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Regenerative Braking Energy

  1. Dec 17, 2009 #1
    Hello, I'am new to this forum and this is actually my first post. And I have some questions about regaining braking energy.

    We are thinking of using an electric motor to speed up and slow down a new and innovative train. The idea is to regain energy while the train is slowing down with the electric motor.

    I've read some articles on the net about braking with electric motors and regaining energy, but I can't seem to find any good formula's for calculating the regained energy. The formula's I found are hard to understand and they differ a lot with other websites.

    Maybe you could help me out. I would like to hear some idea's and formula's from you.

    Thanks,

    some specs of the train:
    20 tons, 20.000kg
    top speed 250 km/h , 70m/s
    it rides on wheels, r=0,5m (max)

    some formula's I found:
    pdf url document from an electric motor manufacturer.
    http://www.reliance.com/prodserv/standriv/appnotes/d7743.pdf
    The formula's who are in this pdf document arent clear to me. Maybe because I'am familiar
    with the metric system

    http://img693.imageshack.us/img693/3609/formule.th.jpg [Broken]
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Dec 17, 2009 #2
    Your kinetic energy is 49 MJ (megajoules) or about 13.6 kilowatt-hours. Where do you plan to store the recovered energy? (Note:the battery in the Toyota Prius holds about 1.2 kWh).
    Bob S
     
  4. Dec 17, 2009 #3

    OmCheeto

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    Gold Member

    greetings trideon
    Energy recoverable from braking equals 0.5 times the mass of the vehicle times the initial velocity squared.

    Until you understand this, all other formula's will only lead to confusion.
     
  5. Dec 18, 2009 #4
    First thanks for responding.

    I haven't really thought about the energy storing. Maybe a lot of big battery's or give it back to the electricity net. And I'am thinking of the usage of multiple electric motors.

    I know that the net regenerated energy is Ek=1/2×m×v2 reduced with friction losses and efficiency losses in the system.

    My question is: Is there any easy or understandable way to calculate the losses. What values/data do I need.
    Oh and the Deceleration is 2m/s^2 = a, in F=m×a. So I know the force F (40.000N) and a = 2m/s^2.
     
    Last edited: Dec 18, 2009
  6. Dec 18, 2009 #5

    OmCheeto

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    In the situation where one comes to a full stop:

    Initial Kinetic Energy = Change in Stored Energy + Energy Lost

    therefore

    Losses = Initial Kinetic Energy - Change in Stored Energy
     
  7. Dec 18, 2009 #6
    Your losses and the behavior of the train doesn't depend on an equation, it depends on the regenerative braking system. There is no way to just simply calculate these losses accurately, or even ballpark the numbers really, unless you have detailed information about the braking system itself.

    Regen braking systems used in cars are very complex and require a lot more than just switching a few resistors on a motor or generator.
     
  8. Dec 18, 2009 #7
    The weight of the train is 20,000 Kg x 9.81 m/sec2 = 196,000 N. The desired stopping force is 40,000 N. So the needed wheel friction coefficient on the rails is about 40,000/196,000 = 0.2. Even with regenerative braking on every steel wheel on steel rails, you will not be able to achieve half of this. So you will need to switch to rubber tires.
    Also, the stopping regenerative power is

    P = Force x velocity = mass x deceleration x velocity = 20,000 Kg*2 m/sec2 x 70 m/sec = 2.8 MW = 3,800 HP.

    So your regenerative motor/generators (brushless dc motor/generators) are significant.

    Bob S
     
  9. Dec 19, 2009 #8
    I was forgotten to put this with it: The train rides frictionless on the rails, because it uses an aquaplaning effect by his advantage. The train is riding on a thin sheet of water. A bit hard to explain everything but the point is the friction caused by wheels touching the rails can be neglected.
    Anyway thanks for the formula's, I will use and apply them.

    Thanks every1 for responding I think I have enough to make some good conclusions about regenerative braking. I was hoping there was a simple way to calculate the regenerated energy but I came to the conclusion that its very hard to do en there are very variables which should not be neglected.
     
  10. Dec 19, 2009 #9

    Ivan Seeking

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    http://www.railway-technical.com/brake1.shtml#DynamicBraking
     
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