As I’ve always said about AWE concepts, rather than starting out chasing megaWatts, try powering a single home first. Or even just a system useful for any purpose. (Charging a phone?)
Maybe the parachute idea could be useful for something - who knows, while not very “efficient”, at least it is relatively simple.
Seems to me it makes more sense to verify a concept at an achievable scale before even worrying about megaWatt versions. I guess it’s easier to dream about giant scale than build at miniature scale. Or for people with misappropriated, unsuitably-large budgets, they can waste money, manpower, and resources, building large models before they have verified the usefulness of the basic concept at a smaller scale.
Betz limit 16/27 is 4 times the limit in pumping mode which is given for 4/27. The paper details the reasons. I see a simple reason: the swept area, in pumping mode, is dragging at 1/3 wind speed which is reel-out speed, keeping 2/3 wind speed. So the force (measured by the square of wind speed) becomes 2²/3² = 4/9. Power is force x reel-out speed: 4/9 x 1/3 = 4/27.
The parachute takes the whole area of power available in the wind, if we do not take into account wind deflections around the parachute, which remains to be seen (but it can be the same for the swept area of a conventional wind turbine). Now if we increase the drag coefficient Cd of the parachute, the drag force will increase. And said drag force will affect the swept area as well as the flying body (in this case the parachute), both being merging. If the Cd is 3, the AWE parachute system could harvest up to 12/27 of the power available in the wind.
I am wondering if drag coefficient is meaningless if a large enough cloth is put in the sky. The Reynolds number maybe converges to something? Seems the cloth would be the ultimate collection of energy in the wind. The wind bending around the edges is not something the Betz limit accounts for?
Probably the answer to your question is yes for a farm of parachutes. But in practice, the deflected wind around a single parachute is not so important, although being possibly a non visible part of the swept area which is subject to the Betz limit.
The Betz limit takes account of energy extraction slowing the wind, whereby a lot of the wind goes around the rotor disc, just as we avoid going into a store if the prices are too high. And even the air that has had energy extracted must still be left with some energy in order to leave the volumetric region of energy extraction, to allow more wind to enter.
I realize the real world is complicated. But I believe the premise of the Betz limit calculation only accounts for the energy of the air passing through the swept area. So if you start talking about the air outside the swept area, you need to calculate in a more sophisticated way.
Also I am suggesting, without much in terms of analysis, that given a big enough swept area, the drag coefficient may converge to a fixed value for any object covering the skies completely, like a parachute. Suggesting the idea, not suggesting that it is true
Hi Doug,
Let’s forget for a moment the questionable terminology of “lift” and “drag” power here.
The power available in the wind for reel-out/in (“lift power”) = yo-yo mode is 4/27, so 1/4 Betz limit (quote 2), so two times more than “1/8th” (quote 1). Indeed the idealized reel-out speed remains 1/3 wind speed (quote 3). The apparent wind speed is 2/3 real wind speed (quote 3) rather than “half the windspeed” (quote 1).
The power is the traction force multiplied by the reel/out speed (quote 3), and can also be reached by using the formula (quote 4) but with simplification, without “CL(CL/CD)²” which is applicable to crosswind kites, but not to our surface, which can be a parachute going downwind. Using the same figures (quote 3), this gives us: 2/27 x 1.2 x 100 x 10³ = 8888 W, so exactly the same result.
Precision about 4/9 (quote 3) of the force (and also the power in the end, the reel-out speed being 1/3 wind speed), resulting from apparent wind speed of 2/3 real wind speed (quote 3):
I get an efficiency of 0.24 for dragging a parachute at 1/3 wind speed and a drag coefficient of 1.62
Which is ~0.4 of Betz efficiency, and approx. half the mechanical efficiency of high end lift wind turbines.
I doubt that the same formulae are directly transferable to a lifting kite because, as in propeller-lift turbines, it has to account for swept area. e.g. doubling the wing’s aspect ratio also doubles the swept area, increases L/D ratio, etc..
Take a sail boat moving in a straight line pulled by a lifting kite, how is the “drag-equivalent efficiency” computed and why would it matter
PS just for reference 1sqm 10m/sec, 1.2kg/mc:
10 ** 3 / 2 * 1.2 -= 600watt raw power in the airflow
600 * .59 = 354 watt remaining since Betz is an asshole.
144 watt from a 1sqm parachute dragging the line at 3.333m/sec and 1.62 drag coefficient
I agree. In my previous comment I preferred to simplify with a drag coefficient of 1, mainly to be able to reuse a previous example. That said, when the Reynolds number increases—that is, when the dimensions and speed of the parachute increase—the drag coefficient tends to decrease and approach 1.
In this case, we use the complete formula, as indicated in the quote 4, including CL(CL/CD)². Unlike what happens with a parachute, the swept area is not the same as the surface area of the kite, being far larger. But the principle is the same, as the power available in the wind within the swept area in yo-yo mode (for crosswind or parachute kites) remains 4/27 (excepted if the drag coefficient Cd is above 1 for a parachute where surface area is the swept area in yo-yo (“lift”) mode, the swept area going downwind?).
If you indicate a 8/27 power because the Cd is 2, you implicitly accept the base value of 4/27 which is half. How do you obtain the 4/27 value if the document is wrong, as you stated only one hour (!) after it was fully available? Perhaps you used my demonstration I put again:
In this demonstration, the 1/4 of power at Betz limit (first formula) is reached by using the formula of force multiplied by reel-out speed. This is not about Cd, only about what remains when the swept area goes downwind, the apparent wind becoming lower.
In the past, I produced a similar line of reasoning three times to arrive at the values I cite below, in a form of question, but without explicitly contradicting the 4/27 limit mentioned in the document, having no sufficient knowledge to challenge the authors, and not having read the complete document at that time.
I would never have hastily asserted that the authors were wrong. I preferred to leave aside the elements that needed clarification, putting forward hypotheses here and there.
If that is assumed for a 100sqm swept area for a propeller (or rotor), with a 10m/s wind orthogonal to the plane of rotation and 1.2kg/m3 air density then the formula either assumes a thrust coefficient of 1 or pretends there is no such thing. It can also be called drag coefficient if you think at a static thing like a parachute.
Yes it’s a formula for dynamic pressure divided by projected disc area but it isn’t an upper limit of force imposed on an object occupying that space in the same way Betz limit is on extracted energy.
Therefore what you show there is a theoretical thrust force for a Ct exactly 1 which you use to derive energy available to a pumping mode device at 1/3, which would be correct if it didn’t start from the false premise of a Ct of exactly 1
Edit: and yes, 8/27 (half of Betz) is a more reasonable power coefficient during line pull for both parachutes and spinning lift devices within a similar swept area than 4/27 . A glider kite spinning in circles and pulling a line axially it’s only the tip of an equivalent autogyro blade.
Whatever you allege about my awe towards AWE is imagined otherwise you would be able to quote were I claimed AWE is in anyway superior to established technologies like HAWTs.
Simply assuming AWE being totally useless in all cases is what I disagree with and all I’m willing to investigate which are those situations - if any - in which it could be sufficiently advantageous.
Why not, falsifiability is a fundamental pillar in science and any theory should hold against empirical evidences. You don’t have to figure out which one’s arguments or assumptions propping a theory are wrong in order to prove it wrong, it is sufficient to provide a concrete example that contradicts (as in produce a different result than predicted by) that theory.
Guys: Is it possible that a high drag coefficient could be a momentary phenomenon, from a steep descent used to spin up the main rotor, to increase its rotational inertia and kinetic energy, for a final flare during landing?
Everything I’ve read is logical and flows smoothly, especially from Figure 5 onwards, which describes the characteristics (curves of the thrust _ with a maximal value of 1 and a value of about 0.9 at Betz limit _ and power coefficients) that “are well accepted in academia and industry as an idealization of a turbine’s behavior, i.e. an upper bound for its efficiency.”
The deductions for AWE lift power systems (“ideal airfoils” exerting a pull (thrust) in their crosswind figure-eight trajectories.) are intelligently issued.
I still have much to learn about this wonderful document.
This document could have positive repercussions by facilitating the choice of a generation method in relation to crucial issues (if AWES develop) such as Power to space use ratio among other possible issues.
Where does this axial induction factor is taken from:
there are three reference values of wind speed in Betz’s assumptions - that wind speed drops from V1 and V2 and the difference is energy that a turbine can extract
undisturbed wind speed in front of the turbine V1
slowest wind in the wake of the turbine V2
wind at turbine level is assumed being an average between V1 and V2: Vt = (V1 + V2)/2
The axial induction factor is defined as how much the wind is slowed down when it reaches turbine level a = (V1 - Vt) / V1 which is the same as a = (V1 - V2)/(2*V1)
Delta V (equation (1), page 1) is V1 - V2. A realistic V2 is between 0 and V1, which means wind can not blow faster after turbine (because that would mean the turbine puts energy in wind instead of extracting it, and greater than 0 means V2 does not end up blowing opposite the original direction - So if a is greater than 0.5 it requires the V2 to become negative.
So Betz assumptions are valid as long as the wind speed at turbine level is no less than half of undisturbed wind speed, and since the maximum Cp is reached at 2/3 (a = 1/3), nobody cared about the a > 0.5 zone because negative wind values are possible only in turbulent regime and that means even more energy lost than predicted mathematically for laminar regime a > 0.5
I think that where the Brazilian paper fail is by labeling regime above 0.5 as “impossible” and ignore it instead of simpler non-solution - Betz assumptions do not apply there therefore we cannot make meaningful conclusions about that regime applying the same maths Betz used for static wind turbines.
