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� Mound Design Worksheet (For flows up to 1200 gpd) <br /> All boxed►ectangles must be entered, the rest will be calculated. <br /> A. FLOW <br /> Estimated 900 gpd(see figure A-1) <br /> or measured x 1.5(safety factor)= 0 gpd <br /> B. SEPTIC TANK LIQUID VOLUMES <br /> Septic tank capaaty 2250 gallons(see figure 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 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= 2 feet <br /> 2. Depth of percolation tests= 12 inches <br /> 3. Texture loam <br /> 4. Soil lo�iing rate(see Figu�D-33) 0.6 gpol ft� <br /> Pe�colation rate 6 MPI <br /> 5. °�L�d Slope 9 °� <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Multiply average design flow(A)by 0.83 to obtain required area of rodc layer. Item A x 0.83= <br /> 900 gpd x 0.83 ft�gpd= 747.0 ft� <br /> 2. Determine rodc layer width =0.83 ft�/gpd x Linear Loadin 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 /> 747 ft/ 10 feet= 75.0 feet <br /> E. ROCK VOLUME <br /> 1. Mulfiply rodc area by rock depth to get cubic feet of rock <br /> 747 X 1 ft= 747.0 ft3 <br /> 2. Divide ft3 by 27 ft31yd3to get cubic yards <br /> 747.0 ft3 / 27 = 27.7 yd3 <br /> 3. Multipty a.ibic yards by 1.4 to get weight of rock in tons; <br /> 27.7 yd3 X 1.4 tonlyd3 = 38.7 tons <br /> F. ABSORPTION WIDTH <br /> 1. Abso�lption 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 />