Author Topic: OUTPUT ? (Read 2594 times)

2windy · « **on:** November 28, 2006, 07:15:14 PM »

I have a 12 volt wind genny that puts out about 25 to 30 amps. When I check open voltage when its windy and putting out 30 amps I get about 40 volts. When I hook it back up to the battery bank it drops to 13 volts of course. My dumb question is am I only generating 13 volts x 30 amps = 390 watts or am I getting 40 volts x 30 amps = 1200 watts ? I can hook up a 800 watt electric heater to my inverter and the genny keeps up with it.

Flux · « **Reply #1 on:** November 28, 2006, 12:37:45 PM »

Confusion here somewhere.

"When I check open voltage when its windy and putting out 30 amps I get about 40 volts."

If it is open circuit, there can be no amps. I presume you mean under the same wind conditions.

It is not wise to open circuit a wind generator in high winds, I am surprised you only get 40V.

Anyway the power out is the current times the battery voltage, that is the only thing that means anything. Sorry you have to live with the 390W, the other figure is dreaming.

Flux

AbyssUnderground · « **Reply #2 on:** November 28, 2006, 01:01:56 PM »

You would be better going for a 24v system since your genny puts out enough to do so. You would make much better use of the power that way.

2windy · « **Reply #3 on:** November 28, 2006, 01:04:27 PM »

I just opened the circuit for an instance to see how many volts it was doing , I do realize it would self destruct. How come I can run the heater with no problem?

Nando · « **Reply #4 on:** November 28, 2006, 01:07:26 PM »

Flux:

Why surprised, reasons ?.

Nando

kurt · « **Reply #5 on:** November 28, 2006, 01:22:14 PM »

"You would be better going for a 24v system since your genny puts out enough to do so. You would make much better use of the power that way."

that depends totaly on what rpm it cuts in and prop size.

Nando · « **Reply #6 on:** November 28, 2006, 01:24:22 PM »

You have two voltages, one with no load: the voltage is the generated voltage .

The second voltage is the voltage clamped by the battery and the difference between both is close to the dissipation that the generator suffers with the standard wind mills being clamped by the battery bank,

With the values You have, One can say that the generator may be dissipating (40 - 14) * 30 amps = 26 * 30 = 780 watts -- the value is not real because the measurements may be wrong.

If they were accurate then the total power at that moment would be 40 * 30 amps = 1200 watts.

(14 volts * 30 ) + ( 26 * 30 ) = 420 + 780 = 1200 watts.

This may be telling you that the generator is dissipating a lot of power ( 65 %) and 35 % into the battery.

Normally, most of the generators that charge batteries directly, have around 53 % efficiency.

Knowing the internal resistance of the windings, then one can say what would be the best harvested power if the power is not clamped by a battery bank.

One can obtain around 80 to 86 % efficiency if one uses a charge controller with MPPT capabilities.

Nando

Flux · « **Reply #7 on:** November 28, 2006, 01:47:09 PM »

I assume you are running the inverter from the battery, if you try it direct without the battery then it won't last long.

The battery will run the load for some time, with the help of the generator it is probably longer than you tried it for.

If it will run continuously then one of your measurements is wrong either the power in or the heater load.

Flux

2windy · « **Reply #8 on:** November 28, 2006, 01:59:14 PM »

I have 8 T-105's hooked to my inverter. The heater [measured with the amp meter amp

and says on it is 800 watts] run for four hours before I shut it off. The battery voltage remained at 12.5 with the heater on and when turned off it went right back to 12.8 and climbed till it started dumping power to my dump load. It was a steady 20mph breeze also.

ghurd · « **Reply #9 on:** November 28, 2006, 02:21:34 PM »

This fits in somewhere.

Might be easiest to use the dump load for heat. Like the AC heater, but automatic.

The dump load will need to be larger than the output at furling.

Maybe a switch to trick the dump load into turning on at a tad lower voltage during windy periods?

(That would depend on the dump controller)

Maybe hack the inverter switch so the dump controller turns on the inverter and VAC heater?

My plan is to do that for an air compressor. Some day.

No baby-sitting. No complex MPPT circuits.

G-

harrie · « **Reply #10 on:** November 28, 2006, 03:00:00 PM »

Hack the inverter switch? This would be nice, can you explain, or give a diagram, maybe even I with little knowage of electronics could do it?

ghurd · « **Reply #11 on:** November 28, 2006, 03:34:20 PM »

Many of the inverters I have seen, and I have not seen them all...

The switch operates some 'stuff' in the inverter. The 'stuff' is of no consequence.

That tiny switch is not handling 500A from the battery!

The 'remote switch' can operated from a phone cord, on most.

The phone cord can not handle 500A either.

The dump controller can operate a very small relay, closing a switch... like the inverter remote control switch closes, turning on the dump load.

The hystersis between the dump load controller On and Off will need to be high.

A homebrew controller can do that easy enough.

I hear some inverters can NOT be turned On with a load connected, but I never saw one that I am aware of. (maybe the very large / expensive / 'sleep mode' style)

Really... What's the point of a remote control switch?

So I can turn the inverter on, then run to the inverter and plug in the load, then run back to where the inverter remote and the load are located, and then turn on the load?

A rapid cycling heating unit on an inverter can't be a great idea.

Hystersis should be 'quite high' I believe, and a 'On' timer (say 30, 60, or 120 seconds minimum per cycle) may be a good idea.

It would be as simple as turning on a LED for 30/60/120 seconds when the battery reaches a cetain voltage.

The LED would simply be replaced with a small relay, the relay would replace the inverter switch.

A basic and simple hack if the inverter and dump load controller are compatable with the concept, and each other.

G-

Ungrounded Lightning Rod · « **Reply #12 on:** November 28, 2006, 06:08:46 PM »

The second voltage is the voltage clamped by the battery and the difference between both is close to the dissipation that the generator suffers with the standard wind mills being clamped by the battery bank,

Actually (unless you measured the open circuit voltage VERY QUICKLY, before the genny had time to speed up) the dissipation in the genny is considerably lower than that. The load slows the genny and reduces the generated voltage.

Part of the "lost" power is in the form of genny heating, part in the form of power than was never converted from air motion to shaft power because the blades slowed down.

bj · « **Reply #13 on:** November 28, 2006, 07:53:24 PM »

Not that he needs the encouragment, I agree with Flux. I only

measure volts open curcuit, as I charge into batteries. That tells

me if I have enough volts at a given RPM to charge the batteries.

Its easy to forget that if you don't have the voltage, nothing

happens.

For my system, I only measure amps into a battery that is

discharged to as close as I can get to 11 volts. (want to protect

my batts) Then measure amps. The only figures I work with are

these. But as I said, I am charging into batteries, and need to

understand what the batteries want.

Flux, am I out to lunch?

bj

bj · « **Reply #14 on:** November 28, 2006, 07:59:36 PM »

Flux, could it be the changing watt load on resistive loads? I

used 12 volt halogens, (300 watt rating) for a while as a dummy load.

The mill would stall, to a point, then the lights would get

kinda bright, and then the mill would take off. Many burnt out bulbs.

Which explains previous post.

Again, am I out to lunch?

bj

Flux · « **Reply #15 on:** November 29, 2006, 01:11:33 AM »

bj in case this turns up somewhere else this is for you.

I am not sure what you are doing.

Are you trying to switch the load from the battery to the halogens when the batteries are charged?. You could use the halogens as a dump load with a controller, but you will not be able to run them directly from the mill. When cold they have a low resistance and will stall the mill. As the wind picks up, they start to heat and the resistance rises, this lets the mill come out of stall and you have a runaway situation and the lamp blows. This is worse than a direct heater and that doesn't work.

Used as a diversion controller, the lamps are ok as the battery determines the voltage.

Now your second point, I really don't follow that. You are right that without enough volts you can't charge the battery, but as long as you have amps then you must have enough volts.

I am not sure what you mean about the 11 volts, it seems as though you may not be treating your batteries very well. Don't wait until the batteries are down to 11v before you charge.

What you need to do is to charge all the time but control the volts so that you don't over charge. If you just have an on/off controller you need to apply a load when you reach about 14v and remove it at a lower voltage ( say 13.6). This load will be taken from the battery. If you want to remove charging current rather than dump in the normal way, then you need to probably dump the mill into a load that will stall it and accept that it will probably stop. You need to be sure that your mill will work safely in that mode ( it needs a powerful alternator)

With pwm control you can keep the volts virtually constant at float volts, but it is still cycling between small limits at high frequency. Again normal practice is to divert from the battery. You can do pwm control before the rectifier but not with a conventional controller unless you adapt it to take its control and power supply from the battery and do the pwm load upstream.

Flux

2windy · « **Reply #16 on:** November 29, 2006, 09:06:27 AM »

Actually I did measure the voltage before the genny had time to speed up. I have a battery disconnect on my control panel, right next to my volt meter. This volt meter is hooked direct to the genny, I have two one for battery voltage and for for this. I know it is not needed but I just thought it would be nat to watch and to see how much wind I needed for cut in. I forgot to mention in my first posting that the amp meter did go to zero when I disconnected the load, but Flux thought that I meant the amps where still there. I have all Simpson anolog meters for amps and voltage except for one digital meter to monitor battery voltage. Could these meters be wrong? One other thing that I did not mention is that if I left the battery disconnected for any amount of time, more that a few seconds the voltage would really climb.

Flux · « **Reply #17 on:** November 29, 2006, 09:30:08 AM »

I am not sure what you are asking.

The open circuit voltage when you pull the switch will be the emf more or less at the stalled operating speed for that wind. If you wait then the prop will speed up to runaway speed and the voltage could well double, that is why I was surprised that you only got 40v open circuit, but everything depends on the type of alternator , prop and wind speed.

The true power into your battery is still Amps times battery volts. The voltmeter seems to be correct as it is reading sensible battery volts.

Whether the ammeter is correct is anyones guess. If it has a shunt then there is potential for errors but without all the details only you can tell. If there is a shunt and even if it is the correct one, an analog meter needs calibrated leads to the shunt, if the connecting leads have a lot of resistance then it could read low.

If you could maintain that 800W load for long periods and still have the battery recover quickly to dump then it does question the accuracy of your ammeter. With 800W going out and 400W coming in you would be using 400W or at least 34A. For your 4 hours that is about 130 AH and would take a fair while to re charge.

You could try feeding a known load from the battery via that ammeter as a rough check. 12V lamps or something would give a fair idea.

Flux

bj · « **Reply #18 on:** November 30, 2006, 05:51:15 PM »

Flux--first of all Thanks. Many of us newbies are floundering

in bad advice. Yours will help the battle.

bj

2windy · « **Reply #19 on:** December 01, 2006, 09:26:58 AM »

I put my clamp on hand held Fluke, it reads the same. I hooked op some 50 watt halogens through that meter and it reads correctly.

Spelljammer · « **Reply #20 on:** December 04, 2006, 02:10:28 PM »

Maybe I missed something, but you said that your windmill was keeping up with an 800 watt heater. If that is a 800 watt / 120 volt heater and you are running 40 volts through it then it may work, but not at 800 watts. More like 88.9 watts. And at that low of a wattage, the windmill would speed up until the wattage load matched the wind power (minus losses) and a steady state is reached. Maybe it stabalized at 60 volts for instance. That would be 200 watt output. Like if I hooked up a 120 volts to a 3000 watt 240 Volt oven heating element, I would only get about 750 watts. My math may be wrong but I thought it was a square of the ratio. For instance, if I use half the volts as intended and square it, that would be 1/4th of the original power.

So the heater would be putting out heat and you would think that it is "keeping up" but it is purely anectdotel. I do agree that you can get more power by using a better matched battery bank voltage though. If you take a gen that can handle 10 amps before burning up and produces 40 volts in a 25 mph wind and hook it to a 12 volt battery, you could only hope for up to 15 volts at 10 amps or 150 watts. If you hook that gen to a 24 volt, then you could max at about 30 volts at 10 amps, or 300 watts.

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Author Topic: OUTPUT ? (Read 2594 times)