Question

mechanically cleaned wastewater bar screen is constructed using 6.5-mm-wide bars spaced 5.0 cm apart center to center; the bars are vertically oriented. The wastewater flow velocity in the channel immediately upstream of the screen will vary from 0.4 to 0.9 m s−1. What is the design headloss for the screens at the two extremes of flow? Assume that the friction coefficient is 0.84.

Answer 1

Answer: The head loss is 1.5 cm

Explanation:  Equation needed to solve is: $h= (vs^2-v^2)/2gC^2$

where C is the friction coefficient. To determine the area of flow, we nee to realize that the area available for flow over an area of 1*1 m, Aₓ= 0.87 m², observed from the correlation that for every width of 5.0 cm, we lose 6.5 mm, roughly about 12.9 or 13%.

For a fresh clean screen, the velocity will tend to increase by a factor of  (1/0.87).

The velocities through the screen equals to (0.4/0.87) = 0.46 m/s and (0.8/0.87) = 0.92 m/s.

Apply the above mentioned equation:

$h= (0.92^2-0.8^2)/2*0.84^2*9.81 \\\\=0.01499\\=0.015 m = 0.015*100 = 1.50 cm$