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f i I LI. <br /> PRESSURE DISTI I13UTION SYSTEM <br /> �G�ross Cover <br /> � epsell, ���?k,"'� <br /> 1. Select number of perforated laterals 3 i ,._--_ MT_-- <br /> Layer of Geoleelile Fabric r- <br /> • Loamy Sand Layer .7Inch layer of hay or evu" <br /> 3 feet . . . with o rosin paper) <br /> 2. Select perforation spacing 44 --ig .__.t.-Perla atlon Or Bled Mont. <br /> /I., Info C p Near Top <br /> :�'/i Pluf •I �-{-Al Leoet I2'to E• <br /> t_Urain Fseld Ro i j( of Rock Layer <br /> 3. Since perforations should not be placed closer than 1 ft. to --` , <br /> -Perforations Located of <br /> the edge of the rock layer (dee diagram), subtract 2 ft. from Clean Sand Layer Bottom of Lmerol <br /> the rock layer length. ....,..,_____„... . : <br /> • <br /> Solt Properly ScarIRed <br /> c0Before Placing Sand Layer <br /> Rock layer length 2 ft. =Alt feet <br /> TABLE OF PERFORATION DISCI FARCES IN C <br /> 4. Determine the number of s aces between perforations. <br /> P Head Perforation diameter(inches) <br /> Divide the length above byperforation spacing and round <br /> down to nearest whole nun ber. 732 '/e <br /> o 1.0a 0.56 0.74 <br /> g = <br /> Length erf. spacing l'0 ft. = 3 f t. = l� spaces1.5 0.69 0.90. <br /> g P P (#3 (#2) 2.0b 0.80 1.04 <br /> 2.5 . 0.89 1.17 <br /> 5. Number of perforations is a ual to one plus the number of a 0 0.98 1.28.13 <br /> 7 <br /> perforation spaces . 5.0 1.26 1.65 <br /> • <br /> aUse 1.0 foot of head for residential systems. <br /> / spaces + 1 = )7 perforations per lateral bUse 2.0 feet of head for othereslablishments <br /> I • <br /> • 6. Multiply perforations per 1 teral by number of laterals to <br /> get total number of perforate ns. <br /> Table 2 <br /> ' X - g� �erfuratiolls Maximum allowable number of minter inch perforations per <br /> laterals peals/lateral 1 lateral to guarantee<10%Discharge variation <br /> perfecalln",peeing 1.25 inch 1.5 inch 2.0 in <br /> (reel) <br /> 7. Determine required flow rat by multiplying 2.5 14 18 28 <br /> number of perforations by fl?w per perforation 3.0 . 13 17_- 26 <br /> 33. 12 16 25 <br /> 4.0 11 15 23 <br /> :I <br /> LI 7[/ 5.0 10 14 22 <br /> pets x gpm/Pert .I <br /> 3�• gpm. <br /> 8. If laterals are connected to h ader I e as shown on u "hil a,a.ac.r,e Imp ter""� ,�'"�,�,'""" <br /> P P <br /> � PPer <br /> example,select minimum required lateral diameter from --4- <br /> table 2; enter table with perforation spacing and number --�-- <br /> perforations per lateral. Select minimum diameter for • �.,. 1.......:-.......>„- <br /> ofperforated lateral = i>iclies ,,,.- >J- <br /> ------ <br /> 9. If perforated lateral system isattached to manifold pipe near <br /> the center, as in lower examp e, perforated lateral length and --•;.��_n.,re--r,..,..,.- <br /> number of perforations per lateral will be approximately one cr" <br /> w= <br /> half of that in # 6. Using these values, select minimum .1,11 E4: -,.,-,-;' ,, <br /> f <br /> ...... .� <br /> diameter for perforated lateral from table 2 cr->»..... <br /> • perforated lateral = ) ,C inches r4�' r' ' <br /> ','. <br />