the Otherpower.com Discussion Board || Windmill Heat

Front Page - [Homebrewed Electricity-- (wind) (solar) (hydro) (steam) (controls) (storage) (mechanical)] - Classifieds - Site News
Everything - Newbies - [Remote Living-- (housing) (heat) (light) (water)] - Rants & Opinion - Diaries - Our Products

Windmill Heat
By ghurd, Section Heat Posted on Thu Jan 12th, 2006 at 06:40:53 PM MST
Move the losses indoors?

Hi everyone,
Lots of interest in heating lately, and there is always interest in efficiency.
The two got me thinking in reverse (as usual), about a combination of the 2.
It started with 100W at 12V will be more lossy than 100W at 24V.
What would be the result if the heating element were in series with the battery?
The battery would still maintain a full charge.
Less wear and tear on the battery.
Less thermal stress on the stator(?).
Less line losses.
The battery would let the blades get going before a load is introduced.
The increased windmill power, feeding a resistance at a higer voltage, will reduce stator and line losses, meaning more watts of heat in the house.
Here is why I say that.
From W=I^2xR, 1A in 1 ohm is 1W, but 5A in 1 ohm is 25W. That 25W is being transmitted at a higher voltage with better efficiency than a battery voltage dump load.
For a 12V system, and a 1 ohm heating element...
With 1A is 13V (12+1), series resistance heat is 1W, 7.7% of the power into the system.
With 5A is 17V (12+5), series resistance heat is 25W, 29.4% of the power into the system.
With 20A is 32V (12+20), series resistance heat is 400W, 63% of the power into the system. Etc.
With a 3 step dump load the first step could be a make-before-break SPST relay removing the heaters short. Maybe the second step could add more resistance too.
A system set to furl a little too early may benefit with some adjustment. Around here the batteries generally would be full a day before it furls anyway.
Just a little different take on the capacitor controlled heating idea.
The whole idea may be better thought of as a "pre-dump load".
I can't do it here anyway, but does anyone have any comments about this line of thought?
G-

Windmill Heat | 11 comments (11 topical, 0 editorial)

Re: Windmill Heat (3.00 / 0) (#1)
by Flux on Thu Jan 12th, 2006 at 12:38:31 PM MST
(User Info)

There is a lot of sense in this under the right conditions.

For less than a 10 ft mill I doubt whether heat is worth bothering about if it adds to the cost of the alternator. Under high wind conditions a bit of water pre heat from a dump load is worth considering. Line heaters are questionable as you will need a more expensive alternator to make it work.

For bigger mills which still have a battery requirement then it makes perfect sense.
The alternator will have to be bigger and more expensive but keeping the losses as heat in line resistors lets you push the stator current a lot higher and you can have a lot of useful heat under normal charging conditions even if the battery is still charging. When charged then you can use the dump load current as well.

I don't quite follow your dump idea, I got the impression that you were proposing doing it on series resistance rather than a shunt load. You will need both.

This is not very practical at low battery voltages, line loss could be too great, heater values are impractically low. I think there may be some possibility at 24V but more useful at 48V.

If you go even larger, mainly for heat I think proper heater control would be better and a bit of battery charging if required could be better done by a separate battery charger but it depends on the ratio of heating power to battery power needed.

In the past with the size of machine most people were building this was not worth considering but now 15 ft and larger are becoming common then the issue needs serious thought.

Flux

Re: Windmill Heat (3.00 / 0) (#2)
by ghurd on Thu Jan 12th, 2006 at 02:47:16 PM MST
(User Info)

Thanks Flux!
I'm glad I'm not entirely crazy.

Yes, I meant this as a substitute as a very early load, maybe in 2 resistive steps a tad before a big dump load is draining the battery.
Then the standard dump load comes on.

But (dosen't everyone hate the 'but' part with horrible numbers), when the first real the first dump load (shuted) dump load is activated with 1 ohm (proposed) series resistance already in place, 1/100 ohm battery resistance, and 1 ohm dump load resistance, the heating efficiency still is much better than a shunted dump load alone. That's what I was getting at.

The 12V part was just a random choice suited to boldly prove the idea. Most of the heating talk of late was related to much higher voltages.
Personally I would not go near a homebrew 480VAC unit. Maybe I am overly afraid of things like death.
G-

[ Parent ]

Re: Windmill Heat (3.00 / 0) (#3)
by farmerfrank on Thu Jan 12th, 2006 at 08:04:58 PM MST
(User Info)

The voltage is proportional to the amperage. Doubling the voltage,doubles the amperage which results in 2x2 or 4 times the power. Using your example of 1 amp at 12 volts would provide 2.5 amps at 30 volts(2.5 times 12)

[ Parent ]

Re: Windmill Heat (3.00 / 0) (#6)
by finnsawyer on Fri Jan 13th, 2006 at 10:00:46 AM MST
(User Info)

He could replace the resistor by the collector to emitter junction of a power transistor. By controlling the base current he can then control the charging current. This would also control the heat dissipated in the transistor, since the collector to emitter voltage Vce will be equal to the rectified alternator voltage minus the battery voltage. Vce can be as low as 0.3 volts. The current can then be adjusted to follow the power curve of the windmill.
GeoM
[ Parent ]

Re: Windmill Heat (3.00 / 0) (#4)
by electrondady1 on Thu Jan 12th, 2006 at 09:54:50 PM MST
(User Info)

my understanding of electricity is in it's infancy, but this sounds like a slightly different concept if i understand correctly , the positive output of the geni is conected perhapse via a diode to the battery which is in series with a resistave load and then back to the geni. ? and the battery remains in a charged state please elaborate if posible.

Re: Windmill Heat (3.00 / 0) (#5)
by ghurd on Fri Jan 13th, 2006 at 05:17:40 AM MST
(User Info)

From the geni positive, through a heater, into the battery (charging current), out of the battery, back to the geni.
The battery is still charging but at a slower rate because of the heater resistance. As the charging current increases a little, the total voltage across the heater AND battery increases a lot, increasing the losses (heat) and a higher percentage of loss (heat) is in the house.

The battery is still charging just at a slower rate, so a standard dump load control is still required.

The heater resistance would need carefully chosen to match the mill. Too low won't make much heat, too high and the mill could overspeed.
G-

[ Parent ]

Re: Windmill Heat (3.00 / 0) (#7)
by farmerfrank on Fri Jan 13th, 2006 at 06:36:26 PM MST
(User Info)

It will be very interesting to see what kind of resistor you sue. Even a 1500 watt/ 120 volt element will only produce 15 watts at 12 volts as the resistance is 10 ohms. If you want to capture heat, have your inverter indoors.

farmerfrank

[ Parent ]

Re: Windmill Heat (3.00 / 0) (#8)
by scottsAI (user name at eml dot cc) on Mon Jan 16th, 2006 at 10:16:36 PM MST
(User Info)

Good idea, lets see how this could work.

To keep this simple, it's just a 12v battery. Rectifiers have zero voltage drop.

The 3 meter wind turbine cut in speed is 6mph to charge the battery.
The available wind power is around 68 watts.

12mph wind the generators output voltage is now 24 volts.
PM generator output voltage is directly proportionally to RPM, twice the wind, twice RPM, twice the voltage. The wind power is a cubic function, the wind power is 545 watts. (this follows the E=NARPB/2 developed in another thread)
The mis-match of the battery and generator will cause a 50% power loss, that shows up as heat in the generator.

A 3m wind turbine will be rated at 1kw around 20mph.The available wind power is 2525w. Generator's output voltage is 40v.

At 24mph wind, the generator output voltage is 48v.
Wind power is 4364 watts.
The battery – gen mis-match has 75% loss as heat in the generator.

With 3kw of heat, why does not the generator's burn out? Actually it can.

30mph, 60v out, with 8525w wind power. Now were smoking!
Most systems will furl the blades before this point...

Looking over 1kw generator output power vs wind speed, the power is sort of a straight line, not a cubic function as you would expect from the available wind power. Why?

Generator output impedance, lets take a look at a few numbers.
The table below is a numbers game, they don't add up correctly, explain below.
Three generator output impedances:
0.576 0.288 0.144
Wind wind bat power bat power bat power
speed power watts loss watts loss watts loss

545 250 250 500 500 1000 1000
1841 500 1000 1000 2000 2000 4000
4364 750 2250 1500 4500 3000 9000
8525 1000 4000 2000 8000 4000 16000

The bat watts is the power going into the battery, notice the linear line of the power vs wind speed. The wind power is not linear. This explains why the plots are a straight line. From this plot the gen impedance can be estimated.

If the sum of the bat watts + power loss is >wind power, then the power is first in the battery any left over goes to the loss. The generators total output power is the sum of the bat and loss power.
If the wind power is greater than the sum, the impedance is limiting the total power the generator can produce.

Now back to what this is all about. Extracting heat from the mis-match of the gen output to battery voltage. Some to most of the power in the loss column can be converted to heat as long as it does not exceed the wind power. More can be used as heat, by reducing power into the battery.
The resistor is added in series between the gen and battery.
Solid state device like several MosFETs could be used. Need to monitor gen frequency and control the voltage across the MosFETs.

If the heat is useful then over all system efficiency can be enhanced greatly.
Have fun,
Scott.

Re: Windmill Heat (3.00 / 0) (#9)
by thunderhead (mail me from my homepage!) on Fri Jan 20th, 2006 at 04:38:00 AM MST
(User Info) http://www.simon.richardson.net/mailme.htm

Surely the power only goes up as a cubic on a fixed-pitch machine if the blades speed up to match the wind. Since battery voltage regulates blade speed, the blades will not speed up, and the power will "only" go up as the square of windspeed.

[ Parent ]

Re: Windmill Heat (3.00 / 0) (#10)
by scottsAI (user name at eml dot cc) on Fri Jan 20th, 2006 at 03:54:40 PM MST
(User Info)

Hello thunderhead, thanks for the reply.
To be true I did not include Blade performance vs loading vs wind speed. Way beyond the points here.
Hopefully I showed a cubic power function. That is the function I used in the spread sheet.

PM generators
Output voltage is V=NARPB/2 (do search for def.) all terms are fixed except R = RPM
The PM generators output voltage is directly proportional to RPM.
The output impedance is fixed. The only way to increase the power is to increase RPM.
Fixed pitch blade speed is proportional to wind speed.

http://users.xplornet.com/~rmanzer/windmill/rotor_calculator.html
Try the wind speed at 5m/s then 10m/s

m blade, 35% eff
= 0.54w
= 4.34w

If square function for 10m/s would be 2.16w. Therefore cubic.

The plots show the power into the battery is a linear function vs wind speed.

This post was about using series resistance in line with the battery to make use of the miss match between the wind power, generator power and fixed voltage of the battery.
The power loss, is in the generator, by using a series resistance this power can be made into useful heat.
Improving overall generator efficiency!
Have fun,
Scott.

[ Parent ]

Re: Windmill Heat (3.00 / 0) (#11)
by thunderhead (mail me from my homepage!) on Tue Jan 24th, 2006 at 04:21:23 PM MST
(User Info) http://www.simon.richardson.net/mailme.htm

The problem is that the impedance is fixed. The current will therefore be proportional to the voltage -- that is, to the rotation speed -- and so the torque loading the blades will also be proportional to the current and so the rotation speed.

The impedance is only matched to the mill at one speed: below this speed too much current is drawn and the mill stalls; above this speed the current is insufficient and the mill overspeeds.

Either way, the efficiency is impaired, and so the power cannot be a cubic.

[ Parent ]

Windmill Heat | 11 comments (11 topical, 0 editorial)

Menu

� create account
� How to use the board
� FAQs
� search the board
� Google search the board
� Old Otherpower Board

Login

Make a new account

Username:

Password:

Total Views

��180 Scoop users have viewed this posting.

Related Links

� Also by ghurd

You must be a registered user to post here. It's easy and free, and the link is on the upper right side of your page.
All trademarks and copyrights on this page are owned by their respective companies. Postings are owned by the poster, but may be deleted or moved at the ADMIN's sole discretion. The Rest � 2003 Forcefield.
You can Email the board ADMIN here. PLEASE include the username you signed up with!