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o.�.� <br /> ae�„�,�,se Job#� <br /> Tw:..T....+r � <br /> PreoMewr <br /> Univers' of Minnesota Mound Design Worksheet <br /> Greaterthan 1%Slopes <br /> A FLOW <br /> Estimated 750 9Pd(see flgure A-1) <br /> I a��� I x 1.5(safety factor)= 0 gpd i <br /> B. SEPTIC TANK LIQUID VOIUMES <br /> Se�ic t�k capaaty 2250 gailons(see figure G1) <br /> Number of t�ks/oompartrnents 0 <br /> Eflluent Filter (yes/no) yes <br /> C-1 Septic Tank Capscily�Galbns <br /> Number of Minimum Capacity with Capadty with <br /> Bedrooms Capacity Garb.Disp. Disp.�d Lift <br /> 2 or I�s 1125 <br /> 3 or 4 1500 <br /> 5 or 6 2250 ' <br /> 7,8 or 9 3000 <br /> C. SOILS(Site eveluation deta) <br /> 1. Depth to restriding layer- 1.3 feet <br /> 2. Depth of percolation tests= 12 inches <br /> 3. Texture bart► <br /> 4. Soil bading rate(see Figure D-33) 0.60 gpol ft <br /> Perodation rate 3 MPI <br /> 5. %L.and Sbpe 11.0 96 <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Mul�ply average design flow(A)by 0.83 to obtain required area of rodc layer:Item A x 0.83= <br /> 750 gpd x 0.83 ft�/gpd= 630 ft2 <br /> 2. Determine rock layer width =0.83 ft`/gpd 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 rodc layer=area divided by width= <br /> 630.0 ftZ / 10.0 feet= 63.0 ft <br /> E. ROCK VOLUME <br /> 1. Muftiply rock area by roc:k depth to get cubic feet of rock <br /> 630.0 X 1.0 ft= 630.0 ft' <br /> 2. Divide ft3 by 27 ft3/yd3 to get cubic yards <br /> 630.0 ft3 / 27 = 23.3 yd3 <br /> 3. MulGply 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 5 <br />