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Mound Desi n Worksheet (For flows up to 1200 gpd) <br /> All boxed rectangles ust be entered,the rest will be calculated. <br /> A. FLOW NI:EsIkmaled Sewage Ram in Gallate per Doi <br /> Estimated 450 gpd(see figure A-1) MIMI01 <br /> or measured 0 x 1.5(safety factor)= 0 gpd00Rs dos I doss II Gas III dos N <br /> 2 300 225 180 60% <br /> B. SEPTIC TANK LIQUID VOLUMES 3 450 300 218 of the <br /> Septic tank capacity ex.2-1000 gallons(see figure C-1) 4 600 375 256 vanes <br /> 5 750 450 294 in the <br /> C. SOILS(S' evaluation data) 6 900 525 332 Goss I. <br /> 1. Depth to r stricting layer= 1.5 feet 7 1050 600 370 I,or II <br /> 8 1200 675 408 coMxrr>s. <br /> 2. Depth of rcolation tests= inches <br /> 3. Texture \ Sandy Loam <br /> 4. Soil loadingrate(see Figure D-33) 0.79 gpd/ft2 <br /> Percolation rate 6 to 15 MPI <br /> 5. %Land Slope 1 8 % D-33: Absorption Wltkh Sizing Table <br /> Percolation Rate Loading Rate <br /> in <br /> M mn per Soil Texture wxe ot <br /> Rw on <br /> C-1: Septic Tan Capacities(In Mons) � >pper <br /> (NIP!) <br /> Lid <br /> Number of Mihimum Liquid Liquid capacity with with dispCapacity <br /> a& F terin.ns M�Sand 1.20 L 00 <br /> Bedrooms Capacity garbage disposal lift inside LoaFinSand <br /> 6 to 15 Sandy Loam 079 t 50 <br /> 2 or less 750 112516 to 30 Loam 0.60 2 00 <br /> 3 or 4 1000 1500 150031 to 45 Silt Loam o.so 2 40 <br /> 5 or 6 1500 2250 3Q ) <br /> 46 w 60 SSSauntys CClay e 0.45 2 67 <br /> 7,8 or 9 2000 3000 400 61 to 120nay 0.24 500 <br /> Sandy Clay <br /> Clay <br /> Slower than 120• <br /> "Symms desiaaad fort dm,soils must to ober or psfasaace <br /> D. ROCK LAVE DIMENSIONS <br /> 1. Multiply avera a design flow(A)by 0.83 to obtain required area of rock layer: Item A x 0.83= <br /> i 450 gpd x 0.83 ft21gpd= 380.0 ft2 <br /> 2. Determine roc layer width =0.83 ft2/gpd x Linear Loadin. Rate(LLR)(see LLR chart) <br /> 0.83 ft2/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 I yer=area divided by width= <br /> 380 ft2 I 10 feet= 38.0 feet <br /> E. ROCK VOLUME <br /> 1. Multiply rock are by rock depth to get cubic feet of rock <br /> 380 X 1 ft= 380.0 ft3 <br /> 2. Divide ft3 by 27 ft/yd3 to get cubic yards <br /> 380.0 ft3 / 27 = 14.1 yd3 <br /> I <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> 14.1 yd3 X 1.4 ton/yd3 = 19.7 tons <br />