CEE 350 - Korshin - Winter 2012 - Homework 8(1)

CEE 350 - Korshin - Winter 2012 - Homework 8(1) - CEE 350...

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CEE 350 Week 8 Homework Solutions Problem 1 (Textbook 6.6) Hor izon ta l sha f t padd le whee ls are used in a tapered flocculation basin comprised of three compartments. The power input to the paddles progressively decreases through the basin with the first compartment paddle driving 186 W of power, the second compartment with 30.0 of power, and the final compartment paddle driving with 7.50 W of power. Each paddle wheel mixes a compartment 4.17 m deep, 3.75 m wide, and 4.17 m long. The normal water temperature is 15ºC and the average flow rate is 16,000 m 3 /day. (a) What is the mixing intensity in each of the three compartment? (b) If the influent water contains a nearly monodisperse (single-sized) particle size distribution of 20 μm particles with a concentration 1.8·10 5 particles/mL, what will be the average number of singlet particles in an aggregate leaving the first compartment? (a) The mixing intensity can be calculated using this formula b V P G P The volume of each compartment is 65.21 m 3 . The viscosity of water at 15ºC is 0.00114 kg·m -1 ·s -1 . Accordingly, the G values calculated for each of the compartments are shown in the table below P (watt) G (sec Ͳ 1) 186.0 50.0 30.0 20.1 7.5 10.0 (b) This can be examined using the equation for the ratio of numbers of particles entering the flocculation basic and exiting from it. This equation is given in the textbook Q GV N N b S D : ± 4 1 0 where G is the mixing intensity, D is the collision efficiency (can be assumed to be close to 1 in these conditions), : is the floc volume defined as 6 0 3 N d p :
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In the above equation, d p is the diameter of the initial particles (20 μm) and N 0 is their initial number per unit of volume (1.8•10 5 particles/mL). For these conditions, the volume of : is ±² ± ² ± ² ± ² 4 3 6 1 5 3 18 3 0 3 10 · 54 . 7 6 / 10 10 · 8 . 1 10 20 14 . 3 6 ³ ³ ³ u u u u : m ml mL m N d p S The flow rate is 16,000 m 3 /day= 0.185 m 3 /sec. Accordingly, the N 0 /N ratio is ± ² ± ² 93 . 17 sec / 185 . 0 14 . 3 21 . 65 sec 50 10 · 54 . 7 1 4 1 4 1 3 3 1 0 4 u u u u u ´ : ´ ³ ³ m m Q GV N N b D So the average number of the particles leaving the first chamber will be 1.8•10 5 /17.93= 1.03·10 4 particles/mL. Problem 2 (Textbook 6.8) A rapid sand filter has a loading rate of 8.00 m/h, surface dimensions 10 m · 8 m and effective filtration rate of 7.70 m/h, and a production efficiency of 96%. A complete filter cycle duration is 52 h and the filter is rinsed for 20 minutes at the start of each cycle. (a) What flow rate (m 3 /s) does the filter handle during production? (b) What volume of water is needed for backwashing plus rinsing the filter in each filter cycle? (a) The loading or filtration rate of the filter is defined by the ratio of the volume of treated water by the filter cross-section: f a A Q v The cross-section of the filter is 80 m 2 . The flow rate is therefore 80·8=640 m 3 /hour.
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This note was uploaded on 03/31/2012 for the course CEE 350 taught by Professor Korshin during the Spring '10 term at University of Washington.

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CEE 350 - Korshin - Winter 2012 - Homework 8(1) - CEE 350...

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