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Potential energy of a 1000 kg weight at 10 meters high

  1. Nov 3, 2008 #1
    What is the most power that can be generated from the potential energy of a 1000 kg mass at a height of 10 meters.

    Suppose you can lift a 1000 kg weight at virtually no cost in energy. How much electricity can be generated either through gear box and ball screw spinning a flywheel, or directly turning high RPM generators. 4 ball screws and gearboxes geared in such a way to control the rate of descent. Is that power limited to the calculated potential energy of the mass.
     
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  3. Nov 3, 2008 #2

    russ_watters

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    Yes, conservation of energy applies, so the potential energy is the limit.

    Power is more difficult because the faster you lower the weight, the more the acceleration due to gravity plays a role. You can't have a constant power output if the speed is high.
     
  4. Nov 3, 2008 #3
    Some control through breaking would be necessary, and if you had a double system I/e
    2 weights tied in so that as you are lifting one the other is in the process of descent you could maintain a somewhat constant rotation. What do you calculate the potential at. I figured it and was surprised at how little energy there was. Potential energy. It seems to me with proper gearing ratios with fine screws and high torques you could manage very slow descent with high RPM generation
     
  5. Nov 4, 2008 #4

    russ_watters

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    If you use one weight to lift another, then there is no energy left over to run a turbine.

    Also note that rpm and torque are inversely proportional to each other when you change a gear ratio, so gearing does not effect power.
     
  6. Nov 4, 2008 #5
    The mass would be lifted from another source at a cost of approximately 30 cents. not with the other mass. So, that being said, is it unreasonbale to believe that the energy generated
    from that potential energy is worth more than the cost of creating it.
     
  7. Nov 4, 2008 #6
    in addition, you would not be fighting the effects torque lifting and only use that torque on descent. This would be done though a ratcheting type mechanism to disengage from the gearbox drag
     
  8. Nov 4, 2008 #7
    Mistake. 30 cent cost was to lift the 1000 kg's 10 feet or approx 3 meters, not 10 meters
     
  9. Nov 4, 2008 #8

    russ_watters

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    Well, did you calculate the energy you could recover from this system? FYI, electricity costs about 15 cents per kWh (retail, residential), so was the answer you got more or less than 6 kWh...?

    [edit] I can answer the question without doing the calculation: you come nowhere close to making money on the deal. If I had to guess, I'd say you are off by two orders of magnitude (ie, you can generate less than a cents' worth of electicity by lowering the weight to turn a generator and spend 90).
    Not exactly sure what you mean there, but like I said, the energy is the energy. The path the mass takes to get to the top only effects the efficiency: the maximum energy is not affected by adding a ratchet.
     
    Last edited: Nov 4, 2008
  10. Nov 4, 2008 #9

    russ_watters

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    Ok, I'll give the answer. 1000 kg, 10 m, is about 100,000 J, which is about .028 kWh or a little under half a cent.
     
  11. Nov 5, 2008 #10

    Redbelly98

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    I have a rather simple approach to this problem. Suppose the energy of the mass is transferred by having it apply a constant force against the energy generating system. Call this force Fgen.

    So the net force on the mass is

    Fnet = Fgen - m g

    and the power transferred is

    (mgh - KE) / t,

    Here KE is the kinetic energy of the mass after falling, i.e. the amount of energy NOT transferred to the generator.
    Also, t is the time it takes to fall given the force Fnet acting on the mass.

    Before solving the problem, there are two extremes to consider:

    1. Fnet = mg (approximately).
    This will transfer nearly all the available energy into the generator, but unfortunately it will take a lo-o-o-o-o-ong time for the mass to fall, so "t" will be large and very little power gets generated.

    2. Fnet = 0
    This will allow the mass to fall in the shortest possible time, which is useful for generating power. However, none of the energy gets transferred to the generator (mgh-KE=0), so there is zero power generation.

    However, for some value of Fgen between 0 and mgh, we can maximize the power expression,

    (mgh - KE) / t

    and that will be the theoretical maximum. Of course there would be efficiency factors to consider in a real system.
     
  12. Nov 5, 2008 #11
    Thank you for your responses. While I understand the theory of conservation of energy, sometimes it just doesn't make sense. To me, it seems with the proper gear ratios that a generator could be run for several hours by the very slow descent of such a mass. .08 KWH may be what the potential energy equates to but I just don't buy that , under any design or configuration,this is the limit to what you could create or generate. Yes, I am a layman, and have only a basic understanding of physics. I have worked in process control for years utilizing robotics, servo controls and PLC's to do very intricate work. I consider myself a logical thinker.
    I'm also very stubborn. Thanks gentleman....
     
  13. Nov 5, 2008 #12
    OK, In your #1 explanation, Yes very slow vertical movement is desired.Lets say I have 4 ball screws that are directly coupled to four 10,000 to 1 gear boxes . as the mass drops 1 inch you get 4 revolutions into each gearbox from the ball screw which gives you 40,000 revolutions out. lets say that takes one minute, could be 10 minutes I believe. so thats 4 generators spinning at 40,000 rpms. In this example, it would take 2 hours to drop 10 feet. So 4 generators in 2 hours can only generate .08 KWH's. This is a rough example. Not sure of the torque of a gear box with that kind of ratio. Should be a great deal
    of resistance to beginning movement. Maybe to much. Anyway, that is my thought as a layman. How can you turn 4 generators at 40,000 RPM's for 2 hours and generate only what is equal to the potential energy I calculated.
     
  14. Nov 5, 2008 #13

    russ_watters

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    Quite true.
    Well, the piece you are missing is what "mechanical advantage" is. It's a common misunderstanding. In short, mechanical advantage multiplies torque while dividing distance and vice versa, which means power is not changed by gearing.

    Try reading the wiki: http://en.wikipedia.org/wiki/Mechanical_advantage

    The calculation is easy: if you have a gear ratio of 3:1, you triple the torque while cutting the rpm by a factor of 1/3. Since power involves multiplying them together, 3*1/3=1.
     
  15. Nov 5, 2008 #14

    russ_watters

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    You can't. If you multiply the rpm by 40,000, you divide the torque by 40,000. You won't have enough torque to overcome the the internal friction of the gears and spin the genereator.

    But even if there were no friction, your generator would not produce much power. The reason is that generators convert mechanical power to electrical power. They don't just spin freely: when you make electrons move, a torque is produced that counteracts the rotation of the generator.
     
  16. Nov 5, 2008 #15

    Dale

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    Out of morbid curiosity, what are you doing that makes it so incredibly expensive to lift the mass? It should take less than 1 cent even with a pretty inefficient system.
     
  17. Nov 6, 2008 #16

    russ_watters

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    Right - a winch would do it a around 1 cent even if its efficiency is pretty low.
     
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