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,ONarre <br /> SewAce Geotextile fabric j <br /> TREATMENT � <br /> PRocrtaM - . <br /> PRESSURE DISTRIBUTION SYSTEM Quarter inch ertorations s aced 3' 12 � <br /> „ <br /> 1. Select number of perforated laterals� ' 9 of rock <br /> �' <br /> 2. Select perforation spacing = � ft Perf Sizing 3/16' - 1/4" � <br /> Perf Spacing 1. ' -3 <br /> 3. Since perforations should not be placed closer than 1 foot to <br /> the edge of the rock layer (see diagram), subtract 2 feet from E�: Maximum allowable number of 1/4-tnch perforattons <br /> per lateral to guarantee<107e discharge variatlon <br /> the rock laye . <br /> perforetfon <br /> Roc�� i.7�/ spacing <br /> length � 2 ft = r ft feet 1 inch 1.25 inch 1.5 inc 2.0 inch <br /> 4. Determine the number of spaces between perforations. e ,a 2a <br /> Divide the len th 3 b erforation s acin 2 and round - 3.o a ,s ,� zs <br /> g ( ) Y P P g ( ) s.s � ,2 ,s 2s <br /> down to nearest whole number. a.o � „ ,s ss <br /> -/ S � �3 �_ /� 5.� 6 ,� ,4 22 <br /> Perforation spacing= spaces <br /> 5. Number of perforations is equal to one plus the number of E-s: Pertoratton o�scnar9e in 9Pm <br /> perforation.spaces(4). Check figure E-4 to assure the number of <br /> perforations per lateral guarantees <10°/o discharge variation. perforation diameter <br /> head inches <br /> � spaces + 1 = � perforations/lateral (fee 3/16 7/32 1/4 <br /> 6. A. Total number of perforations = perforations per lateral (5) �•Oa 0.42 0.56 0.74 <br /> times number of laterals (1) 2.Ob 0.59 0.80 1.04 <br /> ��perfs/lat x � lat= � perforations 5.0 0.94 1.26 1.65 <br /> � B. Calculate the square footag� per perforation. a Use 1.0 foot for single-family homes. <br /> Recommeded value is 6-10 sqft/perf. Does not apply fo at-grades. °use 2.o ree�ro�a� n�� e�se. <br /> Rock bed area= rock width (ft) x rock length (ft) <br /> ��_ft x .Sd it = or� sqft <br /> Square foot per perforation = Rock be area =number of perfs (6) . <br /> � v�d� sqft=�perfs = �� sqft/perf <br /> �O. mandol0 ppe <br /> 7. Determine required flow rate by multiplying'the total number of �'"�� � <br /> pipe Irom Dump <br /> perforations (6A) by flow per perforation (see figure E-6) -''��� <br /> anO cap .l � � <br /> � '/ ` <br /> � perfs x �� m/perfs = � D gpm .r �{D q � <br /> �t „� <br /> J atlemaie iocation <br /> ol p�ps Irpm pump <br /> 8. If laterals are connected to header pipe as shown on upper Figure E-1:Manifold Located at End ot System <br /> example, to select minimum required lateral diameter; enter <br /> figure E-4 with perforation spacing (2) and number of perforations <br /> per lateral (5) Select minimum diameter for Figuro E-2:Manitold Loeated ='l'°""'0 <br /> in tha Cantar of tha Systam " � � <br /> perforated lateral = T_ inches. ;—=%" _�—' <br /> �,.��. i=� <br /> / \�`/y ,—.'' <br /> 9. If perforated lateral system is attached to manifold pipe near � c.f o'�� � .` ;_ —' <br /> the center, lower diagram, perforated lateral length (3) and �o�, �a�, <br /> number of perforations per lateral (5) will be approximately one i >j ,,,.,,�,�„� <br /> half of that in step 8. Using these valu$�, select minimum Z`� <br /> diameter for perforated lateral = l ��'Z- inches. � <br /> �' e e ce tify that I have completed this work in accordance with applicable ordinances, rules and laws. <br /> . (signature) � J Z (litense#) 2� � (date) <br />