A careless person has disposed of arsenic in a well used for drinking water. The arsenic sits at the bottomof the well and diffuses into the water. This diffusion obeys the following equation:Yt = DYxx (1)where Y is the concentration of arsenic. The surface of the well is at x = 0 and the bottom at x = L =20m. At x = 0 the concentration is fixed at Y = 0 and at x = L the concentration is fixed at Y = 1.(a) [2]What is the steady state concentration of arsenic in the well
Question
A careless person has disposed of arsenic in a well used for drinking water. The arsenic sits at the bottomof the well and diffuses into the water. This diffusion obeys the following equation:Yt = DYxx (1)where Y is the concentration of arsenic. The surface of the well is at x = 0 and the bottom at x = L =20m. At x = 0 the concentration is fixed at Y = 0 and at x = L the concentration is fixed at Y = 1.(a) [2]What is the steady state concentration of arsenic in the well
Solution
The steady state concentration of arsenic in the well can be found by solving the given differential equation with the given boundary conditions.
The general solution of the equation Yt = DYxx is Y(x) = Asinh(x) + Bcosh(x), where sinh and cosh are hyperbolic sine and cosine functions, and A and B are constants.
Given the boundary conditions Y(0) = 0 and Y(L) = 1, we can find the values of A and B.
At x = 0, Y(0) = Asinh(0) + Bcosh(0) = 0. Since sinh(0) = 0 and cosh(0) = 1, we get B = 0.
At x = L, Y(L) = Asinh(L) + Bcosh(L) = 1. Since B = 0, we get A*sinh(L) = 1. Therefore, A = 1/sinh(L).
So, the steady state concentration of arsenic in the well is Y(x) = (1/sinh(L))*sinh(x).
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