Energy(work) captured in falling water

  1. I know that there is the equation to capture work of falling water using dams (power= head*flow*gravity constant*efficiency).

    But let's just say that there is an empty basin in the middle of a large body of water that is 100 meters deep. If the basin is 30m * 30m and we have a conduit that is open to the body of water at the top and extends down to the bottom of the basin; is it correct to say that if the water is allowed to fill the basin to the top by allowing water to fall down the conduit and fill the basin that the work done from the falling water is:

    work=Force *distance

    30m*30m*100m *1000kg/cubic meter *9.81m/second square*100m

    =88,290,000,000 newton meters (joule)

    So if there is an energy capturing device (assume 100 percent efficiency), it would capture all the energy of 88,290,000,000 Joules?

    Sorry for the amateur assumptions and calculations :(

    G
     
  2. jcsd
  3. The water falling from 100 m height would convert all of that potential energy to kinetic energy, so you'd have a lot to work with if you could extract it all. But unless you kept removing the water that fell to the bottom, you couldn't really have a steady flow to extract energy from. At best you'd have a progressively shorter drop as the basin filled up. Removing the water would entail raising it back up to the top or pumping it out to the side against the pressure at 100 m.

    The nice thing about a hydro dam is that the water flows out of the reservoir, past your turbines, and just leaves.
     
  4. Thanks olivermsun. I do understand that issue with not really capturing net energy or even being negative if you try to remove the water to restart; but I was curious about not dropping from the surface straight into the basin, but down a conduit that runs down the side of the basin from the surface right to the bottom of the basin (like a pipe or penstock that is open to the surface of the outside of body of water and down into an opening in the bottom of the basin?) The water will fill up higher and higher, but water still will have to flow down and flow past the turbine in the conduit to fill up the basin to the top. The water pressure would still flow until the basin if filled to the top. I know the rate of flow would slow down as the water gets higher in the basin because the water in the conduit would move up the conduit...but the water would still flow and capture work. Would the equation still be force times distance?
     
  5. The water can only convert the potential energy due to its height into kinetic energy (which you harness) because it falls unimpeded. The same problem always arises with this kind of system—no matter where you put the turbine, eventually the "catch-basin" next to your pipe will be filled above the height of your turbine and you either need to expend energy pumping water from low to high, or you have to raise your turbine and accept the ever-decreasing drop height.
     
  6. But there is still flow occuring as water from outside the surface flows down and moves the turbine as the basin gets filled to the level of the outside body of water. I understand the wattage would be lower because the rate of flow would be less. But how can no work be captured?

    For river flow systems, there is no head, but still work is being captured as water flows.
     
  7. Yes that is the total work done by the descending water, but to get the net work done on the turbine you have to subtract the work done raising the water level in the basin. The net work done is therefore the force times the difference in height between the head of water and the height of the water in the basin.

    And of course the force (pressure) is also proportional to the net head.
     
    Last edited: Jul 28, 2014
    1 person likes this.
  8. Maybe you could draw a diagram of your configuration and the expected flow pattern of the the water.
     
  9. @ oliversum - sure...how do i attach a drawing? Can i attach a powerpoint pic and where is the attachment symbol?
     
  10. diagram

    I hope you can see this
     
  11. Not seeing anything.
     
  12. russ_watters

    Staff: Mentor

    Looks to me that all you have to do is divide your total work in half. This accounts for the fact that the average fall height of the water is half the initial.
     
    1 person likes this.
  13. Thank you everyone... the answers are appreciated. Pls if anyone else has another angle I would like to hear :)
     
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