# Flettner Rotors - Calculating Rotating Skin Friction

Discussion in 'Hydrodynamics and Aerodynamics' started by rwatson, Jun 16, 2018.

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### rwatsonSenior Member

This is some good work on the issue. Joakim's questions about the practical issues are of real interest to me.

First, I am not sure that the assumption that high aspect rotors ARE more efficient. On that calculator, a bigger diameter creates more power. There is probably a "sweet spot" where the power required for any particular angular velocity coincides with the most drive.

Second - the question of comparing standard rigs with rotors isn't only about performance. For example, going sailing on a 35ft boat without extra crew on a very windy day would be unwise from a sail handling and safety point of view. But, a close equivalent in Rotors, with "dial up" performance, (no raising , lowering , changing of sails) by one person, can be a very compelling case.

Third - The problem of constant sail tuning during a voyage is always a a nuisance. Constant adjustment of sail sheets, tacking adjustments etc during a voyage are a constant source of problems, especially if the weather is bad.

Fourth - moored windage is reported to be less than standard rigging. Anyone who has put a ping pong ball under a stream of water, and noticed that the flow of water around a perfect sphere actually keeps the ball centred and un-moving. Certainly, the actual drag force would be less of a problem if the wind was coming from the bow or stern.

If I am able to get a test rig built, I would get some remote sensors that would record actual pressure on the rotor under operation. It would automatically record positive and negative pressure at three or four places up the column, the apparent wind direction, wind speed, rotation speed and direction, and boat direction and speed. I bet these figures would vary a bit on individual boats

The ultimate aim would be to develop a databases of ideal configurations, that would be used to automate the selection of ideal rotor speed and direction depending on the desired direction of travel. This is what the commercial systems do, avoiding the need for decisions by the navigation officers.

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### JoakimSenior Member

Bigger diameter has more area and reaches the same velocity ratio and thus lift coefficient at lower RPM. If RPM is limited for some reason, then bigger diameter (lower AR for the same area) can be more efficient.

How do you define efficiency of a Flettner rotor? Higher AR rotors can reach much higher lift coefficients and L/D ratios, thus they can produce more lift at the same area and lower beat angles. I haven't yet looked at the power consumption of different AR rotors. But defining the efficiency as propulsion power out/mechanical power in would lead to infinite efficiency for sails, since they do not need any power input.

Windage is quite easy to calculate and is certainly much higher for a Flettner rotor than for a modern sailboat rig. An historic rig with several thick masts and a lot of wires/ropes may be another case. Windage is also a problem when it is necessary to motor head to wind in heavy weather. A lot of windage makes mooring much more difficult.

I sail quite often alone with my boat. I don't find sailing alone difficult, but manuvering in a harbor can be a challenge. More windage would make it much harder.

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### rwatsonSenior Member

"define efficiency of a Flettner rotor"
The most directional lift for the least power input, for the least capital cost for the least amount of crew effort.

"infinite efficiency for sails, since they do not need any power input."
That's not practical of course, as the higher the wind, the greater the angle of heel, until you are laying over sideways. Also, at maximum efficiency angle (reaching) , sails create maximum heel angle, while Flettners vector is straight ahead. In lighter winds, the Flettner Rotor actually leans INTO the wind.

"I sail quite often alone with my boat. I don't find sailing alone difficult"
Yes you do. Its just that you are young and fit enough to do the sail raising and adjusting and not find it a problem. Try it for 12 hours straight in adverse conditions, and not even you will be immune. You can also look up all the injuries sustained each year by "men on boats" while sail and sheet handling.

"manuvering in a harbor can be a challenge. More windage would make it much harder."
Yes, that makes sense. Sails are actually a major issue in confined or adverse environments. This is where hopefully, the quick and automated functioning of a rotor would make it less problematic - being able to use the wind power for longer more effectively, instead of having to douse sail and have just have the motor at one end of the vessel.

Of course, all this has to be proven in practice. This is the dream

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### JoakimSenior Member

Why would Flettner rotor lean into wind in light winds? Certainly that doesn't happen unless apparent wind is clearly in the rear sector and thus resultant force vector pointing to windward side. In that case there is very little heel from sails. In light winds apparent wind can be clearly in the rear sector only when sailing very close to dead downwind. E.g. my boat it would require TWA>150 at 6-10 knots TWS. At 135 TWA AWA would be below 90 degrees.

Sails produce a lot of heel only when the apparent wind is in the forward sector (or in heavy winds with apparent wind close to abeam). It is easy to avoid excessive heeling by trimming the sails, reefing or sailing with just one sail. In heavy weather there is usually no lack of power and speed is not reduced significantly by this.

I can understand your definition for efficiency for ships looking for fuel saving, but not really for sailboats. For sailboats you need to have enough drive force in light winds as well and L/D ratio is crucial for upwind performance.

I don't find sails a major issue when coming to harbor sailing alone. But I do find it difficult to maneuver alone in the harbour using engine in heavy winds due to windage and I think it would be much harder with added windage from a Flettner rotor.

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### DolfimanSenior Member

In complement of my previous quote, here attached the Pm "Norwood" curves compared with Reid/Naca 1924 measurements + the extension to v/U = 2. I need others first hand data of such power measurements for v/U > 1,2 to consolidate this comparison, many thanks in advance for such precious (but rare) documents.

PS : while this comparison fits quite well with the experiments, I am still not confortable with the approach :
2 U sin(Teta) + v is said the "net air speed seen by the rotor" . It is actually, in my opinion, the net tangential air speed at the surface of the "envelop" of the rotor, itself in rotation v, and is it not more exactly the relative speed to consider when computing a friction force ?
And second remark about the integration of the squared speed on the full perimeter, but without checking before when it is a lower tangential speed / rotor one (which leads to a friction acting on the rotor ) or a higher one (which leads to an entrainement of the rotor).
Your comments welcome on that, as I think there is more to dig ...

#### Attached Files:

• ###### Power to rotate a cylinder in a moving air.pdf
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### rwatsonSenior Member

No, if the wind is abeam ( sideways on ) the rotor actually leans into the wind somewhat if the motor is turning. It "claws" its way to windward. This is why golf balls climb quite a bit after being struck on a drive.

Its true that you also get that same effect when the wind is coming somewhat from dead astern or ahead.

The main point however is, that you don't get the tendency to heel to leeward in heavy winds, when travelling at the optimum wind angle. ie. abeam

Show me a photo of monohull that isn't heeling over to around 10% or more in a decent wind, when running at optimum (for speed) wind direction ie. somewhat abeam.
, reefed or not

This is important for many cargo ships that are not designed for excessive heel due to cargo care.

In a yacht, it makes for a much more pleasurable experience for passengers and crew.

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### JoakimSenior Member

It seems you haven't quite understood how the Flettner rotor works. It doesn't "claw" to windward. It produces lift and drag, just like a sail or a wing. Thus the resultant force vector points always a bit to the leeward side of the apparent wind. With high L/D almost perpendicular to the apparent wind. At L/D of 5 the resultant force vector points 11 degrees (18 deg at 3 and 7 deg at 8 L/D) to the leeward side when the apparent wind is perpendicular to the boat. Thus AWA needs to be 101 degrees for the resultant vector to point straight forward and clearly more than that to get any windward force and thus leaning to wind.

When sailing upwind or close to that L/D and the effective height of the sail or rotor is the crucial factor that decides how much the boat heels, IF the performance is about the same. If you have worse L/D you will have more heel unless that is compensated by lower effective height (or lower force and thus worse performance). For my boat sailing at 5 m/s and 25 deg AWA at 5 L/D the resultant force is pointing 76.3 degrees from keel line. Thus with 3000 N total force the drive would be 710 N and heeling force 2914 N. With 3 L/D at the same operating point the resultant force would point 83.4 degrees from the keel line leading to only 343 N drive and 2980 N heeling force. Thus the total force would need to be more than doubled to get the same drive force and then you would get more than double heeling force and much more heel (and still worse performance due to added hydrodynamic drag and leeway).

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### JoakimSenior Member

Dolfiman: your latest pdf seems to have an error. The Cf computed should be 0.0079, not 0.079.

Using those formulas I get ~1200 W for the 14 m2 5.1 AR rotor at 2.5 velocity ratio and 7.5 m/s apparent wind (beat at 5 m/s). This doesn't take into account the huge endplate, which certainly increases the power needed considerably. It is also far away from the measured velocity ratio range.

Then the 6 m2 13 AR at the same point, but using 5 velocity ratio would take a huge ~3500 W using these formulas (even more out of measured velocity range). This is already about the same power as my boat would need to reach the same speed while motoring!

The effective thrust is about 2000 W thus the high AR rotor would consume more than it produces and also the lower AR one would have poor power efficiency.

I guess that is the reason why high aspect ratio Flettner rotors are not used. The power needed increases to huge amounts when higher velocity ratios and thus higher lift coefficients are used. Already 1200 W is a lot. Would need a huge battery bank or engine running all the time. Not at all practical for a sailboat. And this was calculated only at 5 m/s true wind. The power becomes much higher in heavy winds.

The situation is much better when the wind is abeam or on a reach, since then apparent wind is lower and force is mainly drive instead of heeling force while beating upwind. But then you would like to use high velocity ratio to get high power, but that takes a lot of power.

There seems to be an error in the Jeremy's Excel. The formula used to calculate velocity ratio from apparent wind, diameter and RPM uses u= omega * D, while it should be u = omega *R. Thus the calculated u/V is double or the RPM is half of what it should be. So the all the RPM I gave in the earlier post must be doubled.

E.g. in the case of AR 13 diameter is 0.7 m and RPM 520, which equals to 54.5 rad/s and gives surface speed of 19.05 m/s. The wind speed is 16.8 mph, which equals to 7,5 m/s. So the velocity ratio is 2.54, but the Excel gives 5.075 and all the rest of the values are based on that velocity ratio.

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### DolfimanSenior Member

Yes,it is just a copy/paste error, and 0,0079 of course was used in the computation.
Thanks for the Seifert paper full of info and data. From it and the Norwood one, I try to formulate the Power from the torque formulation proposed by Thom, that gives when rewriting with my notations :
Pm "Thom"= Cf (1/2 * Rho * Sw) * (k*U*v) * v
, the key quadratic function is there in the form of k*U*v, and k ~ 3 ? (rewriting the formulation leads to k = 4 Ct / pi^2/Cf ; when I takes Ct ~ 0,03 from the Seifert paper and Cf = 0,0041 from the Norwood paper, that gives ~ 3).

Here attached the pdf completed with the "Thom" curve and extended to speed ratio of 4 to see the trend.

My next task will be to extend the formulations to take also into account the friction of the end plates.

#### Attached Files:

• ###### Power to rotate a cylinder in a moving air.pdf
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### rwatsonSenior Member

I'm sorry, I didn't understand much of what you were trying to say.

All I know is, I have read operational reports on how the rotors behave, and when a light wind is abeam, and the rotor is spinning in a direction that is directing the boat forward, the rotors tend to lean into the wind a bit. If this boat below were tied up at the dock - the boat would have a slight heel to Starboard, as the excerpt I posted describes.

Under sail, there would be little heel to Port under his configuration.

You can see the Buckau in strong winds here (about 2:30 mins) with perhaps 4-5 degrees of heel, much less than conventional sailboats.

Last edited: Jun 23, 2018
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### JoakimSenior Member

rwatson, the vector diagram of Fig 4. shows a broad reach, not abeam wind. The true wind seems to be around 135 degrees and the apparent wind is also at quite high angle, maybe 120 degrees. The true wind vector is almost three times as long as fair wind vector, thus the vessel would be really slow or the wind extremely heavy (30+ knots for my boat). And still the resultant force is drawn to be just on the keel line, thus no heel at all. Lean to windward would need an even higher true wind angle.

Buckau is a not a sailing vessel. From Steifer paper "The main engine and one screw propeller still served as primary propulsion system. Flettner proposed to use the Magnus effect not as sole but as additional source for propulsion, to reduce fuel consumption".

The rotor area of 87 m2 is not enough to replace the sail area of 883 m2 when AR is low and maximum surface velocity is only 18 or 20 m/s (depending on the source). In 10 ms apparent wind the rotors can only reach velocity ratio of 2 and thus Cl of 4. Even at lower wind speeds the rather low AR (5.6) and very small endplates would limit Cl to about 4.

Sails can easily reach Cl of 1, thus the rotors would have needed at least 2.5 times the area they had to give equal driving force. Obviously with such a little power the vessel designed for much more power doesn't heel much. Just like my boat wouldn't heel much with trysail and storm jib in 10 m/s wind where I would normally sail with full sails with much more heel.

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### JoakimSenior Member

Here is one data point for the power needed: Buckau | Der Flettner Rotor http://flettner-rotor.de/geschichte/buckau/

Buckau's rotor needed 9.6 HP to reach 110 rpm and the installed engine was 11 kW designed to reach at least 120 rpm. On that page 125 rpm is given as maximum. Steifer paper gives 135 rpm. Maybe it depended on the point of sail, since apparent wind has an effect on the power?

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### DolfimanSenior Member

As regard the AYRS / Norwood computation of the friction elementary force on the end plates, I disagree with the use of formulation (12) U = 2 V sin(teta) + w r , i.e. with my notations : local speed at (r,teta) = 2 U sin(teta) + v (r/R).
Not only it is not clear if it is the absolute air speed or the relative speed with regard the rotation of the end plate (I did the same remark for the cylinder and the use of formulation (4), as friction means the use of the relative speed), but anyway the air speed module at (r,teta) is unfortunately a lot more complicated formula, and moreover its orientation is not parallel to the cylinder tangential one in the general case, so you should compute also this angle of orientation alpha and then multiply by cos (pi/2 - teta + alpha). The note attached gives you these formulations and a taste of the complication. And to this air speed tangential component so obtained, in order to have the relative speed, we have to subtract the tangential speed of the end plate rotation at same point (r,teta), then to check the sign of the result in order to know if it is a friction or an entrainement, and double integrate with r and with teta, …., analytically completely crazy to do !!!!
So, a huge but realistic simplification is there necessary for a reliable estimation of the friction acting on the end plates and so their contribution to the power. The good news is that probably the Norwood approach over estimates a lot the power for the end plates, the speeds being a lot less when we move away from the rotor cylinder as shown in the note attached for the example of teta = 90°.

#### Attached Files:

• ###### Speed components, module and orientation.pdf
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### DolfimanSenior Member

For the end plate, I eventually proposes an empirical formula derived from the "Norwood" one for the cylinder, based on the comparison of the air flow speeds at teta 90° and radius from R (the cylinder) to r (the end plate). So I avoid the complication of an integration, and keep it simple and in a form easy for a comparison with experimental results if any (e.g. through ajustements of 3 coefficients a, b and c) :

For the cylinder, the « Norwood » formula :
Pm = Cf * (1/2 * ρ * Sw ) * (a * U^2 + b * v^2) * (c * v)
with :
a = 2 b=1 and c=1

For an end plate, a formula derived from « Norwood » :
Pm = Cf * (1/2 * ρ * Sw ) * (a * U^2 + b * v^2) * (c * v)
with :
a = 1 + [ 2 D / (De + D) ]^2
b = 2 D / (De + D)
c = (De +D) / 2 D

More details and numerical examples in the note attached.

#### Attached Files:

• ###### Power to rotate a cylinder with end plates in a moving air.pdf
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75 KB
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### DolfimanSenior Member

Oupsssss, … with the right document attached (a small copy/paste error in the previous one).

#### Attached Files:

• ###### Power to rotate a cylinder with end plates in a moving air.pdf
File size:
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Views:
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