The ultimate size of particles produced in a flocculation basin is described by
ID: 1844202 • Letter: T
Question
The ultimate size of particles produced in a flocculation basin is described by the equation: Diameter (feet) = 0.000 1 x [1 + (G/50 ?4 X G^2/5,000). Choose a G value to obtain maximum floc size If detention time is 50 minutes and flow is 5 cubic feet per second, determine the size of the basin (length = 5 width = 5 depth) If the flocculation wheel has 8 blades at diameter of width -2 feet and the blades are one foot wide and width -2 feet long (assume CD = 1.5), what is the rotational speed required to attain the needed G value?Explanation / Answer
the size of basin can be determined as
volume of basin = flow * detention time
= 5 * 50*60 = 15000 ft3
also we have L=5B= 5D
volume of basin = L3/25 = 15000 ft3
which gives us L= 72.112 ft
B= 14.42 ft
D= 14.42 ft
now for depth = 14.42 ft we have detention perod = 50*60 = 3000 seconds
therefore setlling velocity of tank = 4.8 * 10-3 ft/sec
to find the maximum size of floc given equation is unclear.
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