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1997-009687 - new septic system
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1997-009687 - new septic system
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Last modified
8/22/2023 5:06:08 PM
Creation date
4/22/2019 1:43:56 PM
Metadata
Fields
Template:
x Address Old
House Number
2520
Street Name
Thoroughbred
Street Type
Lane
Address
2520 Thoroughbred La
Document Type
Septic
PIN
0411723110015
Supplemental fields
ProcessedPID
Updated
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MOUND DESIGN WORKSHEET <br /> (For Flows up to 1200 gpd) <br /> A. FLOW Estimated Sewage Flow in Gallons per Day(gpd) <br /> Estimated 0 gpd Number <br /> of or measured - x 1.5 =_=_gpd. Bedrooms ��I Iyix tt Type III Type IV <br /> 2 300 2Z5 180 so9c <br /> 3 450 300 218 et <br /> B. SEPTIC TANK LIQUID VOLUMES s 7600 3775 294 256 <br /> 6 900 525 332 <br /> i- a Sv J-�DO O gallons 07 1050 600 370 mW <br /> I�L To 16" 8 1 1200 1 675 1 408 <br /> C. SOILS(refer to site evaluation) 1, , Number <br /> �� <br /> 1. Depth to restricting layer = J a') 1 �y inches,syr. a," Bedmoms c <br /> 2. Depth of percolation tests = 21i inches <br /> 3. Percolation rate ?1. 1 mpi 23 cc 4 1less7.050°0 1.50000 <br /> 4. Land slope - % 7 or 9 z� ;;� <br /> over 9 See fig.C-6 (x 1.5) <br /> D. ROCK LAYER DIlvfENSIONS <br /> 1. Multiply flow rate by 0.83 to obtain required area of rock <br /> layer: A x 0.83 = , <br /> _0 6)_gpd x 0.83 sq. ft./gpd sq. ft.-}►o`er=1-%4" <br /> 2. Select width of rock layer (10 feet or less) = /0 ft. <br /> 3. Length of rock layer = area width = Rock Bed <br /> (ogi-l-sq. ft. 20 ft. = L�< ft. ti.,.,.,.,.,.,.�.,.ti...,.�.,...,., <br /> 1•t•1•�•I•f•f•f•t t•f•t•f•t•f•l• <br /> {•ti•\•�H.ti•ti•ti•ti•tiN•ti.ti•ti•ti•ti•ti <br /> r•r•f•r•f•r•r•r•�r•r•r•f•f•r.f• idth 5510 ft. <br /> S•ti•'►•ti.ti•ti•ti•ti•ti•ti•ti•ti•ti•ti•ti•% <br /> f•1•t•r•t•t•il'•f•1 t•t•I•r•t•t•1• <br /> {.ti.�•�•S•S.ti.ti.ti.{•1•Mti•ti•tiH•ti <br /> .j•f.f•I•r.j.r.r•f•I•r•f•r•r.f.f• <br /> E. ROCK VOLUME �-- Length I <br /> 1. Multiply rock area by rock depth to get cubic feet of rock; <br /> (a,tL4 sq. ft. x Lo__�C ft. =015e_cu. ft. <br /> 2. Divide cu. ft.by 27 cu. ft./cu. yd. to get cubic yards; <br /> r?)ee cu. ft. i 27 -;t 17_cu. yd. <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> cu. yd. x 1.4 ton/cu. yd. =,� tons. <br /> F. ADSORPTION WIDTH Z-v-A LO vn a«,width Siting,able <br /> 1. Percolation rate in top 12 inches of soil is ,21,1 mpi pemalati Rate �dy <br /> � 1,0" <br /> `timitrs per inch Soil Texture <br /> ,2. Select allowable soil loading rate from table; Faster than 0.1 oarse Sand 1.20 d/ft2 0.1 toy Sand 1.20 1. <br /> - 0.1 to 5 Fine Sand- 0.60 2.00 <br /> 6 to 15 Sandy Loam 0.79 1.52 <br /> 3. Calculate adsorption width ratio b dividing rock layer 16 to 30 Loam 0.60 2.00 <br /> rp y g }' 31 to 45 Silt Loam 2.40 <br /> loading rate of 1.20 gpd/ft2 by allowable soil loading rate; 46 to 6o ClayLoam .4' 2.67 <br /> -"bT_MM__ ay 4 5.00 <br /> 1.20 gpd/ft2i I�gpd/ft2 Slower than 120 Clav - - <br /> ••Soil having 50%or more of fine or very fine sand. <br /> 4. Multiply adsorption width ratio by rock layer width to get <br /> required adsorption width; <br /> 24 x10ft=a'z ft <br />
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