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septic design - 1992
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1505 Tanglewood Road - 26-118-23-32-0018
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septic design - 1992
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Last modified
8/22/2023 4:17:19 PM
Creation date
4/17/2019 9:56:24 AM
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x Address Old
House Number
1505
Street Name
Tanglewood
Street Type
Road
Address
1505 Tanglewood Road
Document Type
Septic
PIN
2611823320018
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MOUND DESIGN WORKSHEET <br /> (For Flowrup to 1200 gpd) <br /> A. FLOW ea1"„00d SCWW FIM In GORMS Pff&Y <br /> Estimated 1-,�-b gpd(seepages D-7 or I-3,4,5) <br /> rn a (OPQ T <br /> or measured gpd x 1.5 = Of TMr TMn Tm n T99 <br /> eaa0=. <br /> 13. SEPTIC TANK LIQUID VOLUMES 3 50 ZtS 3300 2u <br /> 4 600 375 256 dd e <br /> J-�a�c7 -�Dev gallons'(see pages C-3 or C-5) 294 In <br /> 6 9700 Su 3332 Tr L <br /> 7 1050 600 370 = <br /> C. SOILS(refer to site evaluation) s 1200 675 409 .._. <br /> 1. Depth to restricting layer= � �s o -xy " inches ''" TW*C''""'" b @e% ' <br /> N=bwet bft, t.q.�r ugrWe.p.e1'.w <br /> 2. Depth of percolation tests _ a,, inches SOdnieau a air 4r <br /> 3. Percolation rate mpi 23 1 W 11, <br /> 4. Land slope 3 °Yo ; ' 3M3 <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Multiply flow rate by 0.83 to obtain required area of rock <br /> layer: Daily Flow x 0.83= n <br /> o gpd x 0.83 sq. ft./gpd = (,aa sq. ft-�►d�°-���+ <br /> 2. Select width of rock layer(10 feet or less) <br /> 3. Length of rock layer=Area+Width = <br /> t.,vLi sq. ft.+ 1p ft. = e,, ft. Rock Bed <br /> •r r <br /> r.rti.ti. <br /> r•r•r•r•r• .r•r•r•r•r <br /> • .ti.ti.ti.ti. •ti• ti.ti.ti• <br /> r r•s•s r.r=r.r.rN•r•r•�•I•r <br /> dth S10 ft. <br /> Kry�y;;}ti;�;,rtir�rti}tirtir;stir <br /> E. ROCK VOLUME •r r•r•r• •r•r•r•r• •]'•r.r.r•r <br /> F-- Length <br /> 1. Multiply rock area by rock depth to get cubic feet of rock; <br /> Lam{sq. ft. x I.o5 ft. = r-)vw cu. ft. <br /> 2. Divide cu. ft.by 27 cu. ft./cu. yd. to got cubic yards; <br /> .cu. ft. +27= cu. yd. <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> -2a cu. yd.x 1.4 ton/cu. yd. = &,K tons. <br /> F. ADSORPTION WIDTH <br /> 1. Percolation rate in top 12 inches of soil is mpi Absorption width Sldng Tobit1 <br /> Wan X61100f <br /> 2. Select allowable soil loading rate from table on page E-; in sd1'loon <br /> 10 <br /> L4,< gpd/fF wid►w <br /> 3. Calculate adsorption width ratio by dividing rock layer ttaw*wa1 a anus." .-- <br /> W w s sww 1.20 1100 <br /> loading rate of 1.20 gpd/f t2 by allowable soil loading rate; 0.100300 iMe s.sd M a60 2110 <br /> 1.20 gpd/ft2+ .`- gpd/ft2= ;Jo D 130130 s "°1i:: " aW LM <br /> 31w45 air LIMM 0.10 2.40 <br /> Check this value on page E-16. b o " .,`°M OAS 2A7 <br /> SM <br /> 4. Multiply adsorption width ratio by rock layer width to get s ;;" any -- -- <br /> required adsorption width; <br /> 2.-4-x Z 0 ft= a c.. ft <br />
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