Half the jet speed takes everything

A Pelton wheel is a rim of buckets with a water jet aimed at it. There is no casing, no draught tube and no pressure anywhere in the machine: the water arrives as a free jet at atmospheric pressure, hits a bucket, is turned round, and falls away. Everything the machine does is a momentum exchange in open air.

That makes it the simplest turbine there is, and it makes it the one turbine on this site whose ideal efficiency is exactly one — not a limit approached, not a fraction like Betz’s 16/27, but one, reached at a stated operating point by a machine that could in principle be built.

Getting there needs a change of frame and two lines of algebra.

The frame change #

In the bucket’s own frame the water arrives at V − U and leaves at the same speed, turned through the bucket’s angle. It leaves at the same speed for the reason the previous rung established: the surface is at atmospheric pressure the whole way round, so Bernoulli in the bucket’s frame — which is inertial while the bucket runs at constant speed — allows nothing else.

So the force is

$$F = \dot{m}(V-U)(1-\cos\beta)$$

and the power is that times the bucket’s own speed. The two factors pull in opposite directions and that is the whole of the design problem: a stationary bucket feels the largest possible force and delivers nothing, and a bucket running at the jet speed is never caught up with and delivers nothing either.

$$P = \dot{m}\,U(V-U)(1-\cos\beta)$$

is a parabola in U with roots at 0 and V, so its maximum is at U = V/2 — for every β, every flow rate, every fluid. The site computes it by golden-section search rather than by differentiating, which is not fussiness: it would be entirely possible to write a figure whose curve peaked somewhere else beneath a caption saying a half, and the search knows nothing about the algebra above.

What “efficiency one” actually means #

Substituting U = V/2 into the power and dividing by the jet’s kinetic energy flux ½ṁV² gives

$$\eta_{\max} = \frac{1-\cos\beta}{2}$$

which is one only for β = π — a bucket that sends the water back the way it came.

The temptation is to read that as a formula and move on. It is worth stopping, because what it says is strange: a machine that takes all of the energy out of a moving fluid, leaving nothing. Where does the water go?

It falls out of the bucket motionless. In the bucket’s frame it is travelling backwards at V/2; the bucket is travelling forwards at V/2; in the ground frame those cancel exactly, and the site’s solver returns an exit speed of 6·10⁻¹⁷ of the jet speed, which is the arithmetic’s own noise.

This is the answer to the objection that a turbine cannot take everything because the water has to leave. It has to leave the bucket, which is a statement in the bucket’s frame, and it does. It does not have to be moving in the ground frame, and at the optimum it is not.

The audit, at every speed rather than at the best one #

An efficiency quoted at one operating point is a weak claim. The stronger statement is that the jet’s energy is fully accounted for everywhere: whatever the shaft does not take, the departing water carries away, at every bucket speed including the useless ones.

That the sum is exactly one is not a discovery; it is a check that the velocity triangles are being resolved consistently. It is worth having for the same reason the tangential-force check on the previous rung is: a sign error in the exit velocity produces an audit that fails, and produces a plot that looks perfectly reasonable.

The audit also makes the shape of the curve legible. At U = 0 the water leaves at V backwards, carrying the whole of the incoming energy — the bucket has reversed it and taken nothing. Past U = V/2 the water leaves going forwards again, and the machine is dragging it along rather than stopping it.

Why a real bucket stops at 165 degrees #

A bucket that reverses the jet completely sends the departing sheet straight back along the incoming jet — into the jet itself, and into the bucket following. So real Pelton buckets are split down the middle by a central ridge, throwing two sheets sideways, and turn the flow through about 165° rather than 180°.

1.7 per cent, and that is the entire cost of the most visible feature of the machine. It is a characteristic result of this field: the physics offers a maximum, the geometry of a machine that has to be built out of parts forbids reaching it, and the size of the gap between them is a number rather than a shrug. The sailing polar has the same shape, and so does the actuator disc’s optimum, where the inaccessible point is not a shape but a wake at rest.

Real Pelton wheels reach about 90 per cent. The missing tenth is the bucket angle (1.7 per cent), friction on the bucket surfaces, windage on the wheel — which is considerable, since the rim is travelling at half the jet speed through air, and is a drag problem of the kind the sphere essays compute — bearing losses, and the fact that a bucket is entering and leaving the jet rather than sitting in it. Every one of those is a defect of a particular machine. None of them is in the idea.

A slower bucket and a faster one #

The flatness of the optimum deserves the picture rather than the sentence, because it is what makes the machine usable.

A wheel running ten per cent fast or ten per cent slow is at 99 per cent of its best power. That is what lets a Pelton wheel drive a synchronous generator — the shaft speed is fixed by the grid and the head varies with the season, so the machine is nearly always off design and nearly always fine. The adjustment that matters is the nozzle, which changes the flow rate rather than the speed, and the site’s arithmetic says why: ṁ appears in the power as a simple factor, so throttling the jet scales the whole curve down without moving its peak.

The contrast with a wind turbine is instructive. There the free stream cannot be throttled, the optimum is a tip-speed ratio rather than a speed, and holding the machine at it as the wind changes is the whole of the control problem.

What separates this machine from every other turbine on this site #

A Pelton wheel is an impulse machine: the pressure is atmospheric on both sides of the runner and all the work comes from turning the flow. Euler’s turbomachinery equation covers it as a special case — the work is U ΔV_θ as it is for every rotating machine — and in the language of that essay a Pelton bucket is a stage of zero reaction, where the whole of the kinematic identity’s work sits in the first term.

The contrast with a reaction machine is worth drawing because it explains where each is used. A Francis or Kaplan runner works inside a casing with a pressure drop across it; the water is not free and the machine has to be full. That is the right arrangement for a large flow at a modest head. A Pelton wheel converts the whole head to velocity in a nozzle first, then extracts it in the open, and that is the right arrangement for a small flow at a large head — where the jet velocity is enormous and a runner would have to be impossibly large or impossibly fast to match it.

Which of the two a duty needs is not a matter of taste, and it is not decided by either of these essays: it is decided by a single dimensionless number computed from the flow rate, the head and the shaft speed, and that number is the subject of its own rung.

There is also a pressure question hiding in the choice. A reaction runner works below the ambient pressure over part of its blade, which is where a machine tears the water it is pumping; an impulse wheel never does, because the water is at atmospheric pressure throughout. A Pelton wheel cannot cavitate. Its problems are erosion by silt and fatigue in the buckets, which are materials problems rather than fluid ones.

The straight line hiding under the parabola #

The power curve is a parabola because it is a product, and the other factor is worth separating out, because it is the machine’s characteristic and it is a straight line.

The force is proportional to $(V-U)$ and to nothing else, so the torque falls linearly from a maximum when the wheel is held still to exactly zero when the rim reaches the jet speed. The parabola is that line multiplied by the speed, and every operational property of the machine reads off the line rather than the curve.

Starting torque is twice the running torque. A Pelton wheel held at rest under a full jet is producing double the torque it will produce at its optimum, so it starts against load without any assistance, which is not true of most turbomachines.

And runaway is exactly the jet speed — twice the normal running speed. Take the load off a wheel in full flow and the only thing decelerating it is windage and bearing friction, so it accelerates until the buckets are keeping pace with the water and stops there. That doubling is the design condition nobody wants to meet: centrifugal stress goes as the square of the speed, so the runner has to survive four times the load it carries in service, and the buckets that would fly off are travelling at the speed of a jet driven by a four-hundred-metre head.

Which is what the deflector is really for. Shutting the nozzle would stop the runaway and would cost a pressure transient the penstock cannot take; swinging a plate through the jet removes the driving torque in under a second and leaves the pipe undisturbed. The wheel coasts, the spear valve closes at its leisure, and the two failure modes are separated by putting each on its own timescale.

What the picture cannot show #

No jet is drawn anywhere in this essay, and that is deliberate rather than a gap. There is no solved field here: the analysis is a control volume round a bucket with the faces labelled, and a sketched flow through the bucket would be exactly the decoration this site’s first invariant forbids. What can be drawn honestly is the velocity triangle, which is not a decoration — once the three vectors are to scale the work, the exit velocity and the efficiency are all lengths in the picture.

The relative speed is not quite unchanged. Friction on the bucket surface reduces it by a few per cent, which enters the force as a factor multiplying (1 − cos β) and is quoted in the literature as a bucket coefficient of about 0.9 to 0.98. It is a measurement, this site does not compute it, and no number here depends on it.

The optimum is flat. The power curve is a parabola, so at 0.45 or 0.55 of the jet speed the machine is still at 99 per cent of its best. That flatness is why a Pelton wheel can drive a synchronous generator at fixed speed across a range of heads, and it is a more consequential property than the optimum itself.

Where the head became a jet #

Everything in this essay begins with a jet already made, and the machine upstream of it deserves a paragraph, because it is where the head goes.

A Pelton wheel is fed from a reservoir far above it through a penstock and a nozzle. The nozzle’s job is to convert the whole of the available head into velocity: V = √(2gH), which for a 400-metre head is 89 metres a second. Everything the runner does afterwards is a momentum exchange at atmospheric pressure, which is what makes the machine an impulse turbine rather than a reaction one.

Two consequences follow and both are practical.

The nozzle is the control. Flow is regulated by a spear valve moving in and out of the nozzle, which changes the jet’s area and not its speed — because the speed is set by the head, which does not change. So a Pelton wheel throttled to a quarter of its flow is still running at its optimum bucket speed, and its efficiency barely moves. A reaction turbine throttled the same way is off design everywhere.

Closing the spear valve quickly is forbidden. The penstock is a long pipe full of moving water and shutting it abruptly costs ρaΔV — hundreds of metres of head on top of the four hundred already there. Large installations therefore fit a deflector that swings across the jet in a second while the spear valve closes over half a minute: the runner loses its load immediately and the penstock never notices.

That pairing — a fast deflector and a slow valve — is a design that falls out of two essays in this field which have nothing else to do with each other, and it is the sort of thing the field exists to make visible.

Who found it, and when #

Lester Pelton patented the split bucket in 1880, having noticed — the story goes — that a wheel whose buckets had shifted on their spindles ran better than one whose jets struck them centrally. The underlying result is older and belongs to the same eighteenth-century tradition as Borda’s mouthpiece and Euler’s turbine equation: the statement that the power of a jet on a moving vane peaks at half the jet speed appears in Euler’s own work on hydraulic machines, and Poncelet built undershot waterwheels on the principle decades before Pelton.

What Pelton added was the geometry that makes the ideal nearly reachable — the split bucket that turns the flow through 165° in two sheets and gets them out of the way of the next bucket. The physics gives the maximum; the invention is the arrangement that costs 1.7 per cent instead of throwing the water back into the jet.

What the impulse machine gives up #

Ending with the machine’s limitations rather than its ideal efficiency keeps the account honest, and they are all consequences of the same choice.

An impulse wheel converts the whole head to velocity before the runner, in a nozzle. So the runner has to be large enough that its rim can travel at half of a very large jet speed, and for a modest head that means either an enormous wheel or an impractically fast shaft. Below about fifty metres of head a Pelton wheel is simply the wrong machine, and the number that says so is the specific speed.

It also uses only one part of its runner at a time. A reaction turbine has water in every passage simultaneously; a Pelton wheel has one or two buckets in the jet and the rest of the rim doing nothing but pushing air out of the way. That is why multi-jet machines exist — up to six jets round one wheel, each with its own nozzle — and why the specific-speed band for a Pelton wheel widens when the jets are counted.

And it cannot use a draught tube. A reaction turbine recovers some of the kinetic energy leaving its runner by decelerating it in a diverging passage below the machine, which is the same pressure recovery a Venturi’s diffuser performs. An impulse wheel has nothing to recover — the water leaves at atmospheric pressure and, at the optimum, at nearly no speed at all — so the tailrace level below the wheel is head that is simply lost.

Each of those is a consequence of running at atmospheric pressure in open air, which is also the source of every virtue the machine has.

Where the ladder goes next #

The jet has been made, aimed, split and extracted from. What remains is what happens when a jet is put inside a duct rather than left in the open — where it can drag its surroundings along with it, and where the mixing that would be a pure loss in a pipe fitting becomes the pumping mechanism of a machine with no moving parts at all.