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2010 - 00366 - septic repair
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2010 - 00366 - septic repair
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Last modified
8/22/2023 4:50:52 PM
Creation date
1/21/2020 11:26:18 AM
Metadata
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Template:
x Address Old
House Number
40
Street Name
Wear
Street Type
Lane
Street Direction
North
Address
40 Wear Lane North
Document Type
Septic
PIN
3311823340010
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0x/09/2010 01: 28 7634975011 SPTESTINGINC PAGE 09/13 <br /> MOUND DESIGN WORK SHEET (For Flows u to 1200 1.d <br /> A. Average Design FLOW A-1: tstlmaled Sewage Flows In Gallons per Day <br /> number at <br /> — <br /> Estimated (y,o o gpd (see figure A-1) bedrooms Class I Class II Class III Class IV <br /> or measured — x 1.5 (safety factor) _ — gpd 2 300 225 180 60% <br /> 3 450 300 218 of the <br /> 1'-0*I 4 600 375 256 values <br /> R. SEPTIC TANK Capacity 5 750 450 294 In the <br /> 6 900 525 332 Class I, <br /> a-- i oo 1J gallons (see figure. C-1) 7 1050 600 370 II, or Ill <br /> /C9(7Rnln.1 '�'r-I'n''A cul 1S07t. 4 1200 675 4108 columns. <br /> C. SOILS (refer to site evaluation) C•l: Septic Tnnk Cliwncllles(in Vllongt <br /> Num1. Depth to restricting layer = I. 0 Bcdrler of oolna MI11`aIact1.lrinld Ligllidccdispi with wilhdisposnl& <br /> feet Capacity garbngc disposal lift insillc <br /> 2. Depth of percolation tests = /, a . feet 2orless 750 1125 <br /> 3. Texture /-1.-i r' t-L,1Rrw1 3 or 4 1000 1500 1500 <br /> 5 or 6 1500 2250 2000 <br /> Percolation rate 3S. I mpi cfi-F,.,.-f.� , 7,6 or 9 2000 3000 3000 <br /> 4000 <br /> 4. Soil loading rate r LIS gpd/sqft (see figure D-33) <br /> 5. Percent land slope <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Multiply average design flow (A) by 0.83 to obtain required rock layer area. <br /> .,�1 oo gpd x 0.83 sqft/gpd = 49 sqft ,4-112°'2, ,.<4,r-7 ,1 ' <br /> 2. Determine rock layer width = 0,83 sqft/gpd x linear Loading Rate (LLR <br /> 0.83 sqft/gpd x bz _gpd/sgft = /o ft Mound LLR <br /> 3. Length of rock layer = area width = <br /> S9 r, sqft (Di) i- /o ft (D2) = ii < 120 M P1 < 12 <br /> E. ROCK VOLUME > 120 MPI < 6 <br /> 1. Multiply rock area (D1) by rock depth of 1 ft to get cubic feet of rock <br /> _ ,'Li 9_ sqft x 1 ft = cult <br /> 2. Divide cuft by 27 cuft/cuyd to get cubic yards <br /> _ `'-1 2 cuft = 27 cuyd/cuft = D cuyd <br /> 3. Multiply cubic yards by 1.4 to get wei ht of rock in tons <br /> a o _cuyd x 1.4 ton/cuyd = d{ tons <br /> D-33: AbagrpLi on Width Sizing Tn,ble - J <br /> F. SEWAGE ABSORPTION WIDTH Pereulilion Me Londing ltnle <br /> In MlnuIen per Soil'Iexlurr (Inllonn nbeorpqun <br /> Inch per dny per Rnlin <br /> (MI'r) aqunre loot <br /> rosier Ilion 5 Cnnr•.,e Snnd 1.20 1.00 <br /> Medium Snnd <br /> Absorption width equals absorption ratio (See Figure D-33) L"ntnysnn_' <br /> ..I.50— <br /> times rock layer width (D2) . riny 0_w..__, <br /> bin li„�,�Snhtly.Lpnm _ZU 4 ,ISO— <br /> 1419_35Q Loom Q‘GO Z�00 <br /> JI 1n 45 W$1rConm (f.,(�i�Z.dp <br /> SIU <br /> of.1r'1 x O ft = ,U r r) ft 76 to 60 Slimly C.Iny — <br /> Iny Wong0.45 2.67 <br /> Silly tiny lmnm <br /> 61 lu 120 Silly Cloy 0.2.4 T,D0 <br /> Seedy Clny <br /> . Cr ny <br /> —lower ihnn 71 z -- - <br /> Synlem dnrlrned?inhere sons rum he ether en p7Rennnnen ��—, <br />
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