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Entrance loss from open channel to pipe

  1. Feb 11, 2006 #1

    I have the following equation for the entrance loss from an open channel to a pipe, but I'm not sure how it was derived:

    hloss = K*(V2^2 - V1^2)/2g

    I have always seen entrance losses as: K*(V^2)/2g, but why is the channel flow velocity considered in the equation above.

    Thanks, Michelle
  2. jcsd
  3. Apr 18, 2011 #2
    I think it is because the velocity in the open channel (V1) is assumed to be zero, but I am not 100% sure.
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