Jan 22, 2014, 04:38 AM
homo ludens modellisticus
Netherlands, GE, Nijmegen
Joined Feb 2001
12,831 Posts
Quote:
 Originally Posted by flieslikeabeagle ... The motor constant Io is supposed to represent the constant no-load current, but as I found, it is not a constant at all, and is in fact not even remotely close to being a constant, instead varying quite strongly with rpm. ....
Hysteris losses in the iron go up linear with rpm, eddy current losses go up squared with rpm.

Vriendelijk groeten Ron
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Jan 22, 2014, 09:46 AM
Registered User
Germany, BY, Gräfelfing
Joined Jan 2003
557 Posts
Because of the search of a high efficancy elektric motor i have a little test modelmotor .i yoused this motor to slide different backiron materials outside the windings no need to change any thing on this motor.So i can select + and - of different materials.
So first i slide now some air outside and check the motor running with this freeflux system. Motor has a 2 pol rotor from neodym 22mm dia and 30mm long.
Winding is 5 turn of 245 x 0,1 litzwire ,to get low eddy current in the wire and now i have switcht to parallel wind of this litzwire with 5 turn each phase . Internal resistance is 0,004022 ohm ,no load amps are 3,2 amps.The coils are y terminated to get no circle circuite amps. So kv = 7860 rpm . I run this motor on 1 cell 3,6 volt to mesure no load amps.
Flux calculating show on the surface of the rotor we have 0,560 tesla ,at 0,5mm distenz= 0,52 tesla,at 1,0mm 0,48 tesla,at 1,5mm 0,44 tesla,at 2mm 0,41 tesla,at 2,5mm 0,38 tesla ..... at 5mm 0,27 tesla. So you see the flux is very low.but low flux will give low fluxrelated los.

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Last edited by Christian Lucas; Jan 22, 2014 at 10:03 AM.
Jan 22, 2014, 10:15 AM
Got shenpa?
Los Angeles
Joined May 2004
10,940 Posts
Quote:
 Originally Posted by Ron van Sommeren Hysteris losses in the iron go up linear with rpm, eddy current losses go up squared with rpm. Vriendelijk groeten Ron
Ron, I came to that same conclusion after some head-scratching. Hysteresis losses cause a fixed amount of energy lost every time the magnetic field reverses, so the total energy lost should be proportional to the number of revolutions, so the total power lost (time derivative of energy lost) should be proportional to the rpm.

The same argument seems to hold for bearing friction as well.

So what sort of actual physical motor losses are accurately represented by the constant, unvarying Io in the 3-constant motor model? Are there any, or is the constant Io a purely mathematical convenience with no direct connection to any actual physical energy loss mechanism in motors?

-Flieslikeabeagle
Jan 22, 2014, 10:32 AM
Registered User
London UK
Joined Jun 2006
130 Posts
Quote:
 Originally Posted by flieslikeabeagle So what sort of actual physical motor losses are accurately represented by the constant, unvarying Io in the 3-constant motor model? Are there any, or is the constant Io a purely mathematical convenience with no direct connection to any actual physical energy loss mechanism in motors?
It doesn't represent anything, really. It sort of works in cases where eddy current losses are the minor component.
Jan 24, 2014, 08:17 AM
Registered User
Antony (France)
Joined Sep 2003
3,271 Posts
Hi
The best graph is to represent Io x Vin versus rpm
See the picture for the Scorpion SII 4020 family (3 windings, 3 Kv)
At no load, the copper losses are ridiculously low
BB losses and aerodynamics drags are probably(?) also very low
Conclusion : curve is not linear
A good regression is an order 2 polynomial
Louis

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 Jan 24, 2014, 01:00 PM Dieselized User Chicagoland Joined Feb 2000 7,865 Posts I've found motors to generally fit idle watts is proportional to delta V^(3/2). That is, if a motor is 10W at 10V (1A) no load it will be 2^1.5*10 = 28W at 20V. But I haven't often tested this with junk motors. Greg
Jan 24, 2014, 03:04 PM
Registered User
Antony (France)
Joined Sep 2003
3,271 Posts
Quote:
 Originally Posted by gkamysz I've found motors to generally fit idle watts is proportional to delta V^(3/2). That is, if a motor is 10W at 10V (1A) no load it will be 2^1.5*10 = 28W at 20V. But I haven't often tested this with junk motors. Greg
Hi Greg
with the formula Wo = K V^(3/2) you have only one degree of freedom (K)
with the formula Wo = A + B rpm + C rpm^2 you have three
and the formula is respecting hysteresis losses + eddy currents losses
generally A is very small
Louis
 Jan 24, 2014, 06:36 PM 186mph Funjet Ultra nieuw vennep, the netherlands Joined Feb 2010 544 Posts I have measured a rewound junk motor! Turnigy SK50-55 580kv, phase resistance .011Ohm. On ~11.1V Io ~3A, on 22.2V Io >7A! Laminations are .35mm René
Jan 25, 2014, 05:59 AM
Registered User
Antony (France)
Joined Sep 2003
3,271 Posts
Quote:
 Originally Posted by E-speed buff I have measured a rewound junk motor! Turnigy SK50-55 580kv, phase resistance .011Ohm. On ~11.1V Io ~3A, on 22.2V Io >7A! Laminations are .35mm René
Hi René
As a comparison
Scorpion SII 4025-520
weight 341g
measured Kv 526 rpm/V
Rm 0.0164 ohm
On 11.1V Io = 1.5A
On 22.2V Io = 2.2A
Louis
 Jan 26, 2014, 05:16 AM Registered User London UK Joined Jun 2006 130 Posts Regarding simple modelling: https://www.academia.edu/1360191/Bru...Fixed_Wing_UAV

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