Force & WSA Drag Calculation

Discussion in 'Hydrodynamics and Aerodynamics' started by captncoop, Mar 23, 2022.

  1. captncoop
    Joined: Apr 2014
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    captncoop Junior Member

    First post, and looking for some clarification/help as am confused on the answer.

    My goal is to get a ROUGH estimate for power required on a small HYSWAS. I understand this is complex and opening a can of worms...should probably open a thread on this alone since theres only a handful of information available (I have read HYSWAS Anyone - great info) - but am starting with ONE PIECE - the tube or lower hull.

    Lets assume, the lower hull "tube" is the ONLY piece of this equation (ill worry about the struts later - again, looking for rough back of napkin ballpark estimates).

    From what I understand, the prettymuch only drag on it would be Wetted surface area? And the equation to solve this would be:

    D = .5 rho * V^2 * Cd * S​
    • D = drag (N or lbf)
    • rho = 2 slugs or 1000 kg/m^3 (I know its not exact)
    • V = Velocity (ft/s or m/s)
    • Cd = Coefficient of drag
    • S = Surface Area (ft^2 or m^2)
    Question 1: Is this the only hydrodynamic force to overcome on a submerged "tube" (maybe itd be better to call it a torpedo) - remember, eliminating the struts for now, and trying to get "ballpark"

    Example (I know its in imperial and apologize but units should work out ok):
    • V = 30 knots or 50ft/sec
    • Cd = .5
      • QUESTION 2: I know this is a complete pull-number-out-of-rear, but I feel like im an order of magnitude off.. A NACA 0033 (nice teardrop shape) is .04 Cd...please educate me here (perhaps this is part of Question 3 below?)
    • 18 ft Long, 2 ft Diameter tube displaces 3500lbs
      • assume ends are flat - I know this will effect Cd but trying to simplify for S
      • QUESTION 3: Would S be the frontal, planform, or TOTAL surface area of the tube? In Hydrofoils: Design, Build, Fly by Ray Vellinga he states: "some coefficients are used with frontal area, and some planform. Usually, with parasite drag, the frontal area will be used, but for lift-induced drag the planform area will be used" either way - this is not the total surface area, although I feel like Ive read on the forum here it should be total WSA.
      • I believe it would be total surface area, resulting in WSA of: 2*pi*r^2 (2xends simpified) + 2*pi*r*L (tube) which would equate to ~120 sq feet
    Results in above equation equaling 150,000 lbf

    To get required power, we multiply by the velocity again (50) and divide by 33,000 to get horsepower. This equals ~227HP rounded to 225HP 60% prop efficiency dictates 375HP.

    ~375HP to push a 2' diameter by 18' torpedo through the water at 30 knots? Seems like something is way off...like an order of magnitude...and thanks in advance for your knowledge.
     
    Last edited: Mar 23, 2022
  2. jehardiman
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    jehardiman Senior Member

    Yeah, there is a lot missing. Before starting on something so hydrodynamically complex as a HYSWAS, I would suggest a read through of Heorner's Fluid Dynamic Lift and Fluid Dynamic Drag (both on the web IIRC). The actual drag of the buoyant body is small compared to the appendages and struts required to make a HYSWAS work.
     
  3. captncoop
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    captncoop Junior Member

    I knew I was opening a can of worms mentioning this, instead of simply going for the calculation to move the tube through the water first, and bring that part up in another thread.

    Also, im not sure suggesting reading a 450-page book is really appropriate to answer the above question, though it might be helpful for me going forwards.
     
    Last edited: Mar 23, 2022
  4. Ad Hoc
    Joined: Oct 2008
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    Ad Hoc Naval Architect

    JEH gave you a good point in the right direction.

    Since this is far too simplistic.
    Being close to the surface there will be the residuary and viscous pressure resistance.
    Not insignificant either!

    And that's before you get into the struts/appendages etc as JEH noted.

    Consider this...if what you have proposed, is that simple..anyone could do it! :eek:
     
  5. captncoop
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    captncoop Junior Member

    I was afraid this is the way this would go. I'm trying to establish a ballpark figure here guys. I need somewhere to start, something super simple. I know there's a ton of complexity, that's not what I'm asking for.

    Forget surface, forget all that. And I apologize for even mentioning hyswas. Let's just move a torpedo shaped object through the water, and kindly help me with the questions I listed so I can understand some basic principles I'm hung on, prior to adding in all this other complexity you guys love.
     

  6. jehardiman
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    jehardiman Senior Member

    Then read Hoerner Fluid Dynamic Drag, he shows the approximate equation of drag for any full submerged body of revolution.
     
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