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ICE 203

Chemical Process Computation

Department of Chemical EngineeringFaculty of Industrial TechnologyParahyangan Catholic University

January 2013

Main Aim

• To learn about some numerical methods to be

used in solving Chemical Engineering

Problems.

• To introduce MATLAB as a tool to solve

problems in Chemical Engineering.

• To apply the numerical methods in MATLAB

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Schedule and ContentsWeek Topic

I Introduction

II Solution of Non Linear Equation f(x)=0

III Solution of Non Linear Equation f(x)=0

IV Solution of Linear Equation AX=B

V Solution of Linear Equation AX=B

VI Curve Fitting & Interpolation

VII Numerical Optimization

VIII MID TERM EXAM

Schedule and ContentsWeek Topic

IX Numerical Differentiation/Integration

X Numerical Differentiation/Integration

XI Solution to Ordinary Differential Equation

XII Solution to Ordinary Differential Equation

XIII Solution to Partial Differential Equation

XVI FINAL EXAM

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5

Grading

Mid Term Exam (UTS) 35%

Final Exam (UAS) 35%

Labs+Assignments + quizes 30%

6

Outlines of the Course

• Solution of nonlinear

Equations

• Interpolation

• Numerical Differentiation

• Numerical Integration

• Solution of linear Equations

• Least Squares curve fitting

• Solution of ordinary

differential equations

• Solution of Partial differential

equations

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7

Solution of Nonlinear Equations

• Some simple equations can be solved analytically:

• Many other equations have no analytical solution:

31

)1(2

)3)(1(444solution Analytic

034

2

2

−=−=

−±−=

=++

xandx

roots

xx

solution analytic No052 29

=

=+−− x

ex

xx

8

Methods for Solving Nonlinear

Equations

o Bracketing Method : Bisection and False

Position Method

o Open Method : Fixed Point, Newton

Raphson and Secant Method

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9

Solution of Systems of Linear

Equations

unknowns. 1000in equations 1000

have weif do What to

123,2

523,3

:asit solvecan We

52

3

12

2221

21

21

=−==⇒

=+−−=

=+

=+

xx

xxxx

xx

xx

10

Cramer’s Rule is Not Practical

this.compute toyears 10 than more needscomputer super A

needed. are tionsmultiplica102.3 system, 30by 30 a solve To

tions.multiplica

1)N!1)(N(N need weunknowns, N with equations N solve To

problems. largefor practicalnot is Rule sCramer'But

2

21

11

51

31

,1

21

11

25

13

:system thesolve toused becan Rule sCramer'

20

35

21

×

−+

==== xx

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11

Methods for Solving Systems of Linear

Equations

o Naive Gaussian Elimination

o Gaussian Elimination with Scaled Partial

Pivoting

o Algorithm for Tri-diagonal Equations

12

Curve Fitting

• Given a set of data:

• Select a curve that best fits the data. One

choice is to find the curve so that the sum of

the square of the error is minimized.

x 0 1 2

y 0.5 10.3 21.3

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Interpolation

• Given a set of data:

• Find a polynomial P(x) whose graph passes

through all tabulated points.

xi 0 1 2

yi 0.5 10.3 15.3

tablein the is)( iii xifxPy =

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Methods for Curve Fitting

o Least Squares

o Linear Regression

o Nonlinear Least Squares Problems

o Interpolation

o Newton Polynomial Interpolation

o Lagrange Interpolation

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15

Solution of Ordinary Differential

Equations

only. cases special

for available are solutions Analytical *

equations. thesatisfies that function a is

0)0(;1)0(

0)(3)(3)(

:equation aldifferenti theosolution tA

x(t)

xx

txtxtx

==

=++

&

&&&

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Solution of Partial Differential Equations

Partial Differential Equations are more difficult

to solve than ordinary differential equations:

)sin()0,(,0),1(),0(

022

2

2

2

xxututu

t

u

x

u

π===

=+∂

∂+

∂

∂

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Text Books

• John H. Mathews, Kurtis D. Fink, Numerical Methods

With MATLAB, Prentice Hall, NJ, 1999.

• Steven Chapra, Raymond Canale, Numerical

Methods For Engineers,McGrawHill,2010

• Michael B. Cutlip, Mordechai Shacham, Problem

solving in Chemical Engineering With Numerical

Methods, Prentice Hall,NJ,1999.

• Kenneth J Beers, Numerical Methods for Chemical

Engineers, Cambridge Univ Press, Edinburgh, 2007

• Alkis Constantinides, Navid Mostoufi, Numerical

Methods for Chemical Engineers with MATLAB

Applications,Prentice Hall,NJ,1999

Numerical Method

&

Mathematical Modeling

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Numerical Methods - Definitions

Numerical Methods

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Analytical vs. Numerical methods

Analytical vs. Numerical methods

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Reasons to study numerical Analysis

• Powerful problem solving techniques and can be

used to handle large systems of equations

• It enables you to intelligently use the commercial

software packages as well as designing your own

algorithm.

• Numerical Methods are efficient vehicles in learning

to use computers

• It Reinforce your understanding of mathematics;

where it reduces higher mathematics to basic

arithmetic operation.

Mathematical Modeling

Mathematical Model

• A formulation or equation that expresses the essential

features of a physical system or process in mathematical

terms.

• Generally, it can be represented as a functional

relationship of the form

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Mathematical Modeling

Typical characteristics of Math. model

• It describes a natural process or system in

mathematical way

• It represents the idealization and simplification

of reality.

• It yields reproducible results, and can be used

for predictive purpose.

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Example Simple Process Model

qf

q=kh

h

k

qK

k

AKth

dt

tdh

tkhqdt

tdhA

tqqdt

tAhd

tqqdt

tmd

f

f

f

f

===+

−=

−=

−=

,;)()(

)()(

)]([)]([

)]([)]([

ττ

ρρ

ρ

Total Mass Balance

qf : inlet volumetric flowrate

q : outlet volumetric flowrate

h(0)=0

Analytical solution

Solving the differential equations, it gives

h(t

)

t

hs=K

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Numerical solution

)]([1)(

thKdt

tdh−=

τ)]([

1)()(

1

1i

ii

ii thKtt

thth−=

−

−

+

+

τ

)))](((1

[)()( 11 iiiii ttthKthth −−+= ++τ

New Value Old Value Step Size

Comparison between Analytical vs. Numerical Solution

t

h

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Pre-computer era computer era

Approximations and Errors

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Approximations and Errors

• The major advantage of numerical analysis is that

a numerical answer can be obtained even when a

problem has no “analytical” solution.

• Although the numerical technique yielded close

estimates to the exact analytical solutions, there

are errors because the numerical methods involve

“approximations”.

Chapter 3 34

Approximations and Round-Off Errors

• For many engineering problems, we cannot obtain analytical solutions.

• Numerical methods yield approximate results, results that are close to the exact analytical solution.– Only rarely given data are exact, since they originate from

measurements. Therefore there is probably error in the input information.

– Algorithm itself usually introduces errors as well, e.g., unavoidable round-offs, etc …

– The output information will then contain error from both of these sources.

• How confident we are in our approximate result?

• The question is “how much error is present in our calculation and is it tolerable?”

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Accuracy and Precision

• Accuracy refers to how

closely a computed or

measured value agrees

with the true value.

• Precision refers to how

closely individual

computed or measured

values agree with each

other.

• Bias refers to systematic

deviation of values from

the true value.

Error Definition

Numerical errors arise from the use of approximations

Truncation errors Round-off errors

Errors

Result when approximations

are used to represent

exact mathematical

procedure.

Result when numbers having

limited significant figures are

used to represent exact

numbers.

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Round-off Errors

• Numbers such as π, e, or cannot be expressed by

a fixed number of significant figures.

• Computers use a base-2 representation, they cannot

precisely represent certain exact base-10 numbers

Example:

π = 3.14159265358 to be stored carrying 7 significant digits.

π = 3.141592 chopping

π = 3.141593 rounding

7

Truncation Errors

• Truncation errors are those that result using

approximation in place of an exact mathematical

procedure.

( ) ( )1

1

i i

i i

V t V tdv v

dt t t t

+

+

−∆≈ =

∆ −

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True Error

tε

True error (Et) or Exact value of error

= true value – approximated value

�True error (Et)

�True percent relative error ( )

(%)100

(%)100

×−

=

×==

valuetrue

valueedapproximatvaluetrue

valueTrue

errorTrueerrorrelativepercentTrue tε

Example

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Example

Approximate Error

• The true error is known only when we deal with functions that

can be solved analytically.

• In many applications, a prior true value is rarely available.

• For this situation, an alternative is to calculate an

approximation of the error using the best available estimate of

the true value as:

(%)100×==ionapproximat

erroreApproximaterrorrelativepercenteApproximataε

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Approximate Error

• In many numerical methods a present approximation is

calculated using previous approximation:

aε tε

(%)100×−

=ionapproximatpresent

ionapproximatpreviousionapproximatpresentaε

Note:

- The sign of or may be positive or negative

- We interested in whether the absolute value is lower

than a prespecified tolerance (εεεεs), not to the sign of error.

Thus, the computation is repeated until (stopping criteria):

sa εε <

Prespecified Error

• We can relate (εεεεs) to the number of significant

figures in the approximation,

So, we can assure that the result is correct to

at least n significant figures if the following

criteria is met:

%)105.0( 2 n

s

−×=ε

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Example

The exponential function can be computed using Maclaurin

series as follows:

Estimate e0.5 using series, add terms until the absolute value of

approximate error εa fall below a pre-specified error εs

conforming with three significant figures.

{The exact value of e0.5=1.648721…}

• Solution

2 3

12! 3! !

nx x x x

e xn

= + + + + +L

( )2 30.5 10 % 0.05%s

ε −= × =

� Using one term:

� Using two terms:

� Using three terms:

0.5 1.648721 1.01 100% 39.3

1.648721t

e ε−

= = =

εεεεa%εεεεt%ResultsTerms

---39.31.01

33.39.021.52

7.691.441.6253

1.270.1751.6458333334

0.1580.01721.6484375005

0.01580.001421.6486979176

0.5 1.648721 1.5 1.5 1.01 0.5 1.5 100% 9.02% 100% 33.3%

1.648721 1.5t ae ε ε

− −= + = = = = =

20.5 0.5 1.648721 1.625 1.625 1.0

1 0.5 1.625 100% 1.44% 100% 7.69%2! 1.648721 1.625

t ae ε ε− −

= + + = = = = =

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Taylor Series

48

Motivation

• We can easily compute expressions like:

?)6.0sin(,4.1 computeyou do HowBut,

)4(2

103 2

+

×

x

way?practicala thisIs

sin(0.6)? compute to

definition theuse weCan

0.6

ab

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49

Taylor Series

∑∞0

)(

0

)(

3)3(

2)2(

'

)()( !

1 )(

: writecan weconverge, series theIf

)()( !

1

...)(!3

)()(

!2

)()()()(

:about )( of expansion seriesTaylor The

=

∞

=

−=

−=

+−+−+−+

∑

k

kk

k

kk

axafk

xf

axafk

SeriesTaylor

or

axaf

axaf

axafaf

axf

Taylor Series - Example

Use zero-order to fourth-order Taylor series expansions to approximate the function.

f(x)= -0.1x4 – 0.15x3 – 0.5x2 – 0.25x +1.2

From xi = 0 with h =1. Predict the function’s value at xi+1 =1.

Solution� f(xi)= f(0)= 1.2 , f(xi+1)= f(1) = 0.2 ………exact solution

• Zero- order approx. (n=0) ���� f(xi+1)=1.2

Et = 0.2 – 1.2 = -1.0

• First- order approx. (n=1) � f(xi+1)= 0.95

f(x)= -0.4x3 – 0.45x2 – x – 0.25, f ’(0)= -0.25f( xi+1)= 1.2- 0.25h = 0.95Et = 0.2 - 0.95 = -0.75

)()(1 ii

xfxf =+

hxfxfxf iii)()()(

'

1+=

+

50

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Taylor Series - Example

• Second- order approximation (n=2) � f(xi+1)= 0.45

f’’(x) = -1.2 x2 – 0.9x -1 , f ’’(0)= -1

f( xi+1)= 1.2 - 0.25h - 0.5 h2 = 0.45

Et = 0.2 – 0.45 = -0.25

• Third-order approximation (n=3) � f(xi+1)= 0.3

f( xi+1)= 1.2 - 0.25h - 0.5 h2 – 0.15h3 = 0.3

Et = 0.2 – 0.3 = -0.1

!2

)('')()()(

2'

1

hxfhxfxfxf i

iii++=

+

!3

)(

!2

)('')()()(

3(3)2'

1

hxfhxfhxfxfxf ii

iii+++=

+

51

Taylor Series - Example

• Fourth-order approximation (n = 4) � f(xi+1)= 0.2

f( xi+1)= 1.2 - 0.25h - 0.5 h2 – 0.15h3 – 0.1h 4= 0.2

Et = 0.2 – 0.2 = 0

The remainder term (R4) = 0

because the fifth derivative of the fourth-order polynomial is

zero.

!4

)(

!3

)(

!2

)('')()()(

4)4(3)3(2'

1

hxfhxfhxfhxfxfxf iii

iii++++=

+

5)5(

4!5

)(h

fR

ξ=

52

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Approximation using Taylor Series Expansion

The nth-order Approximation

Taylor Series

• In General, the n-th order Taylor Series will be exact

for n-th order polynomial.

• For other differentiable and continuous functions,

such as exponentials and sinusoids, a finite number of

terms will not yield an exact estimate.

• Each additional term will contribute some

improvement.

54

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55

Maclaurin Series

�Maclaurin series is a special case of Taylor series with the center of expansion a = 0.

∑∞0

)(

3)3(

2)2(

'

)0( !

1 )(

: writecan weconverge, series theIf

...!3

)0(

!2

)0()0()0(

:)( of expansion series n MaclauriThe

=

=

++++

k

kkxf

kxf

xf

xf

xff

xf

56

Maclaurin Series – Example 1

∞.xfor converges series The

...!3!2

1!

)0(!

1

11)0()(

1)0()(

1)0(')('

1)0()(

32∞0

∞0

)(

)()(

)2()2(

∑∑<

++++===

≥==

==

==

==

==

xxx

k

xxf

ke

kforfexf

fexf

fexf

fexf

k

k

k

kkx

kxk

x

x

x

xexf =)( of expansion series n MaclauriObtain

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Taylor SeriesExample 1

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 10

0.5

1

1.5

2

2.5

3

1

1+x

1+x+0.5x2

exp(x)

58

Maclaurin Series – Example 2

∞.xfor converges series The

....!7!5!3!

)0()sin(

1)0()cos()(

0)0()sin()(

1)0(')cos()('

0)0()sin()(

753∞0

)(

)3()3(

)2()2(

∑<

+−+−==

−=−=

=−=

==

==

=

xxxxx

k

fx

fxxf

fxxf

fxxf

fxxf

k

kk

: )sin()( of expansion series n MaclauriObtain xxf =

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59

-4 -3 -2 -1 0 1 2 3 4-4

-3

-2

-1

0

1

2

3

4

x

x-x3/3!

x-x3/3!+x5/5!

sin(x)

60

Maclaurin Series – Example 3

( )

( )

( )

1 ||for converges Series

...xxx1x1

1 :ofExpansion Series Maclaurin

6)0(1

6)(

2)0(1

2)(

1)0('1

1)('

1)0(1

1)(

of expansion series n MaclauriObtain

32

)3(

4

)3(

)2(

3

)2(

2

<

++++=−

=−

=

=−

=

=−

=

=−

=

−=

x

fx

xf

fx

xf

fx

xf

fx

xf

x1

1f(x)

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61

Taylor Series – Example 4

...)1(3

1)1(

2

1)1( :Expansion SeriesTaylor

2)1(1)1(,1)1(',0)1(

2)(,

1)(,

1)(',)ln()(

)1(at )ln( of expansion seriesTaylor Obtain

32

)3()2(

3

)3(

2

)2(

−−+−−−

=−===

=−

===

==

xxx

ffff

xxf

xxf

xxfxxf

axf(x)

62

Convergence of Taylor Series

• The Taylor series converges fast (few terms are needed)

when x is near the point of expansion. If |x-a|=h is large

then more terms are needed to get a good

approximation.

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63

Taylor’s Theorem

. and between is)()!1(

)(

:where

)(!

)( )(

:by given is )( of value the then and containing interval an on

1)( ..., 2, 1, orders of sderivative possesses )( functiona If

1)1(

0

)(∑

xaandaxn

fR

Raxk

afxf

xfxa

nxf

nn

n

n

n

k

kk

ξξ +

+

=

−+

=

+−=

+

(n+1) terms Truncated

Taylor Series

Remainder

64

Error Term

. and between allfor

)()!1(

)(

:on boundupper an derive can we

error, ionapproximat about theidea anget To

1)1(

xaofvalues

axn

fR

nn

n

ξ

ξ ++

−+

=

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65

Error Term - Example

( ) 0514268.82.0)!1(

)()!1(

)(

1≥≤)()(

31

2.0

1)1(

2.0)()(

−≤⇒+

≤

−+

=

=

+

++

ERn

eR

axn

fR

nforefexf

nn

nn

n

nxn

ξ

ξ

?2.00at

expansion seriesTaylor its of3)( terms4first the

by )( replaced weiferror theis largeHow

==

=

=

xwhena

n

exfx

Summary- Taylor Series

• Truncation error is decreased by addition of terms to

the Taylor series.

• If h is sufficiently small, only a few terms may be

required to obtain an approximation close enough to

the actual value for practical purposes.

66