Numerical Methods for Chemical Engineers · Numerical Methods for Chemical Engineers Chapter 4: ... This chapter deals with the case of determining the values ... 20 PM

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Chemical Engineering Numerical Method© copyright PCS- FKKKSA, UTM

Numerical Methods for Numerical Methods for Chemical EngineersChemical Engineers

Chapter 4: System of Linear Algebraic Chapter 4: System of Linear Algebraic EquationEquation

Saharudin Saharudin HaronHaron

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This chapter deals with the case of determining the values x1 , x2 , …, xn that simultaneously satisfy a set of equations:

a11x1 + a12x2 + . . . .+ a1nxn = b1a21x1 + a22x2 + . . . .+ a2nxn = b2

. .

. .an1x1 + an2x2 + . . . .+ annxn = bn

where the a’s are constant coefficients, b’s are constants and n is the number of equations.

System of Linear Algebraic EquationsSystem of Linear Algebraic Equations

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Goals and ObjectivesBe able to solve problems involving linear algebraic equationsAppreciate the usage of linear algebraic equations in any field of engineeringMastering several techniques and their reliability

Naïve Gauss eliminationGauss-Jordan eliminationLU decompositionGauss Siedel

Be able to use a program to successfully solve systems of linearalgebraic equations

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Naïve Gauss EliminationA systematic technique use to solve linear algebraic equations simultaneously with two steps:

a) Forward elimination- the equations were manipulated to eliminate all the elements

below the main diagonal of matrix A

b) Back Substitution- the elimination step result in one equation with one unknown.- the equation could be solved directly and the result back-

substituted into one of the original equations to solve the remaining unknown.

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The procedure of Naïve Gauss EliminationRepresenting the linear algebraic equations in an augmented matrix form.

a11x1 + a12x2 + a13x3 + .. + a1nxn = c1a21x1 + a22x2 + a23x3 + .. + a2nxn = c2a31x1 + a32x2 + a33x3 + .. + a3nxn = c3

: : : : : :an1x1 + an2x2 + an3x3 + .. + annxn = cn

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

nnnnnn

n

n

n

cccc

xxxx

aaaaaaaaaaaaaaaa

3

2

1

3

2

1

3321

3333231

2232221

1131211

........

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

4

3

2

1

321

3333231

2232221

1131211

..

..

..

..

cccc

aaaaaaaaaaaaaaaa

nnnnn

n

n

n Label as row or eqn. (1). - (1)

- (2)

- (3)

- (n)

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Steps of Naïve Gauss Elimination (ex. 3 unknowns in 3 equations).A) Forward elimination

a) To eliminate the first unknown, x1, from the second through the nth row/eqn.- row/eqn (1) is called the pivot equation, and a11 is called pivot element.

)3()2()1(

3

2

1

333231

232221

131211

−−−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

ccc

aaaaaaaaa

(i) (1) x a21/a11 ⇒ (1a)(ii) (2) – (1a) ⇒ (2')(iii) (1) x a31/a11 ⇒ (1b)(iv) (3) – (1b) ⇒ (3')

)3()2()1(

3

2

1

3332

2322

131211

′−

′−−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

′′

′′′′

ccc

aaaaaaa

the prime ' indicates that the elements have been modified.

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Steps of Naïve Gauss Elimination (ex. 3 unknowns in 3 equations).

b) To eliminate the second unknown, x2, from the third through the nth row/eqn.- row/eqn (2') is called the pivot equation.

)3()2()1(

3

2

1

3332

2322

131211

′−

′−−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

′′

′′′′

ccc

aaaaaaa

(i) (2') x a'32/a'22 ⇒ (2'a)

(ii) (3') – (2'a) ⇒ (3'')

)3()2()1(

3

2

1

33

2322

131211

′′−

′−−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

′′′

′′′′

ccc

aaaaaa

the double prime '' indicates that the elements have been modified twice.

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Steps of Naïve Gauss Elimination (ex. 3 unknowns in 3 equations).

B) Back Substitution

From equation (3'') : [a''33 c''33] ⇒ a''33x3 = c''3

∴ x3 = c''3/a''33

the result x3 can be back-substituted into eqn (2') and (1) to solve for x2 and x1.

1131321211

2232322

/)(/)(

axaxacxaxacx

−−=

′′−′=From equation (2') and (1)

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Naïve Gauss Elimination(assignment in class)

Use Naïve Gauss elimination to solve the following equations.

x1 + x2 – x3 = -36x1 + 2x2 + 2x3 = 2-3x1 + 4x2 + x3 = 1

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Gauss-Jordan EliminationSteps of Gauss-Jordan Elimination (ex. 3 unknowns in 3 equations).

a) Change the value of a11 to 1 and eliminate the other elements in the first column.

)3()2()1(

3

2

1

333231

232221

131211

−−−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

ccc

aaaaaaaaa

(i) (1) x 1/a11 ⇒ (1')(ii) (1') x a21 ⇒ (1'a)(iii) (2) – (1'a) ⇒ (2')(iv) (1') x a31 ⇒ (1'b)(v) (3) – (1'b) ⇒ (3')

)3()2()1(

001

3

2

1

3332

2322

1312

′−

′−

′−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

′′′

′′′′′′

ccc

aaaaaa

the prime ' indicates that the elements have been modified.

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Steps of Gauss-Jordan Elimination (ex. 3 unknowns in 3 equations).

b) Change the value of a'22 to 1 and eliminate the other elements in the second column.

)3()2()1(

001

3

2

1

3332

2322

1312

′−

′−

′−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

′′′

′′′′′′

ccc

aaaaaa

(i) (2') x 1/a'22 ⇒ (2'')(ii) (2'') x a'12 ⇒ (2''a)(iii) (1') – (2''a) ⇒ (1'')(iv) (2'') x a'32 ⇒ (2''b)(v) (3') – (2''b) ⇒ (3'')

)3()2()1(

001001

3

2

1

33

23

13

′′−

′′−

′′−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

′′′′′′

′′′′′′

ccc

aaa

the double prime '' indicates that the elements have been modified twice.

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c) Change the value of a''33 to 1 and eliminate the other elements in the third column.

)3()2()1(

001001

3

2

1

33

23

13

′′−

′′−

′′−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

′′′′′′

′′′′′′

ccc

aaa

(i) (3'') x 1/a''33 ⇒ (3′″)(ii) (3′″) x a''13 ⇒ (3′″a)(iii) (1'') – (3′″a) ⇒ (1′″)(iv) (3′″) x a''23 ⇒ (3′″b)(v) (2'') – (3′″b) ⇒ (2′″)

)3()2()1(

100010001

3

2

1

′′′−

′′′−

′′′−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

′′′′′′′′′

ccc

the triple prime ′″ indicates that the elements have been modified three times.

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d) The value of the unknowns can be determined directly without theback substitution step as in the naïve gauss elimination.

)3()2()1(

100010001

3

2

1

′′′−

′′′−

′′′−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

′′′′′′′′′

ccc

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

′′′′′′′′′

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

3

2

1

3

2

1

100010001

ccc

xxx

11 cx ′′′=

22 cx ′′′=

33 cx ′′′=

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The most commonly used iterative method for linear equations solving.The linear equations were derived so that the first equation can be solved for x1, the second can be solved for x2 and the third can be solved for x3.

a11x1 + a12x2 + a13x3 + .. + a1nxn = c1a21x1 + a22x2 + a23x3 + .. + a2nxn = c2a31x1 + a32x2 + a33x3 + .. + a3nxn = c3

: : : : : :an1x1 + an2x2 + an3x3 + .. + annxn = cn

Gauss Seidel Method

22

232312122 a

xaxaxacx nn−−−−=

L

x1 =

c1 − a12x2 − a13x3 −L− a1nxn

a11

33

323213133 a

xaxaxacx nn−−−−=

L

nn

mnmnnnn a

xaxaxacx −−−−=

L3322

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Step for Gauss Seidel

x1 =

c1

a11

x3 =

(c3 − a31x1 − a32x2 )a33

x2 =

(c2 − a21x1 )a22

x1 =

(c1 − a12 x2 − a13x3 )a11

x2 =

(c2 − a21x1 − a23x3 )a22

x3 =

(c3 − a31x1 − a32x2 )a33

Second IterationFirst Iteration

εa ,i =

xij − xi

j−1

xij 100% < εs

convergence criteria where j and j -1 arethe present and previous iterations

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Gauss Seidel (assignment in class)

Use the Gauss Seidel method to solve the following equations (εs = 5 %).

17x1 - 2x2 - 3x3 = 500-5x1 + 21x2 - 2x3 = 200-5x1 - 5x2 + 22x3 = 30

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Gauss Seidel – assignment in classFigure 1.0 shows a chemical process consists of 3 reactors linked by pipes. The mass flowrateof a chemical (g/s) through each pipe is equal to its concentration in each reactor, c (g/m3) multiplied by the volume flowrate (m3/s) of the pipe. Assume the system is at a steady state, so that the transfer into each reactor will balance the transfer out. Develop mass-balance equations for the reactors, and solve the equations simultaneously for the unknown concentrations (c1, c2, c3) using Gauss-Siedel method with εs = 5%.

R1

R2 R3

(5 m3/s)(c1)(2 m3/s)(c2) (3 m3/s)(c3)

(2 m3/s)(c3)

(5 m3/s)(c2)(14 m3/s)(c2)

(7 m3/s)(c1)

200 g/s

(17 m3/s)(c3)30 g/s

Figure 1.0 (5 m3/s)(c1)

500 g/s

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Numerical Methods for Chemical Engineers · Numerical Methods for Chemical Engineers Chapter 4: ... This chapter deals with the case of determining the values ... 20 PM