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2004-P07940 - new septic system
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785 Ferndale Road North - 36-118-23-11-0012
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2004-P07940 - new septic system
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Last modified
8/22/2023 5:00:27 PM
Creation date
9/7/2016 11:50:03 AM
Metadata
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Template:
x Address Old
House Number
785
Street Name
Ferndale
Street Type
Road
Street Direction
North
Address
785 Ferndale Road North
Document Type
Septic
PIN
3611823110012
Supplemental fields
ProcessedPID
Updated
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� M�UND DESIGN WORK SHEET (For Flows up to 1200 gpd) <br /> p p veragQ 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 tl Ciass III Class IV <br /> or measured -- x 1.5 (safety factor) = gpd 2 300 225 t8o c� <br /> 3 . 450 300 218 of the <br /> q 600 375 256 values <br /> B. SEPTIC TANK Capacity 5 750 450 294 in the <br /> 6 900 525 332 Class I, <br /> a,-- � a So gallons (see figure C-1) � ��� � 370 ii, or�t► <br /> g 1200 675 408 column: <br /> C. SOILS (refer to site evaluation) C-]: Se ticTankCa acities(ln allons <br /> Liquid capac; <br /> Numfxr of Minimum Liquid Liquid capacity with W��disposal <br /> 1. Depth to restricting layer = a.o�a .)aa-�feet <br /> gedrooms Capxity garbage disposa] I�ft ioside <br /> 2. Depth of percolation tests = /•O feet zo�►ess �so >>� lsoo <br /> 3 or 4 1000 1500 Zppp <br /> 3. Texture L�.� L o�r�wt 5 or 6 1500 2250 300p <br /> Percolation rate 4�.q mpi �'8°f 9 Z� 3Q00 <br /> 4. Soil loading rate � y 5� 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 /> � O gpd x 0.83 sqft/gpd = (oa , sqft <br /> 2. Determine rock layer width = 0.83 sqft/gpd x linear Loading Rate (LLR <br /> 0.83 sqft/gpd x J� gpd/sqft= / t� ft Mound LLR <br /> 3. Length of rock layer = area+width = <br /> �aa sqft (D1) = �2._ft (D2) = Coz— �t < 120 M PI < 12 <br /> E. ROCK VOLUME � � 2O M PI < C� <br /> 1. Multiply rock area (D1) by rock depth of 1 ft to get cubic feet of rock <br /> (o a� sqft x 1 ft = ��cuft <br /> 2. Divide cuft by 27 cuft/cuyd to get cubic yards <br /> Lo�_cuft +27 cuyd/cuft = �_ cuyd <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons. <br /> �_cuyd x 1.4 ton/cuyd = 3 � tons <br /> D-33: Absorptlon Width Sizing Table <br /> Percol�Gon Rate . Loading Rale <br /> F. SEWAGE ABSORPTION WIDTH inMinutuper So,�T�x� ���o�s Absorpcior <br /> lnch pu day per Rat�o <br /> Mp� s uare foot <br /> Fesler than 5 Coarx Sand 1.20 1•� <br /> Medium Send <br /> � � Lo�my S�nd <br /> Absorption width equals absorption ratio (See Figure D-33) <br /> times rock layer width (D2) 1B`°30 `"°�' °.60 2.� <br /> 31 to 45 Silt Loam 0•5� �•4� <br /> �,�X � _ ^t = ^/ 1 `t 46 ta 60 Sendy C1ay 0.45 2.67 <br /> _�`?_ t o4,o T s�uy c�oy i.o.m <br /> e��o�2o s+ny ciey o.za s.00 <br /> S�ndy Gay <br /> le <br /> � Slower Ih�n I20' <br /> •Sy��em dolEmd!or�hrse wil�rtun be dher or parfomuoce <br />
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