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How to use the hydroelectric power equation

  1. Jun 29, 2016 #1
    I would like to get some help with the power equation that derives the power generated in Watts

    P= density (water) * head (meters) * flow rate (cubic meter per second) * g * efficiency

    I understand how to plug in everything except for the flow rate variable. I would like to get help in deriving 2 things for the following set up: If I have a reservoir that is 20m by 20m by 20m (8000 cubic meters of water) and it is 70 meters high.

    1. How big would my pen stock tunnel diameter half to be achieve a 50 cubic meter per second drop? Or is that a function of the head size and the diameter?

    2. If I achieved the 50 cubic meter per second...would the reservoir empty out in about 160 seconds? (8000 cubic meter / 50 seconds??)
  2. jcsd
  3. Jun 29, 2016 #2


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    Last edited: Jun 29, 2016
  4. Jun 29, 2016 #3
    Is this something you are building yourself? If so, 50 cubic meters per second is a very high flow rate. To get a reasonable flow velocity (say 5 ft/sec or 1.5 m/sec) you'd want a pipe 6.5 meters diameter. Yes, it would drain an 8000 m3 reservoir in 160 seconds.

    when you say 20 x 20 x 20 meters, 70 meters high, do you mean the 8000 m3 tank is on a 70 meter hill, above your turbine?
  5. Jun 29, 2016 #4
    Yes those assumptions are correct. I was just working out some theoretical costs of building a hydro electric dam and calculating payback periods etc.
  6. Jun 30, 2016 #5


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    Yes. If there was no inflow.

    50 cubic meters per second is pretty fast. What's the flow rate into the reservoir?
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