Water heater elements on wild DC voltage

[Titantornado]

So, I'm starting to think about how I want to do my diversion load.  Somewhere on this board, I recall how wattage of WH elements dropped off sharply with lowering voltages.  It got me wondering if this fact would make it ideal as a speed controlling dump load.  

Say I would have a 120v, 1500w water heater element connected to the wild DC out of the rectifier.  During low winds and turbine voltages, wouldn't the resistance of the element prevent current from flowing, and thereby allow the turbine to freewheel?  And, as speed and voltage increases, so would the current to the element?  The faster the wind and higher the voltage, the more "hungry" the element becomes, taking up the energy, no?  Eventually, wouldn't the element overpower the windmill and cause stalling in higher winds, correct?

Below is a chart of my basic calculated output numbers. (disregard the amps and watts figures, as they are intended for battery charging) The voltage would still be a bit low for a 120v element and reach it's wattage potential.  Perhaps paralleling a second element would be a good idea. (if my knowledge of Ohm's Law is correct, that should cut resistance in half)

MPH        RPM        rectified open VDC        Amps      Watts out

7. 6       108             025.71              000.0         0001     (cut - in)
   
10. 1       143             034.04              012.8         0330
    
12. 5       178             042.38              025.7         0659
    
15. 0       213             050.71              038.5         0989
    
17. 4       248             059.04              051.3         1318
    
19. 9       283             067.38              064.1         1647
    
22. 6       321             076.42              078.0         2005  
    
    

(furling/furled by this point. beyond this is speculation)

25. 1       356             084.76              090.9         2334
    
27. 5       391             093.09              103.7         2663
    
30. 0       426             101.42              116.5         2992
    
32. 4       461             109.76              129.3         3321
    
34. 9       496             118.09              142.1         3651
    
    

I don't know, sounds too easy.  I must be thinking wrong here.  OK, awaiting return fire.

Rod

[Opera House]

That chart looks great!  Unfortunantly, your calculator needs new batteries.  Just checking a few numbers I found none of them was right.

[Nando]

Rod:

One can not place a load to the generator expecting that at low winds the wind mill will FREEWHEEL, it is quite the opposite, it will BRAKE the wind mill to a stand still, the same way that one shorts the generator output wires to stop the wind mill.

Reading the cut-in voltage of 25.71 at 108 RPM the V/rpm = 238 mv/rpm, so at 321 RPM the voltage will be 76.42 if we follow your readings or calculations following the cut-in voltage -- the correlation with the capability of a heater will not.

You have a 24 volts system ( basing on the current) and to do maximum harvesting the furling needs to be raised to around 450-500 RPM for good MPPT processing -- your generator is a low voltage that complicates the MPPT a bit.

You need a load controller that presents a high value load, let's say 100 times greater than the load a peak power.

Due to the low voltage the gen produces, it may be necessary to have not one but 2 or 3 loads, the first a PWM load and the 2 & 3 loads to be ON/OFF as defined by the PWM limiting factors, each load to be 1/3 of the total load.

How GOOD are you with electronics ?.

Do you know the generator phase winding resistance?.

Nando

[Titantornado]

"One can not place a load to the generator expecting that at low winds the wind mill will FREEWHEEL"

Really. I would have assumed the high resistance of a 120v heating element would have allowed it.

"How GOOD are you with electronics ?"

Electronics is definitely NOT my strong point. (I guess that's why I'm always trying to find a way to circumvent such)

"Do you know the generator phase winding resistance?"

The phase resistance of my stator is 0.5 ohms, in star configuration.

[commanda]

Titan,

Output voltage is a linear function of rpm.

Power into a resistance is a squared function (P = V*V/R)

RPM is essentially linear with wind velocity assuming constant TSR.

Power from the wind is a cubed function.

Short story is; with a constant resistance it will not stall with increasing wind velocity.

You need zero load at zero rpm so it will start turning. You then need increasing loads with increasing rpm.

It can be done with switched or pwm loads using several methods; all require some electronics ability.

Amanda

[Nando]

ROD:

If you have a 3 KW load for 120 volts the resistance would be 120^2/3000 = 4.8 ohms, would you thing that if you apply such load to the wind mill th at is standing still, the wind mill will start powering such resistor -- NO you are braking the mill.

If at cut-in the generator has, in your case around 25 volts and if you load it with a 4.8 ohms, do you think the wind mill will operate?>

Nando

[Ungrounded Lightning Rod]

Dump loads are not about consuming the power as it's generated.

They're about discharging the batteries - ideally, faster than they're being charged - to keep the level of charge below that which will damage the batteries or cause excessive gassing and require frequent watering.  (essentially:  a squidge short of full, not overcharged).

If a gust causes the mill to charge faster than the dump load can eat it, before the furling takes effect that's just fine.  It means the dump load will stay on a bit longer.

Resistors pull current in proportion to voltage  "Cutin" is zero plus a delta - as soon as the blades are moving if you're hooked up before the rectifiers, when they're generating enough to get past the forward voltage drop of the diodes (about 1.2v) if you're hooked up after them.  This is not what you're after.  You want to use every scrap of power to charge the batteries until they're full, and THEN dump any excess to keep them from overfilling.

The essentially constant-voltage load of the batteries and diode drops, plus the current-dependent voltage drop of the genny and its feed wiring, provide some speed limiting drag on the blade which helps to keep the mill from running away before furling can take effect.

[Titantornado]

Alright, I had the suspicion it couldn't have been so easy.  If it were, everyone would be doing it.  ;-)

Building wind turbines is easy.  Keeping them in one piece is the hard part.

[willib]

soooo you got your stator on?

great news!

what did you end up using to hold it in place?

no pictures?

[Titantornado]

No no, not yet.  I'm still hunting down steel tubing for it.  It's been a lot more difficult finding it than I thought it would be.  Scrapyards around here have the most bizarre hours of operation.  More like, whenever they feel like it, type of hours.  And buying new is ridiculous.  The steel for the head alone, would cost nearly as much as the generator itself.

I got Monday off, so hopefully I can make some headway.