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Mound Design Worksheet (For flows up to 1200 gpd) <br /> ' All boxed rectangles must be entered,the resf will be ca�ulated. <br /> A. FLOW A�I: Eshm°led S�.wwpe Fiana m Gd°�per D°'� <br /> Es6mated 600 gpd(see flgure A-1) �° <br /> or measured x 1.5(safety factor)= 0 gpd �0°"� �� ��� �Itl C�cs IV <br /> 2 3� 225 180 6096 <br /> B. SEPTIC TANK LIQUID VOLUMES 3 450 300 218 of ihe <br /> Septic tank capaaty 2�0 gallons(see figu►�G1) d �0 375 25b vaues <br /> 5 75p �0 294 �tha <br /> C. SOILS(Site evaluafion dafa) e 900 525 332 Class L <br /> 7 1050 b00 370 II,w III <br /> 1. Depth to restricting layer= 2 feet 8 120D e75 d08 caurrru. <br /> 2. Depth of percola6on tests= 12 inches <br /> 3. Texture Sand Loam <br /> 4. Soil loading rate(see Figure D-33 0.79 gpol ft <br /> Percolation rate 15 MPI <br /> 5. %Land Slope 5 °r6 D•33: Absorpllon Widlh Slal�Tabk <br /> Percolanon Raie Load�ng Rwte <br /> in Alinwea per So�l Texcure Gaflona +baa'twon <br /> C-1: Se ic Tank G aciUes Iln allonsl ti�� ���'� K•�� <br /> NumUer of h7i�umurn l.iqwd Liquid ca}r�city W�ith �9�ud capacity Faster than 3 ����a� ��o �o0 <br /> wtth di salc�C <br /> Bcdrnauis Capacity R�'�a8��P�� lift iuis� F�s�e`� <br /> a�v ---iso <br /> 2 or less 750 1125 1� i��o'o �,o.m _ _o.w ___ :oo _ <br /> ii�o�,s s�i�� o.a� i�o <br /> 3 or�i 1 U00 1�00 �� ��o eo �nw i i O.a3 - 2 67 - <br /> �or 6 1� 2250 3� si��v ns��.oam <br /> '?,8 or 9 2d60 3000 <'• , <br /> e��o izo s,��y ci.y o.:a s o0 <br /> Sanay c::�n� <br /> -_ c.�--- ------ -- <br /> Slower tli�n 120, - -- <br /> •5rwm d�rrrd fw Waw wil�rou�W adwr u p+rfamaecv <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Multiply 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 ft21gpd= 498.0 ft� <br /> 2. Determine rock layer width =0.83 ft Igpd x Linear Loadin Rate(LLR)(see LLR chart) <br /> 0.83 ft�/gpd X 12 = 10.0 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 /> 498 f� I � 10 feet= 50.0 feet <br /> E. ROCK VOLUME <br /> 1. Multiply rodc area by rodc depth to get cubie feet of rodc <br /> 498 X 1 ft= 498.0 ft3 <br /> 2. Divide ft�by 27 ft�lyd3 to get cubic yards <br /> 498.0 ft3 I 27 = 18.4 yd3 <br /> 3. Multiply cubic yards by 1.4 to get weight of rodc in tons; <br /> 18.4 yd3 X 1.4 ton/yd3 = 25.8 tons <br /> F. ABSORPTION WIDTH <br /> 1. Abso tion width uals absorption ratio(see Figure D-33)times rodc layer width <br /> 1.5 x 10.0 ft = 15.0 ft <br />