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2004-P08021 - septic
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3640 Jacobs Mill Road - 32-118-23-24-0010
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2004-P08021 - septic
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Last modified
8/22/2023 4:40:10 PM
Creation date
3/15/2017 10:41:30 AM
Metadata
Fields
Template:
x Address Old
House Number
3640
Street Name
Jacobs Mill
Street Type
Road
Address
3640 Jacobs Mill Road
Document Type
Septic
PIN
3211823240010
Supplemental fields
ProcessedPID
Updated
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PRESSURE DISTRIBUTION SYSTEM -Trenches <br /> �x�k«�� <br /> nw� .0 .. .ve a r <br /> /UI bOX6(�lbCfB/i�/93 RIll3t bB M�I6d 1/iB/95�Wl�b9 Cd�Cl1�8�. . . . . � . q'of aoctc � .. <br /> Psrf 3"izing 3/16'-1/�' <br /> Perf� m.ng 1.3•_S• <br /> 1. Select number of perforated laterals: 03 <br /> 2. Seled perforation spacing= �ft F�t�r�.,�a�.��w.aii�r,a,o.�ao�w. <br /> a.r a�w�a a aaarss amc a�enso.•aa�n <br /> 3. Since perforations should not be placed Goser that 1.0 foot to � <br /> the edge of the rock layer(see dlagram),subtrad 2 feet from ' <br /> the roCk la r len z.s e ,e �e 2e <br /> 63 -2 ft= 61 ft ao e ia n za <br /> fOdc 18y6f i@ngth as � i2 ta 2s <br /> ao � i� is xs <br /> 5.0 6 10 14 72 <br /> 4 Determine the number of spaces between perforations. <br /> Divide the length(3)by perforatbn spaang(2)and round dovm to nearest whole number. <br /> Pe�foration spacing= 61 ft/ 3 ft= 20 spaces <br /> 5. Number of perforations is equal to one plus the number of perforation spaces(4). <br /> •Chedc figure E-4 to assure the number of perforations per lateral guarantees <br /> < 1090 dfscha�ge variation. <br /> 20 spaces+1 = 21 perForations/lateral <br /> 6. A.Total number of perForations=perforations per lateral(5)times number of laterals(1). <br /> 21 perfs/lat x 3 laterals= 63 perforations <br /> E� Pe�o�o�on DfscF�pe in ppm <br /> B.Calculate the square footage per pertoration. <br /> Should be 6-10 sqft/perF.Does not appy to at-grades. �d Peffora�d�ameter <br /> 1. Rodc bed area=rocic width(ft)x rock length(ft) �f�t� I 31 t 6 7I� 1/4 <br /> �o ft x ss ft= s3o ft i.o« o.t a o.a2 0.5� o.�a <br /> 2. Square foot per perforation=Rodc Bed Area/numbe�of perfs(6) <br /> 630.0 ft/ 63 peifs = 10.0 ft/perf 2•ob 0.26 0.59 0.80 1.04 <br /> 5.0 0.41 0.94 1.26 1.65 <br /> 7. Detertnine required flow rate by mu�iplying the total number �Us�a 1.G fo7t fot Singla-f-n.��v n�. <br /> Of perforatiOns(6A)by flow pe1' iforations(see figure E-6) b u+e z.o r��ror� tn e�sQ. <br /> 63 perfs x 0.74 gpm/perfs= 46.6 gpm <br /> 8. If laterals are conneded to header pipe as shown ���--_��� ��,,��,m, I <br /> �_=`�� ��<\I <br /> in Figure E-1,to seled minimum required lateral � �_!�_ � , <br /> diameter;eMer figure E�with perforation spacing(2)and �_� f_--���`,, � <br /> number of perforatlons per lateral(5). �l� a�-':��"r� i <br /> wov.e•�:Mmr�d toodw a�erw a A••+.m <br /> Select minimum diameter for perforated laterals= [�inches <br /> 9. If perforated lateral system is attached to manifoid plpe ��-s:M�+� I,l�i-'"""� <br /> In Gn1�t W M ly�Mn <br /> near the cente�,like Figure E-2,perforated lateral length(3) :� <br /> �„y,��, <br /> and number of perforations per lateral(5)wiii be approximately �� �� �- ���-�' <br /> one half of that in step 8. Using these values,select �= �"� ''� �,� <br /> �i/ � - -.����.� <br /> minimum diameter for perforated lateral= 1.25 inches. ��--=� `�,,,,„��,,, <br /> i hereby certiy that I have completed this work in acxordance wfth all applicable oMinances,rules and laws. <br /> (signature) �gZ- (lioense#) - (date) <br />
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