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� Mound Design Worksheet (For flows up to 1200 gpd) <br /> Ail boxed rectangles must be enfered, the rest will be calculated. <br /> A. FLOW <br /> Estimated 750 gpd(see figure A-f) FUTURE SITE SOUTH END <br /> or measured x 1.5(safety factor)= 0 gpd <br /> B. SEPTIC TANK LIQUID VOLUMES <br /> Septic tank capacity 2250 gallons(see fgure G1) <br /> ep c an apaci in a ons <br /> Number of Minimum Capacity with Capacity with <br /> Bedrooms Capaaty Garb. Disp. Disp. and Lift <br /> 2 orless 750 1125 1500 <br /> 3 or 4 1000 1500 2000 <br /> 5 or 6 1500 2250 3000 <br /> 7,8 or 9 2000 3000 4000 <br /> C. SOILS(Sfte evaluation data) . <br /> 1. Depth to restricting layer- 2.1 feet <br /> 2. Depth of percolation tests= 12 inches <br /> 3. Texture LOAM <br /> 4. Soil loading rate(see Figure D-33) 0.6 gpd/ft <br /> Percolation rate 3 MPI <br /> 5. %Land Slope 11 % <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Multiply average design flow(A)by 0.83 to obtain required area of rock layer: Item A x 0.83= <br /> 750 gpd x 0.83 ft gpd= 622.5 ft <br /> 2. Determine rock layer width =0.83 ft/gpd x Linear Loading Rate(LLR)(see LLR chart) <br /> 0.83 ft/gpd x 12 = 10.0 ft <br /> LLR Chart <br /> Perk Rate LLR <br /> <120 MPI <=12 <br /> >=120 MPI <=6 <br /> 3. Length of rock layer=area divided by width= <br /> 622.5 ft I 10 feet= 62.0 feet <br /> E. ROCK VOLUME <br /> 1. Multiply rock area by rock depth to get cubic feet of rock <br /> 622.5 X 1 ft= 622.5 ft3 <br /> 2, Divide ft3 by 27 ft31yd3 to get cubic yards <br /> 622.5 ft3 I 27 = 23.1 yd3 <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> 23.1 yd3 X 1.4 tonlyd3 = 32.3 tons <br /> F. ABSORPTION WIDTH <br /> 1. Abso tion width equals absorption ratio(see Figure D-33)times rock layer width <br /> 2 x 10.0 ft = 20.0 ft <br /> Page 1 of 6 <br />