Chapter1(MBISKKK2133-03 PENGIRAAN KEJURUTERAN KIMIA (CHEMICAL ENGINEERING COMPUTATION))El (1)

7/29/2019 Chapter1(MBISKKK2133-03 PENGIRAAN KEJURUTERAN KIMIA (CHEMICAL ENGINEERING COMPUTATION))El (1)

1/23

CHAPTER 1: INTRODUCTION TO

NUMERICAL METHODS

1.1 Numerical Methods & Computer

NM: technique by which mathematical problems are

formulated so that they can be solved with arithmetic

operations.

Thus, by using computer & NM has significant

influence on engineering problem-solving process.

Dr Mariani Idroas SKF 2133 1

Numerical Methods for

Chemical Engineers


2/23

1.2 Advantages and Disadvantages of NM

Engineers approached to solve problems during

pre-computer era:

a. Using analytical or exact methods only for a

simple linear models or simple geometry and

low dimensionality.

b. Graphical solutions to solve complex

problems but the results are not precise & can

be described using three or less dimensions.

c. Calculators approaches to implement NM manually.

Although perfectly adequate for solving complex

problems, manual calculations are slow, tedious &

consistent results are elusive.


Chemical Engineers



3/23

Nowadays, computer & NM provide an

alternative for complicated calculations.

Since late 1940s digital computers has lead in

the use & development of NMs.

There are several advantages of study NMs:

a. Powerful-solving tools capable of handling

large systems of equations, nonlinearities,and

complicated geometries that uncommon inengineering practice & impossible to solve

analytically.

b. An efficient vehicle for learning to use

computers an efficient way to learn

programming as NMs are the important part

for implementation on computers.

c. Reinforce understanding of mathematics

function of NM is to reduce high

mathematics to basic arithematic operations.


Chemical Engineers



4/23

1.3 Mathematical Modeling

A mathematical modelcan be defined as a formulation

or equation that express aphysicalsystem orprocess in

mathematical terms.

Example 1.1: Very general functional relationship

Dependent = f( independent variables, parameter,variable forcing functions) ---------- Eq. (1)

where:

dependent variable = behavior or state of the system

Independent variable = usually dimensions such as

time & space

Parameter= system properties or composition

Forcing function = external influence

Equation (1) can be ranged from a simple algebraic

relationship to large complicated sets of differential

equations.


Chemical Engineers



5/23

Example 1.2 : Very simple case.

F = ma ------ Eq. (2)

(Newton 2nd law of motion)

F= net force acting on body (N, kg.m/s2)

m = mass of object (kg), a = acceleration (m/s2)

Dividing both sides of Eq. (2) by m to give:

a = F/m --------- Eq. (3)

where,a = dependent variable (or systems behavior)

F= forcing function

m = parameter (or property of system)

No independent variable because we are not yet

predicting how acceleration varies in time or space.


Chemical Engineers



6/23

Equation (3) has characteristics of typical

mathematical models:

a. Describes a physical system or process in

mathematical terms.

b. Represents an idealization & simplification of

reality by ignores details of natural process

& focuses on essential matters.

c. Yields reproducible results can be used for

predictive purposes. For e.g. if the force on an

object and the mass of an object are known, Eq.

(1.3) can be used to compute acceleration.


Chemical Engineers



7/23

Example 1.3: For complex case of falling body

(parachutist)

Newtons 2nd law can be used to determine terminal

velocity of a free-falling body near earths surface.

We expressing acceleration as the time rate of changeof velocity (dv/dt) & substituting into equation (3) to

yield:

dv/dt = F/m --------- Eq. (4)

Rearrange from equation (4); m dv/dt = F

where,

v = velocity (m/s)

F= net force acting on the body

F +ve ------ object will accelerate

F ve ------ object will decelerate

F = 0 ------ object velocity remain at a constant

Now we will express net force in terms of measurable

variables & parameters.

For a falling object, force is composed of 2 opposing

forces;

F = FD + FU ---------- Eq. (5)


Chemical Engineers



8/23

where;

FD = downward force pull of gravity

FU= upward force of air resistance

IfFD is assigned a +ve sign, the 2nd law can be

used to formulate the force due to gravity, as;

FD = mg -------- Eq. (6)where;g= gravitational constant (9.8 m/s2)

Air resistance can be assumed as an upward

direction, thus,

FU= -cv -------- Eq. (7)

where; c = drag coefficient (kg/s)

Net force is the difference betweenFD andFU

Therefore, Eq. (6) and Eq. (7) can be combined

to

yield;

dv/dt = (mg cv)/m -------- Eq. (8)

simplifying the right side,

dv/dt = g (c/m)v -------- Eq. (9)

Equation (9) is a model that relates the

acceleration of a falling object to the force acting

on it.


Chemical Engineers



9/23

Equations (9) is a differential equation because terms

of dv/dt acts as variable that we are interested in

predicting.

However, from equations (3) & (9) the velocity of

falling object cannot be obtained using simplealgebraic manipulation.

More advance techniques such as calculus must be

applied to obtain an exact or analytical solution.

Calculus can be used to solve equation (9) for;

v(t) = gm/c (1-e-(c/m)t

) -------- Eq. (10)

Note: equation (10) is in general form of

equation (1), where;

v(t) = dependent variable

c and m = parameters

g= forcing function


Chemical Engineers



10/23

Equation (10) is called an analyticalorexact

solution because it exactly satisfies the original

differential equation.

Unfortunately, there are many mathematical models

that cannot be solved exactly.In many cases, the only alternative is to develop a

numerical solution that approximates the exact

solution.

As stated previously, numerical methodare those inwhich the mathematical problem is reformulated so

it can be solved by arithmetic operation


Chemical Engineers



11/23

Example 1.4: Analytical/Exact Solution of Falling

Parachutist Problem

Problem statement: A parachutist of mass 68.1 kg

jumps out of a stationary hot air balloon.

Use equation (10) to compute velocity prior toopening the chute. Drag coefficient is equal to

12.5kg/s.

Solution: Inserting the parameter into equation (10)

v(t) = 9.8(68.1)/12.5 (1-e-(12.5/68.1)t

)= 53.39 (1-e-0.18355t)

Then can be used to compute;

t, (s) v, (m/s)

0 0.00

2 16.40

4 27.77

6 35.64

8 41.10

10 44.87

12 47.49

Infinity 53.39


Chemical Engineers



12/23

Thus, according to the model, the parachutist

accelerates rapidly (Fig. 1.1).

A velocity of 44.87 m/s is attained after 10s.

After sufficient time, a constant velocity, called as

terminal velocity of 53.39 m/s is reached.

This velocity is constant because the force of

gravity is balanced with the air resistance.

Thus, net force is zero & acceleration has ceasedor stop to increase.


Chemical Engineers



13/23


Chemical Engineers



14/23

Numerical method approaches to solve mathematical

problem.

From Newtons 2nd law, the time rate of change of

velocity can be approximated by: (Fig. 1.2)

dv/dt=v/t = v(ti+1)v(ti)/ti+1ti --eqn(11)

where;

v and t= differences in velocity and time

computed over finite intervals

v(ti) = velocity at an initial time ti , and

v(ti+1) = velocity at some later time ti+1


Chemical Engineers



15/23


Chemical Engineers



16/23

Note that dv/dt v/ t is approximate because t

is finite.

Equation (11) is called a finite divided difference

approximation at time ti.

Substituted equation (11) into equation (9) to give;

v(ti+1) v(ti)/ti+1-ti = g c/m(v)(ti)

And rearranged to yield;

v(ti+1) = v(ti)+{gc/m(v)(ti)}(ti+1-ti) --- eqn(12)

The term in brackets is the right-hand side of the

differential is provides a mean to compute the rate

of change or slope ofv.


Chemical Engineers



17/23

Thus, the differential equation has been transformed

into an equation that can be used to determine the

velocity algebraically at ti+1 using the slope and

previous values ofv and t.

If you are given an initial value for velocity at sometime ti, you can easily compute velocity at a later time

ti+1.

This new value of velocity at ti+1 can in turn be

employed to compute velocity at ti+2 and so on.

Thus;

New value = old value + (slope)(step size)


Chemical Engineers



18/23

Example 1.5: Numerical Solution to the Falling

Parachutist Problem

Problem Statement: Perform the same computation

as in Example 1.4 but use equation (12) to compute

velocity. Employ a step size of 2 for the calculation.

Solution:

At the start of the computation (ti = 0), the velocity

of the parachutist is zero.

Using this information & parameter values from

Example 1.4, equation (12) can be used to compute

velocity at ti+1 = 2s:

v= 0 + {9.8 12.5/68.1(0)}(2) = 19.60 m/s

For the next interval (from t= 2 to 4 s) thecomputation is repeated, with the result;

v=19.60+{9.812.5/68.1 (19.60)}(2)=32.00 m/s

The calculation is continued in a similar fashion to

obtain additional values:


Chemical Engineers



19/23

t,(s) v,(m/s)

0 0.00

2 19.60

4 32.00

6 39.858 44.82

10 47.97

12 49.96

53.39

Along with the exact solution, the results above are

plotted in Fig. 1.3.There are discrepancy between the

two results.One way to minimize such differences is

to use a smaller step size.

With the aid of computer, large number of

calculation can be performed easily.Thus, you can

model the case without having to solve the

differential equation exactly


Chemical Engineers



20/23


Chemical Engineers



21/23

1.3.1 Other Complicate Engineering Cases

Aside from Newtons 2nd law, there are other major

principles in engineering.

For example the conservation laws of science &engineering:

Change = increases decreases ---- Eq. (13)

If change is zero, equation (13) becomes;

Change = 0 = increases decreases

or

Increases = Decreases -------- Eq. (14)

Thus, if no change occurs, the increases & decreases

must be in balance, which can be called assteady

state computation has many applications in

engineering.


Chemical Engineers



22/23

Table 1 summarizes some of simple engineering

models.

Most of chemical engineering applications will focus

on mass balance for reactor depends on mass flow

in & out.

Both the civil & mechanical engineering applicationswill focus on conservation of momentum.

Electrical engineering applications employ both

current & energy balance to model electric circuits.


Chemical Engineers



23/23


Chemical Engineers


Chapter1(MBISKKK2133-03 PENGIRAAN KEJURUTERAN KIMIA (CHEMICAL ENGINEERING COMPUTATION))El (1)

Documents

Transcript of Chapter1(MBISKKK2133-03 PENGIRAAN KEJURUTERAN KIMIA (CHEMICAL ENGINEERING COMPUTATION))El (1)