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septic info including septic design
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Old Crystal Bay Road North
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0365 Old Crystal Bay Road North- 33-118-23-31-0013
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septic info including septic design
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Last modified
8/22/2023 4:49:23 PM
Creation date
3/7/2018 12:42:33 PM
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x Address Old
House Number
365
Street Name
Old Crystal Bay
Street Type
Road
Street Direction
North
Address
365 Old Crystal Bay Road North
Document Type
Septic
PIN
3311823310013
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MOUND DESIGN WORK SHEET(For Flows up t 1200 d) <br /> A. Average Design FLOW A-1: Estimated Sewage Flows in Gallons per Day <br /> nurroero <br /> Estimated 90 gpd(see figure A-1) bedrooms Class I class II Class III class IV <br /> or measured x 1.5 (safety factor) = gpd 2 300 225 180 60% <br /> 3 450 300 218 of the <br /> 4 600 375 256 values <br /> B. SEPTIC TANK Capacity 5 750 450 294 In the <br /> 6 900 525 332 Class I, <br /> gallons (see re C- ) 7 1050 600 370 Ii,or III <br /> -)o 00 <br /> $ �gu18 1200 1 675 1 408 columns. <br /> C. SOILS (refer to site evaluation) G1: SePdC 7ant capaellies(in M11611s) <br /> aclty <br /> Number of Minaimum liquid Ugidd ip&ity with with dispppoW& <br /> 1. Depth to restricting layer= . feet B�0°1°s i� gmbli <br /> �disposlil Gft inside <br /> 2. Depth of percolation tests= 1-'O feet 2 or less 750 1125 1500 <br /> 3. Texture L1.OK�.19'A V%4 3 or 4 1000 1500 2000 <br /> 5 or6 1500 2250 3000 <br /> Percolation rate 1!�.1 mpi 7,8 or 9 2000 3W <br /> 4. Soil loading rate .4 IC gpd/sqft(see fi re D-33) <br /> 5. Percent land slope % <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Multiply average design flow (A)by 0.83 to tain required rogk layer area. <br /> 7S0 gpd x 0.83 sqft/gpd= A ak <br /> 2. Determine rock layer width= 0.83 sqft/gpd*linear Loading Rate (LLR. <br /> 0.83 sqft/gpd x a-4, gpd/sqft= l0 ft Mound LLR <br /> 3. Length of rock layer= area=width= <br /> ig Q k4 _sgft(D1) in ft (D2) =_Ql 9� jft < 120 MPI <12 <br /> E. ROCK VOLUME > 120 MPI < 6 <br /> 1. Multiply rock area (Dl)by rock depth of 1j ft to get cubic feet of rock <br /> ,meq sgft x 1 ft= t erAq cult <br /> 2. Divide tuft by 27 cuft/cuyd to get cubic yards <br /> �'�cuft +27 cuyd/cult= _cuyd <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons <br /> a S cuyd x 1.4 ton/cuyd = tons <br /> 1D-33: Absorption Width Sizing Table <br /> Percolation Rate Loading Rate <br /> F. SEWAGE ABSORPTION WIDTH in Minutes par Sall Texture ca j.: Absorption <br /> itch per day pa Ratio <br /> (Mpl) square foot <br /> Paster than 5 Coarse Send 1.20 1.00 <br /> Medium Sand <br /> Absorption width equals absorption ratio (S a Figure D-33) 4=1W <br /> times rock layer width (D2) 6 0; 2. <br /> 31 to 43 Silt LAM 2•U <br /> Silt <br /> x 1�ft= ac,, ft 46 to 60 SSa1 ey L=U 0.45 2.67 <br /> 61 to 120 Silry aay 0.24 5.00 <br /> Sand aay <br /> Slower trm 12 <br /> *System designed for the ails amt be adw or pe fonmow <br />
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