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2001-P04627 - new septic
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395 Sussex Lane - 04-117-23-24-0007
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2001-P04627 - new septic
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Entry Properties
Last modified
8/22/2023 5:10:40 PM
Creation date
4/4/2019 2:20:54 PM
Metadata
Fields
Template:
x Address Old
House Number
395
Street Name
Sussex
Street Type
Lane
Address
395 Sussex La
Document Type
Septic
PIN
0411723240007
Supplemental fields
ProcessedPID
Updated
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• PRESSURE DISTRIBUTION SYSTEM - Trenches <br /> c::,-.;e.,.:.�.-r.,ti.,� <br /> �r�.�.�_[.�t�e^��[r��..t�n�^T_'t��rc�ms¢i�tnrr� � _ <br /> n�(._cr._.���_�.�.�,�.,�;�xi-,�.F...�i,.�__ - <br /> AU boxed rectang/es must be entered,the rest wip be calculated. � � � �r ,��.�,k <br /> r.�rr s:;,.�.,K�i�F••-i i r <br /> - t'url.`:i.�.�a'��iK 1.4•�S• <br /> 1. Select number of perforated laterals: 03 <br /> 2. Select perforation spacing= 03 ft �E4: Mnknum utl«ido nunbet uf 1!4-inch W�rlaufas <br /> pw wlaa b sR�a�,tea<lox�Ascr.xWe v«Icflia, <br /> 3. Since perforations should not be placed closer that 1 foot to �"�`�°� <br /> �ia <br /> s <br /> the edge of the rock layer(see diagram), subtract 2 feet from L a��n ����:►� �.�st�r,_ ►.s��,�, ?.o��.:n � <br /> the rock la er len th � zs � e ,d � ,a ,a <br /> 75 -2ft= 73 ft � 3.� s �s n se <br /> rock layer length . i '3 ' '? '6 �s <br /> ' 60 7 17 15 23 <br /> l.�5� -_b�._ �� _1Q � 22.^_ <br /> 4 Determine the numbe�of spaces between perforaGons. <br /> Divide the length(3)by perforation spacing(2)and round down to nearest whole number. <br /> Perforation spacing= 73 ft/ 3 ft= 24 spaces <br /> 5. Number of perforations is equal to one plus the number of perforation spaces(4). <br /> 'Check figure E-4 to assune the number of per/orations per lateral guarantees <br /> < 10'/discharge variation. <br /> 24 spaces+ 1 = 25 perforations/lateral <br /> 6. A.Total number of pertorations=perforations per lateral(5)times number of laterals(1). <br /> 25 perfs/lat x 3 laterals= 75 perforations <br /> 10-13 varies �-b: Pe�tora�or�E�lssricw�o hf s�m <br /> B.Calculate the square footage per perforation. "- "-' <br /> Should be 6-10 sqft/perf. Does not apply to at-grades. p��foratron diarrn3ter <br /> ne-ad (iric;hes <br /> 1. Rock bed area=rockwidth(ft)x rock length(ft) ���e�y i� 31�G �f32 }!� <br /> 10 ft x 75 ft= 750 ft2 1.p� o.l s e.a2 G�.SG o.74 <br /> 2. Square foot per perforation= Rock Bed Area/number of perfs(6) <br /> 750.0 ft� / 75 perfs = 10.0 ft/perf 2•Ob G.25 0.59 0.80 t.04 <br /> 5.0 L�.dl Q.94 1.26 1.65 <br /> 7. Detertnine required flow rate by multiplying the total number -�;;5;�t.�;r,__.I i�_i aiir�I�%�!�tffiYy t„_,-,=5. <br /> of pertorations(6A)by flow per erforations(see figure E-6) t';�,�:-.���������<.,..j�,.:t�i�.: ��T°=.. <br /> 75 perfs x 0.56 gpm/perfs= 42 gpm <br /> _ <br /> 8. If laterals are connected to header pipe as shown _ -- ��"-,';' ^�V <br /> in Figure E-1,to select minimum required lateral - `�� <br /> --.: <br /> diameter,enter figure E-4 with perforation spacing(2)and ' �� ` ` -: =�'• <br /> number of pertorations per lateral(5). � . � �`" , i <br /> ,:,. <br /> � Hqur�E-1:MonHdd LxW�d ot End o1 6Y��m ' <br /> � .. _._...__..._.._._._.._.__...__......--....__...---_.._._._�_...._..._------'-- <br /> Select minimum diameter for perforated laterals= NA inches � <br /> -------------------------------._.__...._____ <br /> 9. If perforated lateral system is attached to manifold pipe Na,�E-z:M�ad��� - - •y�=-� <br /> �n tAo Center d p�e 6�nlam .,..:� <br /> near the center, like Figure E-2, pertorated lateral length(3) . - <br /> and number of perforations per lateral(5)will be approximately � 5^ - ="' <br /> one half of that in step 8. Using these values, select - � - ,=. r <br /> minimum diameter for pertorated lateral= 1.5 inches. •` `�---- , <br /> I hereby certify that I ha completed this work in accordance with all applicable ordinances,rules and laws. <br /> �:�� (signature) �Q� (license#) C� �s�� (date) <br />
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