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Septic info including 1985 design
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0985 Old Long Lake Road - 35-118-23-41-0027
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Septic info including 1985 design
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Last modified
8/22/2023 4:59:31 PM
Creation date
4/12/2018 12:43:26 PM
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x Address Old
House Number
985
Street Name
Old Long Lake
Street Type
Road
Address
985 Old Long Lake Road
Document Type
Septic
PIN
3511823410027
Supplemental fields
ProcessedPID
Updated
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• • t-19 <br /> MOUND DESIGN WORKSHEET <br /> ' (For Flows up to 1200 gpd) <br /> A. FLOW D-7 <br /> Estimated goo gpd (see pages D-7 or I-3,4, 5) ESTIMA,fD 3EWA0E FLOW,116 01.110163 PEP 047 <br /> NMocER TYPt Of R[]10ENCE• <br /> BE0400E6 t C I= 3 <br /> or measured gpd. 2 300 225 1110 00% <br /> 3 430 300 219 0 <br /> 4 GOO 379 206 <br /> B. SEPTIC TANK LIQUID VOLUMES ' 'n 322 `, T. <br /> /��L <br /> 7 1090 600 370 is <br /> ;01,C, gallons (see pages C-3 or C-5) ! 1200 6 <br /> 15 400 Columns <br /> C-3 <br /> C. SOILS (refer to site evaluation) SEPTIC TANK CAPACITIES, IN GALLONS <br /> UONO c0•ACT/ <br /> 1. Depth to restricting layer = + inches 20.t/1 Of 10140300.101 SIMI <br /> CLOY ma <br /> 11141I1001E0 UOWO CAPACITY 01SPOS•L <br /> 2. Depth of percolation tests = /A. inches ,,R,A2A 7E0 2I„ <br /> 3. Percolation rate / 2 mpi f 011 <br /> • 1.00 22110 <br /> 1 04 2500 2220 <br /> 4. Land slope °'o7.14 0R I „e, ,.112 <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Multiply flow rate by 0.83 to obtain required area of rock <br /> layer: A x 0.83 = <br /> SO O gpd x 0.83 sq. ft./gpd = 2..2' sq. ft. <br /> 2. Select width of rock layer (10 feet or less) = /0 ft. <br /> 3. Length of rock layer = area width = <br /> z.S"V sq. ft. i _i 0 ft. = Z.s- ft. Rock Bed <br /> r•r•r•r•r•r•r•r•r•r•r•r•r•r•r�T <br /> tif.f,:pp ti:tif�ftiftiftiftiti.7.1-aWidth s1(I <br /> r•r•r•r•r•r•r•r•r•r•r•r•r•r•rI <br /> rtir�r~r�rtir�rtirtippy_r�r�r'•r11 <br /> E. ROCK VOLUME ►-- Length <br /> 1. Multiply rock area by rock depth to get cubic feet of rock; <br /> 2370 sq. ft. x / ft. =2-.Cb cu. ft. <br /> 2. Divide cu. ft. by 27 cu. ft./cu. yd. to get cubic yards; <br /> 2-Ce, cu. ft. y 27 = 9, 2 cu. yd. <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> 9,L cu. yd. x 1.4 ton/cu. yd. = /3 tons. <br /> F. ADSORPTION WIDTH <br /> 1. Percolation rate in top 12 inches of soil is /Z mpi E-16 <br /> 2. Select allowable soil loading rate from table on page E-16; •LLO•.•ALALOAO••O PATES O.2dAV•02R 11C u•0! <br /> 1 7 ' gpd/fes ...,�.YJ �-w.•._ <br /> 3. Calculate adsorption width ratio by dividing rock layer """" "'" `�•' "� <br /> ...• I IAO 2.00 2.00 l00 <br /> loading rate of 1.20 gpd/ft2 by allowable soil loading rate; 0 12+ 201 Ile 2.11 I f0 <br /> Ie -20 0A0 0.11 2.44 2.00 <br /> 1.20 gpd/ft2 rs� gpd/ft2 = /.S ze.. ,112 ,00 <br /> I0 .00 0 0. 20 2.22 <br /> II .120 0.:. 0.25 II IIII 2.00 <br /> Check this value on page E-16. <br /> 4. Multiply adsorption width ratio by rock layer width to get <br /> required adsorption width; <br /> z- <br /> ,' x /D ft = f. ft <br />
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