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Recovery Test Analysis Calculations <br /> `- RW--1 Parameters <br /> From the graph (following) , the slope of the line over one log cycle <br /> (AS) = 6.5 ft. <br /> Q = 5.0gpm <br /> Conversions <br /> S = 6.5 ft x 114/3.28 ft = 1.98M <br /> Q = wpm (1440 min/day) = 962.6 ft3/day (1M3/35.29 ft3) <br /> 27.28M3/day <br /> 7.48 gal/ft3 <br /> Solving T = 2.30Q <br /> 4v 6s' <br /> 's' = 2.30 (27.28M3/day) <br /> 2.52M2/day <br /> 4n (1.98M) <br /> T = 2.52M2/day [1.0.7584 ft2/M2] = 27.22 ft2/day <br /> MW-4 <br /> Parameters Froin the graph (following) , the slope of the line over one <br /> _ log cycle (6S) = 0.1.72 ft <br /> Q - 5.0 gpm <br /> Conversions <br /> S = 0.172 £t (IM/3.28 ft) = 0.052 M <br /> Q = 27.28 M3/day = 5 gpm <br /> Solvin4 T = 2.30 Q <br /> T = 2.30 (27.28 M3/day) <br /> 96.23 M3/day <br /> 47r (0.052M) <br /> �� T = 96.23M3/day (10.7584 ft3/M2) = 1035.28 ft2/day <br />