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Site B Job#� <br /> s�wwoe <br /> TwtwtMs�rT <br /> P�aoo�ewr <br /> University of Minnesota Mound Design Worksheet <br /> Greater than 1%Slopes <br /> aFLOWEstimated 750 gpd(see figure A-1) <br /> i i <br /> I or measured x 1.5(safety factor)= 0 gpd i <br /> B. SEPTIC TANK LIQUID VOLUMES <br /> Sept tank 3000 gallons(see figure F1) <br /> Number d tanksloomparlments 0 <br /> Effluent f=ilter (yeshw) yes <br /> C-1 SWft To*CqxicKy in Gallons <br /> Number of Minimum Capacity withI=Disp <br /> with <br /> Bedrooms Capacity Garb.Disp. lift <br /> pr <br /> 1125 <br /> 3 or 4 1500 <br /> 5 or 6 2250 <br /> 7,8 or 9 3000 <br /> C. SONS(Site eva/uedon data) <br /> 1. Depth to ms*ft layer= 1.5 feet <br /> 2. Depth d percolation tests= 12 inches <br /> 3. Texdne ban <br /> 4. Soft loading rate(see Figure D-33) 0.60 gpd/ft' <br /> Perooletion rate 14 MPI <br /> 5. %Land Slope 6.0 % <br /> D. ROCK LAYER DI1111115NSIONS <br /> 1. Multiply average design flow(A)by 0.83 to obtain req*W area of rock layer.Item A x 0.83-- <br /> 750 <br /> .83=750 gpd x 0.83 ftp/gpd= 630 ftz <br /> 2. Determine rock layer width =0.83 ft`/gpd x Unear Loading Rate(LLR)(see LLR chart <br /> 0.83 fe/gpd x 1 12.00 j = 10.0 ft <br /> LLR Chart <br /> Perk Rate LLR <br /> <120 MPI <=12 <br /> >=120 MPI <=6 <br /> 3. Length of nx*layer=area divided by width= <br /> 630.0 fe / 10.0 feet= 63.0 ft <br /> E. ROCK VOLUME <br /> 1. Multiply rock area by rock depth to get cubic feet of rods <br /> 630.0 X 1.0 ft= <br /> 630.0 ft3 <br /> 2. Divide ft by 27 ft3V to get cubic yards <br /> 630.0 ft3 / 27 = 23.3 yd' <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> 23.3 yd3 X 1.4 ton/yd 3 = 32.7 tons <br /> h <br /> Page 1 of 5 �= <br />