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PRESSURE DISTRIBUTION SYSTEM - Trenches <br /> C:a-c.ecx�ilo f hnc <br /> ��tW� Ylar� AT <br /> All boxed reCtangles must be entered,the rest will be celCulated. ���,�k <br /> r�r�s:<:.yz�i�b^-i ia' <br /> r«r��..es�s._5. <br /> 1. Select number of perForated laterals: 03 <br /> 2. Select perforation spacing= �ft F���++�o���i¢Khv«b� <br /> par blsd b puQaiea c10%d�et�p�valaNan <br /> 3. Since perforations should not be piaced closer that 1.0 foot to �,� <br /> the edge of the rock layer(see diagram), subtract 2 feet from ' " ' '. "'°" <br /> the rock la er len th z.s e �a �e �e <br /> 63 -2 ft= 61 ft a.o a ts i� za <br /> rock layer length '' � �� 16 � <br /> 4.0 7 17 15 29 <br /> 5.0 6 1C 14 72 <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= 61 ft/ 3 ft= 20 spaces <br /> 5. Number of perforations is equal to one plus the number of perforation spaces(4). <br /> *Check figure E-4 to assur�e the number of perforations per lateral guarantees <br /> < 10%discharge variation. <br /> 20 spaces+1 = 21 perforations/laterai <br /> 6. A.Total number of perforations=perforations per laterai(5)times number of laterals(1). <br /> 21 perfs/lat x 3 laterals= 63 perforations <br /> E6: Perforolfon pscharpe in flpm <br /> B.Calculate the square footage per perForation. pertorotion diameter <br /> Should be 6-10 sqft/perf. Does not apply to at-grades. h�d �,xh� <br /> 1. Rock bed area=rock width(ft)x rock length(ft) ���t� t I 3/16 7132 1/4 <br /> 10 ft x 63 ft= 630 ft� t.oa 0.18 0.42 0.5b 0.74 <br /> 2. Square foot per perforation=Rock Bed Area/number of perfs(6) <br /> 630.0 ft� / 63 perfs = 10.0 ft�/perf 2.Ob o.26 0.59 0.8o t.oa <br /> 5.0 0.41 0.94 1.26 1.65 <br /> 7. Determine required flow rate by multiplying the total number U�9�+�.or��a�..���.�;�-r m��,r,-nkg. <br /> of perForations(6A)by flow per erforations(see figure E-6) C��9�z.or:��r-��.:,.�,r,f, .��.. <br /> 63 perfs x 0.74 gpm!perfs= 46.6 gpm <br /> 8. If laterals are connected to header pipe as shown ( -= ""'r""`� ^' <br /> in Figure E-1,to select minimum required lateral i __- - ``�-� r^ <br /> diameter,ente�figure E�with perforation spacing(2)and i '' - ____ ��•�" ' <br /> number of perforations per lateral(5). , �>,:=_= �<<_- ,,-� <br /> , ._,....., , <br /> L_Rqu�E-1:ManHbld lxal�d af End d SysNm �. <br /> -__--..-_ -- ---___-__J <br /> Select minimum diameter for perforated laterais= NA inches <br /> 9. If pertorated lateral system is attached to manifold pipe �E-Z:M�� _ :- '� s <br /> n c.nwe a�,.MM* -- <br /> near the center, like Figure E-2, perforated lateral length(3) --=_-`�" � _ -_ -�' <br /> and number of perForations per lateral(5)will be approximately __.. � �.��,_ - <br /> one hatf of that in step 8. Using these values,select - " t `� , <br /> minimum diameter for perforated lateral= 1.5 inches. � �- <br /> ,.�., ..�, <br /> I hereby certify that I have completed this work in accordance with all applicable ordinances, rules and laws. <br /> (signature) L g� (license#) Z`Z��� (date) <br />