Question
Download Solution PDFA water treatment plant has a flow rate of 0.6 m3/s. The settling basin at the plant has effective settling volume dimensions of length 20 m, depth 3 m, and width 6 m. What percentage of the particles having a settling velocity of 0.004 m/sec is removed?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFSettling velocity (Vsc) of concerned size particle is called Surface overflow rate (SOR) and it is given by:
SOR = Discharge/Surface area
Or
\(SOR = \frac{{0.6}}{{20\; \times \;6\;}} = 0.005\;m/s\)
Now,
All particles whose settling velocity is greater than or equal to SOR will settle 100 % while particles whose settling velocity is less than SOR will settle partially.
In this case, Settling Velocity of given particles < SOR ⇒ they will settle partially
\({\rm{\% \;Removal}} = \frac{{{\rm{Setlling\;velocity\;of\;given\;particles}}}}{{{\rm{Settling\;velocity\;of\;concerned\;size\;particle\;or\;SOR}}}}{\rm{\;}}\)
\({\rm{\% \;Removal}} = \frac{{0.004}}{{0.005}} \times 100\)
∴ % Removal = 80%
Last updated on Jul 1, 2025
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