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2003-P06605 - new septic system
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2003-P06605 - new septic system
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Last modified
8/22/2023 5:13:24 PM
Creation date
10/21/2016 1:44:37 PM
Metadata
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Template:
x Address Old
House Number
2530
Street Name
Fox
Street Type
Street
Address
2530 Fox St
Document Type
Septic
PIN
0411723410010
Supplemental fields
ProcessedPID
Updated
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. t�IO`[�'ND DESIGN WORK SHEET (For Flows u to 1200 d) <br /> A. Average Design FLOW A-1: Estimated Sewage Flows in Gallons per Day <br /> num er o <br /> Estimated � �� gpd (see figure A-1) bedrooms Class I Class II Class III Class IV <br /> or measured — x 1.5 (safety factor) _ — gpd 2 300 225 180 � <br /> 3 450 300 218 ofthe <br /> 4 600 375 256 values <br /> B. SEPTIC TANK Capacity 5 75o a5o 294 in the <br /> 6 900 525 332 Class I, <br /> a - i a�S� gallons (see figure C-1) � ��� � 3�o u, or ui <br /> 8 1200 675 408 columns. <br /> t - 1'��v 'C'�..1,M� G�.�-,o,-w�'S��• <br /> C. SOILS (refer to site evaluation) C-1: Se ticTankCa acities(in allons <br /> ��"¢>-'1a'S� <br /> s�-�'�'c�• I�S 4►.�d Liquid capacity <br /> Number of Minimum Liquid Liquid capaciry with W��disposal& <br /> 1. Depth to restricting layer = �, I + 3. ) feet Bedrooms Capacity garbage disposa! ►�(t inside <br /> 2. Depth of percolation tests = l.o feet zo�iess �so i�as 15� <br /> 3. Texture ��-�-�5 L.o,a-w� 3ora �000 �soo Z� <br /> 5 or 6 1500 2250 3� <br /> Percolation rate �a•� mpi �,8 or 9 z000 3000 <br /> 4. Soil loading rate _ ��S gpd/sqft (see figure D-33) <br /> 5. Percent land slope '� % <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Multiply average design flow (A) by 0.83 to obtain required rock layer area. <br /> r?S� gpd x 0.83 sqft/gpd = cea� sqft+���,, -ca�ya' <br /> 2. Determine rock layer width = 0.83 sqft/gpd x linear Loading Rate (LLR <br /> 0.83 sqft/gpd x ►a gpd/sqft= �� ft Mound LLR <br /> 3. Length of rock layer = area = width = <br /> ��_ sqft (Dl) = � u ft (D2) = c.� ft < 120 M PI < � 2 <br /> E. ROCK VOLUME > 120 M PI < 6 <br /> 1. Multiply rock area (Dl) by rock depth of 1 ft to get cubic feet of rock <br /> co�y sqft x 1 ft =��cuft <br /> 2. Divide cuft by 27 cuft/cuyd to get cubic yards <br /> lo�s y cuft T 27 cuyd/cuft = _ as cuyd <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons <br /> � cuyd x 1.4 ton/cuyd = 35 tons <br /> D-33: Absorptlon Width Sizing Table <br /> F. SEWAGE ABSORPTION WIDTH Percolation Ra�e Loading Rate <br /> in Minu�es per Soil Tezture Gellons Absorption <br /> Inch per day per Ratio <br /> Mp1 s uarc foot <br /> Fasurthan 5 Coerx Sand 1.20 ].00 <br /> Medium Sand <br /> Absorption width equals absorption ratio (See Figure D-33) LoamySmd <br /> times rock layer width (D2) 16 to 30 Lo�„ o.�0 2.0� <br /> si�o as u�t,�,n oso z.ao <br /> ,lo� x /C� ft = a(o.� f t a6�a 6o s�,ay ca.Y o.as s.6� <br /> s�i� ci.y i.o.m <br /> 61�0 120 Silty Cley 0.24 5.00 <br /> Sandy Clay <br /> I <br /> Slower then 120• <br /> •Syelem deaigned for Nese wi1�m�st be dber or perfomuoce <br />
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