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, ONS�TE <br /> s�r,,,,a� ' Job# <br /> Tw CqTMlKT <br /> Pwoa�can�n ����� <br /> University of Minnesota Mound Design Worksheet <br /> Greater than 1%Slopes <br /> A FLOW <br /> Estimated � 300 gp�(see figure A-1) <br /> or measured ' x 1.5(safety factor)= 0 9Pd <br /> B. SEPTIC TANK LIQUID VOLUMES <br /> ��k qp�y 2000 galbns(see figure C-1) <br /> Number ofi tanks/oompartrnents 0 <br /> Eifluent Filter (yeslno) Y� <br /> C-1 Septic Tank Capacity in Galbns <br /> Number oi Minimum Capacity with Capacity with <br /> Bedrooms Capacity Ga�b.Disp. Disp.and L)ft <br /> 2 orless 750 1125 1500 <br /> 3 or 4 1000 1500 2000 <br /> 5 or 6 1500 2250 300U <br /> 7,8 or 9 2000 3000 4000 <br /> C. SOILS(Site evaluation dafa) <br /> 1. Depth to restricting layer= 1.5 feet <br /> 2. Depth of percolation tests= 12 inches <br /> 3. Texture loam <br /> 4, Soil loading rate(see F'rgure D-33) 0.60 gpd�ft <br /> Percolation rate 17 MPI <br /> 5. %Land SIOpe g•p o/6 <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Multiply average design flow(A)by 0.83 to obt�n required area of rodc layer:Item A x 0.83= <br /> 300 gpd x 0.83 ft/gpd= 250 ft <br /> 2. Detertnine rodc layer width =0.83 ft`/gpd x Linear Loadirg Rate(LLR)(see LLR chart <br /> 0.83 fl�/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 rodc layer=area divided by width= <br /> 250.0 ft� / 10.0 feet= 25.0 ft <br /> E. ROCK VOLUME <br /> 1. Multiply rodc area by rock depth to get cubic feet of rock <br /> 250.0 X 1.0 ft= 250.0 ft3 <br /> 2. Divide ft3 by 27 ft�d3 to get cubic yards <br /> 250.0 ft3 / 27 = 9.3 yd3 <br /> 3. Multiply cubic y�ds by 1.4 to get weight of rodc in tons; <br /> g,3 yd3 X 1.4 ton/yd3 = 13.0 tons <br /> Page 1 of 5 <br />