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How much energy could be collected from such a setup?

  1. Jun 15, 2011 #1
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    Suppose that we have a tower with two compartments. The upper one is filled with mercury and the bottom one with water. The floor can be opened (and so the compartments are abolished), and mercury could flow down to the bottom of the tower. I have calculated the potential energies of the two setup (A and B) and came to the conclusion that there is an energy difference of 14755 kilojoules. My question is that if we modify the setup (C) by adding a channel for the flowing mercury and a waterwheel (or a similar device) then how much energy could be collected of that 14 755 kilojoules?

    If no exact calculation can be made, then any estimate? I have found in Wikipedia that a waterwheel can have an efficiency of 60%, but I suppose that a setup with mercury and water has a different efficiency than a setup with water and air.

    I'm not a professional physicist, nor an engineer, so please no offense if I'm completely wrong at some point.
     
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  3. Jun 15, 2011 #2

    QuantumPion

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    What purpose does the water serve? Why not just drop the mercury over a waterwheel in an empty tank? The only thing having the mercury flow through water accomplishes is reduce the useful energy available due to buoyancy and friction.
     
  4. Jun 15, 2011 #3
    You are right, in this setup water has no purpose, but the drawing is only one part of a system in which water is necessary. I tried to simplify, this is the only part I'm having troubles with.
     
  5. Jun 15, 2011 #4
    Wikipedia says: "If the object is much denser than the fluid, then B (the buoyancy force) approaches zero and the object's upward acceleration is approximately −g, i.e. it is accelerated downward due to gravity as if the fluid were not present"

    So buoyancy isn't reducing the energy available. I don't really feel that because of friction the system would loose "too much" energy in the form of heat. Any ideas on how to calculate it based on the viscosic properties of water and mercury?
     
  6. Jun 15, 2011 #5

    russ_watters

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    60% is as good an estimate as any.
     
  7. Jun 15, 2011 #6
    Maybe you never had the opportunity to walk through water. Otherwise you should feel how much more energy is required to move something through water as compared to moving through air.
    As for calculation, it will depend on the shape and size of the wheel. There is no way to calculate it in general. You may be able to design the wheel in a way that will reduce the drag.
     
  8. Jun 15, 2011 #7

    Q_Goest

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    Hi balaz,
    I'm not going to check your math but it looks like you set it up properly. The amount of potential energy difference between A and B is what you show in your calculations. It is the hypothetical amount of energy that can be extracted given 100% efficiency, which is obviously not possible. Putting a wheel inside the system to extract that energy is one possible way to remove that energy, but it will be much less than 100%. To do any reasonable calculation on it, you'd need to add all sorts of details around the wheel and even then it would only be so accurate. There's just too many variables for a quickie calculation.

    PS: This isn't part of a perpetual motion machine idea is it? Needless to say, perpetual motion is impossible.
     
  9. Jun 16, 2011 #8
    Good point, I was not considering drag and now I have realized that turbulence is another issue to deal with.

    No, it isn't. As I said I'm not a physicist, nor an engineer, but that does not mean that I'm a fool!

    Anyway, thanks for the answers and if someone has any other remarks then please feel free to share!
     
  10. Jun 16, 2011 #9

    sophiecentaur

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    If it's not part of a perpetual motion machine then is it intended as a method of energy retrieval? If it is, then you would be much better to take the mercury through an external tube and through a turbine, (rotating in air) and back into the bottom of the tower.. Your losses would be a lot lower.
    If it's just a 'thought experiment' then change your efficiency to something much lower, to account for the losses of this wheel sloshing around under water. I'd think something like 15% as you'll be stirring up all that water to no avail.
     
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