An ejector's nozzle exit is a condition, not a choice

A choke that belongs to two streams built a steam ejector in one dimension with nothing fitted, found that its entrained gas chokes at Mach 0.886 rather than at one because the supersonic jet beside it is part of the same throat, and then enlarged the mixing tube and watched two quantities move in opposite directions. A tube just large enough to admit the jet compressed by 10.6 and entrained almost nothing; one six times larger compressed by 2.6 and entrained 0.90 kilograms of vapour for each kilogram of motive steam. No area on that curve bought both.

Its last section named the obvious objection. The calculation had one design area, and a real ejector has two. The primary nozzle ends somewhere, and where it ends decides how the jet arrives at the pressure it will share with the entrained gas — still expanding, already expanded, or pushed past it and having to recover. The earlier model could not see that area at all: it let the jet expand isentropically from its throat straight to the shared pressure, as though the nozzle had been cut off exactly where that pressure is reached. The question here is what happens when the nozzle is allowed to end elsewhere, and whether a second area ratio can bend a trade that one area ratio could not escape. The second question is the one the last section also raised, and it turns out to have the larger answer: what happens to all of this when the nozzle, the mixing and the diffuser are allowed to lose something.

How a jet meets a pressure it was not built for #

The machine is the earlier one. A converging–diverging nozzle, choked at its throat, is fed from a boiler at the motive pressure; its jet enters a constant-area mixing tube 60 times the throat’s area; the entrained vapour comes from a suction at a hundredth of the motive pressure. Steam is 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. The jet and the entrained gas first share a static pressure at a section called the hypothetical throat, and there the entrained stream is at the Mach number that makes the flow of the pair a maximum. From there they mix at constant area, and a diffuser takes the mixed stream to its stagnation pressure, which is the discharge.

The new piece is the nozzle’s exit, of area $\varepsilon$ in units of the throat. The jet leaves it at the state that area gives, and in general that state’s pressure $p_e$ is not the shared pressure $p_y$. If $p_e$ is higher the jet is under-expanded and swells outside the nozzle; if lower it is over-expanded and is squeezed back. Either way something happens between the exit and the hypothetical throat that the nozzle’s walls no longer govern.

That adjustment is taken as a control-volume balance across the exit, and it is the balance every rocket designer knows as the effective exhaust velocity. Mass and total enthalpy pass unchanged; momentum is the jet’s own plus the pressure excess acting over the exit area:

$$\dot m V_j = \dot m V_e + (p_e - p_y) A_e .$$

The jet’s temperature then follows from its total enthalpy, $T_j = T_0 - V_j^2/2c_p$, and its area at the shared pressure from its mass flow. When the exit is matched — $p_e = p_y$ — the pressure term vanishes, the jet is the isentropic one, and the machine is the earlier model exactly. With the entrained Mach number, the tube, the suction and the gas all held, the two calculations agree to six parts in $10^{16}$ at four different machines, which is the first check. The matched exit for the 60-throat machine at a hundredth of the motive pressure is 13.65 throats, and the jet leaves it at Mach 3.86.

A rocket’s rule in a vacuum pump #

The first figure is the only one that does not need the ejector. It asks what effective velocity a nozzle with a given exit delivers into a given pressure, and the answer is the oldest result in propulsion: the velocity is largest when the exit pressure equals the pressure outside, and it falls on both sides. A nozzle with half the matched exit leaves 1.6 per cent of the velocity behind; one with twice the matched exit, 2.0 per cent. At a fifth or at five times, the losses are 8 and 14 per cent.

The reason is worth a sentence because it is general. The pressure term $(p_e - p_y)A_e$ is a force the jet receives over a surface where its pressure differs from its surroundings, and it is a real force — but the isentropic expansion the nozzle declined to do would have delivered more. An under-expanded jet has kinetic energy still locked in its pressure and releases only part of it as momentum; an over-expanded jet has spent kinetic energy expanding past the pressure it will meet and pays it back in a compression that is not reversible, and an irreversible compression has a price in entropy that no later expansion refunds. Every mismatch is a loss, and the matched exit is not a compromise but a maximum. The same asymmetry shows up that an over-expanded rocket nozzle makes a practical matter: over-expansion costs more than under-expansion by the same factor, because the pressure deficit acts over a larger exit.

That already predicts the answer to the design question. If the exit area can only lower the jet’s effective momentum, it can only weaken the machine — unless something about the jet’s area at the shared pressure changes in a way that the lost momentum does not undo.

The second area sits on top of a hill #

The second figure asks it. It holds the tube at 60 throats, the suction at each of three pressures, and sweeps the exit area from a quarter of the matched value to four times it, recomputing the whole critical point at every exit: the jet’s effective state, the compound choke, the mixing, the discharge. Both the entrainment and the compression ratio peak at the matched exit, at every suction drawn. At a hundredth of the motive pressure, halving the exit costs 2.4 per cent of the entrainment and 1.5 per cent of the compression; doubling it costs 3.2 and 2.0 per cent; a factor of four either way costs between 9 and 17 per cent of the entrainment.

This is the result that answers the question, and it is stronger than it looked like being. If a mismatched exit had lowered compression while raising entrainment, the exit area would have been a second degree of freedom, and a designer short of capacity could have bought it with an under-expanded jet. It does neither. Both quantities fall together, so the exit area is not a place on a trade at all. It is a condition: the nozzle should end where its jet meets the shared pressure, and every other choice is a loss with no compensating gain.

The mechanism is the effective velocity above, arriving in two places. An under-expanded jet reaches the shared pressure hotter and slower than an isentropic one, so it carries less momentum into the mixing, and for the same mass flow it is wider — at half the matched exit the jet takes 14.5 throats of the tube instead of 13.7 — leaving less annulus for the entrained gas. An over-expanded jet recompresses to the shared pressure through the same kind of irreversible jump and arrives wider still, 15.3 throats at twice the matched exit. Neither the lost momentum nor the lost annulus has anything to trade against.

The frontier does not move #

A peak at one tube does not settle the question for all of them. The earlier essay’s trade curve was drawn by varying the tube’s area, and a mismatched exit might in principle push part of that curve outwards even if it lowers both quantities at the one tube examined. The third figure redraws the whole trade four times: with the exit matched in every tube, and with it held at a quarter, a half and twice each tube’s own matched exit.

Every mismatched curve lies inside the matched one. At a compression ratio of 5 the matched frontier entrains 0.267; the half-exit curve 0.246, the double-exit curve 0.239, the quarter-exit curve 0.190. At a compression ratio of 3.4, 0.552 against 0.523, 0.514 and 0.444. Freeing the exit area adds a dimension to the design space and not a single point to the frontier. The trade that one area ratio could not escape is exactly the trade two cannot either, and a designer who treats the exit as a second knob has one knob and a way of losing.

There is an echo here of the liquid machine. In the nozzle that is best at one thing one area ratio answered three questions differently and none of them jointly; there the second free area — the nozzle, relative to the throat — was a real trade, because an incompressible jet has no expansion to get wrong. The compressible jet has one, and getting it right removes the exit from the list of choices rather than adding it.

A nozzle has one suction it was built for #

The practical form of the matched-exit rule is less comfortable. A matched exit is matched to a pressure, and the shared pressure depends on the suction the machine happens to be drawing. A vacuum ejector pulled down from atmosphere, or one whose load changes, runs at suctions far from the one its nozzle was cut for. The fourth figure fixes the nozzle at the exit matched for a hundredth of the motive pressure and runs it from two thousandths to a tenth.

Above the design suction the jet leaves over-expanded, and the machine loses entrainment and compression by a few per cent — 3 and 5 per cent at five hundredths, 6 and 9 per cent at a tenth. Below it the jet leaves under-expanded, and the loss grows much faster: at two thousandths the fixed nozzle entrains 78 per cent of what a matched one would, and at 1.1 thousandths the jet, swelling outside the nozzle to reach the low shared pressure, fills the tube and entrains nothing at all. The asymmetry is the opposite of the rocket’s, and the reason is the tube. A rocket’s jet has the whole atmosphere to swell into; an ejector’s jet has a fixed tube, and at a deep suction the tube is barely larger than the jet already, so the extra width an under-expanded adjustment adds comes straight out of the annulus.

This is the figure to read with the most caution, and the reason is stated where the model’s limits are. The width an under-expanded jet reaches depends on how it expands outside the nozzle, and a one-dimensional balance gives it a single uniform state where a real jet overshoots, forms a barrel of shock cells and reaches its widest well downstream of the exit. The direction of the effect is robust. Its size near the point where the jet fills the tube is not.

Only the nozzle’s loss reaches the flow #

The exit area was the question the earlier essay asked. The losses were the one it named as more important, and the fifth figure takes them one at a time. A nozzle efficiency applies to the jet’s kinetic energy: at 0.9, the jet reaches the shared pressure with nine-tenths of the kinetic energy the isentropic expansion would have given it. A mixing efficiency applies to the momentum carried into the mixing tube, the form Huang and his colleagues used in 1999 to absorb wall friction and incomplete mixing. A diffuser efficiency applies to the pressure the diffuser recovers from the mixed stream’s velocity.

The left panel has the result the calculation was worth doing for. The diffuser’s efficiency does not move the entrainment at all, and neither does the mixing efficiency over the whole range a working machine occupies — not by a small amount but by nothing, to every digit the calculation carries. The critical entrained flow is set at the hypothetical throat, where the jet and the entrained gas first share a pressure and the compound choke forbids more flow; mixing and diffusion happen downstream of that, and on the flat part of the characteristic nothing downstream of a choke can reach upstream of it. That is the throat that stops listening again, now doing a job in diagnosis. Only the nozzle’s loss is upstream of the choke, and it lowers the entrainment by 4 per cent at an efficiency of 0.95 and by 8 per cent at 0.9, because a slower, hotter jet both drags less gas and takes more of the tube.

The mixing curve has a knee, and the knee is a second choke. As the mixing efficiency falls, the momentum available to the mixed stream falls with it, and the subsonic root of the mixing balance rises towards Mach one: 0.47 in the ideal machine, 0.73 at an efficiency of 0.8, 0.91 at 0.765. That is a stream losing momentum to the wall at constant area, which is exactly Fanno’s line, and it ends where Fanno’s line ends. Below an efficiency of 0.76 the mixed stream would have to be supersonic to carry the flow, the tube itself chokes, and from there the mixing loss does reach upstream: the choke moves from the hypothetical throat to the tube’s exit, as friction moves a nozzle’s sonic point downstream of its geometric throat, and the entrainment falls, 2.4 per cent at 0.75. A real ejector with mixing that bad is broken rather than inefficient, but the knee says where the statement “downstream losses cannot touch the flow” stops being true, and why.

The right panel is the other half. Every loss lowers the critical compression ratio, and the mixing loss lowers it most: at 0.8 the ratio falls from 3.42 to 2.62, where the diffuser’s loss at 0.8 costs only 3.32. The mixing efficiency multiplies the momentum from which the whole pressure rise is made, while the diffuser’s acts only on the dynamic pressure left in a stream already slowed to Mach 0.47.

The practical reading is a diagnosis. An ejector that entrains less than its rating on a test stand, at a back pressure below its break, has a problem upstream of the choke — its nozzle, its steam quality, its supply pressure. An ejector that entrains its rating and breaks early has a problem downstream. The two symptoms point at different parts, and the model says why they cannot be confused.

The sharp edge belongs to the ideal machine #

The earlier essay’s most striking figure was a cliff. The ideal machine entrained its full rating up to a critical back pressure and nothing at all a little above it, and it noted that real machines fall more gradually. The sixth figure puts a number on how much of the difference losses alone account for. The fall is measured from the critical back pressure to the back pressure at which the entrainment reaches zero, with the nozzle fixed at the exit it has at the critical point.

In the ideal machine the fall takes 0.98 per cent of the critical back pressure. At component efficiencies of 0.95 it takes 3.2 per cent; at 0.9, 6.3; at 0.85, 11. The losses lower the critical point further than they lower the shut-off, because the critical point is a stream in full flow paying every loss on every kilogram, while at shut-off the entrained stream is almost stagnant and the mixing and diffusion that the efficiencies tax are nearly absent. So the cliff tilts.

That is the prediction the earlier essay’s last section made about which of its numbers would not survive: the width of the fall exists only in the ideal machine. The losses do not produce the smooth, rounded break of a measured characteristic — that needs the shock train and the separation in a real mixing tube, which a uniform-profile balance cannot contain — but they account for a fall of a few per cent to ten or more, which is the range measured machines occupy.

A four-stage train needs five #

The earlier essay’s train ran from a millibar on a 10-bar supply to the atmosphere in four stages, compressing by 7.8, 6.8, 5.6 and 4.3, each stage sized for an entrainment of a quarter and each following an intercondenser. The seventh figure repeats the design with losses, re-sizing every stage’s tube so that it still entrains a quarter.

With nozzle, mixing and diffuser efficiencies of 0.95, 0.95 and 0.9, the stages compress by 6.6, 5.9, 5.0, 4.0 and 3.0, and the train needs five. With 0.9, 0.9 and 0.85 the first stage manages only 5.6, the stages run 5.6, 5.1, 4.5, 3.8 and 3.0, and the train still needs five, reaching the atmosphere by a smaller margin. The ideal model’s “four” was a lower bound, as that essay said, and the bound is not tight.

The fifth stage costs a quarter more motive steam with intercondensers — five stages of four units each against four — but it costs far more without them. The load compounds fivefold through each stage that is not followed by a condenser, so an uncondensed train uses $4(1 + 5 + 25 + \dots)$ units: 624 for four stages and 3,124 for five. The factor between the two designs grows from 39 to 156. Losses make intercondensers more valuable, not less, because they lengthen the train over which the uncondensed load would compound.

What the calculation was checked against #

The ledger carries three checks. The matched, lossless machine reproduces the earlier model at four machines of different tubes and suctions to six parts in $10^{16}$, which says the new pieces reduce to the old when they should. The exit balance closes mass, momentum and energy to two parts in $10^{16}$ at three exits and three shared pressures, with a nozzle efficiency of 0.93 so that the non-isentropic branch is the one exercised. And the matched exit’s effective velocity exceeds every other exit’s in a sweep from a fifth to five times the matched area, which is the lemma the whole design result rests on.

One observation belongs with the checks rather than the results. With a mismatched exit the critical point no longer satisfies the compound-choking condition written for two isentropic streams — the residual is nearly half the entrained stream’s own term — and that is correct rather than a failure. The condition is derived for streams whose areas respond to the shared pressure isentropically, and a jet that adjusts through an irreversible balance at its exit responds differently. The critical point is still the maximum of the entrained flow, found directly; what changes is only that the closed-form condition for it no longer applies.

What one dimension leaves out of the exit #

The jet’s real expansion. An under-expanded jet does not jump to a uniform state at the shared pressure. It expands through a fan at the nozzle lip, overshoots, and forms a train of shock cells, and it reaches its greatest width some distance downstream. The entrained gas chokes against that widest section, and the classical account of this — Fabri’s, from the 1950s — puts the choke there rather than at a uniform section. The one-dimensional balance gets the effective momentum right and the width only approximately, which is why the off-design figure’s steep end is the least reliable number here.

Efficiencies as constants. A nozzle’s efficiency depends on its expansion ratio and its surface; a mixing efficiency on the tube’s length and the velocity ratio of the streams. Treating each as one number over a sweep is the practice of design methods, not a result, and it is the reason the fall’s width is quoted as a range.

Condensation. Steam expanding to Mach 3.9 supercools and condenses in the nozzle, and the heat released changes the jet’s state in a way no perfect-gas nozzle efficiency represents. It is a loss in the nozzle, and so by the argument above it reaches the entrainment.

The shock train. The one-dimensional machine’s discharge does not depend on where the terminal shock stands; a real one’s does. The losses here do not replace that physics, and a characteristic’s shape between its two ends remains outside the model.

The convention: pressures against the supply #

Pressures are fractions of the motive stagnation pressure; areas are in units of the nozzle throat. Entrainment is the entrained mass flow over the motive; the compression ratio is the critical back pressure over the suction. The matched exit is the one whose isentropic — or, with a nozzle efficiency, lossy — expansion delivers the jet at the shared pressure of the critical point. Nozzle efficiency is on kinetic energy, mixing efficiency on the momentum entering the tube, diffuser efficiency on the isentropic pressure rise.

Who worked it out #

The effective exhaust velocity and the rule that thrust is largest at matched expansion are as old as rocket nozzle theory, stated in the 1920s and 30s by Oberth and by Goddard’s successors. Keenan and Neumann’s constant-area and constant-pressure ejector models of 1950 put the ejector on a one-dimensional footing; Fabri and Siestrunck in 1958 located the entrained stream’s choke against the primary jet’s greatest width; Munday and Bagster in 1977 introduced the hypothetical throat; Huang, Chang, Wang and Petrenko in 1999 attached measured component efficiencies to the model and fitted them against steam and refrigerant ejectors, which is the form the losses take here.

Still open: the jet’s widest section, and the choke that sits against it #

The calculation’s weakest number is the jet’s width outside a mismatched nozzle, and the fix is a known one. An under-expanded jet’s boundary can be computed by the method of characteristics from the nozzle lip to its first maximum, with the entrained stream’s pressure acting on it, and the entrained gas’s choke placed at that maximum as Fabri placed it. The next calculation does that, and asks two things of it: whether an under-expanded jet, whose boundary overshoots before settling, chokes the entrained flow earlier than the one-dimensional balance says — which would steepen the off-design loss below the design suction — and whether, with the choke at the jet’s real widest section, the matched exit is still the one that entrains most, or whether a slightly under-expanded nozzle, whose overshoot is small and whose momentum loss is second order, can edge past it.