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Achievable torque from regenerative braking

  1. Jul 12, 2015 #1
    Hello guys,

    I have been trying to calculate the maximum torque that can be applied to my driveline through regenerative braking, using the equations from this paper: https://smartech.gatech.edu/bitstre...ambamurthy_aravind_201305_mast.pdf?sequence=1

    I have been using the equation T=-(K^2/Ra)*omega , the motor I have has an armature resistance of around 0.2Amps; while also as the magnetic constant is given by K=Em/omega , I have for the sakes of estimation let steady state applied voltage = back EMF voltage, this gave me: K(V/Rad/s)=18/127 = 0.142

    Looking at some physical test data I could achieve 22Nm of torque at 84 rad/sec, but using this equation it shows I should only achieve 8.5Nm (calculated below)

    (T=-(K^2/Ra)*omega = - (0.142^2/0.2)*84=8.5Nm

    Would I be correct in saying that this is because the charge resistance of the batteries increases regeneration torque?

    Kind regards

    Jacob Smith
     
  2. jcsd
  3. Jul 13, 2015 #2

    Lok

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    Hi Jacob,

    By using the : T=V^2/(Omega*Ra) with omega = 84 I get 19.something in T.

    I am not sure but why did you use 127 in one equation and 84 in the other? Where did the 127 come from?
     
  4. Jul 13, 2015 #3
    Hi Lok, and thanks for the reply!

    As K is a constant I used test data from unladed conditions to give me the electromagnetic constant (K), of which returned 18V at 127 rad/sec. I thought using these unloaded conditions was best to find the constant because I assumed supply V to roughly equal back emf in steady-state conditions.

    This gave me the constant, which I then used to estimate the torque at a different speed (84 rad/sec)

    So I know Ra = 0.2 ohms, and K= 0.142

    Therefore the torque should be given by T: -((0.142^2)/0.2)*84 = -8.5Nm

    Comparing this value to some data I actually have from 84 rad/sec, I can see that I actually achieved 22Nm.

    Would I be correct that regen torque is directly proportional to rotational speed?

    Unfortunately there doesn't seem to be many publications which explain regen torque :(

    Thanks for your help
     
  5. Jul 13, 2015 #4

    Lok

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    You should search for a characteristic map of torque vs rpm of your electric motor. If not available of a electric motor of same type AC/DC/BLDC etc.
    What type of motor or generator is it actually?
     
  6. Jul 13, 2015 #5
    Hi Lok,

    Im using a PMDC brushed axial flux motor (Agni 95) , which unfortunately don't have a great deal of data available.

    I understand the torque vs rpm maps and how this is usually higher at lower speed etc. However that is for motoring, where a supplied voltage is greater than the back emf.

    When generating however, the supply voltage is zero for max torque, and the back emf is speed dependent, therefore with a constant armature resistance, surely a higher regen current and thus torque is achieved at higher rpm's.

    The majority of papers I have found suggest the same, with the regen torque crossing though the axis with an intercept of zero.

    The trouble is, I am actually achieving higher braking torques than should be possible from the calculations. I am therefore wondering if any other system constraints (e.g. charge resistance) actually increase motor braking torque?

    Cheers
     
  7. Jul 13, 2015 #6

    Lok

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    Well maybe this link will get some info:
    http://www.jozztek.com/shop/content/9-technical-data-on-agni-motors
    There are some performance graphs for the AGNI not really sure it is exactly the same one.
    While torque and rpm seem to be linear, the performance has a complicated behavior at low speeds. So a high reactive torque vs lower than expected power generated.
    The generating graph will be different but not far from this.

    Honestly it is hard to point the issue, but it seems to be a high loss of power at low rpm range.
     
  8. Jul 14, 2015 #7

    jim hardy

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    all Lok's curves for that motor show it 71RPM per volt
    you report measuring 18 volts at 127 rad/sec

    if my arithmetic is any good, 127 rad/sec is 1212 RPM
    so you should have got 1212/71 = 17.08 volts, and only 5% from published is not bad at all...


    how do you know that ?

    From the 12 volt curve
    as load on motor increases , motor speed drops from , maybe 860 RPM to 570 RPM as armature current increases from 0 to 400 amps.
    That 290 RPM change equates to 280/71 (oops) 290/71 = ~4 volts of the 12 applied are dropped in the armature resistance instead of being counter-emf
    meaning
    at 400 amps Ia X Ra equals ~4
    so Ra looks to me closer to 0.01 ohms.

    What do you think ?
     
    Last edited: Jul 14, 2015
  9. Jul 14, 2015 #8

    jim hardy

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    torque will be in proportion to current.

    generated voltage will be in proportion to speed.

    What determines current ?

    here's a good writeup on DC motors that's brief enough for one sitting.
    http://www.ee.lamar.edu/gleb/power/Lecture%2005%20-%20DC%20motors.pdf [Broken]
     
    Last edited by a moderator: May 7, 2017
  10. Jul 14, 2015 #9
    Hi Jim and thanks for the replies,

    I measured the motor resistance using a multimeter, but maybe this did not have a high enough resolution for the low resistances were dealing with, I will re-check :)

    I see yes, motor torque is proportional to current, so I assume Id be right to say that because during regen there is no applied voltage , but there are high back emf voltages, the potential difference is high. High back emf voltages (which rise with speed), and low resistances create high currents, and thus high torque. Therefore without any applied voltages reducing the back emf advantage, the torque speed curve will be linear, increasing at higher speeds.

    The resistances we talk of are only within the motor, do you know if controller and battery resistances will in-turn reduce current and therefore braking torque?

    A good write-up that is, cheers
     
  11. Jul 14, 2015 #10

    jim hardy

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    I think you've got it, even if not expressed succinctly.

    I learned English units.
    These two formulas describe the external characteristics of a DC motor
    Counter EMF = K X Φ X RPM
    Torque = 7.04 X K X Φ X Iarmature , same K , result in ft-lbs
    i'm foggy this morning - my 7.04 probably needs to be multiplied by n-m per ft-lb for your metric units?


    KVL around the motor and its load(or source)

    Vexternal = Counter EMF +(or -) Iarmature X (Rarmature + Rexternal)
    be sure to pick sign of Iarmature so it adds or subtracts correctly to counter-emf for generator or motoring action

    so yes, an external EMF reduces current

    PM motors are nice because their flux is so nearly constant
    and every single one of Lok's curves shows torque linear with current, reaching ~54n-m at 400 amps
    that's 39.8 ft-lbs

    Using my English units:
    KΦ from published torque and armature current,
    39.8 / (7.04 X 400 ) = .01413
    KΦ from volts per RPM
    1/71 = .01408
    match within 0.3% from torque read off a graph ? Incredible ! You try it using speed off the graph.

    i wonder if those curves are measured or calculated .


    In big machines we assume a couple volts for brush drop irrespective of current.
    Armature resistance you'll have to measure on inside of the brushes. Can you reach commutator?

    If not, try locking rotor and measuring voltage at two different currents, neither of them zero,
    then solve for
    volts = m X current + B, B will be brush drop at standstill an m will be armature resistance

    Have fun !
     
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