Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Loss due to open channel to pipe flow transition

  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. Feb 12, 2006 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    Comes from the energy equation.

    [itex]\frac{p_1}{\rho g}\,+\,\alpha_1\frac{\bar{V}^2}{2g}\,+\,z_1\,=\,\frac{p_2}{\rho g}\,+\,\alpha_2\frac{\bar{V}^2}{2g}\,+\,z_2\,+\,h_{l_T}[/itex] from Robert Fox & Alan MacDonald, Introduction to Fluid Mechanics, John Wiley & Sons, 1978.

    is used with respect to the entrance of the pipe. V is the mean velocity. Perhaps the K's are different (?). Some additional discussion is needed.
  4. Feb 13, 2006 #3


    User Avatar
    Science Advisor

    I can't say that any of my references have this situation. It's either open channel or pipe, not a combination. I will look around to see what I can find as well.
  5. Feb 13, 2006 #4
    Fred and Astronuc:

    Thanks for the help....

    Just to add more info: The case I'm interested in is an open channel discharging into a box culvert and then back into an open channel.

    I actually found a reference that has a similar equation, but I'm still not sure exactly how it was derived. The book is "Hydraulic Engineering" by Roberson, Cassidy and Chaudry (1988 Houghton Mifflin). There is a section on channel-flume transitions where it says:

    *For an inlet transition (contraction): the loss would be KI * V^2/2g, where KI is the loss coefficient for the transition and V is the velocity of the downstream conduit (the highest mean velocity), which would be the pipe velocity in my case of channel to culvert flow.
    *For an expansion, the head loss would be KE (V1^2-V2^2)/2g where KE is the loss coefficient for the expansion, V1 is the mean velocity at the upstream end (culvert in my case) and V2 is the mean velocity at the downstream end (open channel in my case).

    Expansion losses in theory should be greater than contraction losses. However, when I use the 2 equations above I get much greater contraction losses (probably because there is only one velocity term in the contraction loss). I'm really lost here...any clue?

    Thanks, Michelle
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook