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2010-00600 - mound septic system
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2675 Fox Street - 04-117-23-43-0003
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2010-00600 - mound septic system
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Last modified
8/22/2023 5:14:32 PM
Creation date
11/10/2016 12:36:41 PM
Metadata
Fields
Template:
x Address Old
House Number
2675
Street Name
Fox
Street Type
Street
Address
2675 Fox St
Document Type
Septic
PIN
0411723430003
Supplemental fields
ProcessedPID
Updated
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PRESSURE DISTRIBUTION SYSTEM Geotextile fabric <br /> 1. Select number of perforated laterals � <br /> Ouaztt+r in�h narFn��r:.�..�.-....__ � I?-" <br /> a��� <br /> 2. Select perforation spacing = �,� ft 9'•of ro�k <br /> 3. Since perforations should not be laced closer than 1 foot to Perf SPa g13i15�-51/4" <br /> p rerf 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 =—_�_ft <br /> per(oration <br /> 4. Determine the number of spaces between perforations. spacing <br /> Divide the length (3)by perforation spacing (2) and round feet 1 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 = ",,�i�j ft= �'= ft= I v spaces 3.0 8 �3 .17 ;" � <br /> __._____ ___.. �..26� <br /> 5. Number of perforations is equal to one plus the number of 3'3 > 12 �b 25 <br /> perforation spaces(4). Check figure E-4 to assure the number of 4�0 > >> 15 23 <br /> perforations per lateral guarantees <10% discharge variation. 5.0 6 �p �4 22 <br /> �� spaces + 1 =��perforalions/lateral E-6_ Perforation Discharge in gpm <br /> 6. A. Total number of perforations = perforations per lateral (5) — <br /> times number of laterals (1) perfioration diameter <br /> head inches <br /> 1`- perfs/latx u lat= � "�- perforations (feet) 3/16 7/32 1/4 <br /> B. Calculate the square footage per perforation. 1.0a 0.42 0.56 �0.74�) <br /> Should be 6-10 sqft/perf. Does not apply to at-grades. 2�0b 0.59 0.80 1.04 <br /> Rock bed area = rock width (ft) x rock length (ft) 5.0 0.94 1.26 1.65 <br /> 1� ft x ►�� 1 ft = �1 �r� 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 n�� else. <br /> 1.�,�.1�� sqft=_ �L't.. perfs = 4�1, �) sqft/perf <br /> MnMfOlO LOCATED PT ENO OF pqE55URE �ISTRIBUTION y^�STEM <br /> 7. Determine re quired flow rate b y mul t i p l y i n g t h e t o t a l n t u n b e r o f <br /> perforations (6A) by flow per perforation (see figure E-6) <br /> �1�. perfs x �?4 gpm/perfs =_ 31 gpm <br /> �- <br /> 8. If lateraLs are connected to header pipe as shown on upper j�:-�°' <br /> example, to select inuii.mum required lateral diameter; enter �,�..�`"`" ��;'�~`°��,� <br /> figure E-4 with perforation spacing (2) and number of perforations `"""d <br /> per lateral (5) Select minimum diameter for �� <br /> perforated lateral = I��ti- indzes. u•��• �•PERfOMtEO PIPF IaTERnLS,�w <br /> PNESSUIiE OISTpIB�T10H W MO�NO <br /> 0 R�S��C�i'( <br /> . I perforated lateral system is attached to manifold pipe near �,�„ati,�p,��,�o u. <br /> y,�wc <br /> the center,lower diagram, perforated lateral length (3) and �"° "�"" <br /> �Ew o'qc w.e[�;ro���on \ A n�oe�rrw� <br /> number of perforations per lateral (5)will be approximately one r�°"`�� <br /> half of dlat in step 8. Using these values, select miniinum � �s���s��� ���� � <br /> lo�.. '�- <br /> diameter for perforated lateral = �""'-' inches. K:�� <br /> b. <br /> � \ ior��� -n <br /> � o <br /> G M��,,,,,na �.�K- <br /> `f'��rx' <br /> I hereby certify that I have completed this work in accordance with applicable ordinances, rules and laws. <br /> � - t � , <br /> � ,. _ , i <br /> ` _, ,..,,. .. ,,, <br /> -- i.� ,,`` }- , __..m,._ (signature) �`�1 <br /> (license#) ��fo—D`� (date) <br />
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