Volts versus amperes

[lifer]

Hi guys,

I wonder if there's some calculations regarding the maximum power that could be "extracted" from a magnetic field. I'm planning to build an axial flux PMG for a WAVT project and, after I read many kilos of informations about it I still can't decide how many windings the coils should have.

There's no limitations regarding the cut-in voltage as I would use a MPPT charger for the batteries.

Let's suppose the overall dimensions of the coil are fixed, so you can use more windings of smaller diameter wires or less windings of larger diameter.
I understand that you can't go too low or too high with the voltage and current due to copper loses but what is (if there is one!) the ideal ballance betwen voltage and current (the # of windings versus wire diameter), like the maximum power point of a photovoltaic cell?

Thanks in advance for any informations.

PS: I'm new on this excellent forum.. more questions will come soon!

[Flux]

The power you can get out depends on several factors but the size of wire is not one of them. As you say there are limits at each end but over a wide range of wire size the only effect is on the voltage produced. If you can adopt any sensible voltage with the mppt converter the only factor that comes into it is the line loss if the machine is remote from the battery. Higher voltage will reduce this loss.

The limit is mostly dependent on acceptable efficiency and the associated stator heating problem. Unlike iron cored machines the axial flux machine is not usually limited by reactance in the sensible working range if the magnets are adequate.

With wind power the ultimate limit is usually the turbine if the alternator is sized to match the available energy. There is only so much power available form a given turbine size and failure to match the turbine to electrical output will make the output even lower.

Flux

[lifer]

Thanks for your reply, @flux!

I've done some calculations about the copper loss in the coils.
For a given coil leg size (section), if I reduce the wire diameter by 2, I'll get a wire resistance 16 times bigger (the wire length is multiplied by 4 and the wire section is reduced by 4) but the voltage will rise 4 times (theoretically).

So, for a given load, if the applied voltage is multiplied by 4 the current will be reduced by 4 (to obtain the same power). To calculate the wire loses:

P' = R' * I'^2   => P = (16*R) * (I/4)^2 = R*I = P

so the copper loses will be the same for a given load power.

I don't know if dissipating the power has anything to do with number of windings, to decide that smaller/larger diameter is preferable for a "colder" stator. As you mentioned before, the reactance could be negligible in this case so it's not a factor. Also, I'm planning to put the battery and the MPPT charger close to the wind turbine (I'm gonna mount it on the roof) so line loss won't be a factor either. That's why I am so confused..

Anyway.. I think I'll stick with the "common" wire size (2*14 AVG, for an easy coil-winding operation) and I'll put as many windings as I could to fit in the given space.

Any second thoughts?!

[electrondady1]

what are you going to do with this electricity ?
are you storing it ?
or using it as it's being produced?
you need to factor in the rpm range of your mill during your average wind.

[electrondady1]

if you have a voltage in mind you need coils that produce at the right speed and that  dictates  the turn count and  gauge.
 

[lifer]

I plan to live off-grid..

But.. does the rpm affect the power "extracted" from a given magnetic flux, for a different coil wire size?

To answer your questions, I have to tell you about my further plans: I'll die not to use the regular storage devices (lead acid batteries)! I'll struggle to find a way to mechanically store the wind energy (pressurized gases, heated water, lifted weights and so on).
So far I'll use my current setup (few used VRLA batteries) but I need to keep them charged all the time because they pretty lost most of the initial capacity. Thus I'm gonna use some PV panels and a portable gas generator as a backup.
I live in quite a windy (and sunny) area so I can't go wrong but, as I said, I'm looking for a cheap and convenient way of storing the energy.

Regarding my initial question, I'm in the market for some neo magnets and I want to find out if I could "extract" 2-3kW from a 12 poles N40 60x30x15mm or do I have to go with 16 poles/12 coils (that's why I need to optimize the PMG design).

[electrondady1]

"I'm planning to build an axial flux PMG for a WAVT project "
do you mean a vawt?

But.. does the rpm affect the power "extracted" from a given magnetic flux, for a different coil wire size?
yes.
 the faster you spin the alternator the more power you get
until things start to melt.

[lifer]

It's VAWT, sorry (misspelling).

I was not talking about the (obvious) RPM-power raport. You just asked me about RPM range when I was questioning about the importance of wire size for a given condition. So, to reformulate my question: for the same RPM, TSR, torque and coil size (section), one could extract more power if using smaller/larger coil wire size? 

PS: I know it's hilarious to talk about VAWT and off-grid, but I live in a residential area and I am not allowed to build a tower for a HAWT. Actually, I don't like the design(?) of a HAWT neither so the choice it's very simple. I don't want to be off-topic here so I'm gonna start a new thread about my VAWT design (using the edge-flow effect). 

[joestue]

regarding how much power you can get from the magnets, that question is answered here
http://www.fieldlines.com/index.php/topic,147695.msg1023729.html#msg1023729


note however that the only way to get anything close to that for air core machines would require superconducting wire.

[lifer]

Thanks, joestue! That was a good reading - even those calculations were too complicated for me to find a decent answer.

In the meantime, I was playing with this calculation tool and I get the same output power for different coil wire size (for a different ouput voltages/current, obvious). More, seems like a 12 poles/9 coils/3 phases PMG using previously mentioned magnets can output 3kW at wind speed of 10m/s.

As a conclusion, I rather go with a higher voltage output to minimize the line losses and rectifying losses (that's it, you lose the same 1.2V during rectifying for an PMG output of 16V or 96V but for different currents) and because it's cheaper to design electronics (rectifier, charger, inverter) at lower currents and (relative) high voltages.

[john8750]

I like your plans.
With all due respect, 2-3,000 watts from a VAWT in asking a lot.
I am interested in finding ways of storage other than battery.
Would the storage method change from summer to winter?

John Smith

[lifer]

John, I'm thinking of something like this:



My house have a section (9 meters wide) which looks like that so I can mount a large "horizontal" VAWT, taking advantage of the "edge flow" effect mentioned earlier.
Or/and I can mount a bigger (and wider) "true" VAWT in the middle of that flat roof so I can easily get 3-4kW of power.

Regarding storage, I've just opened a new thread to gather some informations about Peltier (actually, Seebeck) devices to convert the thermal energy to electricity.
That water could be heated by an electric heater powered by the wind turbine, so you could store a big amount of energy in some much cheaper "battery" (which don't have the "deep-discharge" effect or overcharging problems).

For now, seems like the actual commercial Peltier modules have very low efficiency so I'll struggle to find a(nother) way to convert that thermal energy (or maybe pressure) into electricity.

The lead-acid batteries are just TOO expensive for their values so I won't put my two cents on those!

[joestue]

The lead-acid batteries are just TOO expensive for their values so I won't put my two cents on those!

It really depends what you want to do with them.

if i recall correctly the 200 amp hour 12 volt cells, or perhaps they were 6 volt, flooded, often for sale at sams club or other places, are still the cheapest at 14 watt hours per dollar. and they supposidly last 500 80% depth of discharge cycles while dropping to 50% capacity.

there's still a bunch of argument about whether 500 at 80% deep discharges would deliver more kilowatt hours per dollar than 2000 cycles at 20% discharge (given that the batteries are below 50% capacity after that and the only option is to chuck them in the blast furnace)

but i do think that Lithium iron phosphate cells do offer cheaper kilowatts hours per dollar life cycle cost at this present time.
(furthermore their 98% electrical efficiency is often important)
that said however it is only rumoured to be that they can in fact withstand 3000 charge cycles before dropping below 80% capacity.
such testing would take 10 years to complete, and there are only a few testing authorities that have delivered charts of the first 500 charge-discharge cycles, and these few datapoints are suspect because they came from a single individual from within the manufacturer, who was acting anonymously.

so the simple fact remains that if energy storage was affordable relative to the grid, well, then if that were the case the grid would be using batteries to provide load management.

Chris has made some very strong statements about which batteries are more cost effective than others but i really wonder if the differences are really worth it when expressed in kilowatt hours per dollar life cycle cost.

if i throw away sealed AGM lead-calcium cells after 2 years of cycleing
or if i maintain 2000 pound flooded lead-antimony cells after 10 years, which are no longer available off the shelf, because about 10 years ago or more now, 20? all the battery manufactuers world wide switched to lead-calcium at the same time..

which one is really cheaper.
and who has the data to back it up?

[lifer]

An AGM 12V/200A battery cost between 500-700$ here, so I must call it "prohibited".

I'll keep searcing for a more adequate alternative to store the energy; I just don't like the limitations of current batteries (under/overvoltage, charging diagram, life cycle).
Anyway, I didn't blame the lead-acid solely - all the technologies have the same limitations.

I'll keep dreaming of my heated (or pressurized) water tank!

[Bruce S]

I like the joke made about pressurized water  .
Since my system in an urban environment is limited by the city's codes  . What I do with the extra once my NiCd battery banks are full; is to use a GHURD controller to keep a small (650CC) free-to-me car battery, that runs a small 100 watt inverter that runs an immersion heater that drives my digester and then my Alky still.
This is then added to petrol used for my home-built 4-stroke genset.
It rarely goes bad and is a good backup not to mention cleaner than straight petrol.
Efficiency of this type of setup is probably too low for most, but when the "Polar Vortex" came through a took out our mains power, I had lights and hot coffee :-)
Just A thought
Then of course there's also less stuff going to a landfill.
Cheers
Bruce S 

[lifer]

Well, Bruce.. I honestly believed in magic!

I've read alot about unconventional energy storage methods.. even about homebrew lead acid batteries (I've watched some movies too).. but all in all it's a dirty job (literally) with uncertain benefits.
My last temptations was a flywhell (the cleaner storage method, I suppose) but I did't find any information about the required size (mass, diameter) for storing 1 kWh of energy, for example.

That being said, I look with more sympathy(!) to my current (old) VRLA batteries and I'm concentrating on energy production.

[xxzxxz]

For volts

e_1 = Blv \sin \omega t
e_2 = Blv \sin ( \omega t +\pi)= - Blv \sin \omega t,

See Wikipedia ))))

For ampers look for "kappa" Greek letter. I do`t remember... May be it`s look B=(m0*N*i)/l
TRhere m0 universal constant (=1,255*10^-6)
N = number of turns of the coil
i = ampers
l = lengh of coils

and....

i= (B*l)/m0*N

Enjoy =)))
Sorry for bad English =)))

[Mary B]

A long time ago someone posted a spreadsheet called number of turns that helped get you close to the coil design needed for a certain voltage and rpm. If you can find that it might help.