uestion 2 (Coagulation &Flocculation; A drinking water treatment plant in centra
ID: 1712032 • Letter: U
Question
uestion 2 (Coagulation &Flocculation; A drinking water treatment plant in central New York relies on one of the Finger Lakes as its main water supply. Unfortunately, the lake has been impacted by cyanobacterial harmful algal blooms (so called "cyanoHABs") over the past few summers, which poses a potential threat to the safety of drinking water. The treatment plant operates a single-compartment, completely-mixed flocculator with a volume of 275 m3 and an average velocity gradient of90 s1. The plant influent flow rate is approximately 5.5 m/min. (a) In summer 2016, the flocculator influent carried at an average number concentration of 109 cyanobacteria/m2 when blooms occurred. What was the diameter of cyanobacteria (in the unit of m), assuming that cyanobacteria were spherical particles with a volume of 10-15 m3? (b) What was the number concentration of cyanobacteria in the flocculator effluent, assuming that cyanobacteria were the only particles present in the flocculator influent and that macroscale flocculation was the dominant flocculation mechanism for cyanobacteria (assuming the collision efficiency factor was equal to 1.0)? (c) Following part (b), what was the average diameter of the cyanobacteria flocs in the flocculator effluent (in the unit of m), assuming that the flocs were also spherical and that the floc volume fraction remained unchanged before and after flocculation? Did flocculation have a significant impact on the size of cyanobacteria flocs? (d) In summer 2017, the treatment plant installed an in-line mixer to dose ferric upstream of the flocculator. Following coagulation, the cyanobacteria floc volume fraction increased to 104 in the flocculator influent. What was the diameter of cyanobacteria in the flocculator influent (in the unitExplanation / Answer
a)
volume of spherical cyanobacteria = 10-15 m3
volume of sphere is 4/3 **r3 = 32/3 **D3 = 10-15
D= (10-15*3 /(32*))1/3 = 3.1017 *10-6 m= 3.1017 m
b)
no concentration = total volume / volume of one cyanobacteria = 1/10-15 = 1015 cyanobacteria /m3
c)
given concentration = 109/m3
volume of one cyanobacteria = 10-9/m3
D= (10-9*3 /(32*))1/3 = 310 m
d)
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