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� PR���,a,Z � <br /> , �YNitT! Job#� <br /> , Scwwae <br /> TRlATM C NT -r <br /> PROQRAM <br /> University of Minnesota Mound Design Worksheet <br /> Greater than 1%Slopes <br /> A FLOW <br /> Estimated 900 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 3000 gallons(see figure G1) <br /> Number of tanks/compartments � <br /> Effluent Filter (yes/no) Y� <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 4�0 <br /> �, SOILS(Site evaluation data) <br /> 1. Depth to restricting layer- 1.3 feet <br /> 2. Depth of percolation tests= �2 ���� <br /> 3. Textu�e clay loam <br /> 4. Soil loading rate(see Figure D-33) 0.45 9P�� <br /> Percolation rate 13 MPI <br /> 5. °k Land Slope 6.0 °� <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Muttiply average design flow(A)by 0.83 to obtain required area of rock layer:Item A x 0.83= <br /> 900 gpd x 0.83 ftZ/gpd= 750 ft2 <br /> 2. Determine rock layer width =0.83 ft`/gpd x Linear Loading Rate(LLR)(see LLR chart <br /> 0.83 ft2lgpd 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 /> 750.0 ftz / 10.0 feet= 75.0 ft <br /> E. ROCK VOLUME <br /> 1. Multiply rock area by rock depth to get cubic feet of rock <br /> 750.0 X 1.0 ft= 750.0 ft3 <br /> 2. Divide ft3 by 27 ft3/yd3to get cubic yards <br /> 750.0 ft3 I 27 = 27.8 yd3 <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> 27.8 yd3 X 1.4 tonlyd3 = 38.9 tons <br /> Page 1 of 5 <br />