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CHEMICAL ENGINEERING COMPUTATION
SKKK 2133
SYED ANUAR FAUA’AD B. SYED MUHAMMADDept. of Bioprocess and PolymerEngineering, Faculty of Chemical and Energy Engineering.
UTM Skudai, JohorN01-257
syed@utm.my
WORLDVIEW KNOWLEDGE,
FRAMEWORK CIVILIZATION & THE
DEVELOPMENT OF A HOLISTIC
PERSONALITY: COMPARATIVE
PERSPECTIVE
11-12 SEPT 2013
PROF. DR. WAN MOHD NOR WAN DAUD
Founder Director
Advance Studies on Islam, Science and Civilisation (CASIS),
UTM Skudai.
• CHAPTER 1 – INTRODUCTION to NUMERICAL METHODS
• CHAPTER 2 – APPROXIMATIONS & ERRORS
• CHAPTER 3 – PART I : MATLAB FUNDAMENTALS
– PART II : PROGRAMMING WITH MATLAB
• CHAPTER 4 – ROOTS OF EQUATIONS
• CHAPTER 5 – LINEAR ALGEBRAIC EQUATIONS
• CHAPTER 6 – CURVE FITTING
• CHAPTER 7 – NUMERICAL DIFFERENTIATION
• CHAPTER 8 – NUMERICAL INTEGRATION
• CHAPTER 9 – ORDINARY DIFFERENTIAL EQUATIONS
(ODE)
COURSE OUTLINES
Chapter 1 : TOPIC COVERS
(INTRODUCTION to NMs)
• Computers and Numerical Methods
• Advantages (NMs) and Disadvantages
(analytical/exact solution)
• Role of Numerical Methods in Engineering
LEARNING OUTCOMES
INTRODUCTION
It is expected that students will be able to:
• Recognize the difference between analytical and
numerical solutions
• Describe how conservation laws are employed to
develop mathematical models of physical systems
• Identify the advantages and disadvantages of learning
numerical methods
Numerical methods
Techniques by which mathematical
problems are formulated so that
they can be solved with
arithmetic/reckoning/calculation
operations – i.e involve large
numbers of calculation
Numerical Methods and
MATLAB operationsLinear algebraic
Differentiation & Integration
Curve fitting Ordinary differential equations
Roots of equation
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
indefinable/elusive/hard to define.
Nowadays, computers & NMs 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 in
engineering 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/support understanding of
mathematics – function of NM is to reduce
high mathematics to basic reckoning/arithematic/calculation
operations.
Mathematical Modeling and
Engineering Problem solving
• Requires understanding of engineering systems:
– By observation and experiment
– Theoretical analysis and generalization
• Computers are great tools, however, without
fundamental understanding of engineering
problems, they will be useless.
A mathematical model is represented as a functional
relationship:
Dependent independent variables,
Variable = f , parameters, forcing functions
• Dependent variable : Characteristic that usually reflects the state of
the system (i.e a = acceleration).
• Independent variables : Dimensions such as time and space along
which the systems behavior is being determined.
• Parameters : Reflect the system’s properties or composition.
(i.e m = mass of an object).
• Forcing functions : External influences acting upon the system
(i.e F = forcing acting on the object).
- - - - (1.1)
Equation (1.1) can be ranged from a simple
mathematical /algebraic/numerical relationship to large
complicated sets of differential equations.
Newton’s 2nd law of Motion
• States that “the time rate change of momentum of
a body is equal to the resulting force acting on it ”.
• The model is formulated as:
F = m a - - - - - - - - - - - (1.2)
F = net force acting on the body (N)
m = mass of the object (kg)
a = its acceleration (m/s2)
Equation (1.2) can be written as:
a = F / m
simple algebraic equation that can be solved
analytically.
a = dependent variable (or system’s 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.
Equation (1.2) has characteristics as typical of
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.
• However, some mathematical models of
physical phenomena may be much more
complex (by adding independent variable, time).
• Most of the cases are complex models may not
be solved exactly/analytically or require more
complex mathematical techniques than simple
algebra for their solution.
• Example of physical phenomena model by
modeling of a falling parachutist:
m
cvmg
dt
dv
cvF
mgF
FFF
m
F
dt
dva
U
D
UD
substituting and
rearrange the equationc = drag coefficient (kg/s)
• This is a differential equation and is written in terms of
the differential rate of change dv/dt of the variable that
we are interested in predicting.
• If the parachutist is initially at rest (v=0 at t=0), using
calculus:
vm
cg
dt
dv
tmcec
gmtv )/(1)(
Independent
variable
Dependent
variable
ParametersForcing
function
- - - - - 1.4
- - - - - -1.3
Equation (1.4) is called an analytical or exact
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 of the exact
solution.
As stated previously, numerical method are those in
which the mathematical problem is reformulated so it can
be solved by arithmetic/mathematics operation
Example 1.1 :
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 (1.4) to
compute velocity prior to opening the chute. Drag
coefficient (c) is equal to 12.5kg/s.
Solution: Inserting the parameter into
equation (1.4):
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 (constant value)
• Thus, according to the model (Fig. 1.2), at initial stage the
parachutist accelerates rapidly.
• Then, a velocity of 44.87 m/s is attained after 10 s.
• 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 (FD ) is
balanced with the air resistance (FU).
• Thus, net force is zero & acceleration has ceased or stop to
increase.
Fig. 1.2 : Parachutist
accelerates rapidly
Apply Numerical methods in theNewton 2nd Law
(Falling Parachutist Problem)
• The time of change of velocity (acceleration, a ) can
be approximated by:
• Equation 1.5 is called finite divided difference
approximation of the derivative at time ti
- - - - - 1.5
dv
dtv
tv(ti1) v(ti)
ti1 ti
• Substitute Eq 1.5 into Eq 1.3 to give:
- - - - 1.6
- - - - 1.7
The term in brackets at the right-hand side of
the differential is provides a mean to compute
the rate of change or slope of v.
Rearrange Eq. 1.6:
Equation 1.7 can be used to determine the velocity at time
ti+1 if an initial value for velocity at time ti is given.
This new value of velocity at ti+1 can in turn be employed to
extend the computation to velocity at ti+2 and so on.
In general:
New value = old value + [(slope) x (step size)]
This approach is formally called Euler’s method (later will be
covered in Chapter 8)
Example 1.2 : Numerical Solution to the Falling
Parachutist Problem
Problem Statement: Perform the same computation
as in Example 1.1 but use equation (1.7) 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 1.1, equation (1.7) can be used to compute
velocity at ti+1 = 2s:
At ti = 0; and vi = 0;
v= 0 + {9.8 – 12.5/68.1(0)}(2) = 19.60 m/s
For the next interval ( from t = 2 to 4 s ) the computation is
repeated, with the result;
v=19.60+{9.8–12.5/68.1 (19.60)}(2)=32.00 m/s
The calculation is continued in a similar fashion to obtain
additional values:
t,(s) v,(m/s)-approx. v,(m/s) – exact
values (Example 1.1) values
0 0.00 0.00
2 19.60 16.40
4 32.00 27.77
6 39.85 35.64
8 44.82 41.10
10 47.97 44.87
12 49.96 47.49
53.39 53.39
Along with the exact solution, the results above are
plotted in Fig. 1.4.There are differences/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 (calculus approaches) .
Other Complicate Engineering Cases
Aside from Newton’s 2nd law, there are other major principles in
engineering.
For example the conservation laws of science & engineering:
Change = increases – decreases - - - - - 1.8
If change is zero, equation (1.8) becomes;
Change = 0 = increases – decreases
or
Increases = Decreases - - - - - 1.9
Thus, if no change occurs, the increases & decreases must be in balance,
which can be called as steady state computation – that has many
applications in engineering.
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 applications
will focus on conservation of momentum.
Electrical engineering applications employ both
current & energy balance to model electric circuits.
For steady-state incompressible fluid flow in pipes:
Flow in = Flow out
Or
100 + 80 = 120 + Flow4
Flow4 = 60
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