portofolio tk3201 jul 11 footer

125
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

Upload: serpihan-hati

Post on 11-Nov-2014

18 views

Category:

Documents


1 download

TRANSCRIPT

Page 1: Portofolio TK3201 Jul 11 Footer

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

Page 2: Portofolio TK3201 Jul 11 Footer

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

Page 3: Portofolio TK3201 Jul 11 Footer

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.

TK3201-II-2010/2011 Chem Eng FTI-ITB 1 of 123

Page 4: Portofolio TK3201 Jul 11 Footer

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

TK3201-II-2010/2011 Chem Eng FTI-ITB 2 of 123

Page 5: Portofolio TK3201 Jul 11 Footer

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.

TK3201-II-2010/2011 Chem Eng FTI-ITB 3 of 123

Page 6: Portofolio TK3201 Jul 11 Footer

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

TK3201-II-2010/2011 Chem Eng FTI-ITB 4 of 123

Page 7: Portofolio TK3201 Jul 11 Footer

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.

TK3201-II-2010/2011 Chem Eng FTI-ITB 5 of 123

Page 8: Portofolio TK3201 Jul 11 Footer

Figure 1. Distribution of Grade.

Figure 2. Cummulative Distribution of Grade.

TK3201-II-2010/2011 Chem Eng FTI-ITB 6 of 123

Page 9: Portofolio TK3201 Jul 11 Footer

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)

TK3201-II-2010/2011 Chem Eng FTI-ITB 7 of 123

Page 10: Portofolio TK3201 Jul 11 Footer

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.

TK3201-II-2010/2011 Chem Eng FTI-ITB 8 of 123

Page 11: Portofolio TK3201 Jul 11 Footer

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

TK3201-II-2010/2011 Chem Eng FTI-ITB 9 of 123

Page 12: Portofolio TK3201 Jul 11 Footer

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.

TK3201-II-2010/2011 Chem Eng FTI-ITB 10 of 123

Page 13: Portofolio TK3201 Jul 11 Footer

APPENDIX A

SYLLABUS, LEARNING SYSTEM AND EVALUATION

TK3201-II-2010/2011 Chem Eng FTI-ITB 11 of 123

Page 14: Portofolio TK3201 Jul 11 Footer

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

TK3201-II-2010/2011 Chem Eng FTI-ITB 12 of 123

Page 15: Portofolio TK3201 Jul 11 Footer

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

TK3201-II-2010/2011 Chem Eng FTI-ITB 13 of 123

Page 16: Portofolio TK3201 Jul 11 Footer

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

TK3201-II-2010/2011 Chem Eng FTI-ITB 14 of 123

Page 17: Portofolio TK3201 Jul 11 Footer

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

TK3201-II-2010/2011 Chem Eng FTI-ITB 15 of 123

Page 18: Portofolio TK3201 Jul 11 Footer

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 -

TK3201-II-2010/2011 Chem Eng FTI-ITB 16 of 123

Page 19: Portofolio TK3201 Jul 11 Footer

APPENDIX B

DOCUMENTATION AND STATISTIC

TK3201-II-2010/2011 Chem Eng FTI-ITB 17 of 123

Page 20: Portofolio TK3201 Jul 11 Footer

B.1. Examination Evidences

TK3201-II-2010/2011 Chem Eng FTI-ITB 18 of 123

Page 21: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 19 of 123

Page 22: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 20 of 123

Page 23: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 21 of 123

Page 24: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 22 of 123

Page 25: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 23 of 123

Page 26: Portofolio TK3201 Jul 11 Footer

B.2. List of Attendance

TK3201-II-2010/2011 Chem Eng FTI-ITB 24 of 123

Page 27: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 25 of 123

Page 28: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 26 of 123

Page 29: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 27 of 123

Page 30: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 28 of 123

Page 31: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 29 of 123

Page 32: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 30 of 123

Page 33: Portofolio TK3201 Jul 11 Footer

B.3. Examination Problems

TK3201-II-2010/2011 Chem Eng FTI-ITB 31 of 123

Page 34: Portofolio TK3201 Jul 11 Footer

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

Page 35: Portofolio TK3201 Jul 11 Footer

1. Theflow2. Steatub3. Assdimradheadurthe in s4. Assis sincthicsymThe3. Tthe AnamatIntecon

the

e higher the wability meaady flow coe length is vsuming the emensional sidial directionat transfer pring cookingentire outespherical coosumptions: steady andce the pipe ickness, andmmetry abouermal conduThere is no pipe. alysis: Notithematical fegrating thenstant. Apply

variation of

T

temperaturans lower vindition, lamvery large wegg to be ronce tempern only becaprocess sincg. Also, the ser surface ofordinates. T1. Heat cond one-dimis long relatd there is ut the centeuctivity is cheat genering that heformulation e differentiaying the foll

|temperature

TK3201 TRSEMEMIDDThursre, the largeiscosity. minar flow owith respect tund, heat trature differause of symme the tempepherical coof a sphericalThe origin is nduction ensional ive to its thermal er line. 2. constant. ration in eat transfeof this prob0 al equation owing boun

is determine

RANSPORESTER II-2DLE EXAMsday, 17 MSOLUT

r the densitof a fluid, coto tube radiransfer to anrences (and metry abouerature at anordinate sysl body can bplaced at th

er is one-dblem can be twice givesndary condit|

0 ed to be T =

RWater

RT PHENO2010/201MINATIOMarch 20TION ty and the benstant densius, no slip cn egg in boithus heat trut the centerny point witstem could bbe describedhe center of

dimensional expressed as: T = C1 ln tions Tf.

R1

2Insula

OMENA 11 N 11 etter the flowsity and visccondition at ling water cransfer) wilr point. Thisthin the eggbe used to sod by a constthe egg.

in the raas r + C2, whe

ated (q=0)

wability of lcosity, Newtthe solid sucan be modell primarily s would be g will changolve this protant value o

adial r direre C1 and C

liquid. Goodtonian fluid,urface. eled as one-exist in thea transientge with timeoblem sincef the radius

rection, theC2 arbitrary

d , -e t e e s

e y

TK3201-II-2010/2011 Chem Eng FTI-ITB 33 of 123

Page 36: Portofolio TK3201 Jul 11 Footer

5. AssMasisotAnamin

mov

stoi

−0.

con

Mak

= 0.

Inte

Wit

the

sumptions: 1ss diffusivitthermal andalysis: There

nus z direct

ving in the

ichiometry o

5NAz, wher

ndition at any

king a mass

. Insertion of

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

Page 37: Portofolio TK3201 Jul 11 Footer

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

Page 38: Portofolio TK3201 Jul 11 Footer

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

Page 39: Portofolio TK3201 Jul 11 Footer

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

Page 40: Portofolio TK3201 Jul 11 Footer

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

Page 41: Portofolio TK3201 Jul 11 Footer

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

Page 42: Portofolio TK3201 Jul 11 Footer

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

Page 43: Portofolio TK3201 Jul 11 Footer

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

Page 44: Portofolio TK3201 Jul 11 Footer

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

Page 45: Portofolio TK3201 Jul 11 Footer

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

Page 46: Portofolio TK3201 Jul 11 Footer

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

Page 47: Portofolio TK3201 Jul 11 Footer

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

Page 48: Portofolio TK3201 Jul 11 Footer

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

Page 49: Portofolio TK3201 Jul 11 Footer

APPENDIX C

EXAMPLE OF STUDENTS WORKS

TK3201-II-2010/2011 Chem Eng FTI-ITB 47 of 123

Page 50: Portofolio TK3201 Jul 11 Footer

C.1. Mid Examination

TK3201-II-2010/2011 Chem Eng FTI-ITB 48 of 123

Page 51: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 49 of 123

Teknik Kimia ITB
Rectangle
Page 52: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 50 of 123

Page 53: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 51 of 123

Page 54: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 52 of 123

Page 55: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 53 of 123

Page 56: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 54 of 123

Teknik Kimia ITB
Rectangle
Page 57: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 55 of 123

Page 58: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 56 of 123

Page 59: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 57 of 123

Page 60: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 58 of 123

Page 61: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 59 of 123

Teknik Kimia ITB
Rectangle
Page 62: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 60 of 123

Page 63: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 61 of 123

Page 64: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 62 of 123

Page 65: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 63 of 123

Page 66: Portofolio TK3201 Jul 11 Footer

C.2. Final Examination

TK3201-II-2010/2011 Chem Eng FTI-ITB 64 of 123

Page 67: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 65 of 123

Teknik Kimia ITB
Rectangle
Page 68: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 66 of 123

Page 69: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 67 of 123

Page 70: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 68 of 123

Page 71: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 69 of 123

Page 72: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 70 of 123

Page 73: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 71 of 123

Teknik Kimia ITB
Rectangle
Page 74: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 72 of 123

Page 75: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 73 of 123

Page 76: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 74 of 123

Page 77: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 75 of 123

Teknik Kimia ITB
Rectangle
Page 78: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 76 of 123

Page 79: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 77 of 123

Page 80: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 78 of 123

Page 81: Portofolio TK3201 Jul 11 Footer

C.3. Quiz 1

TK3201-II-2010/2011 Chem Eng FTI-ITB 79 of 123

Page 82: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 80 of 123

Page 83: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 81 of 123

Page 84: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 82 of 123

Page 85: Portofolio TK3201 Jul 11 Footer

C.4. Quiz 2

TK3201-II-2010/2011 Chem Eng FTI-ITB 83 of 123

Page 86: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 84 of 123

Page 87: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 85 of 123

Page 88: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 86 of 123

Page 89: Portofolio TK3201 Jul 11 Footer

C.5. Quiz 3

TK3201-II-2010/2011 Chem Eng FTI-ITB 87 of 123

Page 90: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 88 of 123

Page 91: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 89 of 123

Page 92: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 90 of 123

Page 93: Portofolio TK3201 Jul 11 Footer

C.6. Assignment 1

TK3201-II-2010/2011 Chem Eng FTI-ITB 91 of 123

Page 94: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 92 of 123

Page 95: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 93 of 123

Page 96: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 94 of 123

Page 97: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 95 of 123

Page 98: Portofolio TK3201 Jul 11 Footer

C.7. Assignment 2

TK3201-II-2010/2011 Chem Eng FTI-ITB 96 of 123

Page 99: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 97 of 123

Page 100: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 98 of 123

Page 101: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 99 of 123

Page 102: Portofolio TK3201 Jul 11 Footer

C.8. Assignment 3

TK3201-II-2010/2011 Chem Eng FTI-ITB 100 of 123

Page 103: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 101 of 123

Teknik Kimia ITB
Pencil
Teknik Kimia ITB
Text Box
(copy paste from solution manual)
Page 104: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 102 of 123

Page 105: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 103 of 123

Page 106: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 104 of 123

Teknik Kimia ITB
Pencil
Teknik Kimia ITB
Pencil
Page 107: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 105 of 123

Page 108: Portofolio TK3201 Jul 11 Footer

C.9. Assignment 4

TK3201-II-2010/2011 Chem Eng FTI-ITB 106 of 123

Page 109: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 107 of 123

Teknik Kimia ITB
Pencil
Page 110: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 108 of 123

Page 111: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 109 of 123

Teknik Kimia ITB
Pencil
Page 112: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 110 of 123

Page 113: Portofolio TK3201 Jul 11 Footer

C.10. Assignment 5

TK3201-II-2010/2011 Chem Eng FTI-ITB 111 of 123

Page 114: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 112 of 123

Teknik Kimia ITB
Pencil
Page 115: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 113 of 123

Page 116: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 114 of 123

Page 117: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 115 of 123

Page 118: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 116 of 123

Page 119: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 117 of 123

Teknik Kimia ITB
Pencil
Page 120: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 118 of 123

Page 121: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 119 of 123

Page 122: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 120 of 123

Page 123: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 121 of 123

Page 124: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 122 of 123

Page 125: Portofolio TK3201 Jul 11 Footer

TK3201-II-2010/2011 Chem Eng FTI-ITB 123 of 123