skf2133-chapter5n

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Chemical Engineering Numerical Method

copyright PCS- FKKKSA, UTM

Numerical Methods forNumerical Methods for

Chemical EngineersChemical EngineersChapter 5: Curve FittingChapter 5: Curve Fitting

SaharudinSaharudin HaronHaron


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Curve FittingCurve Fitting

Data is often given as a discrete values along a continuum and

problems may occur when one need to estimate points between thediscrete values.

This chapter describes techniques to fit curves to such data to obtain

intermediate estimates.

There are 2 general approaches :

a) Least-squares regression

b) Interpolation


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Linear Regression Fitting a straight line that represents the general trend of the data.

y = a o

+ a1 x + e

where

e = error or residual n = number of points

( )

x xn

x x

n

x

x x

x x

of meanarithmetic

yof meanarithmeticy yy

intercepta a-ya

slopea -n

y-yn a

i

i

o1o

12i

2i

iiii1

=∑

=

=∑=

==

=∑ ∑

∑ ∑ ∑=


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Example 5.1:

Fit a straight line from the Table 1 below.

Table 1

xi 1 2 3 4 5 6 7

yi 0.5 2.5 2.0 4.0 3.5 6.0 5.5

Solution: n = 7 Σ xi yi = 119.5 Σ xi2 = 140 Σ yi = 24 Σ xi = 28

x = 28/7 = 4 a1 = 0.8392857

y = 24/7 = 3.42857 a0 = 0.07142857

Therefore, the least-squares fit is

y = 0.01742857 + 0.8392857x

Linear Regression

x

y


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Quantification of Error of Linear Regression Standard deviation, Sy

( )∑ −=−=

2

t

t

y

s

1

S

s

y y

n

i

- the measurement of spread of data around the mean

- the sum of the squares of the residuals between the

data points and the mean

Variance, Sy2 → (standard deviation)2

1

S s t

2

y−

=n

Regression error, Sr

( )∑ ∑= =

==n

1

n

1i

2

i1oi

2

ir xa-a-yesi

- the sum of the squares of the residuals

between the measured and calculated y


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Quantification of Error of Linear Regression Standard error of the estimate

2

S

sr

y/ −= n x- the measurement of spread of data around

the regression line

Coefficient of determination

t

r t2S

SS r −= - shows the efficiency of the estimation of spread of the data points using regression line compare to the mean value

tcoefficienncorrelatio- r r

or

2=

i.e. r 2 = 0.923 - indicate that 92.3 % of the original uncertainty has been

explained by the linear model


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Quantification of Error of Linear Regression

The standard deviation is:

sy

= √[(22.7143 / (7 – 1)] = 1.9457

The standard error is:

s y/x = √ [(2.9911 / (7-2)] = 0.7735

Thus because Sy/x < Sy, the linear regression ,model has merit.

r 2 = (St – Sr ) / St = 22.7143 – 2.9911/ (22.7143) = 0.868

or

r = √r 2 = √(0.868)2 = 0.932

The results indicate that 86.8% of original uncertainty has been

explained by the linear model.


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Interpolation• To estimate intermediate values between precise data points. The most

common method used for this purpose is polynomial interpolation.

• General formula for an nth-order polynomial

• f(x) = a o + a1 x + a 2 x 2 + …….. + a n x

n

• Several methods of interpolating polynomials such as :

•1st order (linear) connecting 2 points.•2nd order (quadratic or parabolic) connecting 3 points.

•3rd order (cubic) connecting 4 points.

• The techniques that will be used are:-•Newton’s Interpolating Polynomials

•Lagrange Interpolating Polynomial

•Splines Interpolation


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Quadratic (or parabola) Interpolation

Connecting 3 points with a curve.

f 2( x) = b0 + b1( x- x0) + b2( x- x0)( x- x1)

where :

b0

= f( x0)

f( x1) - f( x0)

x1 - x0

f( x2) - f( x1) - f( x1) - f( x0)

x2 - x1 x1 - x0

x2 – x0

b1 = f [ x1, x0] =

b2 = f [ x2, x1, x0] =


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Cubic Interpolation (third order)

Connecting 4 points with a curve.

f 3( x) = f( x

0) + ( x- x

0) f[ x

1, x

o] + ( x- x

0)( x- x

1) f [ x

2, x

1, x

o]

+ ( x- x0)( x- x1)( x- x2) f [ x3, x2, x1, xo]

where :

f( x1) - f( x0) x1 - x0

f [ x2, x1] - f [ x1, x0]

x2 - x0

f [ x3 , x2, x1] - f [ x2, x1, x0]

x3

- x0

f [ x1, x0] =

f [ x2

, x1

, x0

] =

f [ x3, x2, x1, x0] =


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Newton’s Divided-Difference Interpolating

Polynomials – assignment in class

Table 1 shows the relationship between applied stress and the time tofracture for a stainless steel:

Fracture time, h 5 10 15 20 25 30

Applied stress,kg/mm2

11.6 10.3 9.1 8.2 7.4 6.8

Estimate the fracture time for an applied stress of 8.5 kg/mm

2

. Use Newtoninterpolation of order 3.


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Lagrange Interpolating Polynomials Is a reformulation of the Newton polynomial that avoids the

composition of divided differences.

It can be represented as :

n

f n( x) = ∑ Li( x) f ( xi)

i=0

where Li( x) = ∏

≠

= −−n

i j

j ji

j

x x

x x

0


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Lagrange Interpolating Polynomials Example :

If n = 1

If n = 2

)()()( 101

00

10

11 x f

x x

x x x f

x x

x x x f

−−+

−−=

)())((

))((

)())((

))(()(

))((

))(()(

21202

10

12101

200

2010

212

x f x x x x

x x x x

x f x x x x

x x x x x f

x x x x

x x x x x f

−−

−−+

−−

−−+

−−

−−=


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Splines Interpolation

3 cases:

Linear Splines Quadratic Splines

Cubic Splines

Linear Splines

- is a simplest method where a straight line is drawn to connect 2 points.

- a group of ordered data points can be defined as a set of linear functions.


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A set of linear functions can be shown as below :-

f ( x) = f ( x0) + m0( x- x0) x0 ≤ x ≤ x1

f ( x) = f ( x1) + m1( x- x1) x1 ≤ x ≤ x2

:

:

f ( x) = f ( xn-1) + mn-1( x- xn-1) xn-1 ≤ x ≤ xn

where

mi = f ( xi+1) - f ( xi)

xi+1 - xi



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Splines Interpolation Example : Fit the data below with first order Splines. Evaluate the

function at x = 5

Solution : For the interval from x = 4.5 to x = 7.0

m = 2.5 - 1.0

7.0 - 4.5

So, f (5) = (1.0) + (0.6)(5.0-4.5)

= 1.3

x f(x)3.0 2.5

4.5 1.0

7.0 2.5

9.0 0.5

= 0.6


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Quadratic Splines

the objective in quadratic splines is to derive a second order polynomial

for interval between data points.

The polynomial for each interval can be represented generally as :

f i(x) = ai x 2 + bi x + ci - general equation for Quadratic Splines

For n+1 data points, ( i = 0, 1, 2,….n) there are n intervals and

consequently, 3n unknowns constants (the a’s, b’s and c’s) to evaluate.


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The function values must be EQUAL at the interior knots. This conditioncan be represented as :

ai-1 x2

i-1 + bi-1 xi-1 + ci-1 = f ( xi-1)

ai x2 i-1 + bi xi-1 + ci = f ( xi-1)for i = 2 to n.

The first and last functions must pass through the end points.

a1 x20 + b1 x0 + c1 = f ( x0)

an x2n + bn xn + cn = f ( xn)

Conditions that are required to evaluate the unknowns


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Cubic Splines

- the objective is to derive a third order polynomial for each interval. f i (x) = ai x

3 + bi x 2 + ci x + d i

For a quadratic splines, 4n conditions are required to evaluate the

unknowns.These are :

The function values must be equal at the interior knots (2n-2 conditions)

The first and last functions must pass through the end points (2 conditions)

The first derivatives at the interior knots must be equal (n-1 conditions) The second derivatives at the interior knots must be equal (n-1 conditions)

The second derivatives at the end knots are zero (2 conditions)


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Splines Interpolation– assignment in class

Given the data:

x 2 2.5 3 4

f ( x) 5 7 8 2

Predict f (3.4) using linear and quadratic splines.

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