No battery, heating circuit

[(unknown)]

Hello

I've almost finished my 8 ft diameneter Hugh Piggot wind turbine. I've constructed it to output nomimally at 12V DC, but I guess it will peak up at 30 / 40 V.

I only want the electricity to heat water in a hot water tank, or heat a storage radiator - haven't decided which yet.

I assume I would need a number of 'heating' resisters which are switched on sequentially as wind speed increases, so as not to stall the turbine.

If possible, I would like to avoid using an intermediate battery - as I assume this is

another component that might lead to energy loss, and it may be weeks or months before

I can check it isn't dry.

Has anyone any circuit diagrams to achieve this ?

Thanks,

Richard, Inverness, Scotland

[tesla man]

you really don't need a complex system (bank of resistors, and switches, etc)

I would use a single, used  hot water heater tank.. it already has (2)-240 Vac heating elements inside it!

just wire them up in series  they then should be able to  EASILY dissipate around 20,000 watts, far, far more than your wind genny will ever provide..very long life..

it will work automatically..no switches needed..

a 240 Vac heating element, will still heat water at 12-vdc - 40 Vdc, (just not as fast..)

whatever the wind genny puts out, is dumped into the water.. and then stored..(as heat)

have some power diverted to a 12vdc water pump, to pump the heated water to a radiator, inside the room of your choice, all controlled with electronic thermostats, etc..

hydronic heating (hot water tubes in the floor) is extremely efficient!

this is a very good idea.I have been wanting to do this for some time!

the cool thing about a used water heater, is that they can be obtained very often for free, already  have the plumbing connection points all built in, and ALSO  be able to easily connect it to a solar hot water heater system as well!

and you  can get good insulation jackets for the hot water tank.. very good overall  "system efficiency" with this plan!

I believe these are called "wind furnaces"..

I hope you share your results..go for it! very good idea indeed!

water, is awesome heat storage medium! (far better than air..)

(no battery needed!)

[cdog]

Please correct me if I am wrong but I understood hooking the load directly to the mill will cause it to act as a brake and not let the mill get up to speed! Also even though I dont know the math I understand that at 12 volts those elements will not work properly with a 240 volt load. Perhaps I am way off on this, someone else care to step in, I`d like to know the answer as well,

Cdog.

[TomW]

tman;

Not to be nasty but, 2 [240 volt] heaters in series connected to a nominal 12 volt load will only dissipate a minute fraction of the 220 volt rating? Probably only 4 or 5 percent. Just so he doesn't waste time on something most of us know will not work.

He should of designed the turbine for higher volts but he put the cart ahead of the horse here.

Just a reality check.

T

[vawtman]

Cant you weld with batteries Tom? Maybe a relay to pulse the power to the elements?

 Crazy has it may be.Shoot me

[TomW]

vman;



Cant you weld with batteries Tom? Maybe a relay to pulse the power to the elements?

Certainly. I have personally spot welded screwdrivers and wrenches to body parts getting across the + terminal on the auto battery and the bodywork.

Seems it would be easier to wire your stator for the voltage of the heater you hope to drive.

You can "decouple" the turbine from the loads using motor run caps. They will let the turbine spin up a ways before they start passing AC to the load. I have seen this done in person.

Cheers.

TomW

[vawtman]

 Run capacitors Hmmm little batteries. Cool

 Thanks Tom

[Flux]

It is going to be very difficult to do anything useful with this. 8ft is too small for serious heating even in Scotland.

Ignoring that minor point, you are forced to load on the dc side with Hugh's 5 phase machine ( if you have built it 3 phase you have a chance with ac)

Conventional 240V heaters that someone suggested will provide negligible load and you will need a tank full of them.

Assuming you can find heaters of suitable resistance you can probably connect the first stage direct as there is no need for the thing to start below 12 mph.

Switching dc with relays is not clever but may work at this low voltage ( below 50V)

Far better would be mosfet controlled pwm .

If it is 12v for battery charging, you will need to be up to about 24v for first loading. If you want significant power and are prepared to live with the noise you had better let the sped go up to at least 600 rpm and you will be up to 40 or 50V which should reduce the effect of line resistance.

If you have wound for 3 phase you can use the controller circuit that Hugh has on his site, but the triacs will waste a lot of power at low volts. If you can mount them on the tank it doesn't matter, you can use the heat.

My choice would be mosfet pwm for this low voltage set up. You need to let volts rise with speed and increase the load to track speed cubed in theory, in practice with alternator loss ans prop characteristics you will likely need to track load with speed squared.

Flux

[(unknown)]

Thanks for all the comments

The plan is for the turbine to hear water during the week while the property is unoccupied, for use at the weekend.

But as cdog understands,  I can't just hook one big load onto the output from the mill, otherwise it will brake it - hence I don't see how I can get away from a number of small loads that are turned on in sequence as wind speed increases.

And I can't use 240V heating elements - the loads are going to have to be matched to the voltage output.

I appreciate Flux's comments about 'mosfet controlled pwm' but I'm really looking for a 'simple' black & white circuit diagram, which I can solder up myself, from someone who has already done something similar,  rather then the theory of how to do it, which is above my head at the moment.

I'll have a closer look at Hugh Pigott's site - I think I came across something close to what I want a while back, and I know he has used a turbine to heat a storage radiator in a school room.

Richard

[Flux]

You didn't say whether you have the 5 phase design or whether you have wound for 3 phase.

For a simple relay control you will do much better with ac, switching dc is a bit specialised.

I would imagine that 3 steps would be enough for this machine, first step at about 12 mph. If you use speed sensing you could connect load at about 300 rpm. At 400 rpm you could add another step and finally a big step at about 600 rpm that could hold it up to about 30 mph.

If you have 3 phase you need to keep things fairly balanced so you need 2 pole relays for each step or electronic relays ( triacs). Hugh's speed sensing circuit and comparators will do this well enough.

With dc you can use single pole relays but even with capacitors and resistors across the contacts the contact life may be rather limited.

Another method that you could try is using a standard charge controller which does the pwm bit for you. Set that to hold 24v into the main heater load and use a series heater in the line to avoid the worst of the stall. This will add to the main heater.

My worry is that a commercial controller will run into trouble if run without a battery, it would be a gamble to try without. The battery could be quite small and would also give you a supply of dc at 24v which you may be able to use.

This does depart from your requirement of no batteries but in many ways I think it will get you there quicker than other methods. A pair of car batteries from the scrap yard would be good enough as long as they are still working and you don't want the dc.

Flux

[CompDoc]

Richard, Flux has good advice. You require a method of 'actively' increasing the load/heater element (by decreasing the effective resistance) to compensate for increased Turbine power as the wind increases speed (RPM increase). You want to extract maximum power as heat, as well as keep your Turbine from over-speeding as the element increases resistance (due to its thermal coefficient)... which would result in less of a load.

To approach this problem from the AC side of your 3hase system would require the use of relays (problematic), or a fair bit of electronics and knowledge.

I have designed a controller that works on the DC side of your rectifiers, making use of Mosfets in-place of relays.  The circuit is adjustable, via a simple potentiometer, to match the load to the turbine characteristics (size).  I am using the circuit with homemade elements (from spent dryer-element NiChrome wire) to convection-heat a living-space.

I posted the proto type circuit previously on this board that has additional info, http://www.fieldlines.com/story/2007/1/16/19336/9986

I have since built the production model with a few changes to add flexibility using the parts many may have on hand (junk-box).  The following circuit and connection diagram is working great with a 10-foot 9/12 turbine.  You can use this to heat water if you have low-resistance elements, or any other means of heating including 'dumping' directly from the battery-bank. The circuit uses less than 50ma.









I built this heater inside a PC-tower case...





All files including PCB layouts and PicAXE code are in this Zip file.  Cheers!

http://www.otherpower.com/images/scimages/6828/DCHeater.zip

[(unknown)]

Thanks very much - that is just what I was wanting.

Before closing the thread, I can tell you it is the 5 phase version that I'm building. Everything is just about built except buying the magnets & putting them on the disks. I've been putting of buying them until last because of their expense... the cheapest I've found for 24 pcs neodym magnets 40H 46*30*10, with postage is £98 = 144 euro = 189 USD !! Does anyone know a cheaper source ?

Richard

[tesla man]

"2 [240 volt] heaters in series connected to a nominal 12 volt load.."

I don't even follow that phrase..

how exactly, is a resistive "load" (the heater elements) - "connected to a nominal 12 volt load.."

[sorry,I don't even (electronically) understand the statement]

when is voltage, "a load" - [unless you are a resistor ?? I guess ]

any voltage, into any resistance, dissipates power..

as long as the rated POWER of the load (the resistive elements) is GREATER than the supplied power, all power will be "absorbed"..(by the water)

10. ,000 watts/220 volts = 45.45 amps (46 amps) (@ rated voltage of a single resistive element)
    
220. v/46amps = 4.78 ohms (per element) -  (lets round it to 4.0 ohms - more of a "load")
     
     

(times two, for being in series = 8 ohms)

the LOWER the resistance, the MORE SEVERE the" load", to any voltage being  supplied

(this principle is important, to full understand..)

i=e/r

12 Vdc  (output of generator) / 8 ohms (load - the resistive elements) = 1.5 amps "load" (to the windmill/generator) - pretty minimal as "a load" to the wind genny (not much of a brake)

P = E x I [power = volts x amps]

12. volts DC- {output of windmill generator} * 1.5 amps  =  18 watts (into the water/resistive elements)
    
18. watts into water (that is insulated)  just keeps warming up, OVER TIME..just keep applying power..continuously..
    
    

(want to load the wind-genny more than 15 watts? put the resistive elements into parallel - thus 4 ohms)

@@@@@

e=IR

i=e/r

r=e/i

p=ei

i=p/e

i=e/r

( i= current,in amps)

(v=voltage - in volts)

(r= resistance, in ohms)

(p=power - in watts)

(current - in amps = volts (of the generator/windmill) {divided by} the resistance of the (2) elements (8 ohms, in series) =

12. vdc/ohms = 1.5 amps  [the "load" to wind-genny]
    
12. *1.5 amps = 18 watts "dissipated" into the water! (with a 12 volt "output" of the wind-genny)
    
    

(a pretty small load -to the wind genny)

(want MORE power to heat the water? put the elements in parallel, for 4 times more power ("load" - if you so wish..)

12 Vdc x 6 amps = 72 watts (dissipated into the water - "more braking force")

(still not much strain to the wind genny)

as long as the voltage output of the "wind-genny" is less than 220 Vdc, the hot water heater elements will EASILY heat any water inside the tank..( they will dissipate ALL power  being supplied (220 volts,or less) - 18, or 60 watts, depending on how the elements are wired)

this "will work", and will not "load down" the wind-genny (put the brakes on it, very much.)

please explain, "how it wont work" ["Just so he doesn't waste time on something most of us know will not work."]

 I am very curious .. (your math?)

who is "us"..

220 volts IS ONLY the "upper limit" of the resistive load (in terms of power dissipation)

anything less  (in terms of voltage) will only give a better safety factor, to the system..(thus the suggestion..)

the wind genny can  thus, easily heat the water

that was my entire point -

that  the restive elements (of a used, easily obtained, hot water heater) can EASILY handle 12-50vdc (output of the wind-genny), no sweat..

you want an optimal wind-furnace design?

measure the resistance of the wind genny coils, and make the heating elements the EXACT same resistance..

remember, if little heat (even at 14 watts) is not removed from the system (hot water removed,or heat radiative loss- if there is no insulation; then the water will only get hotter, over time..

a few days hooked up to the "wind furnace",  one will have a lot of hot water..( time works to the advantage)

I don't think you are "nasty", I just suspect that you just don't have a firm grasp, of the math, of the physics phenomena

.. just reading this, hopefully, will help you learn more of the principles involved..

I hope I have illuminated the subject, to your benefit..

 

[tesla man]

no-one, (that I know of), recommends that the electrical load, TO the generator (windmill), should be used to dampen "over-speed"of the windmill, due, to high wind speeds..

 only use MECHANICAL furling systems..(to correct for over-speed)

as long as you have a minimal load (electrical) to the windmill (less than 100 watts?), all of that power, is available to "220volt heating elements" they will dissipate it all -everything

if it is 12-50 Vdc, then THAT power, will be coupled, (via heat) into the water...

as far as I can tell, everyone has trouble grasping that a "220 volt heating element" ( the upper limit to the power supplied) can, and will, handle ALL the power that 12-50 Vdc supplies..

220 volts is JUST the upper limit of ANY voltage supplied..(nothing else)

so if the wind-genny supplies 220 volts, (or less) the element is rated to SURVIVE..

it will dissipate all the power provided to it  (any voltage, from 0-220 volts)

test this:

.hook 12vdc to a 120 Vac (resistive) heater .it will easily dissipate ALL THE POWER SUPPLIED BY THE 12VDC. voltage

it will just dissipate less power..than the 120 volts..

if the windmill "over-speeds" (produces more voltage) then the 220volt resistive elements will handle it easily..

all one has to do to protect the windmill, is have a "over voltage switch" that switches MORE resistance ( say 10-1000 times more) into (in series with) the heating elements , if the voltage rises to a safe 'trip point" the windmill is thus, easily protected

(more/higher resistance= less current -> less power, even at a higher voltage out..)

but I suspect, that any wind-genny that is rated to produce 12Vdc (at low wind speeds) , if it ever gets close to 220 volts  output ( the upper limit if the restive water heater elements)

  it will have long since disintegrated, due to MECHANICAL stress..(due to wind/over speed)

if you have a goal of a certain power, being coupled into the water, then do the math, and get a proper element of such resistance.. remember you can hook them in series, or parallel..to obtain your goal..

quite a bit of electrical flexibility!

pick your numbers (desired, safe power, to heat the water, at x voltage, and y wind speed..) do the math, and add the right restive elements, and heat the water.. it is pretty simple  (no  complex circuits are needed)  its NOT rocket science.. electrical power equations -  4 equations, tops..

honest, it is, that simple.. water heaters are very cheap, and easy to obtain..

[TomW]

Tesla;

Ok, you caught me. Typed "load" meant "source". I apologize for that confusion.

Lets not forget the simple formula [P]ower equals Current squared times [R]esistance .

If you drive a 120 volt heating resistance with a 12 volt source you will only produce a fraction of the heat it would at 120 volts. Fine to produce a few watts but very inefficient and of no value to control the turbine.

If you rely on these loads to control your wind machine in high winds you will be picking up pieces of it all over the plane of rotation eventually.

Tesla would understand this relationship of current, resistance and voltage. I suspect almost everyone with any clue of how electricity works will agree on the heater point here, despite my flubbing the load and source terms.

As a Tesla Man, you should know this stuff and refrain from passing out bad advice. It is more than simply that the element can "take" the 12 volts. It is that it will not present a safe load to the turbine to control its rotational speed at safe levels. It is an integrated system and all parts need to work together to provide useful and safe power. You can never just look at one piece of the puzzle and design based on one factor.

Mixed terms aside, you might want to go read your basic DC circuits text book again. Maybe they changed it in the 40 years since I studied it but I doubt it.

I hope this clears that up because it could get expensive and dangerous if anyone followed your well meaning but flawed advice.

Thanks.

T

[sdscott]

Richard, Flux is correct.  It is wise to load your turbine sufficiently to avoid over-speeding (run-away)... while staying out of stall of course.  This is where you will get the most efficient harvest of wind-power.

[finnsawyer]

Why in the world did you go through this long winded argument just to arrive at the well known conclusion that a linear voltage source will provide maximum power out when the load resistance is equal to the source resistance?  Whether that's optimal is debatable since half the power generated by the source is then dissipated in the source resistance.  It might, in fact, be more desirable to have the source resistance one tenth as large as the load resistance so most of the power goes into the load resistance.  In the case of a wind mill it then becomes a matter of designing the alternator to deliver as large a voltage as possible with as low a resistance as possible so that the load gets most of the power.  In other words, if you are using a 120 volt heating element, you should shoot for a maximum alternator output of 120 volts with an alternator resistance much less than that of the element.  Admittedly, it could be a tall order.  As you mention you can put several elements in parallel, but then you should lower the alternator resistance by the same ratio.    

[finnsawyer]

What the hell happened here?  I wasn't replying to TomW.  Sorry Tomw.

[(unknown)]

Sorry Tesla man, I think I'm leaning towards Comp Doc and Flux ...

If I use a 2KW, 240 V heating element, resistance = 28 Ohm, that will only produce 5 Watts at 12V  (from Watts = Amps x Volts, and V = I/R).  Which isn't very much.

If I use a 3 Ohm resistance, that will give me a more respectable 48 Watts at 12V. Even if the load on the coils slows the turbine to half the speed so I only get 6V from the turbine, I still get 13 Watts.

If there is a flaw in this logic, please correct me.

Richard

[ghurd]

I figured someone would post this before now.

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

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

http://www.fieldlines.com/story/2006/11/3/211555/017

http://www.fieldlines.com/story/2006/8/7/202620/3156

Just for fun,

http://www.fieldlines.com/story/2006/7/23/224332/349

G-

[(unknown)]

But a lot of the argument will depend on how much different loads slow the turbine down, and influence the output voltage - I've no idea about this, seeing I've not yet built the thing!

Maybe you're right Tesla man with 240V heating element- wind blows, turbine spins like mad, voltage high, watts  =  same as flux/ comp doc with low resistance heating element, turbine spins slower, voltage low.

So I'm none the wiser

[sdscott]

Ghurd, Zubbly has designed his gen to deliver Hi-Voltage. Snippit from his initial post;

"the gen is a 12 pole unit, designed to give 240 volt at 600 rpm unloaded voltage"

I do not believe Richard plans to do this...?

[ghurd]

Still, Caps must be better than relays?

[RP]

You didn't.  You replied to Tesla Man but Tom got there first so yours showed up under Tom's.  According to  the indent structure you both replied to the same comment.  :-)

[tesla man]

it is astonishing to me to see such psychological resistance, to  such a simple idea..

I just wanted to present "the concept", that excess electricity, into a small"load" of a water heater element, can easily heat the water,over the course of a week..(small output power, over a long period of time, CAN result in a lot of hot water available)

since there was no hard numbers  presented (power out put, goal to dampen over-speed,etc, etc) I used  the term "220 volt elements" as an example of utilizing a common, easily obtained, cheap item, to accomplish the goal (of heating water, over a long time - a week..)..

as for the power INTO the system, that desired result, is easily achieved using mathematical analysis, and selecting a proper heating element (again, cheap, easily obtained.. many wiring configurations) for the desired electrical performance expected..

I assumed the windmill was small - (at 100-300 watt output) ..so dissipating 20-80 watts into the water, all week long, would not damage the windmill, or heater system.. (sure, in a hurricane wind-speed burst, any linear load will  eventually result in a damaged windmill.. regardless of the "load" - batteries, air, water)

as the wind speed varies, AND the voltage output would widely vary, and the all the power available (at any given moment) would still be coupled into the water.. this, I saw as an advantage to the:   windmill->electricity-> hot water heating, SYSTEM

if someone want efficiency, design the resistive load for efficiency..

if someone wants to protect from over speed (with an electrical load)  then design the resistive load to accomplish that.. (again, there is a very wide range of resistive heating elements, with various resistances, and then having the ability to wire in series or parallel ( two tanks with different restive circuits?, switched between them (use make, before break, switches) can give the designer a wide range, and choice of electrical performance..

you aren't just stuck..

the concept is widely adaptable (wind->electricity->hot water) depending on how you design the load (the resistive value of the heating elements) there are a wide variety of choices..easily available..

I was just trying to introduce the concept of using a hot water heater, as a  load to a windmill..

your performance will vary..depending on your windmill, and the elements used.. but one can use the water heater, to heat water, over along period of time..(invest in insulation, of the tank, so even a small amount of power input, to the system, can result in a lot of hot water..over a long period of time.. (this, I see as  a critical element to this design)

I would rather dump the excess power, into water - since it can store the thermal power.(better- than) air heating elements..(which a lot of people are happily designing, and using, currently..)

of all the "dump loads", I would prefer to dump into hot water (as opposed to air, for example), as that medium can store, and distribute heat, very effectively..

again I was just trying to advance a simple concept:

 excess wind-genny electricity /power (coupled) -> into water, is a simple system.. no switches, no complex monitoring circuits.. if the time between "output" of the energy system, (vs, a small, constant input) the water can be very hot indeed, with even a small amount of electrical power input.. over a long period of time..

some may find this a useful model..

 others it appears, think it is a very bad idea.. they would appear to have other needs, to be met.. (somehow, a simple, low power resistive load, is insufficient, for their needs)

feel free to accept, and advance the concept,(or reject it, violently...)

I just don't accept that it "wont work", or that it is "dangerous advice".. the overall concept is valid - using hot water heating elements as a "dump load"..

everyone who designs, and builds a windmill, is just as responsible to design the load system..

a windmill, can easily drive a hot heater element (simple, restive load), and heat water.. (that is my simple,  point)..with-out complex switching circuits, monitoring circuits, etc, etc..

I truly hope that whatever solution you find, that works for you, you will continue to share your results, and design, and the your over all performance (meets expectations,etc)

I really wish you the best of luck, with your efforts..

 I tried to advance some (conceptual)knowledge..  my humble apologies, to have wasted so much of everyones time..

[Flux]

Richard I am not sure where this will turn up.

Ignoring internal loss, the voltage of the alternator will rise directly with speed.

Assuming the heater resistance remains constant ( which it probably won't), the power into the load will rise with speed squared. The power in the wind rises as the cube.

If you choose a load to suit low wind you will have less than the ideal load in high winds. If you choose a match for high wind then the blades will stall as the wind drops.

Over a limited wind speed range you will get a reasonable match with just a single resistor but you will need something to switch it in after the blades are out of stall.

All the power you produce will go into the water, but you will not produce as much as you could do unless you lower the resistor with increased speed.

Your small machine will not do much below about 15 mph so loading into a fixed resistor would not be as bad as with a large machine where you may produce useful heat from perhaps 10mph.

There are other benefits in increasing the load with wind speed, an unloaded prop is noisy, prone to tip erosion and more highly stressed. Furling on Hugh's machine was chosen for battery charging and it will furl more effectively if you don't let the prop run away.

There is an additional factor in that props of this type work better with a falling tsr with load and you will produce more power if you load it more heavily than suggested by sticking to a constant tsr.

The mosfet controller will be well worth the effort in terms of extra power gained and more so for peace of mind with better control, less noise and less machine stress.

You may manage with 2 load steps but as it has provision for 3, I would use all three if you can find suitable heaters.

I didn't think of using mosfets just as dc switches, it should work fine.

Flux

[(unknown)]

Thanks everyone, for all your thoughts

I think I am most likely to follow Tesla man idea to start off with, as it is so simple and will get me some hot water soonest.

I might find that works fine for me, and don't need to do anything else.

But, if we arrive at the weekend and find the water tepid, or I spend my time worrying the turbine is going to run-away and disintegrate due to lack of load in windy weather, I'll need to consider flux/ comp doc's suggestions. I'll probably do this anyway, because I like playing with electronics!

Either way, I'll let you know how I get on.

Richard

[finnsawyer]

I don't generally change the title unless Scoop gets nasty, but I am aware of the practice.  It just caught me unawares.

[Dave B]

It works fine, if the blades are turning the water is heating (no matter how little or how much). My past design with my 12' and my current design will be the same with my 18'. (hot water heating, no batteries.) Yes, you can complicate things to make it more efficient by matching the load to the output  but the simplicity of a direct connection by selecting a load for a happy medium of start up vs. high end is a secure feeling. My only suggestion is (from experience) have an overspeed protection of some kind in case of total load loss or stator failure, preferrably mechanical (furling set for unloaded protection ?) As your water heats up your elements resistance will rise, this increases the rpm for a given wind speed so you see what can happen here. Keep us posted,  Dave B.