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2011-00348 - new septic
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2011-00348 - new septic
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Last modified
8/22/2023 4:41:51 PM
Creation date
4/12/2017 11:19:34 AM
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x Address Old
House Number
155
Street Name
Kintyre
Street Type
Lane
Address
155 Kintyre Lane
Document Type
Septic
PIN
3211823430016
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ProcessedPID
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• University of Minnesota Pressure Distribution System Design -10/25/04 <br /> M Dozed rsdanpbs mwt be entered,Me�wY be teladated. <br /> Ow�ra <br /> � <br /> 1. Select number of perforated laterals: � '�""�"'"• <br /> 2. Seleet perforation spacing= �3�ft <br /> ...k.,,.�.-�.�....- <br /> 3. Since perforadons should not be pleced doser that 1 foot to ��� <br /> the edge of Ute rodc layer(see diagram),suMrad 2 feat from ""-�-.�.�,`�,..•.`.....`�`�� <br /> {�w�y v of nx�k <br /> PIo 1{7{ll� <br /> 63 -2R= 61 ft �-..�..�.,,...,.��--��,- <br /> f'E f:Y�a�.rry�I.S.S <br /> 4. Determine tha number of spaces between perforations. <br /> Divide the ler�th(3)by parforation spacing(2)and round down to nearest whole number. <br /> PeAoration spaang= 61 ft/ 3 ft= 20 <br /> 5. Seled pe�foration size 1/4 inch <br /> 6. Number of perfora6ona is equal to one plus the number of pertoratan spaoes(4). <br /> •CI►eWc fg�E-4 do assure the number of pe�Iorations per lateral gua�nGees <br /> <10%G�sd�erye veiietion. <br /> 20 spaoes+1= 21 perforations/lateral <br /> E�4 um ber of 114 iach E-5 Ma�dmwn Nwn at 3M61nch pe <br /> laberal to r�ee<10X.tNscha variaUon la�ral to <10M.d�c va�istion <br /> ration <br /> SPa�nB Pipe Dien►ster Spacin� Ppoe D�ameOer <br /> ft 1 inch 1.25 inch 1.5 inch 2.0 irx;h feet 1 inch 1.25 inch 1.5 inch 2.0 inch <br /> 2.5 8 14 1 28 2.5 1 19 25 39 <br /> ��. t��' 13�: ���;:� _.?��:�' ��,'_�- � _�t` -;� ,�4 , a :��=. <br /> � � <br /> 3.3 7 12 16 � � 25 3.3 . , 10 i 17 23 `, �, �36q <br /> �. �'n 1' �y' `(� . 7{ <br /> 1 f!t' , � $T3�.�..t;!��!si r::a�.':.':'d' � ,t•::` �;:'- . ��'; ��� ;'a7+i'.�.:._�. <br /> 5A � 6 10 14 � 22 5 9 � 15 20 '� 31 <br /> 7. A.Tdai number of perforations=perforations per lateral(5)times number of laterals(1). <br /> 21 perfs/lat x 3 laterals= 63 perforetions <br /> B.Cakxrlate the aquare footage per pe�foration. <br /> Reoarunended value is 6-10 sqNperf.Does not appy to ffi-grades. <br /> 1. Rodc bed area=todc width(ft)x rock length(ft) <br /> 10 ft x 63 ft= 630 (t� <br /> 2. � par perforation=Rodc Bed Areainumber of perfs(6) <br /> 630.0 fl/ 63 perts = 10.0 it�/pe�f <br /> 8. Datamine required flow rate by mulGplying the total number <br /> of peAorationa(6A)by flow per perforations see figure E-6) <br /> 63 perfs x 074 gpm/perfs= 46.6 gpm <br /> E-6 Perforatlon D�cha e q� �PM <br /> Flead rlo ons diameter <br /> i�et inches <br /> 3l16 7/32 1/4 <br /> 1 0.4 0. 0.74 <br /> ' .�" ���0.�' `�,°�Q�R ' � ��t.04 <br /> 5 0.94 1.26 1.65 <br /> a ume�.o roa�«�nome�. <br /> b.Uae 2.0 feet far elae .' _ <br /> ;-�/ , <br /> 9. Detertnine Minanum Pipe Siae _.� ., . <br /> A. Maoifold on End. If laterals are connected to heade�pipe ..... ,, <br /> as shown in Figure E-1,to select minimum requi�d lateral •w�••E-,:M���d�«�•� . <br /> diameter,eMer figure E-4 or E-5 with peAoration spaang and <br /> number of perforations per lateral.Seleat minimum diameter <br /> for perforated Is�rals= �inches <br /> B. Ce�er Manifold. If perfo►ated lateral system�attached to _-F=�� ___ _- -- - <br /> manifold pipe near the oenter,like Figure E-2,perforated lateral length(3) �o�»•�•»•r•~„_ _.. : <br /> and number of perforations per Iateral(�wiil be appro�amatey • <br /> one half of thet in atep A. Using tbese valuea selec;t <br /> ec <br /> minimum diartieter for perfora�ed lateral= 1.5 inches • ` Z_ ,, <br /> I hereby aertify that I have oompieted this work in accordance wit all applicable ordinanoes,rules and laws. <br /> (signature) 810 (license#) 11/03/05 (date) <br />
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