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2012-01197 - septic mound repair
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2012-01197 - septic mound repair
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Last modified
8/22/2023 5:05:58 PM
Creation date
4/29/2016 3:58:44 PM
Metadata
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Template:
x Address Old
House Number
2590
Street Name
Countryside
Street Type
Drive
Address
2590 Countryside Dr
Document Type
Septic
PIN
0411723110010
Supplemental fields
ProcessedPID
Updated
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���N� I��SI�1� �'VOI�.�C S�E�'j' (For Flows u to 1200 d) <br /> A. �ve�age Desig�a �LO� <br /> A-1: Estimaied Sewage Flows in Gallons per�3ay <br /> number of <br /> Estimated � S� gpd (see figure A-1) bedrooms cioss i �IaSS II c��5s ni ceoss Av <br /> or measured x 1.5 (safety factor) = gpd 2 300 225 �so bo% <br /> 3 450 300 218 of the <br /> 4 600 375 256 values <br /> �. SEPTIC'I'ANK Capacaty 5 750 450 294 in the <br /> 6 900 525 332 Class I, <br /> � � 1����:��-? gallons (see figure G1) ��x��.���.��., � �05o boo s�o u, or ni <br /> ���� 9^� .4;.���.��.�;. �ra��� 8 1200 675 408 columns. <br /> C. SOILS (refer to site evaluation) c-�: se ticTankCa acities(in alloos) <br /> , � , `0' 3'� � Number of Minimum Liquid Liquid capacity wilh Liquid capaciry <br /> 1. De th to restrictin la eP = _� � � , (,� ?.c? feet Bedrooms Capaci w�thdisposal& <br /> P g y � ty garbage disposal �ift inside <br /> 2. Depth of percolation tests = ' feet 2orless �so >>zs <br /> 3. Texture ��-.�`� ����r,� -�.v►��vY� 3 0�a �000 isoo �soo <br /> 5 or6 1500 2250 2000 <br /> Percolation rate �• ;�� mpi �:�����Iti�i�t.:- 7,8 or 9 2000 3000 3000 <br /> 4. Soil loading rate .4 � gpd/sqft (see figure D-33) <br /> 5. Percent land slope (o % <br /> D. ROCK LA�CER DIMENSIONS <br /> 1. Multiply average design flow (A) by 0.83 to obtain required rock layer area. <br /> ���� gpd x 0.83 sqft/gpd = i.� �,� sqft -�r������ ���,µ. s��: � <br /> 2. Determine rock layer width = 0.83 sqft/gpd x linear Loading Rate (LLR <br /> - 0.83 sqft/gpd x ! �1. gpd/sqft = )� ft �0��� LLR <br /> 3. Length of rock layer = area =width = <br /> �_sqft (D1) = I Q ft (D2) _ (�� �/ ft G � �o �P� < 1 � <br /> E. ROCK VOLUME � � 2O M P� G b <br /> 1. Multiply rock area (Dl) by rock depth of 1 ft to get cubic feet of rock <br /> (��t�t sqft x 1 ft = �,���u� cuft <br /> 2. Divide cuft by 27 cuft/cuyd to get cubic yards <br /> ,`�''� cuft = 27 cuyd/cuft = ���`l cuyd <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons <br /> ��cuyd x 1.4 ton/cuyd = .`�, � tons <br /> D-33: Absorption Width Sizing Table <br /> F. SEWAGE ABSORPTION �VID'Y'I� Percolation Rate Loading Rale <br /> in Minuios per Soil Tuture Gallons Absorption <br /> Inch per day per Ratio <br /> MP� s uarc foot <br /> Faster than 5 Coarse Sand 1.20 I.00 <br /> Mcdium Sand <br /> Absorption width equals absorption ratio (See Figure D-33) LoamySend <br /> Fne Sand <br /> times rock layer width (D2) 16 to 30 Loar„ o.�o z.00 <br /> 31 io a5 Silt L,oem 0.50 2.40 <br /> c�,��`�� � X '_� ft = ,-���.�r r,.�` f t 46 m 60 Sandy Qay L.o 0.45 2.67 <br /> Silty Clay Loam <br /> 61 l0 120 Silty Clay 0.24 5.00 <br /> � Sandy Clay <br /> Cla <br /> � Slowcrthan]20' <br /> 'Sysiem drsigned for thue soils rtvsi be otber or perfomvnce <br />
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