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Mound Design Worksheet (For flows up to 1200 gpd) <br /> All boxed rectangles must be entered,the rest will be calculated. <br /> A. FLOW <br /> Estimated 750 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 2250 Dgallons(see figure C-1) <br /> c Tank Capacityn Uallons <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 or 4 1000 1500 2000 <br /> 5 or 6 1500 2250 3000 <br /> 7,8 or 9 2000 3000 4000 <br /> C. SOILS(Site evaluation data) <br /> 1. Depth to restricting layer- 1.1 feet <br /> 2. Depth of percolation tests= 12 linches <br /> 3. Texture cia loam <br /> 4. Soil loading rate(see Figure D-33) 0.45 gpd/fe <br /> Percolation rate I MPI <br /> 5. %Land Slope 1 % <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.&3-- <br /> 750 <br /> .83=750 gpd x 0.83 egpd= 622.5 fl? <br /> 2. Determine rock layer width =0.83 If/gpd x Linear Loading Rate(LLR)(see LLR chart) <br /> 0.83 fe/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 fe / 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 ft3/yd3 to get cubic yards <br /> 622.5 ft3 / 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 ton/yd3 = 32.3 tons <br /> F. ABSORPTION WIDTH <br /> 1. Absorption width equals absorption ratio(see Figure D-33)times rock layer width <br /> 2.7 x 10.0 ft = 27.0 ft <br /> Page 1 of 4 <br />