Dual rotor / tiltup furl

[Stewart Corman]

This is a new thread for discussion/info on my redesign of my first turbine assembly.

As a refresher, the original design was displayed with a single rotor, three 10'8" NACA 4425 wooden blades mounted on spars to allow setting pitch angle, 24v DCmotor/generator and had no furling mechanism.

Rotor was placed too far in front of yaw pivot, which caused instability, while tail beam was too whimpy and exhibited wobble.

http://www.greenpowertalk.org/showthread.php?p=2453#post2453

Never got good data, since the leaves were on the trees and tower was only 27 feet high (below treeline and lots of turbulence). Did get it to about 200 rpm in 13mph WS.

New tiltover tower ala Mario's, will be discussed in another thread and is expected to be double the height to around 55 feet

New head design uses similar square tube frame, but a 3/4" pivoting billet to allow tilt up of 45 degrees. A three phase PM 460v/3000 rpm servo motor/generator is installed on uprights, while a chain drive was chosen for an 8.4:1 drive ratio ...at 360 rpm, it will deliver the 460v rating.
With dual rotor and 6 blades total, I expect a TSR of around 5 and a nominal rpm of about 200rpm at 15mph producing about 250volts/1.0KW

first photo shows dual rotors w/chain sprocket, sample spars, and one blade mounted
note that rotors get offset to create jib/mailsail effect

note also that there is a second set of mounting holes to permit all 6 blades to be mounted on single rotor for comparison data

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next photo shows servo mounted above 1 1/2" shaft on centerline
servo has 10 tooth sprocket, mounting plate is 84 tooth sprocket

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next is pivoting head assembly for tiltup furling
aluminum blocks provide bearing point for galv 3/4" shaft
polycarbonate discs provide side clearance
the unpainted slotted member is the 10 foot tail beam

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another view of the shaft vs pivot ..there is a 1.5" offset which provides a lever arm for the 90 pounds of thrust calculated at 20mph WS which is the desired tilt setpoint

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an adjustable mechanical stop is provided to set the normal tilt of the head w/r to the tower ( typically 5 degress off vertical)

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I will continue this thread in the next page

Stew Corman from sunny Endicott

[Stewart Corman]

The total weight of the head assembly incl 6 blades is about 200 pounds.
Tail beam is 10 feet long from pivot point and total tail assembly weighs 30 pounds ..
this is the counterbalance weight for the heavy servo + rotors

entire assembly shown w/o tail visible

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nameplate of servo shows 3.9 amp max, but it is rated at 460v ,
so that equates to 1800watts per phase x 3 = 5KW capacity ...
I will use that as safety factor, running at nominal 1.0KW at 15mph or about 1.3amps/250v
This dictates the requirement of a 62ohm power resistor per leg x 3

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coil windings are 7 ohms, so fairly large resistors will be employed and heat losses in servo will be minimal ie 11%




here is the entire head assembly tilted up about 30 degrees

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I will show how the pivot position can be measured with a digital scale for
determination of furl WS setpoint

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the digital scale is showing 4lb 13.6 oz at the end of the frame which is 17.5 inches from the pivot point ...this equates to a furl wind speed of 15.5 mph whereby the horizontal wind force calculates to be 56lbs ...in order to have it furl at 20mph, the digital scale would have to show 7.9 lbs which equates to 92lbs wind force ..the aluminum pivot blocks have bolt slots which allow +/- 1/4" travel

in the field, a fish scale can measure the force at the end of the frame with everything connected


here is another view of the pivot assembly when tilted up

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note the additional 2 bolts to stabilize the tail boom ...looked to whimpy before ;)


Stew

[Mark Parsons]

Greetings Stew,

This design looks much more robust than your version 1. Looks real good to me. Great photos.

How long is your tail? Looks like you are tilting the whole fuselage assembly? Will that give you enough tilt angle to furl safely?

Best regards,
Mark

[Stewart Corman]

Mark,

did you catch where the pivot point is located?
the entire motor, shaft assembly/bearings and rotors are forward of the pivot

thought you'd like the rigid mounting of the servo ...bottom tapped (4) 3/8" bolts 5/8 " into the aluminum housing on both sides ..
used aluminum blocks to space it out to frame ..the larger servo will fit in nicely at a later time

The 8 sqft sheetmetal tail is 10 feet from the pivot to the outer extremity ..I used (2) 5/8" square steel tubing uprights to hopefully keep it from wobbling (pics later)

reinforcing the tail attachment as we speak ...need two more bolts at beam/frame

those nylock nuts are very thick on rotor, will be cutting them down on lathe to give more clearance from chain to spars by added 1/4" spacer

Note that the front rotor is the original (formerly green), so 5/16" ubolt holes are for the smaller spars, whereby the rear rotor on sprocket is machined for six sets of the heavier 3/8 ubolts/galv 16ga spars. The aluminum tubing (shown only for assembly construction) can slip over to adapt the smaller spars to the larger rear rotor ubolts.
BTW, I will NOT use the 5/16 sheet metal "exhaust clamp" shown, since it creates a stress concentrator, but will reproduce the saddles shown on the rear rotor ( easy to fabricate)

Beefing/fitting up the tower is my next concern ..but winter may delay install until spring

As far as tilt angle ..I was shooting for 45 degrees ..that said I really need to do the experimental model at 45 degrees to see how really inefficient it will run "offwind"....find it hard to believe that TSR>1 and if load is applied, will it stall??

Check out last part of page 2 ..I added some more pics

Stew

[Rob Beckers]

Beautiful work Stew! You're showing how theory is put to practice. Great fabrication work too. I'm looking forward to the the upcoming episodes!

-RoB-

[Stewart Corman]

Looks like you are tilting the whole fuselage assembly? Will that give you enough tilt angle to furl safely?

Well, actually no ...in the process now of flipping the shaft bearings to the underside, then the pivot shaft has to be below that ...this way the square tube frame is on top of pivot instead of underneath, and the whole assembly can then tilt to about 60 degrees rather than 45 (which I wasn't able to attain anyway)...the vertical supports for the servo will be shortened about 6 inches to keep the chain same size ..will make it more solid as well ..the risers on bottom are gone too

I hadn't considered this layout originally, because I had presumed that with all that weight in the rotors and servo, that the pivot point would be centered under the servo ...it is far back enough the frame won't get in the way of the shaft/bearings

new pics will follow shortly

Stew

[Mark Parsons]

Hi Stew,

Great idea on flipping the fuselage above the pivot point to gain more tilt angle and flipping the main shaft below the fuselage tubing allowing shorter servo mounts.

Q. You show measuring the head bias weight at the end of the frame. Shouldn't it be measured at the blade hub? This is where the force is located.

Regards,
Mark

[Stewart Corman]

Q. You show measuring the head bias weight at the end of the frame. Shouldn't it be measured at the blade hub? This is where the force is located.Actually ...NO!

you are correct that if a weight (force) is APPLIED at a longer lever arm,
it will exert more force to the side opposite the pivot fulcrum ..that is basic leverage effect

here however, we have a fixed weight of several components who have a center of mass (somewhere along the frame?) and can be replaced by a single weight vector...but that weight is fixed in space (constant lever arm) w/r to the pivot point so torque is constant (as measured in ft-lbs)

We will presume that the frame is a rigid body ie doesn't bend w/r to the weight applied, then the relationship is simply the weight recorded on the digital scale multiplied by the lever arm at the scale to derive the ft-lbs.

the digital scale is showing 4lb 13.6 oz at the end of the frame which is 17.5 inches from the pivot point If I had measured at the outer rotor at 24 inches, it would measure 3.74lbs and if I moved the scale inward to 12 inches, it would weigh 7.48 lbs

basically, you can measure anywhere along the frame as long as you note the lever arm at the scale ...this makes field measurement with a fish weighing scale very easy

a very simple experiment can confirm this for yourself ...lean a 2 ft piece of angle iron on a short piece upside down as a fulcrum ...weigh at the end and then way half the distance to the pivot ...you'll see 2x the weight closer in
OR in other words ...if you had a six foot piece of angle iron, it would take a heck of a lot less force to lift it off the ground at the end vs only one inch from the pivot

Stew

[Mark Parsons]

Hi Stew,

Thanks for the clarification that you need 7.5 ft-lbs of torque to offset to furl your machine.

This might be a good spot to spell out the formula you are using to calculate torque from wind speed and turbine swept area.

Regards,
Mark

[Stewart Corman]

to calculate how much force a turbine exerts on the tower at any given wind speed had found this on otherpower:

P=(1/2 Cd p A v^2 Fdl) / Gc

where:
P= Pounds of force on the rotor
Cd= Drag Coefficient = 1.57
p= air density = 0.076 Lbm/ft^3
A= Area swept by blade
v= wind velocity in ft/sec
Fdl= Dynamic load factor = 1.4
Gc= sec^2/Lbm-ft (coefficient of gravity?) = 32.2

The equation boils down to:
P= 0.0026 X (Area) X (wind velocity ^2)



so for a 18footer @ 20mph :
P= 0.0026 x 254sqft x 20mph x 20 mph= 264 lbs force
(this number was used to calculate stress in spar)




BTW, DON'T get confused ....in the previous posts where I showed 4lbs 13.6 oz as a measurement to get 7.48 ft-lbs torque ..that was NOT a set point for 20mph furl for 18 footer..it was just an example for the 10 footer I am assembling .. the 10'8" turbine has an area of only 89 sqft so the trust is now 92 lbs ..now multiply that by the 1 1/2" lever arm to get 11.5 ft-lbs of arm torque ...at 17 1/2", I need to get 7.9 lbs offset weight, not 4+lbs which would furl at a much lower wind speed



if you take the above 264lbs thrust for 18footer and couple it to a 1 1/2" lever arm, you get 33 ft-lbs arm torque ..most of that weight is contributed by the additional weight of the much heavier blades, which then requires to reset the balance point +/- from the current 17.5" setting

hope we are not getting too bogged down by bunch of numbers ...point is, we assemble all the hardware and determine where it balances, then move the balance point towards the tail, so that there is extra weight on the forward frame stop ..the amount of weight desired is a function of what turbine diameter (swept area) and what desired furl wind speed.



KISS



Stew

[Mark Parsons]

Hi Stew,

Small correction. Wind Speed in ft/sec, not MPH. 20MPH = 29.3 ft/sec

so for a 18footer @ 20mph :
P= 0.0026 x 254sqft x 20mph x 20 mph= 264 lbs force
(this number was used to calculate stress in spar)

Should be = 0.0026 x 254 ft^2 x 29.3 ft/s x 29.3 ft/s = 568 pounds force.

A not insignificant amount...

Regards,
Mark

[Stewart Corman]

Mark,
Good eyesight!!

I see where you found an error in my calc, but I am not so sure I quoted the right source for the equation.

Also note the very high drag coefficient?

just I found this equation in metric:
P (kg) = D^2 x WS^2 / 24

http://www.thebackshed.com/Windmill/Docs/Furling.asp

so, where 18' = 5.5m 20mph = 8.94m/s

so P = 5.5^2 x 8.94^2 / 24 = 100 kg = 220 lbs

now I feel better ..that is BETTER than the 264lbs I was using

why not find another source that get's it down to say ...95lbs??? <grin>

Seriously tho ..I think I got the equation from another source that I can't identify at this time and it did have 264 lbs as the answer

I will followup to this thread

Stew

[Rob Beckers]

Here is another way to calculate horizontal turbine thrust. This comes from a wind load calculation done by a structural engineer. I'm an electrical one, so I won't comment on the validity of it. The assumption is that maximum wind load is at the wind speed just before furling (which may or may not be valid, depending very much on how it furls and what the survival wind speed is that you shoot for):

Rotor diameter, D = 18 ft = 5.49 m

Swept area, Ar = Pi * D / 4 = 23.66 m^2

Air density (sea level), Rho = 1.225 kg/m^3

Wind speed, V1 = 20 mph = 8.94 m/s

Wind speed downstream of rotor, assuming optimal efficiency, and overestimating thrust, V2 = V1 / 3 = 2.98 m/s

Rotor thrust, Tr = 0.5 * Rho * Ar * (V1^2 - V2^2) = 1030 N

Factoring in wake-induced thrust, Tt = Tr * 1.3 = 1338 N

Convert to pounds, Tt = 1338 * 0.225 = 301 lbs

For what it's worth....

-RoB-

[Stewart Corman]

Rob,
can you please post the link from which this last analysis came from.
The number seems in the ballpark.

I am looking over the specfics and if my perspective is correct, they are taking the total power in the wind for a swept area at the furl wind speed, presuming 2/3 gets absorbed by the turbine and using that for thrust ..

not exactly sure how the wake induced thrust w/ factor of 1.3x adds to the tower load??

However, in the first case I don't see where the removing of energy by the rotation and transfer of mechanical to electrical is accomodated, since that energy is derived in a vector which is rotating in the rotor plane and not directed towards the tower .

(BTW, Still haven't found that missing link I have been looking for with the orig equation I was using)

Here is a link with nice pics and equations :
http://www.eng.fsu.edu/~kroth/eml4450fall04/eml4450L21.pdf

Stew

[Rob Beckers]

Hi Stew,

The analysis comes from an engineering report on rotor thrust for the analysis of a tower. I don't have a source for the equations, they're what the engineer used. I have of course substituted your rotor and wind numbers in them.

Like you I had figured out that what it represents is the force needed to remove 2/3 of the kinetic energy from the air (in essence an approximation of the Betz limit), that's about as far as my knowledge goes I fear. Why there's an additional factor of 1.3x I don't know. I trust this engineer though, he's very knowledgeable when it comes to wind turbines. Keep in mind that for the intended purpose it was not about an exact thrust number, but an approximation that errs on the side of safety so the tower is sure to handle the load. In fact, the tower is designed to handle multiple times the calculated thrust load.

-RoB-

[Stewart Corman]

OK, Rob and Mark,

This total thrust on the tower and then dividings by number of blades to calculate stress on a spar has really been getting to me

I'm game

what is wrong with this math:

KE = 1/2 m v^2
KE = 1/2 rho x area x WS^2
= 1/2 x 1.225 kg/m^3 x 23.66 m^2 x 8.94 m/s ^2 = 1158 kg-m/s^2 = 1158n

multiply by 1n =0.225 lbs and we get 1185 x 0.225 = 260lbs

THINK ABOUT IT ..this is the total force of the wind hitting a solid disc 18 feet in diameter

Do we have the makings of a perpetual motion machine, since one equation gives us a thrust of 568lbs on the tower and Rob's tower expert gets 310lbs ??

I'll post on Otherpower to see if anyone has better links to thrust equation

Stew