Process Model Formulation and Solution, 3E4

Process Model Formulation and Solution, 3E4

Instructor: Kevin Dunn [email protected]

Department of Chemical Engineering

Course notes: © Dr. Benoˆıt Chachuat

Why integral calculation?

Example 1

Calculate theheight of liquid after a given time:

z(t) =

Example 2

Calculate the (constant) water flowin a river/canal:

Numerical calculation of integrals:

quadrature

Problem statement:

Given a continuous function f : [a, b] → IR, calculate itsintegralon [a, b],

I = Z b

a

f (x ) dx

usingfunction evaluations only!

General approach:

Approximate: I = Z b

a

f (x ) dx

by the weighted sum: In= n

X

i =1

ωif (xi)

withnodal abscissas xi andweights ωi

Numerical calculation of integrals:

quadrature

Problem statement:

Given a continuous function f : [a, b] → IR, calculate itsintegralon [a, b],

I = Z b

a

f (x ) dx

usingfunction evaluations only! Two possible scenarios

1. Empirical function: Freeto choose the points at which to evaluate the function f

2. Experimental data: Function values available atdiscrete(fixed) data points only!

Outline and recommended readings

Newton-Cotes formulas The trapezoidal rule Simpson’s rules Romberg integration Gauss quadrature

2-point Gauss quadrature Higher-order Gauss quadrature

(Strongly) recommended readings:

I Chapters21.1-21.3and22.1-22.3in: S. C. Chapra, and R. P. Canale, “Numerical

Methods for Engineers”, McGraw Hill, 5th/6th edition

Newton-Cotes

formulas: principles

I Most commonnumerical integration schemes

I Requireequally spacedabscissas – Most useful if:

I f (x ) already available at equally spaced data points x1, ..., xn; or

I f (x ) can be computed at low cost

Main concept:

Approximate the integrand by a polynomial function, pn(x ) = a0+ a1x + · · · + anxn, that is easy to integrate:

I = Z b a f (x ) dx ≈ In= Z b a pn(x ) dx

The

trapezoidal rule: basics

I First Newton-Cotes formula:

I ≈ I1=

Z b

a

p1(x ) dx

with: p1(x ) = a0+ a1x

I Consider the twonodal abscissas: x1= aandx2= b

 p1(x1) = f (x1) p1(x2) = f (x2) =⇒ p1(x ) = I Trapezoidal rule: I1= Z b a p1(x ) dx = I Interpretation?

The trapezoidal rule:

application

Use the trapezoidal rule to estimate: I = Z π2 0 sin(x ) dx I I = I I1= I E1= |I − I1| =

The

composite

trapezoidal rule

Concept: Apply the trapezoidal rule in apiecewise fashion

I Divide [a, b] into n panels:

h = b − a n , I = n X k=1 Ik

I For each panel k = 1, . . . , n:

Ik = Z xk+1 xk f (x ) dx ≈ I₁k = hf (xk) + f (xk+1) 2 I Overall, I ≈ I1(n)= h 2 n X k=1 [f (xk) + f (xk+1)] = h 2[f (x1) + 2f (x2) + · · · + 2f (xn) + f (xn+1)]

The composite trapezoidal rule:

error

analysis

00_{(ξ), for some ξ ∈ [a, b]}

I The error E₁(n)= |I − I₁(n)| can be decreased by increasing n

I At the limit: lim

n→∞E (n) 1 = 0

How fast does the error go to zero?

E₁(n)= |I − I₁(n)| =O(h2)

The composite trapezoidal rule:

application

Use thecompositetrapezoidal rule to estimate: I = Z π2 0 sin(x ) dx I I = I I₁(2)= I E₁(2)= |I − I₁(2)| = n I₁(n) E₁(n) 1 7.8540e-01 2.1460e-01 2 9.4806e-01 5.1941e-02 4 9.8712e-01 1.2884e-02 8 9.9679e-01 3.2148e-03 16 9.9920e-01 8.0332e-04 32 9.9980e-01 2.0081e-04 64 9.9995e-01 5.0200e-05 128 9.9999e-01 1.2550e-05

I Large number of panels required for high accuracy

The composite trapezoidal rule:

coding

Write a function that estimates the integral I =Rb

a f (x )dx using the

Simpson’s rules: basics

I Concept: Develop Newton-Cotes formulas ofhigher order

I How? Use more points!

n = 3: Simpson’s 1/3 rule I ≈ I2= Z b a p2(x ) dx n = 4: Simpson’s 3/8 rule I ≈ I3= Z b a p3(x ) dx

Simpson’s 1/3 rule:

derivation

p2(x ),uniquequadratic polynomial

such that: p2(x1) = f (a) p2(x2) = f  a + b 2  p2(x3) = f (b)

Use of Lagrange interpolating polynomials:

I ≈ I2=

h

Simpson’s 1/3 rule:

application

Use Simpson’s 1/3 rule to estimate: I = Z π₂ 0 sin(x ) dx I I = I I2≈ I E2= |I − I2| ≈ 2.280 × 10−3

Simpson’s 3/8 rule:

derivation

p3(x ),uniquecubic polynomial s.t.

p3(x1) = f (a) p3(x2) = f  2a + b 3  p3(x3) = f  a + 2b 3  p3(x4) = f (b)

Use of Lagrange interpolating polynomials:

I ≈ I3=

3h

Simpson’s 3/8 rule:

application

Use Simpson’s 3/8 rule to estimate: I = Z π2 0 sin(x ) dx I I = I I3≈ I E3= |I − I3| ≈ 1.005 × 10−3

Composite

Simpson’s 1/3 rule

I Divide [a, b] into n panels, withn even: h = b − a n , I = n/2 X k=1 Ik

I For each pair of panels k = 1, 2, . . . , n/2:

Ik = Z x2k+1 x2k−1 f (x ) dx ≈ I2k = h 3[f (x2k−1) + 4f (x2k) + f (x2k+1)] I Overall, h n/2

Composite

Simpson’s 1/3 rule (cont’d)

(4)_{(ξ), for some xi ∈ [a, b]}

I Simpson’s 1/3 ruleexactfor a polynomial of degree 3 or less Convergence speed (with decreasing h):

E₂(n)= |I − I₂(n)| =O(h4)

I Halving the stepsize, divides the error by sixteen

I One order higher than expected! Case of an odd number of panels:

I Use Simpson’s 1/3 rule for all but last 3 panels

I Use Simpson’s 3/8 rule for last 3 panels

I ≈h 3 n−3 2 X k=1 [f (x2k−1) + 4f (x2k) + f (x2k+1)] + 3h 8 [f (xn−2) + 3f (xn−1) + 3f (xn) + f (xn+1)]

Composite Simpson’s 1/3 rule:

application

Use thecompositeSimpson’s 1/3 rule to estimate: I = Z π₂ 0 sin(x ) dx I I = I I₂(2)= I E₂(2)= |I − I₂(2)| = n I₁(n) E₁(n) 2 1.00227988 2.2799e-03 4 1.00013458 1.3458e-04 8 1.00000830 8.2955e-06 16 1.00000052 5.1668e-07 32 1.00000003 3.2265e-08

I Much smaller number of panels required for high accuracy (compared to composite trapezoidal rule)

Refresher:

Richardson extrapolation

Principle:

Combine two estimatesof a given order, O(hp_{), in order to obtain a new}

estimate of higher-order, O(hp+1₎

I Suppose we want to calculate some quantity Q

I Let Qpbe an estimate of Q of order p,

Q = Qp(h) + O(hp)

I Then, an estimate Qp+1 of order p + 1 is given by: Why?

Q = Qp+1(h) + O(hp+1) =

2p_Q

p(1₂h) − Qp(h)

2p_{+ 1} + O(h p+1₎

I Successfully applied to numerical differentiation

Romberg

integration: principles

Principle:

ApplyRichardson extrapolationto numerical integration, starting with thecomposite trapezoidal rule

Starting point: −→ O(h2₎

I Composite trapezoidal rule: n panels, h = b−a n I1(h) = h 2 n X k=1 [f (xk) + f (xk+1)] = h 2[f (x1) + 2f (x2) + · · · + 2f (xn) + f (xn+1)] I Truncation error: I − I1(h) = O(h2) = α2h2+ α4h4+ · · · + α2kh2k+ · · ·

Romberg

integration: principles

1st extrapolation: −→ O(h4₎

I Combine I1(h) and I1(1₂h) to eliminate theterm in h2, How?

I = 4I1( 1 2h) − I1(h) 3 | {z } I2(1₂h) +α0₄h4+ O(h6)

I Link with the composite Simpson’s 1/3 rule? 2nd extrapolation: −→ O(h6₎

I Combine I2(1₂h) and I2(1₄h) to eliminate theterm in h4, How?

I = 4 2_I 2(1₄h) − I2(1₂h) 42_{− 1} | {z } I3(1₄h) +α00₆h6+ O(h8)

Romberg integration:

algorithm

I

(

h) =

4

I

(

h) − I

(

h)

4

_{− 1}

,

j ≥ 1, k ≥ 1

O(h2) O(h4) O(h6)

I1(h) & I1(1₂h) → I2(1₂h) = 4I1(1₂h) − I1(h) 4 − 1 & & 1 1 4I1( 1 22h) − I1(1₂h) ₁ 42I2(₂12h) − I2(1₂h)

Romberg integration:

application

Use Romberg integration to calculate an O(h8) estimate of: I =

Z π₂

0

sin(x ) dx , starting with h = π₂ (1 panel)

h I1, O(h2) I2, O(h4) I3, O(h6) I4, O(h8)

π 2 0.7853981633 π 4 0.9480594489 1.0022798774 π 8 0.9871158009 1.0001345849 0.9999915654 π 16 0.9967851718 1.0000082955 0.9999998762 1.0000000081

E 3.2148e-03 8.2955e-06 1.2377e-07 8.1440e-09

I I2(π₄) =

I How manyfunction evaluationsare required to get an accuracy of ca. 10−8 for I ?

Romberg integration:

coding

Write a function that estimates the integral I =Rb

a f (x )dx using

Romberg integration; make use of the trapezoidal function (see earlier slide)

Gauss quadrature: principles

Approximate: I = Z b

a

f (x ) dx by the weighted sum: In= n X k=1 ωif (xi) So far, I Newton-Cotes formulas I Romberg integration   

The nodal abscissas xi areevenly

spacedin [a, b]

Only the weights ωi are “optimized”

Gauss Quadrature:

Choose the nodal abscissas and the weightssimultaneously in order to achieve maximal accuracy!

2-point

Gauss quadrature

Find the coefficients ω1, ω2, x1, x2such that:

I = Z b

a

2-point

Gauss quadrature

Z 1

−1

f (x ) dx 0. Condition for aconstant term: f (x ) = 1

Z 1 −1 1 dx = n X k=1 ωi ⇒

1. Condition for alinear term: f (x ) = x Z 1 −1 x dx = n X k=1 ωixi ⇒

2. Condition for aquadratic term: f (x ) = x2

Z 1 −1 x2dx = n X k=1 ωixi2 ⇒

3. Condition for acubic term: f (x ) = x3 Z 1 −1 x3dx = n X k=1 ωixi3 ⇒

2-point

Gauss quadrature (cont’d)

Solution to the nonlinear algebraic equations: Check! ω1= ω2= 1, x1= − 1 √ 3, x2= 1 √ 3

2-point Gauss quadrature on [−1, 1]:

I = Z 1 −1 f (x ) dx ≈ I2= f  −_√1 3  + f_√1 3 

2-point Gauss quadrature on a general interval[a, b]: Why?

2-point Gauss quadrature:

application

Use2-point Gauss quadratureto estimate: I = Z π₂ 0 sin(x ) dx Nodal abscissas: I x1= I x2= Function evaluations: I f (x1) = I f (x2) =

2-point Gauss quadrature estimate:

I = Z π₂

0

sin(x ) dx ≈ I2=

E2= |I − I2| ≈ 1.527 × 10−3

I Error of the same magnitude as with Simpson’s 1/3 and 3/8 rule

Higher-order

Gauss quadrature

The good news: exact same procedure as with 2 points!

Find the coefficients ω1, . . . , ωn, x1, . . . , xn such that:

I = Z b a f (x ) dx ≈ n X k=1 ωkf (xk) 2n conditions required:

I Now, require the formula to beexactfor termsup to order 2n

Coefficients for Gauss-quadrature on [−1, 1]

n = 3 ±xk ωk 0.000 000 0.888 889 n = 4 ±xk ωk 0.339 981 0.652 145 n = 5 ±xk ωk 0.000 000 0.568 889

3-point Gauss quadrature:

application

Use3-point Gauss quadratureto estimate: I = Z π₂ 0 sin(x ) dx Nodal abscissas: I x1= I x2= I x3= Function evaluations: I f (x1) = I f (x2) = I f (x3) =

3-point Gauss quadrature estimate:

I = Z π2

0

sin(x ) dx ≈ I3=