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Creating Energy with Hydraulics

  1. Nov 29, 2009 #1
    I was hoping one of you could look at this idea and tell me if it will produce a positive net-energy supply, with the help of hydraulics? Please ponder it a moment and feel free to capitalize off of it if it does.

    Let me pause for a moment to explain what you already know. I can lift a 5000 lb vehicle with about 25 lbs of force with the help of hydraulics. Now if I connect another hydraulic mechanism to the first I could then lift the same 5000 lb force with about 1 lb of force.

    So if it is true that with some form of advanced hydraulics, you could lift 5000 lbs with 1lb of effort. It seems conceivable you could lift a large weight quite high using very little electricity.

    .Using the same principal as above, imagine a 20,000 lb being lifted up a shaft. This weight is connected to a(n) gear box(s) 1200:1. Then connected to a(n) electric turbine(s). The weight is then released to slowly come to the ground. On its way down it is creating electricity.

    Can I produce more energy with the weight slowly moving down connected to turbine than it takes to hydraulically lift the weight?
     
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  3. Nov 29, 2009 #2

    russ_watters

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    Welcome to PF.
    No, this is not true. In hydraulics, you convert pressure to force. Ie, if you have 25 psi and you want to lift 5000 lb, you need a hydraulic cylinder with a cross sectional area of 200 square inches.

    That said, this is very similar to what happens in a lever, where you do multiply force. The problem is that while you multiply force, you divide distance. Going back to the hydraulics example, rather than a pump, just use two pistons of different sizes and different lengths and you get the same thing. Ie, the piston that you push on with 25 lb of force must be 200x longer than the piston lifting the 5000lb car. And that's how conservation of energy is satisfied - which means that no, you can't use this to create a perpetual motion machine.

    This principle is called mechanical advantage: http://en.wikipedia.org/wiki/Mechanical_advantage
     
  4. Nov 30, 2009 #3
    the principal could be used to accumulate energy over time to do a single action that you wouldn't normally have the power to accomplish (aka slowly lift the weight with solar, water, wind, over a long period, then use the stored pressure to do a job, but heat, fluid leakage and friction all must be paid

    there is no free ride

    dr
     
  5. Nov 30, 2009 #4
    I see what your both saying above. No free ride! Seemingly such a simple idea would being used if it were possible.

    What about these hydraulic wenches that are on the market these days. Some very strong, 30,000lb wenches, also even 50 ton hydraulic hoists. What is the energy conserved by using these devices. for example: with the use of pullies and hydraulics what amount of energy could be saved. 20:1 maybe 50:1.

    Also, with normal water pressure at say your house being 40 psi. Maybe the local water pressure could could be enough energy to push they (for example:2000lb weight) up with the use of various mechanisms that conserve or reduce energy required.

    This is really the answers that I'm seeking.

    How much energy would it take to lift a 20,000 lb weight using the most efficient way possilbe (pulleys,advanced levers, and hydraulics combined)?

    And obviously how much electricity could be created by the energy of the weight moving down with proper gear boxes and turbine create?
     
  6. Nov 30, 2009 #5

    FredGarvin

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    There is not a direct energy conservation between the production of hydraulic power and the power required to perform the task. There are losses involved that mean it ALWAYS takes more energy to make the hydraulic power than what is needed to move the load or whatever you are doing. Some of that energy goes to things like heat, sound, etc... and that is all lost energy. If you take into account all losses in your system then the energy is conserved.
     
  7. Nov 30, 2009 #6

    minger

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    Potential energy is mass times gravity times height.

    See above.

    I think that you have a general misunderstanding of the relationship between forces, work and energy. Check out wikipedia (or a textbook) to try and get a better feel for what these properties actually are.
     
  8. Nov 30, 2009 #7
    I appreciate both of your replies. I'm still not understanding why there isn't an energy surplus with hydraulics. If I can lift the front end of a 5000lb car with 30lbs of pressure, this still leaves 2,500 lbs of car lifted in to the air. Seems like quite a savings of power, force, energy, whatever you want to call it. So with the 30lbs of energy I now have 2500lbs of
    energy in the air that can do a work for me.

    I'm going to research the capabilities and limitations of hydraulics a little bit further.

    I would consider this a little different than perpetual motion, seeing as your using electricity to push the machine into the air.

    God bless
     
  9. Nov 30, 2009 #8

    russ_watters

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    Fred this all seems confusingly worded:
    In an idealized situation (lossless), there is a direct energy conversion between the production and use of hydraulic power: Ein=Eout and regardless of how you model the system, it isn't that hard to figure out where the energy is coming from/going to. I used a double-piston cylinder arrangement for my hydraulic system because it is easier and more direct to show the mechanical advantage, but if you use a hydraulic cylinder and pump, it works the same way. Eout can be force times distance on the cylinder, Ein is flow times pressure on the pump and Ein=Eout.

    If you don't take into account any losses in the system, energy is conserved (Ein=Eout). If you do take losses into account, energy is lost (Ein>Eout).
     
    Last edited: Nov 30, 2009
  10. Nov 30, 2009 #9

    russ_watters

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    Pounds is force, not energy. Energy is force times distance. In my first post, I showed you mechanical advantage which multiplies force and divides distance in direct proportion to each other. Take the following example:

    Say you want to lift 5000 lb with a piston by 1 ft.
    Fout=5000 lb
    Dout=1 ft
    Eout= 5000*1= 5000 ft-lb

    Say you want to do it with an input force on another piston of 25 lb:
    Mechanical advantage= 5000lb/25lb=200
    Dout=1 ft*200=200 ft
    Eout=25 lb*200 ft=5000 ft-lb

    Conservation of energy cross-check:
    Ein=Eout= 5000 ft-lb
    Din=5000 ft-lb/25 lb=200 ft
    By definition, a type 1 perpetual motion machine is a perpetual motion machine that violates the 1st law of thermodynamics (conservation of energy). In other words, it is a machine that has more output than input power/work.
     
  11. Nov 30, 2009 #10

    russ_watters

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    There is no energy savings. The 20:1 or 50:1 is the mechanical advantage which multiplies force (torque) while keeping energy constant.
    Well you're still just describing a hydraulic cylinder. The more weight you want to lift, the more flow rate you need and that's equivalent to more distance in a rope/pully arrangement. It's mechanical advantage.
    Energy (work) is force times distance. So regardless of the method, if you seek 100% efficiency, the energy required to lift a 20,000 lb weight 1 foot is 20,000*1=20,000 ft-lb.
    Assuming 100% efficiency, the energy you can create by lowering a 20,000 lb weight 1 foot is 20,000*1=20,000 ft-lb
     
  12. Dec 1, 2009 #11

    FredGarvin

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    What I was trying to get across is that if you put in 10 HP into a hydraulic system, you will not get 10 HP out of it. I am talking about the larger picture, not just the cylinder itself. I am looking at the pump/driver and all of the associated system. From the standpoint of the OP's statements of creating energy, it seemed that that aspect needed to be impressed. In the end, it will take more energy to create that hydraulic power than what one gets out.
     
  13. Dec 1, 2009 #12
    Thank you for you deatiled explanations and teachings on this science.
     
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