Windmill or Wind Turbine- powered boats: how many are out there, and are they viable?

Discussion in 'Projects & Proposals' started by Duma Tau, Oct 9, 2006.

  1. MPraamsma
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    MPraamsma Junior Member

    I beg to disagree...

    That force of lift is work accomplished against gravity.
     
  2. Guest625101138

    Guest625101138 Previous Member

    The lift force is not doing any work in level flight. A balloon can keep the plane in the sky. It is not doing work. The only work on the plane is that required to move it through the air. There is no work done in the vertical direction against the lift force in level flight.


    Rick W
     
  3. MPraamsma
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    MPraamsma Junior Member

    I'm surprised that you don't see that...

    ...you have defined work as a force times distance, and because you see no distance (what you define as level flight) that the equation says no work is accomplished. Surely you can see the fallacy of that argument. The work in this case is accomplished at the atomic level and is transmitted as a wave of momentum that travels through the mass of the airfoil. Since you eye cannot see displacements of nanometer dimensions you assume nothing is being displaced.
     
  4. Guest625101138

    Guest625101138 Previous Member

    With your system does it work in water?

    Rick W
     
  5. MPraamsma
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    MPraamsma Junior Member

    Water...?

    I'm not sure why you ask that question, because water has no ability to compress since it is virtually a solid. Air is compressible if you can contain it in a closed system (ask any scuba diver). In a dynamic situation, such as between a compressor and turbine air can also be subject to volumetric changes, water cannot.
     
  6. Guest625101138

    Guest625101138 Previous Member

    I am trying to determine why you thing the L/D of a foil has anything to do with work. A foil operates in water as well. The fluid does not need to be compressible for a foil to work.

    Once you start getting into compression that is going to do work you have to have a cycle to extract energy. I cannot find any references to a cycle that works without temperature differences. Hence I cannot see a feasible cycle for you to extract energy from the air.

    Rick W
     
  7. MPraamsma
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    MPraamsma Junior Member

    Rick...Just to elaborate...

    ...on the theory of lift, where you seem to have placed too much faith in formulas, and not enough in the fundamental understanding of the underlying physics. You say there is no work performed, but to show you that that is nonsense, try a little thought experiment. Imagine that gravity was something you could switch on and off instantly like a lightbulb. If a plane were in level flight, what do you think would happen if the gravity switch were suddenly turned off, what would the plane do? Of course it will climb like son-of-a-gun and probably snap the spines of all the passengers. Would that not be Mechanical Work according to your formula?

    Michael
     
  8. MPraamsma
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    MPraamsma Junior Member

    There is no difference...

    ...between a water-foil and an air-foil in the way they develop lift. The collisions between water (/air) molecules on the underside (leading) are more energetic than the topside (receding) therefore there is a net force developed that we have called thrust (lift). On the macro scale there is no visible compression in water, but there is at the atomic level, at the interface between the water and the foil. Those billions of collisions, times the tiny displacement of each atom of the underside of the foil is work.
     
  9. Guest625101138

    Guest625101138 Previous Member

    Michael
    That is not correct. There is huge resistance to it lifting. Nothing to do with gravity. It has to move a huge volume of air out of the way to lift vertically at any great rate. The air compression is very low. In fact it can be treated as incompressible under normal flight conditions. It does not behave any differently to water.

    Rick W
     
  10. MPraamsma
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    MPraamsma Junior Member

    Rick...

    You are letting the formulas get the better of you because you apparently cannot see that level flight is a limiting condition where there are inumerable small impacts against countless atoms causing infinitesmal movements. The sum of these is the work performed, but since this is in perfect equilibrium with the rate at which the airfoil is 'falling' is seems to the naked eye that there is no displacement. Removing the force which opposes lift (gravity in this case) will suddenly reveal the fact that work is being performed.

    Re-read carefully what I said and use your imagination more to visualize this process instead of blindly accepting what some books say.
     
  11. Guest625101138

    Guest625101138 Previous Member

    No formulas just common sense. Air and water foils work much the same other than when at high speed where pressure wave velocity is a problem in the air and cavitation is a problem in the water. Up to those conditions they are essentially behaving as incompressible fluids.

    I accept that if you have very high velocity at low pressure you can get into supersonic conditions and that may be what can be achieved in your venturi but it has nothing to do with the L/D for a foil.

    The operation you describe by switching gravity on and off is effectively the way water gliders work. They can cover huge distances with small power by pumping water in and out to give them positive or negative buoyancy. The same has been proposed for aircraft using helium filled spaces.

    If you switch gravity off on a normal aircraft it will start a climb. It will not be steep because the air will resist it. It has to take an upward glide path.

    Lift is achieved through momentum transfer to the air exactly the same as lift in water. Nothing to do microscopic processes in either medium other than they are both made up of small particles.

    Rick w
     
  12. Windmaster
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    Windmaster Senior Member

    No, I don't have to prove anything with this. If anyone is sufficiently interested they can build it for themselves. The instructions are, I hope, quite clear. The only video is the "pulling on a piece of string" which you say proves nothing. However, if you had seen it in real life you would see that it is pushing a stream of water backwards sufficient that the reaction (Newton's laws of motion) overcomes the drag of the oncoming wind. So it's not a static force, it is work being done. How much work, would be determined by the pull on the string, but, as the natural wind varies all the time you would have to relate the pull to the wind strength in exact moments in time. Only in a wind-tunnel would you get sufficiently stable conditions to get perfect results.
    I don't need to promote this by means of videos, which take my time and effort to make. It's been served up on a plate, if people don't want to take it, then its their problem.
    ...
     
  13. Guest625101138

    Guest625101138 Previous Member

    Peter
    In the pursuit of science I would be interested to see a video of your model boat from side view with some yardstick as a means of determining the speed. An idea of the wind strength would be useful. Maybe from a monitoring station nearby. Typically you can get doppler radar images.

    bil
    Same goes for you if you build one.

    Rick W
     
  14. MPraamsma
    Joined: Sep 2008
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    MPraamsma Junior Member

    Rick...I think I see where you are confused...

    ...and I think if you can answer my previous question of where the missing 19 pound force actually comes from you will see the problem a lot clearer. Your example of the diving and rising subs is not relevent because they are essentially neutrally bouyant nominally, and making them dive or rise (more dense and less dense) creates a glide slope.

    Your other comment about momentum passing from foil to air is also wrong, it is the other way round, and that is the energy used to support a wing. A glider, in tow, leaves a trail of colder air behind it, not a warmer one.
     
    Last edited: Sep 28, 2009

  15. Guest625101138

    Guest625101138 Previous Member

    The missing 19 pound force. This is not sensible. It is essentially the same as saying that if I was in a vehicle with a weight to drag ratio of 100, and it took 1N to move it, there is 99N missing.

    Intuitively I would have thought the glider lost energy to the air. So would warm it but if you have data that shows otherwise I would be interested in it. However this has nothing to do with the L/D of a wing. The glider increases the momentum of the air. If there is cooling it is temporary as there is a net loss of energy to the air due to the momentum put in. If this was not the case the glider would not need thermals or wind shear to gain height. It would just continue to suck energy from the air, which it cannot do.

    Rick W
     
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