The Wind Powered Sail-less Boat

Discussion in 'Boat Design' started by DuncanRox, Oct 20, 2008.

  1. Guillermo
    Joined: Mar 2005
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    Guillermo Ingeniero Naval

    How can you say that? Atonishing.

    That's exactly how I feel about all of you, who do not want to perform the asked for energy analysis :)
  2. spork

    spork Previous Member

    The real life demonstration with the physical cart sure as heck addresses energy - as do many of the analyses presented.

    And yet your analysis above doesn't even attempt to address energy - curious.

    Just as we keep asking of you. It seems we're both going to be out of luck.

    Fair enough. You first.
  3. Guillermo
    Joined: Mar 2005
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    Guillermo Ingeniero Naval

    Of course not. I try to see the system as a whole.
    The vectorial analysis posted by some of you previously for the movement of the blades of the air propeller, proves nothing in my opinion, except that a propeller can go upwind, as is obvious.

  4. spork

    spork Previous Member

    You've already admitted that an ice-boat can outpace the wind by tacking downwind. So you've admitted we can have a vehicle that goes DIRECTLY downwind faster than the wind.

    If all you're arguing about is the mechanics or aerodynamics of the prop cart, that's not interesting enough to keep me engaged in this bizarre debate.
  5. Guillermo
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    Guillermo Ingeniero Naval

    That's rethorical. And please avoid questioning my understanding of sailing physics as I do not question yours.

    Size of sail is not relevant to what I'm talking about.

    If you do not agree with me, please tell me what your energy analysis is for the tacking downwind sailing craft.
  6. Guillermo
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    Guillermo Ingeniero Naval

    You are the ones arguing about the mechanics or aerodynamics of the prop cart, not me.

    Well, debate is what's all these forums are about. Or do you expect everybody agreeing?

  7. Guest625101138

    Guest625101138 Previous Member

    The equation of motion for the DDWFTTW boat is:
    ma= Fp - Fh - Ft

    m is the mass of the hull
    a is the acceleration of the boat
    Fp is the thrust from the propeller
    Fh is the drag from the hull
    Ft is the drag from the turbine

    There solved - it works.


    Rick W
  8. Guillermo
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    Guillermo Ingeniero Naval

    spork, joaquim, etc

    Please perform for me the analysis I've asked you for. I'm honest in saying that if you prove the DDWFTTW thing is feasible from the point of view of energy and mass conservation, I will recognize my mistake and go to the 'Force' side. :)

  9. Guillermo
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    Guillermo Ingeniero Naval

    But not as you asume. Just the contrary I'm afraid.

    Fh + Ft = my previous Fh
    Fp = my previous Fs

    The thing I wanted to highlight is that there is no more energy than the one taken from the wind, whatever the device you use, and it has to overcome all the negative forces from the hull, apendages, turbines or whatever.

    Now I have to leave. I'll be back to all of you at night.

  10. Guest625101138

    Guest625101138 Previous Member

    I can see you are just making a joke of it now Guillermo. No one could be as thick as you are pretending to be on this.

    I provided you exactly what you ask for in post #468 and you make no comment on its validity.

    Rick W
  11. Joakim
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    Joakim Senior Member

    OK. I consider a bit of energy, but not really detailed to a real DDWFTTW craft. I'll do all in the reference frame of water.

    Think of a 100m2 * 100 m = 10 000 m3 air having a velocity of 10 m/s. It has a kinetic energy of 1/2*m*V^2=600 kJ (rho=1.2 kg/m3). If you would be able catch all that, your 100 kg craft would have a speed of 110 m/s, if there were no losses.

    How to catch all that huge amount of energy?

    You could anchor and build up wind mill with 100 m2 disk area. Then you run it for some time and store the energy to a flywheel or a battery and finally drive away with a propeller. Now you would be limited by Betz and thus would only get a maximum of ~350 kJ from the same tube in 10 seconds time. That would still get your 100 kg vehicle a speed of 84 m/s without losses. A well optimized 100 kg craft could do say 12 m/s at 5 kW power and thus it would go for 70 seconds reaching 840 m. In the same time wind has reached 800 m (10 s wind mill + 70 s driving). So now you just need to design a 100 kg craft that has a 100 m2 wind mill build into it (stored while running). Not easy, but nothing physically impossible.


    You could build a craft that doesn't need to stop and run to go faster than the wind. Obviously you can not use a wind mill, since the apparent wind is headwind and thus you would just increase the kinetic energy of air. Thus you need to push air backwards in order to reduce it's kinetic energy. Now there is no Betz limit, since you are using the air rotor as a propeller to slow down air and gain energy to the craft. But since it is a propeller it needs energy to be able to rotate. That energy comes from the shaft and is originally made by the water turbine. How does the wind make the water turbine rotate you may ask. That comes from the thrust of the air propeller. Thus there are two energy paths: The axial force driven and the shaft momentum driven. All the energy is taken from the wind (since water has no kinetic energy as it's velocity is zero) and is put to the craft AND water. Thus water is used as a "tool" to transfer energy from wind to the craft.

    By changing the reference frame the "roles" of water and air are reversed and you could equally well say that you are taking energy out of water and putting it to the craft and air. In the reference frame of wind the kinetic energy of water is much bigger due to density, thus to have that 600 kJ available in 100 m distance you only need an area of 0.12 m2.

    So now you need to add a tube of water to your energy balance (actually it was necessary even before), but that is rather easy as you have a big speed against water. All the drag forces you have against water will increase the kinetic energy of water.

    But the more tricky part comes when W=V and you think of air. Now you are staying stationary to the air, thus you can not consider a long tube of air. What will happen is the propeller will create a local flow field around it. The air will go through the propeller and then find it's way back to the front of the propeller. While doing that a much bigger volume of air is slowed down and some actually sped up, but as a total the kinetic energy of air is reduced. Because of this the propeller is not very efficient at and close to W=V. Actually it works much better when V is increased well above W.

    When thinking all of this doing a control volume analysis is not very straight forward and I don't see how that would really help. When you are doing the propeller, turbine and craft drag analysis in the correct way there is no doubt about energy conservation.

  12. spork

    spork Previous Member

    Rick, what you provide in #468 is just fine, but it's not what G asked for. He specifies a bizarre analysis of a control volume in the transient case where the vehicle begins at rest within the volume and exits the volume at greater than wind speed. He says he wants the energy flux at the inlet and exit of that volume as a function of time.

    The problem with this is that to truly provide what he requests would be quite a task, and the results would be of no use in proving viability of DDWFTTW.

    Whether he's simply yanking our chains isn't clear to me yet. But I have to agree he's pretty quick to dismiss all valid analyses and unwilling to even tell us where he feels they go wrong. He's also completely unwilling to offer us an example of the type of analysis he demands on the simple case of a sailboat on a typical course of sail.
  13. markdrela
    Joined: Jun 2004
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    markdrela Senior Member

    Energy analysis attached. Jeez.

    Attached Files:

    • ddwe.pdf
      File size:
      29.7 KB
  14. spork

    spork Previous Member

    I can't wait to see the response to this. On the one hand, effectively the same analyses have been posted, but on the other hand, they haven't been posted by someone that Guillermo should know to listen to and trust.

    The primary reason I'm curious to see his response is that, while this energy analysis is exactly on point, it's not the specific analysis that he proposed. The problem of course being that the analysis G proposed is just plain silly.

    Waiting to see...

  15. Guest625101138

    Guest625101138 Previous Member

    I believe I provided exactly what he asked for in his post #462. I certainly gave him the proof he asked for.

    For the said boat with 9m prop and 800mm turbine moving through the water at 3.5m/s with wind speed at 3.5m/s the energy balance over one second is:

    Air Stream and Propeller
    The air immediately ahead of the boat is moving at 3.5m/s. In 1 second 1.3m of air is drawn into the prop to be slowed to 2.2m/s immediately behind the prop.
    Ma = pi x 4.5^2 x 1.2 x 1.3 =101kg
    Reduction in air energy in going from 3.5m/s to 2.2m/s:
    Ea = 1/2 x 101 x (3.5^2 - 2.2^2) = 374J

    Work done on the air by the propeller to accelerate it rearwards at 1.3m/s:
    Eaw= 1/2 x 101 x 1.3^2 = 86J

    The water immediately ahead of the turbine is at rest while behind the turbine it will be moving at 0.01m/s. In one second 3.49m of water will have passed through the turbine.
    Mw = pi x 0.4^2 x 1000 x 3.49 = 1754kg
    Energy imparted to water:
    Ew = 1/2 x 1754 x (3.5 - 3.49)^2 = 0.09J

    Work done in one second to overcome the boat drag of 20N moving at 3.5m/s is:
    Eb = 20 x 3.5 = 70J

    The system has so far unaccounted losses as follows:
    Transmission Eml = 2.7J
    Propeller Epl= 195J
    Turbine Etl = 8.9J
    The propeller and turbine losses produce swirl and turbulence in their respective fluids as does the net flow through the discs eventually.

    The energy ballance over 1 second is:
    Net Energy = Ea - Eaw - Epl - Etl - Ew - Eb - Eml
    = 374 - 86 - 195 - 8.9 - 0.09 - 70 - 2.7
    = 11.3J

    There is net energy of 11J available in each second at the Vb = Vw condition to continue to accelerate the boat.


    Rick W
    Last edited: Jan 3, 2009
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