Question: ? Sedimentation Solution (contd.): For 100% removal, U. = Us = 0.0022 m/s (refer slide #18) m3 10,000- 2 = day (...) :. From Eq

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? Sedimentation Solution (contd.): For 100%
Sedimentation Solution (contd.): For 100% removal, U. = Us = 0.0022 m/s (refer slide #18) m3 10,000- 2 = day (...) :. From Eq (7.9), As = Uc = 50.3 m2 0.0022 Picking a L:W ratio of 5 (refer slide # 19: recommended range for L:W- 4:1 to 8:1) Ap = LXW = 5W x W = 5W2 = 50.3 m2 " W = 3.2 m and L = 5W = 16.0 m Basin volume (3.2 m) (16.0 m) (3 m) : HRT = m 3 (...) = 22 min Q 10,000 day b) As in (a), calculate Us for do = 0.025 m. >>>>>>>> >> Us = 0.00056 m/s Again, check validity of laminar flow assumption >>>>>>>>> Ne = 0.014 V Us 0.00056 m/s Calculate removal efficiency using Eq (7.10): n = = 25% Uc 0.0022 m/s

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