensional theory of the ejector as a mixing device was set out by J. H. Keenan and E. P. Neumann in the 1940s. The idea that the entrained stream is choked beside the jet, and that this is what makes the characteristic flat, is due to J. Fabri and R. Siestrunck’s work on supersonic air ejectors in the 1950s. The constant-area model with a hypothetical throat, in the form used here, is J. T. Munday and D. F. Bagster’s of 1977, developed by B. J. Huang and colleagues in 1999 into the standard design calculation. The compound-choking condition for several streams in one duct is A. Bernstein, W. H. Heiser and C. Hevenor’s, from 1967.

The observation that the textbook form of the model and the compound condition disagree — that the sonic assumption sits beside the maximum rather than at it — is small in a large tube and nine or ten per cent in a small one. It is the kind of thing that is invisible until the model is run at every Mach number rather than at the one assumed.

Still open: a shaped nozzle and the shock it moves #

Every stage above has the same nozzle, sized to the throat and left alone. A real design chooses the nozzle’s exit area too, and that choice decides whether the jet leaves the nozzle over-expanded or under-expanded — whether it contracts or swells on its way to meeting the entrained gas. The compound argument above says the jet’s area at the shared pressure is what couples the two streams, so the nozzle exit is the one piece of geometry that can move the choke itself.

The calculation that follows is the ejector with its nozzle exit as a second design variable, and the question worth asking of it is whether the trade between entrainment and compression, which one area ratio could not escape, is a trade two area ratios can bend. Beside it is the question of losses in the form that matters most: a nozzle efficiency and a diffuser efficiency applied to the same model, to see which of the numbers above are robust to them and which — the width of the fall, most likely — exist only in the ideal machine.