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7. <br /> MOUND DESIGN WORKSHEET <br /> ' (For Flows up to 1200 gpd) <br /> A. FLOW E&timatcd Sewage Flows in Gallons per day <br /> Estimated 75 o gpd S&tic-i) (gad) <br /> :urncer I <br /> or measured x 1.3 = gpd• enxio rtta I Type 1 Type a Type m lve <br /> 2 300 223 1150 <br /> 3 450 B. SEPTIC TANK LIQUID VOLUMES (SQ,-,eQszc,5-0,...,1)2,6) 300 218 am� ' , 3775 156 <br /> . " <br /> - 4;2 5-6 gallons � l 6 900 725 332 <br /> 7 1050 600 370s <br /> 8 1200 673 408 .a.•.. <br /> C. SOILS (refer to site evaluation) <br /> 1. Depth to restricting layer = /' inches (.1 - SepumgTore Grimm,.a Wu.. <br /> 2. Depth of percolation tests = N..cer el Mun•.ss Law.t Lepel come!.M <br /> inches B-e.-..1 Ca..ruy wimps dise...t <br /> 3. Percolation rate 'v. mpi 150 rtv <br /> 1>. 1000 ixo <br /> 4. Land slope . 0 % 15;r'9 %COZtlel <br /> d..:9 ; --- <br /> v <br /> v y . <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 = Z <br /> 766-O gpd x 0.83 sq. ft./gpd =(,23 sq. ft. a IG'7c., = e$Q <br /> 2. Select width of rock layer (10 feet or less) = /o ft. <br /> 3. Length of rock layer = area width = Rock Sed <br /> ..(1?e) sq. ft. + /d ft. = Cv . ft. :,:; : :.:;: :\ : ..: )o 'Y ' <br /> • :�.:?.\.,......,.\•\.\.\.,........,�•]] I 1 t 510 ft. <br /> E. ROCK VOLUME ► Length -- <br /> -- 1 <br /> 1. Multiply rock area by rock depth to get cubic feet of rock; to g 9- <br /> G,T <br /> o sq. ft. x / ft. =“10 cu. ft. <br /> 2. Divide cu. ft. by 27 cu. ft./cu. yd. to get cubic yards; <br /> (ago cu. ft. + 27 = a4 cu. yd. <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> 02 G cu. yd. x 1.4 ton/cu. yd. = 37 tons. 4- (G 626 <br /> F. ADSORPTION WIDTH ezAL)74'Ac/ri 60 - /Z•a Absorption Width Sizing Table <br /> 1. Percolation rate in top 12 inches of soil is 7e 2 mpi Pereatauon Rue Gallons Rauoor <br /> in Minutes per Soil Tway' per day per Absorpooe <br /> loch(y.1P1) square foot to Rock L <br /> Wi.1l1 <br /> 2. Select allowable soil loading rate from table; Faster than 0.1• coarsnt Sand -. _._ <br /> 0• 2"47 gpd/ft2 0.1 to 5 Sand 1.20 1.00 <br /> 0.1to5•• Fine Sand•• 0.60 Z00 <br /> 6 to 15 Sandy Loam 0.79 1.52 <br /> 16 to 30 Loam 0.60 2.00 <br /> 31 t.45 Silt .• 0.70 7.40 <br /> 3. Calculate adsorption width ratio by dividing rock layer - .tri:•+-r._ - <br /> loading rate of 1.20 gpd/ft2 by allowable soil loading rate; ., Overt.0 as --.......'s „.�'-00 <br /> 1.20 gpd/ft2+ O'ZYgpd/ft2 = S o 0 . • <br /> 121:1•••• <br /> Soul <br /> Sou too coarse rot installation of a <br /> standard system. <br /> Sce 7080.0170 Subp 2.0.3,�7 age 26. <br /> 4. Multiply adsorption width ratio by rock layer width to get •- Soil having 30%or more of tine sand <br /> plus verytine sand. <br /> required adsorption widt <O width; •••S��ohe' <br /> vyfor installation ofa <br /> ,� x Jo ft = , ft Sue 7080.0210 Subp S.A.page 33. <br />