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device to add additional load as wind increases


By ontfarmer, Section Wind
Posted on Mon Nov 02, 2009 at 01:45:09 PM MST
My email address is carolhanna@quadro.net

Ontfarmer - This is my first posting.  I have followed the Board closely for five or six years.  I built and flew a ten foot three phase 48 volt for three years.  I have now built a 20 foot, 3 phase. 120 volt per phase.  I am running electric heaters with it to supplement heating requirements.  I would like to know if some device can be obtained that would turn on additional loads as the wind speed inreases. I certainly appreciate the knowledge I have gained from the Board Members.  I thank you in advance and I hope somebody will be able to reply and give me information of where something can be obtained or whether it has to be homemade.  An example of what I am trying to say is a voltage regulator or cut-out that was used with generators.

 

device to add additional load as wind increases | 16 comments (16 topical)

Re: information re: voltage regulation device to (3.00 / 0) (#1)
by Ungrounded Lightning Rod on Mon Nov 02, 2009 at 02:01:16 PM MST

Switching a major load causes a mechanical shock on and current spike in the mill.  Not necessarily fatal but may be an issue.  You'd want to switch only part of the load.

There was a discussion earlier about putting motor run capacitors in series with resistive heating loads, so the rising frequency causes the current to go up as the wind picks up.  I don't recall how that worked out.  Maybe somebody who tried it can post experiences.

(Note:  RUN capacitors.  Start capacitors don't have the continuous current rating.)



Re: information re: voltage regulation device to (3.00 / 0) (#2)
by kurt on Mon Nov 02, 2009 at 02:13:58 PM MST

zubbly tried that with large dc caps wired back to back or something like that as he could not get his hands on enough ac motor run caps to do the job if i remember correctly it worked but he stopped using it because the capacitors got very warm under load. you can Google search the board and find out for sure its all here.  

http://www.reresource.org/

IRC
[ Parent ]


add additional load as wind increases (3.00 / 0) (#3)
by wooferhound on Mon Nov 02, 2009 at 02:24:45 PM MST

It seems to me that you could use a Dumpload controller and make that setup work good. You could use a large capacitor instead of a battery and dump all the generators output to your heaters.
W o o f -={(
Huntsville Alabama U.S.A.



Re: add additional load as wind increases (3.00 / 0) (#5)
by SparWeb on Mon Nov 02, 2009 at 09:41:28 PM MST

Use a capacitor instead of a dump load?  Or use a capacitor instead of a battery?  Either way I don't get it, Woof.  Without the battery the system has no reference voltage.  If it's pure AC then the cap makes the power factor crazy and the dump controller can't switch the negative polarity.

Steven Fahey
[ Parent ]


Re: add additional load as wind increases (3.00 / 0) (#9)
by wooferhound on Tue Nov 03, 2009 at 06:44:43 AM MST

Basically, I'm talking about using a dump controller to dump Everything to a heater. That way you would just set the voltage you want the heater to turn on at. the wind turbine will spin freely until it reaches the Dump Voltage. Using this method you do not need a battery since you would be Dumping Everything. But you would need to substitute a large capacitor in place of the battery to make the dump controller work properly. The disadvantage is having to convert everything to DC voltages with large rectifiers, then feeding the DC to Heaters.
W o o f -={(
Huntsville Alabama U.S.A.

[ Parent ]


Re: add additional load as wind increases (3.00 / 0) (#10)
by ghurd on Tue Nov 03, 2009 at 08:29:18 AM MST

Could use a "small" cap for reading the voltage.
As the cap voltage climbs, controllers set at higher and higher DC voltages could trigger progressive dumping on the AC side.
G-


[ Parent ]


Re: add additional load as wind increases (3.00 / 0) (#12)
by Ungrounded Lightning Rod on Tue Nov 03, 2009 at 11:33:17 AM MST

Don't forget that using a capacitor this way requires a rectifier upstream of it.

You can use relays powered by the dump load controller hooked to the capacitor to switch AC loads.

You'll need a BIG hysteresis on the dump load controller if you use a cap rather than a battery.

[ Parent ]



Re: add additional load as wind increases (3.00 / 0) (#13)
by ghurd on Tue Nov 03, 2009 at 11:57:23 AM MST

I don't think the hysteresis would have to be all that big.
It would only be pulling a couple-few ma from the cap, so the cap should maintain some voltage for a short time, sort of effectively making the hysteresis larger (maybe delayed is a better word).
G-

[ Parent ]


Re: device to add additional load as wind increase (3.00 / 0) (#4)
by nibor wind on Mon Nov 02, 2009 at 04:26:12 PM MST

We all know that the ideal situation would be to have the total load watts proportional to the wind speed cubed.  Your generator output voltage assuming a constant tip speed ratio will be proportional to the wind speed.  The power in a resistor is proportional to the voltage squared.  This  means that if you apply the output of your generator to a resistor the power would be proportional to the wind speed squared.  There is not much of a difference at light winds but a very large difference with high winds.  It is not very hard to switch in additional loads at these higher voltages to piece wise approximate the ideal curve.



Re: device to add load as wind increases (3.00 / 0) (#6)
by rossw on Mon Nov 02, 2009 at 11:24:59 PM MST

Just wondering out loud here.... someone point out my logic flaws?

By mounting a small anemometer close to the turbine, we can determine a reasonable estimate of windspeed.

For a given prop, the best operating point is at it's correct TSR

By increasing or decreasing the electrical load, the turbine can be made to operate at lower or higher speeds respectively for a given wind.

TipSpeed / TSR should equal wind speed for "optimum" performance.

So a 10' (3m) prop has a circumfrence of near enough to 10m. At 360 RPM thats 60m/s.
With a TSR of 7, thats about 8.6m/s wind speed (31 kmh).
Turning that all around, it's fairly simple math to find the "required" RPM for a given wind speed.

A simple-ish (almost certainly in a microcontroller/PIC) load control to add or remove load based on the current RPM being too high or too low, should make a nearly universal controller?

A couple of things to add for safety - an absolute maximum speed of turbine (RPM) - at which point the thing tries to shut it all down... The load would ideally be a PWM control, but some of us might be reluctant to DIY high voltage and/or high current controls.

As an example, a simple controller could just have two outputs - one to drive a very small geared motor forwards or reverse to operate (say) a suitable sized commercial dimmer.



Re: device to add load as wind increases (3.00 / 0) (#7)
by Flux on Tue Nov 03, 2009 at 12:04:50 AM MST

The only controller circuit I have seen is one on Hugh Piggott's site and you should be able to adapt that. It uses triacs to switch in heaters as the wind speed rises.

"Nibor wind" sums the situation up perfectly and there is no big issue in designing a controller to track the input power cubed, just that I don't know of one available commercially.

The simplest heater control is phase control with thyristors or triacs but the normal firing circuits work at fixed frequency, I haven't seen anyone do it with wild variable frequency.( I have some ideas but have never actually tested the idea).

The easiest solution is to rectify to dc and use a mosfet or igbt as a buck converter to load the heaters. You could measure prop speed ( frequency) and use a multiplier to get speed squared and track power to this but I still live in the dark ages of analogue control ( and it works). I am sure the digital boys can adapt the idea to digital control but this is probably way too complicated for you.

I think you could adapt Hugh's circuit to do what you want and it should do it well enough with about 3 steps of load increase. You won't get good results with a single fixed load.

The capacitor idea is fine in theory but unless you design a special high frequency alternator the capacitors get out of hand. Motor run capacitors aren't the world's most reliable things and electrolytics ( ac or back to back dc) are something that I wouldn't trust long term at any power level.

In theory heating control is simpler than battery charging but in reality battery charging is easy as long as you don't want good performance. The battery constant voltage reference keeps the mill under control. A fixed resistive heating load doesn't.

Flux

[ Parent ]



Re: device to add load as wind increases (3.00 / 0) (#8)
by electrondady1 on Tue Nov 03, 2009 at 06:12:46 AM MST

hey there ontfarmer ,
i am on the same quest.
i always thought that it would be ok to use a thermostat
 on the first heater to close a parallel circuit on a second heater and so on.
i'm mostly interested in vertical mills with great starting tourque to accomplish this .
a propeller type mill might run away before the increase in voltage caused more heat and triggered the next thermostat.
here are some links to check   out

http://www.fieldlines.com/story/2006/9/14/17359/1190

http://www.fieldlines.com/story/2006/9/16/225354/917

http://www.fieldlines.com/story/2006/10/29/201016/09

[ Parent ]



Re: device to add load as wind increases (3.00 / 0) (#11)
by frepdx on Tue Nov 03, 2009 at 10:25:06 AM MST

"I think you could adapt Hugh's circuit to do what you want and it should do it well enough with about 3 steps of load increase. You won't get good results with a single fixed load."

Why not an infinite number of steps with SCR's or triacs - they're made for it.

"... here are some links to check out; http://www.fieldlines.com/story/2006/9/14/17359/1190 "

http://www.fieldlines.com/comments/2006/9/14/17359/1190/23#23

I made a phase control circuit like this years ago for an automatic lighting display. It was simple and reliable and might lend itself as an infinitely variable dump controller.

Take an isolated voltage signal from the alternator, pass it through a voltage divider and feed to a comparator + reference voltage. By varying the reference voltage you can set the cut-in. The comparator triggers an optoisolator/SCR for the period of the phase that is above the cutin. The greater the alternator voltage, the more time the SCR's are switched on for each part of the phase.

My circuit ran on household current so I'm not sure how to deal with floating alternator voltages - I suppose an isolation transformer for the alternator signal.

[ Parent ]



Re: device to add additional load as wind increase (3.00 / 0) (#14)
by brokengun on Tue Nov 03, 2009 at 07:53:15 PM MST

I think this could be done pretty well digitally as Flux suggested. I really wanted to turn my first turbine into a water heater originally. However, I quickly realized my situation would not allow that for many reasons (mostly inexperience).

However, by reading one of the phases and doing some calculations a load could be nearly perfectly matched to the turbine. I had purchased a solid state relay rated for 40 amps and I was planning on writing some code to basically do just that. It is looking like I might have another opportunity to do that soon. If I do, I will definitely keep everyone updated and log my progress. I'm just scared of having the turbine run away on me if my prototypes fail.



Re: device to add additional load as wind increase (3.00 / 0) (#15)
by anteror on Sat Nov 07, 2009 at 10:29:06 AM MST

This is should be usefull;

http://www.folkecenter.net/mediafiles/folkecenter/pdf/Small_Scale_Renewable_Energy_Control_Systems.p df

Antero
Finland

[ Parent ]



Re: device to add additional load as wind increase (3.00 / 0) (#16)
by ghurd on Sun Nov 08, 2009 at 03:44:03 PM MST

I do not think I trust many of those circuits.

Circuits 1 and 2 both use a relay.  1 looks expensive.  2 has some basic flaws.
Circuit 3 runs mosfets linearly.  And it looks like it would stall most turbines in normal winds, and can not properly load a 400W turbine.
Circuit 4 looks a little over complicated for doing what it does, if I understand what it does.
Circuits 5 and 6 look like they would stall a turbine in normal winds.

The circuits that stall a turbine will probably allow it to run away if they manage to break out of stall.

I don't care to wade through Circuit 7, though the text explanation is the same concept I proposed a few times before.  I think it could be very greatly simplified.
G-

[ Parent ]



device to add additional load as wind increases | 16 comments (16 topical)
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