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drag co-efficient question
By xymox, Section Wind Posted on Mon Sep 05, 2005 at 07:38:23 PM MST
Setting out on a turbine adventure

Hey there, I just bought a round (24) of neo magnets (dual rotor design) and am planning on starting the wind turbine as soon as I get them. I bought a hub assembly from autozone (chevy celebrity), 90 dollars US. I am thinking of a 12 foot prop, and according to my calculations the theoretical limit to the watts it can produce in a 15 mph wind is as follows
(15*15*15)*(area swept = pi * r^2 = 3.14*6^2 = 113.04) * .0052 = 1983.15 watts.
Question, the drag coefficient number (.0052). how is that determined?

drag co-efficient question | 23 comments (23 topical)

Re: drag co-efficient question (3.00 / 0) (#1)
by xymox on Mon Sep 05, 2005 at 01:43:20 PM MST

In case you are wondering why I chose a checy celebrity hub,....well..

I cruised on over to autozone and checked out every hub that they had, I knew I wanted a 5 lug hub and I knew that I wanted something that spun pretty free and was light weight, thus I chose the hub, it also includes a mounting flange so I wont need a spindle, It will bolt right on to the frame(yet to be devised) with the stator holder sandwiched in between.
nough said!
Beat the system, do as much as you can do.

Re: drag co-efficient question (3.00 / 0) (#2)
by Flux on Mon Sep 05, 2005 at 02:25:09 PM MST

I am not sure what drag coefficient has to do with it.

The power in the wind is 1/2(rho)AV^3 where rho is the density of air.

If you use speed in mph and area in square feet then this works out at about

O.0051AV^3 depending on the value of rho that you use.

This is total energy in the wind. Betz limit is 0.59 of this and in reality you will do very well to manage 0.25. Take 1/4 of your figure and you are near the mark.

Flux

Re: drag co-efficient question (3.00 / 0) (#4)
by xymox on Tue Sep 06, 2005 at 05:08:37 AM MST

Ok , not sure if the correct terminology is drag co-efficient of something else
So using your formula what would be the available energy in wind with say a 6 foot radius prop?
Beat the system, do as much as you can do.
[ Parent ]

Re: drag co-efficient question (3.00 / 0) (#5)
by Flux on Tue Sep 06, 2005 at 07:23:16 AM MST

OK I will try again.

The formula that you used is the same as the one I gave, I was trying to explain that air density was the other factor that comes into play.

That formula gives the total power in the wind. You can't extract all of it because the wind would come to a stop and pile up in front the blades.

Betz deduced that you can extract the maximum power of 0.593 of that total when you have reduced the wind speed through the rotor to one third. This is the theoretical maximum. In reality you can not even get this much for various reasons so you are only likely to extract 0.35 to 0.4 of this maximum in mechanical watts. You then need to include your alternator efficiency and then you end up with a figure of between 0.15 and 0.25 that you can use.

So your total power in the wind at 15 mph for a 12ft prop that you calculated at just under 2000W will give you between 15 & 25 % of this so you can expect a bit under 500W.

The complete formula for power out of a windmill is

1/2 (rho)AV^3 x cp where cp is the power coefficient and it has to be less than .59.

I hope this solves your problem.

Flux

[ Parent ]

Re: drag co-efficient question (3.00 / 0) (#8)
by xymox on Tue Sep 06, 2005 at 08:26:21 AM MST

It does, thank you very much for your imput.
Beat the system, do as much as you can do.
[ Parent ]

Re: drag co-efficient question (3.00 / 0) (#9)
by xymox on Tue Sep 06, 2005 at 08:33:40 AM MST

My goal in all of this is to create a 1kw power source for use in a heater. We have an annual average of 13 mph wind velocity around here, (sturgeon bay WI) usually 15 in winter. that may not sound like allot for a heater, but if it is on all the time that will amount to a whole heck of allot of heat! If it gets to hot, open the door (what I'll tell my Wife ha! ha!)
So Im trying to balance out my figures and figure out what size prop I need and what efficiency I need to strive for in my alternator design. Yes I am a newbie! but for what I lack in knowledge I will make up for in skun knuckles!
Beat the system, do as much as you can do.
[ Parent ]

Re: drag co-efficient question (3.00 / 0) (#3)
by jlt on Mon Sep 05, 2005 at 04:54:08 PM MST

ihave used a rear spindle off of a buick regal. it is a sealed bearing set up . i dont know about celebaty hubs but they are probaly very similar.mine has a lot of drag on it .I would check ouy trailer hubs .they make 5 hole models. and a lott cheaper too,

Re: drag co-efficient question (3.00 / 0) (#6)
by kitno455 on Tue Sep 06, 2005 at 07:29:02 AM MST

WARNING:

most tapered bearing front wheel drive hubs (like the celebrity) rely on the outer joint of the CV to hold them together. see that big internally splined hole in the center? if you dont put something in there with a serious bit of tensile strength, the spindle will likely walk apart during gusts and yawing, and that prop is going to fly off. someone is going to be killed, or property severely damaged.

you have been warned.

allan

[ Parent ]

Re: drag co-efficient question (3.00 / 0) (#7)
by xymox on Tue Sep 06, 2005 at 08:18:53 AM MST

Hmmm, That is interesting. So I should definately incorporate a spindle in my design?
I was thinking of boring out that internally splined hole in the middle and welding a piece of 1 inch round steel through it so that I can secure the stator to it.
(avoiding the need for a large perimeter bracket around the outside. Surely there must be a way to keep that bearing where it belongs?
Any ideas?

Beat the system, do as much as you can do.
[ Parent ]

Re: drag co-efficient question (3.00 / 0) (#11)
by kitno455 on Tue Sep 06, 2005 at 02:02:12 PM MST

perhaps we miscommunicate. i said nothing about needing the spindle. i said you need the end of the CV that goes in the splined hole, and has a BIG washer and nut and castle lock AND a cotter pin on the outside. (all that attaching hardware is your first clue that something important is going on, GM would happily leave that stuff off if they could...)

the hub (the part with the splined hole and the 5 wheel studs) is only pressed into the outer bearing, and the spindle mounting flange (three bolt circle) is dropped over that, and then the inner bearing is pressed onto the hub from the inside. you need something serious in tension, with big flanges inside that splined hole, to keep the bearings from sliding off the hub.

allan

[ Parent ]

Re: drag co-efficient question (3.00 / 0) (#14)
by xymox on Tue Sep 06, 2005 at 03:27:25 PM MST

I thought those bearings were pressed on by machined with like 200000 pounds per square inch of pressure.
Beat the system, do as much as you can do.
[ Parent ]

Re: drag co-efficient question (3.00 / 0) (#12)
by xymox on Tue Sep 06, 2005 at 03:20:32 PM MST

In reference to the below picture I was thinking that I could(with a throughbolt and some helacious washers and nuts) I could lock the bearing on, at least I could prevent it from actually coming off. The white section of the diagram represents the throughbolt. Do you think that this would be adequate enough?

Beat the system, do as much as you can do.
[ Parent ]

Re: drag co-efficient question (3.00 / 0) (#13)
by xymox on Tue Sep 06, 2005 at 03:22:23 PM MST

Please note that the lock washer isn't actually putting any pressure on the bearing, its about 1 mm away.
Beat the system, do as much as you can do.
[ Parent ]

Re: drag co-efficient question (3.00 / 0) (#15)
by xymox on Tue Sep 06, 2005 at 07:31:40 PM MST

Allan, Here is a good article on how much force it takes to actually remove the bearings
http://www.rangerovers.net/repairdetails/drivetrain/hubs.html#bearingonly
given that, it seems pretty unlikely that this process could happen on its own.
I still dont feel quite at ease about using this hub design, but given that its fairly easy to replace and will require no welding to attach to a frame(given the 3 bolt flange), I think it is a good trade off.

Also while thinking of it. The force will probably 99% of the time be focused in towards the mounting flange, so I can imagine, if anything it will tend to tighten itself, I say this in jest of course ;)
Beat the system, do as much as you can do.
[ Parent ]

Re: drag co-efficient question (3.00 / 0) (#16)
by kitno455 on Tue Sep 06, 2005 at 08:14:21 PM MST

you are not using an off-road spec hub from a solid axle rover 4.6 (with 32x12 tires), you are using a cheap rebuild from a chevy FWD with some half the tire weight for on-road use :) yes, it takes a big press, yes i have seen presses bent trying to do this. that does not mean that corrosion and vibration and yaw precession will not eventually loosen this thing. a big bolt or the actual end joint of a CV and the appropriate washers are cheap insurance.

oh, and they most DEFINATELY should touch. dont leave a gap, cause that defeats the purpose. follow the axle nut torque specs of the manufacturer, minus a quarter turn since you want a little less friction and have less load.

i think perhaps you are missing the intended location of the inner washer- it should only cover the inner portion of the hub, not the outer, three-bolt flange. take a look an an actual half-shaft at the auto parts store. see how it fits into your hub. replicate that, or just get the outer joint and nut at your local junk yard.

allan

[ Parent ]

Re: drag co-efficient question (3.00 / 0) (#17)
by maker of toys on Wed Sep 07, 2005 at 12:33:16 AM MST

preload is important to the longevity of this sort of bearing system.
prevailing torque is not sufficient to lock the nut.
never rely on a press fit to hold things in place.
things can come unraveled frightfully fast.

anecdotal evidence:

I had occasion to change the wheel bearings twice in one month in a CV equipped car. . . after the mechanic had his try, the nut worked loose because it hadn't been staked over properly. so the new bearings lost their preload and ate themselves by running on the wrong part of the race. the hub also worked its way about 1/3 the way out of the bearing. (this only took about 1000 freeway miles in a very light car. . . I guestimate about 6 months at 250 rpm for a windmill.)
I had caught the lack of staking a couple of days after getting back the car, but hadn't thought to retorque the nut. (it didn't spin by hand, and I didn't have a big enough socket or torque wrench at that point. . . could have bought both for what the job cost. always doublecheck anything someone else has done to your car . . . .)

so, no longer trusting that mechanic, I undertook the task myself.

it turned out that it actually was a pretty light press fit. . . I managed to remove and replace the bearings using some creative bracketry and a 1/2" bolt. the hub was a similar light force-fit in the bearing.
I never got above 40 ftLb torque on the bolt. (I think that's about 60 Nm to those of you who use a real measurement system. . .) I've since done some research, and I figure the 'press fit' was about 6 tons force.

There was a special 'external thread nut' sort of a thing to hold the bearings in place in the hub carrier. . . . seems the engineers at Fiat didn't trust a press fit either. . . .

moral: vibration and impact can undo anything that a press can do.

so: preload those suckers, and at lease use a staked nut or cotter pin if you're going to double-think the original engineer. . . washers alone won't cut it in this application.

It's not a case of 'Save the planet,' it's a case of 'Save the humans.'
[ Parent ]

Re: drag co-efficient question (3.00 / 0) (#18)
by kitno455 on Wed Sep 07, 2005 at 07:57:01 AM MST

preload is important, yes. though more important with ball bearings than tapered roller, though that is mostly due to the 'ramping' effect of the weight of the car in the races. in this case, with no appreciable vertical load, i bet the rear/inner bearing will take most of the force in compression, and the outer/front bearing will be largely unloaded (until vibration or precession starts)

i recommend highly that you use the outer part of a cv, instead of a bolt and washer. it has radiused corners, case hardening, and is cross-drilled for a cotter pin. it is also self-centering, and not fully threaded (less stress raisers). you can cut the birfield cup off, as long as you leave the flange under it.

if you insist on using a bolt and washers to do this, find a way to keep the bolt centered in the hole, and use grade 8 hardware, not the crap from the home improvement store, and try to find a properly made bolt, with threads only at the tip, and a narrowed shank, like a good cylinder head bolt.

allan

[ Parent ]

Re: drag co-efficient question (3.00 / 0) (#10)
by xymox on Tue Sep 06, 2005 at 08:35:24 AM MST

Have you ever noticed any bearing movement movement on the hub?
Beat the system, do as much as you can do.
[ Parent ]

Re: drag co-efficient question (3.00 / 0) (#19)
by xymox on Wed Sep 07, 2005 at 09:23:49 AM MST

Seems this question is worth pursuing, I am starting a new post to get some more feedback about my hub choice.
Beat the system, do as much as you can do.
[ Parent ]

Re: drag co-efficient question (3.00 / 0) (#20)
by IntegEner on Thu Sep 08, 2005 at 08:12:52 AM MST

I've got lots of knuckles myself and in my head so that's why I am a knucklehead. Sturgeon Bay, Wisconsin is near two of the most well known names in home wind turbines - John Hippensteel of Lake Michigan Wind and Sun (www.windandsun.com) and Doug Johnson of Bay Winds (www.baywinds.com). You might find your answers when you pay either one or both of these gentlemen a visit. Hope that tornado didn't mess up the Door Peninsula. Best to you.

"Knucklehead" Chessick
www.integener.com

Re: drag co-efficient question (3.00 / 0) (#21)
by xymox on Thu Sep 08, 2005 at 08:46:31 AM MST

Kewl, jsut found out thatIn Wisconsin it is required by law that the public utilities pay you the same price that you pay them for power from renewable energy systems up to 20 KW

got to like that.
Beat the system, do as much as you can do.
[ Parent ]

Re: drag co-efficient question (3.00 / 0) (#22)
by ghurd on Thu Sep 08, 2005 at 10:19:09 AM MST

Until you get to the red tape, rules and regs, etc.
Not to mention the cost.
G-
Ghurd.info
[ Parent ]

Re: drag co-efficient question (3.00 / 0) (#23)
by IntegEner on Fri Sep 09, 2005 at 07:39:06 AM MST

In Minnesota it is "less than 40 kws." and so the 35 kw. wind turbine was born but it all has been slow. Ask Dan Juhl there. Take a look at www.integener.com and the nine foot tall "wind micro rotator" for the front yard. Get something all your friends and neighbors will want to have for themselves also. The blades will be green with adhesive contact paper and it will look like a tree. People will stop and ask you how to buy one. No one has called me "Kewl" before. It sounds better than "Knucks", my usual name. I am a poverty case but even so am giving an extra dollar or two in church on Sundays for the Katrina disaster relief.

Knucks
www.integener.com

[ Parent ]

drag co-efficient question | 23 comments (23 topical)

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