Friction Coefficient

Discussion in 'Hydrodynamics and Aerodynamics' started by jesdreamer, Sep 30, 2015.

  1. jesdreamer
    Joined: Jan 2013
    Posts: 88
    Likes: 1, Points: 0, Legacy Rep: 10
    Location: Alabama

    jesdreamer Junior Member

    Formulas for both laminar and turbulent skin friction include Reynolds Number for distance from start of contact along the nominal flow to some point in question (such as end of flow contact). This gives a single number for coefficient and thus implies a single level of friction for the whole distance involved. It seems obvious to me that friction varies with distance from the beginning contact point. Does the formula acommodate this "fact" and yield an average friction across the whole distance involved?? It seems to me that to arrive at friction across some distance as defined by Reynolds Number to end point in question, we should be integrating a series of increasing values??
     
  2. Joakim
    Joined: Apr 2004
    Posts: 892
    Likes: 52, Points: 28, Legacy Rep: 422
    Location: Finland

    Joakim Senior Member

    You can find formulas for local (Cfx) and total (Cf) skin friction coefficients. Both decrease with increasing distance from the beginning of the surface, except for transition to turbulent flow, which increases the skin friction (it starts to decrease again after the transition jump).

    The drecreasing nature can be explained be increasing boundary layer thickness. The velocity close to the surface decreases. After transition to turbulent flow the velocity profile will be very different due to much more effective mixing. The transition causes a rapid increase in the velocity close to the surface.
     
  3. daiquiri
    Joined: May 2004
    Posts: 5,372
    Likes: 239, Points: 73, Legacy Rep: 3380
    Location: Italy (Garda Lake) and Croatia (Istria)

    daiquiri Engineering and Design

    As Joakim said.
    For example, for a turbulent flow over a smooth flat plate, the local friction coefficient at a streamwise point x from the leading edge can be described by the equation (Prandtl)
    cf(x) = 0.059 / Re(x)^(1/5)​
    where:
    Re(x)=(V*x)/viscosity < 10e6​

    The mean friction coefficient for a smooth flat plate of length L is then found through integration of the cf(x) over the length L, and is given by the equation
    Cf(L) = 0.074 / Re(L)^(1/5)​
    where:
    Re(L)=(V*L)/viscosity​

    Hence, they are both decreasing with the decreasing length x, or L, from the leading edge.

    Now let's imagine that we have divided the total length L of the plate into a forward part (with length Lfwd) and an aft part (with length Laft), such that
    Lfwd + Laft = L​
    Flat plate.gif

    With few mathematical passages, it can be shown that the contribution of the aft part of the plate to the total friction drag is given by:
    Daft/Dtot = 1 - (1 - Laft/L )^0.8​
    The graph of the above relationship might be easier to comprehend:

    Flat plate turbulent.gif

    If the plate is divided into 2 equal parts, we can see from the above graph (red lines) the contribution of the aft half to the total friction drag is 43%, while the forward half of the plate crates 57% of the overall friction drag.



    In case of a fully laminar flow over a smooth flat plate, the above relationship becomes:
    Daft/Dtot = 1 - sqrt(1 - Laft/L )​
    and the graph is:

    Flat plate laminar.gif

    If the plate is divided into 2 equal parts as before, we can notice that the contribution of the aft half to the total friction drag is 29%, while the forward half of the plate crates a hefty 71% of the overall friction drag.

    The reason for this major importance of the fwd part of the friction area lays in the fact that the shear stress has a maximum value near the leading edge, where the incoming flow undergoes a rapid slow-down when in contact with the plate surface. The transfer of the momentum from the fluid to the plate (which is what we call friction drag) is hence maximum right behind the leading edge. The shear stress then decreases towards the aft part of the plate - rapidly in case of laminar B.L. and less rapidly in case of turbulent B.L. Thus the smaller disproportion between fwd and aft contributions to the friction drag in the latter case.
     
  4. jesdreamer
    Joined: Jan 2013
    Posts: 88
    Likes: 1, Points: 0, Legacy Rep: 10
    Location: Alabama

    jesdreamer Junior Member

    Decreasing friction with length from start

    Formulas I had been using for laminar and turbulent both had Reynolds Number in dividend so I did realize friction would decrease as one would advance along streamlines of flow getting larger Reynolds # -- so I misspoke in original post. But I felt we had to integrate a value which was changing (decreasing) with distance and feared that the formulas were for a specific spot and were not averaging over the length involved. Thanks for the followup and insights --

    But I have more questions -- If we have a flat plate with laminar from beginning at pt A to transition B and ending at C, I feel correct way to determine overall friction would be to calculate via turbulent formula for A to C, then for laminar from A to B, subtract the 2nd from the first to find difference for AB, then subtract this AB difference from turbulent AC to find turbulent BC and finally add this to laminar AB to end up with laminar friction for AB plus turbulent friction for BC -- In this way we are applying applipable Reynolds numbers for the areas involved and their proper location within the flow field -- Is this the right way to apply friction formulas in a flow distance involving transition??
     
  5. daiquiri
    Joined: May 2004
    Posts: 5,372
    Likes: 239, Points: 73, Legacy Rep: 3380
    Location: Italy (Garda Lake) and Croatia (Istria)

    daiquiri Engineering and Design

    That's how I would do it in order to deliver a result, in absence of more accurate tools for numerical integration of BL equations.
    Or, if the extension of the laminar region is small (and it usually is), I would simply assume a turbulent flow all the way from the LE to the TE, and would use the formulae for turbulent Cf. In that way, I would have a safety margin for the calculation of required thrust or power.
     
  6. philSweet
    Joined: May 2008
    Posts: 2,249
    Likes: 147, Points: 63, Legacy Rep: 1082
    Location: Beaufort, SC and H'ville, NC

    philSweet Senior Member

    I'm not sure how far you want to go with this, but the idea is to match the momentum defect across the transition. If you integrate the laminar frictional force from A to B, using something like the Blasius boundary layer solution, then you know that there is a momentum defect in the boundary layer fluid that is proportional to this integral. Just after the transition at B, there is a different velocity profile in the boundary layer, with a different shear at the surface, but the momentum defect of this new profile must equal that of the profile just upsteam at B. This gives you a way to patch things together. Kinetic energy within the boundary layer should also be conserved across the transition. Of course, some of that kinetic energy is now sunk into turbulence.

    So an improvement over your proposed method is to find a new starting position for the turbulent integral, A*, such that the laminar integral from A to B = the turbulent integral from A* to B (edit. meaning from 0 to B - A* in terms of the integral). Then just add the two pieces, Laminar A to B, and turbulent B - A* to C - A*.
     
  7. Joakim
    Joined: Apr 2004
    Posts: 892
    Likes: 52, Points: 28, Legacy Rep: 422
    Location: Finland

    Joakim Senior Member

    Laminar flow and free transitions are a very difficult subject. Search for papers by Ulrich Remmlinger posted on this forum. You can find some Cf formulas, where laminar part is taken into account.

    I think your method would underpredict the friction drag, since it takes away the high local drag in the beginning of the turbulent part. But I haven't checked that.

    You can compare your method to the formula given here: http://www.calpoly.edu/~kshollen/ME347/Handouts/Friction_Drag_Coef.pdf

    Remember to take into account the increased drag due to roughness and fouling, which is another difficult subject, if an accurate drag prediction is needed.
     
  8. jesdreamer
    Joined: Jan 2013
    Posts: 88
    Likes: 1, Points: 0, Legacy Rep: 10
    Location: Alabama

    jesdreamer Junior Member

    Friction decrease with distance and need to integrate

    Just to sumarize, (1) friction increases with distance traveled over a surface so we need to integrate to get total (over a curved surface, I assume this means tangential distance not chordal distance) and (2) basic coef formulas provide an average value for length traveled --

    Now per PhilSweet post above I would like to go further into this. Pressure normal to a convex surface (hull) is max at first contact and decreases until transition, then increases but probably separates due to viscosity before end of surface. On an airfoil in air, this leads to "induced drag" (might we say the plane dominates, and re-directs the fluid??) -- But it seems to me that on a boat, reduced (though increasing) support pressure after transition and dramatic loss of virtually all support pressure after separation just leads to "sinkage or squat" of hull down into equilibrium (Might we say the fluid dominates the boat??) . So with a boat we would be getting no induced drag in same sense as with a airplane -- but the suck-down force would be normal to convex surface and thus provide a component parallel to travel which would be a drag force due to attitude, approach angle, or trim. So we have sinkage due to Bernoulli and a component of the Bernoulli force (reduced pressure) opposite to direction of travel which adds to drag. This component would be "pulling" toward aft while the main component (normal to convex) would be "pulling" down -- (I realize we are talking about pressure differentials not an actual pulling force) -- So if we eliminated sinkage, wouldn't we eliminate this drag??

    But in addition to the above Bernoulli drag effect, isn't there a drag related to pressure ahead of transition being higher than after transition (due to viscosity losses?) and particularly if we have any separation -- (I am going around in circles here since it would seem to me than this pressure differential is equalized by the sinkage in same fashion as the analysis above) -- again as before, I am just trying to understand what is going on and I worry that mys thoughts might be violating conservation laws of physics.

    I would like to hear some of your thoughts per this approach to the situation and/or other ways of looking at what all is going on mechanically --
     
  9. Joakim
    Joined: Apr 2004
    Posts: 892
    Likes: 52, Points: 28, Legacy Rep: 422
    Location: Finland

    Joakim Senior Member

    Induced drag is related to lift force. For a boat induced drag is created when it has non zero angle of attact that it has leeway or it is a planing boat with trim angle. A displacement boat without leeway doesn't create much lift (or sink).

    The same applies to an airfoil (no matter if it is a wing, keel or a rudder). Induced drag is only related to lift. At zero lift there is no induced drag. Induced drag is a 3D phenomen causes by lift while what you are describing applies to 2D as well.

    There are only two possible interactions between a boat hull (or any object) and fluid. Friction, which is always parrallel to to surface, and pressure, which is always perpendicular to the surface. On a boat pressure is very much determined by the wave pattern created by the boat, which makes it very different from airplanes (or submarines). Friction is very similar.
     
  10. daiquiri
    Joined: May 2004
    Posts: 5,372
    Likes: 239, Points: 73, Legacy Rep: 3380
    Location: Italy (Garda Lake) and Croatia (Istria)

    daiquiri Engineering and Design

    The method proposed by Philsweet is formally the most correct one, because the momentum B.L. thickness (indicated with Greek letter Theta) is directly proportional to the integrated friction coefficient over the flat plate. In particular, for a flat plate with no pressure (and hence velocity) gradient, the Von Karman's integral B.L. equation gives:
    Cf(L) = 2 Theta / L​
    So, if we were able to estimate (or to enforce) the position of the transition point Xtrans, in that point the B.L. thickness Theta of the laminar part of the flow (from the L.E. up to the point Xtrans) would have to be equal to the thickness Theta of a fictitious fully-turbulent B.L. which starts at some point A downstream from the L.E. The point A is then determined by solving the condition Theta_laminar = Theta_turbulent at Xtr.
    If the enhanced precision is required, then this method can be used. But if the goal is to make a conservative yet near-the-ballpark estimate of the friction drag, the two methods described in posts #4 and #5 will deliver the result.

    In any case, the formulae seen so far are valid for smooth flat plates with no pressure gradient, set at zero angle of attack to the flow. If they are used for evaluating the friction drag of curved bodies with a pressure gradient, the word "precision" will also indicate the level of optimism and faith of the designer.

    I would say that quite the contrary happens. :)
    As mentioned before, the integrated friction drag is proportional to the thickness of the momentum B.L. at the trailing edge of the flat plate - see the above formula.
    So if in our calculations we use the turbulent B.L. which starts from the L.E. (and not from some downstream point A), then at the trailing edge the Theta will be necessarily higher than a Theta of a turbulent B.L. which originates at some point downstream from the leading edge. Hence, the Cf of the latter case has to be lower. In other words, the subtraction method described in the post #4 yields a higher friction drag than the method in the post #6 and hence gives a conservative estimate.
    The choice of the calculation method depends (as always) on what is the final goal of the calculation, and the required precision.
     
  11. daiquiri
    Joined: May 2004
    Posts: 5,372
    Likes: 239, Points: 73, Legacy Rep: 3380
    Location: Italy (Garda Lake) and Croatia (Istria)

    daiquiri Engineering and Design

    I see a number of issues in what you wrote. Difficult to gather things together and decide where to start sorting them out...
    What is exactly the goal you are trying to achieve here? Are you making all these considerations with some specific design problem in your mind, or just out of personal curiosity?
     
  12. jesdreamer
    Joined: Jan 2013
    Posts: 88
    Likes: 1, Points: 0, Legacy Rep: 10
    Location: Alabama

    jesdreamer Junior Member

    Too Many Questions

    With each post in response to my questions I come up with around 10 more questions. I have no specific design in mind, and in fact would prefer to understand what is going on before getting into any specific design debate --

    I am not interested in precision but do appreciate the expansion of concept understanding that discussion of alternative math approaches do provide --

    I would like to get back to what I last asked -- quote below --
    "But in addition to the above Bernoulli drag (sinkage) effect, isn't there a drag related to pressure ahead of transition being higher than after transition (due to viscosity and/or BL flow losses?) and particularly if we have any separation -- I am going around in circles here since it would seem to me than this pressure differential is equalized by sinkage in same fashion...."

    If you do agree that Bernouli pressure loss yields sinkage which in turn yields a drag "component" parallel to flow, isn't any pressure differential relating to BL thickness or friction also compensated for by additional sinkage which also adds a drag "component"?? And how might we calculate any pressure differential showing up with distance traveled across surface?? Of alternatively, how might we calculate related sinkage or its drag "component" --

    This is starting to sound like doubletalk and it seems that I may be asking the same question in 2 different ways -- I realize that net or integrated pressure differential causes sinkage, and that since the pressure reduction is acting perpendicular to tangent of the convex curve, then with sinkage, we have a component of this force acting parallel to flow as a drag force -- It would appear that that the net sinkage should compensate for all net buoyancy related pressure differentials and would yield some angle of trim due to higher pressures supporting bow -- Is there a way to calculate the related drag?? -- Sorry if I am not making any sense --
     
  13. jesdreamer
    Joined: Jan 2013
    Posts: 88
    Likes: 1, Points: 0, Legacy Rep: 10
    Location: Alabama

    jesdreamer Junior Member

    "Displacement boat doesn't create much lift (or sink"

     
  14. DCockey
    Joined: Oct 2009
    Posts: 4,442
    Likes: 219, Points: 63, Legacy Rep: 1485
    Location: Midcoast Maine

    DCockey Senior Member

    jesdreamer, you may be confusing boundary layer flow effects and "inviscid" flow effects outside of the boundary layer. The pressure in the boundary layer is determined by the pressure outside the boundary layer. The variation in pressure for attached flow around a curved object is due to inviscid effects with the effective shape of the object modified by the boundary layer thickness. To assist with the discussion here have you taken any university level courses on fluid mechanics, aerodynamics or hydrodynamics?
     

  15. markdrela
    Joined: Jun 2004
    Posts: 306
    Likes: 28, Points: 28, Legacy Rep: 324
    Location: MIT Aero & Astro

    markdrela Senior Member

    You can look at drag in two main ways, which are consistent in that if applied correctly they will give the same answer:

    1) "Near-field" drag, defined on the body surface. This is the sum of perpendicular pressure forces which constitute pressure drag, and tangential viscous shear forces which constitute friction drag. This is what's done to calculate drag in Navier-Stokes codes. But this approach is sketchy when used with simpler drag estimation methods, since there's no straightforward way to accurately determine the pressure drag part for general shapes without doing full viscous calculations.

    D = D_friction + D_pressure

    2) "Far-field" drag, defined by the viscous wakes, the trailing vortex wake, and also the wave pattern in the case of a free-surface flow, all examined relatively far behind the body. The overall momentum defect of the viscous wakes is called profile drag. The kinetic energy per streamwise length of the overall inviscid flow about the trailing vortex wake is called induced drag. The kinetic energy of the free surface waves per streamwise length is called wave drag.

    D = D_profile + D_induced + D_wave

    ---------------------------------------------

    Most drag estimation methods assume breakdown 2), and estimate the three pieces separately and add them together. In these estimations the D_profile component is usually estimated as

    D_profile = D_friction' + D_form

    where D_friction' is what you get from a flat-plate skin friction chart. Since this is almost invariably too low for a body with a finite thickness, the D_form is added to make up the missing part. Note that D_friction' above is not the same as D_friction from method 1), although it's comparable. The D_form is typically correlated with thickness/chord ratio, fineness ratio, or similar geometric parameters. But this approach is really just fancy experimental data fitting without much physics behind it.

    I'm not sure how the various types of free-surface codes estimate D_wave, whether it's via the farfield energy outflow or via body surface pressures.
     
    Last edited: Oct 2, 2015
Loading...
Similar Threads
  1. dustman
    Replies:
    53
    Views:
    3,530
  2. rwatson
    Replies:
    34
    Views:
    2,761
  3. Slip
    Replies:
    3
    Views:
    917
  4. designing
    Replies:
    5
    Views:
    1,362
  5. nine6
    Replies:
    5
    Views:
    1,317
  6. Leo Lazauskas
    Replies:
    34
    Views:
    8,239
  7. hprasmus
    Replies:
    11
    Views:
    5,321
  8. Shayne Young
    Replies:
    12
    Views:
    2,539
  9. Leo Lazauskas
    Replies:
    50
    Views:
    7,375
  10. nimblemotors
    Replies:
    47
    Views:
    13,668
Forum posts represent the experience, opinion, and view of individual users. Boat Design Net does not necessarily endorse nor share the view of each individual post.
When making potentially dangerous or financial decisions, always employ and consult appropriate professionals. Your circumstances or experience may be different.