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PRESSURE DISTRIBUTION SYSTEM Geoteati�e t��i�i� <br /> 1. SC'12Ct I1L1LT12J2C Of PC'rfOLdtCC� Ic1feT11S_.�_ Quarter inch �erlorat�ons��.�ced�;' 1� <br /> �"Ot�iOCk <br /> 2. Select perforation spacing =�_tt <br /> PerE Siticig�/16"- I/3'� <br /> 3. Since perforations should not be placed closer than 1 foot to P"`sE"`i"�; 1.`'�-'' <br /> the edge of the rock layer (see diagram), subtract 2 feet h-om <br /> the rock layer length. E-4: Maximum aliowable number of 1 Jd-inch perforcrtions <br /> � <br /> per laferaf to guarantee<10°/o discharge variation <br /> -2 ft =_J�_ft <br /> i;,,,��; „-<<,�„�;,i, perforafion <br /> spacing <br /> 4. Determine the number of s�aces betv��een perforations. (feet) i inch 1.25 inch 1.5 inch 2.0 inch <br /> Divide the length (3) by perforation spacing (2) and round <br /> do�n�n to nearest whole ntimUer. �� $ �q �g Zg <br /> Perforation spacing = .�.� ft=�_ft =j� spaces 3.0 8 13 17 26 <br /> 3.3 7 12 16 25 <br /> 5. Nuinber of perforations is equal to one plt�s the nt�mber of Q o � �; 15 23 <br /> perforation spaces(4). Clteck figure E-4 to rts�i+r�� tii�� �li�>>>���'�'of 5� 6 10 14 22 <br /> perforatio�is per laternt gaiarm�tees <10% discJinr��r���nrinti��;r. <br /> �_�spaces + 1 =�j�erfOrahonS/1lteral E-6: Perforation Discharge in gpm <br /> 6. A. Total number of perforations = perforations per latera( (5) perforation diometer <br /> rimes number of Iaterals (1) head inches <br /> (feet) ��8 3/16 7/32 1/4 <br /> ___�_perfs/lat x_�_lat= �_���erfor�tions <br /> 1.Oa 0.18 0.42 0.56 0.74 <br /> _ ; <br /> B. Calculate the square footage per perforation. 2,0b 0.26 0.59 0.80 1.04 ; <br /> Should be 6-10 sqft/perf. Does itat npply to at-��'1'll[�C�. ' <br /> Rock bed area = rock width (ft) x rock lengttl (Et) 5.0 0.41 0.94 1.26 1.65 '' <br /> 1 f� ft X_��5 ft = ^�SClft ° Use 1.0 foot for singie-family homes. <br /> Square foot per perforation = Rock bed area =nur�iber af perfs (6� b Use 2.0 feei for anythinq else. <br /> ���./� sqft=�_perfs =�S�ll/�7er1 FtqNIFOID LOCATED qT ENO OF PRESSVRE OKTRIBUTION SYSTEM <br /> 7. Determine required flow rate by multiplying the total number of <br /> perforations {6A) by flow per perforation(see figicr� E-6) .,.� <br /> _S� -.�_ / __� �� <br /> a�� .,�. <br /> perfs x ��m perfs � gpm , <br /> ,�i"-- � <br /> 8. IE laterals are coruzectea to header pipe as sho�ti�n on tipper `� �� ,,�,� '��E-�,� <br /> example, to select muumum required lateral diameter;enter G� ,,� <br /> figure E-� with perforation spacing (2) and number af perforations ���`��� <br /> per lateral(5) Select mininlum diameter for <br /> L4TOUI Of PEflFpp4TE0 PIPE L<tERAlS FOH <br /> perforated lateral = �- LLIC�I.S. PHFSSUfiE O�SiR18Up04 �wo�.,o <br /> �rFiaauu:ro R�stK rw[ �G <br /> 9. If perforated lateral systein is attadled to manifolc� pipe near �.,.,o,,.,�.,v�o K. -.-�:� <br /> �rtdii <br /> viw ���`Fw:etrFy� Yia.., li>.A� ,r"� <br /> the center, Iower diagram,perforated latei-al len�th (3) and ,��_�" Y�M� �.s�` <br /> number of perforations per lateral (5)will be ap��roiimately one ��-�:-'���...n.�.�..�,:�.d � P�'" <- <br /> hlSi�(�I�E <br /> �: <br /> half of that in step 8. Using these values, selecl inuiunum \`°� ��`� �""' :�;�,��„,,��«�„�� <br /> � � � � <br /> diameter for perforated lateral =�inches. ,/�.,�b �,�� <br /> �"" ��rr•-'"� <br /> � .s � ��0` �: ,� <br /> '��- �x .,,..,.c�..,,.o�o <br /> cJIC»"�({ <br /> I hereby certify that I have completed tltis work in ac�ordailce �vith applicable ordinances, rules and la�a�s. <br /> (signature) (license#) (date) <br />