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<br /> PRESSURE DISTRIBUTION SYSTEM Geotextile fabr�c
<br /> 3 . � :�,,.: . .;. .. -::: . <.:, -.: ::,- . ::: .2,:;
<br /> 1. Select number of perforated lateraLs�_. arter inch erforatlons aced!3' ,
<br /> . , , . :..,.: � .
<br /> <i , �'r ,�;•�S�� �f',t�c1c:•
<br /> 2. Select perforation spacing=��f{ , ,�. .: ,.. { ` ': . : .. ;:
<br /> Perf Sizing 3/16"-1/4"
<br /> 3. Since perforations should not be placed closer than 1 foot to P�spadns 1.5'-s�
<br /> the edge of the rock layer(see diagram),subtract 2 feet from
<br /> the rock layer length. E-4: Max�rwm dbwabia numbet of 1/4inch pedorafions
<br /> per lateral to�uarraa�tea<10%.discharqe vadatfon
<br /> V�r -2 ft = �.�� � . perioranon
<br /> 4. Determine the number of aces between erforations. `����
<br /> Sp P }ee 1 Inch ].251nch. 1.S•inch 2.O inch
<br /> Divide the length(3)by perforation spacing(2)and�i11d
<br /> ��to nearest whole number. 2;5 � 8 �14 18 28
<br /> Perforation spacing=�_ft+ 3 ft=�_spaces �3.0 � 8 13 17 26
<br /> 3.3 7 12 16 � 25
<br /> 5. Number of perforations is equal to one plus the number of 4 0 � �� t 5 23
<br /> perforation spaces(4)..Check figure E-4 to assure the number of �,p 6 10 14 22
<br /> perforutions per lateral guarantees<1Q%discharge variation.
<br /> ��,_3�7aCeS+1 = �3 . p�rforations/lateral E-6: Po�foratlon Discharqe In�pm
<br /> 6. A. Total number of perfora#ions= perforafions per l�teral(5) perforotlon d(ometer
<br /> times number of laterals(1) inches
<br /> �'S��' (fee) 3/16 7/32 1/4
<br /> �_perfs/lat x�_lat= L`� pe orations �,pa 0.42 0.56 0.74
<br /> B. Calculate the square footage per perforation. 2,pb 0.59 0.80 1.04
<br /> Should be 6-10 sqft/perf.Does not apply to at grades.
<br /> Rock bed area= rock width(ft)x rock length(ft) 5.0 0.94 1.26 1.65
<br /> _�_f{X�ft=�1�_SClft a Use 1.Q foot hx alnple-fam(ly homes.
<br /> Square foot per perforation=Rock bed area+number of perfs(6) b use 2.o reer ror c� r� eige.
<br /> �_sqft+.�—Perfs'_-�--sqft/�perf MMMfOLD LOCATm AT LND 0� NKftURC pSTRIBUTION 4YSTEM
<br /> 7. Determine required flow rate by multiplying the t�otal number of
<br /> perforations (6A).by flow per perforation�see figure E-6) �
<br /> !� erfs x �S m/ erfs= 3`al . gpm ��
<br /> __2.p —�--SP P •
<br /> 8. If laterals are connected to header. i e as shown on u er �M�'r'
<br /> P P PP � �;��-.,,
<br /> example,to select n1inimusri required lateraFdiameter;enter ��,,,,.�'""'�
<br /> figure E-4 with perforation spacing(2)and number of perforations \�
<br /> per lateral(5) Select�miniinum diameter for ,,,,a„„.�,,,,,„,,„��..�,,,�,�,
<br /> NK�ML p�?MWt10M N MOUND
<br /> perforated lateral= inches.
<br /> .�..�.�..�
<br /> 9. If perforated lateral system is attached to manifold pipe near � , �,� ��""
<br /> ��1��'�" M
<br /> tYte center,lower diagram,perforated lateral length{3) and ry�,.
<br /> number of perforations per lateral(5)will be approximately one :•��,�-��«-
<br /> half of that in step 8. Using these values,select minimum , -,�_�.�,;,.�,
<br /> diameter for perforated lateral= �����-- inches. a�, M
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<br /> � w�"'r
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<br /> I hereby certify that I have ompleted this work in accordance with applicable ordinances, rules and laws.
<br /> ��, �j � (si ture) �� (license#) t 1- ) -b� � (date)
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