University:
Massachusetts Institute of Technology
Course:
1.85 | Water and Wastewater Treatment Engineering
Academic year:

2021
Views:

292

Pages:

7
Author:

Henry A

1.85 WATER AND WASTEWATER TREATMENT ENGINEERING HOMEWORK 3 Question 1 (6 points) A wastewater was tested in a settling column test with the following results in terms of suspended sediment concentration: At t = 30 minutes Depth below surface, h (cm) Concentration remaining, c/c0 At t = 60 minutes Depth below surface, h (cm) Concentration remaining, c/c0 At t = 90 minutes Depth below surface, h (cm) Concentration remaining, c/c0 38 0.23 34 0.05 32 0.03 118 0.81 114 0.46 112 0.23 198 0.94 194 0.73 192 0.52 278 0.97 274 0.86 272 0.70 358 0.98 354 0.92 352 0.83 a. What type of settling is indicated by these data? (2 points) b. This particular wastewater is proposed to be treated in a rectangular sedimentation tank having a detention time of 2 hours and a depth of 4 meters. Estimate the percent of suspended sediments that will be removed. (2 points) c. If the wastewater flow rate is 7,500 m3/day, what needs to be the area and volume of the sedimentation tank? (2 points) Question 2 (2 points) The city of Hong Kong has a dual water-supply system that provides freshwater for drinking and bathing, and salt water for flushing toilets. At the Hong Kong wastewater treatment plant, as in most other wastewater treatment plants, wastewater is passed through a primary clarifier to settle out suspended solids. The Hong Kong plant achieves greater removal in their primary clarifier than most plants elsewhere in the world. Why? Question 3 (2 points) A rectangular sedimentation basin is to be designed for a flow of 1.0 mgd (million gallons per day) using a 2:1 length:width ratio, an overflow rate of 0.00077 fps (feet per second), and a detention time of 3.0 hr. What are the dimensions of the basin? Solution to Homework 3, Question 1 a. 0 40 Depth below surface (cm) 97 95 77 90 80 70 54 19 120 77 50 160 200 27 6 40 240 48 30 280 30 14 3 20 10 320 360 17 8 2 400 15 30 45 60 75 Detention time (minutes) Straight removal efficiency lines indicate discrete particle settling. 90 105 120 Solution to Homework 3, Question 1 Detention time (min) t 30 30 30 30 30 60 60 60 60 60 90 90 90 90 90 Depth Conc. remaining (cm) c/c0 0.23 0.81 0.94 0.97 0.98 0.05 0.46 0.73 0.86 0.92 0.03 0.23 0.52 0.70 0.83 z 38 118 198 278 358 34 114 194 274 354 32 112 192 272 352 Settling velocity (cm/min) v 1.3 3.9 6.6 9.3 11.9 0.6 1.9 3.2 4.6 5.9 0.4 1.2 2.1 3.0 3.9 F, Fraction of particles with less than settling velocity v Particle velocity distribution 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 Curve drawn by "eyeball" 0 1 2 3 4 5 6 7 8 v, settling velocity 9 10 11 12 13 Solution to Homework 3, Question 1 Fraction removed from Equation 9.30, pg. 230, of Reynolds and Richards 1 - F0 = 1 - 0.75 0.25 v F 3.3 3.0 2.5 2.0 1.5 1.0 0.5 0.0 Fraction removed = Fraction removed = v - ave 0.76 0.70 0.64 0.49 0.34 0.18 0.04 0.00 ∆F 3.2 2.8 2.3 1.8 1.3 0.8 0.3 v ∆F 0.06 0.06 0.15 0.15 0.16 0.14 0.04 0.19 0.17 0.34 0.26 0.20 0.11 0.01 Sum 1.27 Sum/v0 0.38 0.25 + 0.38 0.63 Alternative calculation using graph and formula from Lecture 6, page 18 R ∆h h 100 90 70 50 40 30 26 0 100 190 230 325 370 400 100 90 40 95 45 30 95 80 60 45 35 28 Sum Percent removed = 64.5 % ∆h/hn * Rave Rave 23.8 18.0 6.0 10.7 3.9 2.1 64.5

1.85 Water and Wastewater Treatment Engineering: Homework 3 Solutions

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