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's�,';a� Site A Job#�� <br /> TRtATMlNT <br /> PROORAM <br /> Univers' of Minnesota Mound Design Worksheet <br /> Greater than 1X Slopes <br /> A. FLOW <br /> � �t� 600 9Pd(see figu�e A-i) <br /> ; a measured i x 1.5(safety factor)= 0 9Pd � <br /> B. SEP'TIC TANK LIQUID YOLUMES <br /> ���k��y 3pp0 galbns(see figure G1) <br /> Number of tankslcompartments � <br /> Effluer�F�7ter (yes/no) Yes <br /> C-1 Septic Tank Capacity in Gallons <br /> Number of Minimum Capaaty with Capacity with <br /> Bedrooms Capaaty Garb.Disp. Disp.and lift <br /> 2 or less 1125 <br /> 3 or 4 �.10Q� 1500 ?000 : <br /> 5 or 6 `1600 2250 3000 <br /> 7,8 or 9 �2000 3000 4000 `. <br /> C. SOIi.S(Site evatua5�n data) <br /> 1. Depth to restriat�nny hayer- 2.0 feet <br /> 2. Depth of peroolation tests= 12 inchas <br /> 3. Texture �►n <br /> 4. Soil loading rate(see Figure D-33) 0.60 9P�� <br /> Percola6on rate 5 MPI <br /> 5. 96 Land Sbpe 4.0 % <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Muttipiy average design fbw(A)by 0.83 to obtain required area of rodc layer.item A x 0.83= <br /> 600 gpd x 0.83 ft�/gpd= 500 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 ChaR <br /> Perk Rate LLR <br /> <�2Q MPI <=12 <br /> >=120 MPI <=6 <br /> 3. Length of rodc layer=area divided by width= <br /> 500.0 ft� ! 10.0 feet= 50.0 ft <br /> E. ROCK VOLUME <br /> 1. Multiply rodc area by rock depth to get cubic feet of rodc <br /> 500.0 X 1.0 ft= 500.0 ft' <br /> 2. Divide ft'by 27 ft�/yd'to get cubic yards <br /> 500.0 ft3 I 27 = 18.5 yd' <br /> 3. Multiply cubic yarYiss by 1.4 to get weight of rock in tons; <br /> 18.5 yd3 X 1.4 tonlyd' = 25.9 tons <br /> Page 1 of 5 <br />