Author Topic: resistive heating loads (Read 4542 times)

fanman · « **on:** April 15, 2006, 01:04:06 PM »

hey fellas

i have a couple of questions for you all

can a person directly hook up electric baseboard heaters to an axial flux wind gen.? of course it needs to be 240 volt output, would it act like a short to the genny? would it act like a battery and let it produce current? i dont really know so i thought i would ask
as i build my axial flux machine once i have all the coils connected together, could i take a big variac transformer and hook it to the coils of one phase, and start pumping the amps to them and watch the temputer rise, measure the heat and record a maximum operating point? does that sound like a good way to know what your output would be? seems if that would work it would be easier than hooking up a motor or tractor to test it, just a thought i guess thanks

Nando · « **Reply #1 on:** April 15, 2006, 08:20:57 AM »

FanMan:

No you can not connect the load directly to the wind mill.

The load needs to "follow" the energy level being produced, so if the mill is producing 50 watts the load HAS to be such that just 50 watts are dissipated, if the mill produces 1000 watts, like in your case of 240 Vac the value would be R= (V^2/P) ; 240^2/1000 = 57.6 ohms which for practical purposes is a short to the 50 watts output.

To use a wind mill for heating purposes a controller is needed to vary, in this case, the baseboard resistance from 0 to 100 % value.

Depending on the wind mill power level, the controller may have to have 2 or more resistance loads, the first one to vary from 0 to 100 %, then the others to be ON/OFF as defined by the controller, and all the loads of equal value, though one can as well use ON/OFF loads in binary value.

Exactly what you are thinking, I have been helping a fellow in the UK with a wind mill, high voltage producer, rated at 5 KW ( 5 meter mill) but really capable of only 2.5 KW, he uses the energy to heat water for house heating, his controller is a small industrial PLC = Programmable Logic Controller, I showed him how to capture the energy available using MPPT = Maximum Power Point Tracking, one of the reports indicates that the wind mill produced 980+ KWH in a month, or 32.6 KWH/day.

Nando

Saidrich · « **Reply #2 on:** April 15, 2006, 09:17:55 AM »

Please correct me if I am wrong but if the resistance exceeds the voltage you have no current?

dinges · « **Reply #3 on:** April 15, 2006, 10:42:09 AM »

Consider yourself corrected.

You might want to check out Ohm's law:

U=I*R

U=voltage (V)

I=current (A)

R=resistance (ohm)

More resistance, at a given voltage, means less current.

More voltage, at a given resistance, means more current.

etc.

Peter.

Saidrich · « **Reply #4 on:** April 15, 2006, 12:13:57 PM »

I think, that is what I was trying to say but... Thanks for the set straight.

electrondady1 · « **Reply #5 on:** April 15, 2006, 12:15:14 PM »

fanman , ive always followed your postings. i admire the generators and towers you build.you are in minesota i believe while i am in ontario. my quest is the same as yours. when the temperature dips to -40c ( oddly enough thats -40f as well, same temp.) and the wind comes off lake huron howling like a demon , it kinda makes a guy think of useing one force of nature against the other. i've posted the same request but never get an anser i can use. hugh piggot has a circuit diagram on his site if your good with electronics. i am not. good luck!

Nando · « **Reply #6 on:** April 15, 2006, 12:44:25 PM »

electrondady1 & fanman:

Now tell me with a good description what type of wind mills each one of you have -- in detail -- also if you want you could email me directly and it would be better to unload the site of many messages.

There are solutions

Nando

WXYZCIENCE · « **Reply #7 on:** April 15, 2006, 01:13:33 PM »

"MPPT = Maximum Power Point Tracking", Nando, can you please explain this a little more. Joe.

Nando · « **Reply #8 on:** April 15, 2006, 02:59:04 PM »

The wind mill follows the wind velocity which has a cube power, or V^3 facto, so if you double the wind velocity the power will be 8 times higher.

If one plots this curve, one will see that every time the wind doubles in velocity the power goes 8 times, so if you have 2 m/s and wind goes up to 4 m/s you have 8 times the energy that is available at 2 m/s and if the wind goes to 8 m/s now the wind mill can produce 64 times the energy available at 2 m/s.

If one has a GIZMO that follows that curve, it needs to follow the peak power generated at every moment of time.

This GIZMO is a Maximum Power Point Tracking controller that detects the peak power and adjusts its load to the wind mill to take that maximum power available at every moment of time and since the wind varies, the energy available varies wild, so the controller with MPPT capabilities varies the load following that wild behavior.

MPPT controllers need high voltage and those that have tried to use them with 1/2 wind mill efficiency find them useless -- which is logical in this case --

For a good MPPT controller to attain its best performance the wind mill needs to have a high voltage capability -- so if the load is 10 times larger than the wind mill internal resistance + its inductance (though careful here) the theoretical capability will be that the controller can obtain 90 % efficiency -- depending on how good is the MPPT circuit this efficiency may run a few point lower.

If the load is 5 times the generator impedance, then the efficiency may run around 80 to 83 %

How the MPPT operates: a good circuit will measure the voltage and the current loading the controller; and the controller starts varying the current up/down finding the behavior of the voltage that may go up/down, if the current is brought up and the voltage still is up or continuing up, then the controller keeps bringing the current up, until the voltage starts to come down, at this point the controller may reduce the current and see the voltage behavior and the UP/DOWN is continuously testing this voltage and adjusting the current.

As a result the energy available is harvested at its maximum peak always following that cube power curve of the wind.

MPPT to work needs to have a generator source that has a limited torque source.

With this statement I made, any one, NOT an electrical or electronic expert knows, why this requirement for a limited torque source.??

Nando

Opera House · « **Reply #9 on:** April 15, 2006, 03:00:30 PM »

You can put less voltage into a resistance heater. If you reduce the voltage to half (110), it will be 1/4 the rated wattage produced.

electrondady1 · « **Reply #10 on:** April 15, 2006, 05:02:23 PM »

nando, what a kind offer, but i don't think i can afford any consultations ( ha ha).

seriously,the little genies i have been constructing up till now have been usefull for experiments in understanding the process of generation but for this next stage it might be prudent to define the load and then build a generator/s to suit.

WXYZCIENCE · « **Reply #11 on:** April 15, 2006, 06:23:25 PM »

Thank you Nando, Joe.

commanda · « **Reply #12 on:** April 16, 2006, 03:22:24 AM »

Nando,

In a true mppt, where it actively seeks the point of operation equal to the maximum harvestable energy for the wind which is currently blowing; surely the actual operating point of applied load versus the generators internal impedance is immaterial.

MPPT controllers need high voltage and those that have tried to use them with 1/2 wind mill efficiency find them useless

I have previously supplied you with a spreadsheet of test results I generated for the Fisher & Paykel 80s, driving a resistive load. Yes, it was driven with a large drill press, where the input torque was not a limiting factor. Never the less, it clearly shows, for this generator, which does have a large inductive reactance component, that the half power equation does hold true in this instance.

I realise this stuff is really not easy for the non-engineering background people to get their head around, and with the typical dual axial machines many people build, your quoted ratio of Rload vs Rz of 8:1 or 9:1 may well be correct; but if the mppt is functioning as it should, this ratio essentially becomes immaterial. Except in defining the maximum input voltage to the converter.

As an aside, for those who are interested. Preliminary testing of the mppt controller detailed recently in my diary; with the toy generator shown, appears to lock on at about 2/3 of the open circuit voltage for any given wind velocity.

Amanda

coldspot · « **Reply #13 on:** April 16, 2006, 09:10:46 AM »

Nando-

Thank you first off-

I am slowly reading every comment of yours-

But,

"and it would be better to unload the site of many messages."

That makes it very hard for us regular types to

"get our heads around" this stuff

You just get a good subject going then-

poof-

Nothing to follow.

Some times maybe it would be of better use to and by

the members here if maybe things didn't go to private

E-mails and did in fact stay here and in doing so

a few more people could then actullay get it

straight and square in the head and gain more.

Not trying to complane or NOTHING along

those lines.

Just $0.02

Thanks

Happy Easter to ALL !!!

%:)

TomW · « **Reply #14 on:** April 16, 2006, 09:50:29 AM »

ColdSpot;

Some times maybe it would be of better use to and by

the members here if maybe things didn't go to private

E-mails and did in fact stay here and in doing so

a few more people could then actualay get it

straight and square in the head and gain more.

Yes, it always strikes me as counter productive to want to take stuff into private channels. Seems to be contrary to the very core of why this site exists.

It also strikes me as kind of creepy like the person or their ideas cannot stand up to peer review but thats another rant.

Here is to keeping it in the public eye so everyone can learn.

Cheers.

TomW

Nando · « **Reply #15 on:** April 16, 2006, 10:35:38 AM »

Amanda:

The XLS file is not longer available it is in a bad PC. Can you send it again ?.

I did not define the operating point of the load versus the internal impedance of the generator

YOU SAID : "Never the less, it clearly shows, for this generator, which does have a large inductive reactance component, that the half power equation does hold true in this instance."

Please send to me the half power equation.

YOU SAID: "but if the mppt is functioning as it should, this ratio essentially becomes immaterial. Except in defining the maximum input voltage to the converter"

MPPT as MPPT, the ratio is immaterial, but MPPT with broad operating range and maximizing the harvested efficiency, the ratio is needed.

And it is not defining the maximum input voltage, it is defining the maximum power to the LOAD because as you say in your diary : voltage times current = power

If the load impedance is fixed and equal to the generator impedance load, the maximum energy that can be harvested is just 50 % and may follow, in this case, the wind mill power curve -- MPPT practically useless --

With a high ratio Rload to Rsource, the MPPT can follow the power curve and maximize the available power at higher efficiency

The relevance here is to harvest the latent power available in the power source with limited torque power source.

Nando

robl · « **Reply #16 on:** April 20, 2006, 12:45:24 PM »

Nando

I think you're still working in the field, so if this request inmpacts your business I'll understand, but I would love an explanation of the way you employ PLC's in wind or (particularly) hydro setups.

I've got a small 240V/300W crossflow turbine that loses all sorts of power on the rectifier/control elements-side of the system. I've talked to industrial PLC vendors but they clearly have no understanding of that aspect of the technology. The closest I've gotten is instructions on activating the back-emf/braking function. Something like a dedicated MX60 or equivalent would be nice, if only the algorithms were convertible from their PV focus. The PLC's seem more site-specific programmable.

Regards

Rob

SmoggyTurnip · « **Reply #17 on:** April 20, 2006, 12:59:06 PM »

Wow - I have to completly disagree with this.

I am currently building a 18' windmill to do

just this. If the generator is matched properly

with the load there should not be a problem.

SmoggyTurnip · « **Reply #18 on:** April 20, 2006, 01:10:13 PM »

The load needs to "follow" the energy level being produced, so if the mill is producing 50 watts the load HAS to be such that just 50 watts are dissipated, if the mill produces 1000 watts, like in your case of 240 Vac the value would be R= (V^2/P) ; 240^2/1000 = 57.6 ohms which for practical purposes is a short to the 50 watts output.

If the internal resistance of the

mill was 5.76 ohms and the blades

delivered 50 watts to the gennie

then 45 wats would be consummed

by the heater and 5 watts would

be consumed in the stator.

In stronger winds where 1000 watts

were delivered to the gennie 100 watts

would be used in the stator and 900 watts

in the heater.

The efficiency of the generator will not

change if the load if fixed.

SmoggyTurnip · « **Reply #19 on:** April 20, 2006, 01:32:29 PM »

With a 3 phase alternator wired in a star configuration

connected to a resistive load that is also wired in star,

say 3 baseboard heaters the electrical efficiency of the system

is:

Efficiency = Re/(3*Ri+Re)

where Re is the load resistance

and Ri is the resistance of of 1 phase.

So by selecting a fixed load it is possible

to set the electrical efficiency to say 90%.

When you add MPPT to this system all it can really

do is effectivly decrease the value of Ri, yes this

will bring the blades closer to design TSR but in

doing so it also reduced the overall electrical efficiency

of the system. The blades are delivering more power

to the gennie but most of the gained power is being

used up in the stator.

So you are just spending alot of time and effort to

slow down the blades and heat up the stator.

This is what I believe to be the

case with resistave loads - I have not

looked at the situation for battery charging

as it is more difficult to analyze and that

is not what I am trying to do anyway.

ghurd · « **Reply #20 on:** April 20, 2006, 03:47:55 PM »

Might want to take a smaller windmill and connect it to a headlight, during low wind.

I think you are going to have a problem somewhere.

G-

oztules · « **Reply #21 on:** April 20, 2006, 04:07:31 PM »

Interesting,... does this "problem" still hold for a drag type blade, or only for those blades where stall kills efficiency?

.......oztules

commanda · « **Reply #22 on:** April 20, 2006, 04:33:51 PM »

Huge over-simplification. Assumes the internal impedance of the alternator is a constant, at all rpm's & current's.

This also ignores the fact that the output power curve of the alternator vs rpm is a straight line; whereas the shaft input power generated by the blades is a cubic power function.

Amanda

ghurd · « **Reply #23 on:** April 20, 2006, 05:06:53 PM »

I don't know for sure. I'm not much of a blade guy.

I 'think' it would be much less severe for a drag type, brute force.

I have hope the drag style will be good for the current project, #3 of 4 billion.

Certainly a lift type blade set under load at startup conditions is severely hindered with much (5%?) resistive load, causing stall before it even starts really spinning.

They just won't go.

Maybe it could be cured with a fast zener, 723, mosfet, couple resistors, but I 'think' when the Voc passes the zener, it'll stall again limiting the output when the potential output is high. Then there is the vibration and blade torque, but didn't try this because of other issues.

Direct connect a car headlight to a large standing still 12V windmill, then wait for the wind. I expect even a 6' 12V lift based wind gennie direct to an #1156 tail light will do.

That is a great demonstration of what I am talking about.

G-

Ungrounded Lightning Rod · « **Reply #24 on:** April 20, 2006, 07:06:21 PM »

Actually the voltage is a striaght line, the power (to a resistive load) is a square function.

Still falls off from the cubic that the blade would be happy to give you. But not as fast as if it were first power rather than square.

A MPPT controller can raise the load at high speeds to pick up some of that shortfall. Yes the efficiency drops, because the current goes up more in the alternator than in the load. But the total power delivered to the load is still greater than without the MPPT controller - the losses in the alternator don't go up fast enough to eat the whole gain.

Downside is that the main limit on your genny is how much heating it can stand - which is proportional to the square of the current through it. As you approach that limit you're better off to let the voltage go up and keep the current down, so you can RAISE the efficiency and delay furling.

commanda · « **Reply #25 on:** April 20, 2006, 07:54:46 PM »

My apologies. You are correct of course. I was thinking of my F&P where the increasing internal reactance vs rpm means the half-power point vs rpm is a straight line.

Amanda

oztules · « **Reply #26 on:** April 20, 2006, 11:36:37 PM »

This all really points back to Flux's proposition of "matching the load".

If you design your alt for late cutin R goes down dramatically, and W= ExE/R (w=watts, e=volts, r=resistance) gives much better control over the stator thermal runaway as the R is maybe 1/4 than for say Danb's direction..(low cut in speed with blade stall to control over revving ...... no offence here Danb, we know why you have done that and it makes sense for you and your needs,..noise, speed, output requirements etc.)

The use of boost at low power levels makes electronic design easier, and of little relevance at higher power levels. MPPT as described above becomes more critical as power levels go up, and as such more prone to failure as power and silicon chips mix. A 500w boost can be cheap and nasty, a 5KW controller is a league above that.

From a practical level, it appears to me that Flux has it nailed from efficiency and designability and above all dependability. MPPT may be the amswer to a mismatched system, but may provide diminishing returns of complexity against reliability and efficiency compared to a well designed "Flux load matched design".

I think 1'm becomming a convert..........oztules

SmoggyTurnip · « **Reply #27 on:** April 21, 2006, 06:31:59 AM »

What size are the blades?

What power is the headlight?

What is the phase resistance of the alternator?

What is the open circuit voltage of the alternator.

Any windmill can be stalled in low winds (or high winds for that matter)

if the load is to large.

SmoggyTurnip · « **Reply #28 on:** April 21, 2006, 06:39:14 AM »

Correction - I ment to say MPPT effectively changes Re not Ri.

For an air core axial flux alternator I think

the internal impeadance can be considered constant

at all rpms.

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Author Topic: resistive heating loads (Read 4542 times)