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COURSE PORTFOLIO
TK3201
TRANSPORT PHENOMENA
SEMESTER II - 2010/2011
STUDY PROGRAM OF CHEMICAL ENGINEERING
Instructor:
Dr. I Dewa Gede Arsa Putrawan
FACULTY OF INDUSTRIAL TECHNOLOGY
INSTITUT TEKNOLOGI BANDUNG
July 2011
CONTENTS
CONTENTS .................................................................................................................................. i
1 Course Description ................................................................................................................... 1
1.1 Syllabus, Goals and Outcomes ......................................................................................... 1
1.2 Learning System ............................................................................................................... 2
1.3 Prerequisites ..................................................................................................................... 2
1.4 Evaluation ........................................................................................................................ 3
1.5 Reading Materials ............................................................................................................ 3
2 Implementation ........................................................................................................................ 3
2.1 Class Meeting ................................................................................................................... 3
2.2 Evaluation ........................................................................................................................ 4
2.3 Grade Statistic .................................................................................................................. 4
3 Reflection ................................................................................................................................. 8
3.1 Advantages ....................................................................................................................... 8
3.2 Disadvantages................................................................................................................... 8
3.3 Feedbacks ......................................................................................................................... 9
3.4 Recommendations .......................................................................................................... 10
A. 1. Syllabus ............................................................................................................................. 12
A.2. Learning System and Evaluation ....................................................................................... 13
A.3. Schedule ............................................................................................................................. 14
A.4. Relationships between Course Objectives and ABET Outcomes ...................................... 15
B.1. Examination Evidences ...................................................................................................... 18
B.2. List of Attendance .............................................................................................................. 24
B.3. Examination Problems ....................................................................................................... 31
B.4. Achievement of Student Outcome ..................................................................................... 44
C.1. Mid Examination ................................................................................................................ 48
C.2. Final Examination .............................................................................................................. 66
C.3. Quiz 1 ................................................................................................................................. 79
C.4. Quiz 2 ................................................................................................................................. 83
C.5. Quiz 3 ................................................................................................................................. 87
C.6. Assignment 1 ...................................................................................................................... 91
C.7. Assignment 2 ...................................................................................................................... 96
C.8. Assignment 3 .................................................................................................................... 100
C.9. Assignment 4 .................................................................................................................... 106
C.10. Assignment 5 .................................................................................................................. 111
TK3201 Chem Eng FTI-ITB i
1 Course Description
1.1 Syllabus, Goals and Outcomes
This course introduces the basic physics and applications of the transport of heat, mass
and momentum. Topics cover: Basic concepts and practical applications of transport
phenomena, Fluid statics (simple transports of momentum, heat and mass), Introduction
to fluid dynamics (equations of motion, energy, continuity, transport phenomena in
complex systems, interphase transport, unsteady transport, multidimensional transport,
simultaneous transport), Fluid dynamic simulator (demo), and Practical applications of
transport equations.
The general objectives of this course are to use the conservation principles of momentum,
heat, and mass to develop models of fluid flow, heat transfer, and mass transfer systems
that can be used to predict the behavior of real-process systems and to explain the
physical properties of a fluid and their consequence on fluid flow, heat transfer, and mass
transfer, expressed in terms of the Reynolds number, the Nusselt number, Sherwood
number and other dimensionless quantities. As measurable outcomes, students who are
successful in this course, i.e. to pass the course with a grade of C or better, should be able
to do the following by the time of the final exam:
1. Students have proficiency in describing and using the basic principles underlying
the momentum, heat and mass transports including Newton’s law, Fourier’s law,
Ficks’s law, non Newtonian fluid and microscopic balances.
2. Students can create conceptual and quantitative models of momentum, heat and
mass transports over simple bodies and in simple channels.
3. Students can estimate momentum, heat, and mass transfer rates in simple
engineering situations including velocity, temperature, and concentration profiles.
4. Students can determine the appropriate dimensionless quantities necessary for
modeling and scaling fluid flow, heat transfer, and mass transfer systems.
5. Students have an understanding and appreciation for the implications of the
science of transport phenomena on society as a whole (in scientific, historical and
economic contexts) and recognize connections between transport phenomena and
other areas of study.
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Among eleven student outcomes recommended by ABET Engineering Criteria 2000, the
students outcomes emphasized in this course are as follows:
Emphasized ABET Student Outcome (SO) Level
An ability to apply knowledge of mathematics, science and engineering (SO a) Mid
An ability to identify, formulate and solve engineering problems (SO e) High
An ability to use the techniques, skills and modern engineering tools necessary
for engineering practice (SO k) Mid
The complete course syllabus and course schedule including the relation between course
objectives and ABET Student Outcomes can be found in Appendix A.
1.2 Learning System
Students are expected to take responsibility for their own learning. Instructor will lead
students through the steps necessary to do and will provide students with opportunities in
every class to test their learning and receive feedback. There will be reading and practice
problems assigned for every class. Students are expected to complete the reading and
make a substantial attempt at the practice problems prior to coming to class. There will be
a mini-quiz at the beginning of each class consisting of a single question from the reading
and practice problems. During class, instructor will spend time checking student’s
understanding of the material (by sampling) with concept questions, discussing difficult
or confusing topics, answering questions, working on group exercises and performing
demonstrations. Some classes will include a short lecture component, depending on the
complexity of the material. Feedback will be collected at the end of every class which
will be used to customize the class as needed. All practice problems will be due once a
week. This is a two unit course. Accordingly, the course has been designed to demand
approximately six hours per week of student’s time, including the class time. It is
expected that each student will prepare for and attend all of the classes.
1.3 Prerequisites
Microscopic balances are the bases of mathematical formulations of transport phenomena
which commonly result in complex deferential equations. Before taking this course,
students are expected to finish TK2201 Material and Heat Balances, TK2102 Chemical
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Engineering Mathematics, and TK2104 Process Computation. Students are also expected
to have experience in attending TK3102 Chemical Reaction Engineering, TK2105 Fluid
and Particle Mechanics, TK2203 Heat Transfer, and TK3101 Separation Process.
1.4 Evaluation
Evaluations include examination, quiz and homework. Examinations consist of midterm
examination and final examination which are open book and open notes. Quizzes are
conducted at the beginning of classes consisting of a single question from the reading and
practice problems. Homeworks consist of all practice problems assigned during a given
week and will be collected in the next week.
1.5 Reading Materials
The primary reading material for the class is the book “Transport Phenomena” written by
Professors R. B. Bird, W. E. Stewart, and E. N. Lightfoot (2002). This textbook is
available at Library and booksellers. Additional readings include “Transport Phenomena:
A Unified Approach” by Professors R. S. Broadkey and H. C. Hersey (1988) and
“Fundamentals of Transport Phenomena” by Professor R. W. Fahien (1983). Handouts
are also available in softcopy and hardcopy.
2 Implementation
2.1 Class Meeting
In this semester, the class meeting time of this course was scheduled on Thursday, 7:00 –
9:00, started on January 27, 2011 and finished on May 5, 2011, total of 14 weeks,
excluding one holiday, i.e., Gong Xi Fa Cai on 3 February 2011. The attendance in
classes of students and instructor are 84% by average and 100%, respectively
(recapitulation from Administration Office). The lowest attendance of student was found
on 28 April 2011 (48%). It was due to student event (Lomba Rancang Pabrik Nasional).
Many students were in charge as Technical Committee at that event. The complete List of
Attendance and Class Minutes can be found in Appendix B.
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2.2 Evaluation
Evaluations consisted of a middle examination, a final examination, quizzes and
assignments/homework. The mid examination was carried on March 17, 2011. The final
examination was carried on May 23, 2011, as scheduled by ITB. The problems of mid
examination covered the simple problems of momentum, heat, and mass transfers,
emphasizing the application of microscopic balances. The problems of final examination
included transport phenomena in complex systems unsteady, multidimensional,
simultaneous), applications of numerical methods in transport phenomena, and the
practical applications of transport equations. Students with class attendance below 80%
were not allowed to take final examination. Student’s grade was determined as 15%-Quiz,
15%-Homework, 25% of Mid Exam, and 45% of Final Exam. The examination problems
are given in Appendix B. Samples of students works are given in Appendix C.
2.3 Grade Statistic
Figure 1 shows the distribution of grade in Semester II-2009/2010. Among 58 students,
14% passed the course with grade A, 24% passed the course with grade AB, 26% passed
the course with grade B, 16% passed the course with grade BC, 16% passed the course
with grade C and 5% did not pass the course. Convertion of numerical grade to alphabetic
grade is as follows: A = [80 – 100], AB = [70 – 79], B = [60 – 69], BC = [50 – 59], C =
[40 – 49], D [30 – 39] and E [20 – 29]. With grade points of A, AB, B, BC, C, D, E equal
to 4.0, 3.5, 3.0, 2.5, 2.0, 1.0 and 0, respectively, it was found that the average grade points
of the whole class is 2.9 (close to the grade point of B). In other words, the average
achievement of the students was good (B). Figure 2 shows that 64% of students passing
the course with grade of minimum B. About 5% (three students) could not pass the
course. Their examinations were not satisfied, they did not understand the fundamentals
of transport phenomena, especially the microscopic balances. Moreover, they lost several
quizes and assignments.
As mentioned earlier, three ABET Student Outcomes are emphasized in this course.Those
three outcomes are Student Outcomes a, e and k which indicate the abilities to apply
mathematics, to formulate problems, and to utilize modern tools, respectively. Figures 3
and 4 show the distribution and cumulative distribution of outcome achievement. The
complete statistic of student outcome is given in Appendix B. Achievement of minimum
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40% for all outcomes emphasized could be achieved by most student (80% to 90%
students). Minimum achievement of 60% for student outcomes a (mathematic), e
(formulation), and k (modern tool) could be achieved by 66%, 50%, and 79% students,
respectively. It can be seen that among the student outcomes emphasized in this course,
student outcome e (an ability to identify, formulate and solve engineering problems) is
most difficult to achieve. Only about 10% of students achieved minimum 80% for this
outcome. This is the reason why the percentage of grade A is low (about 14%). As noted
earlier, the grade was mainly determined by examination results (total contribution of
examinations is 70%). The problems examined can be grouped into three categories: basic
concepts (could be answered by paying attention to classes), review (could be answered
by reviewing examples discussed in classes), and analysis (could be answered by further
studying course materials). Most problems are on the problem identification and
formulation. With the criteria mentioned previously, the students can achieve grade A if
they have good ability in identifying and formulating problems.
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Figure 1. Distribution of Grade.
Figure 2. Cummulative Distribution of Grade.
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Figure 3. Distribution of outcome achievement (ach).
Figure 4. Cummulative distribution of outcome achievement (ach).
Notes on SO (Level):
SO (a): an ability to apply knowledge of mathematics, science and engineering (mid)
SO (e): an ability to identify, formulate and solve engineering problems (high)
SO (k): an ability to use the techniques, skills and modern engineering tools necessary for engineering practice (mid)
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3 Reflection
3.1 Advantages
The advantages of this year implementation could be viewed from the course materials.
Introduction to practical applications of transport phenomena in chemical engineering and
beyond chemical engineering such as civil engineering, medicine, biology, mining, sports,
etc, have added insight on the benefit of mastering transport phenomena. It has also
motivated students to study transport phenomena. Introduction to practical problems
which can be solved by study on transport phenomena at the beginning of the class
meeting is very important as Transport Phenomena is a fundamental course. Without
knowing the practical applications in industrial scales, students will think that transport
phenomena is a course which is only related to laboratory works. Introduction to CFD
simulator at the end of semester further improve the knowledge of students on the various
practical problems that can be solved by studying transport phenomena. In addition, this
year material also covered non Newtonian Fluids which are often found in food
industries. Videos on the unique behavior of non Newtonian Fluids have impressed the
students.
Numerous practical problems, e.g., the dead body, the iceberg, friction coefficients and
heat as well as mass transfer coefficients, boilling eggs, drying polymeric sheet, etc were
discussed in this year. Two weeks were spent to show to the students that many problems
in industries and daily life could be solved by transport equations. The problem allowed
students to utilize various transport solutions found in the textbook and to apply
numerical methods obtained in the Process Computation course on third semester.
3.2 Disadvantages
In this year, three students could not pass the minimum criteria. Moreover, the percentage
of grade A is rather low, only 14% of total students got grade A. Student outcome
assessment showed that the achievement on the ability to identify, formulate and solve
engineering problems is not satisfied.
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3.3 Feedbacks
Feedbacks were obtained by questioner answered online which was managed by the
institute. The recapitulation of questioner was obtained from the Administration Office,
Study Program of Chemical Engineering. Fifty four of sixty four students answered the
questioner. The questioner consisted of eleven questions, covering the competency,
commitment, and attitude of instructor, the implementation and benefit of the course, as
well as the students attendance. The results, shown in Figure 5, were between 2.3 and
3.5, and the average was 3.3 (out of scale 4.0). By average, the responses are very good.
Most questions got answers above 3.0. The lowest was found for question 10 (student
capability after finishing the course), i.e., 2.5.
Figure 5. Results of questioner.
A. Instructor competency
1. Material mastery
2. Communication
B. Instructor commitment
3. Time utilization
4. Attendance
C. Instructor Attitude
5. Course preparation
6. Response and discussion
D. Course implementation
7. Explanation on course objectives, plans, and references
8. Appropriateness materials and credits
9. Grading based on more than one evaluation
E. Course benefit
10. Capability of students after attending the course
F. Students attendance
11. High attendance of students in classes
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3.4 Recommendations
The followings need to consider for the implementation of Transport Phenomena course
in the next academic year:
a. Materials on simple practical problems which are found in industries as well as in
daily life need to be maintained. Such problems are easily found in Chemical
Engineering Education and also can be developed from the problems of standard
transport phenomena textbooks.
b. Group assignment, at least one assignment, is necessary to allow the students to
interact and discuss beyond class meetings in order to improve their
communication skills.
c. Ability to identify, formulate and solve engineering problems needs to be stressed
from the beginning. Students need to know that this outcome is very important in
solving practical problems on transport phenomena.
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APPENDIX A
SYLLABUS, LEARNING SYSTEM AND EVALUATION
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A. 1. Syllabus
Course code:
TK3201
Unit:
2
Semester:
2
Main area:
General
Category:
Compulsory
Course type Class
Course name Transport Phenomena
Short syllabus Introduction (Basic definitions and practical applications); Transport
without flows (one dimensional in steady state of momentum, heat and
mass); Transport with flows (equation of motion, equation of energy,
continuity equation of mixers and components, interphase transport,
unsteady transport, multidimensional transport, simultaneous
transport); Transport in turbulence regime; Introduction to fluid
dynamic simulator.
Complete syllabus This course dealing with phenomena on momentum, heat, and mass
transports. Topics cover : Introduction (basic definitions and practical
applications), Transport without flows (one dimensional in steady state
of momentum, heat and mass), Transport with flows (equation of
motion, equation of energy, continuity equation of mixers and
components, interphase transport, unsteady transport, multidimensional
transport, simultaneous transport), Transport in turbulence regime,
Introduction to fluid dynamic simulator.
General instructional
goals
To give understanding on mechanisms of momentum, heat and mass
transports in stationary and flowing fluids, and to give capability to
derive mathematical formulations for momentum, heat and mass
transports through microscopic balance and to solve them for practical
purposes.
Outcome This course gives skill in understanding and predicting distribution of
velocity, temperature, and composition through study on transport
phenomena.
Prerequisite 1. TK2201 MASS AND ENERGY BALANCES
2. TK2102 ANALYSIS ON CHEM. ENG. MATHEMATICS
3. TK2104 PROCESS COMPUTATION
Related courses 1. TK2202 KINETICS ON CHEM. REACTION AND CATALYSIS
2. TK3102 CHEM. REACTION TECHNIQUES
3. TK2105 FLUID MECHANICS
4. TK2203 HEAT TRANSFER PROCESS
5. TK3101 SEPARATION PROCESS 1
References 1. Bird,R. B., W.E. Stewart, and E.N. Lightfoot, 2002, Transport
Phenomena 2nd ed., John Wiley & Sons, New York
2. Broadkey,R.S. and H.C. Hersey, 1988, Transport Phenomena: A
Unified Approach, McGraw-Hill Book Co., Inc., New York
3. Fahien, R.W., 1983, Fundamentals of Transport Phenomena,
McGraw-Hill Book Co., Inc., New York
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A.2. Learning System and Evaluation
Percentage
Knowledge = 60-80 %
Media
x White board
Skill = 20-40 % x OHP/Projector
Attitude = 0-10 % Computer (lab)
Activity (hour/week)
Course = 2 Courseware
Tutorial = - E-learning
Lab Works = - x Others :
……(simulator)
Others = -
Assessment
Mid Exam = 50 % Grade Score
Final Exam = 40 % A 80 – 100
Homework = 5 % AB 70 – 79
Quiz = 5 % B 60 – 69
BC 50 – 59
C 40 – 49
E 0 – 39
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A.3. Schedule
Week Topic Sub Topic Instructional Goals Act
1 Introduction Basic definitions
Practical cases
Students understand the importance of
studying transport phenomena.
Students understand the wide application of
transport phenomena.
C
2 Basic laws
and transport
properties
Basic laws
Transport properties
Students understand the mechanisms and basic
laws of momentum, heat, and mass transfers.
Students understand the physical meaning of
transport properties and the influences of
pressure and temperature on transport
properties.
C
3 Microscopic
balances Simple momentum transfers
Microscopic balance of
momentum
Students can solve one dimensional
momentum transfer in steady state condition.
C
4 Microscopic
balances Simple heat transfers
Microscopic balance of heat
Students can solve one dimensional heat
transfer in steady state condition.
C
5 Microscopic
balances Simple mas transfers
Microscopic balance of mass
Students can solve one dimensional mass
transfer in steady state condition.
C
6 Equations of
changes Equations of continuity
Energy equations
Equations of motion
Students understand the general equations for
solving transport phenomena problems.
C
7 Equations of
changes Unsteady transport
phenomena
Analytical solution
Students can apply the general equations for
solving unsteady transport phenomena
problems.
Students can utilized analytical solutions
including graphics to solve unsteady transport
phenomena problems.
C
8 Mid exam E
9
10
Iterphase
transport and
turbulence
Interphase transport
Turbulence regime
Students can utilized empirical correlations on
friction factor, heat and mass transfer
coefficients.
Students understand dimensionless groups
commonly used in transport phenomena.
Students understand transport equations for
turbulence regimes.
C
11 Non
Newtonian
Fluids
Behavior of non Newtonian
Fluids
Basic laws and correlations
for non Newtonian Fluids
Students understand the characters of non
Newtonian Fluid and fascinating behavior
showed by non Newtonian Fluids.
Students understand correlations for non
Newtonian Fluids.
C
D
12
13
Finite
difference Applications of finite
difference in transport
phenomena
Students can formulate finite difference
equations for complex momentum, heat, and
mass transfers.
Students can solve complex momentum, heat,
and mass transfers by numerical methods.
C
14
15
Practical
Problems Application of transport
phenomena in industries
Students can solve practical problems found in
the area of chemical, biochemical, and food
industries.
C
P
16 Computation
fluid
dynamic
CFD Simulator Students understand the features and
capability of CFD Simulator.
C
D
17 Final exam E
Notes: C = class E = evaluation P = group presentation D = demo
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A.4. Relationships between Course Objectives and ABET Outcomes
ABET outcomes (ABET Engineering Criteria 2000) are:
a) an ability to apply knowledge of mathematics, science and engineering
b) an ability to design and conduct experiments, as well as to analyze and interpret data
c) an ability to design a system, component or process to meet desired needs within realistic
constraints such as economic, environmental, social, political, ethical, health and safety,
manufacturability and sustainability
d) an ability to function on multidisciplinary teams
e) an ability to identify, formulate and solve engineering problems
f) an understanding of professional and ethical responsibility
g) an ability to communicate effectively
h) the broad education necessary to understand the impact of engineering solutions in a
global, economic, environmental and societal context
i) a recognition of the need for and ability to engage in life-long learning
j) a knowledge of contemporary issues
k) an ability to use the techniques, skills and modern engineering tools necessary for
engineering practice
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No Course Objective Outcomes1)
Evaluation2)
1 Understanding the importance of study on transport
phenomena in solving chemical engineering problems.
e A, Q, E
2 Understanding variations of practical problems that can
be solved by studying transport phenomena.
e A, Q, E
3 Understanding mechanisms and basic laws of momentum,
heat, and mass transfers.
e A, Q, E
4 Understanding physical meanings of transport properties
and the influences of pressure and temperature on
transport properties.
e A, Q, E
5 Ability to solve one dimensional momentum, heat, and
mass transfers in steady state condition.
a, e A, Q, E
6 Ability to apply general equations for solving unsteady
transport phenomena problems.
a, e A, Q, E
7 Ability to utilize empirical correlations on friction factor,
heat and mass transfer coefficients.
e A, Q, E
8 Understanding dimensionless groups commonly used in
transport phenomena.
e A, Q, E
9 Understanding transport equations for turbulence
regimes.
a, e A, Q, E
10 Understanding characters of non Newtonian Fluid and
fascinating behavior showed by non Newtonian Fluids.
e A, Q, E
11 Understanding correlations for non Newtonian Fluids. e A, Q, E
12 Ability to formulate finite difference equations for
complex momentum, heat, and mass transfers.
e, k A, Q, E
13 Ability to solve complex momentum, heat, and mass
transfers by numerical methods.
k A, Q, E
14 Ability to solve practical problems found in the area of
chemical, biochemical, and food engineering.
a, e, k A, Q, E
15 Understanding the features and capability of CFD
Simulator
e, k A, Q, E
1)ABET Student Outcome (SO), 2)Assignment (A), Quiz (Q), Examination (E)
Contribution of Course to Meeting the Professional Component Mathematics and basic sciences Engineering Topics General Education
1 credit 1 credit -
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APPENDIX B
DOCUMENTATION AND STATISTIC
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B.1. Examination Evidences
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B.2. List of Attendance
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B.3. Examination Problems
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TK3201 TRANSPORT PHENOMENA SEMESTER II-2010/2011 MIDDLE EXAMINATION Thursday, 17 March 2011 Time : 75 minutes (Close books/Close notes) 1. Explain briefly why the viscosity of liquid, in general, decreases with temperature. 2. The measurement of viscosity by a capillary viscometer exploits the Hagen-Poiseuille equation: ΔP = 8µLQ/(πR4), where ΔP is pressure drop, µ is viscosity,
L is length, Q is volumetric flow rate, and R is inside radius. Write the assumptions which must be obeyed by the above equation. 3. Consider an egg being cooked in boiling water in a pan. Would you model the heat transfer to the egg as one, two, or three dimensional ? Would the heat transfer be steady or transient ? Also, which coordinate system would you use to solve this problem and where you place the origin ? 4. Consider a chilled water pipe of length L, inner radius R1, outer radius R2, and thermal conductivity k. Chilled water flows in the pipe at a temperature Tf and the film heat transfer coefficient at the inner surface is h. If the pipe is well insulated on the outer surface: (a)-express the differential equation and the boundary conditions for steady one-dimensional heat conduction through the pipe, and (b)-obtain a relation for the variation of temperature in the pipe by solving the differential equation. 5. A dimerization reaction of A is carried out on a flat-surface of catalyst with a large surface area S. The catalyst surface is surrounded by a stagnant gas film through which A has to diffuse to reach the catalyst surface. At the catalyst surface, the reaction 2A → A2 occurs instantaneously, and the product A2 then diffuses back out through the gas film to the main turbulent stream composing of A and A2. The gas-film has thickness d and the concentration of A in the main gas stream is xA0. The reaction occurs in steady condition at constant pressure and temperature, and the ideal gas law can be applied. Derive the concentration profile of A along the thickness of gas film. TK3201-II-2010/2011 Chem Eng FTI-ITB 32 of 123
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TK3201-II-2010/2011 Chem Eng FTI-ITB 33 of 123
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egration twic
th the follow
final result i
1. Mass diffuty and totad isobaric coe is one mole
tion for eve
e plus z
of the reacti
re B stands
y value of z.
1balance on s
f the express
1 0ce with respe
-2 ln (1 – 0
wing boundary
xA = xA0 at
xA = 0 at z
is then
(1 – 0.5 xA
usion is steaal concentraondition. e of A2 mov
ery two mo
direction. F
ion, therefor
s for A2, a
This relation
0.5species A ov
ion for NAz,
.5ct to z gives
0.5 xA) = C1
y conditions
t z = 0
z = d (the rea
A) = (1 – 0.5
ady and oneation are coving in the
oles of A
From the
re: NBz =
at steady
n may be sub
ver a thin slab
developed ab0
z + C2
action occurs
xA0)1-z/d
-dimensionaonstant sinc
bstituted into
b of thicknes
bove, into th
s instantaneou
z=0z=d
edca
al since the ce the syst
the Fick’s la
ss Δz in the g
his equation g
usly)
dge of stagnatalyst surfac
surface aretem is an i
aw II which
gas film lead
gives (for co
AA
nant gas filmce
a is large. 2.deal gas at
gives
ds to dNAz/dz
nstant DAB):
A2 Δzm
. t
z
TK3201-II-2010/2011 Chem Eng FTI-ITB 34 of 123
TK3201 TRANSPORT PHENOMENA SEMESTER II-2010/2011 FINAL EXAMINATION
Monday, 23 May 2011 100 minutes (Open Books/Handout/Notes)
1. What is a non-Newtonian fluid ? What is the power-law model ? [10%]
2. How do numerical solution methods differ from analytical ones ? What are the advantages and disadvantages of numerical analytical methods ? [15%]
3. Consider transient one-dimensional heat conduction in a plane wall with thickness d, no heat generation, and constant thermal conductivity, initially at homogeneous temperature T0. One of the surfaces is then well isolated and the other is kept at temperature T*. Obtain the explicit finite difference formulation of this problem. [20%]
4. Consider a cold canned drink left on a dinner table. Would you model the heat transfer to the drink as one, two, or three-dimensional ? Would the heat transfer be steady or transient ? Also, which coordinate system would you use to analyze this heat transfer problems, and where would you place the origin ? Explain ! [15%]
5. Imagine that you are spending time this morning in a small town near a mount on your way home for holidays. At about 10:00 a.m., the local police officer calls you and asks for your help. He knows that you are a chemical engineer, and naturally assumes you have some knowledge on forensic chemistry. It seems that the body of Mr. Dagdag, a local car dealer, had been found somewhat earlier in a “heavily wooded area” just outside the town. The local forensic doctor was out of office for seminar and there was no one else to estimate the time of death. Mr. Dagdag had been known to deal in “hot” cars and was to be going to the police to confess and name his four accomplices, Mr. Kelor, Mr. Salam, Mr. Bun, and Mr. Sawi. Mr. Kelor had been known to be out of town until 3:00 a.m., this dawn. Mr. Salam had a solid alibi from 1:00 a.m., this dawn. Mr. Bun was with his girl until about 5:00 p.m., early evening yesterday. Mr. Sawi was in jail yesterday for drunkenness, he was not released until 11:00 p.m., last night. When you finally get to the body it is about 11:00 a.m. You measure a rectal temperature (equivalent to core or center temperature) of 80 °F. The air temperature was known to be constant, about 70 °F. Luckily, you brought your text book of Prof. Bird’s Transport Phenomena. Calculate the latest possible time the murder could have occurred and state the possible suspect. For practical purposes, assume Mr. Dagdag to be shaped like a cylinder with diameter of 10 inches. Furthermore, the human body temperature and thermal diffusivity are assumed to be 99°F and 0.0058 ft2/hr, respectively. [20%]
6. Jugs made of porous clay were commonly used to cool water in the past. A small amount of water that leaks out keeps the outer surface of the jug wet at all times, and hot and relatively dry air flowing over the jug causes this water to evaporate. Part of the latent heat of evaporation comes from the water in the jug, as a result, the water is cooled. If the environment conditions are 1 atm, 30 oC, and 35% relative humidity, determine the temperature of the water when steady conditions are reached. [20%]
well insulated
temperature To
d
TK3201-II-2010/2011 Chem Eng FTI-ITB 35 of 123
TK3201 TRANSPORT PHENOMENA SEMESTER II-2010/2011 FINAL EXAMINATION
Solution 1. A non-Newtonian fluid is a material with viscosity at a fixed temperature is not constant, but depends
on shear rate and could also depend on the time of shear and previous shear and thermal history. Power Law model: τ = K (γ)n where τ is momentum flux, γ is velocity gradient, K and n are constants.
2. The analytical solutions are based on (1) deriving the governing differential equation by performing an energy balance on a differential volume element, (2) expressing the boundary conditions in the proper mathematical form, and (3) solving the differential equation and applying the boundary conditions to determine the integration constants. The numerical solution methods are based on replacing the differential equations by algebraic equations. In the case of the popular finite difference method, this is done by replacing the derivatives by differences. The analytical methods are simple and they provide solution functions applicable to the entire medium, but they are limited to simple problems in simple geometries. The numerical methods are usually more involved and the solutions are obtained at a number of points, but they are applicable to any geometry subjected to any kind of thermal conditions.
3. Energy equation: =
IC : T = T0 BC 1 : q = k dT/dt = 0 at x = 0 BC 2 : T = T* at x = d (upper surface as origin). An explicit finite difference formulation is obtained by evaluating space finite difference at lower time level. For example, if time is approached by forward difference and space by center finite difference: −∆ = −2 +(∆ ) = + ∆(∆ ) ( −2 + ) where i and n stand for space and time nodes, respectively.
4. Heat transfer to a canned drink can be modeled as two-dimensional since temperature differences (and thus heat transfer) will exist in the radial and axial directions (but there will be symmetry about the center line and no heat transfer in the angular direction. This would be a transient heat transfer process since the temperature at any point within the drink will change with time during heating. Also, we would use the cylindrical coordinate system to solve this problem since a cylinder is best described in cylindrical coordinates. Also, we would place the origin somewhere on the center line, possibly at the center of the bottom surface.
TK3201-II-2010/2011 Chem Eng FTI-ITB 36 of 123
5. This transf12 ofothersurfaclossewoodAssumfor co
Fromb = 0hoursPossi
6. Assum
and msaturainvolPropeaverasurfacbe 20of 25 W DThe mwaterAnalyequat=where=Note betterthe safrom Pv,∞ =Notin=The aevalu
is an unsteadfer in a cylinf Bird [2002r word, assuce temperatus is neglect
ded” area. ming infinitooling): = 70 − 8070 − 99
m the graph, α0.5·(10 in)·(s 48 minutesible suspects
mptions: 1. Tmass transferated air at 3lved in this aerties: Becauage temperatce temperatu
0°C. Then, th°C and 1 atmWater at 20°Dry air at 20mass diffusivr and air are ysis: The sution as − e the Lewis n= 2.142.5that we cou
r accuracy. Taturation prethe surface i
= Psat,T∞ = (ng that the at30° −(1.accuracy of
uated at (30+
dy one dimennder. Refer t2]. Assume mume no filmure of the bted as the b
Use Figurte length (he09 = 0.34αt/b2 = 0.28 1 ft/12 in)=0s). Murder, are Mr. Saw
The low masr is applicab00 K). 2. Bossumption is
use of low mture of (T∞+Ture Ts. We knhe propertiesm are °C: hfg = 2,450°C: Cp = 1.0vity of water18 and 29 kgurface tempe
⁄ ,number is 4110 /010 /
uld take the The air at theessure of wais determined(0.35) (4.25 tmospheric p2,454007 / °f this result +16)/2 = 18°C
nsional (radito Example 1maximum ram resistanceody is 70 °C
body is foune 12.1.2 oeight),(use th
0.35
. Since α = 0.42 ft, then therefore, ha
wi and Mr. Bu
ss flux condile since the oth air and ws less than 1 p
mass flux conTS)/2 which now that T∞<
s of water at
54 kJ/kg, Pv =007 kJ/kg°C,r vapor in airg/kmol, respeerature of th− ,
= 0.856
Lewis numbe surface is sater at the sud from kPa) = 1.488
pressure is 1 a/)(0.856) /can be impr
C and water p
ial direction)12.1.2 in Chate of cooline in the air C). The radind in a “heof Bird [2he right ord
0.0058 ft2/hn t = 8.8 houad to occur un.
itions exist smass fractio
water vapor percent). 3. Rnditions, we cannot be d
<TS and, for 20°C and th
= 2.34 kPa. P, α = 2.141 1r at 25°C is Dectively. he jug can
ber to be 1 fsaturated, andurface tempe
8 kPa atm = 101.3
/ 18 /29 /roved by repproperties at
) heat hapter ng. In
(the iative eavily 002],
dinate
r and urs (8
around 2:12
o that the Chon of vapor iat specified Radiation effcan use dry
determined athe purpose e properties
Psat, 30°C = 4.210-5 m2/s. Dwater-air = 2.5
be determin
for simplicityd thus the va
erature (2.34
Pa, substitut(2.34 −101peating the t 16.0°C. But
2 a.m. (11:0
hilton-Colbuin the air is lconditions afects are negair propertie
t this point bof property eof dry air at
25 kPa
50 10-5 m2/s,
ned by rearr
y, but we chapor pressure
kPa). The v
ting the know1.488).3 =calculations t the improve
00 a.m. – 8
urn analogy blow (about 2are ideal gasgligible. es for the mbecause of tevaluation, wthe average
and the mol
ranging Chil
hose to incore at the surfavapor pressu
wn quantities= . °
with dry aiement will b
hr 48 min).
between heat2 percent fores (the error
mixture at thehe unknown
we take TS totemperature
ar masses of
lton-Colburn
rporate it forace is simplyure of air far
s gives
ir propertiese minor.
.
t r r
e n o e
f
n
r y r
s
TK3201-II-2010/2011 Chem Eng FTI-ITB 37 of 123
TK3201 Quiz 1 (Close Book, 15 minutes) – 24 Feb 11 Name (NIM): ___________________________ 1. Explain the influence of temperature on the viscosity of liquid in general.
Answer: The higher the temperature, the lower the density of liquid. Lower density results in better flowability so that the viscosity of liquid becomes lower.
2. What are the viscosity of liquid water and the viscosity of air at ambient in cP ? Answer: Viscosity of liquid water (at room temperature) ≈ 1 cP Viscosity of air (at room temperature) ≈ 0.001 cP (for any gas in general)
3. Derive the dimension of the thermal conductivity and write an example of its unit ! Answer: Start from Fourier’s Law for 1D Heat Transfer: q = k dT/dx or k = q/(dT/dx) q [=] J/m2 = N m m-2 = kg ms-2 m m-2 = M T-2 dT/dx [=] °C/m = θ L-1 k [=] M L θ-1 T-2
TK3201-II-2010/2011 Chem Eng FTI-ITB 38 of 123
TK3201 Quiz 2 (Close Book, 15 minutes) – 4 Mar 11 Name (NIM): ___________________________ 1. Heat is flowing steadily in radial direction through an annular wall of inside radius Ri
and outside radius R0. The outer surface of the wall is kept at temperature T0. The inner surface of the wall is in contact with a cooling fluid at bulk temperature Tf and has a film heat transfer coefficient hi. Write two boundary conditions for this problem. Answer: Boundary condition 1: | = Boundary condition 2: | = ℎ ( | − )
2. O2 is diffusing steadily in axial direction at room condition through a circular duct which is filled with stagnant water and has a constant cross area. Explain why, for this problem, the molar flux of O2 can be approached as NO2 = − DO2,H2O dCO2/dz, where z is axial direction. Answer: From Fick’s law II: NO2 = -DO2,H2O dCO2/dz + xO2 (NO2+NH2O). Since the oxygen is insoluble in water, xO2 ≈ 0, then NO2 = -DO2,H2O dCO2/dz.
TK3201-II-2010/2011 Chem Eng FTI-ITB 39 of 123
TK3201 Quiz 3 (Close Book, 15 minutes) – 14 Apr 11 Name (NIM): ___________________________ Write down the forward, central, and backward finite difference approximation for d2T/dz2 with step size ½Δz. Answer: Forward ≈ − 2 ½ +¼(Δz)
Central ≈ ½ − 2 + ½¼(Δz)
Backward ≈ − 2 ½ +¼(Δz)
TK3201-II-2010/2011 Chem Eng FTI-ITB 40 of 123
Assignment 1 24 February 2011
1. For reaction in gas phase on a flat infinite catalyst plate with reaction rate of –rA=k1, prove
that the concentration of A is given as follows: 1 − 0.5 = (1 − 0.5 )( / ) · (1 − 0.5 / ) /
2. Problem 18.A.2 of Bird et al [2002]: Sublimation of small iodine sphere in still air
A sphere of iodine, 1 cm in diameter is placed in still air at 40 C and 747 mmHg pressure.
At this temperature, the vapor pressure of iodine is about 1.03 mmHg. We want to
determine the diffusivity of the iodine-air system by measuring the sublimation rate. To
help determine reasonable experimental conditions,
a. Estimate the diffusivity for the iodine-air system at the temperature and pressure
given above, using the intermolecular force parameters in Table E.1.
b. Estimate the rate of sublimation, basing your calculations on Eq. 18.2.27 (assume that
r2 is very large).
Assignment 2 3 March 2011
From energy equation, derive the energy equation for unsteady axial heat transfer in a
cylinder having constant density, heat capacity and thermal conductivity. Rewrite the
equation in terms of dimensionless groups.
TK3201-II-2010/2011 Chem Eng FTI-ITB 41 of 123
Assignment 3 24 March 2011
1. Problem 6.B.2 of Bird et al [2002]: Friction factor for flow along a flat plate.
An expression for the drag force on a flat plate, wetted both sides, is given in Eq. 4.4.30.
This equation was derived by using laminar boundary layer theory and is known to be in
good agreement with experimental data. Define a friction factor and Reynolds number,
and obtain the f versus Re relation.
2. Problem 6.B.3 of Bird et al [2002]: Friction factor for laminar flow in a slit.
Use the resultas of Problem 2.B.3 to show that for the laminar flow in a thin slit of
thickness 2B the friction factor is f = 12/Re, if the Reynolds number is defined as Re =
2B (vz) ρ/μ. Compare this result for f with what one would get from the mean hydraulic
radius empiricism.
3. Problem 14.A.1 of Bird et al [2002]: Average heat transfer coefficients.
Ten thousand pounds per hour of an oil with a heat capacity of 0.6 BTU/lbm·F are being
heated from 100°F to 200°F in the simple heat. The oil is flowing through the tubes,
which are copper, 1-in in outside diameter, with 0.065-in walls. The combined length of
tubes is 300 ft. The required heat is supplied by condensation of saturated steam at 15.0
psia on the outside of the tubes. Calculate hI, ha, and hln, for the oil, assuming that the
inside surfaces of the tubes are at saturation temperature of the steam, 213°F.
TK3201-II-2010/2011 Chem Eng FTI-ITB 42 of 123
Assignment 4 7 April 2011
Liquid with kinematic viscosity of 2.17 10-4 m2/s is flowing through a slit between two paralel plates, 4 cm away. Both plates move with velocity of 3 m/s. Calculate the velocity profile by numerical methods.
Assignment 5 14 April 2011
1. A wall 1 ft thick and infinite in other directions has an initial uniform temperature of
100°F. The urface temperatures at the two sides are suddenly increased and maintained at
300°F. The wall is composed of nickel steel 40% Ni with a thermal diffusivity
derivatives at lower time level.
2. The same phenomena as in Problem 1. Use central finite difference for time and central
finite difference for spatial derivative at lower time level.
TK3201-II-2010/2011 Chem Eng FTI-ITB 43 of 123
B.4
. Ach
ieve
men
t of S
tude
nt O
utco
me
E
valu
atio
n:
Qui
z A
ssig
nmen
ts
Mid
Exa
m
Fin
al E
xam
A
chie
vem
ent
Wei
ght:
15%
15
%
25%
45
%
Nr
of e
val:
3
quiz
zes
5 as
sign
men
ts
17/0
3/20
11
17/0
3/20
11
AB
ET
SO
: (a
) (e
) (k
) T
otal
(a
) (e
) (k
) T
otal
(a
) (e
) (k
) T
otal
(a
) (e
) (k
) T
otal
(a
) (e
) (k
)
Tar
get:
200
100
300
200
100
200
500
10
90
0 10
0 0
65
35
100
100%
10
0%
100%
No.
N
IM
1 13
0060
56
100
0 10
0 70
90
0
160
5 40
0
45
0 52
,5
25
77,5
44
%
68%
43
%
2 13
0070
41
0 0
0 0
0 0
0 0
0 0
0 0
10
15
25
0%
7%
26%
3 13
0070
97
50
0 50
90
0
180
270
10
20
0 30
0
27,5
25
52
,5
79%
28
%
61%
4 13
0071
08
0 10
15
25
0%
7%
26
%
5 13
0080
01
120
90
210
120
90
180
390
10
50
0 60
0
25
17,5
42
,5
85%
54
%
66%
6 13
0080
03
130
70
200
90
90
180
360
5 65
0
70
0 50
35
85
48
%
76%
92
%
7 13
0080
05
130
50
180
160
90
180
430
10
70
0 80
0
52,5
30
82
,5
93%
79
%
79%
8 13
0080
07
140
50
190
140
70
90
300
10
60
0 70
0
35
35
70
89%
62
%
79%
9 13
0080
09
120
90
210
140
90
180
410
10
55
0 65
0
55
35
90
89%
76
%
96%
10
1300
8011
80
70
15
0 70
0
0 70
5
50
0 55
0
27,5
35
62
,5
44%
39
%
74%
11
1300
8013
12
0 70
19
0 70
90
18
0 34
0 10
45
0
55
0 32
,5
32,5
65
76
%
58%
88
%
12
1300
8015
13
0 90
22
0 17
0 90
18
0 44
0 10
65
0
75
0 60
35
95
94
%
83%
96
%
13
1300
8017
12
0 70
19
0 90
90
18
0 36
0 10
85
0
95
0 52
,5
25
77,5
79
%
82%
75
%
14
1300
8019
12
0 90
21
0 16
0 90
18
0 43
0 10
60
0
70
0 32
,5
35
67,5
93
%
62%
96
%
15
1300
8021
12
0 0
120
70
90
90
250
0 65
0
65
0 42
,5
30
72,5
13
%
70%
60
%
16
1300
8023
12
0 70
19
0 0
90
180
270
5 55
0
60
0 60
32
,5
92,5
31
%
79%
88
%
17
1300
8025
12
0 70
19
0 14
0 90
15
0 38
0 10
70
0
80
0 37
,5
35
72,5
89
%
68%
89
%
18
1300
8027
13
0 70
20
0 16
0 90
18
0 43
0 10
55
0
65
0 25
35
60
93
%
56%
92
%
19
1300
8029
12
0 70
19
0 90
90
18
0 36
0 10
65
0
75
0 50
32
,5
82,5
79
%
75%
88
%
TK3201-II-2010/2011 Chem Eng FTI-ITB 44 of 123
E
val
uat
ion:
Qu
iz
Ass
ign
men
ts
Mid
Ex
am
Fin
al E
xam
A
chie
vem
ent
W
eight:
15%
1
5%
2
5%
4
5%
N
r o
f ev
al:
3 q
uiz
zes
5 a
ssig
nm
ents
1
7/0
3/2
01
1
17
/03
/20
11
A
BE
T S
O:
(a)
(e)
(k)
To
tal
(a)
(e)
(k)
To
tal
(a)
(e)
(k)
Tota
l (a
) (e
) (k
) T
ota
l (a
) (e
) (k
)
T
arg
et:
200
100
30
0
20
0
10
0
20
0
50
0
10
90
0
10
0
0
65
35
10
0
10
0%
1
00
%
10
0%
No
. N
IM
20
13
00
8031
120
50
17
0
90
90
90
27
0
10
70
0
80
0
42
,5
32
,5
75
79
%
71
%
75
%
21
13
00
8033
110
50
16
0
90
90
18
0
36
0
5
35
0
40
0
12
,5
15
27
,5
48
%
40
%
54
%
22
13
00
8035
110
70
18
0
90
90
18
0
36
0
5
70
0
75
0
50
35
85
48
%
76
%
92
%
23
13
00
8037
120
0
12
0
14
0
90
18
0
41
0
10
45
0
55
0
30
32
,5
62
,5
89
%
56
%
74
%
24
13
00
8039
130
70
20
0
14
0
90
15
0
38
0
10
60
0
70
0
25
30
55
89
%
57
%
80
%
25
13
00
8041
130
70
20
0
16
0
90
18
0
43
0
5
45
0
50
0
17
,5
30
47
,5
61
%
48
%
83
%
26
13
00
8043
120
90
21
0
16
0
90
15
0
40
0
5
50
0
55
0
47
,5
35
82
,5
61
%
69
%
93
%
27
13
00
8045
140
70
21
0
14
0
90
18
0
41
0
10
70
0
80
0
57
,5
32
,5
90
89
%
83
%
88
%
28
13
00
8049
120
50
17
0
16
0
90
18
0
43
0
10
60
0
70
0
30
32
,5
62
,5
93
%
60
%
84
%
29
13
00
8051
120
50
17
0
90
90
90
27
0
5
50
0
55
0
15
25
40
48
%
47
%
62
%
30
13
00
8053
110
50
16
0
14
0
90
18
0
41
0
5
45
0
50
0
30
35
65
58
%
55
%
88
%
31
13
00
8055
130
50
18
0
90
90
18
0
36
0
10
70
0
80
0
45
25
70
79
%
74
%
71
%
32
13
00
8057
150
90
24
0
17
0
90
18
0
44
0
5
15
0
20
0
12
,5
7,5
2
0
63
%
38
%
49
%
33
13
00
8059
130
50
18
0
14
0
90
0
23
0
5
70
0
75
0
15
22
,5
37
,5
58
%
53
%
49
%
34
13
00
8061
130
0
13
0
90
90
0
18
0
5
55
0
60
0
30
20
50
48
%
59
%
34
%
35
13
00
8063
120
90
21
0
0
90
18
0
27
0
10
25
0
35
0
15
20
35
63
%
40
%
70
%
36
13
00
8065
150
50
20
0
16
0
90
18
0
43
0
10
65
0
75
0
60
32
,5
92
,5
93
%
84
%
84
%
37
13
00
8067
140
50
19
0
16
0
90
90
34
0
10
70
0
80
0
50
20
70
93
%
78
%
53
%
38
13
00
8069
130
50
18
0
90
90
90
27
0
5
35
0
40
0
47
,5
32
,5
80
48
%
66
%
75
%
39
13
00
8071
110
50
16
0
90
90
18
0
36
0
10
40
0
50
0
47
,5
35
82
,5
79
%
66
%
88
%
40
13
00
8073
70
70
14
0
12
0
90
18
0
39
0
10
70
0
80
0
50
32
,5
82
,5
85
%
73
%
88
%
TK3201-II-2010/2011 Chem Eng FTI-ITB 45 of 123
E
val
uat
ion:
Qu
iz
Ass
ign
men
ts
Mid
Ex
am
Fin
al E
xam
A
chie
vem
ent
W
eight:
15%
1
5%
2
5%
4
5%
N
r o
f ev
al:
3 q
uiz
zes
5 a
ssig
nm
ents
1
7/0
3/2
01
1
17
/03
/20
11
A
BE
T S
O:
(a)
(e)
(k)
To
tal
(a)
(e)
(k)
To
tal
(a)
(e)
(k)
Tota
l (a
) (e
) (k
) T
ota
l (a
) (e
) (k
)
T
arg
et:
200
100
30
0
20
0
10
0
20
0
50
0
10
90
0
10
0
0
65
35
10
0
10
0%
1
00
%
10
0%
No
. N
IM
41
13
00
8075
70
70
14
0
16
0
90
18
0
43
0
10
55
0
65
0
60
22
,5
82
,5
93
%
76
%
71
%
42
13
00
8077
120
70
19
0
90
90
90
27
0
10
65
0
75
0
42
,5
35
77
,5
79
%
70
%
83
%
43
13
00
8079
120
70
19
0
16
0
90
18
0
43
0
10
70
0
80
0
55
30
85
93
%
80
%
83
%
44
13
00
8081
120
70
19
0
16
0
90
15
0
40
0
10
50
0
60
0
20
30
50
93
%
50
%
80
%
45
13
00
8083
110
70
18
0
90
0
90
18
0
5
35
0
40
0
17
,5
17
,5
35
48
%
30
%
53
%
46
13
00
8085
130
50
18
0
90
50
18
0
32
0
5
65
0
70
0
30
35
65
48
%
56
%
88
%
47
13
00
8087
130
0
13
0
0
0
60
60
10
70
0
80
0
10
12
,5
22
,5
63
%
36
%
27
%
48
13
00
8089
0
0
0
70
50
18
0
30
0
0
0
0
0
0
32
,5
35
67
,5
13
%
30
%
78
%
49
13
00
8091
140
50
19
0
90
90
15
0
33
0
10
40
0
50
0
45
27
,5
72
,5
79
%
66
%
72
%
50
13
00
8093
120
70
19
0
14
0
90
18
0
41
0
10
80
0
90
0
50
32
,5
82
,5
89
%
79
%
88
%
51
13
00
8095
110
70
18
0
70
0
90
16
0
5
35
0
40
0
22
,5
35
57
,5
44
%
34
%
83
%
52
13
00
8097
110
70
18
0
16
0
90
18
0
43
0
10
65
0
75
0
17
,5
30
47
,5
93
%
52
%
83
%
53
13
00
8099
60
50
11
0
90
90
90
27
0
10
50
0
60
0
12
,5
25
37
,5
79
%
41
%
62
%
54
13
00
8101
130
70
20
0
14
0
90
15
0
38
0
5
60
0
65
0
65
32
,5
97
,5
58
%
85
%
85
%
55
13
00
8103
120
50
17
0
16
0
90
0
25
0
10
60
0
70
0
32
,5
30
62
,5
93
%
62
%
61
%
56
13
00
8105
110
0
11
0
90
0
90
18
0
5
30
0
35
0
45
10
55
48
%
48
%
26
%
57
13
00
8107
110
50
16
0
16
0
0
90
25
0
10
50
0
60
0
30
35
65
93
%
43
%
79
%
58
13
00
8109
130
50
18
0
16
0
90
90
34
0
10
55
0
65
0
30
15
45
93
%
59
%
45
%
M
inim
um
0,0
0,0
0
,0
0,0
0
,0
0,0
0
,0
0,0
0
,0
0,0
0
,0
0,0
1
0,0
7
,5
20
,0
0%
7
%
26
%
A
ver
age
114,0
54,7
1
68
,8
11
2,6
7
5,6
1
35
,3
32
3,5
7
,8
53
,1
0,0
6
0,9
0
,0
35
,9
28
,1
64
,1
69
%
59
%
72
%
M
axim
um
150,0
90,0
2
40
,0
17
0,0
9
0,0
1
80
,0
44
0,0
1
0,0
8
5,0
0
,0
95,0
0
,0
65
,0
35
,0
97
,5
94
%
85
%
96
%
TK3201-II-2010/2011 Chem Eng FTI-ITB 46 of 123
APPENDIX C
EXAMPLE OF STUDENTS WORKS
TK3201-II-2010/2011 Chem Eng FTI-ITB 47 of 123
C.1. Mid Examination
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C.2. Final Examination
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C.3. Quiz 1
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C.4. Quiz 2
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C.5. Quiz 3
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C.6. Assignment 1
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C.7. Assignment 2
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C.8. Assignment 3
TK3201-II-2010/2011 Chem Eng FTI-ITB 100 of 123
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TK3201-II-2010/2011 Chem Eng FTI-ITB 105 of 123
C.9. Assignment 4
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C.10. Assignment 5
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