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2003-P06591 - septic
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1045 Brown Road South - PID: 10-117-23-24-0002
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2003-P06591 - septic
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Last modified
8/22/2023 3:21:12 PM
Creation date
2/8/2016 1:58:03 PM
Metadata
Fields
Template:
x Address Old
House Number
1045
Street Name
Brown
Street Type
Road
Street Direction
South
Address
1045 Brown Road South
Document Type
Septic
PIN
1011723240002
Supplemental fields
ProcessedPID
Updated
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' PRESSURE DISTRIBUTION SYSTEM <br /> All boxed rectangles must be entered, the rest will be calculafed <br /> 1 Select number of perforated laterals: � � � ' " <br /> -- - -_-�= <br /> ---�---- -o---,,. <br /> 2. Select perforation spacing = Oft � � � w I <br /> � �_ <br /> �-...<<u ,,,,„, _���.. <br /> 3. Since perforations should not be placed closer that 1 foot to �����'`��-��'�� 4'� <br /> the edge of the rock layer(see diagram), subtract 2 feet from <br /> the rock layer len th <br /> 62 -2 ft= 60 ft <br /> 4 Determine the number of spaces between perforations. <br /> Divide the length(3)by perforation spacing(2) and round down to nearest whole number. <br /> Perforation spacing= 60 ft/ 3 ft= 20 <br /> 5. Number of perforations is equal to one plus the number of perforation spaces(4). <br /> 'Check figure E-4 to assure the number of perforations per lateral guarantees <br /> < 10%discharge variation. <br /> 20 spaces+ 1 = 21 perforations/lateral <br /> E�Maximum Number of 1/4 inch perforations E-6 PerForation Dischar e in GPM <br /> r lateral to uarantee<10%dischar e variation Head Perforations diameter <br /> p���� (feet) (inches) <br /> Spacing 3/16 7132 1/4 <br /> feet 1 inch 1.25 inch 1.5 inch 2.0 inch 18 0.42 0.56 0.74 <br /> 2.5 8 14 18 28 2b 0.59 0.80 1.04 <br /> 3.0 8 13 17 26 5 0.94 1.26 1.65 <br /> 3.3 7 12 16 25 a. Use 1.0 toot tor single-family homes. <br /> 4.0 7 11 15 23 b.Use 2.0 feet for anything else <br /> 5.0 6 10 14 22 <br /> 6. A.Tdal number of perforations=perforations per Iffieral(5)times number of faterals(1). <br /> 21 perfs/lat x 3 laterals= 63 perforations <br /> B.Calculate the square footage per perforation. <br /> Recorr►mended value is 6-10 sqft/perf. Dces not apply to at-grades. <br /> 1. Rock bed area=rock width(ft)x rock length(ft) <br /> 10 ft x 62 ft= 620 ftZ <br /> 2. Square foot per perforation=Rock Bed Area/number of perfs(6) <br /> 620.0 ft/ 63 perfs = 9.8 ftZ/perF <br /> 7. Determine required flow rate by muftiplying the total number <br /> of perforations(6A)by flow per perforations(see figure E-6) <br /> 63 perfs x 0.74 gpm/perfs= 46.6 gpm , <br /> 8. If laterals are connected to header pipe as shown �� <br /> in Figure E-1,to select minimum required lateral ` . <br /> diameter;enter figure E-4 with perforation spacing(2)and .. ._ <br /> number of perforations per lateral(5). �,a�..E-':"'a^"o"'�a�'°a'E"'o,9"""" <br /> Select minimum diameter for perforated laterals = 2.0 inches <br /> 9. If perforated lateral system is attached to manifold pipe ���e?,M,�;;„,;,,;e° . ' <br /> near the center, like Figure E-2, perforated lateral length (3) � , ; <br /> and number of perforations per lateral(5)will be approximately � ; <br /> I <br /> one half of that in step 8. Using these values, select ��• `���_ . j <br /> minimum diameter for perforated lateral= 1.25 inches. � ' <br /> � _T <br /> I hereby certify that I have compieted this work in accordance with all applicable ordinances, rules and laws <br /> _ / <br /> � (siqnature) ��U (Gcense#) �L�L�UJ�(date) <br /> -- _ - -- <br /> - - - <br /> f'aqe 1 of 1 <br />
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