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Calculating Blade Loads
By paradigmdesign, Section Wind Posted on Wed Mar 22, 2006 at 08:27:47 PM MST
Does anyone know of a formula?

I was wondering if anyone here knew of a rough formula for calculating the linear loads that will be exerted on a wind turbine blade. I know the way to calculate the amount of wind in a given area, but figuring out how that will load on my blades has been another chalenge.
I am in the process of building some 9 foot composite blades with a TSR of 7, and was going to do some load testing on them before I installed them on a turbine. I saw that a 20k blade, 12+ feet was tested with intervals 100,000, varying from 250-350lbs. The loading was done only a few inches from the tip. I know it can not be an exact nunber because you don't know the profile of the blade, airfoils being used etc, but if anyone can help, it would be greatly appreciated.

Calculating Blade Loads | 16 comments (16 topical)

Re: Calculating Blade Loads (3.00 / 0) (#1)
by ivandenisovich on Wed Mar 22, 2006 at 03:01:00 PM MST

Forces you might find on a wind turbine rotor are basically three different types:
Lateral pushing back on the broad side of the rotor facing the wind, centripetal outward from the center of the rotor toward the tips of the blades and what some refer to as gyroscopic, beter known as gyroscopic inertia or precession.

Lateral forces are fairly easy. Since energy is conserved, it would depend on the rotor efficiency. If your rotor is say 30 percent efficent, then it would be 30 percent of the power available in the area of your rotor. Since most formulas are given in power or watts, you will have to convert that to an appropriate force unit such as Newtons or pound feet.

Centripetal force depends on the weight of the blade material and the speed of
rotation. The formula for that is Fc= (2pi/t) x r where t is the period in seconds (RPM/60 for RPS then 1/RPS for period) and r is the radius.

Precession is much harder to explain. This will be more of a twisting force on the blade. It results from intertia and the greater the RPM of the blade, the more precession it will have.

I don't know if that helps.
Paul, I'd rather spend my money on anything but oil.

Re: Calculating Blade Loads (3.00 / 0) (#2)
by paradigmdesign on Wed Mar 22, 2006 at 03:54:51 PM MST

I guess what I was asking for was the lateral loads. I got that far too, but there is different amounts of force in each part of the area i.e. more power comes from the outside of the radius than the inside. So figuring out what kind of a load is applied where on the blade is the question.
"Don't worry about tomorrow, plan for it."

Re: Calculating Blade Loads (3.00 / 0) (#3)
by windstuffnow on Wed Mar 22, 2006 at 04:50:03 PM MST

You can find the force acting on the blades in flight using the formula;

.00492 x area x windspeed^2 = force in lbs

area is in square feet
windspeed in mph

Area = dia^2 x .7854 ( dia being in feet)

Hope that helps...

.
Have Fun! Windstuff Ed

Re: Calculating Blade Loads (3.00 / 0) (#4)
by willib on Wed Mar 22, 2006 at 05:14:11 PM MST

Ed , i'll be damned , but that works ..the swept area formula..
why not use the conventional formula for area of a circle?
pi is just a key on your calculator
area = pi* r^2 , r being the radius in feet..

Carpe Ventum (seize the wind)
[ Parent ]

Re: Calculating Blade Loads (3.00 / 0) (#5)
by windstuffnow on Wed Mar 22, 2006 at 07:16:01 PM MST

It always seemed simpler for me, I usually know the diameter of the project, if it was an odd size I had to figure out the radius first... seemed like an extra step that wasn't necessary. Besides, Being an old gear head once into race engines the cid formula was a natural...

Bore^2 x stroke x .7854 x #cylinders = cu in

A minor modification of the formula and you have the sq ft of any circle...

.
Have Fun! Windstuff Ed
[ Parent ]

Re: Calculating Blade Loads (3.00 / 0) (#6)
by willib on Wed Mar 22, 2006 at 07:28:51 PM MST

how is your flywheel project comming along.??
have you seen my flywheel , in my diaries ?

Carpe Ventum (seize the wind)
[ Parent ]

Re: Calculating Blade Loads (3.00 / 0) (#7)
by windstuffnow on Wed Mar 22, 2006 at 09:01:32 PM MST

I'm still working on a drive system for it. I did come up with another smaller one since then though. A friend had an old bailer that he was going to scrap out and it had a beautiful 22 inch 225 lb flywheel on it... needless to say I scarfed that up in a heartbeat.

I went through your files but... you have a lot of pictures in there ! You'll have to point me to the proper section to see it.

.
Have Fun! Windstuff Ed
[ Parent ]

Re: Calculating Blade Loads (3.00 / 0) (#8)
by willib on Wed Mar 22, 2006 at 09:51:11 PM MST

there ya go..

http://www.fieldlines.com/comments/2006/3/22/1221/06941/6#6

Carpe Ventum (seize the wind)
[ Parent ]

Re: Calculating Blade Loads (3.00 / 0) (#9)
by windstuffnow on Thu Mar 23, 2006 at 08:11:31 AM MST

Sweet !!! What does it weigh?
Have Fun! Windstuff Ed
[ Parent ]

Re: Calculating Blade Loads (3.00 / 0) (#12)
by willib on Thu Mar 23, 2006 at 10:28:13 AM MST

dont know ...real heavy for it size..
The outer rim is a hair under 7/8" thick x 1.7" by 16" dia..
all i have a 0-20 newton scale , and with a lever i migh be able to weigh it?

Carpe Ventum (seize the wind)
[ Parent ]

Re: Calculating Blade Loads (3.00 / 0) (#14)
by apie on Sat Mar 25, 2006 at 09:50:05 AM MST

Ed, I was trying to put this formula into metric units and I get:

Force (Newton) = 0.0070239 * area (m^2) * velocity (meter/sec) ^2

If I calculate this with a swept area of 1 m^2 and a wind speed of 30 m/s (pretty fast!), I only get 6,32 Newton. So there must be something wrong!

I am sure that a 1 m^2 wind-turbine has something in the order of hundreds (maybe thousands?) of newtons on it in a 30 m/s wind.

Can someone find the error?

BTW: where does this formula come from? That could also clarify a lot!

[ Parent ]

Re: Calculating Blade Loads (3.00 / 0) (#15)
by windstuffnow on Sat Mar 25, 2006 at 04:29:19 PM MST

I'm pretty ignorant when it comes to converting to metric. When someone says "it's 25 mm long" it takes me 4 books and a calculator to figure out its about an inch long.

Hugh's book defines the force using the following formula;

Diameter^2 x Velocity^2 / 24 = Kg

That really didn't seem very accurate but it would tend to make you build the tower and footings stronger than you needed.

Diameter^2 would make a 3meter blade have an area of around 9sq meter when actually it would be around 7.06 sq meter.

Using the Pi x R^2 would be a better approach for an accurate account or as I said before Diameter^2 x .7854 either way.

So I guess I would use Dia^2 x .7854 x Velocity^2 / 24 and convert Kg to Newtons although it seems Kg would be easier to work with than Newtons.

Hope my ramblings helped a bit...
.
Have Fun! Windstuff Ed
[ Parent ]

Re: Calculating Blade Loads (3.00 / 0) (#16)
by apie on Mon Mar 27, 2006 at 07:32:19 AM MST

Thanks for the reply! Yesterday I met someone who had the "feynman lectures" (THE
books of physics) and I found something usefull that I converted into this:

the amount of force applied to a sheet facing the wind is equal to 2 times the mass of air that bounces off it times the velocity of this air.

F = 2 * mass/second * velocity

the factor two is because the velocity of the air is turned around. So it is stopped and then pushed away.

The mass/second is equal to rho * area * velocity.

So that means:

F = 2 * rho * area * velocity^2

where rho is about 1.22 kg/m^3

If I calculate this force, I get higher results than with your formula, but then I think it's better to make the thing stronger...

[ Parent ]

Re: Calculating Blade Loads (3.00 / 0) (#10)
by IntegEner on Thu Mar 23, 2006 at 09:13:03 AM MST

Sounds like something similar to a project being proposed for funding here, with the blade length being 7 feet. I think your figures mean 20 kilowatts (for "k") and 100,000 psi stress load intervals. Yes, some good information is provided here by others and the lateral, or "bend back" force, is the force with which most is to be reckoned. It is proportional, as stated, to the power being delivered and the power being delivered is proportional to the blade rotation rate up to a point. This can put it into a high range if the blade, for example, runs away. Believe it or not, blades still break and such has occurred even here on the big turbines. Send me any updates on this project and we would be glad to respond with anything we have to offer.

Anthony C.
www.integener.com

Re: Calculating Blade Loads (3.00 / 0) (#11)
by paradigmdesign on Thu Mar 23, 2006 at 09:19:41 AM MST

I will be posting a soon a something is completed. Thanks all, for your help, I will let you know how it goes.
"Don't worry about tomorrow, plan for it."
[ Parent ]

Re: Calculating Blade Loads (3.00 / 0) (#13)
by SparWeb on Fri Mar 24, 2006 at 01:13:27 PM MST

Paradigmdesign,

Problems of internal loads and stresses are at the level of the engineering undergraduate. If you are looking for an analytical way of predicting loads and stresses, I suggest you pick up a few engineering textbooks (used) and you will find formulas and examples to guide you. If you are good at math, you will be able to calculate the integrals that are required for detailed stress analysis of a blade.

Furling and weather-cocking are problems of gyroscopic precession, so look for that in whatever books you get. Those are the critical loads in a HAWT.
Steven Fahey
[ Parent ]

Calculating Blade Loads | 16 comments (16 topical)

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