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University of Minnesota Pressure Distribution System Design - 10/25l04 <br /> Al!boxed�ectangfes must be entpred,the rest will be calculated. <br /> ��� vince <br /> s�,�� <br /> 1. Select number of perforated laterals: 0 T"""T"�'" <br /> Pwoo�ewue „�, ��' <br /> 2. Select pertoration spacing= �ft <br /> .h���...-�. <br /> 3. Since perforations shou{d not be placed closer that 1 foot to - <br /> the edge of the rock layer(see diagram},subtract 2 feet from l �� <br /> the rock layer len h � ^ r�W -��r�nk � <br /> 25 -2 ft= 23 ft t•�.._:v,.,,3,�,._,„_ <br /> r�r4:,�,.,y i.s._�. <br /> 4. Determirte the number of spaces between perforations. <br /> Divide the length(3)by pertoration spacing(2)and round down to nearest whole number. <br /> Pertoration spacing= 23 ft! 3 ft= 7 <br /> 5. _. h <br /> 6. Number of perforations is equal to one plus the number of perforation spaces(4). <br /> 'Check figure E-4 fo assure the number of perforations perlaterat guarantees <br /> <10%discharge variation. <br /> 7 spaces+1 = 8 pertoratiorisllateral <br /> E-d Maximum Number of 114 inch pertorations ES Maximum Number of 3N6 inch perforations <br /> er lateral to uarantee<1Q%dischar e variation r lateral to uarantee<10%dischar e variation <br /> Pertoration PertoraUon <br /> Spacing Pipe Diameter Spacing Pipe Diameter <br /> ft 1 inch 1.25 inch 1.5 inch 2.0 inch feet 1 inch 1.25 inch 1.5 inch 2.0 inch <br /> 2.5 8 14 18 28 2.5 12 19 25 39 <br /> �34.:�--� .���: ..�.��' .�. ..���;�• �'�;. �"�. �.s�,..Y`.. �11' ' " ,_t$ 24.;.:;. 37� <br /> 3.3 7. ��� 12 � 16 25 3.3 . 10 17 23 � 36 <br /> 40.._.: _�._.���_ ����.��.__ � .,�,''".1 ���.',;:� gr;"` �^_ ^". '; 10 '��.. .. � �'�' � <br /> ..,. ...�.. . � _ ...: <br /> 5.0 6 10 14 22 5 9 15 20 31 <br /> 7. A.Total number of perforations=perforations per lateral(5)times number of laterals(1). <br /> 8 perts/lat x 3 laterals= 24 pertorations <br /> B.Ca�ufate the square footage per perforation. <br /> Recommended value is 6-10 sqft/pert.Dces not apply to at-grades. <br /> 1. Rock bed area=rock width(ft)x rock length(ft) <br /> 10 ft x 25 ft= 250 ft <br /> 2. Square foot per pertoration=Rock Bed Area/number of perts(6) <br /> 250.0 ft/ 24 perts = 10.4 ft�/pert <br /> 8. Determine required flow rate by multiplying the total number <br /> of perforations(6A)by flow}�er perforations see figure E-6) <br /> 24 perts x 0.74 gpm/perfs= 17.8 gpm <br /> E-S Pertoration Dischar e in GPM <br /> Head Pertorations diameter <br /> feet) inches <br /> 3/16 7t32 1/4 <br /> 1 0.42 0.56 0.74 <br /> 2" 0.5C�3 £3.8C) 1.04 <br /> 5 0.94 1.26 L66 <br /> a. Use 1.D footfor single-family homes. � <br /> b.Use 2.0 feet for anything else ___--"-> "'r""�` � <br /> ___ ;. _` -.�,�-.:'.fr.Y I <br /> .s� '� ��`/� <br /> �',. r-,.''_� -_ I <br /> 9. Determine Minimum Pipe Size ; � �--_°'" __--_-' ! <br /> A. Manifold on End. If laterals are connected to header pipe = '- �,-_ :._, I <br /> ,, . .:.... <br /> as shown in Figure E-1,to sel�t minimum required lateral ;F+a�,s E-,:M�,�����a�E„���.«m ; <br /> diameter;enter figure E-4 or E-5 with pertoration spacing and <br /> number of pertorations per lateral.Select minimum diameter <br /> for perforated laterals= �inches <br /> r---- --- -- - <br /> B. Center Manifold. If pertorated laterel system is attached to � ��EZM��a�a� =_ '°'- <br /> Y�Ib tanler r/F�Syslvrr <br /> manifold pipe near the center,like Figure E-2,perforated lateral length(3) , - - <br /> and number of pertorations per lateral(�will be approximatety � " i <br /> one half of that in step A. Using these values,select I•_-` - r " <br /> minimum diameter for pertorated lateral= Dinches I �-- ; <br /> i - <br /> 1 hereby certify that I have completed this work in accordance with aN applicabte ord�nances,rufes and laws. <br /> _;� (signature) �B� (license#) �l o (date) <br />