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2005-P09267 - new septic system
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175 Golden View Drive - 33-118-23-43-0025
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2005-P09267 - new septic system
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Entry Properties
Last modified
8/22/2023 4:52:25 PM
Creation date
12/30/2016 12:17:08 PM
Metadata
Fields
Template:
x Address Old
House Number
175
Street Name
Golden View
Street Type
Drive
Address
175 Golden View Drive
Document Type
Septic
PIN
3311823430025
Supplemental fields
ProcessedPID
Updated
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� M��[JND DESIGN WORK SHEET(For Flows u to 1200 d) <br /> A.. Average Design FLOW' ,A-1: f:timated Sewope Fl4Ws in Gcllons,per Day <br /> nu er o <br /> "� Estimated '�S� gpd (see fcgure A-1) bedrooms Clau 1 Class II Clau III Class IV <br /> 2 300 225 180 6096 <br /> or measured --� x 1:5 (safety factor) = -- gpd 3 �p 300 2t 8 of tne <br /> _ q b00 375 256 values <br /> B. SEPTIC TANK Capacity 5 750 450 294 in the <br /> 6 qp0 b25 332 Class I, <br /> 7 �p5p 600 370 II,or lll <br /> � - 14a�� �^�f gallons (see fi,gure C-1) g : 12pp. 675 408 columns. <br /> �x�s�'���., �A�� !� � �-�� t��o��.�(. . <br /> C. SOILS (refer to site evaluation) � ci: Se tic7�nkCa acides allons <br /> . ; Liquid capacity <br /> ' Tlqmber o( Minimum liqu.id liquid capaciry wich. .with disposal& <br /> 1. Depth to restricting layer= �/•� 1.S feet B�°°'°s c�'ry B�flg��P� lift inside <br /> 2. Depth of percolation tests = /• � feet za� �s� l�u lsoo <br /> 3«a i000 �soo a000 <br /> 3. Texture ��-+��! �--v,r�v��l s«6 �soo � 2�so 3000 <br /> Percolation rate 3Q.�� mpi � <br /> 7,8 a 9 7A(10 3000 <br /> 4. Soil loading rate .'�S gpd/sqft(see figure D-33) <br /> 5. Percent land slope y_% <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 = �?�. sqft <br /> . 2. Determine rock layer width= 0.83 sqft/gpd x linear Loading Rate(LLR <br /> 0.83 sqft/gpd x � a, gvd/sqft- 1p � ft Mound LLR <br /> 3. Length of rock layer= area+.width= <br /> �_Sqft c��� -� �.ft �Dz> _�ft� � 120 M P I <1.2 <br /> E. Rocx voLu� �. -> 120 M�PI < b <br /> ,_ - — <br /> 1. Multiply rock area (D1)by rock depth of 1 ft to get cubic feet of rock . <br /> �.� sqft x 1 ft= ���� cuft <br /> 2. Divide cuft by 27 cuft/cuyd to get cubic yards <br /> ���? �. 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 = ��1, tons . <br /> �I?�33:.Absprptlon Wldth SWng Ta61e <br /> . . pkrcol�Uon Rale Lwdina Rue' <br /> F. SEWAGE ABSORPI'ION tiVIDTH inMinuwper ShcTextuce c.,,� Absoiption <br /> �h per day per Ratio <br /> � uare(oo <br /> ' Fuwthan5 CwneS�nd 1.20` 1.00 <br /> Medium Sand. , <br /> LwmY Su�d, <br /> AbsorpHon width equals absorption ratio (See Figure D-33) <br /> z <br /> times rock layer width (D2) ' ' ' ,�o. +� � �4 <br /> / ' 46 to 60 S�ndy C1ay 0.45 2.67 .:. <br /> �,, �v�1 )C /� �= e�,�v. f t Silt Qry Lwm <br /> 61 l0 120 5 Iind Q y 0.1A 5.00 <br /> ,. � � . . � . . .. ... . , . . ower - <br /> . � . . . . •Syn�mdulV+dfm�Awdl�rmpAea6�rorperfomrnca � . .. <br />
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