Solution4

# Solution4 - ChE 132B HW4 Solutions 60pt Problem 1. 15pt = 1...

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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 Newton-Raphson method to two nonlinear equations: Lecture 7, pg 25 ) Basic steps required to solve the above equation is similar to Newton-Raphson 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 Newton-Raphson 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.

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Solution4 - ChE 132B HW4 Solutions 60pt Problem 1. 15pt = 1...

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