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2008-P12100 - new septic system
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2010 Colin Drive - 03-117-23-21-0014
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2008-P12100 - new septic system
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Entry Properties
Last modified
8/22/2023 4:34:12 PM
Creation date
4/20/2016 1:13:08 PM
Metadata
Fields
Template:
x Address Old
House Number
2010
Street Name
Colin
Street Type
Drive
Address
2010 Colin Dr
Document Type
Septic
PIN
0311723210014
Supplemental fields
ProcessedPID
Updated
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' ' PRESSURE DISTRIBUTION SXSTEM Geotextile fabric <br /> l. Select nurnUer of perforated laterals ' ouartPr;,,�,,n rf�r,�'� � `l� <br /> _ "u�;�� <br /> 2. Select perforation spacing = �,� ft ' 9 of;rock <br /> 3. Since perforations should not be placed closer than 1 foot to Perf s�ac g si15-51/4" <br /> Perf S g <br /> the edge of the rock layer (see diagram), subtract 2 feet from <br /> the rock layer length. <br /> E-4: Maximum allowable number of 1/4-inch perforations <br /> per lateral to guarantee<10%discharge variation <br /> Rock layer length —2 ft —3�—ft perforation <br /> 4. Determine the number of spaces between perforations. spacing <br /> Divide the length (3) by perforation spacing (2) and round (feet) i inch 1.25 inch 1.5 inch 2.0 inch <br /> down to nearest whole number. <br /> 2.5 8 14 18 28 <br /> Perforation spacing = o� ft= 3 ft= 1,3 spaces 3.0 j�», '�" <br /> ... �, ._...8 .,r]3_,...,.._�...._�:' .`..26� <br /> 5. Number of perforations is equal to one plus the number of 3'3 � 12 16 25 <br /> perforation spaces(4). Check figure E-4 to assure the number of 4�0 > >> 15 23 <br /> perforatians per lateral guarantees <10% dischar�e variation. 5.0 6 10 �q 22 <br /> a 3 spaces + 1 = )�`� perforations/lateral E-6: Perforation Dischar e in <br /> 9 gpm <br /> 6. A. Total number of perforations = perforations per lateral (5) '— <br /> times number of laterals (1) perforation diameter <br /> head inches <br /> 1`~ perfs/lat x 3 lat= �-12 perforations (feet) ��8 3/16 7/32 1/4 <br /> B. Calculate the square footage per perforation. ���° 0.18 0.42 0.56 �,74� <br /> Should be 6-10 sqft/perf. Does not apply to at-grades. 2•0b 0.26 0.59 0.80 1.04 <br /> Rock bed area = rock width (ft) x rock length (ft) 5.0 0.41 0.94 1 ,26 1.65 <br /> 1� ft x v I ft= y�o _sqft <br /> ° Use 1.0 foot for single-family homes. <br /> Square foot per perforation=Rock bed area =number of perfs (6) b Use 2.0 feet for an thin else. <br /> '�t t � sqft= �-i-Z_ perfs = �.5� sqft/perf <br /> 7. Determine required flow rate by multiplying the total number of <br /> mand Id pipe <br /> per orations (6A) by flow per perforation (see figicre E-6) ,� <br /> pipe from pump <br /> _ '-��-_perfs x �7�gprn/perfs = 3 I gpm <br /> end cap <br /> 8. If laterals are connected to header pi�e as shown on upper ��`�, <br /> example, to select minimum required lateral diameter; enter altemotelowfion <br /> figure E-4 with perforation spacing (2) and number of erforations 9 ofpipe6ompump <br /> per lateral (5) Select minimuin diameter for p F� ure E-l;Manifold Located at End of System <br /> perforated lateral= al�. U inches. e�d�op <br /> ��� Figure E-2:Manifold�ocated <br /> 9. If perforated lateral system is attached to manifold pipe near in the Cenferof fhe Sysiem <br /> the center, lower diagram,perforated lateral length (3) and mo�;�o;tlG;� <br /> number of perfarations per lateral (5) will be approximately one <br /> half of that in step 8. Using these values, select miniinuin <br /> C�1dI112t2T f01�2I'{Ol'dteC� IdteT11 = 111C�12S. allemolelocolion <br /> of pipe hom pump <br /> �ipe hom pump <br /> I hereby certify that I have completed this work u1 accardance with applicable ord'uzances, rules and laws. <br /> ! �.Z�^�. '�, ,�� 1��,,1i�' t. <br /> �� (signature) _ �v�y (license#) � -'c�."a —t?S�` (date) <br />
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