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Re-gen efficiency in Prime

Discussion in 'Prime Main Forum (2017-2022)' started by Krzysiek_KTA, Jun 10, 2017.

  1. Krzysiek_KTA

    Krzysiek_KTA Active Member

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    I'm new to hybrid/electric cars, so need some clarification from the experts on this forum.
    The Prime uses electric motors/generators (MEG) to convert the kinetic energy into electricity and charge the traction battery while in re-gen mode. I seek understanding how the system work.

    Is the re-gen efficiency same for all kind of loads on MEG or it varies ?

    In scenario of slowing down to complete STOP how 'gliding' + stronger re-gen compares to just applying continuous slight re-gen all the way to complete STOP?
     
    #1 Krzysiek_KTA, Jun 10, 2017
    Last edited: Jun 10, 2017
  2. Lee Jay

    Lee Jay Senior Member

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    Varies.

    Smooth and steady beats gliding plus harsh application.
     
  3. Krzysiek_KTA

    Krzysiek_KTA Active Member

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    That is what I suspected. It sounds like brake (re-gen) and glide works.
    Nevertheless if the re-gen efficiency curve of the MEG become available it would justify that in more scientific way...

    @Lee Jay
    Thx for reply

    cheers
     
    #3 Krzysiek_KTA, Jun 10, 2017
    Last edited: Jun 10, 2017
  4. SaraBBrown94

    SaraBBrown94 Active Member

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    There's actually a screen on the prius that shows how much braking is going towards Regen vs physical brakes. So if you brake too hard, the physical brakes will be used and the Regen possibly will be wasted. Using B mode is a good way to not waste braking on moderate hills.

    When/if you do get a prime, the screen on the dash that shows where the power is coming from with the car diagram, hitting the down arrow on the steering wheel shows how much brake power is being utilized.

    Posted via the PriusChat mobile app.
     
  5. Tideland Prius

    Tideland Prius Moderator of the North
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    The thing to keep in mind is that friction braking and regen are not mutually exclusive unless you're below 8mph. If you brake hard, it'll use both regen and friction braking. But if you're below 8mph, it'll transition to friction only.
     
  6. CharlesH

    CharlesH CA HOV Decal #5 on former PiP

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    So that means that if the DRCC takes you to a complete stop, that it will apply the friction brakes at 8mph, and presumably also light the brake lights?
     
  7. Prius from Dad

    Prius from Dad Senior Member

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    The DRCC will light the brake lights but B mode will not.
     
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  8. stevepea

    stevepea Senior Member

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    Is this the same screen that also shows your 3 "ECO" scores? If so, while it does show a regen graph (when regen is going on) it doesn't show if (or how much) physical brakes are being used. Unless there's a different screen I'm missing. Any way you could take a picture of it and post it here?

    If it's the screen I'm thinking of, it's indeed a helpful screen, but it only shows how much regen is going on -- not if physical braking is occurring (or not) at the same time along with the regen...
     
  9. Tideland Prius

    Tideland Prius Moderator of the North
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    DRCC *should* illuminate the brake lights if it brakes as you normally would. The braking system in the Prius is by-wire so it's electronically controlled. It doesn't matter if you're sending the signal to the brake ECU from the brake pedal or from the DRCC ECU.

    The brake lights will come on in the usual fashion with DRCC or with you depressing the brake pedal.
     
  10. David Beale

    David Beale Senior Member

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    Driving the guy behind you crazy when using DRCC in the city. ;)
     
  11. DonDNH

    DonDNH Senior Member

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    The Gen-4 will regen down to about 2 MPH.
     
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  12. Krzysiek_KTA

    Krzysiek_KTA Active Member

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    Try to revive the topic:
    Is the re-gen linear (regardless of speed= MEG load) or not as per quoted post?
    Is there a re-gen efficiency curve as function of MEG load available?
    I'd rather stick to some tangible principles/data rather than rely on Toyota's efficiency calculation software ( Eco scores, ev range estimates, etc.)
    cheers
     
  13. Lee Jay

    Lee Jay Senior Member

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    Regen efficiency cannot be linear. Some portion of the losses (probably most of them) are conduction losses which go with current squared.
     
  14. Krzysiek_KTA

    Krzysiek_KTA Active Member

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    Please enlighten me. Does it mean that the losses are higher with the higher MEG load (=higher current)?
    Forgive my indolence in that subject.
    thx
     
  15. Lee Jay

    Lee Jay Senior Member

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    Torque (braking force) goes with current. Losses go with current squared. That means losses get higher with higher braking force.

    To illustrate, let's take two extreme cases:

    Lets say you need 1 unit of braking force for 10 seconds to stop. You could get that two extreme ways:

    1 unit of braking for 10 continuous seconds
    2 units of braking for 5 seconds, 0 units of braking for the other 5 seconds.

    The first one has losses of 1^2 for 10 seconds for total losses of 1*10 = 10 units.
    The second one has losses 2^2 for 5 seconds for total losses of 4*5 = 20 units.
     
  16. Washingtonian

    Washingtonian Senior Member

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    I think it is more complicated than that. You need a certain braking force to stop this heavy car, so some of the force is used to slow the car and some of the force is used to charge the traction battery. So if I barely apply the brakes and take ten seconds to stop the car or at the same speed I take five seconds to stop the car, which scenario has generated the most power? (I know which scenario pisses off the driver behind me the most.)
     
  17. Lee Jay

    Lee Jay Senior Member

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    Unless you push the pedal hard enough to apply the friction brakes, all of the braking force is used to charge the battery.
     
  18. Krzysiek_KTA

    Krzysiek_KTA Active Member

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    That is contradicting your earlier statement::
    Given your calculations the most efficient re-gen is that involving low current therefore smooth and steady beats gliding plus light re-gen application should work better for energy utilization.
     
  19. Lee Jay

    Lee Jay Senior Member

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    No, it's exactly the same as my earlier statement.

    Low current = smooth and steady = lower losses. Gliding plus harsh application = high current = higher losses.
     
  20. Krzysiek_KTA

    Krzysiek_KTA Active Member

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    OK. Got it. It's clear now
    Thanks