Appendix A Numerical Techniques

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A. Numerical Techniques

Lake Michigan—unsalted and shark free.

A.1 Useful Integrals in Reactor Design

$\int_{0}^{χ} \frac{d x}{1 - x} = \ln \frac{1}{1 - x}$ $\int_{0}^{χ} \frac{d x}{1 - x} = \ln \frac{1}{1 - x}$ (A-1)

$\int_{x_{1}}^{x_{2}} \frac{d x}{(1 - x)^{2}} = \frac{1}{1 - x_{2}} - \frac{1}{1 - x_{1}}$ $\int_{x_{1}}^{x_{2}} \frac{d x}{(1 - x)^{2}} = \frac{1}{1 - x_{2}} - \frac{1}{1 - x_{1}}$ (A-2)

$\int_{0}^{x} \frac{d x}{(1 - x)^{2}} = \frac{χ}{1 - x}$ $\int_{0}^{x} \frac{d x}{(1 - x)^{2}} = \frac{χ}{1 - x}$ (A-3)

$\int_{0}^{x} \frac{d x}{1 + ε x} = \frac{1}{ε} \ln (1 + ε x)$ $\int_{0}^{x} \frac{d x}{1 + ε x} = \frac{1}{ε} \ln (1 + ε x)$ (A-4)

$\int_{0}^{x} \frac{(1 + ε x) d x}{1 - x} = (1 + ε) \ln \frac{1}{1 - x} - ε x$ $\int_{0}^{x} \frac{(1 + ε x) d x}{1 - x} = (1 + ε) \ln \frac{1}{1 - x} - ε x$ (A-5)

$\int_{0}^{x} \frac{(1 + ε x) d x}{(1 - x)^{2}} = \frac{(1 + ε) x}{1 - x} - ε \ln \frac{1}{1 - x}$ $\int_{0}^{x} \frac{(1 + ε x) d x}{(1 - x)^{2}} = \frac{(1 + ε) x}{1 - x} - ε \ln \frac{1}{1 - x}$ (A-6)

$\int_{0}^{x} \frac{(1 + ε x)^{2} d x}{(1 - x)^{2}} = 2 ε (1 + ε) \ln (1 - x) + ε^{2} x + \frac{(1 + ε)^{2} x}{1 - x}$ $\int_{0}^{x} \frac{(1 + ε x)^{2} d x}{(1 - x)^{2}} = 2 ε (1 + ε) \ln (1 - x) + ε^{2} x + \frac{(1 + ε)^{2} x}{1 - x}$ (A-7)

$\int_{0}^{x} \frac{d x}{(1 - x) (Θ_{B} - x)} = \frac{1}{Θ_{B} - 1} \ln \frac{Θ_{B} - x}{Θ_{B} (1 - x)}$ $\int_{0}^{x} \frac{d x}{(1 - x) (Θ_{B} - x)} = \frac{1}{Θ_{B} - 1} \ln \frac{Θ_{B} - x}{Θ_{B} (1 - x)}$ $Θ_{B} \neq 1$ $Θ_{B} \neq 1$ (A-8)

$\int_{0}^{W} (1 - α W)^{1 / 2} d W = \frac{2}{3 α} [1 - (1 - α W)^{3 / 2}]$ $\int_{0}^{W} (1 - α W)^{1 / 2} d W = \frac{2}{3 α} [1 - (1 - α W)^{3 / 2}]$ (A-9)

$\int_{0}^{x} \frac{d x}{a x^{2} + b x + c} = \frac{- 2}{2 a x + b} + \frac{2}{b}$ $\int_{0}^{x} \frac{d x}{a x^{2} + b x + c} = \frac{- 2}{2 a x + b} + \frac{2}{b}$ for $b^{2} = 4 a c$ $b^{2} = 4 a c$ (A-10)

$\int_{0}^{x} \frac{d x}{a x^{2} + b x + c} = \frac{1}{a (p - q)} \ln (\frac{q}{p} \cdot \frac{x - p}{x - q})$ $\int_{0}^{x} \frac{d x}{a x^{2} + b x + c} = \frac{1}{a (p - q)} \ln (\frac{q}{p} \cdot \frac{x - p}{x - q})$ for b² > 4ac (A-11)

where p and q are the roots of the equation.

$a x^{2} + b x + c = 0$ $a x^{2} + b x + c = 0$ i.e., $p,$ $p,$ $q = \frac{- b \mp \sqrt{b^{2} - 4 a c}}{2 a}$ $q = \frac{- b \mp \sqrt{b^{2} - 4 a c}}{2 a}$

$\int_{0}^{x} \frac{a + b x}{c + g x} d x = \frac{b x}{g} + \frac{a g - b c}{g^{2}} I n \frac{c + g x}{c}$ $\int_{0}^{x} \frac{a + b x}{c + g x} d x = \frac{b x}{g} + \frac{a g - b c}{g^{2}} I n \frac{c + g x}{c}$ (A-12)

A.2 Equal-Area Graphical Differentiation

There are many ways of differentiating numerical and graphical data (cf. Chapter 7). We shall confine our discussions to the technique of equal-area differentiation. In the procedure delineated here, we want to find the derivative of y with respect to x.

This method finds use in Chapter 5.

1. Tabulate the (y_i, x_i) observations as shown in Table A-1.

2. For each interval, calculate Δx_n = x_n – x_n–1 and Δy_n = y_n – y_n–1.

TABLE A-1

x_i	y_i	Δx	Δy	$\frac{Δ y}{Δ x}$ $\frac{Δ y}{Δ x}$	$\frac{d x}{d y}$ $\frac{d x}{d y}$
x₁	y₁				$(\frac{d y}{d x})_{1}$ $(\frac{d y}{d x})_{1}$
		x₂–x₁	y₂–y₁	$(\frac{Δ y}{Δ x})_{2}$ $(\frac{Δ y}{Δ x})_{2}$
x₂	y₂				$(\frac{d y}{d x})_{2}$ $(\frac{d y}{d x})_{2}$
		x₃–x₂	y₃–y₂	$(\frac{Δ y}{Δ x})_{3}$ $(\frac{Δ y}{Δ x})_{3}$
x₃	y₃				$(\frac{d y}{d x})_{3}$ $(\frac{d y}{d x})_{3}$
x₄	y₄		etc.

3. Calculate Δy_n/Δx_n as an estimate of the average slope in an interval x_n–1 to x_n.

4. Plot these values as a histogram versus x_i. The value between x₂ and x₃, for example, is (y₃ – y₂)/(x₃ – x₂). Refer to Figure A-1.

Graph illustrates equal area differentiation with the help of a histogram and a curve.

Figure A-1 Equal-area differentiation.

5. Next, draw in the smooth curve that best approximates the area under the histogram. That is, attempt in each interval to balance areas such as those labeled A and B, but when th approximation is not possible, balance out over several intervals (as for the areas labeled C and D). From our definitions of Δx and Δy, we know that

$y_{n} - y_{1^{=}} Σ_{i = 2}^{n} \frac{Δ y}{Δ x_{i}} Δ x_{i}$ $y_{n} - y_{1^{=}} Σ_{i = 2}^{n} \frac{Δ y}{Δ x_{i}} Δ x_{i}$ (A-13)

The equal-area method attempts to estimate $d y / d x$ $d y / d x$ so that

$y_{n} - y_{1} = \int_{x_{1}}^{x_{n}} \frac{d x}{d y} d x$ $y_{n} - y_{1} = \int_{x_{1}}^{x_{n}} \frac{d x}{d y} d x$ (A-14)

that is, so that the area under Δy / Δx is the same as that under dy / dx, everywhere possible.

6. Read estimates of dy / dx from this curve at the data points x₁, x₂, ... and complete the table.

An example illustrating the technique is given on the CRE Web site, Appendix A.

Differentiation is, at best, less accurate than integration. This method also clearly indicates bad data and allows for compensation of such data. Differentiation is only valid, however, when the data are presumed to differentiate smoothly, as in rate-data analysis and the interpretation of transient diffusion data.

A.3 Solutions to Differential Equations

A.3.A First-Order Ordinary Differential Equations

See the CRE Web site, Appendix A.3.

$\frac{d y}{d t} + f (t) y = g (t)$ $\frac{d y}{d t} + f (t) y = g (t)$ (A-15)

Using integrating factor = exp $(\int f dt)$ $(\int f dt)$ , the solution is

$y = e^{- \int f dt} \int g (t) e^{\int f d t} d t + K_{1} e^{- \int f d t}$ $y = e^{- \int f dt} \int g (t) e^{\int f d t} d t + K_{1} e^{- \int f d t}$ (A-16)

Example A–1 Integrating Factor for Series Reactions

$\frac{d y}{d x} + k_{2} y = k_{1} e^{- k_{1} t}$ $\frac{d y}{d x} + k_{2} y = k_{1} e^{- k_{1} t}$

Comparing the above equation with Equation (A-15) we note

f(t) = k₂

and the integrating factor $= \exp \int k_{2} dt = e^{k_{2} t}$ $= \exp \int k_{2} dt = e^{k_{2} t}$

Integrating factor $= \exp \int k_{2} dt = e^{k_{2} t}$ $= \exp \int k_{2} dt = e^{k_{2} t}$

$\frac{d (y e^{k_{2^{t}}})}{d t} = e^{k_{2} t} k_{1} e^{- k_{1} t} = k_{1} e^{(k_{2} - k_{1}) t}$ $\frac{d (y e^{k_{2^{t}}})}{d t} = e^{k_{2} t} k_{1} e^{- k_{1} t} = k_{1} e^{(k_{2} - k_{1}) t}$

$e^{k_{2^{t}}} y = k_{1} {\int e}^{(k_{2} - k_{1}) t} d t = \frac{k_{1}}{k_{2} - k_{1}} e^{(k_{2} - k_{1}) t} + K_{1}$ $e^{k_{2^{t}}} y = k_{1} {\int e}^{(k_{2} - k_{1}) t} d t = \frac{k_{1}}{k_{2} - k_{1}} e^{(k_{2} - k_{1}) t} + K_{1}$

$y = \frac{k_{1}}{k_{2} - k_{1}} e^{- k_{1} t} + K_{1} e^{- k_{2} t}$ $y = \frac{k_{1}}{k_{2} - k_{1}} e^{- k_{1} t} + K_{1} e^{- k_{2} t}$

$t = 0 y = 0$ $t = 0 y = 0$

$y = \frac{k_{1}}{k_{2} - k_{1}} [e^{- k_{1} t} - e^{- k_{2} t}]$ $y = \frac{k_{1}}{k_{2} - k_{1}} [e^{- k_{1} t} - e^{- k_{2} t}]$

A.3.B Coupled Differential Equations

Techniques to solve coupled first-order linear ODEs such as

$\frac{d x}{d t} = a x + b y$ $\frac{d x}{d t} = a x + b y$

$\frac{d y}{d t} = c x + d y$ $\frac{d y}{d t} = c x + d y$

are given in Web Appendix A.3 on the CRE Web site (http://www.umich.edu/~elements/5e/appendix/AppA.3_Web.pdf).

A.3.C Second-Order Ordinary Differential Equations

Methods of solving differential equations of the type

$\frac{d^{2} y}{d^{2} x} - β y = 0$ $\frac{d^{2} y}{d^{2} x} - β y = 0$ (A-17)

can be found in such texts as Applied Differential Equations by M. R. Spiegel (Upper Saddle River, NJ: Prentice Hall, 1958, Chapter 4; a great book even though it’s old) or in Differential Equations by F. Ayres (New York: Schaum Outline Series, McGraw-Hill, 1952). Solutions of this type are required in PDF Chapter 15. One method of solution is to determine the characteristic roots of

$(\frac{d^{2}}{d x^{2}} - β) y = (m^{2} - β) y$ $(\frac{d^{2}}{d x^{2}} - β) y = (m^{2} - β) y$

which are

m = ± $\sqrt{β}$ $\sqrt{β}$

The solution to the differential equation is

$y = A_{1} e^{- \sqrt{β x}} + B_{1} e^{+ \sqrt{β x}}$ $y = A_{1} e^{- \sqrt{β x}} + B_{1} e^{+ \sqrt{β x}}$ (A-18)

where A1 and B1 are arbitrary constants of integration. It can be verified that Equation (A-18) can be arranged in the form

$y = A \sinh \sqrt{β x} + B \cosh \sqrt{β x}$ $y = A \sinh \sqrt{β x} + B \cosh \sqrt{β x}$ (A-19)

Equation (A-19) is the more useful form of the solution when it comes to evaluating the constants A and B because sinh(0) = 0 and cosh(0) = 1.0. Also see Appendix A.3.B.

A.4 Numerical Evaluation of Integrals

In this section, we discuss techniques for numerically evaluating integrals for solving first-order differential equations.

1. Trapezoidal rule (two-point) (Figure A-2). This method is one of the simplest and most approximate, as it uses the integrand evaluated at the limits of integration to evaluate the integral

$\int_{X_{0}}^{X_{1}} f (X) d X = \frac{h}{2} [f (X_{0}) + f (X_{1})]$ $\int_{X_{0}}^{X_{1}} f (X) d X = \frac{h}{2} [f (X_{0}) + f (X_{1})]$ (A-20)

when h = X₁ – X₀

2. Simpson’s one-third rule (three-point) (Figure A-3). A more accurate evaluation of the integral can be found with the application of Simpson’s rule:

$\int_{X_{1}}^{X_{2}} f (X) d X = \frac{h}{3} [f (X_{0}) + 4 f (X_{1}) + f (X_{2})]$ $\int_{X_{1}}^{X_{2}} f (X) d X = \frac{h}{3} [f (X_{0}) + 4 f (X_{1}) + f (X_{2})]$ (A-21)

where

$h = \frac{X_{2 - X_{0}}}{2} X_{1} = X_{0} + h$ $h = \frac{X_{2 - X_{0}}}{2} X_{1} = X_{0} + h$

Methods to solve

$\int_{0}^{X} \frac{F_{A 0}}{- r_{A}} dX$ $\int_{0}^{X} \frac{F_{A 0}}{- r_{A}} dX$

in Chapters 2, 4, 12, and

$\int_{0}^{\infty} X (t) E (t) d t$ $\int_{0}^{\infty} X (t) E (t) d t$

in PDF Chapter 17

Graph illustrates the Trapezoidal rule with the help of a curve.

Figure A-2 Trapezoidal rule illustration.

Graph illustrates Simpsons three-point rule with the help of a curve.

Figure A-3 Simpson’s three-point rule illustration.

3. Simpson’s three-eighths rule (four-point) (Figure A-4). An improved version of Simpson’s one-third rule can be made by applying Simpson’s three-eighths rule:

$\int_{X_{0}}^{X_{3}} f (X) d X = \frac{3}{8} h [f (x_{0}) + 3 f (x_{1}) + 3 f (x_{2}) + f (x_{3})]$ $\int_{X_{0}}^{X_{3}} f (X) d X = \frac{3}{8} h [f (x_{0}) + 3 f (x_{1}) + 3 f (x_{2}) + f (x_{3})]$ (A-22)

where

$h = \frac{X_{3} - X_{0}}{3} X_{1} = X_{0} + h X_{2} = X_{0} + 2 h$ $h = \frac{X_{3} - X_{0}}{3} X_{1} = X_{0} + h X_{2} = X_{0} + 2 h$

Graph illustrates Simpsons four-point rule with the help of a curve.

Figure A-4 Simpson’s four-point rule illustration.

4. Five-point quadrature formula.

$\int_{X_{0}}^{X_{4}} f (X) d X = \frac{h}{3} (f_{0} + 4 f_{1} + 2 f_{2} + 4 f_{3} + f_{4})$ $\int_{X_{0}}^{X_{4}} f (X) d X = \frac{h}{3} (f_{0} + 4 f_{1} + 2 f_{2} + 4 f_{3} + f_{4})$ (A-23)

where

$h = \frac{X_{4} - X_{0}}{4}$ $h = \frac{X_{4} - X_{0}}{4}$

5. For N + 1 points, where (N/ 3) is an integer

$\int_{X_{0}}^{X_{N}} f (X) d X = \frac{3}{8} h [f_{0} + 3 f_{1} + 3 f_{2} + 2 f_{3} + 3 f_{4} + 3 f_{5} + 2 f_{6} + \dots + 3 f_{N - 1} + f_{N}]$ $\int_{X_{0}}^{X_{N}} f (X) d X = \frac{3}{8} h [f_{0} + 3 f_{1} + 3 f_{2} + 2 f_{3} + 3 f_{4} + 3 f_{5} + 2 f_{6} + \dots + 3 f_{N - 1} + f_{N}]$ (A-24)

where $h = \frac{X_{N} - X_{0}}{N}$ $h = \frac{X_{N} - X_{0}}{N}$

6. For N + 1 points, where N is even

$\int_{X_{0}}^{X_{N}} f (X) d X = \frac{h}{3} (f_{0} + 4 f_{1} + 2 f_{2} + 4 f_{3} + 2 f_{4} + \dots + 4 f_{N - 1} + f_{N})$ $\int_{X_{0}}^{X_{N}} f (X) d X = \frac{h}{3} (f_{0} + 4 f_{1} + 2 f_{2} + 4 f_{3} + 2 f_{4} + \dots + 4 f_{N - 1} + f_{N})$ (A-25)

where $h = \frac{X_{N} - X_{0}}{N}$ $h = \frac{X_{N} - X_{0}}{N}$

These formulas are useful in illustrating how the reaction engineering integrals and coupled ODEs (ordinary differential equation(s)) can be solved and also when there is an ODE solver power failure or some other malfunction.

A.5 Semilog Graphs

Review how to take slopes on semilog graphs on the Web. Also see www.physics.uoguelph.ca/tutorials/GLP. Also see Web Appendix A.5, Using Semi-Log Plots for Data Analysis.

A.6 Software Packages

Instructions on how to use Polymath, COMSOL, and Aspen can be found on the CRE Web site.

For the ordinary differential equation solver (ODE solver), contact:

Polymath Software
P.O. Box 523
Willimantic, CT 06226-0523
Web site: www.polymathso/tware.com//fogler

MATLAB
The Math Works, Inc.
20 North Main Street, Suite 250
Sherborn, MA 01770
www.mathworks.com

Wolfram
100 Trade Center Drive
Champaign, IL 61820-7237
Tel: +1-217-398-0700
+1-800-WOLFRAM (965-3726)
Fax: +1-217-398-0747
www.wolfram.com

Aspen Technology, Inc.
10 Canal Park
Cambridge, MA 02141-2201
Email: [email protected]
www.aspentech.com

COMSOL, Inc.
1 New England Executive Park
Burlington, MA 01803
Tel: +1-781-273-3322
Fax: +1-781-273-6603
Email: [email protected]
www.comsol.com

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Table of Contents for Appendix A Numerical Techniques

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A. Numerical Techniques

A.1 Useful Integrals in Reactor Design

A.2 Equal-Area Graphical Differentiation

A.3 Solutions to Differential Equations

A.3.A First-Order Ordinary Differential Equations

A.3.B Coupled Differential Equations

A.3.C Second-Order Ordinary Differential Equations

A.4 Numerical Evaluation of Integrals

A.5 Semilog Graphs

A.6 Software Packages

Table of Contents for
Appendix A Numerical Techniques