# 5.3.2: The Pelton Turbine

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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 Atmosphere^{1}, 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.

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.

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

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.

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

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*.*5*V*_{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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^{1}An 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.