Wind Turbines and Betz's Law

1. Jul 10, 2016

gloo

I am trying to understand wind turbine efficiency and the phenomena of Betz's Law. I have a basic grasp of Betz Law which says the maximum efficiency of a wind turbine is 59 % energy capture of the wind swept turbine area (more or less in those words).

I understand that when the wind hits the turbine and energy is transferred to the turbine, the air molecules slow down and impeded the incoming air molecules from getting through and slows them down -- which in turn reduces the kinetic energy captured by the turbine blade.

Is this the only phenomena that contributes to the 59 percent efficiency dictated by Betz? (I am not talking about friction loss in generator etc.. just kinetic energy capture of the wind sweep area). Isn't there some loss of kinetic energy capture - especially in slow winds caused by the air molecules going straight through the sweep area of the blades without touching the blade and is not captured by the turbine at all?

2. Jul 10, 2016

Baluncore

Each blade of a propeller or turbine “slices” the air in a helical path, so the gaps between blades are not as significant as you might at first think. Air pressure couples the air between the “slices” so that if one small parcel of air is slowed down then the one behind it must also slow down. Betz's law is fundamentally based on the requirement that kinetic energy is removed from the air, but the air must still be kept moving so as not to obstruct the turbine flow.
https://en.wikipedia.org/wiki/Betz's_law

3. Jul 11, 2016

gloo

1. So ...if you had to guess..what theoretical efficiency would be increased if some unknown fantasy configuration would be able to eliminate this " untouched air" would be captured in the sweep area

2. In slower speeds, wouldn't this "untouched air " be more of a factor? Is this the reason that electricity can't be produced at lower end wind speeds? The blades spin too slow because it doesn't grab enough air on the blade sweep around?

4. Jul 11, 2016

Baluncore

There is no "untouched air" because the blades rotate and so effectively cut the entire volume of air passing through the swept area. The axial distance between the paths of any two blades in the airflow can be less than the width of a blade. Each blade is an airfoil. An airfoil works best in clean air. It is therefore only necessary that the axial distance between slices be great enough to prevent interference from the disturbance produced by the previous blade's passage. Everything is a trade-off.

Power is proportional to the square of the wind speed. If the wind speed was ½ of the maximum then only ¼ = 25% of the maximum power could be generated. At 10% of maximum wind speed the power will be only 1% of the maximum. At “lower end wind speeds” a turbine will generate some electrical power, but relatively little compared to the maximum possible.

5. Jul 11, 2016

gloo

OK thanks! I had no idea that the sweep was so effective.
What about the amount of water molecules actuated in the sweep area of a tidal generator?? Those spin really slow and lots of water pass through untouched -- they rely more on the density of water to be a driving force for the generator. How much of Betz's Law applies in the efficiency of the turbines here?

6. Jul 11, 2016

Baluncore

Do not think about the path of individual water molecules but about how they push against each other in the one litre parcels of water.

Notice how there is a wide slow flowing eddy downstream of a tidal generator. Water density does not change so it cannot be the density of water that drives a tidal generator. It must be the reduction in water velocity. Betz's law applies because the slower flowing fluid must depart the turbine.

Where the turbine is driven by a significant head such as in a hydroelectric plant, without a significant velocity reduction, Betz's law does not apply because potential energy is being converted from the pressure difference across the turbine.

7. Jul 11, 2016

gloo

Tidal Barrage turbines are big heavy steel blades - they only have a head of around 5 to 7 meters difference between the reservoir and open sea-- yet I hear that they have 90 percent efficiency. How much of a factor is Betz law in those cases?

8. Jul 11, 2016

Baluncore

A tidal barrage turbine with several metres head does not obey Betz's law because it is converting potential energy into electricity.
A tidal turbine in a tidal stream without head does obey Betz's law because it is converting kinetic energy into electricity.
Is that not obvious ?

9. Jul 11, 2016

gloo

No -- not obvious to me. I am not an engineer - just a guy with grade 13 physics with a curious mind.