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- s�:.�� Site B bt 5 biock 1 Job#C� <br /> TREATMEIVT <br /> P�aoo�c�aus � �����^' <br /> University of Minnesota Mound Design Worksheet <br /> Greater than 1°/a Slopes <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 3p00 gallons(see figure G1) <br /> Number of tanks/compartments 0 <br /> Effluent Filter (yeslno) yes <br /> C-1 Septic Tank Capacity in Gallons <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- 1.3 feet <br /> 2. Depth of percolation tests= 12 inches <br /> 3. Te�ure loam <br /> 4. Soil loading rate(see Figure D-33) 0.60 gpd�ft <br /> Percolation rate 9 MPI <br /> 5. °10 Land Slope 9.0 % <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 ft/gpd= 630 ft <br /> 2. Determine rock layer width =0.83 ft`/gpd x Linear Loading Rate(LLR)(see lLR chart <br /> 0.83 ft/gpd x 12.00 = 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 /> 630.0 ft! 10.0 feet= 63.0 ft <br /> E. ROCK VOLUME <br /> 1. Multiply rock area by rock depth to get cubic feet of rock <br /> 630.0 X 1.0 ft= 630.0 ft3 <br /> 2. Divide ft3 by 27 ft3/yd3to get cubic yards <br /> 630.0 ft3 ! 27 = 23.3 yd3 <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> 23.3 yd3 X 1.4 ton/yd3 = 32.7 tons <br /> Page 1 of 5 <br />