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2012-01003 (Septic)
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2380 Abingdon Way - 03-117-23-23-0016
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2012-01003 (Septic)
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Last modified
8/22/2023 4:35:31 PM
Creation date
1/14/2016 11:46:58 AM
Metadata
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Template:
x Address Old
House Number
2380
Street Name
Abingdon
Street Type
Way
Address
2380 Abingdon Way
Document Type
Septic
PIN
0311723230016
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� ` _ _ OSTP Pressure Distribution � � <br /> ,JTa�� L�NIVERSITY `'��`� `'�''�� <br /> MinnesataPollutian Design Worksheet r�¢����-���' <br /> Control A enc OF MINNESOTA :.�s,=;,_.���J <br /> Projett ID: v 11.09Z2 <br /> 1. �Select Number of Perforated Luternts in system/zone: � ^______ <br /> (2 feet is minimum and 3 feet�s mnximum sEmucirtg) _ .. _, ,"""�"N"""'e"` ' _ <br /> - :., - _. .....••�: �-- - - '�=-'r = <br /> ��:,�: �;<_ - ,-.,:� .:::.;•,r:?=r i i <br /> 2. SelectPerforationSpQcing: 3.0 ft 5a' .. , , _ _ ..- '�jZ"sw��o„�. _.- ._� .__"�" "- <br /> -..� � <br /> v.-acr(mac�ons waree 9•a�.an �i�^-�z'�oi`«k ' 1a- <br /> 3. Select Perforatinn Dtometer Size 7/32 in _ _ <br /> G'ot rack <br /> 4. Length of Laterals =AAedia Bed Length-2 Feet ���uo��_v;m v: Pertoralfon spadnp:2'to 3' <br /> 63 - 2ft = 61 ft Perjora�ion con not be closer then i f�t from edge. <br /> 5- Determirte the Number of Perforatron Spaces. Divide the Length of Lnterols (Lirm 4)by the Perforation S�cing (Line 2)and <br /> round down to the rtearest whole number. <br /> Number of Perforation Spaces = 61 ft t Q3 ft = 20 Spaces <br /> 6. Number of PerforoNo�per Lateral is equat to 1.Q pl�the Number of Perforation S�ces (Line 5). <br /> Perforations Per laterol = 20 Spaces + 1 = 21 Perfs. Per Lateral <br /> Chetk table below to verlfy the number of perjorations per lateral guorontees less than a 105�dfscharge varlation. The value is <br /> double if the o center manifold fs used. <br /> Ma�a�t Nun�r a#P�f��t�er Lat�ral to�ke�<1ti����fae�ate�► <br /> '�'a ins P ora�s 7J321�sh Pesforations <br /> Perfarat�ss�S�cing l,Fe�ti ��eter(te�ch�s) Peafarati��ing Pi�D�a�n�.►ter(IR�s► <br /> f 1� 9Yt 2 3 �'��� i 1}� 1Vt 2 3 <br /> 2 90 43 1� 30 b� 2 !1 96 21 34 �e8 <br /> FV4 � 12 15 28 54 2�T 10 14 2Q 32 64 <br /> 3 � 1Z 96 25 52 3 9 14 19 3� 60 <br /> 3996!�h Perf�ati� 9/�Irteh Fe�f�rat€�r�s <br /> PiRe Dia��91�tth�s) Perforatioa�S�ascirsg Pi�[�a��r�tracl�sl <br /> Pe�FQrat�s Spacgng�FePt} 9 1� 9t� Z 3 (Fc�!) t fSa 91� 2 3 <br /> 2 92 18 26 4b S7 t 21 33 44 7�# 149 <br /> ��i 92 17 24 40 �0 Z�e 20 i0 41 69 135 <br /> 3 12 96 22 37 75 3 20 �4 3� 64 42� <br /> 7. Totnl Number of Perforatioru equals the Number of Perforatiorts per Laterot (Line 6)multiplied by the Number of <br /> Perforated L.aterals (Line 1). <br /> 21 Perf.Per Lateral X �Numlxr of Perf.�aterals = 63 Total Number of Perf. <br /> 8. Calculate the Squore Feet per Perforation. Recommended value�s 4 90 ft2 per perforation. °°'�'°"�'°1�'"�°�6P"� <br /> Does not app/y toAt-Grades ,�,r�� ����� , <br /> �/e /�e h: /• <br /> Bed Areo = Bed Width(ft)X Bed Length(ft) �.� o.�a o.s� o.ss o.�s <br /> 1.4 O.YL 0.81 0.69 0.9 <br /> 10 ft x 63 R = 630 fta Z° °.�` °-�° °.H0 ,�°° <br /> 2g �.� 0.65 �.89 ,.,� <br /> SA 0.32 0.72 0.98 1.78 <br /> Square Font per Perforation =Bed Area divided by the Total Number of Perforctiorrs (Line 7). 4.0 0.�� o.e� ,.,3 �.4� <br /> B.O` 0.41 0.93 1.2b 1.88 <br /> t foo[ �Ilfi�vikh 3/161n[h W t/4 StKh <br /> 630 ftZ T 63 perforations = 10.0 ftz/pertaratior�s ��« <br /> Oweliireg with t tB 6xA perfo2ibrm <br /> 9. Select Minimum Average Head: 1.0 ft 2 r�t o�����,��,e�s,�nn s��6 <br /> irch W t/4 trxh peRoratbrm <br /> 5 feet ������and N5T5 wiN 1/8 frxh <br /> 10. Select Perforatfon Discharge (GPM)based on Table Ill: 0.56 GPM per Pertorafion <br /> 11• Determine required Flow Rute by multiplying the Total Mumber of PerforatJons (Line 7)by the Perforation Dfschorge (Vne 10). <br />
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