Physical Treatment Processes Process Purpose Equalization Equalize flow entering wastewater treatment plants Equalize a pumping rate into a storage tank before gravity supply of water Bar Screens Entrance of wastewater treatment plants and some water treatment plants Mixing Rapid Slow: flocculation Settling Type I: Grit removal Type II: Low density flocs removal Type III and IV: high density flocs removal Filtration Removal of low levels of suspended solids Desalination To remove dissolved solids Equalization Tanks Equalization tanks are used to dampen fluctuation in flow rates coming into the WWTP or to pump a constant flow from a WTP Flow, gallon per minute 1200 800 Water pumping out of the plant 400 0 Flow, gallon per minute Water Treatment Plant 1200 800 Water consumption 400 0 10 12 14 16 18 20 22 24 10 12 14 16 18 20 22 24 Time, hr Time, hr Flow, gallon per minute 1200 800 400 Wastewater generation 0 10 12 14 16 18 20 22 24 Time, hr Flow, gallon per minute Wastewater Treatment Plant 1200 800 Wastewater pumping into the plant 400 0 10 12 14 16 18 20 22 24 Time, hr In water treatment plants ET Raw water To consumers Water Treatment Processes Pump In wastewater treatment plants PS GC SS Cl ET BS AT Pump Tank Capacity-Spreadsheet Method Example Consider the data shown for the hourly consumption of water in a typical city Determine the volume of the equalization tank using the spreadsheet method Time Consumption gpm 01-02 02-03 03-04 04-05 05-06 06-07 07-08 08-09 09-10 10-11 11-12 12-13 13-14 14-15 15-16 16-17 17-18 18-19 19-20 20-21 21-22 22-23 23-24 24-01 866 866 600 634 1000 1330 1830 2570 2500 2140 2080 2170 2130 2170 2330 2300 2740 3070 3330 2670 2000 1330 1170 933 Solution Find the volume consumed during each hour (V=Q*t where t is 60 minutes in this example) This is column Find the average of the volume consumed (=111.9 thous gal) Subtract the average from the volume consumed (column 4) Place the positives in one column and the negatives in another (column and 6) Sum either the positive or the negative to obtain the capacity (Capacity= 485.4 thous gal) Add 20% for design purposes: Design capacity= 1.2*485.4 thous gal Time 01-02 02-03 03-04 04-05 05-06 06-07 07-08 08-09 09-10 10-11 11-12 12-13 13-14 14-15 15-16 16-17 17-18 18-19 19-20 20-21 21-22 22-23 23-24 24-01 Consumption Volume consumed Average- volume consumed Positive Negative gpm Thous gal thous gal thous gal thous gal 866 866 600 634 1000 1330 1830 2570 2500 2140 2080 2170 2130 2170 2330 2300 2740 3070 3330 2670 2000 1330 1170 933 Average 52.0 52.0 36.0 38.0 60.0 79.8 109.8 154.2 150.0 128.4 124.8 130.2 127.8 130.2 139.8 138.0 164.4 184.2 199.8 160.2 120.0 79.8 70.2 56.0 111.9 59.9 59.9 75.9 73.9 51.9 32.1 2.1 42.338.116.512.918.315.918.327.926.152.572.387.948.38.132.1 41.7 55.9 59.9 59.9 75.9 73.9 51.9 32.1 2.1 Sum 485.4 42.338.116.512.918.315.918.327.926.152.572.387.948.38.132.1 41.7 55.9 485.4- Tank Capacity-Graphical Method Time Consumption Volume consumed Cummulative gpm thous gal thous gal The y-axis represents the cumulative volume (column 4) 01-02 02-03 03-04 04-05 05-06 06-07 07-08 08-09 09-10 10-11 11-12 12-13 13-14 14-15 15-16 16-17 17-18 18-19 19-20 20-21 21-22 22-23 23-24 24-01 866 866 600 634 1000 1330 1830 2570 2500 2140 2080 2170 2130 2170 2330 2300 2740 3070 3330 2670 2000 1330 1170 933 52.0 52.0 36.0 38.0 60.0 79.8 109.8 154.2 150.0 128.4 124.8 130.2 127.8 130.2 139.8 138.0 164.4 184.2 199.8 160.2 120.0 79.8 70.2 56.0 52.0 103.9 139.9 178.0 238.0 317.8 427.6 581.8 731.8 860.2 985.0 1115.2 1243.0 1373.2 1513.0 1651.0 1815.4 1999.6 2199.4 2359.6 2479.6 2559.4 2629.6 2685.5 For design: The tank capacity is increased by 20% over the one found from the graph or the spreadsheet method Filtration Objective: to remove turbidity (suspended solids) from water Turbidity of water > 4NTU Mixing Flocculation Sedimentation Filter Turbidity of water < 4NTU Filter Granular Media filter Flow through filter = 2- 10 gpm/ft2 Dual Media filter Anthracite (coal) Specific gravity 1.4-1.6 Effective size 0.9-1.1 mm Uniformity coefficient [...]... gpm/sq ft for 12 min Determine the average filtration rate per unit area and the quantity and percentage of wash water used every day Solution Q (2.5 ×106 gal / day )(day / 24 × 60 min) Filtration rate = = = 3.9 gpm / ft 2 As 15 ft × 30 ft Quantity of washwater used daily = q × As × t = (15 gpm / ft 2 )(15 ft × 30 ft )(12 min) = 81,000 gal 81000 gal Percent of wash water = × 100 = 3.2% 6 2.5 × 10