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torque / stress calcs
By scorman, Section Wind Posted on Thu Dec 13, 2007 at 01:14:19 PM MST
throwing some numbers around

In another thread, someone asked if tubing spars were sufficiently strong
to withstand the force of the wind on the blades and admittedly,
I hadn't done any calcs on this and just went by gut feel.
Please feel free to poke holes in any assumptions/equations/conversions I have used.
Here are some specifics and presumptions using some worst case scenarios:
turbine diameter 18 feet
three blades @ TSR=6
furl at 20mph
steel spar tubular dimensions : 1.25"OD, 0.065" wall thickness
soft steel because (hot dipped ?) galvanized
I will now work backwards to get the applied loads:
at 20mph, 100% wind energy = 10KW, so 40% eff = 4KW
for TSR = 6 , rotation @ 20mph = 315rpm
power conversion: 4KW/(746w/hp) = 5.4 hp
torque= hp x 5252 / rpm = 5.4 x 5252 / 315 = 90ft-lbs
since 3 blades, 30ft-lbs exerted by each blade
modulus of elasticity of mild steel is 29,000 Kpsi
yield strength steel = 35,000 psi
from the tubing geometry:
section modulus Z = .068 in^3
maximum stress for cantelevered beam is :
s = weight x length / Z where W x l = torque???
if so, then
s = 30ft-lbs x 12in/ft / .068 in^3 = 5,300 psi
if above are correct, then this tubing is at 15% of the 35,000 yield
How do you apply safety factors??
I have little concern for the first set of 10 foot blades
which obviously have much less power associated with them
note that real tubing is closer to 1.375 diameter (slightly better specs by 23%)
If I choose 6 blades and TSR <= 5,
lowers rotation to 220rpm max but now has more torque,
BUT the force is distributed over 2x more spars (or not exactly?)
I have not accounted for the bending back of the blades towards the tower
( or have I??)
Did I miss anything obvious??
Stew Corman from sunny Endicott

torque / stress calcs | 10 comments (10 topical)

Re: torque / stress calcs (3.00 / 0) (#1)
by Kevin L on Thu Dec 13, 2007 at 07:41:19 AM MST

Impulse loads, (ie wind gusts)placing bending stress on steel spar tubing.
Centrifugal force overloading clamps and placing tension on steel spar tubing.

These acting in tandem where the tubing is stretched to its critical limit coupled with a gust of wind could result in failure. Just some ideas to look at.

Re: torque / stress calcs (3.00 / 0) (#2)
by TomW on Thu Dec 13, 2007 at 08:12:59 AM MST

Stew;

Glad you managed to recovered this.

Cheers.

TomW

The Truth is the Truth, even if no one believes it; and a lie is a lie even if everyone believes it

Re: torque / stress calcs (3.00 / 0) (#3)
by finnsawyer on Thu Dec 13, 2007 at 09:03:43 AM MST

The component of lift that gives the output torque is a small fraction of the total lift. Most of the lift will tend to bend the blades back toward the tower. You should find the bending moment per blade for that force.
GeoM

Re: torque / stress calcs (3.00 / 0) (#4)
by scorman on Thu Dec 13, 2007 at 09:23:10 AM MST

GeoM,

"Most of the lift will tend to bend the blades back toward the tower"
good point ... wasn't thinking along those lines

But we do have some numbers to play with:
I need to incorporate the fact that at 20mph, an 18 foot turbine by calculation exerts 250lbs thrust on the tower.

I can assume a point load in the center of the 6 foot paddle on a 30 inch spar, which would then call for a 5 1/2 foot lever arm or 250lbs/3 blades x 5.5 feet = 460 ft-lbs torque backwards

Or I could presume a distributed load along the length of the 6 foot paddle?

Looks like I have a few more calcs to do

Here is is obvious that 6 blade configuration is of great benefit to distribute the forces over more spars.

OK team ..keep me honest

Stew

[ Parent ]

Re: torque / stress calcs (3.00 / 0) (#5)
by Flux on Thu Dec 13, 2007 at 10:22:52 AM MST

I can't help thinking the biggest force you will face is that due to gyroscopics during a violent yaw. It will be difficult to calculate unless you can make a decent guess at the rotary speed during a sudden gust.

It's that force that causes blades to smack the tower, not the fairly modest and calculable thrust component.

Flux

[ Parent ]

Re: torque / stress calcs (3.00 / 0) (#8)
by finnsawyer on Fri Dec 14, 2007 at 08:26:36 AM MST

I had in the past suggested the possibility of stiffening the blades by placing a ring or rim around them. Since the diameter of the rim can't change, the bending forces on the blades would result in linear stresses along the blades. The rim would also affect the performance of the blades, as it prevents radial airflow off of the tips. The analysis in my latest diary suggests that would be desirable.
GeoM
[ Parent ]

Re: torque / stress calcs (3.00 / 0) (#6)
by harrie on Thu Dec 13, 2007 at 11:54:40 AM MST

Stew. most engineers,do the calulations, than more than double the results if for no other reason than to protect themselves from law suits.

If I were using that method to support the blades, I would use nothing less than solid steel stock, and would also have a metal encloseure completely around the root of the blade instead of having only the thru bolts holding it together. I dont think the wood alone will stand up under the forces.

Re: torque / stress calcs (3.00 / 0) (#7)
by Dave B on Thu Dec 13, 2007 at 12:41:39 PM MST

Hi Stew,
I have to agree with Harrie about your root connection. Just my opinion but you have a very small area of blade attachment at the root as well as another connection of some type at the end of each blade connecting rod besides. There is an incredible amount of total force up there acting on all connections and in all different directions and all wanting to tear things apart. If you have ever tried to bend a 2 x 4 edge wise you can get an idea of what it took to bend 6 glued laminations like this over 11 inches over a span of just over 6'. That is what it took for my blade to hit my tower, it is really quite unbelievable. Just food for thought, Dave B.

[ Parent ]

basic math (3.00 / 0) (#9)
by scorman on Mon Dec 17, 2007 at 11:02:52 AM MST

Someone had pointed out to me that the calculation I used for thrust on the tower was incorrect ..I had used WS in mph rather than ft/s so a 20mph is really 29.3 ft/s

I had been using this equation as a source for my calc:
http://www.fieldlines.com/story/2004/4/26/101410/698

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 , if I plug in the correct WS, I get for a 20ft turbine at 20mph, a thrust of :
T = 0.0026 x 254 ft^2 x 29.3 ft/s x 29.3 ft/s = 568 pounds force

HOWEVER, if I get back to basics:

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 = 260 lbs

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

Can a spinning rotor exert more force on the tower than the wind by itself, AND still produce 4KW energy at the generator??? don't think so!

I found different reference equation in a side furling link at backshed:
http://www.thebackshed.com/Windmill/Docs/Furling.asp

in metric:
P (kg) = D^2 x WS^2 / 24
converting units:
where 18' = 5.5m 20mph = 8.94m/s

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

this is more realistic, BUT, if the turbine is extracting 40% of the wind energy,
why isn't the trust on the tower simply 60% of 260lbs = 156 lbs, not 220??

All the above equations (except KE) make presumptions on how the turbine blades react to the wind and how fast they are spinning. Supposedly lower TSR rotors will have more torque on the shaft and at the same time more thrust on the tower, even though they are more efficient.

If anyone has links to more detailed analysis of thrust on a tower, please post

Stew Corman from sunny Endicott

Re: basic math (3.00 / 0) (#10)
by Dave B on Mon Dec 17, 2007 at 01:47:43 PM MST

Hi Stew,
I'm no math wiz but just something I can relate to. Have you ever done the old trick of spinning a bucket of water around in a circle so it doesn't spill out ? That may have been a gallon or so of water (8 lbs or so). You may have been able to spin it a couple times at around 60 RPM and I doubt very much you thought about turning around while you were doing that. Now think about what just one blade out there even at a light 8 lbs will be doing at say a very modest 2X or 3X (120-200 RPM.) that bucket speed. I realize not all the weight is at the tip but the total force on the hub is incredible even when just spinning with no yaw. The straight on wind force you are calculating is only a very small part of the total forces going on up there. I have seen what it can do and I figured "no way" before building also. Just my opinion but whatever you calculate to be safe, at least double those figures and then chances are it will do something you didn't expect could happen anyway. Furl very early to start ! Good luck. Dave B.

[ Parent ]

torque / stress calcs | 10 comments (10 topical)

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