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Wolf Creek Coal Drewdowns <br />Sege Creek Side <br />T = .221 ft2/day <br />t = 344 days in (2001) <br />g = 1,363,334 pals/344 days = 3,9b3 pal/day = 2.75 gpm = 530 ft3/day <br />S is calculated to match a t of 344 days and a drewdown of 22.4 feet <br />Pit area in water = 13 acres/2 pi t:> = 283,140 ft2 <br />s = total potential head = 22.4 fret <br />1. Area of equivalent well t7 r2 = 283,140 ft2, r 300 feet <br />2. Calculate u for 22.4 ft of drawdown <br />4(u) = 4 xTi x .221 x 22.4 = .11737 u = 1.39 <br />530 <br />3. Calculate S that will give 22.4 ft of drawdown in 344 days at a radius of 300 ft. <br />5= 4 x .221 x 344 x 1.39 = .0047 <br />300 x 300 <br />4. Calculate u for 1 ft, drawdown <br />w(u) = 4 xTT x .221 x 1 = .005240 u = 3.73 <br />530 <br />5. Calculate distance to 1 ft. drewdown using S = .0047 <br />r1 = (4 x .221 x 344 z 3.73)1/2 = 491 ft to the 1 ft. drawdown contour <br />.0047 <br />6. Caleulate u for 5 ft. drewdovn <br />w(u) = 4 x 1T x .221 x 5 = .02620 u = 2.46 <br />530 <br />7. Calculate diatence to 5 ft. drewdown using S = .0047 <br />rs = (4 x .221 x 344 x 2.46)1/2 = 399 ft. to the 5 f[. drewdown contour <br />.0047 <br />Revised 03/01/95 <br />