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A. Determine pump capacity: <br />Gravity Distribution <br />1. Minimum suggested is 600 gallons per hour (10 gpm) to stay ahead of <br />water use rate. <br />2. Maximum suggested for delivery to a drop box of a home system is 2,700 <br />gallons per hour (45 gpm) to prevent build-up of pressure in drop box. <br />Pressure Distribution <br />3. a. Select number of perforated laterals - <br />b. Select perforation spacing - --- ft. <br />c. Subtract 2 ft. from the rock layer length. <br />so up WW -2ft. = ft. <br />d. Determine the number of spaces between perforations. <br />Length perf. spacing i. ft. t__ ft-= --- @paces <br />e. spaces a 1 = . perforations/I. teral <br />f. Multiply perforations per lateral by number of laterals to <br />get total number of perforations. <br />t .. . x Fw = perforations. <br />g, -;= xIPW?PW . gpm. <br />SELECTED PUMP CAPACITY L" gpm 1' ) 4A- <br />S. Determine head requirements: <br />I. Elevation difference between pump and point of discharge. <br />_(` feet <br />2. If pumping to a pressure distribution system, add five feet for pressure <br />required at manifold <br />- feet <br />3. Friction loss <br />a. Enter friction loss table with gpm and pipe diameter. <br />Read friction less in feet per 100 feet from table. <br />F.L. a d ft./100 ft of pipe <br />b. Determine total pipe length from pump to discharge <br />point. Add 25 percent to pipe length for fitting <br />loss, or use a fitting loss chart. Equivalent pipe <br />length -1.25 times pipe length • <br />x 1.25 - feet <br />c. Calculate total frictiot lose by multiplying <br />friction loss in ft/100 ft by equivalent pipe length. <br />Total friction loss x •100 • feet <br />4. Total had required is the sum of clevation difference, <br />special head requirements, and total fnction loss. <br />ll) (2) + (3c) <br />TOTAL HEAD feet <br />C. Pump selection <br />1. A pump must be selected to deliver at least <br />with at least feet of total head (Step B). <br />so PW WArtetl a A 10- w aTf1. 1AIOIM. <br />-.0-a1M am <br />M <br />10%V; Old � <br />TAnA Oe rwaMATM ONCHAOt".t• M C M <br />And PiNgs lua Nenrtr ondma <br />'/a <br />'/• <br />lam <br />096 <br />ON <br />13 <br />so <br />too <br />2Ab <br />M <br />1AI <br />23 <br />so <br />1.17 <br />3.0 <br />OA <br />Im <br />4.0 <br />1.13 <br />to <br />9.0 <br />I m <br />US <br />arts 1.0 hat d hnd hr wdrarlt rl�a <br />(bust 2.0 hltdlutl ograttisraslrMrtianar <br />gpm (Step A) <br />PII)O LAMO <br />Pais( of Db*Awp <br />Ekvatkm DiRatem <br />K� <br />F-1!b <br />1.5 iasti 2.0 iadt 10 IN A <br />i1� 111rt�lrrp►tMaf/� <br />10 <br />0.69 <br />0.20 <br />12 <br />0.96 <br />0.28 <br />14 <br />1.n <br />0.38 <br />16 <br />1.63 <br />0.48 <br />11 <br />2.03 <br />0.60 <br />20 <br />2.47 <br />0.73 <br />all <br />25 <br />3.73 <br />1.11 <br />US <br />30 <br />3.23 <br />1.33 <br />0.23 <br />35 <br />7.90 <br />2.06 <br />0.30 <br />43 <br />0.w <br />14.73 <br />. j& , <br />so <br />3." <br />0-W <br />Ss <br />4.76 <br />0.70 <br />60 <br />3.60 <br />0.82 <br />