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septic info-including '99 design
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septic info-including '99 design
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Last modified
8/22/2023 4:19:37 PM
Creation date
1/17/2019 3:18:08 PM
Metadata
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Template:
x Address Old
House Number
2135
Street Name
6th
Street Type
Avenue
Street Direction
North
Address
2135 6th Avenue North
Document Type
Septic
PIN
2711823310005
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ProcessedPID
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• <br /> � 4 <br /> . �:, <br /> • � � <br /> E`� <br /> Rich Valiton <br /> 2135 County Rd 6 <br /> Long Lake MN. <br /> 55356 <br /> Hennpin County <br /> PID # 271182331 <br /> A. FLOW ' <br /> Estimated Gallons per Day: �OU <br /> B. SEPTIC TANK LIQUID VOLUMES <br /> ' Septic Tank Volume: 2000 gallons <br /> C. SOILS {Refer to Site Evaluation) <br /> C1. Depth to Restricting Layer: 30 inches <br /> C2.� Depth of Percolation Test : 12 inches <br /> C3. Texture: Loam Percolation Rate: 30 mpi <br /> C4. Land Slope: 3$ or Rise per 100 foot Run <br /> D. ROCK LAYER DIMENSIONS <br /> D1. Multiply flow by 0.83 to obtain required area of rock layer: <br /> Flow (A} gpd x 0.83 sq.ft/gal/day = <br /> 600 x 0.83 = 498 square feet+ lOg = 548 square feet. <br /> D2. Selected Rock Layer Width: 10 feet <br /> (max 10' if <120 mpi max 5' } <br /> D3. Length of Rock Layer = Area / Width <br /> 598 ft divided by 10 ft = 54.8ft <br /> E. ROCK VOLUME <br /> E1. Multiply rock area by rock depth to get cubic feet of rock <br /> D1 x 1 ft = 5�8 x 1 ft = 548 cubic ft <br /> E2. equals E1 divided by 27 cubic feet per cubic yard <br /> 548 / 27 = 20.29 cubic yards <br /> E3. equals E2 multiply by 1.4 tons per yard <br /> 28.4 tons = 1.4 x 20.28 cubic yards <br /> F. ABSORPTION WIDTH <br /> F1. Percolation Rate: 30 mpi Texture: Loam <br /> F2. Selected Soil Loading Rate: .6 <br /> F3. Calculated Adsorption Width Ratio <br /> Rock Layer Loading Rate Divided by Soil Loading Rate <br /> 1.20 gal/day/ft^2 divided . 6gal%day/ft^2 <br /> Adsorption Width Ratio = 2 <br /> F4. Equals Adsorption Width Ratio, I��zltiplied by Rock Layer Width <br /> 2 x 10 ft = 20 ft <br /> G. MOUND DIMENSIONS FOR SLOPE GREATFR THAN OR EQUAL TO 1$ <br /> G1. Minimum Downslope Berm equals Rock Layer Width minus Adsorption Width ' <br /> G1 = D2 - F4 == lOft = 2p - 10 <br /> G2. Minimum Mound Si�e Calculation <br /> G2a. Depth of Clean Sa:�d Fill Required at LTpslope Edge of Rock Laysr <br /> 3 foot Separation - (Ci incnes i t i� ir�ci�esi zootj j <br />
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