# Flettner Rotors - Calculating Rotating Skin Friction

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

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### rwatsonSenior Member

For the terminally bored, a bit of mental (mathematical) exercise.

Over in the Flettner thread, we've been discussing the challenge of calculating the power required to rotate cylinders in moving air.

The best discussion I have come across is on page 29 of this copy of the Amatuer Yacht Association publication (attached) , but is by no means a "turnkey solution.

If anyone feels inclined to explain the calculation, and /or suggest other strategies for solving the problem, that would be very much appreciated.

The ultimate solution would be an spreadsheet with plug in variables, but this might be a ways off yet.

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• ###### Catalyst_N05_Jul_2001.pdf
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### DolfimanSenior Member

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### rwatsonSenior Member

Damn - pages past 41 are missing

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### DolfimanSenior Member

Can you try again, that's work for me, the pages download speed is variable, sometimes you have to wait a bit to get the following ones.

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### DolfimanSenior Member

In this one, we have the basic equations of a 2D cylinder in a wind flow, without and with rotation, and in particular this gives the origin of the U = 2 V sin(téta) + v
which lacks in the Norwood paper :
http://air-et-terre.info/aerodyn_theorique/cylindre_2D.pdf

Well, we will have all the pieces step by step ....
P.S. : each authors use different notations, not easy and can be confusing

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### rwatsonSenior Member

Thanks for that. I was going to say "its all Greek to me:, but in actual fact its all French
Luckily, it seemed translate to English through Google, and I have attached the English version here.

After a quick, inexpert review, I cant see where any calculations for the power to turn the Cylinder are covered.

I await any competent review of the paper with great anticipation.

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• ###### Flow_English.doc
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### thiseasmJunior Member

Both are very helpful. @rwatson you are not lucky because you speak with a Greek and French.
Because this thread started as a way to evaluate the power needed to rotate Flettner Rotors, from my research I found 3 ways that previous researches have done it (attached files).
In the ''Net Power Generated by Flettner Rotor'' they used only a skin friction formula. They neglect bearing and mechanic friction.
In the ''Flettner rotor power contribution'' they just use a formula with a moment coefficient. Nothing more.
Lastly, in the ''The Use Of Flettner Rotors In Efficient Ship Design Conference paper'' he uses two equations for the flow around the End Plates and the rotor from the book Rotating Flow of Peter Childs (not available online) and one equation for the bearings.

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• ###### The Use Of Flettner Rotors In Efficient Ship Design Conference paper.pdf
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### DolfimanSenior Member

I am myself a "half-expert" in this domain as I was in charge of the refit of this towed submarine vehicle (photo attached, copyright Ifremer) equiped with 4 rotating cylinders so to move the vehicle vertically or transversaly when towed at speed. This vehicle is used to film the entry of a trawling net : EROC - Station Ifremer de Lorient https://wwz.ifremer.fr/lorient/Equipements/Outils-de-mesures-et-d-observation/EROC
The first set of cylinders were too small and especially their axis were too flexible and the capacity of y or z moves was poor. So we increased a bit the cylinders diameter and resized the steel axis and the motors to have both a better rigidity and more power. But without going deep into the theory (as the power optimisation was not at stake) and without time to make measurements in the towing tank other than the ones to check that the new set worked well. An in situ tests were ok too, with a wider range of possible move in both y and z directions.

I have got also this one https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19930080991.pdf

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### rwatsonSenior Member

That NASA paper looks good. I have read part of it. This is interesting.

"The fact that the power input is smaller with moving than stationary air indicates a reduction of air friction . This would be expected as the relative velocity of air to cylinder is reduced, around most of the circumference, by the rotation."

Currently I am toying with the idea of doing a test concept with these components

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### DolfimanSenior Member

Dear all, thanks for this thread and your documents which promises to be very informative and perhaps to lead to clarifications, not only on power input but on all the basics of these rotating cylinders : lift , drag, aspect ratio with end plates, fore stagnation point, ...
So, my first tentative of review, in various posts :
The French paper is about the perfect fluid theory on a 2D cylinder, steady and rotating,
It is a classic academic approach using potential theory, you can find a lot of similar papers, like this one for example :
https://www.springer.com/cda/conten...798242-c2.pdf?SGWID=0-0-45-1500783-p177271608
From this approach, 2 simple things are interesting to confront with experimental results : fore stagnation point, lift
1) the fore stagnation point : the theory leads to a simple formulation for the flow tangential speed at the skin of the cylinder : Vt = 2 U sin(teta) + v (I take U for the incident upstream flow and v for the tangential speed of the rotating cylinder). From this , Vt = 0 leads to 2 symmetrical value of teta, i.e the position of two stagnation points (as long as v < 2 U) , the fore one can be realistic, the rear one has poor chance to be realistic due to the viscosity of the real fluid. Here we can compare with photos of the fluid as in the NASA paper, given for v/U = 0,74 ; 1,48 ; 2,96
0,74 >> teta = 21,7° and 1,48 >> teta = 47,7 these two values seems effectively in coherence with the stagnation points that we can estimated from the photos.
On the other hand, 2,96 leads of course to no stagnation point (sin(téta) > 1) , here the theory completly diverges with the reality.
>>> to collect more photos of such flow around a 2D cylinders with various v/U so to consolidate the comparison with fore stagnation point from Vt formulation, up to now one can say that can match for v/U 0,5 to 1,5.
From the photos, one can also see that the rear flow is of course different from the theory wich leads to a perfect symmetry fore/aft and so no drag (D'alembert paradox) .
In my next post about the lift ...

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### rwatsonSenior Member

Just in case this might be handy, a spreadsheet that Jeremy created some years ago

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• ###### Calculator6mx12.xls
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### DolfimanSenior Member

Another document, the Hoerner Fluid-dynamic lift, see pages 487 and 488 for the Magnus effect :
https://s3-ap-southeast-1.amazonaws.com/erbuc/files/4099_22ac3a91-02b3-4e33-9e95-4caf350b518d.pdf

I am in trouble when Hoerner says that for the speed ratio "U/V = 4" the two stagnation points coincide and that gives a theoritical lift coefficient Cl = 4 pi , I think it is for v/U = 2 , for which the theoritical Cl = 2 pi (v/U) becomes effectively 4 pi (pages 6 and 7 of the French document).

To note that this theoritical (perfect fluid) Cl is anyway a lot higher and unrealistic when compared with the experimental one, which is approximately linear with v/U only when v/U is 0,5 to 2,5 (but probably that it is the domain of practical design).

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### DolfimanSenior Member

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### DolfimanSenior Member

Here is my tentative of review of this AYRS / Varvill & Norwood paper and of comparison with the power measurements available in the Naca/Reid 1924 paper :

The notations : At first, the notations that I have adopted to a better understanding and leading to more uniformed formulations, easier for further comparison :
U is the upstream flow (= the apparent wind when a sailing ship is considered)
v is the tangential speed of rotation at the surface of the rotor.
D and H are the diameter and the height of the rotor.
S the "wetted" surface of the rotor is then : pi D H
>>> Norwood uses rotor radius a , rotor height 1,5 a and the rotational speed w, so the correspondance are : D = 2 a , H = 1,5 a and v=a*w
Rho is the mass volumic of air
Cf is the friction coefficient
Teta is the azimuth angle, 0° is for the rear point of the rotor diameter, and then, following the usual trigonometric convention, 90° is the up point, 180° is the fore point, 270° is the down point.

The approach : Norwood considers the power resulting from the friction drag on the cylinder surface crossed the tangential speed of rotation.
The tangential speed of rotation v is a constant.
The friction drag is formulated using the classic Cf (1/2 rho Speed^2) dS , where "Speed" is the tangential component of the air flow at the surface dS of the rotor, so an integral is necessary as long as we consider a "Speed" function of the azimuth teta. It is the key point, and Norwood adopts the "Speed" directly derived from the 2D perfect fluid theory,
Speed = 2 U sin(teta) + v

Before going further, some physical considerations to assess a priori the range of validity :
Using this formula means that the rotor faces v at teta 0°, 2 U + v at 90°, v at 180°, 0 at the stagnation fore point positioned somewhere in the third quadrant, - 2 U + v at 270°.
We can estimate that with the real fluid we will have flow detachments, roughly at the frontier of the fourth quadrant, at teta ~ 270° and ~ 0°. An alternative could be then to consider only v in the fourth quadrant, the integral computation is not much complicated, I called this alternative Norwood bis.
The speed formulation can be consider valid as long as :
1) we have effectively a flow pattern corresponding to the existence of a lift, needing v/U > 0,5 according to experiments.
2) the stagnation fore point is in the 3rd quadrant and in good coherence with the one observed, meaning v/U < ~ 1,5
When U =0 (no incident flow, just rotation), we are in a complete different aerodynamic context, probably that a thicker boundary layer is then entrained and could causes a higher drag. This could explained the higher power measured by Reid and al. in the Naca paper. Anyway, that situation is not of design interest as no thrust can be expected from a rotation without wind.

The resulting formulations for the power, after integration on the rotor perimeter (for the rotor only at this stage, end plates issue will be adresses later on) :
Norwood ( with my proposed notations) :
Pm = Cf * (1/2 * rho * S ) * (2* U^2 + v^2) * v
Norwood bis (with v in the 4th quadrant) :
Pm = Cf * (1/2 * rho * S) * (3/2 * U^2 + 2/pi * U * v + v^2) * v

Comparison with the Naca measurements :
Data of the experience :
D = 0,1143 m ; H = 1,5232 m
Cinematic viscosity nu= 1,429 E-5 m2/s (estimated)
Rho = 1,23 kg/m3 (estimated)
Series with U = 15 m/s and 12 values of v/U from 0,5 to 1,2
Cf computed with Re = U D / nu and Cf = 0,075 / (log10(Re) - 2)^2 >> Cf = 0,079
>>> we obtain for the power Pm :
Pm measured = 1,21 to 1,87 Pm Norwood = 1,12 to 1,97 Norwood bis
(1,21 or 1,12 being for v/U 1,2. ; 1,87 or 1,97 being for v/U 0,6)
To note that if we consider only the values v/U 0,89 to 1,2 (dropping the results for the lowest values of v/U), we have a lot better stability :
>>> Pm = 1,21 to 1,29 Pm Norwood = 1,12 to 1,26 Pm Norwood bis
Norwood bis is not significantly better than Norwood, actually it lacks measurements for v/U > 1,2 to extend and consolidate the comparison.

So a preliminary conclusion at this point would be to consider the Pm Norwood formulation with a coefficient of ~ 1,25, when v/U in the range of 0,9 to 1,2.

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### JoakimSenior Member

There seems to be still a lot not well known about Flettner rotors. I tried to calculate how a Flettner rotor would perform on my sailing boat compared to its normal rig. Just out of technical interest, no plans to actually use one. It's a rather modern 35' boat from 2005 with 70 m^2 upwind (including roach) and 130 m^2 downwind sail area. I have my own VPP, so I can calculate the lift and drag of the sails and predict the speed of the boat rather accurately.

In 5 m/s wind on a beat (~41 deg TWA ~6 knots BSP) the current sails have about 3000 N total force with a bit over 5 L/D ratio. If I try to match that with a Flettner rotor, I need to find the same or better L/D and total force. Jeremy's Excel gives maximum L/D about 7.5, but it is for a 2D thus infinite AR rotor. This paper gives only 2.8 L/D for 5.1 AR Flettner rotor without an endplate and for better than 5 L/D it would need a huge 2.5*D endplate. And for a good L/D the velocity ratio can only be ~2.5 leading to leading to Cl of ~6 and Cd ~1.2. So for 7.5 m/s apparent wind we would need 14 m^2 rotor area. With 5.1 AR that would be 1.65 m * 8.4 m at 110 RPM. The endplate would then be 4.1 m diameter, thus much wider than the boat (3.28 m) and quite unpractical. If the rotor would be installed on top of the coach roof, it would be about 10 m above waterline. The current mast top is about 16.5 m above waterline. Due to lower height the rotor would have lower AWS and AWA at the same TWA and TWS due to vertical wind profile. It would also suffer more from the hull. Thus it would probably need to be even bigger to reach the same performance. But due to lower heeling arm the boat would heel less at the same side force.

At 90 TWA the sails (no spinnaker, ~7.5 knots BSP) work at L/D of 2.8 and total force of 3000 N. The apparent wind is about 6.2 m/s. Now the rotor needs to have a Cl of almost 9, which needs about 3.5 speed ratio (~130 RPM). With 2.5x endplate Cl could go to about 11 and at 9 the L/D would still be ~4.5 so it should outperform the sails. However having a more reasonable endplate of say 1.5*D, the maximum Cl is only ~6 and L/D lower than the sails, thus it would not be adequate.

At 150 TWA the sails (spinnaker, ~6.2 knots) work at L/D 1.3 and total force of 680 N. The apparent wind is about 2.6 m/s. Now Cl needs to be 11, this 2.5x endplate rotor could just match. However at such a low AWS the vertical wind profile is crucial and the actual Cl needs to be above 12. So sails would outperform this rotor.

To make the comparison fair, power needed to rotate should be included as well. How much power would it take? How much faster would the boat be with sails + equal motor assistance?

High AR rotors seem to be much more efficient (higher maximum Cl and L/D), but not seen in use. The ones used by Flettner, Norsepower etc. seem to have AR around 5 and have rather small endplates, thus quite low L/D. I guess the mechanical problems become too bad at high AR, which also leads to much higher RPM, especially when high Cl is targeted. Also high L/D would be a problem when AWA is high and lift produces mostly side force.

If e.g. AR 13 with endplates is used from this paper, a Cl of ~15 could be used at L/D of ~8. Then the beat case would need only ~6 m^2 (0.7 x 9.1 m at 520 rpm). So about the same height, but smaller diameter and much higher RPM. Should easily outperform the sails with much better L/D. But at other points of sail it could not produce as much force. In order to have the same performance even at 150 TWA, it needs to have about 11 m^s area thus 0.9 * 11.7. Due to bigger diameter the RPM would be more reasonable.

I just slept in my boat during heavy winds in the marina. The boat heeled quite a lot during each gust and mooring lines had a lot of tension. What would happen if I replaced my 0.11 x 0.21 m mast section with a Flettner rotor? Although it would be 3-6 m lower, it would have 7-8 times the frontal area or 3.5-4.5 times the side area of my current mast + endplate. Shrouds and stays have also some frontal area, but rather minimal compared to the rotor. Shrouds are 5-7.1 mm rods.

And the 2.5*D endplate would have about the same area as the rotor. With some initial heel it would catch wind as well and have a high heeling arm.

Why are high AR rotors not in use? Are there some real data of sailboats with Flettner rotors compared to modern effective rigs? Buckaus's original rig was really inefficient compared to modern sail plans.

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