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FLUID MECHANICS - CASE STUDY SOLUTION


Problem Diagram I

 

The upstream Froude number (Fr1) can be determined from the relationship between the upstream and downstream depths across the hydraulic jump as follows:

     

The upstream velocity is obtained from the Froude number:

     

The continuity equation,

     Q = V1by1 = V2by2

gives
     V2 = V1 (y1/y2) = (21.23)(0.5/3.5) = 3.03 ft/s

The downstream Froude number is then calculated to be:

     

     


Specific Force Diagram


Specific Energy Diagram

 

The momentum function is defined as

     M = Q2/gA + zA

where A = by and z = y/2 for a rectangular channel, and the volumetric flow rate Q is determined as follows:
     
     Q = V1by1 = (21.23)(10)(0.5) = 106.15 ft3/s

The momentum function thus becomes

     M = Q2/gby + by2/2     
         = (106.15)2/(32.2)(10)y + (10)y2/2
         = 34.99/y + 5y2

The plot of y versus M is given in the specific force diagram. Note that the momentum functions are the same at sections 1 and 2 before and after the hydraulic jump (for a flat-bottomed channel).

The specific energy is given by

     E = y + (1/2g)(Q/by)2
          
= y + (1/64.4)(106.15/10y)2 = y + 1.75/y2

The plot of y versus E is given in the specific energy diagram. The hydraulic jump process is indicated by the dashed line. Note that the specific energy lost across the hydraulic jump equals the change in E between points 1 and 2. It is given by

     ΔE = E1 - E2
           = 0.5 + 1.75/0.52 - 3.5 - 1.75/3.52
           = 3.86 ft

     
   
 
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