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B Can we transfer 100% Potential energy to electricity?

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  1. Jun 3, 2018 #1
    I read somewhere PE=KE, how? It mean we can get transfer 100% PE to KE and KE to Joule
    Please anyone can explain?

    If I throw mass of 1000kg from 50 meters height can I get energy equal to Pe=mgh = 1000*9.81*50=490500 Joule? It mean 490500Kn = 490500watt= 490.50Kw.
    Is it possible to to get this energy if I throw a round 1000kg sphere from 50 meters height?
     
    Last edited by a moderator: Jun 3, 2018
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  3. Jun 3, 2018 #2

    mfb

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    Where did you read that and in which context?
    In practice you'll always have some losses, but letting something drop in a good vacuum achieves a nearly 100% conversion of potential energy to kinetic energy. "to Joule" doesn't make sense, Joule is a unit of energy, not a type of energy.
    The units don't match, this equation makes no sense.
    That is the initial potential energy of the object (apart from some missing units).
     
  4. Jun 3, 2018 #3

    sophiecentaur

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    That is not correct in general but there are many times where it allows you to calculate a 'good', near enough answer if you assume it's true. The statement implies that (as an example)all the PE (where an object starts) is converted to KE as it falls to your chosen height=zero point.
    In any real situation, some of the PE will be transferred into heat by stirring up the air (or water etc.) on the way through.
    Note: this KE that has been obtained by dropping from a height is not a lot of use as it is and would need to be transferred into another form, say, Electrical Energy in a Hydroelectric generator, which would have an efficiency of a lot less than 100%.
     
  5. Jun 3, 2018 #4
    When you find any real process with zero losses, be sure to let me know first, please. They are truly rare!
     
  6. Jun 4, 2018 #5
    The statement KE = PE is very misleading. It is not correct all the time, as Sophiecentaur pointed out. I have seen this happen many times, typically where an object slides a frictionless plane or roller coaster, starting from rest at the top. In this case, it is true that:

    PE at the top = KE at the bottom
    It is definitely not true that
    PE at the top = KE at the top!

    The correct statement, which is always true in the absence of friction, is that

    PE + KE = constant throughout the motion.
     
  7. Jun 4, 2018 #6

    sophiecentaur

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    A much better way of stating the situation.
     
  8. Jun 4, 2018 #7

    russ_watters

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    Part of that confusion can be avoided by adding numbered subscripts to identify the two states.
     
  9. Jun 4, 2018 #8

    berkeman

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    @Deep Trivedi -- To answer the question you asked in your thread title, hydroelectric power generation comes pretty close to 100% efficiency for large installations...

    https://www.mpoweruk.com/hydro_power.htm
     
  10. Jun 4, 2018 #9
    I'm amazed that a water turbine is 90+ efficient, something I learned a few years back. It just seems that to capture that much energy, that 95% of the water must be utilized, with less than 5% 'leaking' past the blades (there will be some loss due to friction, even some from emitted noise/sound). And I would think that a fit that tight would create all sorts of turbulence, which would react against the turbine. My thinking is obviously off, since those are the numbers, it's just a little hard to visualize it.
     
  11. Jun 4, 2018 #10

    sophiecentaur

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    Hydroelectric Generation.
    I'm not too sure that the energy is actually all Kinetic at the turbine. I think it must be partly 'Pressure times Water Velocity' based. A bit like the bicycle chain argument. What do you all think?
    This has nothing to do with the overall efficiency which relates Potential to Electrical Energy.
     
  12. Jun 5, 2018 #11

    RonL

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    It might help to consider the first test of heat transfer from a paddle wheel to water, on turbine blades being driven by vast amounts of water the change being almost undetectable should be quite large. Just a thought.
     
  13. Jun 5, 2018 #12

    russ_watters

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    What I have always found counter-intuitive is that most turbines do not, in fact, harness the pressure at the turbine, but rather convert the high pressure, low speed flow to high speed, low pressure flow and then harness the kinetic energy of the flow.
    https://en.m.wikipedia.org/wiki/Turbine#Operation_theory
     
  14. Jun 5, 2018 #13
    If you consider the electron in a Hydrogen atom as having a constant energy which varies in form, that is a lossless yet quite real process.
     
  15. Jun 5, 2018 #14

    sophiecentaur

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    Yes but is not a 'process'. It is a Quantum 'state'. No Energy is transferred.
    But all these things are worth thinking about to find where they fit in.
     
  16. Jun 6, 2018 #15
    I forgot, were not supposed to conjecture about what really happens in a physical atom with physical charges and particles. We call it a quantum state and pretend nothing knowable happens inside.
     
  17. Jun 6, 2018 #16

    russ_watters

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    An electric heater.

    ...but this is getting off topic, so let's let that go.
     
  18. Jun 6, 2018 #17

    sophiecentaur

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    It’s easy to ignore the inevitable loss of energy when an actual supply is involved. There is always a source resistance involved. Having said that, we (officially) only get charged for the electrical Energy used. There’s thermal loss too.
     
  19. Jun 7, 2018 #18
    If one considers a single photon at just the correct energy to jump the band gap in a silicon solar cell junction and you consider the electricity as the immediate instantaneous electron current thus generated before they begin bumping into things, one can get very close to 100% according to the original question.
     
  20. Jun 7, 2018 #19

    berkeman

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    So I can buy a 100% efficient solar cell? Very cool. Can you give me a link? I've got my VISA card ready to go!
     
  21. Jun 8, 2018 #20

    sophiecentaur

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    No. The Mechanical Equivalent of Heat was what Joule and others called it. There is no reason why the temperature change should be easily detectable because the MEH is actually a much higher value than you might think (mechanical Joule per heat calorie). You seem to be implying there was something wrong with the understanding of the effect you describe. Water has a massive specific heat capacity.
    To be fair, that wasn't the implication. If we received monochromatic light at the appropriate frequency from the Sun, we could have PV cells that are a lot more efficient. But the probability of a given photon producing a photo-electron would still not be 1.
     
  22. Jun 8, 2018 #21
    Hey, give me a break. :)

    I specifically stated the situation was for one specific energy photon and before the losses as current flow were considered for an external usable current. Even so, a solar cell operating with fixed monochromatic light of the exact right frequency would be extremely efficient. Even in the real world, with real solar spectra, multifunction cells could get up into the 80% range. Why aren't they doing this now? Simple economics. Why spend a fortune making 85% multi-junction cells when you can make a killing selling 15% cells.
     
  23. Jun 8, 2018 #22

    sophiecentaur

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    The space industry would snap them up if they could be made at even a high price.
     
  24. Jun 9, 2018 #23
    Unless we start building really big in space, the market is too small.
     
  25. Jun 9, 2018 #24

    sophiecentaur

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    The concept of Efficiency for a PV array is questionable. In many cases, it doesn't matter too much what area of PV array is used so energy out / energy in is not necessarily relevant unless available area is limited. Add to that the cost of cells and land and the choice is even less obvious.
     
  26. Jun 9, 2018 #25
    In your simple economics, I think you may be mixing up the supply side and the demand side.

    I think it is more useful to look at it from the other way. To paraphrase your comment:

    Why spend a fortune making buying 85% multi-junction cells when you can make a killing selling buy 15% cells at far less/watt?

    And if there is no demand, there won't be a supply.

    As @sophiecentaur points out, there are other factors. The PV is coming down in price, I don't think it is even the majority of cost in an installation - that is divided up between mounting hardware, labor, inverters, etc. In that regard, more efficient cells mean lower installation costs (2x the efficiency means half the panels to install - same amount of inverter though).
     
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