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How do hydroelectric dams generate power?

  1. Sep 15, 2011 #1
    I don't quite understand how the law of conservation of energy applies to a hydroelectric dam, such as the one pictured below. My understanding is that the gravitational potential energy of the water in the reservoir will be converted to kinetic energy as it moves through the penstock. Where I'm confused is what happens to the energy of the water as it passes through the blades of the turbine? I see two possibilities.

    1. Some of the kinetic energy of the water is transferred to the turbine causing it to spin. But if this is the case the water must slow down, meaning it would get backed up in the pipe? How can water have two different velocities when flowing through a pipe?

    or

    2. The water has the same speed both before and after it passes through the blades of the turbine. But if this is the case the water must have the same kinetic energy on both sides of the turbine, meaning no energy was transferred and thus no power can be generated.

    Is there something in play here that I'm not seeing?

    Your thoughts would be much appreciated. Thanks!

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  2. jcsd
  3. Sep 15, 2011 #2

    Pengwuino

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    Why would the water go back up the pipe if it slows down? When a car decelerates, it doesn't move backwards. Why would water?
     
  4. Sep 15, 2011 #3

    rcgldr

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    For a reaction turbine, the energy transferred to the turbine corresponds to a drop in pressure of the water as it passes through the turbine. Water is almost incompressable, so since the mass flow rate is the same at every cross section of the penstock, turbine, and outlet (otherwise water would be accumulating), there's almost no change in speed or kinetic energy.

    http://en.wikipedia.org/wiki/Reaction_turbine#Reaction_turbines

    http://en.wikipedia.org/wiki/Francis_turbine

    An impulse turbine involves a nozzle that directs a high speed flow at the blades of a spinning turbine, which change the direction and speed of the flow with little change in pressure, but that is not the type of turbine shown in the diagram of the first post of this thread.
     
    Last edited: Sep 15, 2011
  5. Sep 15, 2011 #4
    Hmm. I guess what I still don't understand is where the energy to turn the turbine comes from if the water doesn't loose any energy as it passes through the turbine. If the energy of the water remains unchanged, but the energy of the turbine increases, it seems like you're creating energy, which is not possible.

    If the water never slows down it seems you could, in theory, put infinite turbines on a river, and produce an infinite amount of power. However from what I understand about energy it seems there must be an upper limit on the amount of electric power produced by a single river.
     
  6. Sep 15, 2011 #5
    The individual water molecules do slow down and loose kinetic energy as they strike the turbine, so the total energy of the system is conserved. But the average flow velocity of billions of interacting water molecules does not just depend on the individual molecules' velocities, but also on the pressure, which in turn depends on the pipe diameter and what is going on upstream. So as the individual molecules loose energy, this translates to a drop in pressure for a large volume of water instead of a drop in average flow velocity. Put another way, the entire system is filled with liquid water all bound together, so that is is better to think of a total column of water being heavy due to gravity and exerting pressure on the turbine, than on individual molecules striking the turbine.
     
  7. Sep 15, 2011 #6

    A.T.

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    No problem, if the pipe changes in diameter.
     
  8. Sep 15, 2011 #7

    russ_watters

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    It loses pressure energy, not kinetic energy. You're just confused about which type of energy you're dealing with.
     
  9. Sep 15, 2011 #8
    It is not the kinetic but potential energy that matters. The water that goes through the turbine does not lose kinetic energy, but the water in the reservoir loses its potential energy.

    Consider a cross-section of a pipe. The water on each side acts on another side with a force F = P S (P-pressure, S-area of the cross-section). If the water in the pipe is moving, this force produces work at a rate dW/dt = Power = P S V = P F (V-velocity, F-flow (in m^3/s). In an ordinary (horizontal) pipe with the same pressure at both ends, the work at both ends cancels out, but the pipe effectively transfers energy from one end to another (hydraulics). In case of turbine, the pressures before and after are different, so there is a net power = delta P x F. Pressure diferential is the usual delta P = rho g H.
     
  10. Sep 15, 2011 #9
    In addition to the excellent points made by some of the previous posters, you also seem to not be taking into account that the outflow pipe is at a lower elevation than the turbine. In other words, gravity continues to pull on the water after it has gone through the turbine, slowed down, but then accelerates again as it 'falls' through the outlet pipe, yes?
     
  11. Sep 15, 2011 #10
    Interesting question!!


    PE plus KE at the top of the penstock seems greater than the similar total at the bottom.
    The difference is what is transferred to the turbine.
     
    Last edited: Sep 15, 2011
  12. Sep 15, 2011 #11
    The water can't slow down.....unless the outlet pipe is bigger......otherwise water would accumulate....If the velocity at the bottom of the penstock is lower than at the top, where does all that volume go??
     
  13. Sep 16, 2011 #12
    wikipedia says:

    which at superficially seems ok,,,

    but if the turbine is removed, what happens to the energy otherwise used to power the turbine....it's obviously significant....

    and if the pressure drops, how is the volume of water flow maintained??
     
  14. Sep 16, 2011 #13

    rcgldr

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    With the restriction of the turbine removed, the overall rate of flow through the penstock from the upper reservoir to the lower reservoir would increase.
     
  15. Sep 16, 2011 #14

    russ_watters

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    If the turbine is removed, there is less restriction to flow and the entire column of water flows faster.
    What do you mean? Flow is maintained the way it always is: as a function of pressure drop from point A to point B. In the case of the hydroelectric turbine, there is virtually no restriction anywhere except the turbine, so the pressure loss everywhere else in the piping system is near zero.
     
  16. Sep 17, 2011 #15

    I had the same thought too....but could not reconcile it with this idea:

    we INCREASE pressure at the bottom of the penstock,,,,, when the turbine is removed......and water flows faster??? that seems counter intuitive....I am missing something here....
     
    Last edited: Sep 17, 2011
  17. Sep 17, 2011 #16

    A.T.

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    I have a related question regarding "pressure energy" (as described http://hyperphysics.phy-astr.gsu.edu/hbase/press.html#ed") and the assumption of incomprehensibly:

    Let's have water in a cylinder, with an increasing external force applied to the piston. If we assume incomprehensibly, the piston is not moving, so we are not doing any work. But the pressure and thus the pressure energy increases.

    Is this where the assumption of incomprehensibly fails? If not: Where does the energy come from?
     
    Last edited by a moderator: Apr 26, 2017
  18. Sep 17, 2011 #17

    russ_watters

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    Who said removing the turbine increases the pressure at the bottom of the penstock?
     
  19. Sep 17, 2011 #18

    russ_watters

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    No, that concept assumes incompressibility.

    The energy is potential energy. You haven't done any work on an incompressible fluid by pressurizing it; you've only gained the potential to do energy. That's what a reservoir above a hydroelectric dam is for! It only releases the energy when you open a valve and let the water flow out the bottom.
     
    Last edited by a moderator: Apr 26, 2017
  20. Sep 17, 2011 #19

    A.T.

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    But usually you have to do work to increase the potential energy of something.

    Just to make sure: I'm asking about pressure energy as defined here:
    http://hyperphysics.phy-astr.gsu.edu/hbase/press.html#ed
    Not about potential energy as defined here:
    http://hyperphysics.phy-astr.gsu.edu/hbase/press.html#fpe

    Not sure what "potential to do energy" means. My question is: Does the total energy of the fluid increase, when the force on the piston increases?


    That is potential energy due to position in a gravity field, which doesn't change for the fluid in my cylinder. I'm interested in the pressure energy.
     
    Last edited: Sep 17, 2011
  21. Sep 17, 2011 #20

    russ_watters

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    Sorry, that should have been the potential to do work.
    I think it would have to depend on how the force is applied. If the force is applied by a weight being lifted onto the top of the cylinder, for example, then the total energy of the system has certainly increased. In the case of the hydroelectric dam, the water at the bottom gains the potential to do work by piling more water on top of it, just as if you had lifted a weight on top of a piston.
    In this case, they are one and the same. Or if you prefer, one causes the other: At the bottom of a hydro reservoir, the pressure is equal to the weight of the fluid. That's why the equations are identical to each other. In your piston-cylinder example, the piston has to do the work, so only by seeing how it applies the pressure can we know if the whole system has gained potential energy. It could be a spring, compressed air, a weight, etc. The potential energy could be different for the same pressure, depending on how the force is applied. In a hydro dam, the force is applied by piling water on top of more water.
     
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