The nozzle that is best at one thing

Mixing is a pump drew a box round a mixing tube and found that two streams at different speeds, forced to become one, raise their pressure while destroying energy. The box had nothing else in it. The motive jet arrived from a perfect nozzle, the entrained stream from a perfect inlet, the mixed stream left through a tube with no wall friction, and nothing recovered its velocity afterwards. That model had one loss — the mixing — and a design sweep with a peak in delivered power at about half the tube’s area.

A real water jet pump has four more losses, each small, and they change what the design question is. The nozzle does not deliver all of the head it is given as speed. The entrained stream loses a little on its way round the nozzle into the throat. The throat has a wall, and the wall has friction. And the mixed stream leaves through a diffuser, which recovers most of its velocity head as pressure and not all of it.

With those four back, the machine has a reason to prefer one nozzle over another, and it turns out to have three different reasons that point at three different nozzles.

The machine, written down #

The quantities that specify a jet pump are two ratios. The flow ratio $M = Q_s/Q_m$ is how much fluid is entrained for each unit of motive fluid. The head ratio is what that entrained fluid gains against what the motive fluid gives up:

$$N = \frac{P_d - P_s}{P_m - P_d},$$

where $P_m$, $P_s$ and $P_d$ are the total pressures of the motive supply, the suction plenum and the discharge. The design variable is the area ratio $R$, the nozzle’s exit area as a fraction of the throat’s.

Everything else is the momentum balance from the essay before, with a loss coefficient attached to each stream’s dynamic head: $K_n$ on the nozzle jet, $K_s$ on the suction inlet, $K_t$ on the mixed stream for throat friction, and $K_d$ on the mixed stream for the diffuser. The values used here — 0.05, 0.10, 0.10 and 0.15 — are borrowed. They are the representative figures jet-pump design practice quotes for a well-made machine with a smooth nozzle, a bell-mouthed suction and a throat about seven diameters long, and they are the only numbers in this essay that were not computed.

The family has the shape anybody who has read a centrifugal pump’s data sheet expects: high head at no flow, falling to no head at the most flow. What differs from nozzle to nozzle is the trade between the two ends. A nozzle filling four-fifths of the throat makes a head ratio of 3.2 with nothing entrained and stops entraining altogether at a flow ratio of 0.20. A nozzle a tenth of the throat makes a head ratio of 0.22 and keeps entraining until the flow ratio reaches 4.45.

What efficiency is here, before it is anything else #

The efficiency of this machine is the hydraulic power the suction stream receives over the hydraulic power the motive stream gives up. The two flows are $Q_s$ and $Q_m$, the two pressure differences are the numerator and denominator of $N$, and so

$$\eta = \frac{Q_s (P_d - P_s)}{Q_m (P_m - P_d)} = M\,N.$$

That line looks like bookkeeping, and it says something unusual about the device. The efficiency is fixed by the duty, not by the machine. Any jet pump that delivers a flow ratio of 0.9 at a head ratio of 0.37 is 33.3 per cent efficient, whatever its nozzle, its throat and its losses. What the construction decides is whether it can reach that duty, and which nozzle it needs to.

The picture makes the identity concrete. Two machines were solved for the area ratio that puts their characteristic through $M = 0.9$, $N = 0.37$: an idealised one with every component loss set to zero, and a realistic one. The idealised machine needs a much narrower nozzle — 0.172 of its throat against 0.267 — because it wastes nothing and so can afford a larger velocity mismatch. The two efficiencies at the duty agree to $10^{-16}$, which is what an identity looks like computed.

The consequence is that “a more efficient jet pump” is an ambiguous phrase. At a fixed duty there is no such thing. There is only a jet pump that can reach a more efficient duty — a larger product of flow ratio and head ratio — and the question worth asking of a design is which duties it can reach and which of them the application actually wants.

A centrifugal pump does not behave like this, and the reason is worth a sentence. Its efficiency is shaft power into hydraulic power, and the shaft power is not one of the two duty numbers; a better impeller delivers the same flow and head for less torque. A jet pump has no shaft. Its only input is another stream of the same fluid, measured in the same units as the output, so the efficiency collapses into the product of two ratios the duty already names.

The head, and the one optimum that has a closed form #

At shut-off nothing is entrained, the suction stream is stationary, and the mixed stream in the throat is just the motive jet spread over the throat’s area, at speed $R$ in units of the jet. The discharge pressure then comes out of the momentum balance in one line:

$$P_d = R - \tfrac{1}{2}(1 + Kt + Kd)\,R^2,$$

which is a parabola in the area ratio. The head ratio rises with $P_d$, so it peaks where the parabola does:

$$R_\text{head} = \frac{1}{1 + Kt + Kd}.$$

With the borrowed losses that is exactly 0.800, and a golden-section search over the full model lands on 0.800 to eight figures.

Two things about that expression are more interesting than its value.

The first is what is not in it. The nozzle loss does not appear. It scales the motive head and so moves the height of the peak — from 3.2 to 4.0 if the nozzle were perfect — without moving its location at all. A designer tuning a nozzle’s finish is tuning how much head the machine makes, not which machine makes the most.

The second is what the parabola is. Its linear term, $R$, is the momentum the jet brings into the throat; its quadratic term is the velocity head the mixed stream must carry out, charged at one plus whatever the throat and diffuser take of it. A wider nozzle brings more momentum and forces a faster mixed stream, and the downstream losses charge for speed. With no downstream losses the charge is the bare velocity head, and the peak is at $R = 1$ — a machine that is all nozzle and no suction, which makes the most head by having nothing to lift. The losses are what bring the optimum back inside the range where a jet pump is a jet pump.

The efficiency, which peaks somewhere else entirely #

The efficiency along each characteristic rises from zero at shut-off, where nothing moves, to a peak, and falls to zero again at free delivery, where nothing is lifted. Every area ratio has such a peak, and the peaks themselves form a curve with its own maximum.

The best efficiency found by searching over both the area ratio and the flow ratio is 33.4 per cent, at $R = 0.275$, $M = 0.92$ and $N = 0.36$. That figure is roughly half of what a good centrifugal pump reaches and is in the band jet pumps are known to reach in practice. It is not a claim about any particular machine, because the loss coefficients are borrowed; it is what those coefficients imply.

The peak is broad. Anything between a fifth and two-fifths of the throat is within a couple of points of it — 32.8 per cent at $R = 0.2$ and 32.2 per cent at $R = 0.4$ — which is why published recommendations for the area ratio range from 0.2 to 0.35 without contradicting each other.

Three questions, set side by side #

Putting the two curves on one axis makes the separation hard to miss.

A machine built at 0.800 makes the most head and wastes most of its motive power doing it. A machine built at 0.275 is the most efficient there is and makes a quarter of the head. There is no compromise between them that is good at both: the curves cross at about 0.6, where each is at three-quarters of its best.

The third question, the flow, does not produce a third peak. The flow ratio at free delivery rises steadily as the nozzle narrows — 0.20 at the head optimum, 1.72 at the efficiency optimum, 4.45 at a tenth of the throat and 13.5 at a fiftieth — and nothing in the machine turns it over. Entrainment is bought with head at every area ratio, and the purchase has no natural end. The entrainment optimum is not an optimum; it is the smallest nozzle anybody is prepared to make, and in practice that is decided by manufacturing, by how clean the motive fluid is, and by how little head the application can tolerate.

So the claim that a jet pump has “an optimum area ratio” is true of one of three questions and silent on the other two. A deep-well pump lifting water from a borehole wants head and is built wide. A slurry or fume eductor wants volume and is built narrow. A jet pump used as a booster in a system where the motive fluid is expensive wants efficiency, and only that one is built at a quarter.

What pins the efficiency optimum #

The head optimum was pinned by the downstream losses, in closed form. The efficiency optimum has no closed form, and the most direct way to find what holds it in place is to remove the losses one at a time and watch it move.

The inlet-side losses barely matter. Removing the nozzle loss adds three points of efficiency and moves the optimum by two hundredths; removing the suction loss adds two points and moves it the other way by the same amount. The downstream pair are the ones that hold the optimum where it is: without them it moves two-thirds of the way towards the middle of the range and the efficiency nearly doubles.

The reason is the same one that placed the head optimum. A wider nozzle brings the two streams closer in speed, which is good for mixing, and makes the mixed stream faster, which is bad for the throat and the diffuser. The efficiency optimum is where those two effects balance, and the next figure shows them doing it.

The mixing loss falls steadily with the area ratio, because the velocity mismatch it is made of shrinks. The diffuser and throat losses rise, because they charge for the mixed stream’s velocity head, and that velocity is proportional to $R$. The crossover of mixing and diffuser is near 0.6, but the optimum is not at the crossover: it is where the marginal saving in mixing loss from widening the nozzle equals the marginal cost in the downstream pair, and at a quarter of the throat the mixing loss is still falling fast enough to be worth chasing. The same balance, run with the throat and diffuser made perfect, has only the rising suction and nozzle shares to stop it, and they rise slowly — which is exactly why the optimum walked to 0.46 when they were all that was left.

Where the power goes at the best point #

The budget closes, and it closes because two independent calculations agree. The motive power given up and the suction power gained come out of the momentum balance and the pressures it produces. The five losses come from five separate expressions — four loss coefficients times their streams’ dynamic heads, and the two-stream Borda–Carnot expression for the mixing. The difference between the first pair and the sum of the second is $1.8 \times 10^{-15}$ of the motive power at the worst of twenty operating points, which is rounding.

Mixing takes more than the four component losses together — 37.4 per cent against 29.2. That is the number to hold on to when a jet pump is being improved, because it says where improvement cannot come from. A perfect nozzle, a perfect inlet, a frictionless throat and a lossless diffuser would, between them, recover less than the mixing destroys, and the mixing is the machine’s mechanism. It is the Borda–Carnot loss that makes the pressure rise in the first place, and a jet pump that mixed without loss would also mix without lifting.

The diffuser is the largest of the component losses, which is worth a remark because a diffuser is usually thought of as the part that recovers something. It does: it turns 85 per cent of the mixed stream’s velocity head into pressure. But that head belongs to the whole combined flow, moving at the throat’s speed, and fifteen per cent of it is a large number in a machine whose useful output is a third of its input.

The same argument in a different machine #

The pattern here — one geometric ratio, several things worth optimising, a different optimum for each — is not special to jet pumps. A rotating machine’s specific speed sorts impellers by what they do best, and the axial pump that is best at moving a large flow against a low head is a poor machine for a high head at a small flow. The difference is that a rotating machine has enough free geometry to be redesigned for each duty — blade angles, number of blades, diameter — while a jet pump has one ratio and a length, so its whole design space is visible on one axis.

What is surprising is how much of the jet pump’s behaviour is carried by the one closed form. The head optimum at $1/(1 + Kt + Kd)$ has the same structure as a Pelton wheel’s half-speed rule: a product of a quantity rising linearly and a penalty rising as its square, peaking where the two balance. In the wheel it is force times bucket speed; in the jet pump it is momentum brought in against velocity head charged out. Both are the same parabola, and both put the answer at a ratio of losses rather than at anything to do with size.

What the picture cannot show #

No mixing length. The model assumes the streams are fully mixed by the throat exit. A throat shorter than about six diameters does not achieve that, and the whole characteristic falls; a throat much longer pays more friction. The value of $K_t$ borrowed here belongs to a throat of about seven diameters, and a different length is a different $K_t$ and a different machine.

No nozzle-to-throat spacing. Real jet pumps are sensitive to how far the nozzle sits upstream of the throat entry, which changes how the suction stream is accelerated round it. That enters the model only through $K_s$ and is otherwise invisible.

No viscosity dependence in the losses. The four coefficients are constants. In a real machine they rise at low Reynolds number, and a jet pump handling a viscous oil has a lower, narrower efficiency peak at a wider nozzle. The loss coefficients are where that would enter, and they are borrowed rather than modelled.

No cavitation. The lowest pressure in the machine is at the throat entry, and at a low suction pressure it can reach vapour pressure before any of the optima above are reached. That limit is a wall rather than a curve, and it is large enough to need its own argument.

Uniform profiles on every face. The motive jet is a top-hat and the mixed stream at the throat exit is uniform. Both are idealisations, and the second is the same one the mixing-tube essay flagged as its largest single gap.

Who worked it out #

The one-dimensional theory of the water jet pump was set out by J. E. Gosline and M. P. O’Brien in 1934, at the University of California, for lifting oil from wells, and it is essentially the momentum balance with loss coefficients used here. R. G. Cunningham’s work at Pennsylvania State University in the 1950s and 1960s extended it to viscous fluids and to cavitation, and established the loss values and the 0.2-to-0.35 range of area ratios that design practice still quotes. The Borda–Carnot loss at its centre is from 1766, and the momentum balance is Newton’s.

What none of them needed, and what makes the separation of the optima easy to see now, is the ability to sweep every area ratio, every flow ratio and every combination of losses in a fraction of a second. The closed form for the head optimum was always there to be written down; the efficiency optimum’s dependence on the downstream losses, rather than on the nozzle, is the kind of thing that is clear only once the losses can be switched off one at a time.

Still open: where the machine stops being allowed to run #

Every optimum above assumes the machine can operate wherever its characteristic says. It cannot. The entrained stream reaches its lowest pressure where it enters the throat, beside the jet, and in a liquid that pressure has a floor at the vapour pressure. At a low enough suction pressure the floor arrives before the best-efficiency point does.

What happens then is not a gentle loss of performance. The next calculation is the flow ratio at which the throat entry reaches vapour pressure — a closed form in the area ratio, the suction margin and the two inlet-side losses — and what it does to the characteristic, which is to put a vertical wall in it past which no lower discharge pressure moves any more fluid. The questions it answers are how much suction margin the efficiency optimum needs, and what the best reachable efficiency becomes when the margin is short.