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Assignment_5 sol

# Assignment_5 sol - CIVL242 Assignment5/Sedimentation...

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CIVL242 Assignment5/Sedimentation Questions and Solutions 1. Assume that the water temperature is 10 and the depth of the settling tank is 3.0 m (9.80ft). Calculate the theoretical settling velocity (in cm/s), tank loading (m 3 /m 2 /d) and detention time (h) required for the removal of the following particles in a settling tank. Assume discrete particle settling. (a) Sand particles with a relative density of 2.65 and diameter of 0.001 cm to be removed in a presettling tank. (b) Alum floc with a relative density of 1.002 and diameter of 1.0 mm to be removed in a settling tank after coagulation. (c) Calcium carbonate precipitates with a relative density of 1.20 and diameter of 0.10 mm to be removed in a settling tank after softening treatment. (d) What will be the corresponding design loading rate and detention time for the above three cases if the efficiency of the settling basins is 50%, assuming that removal is linearly related to efficiency? Solution: At 10 : ρ= 0.9997 g/cm 3 = 999.7 kg/m 3 , μ = 1.307 × 10 -3 N-s/m 2 . Depth, D = 3.0 m (a) Sand with s.g. = 2.65, d = 0.001 cm = 1 × 10 -5 m The particles are assumed to be spherical. v = ρ ρ - ρ p D C 3 gd 4 Assume Stoke’s range (laminar range), i.e., R e 1, then C D = Re 24 = μ ρ vd 24 Substituting this into the settling velocity equation: v 2 = ρ ρ - ρ ρ μ p vd 24 gd 3 4 , v = ( 29 ρ - ρ μ p 2 18 gd v = m s / kg 10 307 . 1 ) m / kg 10 )( 9997 . 0 65 . 2 ( ) m 10 )( s / m 81 . 9 ( 18 1 3 3 3 2 5 2 - × - - - = 0.688 × 10 -4 m/s = (0.688 × 10 -4 m/s)(100 cm/m) = 0.00688 cm/s Check the Reynold’s number: 1 0005 . 0 m s / kg 10 307 . 1 ) m 10 )( s / m 10 688 . 0 )( m / kg 7 . 999 ( μ ρvd Re 3 5 4 3 = - × × = = - - - Re is OK Loading rate: Q/A = v = ((0.688 × 10 -4 m/s)(86 400 s/d)(m 2 /m 2 ) = 5.94 m 3 /m 2 /d 1

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CIVL242 Assignment5/Sedimentation t d = V/Q = AD/Q = D/v = ) d / h 24 ( d / m 94 . 5 m 0 . 3 = 12.1 h (b) Alum floc with s.g. = 1.002, d = 1.0 mm = 0.001 m Assuming Re < 1, v = ( 29 ρ - ρ μ p 2 18 gd v = m s / kg 10 307 . 1 ) m / kg 10 )( 9997 . 0 002 . 1 ( ) m 10 )( s / m 81 . 9 ( 18 1 3 3 3 2 3 2 - × - - - = 9.60 × 10 -4 m/s = (9.60 × 10 -4 m/s)(100 cm/m) = 0.096 cm/s Check the Reynold’s number: Re = μ ρ vd = m s / kg 10 307 . 1 ) m 10 )( s / m 10 60 . 9 )( m / kg 7 . 999 ( 3 3 4 3 - × × - - - = 0.73 1 Re is OK Loading rate: Q/A = v = ((9.60 × 10 -4 m/s)(86400 s/d)(m 2 /m 2 ) = 82.9 m 3 /m 2 /d t d = V/Q = AD/Q = D/v = ) d / h 24 ( d / m 9 . 82 m 0 . 3 = 0.9 h (c) Calcium carbonate particles with s.g. = 1.20, d = 0.1 mm = 0.0001 m Assuming Re < 1, v = ( 29 ρ - ρ μ s 2 18 gd v = m s / kg 10 307 . 1 ) m / kg 10 )( 9997 . 0 20 . 1 ( ) m 10 )( s / m 81 . 9 ( 18 1 3 3 3 2 4 2 - × - - - = 8.35 × 10 -4 m/s = (8.35 × 10 -4 m/s)(100 cm/m) = 0.0835 cm/s Check the Reynold’s number: Re = μ ρ vd = m s / kg 10 307 . 1 ) m 10 )( s / m 10 35 . 8 )( m / kg 7 . 999 ( 3 3 4 3 - × × - - - = 0.64 1 Re is OK Loading rate: Q/A = v = ((8.35 × 10 -4 m/s)(86400 s/d)(m 2 /m 2 ) = 72.1 m 3 /m 2 /d t d = V/Q = AD/Q = D/v = ) d / h 24 ( d / m 1 . 72 m 0 . 3 = 1.0 h (d) Assume efficiency is linearly related to loading rate. The efficiency of a basin is 50%, it is equivalent to assuming that the effective settling velocity is 50% of the above calculated 2
CIVL242 Assignment5/Sedimentation velocity.

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Assignment_5 sol - CIVL242 Assignment5/Sedimentation...

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