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' PRESSURE DISTRIBUTION SYSTEM Geotextile fabric <br /> 1. Select number of perforated laterals �� � <br /> OUarter inc-h nPrfnr�N., 12� <br /> --- --'�3' <br /> 2. Select perforation spacing = ?;��:? ft 9"oE.=o�k <br /> 3. Since perforations should not Ue placed closer than 1 foot to Perf Sp�g,3 115:'S1/9" <br /> Perf S a g <br /> the edge of the rock layer (see diagram),subtract 2 feet from <br /> the rock layer length. <br /> E-4: Maximum allowable number of 1/4-inch pedorations <br /> 5� �'� � per lateral to guaroi�tee<10%discharge variation <br /> Rock layer length -2 ft --�ft <br /> per(oraBon <br /> 4. Determine the number of spaces between perforations. spocing <br /> Divide the length (3)by perforation spacing (2) and round feef 1 inch 1.25 inch 1.5 inch 2.0 inch <br /> down to nearest whole number. <br /> 2.5 8 14 18 28 <br /> PerforaHon spacing= �'t,� ft= 3.O ft=�spaces 3.0 g 13 �� ��26�� <br /> 5. Number of perforations is equal to one plus the number of 3�3 � 12 �b 25 <br /> perforation spaces(4). Check figure E-4 to assure tlie number of 4�0 � >> 15 23 <br /> perforations per lateral g-uarantees <10% discharge variation. 5.0 6 �p �4 22 <br /> _�_spaces + 1 = � �? perforahons/lateral E-6: Per(oration Dischar e in <br /> 9 9pm <br /> 6. A. Total number of perforations = perforations per lateral (5) <br /> times number of laterals (1) perforation diameter <br /> head inches <br /> r_�s�perfs/lat x ? lat= G�-U perforations �feet) 3/16 7/32 1/4 <br /> B. Calculate the square footage per perforation. 1.0� 0.42 0.56 �0.74 1 <br /> '��-�:� <br /> Should be 6-10 sqft/perf. Does not apply to at-grades. 2•0b 0.59 0.80 1.04 <br /> Rock bed area = rock width (ft)x rock length (ft) 5.0 0.94 1.26 1.65 <br /> l� ft x�_ft =_ �M�'v sqft <br /> Square foot per perforation = Rock Ued area =number of perfs (6) b Use 2.0 feei for nglhino ellsehomes. <br /> .;''�c� sqft= (,�0 perfs = ' ,�) sqft/perf <br /> � MANIFOID LOCATEO 47 END OF pqE55VRE DISTFIBUTION SYSfEM <br /> 7. Determine required flow rate Uy multiplying the total number of <br /> perforaHons (6A) by flow per perforation (see figure E-6) <br /> ��'-#��;a��1, .,.� <br /> ��� perfs x � ��� gpm/perfs =�-�m <br /> �- <br /> 8. If laterals are connected to header pipe as shown on upper j�:,� <br /> example, io select miniinum required lateral diameter;enter �,�����°`"� ��:wp� <br /> figure E-4 with perforation spacing (2) and number of perforations ���`"""d <br /> per lateral (5) Select minimum diameter for <br /> perforated lateral= � ��e8. �..o�,o.renronnreo PiPE LnTERnLS�na <br /> PRESSURE OiSTPIBUIIDN W MOUNO <br /> 9. If perforated lateral system is attached to manifold i e near � � ��~� � �� <br /> P P �..,a.. �� „�y,�w� <br /> the center, lower diagram,perforated lateral length (3) and �Ew =A"%:�""^`�E"��'"" ;-�K"'°"� <br /> numUer of perforations per lateral (5)will be approximately one k�,�:o�,,�,.o„�,„ r�'16`�� <br /> half of that in step £3. Using these values, select mininlum '�• � � <br /> � �._ <br /> diameter for perforated lateral = -�'"�° inches. -'�`^�^�`^-"w <br /> b. <br /> . � 0��![��� n <br /> K"�°�irt r�.,.':"� <br /> `� �[N'M d <br /> I he�eby certify that I have corr►pleted this work in accordance with applicable ordinances, rules and laws. <br /> f , <br /> ��#- ' ,/1 � �,.-' <br /> �-� �_� - (si nature "��, c, � � � <br /> g ) v . � (license#) -1 � �.� _L��} (date) <br />