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Another matching question
By Titantornado, Section Mechanical Posted on Tue Mar 14, 2006 at 10:34:25 PM MST
In regards to losses . . . .

Just when I thought I had a handle on this matching thing, I read more past posts and only confuse myself further. I have tried to narrow my information I gather from a single person, to perhaps help myself from becoming overloaded with mixed opinions. Flux, I choose you! ;-)
From one post, you wrote:
> Knowing the open circuit voltage, battery voltage and winding resistance
> you can then predict the output and see how it compares with the prop power.
>
> If you cut in at 200 rpm for 12v, at 600rpm the o/c voltage will be 36v. You
> have 24v in excess of the battery so the current into the battery would be 24
> divided by the winding resistance. This current x 12v gives you power out.
>
> Power lost in the alternator will be current^2 x winding R. From this you can
> find the input power ( output + losses) and compare this with your blade power.
Then in another post, you wrote:
> 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)
I believe the second quote is probably a more complete explanation for finding losses, unless I'm totally off and comparing two different formulas for different things.
I guess whichever it is, I should add power out and losses, and compare the total wattage to propellor wattage available, (like that found from Alton's Propeller Carving And Power Calculator) and the two numbers should be close to the same? Yes?
Sorry I keep bringing this subject up, but I feel it is my most misunderstood part of building these things, and most delicate area as far as getting things right. I don't want to build a stator for a given RPM/volt/TSR, only to find that the prop size I choose won't work with the alternator, and I'll end up with a mismatched RPM/TSR of a smaller/larger prop than planned.

Another matching question | 13 comments (13 topical)

Re: Another matching question (3.00 / 0) (#1)
by wpowokal on Tue Mar 14, 2006 at 07:55:56 PM MST

Oh I wish I had the time to describe voltage drop across the stator, but then.......

Take it away Flux.

"Dream as if you'll live forever, live as if you'll die today." James Dean

Re: Another matching question (3.00 / 0) (#2)
by Nando on Tue Mar 14, 2006 at 11:09:51 PM MST

Titantornado:

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

I believe is not accurate, the winding phase-resistance in a Star configuration, feeding a charge controller, needs to be multiplied by 1.73 to get the dynamic generator resistance.

The power dissipated will be in this the Dynamic-resistance * average-current squared and it does not need to be multiplied by 3, because the Star windings are full wave rectified and only the peaks of the sine wave are loaded during 60 degrees of their peak.

One needs to think how the current is drawn, taking in consideration the angle of conduction as well as remembering that normally there is 60 degrees peak conducting and the opposite side there are two crossing waveforms that affect the value of the phase resistance, lowering the value of 2 * Phase-resistance to phase-resistance * 1.73 or Dynamic-phase-resistance.

I have tried to find one reference book, for better explanation, but I see it is in storage.
I hope I am not wrong

Nando

Re: Another matching question (3.00 / 0) (#3)
by Flux on Wed Mar 15, 2006 at 01:57:02 AM MST

Oh dear!

I don't like the situation where the idiot has to argue with the engineer, I respect Nando's experience and he is certainly better qualified to deal with maths than I am, but his conclusions don't match my experience.

None of this is easy and my conclusions are based on what I have found to work.

For a 3 phase rectifier without overlap (justified in this case), current is carried by a pair of wires at any moment. For star at any moment current is flowing in 2 phases that have the highest voltage and are feeding the active 2 lines at that instant. There is no contribution from the other phase. Each phase carries current for 120 electrical degrees but rectifier diodes change over at 60 deg.

I often have problems with Nando's terminology, maybe because I am not in North America, but I assume the dynamic resistance is the resistance that the alternator appears to have to the outside world.

With a star alternator the "line" resistance is 2 x phase resistance and as the current is carried by 2 coils in series it would seem that this would be the determining value. Nando seems to suggest it should be 1.73 x phase resistance.

From practical observation it is neither. For predictions of current into a battery it seems to behave as "line" resistance x 1.3 ( 2 x phase R x 1.3 )

This gives a good prediction of current into a battery at any speed.

Now for losses, that is a different issue. If we use the same resistance figure to predict losses the figures are way too high. Most people here use the line resistance value x the dc current squared and this gives a more reasonable figure but is still high.

From first principles as far as I can see, the power loss in a winding is I^2 x R, where the current is the rms current.

For the star alternator it is easiest to consider one phase and multiply by three to get the total loss.

The current in a phase is the same as line current for star. The line current is the rms current into any terminal of the rectifier. Without overlap this is roughly
0.76 of the dc mean current.

From loss measurements on test rigs I have found this gives by far the nearest answer.

Titantornado , I am sorry this has become so complicated, but Nando has questioned the thing and I can only say that using the power per phase method x 3 has given me the best results.

You are correct that you need to add output power and losses to get your input power.
Flux

[ Parent ]

Re: Another matching question (3.00 / 0) (#4)
by Nando on Wed Mar 15, 2006 at 08:19:13 AM MST

FLUX:

Let's un-complicate this case !!!

No body here is an IDIOT, not You, nor Me, not anyone, in a semi-technical or technical discussion there is no space for such definition. Just Opinions !!!!.

I can be wrong, you could be wrong, either You or I took a wrong reference for the supplied statement.

With your remark, I had to seat down and think what I have been saying regarding the GENERATOR internal resistance presented to the external world.

FLUX YOU SAID:
>With a star alternator the "line" resistance is 2 x phase resistance and as the >current is carried by 2 coils in series it would seem that this would be the >determining value. Nando seems to suggest it should be 1.73 x phase resistance

I unhappily took a wrong reference from another work and the "line" resistance is 2 X the phase resistance, FOR SOME REASON, I dropped the 2 multiplier. I AM WRONG HERE!!.

The 1.73 was the multiplier that may vary from 1.2 to about 2, THIS DUE TO THE magnetic field and the space between the rotor and stator that we can represent by a resistor with a power dissipation defined by the current.

LOWER the magnetic field and wider Rotor-Stator Space closer to 2 is the multiplier.

You have found 1.3 factor, this factor is lower because the magnetic field may be quite greater than the ones I have used in my work and, as well, the rotor-stator spacing closer.

FLUX YOU SAID:
>The current in a phase is the same as line current for star. The line current is >the rms current into any terminal of the rectifier. Without overlap this is roughly
>0.76 of the dc mean current.

I do not follow You here.

And can you define 0.76 ?.

In a rectified 3 phase, one can look the phase current as a square pulse around 60 degrees wide for each half pulse for a total of 120 degrees.
The square pulse with a top sine wave, with 60 degrees width, that has about 1.2 times the RMS of a sine wave which extends constant because is a 3 phase rectification, so it gives about RMS X 1.2 Voltage average as the output voltage

I am not trying to be precise here just a general workable value.

Nando

[ Parent ]

Re: Another matching question (3.00 / 0) (#5)
by Flux on Wed Mar 15, 2006 at 08:50:14 AM MST

Nando
FLUX YOU SAID:
>The current in a phase is the same as line current for star. The line current is >the rms current into any terminal of the rectifier. Without overlap this is roughly
>0.76 of the dc mean current.

I do not follow You here.

And can you define 0.76 ?.

I more or less agree with you about the rectifier conduction. At any instant the dc current flows through a pair of leads of the 3 phase rectifier and that current flows as a phase(and line) current in the alternator.

This current flows for 2/3 of the time in any wire or phase winding.

If we have a dc current of say 10A, then the phase current is 10A flowing for 2/3 of the time. Rms current will be 10x(2/3)^.5 or about .82A. Figures I have measured have been a bit lower and the .78 is what I measured in a typical case.

You are probably used to machines with slotted iron cores and a significant leakage reactance, in which cases overlap occurs in the rectifier and the rms value is significantly lower than the case of these air gap alternators, that show no leakage reactance effects in the working range.

These type alternators are rather peculiar and exhibit very different characteristics from the conventional radial field slotted core alternators and fortunately seem to respond to fairly simple analysis.

You may not agree with me about the rms current in the line of one of these alternators but the figure is repeatable and easliy measured with a rms meter in the ac line.
Flux

[ Parent ]

Re: Another matching question (3.00 / 0) (#6)
by Nando on Wed Mar 15, 2006 at 10:00:51 AM MST

FLUX:

YOU SAID:

>This current flows for 2/3 of the time in any wire or phase winding.

>If we have a dc current of say 10A, then the phase current is 10A flowing for 2/3 of the time. Rms current will be 10x(2/3)^.5 or about .82A. Figures I have measured have been a bit lower and the .78 is what I measured in a typical case.

In a rectified 3 phase system, the current in any phase runs for 120 degrees out of 360 degrees, therefore 120/360 = 1/3 of the time

Here is where I do not follow the 2/3 of the time -- AM I LOOKING in the wrong place ?.

You meant 8.2 A not 0.82 A DO YOU ?

I can not disagree with an RMS voltmeter.

I try to think about 8.2 and 7.8 differences and I do not find a reason.

Regards

Nando

[ Parent ]

Re: Another matching question (3.00 / 0) (#7)
by Flux on Wed Mar 15, 2006 at 10:26:48 AM MST

In a rectified 3 phase system, the current in any phase runs for 120 degrees out of 360 degrees, therefore 120/360 = 1/3 of the time

Not for full wave, there is 2 conduction periods when the voltage is positive and when it is negative. 120 deg conduction requires half wave and connection to neutral.

sorry about the decimal point it would have been better to consider dc as 1A.

I am not surprised that there is a difference between the measured rms and the simple theory. The real theory is far too complex for most machine design even in this computer age.
Flux

[ Parent ]

Re: Another matching question (3.00 / 0) (#9)
by Nando on Wed Mar 15, 2006 at 12:05:46 PM MST

FLUX:

Yahoo is killing me, I tried to insert a file and erased my message -- quite often!!

YOU SAID:
>Not for full wave, there is 2 conduction periods when the voltage is positive and when it is negative. 120 deg conduction requires half wave and connection to neutral

I say that in full wave 3 phase rectification each phase conducts for 60 degrees for the positive and 60 degrees for the negative, for a total of 120 degrees . which gives a ratio of 120/360 = 1/3 conduction and each phase conducts 1/3 so X 3 for 360 degrees or 100 % conduction.

AM I RIGHT ?.

I am inserting a 3 phase display to have bases for the discussion.

Nando

[ Parent ]

Re: Another matching question (3.00 / 0) (#11)
by Flux on Wed Mar 15, 2006 at 01:56:52 PM MST

Nando
See if this works.

If it is legible it should show the diode currents and line current in a 3 phase bridge.
Flux

[ Parent ]

Re: Another matching question (3.00 / 0) (#12)
by ghurd on Wed Mar 15, 2006 at 02:43:22 PM MST

I have a feeling this is somehow related to the star / delta / Jerry rigged discussion.

One way may have a higher peak voltage, or higher peak amps.
Another way may have the amps flowing for a longer duration (duty cycle?) per cycle per phase.

ie. One PMA can make 100VAC peak at X RPMs, but only 1A shorted (Don't laugh, I have one). Another can make only 10VAC peak, but 500A shorted. Another can make 200VAC open and 200A short.

A poor attempt at a graph charting amperage follows.

This for AC amps ---n---u---n---u---
This for DC amps ---n---n---n---n---

No current, then a spike, then no current...
The wider the amps spike, the more amp-hours?

I couldn't say it before. Maybe this try is better.
G-
Ghurd.info
[ Parent ]

Re: Another matching question (3.00 / 0) (#13)
by Nando on Wed Mar 15, 2006 at 10:28:36 PM MST

FLUX:

You are right, I have put my big four feet in my mouth.
The same mistake I made in my first semester of electrical and I have been repeating it for sometime

Same mistake made several decades back in college -- it must be old age or that German man, his name ?? Alz.... what is his name ??>

I am going to hibernate to see if I recover my youth !! NOT !!

Nando

[ Parent ]

Re: Another matching question (3.00 / 0) (#8)
by Titantornado on Wed Mar 15, 2006 at 11:04:55 AM MST

Ummmm, right. Now it's all clear.

I just knew I was gonna stir up the hornet's nest, but I just cringed and put it out there. Hey, at least we can all agree that we don't always agree.

LOL, maybe I best just stick to multipling the forcing volts by the dc current. (seems to figure conservatively on the high side) At least I know why this confuses me so. There doesn't seem to be a clear answer to the question. Alot must come from trial and error, I was just hoping to avoid some of the latter. :o)

Hmmmmm, perhaps a side-by-side example comparisons might be an idea?

Rod

Re: Another matching question (3.00 / 0) (#10)
by wpowokal on Wed Mar 15, 2006 at 01:49:15 PM MST

Oh dear such postering, this is a very valid question that Rod posed, so a simple answer. Now to be fair there are "things " that happen within a generator that I will not take into account for simplicity.

Concider one phase, for if it is three it is happening proportionally to the other 2, or 5 phase other 4 etc.

E=IR

Where E=volts
I=current in amps
R= resistance in ohms.

Let us assume your resistance of one phase is 1.5 ohm.

Now when one amp is flowing from the generator there is a voltage drop of 1x1.5 (amps x resistance)or 1.5v across the generator. This also is 1.5 watts of heating.

Now for 10 amps that is 10x1.5=15 volts or 150 watts of heating if you will.

Now I know this is an Ac machine and there are things like back EMF etc but I can not measure that nor I suspect most home constructors so I have applied the KISS principal.

regards Allan
"Dream as if you'll live forever, live as if you'll die today." James Dean
[ Parent ]

Another matching question | 13 comments (13 topical)

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