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1997-00943 - new septic system
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2995 Deer Run Trail - 04-117-23-23-0033
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1997-00943 - new septic system
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Last modified
8/22/2023 5:10:36 PM
Creation date
6/20/2016 11:44:05 AM
Metadata
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Template:
x Address Old
House Number
2995
Street Name
Deer Run
Street Type
Trail
Address
2995 Deer Run Tr
Document Type
Septic
PIN
0411723230033
Supplemental fields
ProcessedPID
Updated
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� MOUND DESIGN WORKSHEET <br /> ' � ' (For Flows up to 1200 gpd) <br /> A. FLOW Estimatcd Sewage Flows in Galluns per day <br /> �gPd) <br /> Estimated '� c� gpd (see pages D-7 or I-3,4,5) �m,�� <br /> or measured -- gpd x 1.5 = - ���5 Type I Type lI Type III �� <br /> 2 300 2?S 180 <br /> B. SEPTIC TANK L:IQUID VOLIJMES 3 aso 30o Z�s � <br /> 4 600 375 256 „��� <br /> ) -)�_�0 d� �-100o gallons (see pages C-3 or C-5) s �s� 4so z�o ;o <br /> 6 900 525 332 TIl Q�• <br /> 7 1050 600 370 m <br /> 8 1200 675 408 �d�d <br /> C. SOILS (refer to site evaluation) �� �� u 5� ,��''.�.},►a <br /> 1. Depth to restricting layer = ) 4 }oy� inches �P�rT.nkC.pxitr�,;,,K.��,,,, <br /> \umbct of Stinimum Liquid I.iquid c:.px�ty wiih <br /> 2. Depth of percolation tests = i � inches �Wm= �:�y�,y rub��=���„�.� <br /> 3. Percolation rate �. ) mpi z a�S� ,� "u <br /> 3a.a ian tsoo <br /> 4. Land slope (o % �,`aa9 `� ;c o <br /> over 9 .._.. <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Multiply flow rate by 0.83 to obtain required area of rock <br /> layer: Daily Flow x 0.83 = <br /> �S� gpd x 0.83 sq. ft./gpd = � sq. ft.� ���`%� <br /> 2. Select width of rock layer (10 feet or less) = J o ft. <br /> 3. Length of rock layer = Area = Width = <br /> �y_sq. ft. _ �ft. _ �� ft. Rock Bed <br /> r-r•r•r•r•r�r•r�r�f•r�f�r:r�: <br /> �.�..•..�.ti.�.�.�.�..�........�.ti.... T <br /> ti�`�{ftiftiftiftirtif:ftiftftiftif:f�.� tdth 510 ft. <br /> tif,f,r,r,f,f,r,f,f,f,f,f,f,f,f <br /> .� ti.�.�.ti �.�.... ............... 1 <br /> E. ROCK VOLUME •�f : :.:::.:.:.:.:.: <br /> Length � <br /> 1. Multiply rock area by rock depth to get cubic feet of rock; <br /> ��s�sq. ft. x .1 ��ft. =2�cu. ft. <br /> 2. Divide cu. ft. by 27 cu. ft./cu. yd. to get cubic yards; <br /> �cu. ft. 127 = �_cu. yd. <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> �cu. yd. x 1.4 ton/cu. yd. _ � tons. <br /> F. ADSORPTION WIDTH L`--�`� L°��''''� <br /> 1. Percolation rate in top 12 inches of soil is L . I mpi Absorption Width Sizing Table <br /> Percolation Rate Galbns Ratio of <br /> 2. Select allowable soil loading rate from table on page E-; in Minutes per so��T�x��� �.r day pet �,b���o„W;�,�h <br /> �� gpd/f� r�n cMPn �u�r� �o Rwk�a�n y�. <br /> 3. Calculate adsorption width ratio by dividing rock layer Facttr than 0.1• c�u s� ----- _-_-_ <br /> 0.1 w 5 Sand 1.20 1.00 <br /> loading rate of 1.20 gpd/ft�by allowable soil loading rate; o.�a s•• Fine Sand•> o.�o z.no <br /> 1.20 d/ft� . ��-1 d/ft - � ,(� � 6`°�s s�cy� o.�v i,sz <br /> p�p ��_�» Z- 16 to 30 Loam 0.60 2.W <br /> or� r� 31 to 45 Silt Loam 0.50 2.40 <br /> Check this value on page E-1 b. ���o Clay Lavn o.4s 2.�� <br /> 60 w I20 Clay 0.24 S.W <br /> 4. Multiply adsorption width ratio by rock layer width to get Sl;���an c�aY ----- __--- <br /> required adsorption width; <br /> �. � x � ft = � ft <br /> � <br /> 1 <br />
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