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2010-00294 - septic
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650 Big Island - PID: 22-117-23-31-0035/0028
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2010-00294 - septic
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Last modified
8/22/2023 4:12:11 PM
Creation date
4/18/2016 2:04:36 PM
Metadata
Fields
Template:
x Address Old
House Number
650
Street Name
Big Island
Address
650 Big Island
Document Type
Septic
PIN
2211723310035
Supplemental fields
ProcessedPID
Updated
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�� ���-1�c�l�( <br /> PRESSURE DISTRIBUTION SYSTEM <br /> � Geotextile fabric <br /> 1• Select number of perforated laterals �.'� <br /> ---- Quarter inch perforaHons s aced�3' lZ� <br /> 2- Select perforation spacing - ft y„of'ro�k <br /> 3. Since perforations should not be placed closer than 1 foot to I'erf Sizing 3/16•�_l�q•• <br /> the edge of the rock layer (see diagram),subtract 2 feet from Perf Spacing 1.5�-s• <br /> the rock layer length. <br /> � E-4: Maximum allowable number of t/4-inch perforations <br /> T `' -2 ft - i per lateral to guarantee<10%discharge variation <br /> Rockl ayer lenglh �' - f( <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 /> Perforation s acin "� • -s 2'S 8 14 18 28 <br /> P g = �-� ► ft— ft= �{.� spaces 3.0 8 13 17 <br /> 5. Number of perforations is equal to one plus the number of 26 <br /> 3.3 7 12 16 25 <br /> perforation spaces(4). Check figure E-4 to assure the number of 4'0 � >> 15 <br /> perforations per lateral guarantees <10% dischqrge variation. 23 <br /> 5.0 6 10 �q 22 <br /> _ 1� spaces + 1 = 1�perforations/lateral <br /> E-6: Pertoration Discharg�in gprn <br /> 6. A. Total number of perforations = perforations per lateral (5) <br /> times number of laterals (1) perforation diameter <br /> head inches <br /> �_perfs/lat x " lat= ;_"�; perforations (feet) 3/16 7/32 1/4 <br /> B. Calculate the square footage per perforation. �•�� 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) <br /> '�=? ft x .�`r: ft= :�.;. �-,� 5.0 0.94 1.26 1.65 <br /> ______sqft <br /> Square foot per perforation = Rock bed area =number of perfs (6) b Use 2.0 feet for ong'hina ellsehomes. <br /> '� `�' sqft— erfs = � <br /> P �" sqft/perf <br /> 7. Determine required flow rate b j]ZL]itl 1 ]� MGNIFp�D IOCATEO n7 END OF pqESSUqE OISTRIBUTION SYS(EM <br /> perforations (6A) by flow per perforat on (s�ee figuretE-6�ber of <br /> �1 �<< <br /> A ,�,,Y`� W„„<„ <br /> ,;,, r ., <br /> <� �� perfs x � �.... �m/perfs = �.L� m <br /> gP �. <br /> 8. If laterals are connected to header pipe as shown on upper °' <br /> example, to select minimum required lateral diameter;enter /�'-� <br /> d��,��m�"`"� ;;�'�wa...� <br /> figure E-4 with perforation spacing (2) and number of perforations <br /> per lateral (5) Select nunirrium diameter for ����TM <br /> perforated lateral= "�. ���eS. ur��r o�Pea�o�.Eo PIpE LCTEfl4�3 fDH <br /> PNESSURE D�STpIB�T10N W MOUNp <br /> 9. If perforated lateral system is attached to manifold i e near � � �� ����� <br /> the center,lower diagram,perf P p �•`"`p^;� �°.A�,� �, ,,,�y,°^� <br /> orated lateral length (3) and �Ew o;��;`�'�"�°£�`°-�' ,- �� °^ <br /> numUer of perforations per lateral (5) will be approximately one Y"°°�`° <br /> half of t�lat in step 8. Using these values, select mi.n_unum n '"''"`''�"°°"�" <br /> b. � a_ � <br /> diameter for perforated lateral = --------�-�nches. � ,,,,.,, <br /> b.. <br /> � ,,,,�.•� � <br /> Kn,e„✓° °,,,;,,nn- <br /> ���N6M� <br /> I hereby certify that I have completed this work in accordance with applicable ordinances, rules and laws. <br /> /``���� (/� � ��J �� ......_..-_._.._.. <br /> • (signature) `"'`���-} <br /> �� � (license#) �'l--o��-}_�,^� (date) <br />
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