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DC motor theory and model

42K views 115 replies 14 participants last post by  gor 
#1 · (Edited)
Hello All,
You may have seen my posts on other threads about a motor model I am developing. Well, I am pretty happy with it and so let me explain.
First, the torque generated by a motor is given by the armature current times the "field" times some constant depending on the motor structure and materials.
Second, the speed of the motor is given by the armature voltage applied to the magnetics [commonly called back emf or speed-voltage] divided by the "field' times some constant again depending on the motor structure and materials.
Lastly, the really *neat* thing is that, if we use SI units, the "field" and the constants are identical... So, we use Volts, Amps, Newtons-meter and radians per second to extract a single field function [or map] which depends only on the field coil ampere-turns and which implicitly contains all the mechanical and material constants. This motor seems to have very little hysteresis in the fields, if it did some other complications would arise.
Next we need to address losses:
First, the so called copper or resistive losses are easily understood with Ohm's Law in that is the speed-voltage is the applied voltage minus the current times the resistance minus brush voltage. So we have:
Speed=(V-IR-Vb)/field, which is field*speed=V-IR-Vb or we can solve

field=(V-IR-Vb)/Speed

Second, the so called iron or magnetic torque losses can be modelled as field squared times speed times a constant. There is also torque loss due to brush friction (Tb) I prefer the form:
Torque=field*Ia*(1-field*Speed/Ia*Rm) - Tb. This is a quadratic equation that can be solved for the field:

field = (Rm/2*speed)*(Ia - SQRT(Ia^2 - 4*speed*(Torque + Tb)/Rm))

When you have dynamometer data, you can extract the loss factors by doing a leastsquares minimization of the difference between the values of the field derived from torque and speed.

I will change this section soon and update the curves:
[I have taken the published dynamometer data for the WarP 9 at 72 volts and performed a least squares fit to the RPM vs. torque data using a four-parameter model for the field plus the resistances. The results for the field map and data fit are in the first attachment.]
I then used those parameters for currents up to 1000 Amps and voltage up to 150 Volts to generate conventional torque vs. RPM curves as in the second attachment.
The third file shows my preferred display of net power generated vs RPM for various voltages and currents. Efficiencies are also shown.
The model allows for separate handling of armature and field resistances, but the dataset does not allow that separation. I hope someone can tell me the those values. I believe this motor has 13 turns per field coil.
It is not possible to get brush losses from the data either.
The model does not account for "armature reaction".
The model is implemented as an excel workbook, not friendly enough for distribution yet. I may model other common motors if you make some specific measurements.
[12/30] At "Gor's" suggestion, I have added the current lines to the torque vs rpm curves posted.
[1/3] Added nominal torque values for each current line on power vs. rpm
[1/3] Added Torque vs. RPM chart
[2/6] edit formulas
 

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#96 ·
Nope never did a coast down. The controller right now is in Alaska racing in a Junior Dragster so as soon as it comes back I can do that test.

Here is the tranny and rear end gearing:

Mitsubishi KM145-8 Manual 4WD Transmission
Gear ratios:
First ………………….…………………. 3.967 : 1
Second …………….……………………. 2.136 : 1
Third …………….……………………….1.360 : 1
Fourth ………….………………………... 1.000: 1
Fifth ………….…………………………. 0.856 : 1
Reverse …….…………………………… 3.578 : 1

Rear Differential .................................... 3.909:1

Tires 225-75R15

Mike
 
#97 ·
Nope never did a coast down. The controller right now is in Alaska racing in a Junior Dragster so as soon as it comes back I can do that test.

Mike
Actuall the typical drive dataset includes a couple of coasting segments. Question is are they on level ground?
The BEMF/Speed from this data set is attached.
Gerhard
 

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#100 · (Edited)
Gerhard, thanks for posting that data! What unit is omega supposed to be expressed in within that pdf? Normally, it is 2pi/T, but I'm not sure if that is what you mean with that symbol.

At 5500 rpm, you get 576.66 for omega when expressed in radians per second. At ~0.32 BMEF per omega at 600A, then the torque generated at 600A should be produced at 5500 rpm at 184V. But, 184V and 600A into the motor is 110.4 Kw electrical, and given that at 600A the motor will produce about 120 lb-ft, at 5500 rpm, this is ~94 kW mechanical, giving an efficiency of 85%.

Provided this is correct, that's not too bad! But it seems quite optimistic for power to be made that high in the RPM range, given the dyno results for cars like Otmar's 914 and his 8" motors, and it would also seem optimistic for efficiency to be that high with that amount of current given the amount of losses from heat alone at that current, let alone losses from the high rotational speed of the motor, brush losses, ect.
 
#105 ·
Gerhard, thanks for posting that data! What unit is omega supposed to be expressed in within that pdf? Normally, it is 2pi/T, but I'm not sure if that is what you mean with that symbol..
I prefer to use SI units: that means rotational speed, omega is in radians/second.
At 5500 rpm, you get 576.66 for omega when expressed in radians per second. At ~0.32 BMEF per omega at 600A, then the torque generated at 600A should be produced at 5500 rpm at 184V. But, 184V and 600A into the motor is 110.4 Kw electrical,
None of the data that was used to derive the saturation curve was at higher than 4K RPM, so your 5500 calculation is an extrapolation ...in an earlier post I noted that This system has a resistance of 16.1 milli ohms and brush loss of 1.18 volts. At 600 amps that is 10.8 volts and 6.5 kW loss, We would be asking for 117 KW at 195 volts which is beyond the battery pack capability: it sags to 130 volts at that kind of power demand.
and given that at 600A the motor will produce about 120 lb-ft, at 5500 rpm, this is ~94 kW mechanical, giving an efficiency of 85%.
I don't have any dyno data for these motors other than that from net-gain which don't match these curves as noted in an earlier post in this thread. Without dyno data, I can't model The iron losses in the motor. Calculateing raw torque from my saturation curve gives 600x0.32 =192 Nxm or 142 ft-lbs, correspounding to 110 kW.
Provided this is correct, that's not too bad! But it seems quite optimistic for power to be made that high in the RPM range, given the dyno results for cars like Otmar's 914 and his 8" motors, and it would also seem optimistic for efficiency to be that high with that amount of current given the amount of losses from heat alone at that current, let alone losses from the high rotational speed of the motor, brush losses, ect.
 
#103 ·
If you really want to get technical you can look up terrain data between 10121 Marmot Ct, Anchorage 99515 and 2550 Denali St, Anchorage 99503. The highway route with speeds over 45 mph were along Minnesota Drive. Any full commutes 45 and under were along C St (which is the way I predominately travelled).

Mike
 
#107 · (Edited)
I applied the model to some Kostov 11 data found here http://kostov-motors.com/files/productattachments/04a1d5e7278db48475f2aa409b5df5ce_S192F01.1.pdf Which includes data for the same motor with the fields connected in parallel. In the first plot I show the saturation curves derived plotted vs. motor current. I also show the parallel results vs. one half the armature current. The results do not overlap as expected:confused:.
I made a guess that the fields were miswired. Instead of:

/--->1/2 current-->coil1-->coil2-->\
\-->1/2 current-->coil3-->coil4--->/

We have
/-->1/4 current-->coil1-->coil2---->coil3-->\
\-->3/4 current-->coil4---------------------->/

In the lower configuration, the sum of the number of poles times current is 1.5 compared to 2.0 for the actual parallel hookup and 4.0 for the series setup. The second graph shows the "parallel" data vs. 3/8 times the armature current which makes the curves coincide. Actually, the lower configuration gives a much better field weakening.
I also display my recently posted Warp11-HV curve.The Kostov total series resistance is 68 mOhms. Compared to 27 for the Warp.
Notice that if you divide the Warp current by two you get close to the Kostov curve.
 

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#110 ·
Have you/anyone found the Piece wise resistances for each field and the Rotor?

I'd like to work out the modelling of a DC motor like the Kostov 11" with interpole/commutating-poles.

I've been try to come up with the details using the Kostov Spec too.
I'm not sure if it's the saturation or the armature reaction or the size of the interpoles that makes it so difficult to analyse.

I just want to know what voltage to apply to the field and armature to make a direct drive setup go in reverse at a low 200RPM.:confused:
 
#116 ·
Gerhard, can you use your model to estimate warp 11hv values 0- 2000a at v=72;144;288? (regular torque-amps-rpm chart, only extended)
attachment - data from netgain site in text file
thank you
and big thanks on behalf of diy community and all people for sharing your knowledge
 

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