Windmill Heat

[ghurd]

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-

[Flux]

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

[ghurd]

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-

[farmerfrank]

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)

[electrondady1]

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.

[ghurd]

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-

[finnsawyer]

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.

[farmerfrank]

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

[scottsAI]

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

12.     545  250    250         500   500        1000  1000
    
18.    1841  500   1000        1000  2000        2000  4000
    
24.    4364  750   2250        1500  4500        3000  9000
    
30.    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.

[thunderhead]

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.

[scottsAI]

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

3. m blade, 35% eff
   
5. = 0.54w
   
10. = 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.

[thunderhead]

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.