Nice idea. Going to be some interesting engineering problems in scaling that up. I don't know much about fluid flow, but it's got to be turbulence that makes that mylar ribbon flap, which I presume would be induced by the slot the wind is forced to pass through and the ribbon itself. It might have to be adjustable in several ways--the gap between the slots and the tension on the ribbon come to mind--to get it to flap at different wind speeds. I wonder what the relationship is among wind velocity, tension, gap size, and frequency. Be a bit of a nuisance if the thing produces a variable frequency AC current. I can see problems with materials too, as in, how long would it take for that ribbon to flap itself to pieces. But those are engineering problems, there's no new science involved, I'm sure that guy's bright enough to figure it out. I wish him well.
have you stood near the large wind turbines? I've been told they're far from quiet.
In retropect, if they can build the Tacoma Narrows bridge then this concept should be doable.
I should imagine that the Tacoma Narrows bridge could have been made a lot stronger and still collapsed. A way has to be found to limit the destructive resonance to the point where the forces could be harnessed. I've thought a bit about this and I can't help feeling that more conventional wind turbines, adapted for varying wind velocities would be the way to go.
There is a lot of kinetic energy there and all kinetic energy eventually turns to heat at some point, but I have no idea how it might be harnessed.
n dimensional analysis, the Strouhal number is a dimensionless number describing oscillating flow mechanisms. Often, it is given as: where S t is the dimensionless Strouhal number, f is the frequency of vortex shedding, L is the characteristic length (for example hydraulic diameter) and V is the velocity of the fluid.
For spheres in uniform flow in the Reynolds number range of 800 < Re < 200,000 there co-exist two values of the Strouhal number. The lower frequency is attributed to the large-scale instability of the wake and is independent of the Reynolds number Re and is approximately equal to 0.2. The higher frequency Strouhal number is caused by small-scale instabilities from the separation of the shear layer (Kim and Durbin, 1988 and Sakamoto and Haniu, 1990).
Hydraulic diameter
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It has been suggested that this article or section be merged into Frictional pressure drop . (Discuss) The hydraulic diameter, D h , is a commonly used term when handling flow in noncircular tubes and channels. Using this term one can calculate many things in the same way as for a round tube.
Definition:
where A is the cross sectional area and U is the wetted perimeter of the cross-section.
For a round tube, this checks as:
Frequency of the wave
Once the speed of propagation is known, the frequency of the sound produced by the string can be calculated. The speed of propagation of a wave is equal to the wavelength λ divided by the period τ, or multiplied by the frequency f :
If the length of the string is L, the fundamental harmonic is the one produced by the vibration whose nodes are the two ends of the string, so L is half of the wavelength of the fundamental harmonic. Hence:
where T is the tension, μ is the linear mass, and L is the length of the vibrating part of the string. Therefore:
- the shorter the string, the higher the note
- the higher the tension, the higher the note
- the lighter the string, the higher the note
For the common case of a homogenous substance of length L and total mass m, this simplifies to:
The term linear density also refers to how dense a line is drawn.
I really don't think this concept will scale up well. I see it more as an application for remote areas with low power requirements. That said if it is going to be scaled up the tension probably needs to be increased because if the flapping part is in greater tension then it should be able to lift a greater load.
Indeed, the concept wasn't intended for something which can be scaled up. It was designed for micro applications, because the turbines don't scale down very well.
I have a very small axial flux turbine that I'm building from a modified hard drive. What's the problem you're talking about with scaling turbines down?
Conventional wind turbines don’t scale down well—there’s too much friction in the gearbox and other components. So poor, remote communities don’t have any way to harness the power of the wind. Till Shawn Frayne, a 28-year-old inventor based in Mountain View, Calif., saw the need for small-scale wind power to juice LED lamps and radios in the homes of the poor. Frayne’s device, which he calls a Windbelt, is a taut membrane fitted with a pair of magnets that oscillate between metal coils. Prototypes have generated 40 milliwatts in 10-mph slivers of wind, making his device 10 to 30 times as efficient as the best microturbines. Popular mechanics has a good article about the device here.
105. RE: Windbelt, Cheap Generator Alternative, Set to Power Third World
I'm not that impressed. 50 milliwatts is nothing. I have a propeller generator that will produce more power in low wind. There is not enough room fot the magnet to travel to develope usable power. Nothing will become of this. Lets have some details, how many volts at how many amps, under what conditions.
78. some really stupid commenters
Do people have any idea what 0.04 Watts at 10 mph means? It's a *micro*power device! When you need 100 Watts, get a wind turbine for $500, not several thousand of these. And fools who propose sticking these on cars (which already have huge alternators and 12V outlets) or next to laptop fans (which already have batteries and powered USB ports)... learn about conservation of energy.
67. Great!
This is a vary cool idea! Not only that but its well thought out. (All credit to Wikipedia for without it I could not argue a valid point) The point "it doesn’t generate much energy" keeps popping up, so ill see if I can put some facts into the fight. The power available in the wind is given by: (for a Turbine) P = 0.5A Rho r^2 V^3 Where: P = power in watts, A = efficiency constant, Rho = mass density of air in kilograms per cubic meter, (Mass density is worked out as – Rho = Mass / volume) r = radius of the wind turbine in meters, V = velocity of the air in meters per second. So for R=30 Meter's (average inland turbine blade radius) V = 10 mph (16 km/h or 4.5 m/s) (The average speed at wind farms) Rho = 1.225 (At 15*C at sea level) A = 30% (because I don’t know the true figures here I am using a figure someone mentioned earlier) P = 0.15*1.225*900*91.125 = 15069 watts (15Kw or 15,069,000 milliwatts) For the Wind Belt Looks to measure about 80 CM * 10 CM * 10 CM or there about’s Apparently generates 40 milli watts at 10 mph So in its current state if you built a wall out of them 80 meters across and 60 meters high (to match the un-usable area of a turbine) you would generate Number of units = 60000 (80 meters / 10 CM) * (60 meters / 80 CM) 40Milliwatts * 60000 units = 2,400,000 milli watts So to conclude, no it doesn’t generate as much power as a turbine. But Its easier to repair, and requires less land to be dedicated to its use, its cheaper and its user friendly. A good product for an unexplored market
19. RE: Windbelt, Cheap Generator Alternative, Set to Power Third World
Website: http://ww.com/
I'm sorry but this device is not 10 to 30 times better than then the best microturbines, which are over 20% efficient. Betz' limit is not about to be crossed, no matter whether it is with rotating or vibrating devices. I predict this invention will not 'break' because it it tuned to only a narrow range of windspeeds and wildly inefficient compared to modern rotary machines. Nice try though !