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2005-P08997 - septic
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2005-P08997 - septic
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Last modified
8/22/2023 5:02:07 PM
Creation date
3/6/2019 12:25:38 PM
Metadata
Fields
Template:
x Address Old
House Number
745
Street Name
Spring Hill
Street Type
Road
Address
745 Spring Hill Road
Document Type
Septic
PIN
3611823210003
Supplemental fields
ProcessedPID
Updated
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1 � MOUND DESIGN WORK SHEET(For Flows u to 1200 d) <br /> , A. A;verage Design FLOW A-�: Eatimaled Sewage Flows in Gallona per Day <br /> L�� � 1�6� Vu�st 1�Oa s� c� <br /> L-,y�¢�, num er o <br /> Estimated /US� gpd (see figure A-1) bedrooms Class I Claas II Claas III Class IV <br /> or measured -- x 1.5 (safety factor) = gpd 2 � 225 �$o � <br /> 3 . �o soo 2�a of ine <br /> 4 600 375 256 values <br /> B. SEPTIC TANK Capacity 5 750 450 294 in the <br /> 6 900 525 332 Class I, <br /> aS� � /�ov allons (see re C-1)A'� N � » � 3�o u,or m <br /> � $ .�$� �S� 8 1200 675 408 columns, <br /> a,—/t'1d� o,a� 14'S la�-1'�`sC �OaS�G. <br /> C. SOILS (refer to site evaluation) c-i: s ticTankCa acities ib� <br /> Liquid cap�iry <br /> Number of Miaimum Liquid liquid capaciry with W��disposal& <br /> 1. Depth to restricting layer= /'-3 . 4�I. feet '�S,�� B�°°"15 �P��ry B�B��P� lift inside <br /> 2. Depth of percolation tests = /• � � ��O� <br /> c� feet 2 a less 750 1125 �Spp <br /> 3. Texture L l�� L01�-Y�I� 3�a �� �� 20°° <br /> 5 or 6 1500 2250 300p <br /> Percolation rate 0.c� mpi ��8�9 20°° 30°° <br /> 4. Soil loading rate -4� gpd/sqft(see figure D-33) <br /> 5. Percent land slop� % <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Multiply average design flow (A)by 0.83 to obtain required rock layer area. <br /> � � _gpd x 0.83 sqft/gpd = 4�� I sqft � <br /> 2. Determine rock layer width= 0.83 sqft/gpd x linear Loading Rate (LLR <br /> 0.83 sqft/gpd x 12 gpd/sqft= `� ft Mound LLR <br /> 3. Length of rock layer = area+width = <br /> 4��! sqft(D1) + �ft (D2) _�ft < 120 M PI <� 2 <br /> E. ROCK VOLUME � � 2O M PI < 6 <br /> 1. Multiply rock area (D1)by rock depth of 1 ft to get cubic feet of rock <br /> �;L sqft x 1 ft=�2.�cuft <br /> 2. Divide cuft by 27 cuft/cuyd to get cubic yards <br /> � �l cuft +27 cuyd/cuft = 3 Z cuyd <br /> 3. Multiply cubic yards by 1.4 to get wei ht of rock in tons <br /> ��_cuyd x 1.4 ton/cuyd =�_tons <br /> D-33: Absorption Width Slzing Table <br /> F. SEWAGE ABSORPTION WIDTH ���0°�u I.�d�naRa�e <br /> in Minuw pa Shc Teacnue Gdlons Absorption <br /> Wch pu day per Ratio <br /> ►y{pl u foot <br /> Faater than S Cosne Sand 1.20 1•00 <br /> Medium Sand <br /> Absorption width equals absorption ratio (See Figure D-33) `'°°"'y S'nd <br /> times rock layer width (D2) ,,o, ;,, . 2: <br /> �•� X �Q lt = V� lt 46io60 SiNItdCQyl.am 0•45 2.67 <br /> 61 l0 120 5 ilty q�,�a� 0.24 5.00 <br /> and <br /> Slowa than 1 0' <br /> •Syaam dai�ned tm Wae wib m�a be ahu or psRortnuec <br />
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