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05/20/2001 23:07 FAX 3982714 msts la 003 <br /> Mound Design Worksheet (For flows up to 1200 gpd) <br /> All boxed rectangles must be entered,tate rest wilt be cakulated. �-- <br /> A. FLOW 1 A,1:St;rrated Sewage Ham in Gallons per Day <br /> Estimated 900 gpd(see figure A-1) 1 <br /> or measured x'1.5(safety factor)= 0 gpd bedrooms f Class I Uns 1`1t Coss 01 CknIV <br /> 2 3X 225 i80 t 61% <br /> B. SEPTIC TANK LIQUID VOLUMES 3 450 A 218 of the <br /> Septic tank capacity 2 1300 gallons(see figure C-1) 4 61 1 375 256 values <br /> p T511 V 244 t inihe <br /> C. SOILS(Site evaluation data) 6 °Ga 525 332 r cbzf, <br /> 1. Depth to restricting layer- 1.5 feet 1 151 1 600 370 11,03( 8 t <br /> 2. Depth of percolation tests= 12 inches 1104 t 67-D 0 cciumra. <br /> 3. Texture ILoam(Silt Loan <br /> 4. Soil loading rate(see Figure D-33) 0.5 gpd/ftz <br /> Percolation rate 11.6-24.3 MPI <br /> 5. Q6 Land Slope 9 % D-33, AbsGrpUon Wkxh SkAng Tabie <br /> pc+c gmn Rate Loading - <br /> .....,�......�.�,rn.. in).Siau=per Soil Texture CratkXn Ab9ap " <br /> Se <br /> C-1: ike'i'ankCe aeltits(in xlloms t=b <br /> rw�. r...rr�r th(� arU:our <br /> ukvid c+1pcity Faaor darn S COW <br /> veJeod t.24 1.00 <br /> Nuuilxr cs Xiinimurn Liq)ud t iclWi capaci y with �;lulu dis�sa!i� J� +�r Medium�� <br /> Badrooins Capacity P-bflgt;dispasal lift inside FJW s <br /> &adYLIAM- 0.79 <br /> 2 <br /> or less -50 ]125 <br /> r <br /> _ _ t,ran.-- -_-zoo <br /> 1 ]i� to4� Jih +ro USO2.40 <br /> 3 or4 I Wo 150 > 60 ct� a.4s z <br /> Sof Int 2� ? si1-`Lt,i ,l <br /> 8 0 9 2000 - <br /> b11w120 $(tyaxy a._4 Soo <br /> +vim r!' w } Sand r Ca17 <br /> a <br /> 3. - stowexwan:=0' is <br /> per. 'Sys+cm dr'Face fmritie alis uxv Re+.ler n t�faawre <br /> L <br /> D. ROCK LAYER DIMENSIONS <br /> St Multiply average design flow(A)by 0.83 to obtain required area of rock layer.Iters A x 0.83-- <br /> 900 <br /> .83=900 gpd x 0.83 egpd= 747.0 ft <br /> 4 ., <br /> 2. Determine rock layer width =0.83 IF/gpd x Linear Loading Rate(LLR)(see LLR chart) <br /> ,., 0.83 f?/gpd X 12 = 10.0 ft <br /> rd <br /> LLF:Chart <br /> Pert;Rate LLR <br /> <120 MPI <=12 <br /> >=120 MPI <=6 <br /> 3. Length of rock layer=area divided Jy width= <br /> 750 fiz 1 10 feet= 75.3 feet <br /> E. ROCK VOLUME <br /> 1. Multiply rock area by rock depth to get cubic feet of rock <br /> 750 X 1 ft= 750.0 ft3 <br /> 2. Divide ft3 by 27 fe/yd3 to get cubic yards <br /> 750.0 fe / 27 = 27.8 yd3 <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in bars; <br /> 27.8 -yd' X 1.4 ton4d3 = 38.9 tons <br /> F. ABSORPTION WIDTH <br />