The duty that had no machine

The rung below this one has a section called a duty with no machine. It puts a flow rate, a head and a shaft speed into the one group with no diameter in it, finds the answer outside every band, and observes — correctly, and it is the whole point of the group — that making the machine bigger cannot help, because the number contains no size.

The essay stops there, and stopping there gets the situation exactly half right. A duty outside every band cannot be met by one runner. It can be met, and the same group says how.

The two exponents, and where they come from #

The group is

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

and everything in this essay is a consequence of the two exponents in it.

Split the head between $n$ stages in series. Each stage passes the whole flow and delivers $H/n$, so each stage’s own specific speed is the duty’s multiplied by $n^{3/4}$. Staging moves a machine up the axis.

Split the flow between $k$ units in parallel. Each unit delivers the whole head and passes $Q/k$, so each unit’s specific speed is the duty’s divided by $\sqrt{k}$. Paralleling moves it down.

Neither exponent is remembered and neither is a rule of thumb. They are the powers of $H$ and $Q$ in the group, which came out of the null space of a dimension matrix — and the group itself is not the only group the variables permit, only the one with no diameter in it and which the previous rung checks survives an arbitrary rescaling to nine decimal places.

And the two are not mirror images, which is the first useful consequence. Three stages buy a factor of 2.28; three units cost a factor of 0.577. To move a duty by the same factor takes fewer stages than units, because three quarters is more than a half — so a designer reaching upward has a cheaper tool than one reaching downward, and that asymmetry is in the group rather than in any engineering preference.

The duty from the rung below, resolved #

Take the case that ended the previous essay: a small flow against an enormous head. Twenty litres a second at three thousand metres, on a 1,720 rpm shaft, is a specific speed of 0.0113 — below the 0.02 where the impulse-wheel band starts, and therefore a duty with no runner.

Three stages divide the head into three of a thousand metres each. Each stage then sees $n_s = 0.0113 \times 3^{3/4} = 0.0258$, which is inside the Pelton band, and the answer is three impulse wheels on one shaft.

The search behind that figure is exhaustive over the same group — every combination of stages and units up to a bound, each evaluated with the same expression, with the cheapest taken as fewest runners times fewest stages and ties going to fewer units. The tie-break is the only judgement in it, and it is a judgement about cost rather than about fluid mechanics: parallel units need their own inlets, valves, controls and foundations, and stages share a shaft.

What the figure shows that a single answer would not is how many arrangements work. The duty is not on a knife edge; a whole region of the grid is feasible, and the choice within it is made on grounds this group knows nothing about.

What splitting does not change #

It is worth being explicit about what has and has not moved, because the arithmetic invites a misreading.

The duty has not changed at all. Twenty litres a second at three thousand metres is still what the installation must deliver, and its specific speed is still 0.0113. What changed is that no single machine is being asked to do it.

No machine has been made bigger. That was the previous rung’s whole point and it survives. Size never enters, and a Pelton wheel of any diameter at that duty is the same impossible machine.

And the shaft speed has not changed either, in this arrangement. It could have: raising $\omega$ raises $n_s$ directly, and a fast enough shaft would put a single runner into the band. That is the third tool and it is the one with a hidden cost, which is what the rung above this one is about.

Where this shows up as real hardware #

The arithmetic is not academic and the machines it describes are ordinary.

A boiler feed pump raises water a couple of thousand metres of head at a modest flow, at a pressure where the margin against tearing the water is the other constraint, which is the same corner as the example above with the sign of the work reversed. They are built with six to twelve stages on a shaft, and the number of stages is chosen exactly this way.

A hydroelectric plant on a high head, where the head is set by the section the river makes available, does the same in reverse with the flow rather than the head: a large flow at a large head gives a specific speed above the Francis band, and the resolution is several runners in parallel, which is why a large station has a row of identical units rather than one enormous one. Paralleling divides by $\sqrt{k}$, and the row exists to buy that division.

A multi-jet Pelton wheel is the same operation done inside one machine, and it is the reason the bands in the rung below list one-jet and multi-jet separately. Splitting the flow between $j$ jets on one runner divides each jet’s specific speed by $\sqrt{j}$ without adding a shaft — which is paralleling with the units stacked on top of each other, and it is why the multi-jet band sits above the single-jet one by very nearly a factor of $\sqrt{4}$ for a four-jet wheel.

And a gas turbine’s compressor is the extreme case, with each stage doing the same work-out-of-swirl the Euler equation prices: fifteen or more stages, each doing a small fraction of the pressure ratio, because a single axial stage that tried to do the whole thing would be at a specific speed no blade can be shaped for. The stage count of a compressor is very largely this arithmetic.

Why a duty falls outside a band at all #

It is worth asking why the bands have edges, since the whole essay treats them as a target to be reached. The answer is geometric and it makes the staging arithmetic feel less like a trick.

The specific speed is, in effect, a measure of a runner’s shape — how much flow it passes for how much head it develops, with the size divided out. A machine at a very low specific speed is passing very little flow against a very large head, which means the fluid must be turned through a large change of swirl in a passage that carries almost nothing. Geometrically that is a large-diameter runner with tiny buckets on it, and beyond some point the buckets are too small to be made, too small to admit the flow without the boundary layers filling them, or too small relative to the disc they are attached to for the disc friction not to dominate.

At the other end, a machine at a very high specific speed passes an enormous flow against almost no head, which is a small-diameter runner with blades that are barely turning the flow at all — an axial propeller, and past some point a propeller with no blade angle left to reduce.

So the bands’ edges are where a shape stops being makeable, which is why they are practice rather than a result and why they have moved over the last century as manufacturing has. What has not moved is the arithmetic: whatever the edges are, the group says where a duty sits relative to them, and it says how splitting moves it.

That also explains why the fix for a duty below the bands is staging and not paralleling. Below the band the runner’s buckets are too small; dividing the head between stages gives each stage less work and therefore, at the same flow, a fatter passage. The arithmetic and the geometry are saying the same thing, which is a good sign that neither is an accident of the units.

The limit of the tool #

There are three things this argument does not license, and they matter because the arithmetic is so easy that it invites over-application.

Stages are not free and they are not identical. The expression treats each stage as seeing $H/n$, which assumes the head divides equally. In a real multistage pump it very nearly does and not exactly: the first stage sees a different inlet condition from the others, the last one discharges into a different volute, and the stages are usually identical castings anyway, so the split is a design target rather than an outcome.

Efficiency has not been mentioned once. The group says which kind of machine a duty asks for and says nothing about how well it will do it. A three-stage arrangement at $n_s = 0.026$ and a seven-stage one at $n_s = 0.021$ are both impulse wheels, and which is more efficient is a question the group cannot reach — it needs the efficiency-against-specific-speed correlations, which are measured rather than derived and which are therefore an exponent dimensions cannot give and which every manufacturer keeps its own version of.

And the bands are borrowed. Nothing in conservation says a Francis runner stops at $n_s = 2.2$; that number is a summary of what has been built and what has worked. Every figure in this ladder draws the bands in the colour this site reserves for a claim taken from elsewhere, and a duty landing two per cent outside one is not a duty with no machine — it is a duty near the edge of somebody’s experience.

The third variable, and why it is not free #

There is a symmetry in the group that the two operations above break, and noticing it is the cleanest way into what comes next.

The specific speed is built from three quantities and every one of them can be moved. Splitting the head moves $H$; splitting the flow moves $Q$; and the third, $\omega$, can simply be turned up. It appears to the first power, so doubling the shaft speed doubles the specific speed — a bigger lever than either of the others, needing no extra hardware at all.

It is the cheapest move in the group and it is the one with a constraint outside it. Three constraints, in fact, and the first two are mechanical rather than fluid: a runner at twice the speed carries four times the centrifugal stress, and its blade tips may reach a Mach number where the flow inside it is no longer the incompressible one the whole analysis assumed.

The third is the one that belongs to this subject, and it is the rung above. Raising the shaft speed also raises the pressure the impeller must drop the liquid to at its own inlet, and there is a floor under that pressure — the liquid’s vapour pressure — below which the flow stops being a single phase and the machine stops being the machine that was analysed.

So the free variable is not free, and the constraint on it is a second dimensionless group formed from the same six quantities. That group has no delivered head in it at all, which is the surprising part and the reason it is worth its own rung.

Reading the grid rather than the answer #

One more thing about the arrangement figure, because the habit it encourages is more useful than the number it reports.

A selection tool that returns one answer invites the question is this right?, which is usually unanswerable. A tool that returns the whole feasible region invites a better question: how much room is there? — and the answer to that is legible at a glance and is what a designer actually needs.

A duty with one feasible cell is fragile: a small change in the head, the flow or the shaft speed puts it outside every band again, and it will do so during commissioning. A duty with a dozen feasible cells is comfortable, and the choice among them can be made on grounds that have nothing to do with fluid mechanics — a standard frame size, an existing spares inventory, a maintenance access, a partial-load schedule.

That is the honest use of a selection group. It narrows a field, it does not choose, and a tool that reported one arrangement would be concealing the width of the choice it had made on the designer’s behalf. The same argument applies to every band in this ladder, which is why they are drawn as borrowed claims rather than as boundaries.

Why the exponents are what they are #

The three-quarter and the one-half have been used throughout as facts read off the group. It is worth spending a paragraph on where the group’s own exponents come from, because they are not arbitrary and the reason explains why the arithmetic is so robust.

The specific speed is the one combination of a machine’s six variables with no diameter in it. Set that requirement and the exponents follow: the flow coefficient carries $D^{-3}$, the head coefficient $D^{-2}$, and eliminating $D$ between them forces the flow to appear as a square root and the head as a three-quarter power. Nothing was chosen.

That is why the two operations’ effects are exactly $n^{3/4}$ and $k^{-1/2}$ rather than approximately. They are not fits, they are not correlations, and they do not depend on the machine being efficient, well-designed or even sensible. Any device that takes a flow and a head and turns a shaft obeys them, because they are consequences of the dimension matrix rather than of anything the device does.

Which is the difference between the exponents and the bands. The exponents are results; the bands are practice; and the figures draw the second in a different colour for that reason. A reader who remembers one thing from this essay should remember which half of it is which.

When the duty is not one duty #

Every calculation above evaluates the group at a single flow and a single head, and a real installation has a range. It is worth saying what happens to the arithmetic, because it is where the tool’s honest limit is.

A pump serving a system does not sit at a point. It sits where its own characteristic crosses the system’s — and the system’s curve rises with flow, because the friction in the pipework does. Throttle a valve and the crossing moves; the flow falls, the head rises, and the specific speed of the duty changes.

That has a consequence for staging that is not obvious. An arrangement chosen at the design point may put each machine outside its band at part load, and the machine’s efficiency falls away accordingly. A multistage pump throttled to half flow is running its stages at a specific speed they were not chosen for, and the loss is not the throttle’s alone.

So the grid of feasible arrangements in the figure above is a grid at one condition, and the honest version of the question is which arrangement keeps every machine inside a band across the whole operating range. That is a harder search, it has a different answer, and it is what a selection actually is — which is why the tool narrows a field rather than choosing from it, as the section above insists.

The extreme form of it is a duty that moves seasonally or with demand, where no fixed arrangement is right and the answer is a variable-speed drive: change $\omega$ to follow the duty, keeping the specific speed where it was chosen. That is the third lever of this essay used continuously rather than once, and it is why variable-speed pumping is worth its cost wherever a duty varies.

What the picture cannot show #

No machine was designed. Nothing here computes a blade angle, a diameter, a runner shape or an efficiency. The whole of the content is dimensional analysis on six variables, and its output is the kind of machine and nothing more.

The search is over a bounded grid. The arrangement figure examines up to eight stages and six units, which covers every ordinary case and would miss an exotic one. It is a search rather than a proof of optimality.

No losses anywhere. Head is treated as delivered head with no allowance for the friction in the piping, the mechanical losses or the volumetric leakage, all of which move the duty a real machine sees.

And the tie-break is an assertion. Ranking arrangements by stages times units is a cost model with no costs in it. A real selection weighs a spare-parts inventory, a maintenance access, a partial-load requirement and a failure mode, and any of those can beat the arithmetic here.

The assertion behind the figures is the one that could reject: staging must multiply the group by exactly $n^{3/4}$ and paralleling divide it by exactly $\sqrt{k}$, checked against the group at six stage counts and five unit counts to twelve decimal places; the arrangement search must land inside a band and its per-unit number must agree with a direct computation; and a request for two and a half stages must be refused rather than rounded.

Who did it, and when #

Specific speed as a selection tool dates from the turn of the twentieth century and the number’s various forms — the American, the European and the dimensionless radian one used here — reflect the different unit systems of the people who standardised them. The staging arithmetic is older than the group in the sense that multistage machines were built before anybody wrote $n^{3/4}$ down; what the group supplied was the ability to decide the stage count before building anything.

The interesting historical point is that the two operations were discovered from opposite ends. Staging came from steam turbines, where Parsons’ whole insight in the 1880s was that expanding a large pressure ratio in many small steps gives blade speeds a material can survive — a velocity argument rather than a specific-speed one, arriving at the same answer. Paralleling came from hydro plants, where the driver was availability and maintenance rather than any dimensionless group.

That both practices turn out to be the same arithmetic on one number, moving it in opposite directions by two exponents in the same expression, is the kind of unification dimensional analysis is for and is not how either was found.

Where the ladder goes next #

The third tool was named above and set aside: raise the shaft speed. It raises the specific speed directly, it needs no extra hardware, and it is the first thing anybody reaches for.

The rung above is what it costs. A second dimensionless group is formed from the same variables with the delivered head replaced by the margin available at the inlet, and the delivered head leaves the expression entirely. That group rises with shaft speed exactly as this one does — so the move that finds a machine is the move that drowns it, and the two constraints together make the shaft speed a window rather than a free choice. A duty whose window is empty needs something that is not a different machine.

The one beside it is the efficiency correlation this essay refused to use. It is measured rather than derived, every manufacturer has its own, and setting several of them side by side to see how much they agree would say how much of “specific speed picks the machine” is a result and how much is a summary of what has been built.