What kind of turbine is a Pelton turbine?

A Pelton turbine is an impulse turbine: a nozzle produces a high-speed jet of water, which strikes the cup-shaped buckets arranged around the rim of the runner and drives it. Its defining feature is that the pressure energy of the water is fully converted to kinetic energy in the nozzle before being used. It suits sites with a high head and a small flow rate, and is widely used in mountain hydropower plants.

Why is power maximized when the bucket speed is half the jet speed?

The force transferred to the bucket is proportional to the relative velocity (jet speed minus bucket speed), and the power is that force times the bucket speed. So the power has the form (V minus u) times u, which is maximized when u equals V/2. With the bucket stopped the force is large but it does not move, so the work is zero; with the bucket moving as fast as the jet the relative velocity is zero so the force is zero. The optimum is exactly halfway.

Why is the speed ratio of a real machine slightly below 0.5?

The theoretical optimum speed ratio is 0.5, but real machines are set at about 0.44 to 0.48. This accounts for friction losses on the bucket surface, the timing of the jet striking successive buckets, and care to keep water leaving one bucket from interfering with the next. The efficiency curve is a gentle hump near 0.5, so a small deviation costs only a little efficiency.

Why is the bucket deflection angle not 180 degrees?

In theory, reversing the jet through 180 degrees gives the largest force, but real buckets are designed at around 165 degrees. A full 180-degree reversal would make the rebounding water collide with the incoming jet or the bucket itself. Leaving a small angle steers the departing water in a direction that does not obstruct the rotation of the runner. The power loss is only about 1 to 2 percent.

Pelton Turbine Simulator — Free Online Calculator

Parameters

Net head H

Flow rate Q

m³/s

Speed ratio φ = u / V

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Bucket deflection angle β

A nozzle velocity coefficient Cv = 0.98 and a bucket friction coefficient k = 0.9 are assumed.

Results

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Jet velocity V

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Power output P

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Hydraulic efficiency η

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Force on the runner F

Pelton Runner and Jet

Cyan = jet from the nozzle / cups around the runner = buckets / arrow = bucket speed u

Hydraulic Efficiency versus Speed Ratio η(φ)

X axis = speed ratio φ = u/V / Y axis = hydraulic efficiency η (yellow dot = current φ, dashed = optimum φ = 0.5)

Theory & Key Formulas

A Pelton turbine is an impulse turbine in which the nozzle converts all the pressure energy of the water into kinetic energy, and the resulting jet strikes the buckets of the runner.

Jet velocity V. Cv is the nozzle velocity coefficient, g is gravity, H is the net head:

$$V = C_v \sqrt{2 g H}$$

Force F on the runner. u is the bucket speed, k is the bucket friction coefficient, β is the deflection angle, ρ is the density of water, Q is the flow rate:

$$F = \rho Q (V - u)\,(1 + k\cos(180^\circ - \beta))$$

Power P and hydraulic efficiency η (relative to the available power ρgQH):

$$P = F\,u, \qquad \eta = \frac{P}{\rho g Q H}$$

Since the power has the form (V−u)·u, it is maximized when the bucket speed is exactly half the jet speed (speed ratio φ = u/V = 0.5).

What is the Pelton Turbine Simulator

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There are lots of kinds of water turbines, right? What is special about a Pelton turbine?

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A Pelton turbine is the type called an "impulse turbine". Roughly speaking, a nozzle shoots water out at high speed, and that jet strikes the cup-shaped buckets arranged around the runner and drives it. The key trick is that the pressure energy of the water is fully converted to kinetic energy in the nozzle before it hits anything. Press "Spin the Runner" in the simulator above and you can see the jet from the nozzle hitting the buckets.

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What sort of site is it suited to?

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High head, small flow rate. In the mountains, where there is little water but a drop of hundreds of metres — that is where it shines. Increase the "net head H" in the simulator and the jet-velocity card shoots up. From $V = C_v\sqrt{2gH}$, it scales with the square root of the head. At 800 m of head the jet velocity is over 400 km/h. That is a tremendous force.

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When I move the "speed ratio" slider, the efficiency graph makes a hump. It is highest at 0.5 in the middle.

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That is the most interesting part of this turbine. The force transferred to the bucket is proportional to "jet speed minus bucket speed", the relative velocity. And the power is force times bucket speed. As a formula, the power has the form $(V-u)\cdot u$. If the bucket is stopped the force is largest but it does not move, so the work is zero. If the bucket runs away as fast as the jet, the relative velocity is zero so the force is zero. Halfway between — speed ratio 0.5 — the power is maximized.

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But the default for a real machine is 0.46. Why not 0.5?

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The theoretical optimum is 0.5, but real machines are set at about 0.44 to 0.48. There is friction on the bucket surface, and care has to be taken so water leaving one bucket does not hit the next. But look closely at the efficiency curve — near 0.5 it is a gentle hump. So a small deviation costs only a little efficiency. The "margin" you get in design comes from this gentle hump.

Frequently Asked Questions

You choose by head and flow rate. The Pelton turbine is an impulse turbine suited to sites with a high head (hundreds of metres and up) and a small flow rate. The Francis turbine is a reaction turbine for medium head and medium flow that uses both the pressure and the speed of the water. For low head and large flow, the Kaplan (propeller-type) turbine is used. A dimensionless figure called the "specific speed" guides which turbine type is best for a given site.

The runner speed is fixed by the generator frequency, so the speed ratio stays roughly constant. Power is adjusted with the "needle valve (spear valve)" at the tip of the nozzle, which changes the thickness of the jet, that is, the flow rate Q. Since the jet velocity is set by the head and stays constant, throttling the flow reduces the power. On a sudden load rejection, a "deflector" diverts the jet away from the runner to prevent overspeed.

A single runner may be fed by two to six nozzles. Splitting the same total flow into several thinner jets makes each jet thinner, which raises the efficiency of the energy conversion at the buckets. It also makes it easier to adapt to changes in flow rate. If an even larger flow is needed, a horizontal-axis arrangement with several runners side by side can be used.

Because high-head water leaves the nozzle at hundreds of km/h, fine sand grains carried in the water scour the bucket surface at high speed — "sand erosion". It is especially severe in rivers fed by glaciers, which carry a lot of sediment. Countermeasures include removing sand with a settling basin at the intake, hard coatings on the bucket surface, and periodic weld repair. Because the precision of the bucket shape directly affects efficiency, wear management is an important maintenance item.

Real-World Applications

Mountain hydropower: The Pelton turbine's main stage is the high-head hydropower plant. Water is led from a dam or intake weir through a long penstock and a drop of hundreds of metres is converted in one go into the kinetic energy of a jet to generate electricity. In countries with steep terrain — the mountainous regions of Switzerland, Norway and Japan — it forms a backbone of the electricity supply.

Pump-turbines of pumped-storage plants: Some high-head pumped-storage plants use a Pelton-type configuration. Surplus power at night pumps water up to a high reservoir, and it generates during the daytime demand peak — as the heart of this large-scale "battery", the Pelton turbine, strong at high head, is chosen.

Small-scale and micro hydropower: Pelton turbines (and their relative, the Turgo turbine) are also used at small high-head sites, such as drop structures on agricultural canals or small mountain streams. Because the structure is relatively simple, it tolerates flow variation, and it resists sediment to some extent, it is attracting attention as a distributed renewable energy source.

Education and the fundamentals of fluid machinery: The relationships "force proportional to relative velocity" and "(V−u)·u is maximized at u=V/2" are among the most basic ideas for learning energy conversion in fluid machinery. Because the interaction between the jet and the buckets is visible and easy to grasp, the Pelton turbine is widely used in textbooks as an introductory example of turbomachinery.

Common Misconceptions and Cautions

The most common misconception is to think that "the faster you spin the bucket, the more power you get". The power has the form $(V-u)\cdot u$, and beyond a speed ratio of φ = u/V = 0.5 it actually falls. When the bucket gets too fast, its relative velocity against the jet becomes small and the transferred force drops. In the simulator, move the speed-ratio slider from 0.5 toward 1.0. The efficiency curve crosses the hump and goes downhill, and at φ=1.0 (bucket as fast as the jet) the efficiency is zero. The answer is not "faster is better" but "exactly half".

The next most common error is to assume that the net head H and the jet velocity V are proportional. In reality $V = C_v\sqrt{2gH}$, proportional to the square root of the head. Quadrupling the head only doubles the jet velocity. And since the power depends on both the flow rate and the jet velocity, the power gain from raising the head is gentler than intuition suggests. Moving the head slider in equal steps while watching the jet-velocity card shows how the increase slows down toward the higher end.

Finally, take care that what this simulator shows is the "hydraulic efficiency", not the overall efficiency at the generator terminals. What is computed here is the fraction of the available water energy ρgQH that reaches the runner. In a real plant, on top of this come the friction losses of the penstock, the losses in the mechanical bearings, the electrical losses of the generator, and so on. The overall efficiency of a large Pelton plant is typically about 85 to 90 percent, and the simulator's hydraulic efficiency represents only the runner stage of that. Understanding each loss separately is the starting point of performance evaluation.

Pelton Turbine Simulator — Power and Efficiency of an Impulse Turbine

What is the Pelton Turbine Simulator

Frequently Asked Questions

Real-World Applications

Common Misconceptions and Cautions

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