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" s�.:�� Job#� <br /> T�wrMswrr <br /> P�eoa�ei►�r <br /> Univers' of Minnesota Mound Design Worksheet <br /> Greater than 19�Slopes <br /> a F�ow <br /> Es�,ar�a � soo �d(see r��►�a-�� <br /> or measured x 1.5(s�ety factor)_ �0 9Pd <br /> B. SEPTIC TANK LIQUID VOLUMES <br /> Septic tank capacity 20p0 gallons(see figure G1) <br /> Number oi tanks/compartments 0 <br /> Eifluent Filter (yes/no) yes <br /> C-1 Septic Tank Capacity in Galbns <br /> Number of Minimum Capaaty with Capacity with <br /> Bedrooms Capacity Garb.Disp. Disp.and Lift <br /> 2 a less 750 1125 1 <br /> 3 or 4 1000 1500 2000 <br /> 5 or 6 1500 7150 3000 <br /> 7,8 or 9 2000 3000 4000 <br /> �, SOILS(Site evaluation dafaJ <br /> 1. Depth to restriding layer- 2.0 feet <br /> 2. Depth of pe�oolation tests= 12 inches <br /> 3. Textur� loam <br /> 4, Soil loading rate(see F'rgure D-33) 0.60 gpd/ft2 <br /> Peroolaation rate 8 MPt <br /> 5. %Land Slope 11.0 °� <br /> D. ROCK IAYER DIMENSIONS <br /> 1. Multiply average design flow{A)by 0.83 to obtain requir�ed area of rock layer:Item A x 0.83= <br /> 600 gpd x 0.83 ft�/gpd= 500 ft <br /> 2. Determine rocic layer width =0.83 ft`/gpd x Linear Loading Rate(LLR)(see LLR chart <br /> 0.83 ft Igpd x 12.00 = 10.0 ft <br />�`�� �ut cnart <br />. Pe�1c R�e LLR <br /> <120 MPI <=12 <br /> >=120 MPI <=6 <br /> 3. Length of ra:k layer=area divided by width= <br /> 500.0 ft/ 10.0 feet= 50.0 ft <br /> E. ROCK VOLUME <br /> 1. Mutdply rodc area by rock depth to get cubic feet of rock <br /> 500.0 X 1.0 ft= 500.0 ft3 <br /> 2. Divide ft tiy 27 ft tyd3 to get cubic yards <br /> 500.0 ft3 / 27 = 18.5 yd3 <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> 18.5 yd3 X 1.4 ton/yd3 = 25.9 tons <br /> Page 1 of 5 <br />