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The hydroelectricity plants extract the potential energy of highly deployed massive object (water) as it falls down. Without turbine, all that energy would be converted into speed (kinetic energy) at the bottom of the waterfall and further into heat. The turbine produces energy by slowing water down.

The efficiency of turbine, how much energy is extracted by turbine, can be charactarized by the exhaust speed: the faster is the output stream, the less efficient our turbine is since not all speed/energy is extracted. So, slower the turbine spins, the higher is its the efficiency. The extraction is 100% when turbine does not spin and no electricity is produced at all. So, there must be a trade-off between the efficiency and amount of the output, the trade-off determined by the turbine spinning speed (exhaust speed). How is it decided?

I read that large modern water turbines operate at mechanical efficiencies greater than 90%. Since couple of percent losses are inevitable whatever you do, it seems that they say that theoretical efficiency is 100%. Identical efficiency is provided by switching power supply converters, which are 100% efficient in theory. I understand the secret exploited by SMPS. My question is how similar, 100% energy extraction, is achieved through the turbines, which seem to operate linearly (spinning at the same pace) rather than switching mode pumping. What is the water release speed when 100% energy extraction is achieved?

This question is actually is not limited to water wheels. Today wind turbines are becoming more popular and I am curious how do you extract all power from the wind flow. If turbine spins quckly, the air is realased at high speed, which means that you do not slow down the flow, which means that it makes no work. On the other hand, if if you stop the flow completely, your turbine stops and no power is extracted either. What is the optimal turbine speed?

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  • $\begingroup$ It seems you've answered your question mostly in your introduction of the problem. The release speed would be 0 if it wasn't for some contingencies: the turbine cannot be of 0 size, you cannot increase its torque in such a way that electricity production would remain finite as the turbine rotation rate actually goes to zero, etc. $\endgroup$ – Joce Jun 21 '14 at 14:54
  • $\begingroup$ @Joce torque != efficiency. In my introduction, I have explained that losses are inevitable, yet, there are theoretically 100% efficient converters in electronics. I guess smart designs are possible for the turbines also. This does not mean that their size must be zero. You loose not because your turbine is large. You loose if you let the water go without extracting all its kinetic energy. $\endgroup$ – Val Jun 21 '14 at 16:51
  • $\begingroup$ Keep in mind that there's a lot of science that goes into the shaping of the sluices feeding the wheel and the shaping of the vanes/blades on the wheel itself. Generally one wants the energy in the water to be expended at relatively a steady rate, from the point of first impingement on the wheel until the water leaves the wheel. As the water loses energy it necessarily slows down (though don't ask me to define the frame of reference), meaning that the "wheel" must somehow accommodate larger volumes of slow water as it "builds up". This clearly calls for a wheel that gets larger near the exit. $\endgroup$ – Hot Licks Feb 11 '15 at 23:09
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There is indeed an optimal arrangement for extracting the most usfeul energy from moving Newtonian fluids.

The highest efficiency possible is $\frac{16}{27}$ - this is known as the Betz Limit or Betz Coefficient, after Albert Betz, who rediscovered it in 1920 after Fred Lanchester first calculated it in 1915.

The ratio of exit speed to entry speed is 1:3 at the Betz limit.

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There is a simple trick to achieve really high (>= 90% )efficiency: Just before the flowing (falling) fluid hits the propeller, make it rotate using guides, i.e., give it some angular speed beside the linear speed in a closed tube. To benefit from the fluids kinetic energy (in this case it is saved in both the rotational kinetic energy and linear speed kinetic energy form) what you would do is clear: Slow down the rotational speed (till down to almost zero) and leave only very little linear speed sufficient for escape of the water from around the propeller. In this setup, where you have chance to "first" convert the fluid's kinetic energy to "mostly" rotational kinetic energy (Er = (1/2)ww*I), you have a fine range of design options to adjust the angular speed of the blades, their size, and the blade angles. You can just search for "Kaplan Turbine design" for technical details.

For the wind case, you don't first give an early rotation to the air flow. It directly and linearly hits to the propeller blades on the open air. Then you have the Betz'z limit. If you could have closed tube where you would have more play area as in , for instance, the Kaplan turbine case, you would achieve higher efficiency values.

Additional note on water turbines: In some (most) cases, the angular speed of the blade is adjusted to fit into system frequency (taking into account also the generator's # of poles). Don't be mislead that having a non-zero w, angular speed means you waste some energy of fluid. The power generated involves also the torque produced by the blade (as the entering & rotating fluid's rotation almost stops as it leaves the blades and producing torque). It is in a way like this: Rotating water hits the propeller blades, makes it rotate, and finally as the water just leaves the tip of the blades its rotational speed (not the blade's!!) drops almost to zero. As the blades keep rotating at a constant non-zero angular speed, almost all of the rotational and a large percentage of linear kinetic energy of the fluid are left in the system (by generating torque; and the produced power is, P = Torque*AngularSpeedOfBlades).

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This is actually quite simple. You measure the maximum no-load velocity of the Turbine and divide it with 2. At this speed you have the highest power. Practically the optimal value is between 1.8-2.3, depending on turbine type. This all is clearly explained in the book of Carl Pfleiderer, Strömungsmachinen, 1952. Page 248.

Carl Pfleiderer, Strömungsmachinen, 1952. Page 248

The diagramm has the rpm on x-Axle and the Mx is Moment, Px is Power, Vx is Mass flow and ηx is efficiency. At the text is the clear simple statement; theoretischen wert 2; theoretical value is 2.

It should be noted that this principe works with all turbines. Even with wind turbines. The Betz-law is not be mixed with this. It has absolutely nothing to do with the Turbine rpm; you can slow down the wind with 1:3 ratio with many blades rotating slowly, or with few blades rotating fast.

So if you want an easy setting; let the turbine run without load, and regulate it rotate with 1/2-speed, and then seek the optimum be reducing the rpm as long as the power output rises. And it's done.

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Turbines (impellers) have fins that work more like aircraft wings and sails than the buckets of a old water wheel. They have have very high lift/drag ratios which translate the high efficiency. It is friction and heat loss that kill efficiency not the work of the turbine (how much it spins).

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  • $\begingroup$ I am asking how the airoplane-style turbines achieve the 100% theoretical efficiency rather than anything else. I cannot even think of a question that you are responding to. $\endgroup$ – Val Jul 21 '14 at 19:55
  • $\begingroup$ Downvoted. Sorry, but that's just wrong. Friction and heat are not major factors, let alone dominant. $\endgroup$ – WhatRoughBeast Oct 2 '15 at 15:26

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