Assignment_4 sol

Assignment_4 sol - (a) Total Power Paddle velocity for...

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Assignment 4 Solutions 1. Q=8.0×10 3 m 3 /d, G=1000 S -1 T= 15 , square basin with a depth equal to 1.25 times width, t d =30 s t d = V/Q V = Qt d = (8.0×10 3 m 3 /d)(30 s)(1d/86400 s) = 2.78 m 3 L = w, D =1.25w V = LwD = w 3 (1.25) = 2.78 m 3 w = 3 3 (2.78m )/1.25 1.31 = m L = w = 1.31 m, D = 1.25 (1.31 m) = 1.64 m G = 1/ 2 P V μ , P = μVG 2 At 15 : μ = 1.139 × 10 -3 N-s/m 2 (kg/m-s), ρ = 999.1 kg/m 3 P= (1.139 × 10 -3 kg/m-s)(2.78 m 3 )(1000 s -1 ) 2 = 3156 kg-m 2 /s 3 = 3156 J/s =3156 W Impeller diameter, d = 0.50w = (0.50)(1.31 m) = 0.66 m 2. A flocculator paddle of the design and dimensions shown below is rotated through water at 20 o C with an angular speed of 4.0 r/min. (a) How much power is dissipated into the water? (b) If the tank in which this paddle is rotating has the dimensions of 4 × 4 × 4 m and the flow through the tank is 5000 m 3 /d, determine the Gt value for the flocculator.
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Unformatted text preview: (a) Total Power Paddle velocity for Inner board = x 2 x 4 x 1/60 = 0.419 m/s Paddle velocity for outer board = x 3 x 4 x 1/60 = 0.628 m/s Power for each paddle Paddle tip speed (V p ) = 0.75 x paddle velocity Co-efficient of drag (Cd) = 1.8 for flat blade Density of water ( ) = 998.2 kg/m 3 Viscosity, = 1.002 x10-3 N.s/m 2 P = (C d A p V p 3 )/2 Power, P1 for inner board = [1.8 x (0.1 x3 x 4) x 998.2 x (0.75 x 0.419) 3 ]/2 = 33.46 watt Power, P2 for inner board = [1.8 x (0.1 x3 x 4) x 998.2 x (0.75 x 0.628) 3 ]/2 = 112.64 watt Total power P = P1 + P2 = 33.46 + 112.64 = 146.1 watt (b) Gt Retention Time, t = V/Q = 4x4x4/5000 = 0.0128 days = 1106 sec G = [P/(Vx )] 1/2 = [146.1/(4x4x4x1.002x10-3 )] 1/2 = 47.75 s-1 Therefore, Gt = 1106 x 47.75 = 52811.5...
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This note was uploaded on 09/13/2011 for the course CIVL 242 taught by Professor Gh during the Fall '09 term at HKUST.

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Assignment_4 sol - (a) Total Power Paddle velocity for...

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