[Faller] 03FALL - HWK4 SOLN

[Faller] 03FALL - HWK4 SOLN - ECH 159, Fall 2002, Solution...

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Unformatted text preview: ECH 159, Fall 2002, Solution for Homework 3, October 16, 2002 Problem 1 The heat equation in a one-dimensional rod with heat source or sink proportional to temperature reads t u = k 2 x u- u. The sign of decides between two different scenarios: > 0 The rod loses heat. This corresponds to a wire in lateral thermal contact with a big, cold background. The wire radiates heat out to the side < 0 The rod picks up heat. Either the same as above with a hot background or a exothermal chemical reaction takes place with a reaction rate propor- tional to temperature. Boundary conditions: u (0 ,t ) = u ( L,t ) = 0 (a) Determine the steady state solution for all three possible cases (including = 0). (b) Solve the full problem for > 0 and the initial condition u ( x,t ) = f ( x ). (c) Same as (b) but < 0, Hint: The two problems are not symmetric. Solution (a) Steady state u = 0 k 2 x u- u = 0 Case 1: = 0 2 x u = 0 u = c 1 x + c , Boundary conditions u (0) = 0 c = 0 ,u ( L ) = 0 c 1 = 0 u ( x ) = 0 Case 2: > k 2 x u = u 2 x u = k u As > 0 we come to hyperbolic functions: u = c 1 sinh( p k x ) + c 2 cosh( p k x ) Boundary conditions: u (0) = 0 c 2 = 0, u ( L ) = 0 c 1 = 0, or k = 0. As > c 1 = 0 u ( x ) = 0 Case 3: < k 2 x u =-| | u 2 x u =- | | k u This leads to u = c 1 sin( q | | k x ) + c 2 cos( q | | k x ) Boundary conditions u (0) = 0 c 2 = 0, u ( L ) = 0 c 1 = 0, or q | | k L = n This means that if the length of the rod is L = n q k | | there exist non-zero solutions. These are the eigenmodes or standing waves of the problem. (b) > Solve the problem by separation of variables u ( x,t ) = X ( x ) T ( t ) leading to d t T ( t ) T ( t ) = k d 2 x X X- = 2 Solving the time equation d t T = 2 t , we exclude = 0 as we dealt with the steady state already, leading to T ( t ) = T exp 2 t For > 0 we saw that the steady state is u = 0 and physically we expect that if we have a negative source, i.e. heat is taken out of the system, that we have- 2 So we deal with the space problem k d 2 x X- X =- 2 X leading to d 2 x X =- 2- k X , lets define 2 = 2- k . For > 0 we find trigonometric solutions X ( x ) = c 1 sin( x ) + c 2 cos( x ) and for < 0 we find hyperbolic solutions X ( x ) = c 1 sinh( x ) + c 2 cosh( x ). In this latter case the boundary conditions enforce X ( x ) = 0. In the trigonometric case we use X (0) = 0 to find c 2 = 0 and X ( L ) = 0 to find L = n . Remembering the definition of we find n 2 2 L 2 = 2- k . So our eigenvalues are = q kn 2 2 L 2 + and the general solution is u ( x,t ) = X n =1 A n sin( n L x )exp- ( kn 2 2 L 2 + ) t ....
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[Faller] 03FALL - HWK4 SOLN - ECH 159, Fall 2002, Solution...

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