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a-OEa <br /> The general agreement between the data from the two plots <br /> supports the interpretation, however these calculations should be <br /> considered rough estimates only. <br /> The storativity as well as the hydrogeologic setting clearly <br /> indicate unconfined groundwater conditions at the site. <br /> -` Calculations of Capture Zone & Recovery <br /> Well Performance Requirements <br /> E <br /> Assumptions: <br /> Gradient = 0.005 (from McClaren Report) = i <br /> Saturated Thickness = 17 ft. = b <br /> F_a Transmissivity = 10000 gpd/ft. = T <br /> Hydraulic Conductivity = 588 gpd/ft2 K = T/b <br /> Natural Groundwater Velocity = Vn = Ki/9 <br /> 1.12 ft./day <br /> This calculated groundwater velocity indicates a contaminant <br /> travel distance of over 365 feet would have occurred since the <br /> time of tank removal. The drilling evidence that shows a zero <br /> line at less than 100 feet suggests one or a combination of the <br /> following: <br /> -- 1. Extensive dispersion and biodegradation is occurring as <br /> the plume moves downgradient. <br /> 2. The contaminants are severely retarded by aquifer <br /> materials. <br /> 3. The calculated natural groundwater velocity is too high <br /> ti. because either the gradient is usually lower, or the <br /> aquifer hydraulic conductivity is lower than calculated. <br /> A capture zone has been calculated first assuming condition 1, <br /> and then condition 3. <br /> Case 1. <br /> Vn= 1.12 ft/day, <br /> For a pumping system in well one to be able to capture <br /> contaminants at well 6 a capture zone radius of <br /> approximately 70 ft. is required in the down-gradient <br /> direction. <br /> Therefore, because velocity due to pumping (Vp) equals the <br /> natural velocity at the edge of the capture zone: <br /> Vn = Vp = 1.12 ft/day or since <br /> Vp = Q/2 rb9 where Q = pumping rate and r = capture zone <br /> radius, <br />