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ONa�T� <br /> s�,,,�,�� Job#� <br /> T/t'lATMeNT <br /> P�aoa�ewM <br /> Universi of Minnesota Mound Design Worksheet <br /> Greater than 1%Slopes <br /> a F�ow <br /> Estimated 450 gqd(see figure A-1) <br /> or measured ! �C 1.5(safety factor)= 0 gpd <br /> B. SEPTIC TANK LIQUID VOLUMES <br /> Septic tank capaciry 2000 gallons(see 8gure G1) <br /> Number of tanks/compartments 0 <br /> Effluent Filter (yeslno) yes <br /> C-1 Septic Tank Capacit�r in Gallons <br /> Number of Minimum Capacity with Capacity with <br /> Bedrooms Capacity Ga�b.Disp. Disp.and Lift <br /> 2 or less � �f= � 1125 �'l � � <br /> � � <br /> 3 or 4 '�°'S;'����i00 r 1500 � �4��� <br /> 5or6 ;�'�t�'��OO,ti� 2250 , .0 kr�;,, <br /> 7,8 a 9 ,'%��� ' �F2� 3000 �, ,��:, <br /> �. SOILS(Site evaluation data) <br /> 1. Oepth 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 9P�� <br /> Percolation rate 3 MPI <br /> 5. %Land Siope 11.0 96 under rock <br /> overa s ope <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 /> 450 gpd x 0.83 ft/gpd= 380 ft <br /> 2. Determine rodc 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 rock layer=area divided by width= <br /> 380.0 ft2 / 10.0 feet= 38.0 ft <br /> E. ROCK VOI.UME <br /> 1. Multiply rock area by rock depth to get cubic feet of rock <br /> 380.0 X 1.0 ft= 380.0 ft3 <br /> 2. Divide ft3 by 2�ft3lyd3 to get cubic yards <br /> 380.0 ft3 I 27 = 14.1 yd3 <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> 14.1 yd3 X 1.4 tonlyd3 = 19.7 tons <br /> � Page 1 of 5 <br />