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1994-006381 - septic system
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Old Crystal Bay Road South
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0825 Old Crystal Bay Road South - 09-117-23-21-0018
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1994-006381 - septic system
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Last modified
8/22/2023 3:18:19 PM
Creation date
3/27/2018 1:46:28 PM
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x Address Old
Address
0825 Old Crystal Bay Rd S
Document Type
Septic
PIN
0911723210018
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ProcessedPID
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• <br /> MOUND DESIGN WORKSHEET <br /> I <br /> (For Flows up to 1200 gpd) <br /> A. FLOW Estimated Sewage Flow in Gallons per Day(gpd) <br /> Estimated 6,oc.) gpd N ofd Type I Type II Type III Type IV <br /> or measured x 1.5 = - gpd. Bedrooms <br /> 2 300 225 180 60% <br /> 3 450 300 218 of <br /> B. SEPTIC TANK LIQUID VOLUMES 54 7750 iso 2294 5 256 "in <br /> -/000 gallons 6 <br /> 9007 525 332 c:1- <br /> „ 8 1200 675 408 <br /> C. SOILS (refer to site evaluation) t Numb., � :,, <br /> 1. Depth to restricting layer= ' .o inches Bedrooms ct•ooky t>ispo•.t <br /> 2. Depth of percolation tests = 1 a. ` inches ` ' . t`�' <br /> 3. Percolation rate /0. 1 mpi 2ess <br /> 3«l4 i o0o 1.5000 <br /> 4. Land slope L, s7«6 1.500 <br /> 000 3,000 <br /> % 5«8 2.000 3,000 <br /> over 9 See fig.C-6 (x 1.5) <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 /> (.00 gpd x 0.83 sq. ft./gpd = 49 v. sq. ftp-logo= Sti')Q <br /> 2. Select width of rock layer (10 feet or less) = /0 ft. <br /> 3. Length of rock layer= area i width = Rock Bed <br /> ,Cy ') sq. ft.+ /0 ft. = < ft. e�::..•.•.•,wa.•.•;:a.wa.•+•.;••• 1 <br /> rti.rtirti .r♦r•rr'r;r;rtiN�.tirtrrr.V <br /> . r•r•r•r•r•r•r•r•r•rr•r•r•r•r•r•r <br /> Vidth <10 ft <br /> r•rr•r•r•rr•r•!•tr•I•r•rr•r•r•I _ .✓7;47%/1...14.74:titti74... ti•rtirtir•tif•ti• <br /> - <br /> E. ROCK VOLUME 1-- Length <br /> 1. Multiply rock area by rock depth to get cubic feet of rock; <br /> . 1 sq. ft. xjos ft. =.0t/ cu. ft. <br /> 2. Divide cu. ft. by 27 cu. ft./cu. yd. to get cubic yards; <br /> 5')L1 cu. ft. -t-27= a ) 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. = a-7 tons. <br /> F. ADSORPTION WIDTH LA-Ay L0'1>> 1 Absorption Width Sirng Db.e <br /> 1. Percolation rate in top 12 inches of soil is jo. / mpi Percolator Rate Callool Ravi*of <br /> Soil Texture p`r°•• ''a°e.", <br /> Minutes per inch pelt_ : � <br /> (mpi) .Hath <br /> 2. Select allowable soil loading rate from table; Faster than 0.1 Coarse Sand 1.20 1.00 <br /> 1.20 1.00 <br /> •L►Lr. gpd/f t' 0.1 to 5 Fine Sand" 0.60 2.00 <br /> 6 to 15 Sandy Loam 0.79 1.52 <br /> 16 to 30 Loam 0.60 2.00 <br /> 3. Calculate adsorption width ratio by dividing rock layer 31 to 45 Silt Loam 050 2.40 <br /> loading rate of 1.20 gpd/ft2 by allowable soil loading rate; to 60 Clay Loam 0.45 2.67 <br /> 61 tO 120 Clay 0.24 5.00 <br /> 1.20 gpd/ft2- .Ll< gpd/ft2 _ D - /7 Slower than 120 Cray -- - <br /> ••Soil laving 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 /> ..,") x /O ft = .' ft <br />
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