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• Abdifatah Muhumed

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BAHIR DAR UNIVERSITY BAHIR DAR INSTITUTE OF TECHNOLOGY School of Research and Graduate Studies Faculty of Civil and Water Resources Engineering Wastewater treatment process and plant design1 assignment Chapter three assignment Submitted by

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Abdifatah Muhumed Abdi

BDU1209509 Submitted to: - Dr. Andinet K. June 2020

Problem 6.4 The water elevations upstream and downstream of a bar screen are 0.89m and 0.85 m. If the approach velocity is 0.45 m/s, what is the flow velocity through the screen? The discharge coefficient of the screen is 0.7. Given Depth of u/s of a bar screen= 0.89 Depth of d/s of a bar screen= 0.85 Approach Velocity= o.45m/s The discharge coefficient of the screen= 0.7 Vs=? Solution Hl= depth of U/s- depth of D/s Hl= 0.89-0.85 Hl= 0.04 Hl=0.74(V-U) ^2/ (2*g) 0.04=0.74(V-0.45) ^2/ (2*9.81) V=√0.752 V= 0.87m/s

Problem 6.5 A mechanically cleaned bar screen is used in preliminary treatment for the following conditions: Incline from vertical = 30° Wastewater flow rate = 150,000 m3/d Approach velocity = 0.6 m/s Open area for flow through the screen = 1.6 m2

Head loss coefficient for clean screen = 0.74 Head loss coefficient for clogged screen = 0.60 a. Calculate the clean water head loss through the bar screen. b. Calculate the head loss after 50% of the flow area is clogged with solids. Solution Flow rate (Q) = 150,000 m3/d/ (24*60*60s/d) Q= 1.74 m3/s Vs= (1.74 m3/s)/ (1.6 m2) Vs= 1.09m/s A. Determine the clean water head loss Hl= 1/CD (V-U) ^2/ (2*g) Hl= 1.35 (1.09-0.6) ^2 / (2*9.81) Hl= 0.057m= 57mm B. Calculate Vs for clogged screen  We know in the case of partially clogged screen Area is reduced to half and the velocity is doubled Area (A) = 1.6/2= 0.8m2 Velocity (Vs) = 2*1.09= 2.18m/s Head loss for clogged screen Hl= 1/CD (V-U) ^2/ (2*g) Hl= 1.67(2.18-0.6) ^2 / (2*9.81) Hl= 0.374m= 374mm  Vs for clogged screen exceeds maximum suggested value of 0.9 m/s, therefore Screens should be cleaned either at regular time intervals or when a specified maximum head loss value is reached.

Problem6.6 Estimate the head loss for a bar screen set at a 30° incline from the vertical. The wastewater flow rate is 90,000 m3/d. The bars are 20 mm in diameter, with 25 mm clear spacing in between bars. The water depth is 1.2 m, channel width is 1.5 m, approach velocity is 0.65 m/s, and the head loss coefficient is 0.65. Solution A. Estimate the number of bars and openings Approximate number of bars= (1500mm-25mm) / (20mm+25mm) = 33 # of bars space= 33+1= 34. B. Determine the effective area of the screen openings by calculating the projected total open area. Area= (34spaces) (25mm/space) (10-3 m/mm) (1.2 m) Area (A) = 1.02m2 C. Velocity through the bar rack Vs = (90,000 m3/d) / (24*60*60) Vs= 1.04m/s D. head loss through the clean bar rack Hl= 1/CD (V-U) ^2/ (2*g) Hl= 1.54 (1.02-0.65) ^2/ (2*9.81) Hl= 0.0485m= 49mm

Problem6.9 Design a grit chamber for a wastewater treatment plant with an average flow rate of 25,000 m3/d and a peak flow rate of 55,000 m3/d. The detention time at peak flow is 3.0 min. The width to depth ratio is 2:1. Use a depth of 2 m. The aeration rate is 0.4 m3/min per m of tank length. Determine the total air required and dimensions of the grit chamber

Solution  If we Use two chambers in parallel and designed it for peak flow rates Step1. Determine tank dimensions. Peak flow rate (Q) = 55,000 m3/d Flow in each tank at peak flow (Q) = 55.000/2 Q= (27,500 m3/d)*(1/24*60min/d) Q= 19.1 m3/min Volume of tank (V) = 19.1 m3/min*3min V= 57.3 m3 Width to depth ratio is 2:1 Depth= 2m, therefore Width= 4m Length= (57.3 m3)/ (4*2m2) = 7m Tank dimensions are= 7m*4m*2m Step2 determine total air required Air required= (0.4 m3/min per m)*(7m) Total air required= 2.8 m3/min