Computational Methods in the Introductory Courses to ... · Computational Methods in the Introductory Courses to Chemical Engineering (KETA01) and Biotechnology (KKKA05) Carmen Ar

Computational Methods in theIntroductory Courses to

Chemical Engineering (KETA01) andBiotechnology (KKKA05)

Carmen Arevalo

Lund University

[email protected]

Lecture 1

Arevalo KETA01 & KKKA05

Anonymous functions in matlab

Anonymous functions can be written directly in the Command Window.

Structure of an anonymous function:

”name of function” = @(”name of variable”) ”formula”

Example: To construct the anonymous function for g(x) =( x2.4

)3

− 2x+ cosπx

12write in the Command Window

g = @(x) (x/2.4)^3 - 2*x+cos(pi*x/12);

To evaluate this function at x = 4.8, write

y = g(4.8)

Arevalo KETA01 & KKKA05 1

M-file functions

• Always starts with the word function

• Must be saved in an M-file

• The names of the M-file and of the function must be the same

Structure:

function ["output param"] = "function name"("input param")"body of the function" (can consist of several statements)

If only one output parameter, brackets may be ommitedIf several input or output parameters, separate by commas.


Example of a simple M-file function

fun(x) → p

function p = fun(x)% Evaluates a function p at x% Input: x, any real number% Output: pp = (x/2.4)^3 - 2*x + cos(pi*x/12);

This is written using the matlab editor and saved in a file called fun.m

To evaluate the function at 4.8 and name the result y we write the command

y = fun(4.8)


Example of an M-file function with 2 output parameters

The following function finds the roots of a quadratic polynomial. It mustbe saved in a file called quadroot.m

function [x1,x2] = quadroot(a,b,c)% Solves equation ax^2+bx+c=0% Input: a,b,c, coefficients of the quadratic polynomial% Output: x1,x2, the complex roots of the polynomialt = -b/(2*a);d = sqrt(t^2-c/a);x1 = t + d;x2 = t - d;

To name the roots of 3x2 − 5x+ 7 root1 & root2, write the command

[root1, root2] = quadroot(3,-5,7);


The matlab for-loop

Used to repeat some commands.

Always starts with the word for and ends with the word end

Example: construct the row vector x with 50 elements

x(j) =1

2j + 3j, j = 1, . . . , 50

We construct a script with the commands

for j=1:50v(1,j) = 1/(2^j+3^j);

end


Examples of for-loops

for j = 1:3a(j,1) = j^2;

end

produces vector a =(

12 22 32)T =

(1 4 9

)Tw = 1:10;for k = 1:2:10

w(k) = 0;end

produces w =(

0 2 0 4 0 6 0 8 0 10)


Constructing a for-loop

Let’s add −2 times the first row to the second and third rows of

A =

1 3 22 2 12 1 3

for j = 2:3

A(j,:) = A(j,:) - 2*A(1,:);end

The result is

A =

1 3 20 −4 −30 −5 −1


Gas laws

Ideal gas law: pV = nRTp is absolute pressure, V is volume, n is number of moles, R is universalgas constant, T is absolute temperature.

van der Waals equation:(p+

a

v2

)(v − b) = RT

v = V/n is molar volume, a and b constants which depend on the gas.

Project: Estimate v for carbon dioxide and oxygen for all combinations ofT = 300, 500, 700 K, and p = 1, 10, 100 atm, in order to choose appropriatecontainment vessels.

Data: R = 0.082054 L atm/(mol K), carbon dioxide (a = 3.592, b =0.04267), oxygen (a = 1.360, b = 0.03183)


Using the Ideal Gas Law

Molar volumes for both gases: v =V

n=RT

p

Temperature, K Pressure, atm Molar Volume, L/mol300 1 24.6162

10 2.1416100 0.2462

500 1 41.027010 4.0127

100 0.4103700 1 57.4378

10 5.7438100 0.5744


Using the van der Waals equation

We know a, b, R and are given P , T . The unknown is V . To solve(P +

a

V 2

)(V − b) = RT

is the same as to solve(P +

a

V 2

)(V − b)−RT = 0

or to find the zeros (roots) of the function

f(V ) =(P +

a

V 2

)(V − b)−RT

We need a numerical method that solves (finds the zeros of) non-linearequations of the form f(x) = 0.


The graphical method: where f crosses the x-axis

Let’s solve in the interval V ∈ [0, 50] for the case P = 1, T = 300.

To plot the function, we first define the function by constructing an M-function

function f=myvanderW(v)% Input: v, vector of volumes% Output: f, value of function at vR=0.082054;a=3.592; b=0.04267; %carbon dioxideP=1;T=300;f=(P+a./v.^2).*(v-b)-R*T;


Plotting the van der Waals function

>> v=linspace(0,50);>> y=myvanderW(v);>> plot(v,y)>> grid

0 5 10 15 20 25 30 35 40 45 50−30

−20

−10

0

10

20

30van der Waals equation for carbon dioxide

volume

nonl

inea

r fu

nctio

n of

vol

ume


Matlab command ginput

enables you to select points from the figure using the mouse

>> v=linspace(0,50);>> y=myvanderW(v);>> plot(v,y,v,zeros(size(x)))>> grid>> v0=ginput(1)v0 =

2.447076612903226e+001 0


The Newton-Raphson method: a graphical understanding

We start by guessing the zero to be x0. The point were the tangent of thecurve at x0 crosses the x-axis is usually an improved estimate.


The Newton-Raphson idea applied to the gas problem

The slope of the tangent at x is f ′(x).

f(V ) =(P +

a

V 2

)(V − b)−RT

f ′(V ) = P +a

V 2− 2aV 3

(V − b)

Guess V = 30: f(30) = 5.4607, f ′(30) = 0.9960

Straight line with slope 0.9960, passing through (30, 5.4607):

y = 0.9960(V − 30) + 5.4607

crosses the X-axis at 0 = 0.9960(V − 30) + 5.4607⇒ V = 24.5175.

Note that f(24.5175) = 0.0049.


The Newton-Raphson method

Also called Newton’s Method, approximates a solution to f(x) = 0.

An approximation to the derivative at xi is

f ′(xi) =f(xi)− f(xi+1)

xi − xi+1

and if we suppose that we reached the zero of f at xi+1,

f ′(xi) =f(xi)− 0xi − xi+1

⇒ xi+1 = xi −f(xi)f ′(xi)


Newton-Raphson is an iterative formula

Newton-Raphson’s formula

xi+1 = xi −f(xi)f ′(xi)

starts by guessing the value of the zero of f . This starting value, x0, isthen used iteratively to produce

x1, x2, x3, x4, x5, . . . .

The last calculated value will be an approximation to the zero of thefunction.


Stopping criterion

We can repeat this process until one (or more) of these occur

• f(xi+1) ≈ 0

• absolute error = |xi − xi+1| ≈ 0

• relative error = |xi − xi+1|/|xi+1| ≈ 0

• we have repeated the process too many times


Error after each Newton iteration

The number of correct figures approximately doubles with each iteration.

i 0 1 2 3 4xi 0 0.500000000 0.566311003 0.567143165 0.567133290

In some special cases Newton’s Method converges slowly or not at all, andthat is why we need to set a maximum number of iterations.

If the starting point is not close enough to the solution, Newton’s methodmay not converge.

Use a graphic approach to get a good starting point.


Matlab solvers for f(x) = 0

• Polynomials: r = roots(p)The van der Waals equation can be written as

PV 3 − (RT + bP )V 2 + aV − ab = 0

>> p=[1 -0.082054*300-.04267 3.592 -3.592*.04267];>> r=roots(p)r =24.51260.0731 + 0.0301i0.0731 - 0.0301i


• Other nonlinear functions:

– if fun is an M-function:>> x = fzero(@fun,x0)

– if f is defined as an anonymous function:>> f=@(t)exp(t/5)-0.3;>> x = fzero(f,2)x = -6.0199

– f can also be defined inside the call to fzero:>> x = fzero(@(t)exp(t/5)-0.3,2)x = -6.0199


Residual

The residual for f(x) = 0 with approximated solution x is f(x).

We want the residual to be as close to 0 as possible.

>> [x,residual] = fzero(@fun,x0)

>> f=@(t)exp(t/5)-0.3;>> [x,residual] = fzero(f,2)x = -6.0199residual = -5.5511e-017

>> [x,residual] = fzero(@(t)exp(t/5)-0.3,2)x = -6.0199residual = -5.5511e-017


Solution with fzero

>> v0=ginput(1)

v0 =2.497031112812236e+001 0

>> [x,res] = fzero(@myfunction,v0(1))

root =

2.451258812844150e+001

res =

3.552713678800501e-015


Failing Newton-Raphson method: dependence on initialguess

Apply the method to find the root of f(x) = −x4 + 3x2 + 2 with initialguess of x0 = 1.

xi+1 = xi −−x4

i + 3x2i + 2

−4x3i + 6xi

x1 = −1

x2 = 1

x3 = −1

The method alternates between the two non-roots -1 and 1, and fails tofind a root. The actual roots are -1.8872 and 1.8872.


Good choice of initial guess

The Newton-Raphson method is a local method. This means the initialguess should be chosen close enough to the solution. Taking as initial guessx0 = 2.5, we get

x1 = 2.1145

x2 = 1.9323

x3 = 1.8895

x4 = 1.8872


Newton-Raphson method for systems of equations

For a single equation use fzeroFor a system of equations use fsolve

Example: to solve the system

x2 − 3y + 2 = 0

x3 − 4x2 − xy + 1 = 0

function f=newf2(x)f1=x(1).^2-3*x(2)+2;f2=x(1).^3-4*x(1).^2-x(1).*x(2)+1;f=[f1;f2];

In command window: [x,fval] = fsolve(@newf2,[0,0])


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Computational Methods in the Introductory Courses to ... · Computational Methods in the Introductory Courses to Chemical Engineering (KETA01) and Biotechnology (KKKA05) Carmen Ar