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2011-01231 - new septic
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315 Hollander Road - 25-118-23-43-0014
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2011-01231 - new septic
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Entry Properties
Last modified
8/22/2023 4:15:52 PM
Creation date
2/24/2017 3:20:10 PM
Metadata
Fields
Template:
x Address Old
House Number
315
Street Name
Hollander
Street Type
Road
Address
315 Hollander Road
Document Type
Septic
PIN
2511823430014
Supplemental fields
ProcessedPID
Updated
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`�'a;j S,�. � 315 t���cr�a����'. .0 2�a�a�� V Pf''�'u � ��.I�Dc� . <br /> '- - �o�'y m�-�Q c3�- �� s`'c s�-�A <br /> �MOtJ�ID DE GN W�K SHEET �or' lows u "ta 1200 d <br /> A:: Average Design FLOW <br /> • �R�� A•i: Estimated Sewage Flows In Gatlons per Day <br /> v d(see ure A-1} num r o <br /> . •Bstimated�_gp �$ bedrooms C)ass I Ctass li Class!!i Class IY ' <br /> or measured x 1.5(safeey f�ctor) Z aoo zzs �so so��, <br /> ' _ •g�d 3 450 300 2S8 of the <br /> � li. SBPTiC TANK Ca acir 4 Bao 3T5 zss �awes <br /> 1} � 5 750 450 294 in Ihe <br /> 8 900 525 332 C�ass i, <br /> ��6 atO gallons{see figure C-1)1- � 1450 800 37U 11.�IN <br /> ��it. �(� <br /> �_/000�� '�"U.v1�/� 8 12A0 875 408 tolumns. <br /> G SOTT.S(refer to sife evaluation ���'�"��.st.�t+3`J G�:g� Uc7�w�CCa dlhs in albns. <br /> cn�����, r�r� .�r��� �� �� ��;� ���`r <br /> 1: Depth to restr�c g ayer= ��a feet e�a� Gww�r �w�v�� ���=b� <br /> 2. Depth of ercaiation tests= �1.c, f�et Z3 a 4 i� i'soo � <br /> 3. Texture �/.��,�' U S r ,se 6g 'zoo50°0 0� <� <br /> : Percolarion rate�t.a mpi `��,-c�o ,►�pr � u�..�.�. �� � <br /> 4. Soit toading rate.�_p,p�lsqft{see figure D-33? ' � <br /> 5. Percent Iand slope ��,d -% <br /> D. It{}CI<tAYBR DTMENSIONS <br /> l. Multiply average design flow(l�.)by 0.�3 ta oUtain req��ired rock layer area. <br /> Z Sci g�d x A.83 sqft/gpd = 3c� sqft � <br /> 2. Determine rack layer width=�.83 sqft/g�d x linear I..�ading Rate(LLR) _ <br /> � 0.83 sqft/gpd x j Z Qnd/sqft= � �° ft �ou nd LLR <br /> 3. Length of rock layer=area=width =�; <br /> �c�sqft(Dl}� io ft(C2}_��ft < 120�MPI <12 <br /> . , <br /> ,, <br /> �- �a.."�Roc�c voL�� > '�20 M P 1 < 6 <br /> � � 1. Multiply rock area{Dl)by rock depth of 1 ft to get cubic feet of rock <br /> !0 3o sqft x 1 ft=�03 cr—cuft • � <br /> . .2. Divide cuft by 27 cuftJcuyd eo get cubic yards <br /> (0 3 v cuft �27 cuyd/cuft=�_cuyd <br /> , 3. Multiply cubic yards by 1.4 to get weigllt of rock in to�ts , <br /> �, '?1� cuyd x�1.4 ton/cuyd=_��tons f /a��o <br /> . . D�3k AhawytMn YdldtA Sfelrm Tabb <br /> �.. SEWAG�ABSORPTION WIDT'H ..�++•� �o�� <br /> . h�w aw r.wn a�� "°f0'a.wm <br /> r <br /> i - <br /> � .� FawrNnS �� t.t0 �.Oo . <br /> Absorption width equals absorp�ion rakio(See Figure D-33) �6 _ _ <br /> , dmes rock layer width{D2) . <br /> �X�ft�,�U ft �b � .. �..� <br /> ,a, �w+n .� aoo <br /> t � . <br /> Orestrs - y <br /> Sewwoe ; <br /> TNEAT�f�NT w.e.+rwarau�armma.wa. <br /> PRoc�N �� <br /> > . . . <br /> • ri <br />
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