Ch E 310 - Fall 10 - Lecture 21

Ch E 310 - Fall 10 - Lecture 21 - Lecture 21 November 9,...

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Lecture 21 – November 9, 2010 Agenda: Numerical differentiation Derivatives of unequally spaced data Numerical differentiation in MATLAB Integrating ordinary differential equations (ODEs) Euler’s Method In-class 9 (due Friday)
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Differentiating Unequally Spaced Data If the spacing between data points is not equal, we can fit a Lagrange interpolating polynomial to adjacent points and differentiate analytically; second order form: where x is where you choose to evaluate the derivative This works anywhere in the range of x 0 to x 2 It has the same accuracy as the centered finite difference equation regardless of our choice of x            1 2 0 2 1 0 2 2 1 0 1 2 0 1 2 0 1 0 2 1 0 2 2 2 x x x x x x x x f x x x x x x x x f x x x x x x x x f x f
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Differentiating Unequally Spaced Data Example: Fourier’s law of heat transfer in a soil system q = heat flux (W/m 2 ), k = thermal diffusivity (3.5×10 -7 m 2 /s), r = density (1800 kg/m 3 ), C p = heat capacity (840 J/kg· C) Calculate the heat flux into the soil at the air-soil interface given the following temperature data measured below ground ( z = soil depth): z = 0.00 cm ( x 0 ) T = 13.5 C ( f ( x 0 )) z = 1.25 cm ( x 1 ) T = 12.0 C ( f ( x 1 )) z = 3.75 cm ( x 2 ) T = 10.0 C ( f ( x 2 ))   0 0 p z dT q z k C dz   soil air z
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Differentiating Unequally Spaced Data 8 9 10 11 12 13 14 15 0 0.5 1 1.5 2 2.5 3 3.5 4 temperature (C) depth (cm) Temperature profile: Use a Lagrange interpolating polynomial to estimate the thermal gradient: q ( z = 0) = (-3.5×10 -7 m 2 /s)(1800 kg/m 3 )(840 J/kg· C)(-133.3 C/m) = 70.56 W/m 2            C/cm 333 . 1 25 . 1 75 . 3 0 75 . 3 25 . 1 0 0 2 0 . 10 75 . 3 25 . 1 0 25 . 1 75 . 3 0 0 2 0 . 12 75 . 3 0 25 . 1 0 75 . 3 25 . 1 0 2 5 . 13 1 0 x f dz dT z
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Numerical Derivatives in MATLAB MATLAB has two key built-in functions for evaluating derivatives diff gives differences over intervals gradient centered version of diff If you use diff to operate on a vector of length n , the output is a vector of length n – 1 with differences between elements; this can be used with finite difference formulas Example: use diff to estimate the derivative of f ( x ) = 0.2 + 25 x – 200 x 2 + 675 x 3 – 900 x 4 + 400 x 5 from x = 0 to 0.8
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Ch E 310 - Fall 10 - Lecture 21 - Lecture 21 November 9,...

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