PGE 310 - HW9 - Solution

PGE 310 - HW9 - Solution - The University of Texas at...

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1 The University of Texas at Austin PGE 310: Formulation and Solution in Geosystems Engineering Homework #9: Spline Interpolation and Numerical Integration By Hossein Roodi 1. Cubic Splines Interpolation: (By HAND and PLOTTING IN MATLAB) Thermal stratification of a reactor is shown in the below table. ! " !" 10 13 22 55 70 ) ( 5 . 2 2 5 . 1 1 5 . 0 ) ( # $ # C in e Temperatur T m in depth z a) Use cubic spline to find Temperature as a function of Depth. Note that you need to find 4 functions for the four intervals between the data points. You can use backslash operator of MATLAB to solve system of equations. Step 1: System of Equations for deriving Second Derivative of the midpoints There are 5 points, so a system 3 eqns 3 unknowns will be derived in Cubic Spline Solution )] ( ) ( [ 6 )] ( ) ( [ 6 ) ( ) ( ) ( ) ( 2 ) ( ) ( 1 1 1 2 1 2 2 1 2 1 2 1 i i i i i i i i i i i i i i i i i x f x f x x x f x f x x x f x x x f x x x f x x % % % % % # & & % & & % & & % For points 1, 2, and 3: 216 ] 70 55 [ 12 ] 55 22 [ 12 ) 5 . 1 ( 5 . 0 ) 1 ( 2 ) 5 . 0 ( 5 . 0 )] 5 . 0 ( ) 1 ( [ 5 . 0 1 6 )] 1 ( ) 5 . 1 ( [ 1 5 . 1 6 ) 5 . 1 ( ) 1 5 . 1 ( ) 1 ( ) 5 . 0 5 . 1 ( 2 ) 5 . 0 ( ) 5 . 0 1 ( % # % % % # & & & & & & ( % % % % % # & & % & & % & & % f f f f f f f f f f For points 2, 3 and 4: 288 ] 55 22 [ 12 ] 22 13 [ 12 ) 2 ( 5 . 0 ) 5 . 1 ( 2 ) 1 ( 5 . 0 )] 1 ( ) 5 . 1 ( [ 1 5 . 1 6 )] 5 . 1 ( ) 2 ( [ 5 . 1 2 6 ) 2 ( ) 5 . 1 2 ( ) 5 . 1 ( ) 1 2 ( 2 ) 1 ( ) 1 5 . 1 ( # % % % # & & & & & & ( % % % % % # & & % & & % & & % f f f f f f f f f f For points 3,4 and 5: 72 ] 22 13 [ 12 ] 13 10 [ 12 ) 5 . 2 ( 5 . 0 ) 2 ( 2 ) 5 . 1 ( 5 . 0 )] 5 . 1 ( ) 2 ( [ 5 . 1 2 6 )] 2 ( ) 5 . 2 ( [ 2 5 . 2 6 ) 5 . 2 ( ) 2 5 . 2 ( ) 2 ( ) 5 . 1 5 . 2 ( 2 ) 5 . 1 ( ) 5 . 1 2 ( # % % % # & & & & & & ( % % % % % # & & % & & % & & % f f f f f f f f f f And also: 0 ) 5 . 2 ( 0 ) 5 . 0 ( # # & & f f So, the system of equations is: ) ) ) * + , , , - .% # ) ) ) * + , , , - . & & & & & & ) ) ) * + , , , - . 72 288 216 ) 2 ( ) 5 . 1 ( ) 1 ( 2 5 . 0 0 5 . 0 2 5 . 0 0 5 . 0 2 f f f Using the backslash operator of MATLAB, the solution will be: ) ) ) * + , , , - . % % # ) ) ) * + , , , - . & & & & & & 2857 . 10 1429 . 185 2857 . 154 ) 2 ( ) 5 . 1 ( ) 1 ( f f f
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2 Step 2: Using the result of Step 1 to determine cubic spline functions at each interval The formulation of cubic functions at each interval will be: ) ]( 6 ) )( ( ) ( [ ) ]( 6 ) )( ( ) ( [ ) ( ) ( 6 ) ( ) ( ) ( 6 ) ( ) ( 1 1 1 1 1 1 1 3 1 1 3 1 1 i i i i i i i i i i i i i i i i i i i i i i i x x x x x f x x x f x x x x x f x x x f x x x x x f x x x x x f x f % % & & % % % % & & % % % % & & % % & & # There will be 4 functions for the four intervals between the points: f1(x)for interval between points 1, 2: ) 5 . 0 ( 9021 . 122 ) 1 ( 140 ) 5 . 0 ( 4286 . 51 ) ( ) 5 . 0 ]( 6 ) 5 . 0 ( 2857 . 154 5 . 0 55 [ ) 1 ]( 0 5 . 0 70 [ ) 5 . 0 ( 3 2857 . 154 0 ) ( ) 5 . 0 ]( 6 ) 5 . 0 1 )( 1 ( 5 . 0 1 ) 1 ( [ ) 1 ]( 6 ) 5 . 0 1 )( 5 . 0 ( 5 . 0 1 ) 5 . 0 ( [ ) 5 . 0 ( ) 5 . 0 1 ( 6 ) 1 ( ) 1 ( ) 5 . 0 1 ( 6 ) 5 . 0 ( ) ( 3 1 3 1 3 3 1 % % % % # ( % % % % % % % # ( % % & & % % % % & & % % % % & & % % & & # x x x x f x x x x f x f f x f f x f x f x f 55 . 78 10 . 17 ) 5 . 0 ( 43 . 51 ) ( 3 1 % % % # ( x x x f f2(x)for interval between points 2, 3: ) 1 ( 5714 . 28 ) 5 . 1 ( 8571 . 122 ) 1 ( 7143 . 61 ) 5 . 1 ( 4286 . 51 ) ( ) 1 ]( 6 ) 5 . 0 ( 1429 . 185 5 . 0 22 [ ) 5 . 1 ]( 6 5 . 0 2857 . 154 5 . 0 55 [ ) 1 ( 3 1429 . 185 ) 5 . 1 ( 3 2857 . 154 ) ( ) 1 ]( 6 ) 1 5 . 1 )( 5 . 1 ( 0 . 1 5 . 1 ) 5 . 1 ( [ ) 5 . 1 ]( 6 ) 1 5 . 1 )( 1 ( 1 5 . 1 ) 1 ( [ ) 1 ( ) 1 5 . 1 ( 6 ) 5 . 1 ( ) 5 . 1 ( ) 1 5 . 1 ( 6 ) 1 ( ) ( 3 3 2 3 3 2 3 3 2 % % % % % # ( % % % / % % % % % # ( % % & & % % % % & & % % % % & & % % & & # x x x x x f x x x x x f x f f x f f x f x f x f 71 . 155 29 . 94 ) 1 ( 71 . 61 ) 5 . 1 ( 43 . 51 ) ( 3 3 2 % % % % # ( x x x x f f3(x)for interval between points 3, 4: ) 5 . 1 ( 8571 . 26 ) 2 ( 5714 . 28 ) 5 . 1 ( 4286 . 3 ) 2 ( 7143 . 61 ) ( ) 5 . 1 ]( 6 ) 5 . 0 ( 2857 . 10 5 . 0
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This note was uploaded on 04/12/2011 for the course PGE 310 taught by Professor Klaus during the Spring '06 term at University of Texas at Austin.

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PGE 310 - HW9 - Solution - The University of Texas at...

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