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Unformatted text preview: ChE 132B HW4 Solutions 60pt Problem 1. 15pt = 1 , = 3 , = 2 , and = 1 + 2 ↔ 2 1 = 0.0100 / 2 + ↔ 2 = 0.0512 2 / 2 First, we need to write the reaction as an equation. 1 = [ ] 2 [ ][ ] 2 2 = [ ] [ ] 2 [ ] Let the extent of the first reaction as x, and the extent of the second reaction as y. / − [ ] = 2 / − [ ] = Then, + 2 ↔ 2 − − 2 + 2 2 + ↔ − 2 − + Thus, [ ] = − − 2 = 3 − − 2 [ ] = 2 − 2 [ ] = 2 + [ ] = 1 − Therefore, at equilibrium, 1 = (2 + ) 2 (3 − − 2 )(2 − 2 ) 2 = 0.0100 2 = (2 + ) (3 − − 2 ) 2 (1 − ) = 0.0512 Let these two equilibrium equations as function f(x,y) and g(x,y) . ( , ) = 0.0100(3 − − 2 )(2 − 2 ) 2 − (2 + ) 2 = 0 ( , ) = 0.0512(3 − − 2 ) 2 (1 − ) − (2 + ) = 0 Taylor expand these functions, ( , ) = ( , ) + ¡ ¢ , £ ¢ ( − ) + ¡ ¢ , £ ¢ ( − ) + ⋯ ( , ) = ( , ) + ¡ ¢ , £ ¢ ( − ) + ¡ ¢ , £ ¢ ( − ) + ⋯ In matrix notation, ¤ = ¤ + ⎣ ⎢ ⎢ ⎢ ⎡ ⎦ ⎥ ⎥ ⎥ ⎤ ¥ − − ¦ = ¥ 0 ¦ ⎣ ⎢ ⎢ ⎢ ⎡ ⎦ ⎥ ⎥ ⎥ ⎤ ¥ ¦ = ⎣ ⎢ ⎢ ⎢ ⎡ ⎦ ⎥ ⎥ ⎥ ⎤ ¥ ¦ − ¤ [ ]{ +1 } = [ ]{ } − { } = { } { +1 } = [ ]\{ } [Z] = “Jacobian matrix” (See Generalized NewtonRaphson method to two nonlinear equations: Lecture 7, pg 25 ) Basic steps required to solve the above equation is similar to NewtonRaphson method that we used before, but here, we need to calculate matrix inverse to get {x}. Instead, we will use the backslash operator. MultiNewt.m >> MultiNewt(0.5,0.5,0.001,20) iteration: 7 [A]: 2.6249 [B]: 1.8504 [C]: 0.2998 [D]: 0.8498 Problem 2. 15pt ( ) = 4 − 1.8 2 + 1.2 3 − 0.3 4 ′ ( ) = 4 − 3.6 + 3.6 2 − 1.2 3 ′′ ( ) = − 3.6 + 7.2 − 3.6 2 The first step to solve this problem is to write code to solve the system using the NewtonRaphson method. Since we already have the code, we just need to change ‘eps’ to a fractional relative error. method....
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This note was uploaded on 02/12/2012 for the course CHE 132b taught by Professor Ceweb during the Fall '09 term at UCSB.
 Fall '09
 Ceweb
 Reaction

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