|Feb11-06, 04:30 PM||#1|
Entrance loss from open channel to pipe
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.
|Apr18-11, 10:01 AM||#2|
I think it is because the velocity in the open channel (V1) is assumed to be zero, but I am not 100% sure.
|Similar Threads for: Entrance loss from open channel to pipe|
|Transition from pipe flow to open channel flow||Mechanical Engineering||6|
|Question on open channel flows||Advanced Physics Homework||0|
|Heat Loss in a Hot Water Pipe System||Engineering Systems & Design||35|
|Loss due to open channel to pipe flow transition||General Engineering||3|
|Open pipe problem||Introductory Physics Homework||2|