Open Channel Flow - More complicated application

 

Example:   Water flows in a wide rectangular channel at a flow rate, q = 16 ft2/sec.  The channel

bed has a slope, So = 0.0009 and is lined with brick.  The flow enters the channel by flowing

under a gate with depth of 1.25 ft.  Determine the depth for equilibrium flow and the critical depth.  Classify the slope as mild or steep and establish that a hydraulic jump will occur downstream. 

Sketch the surface profile.  Find the energy dissipated across the hydraulic jump.  

Also determine the depth just upstream of the hydraulic jump by determining the downstream

depth of the hydraulic jump.  Show a specific energy plot that locates each of these depths.

 

                                            

                                      

 

Strategy:  Apply Manning's equation to find the depth for equilibrium flow.

 

               Q  =  (K/n)  A RH2/3 So 1/2              K  =  1.49 for English units     g = 32.2 ft/sec2

 

Data:  From tables:  n  = 0.016 for brick     q = 16 ft2/sec,   So = 0.0009

 

A  =  w yN ,      wetted perimeter  =  w + 2 yN,        RH  =  w yN / ( w + 2 yN )  ≈  yN    

 

                                        q  =  Q/w  =  (K/n) yN  (yN) 2/3 So 1/2

 So            (yN) 5/3  =  q n / K So 1/2    and for the data    yN  = 2.85 ft   (result)

 

 

Summary so far:   y1  =  1.25 ft,     yc = (q2/g)1/3 = 2 ft,       yN = 2.85 ft 

 

Since   y1  <  yc   the channel slope is mild.  Also the Froude Nb =  Fr1  =  V/√gy1

 

And   q  =  y1 V1  so  V1  =  16 / 1.25  = 12.8 ft/sec   and  Fr1  =  2.0   supercritical flow

 

Since the flow under the gate is supercritical and the slope is mild a hydraulic jump

will occur with sufficient reach of the channel.   (result)

                                                                                     Click here to continue with this example.

 

 

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