Battery resistance - Whats the mystery ?

[Dave B]

 I'm wanting to set up my mill to charge batteries as an option to my current set up of heating direct. I am conservatively rating my 16' at 3 KW @ approx. 100 VAC 3 phase star. I'm planning on a 48 VDC bank with a nothing cheap inverter and dump load at full charge.

 My questions are : Is there a minimum safe capacity for my battery bank ? If the rectified voltage is 1.4 x the AC this means it's possible I could see 140 VDC ? Am I figuring this right ? As is now I measure approx. 35 VAC per phase @ 105 RPM. If this is times 1.4 this gives me approx. 50 VDC. So if I'm figuring this right I would hit cut in at near 105 RPM.

 Fully furled now is limiting output to around 100+- VAC @ approx. 300 RPM with my direct load of (3) 4500 watt 240 VAc heating elements wired delta, (alternator is star) This figures to around 3 KW output.

 Back to my original title question : Other than an ammeter in series with the charging circuit and using ohms law is there any way to know what the battery resistance is at different voltages ? Any help or suggestions would be appreciated. Thank you Dave B.  

[electronbaby]

in your case, ohms law would be the best way to find the battery resistance. I would try and make a chart, that way, it would be easy to figure out what the internal resistance does at different charging currents.

Battery resistance is a factor of many things. It has to do with the charging rate, the voltage of the bank, and other physical characteristics of the cells (actual plate surface area, material makeup, and specific gravity at the time you measure). Every battery will be different. I suggest working with the battery manufacturer to determine what the charging characteristics that they recommend.

The resistance will always change. This is a factor of the SOC of the bank, so dont plan on having one simple answer. You need to understand the resistance curve.

The biggest difference you will see if you change out your heating elements for batteries, is a much higher current from the alternator at times. Although the ohms law calculations might work out to a similar wattage (voltage and current are inversely proportional), what you are doing is changing the power curve of the alternator. This will in turn change the way the machine furls and behaves in different wind speeds. It will definitely change your cut in speed from what you are used to because heating elements are for the most part a constant resistance. Batteries behave in a whole different fashion.

[Flux]

I am reluctant to reply to this, I have tried at least twice and the concept doesn't seem to register.

Battery resistance is meaningless in this case. If you short a large 48v battery bank the potential current is in the thousands of amps. The actual battery resistance is in the milliohm region and not relevant to the charging requirements.

There is probably a minimum  safe ( prefer to call it practical ) battery capacity as you would probably need something in the region of 100Ah to safely absorb your charging current in the bulk charge region without damaging the battery.

"If the rectified voltage is 1.4 x the AC this means it's possible I could see 140 VDC ? "

No this is not going to happen, that is probably where you are struggling. If it is a partly charged 48v bank then you are going to see something a bit over 48v and it is not going to rise ( except very slowly as the battery charges).

Just above cut in your current will try to rise far more rapidly than with the heating load and if you had an engine driving this then you would have serious currents to contend with. With wind, you will just drive your prop into stall and you will likely get less current than with the better matched heating load. Unless you have some serious line resistance or you deliberately add some I doubt that you will get the speed up to anywhere near the 300 rpm. If you cut in at 105rpm I would guess that you will hold hard in stall and be surprised indeed if you can get it up to 200 rpm.

Whatever you do you will need to limit the alternator ac line current to the same value as your heating scheme to keep the stator safe.

Your load voltage will be the charging battery volts and the load power will be this voltage x the dc charging current, if you are trying to think in terms of an equivalent load resistance you will get seriously confused as it is a non linear load. If you try to base things on Current squared x true battery internal resistance your figure will be in watts or fractions of and the result will be meaningless.

This is an energy balance situation, you are balancing the mechanical input of the prop against the electrical battery charging load. You don't have unlimited power from the prop and the power you have will be a function of the wind speed and the operating tsr. Your tsr will drop like a stone above cut in unless the alternator is seriously inefficient.

Not going to try again, I can't explain it any other way.

Flux

[tecker]

The batteries and rectifiers will limit the voltage   . I believe your at 103 turns of 16 wire total amperage considerations for around 140 feet of 16 gage. I think your going to be getting hot  at 15 amps and above. You may find the voltage limiting to your elements will push the current up.

[Dave B]

Flux,

  What you write makes sense of what you explain. Ohms law I understand. Yes, if you short any battery you will have I = E/R.

"Battery resistance is meaningless in this case. If you short a large 48v battery bank the potential current is in the thousands of amps. The actual battery resistance is in the milliohm region and not relevant to the charging requirements."

  How is the battery resistance not relevant ? If say someone measures 30 A charging doesn't ohms law state that there must be a resistance and voltage to measure current ? And, won't this battery resistance change as the state of charge changes ?

  "If the rectified voltage is 1.4 x the AC this means it's possible I could see 140 VDC ? "

"No this is not going to happen, that is probably where you are struggling. If it is a partly charged 48v bank then you are going to see something a bit over 48v and it is not going to rise ( except very slowly as the battery charges)".

  Yes, I understand this also (if charging) I just wondered if I had the 1.4 x the AC voltage correct for figuring the (open) DC voltage. If so then I could expect to see approx. 140 open volts DC from 100 VAC. Correct ?

   "This is an energy balance situation, you are balancing the mechanical input of the prop against the electrical battery charging load. You don't have unlimited power from the prop and the power you have will be a function of the wind speed and the operating tsr. Your tsr will drop like a stone above cut in unless the alternator is seriously inefficient."

  This makes perfect sense to me also, only though if in your first sentence ... battery charging LOAD is a "relevant" resistance.

  I am hoping to understand battery charging, IE: current, resistance, charge rate, amp hour rating etc. with regard to a constantly changing DC charge voltage such as with a wind generator.

  I am seriously trying to understand this and you are correct that if the resistance component is "meaningless" then the other calculated relationships I am trying to figure will be impossible to figure and therefore will be "meaningless" also.

  Going back a few postings under controls maybe someone can answer my following question : Let's say I have a constant wind for as long as I want and this produces a constant rectified 70 VDC voltage. I have an ammeter properly connected to the charging circuit to a 48 VDC battery bank that is at 50% fully charged. What will I read at the ammeter and what components of the circuit are responsible for this reading ? As charging time passes, what will I notice for  the ammeter reading ? I'm not looking for exact figures just : what happens as the battery bank charges ? When during this process does the resistance of the stator + line resistance become a concern for stator heat and or burnout ?

  I think understanding these relationships will help me and possibly others design safe DC battery charging set ups for our specific systems. I appreciate any information in regards to charging batteries from devices whose DC output constanly changes from well above and well below the rated battery bank voltage. Thank you,  Dave B.

[ghurd]

I'll have a crack at it.

The battery resistance is almost 0.00 ohms.

'Nothing' compared to the ohm or 2, or 10 ohms, of the rest of the circuit.

Not enough to bother figuring in any calculations.

Maybe consider the battery the same as a perfect 1Giga Watt 48V Zener diode.  Or 56V if you like.  Same idea.

G-

[Dave B]

Hello Ghurd,

  Thanks for your input also. If say someone here says they saw 50 Amps charging into their battery bank, what would this mean ? Please correct me if I'm thinking wrong ...

  OK, lets say the battery bank is .01 Ohm (maybe this is a high resistance for a battery bank) I don't know and is a big part of my questions. Maybe this is a poor example but if my math is correct and if the ammeter is measuring 50 Amps charge then I figure the DC charge voltage that the battery bank sees must be 1/2 a volt ? I=E/R For a 48 VDC bank does this mean that it is nearly charged at 47.5 VDC and can only take another 1/2 VDC but at 50 Amps and 25 watts ?

  Then say the alternator was putting out 70 VDC at the time, does this mean that 69.5 VDC is what the stator resistance + the line resistance sees ? So say this total is 2 ohms, would this be 69.5 /2 or 34.75 Amps. Then 34.75 x 69.5 = 2415 Watts the stator and line must dissapate ?

  If this is the general idea behind charging the battery bank then I can very easily see why people are concerned with burning up their stator. It's not the stator it's back to the load balance thing again and my original question.

  Is there any way you can measure or know the battery bank resistance at different charge voltages ? Like I had always suspected, this then is not meaningless but very relevant to understanding the charging circuit and applying ohms law throughout charging from needing a charge to fully charged.

  If the battery bank resistance changes as it charges then the performance of the wind generator would be directly related to these changes throughout the charging process. Different blade designs of the same diameter for the same charging system then would also perform differently which again brings us full circle ... matching the output, load and blades. Am I getting it ?

  Thanks Ghurd for your help and anyone else who wants to chime in.  Dave B.  

[Dave B]

Electronbaby, (Roy)

  Thank you for your explanation also, for some reason it didn't show up until after I refreshed my screen and this was after my last posting and reply to Ghurd.

  It sounds like I'm getting it and if in fact the resistance of the battery bank changes during charging then yes, I understand the effects this can have on the overall performance of the wind generator too.

  Sometimes it's hard to ask the questions in such a way that you connect with others. I think this is a very important bit of information and a big reason why running an inefficient alternator along with basically running blades stalled is a fairly safe way to charge batteries. This of course as long as your furling kicks in before the blades come out of stall and really start to fly when you near fully charged.

  Fried stator if all this is not tweaked just right. Thank you again for your help, any other replies are always welcome.  Dave B.

[ghurd]

OK, your numbers are all mixed up.  It's late so maybe it's me.

Seems like you started at both ends of different candles. (I'm rejecting some of your numbers, and took some other liberties)

Your numbers... 50A charging current, 47.5V battery, 70VDC open circuit PMA voltage.

Start with 70V - 47.5V = 22.5V.

Kirchoff's Law.  That 22.5V went somewhere.

Ohm's Law.  It was used up in the resistance.

The total circuit resistance is 22.5V / 50A = 0.45 ohms.  (one Damn Big^2 windmill!)

Lets say the battery is 0.010 ohms, because that's kind of high (I think- not sure).

The 70V open, and 47.5V battery are the same.

The current, if the battery resistance is ignored = 22.5 / (0.45 - 0.010) = 51.1A.

The difference in watts is 47.5V x 1.1A = 52.25W.

Seems like a lot?  Think about it this way...

A "48V" windmill should cut-in at about 52V open, and about 6MPH?

And 70V/52Vx6MPH=8.08MPH.  Those numbers are for a windmill making 2,375W in an 8MPH wind.

Think about it this way...  Cube square, furling wind speed, percentage of losses at 8MPH, ya-da, ya-da, ya-da...

That's like a 50KW machine worried about 52.2W.

Think about it this way...  That 0.01 ohms is high, super high for a battery bank large enough to take those kinds of peak amps.

(From working with batteries, I believe the calculated resistance of a battery would be related to the current flowing through it. There is probably a name for it.  It is late.)

If a suitable sized battery is used, the resistance is more realistic for the system battery, and the math is done again, the battery resistance is "Nothing".

Not enough to bother figuring in any calculations.

If you care to get a migraine over it, there is more to think about that may be a far larger margin of error.

Elevation, humidity, temperature, 101% perfect blades...

Loaded blades not actually running at their perfect design TSR, load, and wind speed all at the same time.

Even the diodes.  The forward voltage of a diode is related to the current, and temperature of the diode.  

I have tested the same diode at the same temperature, at ultra low current (as in the above example) from 0.3V up to 0.8V (full rated current).

So much for tha "0.7Vf" in the math.

It all comes down to the battery resistance is 0 ohms.

Gosh.  All my typing fingers have blisters. All both of them.

Hope that made some kind of sense.

G-

[zeusmorg]

 Battery measured resistance is not an important factor (nor accurate), really. batteries will vary as to what their resistance actually is. This depends on factors such as size and thickness of the plates, connections etc. Resistance does change with state of charge, however it is not a good way to determine the state of charge.

 The measured battery resistance changes so slightly between discharged at 75% and 100% it would not be an accurate way to determine your SOC.

 Battery Resistance does increase with age,and abuse, a high internal resistance can mean it is near or at end of life.

 It may be a good idea to read the lead acid battery FAQ to understand how to properly test your batteries, and don't use cheap equipment to do so. Why rely on a 5 dollar hydrometer when you have a fortune tied up in your batteries?

  A battery will not charge below the voltage at which it currently is, for lets say a 12v system, the charging voltage will remain just above the current SOC voltage, and rise as the battery charges up.

 For example, let's say your battery was down to 10v. (bad btw) so at around 10.2 or so it would start accepting a charge,(this is regardless of the amperage) and this would increase to 14.8 (about) for a 100% charge. Putting in a higher voltage, let's say in this case 25 v., the battery "pulls down" that to near it's current voltage. So the measured voltage across the terminals while charging would be at 14.8.

 To ACCURATELY determine battery resistance, a load test must be done. Applying a known load at a measured voltage drop.

 I hope this clears up a few misconceptions.

[Flux]

Ohm's law only applies to linear resistors. Any other form of Load doesn't follow Ohm's law. If you connect an alternator to a capacitor or a perfect inductor you will have current but there will be no power involved. You can sort of extend ohm's law at least to the extent where you can deal with voltages and currents if you substitute impedance for resistance, but the power calculations don't work.

For any non linear impedance you can't even go this far. A battery has most of the characteristics of a zener diode and in some ways it looks to a rectifier as a capacitor when charging batteries but this analogy is not exact.

 " Going back a few postings under controls maybe someone can answer my following question : Let's say I have a constant wind for as long as I want and this produces a constant rectified 70 VDC voltage. I have an ammeter properly connected to the charging circuit to a 48 VDC battery bank that is at 50% fully charged. What will I read at the ammeter and what components of the circuit are responsible for this reading ? As charging time passes, what will I notice for  the ammeter reading ? I'm not looking for exact figures just : what happens as the battery bank charges ? When during this process does the resistance of the stator + line resistance become a concern for stator heat and or burnout ?"

This can't even happen directly without big problems. You just can't apply 70v dc to a 48v battery, the current would be enormous.

I assume you mean 70v dc open circuit ( emf). If you connect this to a 48v battery the volts will fall instantly to battery volts ( say 50v). If the speed doesn't drop ( which it will in real life) your current will simply be 70 - 50 =20v divided by the circuit resistance. If the alternator was 1 ohm internal resistance and the line was 1/2 ohm for example then the circuit current would be 20/ 1.5 = 13.3A.

The power dissipated in the alternator would be 13.3^2 x 1 Watts and the power lost in the line would be 13.3^2 x 0.5 Watts. Power into battery is 50 x 13,3 Watts.

Adding a milliohm for the battery internal resistance is not going to change things.

As the battery bank charges the nominal 50v that we chose will rise to about 56v at full charge so the forcing voltage will fall from 20 to 70 -56 = 14v. At true constant speed the charging current will fall. Under normal wind conditions you couldn't even assume that, many machines would come out of stall to some extent and the charging current could rise rather than fall. In most cases there will be very little change in current.

The line resistance should never be a concern for burn out, it could be sensible to use the cable resistance if you need extra circuit resistance but it should be capable of handling the full charging current within its thermal rating.

The stator will be in trouble when the charging current squared times its internal resistance produces sufficient heat to cook it.

Roughly the rms ac into a 3 phase rectifier charging a battery is 10/13 of the dc charging current. In the case we chose to look at you would see the same heating in the stator with your 13.3 A charging current as with 10A rms per line into a resistor connected as 3 phase load. If you rectify the dc to a heater the mean to rms factor will be a bit different into a heater but not a lot.

Hope this helps.

Flux

[tecker]

If the elements are Delta connected with the rectifiers parallel the elements won't react . Series connecting the elements to the rectifier is good for both battery and heater the heaters will react to the 7  to 10 amps your putting into the batteries per phase and keep the voltage up a little  .You won't be wasting  power across the stator and in fact storing power that is normally sinked . Batteries in good condition look like a short regardless of sate of charge  resistance wise to a voltage 5 to 10 volts over rated battery volts .If your stator can maintain the current .

I don't use three phase because I don't have the wind to drive it . I don't think I'd use it even If did because I like shifting power to multiple loads and run not even worrying about furling out (except a few protos that need to be shut down ).

 My point is  I don't have any star data that would help .All the wind units are feeding through some load before they hit charging rectifiers . This of little use now but having potable water and heating fluid available on cold windy days works .

The bottom line is the devices that are power hogs can be series circuits with a little planning .

 You can dump to more elements when the batts are not in use . Or turn on lights etc.

[bob g]

from an ignorant bystanders viewpoint

seems to me that trying to get to a definitive answer is much like holding water in a sieve or shooting at a moving target. (a very irratically moving target)

and i am not at all convinced that ohm's law is the last word on figuring where the power goes.

considering ohm's law is most useful in purely dc circuits, and somewhat less useful in ac circuits,, the question then becomes

is an AC generator/alternator charging a battery via rectifiers more like a purely dc circuit?   more like an ac circuit?     something in the middle?   or worse ... something in the middle that moves around acting more like a dc circuit at one point in the charging process and then more like an ac circuit at another point?

and is this relationship linear?

seems like a pretty good case for more effort in development of controlled rectification  or some form of switch mode conversion scheme.

there is just so many friggin variables, i can't see how anyone is going to do all th math needed to get a load match between generator and battery that suits a broad spectrum of windspeeds, and the broad spectrum of power that goes into charging a battery.

sincerely,, what am i missing here?

bob g

[valterra]

From another layman's point of view -

"Battery resistance" is one of those things that you never worry about because you never see it manifest itself in your design.  It has no effect on anything.

Then one day you hear that batteries have internal resistance.

Suddenly it is a factor in all your calculations, and you're having heart palpitations and cold sweats over it.

[Dave B]

It's pretty obvious now that not too many understand what is happening both on the AC side and the DC charging side of their wind generator battery charging circuits. Copying nearly exact to what has worked is the safest bet but change this a bit and you see the burnouts, overspeeds, poor output and stalling etc. I think with a complete understanding of the battery charging circuits most of these problems could be avoided as well as quite possibly designing a more efficient and safe system overall for charging. (this is what I am attempting to do). If I'm not mistaken it seems to me that there is a tremendous waste of power in most of the charging set ups I see posted here. Sooner or later with new comments showing I'll get a better idea of what's going on. Thanks again to all who have commented, I'll be reading as we move forward.  Dave B.

[Flux]

There is no mystery about conventional charging circuits at least as far as I am concerned.

The conventional compromise works and is perfectly easy to understand, there is little trouble in calculating what is going to happen. Some degree of stall is inevitable and is also the low efficiency in high wind.

years ago I covered all this issue under a diary entry about matching the  load, where I proposed virtually every method I have come across to improve this load matching. Most of the schemes work in some respects and add additional complications. The only schemes that seem to offer much benefit without introducing other issues are electronic converters that change the voltage ratio with speed. These really do let you match the variable alternator voltage to the fixed battery voltage. They significantly improve the efficiency and let you reach the full charging potential in high winds without frying the stator.

Even these schemes come at a price, firstly the efficient alternator needed is far more costly than the low efficiency things used for normal matching. The electronics adds complexity and although it can be made adequately reliable, every complication adds to the chance of failure. As wind energy is free most consider the present compromises adequate and most of the power is produced in lighter winds where the efficiency is not so bad. Unless you have a large battery bank the thing is usually fully charged on the very windy days and power is being dumped so worrying about higher efficiency will likely only increase the problems of the charge controller.

Stator burn out is completely unnecessary and if it occurs then it is simply because the furling is wrong or in some cases just doesn't work because basic ideas again are not understood or sometimes people are plain greedy and try to get more than is realistically safe.

I have had converter schemes running for years, but few others seem to have managed it.

Equally true, some others have had machines running for years the conventional way and achieved acceptable results and not burnt out. It is only a case of understanding the compromises and working within the known limits.

The solution is not going to be magically be found in some magic battery characteristic that doesn't exist. You accept the the electronic converter route and have all the benefits with some of the added cost and possible risk or you settle for the tried and tested ( or should be) conventional way. If you go the simple route some stall and considerable loss of efficiency is inevitable but if you burn the machine out is is just your own fault, that is not inherent.

Flux

[Dave B]

  Great comments and an interesting thread for sure. Although I may not understand all of it the one thing it does explain is why I have had to tweak here and there, rewind stators, change furling angles and experiment with different blade profiles and length etc. for me to adapt the "norm" axial plan for direct heating. It also could explain to others who have tried direct heating with the "norm" axial plan here and have had problems. Possibly burning up a stator, very bad stall with little output and or the opposite of furling way too late and overspeeding or even worse a blade strike or self destructing.

  I guess I'm kind of on my own again as I was with very few out there direct heating. Now I want to take a modified machine and safely charge batteries with it, this is why I am trying to learn about battery charging and specifically how to analyze the AC and DC circuits of a variable AC source voltage.

  It's a challenge and one I will work with until I have a satisfactory set up that I can switch from direct AC heating to DC battery charging without changing the machine. Life would be so much easier if there were battery resistance charts for specific batteries at certain voltages or percent of charge. Maybe there are, you can be sure I'll be searching and experimenting too.

  For those out there who say "just do this, or try that it should work" I say thank you for the suggestions but you can't compare apples to oranges. I have worked hard to have an orange, now I'm trying to make my orange charge a battery bank. My persistence to thouroughly understand the AC and DC circuits of battery charging will help me to more confidently design my own unique system to do so.

  Thank you for everyone's patience to help me understand battery charging. I have been through the overspeed (twice), burnt stator (twice) and blade crash experience (once) all by my choice to find the design limits of my previous machines for direct heating. I'm not in a hurry for a re-build as my machine now is a solid performer. I am walking into this battery charging system very carefully, hence asking so many questions. I'll be posting as I learn and my progress continues. Thank you,  Dave B.

[Flux]

Dave, Just one final comment. You can build a machine that works well for heating and battery charging and in fact you can combine the two functions to good effect.

Build an alternator with the correct cut in speed for your battery and build it with more magnet and copper than the normal battery charging machine ( it needs high efficiency for this).

Directly into the battery it will stall hard and the performance will fall off badly in winds not much above cut in. Now include a heater as a series resistance ( simplest will be one strange value one in the battery positive lead). Keep increasing this resistance until you get out of stall and the prop is on the peak of the power curve at your rated full load. Get it proportioned right and nearly half the power will go into the heater and nearly half into the battery. You will have an excellent battery charger in low winds, where heating is virtually useless. You will have a better battery charger than the normal stalled machine in high winds and you will have a heating machine of nearly half the capacity of the direct heating scheme.

If you can afford the magnet and copper to do it well enough you will have very little loss in the machine stator and heating will not be an issue. You will ultimately have to furl but you may be able to hold it off to quite high winds, where the heating could almost equal a heating alone scheme.

You will also have far more battery charging capability and unless you have monster batteries you can use an extra heater on the dump load to capture most of the energy again as heat once the batteries are charged.

Net result a better battery charger, good heating scheme, well behaved without complicated heating or battery charging controllers, predictable furling if you base it on your heating machine and no stator burnout issues. Downside large and expensive alternator costing perhaps 3 to 4 times a cheap pushed to the limit stinking hot battery charger machine produced the usual way.

Flux

[DamonHD]

Very counter-intuitive Flux, and very interesting.  Load matching is bizarre, along with transmission-line effects, RF ground-planes, and D-type flip-flops: all everyday bits of electronics but with a fair amount of black magic in the mix IMHO.

Rgds

Damon

[bob g]

another thing to consider is that batteries as they age change in internal resistance,, not sure how much, but again another variable to have to account for.

good luck with your project

bob g

[tanner0441]

Hi

To add a few more variables, how about reactance, inductance, and AC ripple passing the diodes.

It sounds like you have a package unit from plans or a kit. The maths will have been worked out, and I think you said you had 700 AH of battery power, that should sink any surplus power from the PMG and the cable run will heat up a bit if it is pushed.

Interesting reading though and it shows there is a place for a very simplyfied manual for people starting to be interested in RE.  All the sites I have looked at assume pre knowledge.  I am sure if the knowledge was readily available it would make RE less of a black art followed by geeks and strange people.  It would also stop some of the companies advertising turbines that defy the laws of physics.

good luck and don't give up, and circuit breakers work out cheaper than fuses in the long run....

[Dave B]

Tanner,

  Yes, there are many variables for those charging batteries and for those satisfied with what works and want to build their own they should follow the plans to the letter.

  I think you may have possibly read a different post thinking it in regards to my project. You mention me possibly using a "package unit" from plans or a kit. Nothing could be further from the truth as I have designed my system for hot water heating direct.

  Also, something about a 700 AH battery. Currently I do not use any batteries but plan to wire an option to do so. And as far as the good luck and don't give up ? Thanks for that.  Dave B.

[chubbytrucker01]

I heard a guitar instructor say you can't be really good at something until you have had to teach it. By now flux and ghurd should be the best teachers that ever lived. But I guess there is always someone who can come up with a better "What if?".

 If flux AND ghurd told me to ignore battery resistance.

 Just my $.02