Wind pressure on tower - how calculated

[domwild]

Scenario:

By the looks of it my PVC prop will never overspeed as it stalls at some max. RPM and the experts on this board have suggested that this is a feature of PVC pipe props.

The question is:

1. If overspeeding is no problem is the bending moment on the tower the deciding factor if I need furling or not to reduce the swept area?
   
2. Can I easily calculate the bending moment by using the swept area given by the dia. of the blades, the height of the tower and max. wind speed?
   
3. I take it the force on the tower is reduced by Betz's limit or not??
   
   

Pls do not answer "You need furling" without explaning your reason(s) as overspeeding is not possible but a strong wind may topple the tower.

Thanks.

[RP]

You need furling.

:-)

Seriously, how big is your turbine? If it's 8" in diameter, then I wouldn't worry about it.  If it's 2 feet in diameter then you may not need to worry about it.  If it's 6 feet in diameter then you need to worry about it.

Even if the blades will not speed up past a max rpm, they will be pushed further and further back into the tower leg possibly hitting it in a gust.

Others here can describe the wind loading on the tower part better than I.

[Flux]

You need furling.

Without furling your prop will have a thrust not much lower than a solid disc. The wind loading will increase with wind speed squared and will require a monster of a tower to survive long enough to see your pvc prop disintegrate spectacularly ( which it will in a good storm)

With furling the tower loading from the prop stays fairly constant, there is still an increasing force from the area of the tower itself but that is small compared with the prop thrust.

Anything greater than about 3ft diameter needs some form of furling or protection if it is ever to see a real storm.

The tower loading on the little Marlec is very high in high winds and that is only 3 ft.

Flux

[Slingshot]

Is this really true?  It seems that they would have to be quite different, as the "solid disc" absorbs all the forward momentum of the wind, while the prop only slows the wind down.

Flux wrote: "Without furling your prop will have a thrust not much lower than a solid disc."

[Flux]

Yes you are right, but for tower loading I would take the solid disc value, there is no point in taking risks. If the prop only gives you half the loading of a disc then you have a safety factor.

The actual loading seems to depend on the power absorbed from the wind by the prop and I have found that typically the thrust is a bit over half the disc value.

The main point is that without furling this force continues to rise rapidly even if it is less than the ideal disc value. With furling the thing effectively settles at constant force.

Flux

[SparWeb]

I wonder if this is where some of the concern about furling comes from?   On the one hand, we talk about the load on a solid disk area to design the tower, which really high-balls the number.  On the other hand, we talk about inches of offset and pounds of ballast on the tail to design the furling system.  

To put it succinctly, you OVERESTIMATE the thrust load to make the tower strong,

and you UNDERESTIMATE the thrust to prevent overspeeding before the mill furls itself.

[Flux]

Interesting point. Some of this has come up in discussion about furling on a 10ft machine elsewhere recently.

I think it is reasonable to take the disc loading when considering towers as there other loads than the prop thrust that we really don't take into consideration.

Furling is a much more difficult issue. The actual thrust on the prop is very different from the force that is available to do the furling.

If there was no seeking force from a prop then most of these things would furl very early and it would be possible to use any value of offset and alter the tail weight to suit.

An example I quoted in the other discussion was for a 10ft machine of mine. The thrust at 25 mph is about 90lbs. Without furling it would rise to about 9 times that value at 75mph. With furling it will rise very little.

In my case the seeking force of the prop reduces the effective thrust from 90 to about 67 lbs and the tail moment has to support a torque of about 67 x 5/12 or about 28lb-ft.

Somewhere Dan claims that it needs only 25lb thrust to furl his 10ft machine at 1kW so an even larger seeking torque is present.

We are looking at a thrust of 90lb when only 25lb is available to actually do the furling.

Changes to the prop or alternator loading can make a drastic change to this seeking torque and so it is not surprising that some of these things furl perfectly safely and others furl at speeds likely to burn the alternator.

Anyone who does calculations based on thrust without considering the seeking force will furl at a very high wind speed.

Flux

[domwild]

Thanks for the many answers. The prop has a dia. of 3m or about 10' as the Fisher and Pykel suffers from cogging.

So furling will be implemented!

As an additional safeguard I intend to lift the yaw plate a few degrees from the horizontal to ensure any storm-bent prop does not hit the mast as the distance from the mast will then be greater.

Regards,

[Spelljammer]

Furling definitely.

Here is a table for disk loading.

[DanB]

I think this chart keeps popping up - I wish it would go away.  Unless I'm wrong - it's off by a factor of more than two - the wind load on a solid disk is much less than this chart suggestions.

The formula P=.00246AV^2 is correct but V needs to be in miles per hour - not feet per second.  Whoever made this chart got that wrong I think.  I figure the thrust on a 10' solid disk in a 20mph wind to be about 80 pounds (not 176) - and keep in mind the thrust against the blades will be much less than a solid disk.  It's not nearly as scary as some people would think - although without furling things do get out of hand quickly.  I guess if you go by it...  your tower should never fall down!

[Flux]

Dan I agree.

I have never found a figure for air density in imperial units that I can understand so I have to work in metric.

Converting back I agree that your 80 lb is pretty near the mark.

I can't accept equations that use numbers that I don't recognise. If someone has derived them and they are correct then fine but I would rather get there from figures that I understand. I was brought up on imperial units but for wind power metric seems to make much more sense.

If you use air density of 1.1 ( or 1.2 ) kg/M^3 and you use area in M^2 and wind speed in M/s then the answer comes out in Newtons, without any unknown fiddle factors.

Similarly if you use the metric units to calculate power with the correct equation the answer comes directly in watts.

Use these strange formulae as long as you are sure they are correct and you know what units to put into them.

Flux

[thefinis]

Not sure what formula is right as some times the Cd ratio is included and other times it is not. Here is a quote and a link to a page that discusses it and presents the formulas for different wind loading on flats rounds etc on an antenna tower.

http://www.arraysolutions.com/Products/windloads.htm

The Generic Formula

For using the actual sustained wind speed expected (were we to actually determine it) :

Force, F = A x P x Cd

A = The projected area of the item

P , Wind pressure (Psf), = .00256 x V^2  (V= wind speed in Mph)

Cd , Drag coefficient,  = 2.0 for flat plates. For a long cylinder (like most antenna tubes), Cd = 1.2.

Note the relationship between them is 1.2/2 = .6, not quite 2/3.

Hope this next link muddies the water even more. It seems there are two formulas floating out there and many use this next formula which is from Bernoulli.

http://mohandes.net/eng/lds/index.php?pg=wloads

The cart below taken from a link on the above website would seem to go with DanB and gives about half of the psf of the chart posted in an earlier post.

Wind Stagnation Pressure (qs) at Standard Height of 33 Feet

Basic Wind Speed (mph)1 (X 1.61 for km/hr)     70     80     90     100     110     120     130

Pressure qs (psf) (X 0.0479 for kN/m2)     12.6     16.4     20.8     25.6     31.0     36.9     43.3

Does the main formula from Bernoulli take into account the drag (reduced air pressure)seen behind the object or just the air pressure on the front?

Finis