In the initial post there is a link to the document. I put it again:
Got me thinking: could the earths rotation be used to make energy with such a large flywheel? Not that it seems feasibleā¦
It powers storms and weather. Coriolis effect. Good idea!
http://ww2010.atmos.uiuc.edu/(Gh)/guides/mtr/fw/crls.rxml
Is there some diagram? Iām still confused when I read this. The cone of cables?
The hub is at the centre on the surface of the water ( Ocean version) . The position of this hub is secured by a number of cables forming a cone to the sea bed. For the simplified Couette flow analysis assume these cables are secure to the sea bed under the outer diameter of the system. The water at and under these securing cables can be considered as stationary for analysis.
Iām not good enough at fluid dynamics to challenge this. We could ignore this for now and look at the thing from other, equally important, perspectives. As in if one of them doesnāt compute the idea in its current form doesnāt make sense.
Some that come to mind: environmental impact, structural analysis, material cost, maintenance, time to market, LCOE, ROI, comparison to alternatives, public acceptance and funding.
Maybe the first step is to make more drawings. I can understand a thing better when I can look at it and donāt just have to visualize it. And it is a way of rubber ducking. The other first step is to realize ideas always evolve.
Yeah Definitely think more drawings would help conceptualization.
Iām concerned
1
if the mass of rotary water isnāt contained, the energy dissipates to the outsides very easily.
2
is it just the hub bearings taking all of the lateral non-torque force? e.g. resistance against oncoming wind
1
if the mass of rotary water isnāt contained, the energy dissipates to the outsides very easily.
2
is it just the hub bearings taking all of the lateral non-torque force? e.g. resistance against oncoming wind
1: It has been assumed that the shear rate will be the same horizontally to that vertically. If the water rotates further out than assumed ( dissipating energy) then the shear rate will actually be lower resulting in less drag on the system.
Within the original patent application is the option to enclose the system with a floating dam with the option of generators mounted on such a dam, which could be driven by the rotating flow.
Careful design of such a dam could result in wave energy being converted into rotational energy within the dam.
Note this ādamā does not contain the water volume so does not need have to be a continuous wall, nor does it need to be full depth.
2: The central hub does take the load, which requires many cables ( the cone of cables ) , which also protects the sea floor from a rotating current.
It will be impossible to fish within the systems diameter, providing a safe haven for fish stock.
I would have to say the starting point for energy-storage-by-uncontained-whirlpool/vortex should be skepticism, and running some initial numbers. (Maybe someone already has run some numbers), and as a start, Iād think youād want to calculate:
- how much energy is taken away from generation, and lost to creating a giant whirlpool;
- how fast the whirlpool dissipates (loses) (wastes) the āstoredā energy;
- how much energy you could expect to get back from the whirlpool and under what conditions, how exactly, and for how long.
As to question 1, is it even realistic to think you could add enough energy to create such a whirlpool in the first place? How much energy would such a whirlpool contain, versus how much energy can be added from wind-power and spoke-drag? In other words, is it even possible to create such a whirlpool using ambient wind?
As to question 2, would the whirlpool lose energy so fast that it never actually stores much that could be retrieved?
As to question 3, how efficient would whatever energy-retrieval mechanism is imagined be? How much energy could you really get back, in proportion to how much energy was put in.
Personally, the first thought I have is the turbine sounds interesting and the low-friction possible by floating on air is intriguing, although untested and possibly fraught with āissuesā.
The whirlpool energy storage aspect is hard to swallow, and I suspect might turn out to be an interesting idea whose actual numbers just donāt hold water, pun intended.
If the steady state losses ( Couettte flow analysis example ) indicates a drag of around 50 KW at a Tip speed of 25 m/s. We could expect it to be more at 33 m/s but compared to the system power of 375 MW in 11 m/s winds that loss is tiny.
The energy dissipated can not be more than the drag on the system.
The rotational speed is around 1/4 RPM at the surface and the depth of and within the systems diameter.
It is slower with increasing depth.
It is also slower the further out from the systems diameter.
It is most unlikely that it will create a whirlpool.
Well, whatever you want to call the rotating water, that is just a word. Call it a non-whirlpool. The whole thing sounds pretty doubtful to me, but that is just a feeling, not a proper analysis.
Has anyone calculated the amount of energy required to create the non-whirlpool? What is the total amount of energy contained in such a rotating mass of water? Could the wind provide enough energy to rotate that amount of water?
From your indications:
A 70 cm diameter bicycle wheel has an axle of about 7 cm. We can say that such a wheel is composed of two symmetrical cones with the same base and two summits. If we cut it in two parts then keep only one part, there will remain a 3.5 cm high cone with only one summit. Such a cone would have substantially the same configuration (in a reduced model) while being relatively twice as deep: 70 cm /3.5 cm VS 2500 m / 52 m. Everything is there including the spokes instead of the cables.
I doubt that such a wheel spinning in water generates little resistance. Indeed I do not think that it is correct to count only the resistance of viscosity (which is generally very low): it is also necessary to count the resistance of form as soon as a significant displacement of the fluid is induced. In the present case, this resistance would be rather tangential. Perhaps a more correct and complete explanation would be possible with a complete simulation.
If what I said is not correct, the cyclists should go almost as fast in the water (considering only the resistance of the wheels), and by extension the swimmers compared to the runners. So the storage concern would even not be able to be evoked.
You can actually build a small scale version of this to test some aspects of it. You could build a circular water tunnel, or two concentric ones, and different types and weights of floats. To propel the floats, instead of using the wind, you could attach them to an arm connected to a motor in the center of the circle. Then it is a game of looking at the system and trying to see what you can improve.
I personally wouldnāt want to deal with the extra losses from trying to extract energy from moving water, and the turbulence that then would again introduce, so I would fix any possible flywheel/weight to the float or hub.
I doubt that such a wheel spinning in water generates little resistance. Indeed I do not think that it is correct to count only the resistance of viscosity (which is generally very low): it is also necessary to count the resistance of form
You need to count form resistance if you drive the system though stationary water like a ship.
The VAWT is not moving its location and there is no bow or stern. The only relevant resistance at steady state is the viscose resistance which depends on the shear rate.
There is a mechanism of transferring power, during a speed change, from the wheel to the water or from the water to the wheel:
For this It may be relevant to consider form resistance of the the cables / spokes at a relevant velocity difference.
A hybrid approach that may also be considered is to keep the structure light but add āpivoting flapsā within the lattice structure to act as valves so that the ring can pick up speed quicker, allowing the water to catch up but when the wind drops the structure is driven with more torque by the water.
The alternative, I prefer, is to consider an increase in the shear rate while the system is changing speed, with some of the water considered as trapped within the lattice structure of the ring.
Depending on how much of the water is trapped depends how much of the mass is considered as a solid flywheel and how much of the water within the rotating structures depth is viscose coupled to it.
Note all the water below that depth is only viscous coupled.
With a wide beam putting buoyancy chambers and also containing water in sealed chambers each side ( radially) of each aerofoil, increases the torsional stability of the lattice structured ring while minimising the required torsional strength.
Flettner & Ballon:
The Flettner case is the same as moving a systems location or operating in a current. That is not the case for the VAWT.
The ballon was not rotating with a constant radius effectively pumping the air radially.
A man can quite easily push or pull a fully loaded barge out from the side of a canal and pull it along.
Pulling it includes form drag and the depth under the barge can be minimal leading to a high shear rate.
The whole concept of using Couette flow analysis is to find an approximate resistance of the steady state condition. The time it takes to get to steady state will depend on many factors, including depth, wind velocity and contained mass. When a system starts energy goes into accelerating the ācontainedā flywheel and the liquid flywheel via the viscose drag which is dependent on the shear rate. Due to the mass of the system the rate of acceleration will be low but that enables a lot of energy to be stored for a small change in velocity.
The critical parameter for modelling is the shear rate, because the shearing stress = dynamic viscosity x shear rate. The density is relevant only to the inertia within a fluid flywheel but not the viscose connection.
Referring to starting tangential force:
consider just 1 of several aerofoils being close to the 90 degree position such that the lift force acts tangentially:
300 metres high with an average chord of 15 metres at an angle of 16 degrees in 11 m/s wind .
At that Chord the Re will be around 11.6x10^6 and a Coefficient of lift for a symmetrical aerofoil will be around 1.682
The lift force will be:
0.5x1.225x4500x11^2x1.682 Newtons = 560.9 KN
The torque = 560.9 x 1250 KNm = over 700 MNm from just 1 aerofoil.
For 375 MW at a tip speed of 33 m/s the tangential force ( from all the aerofoils) is 11.36 MN
At this stage the apparent wind speed on an aerofoil will vary between 22 and 44 m/s dependent on its position. The Re number is dependent on velocity and coefficient of lift changes with Re. The combination of increase in coefficient and velocity^2 causes a dramatic increase in the lift force, but this will be acting at various angles to the tangential direction.
The Flettner rotor and Magnus effect balloon generate lift and drag by the stream as they rotate. Here the stream generating Magnus effect is not taken into account.
The comparison with VAWT flywheel stands because they expense power consumption in order to rotate respectively in the air (Flettner and balloon Magnus effect rotors) or in the water (VAWT flywheel).
The power consumption of a Flettner rotor is indicated in the figure 13 as I mentioned above.
For the Flettner rotor or Magnus effect balloon, the rotation is provided by a motor. For the giant carousel VAWT, the rotation of the flywheel in the water is provided by wind power via the blades. Power consumption is realized for both.
So extrapolations for power consumption of the flywheel as it rotates in water can be done.
If anyone were truly interested in the amount of kinetic energy in a spinning mass of water, it would not be difficult to calculate or at least estimate using simple math and a few approximations.
Perhaps these inflatable sails (each sail including blowers and a telescopic mast) could be acceptable blades for the giant VAWT carousel:
Clever idea. All renderings though. The only real one they left on a trailer. There must be a reasonā¦
Perhaps not this:
The same principle for a wing, being also inflatable and comprising also blowers and a spar (instead of a telescopic mast for the sail):
If you want pilot it, ask to Laurent de Kalbermatten, co-inventor of both sail and wing.
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The Couette flow I referred you to is an example of a solution using the Navier Stokes equations
for steady state conditions in which there is no pressure gradient.
The Force = Dynamic Viscosity of water x Velocity x Area / Gap to stationary surface.
The power = Force X Velocity
Note
The velocity / Gap = the shear rate
This approach is also used in journal bearing but in that case the gap is very small leading to much higher shear rates
Example bearing 50mm diameter at 1000 RPM
Surface speed = 1000 x Pi x 0.05 / 60 m/s = 2.618 m/s
If the Gap is 1 mm the shear rate is 2618 per sec
which is 5000 x greater than the example on the VAWT