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Mound Design Worksheet (For flows up to 1200 gpd) <br /> AU boxed rectangles must be entered,the rest will be calculafed. ���ti►AR y Si T� <br /> A. FLOW <br /> Estimated 6(?0 gpd(see figure A-1) <br /> or measured x 1.5(safety factor)= 0 gpd <br /> B. SEPTIC TANK LIQUID VOLUMES <br /> Septic tank capacity 2000 gallons(see figure G1) <br /> ep ic an apaci m a ons <br /> Number of Minimum Capacity with Capacity with <br /> Bedrooms Capacity Garb.Disp. Disp.and Lift <br /> 2 or less 750 1125 1500 <br /> 3 or4 1000 1500 2000 <br /> 5 ar 6 1500 2250 3000 <br /> 7,8 or 9 2000 3000 4000 <br /> C. SOtLS(Site evaluation data) <br /> 1. Depth to restricting layer- 2.1 feet <br /> 2. Depth of percolation tests= 12 inches <br /> 3. Texture. sand loam <br /> 4. Soil loading rate(see Figure D-33 0.79 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 /> 600 gpd x 0.83 ft'�gpd= 498.0 ft <br /> 2. Determine rock layer width =0.83 ft�/gpd x Linear Loadin Rate LLR)(see LLR chart) <br /> 0.83 fllgpd x 12 = 10.0 ft <br /> LLR Chart <br /> Perk Rate LLR <br /> <120 MPI <=12 <br /> >=120 MPI <=6 <br /> 3. Length of rodc layer=area divided by width= <br /> 498 ft� / 10 feet= 50.0 feet <br /> E. ROCK VOLUME <br /> 1. Multiply rodc area by rock depth to get cubic feet of rock <br /> 498 X 1 ft= 498.0 ft3 <br /> 2. Divide ft3 by 27 ft3/yd3 to get cubic yards <br /> 498.0 ft3 / 27 = 18.4 yd3 <br /> 3. Mulfiply cubic yards by 1.4 to get weight of rodc in tons; <br /> 18.4 yd3 X 1.4 ton/yd3 = 25.8 tons <br /> F. ABSORPTION WIDTH <br /> 1. Absor tion width uals absorption rabo(see Figure D-33)times rodc layer width <br /> 1.5 x 10.0 ft = 15.0 ft <br /> Page 1 of 6 <br />