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ONs�n <br /> Scw.►oc � future lower rock bed JOb# <br /> TRlATMlN'T <br /> P�coo�ci►iw _, <br /> University of Minnesota Mound Design Worksheet <br /> Greater than 1%Slopes <br /> A FLOW <br /> Estimated 750 gpd(see figure A-1) <br /> or measured ; x 1.5(safety factor)= i 0 gpd <br /> B. SEPTIC TANK LIQUID VOLUMES <br /> Septic tank capacity 2250 gallons(see figure C-1) <br /> Number of tankslcompartments � <br /> Effluent Fitter (yes/no) yes <br /> C•1 Sepdc 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 ;10Q0 1500 �'2000 <br /> 5 or 6 '1500 7250 '3000 <br /> 7,8 or 9 2000 ' 3000 4000 <br /> �, SOILS(Srte evaluation dafa) <br /> 1. Depth to restricting layer= 2.0 feet <br /> 2, Depth of percolation tests= 12 inches <br /> 3. Texture loam <br /> 4. Soil loading rate(see Figure D-33) 0.60 9Pd�� <br /> Percolation rate 10 MPI <br /> 5. °/a Land Slope 12.0 % <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 /> 750 gpd x 0.83 ft�lgpd= 630 ft2 <br /> 2, Determine rock layer width =0.83 ft`/gpd x Linear Loading Rate(LLR)(see LLR chart <br /> 0.83 ft2/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 /> 630.0 ftZ / 10.0 feet= 63.0 ft <br /> E. ROCK VOLUME <br /> 1. Multiply rock area by rock depth to get cubic feet of rock <br /> 630.0 X 1.0 ft= 630.0 ft3 <br /> 2, Divide ft3 by 27 ft3/yd3 to get cubic yards <br /> 630.0 ft3 I 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 tonlyd3 = 32.7 tons <br /> Page 1 cf 5 <br />