"Peak Efficiency" Control Mode?

General topics and discussions about the VESC and its development.
devin
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Re: "Peak Efficiency" Control Mode?

Postby devin » 29 Aug 2017, 06:40

devin wrote:Source: https://en.wikipedia.org/wiki/Drift_velocity
Image

^We can see halving the conductor length doubles the drift velocity... & doubling the voltage doubles the drift velocity... & and doubling the wire cross section has no effect on drift velocity... behaviors also describing changes to the rotor rpm at no load rpm (halving conductor length doubles no load rpm, and doubling voltage doubles no load rpm, and changing the thickness of the conductor has no effect on no load rpm).


rew wrote:A change from the pattern. Seems you're right.

If you keep the voltage constant, and double the wire (e.g. a second wire), you get twice as much current, giving the same "flow of electrons" in each wire. But I fail to see how this relates to rotor RPM and double the wire thickness. In that case, even though the rotor RPM ALMOST stays the same, the total current will rise "a little", and the drift speed will lower a lot.

There are lots of cases where you can find a "change this, and that other parameter doubles, change that and it still doubles, but change something else and it doesn't, but stays the same". I have a vat of water. Double the water height, and the pressure at the bottom doubles. Double the density of the fluid and the pressure doubles, but double the area of the vat, (keeping the water level the same) the pressure stays the same. So what?


@rew are we 100% sure it's the number of "turns" that determines kv or do you think it may possibly be the drift velocity?

rew
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Re: "Peak Efficiency" Control Mode?

Postby rew » 29 Aug 2017, 18:38

Yes.

devin
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Re: "Peak Efficiency" Control Mode?

Postby devin » 29 Aug 2017, 19:42

This is the way I conceptualize what is happening at no load rpm and how it involves the drift velocity.

At no load rpm, if we choose the spinning rotor magnets as the "stationary" frame of reference, then both the stator winding copper & the copper's free electrons have kinetic energy relative to the "stationary" rotor magnetic moment reference frame at no load rpm. At this RPM, if the voltage applied by the battery is 50V, the back voltage (Back EMF) caused by the relative motion of the free electrons through the rotor magnetic moment is also 50V in the opposite direction, leading to 0w electrical in the copper winding and no net rotor acceleration.

At stall @ 100% duty cycle in a copper wye motor, if the battery voltage is doubled, the drift velocity of the electrons is doubled, but their kinetic energy (& electrical wattage) is quadrupled. As a result of this quadrupled kinetic energy & doubled drift velocity at stall, the windings must have 4 times as much kinetic energy relative to the "stationary" reference frame of the rotor magnetic moment in order for the battery voltage to equalize with the back emf voltage resulting in net 0 watts and net 0 acceleration at no load rpm (& exactly 4 times as much rotor kinetic energy equates to exactly 2 times as much rotor velocity).

^Thus, doubling the battery voltage (or halving the number of turns) doubles the rotor's rpm at no load rpm.

Addy
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Re: "Peak Efficiency" Control Mode?

Postby Addy » 30 Aug 2017, 00:22

Not sure why you're trying to think of this in terms of drift velocity. Current is a simpler concept to think about instead of drift velocity.

To maximize efficiency like you're talking about you would have to keep the motor at the speed at which it is the most efficient, and that speed is determined by how the motor is wound / constructed. I don't think you could dynamically change that. You could perhaps limit the motor speed to keep the efficiency high, but a lot of people do this manually by not blasting around at full throttle all the time.

For the question of aluminum vs copper windings, for the same wire gauge and number of turns, aluminum would have a higher resistance, which would reduce the maximum current of the motor and reduce the maximum speed, so the kV rating would not be the same.

rew
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Re: "Peak Efficiency" Control Mode?

Postby rew » 30 Aug 2017, 07:46

If you keep talking about kinetic energy in electrons, one day I'll calculate that kinetic energy for you. It's going to be less than you expect. For all practical purposes the kinetic energy in the electrons is zero. You should also stop talking about 100% duty at stall.

The hobbyking motor I linked simply allows 1.5V of effective voltage above the BEMF. So if the rotor is stalled, you're allowed 1.5V or 3% duty cylce (at 50V battery voltage). If the rotor is doing 150RPM, (1V BEMF) you're allowed to do 2.5V or 5% dutycycle.

devin
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Re: "Peak Efficiency" Control Mode?

Postby devin » 30 Aug 2017, 09:38

rew wrote:If you keep talking about kinetic energy in electrons, one day I'll calculate that kinetic energy for you. It's going to be less than you expect. For all practical purposes the kinetic energy in the electrons is zero. You should also stop talking about 100% duty at stall.

The hobbyking motor I linked simply allows 1.5V of effective voltage above the BEMF. So if the rotor is stalled, you're allowed 1.5V or 3% duty cylce (at 50V battery voltage). If the rotor is doing 150RPM, (1V BEMF) you're allowed to do 2.5V or 5% dutycycle.


@rew that's a kind offer, these tables might help showing the fermi velocity and free electron density of copper:

http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/fermi.html#c1

Copper Fermi Velocity: 1570000 meters per second
(about 1570 kilometers per second or about 0.5% the speed of light)

Copper Free Electron Density: 8.4*10^28 per m^3

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/ohmmic.html

Lastly, the drift velocity formula:

Image

Simply the copper valence electrons have a tremendous intrinsic velocity at room temperature (with the cummulaive random motions resulting in overall net 0 velocity) when there is no current, superimposed with a slight net drift velocity in one direction when current is flowing.

devin
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Location: San Francisco, California, US

Re: "Peak Efficiency" Control Mode?

Postby devin » 31 Aug 2017, 16:36

rew wrote:If you keep talking about kinetic energy in electrons, one day I'll calculate that kinetic energy for you. It's going to be less than you expect. For all practical purposes the kinetic energy in the electrons is zero. You should also stop talking about 100% duty at stall.

The hobbyking motor I linked simply allows 1.5V of effective voltage above the BEMF. So if the rotor is stalled, you're allowed 1.5V or 3% duty cylce (at 50V battery voltage). If the rotor is doing 150RPM, (1V BEMF) you're allowed to do 2.5V or 5% dutycycle.


@rew if we use this motor as an example:

http://alienpowersystem.com/shop/brushless-motors/alien-5065-outrunner-brushless-motor-270kv-2200wa/ wrote:
MOTOR: 5065
KV: 270
MAX POWER: 2200W
WIRE WINDS: 9
MAX AMP: 60A
ESC: 80/120A
MAX VOLT: 8S
RESISTANCE (Ohm): .42
NO LOAD CURRENT: 1,5
SIZE: 50 x 65 ( without shaft )
WEIGHT (g): 0,380
SHAFT: 8mm with 3mm keyway


It says it allows 60A @ 0.42ohm, and I will assume this equates to 60A Motor amps.

Stall 60A @ 0.42ohm = 1512W Electrical at Stall = 25.2V @ Stall = 50V @ %50.4 duty cycle

The 270KV motor is capable of turning @ 50V Battery * 270KV = 13,500rpm at no load rpm @ 100% duty cycle

If we use the "peak efficiency" algorithm, and the motor is turning 6000rpm:

A = (20B)/(17C)

A = XX.XXV = Full Throttle Effective Voltage
B = 6000rpm = Present RPM
C = 270kv = Motor KV

therefore

A = (20B)/(17C)

A = ((20)(6000))/((17)(270))

26.14V = ((20)(6000))/((17)(270))

Simply according to the algorithm, in theory, for peak electrical to mechanical conversion efficiency, when the rotor is turning 6000rpm, the applied effective voltage should be 26.14V.

1/270=0.003703volts per rpm <---- simply 270 rpm per volt is 0.003703 volts per rpm

6000rpm*0.003703v=22.218V <---- present back emf voltage at 6000rpm

26.14V-22.218V=3.93V <------- Effective Voltage minus back emf voltage equals 3.93 net volts

3.93V=9.35A*0.42ohm <---- 9.35A Motor Amps @ 6000rpm

270*2=540
540*pi=1696.45
1696.45/60=28.27r/vs
0.3377Nm/A=60/((2*pi)*28.27) <--------270KV = 0.3377Nm/A

9.35A * 0.3377NM/A = 3.15Nm <-----3.15 Newton Meters Torque at 9.35A Motor Amps

Power (kW) = Torque (N.m) x Speed (RPM) / 9.5488
Source: http://www.wentec.com/unipower/calculators/power_torque.asp

1.98kW = 3.15Nm*6000rpm/9.5488 <------ 1980W Mechanical Output @ 6000rpm @ 26.14V Effective Volts

Assuming 50V battery pacK

(26.14V Effective V/50V Pack V)*100= 52.28% duty cycle

&:

Battery Amps = Motor Amps x (Duty Cycle/100)

therefore:

4.88 battery amps = 9.35 motor amps * (% 52.28duty cycle/100)

4.88battery amps * 50V = 244W Electrical <-------- 244W Electrical In

1.98kW = 3.15Nm*6000rpm/9.5488 <------ 1980W Mechanical Output??? @ 6000rpm @ 26.14V Effective Volts

----------------------------

So long story short I was trying to analyze whether this algorithm would work and got some strange results, not sure whether or where my math is wrong...

rew
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Location: Delft, Netherlands.

Re: "Peak Efficiency" Control Mode?

Postby rew » 01 Sep 2017, 12:17

I searched for a motor that had 5065 in the name. I found the Turnigy Aerodrive SK3 5065. However that is not available in 270KV version. It is however in a 236kV version. https://hobbyking.com/en_us/turnigy-aer ... motor.html

Some parameters are so wildly different from your example, that your example is totally irrelevant. So where your motor produces 2200W in mechanical output (we hope), it can tolerate 1500W in losses. That is outrageous and not a real example. The Turnigy I linked will tolerate (according to specifications) 68W of losses. A believable number. At standstill/stall (i.e. no selfcooling due to turning) it won't manage that indefinitively. But given some external cooling or allowing the motor to do reasonable RPMs this is a believable number.

So you may start again.
Motor: Aerodrive SK3 5065-236KV
KV: 236
max power: 1850W
MAX AMP: 60A
resistance: 0.019 Ohm
(I've done the "Watts at stall" calculation for you already, see above)

If you want me to consider your numbers you must provide me with a link to where you found those unrealistic numbers. I'll then tell you why those are bullshit.

devin
Posts: 225
Joined: 08 May 2017, 01:55
Location: San Francisco, California, US

Re: "Peak Efficiency" Control Mode?

Postby devin » 01 Sep 2017, 18:42

@rew i think this is where I went wrong....

270*2=540
540*pi=1696.45
1696.45/60=28.27r/vs
0.3377Nm/A=60/((2*pi)*28.27) <--------270KV = 0.3377Nm/A


i think it should have been....

270*2=540
540*pi=1696.45
1696.45/60=28.27r/vs
0.03537Nm/A=1/28.27r/vs <--------270KV = 0.03537Nm/A


therefore:

http://alienpowersystem.com/shop/brushless-motors/alien-5065-outrunner-brushless-motor-270kv-2200wa/ wrote:
MOTOR: 5065
KV: 270
MAX POWER: 2200W
WIRE WINDS: 9
MAX AMP: 60A
ESC: 80/120A
MAX VOLT: 8S
RESISTANCE (Ohm): .42
NO LOAD CURRENT: 1,5
SIZE: 50 x 65 ( without shaft )
WEIGHT (g): 0,380
SHAFT: 8mm with 3mm keyway


It says it allows 60A @ 0.42ohm, and I will assume this equates to 60A Motor amps.

Stall 60A @ 0.42ohm = 1512W Electrical at Stall = 25.2V @ Stall = 50V @ %50.4 duty cycle

The 270KV motor is capable of turning @ 50V Battery * 270KV = 13,500rpm at no load rpm @ 100% duty cycle

If we use the "peak efficiency" algorithm, and the motor is turning 6000rpm:

A = (20B)/(17C)

A = XX.XXV = Full Throttle Effective Voltage
B = 6000rpm = Present RPM
C = 270kv = Motor KV

therefore

A = (20B)/(17C)

A = ((20)(6000))/((17)(270))

26.14V = ((20)(6000))/((17)(270))

Simply according to the algorithm, in theory, for peak electrical to mechanical conversion efficiency, when the rotor is turning 6000rpm, the applied effective voltage should be 26.14V.

1/270=0.003703volts per rpm <---- simply 270 rpm per volt is 0.003703 volts per rpm

6000rpm*0.003703v=22.218V <---- present back emf voltage at 6000rpm

26.14V-22.218V=3.93V <------- Effective Voltage minus back emf voltage equals 3.93 net volts

3.93V=9.35A*0.42ohm <---- 9.35A Motor Amps @ 6000rpm

270*2=540
540*pi=1696.45
1696.45/60=28.27r/vs
0.03537=1/28.27 <--------270KV = 0.03537Nm/A


9.35A * 0.03537NM/A = 0.3307Nm <-----0.3307 Newton Meters Torque at 9.35A Motor Amps

Power (kW) = Torque (N.m) x Speed (RPM) / 9.5488
Source: http://www.wentec.com/unipower/calculators/power_torque.asp

207.79w = 0.3307Nm*6000rpm/9.5488 <------ 207.79W Mechanical Output @ 6000rpm @ 26.14V Effective Volts

Assuming 50V battery pacK

(26.14V Effective V/50V Pack V)*100= 52.28% duty cycle

&:

Battery Amps = Motor Amps x (Duty Cycle/100)

therefore:

4.88 battery amps = 9.35 motor amps * (% 52.28duty cycle/100)

4.88battery amps * 50V = 244W Electrical <-------- 244W Electrical In

207.79W = 0.3307Nm*6000rpm/9.5488 <------ 207.79W Mechanical Output @ 6000rpm @ 26.14V Effective Volts

(207.79W/244W)*100=86.57% <----------- 86.57% conversion efficiency of electrical to mechanical watts.

The question then becomes, does any other duty cycle besides 52.28% provide a higher electrical to mechanical wattage conversion efficiency than 86.57% at 6000rpm?

devin
Posts: 225
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Location: San Francisco, California, US

Re: "Peak Efficiency" Control Mode?

Postby devin » 01 Sep 2017, 22:35

devin wrote:The question then becomes, does any other duty cycle besides 52.28% provide a higher electrical to mechanical wattage conversion efficiency than 86.57% at 6000rpm?


-------------------------------

what if I randomly select 75% duty cycle at 6000rpm? is it more or less efficient than 52.28% duty cycle at 6000rpm?

---------------------------

1/270=0.003703volts per rpm <---- simply 270 rpm per volt is 0.003703 volts per rpm

6000rpm*0.003703v=22.218V <---- present back emf voltage at 6000rpm

(37.5V Effective V/50V Pack V)*100=75% duty cycle <--------- 75% duty cycle is 37.5V Effective Volts

37.5V Effective V - 22.218V Present Back EMF = 15.282V Net V <----- Net voltage @ 6000rpm @ 75% duty @ 50V Battery is 15.282V

15.282V=XX.XXA*0.42ohm

15.282V=36.385A*0.42ohm <------ Motor amps @ 6000rpm @ 75% duty @ 50V Battery is 36.385A motor amps

270*2=540
540*pi=1696.45
1696.45/60=28.27r/vs
0.03537=1/28.27 <--------270KV = 0.03537Nm/A ...simply 270KV is 0.03537 newton meters of torque per motor amp

36.385A * 0.03537NM/A = 1.2869Nm <-----1.2869 Newton Meters Torque at 36.385A Motor Amps

808.625w = 1.2869Nm*6000rpm/9.5488 <------ 808.625W Mechanical Output @ 6000rpm @ 37.5V Effective Volts

&:

Battery Amps = Motor Amps x (Duty Cycle/100)

therefore:

27.262 battery amps = 36.385 motor amps * (% 75 duty cycle/100) <-------- 36.385 motor amps at 75% duty cycle is 27.262 battery amps

&

1363.1W Electrical = 27.262 battery amps * 50V <-------- 1363.1W Electrical In

808.625w Mechanical = 1.2869Nm*6000rpm/9.5488 <------ 808.625W Mechanical Output @ 6000rpm @ 37.5V Effective Volts

(808.625W/1363.1W)*100=59.32% <----------- 59.32% conversion efficiency of electrical to mechanical watts @ 75% duty cycle @ 6000rpm.

-----------------------------

devin wrote:The question then becomes, does any other duty cycle besides 52.28% provide a higher electrical to mechanical wattage conversion efficiency than 86.57% at 6000rpm?


devin wrote:what if I randomly select 75% duty cycle at 6000rpm?


Nope! 75% duty cycle is not as efficient as 52.28% duty cycle @ 6000rpm. At 52.28% duty cycle at 6000rpm @ 50V battery @ 270kv, the electrical to mechanical conversion efficiency is 86.57% -- compared with only 59.32% conversion efficiency at 75% duty cycle.

86.57-59.32=27.25 <---- Loss of 27.25% conversion efficiency

...a loss of 27.25% of the conversion efficiency by not using the peak efficiency algorithm & choosing 75% duty cycle @ 6000rpm!


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