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r- <br /> . �p �� <br /> University of Minnesota Pressure Distribution System Design - 10/25l04 <br /> All boxed rectangles must be entered,the rest wil!be cakulated. <br /> oRas�re � <br /> 8eww�e ��������'� •- <br /> 1. Se�ect number of perforated Iaterais 03 T�RT+�++�►+* <br /> Prtoa�tww `y�r <br /> 2. Select pertoration spaci�g= � 3�ft <br /> ����..<<�,..;. <br /> -- .., .. . . . . <br /> 3. Since perforations shoutd not be placed doser that 1 foot to - -_ �_. .- <br /> the edge of the rock layer(see d"ragram),subtract 2 feet from i �� �W x rr.w,k- <br /> the rock layer len h ! " ' <br /> 40 -2 ft= 38 ft r���,.�.�4`.5�5•'"" <br /> 4. Determine 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= 38 ft/ 3 R= 12 <br /> 5. Select pertoration s¢e 1 4 inch <br /> 6. Number of perfora6ons is equal to one plus the number of pertoration spaces(4}. <br /> "Check figure E-4 fo assure the number of perfora6ons perlaferaf guarantees <br /> <10%discharge variation. <br /> 12 spaces+1 = 13 perforationsAateral <br /> E-4 Maximum Number of 1I4 inch perforations ES Maximum Number of 3116 nch pertorations <br /> er lateral to uarantee<10°k discha e variation lateral to uarantee<10%dischar e variation <br /> Pertoration Perforation <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 /> .€��' ;::"" � <br /> 3.3 7 12 1& 25 3.3 10 17 23 36 <br /> '� � �z�;�"� �-� �~ �s� �': '� +-.:�*' <br /> ��.� _�}'��s�' �`�fl"�`.�` ��.: <br /> 5.0 6 10 7 4 22 5 � 9 15 20 ^ 31 <br /> 7. A_Total number of perforations=perforations per tateral(5}times number of laterats(1). <br /> 13 perts/lat x 3 laterals= � pertorations <br /> B.Calculate 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 /> �ft x 40 ft= 320 ft <br /> 2. Square foot per pertoration=Rock Bed Area/number of perts(6) <br /> 320.0 ft/ � perts = 8.2 ft/pert <br /> 8. Determine required flow rate by mufGplying the total number <br /> of perforation,s(6A)by flow per perforati�s see figure E-6) <br /> � perfs x 0.74 gpm!perfs= 28.9 gpm <br /> E�Pertoration Dischar e in GPM <br /> Head Pertorations diameter <br /> (feet} inches <br /> 3/16 7/32 1/4 <br /> 1 0.42 0.56 O_74 <br /> �"2" �.��,�` �� ��,.��`� ..:� �.:��=� <br /> 5 . 0_94 126 1.6'5 <br /> a. Use 1.O foot for single-family homes. � <br /> b.Use 20 feet for anything etse � �`°� I <br /> i _ _ _ _ - ' ,J�;��..�,-� <br /> 9. Detertnine Minimum Pipe Size i '.°'�^�_-_=_===`� � __-=_`==' I <br /> A. Manifold on End. If laterals are connected to header pipe ..,,_=="-'��~ ,,<< ,-�,-� � <br /> as shown in Figure E-1,to select minimum required Iateral i��*e E-1:Manibltl lx�ecf at Erb of Sy�tam c• __I <br /> diameter,e�ter figu�e E-4 or ES with perforation spacing and , <br /> number of perforatiorts per lateral.Select minimum diameter <br /> for pertorated�aterals= 125 inches <br /> B. Center Manifold. If perforated fateral system is attached to ! ��E2M��a�� _-- =- �'-- � <br /> i ti�roc�a.uh.snam _^.'" -- I <br /> manifold pipe near the center,like Figure E-2,perforated lateral length(3) i _�,�„„y_ ' _ ; <br /> and number of pertorations per latetal{�will be approbmately _- - " _ - - __-_ -�= t <br /> _ _' !,� ; <br /> one half of that in step A. Using these values,select - ; <br /> = <br /> >.: <br /> minimum diameter for pertorated lateral= �inches j • __--- �_ �'�� ! <br /> , _:- - �_ �._...._ ; <br /> I hereby certify that I have completed this work in accordance with all applipble ordinances,rufes and laws. <br /> (signature) �ol (Cicense#) ��1�T)�(date) <br />