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2005-P09293 - new septic system
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Old Crystal Bay Road North
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0160 Old Crystal Bay Road North - 33-118-23-43-0006
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2005-P09293 - new septic system
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Last modified
8/22/2023 4:52:01 PM
Creation date
3/7/2018 9:39:33 AM
Metadata
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Template:
x Address Old
House Number
160
Street Name
Old Crystal Bay
Street Type
Road
Street Direction
North
Address
160 Old Crystal Bay Road North
Document Type
Septic
PIN
3311823430006
Supplemental fields
ProcessedPID
Updated
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f / e <br /> a <br /> MOUND DESIGN WORKSHEET <br /> (For Flows up to 1200 d) <br /> �. FLOW EpUluttcd se+�e F1°..e la calleln Va dq <br /> too <br /> Estimated 75-0 gpd r�rll �"�p �` <br /> or measured x 1.5 = gpd. 2 300 225 lap <br /> 3640300 300 216 eon <br /> 37S 256 .ran <br /> in <br /> B. SEPTIC TANK LIQUID VOLUMES 6don <br /> 9 b o r Tia <br /> Two- Soo gallons a 1700 6673 401 m <br /> w:iM, owa. Ieoo YaAI e•-% dos` �o•hh 'dun° <br /> C. SOILS(refer to site evaluation) SeNk Tank r-s des On <br /> 1. Depth to restricting layer= f S inches feet Nmrkaot M�u,.°„6ge1a 6tynta�o�ln.�In � <br /> 2. Depth of percolation tests = 2 inches ' rdk <br /> 3. TextureS��!k laom ckj�MPercolation rate 5710 mpi 23or i o �'3 0 2M <br /> 4. Land slope 3- (o % ,'s 9 2M 30 0 4 <br /> D. ROCK LAYER DIMENSIONS Mound LLR <br /> 1. Multiply flow rate by 0.83 to obtain required area of rock layer: A x 0.83 = Perc Rate LLR <br /> '75o gpd x 0.83 sq. ft./gpd = Cp3d sq. ft. <120 MPI :512 <br /> 2. Determine width of rock layer =0.83 sq. ft./gpd x Linear.Loading Rate (LLR) >120 MPI < 6 <br /> 0.83 sq. ft/gpd x I z- _ 10 ft _ <br /> 3. Length of rock layer=area-1-width= <br /> 4 e+r. r: . ��. r• v.trtir <br /> S-31) sq. ft.+ co ft. e �c,:t~7:9< , b <br /> A <br /> i:pbpNQ6'wtrr.:r*. <br /> Width -I* ft 6 <br /> <120mpi <10' Length 63 ft <br /> E. ROCK VOLUME >120mpi <5' <br /> 1. Multiply rock area by rock depth to get cubic feet of rock;e3 o sq. ft.x I-os ft._ <br /> (o eo.4 cu. ft. <br /> 2. Divide cu.ft.by 27 cu.ft./cu. yd. to get cubic yards; <br /> (,,eo.g0 cu. ft. +27 cu. yd. <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; 2s.i cu.yd. x•1.4 ton/ 35.3 '10 4-e. <br /> F. ABSORPTION WIDTH <br /> 1. Percolation rate in tole 12 inches of soil is S o mpi Absorption Width Sla''"76le <br /> Texture sr Loaw, - u�r GIS raFtfsa1r .. oa�rw► AWN •-1�. <br /> QoM OPIUM <br /> Rotadm0.1 CerwSna 610 600 <br /> 2. Select allowable soil loading rate from table; 0•'a3 �� 10 600 <br /> °.ins 19a. is aa° 7.00 <br /> d �'S gpd/ft2 dais amu.. ON 1.1 <br /> 16 to" � in <br /> • SIM <br /> 3. Calculate absorption width ratio by dividing rock layer "16D Q, °" '" <br /> loading rate of 1.20 gpd/fF by allowable boil loading rate; I 91�• <br /> 1.20 gpd/ft2 + o•'9r gpd/fe = Z.1-- <br /> 4. Multiply absorption width ratio by rock layer width to get <br /> required absorption width; <br /> 1 0 x z.6-7 ft= z-7 ft. <br />
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