Miles' 90mm inrunner build thread

I've put in a request to Stephan to add an indication of whether the analysis is linear or non-linear to the header of the PDF report. It's easy to press the wrong button when you are doing a series... Also, I wasted one credit by forgetting to download the report :oops:
 
Interesting that the 27t 30p motor simulations used up one credit each, on average. The 24t 28p simulations used over 2 credits each. So, it's a lot cheaper to design a motor with 3 winding symmetries than it is to design one with only two :mrgreen:
 
Miles said:
How much heat can an outrunner 92mm in diameter, 56mm long and 1.1kg in weight, disperse continuously? No venting (sealed case).
Click to expand...

Are those the outer dimensions of the shell? What kind of magnet temperature and stator temperature are you willing to accept...what ambient temp? That can get you to a reasonable estimate with smooth interior and exterior. Then some relatively minor changes can double it without opening it to the elements.
 
Miles said:
I don't want the magnets to get above 75 deg. C. Of course, they'll be rated at 120 deg. C. , or more.

Stator around 90 deg. C. max., I guess.
Click to expand...
Looking at the table above, the maximum heat that I'll need to disperse continually will be about 150Watts, hopefully.... :)
 
What are the unknowns?

- My ability to wind to a 0.6 fill factor.

- Losses/heat from eddy currents in the magnets. (this is the one that worries me the most)

- Construction tolerances.

- ?
 
At what voltage have you calculated these efficiencies? Is it at 44.04V as stated earlier?
What No-Load current did you estimate as this highly determines the efficiency?
 
Honk said:
At what voltage have you calculated these efficiencies? Is it at 44.04V as stated earlier?
Click to expand...
The simulations were all done at a constant speed of 3000rpm.

Honk said:
What No-Load current did you estimate as this highly determines the efficiency?
Click to expand...
The parasitic torque determines the no load current. This is calculated in the Emetor simulations.
 
OK, good info but I'd like to know a bit more if possible.

Is the constant speed simulation of 3000rpm being equal to 44V input?
What level of No-Load current did Emetor simulations estimate?

If I know these parameters I can calculate the motor using another program I have access to.
1) Input Voltage
2) No-Load Current
3) Motor Phase Resistance

The program is highly accurate and I would like to help you out if I can be of any assistance. :mrgreen:
 
With only those three inputs, your program will only give a simple estimate of the motors performance, far less accurate than the output of Emetor.
 
Thanks for the numbers, even though the No-Load number is lacking.
Iron losses is only one part of a motors No-Load current.

I translated the Iron losses into 0.462amp at 48.52V and then trippled that to match the rest of the drag losses.
Like bearing drag and other stray losses that is perhaps not handled by Emetor.
A No-Load of 1.4amps is problably a lot more realistic than 0.462amps even thoug it still feels a tad to low.

Well here goes, the attached file shows the possible efficency and losses of the motor up to 4000W.
The losses shown is the total loss of the motor, all combined to one post. Losses = heat no matter how it is generated.
Here's a pick of the list (trying to match the amps of the tabulated 27T 30P analysis), just for a quick comparison.
Please see the attachment for a complete list from 0W to 4KW.

48.52V Efficiency = 63.80% at 120W output and 3.88A input at 68.08W loss
48.52V Efficiency = 91.78% at 840W output and 18.86A input at 75.27W loss
48.52V Efficiency = 93.53% at 1200W output and 26.45A input at 83.02W loss
48.52V Efficiency = 93.96% at 1360W output and 29.83A input at 87.38W loss
48.52V Efficiency = 94.54% at 1720W output and 37.50A input at 99.28W loss
48.52V Efficiency = 94.80% at 2080W output and 45.22A input at 114.14W loss
48.52V Efficiency = 94.87% at 2400W output and 52.14A input at 129.88W loss
48.52V Efficiency = 94.83% at 2760W output and 59.99A input at 150.52W loss

Something is strange though.....Have a look at the earlier 27T 30P analysis, e'g the 60amps row.
Input 60 amps x 48.52V = 2911W
Output is claimed as 2210W and heat loss 157W = 2367W
2911W input - 2367W used = 544W missing ????
Why are these hidden 544 watts not reported in the Emetor simulation? They can only be losses not given an account of.
 
Hi Honk,

I think tripling the iron losses to arrive at total parasitic losses is a bit unfair.... :)

That other simulation was done using AC.

I gave you the figures for a DC simulation.

The figures I gave you make sense in themselves.
 
Something is strange though.....Have a look at the earlier 27T 30P analysis, e'g the 60amps row.
Input 60 amps x 48.52V = 2911W
Output is claimed as 2210W and heat loss 157W = 2367W
2911W input - 2367W used = 544W missing ????
Why are these hidden 544 watts not reported in the Emetor simulation? They can only be losses not given an account of.
Click to expand...

I'm pretty sure it's a 60A sine through the motor windings. Not the same as 60A from a 48VDC source.

You are going about this the wrong way, IMO. You get some rough data, interpret it wrong, and then expect it to be the truth, and goes on to claim that there is something wrong with the source.
 
I agree, I don't have enough data for this calculation but I have used my program on lots
of different motors and I know it's correct when having proper input data.

OK, 60Amps sine.....but still, there's 544W of loss is missing in the Emetor simulation.
Not even Sine Amps can explain that large difference.....Peak-Peak or RMS.
Could you perhaps clarify whats going on here?
 
The problem is that you used put P = U * I as input, and Emetor calculated the input as something like P(w) = (EMF1(w) + R1 * I1(w)) * I1(w) + (EMF2(w + 120°) ... and so on, where EMF(w) is the waveform of a phase without connection to other phases. At least, that's what I think.
 
Current
Peak phase current 60A
Peak phase current, q-axis 60A
Peak phase current, d-axis 0A
Current density RMS 15.01A/mm²
Peak stator current loading 80.57A/mm
Phase resistance 0.02406W
Conductor losses 129.9W
Slot area 47.12mm²
Conductor area 2.827mm²

Torque
Mean airgap torque (by flux linkage and current) 7.121m*N
Mean airgap torque (by Maxwell stress tensor) 7.123m*N
Torque reduction due to iron losses 0.08533m*N
Torque ripple 2.062%

Flux density
Maximum stator iron flux density: 2.22T
Maximum rotor iron flux density: 1.447T
Fundamental airgap flux density: 1.191T
Iron losses: 26.81W

Voltage, Power, Flux linkage
Peak phase voltage 36.23V
Peak phase voltage, q-axis 26.3V
Peak phase voltage, d-axis -24.92V
Power factor cos(f) 0.7259
Electrical input power 2367W
Mechanical output power 2210W
Efficiency 93.38%
Peak flux linkage, q-axis 0.005288V*s
Peak flux linkage, d-axis 0.005274V*s
q-axis inductance 8.813e-05H
 
bearing said:
The problem is that you used put P = U * I as input, and Emetor calculated the input as something like P(w) = (EMF1(w) + R1 * I1(w)) * I1(w) + (EMF2(w + 120°) ... and so on, where EMF(w) is the waveform of a phase without connection to other phases. At least, that's what I think.
Click to expand...
That's possible.

The losses in all motors consist of a few well known factors.
Eddy currents and Hysteresis losses, Bearing and other drag losses, and of course I2R losses.
All of these losses except I2R is shown as the No-Load current at free RPM.
It peaks at its highest level at free RPM without any shaft load.
Loading the motor decreases No-Load as the RPM goes down.
Mainly I2R losses goes up at load due to increasing currents.

I guess we will have to wait som some "real world tests" as any simulation is 100% dependent on correct parameters.
 
Honk said:
bearing said:
The problem is that you used put P = U * I as input, and Emetor calculated the input as something like P(w) = (EMF1(w) + R1 * I1(w)) * I1(w) + (EMF2(w + 120°) ... and so on, where EMF(w) is the waveform of a phase without connection to other phases. At least, that's what I think.
Click to expand...
That's possible.

The losses in all motors consist of a few well known factors.
Eddy currents and Hysteresis losses, Bearing and other drag losses, and of course I2R losses.
All of these losses except I2R is shown as the No-Load current at free RPM.
It peaks at its highest level at free RPM without any shaft load.
Loading the motor decreases No-Load as the RPM goes down.
Mainly I2R losses goes up at load due to increasing currents.

I guess we will have to wait som some "real world tests" as any simulation is 100% dependent on correct parameters.
Click to expand...
If you know that, then you should also understand that using a simple constant no load current will not be very accurate, compared to reality or the output of an advanced simulation program like Emetor or FEMM.

Bearing losses at no load should be practically nothing, unless you have axial tension in a bearing made for radial load or something. Viscous drag is unknown so far.
 
As far as I'm concerned, bearing losses will be considered within the total losses of the transmission that the motor is incorporated into.

Windage losses on a 63mm dia rotor @ 3000 rpm?
I'll be putting a smooth cover over the spokes and the gaps between the magnets will be filled with resin.
 
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