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f'Ri SSURI: O1SI_R(UU f fON SYSJiM cHHO r•crtr-ort.,rlcx, or
<br /> r'CRrgrtnrro I Al,r,.
<br /> I Select number of perforated laterals �v;
<br /> ,:IfGrata c.,„„,
<br /> loos; f
<br /> • t.00mr Son4 L �r<, °,
<br /> 2 Select perforation spacing _ ?i 5 feet. �r< .,,,, 0,+1<<m< roo,i< ...
<br /> � w _
<br /> ,r..-.r- --I •ill"r<md rnln ... , a- tr.-..
<br /> L ^I r•O�<r 1 r,
<br /> •000000 Lon Lae
<br /> 1 • I 1 C-o° taco,IiIrJ .rn..:'.
<br /> 3. Since perforations should not be placed closer than I ft. to C, . f_ _"° °
<br /> �_=i
<br /> the edge of the rock layer (see diagram), subtract 2 ft. from �c244, `- _� <- ;; Ro°,'r,;c;°
<br /> the rock layer length P«ro<°noni uscoi n 0,
<br /> CI<°n sone Lo, Oonom °, 1oa„,,,
<br /> ft. _ -70, 5 feet.
<br /> 7O<,«,r.ni Soq Prop«I, Scour,<J
<br /> F.a_t layer h%th -- - B<r a PIOU"p Sond l°,<,
<br /> 4 Determine the number of spaces between perforations '-"b;"r°at:on°art:
<br /> C Pu mn.,,<(dVmt
<br /> Divide the length above by perforation spacing and round °H<.A=`
<br /> down to nearest whole number. _ __
<br /> 1.0a 0.56 0 7,1
<br /> Length pert- spacing = 7R_Sft. _;,5 ft = 29 spaces
<br /> 2.0b 0.80 l.04
<br /> (3) (2) a Usc for single family homes
<br /> b Use for all otherappllcel.ons
<br /> 5. Number of perforations is equal to one plus the number of
<br /> perforation spaces
<br /> O Maximum number of quarter inch t�erfc�rahot 6
<br /> . 2_._ spaces 4 1 = 3- perforations/lateral Lateral toguarnantee < 10%discharge
<br /> Perforation
<br /> 6 Multiply rperforations lateral number of laterals to spacing i
<br /> P } perby (feel) ( < (2 2
<br /> get total number of perforations. - 2.5 14 18 28
<br /> 303.0 13 17 26
<br /> t,lcrals x rc,rs/ cral A 0 perforations.
<br /> 3.3 12 16 25
<br /> 7. Determine required flow rate by multiplying 4.0 11 15 23
<br /> number of perforations by flow per perforation 5.0 10 14 22
<br /> 9ox 0,7y �Gt°'(0 gpm. .•-,-moo s«„,m .r c.o a •�rn,<e asrnr°v<a.r M.t-
<br /> ric gpm/pe
<br /> 8. If laterals arc connected to leader p )e as shown on upper --'��
<br /> example, to select minima 1 re red lateral diameter; enter
<br /> table with perforation spac�n„ nd number of perforations <--"• ` "
<br /> per lateral. Select minimur• diameter for
<br /> perforated lateral -: _ _. ,aches.
<br /> •
<br /> U.a,< M KNO".<O2 Nt....e."..Ni
<br /> ...C..,a tr�.,c.o.. w
<br /> 9 (f perforated lateral system is attached to manifold pipe near ni= =�'7"" �'�M
<br /> the center, lower diagram, perforated lateral length and - ---t-:.77.:-A—- . ---
<br /> number of perforations per lateral will be approximately one
<br /> half of that in step 8 Using these values, select minimum
<br /> diameter for perfoiatrd lateral =. rc? . _ inches. \- ,,.--
<br /> N
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