Everything Old is new again - Flettner Rotor Ship is launched

See also Post #43 with the math equations in the attachment

 

thank you very much, Jeremy and rwatson! I will read your files very carefully and redo your calculations.
Pierre-Yves

 

Dear rwatson,
I don't find the attachment, in your post 91: <<you will find that the "tacking angle" is extremely good. Figure 53 in the attachment "performance02.jpg" shows that downwind performance is *not* the same as upwind performance. This is based on actual results from the original Flettner Ships.
yes, for optimum performance, you may "tack" downwind, and this is what 80% of high speed yachts (particularly multihulls) do when racing.>>

Thank you!!!
Pierre-Yves

 

The attachments are in Post #43


Post 43 "Wow Dennis, thank you so verrrrry much for that link." Rwatson


Heres another useful example

Post 105 "Rotor geometry - it's a cylinder" EuroCanal

 

It is OK for L formulae with the different values of V and s, but not for D!

By example, for 100 kph, I find:
D = (rho . V^2 . d / 4) . 5 = (1.224 . 27.78^2 . 2 / 4) . 5 = 2361 N, instead of 3,717 N

I'm searching the explanation...I have found (http://www.ehow.com/how_7991933_lift-drag-cylinder.html):
D = rho . V^2 . d . (drag coefficient of the cylinder) / 2 (per unit length)
and the comment: "For a long cylinder, the drag coefficient is about 0.87) but not the way to obtain the drag coefficient!!

Pierre-Yves

 

Why do you think the Drag formulae is wrong ?

 

I found that the FoilSim data didn't seem that good.

If you look at the spreadsheet model I put together, which was verified against real test data from two sets of wind tunnel tests, then neither the wind tunnel test data or the simple model I put together match up well with the FoilSim predictions.

The drag calculation for a stationary cylinder is straight forward, just 0.5 x rho x Cd x A x V²

where:

rho = fluid density in kg/m³
Cd = drag coefficient for the shape of the object
A is the projected area of the object in m²
V is the relative velocity of the fluid past the object

 

thank you, rwatson and Jeremy!

applying the drag formulae of EuroCanal, my values are not the same, see post 105. The lift values are right. We have:

EuroCanal: D = 0.25 rho A V^2
Jeremy: D = 0.50 rho Cd A V^2

How to compute the drag coefficient, Cd? I have only Cd is about 0.87 "for a long cylinder"!

 

Like I said before, the formulae is in the attachments to Post #43

I attach them again again

 

Cd of a long cylinder (i.e. a cylinder that is at least 5 times longer than its diameter) will be between 0.8 and 0.9, depending on Re. If you assume that Re is fairly high (above the laminar flow velocity relative to the cylinder diameter) which is almost certainly likely to be the case, then I would use a figure of around 0.85 as a fairly good approximation.

I note that the copied document above assumes a Cd of 1, which is too high. I did a lot of work on determining the drag coefficient of various bodies many years ago and concluded that Hoerner had it pretty much spot on in his book Fluid Dynamic Drag (now out of print, I believe). He gives a figure of around 0.82 for the Cd of a long cylinder IIRC.

Here's a worked example using a Cd of 0.85 for a cylinder of length 5 metres and diameter 0.5m (so projected area of 2.5m²) at a relative air flow velocity of 5m/S that may help:

Drag = 0.5 x 1.226 x 0.85 x 2.5 x 5² = 32.57 Newtons

 

I had already read these pages. I find it hard to read the equations and calcs02 calcs03.jpg. Perhaps with a higher resolution?

Thank you!!!

P.Y.

 

I'll try this value of Cd. Thank you!

P.Y.

 

Emden

Hi RWatson,

After a few times turning in the wrong direction, having asked lots of people, who just glared at me, we suddenly found the ship berthed in a small harbor. The gates were closed, the fencing was high and the building was deserted. I walked over to the other side and asked somebody for information. All what he could tell me was that the ship is already for over a year there and was being modified.
Pick your own scenario.
1) Scenario 1. Due to the slump in the windmill business, it was cheaper to use external transporters and negotiate the best price.
2) Scenario 2. The ship has mechanical design problems, and it has to be mechanical modified. However, I feel the ship would then be lying in a dry -dock.
3) Scenario 3. The ship has software and electrical design problems, and there is a dispute between owner and contractor and it will take more time to come up with solutions. Sometimes software problems take months and months to solve.
4) Scenario 4. The owners have a legal dispute with contractors, which could take years.

I will still follow this up with some phone calls.
Bert

 

Great detective work and an interesting saga, Bert!
(IMO, Scenario 5: They have concluded that Flettner rotors are not efficient
enough to justify the effort of putting them on ships but are too
embarrassed to admit it.)

 

Indeed, well done Bert - closer to the mystery. Thanks for keeping us in the 'loop'