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Sep 12 06 10: 53a Josh Swedlund (9521873-3292 p. 8 <br /> � , � <br /> � PRESSURE DISTRIBUTION SYSTEM - Trenches <br /> GootezUlc Iabnc <br /> �Ftar1�W\%IAIY I�VI +Mrnt1U':i' <br /> All boxed necfangles must be entergd,the�est will be celculated. 9..��„M.,� <br /> ��rt�z�ng../�6•'-i/r.. <br /> P�.i sF.nc�ng i.S._s• <br /> 1. Select number of perforated laterals� �3� <br /> 2. SeleCt perforation spacing= Oft Fa�ti.,T,��+.n.,�or�ie-�,P«ro.�«. <br /> par la�a b ouQarAee aox�ccf�roe vcrlalfo�, <br /> 3. Since perforations should rtot be placed closer that 1 foot to � <br /> the edge of the rock layer(see diag�am), subtract 2 feet from ��n �. r,a, �s o <br /> the rock la er ien <br /> 50 -2 ft= 48 ft 3o e ia i� � <br /> rock layer length a o � �i ,s i <br /> 5.0 6 10 14 72 <br /> 4 Oetermine the number of spaces between pe�fora6ons. <br /> Divide the length (3)by perforation spacing(2)and round down to near�est whole number. <br /> Perforation spacing= 48 ft/ 3 ft= 16 spaces <br /> 5. Number of pe�forations is equal to one plus the number of pertoration spaces (4). <br /> 'Check fiqure E-4 to assure the number of perforations per lateral guarentees <br /> < 10%discharge variaGon. <br /> 16 spaces+ 1 = 17 perforations/lateral <br /> 6. A. Total number of perforations=perforations per lateral (5)times number of laterals(1). <br /> 17 perfs/lat x 3 laterals= 51 pertorations <br /> E-6: Perfarolfcn Ofscharoe fn tipm <br /> B. Calculate the square footage per per�oration. """ <br /> Should be 6-10 s ft/ erf. Does not a I to at rades. perioration diameter <br /> q p PP Y -9 h�� Inches <br /> 1. Rock bed area= rock width(ft) x rock length (ft) �f�t� 1! � 3 J 16 7132 1/4 <br /> 10 ft x 50 ft= 500 ftZ t.o� 0.18 0.42 0.56 0.74 <br /> 2. Square foot per perforation= Rodc Bed Area/number of perFs(6) <br /> 500.0 ft2 l 51 perfs = 9.8 ft2/perF 2•04 0.26 0.59 0.80 1.04 <br /> 5.0 O.d 1 0.44 1.26 1.65 <br /> 7. Determine required flow rate by multiplying the total number �tJSA 1.OlO:�t l��r 5irr,�E•f,��+�Y���Y��. <br /> of pertorations(6A)by flow per erforations(see figure E-6) � u z.o���r<,�.��� �.��,_ �9;.�. <br /> 51 perfs x 0.56 gpm/perFs= 28.6 gpm <br /> ---... _----�-1 <br /> 8. If Iatecals are connected to header pipe as shown � , ��"` <br /> I ___--_-�_- .�� •Y�.�,: ...w• � <br /> in Figure E-1, to select minimum required lateral .-_- --'-" __:---,��=�'��`m� � <br /> � ,�°' �� I <br /> diameter; enter figure E-4 with perForation spacing (2)and � ' �` --"---- � --`�t•• <br /> ` - - � I <br /> number of perForations per lateral (5). �_��=--`"'�"-� � :`=,^�'::::�'.�^ <br /> 1I flpun E•1:fdmiMfd tooet�ei m End o1 BysMm . . .. i <br /> I�_--- -_ --------......._--------� <br /> Select rninimum diameter for perforated laterals= �inches <br /> 9. If perforated lateral system is attached to manifold pipe Flp�r�E•2MoMlddlooul�d ,___:.�s%'��°`�""'�' <br /> In1M CMMr of Ik�6yNm �_.- <br /> near the center, like Figure E-2, perforated lateral length (3) -<=��"�� _ - <br /> .�:.���;,-�..�:<,_._�-'"" _.,. <br /> and number of perforations per fateral(5)will be approximately �,_.�,-�-= ::,�,.�---� _ _ -- <br /> one half of that in step 8. Using thess vatues, select -���- _- ' � `<--<--";�.;».:...:Y <br /> minimum diameter for perforated lateral= C�inches. �`'�� _ . _ ---� �--���� �4 <br /> �=-=_ ..,.,,....., <br /> I here rti at I hav mpleted this work in accordance with aN applicable or8inances, n�les and laws. <br /> (signature) ��02 (license#) / (date) <br />