Froude and planing

Discussion in 'Hydrodynamics and Aerodynamics' started by sandhammaren05, Feb 26, 2017.

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  1. Rastapop
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    Rastapop Naval Architect

    I don't think anyone's arguing otherwise.

    The same isn't the case for a hull on the water surface. The Kutta condition becoming satisfied, and lift becoming non-zero, are not (necessarily) the same event here.
     
  2. Joakim
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    Joakim Senior Member

    I don't know if there is a general definition for the reference point of trim. For prismatic hulls it is typically referenced to keel line, not static LWL.

    In the paper displacement and high speed displacemet hulls have zero trim at low speeds while the prismatic planing hull has about -0.5 degrees. I'm guessing that the displacement hull trim is referenced to LWL and planing to keel line (as shown in Figure 6). High speed displacement seems to have straight keel line, so it could be referenced to either one.

    Here is another paper showing sinkage and pressures both from simulations and measuruments: http://www.scope.unina.it:8080/docu...0122.pdf/fe04f7c3-b661-47f9-915b-c61fe2a59275

    I wonder how much of the sinkage is caused by lower pressure due to accelerated flow under the hull. The sinkage could be caused also by the wave pattern. This would also show lower pressures under the hull, but the reason would be lower local hydrostatic pressure due to lower local water surface.

    Also I wonder how accurately measured heave describes the volume under static water level. It seems to be measured as the change of VCG. But a change in trim could change the volume without a change in VCG.
     
  3. sandhammaren05
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    sandhammaren05 Senior Member

    What other lift do you then have in mind?
     
  4. Rastapop
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    Rastapop Naval Architect

    Safest just to assume the hull is drawn at level trim (for "semi-displacement" hulls at least), and that all trims are relative to that.

    To be honest Joakim I'm too lazy (busy?) to look through a thesis to try and find what you're referring to ;)
    If you provide numbers for pages of interest I'd look at those...

    I couldn't say how much without data to look at, and I don't have any that's applicable.

    Yes, pretty standard to describe heave in terms of the CG. Together with trim you can calculate immersed volume.


    It's what I don't have in mind when talking about sharp-edged transom hulls on the water surface that is relevant: lift conditions for foils immersed in a single fluid.
     
  5. sandhammaren05
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    sandhammaren05 Senior Member

    Corrected. With a small trim angle alpha and wet length L the submerged depth is d≈Lxalpha. Alpha is the boat's trim angle measured relative to the horizontal.
     
  6. sandhammaren05
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    sandhammaren05 Senior Member

    I have benefited from several of the comments and have revised both my ms and my original post. In light of the comments and subsequent analysis, here would be what I would have posted today, with hindsight (thanks to several stimulating comments):

    Faltinsen states.....

    "The pressure carrying the vessel can be divided into hydrostatic and hydrodynamic pressure. The hydrostatic pressure gives the buoyancy force, which is proportional to the submerged volume (displacement) of the ship. The hydrodynamic pressure depends on the flow around the hull and is approximately proportional to the square of the ship speed. Roughly speaking, the buoyancy force dominates relative to the hydrodynamic force effect when Fn is less than approximately 0.4. Submerged hull-supported vessels with maximum operating speed in this Froude number range are called displacement vessels. When Fn > 1.0-1.2, the hydrodynamic force mainly carries the weight, and we call this a planing vessel. Vessels operating with maximum speed in the range 0.4-0.5 < Fn < 1.0-1.2 are called semi-displacement vessels."

    I disagree strongly in part. For planing, I use a Froude nr. based on submerged depth of hull. With the boat plowing and the transom submerged at a depth d then the only force is full buoyancy FB=rVg, so long as there is backflow up the transom (the transom is partly wet). Here, the displaced water volume is V=Lbd with the beam b corrected for deadrise of a V-bottom. As the speed is increased slightly the bow rises and the transom drops while the boat still plows, due to the bow's riding up the bow wave. There is still no lift, the backflow at the transom is still there. The boat speed is again increased slightly with the boat still plowing bow-high. Finally, at a critical speed the flow suddenly separates from the bottom at the transom and the entire transom is dry: the trailing vortex creating the backflow was washed downstream (Kutta condition satisfied). I have verified this scenario scientifically by leaning over the transom to look whilst my son piloted the craft and read the GPS. At the speed where the transom is suddenly completely dry, this is exactly the point where lift sets in, with the running surface trying to plane. From this point on the buoyancy is much reduced, FB≈bLd/2 with d≈Lxalpha. From this point on lift and buoyancy both contribute to carrying the weigh but buoyancy quickly becomes neglible as U is increased. With the lift coefficient nonzero (U≥8 mph for our Glastron v153), the lift to (reduced) buoyancy is FL/FB=cLxF^2 where F^2 = U^2/gd. The details are presented in the attachment.

    A Froude nr. for a particular boat has no absolute significance. A hull operates at a range of Froude nr. depending on how wet the boat runs. Submerged depth d must be used in the Froude nr. to describe both lift and buoyancy.
     

    Attached Files:

  7. Rastapop
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    Rastapop Naval Architect

    Which part do you disagree with?

    Your Froude numbers cannot be compared directly with Faltinsen's Froude numbers - they don't use the same definition.

    The transom doesn't instantaneously become dry at some point - there is a transition speed range. Short perhaps, but not instantaneous.

    Lift does not (necessarily) suddenly become non-zero when, and only when, the Kutta condition is satisfied, except when you're talking about a foil immersed in a single fluid.


    You're right about no absolute significance, although when the Froude number definition is known we can reliably associate certain conditions and events with certain Froude numbers.

    The topic that Froude numbers are most often used for is resistance (what it was invented for, after all), and in that case the most useful parameter to use is typically length. When no other definition is supplied, length is the parameter assumed.
     
  8. sandhammaren05
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    sandhammaren05 Senior Member

    Well, of course nothing is literally instantaneous but I watched the flow separate and I would call it instantaneous. The boat used is 15' long. With a bigger boat and a lot more backflow it may well take longer for the starting vortex to be shed.

    I think you're wrong about the Kutta condition. If you'll check my preprint you can see how I've used the Froude nr. to formulate boat's the form drag after lift has developed. That leads directly to my speed-power-weight formula, which I've tested against data on established speed records, as well my data on some classic motorboats.
     
  9. Rastapop
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    Rastapop Naval Architect

    I didn't read your entire paper, but I notice it still uses wetted length to calculate Froude numbers, and still apparently claims Kutta–Joukowski is responsible for lift of a hull despite the fact that there is (obviously) no circulation around the leading edge.

    I found your Cf for turbulent flow over a flat plate to be too large. Multiple other sources gave Cf=0.027*Re^(-1/7) which matched my data reasonably well.
     
  10. Joakim
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    Joakim Senior Member

    That is local Cfx not Cf to be used for the whole area: http://www.mne.psu.edu/cimbala/me320web_Spring_2015/pdf/Flat_plate_turbulent_BL.pdf

    The curve used by sandhammaren gives lower friction than ITTC friction line. It is about 5 % lower at 1e7 and 12% at 1e8. It assumes very smooth surface, especially when used for exceptionally high speed boats.
     
  11. Joakim
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    Joakim Senior Member

    sandhammaren I really tried to read your paper, but it is just impossible to keep track of what you are trying to say.

    You really seem to make your own definition and not at all follow the well established notations used by all published papers.

    You seem to use form drag as a basis of the drag of a planing boat and you are (maybe) assuming drag depends on the square of the speed and thus needed power on the cube of the speed. This is not at all true for most planing boats. Look at the top speeds of the same boat with different power. Or at the measured drag curves for any planing boat.

    The drag of a planing boat have many components and not all of them scale by the square of speed. Aerodynamic drag would, but trim changes with speed typically making the boat more aerodynamic at higher speed (lower trim). Friction would also scale with the square of speed, but wetted area is reduced with increasing speed and friction coefficient may get lower or higher depending on how Re changes. Drive unit (or shaft and its support) have both form drag and friction drag that could scale with square of speed, but drive may be higher at high speed. Then there is whisker spray drag, which can be very high for a high speed light boat. It depends on many things that change with speed and trim. And finally the lift force is perpendicular to hull and thus its drag component depends on trim angle and on the amount of lift.

    Your approach seems quite complex. Why not just use Savitsky method, which is known to be accurate and is well tested for all kinds of planing boats?
     
  12. sandhammaren05
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    sandhammaren05 Senior Member

    I do not use wet length L to get F, I use the depth d of the submerged wedge at the transom: d≈Lxalpha where alpha ≈1/10 at high speed. Alpha is the trim angle of the boat. This is quite different than using L to define F.

    If you integrate over both sides of a vortex sheet (the mean camber line of a thin wing) then you get the circulation. Integration over half the sheet gives half the circulation and therefore half the lift coefficient. A planing surface is half a vortex sheet once the flow separates from the bottom at the transom. The lift coefficient for 12-13 degrees deadrise is roughly equal to alpha, more than a third down in magnitude from a sheet of infinite span.

    I got the turbulence estimate for a flat plate from Prandtl and v. Karmann. I think that's still taught in Landau-Lifshitz, e.g., but will check.
     
  13. Rastapop
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    Rastapop Naval Architect

    Yes, my apologies, you're now going from wetted length to depth before Froude.
    Like I said, I haven't read everything - I skimmed through to find things that caught my eye, and "wet length" was one.

    A hull on the water surface as a vortex sheet does not have equal and opposing pressure on its upper side, and therefore doesn't require circulation for lift.

    A downstream counter-rotating votex cannot provide lift at the interface of two different fluids like water and air: they separate.

    I can only assume you're taking what's familiar to you, and trying to apply it in places it doesn't belong.
    I'm reminded of the idiom: "To a hammer, everything looks like a nail."

    As Joakim pointed out, I may have picked an incorrect equation in my haste - it did match viscous drag from my CFD data reasonably well though, whereas the one you used didn't. It isn't something that would affect your paper much either way, as you only used it to show that viscous drag was dominated by pressure drag in the case you were examining.
     
  14. Joakim
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    Joakim Senior Member

    That's one of the very bad assumptions made. I just run a simulation of a boat I used to have at 52 knots (60 mph), which it would have made with 50 HP (did 48 knots with Yamaha 40 700 cm3 3 cylinder).

    The total thrust 758 N. Hull frictional drag 332 N, drive drag (mostly frictional, propeller shaft at keel level) 92 N, pressure drag on hull 174 N, Aerodynamic drag 152 N and whisker spray drag (frictional) 9 N.

    Pressure drag on the planing hull is not really form drag. It doesn't depend on square of speed. It depends on the trim angle, since it is the force supporting the hull so vertical component of it is always supporting the weight of the boat (with possible help from propeller thurst vertical component).

    So actually aerodynamic drag and part of the drive drag are the only form drag components. They are about 1/4 of the total drag.

    The comparison used only one square foot of wetted area for a boat needing 75 hp for 52 knots. That's not at all realistic. For my boat it would require 7.5 degree trim angle, which is way above porpoising limit (about 3.5 degrees) and would require CG to be only 30 cm from transom. This boat weighs 300 kg with driver and fuel. Your boat is likely much heavier.
     

  15. sandhammaren05
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    sandhammaren05 Senior Member

    My only hammer is brass, for prop work. Recall D'Alembert's Theorem: the force on an object in a steady nonviscous flow is zero. Unless there is circulation. The vortex sheet is necessary, otherwise there is no transverse force. Including viscosity for the thin boundary layer will provide lift without circulation. I'm open to a full theoretical calculation showing that I am wrong.

    To get lift from flow past a flat plate you need two conditions: (i) there is a finite angle of attack, (ii) the flow separates sharply from the trailing
    edge of the plate. For a half-submerged plate this gives a finite lift coefficient proportional to half the circulation about a fully submerged plate. If the flow doesn't separate sharply at the trailing edge then the lift coefficient is zero. This is the Kutta condition, there is no backflow to the top of the plate (or up the transom).

    A vortex sheet is a velocity discontinuity. The bottom of a boat is a vortex sheet: the water speed beneath the bottom is U, the water speed above the bottom is zero. To describe this mathematically you must treat the sheet as a continuous distribution of line vortices. There is no other way. Therefore no hammer was used.
     
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