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� oNs�� �_ Job#� <br /> s�w.w� Future Site <br /> Tawrw��*rr <br /> P�eoo�ewM <br /> Universi of Minnesota Mound Design Worksheet <br /> Greater than 1%Slopes <br /> A FLOW <br /> Estimated 900 9Pd(see'6gure A-1) <br /> or measured � x 1.5(�afety fador)_ � 9Pd <br /> B. SEPTIC TANK LIQUID VOLUMES <br /> ��k���, 2250 gallons(see(�gure G1) <br /> Number of tanks/compartments � � <br /> Effluent Ffter (yes/no) Y� <br /> G1 Septic Tank Capacity in Gallons <br /> Number of Minimum Capacity with Capacity with <br /> Bedrooms Capaaty Garb.Disp. Disp.and L'rft <br /> 2 or less 75p� >. 1125 �1500 1� :> <br /> 3or4 �� :� K 1500 �'��`-,204p ��"'�� <br /> s x ;. .€�. �. �. <br /> 758069 �:����%; � �,� '�Y'r. <br /> �' p; <br /> �. SOILS(Site evaluation data) <br /> 1. Depth to restriding layer= 2.3 feet <br /> 2. Depth of percolation tests= 12 inches <br /> 3. Texture loam <br /> 4. Soil bading rate(see Figure D-33) 0.60 9P�� <br /> Percolation rate 3 MPI <br /> 5. 9b Land Slope 3.0 % <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 /> 900 gpd x 0.83 ftZ/gpd= 750 ft2 <br /> 2. Determine rock layer width =0.83 ft`/gpd x Linear Loading Rate(LLR)(see LLR chaR <br /> 0.83 ftZ/gpd x 12.00 = 10.0 ft <br /> LLR Cha�t <br /> Perk Rate LLR <br /> <120 MPI <=12 <br /> >=120 MPI <=6 <br /> 3. Length of rock layer=area divided by width= <br /> 750.0 ftZ / 10.0 feet= 75.0 ft <br /> E. ROCK VOLUME <br /> 1. Multiply rock area by rock depth to get cubic feet of rock <br /> 750.0 X 1.0 ft= 750.0 ft3 <br /> 2. Divide ft3 by 27 ft3/yd3 to get cubic yards <br /> 750.0 ft3 / 27 = 27.8 yd3 <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> 27.g yd3 X 1.4 tonlyd' = 38.9 tons <br /> Page 1 of 5 <br />