Laserfiche WebLink
� Mound Design Worksheet (For flows up to 1200 gpd) <br /> All boxed►ectangles must be entered,the rest will be calculated. <br /> A. FLOW <br /> Esbmated 750 gpd(see figu►�A-1) <br /> or measured x 1.5(safety factor)= 0 gpd <br /> B. SEPTIC TANK LIQUID VOLUMES <br /> Septic tank capaaty 2000 gallons(see figune G1) <br /> c an apac n a ons <br /> Number of Minimum Capacity with Capaaty with <br /> Bedrooms Capacity Garb.Disp. Disp.and Lift <br /> 2 orless 750 1125 1500 <br /> 3 or 4 1000 1500 2000 <br /> 5 or 6 1500 2250 3000 <br /> 7,8 or 9 Z000 3000 4000 <br /> C. SOILS(Sfie evaluation data) <br /> 1. Dep1h fio restricting layer- 1.8 feet <br /> 2. D�th of peroolation tests= 12 inches <br /> 3. Texture bam <br /> 4. Soil loading rate(see F'gwe D-33 0.6 gpd/ft� <br /> Pe�olation rate 8 MPI <br /> 5. %Land Slope 5 96 <br /> D. ROCK LAYER DIMENSIOt�IS <br /> 1. Multiply average design flow(A)by 0.83 to obtain required area of rodc layer.Item A x 0.83= <br /> 750 gpd x 0.83 ft�gpd= 620.0 ft� <br /> 2. Determine rodc layer width =0.83 ft�/gpd x Linear Loadin Rate LLR)(see lLR chart) <br /> 0.83 ft�/gpd x 12 = 10.0 ft <br /> LLR Chart <br /> Pe�C Rate LLR <br /> <120 MPI <=12 <br /> >=120 MPI <=6 <br /> 3. Length of rodc layer=area divided by width= <br /> 620 fi� / 10 feet= 62.0 feet <br /> E. ROCK VOLUME <br /> 1. Multiply rodc area by rodc depth to get cubic feet of rock <br /> 620 X 1 ft= 620.0 ft3 <br /> 2. Divide ft3 by 27 ft3lyd3 to get cubic yards <br /> 620.0 ft3 / 27 = 23.0 yd3 <br /> 3. Multiply cubic yards by 1.4 to get weight of rodc in tons; <br /> 23.0 yd3 X 1.4 tontyd3 = 32.1 tons <br /> F. ABSORPTION WIDTH <br /> 1. 6on width uals absorption ratio(see Figure D-33)times rock layer width <br /> 2 x 10.0 ft = 20.0 ft <br /> Page 1 of 6 <br />