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Sep 12 06 10: 52a Josh Swedlund (9521873-3292 p. 5 <br /> � � Mound Design Worksheef (For flows up to 1200 gpd) <br /> All boxed recfangles must 6e ente�d,the rest wrlf be calculated. A•I:Esiiraled SewaOe Flows in Gdons P��V <br /> A. FLOW <br /> Estimated 600 gpd(see figurB A-1) �° <br /> or measured x 1.5(safe factor - 0 gpd � �I doss u Cbss ul aass IV <br /> ty )- 2 3p0 225 I80 60% <br /> B. SEPTIC TAWK LlQUID VOI.UMES 3 450 �0 2�6 of the <br /> SepBc tank capaaty 2000 gallons(see figur�G1) 5 � � ��94 n q�e <br /> C. SOILS(Site evaluation dafa) d �00 525 332 C�t, <br /> 1. Depth to restricting layer- 1 feet 7 1050 600 3)0 Q.a NI <br /> 9 1200 6�5 408 cohxmu. <br /> 2. Depth of percalation tests= 12 inches <br /> 3. Texture loam <br /> 4, Soil ioading rate(see Figure D-33 0.6 gpd/fi� <br /> Percola6on rate 20 MPI <br /> 5. °k Land Slope 3 �o D-33: AbsotpUon�Yldth Slzln�Trbk <br /> Pe/caaioa Kxe Loadiug R��c <br /> in Atinv�er per Sail Texcurr Galloas AbsoRwon <br /> C-lt Se feTsnkC.a tcides(In alions► �"" �"`d'Y R�'� <br /> � �� <br /> Liqttidcapacity P�u,ms c��.�w �.zo ioo <br /> Nutuber of Minimun�Liq�id l..iqwd capacity u�iih ����& ��Y� <br /> Bedroouis Ca�aaty g�'�8C�1� tift insidc •�� <br /> .t4_._ _ <br /> 2 or less 750 1125 0� _ �� <br /> jSQQ a i�as sa�t.o«n, o. o �40 <br /> 3 or-i 1000 I 500 200c) aa�o Qo s�ay c7�•to.� u.,s i a� <br /> 5 or 6 l 500 2250 � s��q a.r� <br /> 7,8 or 4 2000 3000 6,�,�20 ��ry a.y o.+, s o0 <br /> S.no Gq- <br /> Slower Ui�n 12 ' — <br /> �y'Ma 6�wrd fa tlrw wi1R nua M o�.r ot p.fo�m�.a <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Mul6ply average design flow(A)by 0.83 to obtain required area of rodc layer:Item A x 0.83= <br /> 600 gpd x 0.83 fi�gpd= 498.0 fi� <br /> 2, Detettnine ra�C layer width =0.83 f'�/gpd x Linear Loadin Rate(LLR)(see LLR chartj <br /> 0.83 ft�lgpd X 12 = i 0.0 ft <br /> LLR Chart <br /> Perk Rate LLR <br /> <120 MPI <=12 <br /> >-120 MPI <=6 <br /> 3. Length of rock layer=area divided by width= <br /> 4g8 ft I � 10 feet= 50.0 feet <br /> E. ROCK VOLUME <br /> 1. Multiply rock area by rock depth to get cublc feet of rock <br /> 498 X 1 ft= 498.0 ft3 <br /> 2. Divide ft3 by 27 ft31yd3 to get cubic y�rds <br /> 498.0 ft I 27 = i 8.4 yd3 <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> 18.4 yd3 X 1.4 toNyd3 = 25.8 tons <br /> F. ABSORPTION WIDTH <br /> 1. Abso 6on widkh e uals abso�ption ratio(see Figure D-33)times rock layer width <br /> 2 x 10.0 ft = 20.0 ft <br />