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ps1sol

ps1sol - ACM 95/100c Problem Set 1 Solutions Lei Zhang...

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ACM 95/100c Problem Set 1 Solutions Lei Zhang April 3, 2006 Each problem is worth 10 points - each part of a multi-part problem is weighted equally Problem 1 (Collaboration allowed) As a result of runoff from the surrounding land following a rainstorm, a large lake of depth h receives a sudden influx of a soluble phosphate compound in its upper layers. The phosphate then diffuses downward into the rest of the lake. Assuming that the phosphate concentration c is a function of depth z (measured positively downward from the lake surface) and time t only, and that the rate at which phosphate diffuses across any horizontal cross section is proportional to ∂c/∂z , show that c satisfies an equation of diffusion type. Check that the equation is dimensionally consistent. Explain why reasonable additional conditions to be satisfied by c are c z = 0 at z = 0 and z = h . Solution 1 By the law of conservation of matter, we have: rate of change of amount of phosphate = phosphate diffuse across boundaries per unit time+ phosphate generated inside per unit time (1) Let f ( z, t ) be the amount of phosphate diffuses downward through the unit area of horizontal cross section at z per unit time, f is proportional to the ∂c/∂z , i.e., for constant k f = - k∂c/∂z (2) By the above conservation law (1), suppose S is the (constant) area of horizontal cross section, since there is no phosphate generation or loss inside, we have ∂t ( c ( z, t ) S Δ z ) f ( z, t ) S - f ( z + Δ z, t ) S (3) let dz 0, we have ∂c ∂t = lim Δ z 0 f ( z, t ) - f ( z + Δ z, t ) Δ z (4) = - ∂f ∂z (5) = k 2 c ∂z 2 (6) 1

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i.e. ∂c ∂t = k 2 c ∂z 2 (7) check dimensional consistency: to be specific, consider the molarity concentration c of phosphate, which has the dimension mol · m - 3 , dimension of left hand side of equation is mol · m - 3 · s - 1 . k∂c/∂z is the amount of phosphate diffuses across unit area of horizontal cross section in unit time, has dimension mol · m - 2 · s - 1 , the dimension of right hand side is also mol · m - 3 · s - 1 . Since no phosphate can go through the lake surface and lake bottom, at z = 0 and z = h , we have insulated boundary condition, i.e. c z = 0. Note: If the area of horizontal cross section is S ( x ), and k = k ( z ), the equation will becomes S ∂c ∂t = ∂z kS ∂c ∂z (8) Problem 2 (Collaboration allowed) Modify the derivation leading to the heat equation, T t = a 2 T xx , so that it is valid for diffusion in a nonhomogeneous medium for which the specific heat c and the conductivity K 0 are functions of x and so that it is valid for a geometry in which the area A is also a function of x . Show that in this case we get cρAT t = ∂x K 0 A ∂T ∂x . Solution 2 We follow the notation of the class notes. Given the material’s “specific heat” c ( x ) = c (constant), the material’s density ρ ( x ) = ρ (constant), the rod’s temperature T ( x, t ) and the rod’s cross-sectional area A ( x ), the total thermal energy inside the rod between x = a and x = b at time t is Z b a cρT ( x, t ) A ( x ) dx. (9) The amount of energy per unit time per unit area flowing to the right is given by φ ( x, t
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