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JUN-1�-1998 08:19P FROM: T0:9522494616 P:11�13 <br /> � . <br /> �`� PRESSURE DISTRIBUTION SYSTEM G�otext�le fabric I <br /> - � _,r.,� � .<:{,-,;:�,_>� ;:. �;: - :,,: 7:, - . . <br /> -f.G??JeiC:..�i l� ,q y.�'.� <br /> 1. Select nu.mber of perEorated laterals� uarter inch foratiuns s_aced�3' •j , <br /> ��:,� �n„�:,�.���r�..,;.:s�; �:�:. .. ....x:.. .. . : �: <br /> �:> y�i.S f��,..,i�M1.�,a c� ,<'� '' ��v°WY Il'�C�{L'.�,s, ; <br /> —� � or o�.�So.i� ;o �•:?s 4 +•' ''t� .'C , . <br /> 2. Select perforation spacing 4xr���" .;7�'"• :�;::.n `�_ - �_r°'?,� . <br /> ,� � „ o -�=r<. -. r} ,. <br /> Perf Sizin�3/16"-1/4" , <br /> 3. Since perforations should not be placed closer than 1 foot to PerE Spacut�1. ' <br /> d�e edge uf the rock layer(see diagram),subtract 2 feet from — <br /> llte rock layer length. E-4: Maxlmum albwable number ot i l4•inch perEorafions <br /> �� per lateral b guora�ee<10%discha�ge voriafion <br /> ~ rforallon <br /> �Kkz�i��t.�ti -2 ft — .. .� ft � <br /> spacing <br /> 4. Deternnine the nwmber of spaces between perforations. 1 inch �.25 in 1.5 inch 2.o inch <br /> Divide the length(3)by perforation spacing(2) and round <br /> down to nearest whole number. g �q 18 28 <br /> Perforation spacing=��= ft=�spaces 3.0 8 l3 17 26 <br /> 3.3 7 12 16 25 <br /> 5. Ntunber of perforations is equal to one plus the number of 4 0 7 11 15 23 <br /> perforation spaces(4). Check figure E-4 to assure the number of 5.0 6 10 14 22 <br /> perforatians per lateral guarantees <1 % discharge variation. <br /> ��spaces+1 = perforations/lateral E-6: Perforation Discharge,in gpm <br /> 6. A. Total number of perforations= perforations per lateral(5) perforotion dt �, ter <br /> times n ber of lateraLs (1) head inches <br /> �� � �P a 3/16 7/3 1/4 <br /> perfs/lat x lat= � erEorations <br /> 1:0 0.42 0.56 0,74 <br /> B. Calculate the square#ootage per perforation. 2,0b 0.59 0.80 1.04 <br /> Recommeded value is 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 /> �d _{�t X��ft=�sqft ° Use 1.�fOOt for singls-fom(ly homes. <br /> Square foot per perfora�ion=Rock bed area+number of perEs(6) �Use 2.0 feet for on n�, eise. <br /> •��? sqft=_.�'`f_perfs= /(� sqft/perf <br /> 7. Determine required flow rate by multiplying the total number of .._ <br /> perforations(6A) by flow per perforation(see figure E-6) ,,,o,,,, �„ <br /> `� \ �nPAV <br /> �J��r ���� <br /> perfs x 1�m/Pe.rfs= � gpm �„� <br /> �'~ '�,. ' <br /> 8. Tf laterals are connected to header pipe as shown on uppez �, �� d�•� <br /> a�a»Iwm prm <br /> e�acample,to select minimum required lateral diameter;enter py��o E.�;�►oecfad ot End of Syttwn <br /> figure E-4 with perforation spacing(2) and number of perforations <br /> per lateral(5) Select minimum cliameter for <br /> perforated lateral= inches. <br /> 9. Tf perforated Iateral system is attached to manifold pipe near �E'�a m:s,� =`�� <br /> the center,lower diagram,perforated lateral length(3)and �+�� <br /> number. of perforations per lateral(5)will be approximately one '— � l� <br /> ha1F oE that i�n step 8. Using these valu�s,select minimum ryfa y ' �� <br /> diameter for perforated lateral=__�_4_inches• �-�^ <br /> �.' � J � ,� �_ <br /> ( c tify that I have completed this work in accordance with applicable ordinances, rules and laws. <br /> � (signature�1 (license#) 2 (date) <br />