Calculating power curve for my alternator

[DanB]

Perhaps someone can help me here - I get bogged down in simple mathmatics.

Regarding my large new alternator for the 20' machine.   As it's currently setup we cutin (48VDC after the rectifiers) at 65 rpm.

This gets a little goofy, but I weighed the stator (Before we cast it) and basicly there's between 21 and 22 pounds of copper in it.  It's wound with 3 strands of #15 gage wire and based upon that I figure each phase comes in around .6 ohms.

Am I correct in thinking that I can use the resistance of 1 phase (10 coils since we each phase is 5 in series and we're wired in Star) to figure the efficiency of the alternator at different output levels?  If so I figure it'll be right about 50% efficient at 4KW output...  which is about what I was hoping for.

By the same calculation my 17' machine is just about exactly 50% efficient @ 3KW output.  (I figure based on the weight of copper that 1 phase is around .84 ohms) I was seeing 96 amps from it a couple weeks ago - the stator must've been disapating around 7.5KW in heat!  (yikes)

At any rate - looking at my new one, figuring the resistance of 1 phase is about .6 ohms and figuring the cutin speed to be 65 rpm, how can we roughly calculate the power out over a range of rpm?  I'd like to actually test this with an engine but till then it would be fun to have a good guess.

[DanB]

Is it as simple as figuring open voltage at any given rpm, subracting battery voltage and then multiplying that times resistance?

so at 130 rpm we have 96 open volts.  Subtract battery voltage to get 48V.  48V x .6 =28.8 amps = about 1382 watts.  Is that a reasonable way to estimate what might happen?

[Flux]

Dan

Let's assume your resistance figure is right and I assume that by resistance of one phase you mean the resistance of 2 phases in series as you say 10 coils.

I prefer to call the resistance of one phase that of  an actual phase and for star I would call the resistance between a pair of terminals line resistance.

Assuming the resistance between a pair of leads is 0.6 ohms, I have found by experiment that it will behave as if it had an effective resistance of 1.3 times that value( lets call it Flux's constant and not worry where it comes from)

You are working at 48V ?.

Cut in is 65 rpm.

At any speed you can find the voltage open circuit by proportion. Say 130 rpm the emf is 96 volts. forcing volts above battery is therefore 48v.

Current = forcing volts/ effective resistance

I = 48 / 0.78  or 61.5A

For  195 rpm   emf = 3 x 48v = 144v   Forcing v = 144 - 48 = 96V therefore

Current = 96 / .78 = 123 A

You can work out different values to find your characteristic.

Basically to find watts lost in the alternator you will be close if you multiply the forcing volts by the dc current.

A more accurate way is to multiply the rms current squared by phase resistance to find power per phase and multiply this by 3.

Near enough for this type of alternator, the rms current is 10 / 13 of the dc value.

Remember in this case the resistance you use will be for one phase ( half that at the terminals).

Hope this is near enough for what you want, it works near enough for mine.

Flux

[Flux]

Dan

Second part of the question, 50% efficiency at about 4.3 kW

At 160 rpm  E=118v   forcing volts = 70  current is 70/.78 = 90A

P out = 90 x 48 = 4.3 kW.

RMS current is 69.2A   power per phase lost is 69.2^ x .3 or 1438W

Total loss is 3 times this 4.31kW

[Nando]

DanB:

A 4 KW generator utilized as a direct charging source, the generator will dissipate 1/2 of the energy, so if 4 KW out about, 4 kw being dissipated in the generator, this % is not well defined because I do not know enough generator parameters.

BUT you could use the same generator to produce into the load a power level that could reach 85 to 90 % or 6.8 to 7.2 KW --INSTEAD OF THE 4 KW You are expecting.

If each phase has 0.6 ohms and you may want to maximize the generated power, it is necessary to know the OPEN ( UNLOADED) Voltage generated at maximum RPM to determine its highest efficiency point.

Also, it is necessary to measure the voltage with a light load ( around 10 % of maximum power) @ the Maximum RPM, to define the voltage drop caused by the stator-rotator GAP.

With these parameters ( phase ohms, peak unloaded output voltage and peak lightly loaded output voltage @ PEAK RPM) the efficiency curves can be plot, as well as, the power output curves

ONCE AGAIN if you connect directly with me and supply the info, I can show you how to optimize such generator AND MAXIMIZE ITS OUTPUT to 85-90% efficiency.

Here you have a case that for once you could test what I say is a reality with a very simple set up, instead of dumping the raw energy into a battery load and wasting half of the energy in the generator, which under certain conditions will burn the coils, the same way that happened recently with one of the larger size generators that one of your friends fried.

With a simple resistive load you could harvest 85 to 90 % of the energy, then is easy to see what the wind mill will do with the proper WIND MILL-TO-CHARGER CONTROLLER-TO-BATTERY setup for 7 KW wind mill.

Nando

[willib]

I have noticed a somewhat dramatic decrease in physical resistance to turning the rotors of a machine when diodes are in place . this decrease is caused by the diodes.

what bridge rectifier will you be using ? make , model # ?

so i can look it up

[DanB]

Hi Nando -

"A 4 KW generator utilized as a direct charging source, the generator will dissipate 1/2 of the energy, so if 4 KW out about, 4 kw being dissipated in the generator, this % is not well defined because I do not know enough generator parameters."

At this point untill I actually test it with an engine Im not sure we can do better than to make predictions based on the information we do know.

"BUT you could use the same generator to produce into the load a power level that could reach 85 to 90 % or 6.8 to 7.2 KW --INSTEAD OF THE 4 KW You are expecting."

I don't see how I could attain that efficiency with this alternator without increasing the speed.

"If each phase has 0.6 ohms and you may want to maximize the generated power, it is necessary to know the OPEN ( UNLOADED) Voltage generated at maximum RPM to determine its highest efficiency point."

I suppose maximum rpm will depend on the nature of the blades.  If I want to furl early (Say 20mph) and if the alternator leans towards stalling the blades -  perhaps we can figure TSR of 5 and figure maximum rpm to be around 140 rpm.  So open voltage would be around 103 Volts and output would be around 70 amps (3.5KW) based upon what Flux has said - which seems a bit optimistic considering it is a homebrew machine, the energy available in the and the losses in the alternator.  Of course the line will add a touch of resistance but I don't think that will be too significant.  I have a feeling I may need to add more resistance in order to prevent the blades from stalling.

"Also, it is necessary to measure the voltage with a light load ( around 10 % of maximum power) @ the Maximum RPM, to define the voltage drop caused by the stator-rotator GAP."

It would be fun to measure it - but can't we work from the calculations above to some degree?

"With these parameters ( phase ohms, peak unloaded output voltage and peak lightly loaded output voltage @ PEAK RPM) the efficiency curves can be plot, as well as, the power output curves"

Can't we roughly calculate all these parameters with the information we know allready?

"ONCE AGAIN if you connect directly with me and supply the info, I can show you how to optimize such generator AND MAXIMIZE ITS OUTPUT to 85-90% efficiency."

Again - I don't understand how we can have such high efficiency with an alternator which has perm. magnets and must produce 4KW or so at such low rpm.

"Here you have a case that for once you could test what I say is a reality with a very simple set up, instead of dumping the raw energy into a battery load and wasting half of the energy in the generator, which under certain conditions will burn the coils,"

Id be curioius to hear details about your thoughts.  I don't worry a great deal about it - my goal is to furl early.  I expect more than 90% of my 'energy' comes in low winds at or below 10mph where the machine is going to be very efficient so picking up lots of extra efficiency in high winds will make a very small change in overall energy harvest.  It would be nice to reduce heat in the stator and if it could be done... interesting to understand how.  

 "the same way that happened recently with one of the larger size generators that one of your friends fried."

His was not furling correctly - he was having sustained output of 1800 watts from a machine that should be rated more like 700 -  1000 watts.

"With a simple resistive load you could harvest 85 to 90 % of the energy, then is easy to see what the wind mill will do with the proper WIND MILL-TO-CHARGER CONTROLLER-TO-BATTERY setup for 7 KW wind"

Yes- Im curious to hear your thoughts.  If it is not too difficult or time consuming I would be very happy to perform any tests on this that you would suggest.  Im hoping to build a hub that will let me hook it to an engine soon - that should be interesting.  Thanks for your thoughts on it all!

[windstuffnow]

  Dan,

    I discovered some time back while trying to figure out efficiency that ohms law seems to dictate where the 50% point would be given a set of numbers.

So... your 48 volt system divided by .6 ohms would dictate a 50% efficiency at around 80 amps or 3840 watts.   ( 48 v / .6 ohm = 80amps )

Ever since I discovered this I model my expected output around this so when its producing it's "rated" output it is 50% efficient at that point.   Sort of keeps it simple...

.  

[willib]

just thinking out loud , so to speak.

Dan , since efficiency depends on voltage output of the machine , and on the voltage of the battery you wish to charge .ie efficiency drops when charging a battery with much more voltage than the battery voltage.

What if, you could change the load voltage to match the ouput of the machine?

example : your 48V @65 RPM is perfectly matched to your 48V  load.

but at double the RPM you have 96V driving a 48V load. the current is through the roof.But if at 130 RPM your load was now 96V your overall efficiency would improve quite dramatically, because of the power lost in the stator would be a lot less.

[scoraigwind]

Hi Dan,

I calculate the resistance at 0.45 ohms (between two line wires) but this will increase with temperature, so 0.6 is probably good.  I agree with Flux's comments and the 1.3 constant.  That usually works well.  I agree that you would probably need a little line resistance to keep out of stall, especially is the battery is low (but that seems an unlikely thing to happen with this beast on the end of a wire).  If you already have a big thick wire then it might be time to consider a low resistance heating element of some sort, to warm your house.

Nando is right that you can get better efficiency by using a variable voltage machine (rather like maximum power point tracking for a PV panel).  Flux has built some successful devices to this end himself I suspect.  It's heavy electronics and it works but there is always the question did it work well enough to pay its way?  Sooner or later the answer will be yes.  Several commercial turbines use some sort of controller that boosts or bucks the voltage.  Bergey XL.1 is probably the best known.

The simplest way to get variable voltage is to kick up a bit of volt dop in the line and although it is lost power (as far as the battery is concerned) it can be gained heat and with a lot less cleverness required than an electronic controller.

I am waiting to see how you test this brute and the dump loads you will use :-)))  Mind out for funny smells!

[Slingshot]

I believe that is because there is no current drawn until the generator spins fast enough for it's voltage to be to forward-bias the diodes (including the battery on the other side of the diodes.  In other words, an air-core alternator would almost free-wheel until it reaches cut-in.

[Nando]

DanB:

Two messages to your message, the second will respond your questions more directly.

To harvest certain KW, the design of the generator has to be defined at the peak operating RPM.

Then, once the power and voltage is defined, the winding resistance needs the attention to fix the generator dissipated power at that output power.

So we could say 10 % in the generator and 90 % to the outside world.

I do not know the parameters of your 4 KW unit, it may not be possible to attain the 85 to 90 % efficiency.

Assuming a 5 KW generator producing 380 volts @ let's say 300 RPM

Load resistance = (V^2) / power

(380^2)/ 5000 = 28.88 ohms

The generator dynamic ohms is 10 % = 2.888 Ohms

phase resistance = 2.888/ 1.73= 1.67 ohms per phase winding resistance

Current @ 300 RPM

380/ 28.88 = 13.16 Amps

Generator dissipation = 38 volts * 13.16 amps = 500 watts

SO the design has to be defined from the beginning to attain desired characteristics.

Furling needs to be defined and not guessed.

Nando

[Gary D]

A strange thought keeps rolling thru my mind. On a post about a center tap on the star winding, Flux(?) stated the voltage would be approx 1/2 the peak of the regular voltage (I think). Since Dan B. has a duel voltage system, would it be possible to center tap with the unused ground wire and charge his 12 volt bank simultaniously? It could save running a battery charger off his inverter. The banks would need to be tied ground wise to complete the circuit... Probably foolish...  Gary D.

[Nando]

To calculate what is your generator capable to produce in a realistic way, the high RPM voltage is needed.

The generator has 2 losses, one is the resistive loss due to current and the other is the band gap stator-rotor that needs definition to know the dynamic losses the Gen has due to its gap -- which can be represented as a resistor value since this is power dissipated internally in the generator.

The calculated values I have presented before do not show the Gap loss mostly to avoid confusion BUT they do exist that may lower the over all efficiency.

In your case, yes the efficiency is obtained by increasing the output voltage to a level of better power transfer within the generator source impedance, critical value to obtain the desired high efficiency.

There is the need to maintain a high ratio of Load to Gen source impedance of at least 10 to 1 for the desired 90 % efficiency.

LOW RPM indeed may generate low voltage if the gen is designed as to produce low voltage - so your gen may not be usable for high efficiency loading.

Furling point is done mostly to limit the maximum power generated to "save" the generator from burning itself up due to over heating as you stated in your message.

~~~~~~~~~~~~~~~~

BY THE WAY one could install [sometimes] a maximum current detector to trip the furling point on maximum desired power ( once again; electronics that most people do not want because it costs money and may fail )

~~~~~~~~~~~~~~~~~~~~~~~

In my previous message I showed a way to obtain a high efficient output, also several months ago I presented the ROUGH calculations done on Chinese wind generator that has been sold in Europe to obtain much higher efficiency.

So finishing: high RPM ( 2-300+ ?), high output voltage and low phase resistance are most of the parameters for a high efficiency generator loading.

I may attach the power curves of the generator ( I found the power curve but I could not find what I wrote and the calculations).

Using the curve values, I think that, I showed how to obtain around 85 % efficiency.

Nando

[Flux]

Gary

Yes it is perfectly possible to take a 24 v load from the star point and it could charge 12v with a resistor or pwm controller.

[Gary D]

Thanks Flux, wasn't sure if I was totally insane or not! The ground wire resistance would be higher just due to being a different wire gauge. Don't know if this fourth wire use would be of interest to Dan, just tossed it out there. A knife switch could be put in line to cut power into the 12 volt bank when full, depending on bank size and use. Sorry to have hyjacked your post Dan..... Gary D

[willib]

This is true , but not what i was talking about.

i have done a lot of hand crank testing , and in a resistive load , at similar RPM, an alternator will present much more of a physical load to cranking , when no diodes are in the circuit..

this is because diodes have a forward resistance , i would imagine to be in the range of a hundred milli ohms or so,

this is why i was asking for the data on Dans bridge rectifier

[willib]

"The generator has 2 losses, one is the resistive loss due to current and the other is the band gap stator-rotor that needs definition to know the dynamic losses the Gen has due to its gap -- which can be represented as a resistor value since this is power dissipated internally in the generator."

when i dont understand something i question it.

I seriously question your thinking on this .specifically the second loss ? the first one is a given.

i have never heard of such a thing.

Band gap stator rotor?

as i see it the only loss in a generator is the first , the power lost in the stator , due to its internal resistance.

if you could elighten me , i would be very appreciative.

[johnlm]

Dan,

I plugged your info into my spreadsheet and came up with the following power curve for your alternator assuming you use a 20 ft dia prop with a TSR of 4.4. If you use a faster prop the unit will go into stall earlier than my info shows, but you might also hit cutin a wind speed that is less than desireable. (less than 5 mph). I also tried to calculate your phase and line resistance and came up with around 0.25 Ohms (phase) and 0.5 Ohms (line). I used the 0.5 Ohm number in my calcs for the internal resistance of the alternator and used 0.65 Ohms as the total resistance through the alternator and rectifiers and into the battery. Note that in my calculations the prop drops to running at design TSR at 18 mph and is running slightly less than design TSR at the 20 and 22 mph entrys. The effeciency column is for alternator effeciency only, and the calculations assume the prop to run at 70% of Betz for its effeciency. Maybe someone has better numbers but I think these will be close.

Johnlm

[RP]

A diode won't even conduct until you put about 0.6 volts across it.  Since your rectifier is likely a bridge, you have 2 diodes in the circuit so you won't pass any current through your load at all until you reach 1.2volts.  At hand cranking speeds you are likely to be in this voltage realm so the effect is more noticable.

In effect you raise your 'cut-in" speed into a resistive load from 0 volts to 1.2V.

[willib]

"you have 2 diodes in the circuit so you won't pass any current through your load at all until you reach 1.2volts.  At hand cranking speeds you are likely to be in this voltage realm so the effect is more noticable.

In effect you raise your 'cut-in" speed into a resistive load from 0 volts to 1.2V"

this is all true ,but it's not only that , i am saying that at all RPMs turning by hand is easier when diodes are in place. due to their forward resistance.

The forward resistance of the diodes is something that should  be considered when estimating the power output of a generator.

Judging by the size of the heatsinks that dan uses, i think that a considerable amount of power is lost in rectification.

[Nando]

The bad gap in a generator periodically is talked about but never analyzed.

Sometimes one says that the stator rotor spacing was modified, either to increase or decrease the cut in RPM.

This gap can be represented as a loss since with loading the voltage drop is reduced by the current in addition to the loss caused by the resistance value of the winding.

Close gap produces the highest coupling and the highest power level and possibly the highest COGGING.

Measure the phase winding resistance and multiply by 1.73 to get the dynamic Star winding resistance value = Rdyn

A way to define this gap loss is to set a fixed RPM, measure the unloaded output voltage, then load the generator with a light load, like 10 % of its capacity, measure the output voltage again, and obtain the current = I

Vunloaded - Vloaded = Rdyn* I + Vgap

In generators producing low voltages is sometimes difficult to measure Vgap.

In the electrical Industry this gap is represented by magnetic values when calculating the motor-generator.

I presented this effect here in a simplistic manner.

I hope that this helps

Nando

[Nando]

Forward resistance of a diode.

The resistance of a diode depends on the size which represents the current capability.

let's say a diode capable of carrying 100 amps with a voltage drop of 1.4 volts.

The equivalent forward resistance is 1.14 / 100 = 0.014 Ohms.

Simple mathematics will give you this FORWARD EQUIVALENT RESISTANCE.

A diode should be used at a current where the intrinsic resistive value of the diode is approaching the non-log section of the voltage versus current curve.

So a 100 amps rated diode should not be used in a circuit drawing more than around 40 to 50 % of its rated value.

Nando

[willib]

ok , lets say that the diode below is used.

this is the data sheet..

http://www.irf.com/product-info/datasheets/data/70hf.pdf

the graph of I vs V

http://www.otherpower.com/images/scimages/2965/Vf_If.JPG

even though the diode is rated for much more current lets use 30A @ 1Volt , and you need two , so , 2/30 = 0.0666...ohms .

what i was getting at is , 66 milli ohms is 66 times the internal resistance (0.001 ohms) of a typical lead acid battery,(i think that is the value that has been used 0.001 ohm ?)

so instead of the generator seeing 0.001 ohms,  as the load.

It is really seeing 66 milli ohms per phase as a load , plus the resistance of the line .

[willib]

"This gap can be represented as a loss since with loading the voltage drop is reduced by the current in addition to the loss caused by the resistance value of the winding."

i'm not following your logic on this, how is the voltage drop reduced by the current?

as i see it , the gap is adjusted out , to increase blade rpm ,to prevent stall .

i think this is not a loss

[Nando]

ERROR DOES NOT COMPUTE --format the hard disk --

"This gap can be represented as a loss since with loading the voltage drop is reduced by the current in addition to the loss caused by the resistance value of the winding."

CORRECTING

"This gap can be represented as a loss since with loading the voltage drop is INCREASED by the current in addition to the loss caused by the resistance value of the winding."

The GAP is adjusted to adjust the magnetic field, either weakening it or making it stronger to produce less voltage (less power ) or more voltage (higher power) which by your point of view more or less RPM --that is affected by the loading of the generator.

It is a loss that depends on the gap magnetic field -- common problem in the manufacture of electric motors that affect the motor over all efficiency.

Nando

[Flux]

When charging, the battery is effectively 12/ current   ohms plus any internal resistance. At 30A it looks like 400 m ohms.

The rectification loss is significant for a diode drop of 2v in a 12v system the diode loss is one sixth of the total power.

Flux

[willib]

That says a lot for useing a higher voltage battery system , i see why dan uses a 48 v system..

it is overall a better way to go.

[willib]

I have an educated guess.

at cutin 48V/ Sqrt(3)= 27.7 Vrms/ phase

i have to make an assumption of .2ohms for your diodes.

(Vrms/phase)/ R(total) 27.7/.8 = 34.64A at at 65 RPM.

I^2(R(internal))= (34.64)^2(0.6)= 720W/ phase

      720/5(coils)= 144W/ coil

I^2(R(Load))= 34.64)^2 (.2)= 240W/ phase

total Power per phase= 720+240=960W/ phase and P(Tot)/ phase = 27.7(34.64)= 959.56 W/ phase

power Tot/gen=48(34.64)= 1660W at 65 RPM..

close ? or way off?

[SmoggyTurnip]

I am not sure you can consider this a loss either - I guess it depends on how you define "loss".  If you define it as mechanical power that went into the alternator but did not come out as electrical power then I don't think you can say the GAP contributes a loss.

[willib]

Thank you. i agree..