Towing resistance – surface friction

Discussion in 'Boat Design' started by kolloff@get2net, Mar 1, 2011.

  1. DCockey
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    DCockey Senior Member

    Would the tugboat wake/propeller have more effect on transom flow than flow separation itself, at such low Fn?
     
  2. DCockey
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    DCockey Senior Member

    If overall wavemaking is also negligable at the Fn in question, then the only mechanism for drag will be skin friction.

    If skin friction is the only non-negligable drag mechanism then a tow tank test is not needed unless it is for a Prohaska plot adjustment.
     
  3. daiquiri
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    daiquiri Engineering and Design

    You guys are discussing a transom drag, but I see no transom in the photo attached to the post #11, just an aft hull bottom exiting the water with an inclination.

    And, besides that, I agree wih Alik here:
    So, if no other data is available apart the table in post #4, then a consultation with MARIN staff is necessary in order to understand the origin of the unusual test data. Imho.
     
  4. MikeJohns
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    MikeJohns Senior Member

    Before the non linear differential equations kick in in the chaotic boundary layer. ;) ..... [edit] suppose I should say Turbulent not chaotic too.
     
  5. MikeJohns
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    MikeJohns Senior Member

    That's essential given the nature of the study, and I expect that it is corrected data. You just need to know what corrections they applied.

    Initially I thought they were looking for a Form factor too. But it's just a big slow platform designed to be towed into position.

    The reality of a towed hull like this at low Froude Numbers is that there will be a significant 'added' resistance component, look to the RAO of a similar vessel if this hasn't been performed. Towing is seldom in calm water and you will probably be surprised how much added resistance applies.
     
  6. Ad Hoc
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    Ad Hoc Naval Architect

    Kolloff

    I am not at my own pc (im on my hols..just checking emails).
    The figures you posted looks like the "M" is for model....anyways..

    The values presented are clearly in error..massively. I don't have my tables to hand to give you the correct solution..but can upon my return next week.

    They have even got the Fn calculation wrong...for example, a the last entry of 9.04 knots, the model Fn is given as 0.087...this simple calculations shows the correct figure to be 0.075.

    If you have time to wait i can run through these numbers when i return. For now...don't trust them one bit.

    As for MARIN...well..suffice to say..not impressed.
     
  7. Leo Lazauskas
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    Leo Lazauskas Senior Member

    0.6/sqrt(9.81*4.885) = 0.0867 on my super-computer.
    Assuming the length of the model is 4.885m as in the footnotes.
     
  8. daiquiri
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    daiquiri Engineering and Design

    Even the best ones can make mistakes sometimes. :) Ad Hoc is writing from a place where he can have just a short access to internet, so his mistake is clearly do to having to read the data and to reply in haste.

    AH, the length of the model is 4.885 m, so the Fn=0.087 is correct. The 6.469 figure you have used for length is really the wetted surface of the model.

    But we do agree that numbers are not reasonable, unless there's something else making a huge pressure (form) drag, which we ignore at this point. And also that the name MARIN alone is not sufficient to blindly trust the test data.

    Cheers!
     
  9. Tackwise
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    Tackwise Member

    I did a quick check of the figures myself and I seem to arrive at a different Reynolds number...?

    RN_model being:
    v_model *L_model / kinematic viscosity = v_model*4.885/0.00000110438

    There seems to be a factor in place of
    RN = RN_model * (density seawater)/(g*100).

    I cannot place this factor, anybody else can?
     
  10. daiquiri
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    daiquiri Engineering and Design

    Tackwise,
    The Rn value depends on viscosity, which in turn depends on water temperature in the towing tank. If you use the value of 1.058E-6 (water at around 18 °C) you will get the same Rn.
    Cheers!
     
  11. Tackwise
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    Tackwise Member

    mmmm you are correct, I was already puzzled that I needed to use Seawater density & viscocity :confused: for the Reynoldsnumber of the model. I should have stayed with my first and most logical assumption, which was that they alway use fresh water in the towing tanks....
     
  12. fastwave
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    fastwave Senior Member

    Dear gentlemen,
    Leo already posed the question about form factor. The form factor can easily be in the region of 1.15 for a streamlined vessel. In this case this can be significanlty higher.
    In addition the CFM which is calculated using the ITTC57 assumes no separation. Is this the case here?? I would be surprised. There are quite some tight corners and a steep stern.
    You cannot draw any conclusions. You need to ensure that the test was done correctly and and that it relflects something close to reality and just live with it.

    If you are seriously concerned, then perform a CFD simlulation to compare the drag with the ITTC57 prediction. Before people start saying CFD is **** and it is not possible, the differences in numbers here are big, so even if the simulation is not perfect you will get an indication.

    CFD can be viewed liked a gun. Give it to an experienced person, he can put it to good use. Give it to a child and he will kill half his school.
     
  13. Alik
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    Alik Senior Member

    ... give it to boat designer, he will shoot his foot.

    :D
     
  14. DCockey
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    DCockey Senior Member

    Along the lines of what Alik suggested earlier a starting point for the drag (assuming wave drag is negiligable) might be the drag of a similar bluff body with the water surface treated as rigid. My estimate, based on experience in bluff body aerodynamics, would be a CD based on frontal area of 0.4 to 0.55. It might be a little lower, but I would be very surprised if it lower than 0.2, or it might be higher but certainly below 1.0 That's a large range but it will provide an estimate of the magnitude of the drag as a starting point.
    Drag = CD (based on frontal area) * frontal area * dynamic pressure
     
    Last edited: Mar 3, 2011

  15. daiquiri
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    daiquiri Engineering and Design

    LOL!!! :D :D :D
     
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