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Q�sr+e <br /> ���� �� Job# Vnce <br /> -r�,►,��,� <br /> v��.r...,. �. <br /> Universit of Minnesota Mound Design Worksheet <br /> Greater than 1%Siopes <br /> A FLOW <br /> Estimated � � lodrHc-� �y�p�j�) 300 g�d(see BgureA-1) <br /> or me�ured x 1.5(safety factor)= 0 �d <br /> B. <br /> Septic tank c�acity :24t10 gailqt�see figure G1) <br /> Number of tankslcomp�trnents 2 <br /> Effluent Fiiter (yesJno) no <br /> C-1 Septic Tank Capacity in Gallons <br /> Number of Minimum C�acity with Capacity with <br /> Bedrooms Capacity Garb.Disp. Disp.and Lift <br /> 2 or less 750 ' 1125 15{30 <br /> 3 or 4 1� - 1500 2000 <br /> 5 or 6 tSU� 2250 3000 ` <br /> 7.8 or 9 20(10 3000 40fl4 <br /> C. SOILS(Site evaluation datal . <br /> 1. , _ 1:2 feet;� <br /> 2. Depth of percolafion fests= 12 inches <br /> 3. Texture loarn <br /> 4. Soii loading rate(see Figure D-33) 0.60 �dl ft� <br /> Percola4on rate 3 MPI <br /> 5. %Land Slope 23.0 % <br /> ,_ :_,,.._. . . <br /> D. <br /> 1. Multipty average design flow(A)by 0.83 to obtain required area of rock layer:Item A x 0.83= <br /> 300 �d x 0.83 ft2tgpd= 250 ft <br /> 2. Determine rock lay�width =0.83 ft`J�d x Linear loading Rate(LLR)(see LLR chart) <br /> 0.83 ft�l�d x 12.00 = ft <br /> LLR Chart <br /> Perk Rate LLR <br /> <120 MPI <=t2 <br /> >=120 MPI <=6 <br /> 3. Length of rock layer=�ea divided by width= <br /> 250.0 ft� / 10.0 feet= �ft <br /> E. <br /> 1. Multiply rock area by rock depth to get cubic feet of rock <br /> 250.0 X 1.0 ft= 250.0 fY� <br /> 2. Divide ft3 by 27 ft3tyd3 to get cubic yards <br /> 250.0 ft3 t 27 = 9.3 yd3 <br /> 3. Mul6ply cubic yards by 1.4 to get vueight of rcek in tons; <br /> 9.3 yd3 X 1.4 toNyd3 = ons <br /> Page 1 of 6 <br />