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Reepicheep
| | Posted on Friday, November 18, 2011 - 08:57 am: | 
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Huh. I learn something new every day. I thought US AC voltage (120V) was peak to peak. Turns out it's actually effective power, and that actual peak to peak is closer to 170V.
So by measuring Volts AC, you may actually have already accounted for that .707 factor (corrected from .606 as I previously was recalling). Provided the different frequency didn't confuse your meter.
That still leaves the "saturation" leveling off issue though, if such a thing even exists. I assume it does on a permanent magnet system... there must be some point where the windings are simply as full of magnetic field as they can ever get... a point where the magnetic field can no longer get traction, so spinning it faster gains nothing.
This showed up and suprised me on a mythbusters episode when Jamie built the magnetic wall climber. His magnets were plenty strong, but when he put them on an air duct, there simply wasn't enough ferrous material for them to stick too. It never occured to me before then that this could be an issue, but made perfect sense in hindsight (story of my life  ).
But this is pure speculation on my part, I'd love to be enlightened by facts. |
Alaskacr
| | Posted on Friday, November 18, 2011 - 10:20 am: |
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Not to nit-pick, but 120v rms is about 340vpp (peak to peak) or 170vp (peak). |
Reepicheep
| | Posted on Friday, November 18, 2011 - 10:32 am: |
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Correct, thanks! Too many peaks. |
Nightsky
| | Posted on Friday, November 18, 2011 - 11:25 am: |
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There are no factory specs relating to wasted power and heat that burn up the stator.
It's a function of stator, regulator type, and RPM.
There is no saturation of the stator iron core here. The core's B field is from the permanent magnet rotor, which is fixed.
The stator core material has a magnetic permeability and cross sectional area that by design can handle the rotor.
The output current of the stator is limited by the B field of the permanent magnet in the rotor.
There is no "saturation" in this design.
So says Ampere's law.
Output voltage of the stator is proportional to the rate of change of magnetic flux i.e. rotor RPM.
So says Faraday's law.
Shunt regs short the stator, making the rotor mechanically hard to turn.
The bike's engine must overcome this torque and expend energy doing so.
This results in several lost horsepower. |
Reepicheep
| | Posted on Friday, November 18, 2011 - 01:02 pm: |
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Not trying to be difficult, this is a good discussion...
quote:The stator core material has a magnetic permeability and cross sectional area that by design can handle the rotor.
Would it be sized to handle the rotor? Or sized to produce the desired power? I could (unburdened by facts) see designing in "overkill" on the magnets relative to the stator in order to eek out more power at lower RPM's. Kinda like de-tuning an internal combustion engine for more low RPM power at the cost of peak RPM power.
Good point about the stator losses not being part of the stator rating.
The manual lists the DC resistance of the stator windings, right? .6 ohms?
So lets do a hypothetical... say the stator is "loosing" 60 watts. (P = I*I * R), so 60 watts = (X amps * X amps) * .6 ohms.
Solve for X, and that means the stator is outputting 10 amps. Assuming that at that same condition, we are putting 35 volts out the stator, that's outputting 350 watts. That seems reasonable.
Assuming that same 10 amp current, but at 60 volts, that's outputting 600 watts, which also sounds about right. The stator would still be "loosing" about 60 watts.
Of course that assumes constant current, which seems like a dangerous assumption given we have a regulator in the mix. Though maybe it works for a shunt regulator.
Fun to think about. |
Timebandit
| | Posted on Friday, November 18, 2011 - 01:35 pm: |
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Yes, this stuff is fun to think about. I'm going to go backward and address some of your questions/comments to try to clarify things a little:
> "Efficiency would not come into it, as the efficiency loss is converting the gasoline into kinetic energy."
In other words you're saying:
Input Energy * k = Output Energy, where k represents the conversion loss factor.
Or more directly:
Chemical Combustion Energy * k = Rotational Kinetic Energy
Efficiency always matters; the loss factor of each subsequent energy conversion has to be multiplied by the loss factor of every stage that preceded it to yield the right answer for the net system loss.
You're right, there is a HUGE loss in efficiency when internal combustion converts chemical energy from gasoline into rotational energy of the crankshaft. Let's pick an arbitrary value of “k” and say that the conversion loss = 0.30. To really understand the situation, it's worth understanding what factors go into deriving the value of “k”. I'll limit this discussion to one of them. Parasitic load from the charging system.
The electrical charging system contains mass that needs to rotated for the 1125 engine to turn. How much torque? This depends upon the way in which the charging system is energized, ie: whether it is dissipating zero power (open circuit), maximal power (shunt regulation) or something in between (series regulation or switching regulation). Each of these regulation paradigms will change the amount of torque that the engine will be required to impart on the charging system to cause it to rotate. These differing electrodynamic states each require a different amount of torque to rotate the charging system, and this amounts to differing levels of parasitic loss that are imposed on the engine as it performs work.
The result is that these different electrodynamic paradigms have the ability to modify the value of “k” in the above equation. Ultimately, this can result in one of two possible outcomes: a) the motor develops more power, or b) the motor consumes less fuel to produce the same amount of power. |
Timebandit
| | Posted on Friday, November 18, 2011 - 01:42 pm: |
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> " If you "gain back" the power generation loss, you only gain back the part of it that is post gas-to-motion conversion."
The example shows how it's possible to change the value of “k”, which represents the amount of useful work energy yielded per unit of fuel energy from internal combustion. How much will it matter? It's hard to say -- without very complicated direct measurement we can only make mathematical predictions. The important take home point is that every loss stage matters in reaching the final answer, and anything that effects the first loss stage has the most significant impact on total system loss, because several subsequent loss factors all use it in their calculations.
> "Which, as per the spec, is 400 or 500 watts."
This seems to be the point that you were missing at first. Just in case anyone reading this is still confused, 400W to 500W is the charging system's OUTPUT power specification. In order to determine how much heat the stator imparts into the engine, you need to calculate the charging system's INPUT power requirement, then the difference between input and output power. As Nightsky noted, the Buell manuals contain an OUTPUT power specification for the charging system, but no INPUT power specification. Input power has to be calculated. It has nothing to do with a 400W or 500W specification in the system manual. Fortunately, input power is easily calculated. Nightsky did it accurately in his initial post.
(edit: fixed a typo: changed "300W to 400W" to read "400W to 500W".)
(Message edited by timebandit on November 18, 2011) |
Timebandit
| | Posted on Friday, November 18, 2011 - 01:53 pm: |
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Back to Nightsky's calculations:
> "The original errors are, I think, two fold. First, I don't think the original measurement of power was right. It would need to measure both volts and amps simultaneously, and that's really hard."
There wasn't an error. What Nighstky did was to calculate the difference between two power states, open-circuit and fully-loaded. By definition, there is no current flow in the open-circuit measurement. The voltage is easily measurable and the current is zero. Ohm's law tells us that the electrical power dissipation in the open-circuit state is zero. Then he measured voltage and current in the shunt-regulated state to calculate power. The difference of power in the first and second states is the net power difference.
Ohm's Law: Power = Voltage * Current * 1.5; where 1.5 is the 3 phase conversion factor
Nightsky:
Open-circuit: 38 V * 0 A * 1.5 = 0W
Shunt-regulated: (38-11) V * 23.7 A * 1.5 = 960W
Net Power = (38-11) * (0-23.7) * 1.5 = -960W
There is one trivial mistake in his numbers: the magnitude of the current in the final calculation should be negative; it should be -23.7 instead of 23.7. The result is that the calculated power differential is negative, meaning that it is a system power loss. I know it's nit-picking, but the minus sign is a useful piece of information that needs to be carried through in the calculations.
Nighstky's measurements are right. He's measuring the total change in voltage across the resistance of the stator coils, and the total change in current flow in the stator to calculate the total change in power between the two test states.
> "Secondly, I don't think whatever value you get at low RPM will scale linearly with higher RPM's. I think the magnetic field parts saturate, and the whole thing "tops out" and just holds there above a particular RPM (that I don't know where it is). "
Nightsky correctly quoted Faraday's Law, which defines the voltage on an unloaded rotating stator to vary as a function of RPM.
Nightsky also addressed the popular misconception that the “core saturates”. People often quote “core saturation” as a method of explaining why the power and current curves level off asymptotically. The correct answer, as he pointed out, is that the B-field of the rotor limits the total current output that can be produced by the system. Ampere's Law applies here.
The bottom line is that Nightsky's calculations are right. Simple, back of the envelope math tells us that the 2008 charging system throws away ~1 HP as heat at 3,000 RPM and ~5.5 HP as heat at 10,000 RPM.
I think this pretty much wraps up the discussion.
Hope this helps.
(Message edited by timebandit on November 18, 2011) |
Reepicheep
| | Posted on Friday, November 18, 2011 - 02:17 pm: |
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That does really help, thanks for taking the time to go into such detail. Very enlightening.
And to further correct myself, the real world case where I corresponded with somebody who did actual dyno testing was on a tuber, which means, what, a 6800 RPM redline? Maybe less, my M2 didn't have a tach.
So the losses would be approaching half that 5.5 HP, which at 2.x HP is probably is getting very close to the "measurement noise threshold" on a dyno.
That loss would scale with RPM... so an 1125 would have more to gain.
Cool discussion, thanks!
(Message edited by reepicheep on November 18, 2011) |
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