efluent rate controller and works at a constant rate of 28.3 1/s. 3. A filter op

Question

efluent rate controller and works at a constant rate of 28.3 1/s. 3. A filter operates with a venturi-tube effluent rate controller and w The gravel, sand, and water (above de sand surface) depths are respectively. The water levels at both the filter m and 0.0 m, respectively. The head loss through the sand, the cm at the beginning of the filter run, and 2.4 m at the end. .6 m, 1.0 m, and 1.5 m. box and the effluent channel remain constant at 3.1 ly. The head loss through the sand, the gravel, and the fiter bottom is 30 a) What is the available head at the beginning and at the end of the filter run, and what is the head lost at the rate controller? Assuming the rate controller works as an orifice, and that the head loss in the effluent pine is negligible, calculate the ratio of the areas of the rate controller opening at the end and at the beginning of the run. b)

Explanation / Answer

Water level at filter box =3.1 m

Head loss at beginning of filter run = 30 cm =0.3 m

Head loss at end of filter run = 2.4 m

Thus available head at beginning of filter run = 3.1- 0.3= 2.8 m

Available head at end of filter run = 3.1- 2.4 = 0.7 m

Thus head lost at rate controller = 2.8- 0.7 = 2.1 m

Ratio of area of orifice at beginning and end of filter run is given as square root of ratio of head available at end to head available at beginning and thus

A1 / A2 = sqrt( h2 / h1 )

Where A1 and A2 is area of orifice at beginning and end of filter run respectively and h1 and h2 are the available heads at beginning and end of filter run

Thus A1/ A2 = sqrt( 0.7/ 2.8 ) = 0.5

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