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Qtist-<ie �R tY _.J�ti�: � <br /> s�r„A�� Job#� <br /> TRCATM�.NT ��-� <br /> PR06�i��1/ ti"���,` <br /> University of Minnesata Mound Design Worksheet <br /> Greaterthan 1%Slopes <br /> A FLOW <br /> Estimated 750 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 2-1000 existing gallons(see figure C-1) <br /> Number of tunkslcompartments <br /> Effluent Filter (yes/no) �a <br /> C-1 Septic Tank Capacity ia Gallons <br /> Number of Minimum Capacity with Capacity with <br /> Betlrooms Capacity Garb.Disp. Disp.and Lift <br /> 2 or less 750 �`:�.`,; 1125 � 15U0 <br /> 3 or4 1000 1500 2000 <br /> 5 or 6 15G0 . 2250 30G0 <br /> 7,8 or 9 20pQ ,, 3000 4400 . ` <br /> �. SOILS(Sita evaluation data) <br /> 1. Depth to restricting layer= 1.5 feet <br /> 2. Depth of percolation tests= 12 inches <br /> 3. Texture clay loam-Loam <br /> 4, Soil loading rate(see Figure D-33) 0.45 9Pd�ft2 <br /> Percolation rate 31�5 MPI <br /> ' 5. %Land Slope 4.0 % <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 ft2/gpd= 630 ftz <br /> 2. Determine roc;k layer wldih =0.83 ft`Igpd x Linear Loading Rate(LLR)(see LLR chart <br /> 0.83 ftZ/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 ft2 / 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 h'by 27 ft31yd3to get cubic yards <br /> 630.0 ft3 I 27 = 23.3 yd3 <br /> 3. MulUply 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 of 6 <br />