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ONarrs <br /> S�w�ao� ��__ Slte B lOt 1 bIOCk 2 Job#� <br /> T/tE4TMENT "` _---_- <br /> Pw06RAM �'�-�`'- <br /> University of Minnesota Mound Design Worksheet <br /> Greaterthan 1�Slopes <br /> A FLOW <br /> Estimated 750 gpd(see figure A-1) <br /> or measured x 1.5(safety factor)= 0 gpd <br /> 8. SEPTIC TANK LIQUID VOLUMES <br /> Septic tank capacity 3ppp gallons(see figure G1) <br /> Number of tanks/compartments p <br /> Effluent Filter (yeslno} y� <br /> C-1 Septk Tank Capacity in Galbns <br /> Number of Minimum Capacity with Capacity wiUi <br /> Bedroans Capacity Garb.Disp. Disp.and Lift <br /> 2 orless 750 1125 1500 <br /> 3 or 4 10(}0 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 resVicting layer- 1.0 feet <br /> 2. Depth of percolation tests= 12 inches <br /> 3. Texture loam <br /> 4. Soil loading rate(see Figure D-33J 0.60 gpd/ft2 <br /> Percolation rate 13 MPI <br /> 5. %Land Slope 6.0 % <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Multiply average design flow(A)by 0.83 to obtain required a�ea of rock layer:Item A x 0.83= <br /> 750 gpd x 0.83 ft�/gpd= 630 ft2 <br /> 2. Determine rocfc layer width =0.83 ft`lgpd x Linear Loading Rate(L�R)(see LLR chart <br /> 0.83 ft2lgpd x 12.00 = 10.0 ft <br /> LLR Chart <br /> Pertc 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 rodc depth to get cubic feet of rock - <br /> 630.0 X 1.0 ft= 630.0 ft3 <br /> 2. Divide ft3 by 27 fl31yd3 to get cubic yards <br /> 630.0 ft3 / 27 = 23.3 yd3 <br /> 3. Mul6ply cubic yards by 1.4 to get weight of rock in tons; <br /> 23.3 yd3 X 1.4ton/yd3 = 32.7 tons <br /> Page 1 of 5 <br />