One way to treat acid mine drainage is to add limestone at the beginning of a na
ID: 1818308 • Letter: O
Question
One way to treat acid mine drainage is to add limestone at the beginning of a narrow AMD stream to precipitate the metals. The first order rate constant is 10/day. Assume the stream is 3 feet wide and 1 foot deep (assume a rectangular cross section). If the flow through the stream is 1000 gallons/day and the concentration at the beginning of the stream is 250 ppm, at what distance will the concentration of the metals be 5 ppm?
The "correct" answer provided by my professor is 17.4 ft, but I keep getting 130 ft as my answer. This is the equation i'm using:
I have C(d) = 5ppm, Co (d=0) = 250ppm, K = 10/day, As = surface area = 3ft x 1ft = 3 ft2 , Q = 1000 gal/day, and d = unknown
One way to treat acid mine drainage is to add limestone at the beginning of a narrow AMD stream to precipitate the metals. The first order rate constant is 10/day. Assume the stream is 3 feet wide and 1 foot deep (assume a rectangular cross section). If the flow through the stream is 1000 gallons/day and the concentration at the beginning of the stream is 250 ppm, at what distance will the concentration of the metals be 5 ppm? The ''correct'' answer provided by my professor is 17.4 ft, but I keep getting 130 ft as my answer. This is the equation i'm using: C(d) = Co (d=0) e^[(-K)( (d)(As)/Q] I have C(d) = 5ppm, Co (d=0) = 250ppm, K = 10/day, As = surface area = 3ft x 1ft = 3 ft^2 , Q = 1000 gal/day, and d = unknownExplanation / Answer
C=Co*e^-kt where t=v/Q C=Co*e^-k(v/Q) 5ppm=250ppm*e^-10/day((3ft(x))/1000gal/day) 100gal/day=133.68ft^3/day so, 5ppm=250ppm*e^-10/day((3ft(x))/133.86ft^3/day) solve: ln(.02)=-10/day(3ft(x)/133.68ft^3/day) x=17.4ft
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