5.3.2: The Pelton Turbine

Sometimes there is much less water in a stream than in a big river – but the stream flows at high altitude, and at some point cascades down a steep mountain side, or even forms a waterfall. It’s a good situation for building a power plant of another type. Namely, the water is “sent downhill” through a long penstock. For each 10 meters (33 feet) of vertical drop the pressure in the pipe increases by one Atmosphere¹, and if the overall vertical fall is large, e.g., several hundred meters, then the pressure at the bottom of the pipeline becomes really high, and water squirts from the nozzle at the pipeline’s end with a very high speed. The turbine type appropriate for using in such a situation is the Pelton Turbine.

The Pelton Turbine has one considerable advantage for an instructor. Namely, when talking about the Kaplan and the Francis turbine, the need instructor tells you that both have high efficiency, they can “intercept” over 90% of the energy carried by flowing water. But if you start asking indepth questions, e.g., What specifically makes the efficiency so high?, then the answer may be either: It’s what the measurements show – in other words, the instructor honestly admits that he/she doesn’t know much more – or, if the instructor does know, you may expect an answer: Well, I can tell you, but it may help if you first read a book about fluid mechanics. And fluid mechanics, you must know, is a highly sophisticated branch of Newtonian mechanics, loaded with pretty high level calculus.

The Pelton Turbine also has a very high efficiency – even higher that the Kaplan and Francis – and the explanation is pretty simple. No need to delve into the intricacies of Newtonian fluid mechanics! Everything will become clear if we first consider two really simple problems: – so that if a pressure is expressed in the units of atm, one quickly gets the idea of what is being talked about.

Problem \(\PageIndex{1}\)

A big U-HAUL truck is parked on a street. A boy throws a basketball at the truck’s back door with the speed of v, a. How much of the ball’s kinetic energy will be transferred to the truck? Assume that the ball-door collision is perfectly elastic.

Solution

if the collision is perfectly elastic, the ball is bounced back with the same speed v, so it’s kinetic energy remains unchanged – ergo, no energy is transferred to the truck.

Figure \(\PageIndex{1}\): Typical arrangement of a power plant using Pelton turbines: the reservoir is located high above the turbine house. Water is brought down through a pipeline (“penstock”). Due to the specific geography of Norway, such powerplants are popular in this country. Pipelines running down on mountain slopes are a common view in the Norwegian landscape (aod).

Problem \(\PageIndex{2}\)

Now, assume that the truck started moving, and when the ball hits the door, its speed is exactly v/2. How much of the ball’s kinetic energy will be transferred now?

Solution

For an observer moving together with the truck, the ball hits the door with a v − v/2 = v/2 speed. After the collision, it moves with the speed of v/2 relative to the truck, meaning that its speed relative to the street is now zero – ergo, it has no kinetic energy, meaning that all kinetic energy it had before the collision has been transferred to the truck

Figure \(\PageIndex{2}\): Left side: a measuring spoon acting as a “water jet reverser”. The same effect sometimes gives a person washing dishes an unexpected and unpleasant “shower” (aoph)

Figure \(\PageIndex{3}\): The Pelton turbine also uses elements that act as “turnarounders” of water streams hitting them. They are called “buckets” and a number of them is mounted at the perimeter of the runner. The water jet hits the “ridge” in the bucket’s center and is split into two streams that are both turned around (source: Wikimedia commons).

Now, let’s think: is it possible to do the same not to a ball, but to a jet of water? Is it possible to find a gizmo that would “backreflect” the jet so that it would retain its kinetic energy? The answer is yes and, most probably, everyone who was ever washing dishes in a kitchen sink did have a “bad experience” with such a gizmo: a spoon, if carelessly placed under a running faucet, can give one an unexpected unpleasant “shower”. In Fig. \(\PageIndex{2}\) one can see the results of a “kitchen sink experiment” carried out by the author: a stream of water from the faucet falls on a measuring spoon of spherical shape (a blown-up picture of this spoon is in the right-side picture) and is almost completly reversed and runs upwards.

Figure \(\PageIndex{4}\): Left side: a complete “runner” with a full set of “buckets”. atteched. The Pelton turbine runners do have a lot of charm, don’t they? Therefore those from decomissioned turbines are often used as decorations in parks or outdor art galleries. The one in the photo is such a decoration in the front of a building in Barcelona, Spain. Right side: a scheme of a Pelton turbine assembly with a single nozzle; usually, more one nozzle is used in order to increase the power output from the runner (source: Wikimedia Commons).

A Pelton turbine uses a number of “buckets” attached to a wheel called a “runner”(Fig. \(\PageIndex{4}\), left). In a working assembly (Fig \(\PageIndex{4}\), right), the water jet is sent to a bucket center. The “reversal” of the water jet is equivalent to the elastic collision of the ball with the truck door that was discussed earlier. So, if the turbine rotates with such a speed that the bucket moves away from the nozzle with a speed equal to one-half of the water jet speed, then the water flows out of the recessing bucket with zero speed – meaning that all its kinetic energy has been passed to the turbine.

The reasoning that has been used in the problem with the basketball hitting a receding truck remains perfectly valid for the water jet reversal in a Pelton bucket – but we can redo the problem in a more formal way. Let’s use the symbols from Fig. \(\PageIndex{5}\): velocities observed in the static reference frame are marked with the subscript “S”, and those observed by a “virtual person” sitting in the bucket are marked with the subscript “B”.

Water is squirting from the nozzle with velocity V_S, but because the bucket is receding from the nozzle with velocity U_S, the virtual observer sitting in the bucket sees a jet flowing in with velocity V_B:

\[V_{B}=V_{S}-U_{S}\notag \]

The jet is reversed inside the bucket, and the virtual observer sees it flowing out with velocity \(V_{B}^{t} = -V_{B}\), so that

\[V_{B}^{t} = -V_{S} + U_{S}\notag \]

Figure \(\PageIndex{5}\): An idealized analysis of a water jet velocity reversal in the Pelton bucket. All “red” velocity vectors are those observed in the static reference frame and the “blue” ones are those seen by a virtual observer moving along with the bucket (aop).

To get the speed \(V_{S}^{t}\) of the flowing out water seen in the static system, we have now to add the recession velocity to \(V_{B}^{t}\) :

\[ V_{S}^{t} = V_{B}^{t} + U_{S} = -V_{S} + 2 U_{S} \notag \]

From the above, one gets immediately that V_S is zero when the bucket recedes with a speed equal to one-half the speed of the water jet squirting out from the nozzle. And \(V_{S}^{t} = 0\) means that the water flowing out of the bucket carries zero energy –ergo, all energy carried by the water jet entering the bucket is “left inside the bucket”, i.e., is transferred to the turbine wheel. It implies that with the condition U_S = 0.5V_S the efficiency of the Pelton turbine id 100% – but one has to remember that “nothing is perfect”, there are always some deviations from idealized assumption, so in practice the efficiency will be always lower than 100%.

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¹An Atmosphere (symbol: atm) is a pressure unit that was widely used in the old days, but now is barely tolerated by the official SI system. But sometimes it is convenient to use it, because the pressure of 1 atm is approximately equal to the pressure of ambient air.