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• + 11-3 SINGLE-PASS (INTERMITTENT) PACKED-BED FILTERS 735 <br /> EXAMPLE 11-5. DESIGN OF AN INTERMITTENT SAND FILTER FOR A SINGLE DWELLING. Size <br /> and lay out an intermittent sand filter and distribution system for an individual three- <br /> bedroom residence in a nonsewered area. Determine organic loading rate, the flowrate,. <br /> the duration of an individual dose,and the discharge per orifice.Check the difference in <br /> the discharge between the orifices.If the difference in the discharge between the orifices <br /> is greater than 5 percent [(l - m) x 100] in either case, the distribution system should <br /> be resized. The septic tank is sized to provide a storage capacity of 2.5 times the av- <br /> erage flow. Finally, estimate the total dynamic head. Assume the following conditions <br /> I! apply: <br /> E <br /> 1. Average occupancy.= 3.0 personsld <br /> 2. Average wastewater flow = 50 gal/capita•d <br /> 3. Assumed peaking factor = 2.5'(see Table 4-20) <br /> 4. Size of septic tank = 1500 gal (4.5 m3) <br /> 5. Effluent BOD from septic tank = 130 mg/L <br /> 6. Effluent TSS from septic tank = 30 mg/L <br /> 7. Hydraulic application rate = 1.25 ga1/ft2•d(51 Um'---d) based on peak design flow <br /> 8. Sand filter dose rate per day =;24 times/d <br /> 9. Distribution system orifice size= $ in(3.2 mm) <br /> 10: Orifice discharge head = 5 ft minimum (1.5 m) <br /> 11. Distance from septic tank to sand filter = 60 ft <br /> 12. Difference in elevation between septic tank and sand filter = 10 ft <br /> Solution <br /> 1. Determine the size of the sand filter. <br /> a. Determine the average and peak flow rates: <br /> Qa,,e = (3.0 persons) x 50 gal/capita-d = 150 galld <br /> I Qpeak =.150 galld x 2.5 (see Table 4-20) = 375 galld <br /> i <br /> b. Determine the required area of the sand filter: <br /> Area = (375 gal/d)/(1.25 gallft2•d) = 300 ft2 <br /> C. Use a filter 17 ft x 18 ft. Check area: (17 x 18) = 306 ft'-, OK. <br /> i <br /> 2. Determine the organic loading rate. <br /> (1.25 gal/ft2•d)(3.785 L/gal)(130 mg/L) <br /> L°'� _ (1000 mg/g)(454 gllb) <br /> = 0.00135 lb BODIft2-d <br /> . 3. 'Lay out sand filter.and the effluent distribution systein.Use a spacing of 1.5 ft between <br /> •distribution pipes and orifices, and a wallspace of 1 ft. <br /> a. Determine the number of laterals fora nominal spacing of 1.5 ft: <br /> Number of laterals = (17 - 2)11.5 + 1 = 11 laterals <br /> b. Determine the number of orifices fora nominal spacing of 1.5 ft: <br /> Number of orifices = (18 - 1.5)11.5 11 orifices/lateral <br /> c. The layout of the sand filter and distribution system is shown below. <br /> I <br /> I <br />