Alex hibbert originals / 3D printed micro WT

[Mr Gus]

Enjoy



Professional review, no holding back.

[Oldgreybeard]

Knowing what we have learned over the years about small wind turbines, and their often wildly over-optimistic claims, this does not surprise me one bit. Many of them seem to be produced by people that still believe in perpetual motion, or who, at least, have a very tenuous grasp of the physics involved.

Leaving aside the poor standard of construction and the probably inadequate choice of 3D printed materials (the most common of which are actually biodegradable) there is always a very simple check that can be done on any claim by the sellers of these things. Just check the average wind speed in the area where it is to installed, determine the swept blade area from the specification or any dimensions given, the use this formula to find the absolute maximum performance possible with zero losses and 100% efficiency (i.e.up to the Betz limit that defines the maximum):

Power (W) = 0.5 x Rho x A x V³ x 0.593

Where:

Rho is the air density (usually 1.204kg/m³ at sea level and 20°C)

A is the swept area of the blades, in m²

V is the mean wind speed (for the UK the mean wind speed is about 4m/s, but varies a very great deal from place to place)

0.593 is the Betz constant, and is the factor for the absolute maximum amount of power that can be extracted from the wind by a machine that is 100% efficient.

The "Cutting Edge Power" bit of crap reviewed in that video has an 18" diameter (taken from their advert, here: https://cuttingedgepower.com/products/2022-micro). Let's do some arithmetic and see how their claims stack up, using the (pretty generous) UK mean wind speed of 4m/S and assuming that it is 100% efficient (in reality I doubt it's much better than 50% efficient, relative to the Betz limit). Putting the numbers into the formula above, and assuming the whole 18" diameter is the active swept are (it isn't, that's again being generous to them) then the are aworks out to be 0.164m² in real units.

Therefore:

Power (W) = 0.5 x 1.204 x 0.164 x 4³ x 0.593 = 3.747W

To get 15W from this bit of crap, again assuming that it is 100% efficient relative to the Betz Limit, would need a wind speed of about 12.795kts. More realistically, with an optimistic efficiency of 50% (and I believe that is VERY optimistic) to get 15W is going to need around 16kts, and that's a pretty stiff breeze, about double the mean wind speed for the UK, plus it assumes no obstacles or turbulence.

My guess is that at average UK wind speeds, allowing for the likely overall efficiency, plus allowing for turbulence and the fact that the true effective swept area is less then the full diameter would imply, someone would be lucky to get about 1W out of this thing most of the time.

[Mr Gus]

Nicely done OGB, ..archival elements of this could not be clearer (except maybe when done in led's & powered up in a flash mode)