Two small pumps carry two big discs

One fast pump or two slow ones worked out how a pumping station of several identical pumps should run them: how many at once, at what shared speed, and when to start the next. Its last comparison put one large pump under a variable-speed drive against two half-size pumps staged perfectly, over a year of flows, and found that the pair wins when the system has a lot of static lift and loses when it has little. Where the two crossed depended on a number the essay stated and did not derive: a penalty of two points of best efficiency for the smaller pumps. With it, the crossing fell at 44 per cent static lift. The essay said plainly that the penalty was a guess with a physical source, and that the source was the losses that grow as a share of the power when the specific speed falls.

This builds that source, and the first thing it finds is that the smaller pumps are not smaller in the way the argument assumed.

Half the flow is not half the specific speed, and not half the pump #

The specific speed of a duty is $\Omega_s = \omega\sqrt{Q}/(gH)^{3/4}$. It goes as the square root of the flow, so a pump taking half the flow at the same head and shaft speed has $1/\sqrt2$ of the specific speed, not half — 0.385 against 0.545 for the station’s duty of a tenth of a cubic metre a second at 40 metres and 1450 revolutions a minute. The essay below wrote “half its specific speed” in two places; the factor is 0.707, and it has been corrected there.

The more important consequence is the impeller’s size. A centrifugal pump’s head is set by its tip speed: $gH = \psi u_2^2$ with a head coefficient $\psi$ near a half for radial impellers. Same head, same coefficient, same tip speed; same tip speed and same shaft speed, same radius. The half-flow pump has the full-flow pump’s impeller diameter, 369 millimetres here. What shrinks is everything the flow passes through: the outlet width, which carries the flow at a fixed fraction of the tip speed and so goes as $\Omega_s^2$, and the eye, whose area carries it at a fixed fraction of the eye’s own speed and whose radius therefore goes as $\Omega_s^{2/3}$. Two half-flow pumps are two narrow pumps of full diameter.

The four losses, built on that geometry #

Three stated design coefficients fix the impeller from its duty — a head coefficient of 0.5, an outlet flow coefficient of 0.12 and an eye flow coefficient of 0.25 — and four losses are built on it. Disc friction is the torque on the two shrouds spinning in the casing, treated as enclosed discs in the turbulent regime with Daily and Nece’s torque coefficient, $C_M = 0.051(s/r)^{0.1}Re^{-1/5}$. Leakage is the flow back through the wear ring at the eye, an orifice of discharge coefficient 0.6 across 70 per cent of the head, with a radial clearance of an eighth of a millimetre plus a thousandth of the ring’s radius — a floor the ring cannot be made below and a part that grows with it, as running clearances are specified. Passage friction is the impeller channel and the volute treated as ducts, with friction factors from Haaland’s formula for a cast surface of 25 micrometres, a roughness a viscous sublayer can hide when it is thick enough. And a fixed share of seven per cent of the head, plus one per cent for the bearings and seal, stands for mixing, incidence and diffusion, which do not change with the size or the specific speed in this model.

The coefficients are typical rather than measured on any one pump, and the claim is the shape. One number picks the machine set the efficiency side of the specific speed aside as an empirical envelope; this is a way of seeing where the envelope’s low side comes from. At a specific speed of 1.0 the efficiency is 0.902 and almost all of the loss is the fixed share. At 0.3 it is 0.777, with disc friction at 7.7 per cent of the shaft power and passage friction at 5.3. At 0.15 those two take 38 per cent between them and the pump is barely half efficient — which is why duties of very low specific speed go to multistage pumps, and to positive-displacement machines below that.

What the discs cost, and why it is the inverse square #

The disc term makes the whole argument in one line. The shrouds’ power is $C_M\rho\omega^3 r_2^5$ and the useful power is $\rho g Q H$, and with $r_2 = u_2/\omega$ and $gH = \psi u_2^2$ their ratio is

$$\frac{P_{\text{disc}}}{\rho g Q H} = \frac{C_M}{\psi^{5/2}\,\Omega_s^2},$$

the inverse square of the specific speed, with the size entering only through the Reynolds number’s fifth root in $C_M$. A pump of lower specific speed is one that spins the same discs for less flow.

In kilowatts the station’s case is stark. The single pump spends 1.17 kilowatts on its discs to deliver 39.1 kilowatts to the water. Each half-flow pump has the same impeller and spends the same 1.17, so the pair spends 2.34 for the same 39.1. Its leakage rises from 0.84 to 1.21 kilowatts, because two eyes have more circumference than one, each fitted with a clearance whose floor does not shrink. Its passage friction rises from 0.87 to 1.64, because each outlet channel is half as wide for the same tip speed. The fixed share is the same. The pair is 0.825 efficient against the single pump’s 0.867: 4.2 points, more than twice the two the station comparison assumed.

The discs are boundary layers #

The torque on a shroud is not a mysterious machine loss; it is skin friction on a spinning surface. A disc turning in open fluid drags a layer round with it and flings it outwards, and von Kármán’s solution for the free disc is one of the few exact solutions of the full equations with the nonlinear terms kept. Enclosed in a casing a few per cent of its radius away, the disc and the casing each carry a layer and the fluid between them turns at about half the disc’s speed. Daily and Nece’s measurements sorted the enclosed disc into four regimes by gap and Reynolds number; a pump’s shrouds, at Reynolds numbers of millions and gaps of a fiftieth of the radius, sit in the turbulent regime with separate layers, where the torque coefficient falls as the inverse fifth root of the Reynolds number for the same reason a turbulent plate’s friction does.

That makes the size dependence of the disc term weak and its specific-speed dependence strong. A pump twice as large at the same specific speed has a torque coefficient 13 per cent lower; a pump of half the flow at the same size has a disc share twice as large. The ring leakage is the other classical term, an orifice whose discharge coefficient is only the shape of the hole, and it is the one whose clearance does not scale, which matters more for size than for specific speed.

Three powers of the specific speed #

On logarithmic axes each loss is close to a straight line, and the slopes are the geometry. Disc friction falls with a slope of exactly −2. Passage friction falls almost as steeply, −1.97, because its narrowest dimension is the outlet width, which goes as $\Omega_s^2$: the friction loss of a channel at fixed velocity goes as its length over its hydraulic diameter. Leakage falls more slowly, −1.09 here, because the eye radius goes only as $\Omega_s^{2/3}$, and a clearance with a floor is relatively wider on a smaller ring, which puts the slope between −2/3 and −4/3.

That ordering is the reason a low-specific-speed pump is inefficient for the reasons it is. It is not mainly leakage, though leakage is what pump handbooks list first. At the station’s 0.385 the discs and the passages together are three times the leakage, and both of them grow as the inverse square. A multistage pump, which divides the head between stages so that each stage sees a higher specific speed, is attacking exactly that inverse square.

The penalty, at every duty #

The penalty for splitting depends on where the duty starts, and steeply. At a specific speed of 1.2 two pumps give up 1.25 points; at the station’s 0.545, 4.2; at 0.3, ten. Three pumps, each at $1/\sqrt3$ of the duty’s specific speed, give up eight points at the station’s duty. The two-point penalty the earlier essay assumed is right for a duty of specific speed near 0.9 and too small for anything below it.

The station, re-run #

The station comparison is run again with nothing changed but the penalty, applied as the budget’s ratio of efficiencies to the earlier model’s best efficiency of 0.82 — four points on that scale. Two staged pumps with no penalty beat one pump at every lift, because staging is a second control. With two points they lose below 44 per cent lift, as before. With the derived four they lose below 65 per cent: at 60 per cent lift, where the earlier essay found the pair ahead, 0.781 against 0.769, the pair is now behind, 0.761 against 0.769. Three pumps, giving up eight points, never catch one pump below 95 per cent lift.

The station’s own rule from the essay below — that every running pump sits at its best point when the flow per running pump equals the square root of the system head, $qN/n = \sqrt h$ — is untouched. What has changed is the price of having the second control, and for a duty of this specific speed it is a price most systems do not have enough lift to pay.

Where splitting pays #

Swept across the duty’s specific speed, the crossing becomes a boundary. Above the curve the split station saves energy; below it one large pump with a drive does, and the case for several pumps rests on standby — the ability to lose one and keep supplying — rather than on the power bill. Two pumps never save energy below a duty specific speed of 0.46 at any lift up to 95 per cent. At 1.2 they save above a third of the head as lift, and three pumps above 39 per cent.

The surprising part is which way the boundary runs. The intuition behind staging is that low flows are where a single large pump suffers, so a duty that is small for its head — a low specific speed — looks like the natural candidate for splitting. It is the worst candidate: its pumps are already on the steep part of the inverse square, and splitting moves each of them further up it. The specific speed a pump spends its life at found that speed control drags a pump with static lift towards low specific speeds over its duty; splitting does the same thing to the pump’s design, and does it permanently.

In a year’s electricity #

The station’s duty, run for a year — a tenth of the time at full flow, a quarter at 80 per cent, 35 per cent at 60 and 30 per cent at 40 — delivers 142 megawatt-hours to the water against a system with 30 per cent static lift. One pump under a drive draws 177 megawatt-hours to do it. Two staged half-flow pumps, priced at the derived penalty, draw 185: seven megawatt-hours a year more, about four per cent, for the privilege of having a second control the system did not need. Under the assumed two points the same comparison came out nearly even. The direction of the decision has changed, and so has its size: at an industrial tariff the difference is a recurring cost of the same order as the drive that would have avoided it.

None of that argues against several pumps. A water supply needs standby, and a station that can lose a pump and still meet its peak buys that with the second unit whatever the energy says. What the budget changes is the reason given for the second unit. On a duty of this specific speed with ordinary lift, it is redundancy, paid for in energy, not a saving.

The loss budget in a sentence, and a check #

The geometry was checked to close. The outlet built from the outlet flow coefficient passes the design flow to $10^{-16}$, the eye passes it to $10^{-16}$, and the specific speed recovered from the built duty is the one asked for. The disc share agrees with its closed form, $C_M/\psi^{5/2}\Omega_s^2$, to $10^{-16}$ at three specific speeds. Haaland’s friction factor is within two per cent of Blasius’s in the smooth range where Blasius holds, and the passage friction switches to $64/Re$ below 2300. And the half-flow pump’s specific speed divided by the full pump’s is 0.707107, $1/\sqrt2$.

The convention: best points, one head, one speed #

Every efficiency here is a best-point efficiency, the peak of a pump’s curve at its design duty; the station comparison carries the shape of the curve away from that point unchanged from the essay below. The specific speed is dimensionless, with $\omega$ in radians a second; the metric type number used in pump catalogues is 52.9 times it. Splitting keeps the head and the shaft speed, so the smaller pumps run at the same synchronous speed on the same motors’ frame size; a station that also chose a faster speed for its smaller pumps would raise their specific speed back, and that option is real and not priced here.

What the budget leaves out #

The rising side. The model has no loss that grows with specific speed, so its efficiency keeps climbing past 1.0 where real pumps level and then fall, as mixed-flow and axial impellers trade disc friction for tip clearance and secondary-flow losses. Everything argued here is on the low side, where the budget’s terms are the ones that matter. The coefficients. A head coefficient, two flow coefficients, a clearance rule and a fixed share are stated, and a real design adjusts all of them with specific speed — a low-specific-speed impeller is usually given a smaller head coefficient and more blades. The slopes survive that; the levels do not. Recirculation and part load. A narrow impeller recirculates at part flow sooner than a wide one, which the group with no head in it would read as a suction problem; nothing here models it. And the motor and drive, whose own efficiencies fall at part load and favour running fewer machines harder, sit outside the pump.

A penalty pump designers already priced #

That a pump’s best efficiency falls with its specific speed is in every pump handbook as a family of measured curves, one per flow, and the attribution of the fall to disc friction, leakage and narrow passages is as old as Stepanoff’s and Pfleiderer’s textbooks of the 1940s and 1950s. Daily and Nece measured enclosed-disc torque across its four regimes in 1960. Modern loss-budget methods, of which Gülich’s is the most complete, assemble the same terms with fitted coefficients and reproduce the handbook curves. What the budget adds here is only the reading of a split station through it: that the pair keeps the large pump’s discs, and that the price of staging is therefore set by the specific speed of the duty it serves.

Still open: whether a smaller pump should spin faster #

Every split here kept the shaft speed, because a station’s pumps usually share a motor speed. They need not. A half-flow pump run at $\sqrt2$ times the speed has the original specific speed and a smaller impeller — $1/\sqrt2$ of the diameter at the same tip speed — and its discs then cost what the large pump’s did as a share of its power, with a Reynolds-number penalty from being smaller and a clearance penalty from a smaller ring. The next calculation asks how much of the four points a faster, smaller pump recovers, and therefore whether the right way to split a duty is two pumps of the same specific speed rather than two of the same size; that turns on how efficiency changes with size at a fixed specific speed, which is the question Moody’s step-up formula claims to answer.