Tema30 - CHAPTER 30 30.1 The key to approaching this problem is to recast the PDE as a system of ODEs Thus by substituting the finite-difference

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CHAPTER 30 30.1 The key to approaching this problem is to recast the PDE as a system of ODEs. Thus, by substituting the finite-difference approximation for the spatial derivative, we arrive at the following general equation for each node dT dt k T T T x i i i i = - + - + 1 1 2 2 By writing this equation for each node, the solution reduces to solving 4 simultaneous ODEs with Heun’s method. The results for the first two steps along with some later selected values are tabulated below. In addition, a plot similar to Fig. 30.4, is also shown t x = 0 x = 2 x = 4 x = 6 x = 8 x = 10 0 100 0 0 0 0 50 0.1 100 2.04392 0.02179 0.01089 1.02196 50 0.2 100 4.00518 0.08402 0.04267 2.00259 50 3 100 37.54054 10.2745 6.442321 18.95732 50 6 100 53.24295 24.66054 17.46032 27.92252 50 9 100 62.39033 36.64937 27.84901 34.34692 50 12 100 68.71329 46.03496 36.5421 39.53549 50 0 20 40 60 80 100 0 5 10 t = 0 t = 3 t = 6 t = 9 t = 12 30.2 Because we now have derivative boundary conditions, the boundary nodes must be simulated. For node 0, T T T T T l l l l l 0 1 0 1 0 1 2 + - = + - + λ ( ) ( i ) This introduces an exterior node into the solution at i = - 1. The derivative boundary condition can be used to eliminate this node, dT dx T T x 0 1 1 2 = - - which can be solved for dx dT x T T 0 1 1 2 - = - which can be substituted into Eq. (i) to give
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- - + = + dx dT x T T T T l l l l l 0 0 1 0 1 0 2 2 2 λ For our case, dT 0 / dx = 1 and x = 2, and therefore T - 1 = T 1 + 4. This can be substituted into Eq. ( i ) to give, ) 4 2 2 ( 0 1 0 1 0 + - + = + l l l l T T T T λ A similar analysis can be used to embed the zero derivative in the equation for the fifth node. The result is T T T T l l l l 5 1 5 4 5 2 2 + = + - λ ( ) Together with the equations for the interior nodes, the entire system can be iterated with a step of 0.1 s. The results for some of the early steps along with some later selected values are tabulated below. In addition, a plot of the later results is also shown t x = 0 x = 2 x = 4 x = 6 x = 8 x = 10 0 50.0000 50.0000 50.0000 50.0000 50.0000 50.0000 0.1 49.9165 50.0000 50.0000 50.0000 50.0000 49.9165 0.2 49.8365 49.9983 50.0000 50.0000 49.9983 49.8365 0.3 49.7597 49.9949 50.0000 50.0000 49.9949 49.7597 0.4 49.6861 49.9901 49.9999 49.9999 49.9901 49.6861 0.5 49.6153 49.9840 49.9997 49.9997 49.9840 49.6153 200 30.00022 31.80019 33.20009 34.19992 34.79981 34.99978 400 13.30043 15.10041 16.50035 17.50028 18.10024 18.30023 600 -3.40115 -1.60115 -0.20115 0.798846 1.398847 1.598847 800 -20.1055 -18.3055 -16.9055 -15.9055 -15.3055 -15.1055 1000 -36.8103 -35.0103 -33.6103 -32.6103 -32.0103 -31.8103 -40 -20 0 20 40 0 2 4 6 8 10 0 200 400 600 800 1000 Notice what’s happening. The rod never reaches a steady state, because of the heat loss at the left end (unit gradient) and the insulated condition (zero gradient) at the right. 30.3 The solution for t = 0.1 is (as computed in Example 30.1),
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t x = 0 x = 2 x = 4 x = 6 x = 8 x = 10 0 100 0 0 0 0 50 0.1 100 2.0875 0 0 1.04375 50 0.2 100 4.087847 0.043577 0.021788 2.043923 50 For t = 0.05, it is t x = 0 x = 2 x = 4 x = 6 x = 8 x = 10 0 100 0 0 0 0 50 0.05 100 1.04375 0 0 0.521875 50 0.1 100 2.065712 1.09E-02 5.45E-03 1.032856 50 0.15 100 3.066454 3.23E-02 0.016228 1.533227 50 0.2 100 4.046528 6.38E-02 3.22E-02 2.023265 50 To assess the differences between the results, we performed the simulation a third time using a more accurate approach (the Heun method) with a much smaller step size ( t = 0.001). It was assumed that this more refined approach would yield a prediction close to true solution. These values could then be used to assess the relative errors of the two Euler solutions. The results are summarized as x = 0 x = 2 x = 4 x = 6 x = 8 x = 10 Heun (h = 0.001) 100 4.006588 0.083044 0.042377 2.003302 50 Euler (h = 0.1) 100 4.087847 0.043577 0.021788 2.043923 50 Error relative to Heun 2.0% 47.5% 48.6% 2.0% Euler (h = 0.05) 100 4.046528 0.063786 0.032229 2.023265 50 Error relative to Heun 1.0% 23.2% 23.9% 1.0% Notice, that as would be expected for Euler’s method, halving the step size approximately halves the global relative error.
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This note was uploaded on 03/13/2012 for the course AEROSPACE 301 taught by Professor Pfchang during the Spring '12 term at Shandong University.

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Tema30 - CHAPTER 30 30.1 The key to approaching this problem is to recast the PDE as a system of ODEs Thus by substituting the finite-difference

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