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MOUND DESIGN WORKSHEET <br />(For Flows up to 1200 gpd) <br />A. FLOW <br />Estimated -) f ,) gpd <br />or measured - x 1.5 = gpd. <br />B. SEPTIC TANK LIQUID VOLUMES <br />-;^oo gallons <br />C. SOILS (refer to site evaluation) <br />1. Depth to restricting layer = '_ 3,4 inches <br />2. Depth of percolation tests = 11 " inches <br />3. Percolation rate mpi <br />4. Land slope S % <br />Bummed Sewdre Flow in GaUms pen pay (z;4 <br />Nurned <br />of Type I II Typ< III Type ryJBC&Oorm <br />2 300 223 I b as <br />130 100 211 •r <br />1 600 175 236 <br />3 730 430 241 <br />6 9W 323 73: e. <br />7 IO:SO 600 170 I �� <br />>! 1 2 <br />D. ROCK LAYER DIMENSIONS <br />1. Multiply flow rate by 0.83 to obtain required area of rock <br />layer: A x 0.83 = <br />sO gpd x 0.83 sq. ft. / gpd = x x sq. f t.+ )o% <br />2. Select width of rock layer (10 feet or less) _ I ft. <br />3. Length of rock layer = area + width = Rock Bed <br />sq. h. + 10 ft. = _(pisr_ ft. <br />tdtF. 510 ft. <br />E. ROCK VOLUME <br />1. Multiply ruck area by rock depth to get cubic feet of rock; <br />(QY4 sq. ft. x 11 ft. = -" cu. ft. <br />2. Divide cu. ft. by 27 cu. ft./cu. yd. to get cubic yards; <br />2L!�' cu. ft. + 27 = �,Z cu. yd. <br />3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br />�') cu. yd. x 1.4 ton/cu. yd. = _-zz tons. <br />F. ADSORPTION WIDTH C-,.,#" >rori-w1 <br />1. Percolation rate in top 12 inches of soil is ice, lnpi <br />2 Select allowable soil loading rate from table; <br />. tiS gpd /ft2 <br />3. Calculate adsorption width ratio by dividing rock layer <br />loading rate of 1.20 gpd /ft2 by allowable soil loading rate; <br />1.20 gpd/ft2+ 1-i�, gpd/W = .1-L-1 <br />4. Multiply adsorption width ratio by rock layer %vidth to get <br />required adsorption width; /� <br />7-L'1 x �_ ft =fitr / ft <br />wtaih <br />1a►ia <br />rerteiadonRaw <br />.4. p.. roti <br />IIeArioma <br />�. <br />(a.r <br />�... <br />tam <br />(akpu <br />v <br />2 ar ha <br />730 <br />1.123 <br />3W4 <br />Low <br />1300 <br />3 or 6 <br />1300 <br />1.00 <br />7 or 92.000 <br />Fine Sand" <br />3.OW <br />o.w 1 <br />Sat f4 C4 <br />(a 13) <br />0.79 <br />1.52 <br />16 to 30 <br />Lorm <br />0.60 <br />2.00 <br />31 to 43 <br />673 <br />Ia <br />46 to 60 <br />D. ROCK LAYER DIMENSIONS <br />1. Multiply flow rate by 0.83 to obtain required area of rock <br />layer: A x 0.83 = <br />sO gpd x 0.83 sq. ft. / gpd = x x sq. f t.+ )o% <br />2. Select width of rock layer (10 feet or less) _ I ft. <br />3. Length of rock layer = area + width = Rock Bed <br />sq. h. + 10 ft. = _(pisr_ ft. <br />tdtF. 510 ft. <br />E. ROCK VOLUME <br />1. Multiply ruck area by rock depth to get cubic feet of rock; <br />(QY4 sq. ft. x 11 ft. = -" cu. ft. <br />2. Divide cu. ft. by 27 cu. ft./cu. yd. to get cubic yards; <br />2L!�' cu. ft. + 27 = �,Z cu. yd. <br />3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br />�') cu. yd. x 1.4 ton/cu. yd. = _-zz tons. <br />F. ADSORPTION WIDTH C-,.,#" >rori-w1 <br />1. Percolation rate in top 12 inches of soil is ice, lnpi <br />2 Select allowable soil loading rate from table; <br />. tiS gpd /ft2 <br />3. Calculate adsorption width ratio by dividing rock layer <br />loading rate of 1.20 gpd /ft2 by allowable soil loading rate; <br />1.20 gpd/ft2+ 1-i�, gpd/W = .1-L-1 <br />4. Multiply adsorption width ratio by rock layer %vidth to get <br />required adsorption width; /� <br />7-L'1 x �_ ft =fitr / ft <br />wtaih <br />1a►ia <br />rerteiadonRaw <br />.4. p.. roti <br />IIeArioma <br />�. <br />(a.r <br />�... <br />tam <br />(akpu <br />v <br />2 ar ha <br />730 <br />1.123 <br />3W4 <br />Low <br />1300 <br />3 or 6 <br />1300 <br />1.00 <br />7 or 92.000 <br />Fine Sand" <br />3.OW <br />o.w 1 <br />Sat f4 C4 <br />(a 13) <br />D. ROCK LAYER DIMENSIONS <br />1. Multiply flow rate by 0.83 to obtain required area of rock <br />layer: A x 0.83 = <br />sO gpd x 0.83 sq. ft. / gpd = x x sq. f t.+ )o% <br />2. Select width of rock layer (10 feet or less) _ I ft. <br />3. Length of rock layer = area + width = Rock Bed <br />sq. h. + 10 ft. = _(pisr_ ft. <br />tdtF. 510 ft. <br />E. ROCK VOLUME <br />1. Multiply ruck area by rock depth to get cubic feet of rock; <br />(QY4 sq. ft. x 11 ft. = -" cu. ft. <br />2. Divide cu. ft. by 27 cu. ft./cu. yd. to get cubic yards; <br />2L!�' cu. ft. + 27 = �,Z cu. yd. <br />3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br />�') cu. yd. x 1.4 ton/cu. yd. = _-zz tons. <br />F. ADSORPTION WIDTH C-,.,#" >rori-w1 <br />1. Percolation rate in top 12 inches of soil is ice, lnpi <br />2 Select allowable soil loading rate from table; <br />. tiS gpd /ft2 <br />3. Calculate adsorption width ratio by dividing rock layer <br />loading rate of 1.20 gpd /ft2 by allowable soil loading rate; <br />1.20 gpd/ft2+ 1-i�, gpd/W = .1-L-1 <br />4. Multiply adsorption width ratio by rock layer %vidth to get <br />required adsorption width; /� <br />7-L'1 x �_ ft =fitr / ft <br />wtaih <br />1a►ia <br />rerteiadonRaw <br />.4. p.. roti <br />Soil Texture <br />ca.. <br />P <br />law <br />-m: <br />(akpu <br />v <br />Faster than 0.1 <br />Coarw Sand <br />120 <br />lbo <br />0.1 to 3 <br />Sand <br />120 <br />1.00 <br />0.1 to 5 <br />Fine Sand" <br />0A0 <br />2.00 <br />6 to 15 <br />Sandy Lora► <br />0.79 <br />1.52 <br />16 to 30 <br />Lorm <br />0.60 <br />2.00 <br />31 to 43 <br />silt Loam <br />050 <br />2-40 <br />46 to 60 <br />Cla� Loam <br />0.45 <br />2-67 <br />61 t0120 <br />Clay <br />024 <br />3.00 <br />Slower than 120 1 <br />Clav I <br />-S <br />- <br />" W ha." 304 w ware of &W w .oy My -,t <br />C <br />