MATLAB for Chemical engineer

University of Baghdad

College of Engineering

Department of Chemical Engineering

2016

Lecture No. 9

Dr. Mahmood Khazzal Hummadi

Samar Kareem

MATLAB for Chemical engineer

Basic and Applications

Department of Chemical Engineering Baghdad University

2 MKH

Lecture No. 9

Looping statements and descision statements:

The conditions are given below. The column of the left hand side is to compare two values and those on the right

hand side are to add multiple conditions:

1. LOOPING: Looping or iterative functions are extremely useful in engineering problem solving.


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1. For statements loop a specific number of times, and keep track of each iteration with an

incrementing index variable.

A loop statement allows us to execute a statement or group of statements multiple times and following is the general form of a loop statement in most of the programming languages.

For example,

preallocate a 10-element vector, and calculate five values:

x = ones(1,10);

for n = 2:6

x(n) = 2 * x(n - 1);

end

2. While statements loop as long as a condition remains true.

For example, find the first integer n for which factorial(n) is a 100-digit number:

n = 1;

nFactorial = 1;

while nFactorial < 1e100

n = n + 1;

nFactorial = nFactorial * n;

end

Each loop requires the end keyword.

http://www.mathworks.com/help/matlab/ref/for.html

http://www.mathworks.com/help/matlab/ref/while.html


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It is a good idea to indent the loops for readability, especially when they are nested (that is, when one loop contains another loop):

A = zeros(5,100);

for m = 1:5

for n = 1:100

A(m, n) = 1/(m + n - 1);

end

end

You can programmatically exit a loop using a break statement, or skip to the next iteration of a loop using

a continue statement. Exit Loop before Expression Is False.

Example: Sum a sequence of random numbers until the next random number is greater than an upper limit. Then, exit the loop using a break statement.

limit = 0.8;

s = 0;

while 1

tmp = rand;

if tmp > limit

break

end

s = s + tmp;

end

http://www.mathworks.com/help/matlab/ref/break.html

http://www.mathworks.com/help/matlab/ref/continue.html

javascript:void(0);


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2. Decision

Decision making structures require that the programmer should specify one or more conditions to be evaluated or tested by the program, along with a statement or statements to be executed if the condition is determined to be true, and optionally, other statements to be executed if the condition is determined to be false.

1. if ... end statement

If the expression evaluates to true, then the block of code inside the if statement will be executed. If the expression evaluates to false, then the first set of code after the end statement will be executed.

Example: Create a script file and type the following code :

a= 10;

% check the condition using if statement

if a < 20

% if condition is true then print the following

fprintf('a is less than 20\n' );

end

fprintf('value of a is : %d\n', a);

http://www.tutorialspoint.com/matlab/if_end_statement_matlab.htm


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2. if...else...end statement An if statement can be followed by an optional else statement, which executes when the expression is false.

If the boolean expression evaluates to true, then the if block of code will be executed, otherwise else block of code will be executed. If the boolean expression evaluates to true, then the if block of code will be executed, otherwise else block of code will be executed.

Example

Create a script file and type the following code −

a = 100;

% check the boolean condition

if a < 20


fprintf('a is less than 20\n' );

else

% if condition is false then print the following

fprintf('a is not less than 20\n' );

end

fprintf('value of a is : %d\n', a);

3. if...elseif...elseif...else...end Statements.

An if statement can be followed by one (or more)

optional elseif... and anelse statement, which is very useful to test various conditions.

When using if... elseif...else statements, there are few points to keep in mind:

An if can have zero or one else's and it must come after any elseif's.

An if can have zero to many elseif's and they must come before the else.

Once an else if succeeds, none of the remaining elseif's or else's will be tested.

if <expression 1>

% Executes when the expression 1 is true

<statement(s)>

http://www.tutorialspoint.com/matlab/if_else_statement_matlab.htm


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elseif <expression 2>

% Executes when the boolean expression 2 is true

<statement(s)>

Elseif <expression 3>


<statement(s)>

else

% executes when the none of the above condition is true

<statement(s)>

end

Example:

a = 100;

%check the boolean condition

if a == 10


fprintf('Value of a is 10\n' );

elseif( a == 20 )

% if else if condition is true


elseif a == 30

% if else if condition is true


else

% if none of the conditions is true '

fprintf('None of the values are matching\n');

fprintf('Exact value of a is: %d\n', a );

end

4. The Nested if Statements

It is always legal in MATLAB to nest if-else statements which means you can use one if or elseif statement inside another if or elseif statement(s).

Syntax


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The syntax for a nested if statement is as follows −

if <expression 1>


if <expression 2>


end

end

You can nest elseif...else in the similar way as you have nested if statement.

Example Create a script file and type the following code in it −

a = 100;

b = 200;

% check the boolean condition

if( a == 100 )

% if condition is true then check the following

if( b == 200 )


fprintf('Value of a is 100 and b is 200\n' );

end

end

fprintf('Exact value of a is : %d\n', a );

fprintf('Exact value of b is : %d\n', b );

5. The switch Statement

A switch block conditionally executes one set of statements from several choices. Each choice is covered by a case statement.

An evaluated switch_expression is a scalar or string.

An evaluated case_expression is a scalar, a string or a cell array of scalars or strings.

Answer Value of a is 100 and b is 200 Exact value of a is : 100

Exact value of b is : 200


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The switch block tests each case until one of the cases is true. A case is true when −

For numbers, eq(case_expression,switch_expression).

For strings, strcmp(case_expression,switch_expression).

For objects that support theeq(case_expression,switch_expression).

For a cell array case_expression, at least one of the elements of the cell array matches switch_expression, as defined above for numbers, strings and objects.

When a case is true, MATLAB executes the corresponding statements and then exits the switch block.

The otherwise block is optional and executes only when no case is true.

Syntax

The syntax of switch statement in MATLAB is −

switch <switch_expression>

case <case_expression>

<statements>

case <case_expression>

<statements>

...

...

otherwise

<statements>

end


grade = 'B';

switch(grade)

case 'A'

fprintf('Excellent!\n' );


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case 'B'

fprintf('Well done\n' );

case 'C'

fprintf('Well done\n' );

case 'D'

fprintf('You passed\n' );

case 'F'

fprintf('Better try again\n' );

otherwise

fprintf('Invalid grade\n' );

end

When you run the file, it displays −

Well done

6. The Nested switch Statements

It is possible to have a switch as part of the statement sequence of an outer

switch. Even if the case constants of the inner and outer switch contain common values, no conflicts will arise.

Syntax

The syntax for a nested switch statement is as follows −

switch(ch1)

case 'A'

fprintf('This A is part of outer switch');

switch(ch2)

case 'A'

fprintf('This A is part of inner switch' );

case 'B'

fprintf('This B is part of inner switch' );

end


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case 'B'

fprintf('This B is part of outer switch' );

end


a = 100;

b = 200;

switch(a)

case 100

fprintf('This is part of outer switch %d\n', a );

switch(b)

case 200

fprintf('This is part of inner switch %d\n', a );

end

end

fprintf('Exact value of a is : %d\n', a );

fprintf('Exact value of b is : %d\n', b );

When you run the file, it displays −

This is part of outer switch 100

This is part of inner switch 100

Exact value of a is : 100

Exact value of b is : 200


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N

ZY

Q

N

i

i

N

i

i

Problems

Q1) Find the value of y . Use function .

A)

B)

Q2)Write the code required to determine the value of Q. The arrays Z and Y are generated as following:

Zi=i+1 , Yi=i2-i .Use Subroutine

Q3)Solve the second order algebraic equation with one variable ax2 + bx + c =0 . The solution . Note use

Function.

Q4) Calculate the kinetic and potential energies of an object at different velocities and heights respectively. . Use velocities and heights ranged from 1 to 1000 m ( step 1). Q5) Make an array for the numbers from 0 to 100 , then make a second one represents the results of multiplying each sequential two numbers in the first array . Q6) The one dimension arrays A and B ,and the number for the each array are 20 value . calculate the summation of array A and the summation of array B and find the y value , where A(i)=i and B(i)= i^2 , where the value of y is calculate as following :

Where MB is the average of B array.

a

acbbx

2

42


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Q7) Design windows application to find the densities of different six gases ( O2 ,H2 , N2 , CO2 , CH4 ,C3H8 , C2H2) at temperatures ranged from 0 to 300 oC (step 10 ) . The user should enter pressure in atm . Q8)Use loop to find the bubble point of ternary system (Ethanol 40 mol%, Water 20 mol% and Benzene 40 mol%). Knowing that the vapor pressure for three components are calculated by: Ethanol Po e=exp(18.5242-3578.91/(T-50.5)) Water Po w=exp(18.3036-3816.44/(T-46.13)) Benzene Po b=exp(15.9008-2788.51/(T-52.36)) Where Ki= Po

i /Pt ,Pt=760 ,yi =Kixi , At Bubble point Σyi=ΣKi×xi =1

Q9)Given that the vapor pressure of methyl chloride at 333.15 K is 13.76 bar, write a code to calculate the

molar volume of saturated vapor at these conditions using Redlich/Kwong equation. Knowing;

a=0.42748*R2T

c

2.5 /P

c

b=0.08664*RTc /Pc

Vi+1

=(RT/P)+b- (a*(Vi-b))/(T

1/2PV

i(V

i+b))

R=83.14 , Tc= 416.3 k , Pc= 66.8 bar Q10) A simple force balance on a spherical particle reaching terminal velocity in a fluid is given by; Where: V

t : Terminal velocity in m/s

g : Acceleration of gravity p

p: Particle density

Dp: The diameter of the spherical particle in m

CD: Dimensionless drag coefficient.

The drag coefficient on a spherical particle at terminal velocity varies with Reynolds number (Re) as followings: C

D=24/Re for Re< 0.1

CD=24*(1+0.14* Re^0.7)/Re for 0.1=<Re=< 1000

CD=0.44 for 1000<Re=< 350000 CD=0.19-8*10^4/Re for 350000 <Re Q11)Write a computer program to calculate molar volume and compressibility factor for gaseous ammonia at a pressure values (1-30 atm) by using of the Redlich-Kwong equation of state. The equations and variables are listed below. R= gas constant (R = 0.08206 atm·L/g-mol·K) ,Tc= the critical temperature (405.5 K for ammonia) Pc= the critical pressure (111.3 atm for ammonia) Compressibility factor is given by Z= PV/RT


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Q12)Calculate pressure drop (N/m2) across a pipe of diameter d (m) and length L (m). The fluid properties

are velocity u (m/s), density ρ (kg/m3), and viscosity µ (kg/m.s).

duRe

For laminar flow: Re < 2100: Re

8fj

For turbulent flow: 105 > Re > 2100: 4/1Re

0396.0fj

24 ud

LjP f

Q13) write the code required to calculate the density in g/lit of 50 mol% propane (C3H8)-50 mol% isobutene

(i-C4H8) gaseous mixture.The pressure in atm, temperature in K and critical constants Pc and Tc for propane

and iso-butane.(Note: Use function ).

RTz

PM

zyz

pTz

T

TT

p

pyp

MM

mix

iimix

ririi

ci

ri

c

i

ri

i

019.02.0

,

y i

Q14) Estimate the heat capacity of difference gases (O2 ,H2 , N2 , CO2 , CH4 ,C3H8 , C2H2) at different temperatures ( from 0 to 1000 oC step 10 ) .

, Cp J/mol K , T deg K.

component a b c d

O2 28.106 -3.680E-06 17.459E-06 -1.065E-08

H2 27.143 92.738E-04 -1.381E-05 76.451E-10

N2 31.150 -1.357E-02 26.796E-06 -1.168E-08

CO2 19.795 73.436E-03 -5.602E-05 17.153E-09

CH4 19.251 52.126E-03 11.974E-06 -1.132E-08

C3H8 -4.224 30.626E-02 -1.586E-04 32.146E-09

C2H2 26.821 75.781E-03 -5.007E-5 14.122E-09

Comp. Pc, atm Tc, K M, g/gmol

C4H8 36 408 56

C3H8 42 370 44 kgmol

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MATLAB for Chemical engineer