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Unformatted text preview: Introduction to Numerical Methods I. Introduction 1.1 Why Numerical Method? T=T 1 T=T > T 2 1 ∂ T/ ∂ y=0 ( insulated ) ∂ T/ ∂ x =0 x y Heat flow Example 1 . Steady state heat conduction ∇ T= 0 2 Governing equation: How do we determine the heat flow from wall AB to wall AD? A B C D Possible solutions: 1. Experiment 2. Analytical solution 3. Numerical solution T=T 1 T=T > T 2 1 ∂ T/ ∂ y=0 ( Insulated ) ∂ T/ ∂ x =0 x y Heat flow Example 2 . Steady state heat conduction in a nonsimple geometry ∇ T= 0 2 Governing equation: How do we determine the heat flow from wall AB to wall AD? A B C D ∂ T/ ∂ x=0 ∂ T/ ∂ y=0 F G E Example 3 Unsteady state heat conduction in a nonsimple geometry: t T ∂ ∂ = α ∇ 2 T 1 Experimental approach : Design the experiment S et up a facility to satisfy boundary conditions (insulation and constant temperature) Prepare instrumentation Perform experiment & collect data (measure heat flux on wall AB, for example) Analyze data and present data Develop a model (say, to describe the effect of L AB /L CD on heat transfer) O.K. for all three examples; no knowledge on T(x, y) inside the cavity; relatively time consuming & tedious. Analytical approach : * Solve mathematical equation ( ∇ 2 T=0 or t T ∂ ∂ = α ∇ 2 T) based on a physical model. Use method of separation of variable? Green's function? * O.K. for simple geometry in Example 1; unlikely for Example 2 & 3. 2 Numerical approach : Solve equations that can be much more complicated than t T ∂ ∂ = α ∇ 2 T using a computer. Solution is discrete, approximate, but can be close to exact. t 1 t 2 t n t n+1 ∆ t ... t y n y n+1 Computer program can handle more complicated geometries (as in Example 2 & 3). Gain insight into the temperature distribution inside the cavity. Cautions: † NO numerical method is completely trouble free in all situations. † NO numerical method is completely error free. † NO numerical method is optimal for all situations. † Be careful about: ACCURACY , EFFICIENCY , & STABILITY . 3 * Example 4 . Solve a simple ODE: dt dy =  10y, y(0) = 1 First note: exact solution is: y exact (t) = e10t . Approximating: LHS ≡ dt dy as t y y n n ∆ + 1 & treating: RHS ≡10 y as 10 y n (n=0, 1, 2, 3,…) i.e. t y y n n ∆ + 1 = 10 y n => y n+ 1 = y n10 ∆ t y n = y n ), with y =1. y n+ 1 = y n ) = y n 1 ) 2 = y n 2 ) 3 = ... = y ) n Choose ∆ t = 0.05, 0.1, 0.2, and 0.5, = 10 ∆ t =0.5, 1, 2, and 5. See what happens ! n y n ( ∆ t=0.05) y n ( ∆ t=0.1) y n ( ∆ t=0.2) y n ( ∆ t=0.5) 1 1 1 1 1 0.514 2 0.25 1 16 3 0.1251 64 4 0.0625 1256 5 0.031251 1024 6 0.0156250 12048 7 0.00781251 … Comments: ok inaccurate oscillate blow up 40.1 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 0.5 1 1.5 y_numerical y_exact y t ∆ t=0.05 (This graph compares the exact solution with the stable numerical solution for ∆ t=0.05.) Questions : Why the solution blows up for ∆ t=0.5?...
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This note was uploaded on 04/23/2009 for the course EGM 6341 taught by Professor Mei during the Spring '09 term at University of Florida.
 Spring '09
 MEI

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