Field

Fluids at work

Turbines, pipes, weirs, balls, sails, arteries and blades. What the conservation laws say about machines, which is more than a designer expects and less than a brochure claims.

The tube widens because the air slows. The streamtube through an actuator disc at an induction factor of 0.333. The three radii are not drawn to taste: each is fixed by requiring the same mass to pass every station, and the slowest station is therefore the widest. The tube widening in front of a wind turbine is why some of the wind goes round it rather than through it, and it is the whole reason a disc cannot take everything.

The most a disc can take

A wind turbine cannot extract more than sixteen twenty-sevenths of the energy passing through the circle its blades sweep. That is not a limit on turbines — it is a limit on anything at all, and it follows from three conservation laws and no engineering.

Efficiency is decided before the engine is chosen. Froude's propulsive efficiency against the ratio of jet speed to flight speed. It is 2/(1+σ) and nothing else — no engine, no fuel, no combustion. A turbojet with a jet at three times flight speed cannot exceed 50% however good its core is, and a propeller moving a great deal of air slowly is above 90% before anybody has designed anything.

A big slow push

The same thrust can be had from a lot of air moved a little or a little air moved a lot, and the two are not equivalent. One number decides which, it contains no engine, and it is why every airliner built since 1970 has a fan far larger than the machine driving it.

Betz's ceiling, and the rotor that cannot reach it. The power coefficient of Glauert's optimum rotor against tip-speed ratio, with Betz's 16/27 drawn as the ceiling it is. The gap is wake rotation: a rotor that extracts power applies a torque, a torque leaves the wake spinning, and that rotational energy never reaches the shaft. It falls as the rotor is geared up and is never zero — which is why large wind turbines turn so slowly and yet have such fast tips.

The wake that has to spin

A rotor that takes power out of the wind must apply a torque to it, and a torque applied to air is angular momentum left behind. The axial theory has nowhere to put that energy, so Betz's ceiling is unreachable at every finite tip-speed ratio — and the gap is computable.

The chart, with one exact line on it. The friction factor of a pipe against Reynolds number, for five relative roughnesses. Every curve here except one is Colebrook's correlation, solved by iteration rather than read off a chart. The exception is the short straight line at the left: f = 64/Re is the laminar solution and it is exact. The curves flatten to the right because once the roughness pokes out of the viscous layer the Reynolds number has nothing left to change.

The roughness a wall cannot feel

A rough pipe and a polished one carry the same flow for the same pressure over three decades of Reynolds number, and then suddenly they do not. What changed is not the pipe. It is the thickness of the film of fluid at the wall, which is the only part of the flow that can see the roughness at all.

The box, and the one thing assumed about it. The control volume across a sudden enlargement. Mass and momentum crossing the two ends are known exactly. The only modelling statement in the whole derivation is written on the annular step: the pressure there is taken to be the upstream pressure, because the fluid in the corner is nearly stationary. Measurement supports it well. Nothing else is assumed, and in particular nothing at all is assumed about the eddy that lives in that corner — which this figure therefore does not draw.

A loss with no viscosity in it

Where a pipe suddenly widens, energy is destroyed. The amount is exact, it has been known since 1766, and the derivation never mentions viscosity, Reynolds number or roughness — because momentum does not care where the energy went, only that it left.

The shock in a river

Shallow water is a gas whose ratio of specific heats is two. The white water below a weir is a shock wave, momentum is conserved across it exactly, energy is not, and one direction is forbidden for the same reason an expansion shock is forbidden — which makes the analogy exact to first order and wrong at the second.

The depth that costs least

For a given flow there are two depths that carry it at any energy above a floor, and exactly one at the floor. That one depth is where the Froude number is one, the least energy is exactly three halves of it, and a bump in the bed that asks for more than the flow has does not thin the water — it backs it up.

The cliff a rough ball reaches sooner. The drag coefficient of a sphere against Reynolds number, on log axes. The smooth curve is Morrison's correlation, which is a fit to measurements and is drawn in the colour this site reserves for a borrowed claim. The other is the same curve shifted along the Reynolds axis by a factor of 6 — a stated model of what roughness does, which is to trip the boundary layer early, and not a measurement of any real ball.

The drag that falls as it speeds up

There is a band of speeds in which a smooth ball experiences less drag the faster it goes. Not a smaller coefficient — a smaller force. Dimples move that band down to where a golf ball actually flies, and they do it by making the friction worse.

Two sides of one ball, at different pressures. The surface pressure coefficient round a ball, measured from the front stagnation point, on the side the seam trips and on the side it does not. Up to separation both follow the exact potential-flow distribution 1 − (9/4)sin²θ. After it both take the same wake pressure, which is what a manometer measures rather than what the ideal theory predicts. The asymmetry is the shaded area between them, and integrating it gives a side force of 0.2949 towards the later-separating side.

A ball that swings without spinning

A cricket ball curves in flight with no spin about any useful axis. The mechanism is not the Magnus effect; it is a seam tripping the boundary layer on one side so that side lets go later. Which way the ball then goes depends on one borrowed number, and this site's own inviscid solver supplies the value that gets it wrong.

Three times the wind, on a reach. The polar diagram: boat speed in every direction, as a multiple of the true wind speed, for drag angles of 14 and 6 degrees. The shaded wedge at the top is the no-go zone, whose half-angle is exactly the sum of the two drag angles. Everywhere outside about twice that angle the boat is faster than the wind, and the maximum is 2.92 times the wind at 110 degrees — which is 1/sin λ at 90° + λ, both checked.

Faster than the wind that drives it

An ice yacht in a fifteen-knot breeze does forty. That is not a trick and it does not need a special sail — it follows from two drag angles and a triangle, and the best speed a boat can reach is one over the sine of their sum.

The fastest way is not the straight one

A boat racing to a mark dead upwind sails seventy per cent further than the distance to it, and arrives first. The best angle is forty-five degrees plus half the apparent wind angle, it comes out of one line of trigonometry, and the same argument says to gybe downwind rather than run.

The radius that costs least

A vessel that carries a flow costs two things to own — the power to push fluid along it and the price of the tissue itself. Minimising the sum gives a best radius, the best radius makes flow proportional to radius cubed, and the rule that follows is a statement about a photograph that came out of a cost function.

The stretch of surface that is boiling. The pressure coefficient along both surfaces of a section at 4 degrees, computed from the same potential-flow solution as the other ideal-flow aerofoil figures. The horizontal line is the vapour pressure at a cavitation number of 1: wherever the suction curve is above it, the liquid there has been pulled below its vapour pressure and is boiling at whatever temperature it happens to be. This section cavitates at any σ below 1.428.

When a body tears the water

A propeller blade moving fast enough pulls the pressure at its own surface below the vapour pressure of the liquid, and the water boils at whatever temperature it happens to be. Where that happens is decided by an inviscid calculation of the pressure along the blade.

$Ninety microseconds, and most of it spent barely moving. The radius of a collapsing cavity against time, both as fractions of their own totals. The bubble spends most of the collapse near its original size and the last few per cent of the radius in the last fraction of a per cent of the time. The total is 91.47 microseconds for a millimetre cavity at one bar, computed by quadrature and agreeing with the closed form in gamma functions to a part in 10⁹.$

The bubble that hammers

A vapour cavity swept into higher pressure does not deflate. It collapses, in ninety microseconds for a millimetre bubble, and the model that describes the collapse predicts a wall speed that reaches the speed of sound in water at three per cent of the original radius — which is to say it predicts its own failure, and locates it.

Work out of a change of swirl

The work a rotor does per unit mass is the blade speed times the change in swirl, and that is all of it — no blade shape, no pressure, no efficiency, no gas properties. It is the same equation for a pump, a compressor, a turbine and a fan, and it follows from angular momentum on a box with nothing assumed about the inside.

The jet is 61.1% of the hole. Flow out of a slot in a plane wall, solved by Kirchhoff's free-streamline method. The outer curve is not a wall and not a guess: it is the streamline on which the pressure is ambient, and where it goes is part of the solution. It leaves the edge of the slot travelling straight down the wall and turns through ninety degrees, settling to a jet whose width is π/(π+2) = 0.6110 of the opening. Every streamline drawn is a level set of the streamfunction the conformal map supplies.

The hole that halves the flow

A jet leaving a sharp-edged hole is narrower than the hole, and by an amount that is not measured but computed. One geometry gives exactly one half from momentum alone; another gives exactly π/(π+2) from a conformal map in which the shape of the free surface is part of the answer.

The jet divides 75% to 25%. A jet striking a plate at 60 degrees. Both sheets leave at the jet's own speed, because their surfaces are at ambient pressure and Bernoulli allows nothing else, and the plate can exert no force along itself because the fluid has no viscosity. Momentum along the plate then fixes the split at (1 + cos β)/2 = 0.7500, and the normal force at ṁV sin β = 0.8660. Nothing about the plate's material, size or roughness enters either.

What a jet cannot push sideways

A jet striking a plate divides in two, and how it divides is fixed by a single sentence — an inviscid fluid exerts no force along a surface. That one statement, plus mass, gives the split exactly — and the same sentence turns a flat plate into a bucket worth twice as much.

Half the jet speed takes everything

A bucket standing still feels the largest force and does no work; a bucket running with the jet does no work either. Between them the power peaks at exactly half the jet speed, for every bucket shape and every flow rate — and at that speed a perfect bucket leaves the water motionless.

The same reading, and only one of them gives it back. A Venturi and an orifice plate at the same diameter ratio, with the pressure along the axis drawn beneath each. Both narrow the flow by the same amount, both read the same difference between the pipe and the narrowest section, and both infer the same flow rate from it. Downstream they part company: the Venturi's diffuser turns the throat's speed back into pressure, and the orifice's jet expands into the pipe and destroys 73% of the reading. The picture is a section rather than a solved field: nothing here computes the jet, and the recirculating corner is not drawn.

The price of knowing the flow rate

Two flowmeters can narrow a pipe by the same amount, read the same pressure difference and infer the same flow rate, and cost pressures that differ by an order of magnitude. What separates them is not viscosity, and not workmanship — it is whether the flow is decelerated or abandoned.

The tunnel makes the stream 7.4% faster. The flow past a cylinder between two walls, solved from the closed form of an infinite image row rather than from a truncated sum. The walls are streamlines exactly — that is what the images are for, and the normal velocity on them is zero to the last bit — and the flow beside the body is squeezed between the body and the wall, which is the whole of the blockage effect. The stream at the model is 7.40% faster than the speed the tunnel's own instruments report far upstream.

The instrument in the answer

A model in a wind tunnel is not a model in the sky. The walls are supplied by an infinite row of reflections, the stream at the model is faster than the tunnel's own instruments report, and the correction is not an empirical fudge — it is a series with a closed form.

12 bar from stopping one metre per second. The head at the valve after it shuts, computed by the method of characteristics on a 600 m pipe. The rise is 122.4 m of water, which is ρaΔV/ρg to 1.4e-14 m — and the scheme was told neither ρaΔV nor anything else about the answer. The wave then runs to the reservoir and back every 2.000 s, and with no friction in the model it never decays: a real pipe damps this out in a few tens of cycles.

Stopping water costs more than moving it

Shut a valve on water running at one metre per second and the pressure that appears is twelve bar — not because the water was pushing hard, but because the only way to stop a column of fluid is to send a message back along it, and the message travels at the speed of sound in the pipe.

$Nothing can deliver more than h/H, and here that is 10.0%. The fraction of the supply a ram can deliver, against the height it is asked to deliver to, from a supply falling 2 m. The upper curve is the exact ceiling h/H, which follows from the energy audit with every loss set to zero and can be reached by no real machine; the lower one is what a ram at 65% efficiency actually sends. Asking for twice the height halves the delivery, exactly, and there is no design that escapes it.$

A pump with no engine

A hydraulic ram lifts water uphill using nothing but the water that is already falling. It has one moving part and no power supply, and everything it can and cannot do follows from an energy audit that fits on one line — including a ceiling nothing about its design can move.

A pump curve out of the momentum theorem. The pressure an ejector delivers, against how much it is entraining, at a fixed nozzle. It has the shape of every pump characteristic ever measured — a shut-off pressure with no flow, falling to no pressure at free delivery — and it was obtained from a momentum balance on a tube with nothing in it. The shut-off value here is 36.0 kPa and the machine at its best power runs at 4.56 times its own motive flow.

Mixing is a pump

Two streams at different speeds mixing in a tube destroy energy — the same Borda–Carnot expression a handbook prints beside a sudden enlargement, with two streams in it instead of one. And while they destroy it the pressure rises, which makes the loss the mechanism of a machine with no moving parts.

1.139 asks for a Francis. The specific-speed axis, with the four machine types on it and one duty marked: 3 m³/s at 60 m, on a shaft turning at 750 rev/min. The number is 1.1387, and the choice of runner follows from it before any blade has been drawn. What the number contains is a ratio of flow to head; what it does not contain is any size at all, which is why one axis serves a garden pump and a gigawatt turbine.

One number picks the machine

A flow rate, a head and a shaft speed contain exactly one dimensionless combination with no size in it. That combination decides whether a duty wants an impulse wheel, a Francis runner or a propeller — before anything has been drawn, sized, or costed.

$Two velocities, 2.50 apart, and only one of them is anybody's. The velocity in Darcy's law is the flow rate divided by the whole cross-section, solid included: a speed no fluid particle ever has, since the fluid occupies only the fraction ε of that area. The speed the fluid actually averages is larger by exactly 1/ε — 2.50 times here — and it is the one that belongs in a residence time, in a pore Reynolds number and in any statement about when a tracer arrives. The grains are drawn to say that the pore-scale flow is not computed anywhere: no figure on this site claims to resolve it.$

A velocity nobody has

The velocity in Darcy's law is the flow rate divided by the whole cross-section, solid included — a speed no fluid particle in the bed ever has. Averaging buys a linear law and charges for it in exactly this coin, and the constant that comes with it is an area of about a square micron.

Two terms, one with viscosity in it and one without. Ergun's two contributions to the pressure gradient, against the pore Reynolds number, on logarithmic axes. The viscous term rises with the first power of the velocity and the inertial one with the square, so on these axes they are straight lines of slope one and two and there is exactly one crossing. The second term contains no viscosity at all — it is the price of accelerating fluid into every pore and out again — which is why a linear resistance law has to fail eventually whatever the fluid is.

Where Darcy stops

A linear resistance law has to fail eventually, because pushing fluid into a pore and out again costs energy that has nothing to do with viscosity. Where it fails is one dimensionless number, and that number is 150/1.75 — read out of a correlation's own constants rather than measured.

The pressure drop stops rising at 0.213 m/s. The pressure drop across a bed of 500 µm sand, against the velocity through it, in units of the fluidisation velocity. The rising branch is Ergun's resistance and the flat one is the bed's buoyant weight, which the flow cannot exceed however hard it is pushed: past the corner the bed expands rather than resisting more. That flat line is the reason fluidisation is unmistakable in practice — the corner is a crossing of two curves rather than a gradual departure, and it can be read off a gauge.

The bed that weighs itself

Blow hard enough through a pile of sand and the pressure drop stops rising. It cannot rise — a control volume round the bed says the drop can never exceed the buoyant weight of the solid in it, and at the velocity where the two meet the bed stops being a structure and starts being a fluid.

A factor of 1.8 between the extremes. The same numbers as a bar. The spread is exactly 96 over 160/3, which is 1.8 — and it is a spread in a quantity that a single correlation using the hydraulic diameter treats as a constant. A rectangle twenty times as wide as it is deep is 78 per cent more resistant than a circle of the same hydraulic diameter.

A number that is only the shape of the hole

The friction factor times the Reynolds number for laminar flow in a duct is a pure number that depends on the cross-section's shape and on nothing else — not the fluid, not the size, not the flow rate. It is exactly 64 for a circle and exactly 96 for a slot, and the correlation that treats them as the same is wrong by a third.

The section that decides the river

A reach of open channel has a normal depth and a critical depth, and the water has neither. What it has is a profile obeying a first-order equation, which needs exactly one condition — and whether that condition belongs at the upstream end or the downstream end is not a choice, because the equation is stable in one direction and unstable in the other.

Two different moments of one distribution. The permeability and the specific surface of a log-normal bundle, against the width of its pore-size distribution at a fixed median. The permeability rises by five orders because it is a fourth moment and the widest tubes dominate it; the surface falls because it is a first moment and the narrowest tubes dominate that.

A permeability that is only the geometry

Kozeny–Carman says that a porous medium's permeability follows from its porosity and its specific surface. Both are real, both are exactly measurable, and they do not determine the answer: forty-nine tube bundles built with identical values of each span a factor of eighty-two in permeability.

The junction the minimisation is over. A parent vessel entering from the left and two daughters leaving to fixed points. The radii are settled by Murray's law; what is left free is where the branch point sits, and the cost of the junction depends on it. The point drawn is the one the minimisation finds.

The angle a junction chooses

Murray's law fixes the radii at a branching vessel and is where every account of it stops. The same minimisation fixes the angles completely — 74.93 degrees for a symmetric bifurcation, a right angle for a vanishing side branch — and it does so as a triangle of forces, with tensions proportional to the squares of the radii.

Five approach profiles a meter might be looking at. The velocity across the pipe upstream of a contraction, for a uniform flow, fully developed laminar flow, two turbulent power laws and an annular jet of the kind a bend or a partly open valve leaves. All five carry the same volume flow. The meter reads a pressure difference and cannot see any of this.

The profile a meter cannot see

A differential-pressure flowmeter measures a force balance and reports a flow rate. The step between them needs two integrals of a velocity profile the instrument has no access to — and two profiles differing by half the mean velocity across the pipe give identical readings, which is why the standards specify straight pipe rather than a correction.

How much the peak cares about the valve at each instant. A small bump is added to the valve's opening at one moment and the peak is recomputed; the difference, divided by the bump, is plotted against the moment. It is flat and zero for the first 0.45 seconds — that part of the closure is invisible to the peak — rises through the reflection arrivals, and falls to nothing after the peak has happened, which is the one part of the shape that needs no fluid mechanics.

The part of the closure a pipe cannot see

A surge specification says how long the valve takes to shut. The line does not integrate the duration, it integrates the shape — and it is measurably blind to the first half-second of an eight-second closure, while three closures of identical length give peaks 97 per cent apart.

When the tracer leaves, given when it went in. The residence-time distribution of three beds with the same mean residence time and different Peclet numbers. Every one of them has its mean at exactly one, and they are nothing alike: the loosest lets a tenth of the tracer out before a third of the mean time has passed, and the tightest is nearly the spike a plug-flow calculation assumes.

The outlet is the inlet, a while ago

A bed has a mean residence time and everybody quotes it. Six beds with the same mean let their first hundredth through at 0.20 and at 0.85 of it, mix a window of inlet history between 1.53 and 0.18 wide, and convert a first-order reaction by amounts the mean cannot distinguish.

A threshold that is a curve, not a pressure. The tension a rectangular pulse has to reach to make a five-micron bubble run away, against how long the pulse lasts. At a tenth of a microsecond it takes forty bar; at a hundred microseconds it takes 1.05, which is within about one per cent of the static threshold. There is no such thing as the cavitation pressure of this bubble on its own.

A threshold that is also a duration

The cavitation number treats inception as a pressure: below it the liquid tears, above it does not. A five-micron bubble asked to grow in 0.3 microseconds needs 43 bar of tension and the same bubble given a hundred needs 1.05, because it has to make a journey and not merely respond.

$Spin against distance flown, not against time. The spin of a struck golf ball as a fraction of its launch spin, against how far it has flown. The integrated flight sits exactly on an exponential in distance, because the spin-down torque is proportional to speed times spin and the speed cancels. The constant is 1,202 metres and the drive is 200, so the ball arrives with five sixths of the spin it left with.$

The ball that never forgets its spin

Commentary explains a late-swerving ball by saying the spin is dying away. The spin-down torque goes as speed times spin, so spin decays over a distance rather than a time — 1,202 metres for a golf ball against a 200-metre drive — and what dies away is the speed, which makes the swerve stronger.

Two ways to move a duty, along one axis, in opposite directions. A duty at a specific speed of 0.01 and what splitting it does. Dividing the head between stages in series multiplies each stage's specific speed by the number of stages to the three-quarter power, because the group carries (gH)^(−3/4); dividing the flow between units in parallel divides it by the square root of the number of units, because the group carries √Q. Both exponents are read off the group rather than remembered, and they are why the two operations are not interchangeable: three stages buy a factor of 2.28 and three units cost a factor of 0.58. The bands are drawn in the colour this site reserves for a borrowed claim, because where a Francis runner stops is practice rather than a result.

The duty that had no machine

Some duties have a specific speed outside every band, and no runner will do them at any size because the number contains no size. That is true and it is not the end. The same group says how to split the duty until it fits, and the two ways of splitting move it along the same axis in opposite directions, by exponents read straight off it.

The shaft speed is a window, and cavitation closes the top of it. Two groups against shaft speed for the same duty: the specific speed, which must be inside a band for a runner to exist, and the suction specific speed, which must be below about 3 for the impeller not to cavitate. Both rise with the shaft speed, so raising it to reach a band is also raising it towards the cavitation limit. The window here runs from 274.98 rpm to 962.31 rpm and cavitation sets its top. A duty whose window is empty needs something other than a different machine — a booster, a lower installation, or an inducer.

The group with no head in it

The number that picks a machine says nothing about whether the machine can exist. A second group formed from the same variables, with the delivered head replaced by the margin available at the inlet, decides that — and the delivered head has left the expression entirely, so how far a pump lifts is irrelevant to whether it tears the liquid apart at its own entrance.

Where a rotor's pressure rise comes from, as the radius moves. The static pressure rise across a rotor, split into the two terms rothalpy gives it. The diffusion term is held at the de Haller limit throughout — the blade is being asked to slow the relative flow as hard as a boundary layer will allow — so it is a flat 11558.4 Pa at every radius ratio. Everything above that line is the centrifugal term, which costs no diffusion and has no limit of its own. At a radius ratio of 2 it supplies 49.92 per cent of the rise and at 3, 72.66 per cent. An axial machine, at a ratio of exactly one, gets none of it.

What a turning frame keeps

Euler's equation prices the work and says nothing about where the pressure comes from. In the frame turning with the blades — which is accelerating, and carries two fictitious forces — a Bernoulli-like quantity survives both of them, and it splits the pressure rise into a term a boundary layer limits and a term that is free if the radius moves.

A blade root at 412.51 megapascals, and the chord is not in it. The centrifugal stress at a blade root against shaft speed, for one annulus of 0.36 m², in four materials. Integrating the blade's own weight outward gives σ = 2π ρ_b A N² with a taper relief — and the chord has cancelled, the blade count has cancelled, and every property of the gas has cancelled. What is left is an area times the square of a speed, capped by a material. The horizontal lines are each material's allowable stress, and where a curve crosses its own line is the fastest that annulus may be turned in that metal. At 12000 rpm this blade carries 412.51 MPa with a tip speed of 439.82 m/s.

The stress that picks the aerodynamics

The free term in a rotor's pressure rise wants radius and speed, and both are capped by something with no fluid in it. Integrate a blade's own weight outward and the root stress comes out as the annulus area times the square of the shaft speed — with the chord cancelled, the blade count cancelled, and every property of the gas cancelled.

A 204 m hammer traded for a 8.57 m swing over 299 seconds. The water level in a 10 m surge tank at the end of a 2 km tunnel 3 m across, carrying 2 m/s, after the turbine is shut off at once, with the level measured from the reservoir's. Without friction it rises to V₀√(L Aₜ/g Aₛ) = 8.57 m and swings with a period 2π√(L Aₛ/g Aₜ) = 299.1 s, the integration agreeing with both closed forms. With the tunnel's 5 m of friction the level starts 5 m below the reservoir, peaks at 5.61 m after 98 s, falls to −3.70 m, and decays. The same tunnel shut at its end with no tank would take the Joukowsky rise of 204 m. The tank does not remove the column's momentum; it gives it a free surface to push against, slowly.

A tank that turns a hammer into a swing

Shut a turbine at the end of a two-kilometre tunnel in two seconds and the valve takes a rise of 256 metres of head. Put a shaft open to the air beside it and the rise is 51, the tunnel never carries the closure as a wave at all, and its water slows instead against a level that climbs for a minute and a half — to a height that is a closed form with the tank's area under a square root.

Below Thoma's 6.07 m² the governed tank's swing grows; above it, it dies. The tank level after the turbine's power demand drops by two per cent, with a governor holding the power constant, for tanks of 0.7 and 1.3 times Thoma's area of 6.07 m² — a tank 2.78 m across. The smaller tank's swing grows by a factor of 1.47 every 74 s cycle and has reached −12.20 m by 427 s; the larger one's keeps 0.75 of itself every 100 s and is barely visible. Carried on, the smaller tank's run is refused at 794 s, where the head at the turbine has fallen below a quarter of its design value and the governor would be asking for a flow no turbine passes. The instability has nothing to do with the tank's height: it is the governor drawing more water as the level falls, which feeds the swing, against the tunnel's friction, which is the only thing damping it.

The better tunnel needs the bigger tank

A turbine governed to hold its power opens further when the head at it falls, and draws the tank down harder. That makes it a negative resistance, the tunnel's friction is the only thing damping the swing against it, and so the smallest stable tank grows as the friction shrinks — 2.78 metres across for five metres of friction, 6.09 for one.

$A force ceiling ends the similarity: a breeze makes the boat slower as a share of the wind. Speed made good to windward as a fraction of the true wind, on the best course for each wind, for a rig that is flattened once the righting moment binds and for one that is reefed, with the hull's drag angle held at 6°. Below 3.70 m/s the two are the same boat, and the fraction does not depend on the wind — 1.122 at every speed, which is the similarity the two-angle polar rests on. Above it the flattened rig's fraction falls: 0.918 at 6 m/s, 0.572 at 10 m/s, 0.321 at 15 m/s and 0.169 at 20 m/s. The reefed rig's stays at 1.122, because a reefed sail keeps its least drag angle and the hull's angle is held. Once a force is limited, the triangle is no longer the same shape in every wind.$

A breeze the boat cannot use

The two-angle polar makes a boat's speed a fixed fraction of the wind, in any wind. A righting moment ends that at 3.7 metres a second on the beat. Past it the crew must spill force, a flattened sail's drag angle climbs, and the best course to windward moves closer to the wind rather than away from it — which 45° + λ/2 cannot say, because λ now depends on the course.

$A keel pulls the best beat 12 degrees closer to the wind. Speed made good as a fraction of the wind against the course, in 3 m/s: for the boat with its keel solved as a wing, and for a boat whose λ is fixed at 25.69°, the value the keeled boat has on its own best course. The search puts the keeled boat's best beat at 45.83°; the fixed-angle rule puts it at 45° + λ/2 = 57.84°. Pointing higher loads the keel towards its best lift coefficient — 0.130 at 40° against 0.076 at 60° — and lowers its drag angle, so the curve peaks early and falls away faster on the far side. The fixed-angle curve promises 0.653 of the wind; the keeled boat can make 0.553, and only by sailing twelve degrees higher than the rule says.$

A keel flies wherever the course puts it

A keel is a wing whose lift is whatever side force the rig happens to make, so its lift coefficient is chosen by the course and the wind rather than by its designer. Solved that way, the hull's drag angle stops being a property of the boat: it pulls the best beat twelve degrees closer to the wind, halves the reaching speed a fixed angle predicts, and finally charges for reefing.

121 m after the cavity closes, against 69 m from the closure. The head at a valve shut instantly on water flowing at 0.36 m/s through 600 m of 100 mm pipe, a = 1200 m/s, with a steady head of 25 m. The closure raises it to 69.1 m, the Joukowsky head; the reflection returns at one round trip, 1.00 s, and takes the head down to the vapour head, −10.1 m, where a cavity opens (shaded). It closes 2.146 round trips after the closure, and the first pulse after it reaches 121.3 m — 52.3 m above the Joukowsky head — for 146 ms. The step line is the exact solution between events; the thin line is a 240-reach grid solver that was told nothing about it and agrees with its first pulse to better than a millimetre.

Twice the margin, on top of the hammer

Shut a valve on a line whose pressure is low and the returning wave boils the water beside it. When that cavity closes, the head at the valve can pass the Joukowsky rise — by up to twice the margin that let the water boil, in a sawtooth that jumps each time one more round trip fits into the cavity's life, and for a time that is shortest exactly when the pulse is tallest.

Matched at one speed and at no other

Every stage of a compressor passes the same mass flow, and the annulus behind each one is cut for the density the air will have reached there at design speed. Slow the shaft and the air is less dense than the metal expects, so the rear stages carry more volume than they were drawn for while the front ones starve — and below a definite speed no throttle setting keeps all of them working at once.

Between hover and twice the hover inflow

A rotor's momentum balance has an answer for every climb and for every fast descent, and none for descending at anything between zero and twice its own hover inflow. There one root sends air out of both ends of its streamtube and the other root is not a real number, and at each edge of the band one end of the tube stops moving — which is where the vortex ring state lives, and where a wind turbine's thrust coefficient of one sits.

$A pump slowed into a system with a static lift leaves its own specific speed. The specific speed of the operating point, as a share of its value at the best point, against the fraction of the design flow delivered, for a pump drawn for 0.1 m³/s against 40 m at 1450 rpm, specific speed 0.545 at its best point. Under speed control into a system whose static lift is 0 per cent, 30 per cent, 60 per cent, 90 per cent of the design head (lines), and under a throttle at design speed into the 60 per cent system (dashed). With no static lift the speed-controlled pump stays exactly at its best point and its specific speed never moves. At half the design flow it has fallen to 1.000, 0.800, 0.708, 0.652 of the design value as the static share rises, and to 0.598 under the throttle.$

The specific speed a pump spends its life at

A pump is chosen by its specific speed at its best point and then run somewhere else. Written in the pump's own coefficients the number is √φ/ψ^¾, a position along its characteristic, and a variable-speed drive keeps it there only when the system it pumps into has no static lift. Every metre of lift moves a slowed pump along its own curve, towards shut-off, and a throttle moves it further.

Six nozzles, six pump curves, and where each one is best. The head ratio a water jet pump delivers against the flow ratio it entrains, for six area ratios from a narrow nozzle to one filling four-fifths of the throat. A wide nozzle makes a tall, steep curve that is finished at a small flow; a narrow one makes a low, long curve. The dots are each curve's best-efficiency point. Nozzle, suction, throat-friction and diffuser losses are included at borrowed representative values, and the mixing loss is computed.

The nozzle that is best at one thing

Put the four losses back into a jet pump and three questions get three answers. The most head comes from a nozzle four-fifths of its throat, in closed form; the best efficiency from one a quarter of it; and the most flow from whichever nozzle is smallest, because flow has no optimum at all.

The characteristic gets a wall, and the wall does not care about the discharge. One jet pump's characteristic, at an area ratio of 0.275, with the flow ratio at which its throat entry reaches vapour pressure drawn for three values of the cavitation parameter σ = (Pₛ − pᵥ)/(Pₘ − Pₛ). Left of a wall the machine runs on its curve. At the wall no lower discharge pressure raises the flow: the head ratio can fall to zero along the vertical and the flow ratio stays where it is. The wall's position contains the nozzle and suction losses and nothing downstream of the throat entry.

The wall the suction puts in the curve

A liquid jet pump's lowest pressure is where the entrained stream enters the throat, and when that reaches vapour pressure the pump curve stops being a curve. The flow ratio freezes at a value no lower discharge pressure can move — and raising the motive pressure, the obvious cure, brings the wall closer.

Flat until the back pressure reaches a value, then gone. Entrainment ratio against back pressure, as a multiple of the suction pressure, for three mixing-section sizes of one steam ejector driven from a motive supply a hundred times its suction pressure. Each is exactly flat while the entrained stream is choked beside the jet, up to its critical back pressure. The dashed lines join that point to the back pressure at which the entrainment has fallen to nothing, 1.4 per cent higher for the middle machine. The model fixes those two ends and not the path between them. A larger mixing section entrains more and breaks at a lower back pressure.

A choke that belongs to two streams

A steam ejector entrains a fixed amount of gas whatever its back pressure, up to a pressure where it stops. The flat part is a choke, and the entrained gas is not at Mach one when it happens: it is at 0.886, because the supersonic jet beside it is part of the same throat. One area ratio then trades that entrainment for compression, and the trade decides how a vacuum train is built.

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