Open Channel Flow - Combination Application

 

Example:   (Continued)                The surface profile looks approximately:

 

                                                       

                                                                           

 

For the case of a mild slope,   y3  =  yN  =  2.85 ft     Also  q  =  y3V3,  so  V3 = 16/2.85 = 5.6 ft/sec

 

Fr3  =  V3 / √ gy3  =  5.6 / √ (32.2)(2.85)  =  0.34

 

Now for a hydraulic jump,   y2 / y3  =  [ √ ( 1 + 8 Fr32 )  ˗ 1 ] / 2       So  y2  =  1.3 ft   (result)

 

 

The specific energy plot is as follows:

                                           

 

 

The energy dissipated across the hydraulic jump is  E2 ˗ E3  =  0.3 ft    (result)

 

Click here go to a summary of equations for open channel flow.

 

 

Return to Notes on Fluid Mechanics


Copyright © 2019 Richard C. Coddington
All rights reserved.