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2002-P05491 - repair septic system
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2002-P05491 - repair septic system
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Last modified
8/22/2023 4:16:38 PM
Creation date
4/10/2017 3:12:21 PM
Metadata
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Template:
x Address Old
House Number
1100
Street Name
Knoll Manor
Street Type
Road
Address
1100 Knoll Manor Road
Document Type
Septic
PIN
2611823310012
Supplemental fields
ProcessedPID
Updated
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♦ ' r' <br /> PRESSURE DISTRIBUTION SYSTEM Geotextile fabric <br /> 1. Select number of perforated laterals � .arter inch erfornHona, seed�3' . 12�� <br /> . ..�.. . <br /> . .•r'' •�'•.�••�,L�7C�C <br /> 2. Select perforation spacing= 3• � ft .: � ' " ��~�'=e'�� - � <br /> Perf S 3/16"-1/4" <br /> 3. Since perforations should not be placed closer than 1 foot to p�Sp�ns 1.5'-5' <br /> the edge of the rock layer(see diagram),subtract 2 feet from <br /> the rock layer length. E-4: Maxirrxim anowable number of 1/4-Mch pedorations <br /> Z- -2 ft =�0 _ft per lateral to puorantee<10%dlscharge vadation <br /> c �y�r �ns pertaa8on , <br /> 4. Determine the number of spaces between perforations. � <br /> Divide the length(3)by perforation spacing(2)and�y� � ��h � 1.25 inch 1.5 inch 2.0 inch <br /> � �own to nearest whole number. <br /> 2.5 8 14 18 28 <br /> Perforation spacing= L�n ft+�ft= �-d spaces �3.0 8 13 17 . 26 <br /> 5. Number of perforations is equal to one plus the number of 3.3 7 12 16 25 <br /> 4.0 7 11 15 23 <br /> perforation spaces(4). Check�gure E-4 to assure the number of <br /> 5.0 6 10 14 22 <br /> perforations per lateral guarantees <10%discharge variation. <br /> o�O 5paces+1 = a, l PerfOratiOnsJlateral E-6: Perforation Discharge in gpm <br /> 6. A. Total number of perforations= perforations per lateral (5) perforation diameter <br /> � times nwnber of laterals(1) head Inches <br /> �_perfs/lat x 3 lat=�o�perforations (feet) 3/16 7/32 1/4 <br /> 1.Oa 0.42 0.56 0.74 <br /> B. Calculate the square footage per perforation. , 2,Ob 0.59 0.80 1.04 <br /> Should be 6-10 sqft/perf.Does not apply to at grades. <br /> Rock bed area= rock width(ft)x rock length(ft) 5.0 0.94 1.26 1.65 <br /> I 0 ft x�ft= c�a o sqft a Use 1 A foot f�single-family homes. <br /> Square foot per perforation=Rock bed area+number of perfs (6) b use 2.o r�et to�a� i� Qise. <br /> („�c� sqft+ (�3 perfs= � .� sqft/perf M.�«�, �,u„m AT � ��E �R��, �EM <br /> � . <br /> 7. Determine required flow rate by multiplying the�otal number of <br /> perforations (6A) by flow per perforation(see figure E-6) � <br /> !I <br /> ,� <br /> � perfs x��y_.gpm perfs= �So gpm <br /> 8. If laterals are connected to header pipe as s�own on upper �.�1 <br /> example,to select minimum required latera�diameter;enter ,,.�"'"`� '�''��'�� <br /> figure E-4 with perforation spacing (2) and number of perforations ��``'"� <br /> per lateral(5) Select minimum diameter for <br /> urart a rc�roiurro r�rE uTceus roa <br /> perforated lateral= inches. �c wsrwurwx w wwNo <br /> �[A'OIY1l0 RYiK rI[ <br /> 9. If perforated lateral system is attached to manifold pipe near �,,,,,�„ ,� <br /> the center,lower diagram,perforated lateral length(3)and � �""` ��' � x"� <br /> r�ao <br /> number of perforations per lateral(5)will be approximately one K�,n.�,�.. <br /> half of that in step 8. Using these values,select minimum �- _ <br /> � •-��.��,;::� <br /> diameter for perforated lateral = � � U inches. �,�, ,� <br /> o�„" <br /> . �� <br /> s �. .� <br /> �� <br /> I hereby certi that I e completed this work in accordance with applicable ordinances, rules and laws. <br /> ' � l�- (signature) 3�y (license#) � -�-4�-- (date) <br />
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