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1997-008935 - repair septic
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1125 Spring Hill Road - 26-118-23-43-0004
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1997-008935 - repair septic
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Last modified
8/22/2023 4:18:54 PM
Creation date
3/7/2019 12:46:13 PM
Metadata
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Template:
x Address Old
House Number
1125
Street Name
Spring Hill
Street Type
Road
Address
1125 Spring Hill Road
Document Type
Septic
PIN
2611823430004
Supplemental fields
ProcessedPID
Updated
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MOUND DESIGN WORKSHEET <br /> � � (For Flows up to 1200 gpd) <br /> ' A. �..�w Estimaced Sewage Flow in Gallons pa D�y(g,pQ) ..•.-� <br /> Estimated Co gpd `�� ; <br /> or �ya t �[y�tt zya t�t �tya rv <br /> or measured x 1.5 = - gpd. s��� <br /> Z 300 27S 180 6oR <br /> 3 450 300 218 et <br /> B. SEI''ITC TANK LIQUID VOLUMES � ° � 3�s u� � <br /> s �so 4so Zsa `v°" <br /> �-��$� � l-1 C D O gdl1011S 6 900 525 332 � <br /> 7 1050 600 370 ��„ <br /> 8 1200 675 40� <br /> C. SOILS(refer to site evaluation) � � ,;,�� � �� <br /> 1. Depth to restricting layer= ����� a a�'� inches �,� �, �,, <br /> 2 Depth of peroolation tests = ) a" inches `"°"' `"�'' <br /> i z«� �so �.iu <br /> 3. Percolation rate 9.� mpi s«a �,000 �.soo <br /> 4. Land slope L % �;a s zo�oo 3,0�00 <br /> wv 9 See fig.C-6 (x 1.5) <br /> D. ROCK LAYER DIlvIIIVSIONS <br /> 1. Multiply flow rate by 0.83 to obtain required. area of rock <br /> layer:A x 0.83 = <br /> g o o . gpd x 0.83 sq.ft./gpd = �.0� sq. ft. <br /> 2 Select width of rock layer(10 feet or less) _�ft. <br /> 3. Length of rock layer = area=width = Rock Bed <br /> ��-l7 sq. ft.i 1 o ft. _�ft. ,.�...ti...�.,.�...�.,.,.,...,.... <br /> 1.1•I•�•J�1.!•J•r r•f•�•d•t•t•t• <br /> 1,�fti+�fti fti�1 f�f�ft fti��fti fti fti fS jt <br /> ti.t•ti.L•ti•ti•ti•ti•Mti.�.ti.ti.ti.ti•ti•y idth 510 ft <br /> !•1•t•r•�•1•t•f•t t•r•r•t•l•f•f• <br /> . ti•tiH•ti•ti•�•�•�•1•tN•�•1�•ti•ti•1•1 <br /> .r.f.f.f•r.f.r•f•f•f.f• •f•f•f.f• <br /> E. ROCK VOLIJME ~- �'g�' -� -��-� <br /> 1. Multiply rock area by rock depth to get cubic feet of rock; ``� <br /> �sq.ft. x .�ft. _�4s 4 cu. f t. <br /> 2 Divide cu. ft.by 27 cu. ft./cu. yd. to get cubic yards; <br /> �cu. ft. s 27=�_cu. yd. � <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> �g_cu.yd. x 1.4 ton/cu. yd. _ �1I tons. <br /> F. ADSORP'IION WIDTH L l�Y`C Loz}n�l ,e,b� �«,w�a�;s�, vb�� <br /> 1. Percolation rate in top 12 inches of soil is �..�' mpi ,��o,,,�,�,p �� R.md <br /> �„s;�,,,u,pR;,,�y, SoilTexture �,�;� �°` <br /> cmpil as .e.o.�a <br /> M <br /> 2. Select allowable soil loading rate from table; Faster than 0.1 oarse Sand 1.20 l.00 <br /> .ys gpd/ffi� O.lto� Sand 1.20 1.00 <br /> 0.1 to� Fine Sand" 0•� Z•� <br /> 6 to 15 ndy Loam 0.79 1 S2 <br /> 3. Calculate adso tion width ratio b dividin rock la er 16 to�o [„o�m o.bo 2.00 <br /> � }' $ }� 31 to 45 Silt Loam OSO 2.40 <br /> loading rate of 1.20 gpd/ft2 by allowable soil loading rate; �to bo cta Lo� 0.45 2.6� <br /> 1.20 gpd/ft2y .NS gpd/ft�_ �. �� Slower�than120 ciay o.z4 �.00 <br /> �s«t tv.y,g sox a mae d ru,�a�ay fu�e su�a <br /> 4. Multiply adsorption width ratio by rock layer tividth to get <br /> required adsorption width; <br /> �.�� x Jo ft= a�.� ft <br /> `�.,:.� <br />
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