Laserfiche WebLink
� Mound Design Worksheet (For flows up to 1200 gpd) <br /> All boxed�ctangles must be entered, the rest will be calculated. <br /> A. FLOW <br /> Estimated 750 gpd(see figur�A-1) <br /> or measured x 1.5(safety factor)= 0 gpd <br /> . B. SEPTIC TANK LIQUID VOLUMES <br /> Septic tank capacity 2250 gallons(see figu�t G1) <br /> ep ic an apaci in a ons <br /> Number of Minimum Capacity with Capacity with <br /> Bedrooms Capaaty 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 evaluation data) <br /> 1. Depth to restricting layer= 1.6 feet <br /> 2. Depth of percolation tests= 12 inches <br /> 3. Texture loam <br /> 4. Soil loading rate(see Figure D-33) 0.6 gpd/ft <br /> Percolation rate 15 MPI <br /> 5. °k Land Slope 9 % <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Multiply average design flow(Aj by 0.83 to obtain required area of rodc layer: Item A x 0.83= <br /> 750 gpd x 0.83 ft gpd= 622.5 ft <br /> 2. Determine rodc layer width =0.83 ft/gpd x Linear Loadin Rate LLR)(see LLR chart) <br /> 0.83 ft2lgpd 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 ft I 10 feet= 62.5 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 I 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. Abso tion width equats 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 />