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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-i) <br /> or measured x 1.5(safety factor)= 0 gpd <br /> B. SEPTIC TANK LIQUID VOLUMES <br /> Septic tank capacity 3000 gallons(see figure C-1) <br /> ep c an apaci in a ons <br /> Number of Minimum Capacity with Capacity with <br /> Bedrooms Capacity 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(Site evaluafion data) <br /> 1. Dept�to restricting layer 2 feet <br /> 2, Depth of percolation tests= 12 inches <br /> 3. Textu�e loam <br /> 4. Soil loading rate(see Figure D-33) 0.6 gpd/ft <br /> Percola6on rate 13 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 0.83= <br /> 750 gpd x 0.83 ft2Jgpd= 622.5 ft <br /> 2. Determine rock layer width =0.83 ft/gpd x Linear Loadin Rate(LLR)(see LLR chart) <br /> 0.83 ft Igpd 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 ft2 1 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 ft3lyd3 to get cubic yards <br /> 622.5 ft3 1 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. Absorption 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 />