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Mound Design Worksheet (EXISTING MOUND ADDITION) <br /> All boxed rectangles must be entered,the rest will be calculated. <br /> A. FLOW �r 1' Fshn°icd Sewape Rans�Gd°ro Per Day <br /> Esamated 750 gpd(see figure A-1) �,� q�I C�Ss II C�ss ill q�s IV <br /> or measured 0 x 1.5(safety factor)= 0 gpd 2 �p 225 �80 b0� <br /> B. SEPTIC TANK LIQUID VOLUMES 3 450 300 218 of the <br /> Septic tank capaaty 2300 gallons(see f�ure G1) 4 � 375 256 values <br /> 5 750 �i0 294 in 1he <br /> C. SOILS(Site evaluafron data) a 900 525 332 qa�I, <br /> 1. Depth to restricting layer= 1.5 feet > >�� � 370 I,or ql <br /> 8 1200 6�5 d08 cd�ru, <br /> 2. Depth of percolation tests= inches <br /> 3. Texture LOAM _ <br /> 4. Soil loading rate(see Figure D-33 0.6 gpol ft� <br /> Percolation rate 6.7 MPI <br /> 5. %Land Slope 1 % D-33: Absorptbn R'Itph SIaJ�Tabk <br /> Percolmroo Rate Loading Rate <br /> in Atinute�Fer Swl Texnne Gelbro Absory+uon <br /> C-l: Se k Tank C cftks I in tllonsi �`° �'e`d'y R'"° <br /> n�r� �f� <br /> �.iqilld Cd�1,9t� Fauer tl�au 5 Coarse Sand 1.20 l 00 <br /> Number of :�linimwn L�qwd L�qwd capacity w�ith with d�sposal& "'�,;,y�ad <br /> Bedrooms Capaat�j gart�age d�s�sal ��ft instde Fme Sand <br /> �- <br /> 16�0 30 Loam O.eO 2 00 <br /> 2 OC�eJS �� 1��5 ]i()() 31 to 45 Silt Lwm � 0.30 2 40 <br /> 3 or�3 l OGi) 1�00 :t�) ,6�0 6o s�,ay<a.> o.as a s� <br /> S Of b ��} �.�.� ���) Silc?•Clay Lo�m <br /> '',8 ar 9 200� 3000 � <br /> 61 to 120 Salty l:�ay O.:a 3 00 <br /> Sandy<:la�. <br /> ��- <br /> Slower tban 130' <br /> `S��sWm d�tipM for tles�aoilf mua Ee Mr�a prfamaoc� <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Multiply average design flow(A)by 0.83 to obtain required area of rock layer: Item A x 0.83= <br /> 750 gpd x 0.83 ftygpd= 630.0 ft� <br /> 2. Determine rock layer width =0.83 ft�/gpd x Linear Loadin Rate(LLR)(see LLR cha�t) <br /> 0.83 ft�lgpd 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 /> 630 ft I 10 feet= 63.0 feet <br /> E. ROCK VOLUME <br /> 1. Multiply rock area by rock depth to get cubic feet of rock <br /> 220 X 1 ft= 220.0 ft3 <br /> 2. Divide ft�by 27 ft31yd3 to get cubic yards <br /> 220.0 ft3 / 27 = 8.1 yd3 <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> 8.1 yd3 X 1.4 ton/yd3 = 11.4 tons <br />