ECH140FinalExam_2013Soln - HANOI UNIVERSITY OF MINING and...

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HANOI UNIVERSITY OF MINING and GEOLOGY Advanced Program in Chemical Engineering ECH140: Mathematical Methods for Chemical Engineering Final Exam Solution September 17, 2013 (9:00-11:00 ) (Closed book and notes no cell phones; no questions asked or answered) Prof. Higgins Hanoi, September 2013 Problem 1 (20 points) Consider the following PDE ¶∂ u ¶∂ t = - au + D ¶∂ 2 u ¶∂ x 2 IC: u H x, 0 L = f H x L BC1: ¶∂ u ¶∂ x H 0, t L = 0 BC2: u H H, t L = U 0 (i) Determine the steady state solution(s) for a > 0. (ii) Determine the transient solution. Solution Problem 1 (i) The steady solution satisfies D 2 U s x 2 - aU s = 0 BC1: U s ¶∂ x H 0 L = 0 BC2: U s H H L = U 0 The general solution of (5) for a > 0 U s H x L = C 1 Cosh B a D 1 ê 2 x F + C 2 Sinh B a D 1 ê 2 x F We apply BC 1 to get U s x = C 1 a D 1 ê 2 Sinh B a D 1 ê 2 x F + C 2 a D 1 ê 2 Cosh B a D 1 ê 2 x F Evaluating at x=0 gives This study resource was shared via CourseHero.com
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C 2 a D 1 ê 2 = 0 ï C 2 = 0 Apply BC2 gives U 0 = C 1 Cosh B a D 1 ê 2 H F ï C 1 = U 0 Cosh BI a D M 1 ê 2 H F Thus the steady state solution is U s H x L = U 0 Cosh BI a D M 1 ê 2 H F Cosh B a D 1 ê 2 x F (ii) To determine the transient solution we make the substitution u H x, t L = U s + w H x, t L The resulting PDE for w H x , t L is ¶∂ w ¶∂ t = - aw + D ¶∂ 2 w ¶∂ x 2 IC: w H x, 0 L = f H x L - U s H x L BC1: ¶∂ w ¶∂ x H 0, t L = 0 BC2: w H H, t L = 0 To solve this PDE we make use of separation of variables. We seek a solution of the form w H x, t L ~ G H t L f H x L Since w H x , t L Ø 0 as t Ø 0 , we have G t = -l 2 G ï G H t L = A 1 Exp A -l 2 t E For f we have the following eigenvalue problem -l 2 f = - a f + D 2 f x 2 BC1: „f x H 0 L = 0 BC2: f H H L = 0 We can reorganize the ODE as 2 f x 2 + b 2 f = 0, b 2 = l 2 - a D The general solution is f H x L = C 1 Cos H b x L + C 2 Sin H b x L 2 ECH140FinalExam_2013Soln.nb This study resource was shared via CourseHero.com
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BC1 gives „f x H 0 L = 8 - C 1 b Sin H b x L + C 2 b Cos H b x L< x = 0 = C 2 b ï C 2 = 0 BC2 givesA n f H H L = C 1 Cos H b H L = 0 ï b = H 2n + 1 L p 2H , n = 0, 1, 2, … Thus b 2 = H 2n + 1 L 2 p 2 4H 2 = l 2 - a D Solving for l 2 we find l 2 = H 2n + 1 L 2 D p 2 4H 2 + a Hence the general solution is w H x, t L = n = 0 A n Exp B H 2n + 1 L 2 D p 2 4H 2 + a t F Cos KH 2n + 1 L p x 2H O Since the eigenvalue problem is a SLEP, we have the following orthogonality relationship 0 H f n H x L f m H x L x = 0, ifn ¹≠ m To determine the coefficients A n we make use of the initial condition w H x, 0 L = f H x L - U s H x L = n = 0 A n Cos BH 2n + 1
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