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�oN:�n Site e Job#� <br /> s�.,...oa <br /> Trc��►�Er�rr <br /> Pnoo�ewN ' <br /> Universi of Minnesota Mound Design Worksheet <br /> Greater th�19G Slopes <br /> A. FLOW <br /> Estirr►ated 750 9P�(�fi9u�e A-1) <br /> or measured� �c 1.5(safety factor)_ � 9Pd <br /> 8. SEPTIC TANK LIQUIO VOLUMES <br /> Sep6c tank capac�ty 3000 gallons(see fgure G1) <br /> Number of tanks/compaM�ents � <br /> EfflueM Fitter (yes/no) Ye3 <br /> G1 Septic Tank Capacity in Gailons <br /> Number of Minimum Capaaty with Capaaty with <br /> Bedrooms Capacity Garb.Disp. Disp.and Lift <br /> 2 or less 1125 1 <br /> 3 or 4 1000 1500 2000 <br /> 5 or 6 15U0 2250 3WQ <br /> 7,8 or 9 2000 � '� <br /> C. SOILS(Sihe eveluetion data) <br /> 1. Depth to restricting layer- 2.0 feet <br /> 2. p��{�a{�t�= 12 inches <br /> 3. Texture loam <br /> 4. So�bading rafe(see Figure D-33) 0.60 9P�� <br /> p��e 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 ftlgpd= 63� ft <br /> 2. Detennine rodc layer width =0.83 ft`/gpd x Linear Loading Rate(LLR)(see LLR chart <br /> 0.83 ft�/gpd x 12.00 = 10.0 ft <br /> LLR Chart <br /> Per1c Rate LLR <br /> <120 MPl <=12 <br /> >=120 MPI <=6 <br /> 3. Length of rodc layer=a�ea divided by width= <br /> 630.0 ft 1 10.0 feet= 63.0 ft <br /> E. ROCK VOIUME <br /> 1. MulUply rodc area by rock depth to get cubic feet of rock <br /> fi30A X 1.0 ft= 630.0 ft' <br /> 2. Divide ft'by 27 ft'tyd'to get cubic yards <br /> 630.0 ft' 1 27 = 23.3 yd' <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> 23.3 yd' X 1.4 ton/yd' = 32.7 tons <br /> Page 1 of 5 <br />