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11-3 SINGLE-PASS (INTERMITTENT) PACKED-BED FILTERS 727 the field capacity, the applied liquid will flow over the filter medium in a thin layer, maximizing the opportunity for absorption of constituents from the applied liquid and the transfer of oxygen from the air in the interstices. It should also be noted that the field capacity of a filtering medium will increase as the film thickness within the filter increases. I Organic loading rate. The organic loading rate is composed of the soluble and particulate organic matter applied to the filter. The organic loading rate is typi- cally expressed, on an area basis, as lb BOD or COD/ft2•d (kg BOD or COD/m2•d). In some references, the organic loading rate is expressed, on a volumetric basis, as lb BOD or COD/ft3•d (kg. BOD or COD/m3•d). The amount of organic material added to the filter with each dose must be such that the microorganisms in the bi- ological film can process the mass of organic matter added without accruing addi- tional mass between doses (see Example 11-4). Although the appropriate organic loading rates are not well defined, typical values are in the range from 0.0005 to 0.002 lb BOD/ft2•d (0.0025 to 0.01 kg BODIm2•d). . I I EXAMPLE 11-4. ORGANIC LOADING OF INTERMITTENT SAND FILTER. Estimate the amount of material that can be processed in an intermittent sand filter with a porosity of 0.40. Assume the bacterial filen occupies 5 percent of the void space in the filter bed and that the specific gravity of the biological film is 1.125. Solution I 1. Determine the mass of the biological film in the sand filter,based on 1.0 ft3 of filter bed. a. Volume of bacterial film: Volume of film = (1.0 ft3/ft3)(7.48 gal/ft3)(3.785 Ugal)(0.4)(0.05) = 0.57 L/ft3 - b. Mass of bacterial film: j Mass = (0.57 LIft3)(1.125)(1.0 kg/L) = 0.64 kg/ft3 = 640 g1ft3 2. Assuming the endogenous respiration coefficient is 0.05d-I (see Chap. 7) and that active microorganisms make up 10 percent of the bacterial film,the amount of waste j that can be processed per day due to endogenous respiration is: a. Mass of active bacteria in the bacterial film: Mass of bacteria = (640 g/ft)(0.1) = 64.0 g/ft3 b. Mass of organic matter that can be processed to maintain the bacteria is: Mass of organic matter = (64.0 g/fO)(0.05/d) = 3.2 g/ft3-d, 3.. Convert the amount of organic matter processed per day to an organic loading rate applied per square foot: Organic loading rate = (3.2 g/ft3•d)/(1.0 ft2 x 454 g/lb) 0.007 lb/ft�•d = 0.034 kg/M2 •d r Comment. The above computation illustrates the point that the performance of inter- mittent sand filters is extremely sensitive to the active mass of microorganisms in the film within the filter. i I