The group with no head in it

The rung below ends by naming the cheapest way to move a duty into a band and then declining to use it. Raise the shaft speed: the specific speed goes up in direct proportion, no hardware is added, and a duty that needed three stages needs one.

The reason for declining is that the shaft speed is constrained by something the first group cannot see — and the constraint is a second dimensionless group, formed from the same six variables, with one substitution.

The substitution #

The first group answers which machine, and it is

$$n_s = \frac{\omega\sqrt{Q}}{(gH)^{3/4}}$$

with $H$ the head the machine delivers. Replace that head with the net positive suction head — the margin between the absolute pressure at the inlet and the liquid’s vapour pressure, expressed as a height — and the same combination becomes

$$S = \frac{\omega\sqrt{Q}}{(g\,\mathrm{NPSH})^{3/4}}.$$

That is the suction specific speed, and the important feature of it is what is absent. The delivered head is not in it. A pump raising water two metres and one raising it two thousand, at the same flow and the same shaft speed with the same inlet condition, have the identical suction specific speed and the identical risk of cavitating.

That is not obvious and it is worth stating why it is true. Cavitation happens at the impeller’s inlet, where the liquid is accelerated onto the blades and its pressure falls below the ambient before any work has been done on it. What happens downstream of that — how much pressure the machine subsequently adds — is a different part of the machine and a later part of the process. The inlet does not know.

What the group is about #

The physical content is the cavitation number of the applied field, rearranged into the machine’s variables.

A liquid cannot be pulled. Its absolute pressure has a floor at the vapour pressure, and below that floor it boils — which is the whole of what “suction” can mean. At an impeller’s inlet the liquid is accelerated from the approach pipe onto the blade, and the acceleration costs pressure. How much it costs scales as the square of the blade speed, so it scales as $\omega^2$ at fixed geometry.

The margin available is the NPSH, set by the plant: the pressure at the free surface, minus the elevation lift, minus the friction to the inlet, minus the vapour pressure. It has nothing to do with the pump.

So the whole question is one number against another — the pressure the impeller’s own inlet takes away, against the margin the plant supplied — and forming the dimensionless ratio of the two gives the group above.

The window, and why it has two ends #

Put the two groups side by side against shaft speed and the structure of the choice appears.

The specific speed rises linearly with $\omega$, and it must be above the bottom of the lowest band for any runner to exist at all. That is a floor on shaft speed.

The suction specific speed also rises linearly with $\omega$, and it must be below the practical limit. That is a ceiling.

So the shaft speed is a window rather than a free variable, and the two ends come from completely different physics: the floor from what shapes can be built, the ceiling from what a liquid can be asked to withstand. In the case drawn, the window runs from 275 rpm to 962 rpm and the top is set by cavitation.

Which end binds is worth reading rather than assuming. At a large NPSH — a flooded suction, a tank above the pump, a cold liquid — the cavitation ceiling can be above the top of the band structure, and then the first group is the only constraint and this essay is unnecessary. At a small NPSH it binds hard, and the pump has to be run slower than the machine-selection arithmetic would like, which sends the designer back to the staging of the rung below.

That is the loop this ladder closes. A duty too low for a band is staged, or it is speeded up. Speeding it up may be forbidden by the suction group. If it is, staging is the remaining move — and a multistage pump exists partly because its shaft speed is capped at the inlet.

The duty with no window #

The case that makes the point is the one where the window closes.

Three cubic metres a second against two thousand metres, with six hundred millimetres of margin at the inlet, gives a floor at 183 rpm and a ceiling at 62 rpm. No shaft speed satisfies both. The duty is not merely awkward; the two constraints are inconsistent.

And the resolutions are not different machines. They are all changes to the problem:

Supply more margin. Raise the vessel, lower the pump — which is the same absolute-pressure accounting a siphon obeys — cool the liquid, or fatten the suction pipework. All four raise the NPSH, all four are plant changes rather than machine changes, and the first two are the reason a feed pump sits in a basement.

Add a booster. A low-speed, low-head first machine that raises the pressure before the main pump sees it, so the main pump’s NPSH is what the booster delivered rather than what the plant supplied. This is a very common answer and it is why a feed system is often two pumps rather than one.

Or change the inlet geometry. An inducer is a slow-turning axial screw in front of the impeller, designed to tolerate some cavitation itself while raising the pressure enough that the impeller behind it does not. It buys a factor of two or three in the group, and it is a machine analysed by drawing a box round it like every other in this field. It is why rocket turbopumps — the most extreme case of this arithmetic anywhere — all have one.

None of those is a different runner, and that is the essay’s point restated. The first group’s answer to which machine was never wrong; the second group’s answer was that the machine cannot be given the conditions it needs.

Why the limit is borrowed and what that costs #

The limit near $S = 3$ in the radian form is not a result and this site draws it accordingly.

It is a summary of what impellers have been built to. Conventional centrifugal impellers reach it; inducers exceed it substantially; and the number has drifted upward over decades as inlet geometry has improved. There is no conservation law behind it and no derivation this site could offer.

What is a result is the group itself, and the distinction is the same one the machine bands rest on. That a machine’s cavitation behaviour depends on $\omega\sqrt{Q}$ and on the NPSH in the combination above, and on nothing else at geometric similarity, follows from dimensional analysis exactly as the first group does. Where the threshold sits is measured.

That has a practical consequence worth stating. A duty at $S = 2.8$ and a duty at $S = 3.2$ are not on opposite sides of a physical boundary; they are on opposite sides of somebody’s experience, and the correct response to the second is to find out whose experience and under what conditions rather than to declare it impossible.

The two groups are the same group #

There is an observation worth making explicitly, because it explains why the substitution works rather than merely noting that it does.

Both expressions are the same combination of the same variables. Take the six quantities a rotating machine has — shaft speed, diameter, volume flow, specific energy, density and power — and the dimension matrix has rank three, so three independent groups exist and exactly one combination of them has a diameter exponent of zero. That combination is $\omega\sqrt{Q}/(g\cdot\text{energy})^{3/4}$, and which specific energy is substituted into it is a choice about which question is being asked.

Put the delivered head in and the group is about the machine’s shape. Put the inlet margin in and it is about the machine’s inlet. The arithmetic did not change; the physical quantity playing the role of “the specific energy this machine works against” did.

That is a general habit rather than a trick of this subject. A dimensionless group is a template with slots, and a second use of the same template is available whenever a second quantity of the same dimensions is doing a different job in the same problem. It is why the same $\omega\sqrt{Q}$ appears in both, and why the exponent on the energy is three quarters in both — those come from the matrix rather than from anything about heads or margins.

And it explains why the two constraints move together. Both groups carry $\omega$ to the first power, so a change in shaft speed moves them in the same direction by the same factor, and the window between them is fixed in width by the duty rather than by the shaft. Raising the shaft speed never widens the window; it only slides the operating point along it. Widening it means changing $Q$, $H$ or the NPSH, which is why the resolutions listed above are all plant changes.

Reading the constraint backwards #

The group is usually used to check a proposed machine, and it is more useful used the other way round.

Given a duty and a shaft speed, the group returns the NPSH the installation must supply — 10.8 metres in the case above. That is a number a plant designer can act on: it is a height, it is a temperature, or it is a pipe diameter, and all three are decisions made before any pump is ordered.

Given a duty and an available NPSH, it returns the fastest shaft the installation will tolerate — 962 rpm in the same case. That is a number a machine designer can act on, and it is a much harder constraint than it looks: standard motors run at synchronous speeds set by the supply frequency and the pole count, so “962 rpm” in practice means “the next standard speed below it”, which is 750.

Both readings are the same expression solved for a different unknown, and which one a project uses says who is being told what. A specification that quotes a required NPSH has put the constraint on the plant; one that quotes a maximum shaft speed has put it on the machine. They are equivalent statements and they are not equivalent contracts.

What the two numbers cost between them #

A last piece of arithmetic, because it puts a price on the window and makes clear what the constraint is actually forbidding.

Suppose the window’s top is binding, as it is in every case in this essay. The shaft must run at the cavitation ceiling rather than wherever the machine-selection arithmetic would have liked, and the consequence is that the duty’s specific speed at that shaft is lower than it could be. That in turn means more stages, by the arithmetic of the rung below: each stage buys $n^{3/4}$, so the number of stages needed rises as the shortfall in specific speed to the four-thirds power.

So the inlet margin buys stages, at a computable rate. Doubling the NPSH available raises the permitted shaft speed by $2^{3/4}$, which is 1.68, which raises the achievable specific speed by the same factor, which reduces the stage count by $1.68^{4/3}$ — a factor of about two. Two metres of elevation in a basement is one stage fewer in a feed pump, and the trade between civil works and rotating machinery has a number in it.

That is the kind of statement dimensional analysis is unusually good at producing. It contains no efficiency, no blade shape and no manufacturer, and it holds for any liquid at any scale in geometric similarity — which is exactly the class of statement the whole ladder has been assembling.

What the picture cannot show #

No impeller was solved. Nothing here computes the pressure at any point of any blade. The whole argument is dimensional, and it identifies the group without saying what value of it any particular geometry can survive.

NPSH is treated as a given. In practice it is computed from the plant — the surface pressure, the static lift, the friction in the suction line and the vapour pressure at the operating temperature — and each of those carries its own uncertainty. The friction term in particular changes with the flow, so the available margin falls as the duty rises, which makes the constraint tighter exactly where it is being pushed.

Cavitation is treated as a threshold and it is not one. Inception, a measurable drop in delivered head, and damage to the impeller happen at three different values of the group, in that order, and the gap between them is where a great deal of industrial practice lives. A pump running with some cavitation is normal; a pump running with enough to erode a blade is not; and nothing here distinguishes them.

And the liquid is treated as clean. The tensile strength of pure water is enormous and that of tap water is essentially zero, because what decides the threshold is whether there is anything for a bubble to nucleate on. Dissolved gas content moves cavitation inception substantially, and it is not in the group.

The assertion behind these figures is the one that could reject and does the work: the suction group must be bit-identical when the delivered head is changed by a factor of ten, while the specific speed must move by exactly $10^{3/4}$. A group that moved when the head moved would not be the group this essay is about, and the check is the cheapest way to notice.

Who found it, and when #

Suction specific speed as a design number dates from the 1930s and 1940s, and its association with a limiting value comes from Wislicenus and co-workers in the United States in the 1940s — from a correlation of machines that had cavitated against machines that had not, which is exactly what a borrowed limit is — and it sits in the same category as a discharge coefficient, a number that summarises measurements rather than deriving them.

The extreme application arrived with rockets, and it is worth naming because it is where the constraint became the design driver rather than a check on it. A liquid-propellant turbopump handles a cryogenic liquid at its own boiling point, so the NPSH available is nearly nothing by any ordinary standard; and it must run at enormous shaft speed, because the whole assembly’s mass is what is being minimised. Those two requirements are in direct opposition in the group above, and the inducer exists because of it — a component invented to buy a factor in one dimensionless number.

That is a good illustration of what a dimensionless group is for. Nobody had to be told that a turbopump might cavitate; what the group supplied was the statement that the difficulty depends on $\omega\sqrt{Q}$ and the inlet margin, in a fixed combination, so that a component which buys a factor of two in it buys the same factor for every propellant, every flow rate and every machine size.

Where the ladder goes next #

This anchor now has the number that picks a machine, the arithmetic that splits a duty until one exists, and the second number that can forbid the result. Two directions are open.

The rung above is the efficiency correlation, which both of the rungs below refused to use and which is where the ladder’s remaining content is. Specific speed picks the kind; the peak efficiency attainable at each specific speed is a measured curve with a maximum in the Francis range, and the shape of that curve is why the bands sit where they do. Setting several published versions of it side by side would say how much of the selection rule is a result and how much is a summary of practice — which is a question this ladder has raised three times and answered none.

The one beside it is the partial-load problem the whole ladder has ignored. Every number here is evaluated at one duty, and a real machine runs across a range: a pump throttled to half flow is at a different specific speed, and a machine selected at its design point can be badly wrong at the condition it spends most of its life at. That is the same group evaluated along an operating line rather than at a point, and the operating line is what a control system chooses.