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PRESSURE DISTRIBUTION SYSTEM <br /> �b��.� <br /> : �� <br /> Atl boxed redang/es must be entered,the iest wil!be calculated. '""a"""'�" ' <br /> . � : . :.. 9'.�of aacl� . .. �� . <br /> f+aaf SiainR 1/I6'-'1/4'. .� <br /> 1. Select number of perForated laterals: � p"=�s`"`�"'g'.S•_s� <br /> 2. Select perforation spacing= 2.5 ft E;���Q�����4��� <br /> . par fabd b fltaranbs d0%Weiwye variabn <br /> 3. Since perforations should not be placed cioser that 1 foot to i°�Q1O�O" �r� i. - <br /> the edge of the rock layer(see diagram), subtract 2 feet from �� � � �, <br /> the rock la er len th <br /> 75 -2.5 ft= 72.5 ft ��� 8 ia ia �a <br /> ao s » » � <br /> roCk layer length a.a � i2 ib as <br /> ao � i> >s r3 <br /> 5.0 b 10 74 � <br /> 4 Determine the number of spaces between perforations. <br /> Divide the length(3)by perforation spacing(2)and round down to nearest whole number. <br /> Perforation spacing= 72.5 ft/ 2.5 ft= 29 spaces <br /> 5. Number of perforations is equal to one plus the number of perforation spaces(4). <br /> 'Check figure E-4 to assure the number of perforations per lateral guarantees <br /> < 10%discharge variation. <br /> 29 spaces+ 1 = 30 perforations/lateral <br /> 6. A.Total number of perforations=perforations per lateral(5)times number of laterals(1). <br /> 30 perfs/lat x 3 laterafs= 90 perforations <br /> ..,.__._.__. <br /> E-� Aettbro�on D�rqe h ppm <br /> B. Calculate the square footage per perForation. ..�_ <br /> Should be 6-10 sqft/perf. Does not apply to at-grades. i�f���d�neter <br /> 1. Rock bed area= rock width (ft)x rock length(ft) �� <br /> hec�d ! 3Jlb 7j32 1/4 <br /> 10 ft x 75 ft= 750 ft{��� <br /> 2. Square foot per perforation=Rock Bed Area/number of perts(6) �'� �•�$ O.d2 0.5b 0.74 <br /> 750.0 ft/ 90 perFs = 8.3 ft/perF 2� 0.26 0.59 v.80 1 A4 <br /> 5.0 O.dl 0.94 l2b I.bS <br /> 7. Determine required flow rate by multiplying the total number <br /> a u�e a.aroo�rcrsrn�ta.r�r�r rkxne,. <br /> of perforations(6A)by flow per erforations(see figure E-6) b raQ ,�, �. <br /> 90 perfs x OJ4 gpm/perfs= 66.6 gpm <br /> 8. If laterals are connected to header pipe as shown J _,_.s.��„d,�p,� <br /> in Figure E-1, to select minimum required lateral j __�.��_--'""�'"����� ` """""""� <br /> diameter, enter figure E-4 with perforation spacing(2)and � °"=� =� � <br /> -.a �.,..:..- <br /> number of rforations `�J� - �----"� �_,• <br /> pe per lateral(5). � sr_�;-- „r„�,�,�, <br />� vr-na1.:c��rip <br /> Rpu�E 1:ManpoW iooabd ot ErfO W Sysf�n <br /> Select minimum diameter for pe�Forated laterals= C�inches <br /> 9. If perforated lateral system is attached to manifold pipe �„E.rM�,,b,,,�,«,., ,,,v�„ <br /> M m.c.n�«a Mf.av.�«» �-�� <br /> near the center, like Figure E-2, pe�forated lateral length(3) '" _,_...--�.-_--� �� <br /> and number of perForations per lateral (5)will be approximately _....---�:�'~""""°`"° �� �. <br /> _.._- , �- <br /> one half of that in step 8. Using these values, select (%���JJ~~�J'~ „��`���,,, �� <br /> _:...�-- ��---;�.a�.� <br /> minimum diameter for perforated lateral= 1.5 inches. �='�'" �,.f1�=""�_w'r�`^""` <br /> -�" r�c+x.w.nn» <br /> �:r <br /> I hereb certify that I have completed this work in accordance with all applicable ordinances, rules and laws. <br /> (signature) 3t�3� (license#) ��y'�3 (date) <br />