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OSTP Pressure Distribution
<br /> ��/ �` U�IIVERSI'T'Y
<br /> Minnesota Poilution �eSl�n 1►�Ort'�$hee� OF ���INNESO"fA �"' ��'`' ��
<br /> Control Agency ��'��
<br /> Geutc t e
<br /> 1. Select Number of Perforated Loterats in systemlzone: 3 ��,�A,�,�s�,,,�eG��r�o�.,�bc�,��a,o�t�M��,r �,��r': � �,o��d ;n
<br /> c�t p�r{oraurns spaced 3 apart,�i�� :'of r>d� '�d �f�
<br /> (2 feet is minimum and 3 feet is maximum spacing) �n� � :c,�y
<br /> � ,v r �r�-�mJs- �os���r-��ia���cr�r --s �
<br /> 2. S e t e c t P e r f o r n t r o n S p a cin g: 3.0 fL ��,�'� � n��✓ o o �
<br /> ����r,��0(�t�u�T BC�C��l;�a�g o��k�sD� v������:-v.��D�j�� �o�
<br /> ��,,,-��� ���f���� � p�a �Yi�`�Y��F�O� �� �;-�„ar� ra�Ch°�
<br /> 3. Setect Perforation Diameter Size 7/32 inch i �,e1f��,,,�„y,.;��: , ,�, ,,.
<br /> �--._..__.__ Pertorenori spaonc�2'r.�3'
<br /> 4. Length of Laterafs =Media Bed�ength-2 Feet. Perforation con not be cioser then 1 foot from edge.
<br /> 63 - 2ft - 61 ft
<br /> 5. Determine the Nurnber of Perforation Spaces. Divide the Len�th of Loterats (Line 4)by the Perforotion Spocing (Line 2}and
<br /> round down to the nearest whole number.
<br /> Number o(Perforotion Spotes = 61 ft = C 3 �ft = z0 Spaces
<br /> 6. Number of Perforotions per Loterot is equal to 1.d ptu5 the Number of PerJoration Spaces (Line 5).
<br /> Perforations Per Loterat = 24 Spaces + 1 = 21 Perfs_ Per Laterat
<br /> Check Tabie 1 to verifry the number of perfo�ations per tateral guarantees tess than a 1090 dischorge variation. The value is doub(e i f
<br /> the a center monijotd is used.
<br /> 7� Totol Number of Perforatrons equats the Number of Perforotions per LaterQl (�ine 6)multiptied by the Number of
<br /> Per/orated L4terats (Line 7).
<br /> 21 Perf. Per Lateral X �J.�Number of Perf. Laterals = 63 Total Number of Perf.
<br /> 8. Catculate the Squore Feet per Perforotion. Remmmended va(ue is 4-10 ft1 per perforation. — """- -- ''
<br /> . r rtor:no�as�,ar¢e�cv�xi
<br /> Does not a t to At-Grades `
<br /> _--
<br /> PP Y �— vnra,non r,�,mec�
<br /> —- ---r— �—`--
<br /> n,a ifp ,� I � � ,; �.
<br /> Bed Area = E3ed Width (fC)X Bed Length (ft) ��
<br /> 1 0� 0 i6 D{7 � Q 56 i 0.74 _
<br /> 1 S 0 22 0 51 '�. 0 69 �.9
<br /> 10 ft x 63 ft = 630 ftz � z a��-� o�a _ o s9 ' 0 80� +.o<�
<br /> 2.3 O.I9 � 0.65 � 0.39 7.17
<br /> __
<br /> Square Foot per Perforation = Bed Area divided by the Total Number of Per(orations (Line 7j. �'o , 0 3� o�2 0 98 ,�E
<br /> a o... � 0 3i o a� ,,Z -,a�—
<br /> � _.... -- .._- —'.
<br /> i 5 0` 0�� 0 9? 1 26 t_65 �
<br /> (� _ __.. . _:�.
<br /> �� Z _ �� 1 _ "___ i S t arv.'tl(6� h Peri t —
<br /> 630 ft 63 perforations _ i0.0 ft /perforations , ,,,�, �z �°
<br /> — --
<br /> 9 i :-,-1� ���iw .
<br /> 9. Selec.t Minimum Averoge Nead: �.a ft """ ` �� �
<br /> la 1 ;� /ic chpelo cb � ,MS�
<br /> .." __'__..
<br /> 5 ikvc. � l"inch pcR�o!'a'n on NS'�
<br /> 10. Setect Perforation Discho�ge (GPM} based on Table IIi: 0.56 GPM per Perforation
<br /> 11. Determine required F1ow Rate by muttiptying the Tota!Number of Perforations (Line 7)by the Perjoratian Discharge (Line 10).
<br /> 63 Perfarations X 0.5b GPM per Perforation = 36 GPM
<br /> 12. Select Type ojManifotd Conneciion (End or Center): �rnd ( j Centcr
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