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• wolf Creek Overburden Drawdowns <br />Sege Creek Side <br />2 <br />T = .0844 ft /day <br />t = 300 days in (2001) <br />0 = 1,670,651 gels/300 days = 5,569 gel/day = 3.87 gpm = 745 ft3/day <br />5 = is calculated to match a t of 300 days and a drawdown of 42.2 feet <br />2 <br />Pit area in water = 11.34 acres/2 pits = 246,985 ft <br />s = total potential head = 42.2 feet <br />1. Area of equivalent well 1{r2 = 246,985 ft2, r = 280 feet <br />2. Calculate u for 42.2 ft of dra wdown <br />Y(u) = 4 x ~Tf x .0844 x 42.2 = .06008 u = 1.85 <br />765 <br />3. Calculate S that will give 38 ft of drawdown in 207 days at a radius of 234 ft. <br />S= 4 x .0844 x 300 x 1.85 = .00239 <br />280 x 280 <br />4. Calculate u for 1 ft. drawdown <br />W(u) = 4 x fi x .0844 x 1 = .001424 u= 4.82 <br />745 <br />5. Calculate distance to 1 ft. drawdown using S = .00239 <br />r1 = (4 x .0844 x 300 x 4.82)1/2 = 452 ft to the 1 ft. drawdown contour <br />.00239 <br />b. Calculate u for 5 ft. drawdoun <br />W(u) = 4 xTT' x .0844 x 5 = .007118 u = 3.48 <br />745 <br />7. Calculate distance to 5 ft. drawdown using S = .00239 <br />. rs = (4 x .0844 x 300 x 3.48)1/2 = 384 ft. to the 5 ft. drewdown contour <br />.00239 <br />Revised 03/01/95 <br />