Hydrofoil propulsion

Discussion in 'Boat Design' started by ace.maverick, May 19, 2009.

  1. ace.maverick
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    ace.maverick New Member

    Hi,

    I'm designing a hydrofoil boat with fully submerged foils which will be attached to a torpedo like structure which I plan to house a water jet turbine.

    I've carried out CFD analysis on my foil designs and have a lift of around 178kn with a NACA0018, and a drag of around 520kN for the whole structure at 20m/s which i'm taking as maximum resistance.

    Does anyone know how I'd go about specifying what size engine / output I'd require from a water jet turbine to power the boat along at this speed...?

    Thanks
     
  2. eponodyne
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    eponodyne Senior Member

    Big. That's what size.
     
  3. mydauphin
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    mydauphin Senior Member

    Double it.
     
  4. ace.maverick
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    ace.maverick New Member

    It's about 8.5m * 3.3m & about 2m deep, I also shelled the hull 140mm thick with S-glass as the chosen material which gives a weight of 12.5 tonnes for the hull and around 18.7tonnes for the entire structure. There is alot of parasitic drag and needless weight which can be optimised later.

    The volume of the hull is about 25.7 m^3 displacing about 18.2m^3 water fully loaded.
     
  5. mydauphin
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    mydauphin Senior Member

    Ace, I am not a expert on this but I have some ideas. To bring the boat to flying speed requires one set of power numbers. After it is flying then power needs are going to be dictate by speed desired. So you have to calculate all of surface area underwater including fins and torpedo, then factor in hydrodynamics of it all. Then lift causes drag. Very very complicated. If I where you the question should be, does anyone have computer program that can do this on such a hull design. Apart from that -try working on scale models of hull without appendages, then use program to see how it matches. Then build appendages and see how it affects total resistance and etc..
     
  6. Ad Hoc
    Joined: Oct 2008
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    Ad Hoc Naval Architect

    ace.m

    do you have a drag curve from your CFD, at various speed and beyond your target speed?
     
  7. Guest625101138

    Guest625101138 Previous Member

    The drag of 520kN looks silly. You have a lift to drag ratio of 178/520 = 0.34. That is a silly number. It means the boat would need enough thrust to accelerate at 2g vertically - that is better than many rockets achieve.

    Anyhow the power is a simple calculation once you have the drag.
    Power = Force X Velocity
    Power = 520,000 x 20 = 10,400,000W or 10.4MW

    That is a lot of power even before you allow for the jet losses. You will need good design to get 60% efficiency from the jet so installed power will be 17.3MW. A VERY LARGE NUMBER. I will give you the optimum nozzle size if you like but it is going to be a very big hole. I doubt that you will get the propulsion system in the 8.5m length. I am also doubtful that you can get any 17.3MW engine within your weight target.

    I would check the CFD.

    Rick W
     
  8. mydauphin
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    mydauphin Senior Member

    Rick, can you explain what kN is? I am not familiar with that unit. I never learned physics in Metric. If by 10.4 MW, do you mean 10 million gallons or 10 million watts? Either number seems like Saturn V powerplant to power this guys boat. thanks
     
  9. Guest625101138

    Guest625101138 Previous Member

    1kN = 1000N

    N means Newton. This is a unit of force. 1N will accelerate 1kg at 1m/s.

    Gravitation pull is 9.8m/s/s.

    The boat in the example weighs 18.7t which is 18,700kgf or 181.4kN. So a thrust of 181.4kN would be enough to lift it vertically. With 520kN there would be 338.6kN thrust to accelerate it vertically in addition to that required just to support.

    W means watt. It is a unit of energy. It is the energy required to move against a force of 1N at 1m/s.

    1MW = 1,000,000W = 1000kW

    The 17.3MW is a very large power plant. In Imperial units it is 23,190HP.

    A good hydrofoil would have an L/D ratio of 20. So an 18.7t craft should require about 9kN of thrust. Hence you get an idea of why I consider the 520kN unrealistic.

    Rick W
     
  10. mydauphin
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    mydauphin Senior Member

    I guess Kn is like a lbs of torque, what imperial measure on it.
     
  11. daiquiri
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    daiquiri Engineering and Design

    :eek:
    Units are very important, guys... Wrongly calculated unit conversions can lead to a big mess:

    http://www.techexpo.com/techdata/si-units.html

    http://www.themeter.net/index_e.htm

    Mydauphin, torque (M) and force (F) are two very different quantities, though they can be related by a formula M = F*L, where L is the lever length.
    So, in SI system, a force is measured in N (Newtons), a torque in Nm (Newton-meters, because lever is measured in meters).

    Lbs of torque is a non-existant quantity. Lbs is a force, but you need a lever too (in feet) in order to have a torque. That would make a foot-pound, which is the English unit for a torque.
     

  12. mydauphin
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    mydauphin Senior Member

    Thanks daiq and Rick. Sorry but when I went to school we used feet, gallons, etc... and it is hard to break the habit or picture other units sometimes.
    Show me a boat and I can tell you it displacement, and horsepower needs almost by magic, but don't ask me to do calcs in metric....lol
     
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