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2009-00269 - mound system
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2009-00269 - mound system
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Last modified
8/22/2023 5:06:41 PM
Creation date
7/10/2017 11:27:01 AM
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x Address Old
House Number
145
Street Name
Manor
Street Type
Circle
Address
145 Manor Cir
Document Type
Septic
PIN
0411723110024
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University of Minnesota Pressure Dist�ibufion System Design-10/25/04 <br /> a,�boxea re�angles nwwsrme enr�ea,a�msr w�ne ca�. <br /> owsna <br /> Serrwae <br /> 1. Select numbe�of perforated laterals: 0 z'��� �� <br /> 2. Select pertoration spacing= 03 ft <br /> <:;��k��� <br /> 3. Sic�ce pe�Forations shou{d not be piaced cbser that 1 foot to <br /> the edge of the rock layer(see diagram),subtract 2 feet from ""' ;���k <br /> the rock layer ten <br /> P��:.:..�,���-_�...- <br /> 50 -2 ft= 48 ft ^g��,��.�-S <br /> 4. Determine the numbe�of spaces between pe[forations. <br /> Drvide the length(3j hy perfocation spacing(2)and roufld drnim ta ne�arest whole number <br /> Perforafion spacang= 48 ft/ 3 ft= 16 <br /> 5. Salec�pe[forafion sae t!4 inch <br /> 6. Number of perforations is equal to one plus the number af pe►foration spaces(4). <br /> 'Checic frg�ne Eft to assuie fhe number of prerfaua6ons per/ai�l guaranGees <br /> <10%drsChsrge va►ieEiort. <br /> 16 spaces+1= 17 perforafio�s/lateral <br /> E-4 Maximr�m Namber of 7/4 inch perForatior�s E-5 Maximum Number of 3H6 ir�ch perFora�ions <br /> r lateral to uararrb�<1U9e d'es variation r latera[to uarantee<70°Yo disch variation <br /> PerForation Perfocation <br /> Spacing Pipe Diameter Spaang Pipe Dismeter <br /> ft 1 inch 125 inch 1.5 inch 2.0 inch feet 1 inch 1.25 inch 1.5 inch 2_0 inch <br /> 2.5 8 14 18 28 2.5 12 19 25 39 <br /> 3.0' 8> 13- 17 26'; 3 . '!�t i8 24; 37 <br /> 3.3 7 12 16 25 3.3 10 17 23 36 <br /> 4_0 7 44 ` -15 ; 23; 4 i 1C�. i6 2't ':33 <br /> 5A 6 10 _ 14 22 5 ,-g 15 20. 31 <br /> 7. A.Totat number of perforations=perforatio�per IaFeral(s�times number of late�als(7). <br /> 17 pe►fs/lat x 3 laterals= 51 perforations <br /> B.Calculate the square t�age per perioration. <br /> Recommertded vatue is 6-10 sqftfperf.Does not apply to at-grades. <br /> 1. Rodc bed area=rodc width(ft)x rock tength(�t) <br /> 90 ftx SQ ft= 500 fl <br /> 2. Squa[e foot per perfora6on=Rodc Bed Areah�umber of Pe►fs¢6) <br /> 500.0 ft/ 5t perFs = 9.$ f1�7 perf <br /> 8. Determine required flow rate by maittiplyiog the total number <br /> c�f�erforations(6A)L'�Y�P�P�rations see figure E-6) <br /> 51 perts x 0_74 gpm/perfs= 37.7 gpm <br /> E-6 Perfrora4ion Discha e in GPM <br /> Head Perh>rations diameter <br /> feet inches <br /> 3/i$ 7/32 1!4 <br /> 1 0.42 0.56 0.74 . <br /> 2° 0:59 . O.SQ 4:04` <br /> 5 0.94 _ 126 . 1.65 <br /> a_ Use 1.0 foot for single-fampy homes. <br /> b.Use 2.0 f�for else i �=�;^•�,,a°>�.. <br /> l m�c. <br /> 1 .�� �.�� <br /> 9. Defertnine Minimum Pipe Size }w��~��� __». <br /> A. Manifold on End. if taterals are connected to header pipe I `� � � ��.���-m� <br /> as shown in Fgure E-1,to seled minimum required lateral i��..E-,:M�w����a� a <br /> diamefer,enter figure E-4 or E-5 with perforation spaang and <br /> number of perfaratio�s per laberai.Select minimum diame�er <br /> f�r perforaEed laterals= 2.0 inches <br /> B. Cemter Manifold. If perForated fateral system is attached to Mg�E2Nm�NotltoealW _ �;-•""i9C -� <br /> inlhaCwNv�fsSY+�n .�:== � f <br /> manifold pipe near the center,Gke Figure E-2,perforated latera!length(3) __=� �-�-" ' � <br /> and number of perforations per lateral(5)wiil be apprwdmately ���� �`""�-`'�� _ __: <br /> �=="` - =' <br /> one half of that in step A. Using these values,select - �' ___== �,=� <br /> minimum diameter for pertorated lateral= 1.5 inches `M -��� --�`��_�'""`"`` <br /> .:-� � �.«,.��.n <br /> I herety certify that I have campleted this work in accondanee with al[appGpbie oMinances,rules and laws. <br /> ,/� <br /> (signature) 810 (licertse#) Q5/11/08 (date} <br />
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