James Newtons Massmind wrote: >> I worked thru the equations of getting >> power from pure drag, and if I didn't mess up it shows that >> maximum power from an object moving downwind results from the >> object's speed being 1/3 the wind speed. If you apply this >> to the wind power equation for a square flat plate >> (coefficient of drag about 1.17), then you get about 17% >> efficiency (again if I didn't mess up). > > I would love to see how you made that calculation. I don't question > it, I'm just curious how you calculated it. I got your message first thing after coming back from the recycling center. My notes are gone, but I'll try to reconstruct the logic. I'm sure everything I did is on some professor's or NASA web site, but I sometimes like to derive the answer for myself to get better insight. If nothing else it tells you the limits beyond which the stock answer doesn't apply. Unfortunately this is all too often missing from descriptions and the "well known" answer gets applied in circumstances where it is no longer correct. I think a little of that is going on in the discussion of lift here. First we'll derive the optimum speed for getting the most power from an object moving directly downwind. The force (F) on an object is proportional to the wind speed (W) squared. For simplicity we'll asume we are using units where the proportionality constant is 1, so we have: F = W**2 If the object is moving downwind with speed S, then the force on it is F = (W - S)**2 to compensate for the reduced apparent wind speed by moving downwind. The power (P) that can be taken from the object's motion is the force on the object times it's speed: P = SF = S(W - S)**2 As a sanity check we can see that the power is zero both when S=0 (not moving) and when S=W (moving exactly with the wind) as expected. In other words, to get power you want to move fast, but if you move too fast your apparent wind goes down and you lose power. Since we only care about the optimum ratio of S/W for maximum power, we can pick an arbitrary value for W in the equation. I pick 1: P = S(1 - S)**2 Expanding out to show the individual S terms yields the third order polynomial: P = S**3 - 2S**2 + S Again, note that P goes to 0 when S=0 and S=1 as expected. We are looking for the maximum point between S=0 and S=1. The derivative of this equation must be zero there: dP/dS = 3S**2 - 4S + 1 = 0 Solving the quadratic (A=3, B=-4, C=1): -B +- sqrt(B**2 - 4AC) 4 +- sqrt(16 - 12) 4 +- 2 ---------------------- = ------------------ = ------ = 1/3 and 1 2A 6 6 So the two inflection points of the P cubic equation are at 1/3 and 1. Clearly the one at 1/3 is the maximum we are looking for. We already know the power at 1 is 0. Therefore, wind does the most work on an object moving downwind when its speed is 1/3 the wind speed. Now let's look at the maximum power we can extract from the wind using pure drag (which is the case we calculated the optimum speed ratio for above). According to several sources I looked at, the power in the wind (Pw) is Pw = .5 p A V**3 where p is the air density, A is the area, and V the wind speed. Again according to what looked like reputable sources, the drag force on an object is F = .5 p A Cd V**2 where Cd is the coefficient of drag. It seems this is about 1.17 for a thin square plate oriented flat into the wind with the wind able to flow around all sides. The power produced by the object (Po) is the force on it times its speed (S). We already know the optimum speed for maximum power output is V/3, so the maximum power (Po) that can be taken from an object moving downwind can be calculated. In this case the apparent wind on the object is 2V/3 and its speed (S) is V/3: Po = F * S = .5 p A Cd (2V/3)**2 V/3 The total effeciency at this point is: Po .5 p A Cd (4/27)V**3 Cd * 4 -- = -------------------- = ------ = 0.148 Cd Pw .5 p A V**3 27 For the case of the flat thin square with Cd=1.17 this comes out to 17%. The equation above should be true for any rigid object moving downwind. The only way to change the theoretical maximum efficiency is to increase the coefficient of drag. The rest is physics that you don't get to mess with. For example, a cup shape has a higher coefficient of drag, and therefore has a higher maximum possible efficiency. So you could get around 20-25% with a well shaped object moving linearly down a track directly downwind at 1/3 the windspeed. For an operating wind turbine you generally want your materials to be reusable, so you have to figure out how to get them back upwind at less drag then you got out of them going down wind. This is where my idea of rotating flat plates came from. I get the 1.2 or so at the full area going downwind, and a much much smaller area going upwind. In my design it is possible to prevent the wind from flowing around three of the sides of the plate when going downwind, thereby creating a much larger stall area. In the terms of the equation above, that means the effective Cd of my paddles on the right side can be higher than 1.17. I don't have a good feel for how much higher, but I do think that my design is pretty efficient compared to all other drag types. I haven't seen this analysis done for airfoils, but I suspect it gets a lot more complicated. Several people have pointed out that drag turbines can't be as efficient as airfoil types, and that may well be true. However there are a number of factors that decide on the relative merit of wind turbines for particular circumstances. In most cases total wind efficiency is less important than other attributes. Drag turbines have some inherent advantages too. Since they operate at low speeds, they are much quieter and don't kill birds which can handle things moving at 1/3 the wind speed. The vertical axis topology also leaves more options for using the mechanical output since it doesn't have to be transferred over a rotating interface. If higher speed is desired, such as for generating electricity, the heavy transmission can be on the ground where it's weight is pretty much irrelevant. Also there are direct mechanical uses, like pumping water, running a compressor, etc, where slow and high torque is desirable. One of my other ideas is a stand alone device by the seashore that uses mostly solar heating to make water for drinking or irrigating crops. The device itself has no moving parts but requires a partial vacuum to operate efficiently. Air that was dissolved in the seawater builds up inside and needs to be occasionally removed. I've thought of having a simple wind turbine for this purpose. Think of a device that just sits there next to the ocean with a small wind turbine on top with a tank and spiggot at the bottom. The tank gets automatically filled with clean water at no incremental cost other than maintenence of the apparatus. Anyway, back to wind turbines. If you see any error in my calculations, please let me know. ****************************************************************** Embed Inc, Littleton Massachusetts, (978) 742-9014. #1 PIC consultant in 2004 program year. http://www.embedinc.com/products -- http://www.piclist.com PIC/SX FAQ & list archive View/change your membership options at http://mailman.mit.edu/mailman/listinfo/piclist