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' Univ�rsity of Minnesota Pressure Distribution System Design - 10/25/04 <br /> an nwr.a rerxangks m�sr be e�oa nb rat wa►ae�,,a�ea <br /> Ows�s� <br /> sewwse <br /> 1. Sele��tu�ilbet Of perfof8ted 12tt6rels: � r�wa�i`� <br /> 2. Select perforation spacing= 03 ft <br /> �.,.k�x,�. <br /> 3. Since pe�orations should not be plaoed doser tl�at 1 foot to <br /> the edge of the rock lay�er(see diagram).subtract 2 teet irom r"'�'�."".�"•�•"...�'m""T <br /> M� !/.+f mck <br /> �,�Jl�� <br /> 63 -2 ft= 61 ft a'..�:''.'" ��M_��.- <br /> I'�(sfwc� /.5•_5• <br /> 4. Determine the number of spaces be4vMeen perforations. <br /> Divide the length(3)by perforation spacing(2)and round dawn to nearest whole number. <br /> Perforation spacing= 61 ft/ 3 ft= 20 <br /> 5. Seled perforation size 7�32 inch <br /> 6. Number of perforations is equal to o�e plus the number of perfwation spaoes(4). <br /> 'Chedc figure E•4 do assure Hie number of pe�fioratiorts per laterai gua�antees <br /> <10%�Sd�e/ge ve/iedon. <br /> 2p spa�s+1= 21 perforations/lateral <br /> E�Yaxitnum Number of 1l4 inch perforations E�b Maxtrnum Number of 3/16 inch pe�torallons <br /> labe�d to c10X d variatlon lateral to <10X�scha variation <br /> Perforadon Perforation <br /> SPacin9 P�pe D�me�r Spaa�9 P�e Diamed�r <br /> ft 1 inch 1.25 inch 1.5 i�h 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 11 18 24 37 <br /> 3.3 7 12 16 25 3.3 10 17 23 36 <br /> 4.0 7 11 15 23 4 10 16 21 33 <br /> 5.0 6 10 14 22 5 9 15 20 31 <br /> 7. A.Total number of perforations=perforations per lateral(5)times number of laterals(1)- <br /> 21 pertsJ tat x 3 laterals= 63 perforations <br /> B.Ca�ulate the square footage per perfaation. <br /> Reoanmended va�ue is s-�o sywperf.ooes na apply to at-grades. <br /> 1. Roac bed area=rodc width(Rj x rodc ler�gth(R) <br /> 10 ft x 63 ft= 630 ft� <br /> 2. Square fo�per pertoration=Rodc Bed Area/number of perfs(6) <br /> 630.0 ft! 63 perfs = 10.0 It�/perf <br /> 8. Detertnine required flow rate by muitiplying the to4�al number <br /> of perforatio�s(6A)by flow per perfaations�_see figure E-6) <br /> 63 perfs x 0.56 gpm/perfs= 35.3 gpm <br /> E�'i Perforatlon Discharge In GPM <br /> Head Perforations diameter <br /> feet inc�es <br /> 3/16 7J32 1/4 <br /> 1 0.42 0.56 0.74 <br /> 2° 0.59 0.80 1.04 <br /> 5 0.94 1.26 1.65 <br /> a Use t.o tooe tor angladmniy homes. ----- --� <br /> b.Use 2.0 teet for else � _=�� �„`." � � <br /> �-��_ '=!t�'� <br /> 9. Determine Minimum Pipe Size �"'.,`"' ._---_�-� _--'• <br /> .� <br /> A. Manifdd on End. If laterals are connected to header pipe � _ -- �-=-_, �.,- �r. <br /> .r- <br /> �`-,��_��.,�.. <br /> ,. ..�. . . <br /> as shown in Figure E-1,to selec!minimum required lateral _f�..F�_M����aE�«,,...M, <br /> diameter,enter figure E-4 or E-5 with perfaation spacing end <br /> - ------------ ___----- _ _ <br /> number of perforations per lateral.Sefect minimum diameter <br /> for perforated taterals= 2.0 inc�es <br /> B. Cenber NanifOld. If perfOrefsd Hdberel systeln is ettached to a�c-:..a�.,dwoa..a ----- � �� - <br /> In CrMr ef M MI�m ,.--'' --..� <br /> manifold pipe near fhe oenter,like Figure E-2�Perforatad lateral len9th(3) ,=s-�:;.;;,,_ _.._� `��f <br /> and number of perforations per Iateral(�wi11 be approbmatey ,_- _� �= _ - -= <br /> one haH of that in sbep A. Using these values seiec.t - _ � __ <br /> T� <br /> minimum diameter for perforated lateral= 2.0 inches : �-�..__�v� <br /> I here that I have completed this work in accordance wfth all applicable ordinances,rules and Iaws. <br /> (signature) 810 (license�) 05/18/07 (date) <br />