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. ✓ <br /> ' j--lf�i.3 f:l.=, <br /> • Ow�irr� <br /> • s�.�,...o� ' Job#� <br /> TRlATMlN'T <br /> P�coO�cwM c�^ <br /> University of Minnesota Mound Design Worksheet <br /> Greaterthan 1°/a Slopes � <br /> A FLOW <br /> Estimated 900 gpd(see 6gure A-1 J <br /> or measured x 1.5(safety factor)= 0 gpd <br /> B. SEPTIC TANK LIQUID VOLUMES <br /> SepGc tank capacity 3000 gallons(see figure G1J <br /> Number of tanks/compartments <br /> Effluent Fifler (yes/no) yes <br /> C-1 Septic Tank Capacity in Gallons <br /> Number of Minimum Capacity with Capacity with <br /> Bedrooms Capacity Gar�.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- 1.5 feet <br /> 2. Depth of percolation tests= 12 inches <br /> 3. Texture loam <br /> 4. Soil loading rate(see Figu�D-33) 0.60 gpd/ftz <br /> Percolation rate 13 MPI <br /> 5. %Land Siope 7.0 % <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 /> 900 gpd x 0.83 ftZ/gpd= 750 ftZ <br /> 2. Determine rock layer width =0.83 ft`Igpd x Unear 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 /> 750.0 ft2 / 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 ft3lyd3 to 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.g yd3 X 1.4 ton/yd3 = 38.9 tons <br /> Page 1 of 5 <br />