eviDences foR classRoom innovations, leaRning gains ...€¦ · Rankine Cycle Gas Mixtures...

eviDences foR classRoom innovations,

leaRning gains & teaching-leaRning

mateRials

“The great aim of education is not knowledge, but action”

- Herbert Spencer -

1. thermodynamics innovations pefalec course code : cmt251 & cmt4082. philosophy course code : fsg5003. Basic physics ii course code : phy4074. scholarship of teaching & learning

- Jean Piaget-

“The principle goal of education is to create men who are capable of doing new things, not simply of

repeating what other generations have done -- men who are creative, inventive and discoverers”

theRmoDynamics innovations pefalec

couRse coDe : cmt251, cmt408

website for thermodynamics course

Main Menu FSG500 PHY407 RiHanum Thermodynamics July 05 Karachi pics ACES Sports 2003 Pics

Thermodynamics - For Students

Download Adobe Acrobat Reader Free. Acrobat Reader 5.0 for Win 95/98. An alternate site can be found at University Teknologi MARA website

http://www.itm.edu.my/ciis/download/acrobat/.

Once you are asked to save the file, save it to the desktop. When downloading is finished, you will see an ar50 icon on the desktop. Double click on that icon and the Acrobat Reader will be installed on the computer. Just follow the on-screen instructions during the installation process. As soon as the installation is completed, you are ready to read all of my pdf files by clicking on the files. Be sure to enter the password when prompted. If the document is not previewed on the screen, even after the downloading is done, click the Refresh button on the toolbar. You may need to do this once or twice. Don't ask me why, I think it may be related to how the cache of the computer is configured. Contact me at [email protected] or [email protected] or

[email protected] for the password.

Lectures (in PDF)

Lectures (in pps)

Students' Obligations!

1. Buy a copy of the textbook; "Thermodynamics An Engineering Approach" by Cengel & Yunus; Edition 4. 2. Download the specific operational objectives available on this site, for the purpose of self-assessment. 3. Use the course outline (details of what to do and when to do) along with the textbook and the operational objectives to prepare for your collaborative (group) learning sessions. 4. Mark the options on the operational objectives after doing your reading and preparation before you attend the collaborative learning on Tuesdays. Then save your options in

Microsoft Excel with the filename "myname.xls". An example of a filename could be jaafar.xls 5. Send an email with your excel file as an attachment to the email address [email protected] to be received by Tuesday at 3 pm. It is best that you send it over the

weekend along with your reading assignment/concept map. I need to compile and review your responses before Tuesday's class. Failure to send the responses results in penalty not

more that 1% (for each week) of the total course percentage. 6. Once the collaborative session is over, you must send your post-responses to the email address [email protected] to be received no later than 3pm on Fridays.

Failure to send the responses results in penalty not more that 0.5% of the total course percentage. I will make comparison between your pre and post responses to the operational

objectives.

Important Dates:

Self Assessments sent via email with your excel file attached. Get Your Files By Clicking Here

Pre-assessment: Due Tuesday by 3 pm. Post-assessment: Due Fridays by 3pm.


Course Objectives

Accumulated (Carry) Marks (pdf)

(Excel)

Self Assessment Responses Excel Files

More Links to Thermodynamics Material

Important Dates

Faclitator's Name List

Example of Self Assessments

Introduction Pure

Substance part1

Pure Substance

part2

Heat and Work

First Law Control Mass

First Law Control Volume

Second Law_1

Second Law_2

EntropyKinetic Theory

Rankine Cycle

Gas Mixtures

Combustion

Introduction Pure

Substance part1

Pure Substance

part2

Heat and Work

First Law Control Mass

First Law Control Volume

Second Law_1

Second Law_2

EntropyKinetic Theory

Rankine Cycle

Gas Mixtures

Combustion

Months Pre & post assessment

Reading Assignment Due Dates Tuesdays

Quizzes Tests

July

26 & 29 26 Quiz 1: 3rd or 4th Quiz 2: 10th or 11th

Aug 2 & 5

9 & 12, 17 &19 4 & 25

Quiz 1: 3rd or 4th Quiz 2: 10th or 11th

Test 1:

24th 3:30PM OR

25th 2:30pm

September 6 & 9

13 & 16, 20 & 23 13, 20, 27

Quiz 3: 7th or 8th

Quiz 4: 21st or 22nd

Quiz5: 28th or 29th

Test 2:

19th Monday night 8:30pm.

Oct 4 & 12

Test 3:

19th 3:30PM

Page 7 of 11Dr JJ or Dr Jaafar Jantan Homepage

13/05/2011http://drjj.uitm.edu.my/

Facilitators Name List:

All Groups


Course Objectives:

On completion of this course, students should be able to:

1. Use the property table and simple mathematical computation to determine phases of pure substances when given 2 independent

intensive properties. 2. Represent phases and states of pure substances and thermodynamics processes on property diagrams such as the temperature-volume

and pressure –volume property diagrams. 3. Identify the forms of dynamic or interacting energies and forms of energies within a system. 4. Write the general energy and mass balances and specific energy balances in any units, for the many different devices such as nozzles,

piston-cylinder devices, turbines, and compressors. 5. State the characteristics of heat engines, draw energy flow diagrams and schematic diagrams for steam power plants, heat pumps and

refrigerators. 6. Determine thermal efficiencies for real and Carnot engines. 7. Identify, and discuss causes of energy degradation and the use of Increase of Entropy Principle and entropy balance to quantify energy

losses in a device. 8. Sketch a temperature-entropy property diagram representing Carnot cycles, Rankine cycles and other modified Rankine cycles. 9. Solve conceptual and computational problems related to energy balances, mass balances, mass flow rates, power generated by

turbines, power input to compressors or air conditioners, heat exchangers and thermal efficiencies of Carnot and Rankine cycle engines. 10. Convert from gravimetric (by weight) to volumetric (by mole) and vice versa and to write equation of states for gas mixtures. 11. Explain the idea of combustion, to state the sources of combustion and hence able to determine the amount of oxygen and air

required for combustible materials to burn completely.


Links to Thermodynamics Materials This Semester

"What we have to learn to do, we learn by doing." "What we have to learn to do, we learn by doing." "What we have to learn to do, we learn by doing." "What we have to learn to do, we learn by doing." ----AristotleAristotleAristotleAristotle

Course Outline Peer Facilitating Learning-Cycle Instruction

Specific Operational Objectives used as Pre & Post assessment Facilitators Notes

Distribution of Previous Course Grades Students Evaluate Dr. J.J.'s Instructional Model




Course Outline & Instructional Methods


Self-Assessment & Notes for Facilitators


Self-Assessment Responses Excel Files

� Save each of the files below onto your diskette or hard drive. � Fill-in your responses. � Then send the file containing your responses, to me, by attaching it with your email . � ALWAYS COPY the email to yourself to ensure what I shall receive is the same as what you had sent. � ALWAYS SAVE your responses on a diskette and bring the diskette containing your responses, to class, in case I did not receive your email with the attachments.

"One who learns by finding out has sevenfold the skill of the one who learned by being told." Arthur Gutterman

Course Outline (pdf) Peer Facilitating Learning-Cycle Instruction (pdf)

Specific Operational Objectives (pdf) Facilitators Notes (pdf)

Pure Substance-Part 1 Pure Substance

Pure Substance-Part 2 Heat and Work

Heat and Work First Law - Control Mass

First Law - Control Mass First Law - Control Volume

First Law - Control Volume Second Law-Heat Engines

Second Law-Heat Engines & Carnot Engines Second Law - Carnot Engines

Second Law-Heat Engines & Carnot Engines Entropy

Entropy Rankine Cycle

Rankine Cycle

Pre Assessment Must Receive on Tues before 3 pm.

Send your Email to: [email protected]

Post Assessment Must Receive on Friday before 3pm

Send your Email to: [email protected]

Pure Substance-Part 1 Weds Thurs Pure Substance-Part 1 Weds Thurs

Pure Substance-Part 2 Weds Thurs Pure Substance-Part 2 Weds Thurs

Heat and Work Weds Thurs Heat and Work Weds Thurs

First Law - Control Mass Weds Thurs First Law - Control Mass Weds Thurs

First Law - Control Volume Weds Thurs First Law - Control Volume Weds Thurs

Second Law-Heat Engines Weds Thurs Second Law-Heat Engines Weds Thurs

Entropy Weds Thurs Entropy Weds Thurs

Rankine Cycle Rankine Cycle




Grades Distribution


Student Assessing Dr. J.J.'s Instructional Model

Some of the pdf files above are encrypted with password. Send me an email at [email protected] or [email protected] to obtain the password to open and print the documents.


Samples of Pre and Post Self Assessments

Analysis of Self-Assessments - Chap 2 [pdf] Sample of self assessments - Chap 2 [xls] Sample of self assessments - Chap 6 [xls]


Picture Album for ACES 2003 Sports Day

Distribution of Course Grades May 02 - Jan 01 [pdf] Distribution of Course Grades June 99 - May 00 [pdf]

End of semester Course Evaluation Jun 01 - Oct 01 End of semester Course Evaluation Nov 01 - March 02

End of Semester Course Evaluation May 02 - Oct 02 End of Semester Course Evaluation Dec 02 - April 03 DICBSAC



syllabus

Semester Dec 2002 – May 2003 @Hakcipta Fakulti Sains Gunaan, UiTM, Shah Alam

Thermodynamics CMT 251/CMT 408 Lecturer: Dr. J.J. Page of 5

APPENDIX 3 - COURSE OUTLINE

THERMODYNAMICS TEXTBOOK Thermodynamics: An Engineering Approach 4

th Edition International Edition By

YUNUS A. CENGEL & MICHAEL A. BOLES

This class meets for 2 hours on Tuesdays and on Thursdays for the Dec 2002 - Apr 2003 session. In addition, peer facilitators will see me on Saturdays (2 hrs) & on or Sundays (3 hrs) (only twice in a semester) for consultation (mentor-mentee) to become facilitators.

Lecturer: Associate Prof. Dr. Jaafar Jantan a.k.a. Dr. JJ Room: 516, Level 5, FSG Phone: 5544-4593 or 019-355-1621 Email: [email protected] or [email protected] Website: http://www.uitm.edu.my/faculties/fsg/drjj1.html Raja Razuan Raja Deris Room:614, Level 6, FSG Phone: 5544-4607 Email: [email protected] or [email protected] WEEK 1 (Dec 26

th – Dec 28

th)

Welcome Back to Campus – Time to ponder WEEK 2 (Dec 30

th – Jan 4

th)

Getting acquainted & Introduction How to do Concept Map Index of learning styles & Force Concept Inventory WEEK 3 (Jan 6

th – Jan 11

th)

Getting acquainted & introduction to the course Format: Lecture/Group Discussion/Concept Map 1.0 Introduction to thermodynamics

1.1 Thermodynamics and energy 1.2 A note on Dimensions & Units 1.3 Closed and opened systems 1.4 Properties of a system 1.5 State and equilibrium 1.6 Processes and cycles 1.7 Forms of energy 1.8 Energy & Environment (Sugg. Reading) 1.9 Temperature & Zeroth Law of Thermo

1.10 Pressure 1.11 The Manometer (Suggested Reading) 1.13 Problem-Solving Technique

First Consultation: Group 1 – Sat & Sunday (Jan 11th & Jan 12

th)

Read Assgn/Concept Map 1: CHAPTER 2 (Part 1) due by 3:00pm Jan 13th.

WEEK 4 (Jan 13

th – Jan 18

th)

Format: The Learning Cycle (Consultation/Concept Introduction, Collaborative Learning/ Reinforcement/Lecture-Discussion) by Group 1.

2.0 Properties of pure substances 2.1 Pure substance 2.2 Phases of pure substance 2.3 Phase change processes of pure substances

2.4 Property diagrams (T-ν) for phase change processes 2.5 Property Tables (Compressed Liquid & Superheated Vapor)

2nd

Consultation Group 2– Sat & Sun (Jan 18th

& Jan 19th

)) Read Assgn/Concept Map 2: CHAPTER 2 (Part 2) due by 3:00pm Jan 20th. WEEK 5 (Jan 20

th – Jan 25

th)

Format: The Learning Cycle (Consultation/Concept Introduction, Collaborative Learning/Reinforcement/Lecture-Discussion) by Group 2.

2.5 P-ν diagrams & Property Tables (Wet Mix Phase & more) 2.6 The Ideal-Gas Equation of State 2.9 Specific Heats

Problems: 1C till 7C, 17C, 19C till 24C, 26, 30, 32, 50, 57, 68. QUIZ 1

3rd

consultation Group 3 –Sat & Sunday (Jan 25th

& Jan 26th

)

Read Assgn/Concept Map 3: CHAPTER 3(1) Heat & Work due by 3:00pm Jan 27th. WEEK 6(Jan 27

th– Feb 1

st)


3.0 Energy Transfer by Heat and Work & Mass 3.1 Heat as a form of energy (Also read methods of heat transfer). 3.2 Energy Transfer by Work 3.3 Mechanical forms of work (moving boundary) 3.4 Non-mechanical Forms of Work 3.5 Conservation of Mass Principle 3.6 Flow Work and the Energy of a Flowing Fluid



PROBLEMS: 1C till 7C, 10C till-13C, 14, 15, 18, 19, 29, 35C, 36C 37, 45C till-48C, 50, 53, 57C till-59C, 62 QUIZ 2 WEEK 7(Feb 3

rd– Feb 8

th) Feb 3

rd Chinese New Year

4th

consultation Group 4 –Tue & Thurs (Feb 4th

& Feb 6th

)

WEEK 8 (Feb 10

th – Feb 15

th)

Raya Haji & Mid-Semester Break

Read Assgn/Concept Map 4: CHAPTER 4(1) Closed system-due by 5:00pm Feb 17th. WEEK 9(Feb 17

th– Feb 22

nd)


4.0 First Law of Thermodynamics 4.1. The First Law of Thermodynamics 4.2. Energy balance for closed systems

PROBLEMS: 1C, till 4C, 5, 7, 11, 12, 18, 20, 21. 27, 28 TEST 1 (Mon Feb 24

th, 8:15pm till 9:30pm) at DKA or DKB.

5th

consultation Group 5 – Sat & Tue (Feb 22nd

& Feb 23rd

)

Read Assgn/Concept Map 5: CHAPTER 4(2) - Control Volume due by 5:00pm Feb 24th WEEK 10 (Feb 24

th – Mar 1

st).


4.3. Energy Balance for Steady-Flow systems 4.4. Some steady-flow engineering devices 4.5. Internal energy, enthalpy, and specific heats for solids and liquids

PROBLEMS: 56C till 60C, 61, 74, 75C till 77C, 79, 81, 91, 92C till 95C, 96, 97, 101C till 103C, 105, 110. QUIZ 3

6th

consultation Group 6–Sat & Sunday (Mar 1st

& Mar 2nd

)

Read Assgn/Concept Map 6: CHAPTER 5(1) Second law–Heat engines - due 3:00pm Mar 3rd. WEEK 11 (Mar 3

rd – Mar 8

th)

Format: The Learning Cycle (Consultation/Concept Introduction, Collaborative Learning/ Reinforcement/Lecture-Discussion) by group 6.

5.0 The Second Law of Thermodynamics 5.1 Introduction to the second law

5.2 Thermal energy reservoirs 5.3 Heat Engines

5.31 Thermal Efficiency 5.32 Can we save Qout ? 5.33 The Second Law: Kelvin –Planck Statement

5.4 Energy Conversion & Efficiencies (Suggested Reading) 5.5 Refrigerators and Heat Pumps

5.41 Coefficient of Performance 5.42 Heat Pumps 5.43 The Second Law: Clausius Statement 5.44 Equivalence of the two statement

5.6 Perpetual-Motion Machines: Suggested reading PROBLEMS: 5C, 9C, 10C, 13C till 16C, 17, 18, 40C till 47C, 50, 54, 62 QUIZ 4

7th

consultation Group 7–Sat & Sun (Mar 8th

& Mar 9th

)

READING ASSIGN. 7:CHAPTER 5(2) Second law–Reversibilities- due by 3:00pm Mar 10th.

WEEK 12 (Mar 10th

– Mar 15th

) Format: The Learning Cycle (Consultation/Concept Introduction, Collaborative

Learning/ Reinforcement/Lecture-Discussion) by Group 7. 5.7 Reversible and Irreversible Processes

5.71 Irreversibilities 5.72 Internally and externally reversible processes

5.8 The Carnot Cycle 5.81 The Reversed Carnot Cycle

5.9 The Carnot Principles 5.10 The Thermodynamic Temperature Scale 5.11 The Carnot Heat Engine

5.111 The Quality of Energy (Suggested Reading) 5.112 Quantity Versus Quality in Daily Life (Suggested Reading)

5.12 The Carnot ReSatgerator and Heat Pump TEST 2 Mar 17

th 8:30pm till 10:00pm)

PROBLEMS: 67C, 73C till 75C, 80, 82, 87, 89C, 90C, 93C, 95, 98 104, 108

8th

consultation Group 8–Sat & Sunday (Mar 15th

& Mar 16th

)

Read Assgn/Concept Map 8: CHAPTER 6 - Entropy - due by 3:00pm Mar 17th.

WEEK 13 (Mar17

th – Mar 22

nd)


6.0 Entropy 6.1 Entropy



6.2 The Increase of entropy Principle 6.3 Entropy change of pure substances 6.4 Isentropic Processes 6.5 Property diagrams involving entropy 6.6 What is entropy? 6.7 The T-ds relations 6.8 Entropy change of liquids and solids (Read) 6.9 Entropy change of ideal gases (Read) 6.13 Entropy balance

Problems: 1C till 23C, 24, 26, 29, 32, 38, 113. QUIZ 5

9th

consultation Group 9–Sat & Sun (Mar 22st & Mar 23

rd)

Read Assgn/Concept Map 9: CHAPTER 9 – Vapor Cycles - due by3:00pm Mar 24th WEEK 14 (Mar 24

th – Mar 29

th)

Format: The Learning Cycle (Consultation/Concept Introduction, Collaborative Learning/ Reinforcement/Lecture-Discussion) by group 9.

9.0 Vapor and Combined Power Cycles 9.1 The Carnot Vapor Cycle 9.2 Rankine Cycle-The Ideal Cycle for Vapor Power Cycle

9.21 Energy analysis of the Ideal Rankine Cycle 9.3 OMIT 9.4 How can we increase the efficiency of the Rankine Cycle

9.41 Lowering the Condenser Pressure 9.42 Superheating the steam to high temperatures 9.43 Increasing the boiler pressure

9.5 The Ideal Reheat Rankine Cycle

PROBLEMS: 1C, 2C, 6, 7C till 10C, 15, 22, 24C, 32 QUIZ 6 Read Assgn/Concept Map: CHAPTER 12 and Chap 14 (DO NOT SUBMIT) WEEK 15 (Mar 31

st – Apr 5

th)

Format: Lecture-Discussion 12.0 Gas Mixtures

12.1 Composition of a gas mixture - Mass and mole fractions : Gravimetric and volumetric analyses

12.2 P-v-T behaviour of gas mixtures – Dalton’s Laws of additive pressures and Amagat’s Law of additive volumes

14.0 Properties of pure substances

14.1 Introduction to combustion

14.2 Proximate and ultimate (gravimetric and volumetric) analyses of solids fuels

14.3 Combustion of fuels in excess air supplied 14.4 Conversion of volumetric analyses into gravimetric analyses and vise-

versa WEEK 16 (Apr 7

th – Apr 12

th)

TEST 3 (Tuesday Apr 8th

2:15 pm till 3:30 pm)

Apr 14

th – Apr 19

th

DEAD WEEK –HELP AVAILABLE DURING CLASS HOURS

FINALS (Apr 20th – May 9th)

May 12th – June 15th SEMESTER BREAK

NOTE: Since 50% of all the tests and the final will incorporate problem-solving

skill based on conceptual understanding, your comprehension of the subject matter is highly significant before you try to solve any problems. Hence, you need to practice doing the assigned problems while brushing your understanding of the subject matter. The other 50% will concentrate on sketching, drawings, labelling, and writing down concepts both in your own words and its mathematical representation. Your ability to sketch, label and draw & concept mapping will determine how much you understood the subject matter and in most cases the questions that are asked of you and will also indicate your success in the course.

Semester May 2001 – Oct 2001 @ Hakcipta Fakulti Sains Gunaan, UiTM, Shah Alam

Thermodynamics CMT 251/CMT 408 Instructor: Dr. J.J. Page 4 of 5

ASSESSMENT

TESTS 3 x 10% = 30%

Read Assgn/Concept Map 9 x 1% = 9%

QUIZ 5 x 2% = 10%

PROJECT (Peer-Facilitating) (2+4.5+4.5)% = 11%

FINAL EXAM 1 x 40% = 40%

Read Assgn/Concept Map

ALL the assignment must be submitted no later than 3:00pm on Mondays. Your handwriting must be legible. Remember that this is only a summary of the chapter. Hence, I anticipate you to draw concept maps, draw, label, draw again and write down conceptual ideas. In addition, the mathematical representation for how the concepts are related and maybe some derivation on how to represent the concepts mathematically, must also be included, where applicable. DO NOT COPY ALL THAT YOU SEE IN YOUR TEXTBOOK. SUMMARIZE THEM. A good summary should not be more than 3 page long with lots of drawings and labeling or concept/mind mapping. You receive credit for each assignment that you submit.

PROJECT (PEER FACILITATING)

Assessment Components include the following: i) Preparation before and after consultation (must sent self-assessment via email) and

attendance whether as a facilitator or as a peer. – 4.5% + 4.5% ii) Your presence as facilitators on Saturday, Sunday and Tuesday –2 X 1% = 2% FORMAT: 1) Ten (DIC) and five (BSAC) groups will be formed and depending on the class size, a group

may consist of between three and four members. 2) Each student will be assigned to do peer tutoring (facilitating) ONLY TWICE during the

semester. This works really well if there are four members in each group but we will make do with whatever numbers we have.

Procedure: a) Group members are randomly selected by me. b) Group coordinators will be selected by me based on CGPA. c) Each group will receive the assigned topic as listed in the outline. d) Group leaders will then discuss and distribute reading responsibilities to group members. e) Group members will then meet to put together materials and methods they will use to facilitate

discussion with peer students. f) On Saturdays & Sundays, groups doing the facilitating the following week will see me to

discuss what needs to be prepared. During this hour, we will go through your preparation (the concepts involved, the graphs or drawings and labeling that are required, the reasoning for a process or relationship between concepts, and the mathematical representations of the relationships), your understanding, your needs and your lacking. Group members are to read the materials before seeing me. We will be using the operational objectives & facilitator’s notes that can be downloaded from my website at http://www.uitm.edu.my/faculties/fsg/drjj1.html. A password is required to open the PDF files. You will need to install Acrobat Reader 5.0 (free) on the computer you are using. Bring the operational objectives when you see me on Saturdays & Sundays.

g) Facilitating will be on Tuesdays (DIC) for two hours. Your role is to start and guide discussion among members and NOT TO TEACH. In addition to sharing concepts with your peers, you will also need to work on some examples and exercise problems during this session. Before the session begins, your students would have already evaluated themselves by using the operational objectives (self-assessment). All sessions will begin on time (2:10pm).

h) Wednesdays/Thursdays are designated for reinforcement (lecture/discussion) led by me with all students. In addition, there will be quizzes (as designated on pages 1 – 3 of this outline).

Semester May 2001 – Oct 2001 @ Hakcipta Fakulti Sains Gunaan, UiTM, Shah Alam

Thermodynamics CMT 251/CMT 408 Instructor: Dr. J.J. Page 5 of 5

Important Dates:

Months Consultations Saturdays &

Sundays

Reading Assignment

Mondays

Quizzes Thursdays

Tests

Jan 11 &12 18 & 19 25 & 26

10, 20 & 27 23 & 30 -

Feb 4 & 6 22 & 23

17,24 27 24 Monday night

March 1 & 2 8 & 9

15 & 16 22 & 23

3, 10, 17, 24 6, 20 & 27 17 Monday night

April 8 Tuesday class

Self Assessments sent via email with your excel file attached. Pre assessment: Received Mondays by 3 pm. Email [email protected] Post assessment: Received Wednesdays by noon. Email: [email protected]

Efficient learning = Active learning = Getting involved in the learning process

Pefalec cycle, leading questions

&notes

Thermodynamics Semester Jun – Dec 20000

Lecturer: Dr. J.J. Page 1 of 1

Peer-Facilitating Session for Thermodynamics CMT 251/408: DIC/BSAC/CHEM ENG

Concept Introduction - Friday• Facilitators (FC) (between 10-15 pupil) meet me for about 3

hours.• A coordinator is appointed – those with highest CGPA.• FC gets to read a set of guided questions or simplified notes

and very specific operational/outcome objectives on the week’stopic.

• Review of previous week’s topic. Identify weakness & remedy.• New concepts are introduced via talk & discussion method with

lots of pauses and using many diagrams.

Concept Reinforcement• FCs meet me again for

about 3 hours on Sat/Sun.• Reinforcement of concept.

Identify weakness and findremedy.

• Solve examples andproblems.

• Discuss strategy to do peer-facilitating.

• Reading assignment on thetopic is due – all students.

Peer Facilitating - Tuesday• FCs are given between 3 to 4 students.• Students evaluate their state of content preparedness by

reviewing specific outcomes for the topic before session starts.• Review of previous topic by making corrections to their quiz.• Review of previous week’s topic. Identify weakness & remedy.• New concepts are introduced via show & tell & discussions

with lots of pauses and using many diagrams.• FCs are evaluated.• Students reevaluate themselves on the content preparedness

to check for any conceptual gains.

Concept Reinforcement &Summative EvaluationThursday• Summary of objectives and

outcome for the week’s topic• Reinforcement of concepts.

Identify weakness and findremedy- lots of diagrams.

• Solve examples and problems– more diagrams.

• Quiz..

The Peer-Facilitating Learning CycleThis learning format is employed in this class in view of researchfindings that lecture is the least effective mode of learning. Bestlearning can be done in very small groups and throughcollaborative methods. Hence, peers-teaching-peers isimplemented in the hope that more learning than teaching istaking place. Furthermore, students are found to lack the abilityto do verbal reasoning due to lack of exposure. This learningformat provides the opportunity to pick up on that lacking skill.

Thermodynamics CMT 251/408 Notes for Peer Leaders Page 1 of 4

Topic: Heat & Work Lecturer: Dr. J.J. Semester Nov 01 – March 02

Heat & Work

No Question on Heat & Work Suggested Response

1. What do I know so far about the study of thermodynamics?

Thermodynamics means thermal energy in motion. Hence the study involves harnessing (getting) thermal energy that exists in material substances into useful energy such as electrical energy. The energy conversion requires a medium as an agent. Hence we need to look into properties of pure substances, otherwise known as the system, which are commonly used as the medium. Then we need to look at the process that the system undergoes and how the properties of the system change when a process takes place. Following that we will observe the causes of change that usually involves the energy that is tapped from material substances. Processes that systems undergo are caused by energy crossing the system’s boundaries. In this chapter, we will look at the types of energy that crosses a system’s boundary.

2. What is an energy interaction and in what forms do they exist?

An energy interaction or dynamic energy is the energy that crosses a system’s boundary and causes the system’s state or properties to change. There are 3 forms of interaction energy namely heat transfer, work done and energy of mass transfer. The energy of mass flowing across a system’s boundary is only discussed for an open system (control volume) while heat transfer and work done are discussed in both a closed system (control mass) and an open system.

3. What is heat? Heat is short for heat transfer & is thermal energy in motion by virtue of temperature difference between the system and the surrounding. This energy is recognized only as it crosses the system’s boundary.

4. Will there be any heat transfer if the system’s temperature and the surrounding’s temperature is the same?

Generally, no heat is transferred since there is no temperature difference. At this point, both the system and the surrounding are in thermal equilibrium. But in an ideal situation, heat can be transferred isothermally in which case the temperature difference is always zero. For this to happen, the transfer process has to be done very slowly.

5. If heat is received by a system, does it mean that the system has more heat than before?

NO. Substances or matter do not possess heat. They

have energy in the form of thermal or internal energy. Hence when a system receives (rejects) energy, its thermal or internal energy will increase (decrease). Remember that heat is only recognized at the boundary. Before crossing, it is part of the surrounding’s (system’s) internal energy and after crossing, it becomes part of the system’s (surrounding’s) internal energy.

6. How do I physically measure the increase in thermal or internal energy of a system?

A system’s internal or thermal energy is physically measured by measuring its temperature. An increase (decrease) in temperature indicates that the energy has increased (decreased).

7. What symbols are used to represent heat transfer and is there a certain sign conventions that are generally used? Note that 1 kilo = 1000. Hence, 1 kilojoule = 1000 joule or abbreviated as 1 kJ = 1000 J.

The letter Q is used to designate the total heat transfer in or out of a system and the unit is Joule (J) or better yet, kilojoule (kJ). But in thermodynamics, we usually specify quantities for one kilogram of the substance. Hence the letter q is used to designate the amount of heat transfer for each kilogram of the substance. Q

and q are related by the relation, kg

kJ ,

m

Qq . Often

times, we also specify the rate of heat transfer,

Q ,

which is the amount of heat transferred in one second.

System

Combustible

Materials

Work

done




So, t

QQ

, in units of kJ/s or sometimes referred to

as kilowatts, kW. Note that t is the time interval during which Q is transferred. The convention chosen is positive for Q into a system and negative for Q out of a system. But we will use the intuitive approach that is by using Qin and Qout for heat transferred into and out of the system respectively. If we are not sure whether it is in or out, we will just assume and will only ascertain its direction after performing a calculation.

8. What are the methods of heat transfer to a system?

There are 3 methods of transfer namely conduction, convection and radiation. Refer to the textbook for the discussion.

9. What is the other form of energy interaction for a closed system?

Heat is energy transfer across a system’s boundary by

virtue of temperature difference, T. If the energy interaction is not Q, then its has to be work for any closed system.

10. What is work? Work usually involves a force and a position change of an object or matter due to the action of the force. Energy is required to do work and work that is done can be stored in any form of energy. Hence, work is a form of energy transfer.

11. Is there a certain symbol and convention used to represent work done?

The symbol W represents the total work done while

the symbol is used to represent work done onto (by)

a 1kg substance. Hence, kg

kJ,

m

W . Another

symbol is the power, kW ors

kJ ,

t

WW

, which

represents the amount of work done in 1 second. Note

that t is the amount of time that the work is done. If work is done by a system, designate this as Wout or positive, and vice versa.

12. What are the similarities or differences between Q and W?

Both are recognized at the boundaries of a system. Hence both are boundary phenomena. Both are energy interactions and cause the energy within a system to change. Both are not states, but rather causing states of a system to change, i.e. both will cause a process to happen. Both do not have values at a state.

13. Can we discuss some examples to see what kind of energy interaction is involved?

See examples 3-2, 3-3, 3-4, and 3-5.

14. Can you elaborate more on the work done by an electrical resistor?

As shown in e.g. 3-5, when the filament is included as the system, then electrons are accelerated across the system’s boundaries due to a potential difference, V, across the battery terminals. Since the electric force, Fe, has moved the charges, q = Ne, where N is the number of electrons and e is the charge of an electron, across the boundary, hence the work done is

qVs qE moved distanceFW e

then rate of work done or electrical power is

kW ors

kJ ,iV

t

qVW

. Note that, i, is the

electrical current or the number of charges passing through a point in one second.

15. Can we try to do some examples involving both Q and W?

Try reading & comprehend the solution to example 3-6 in the textbook.

16. Besides the electrical work, what other forms of work will we consider in thermodynamics?

An important form of work is the mechanical work which involves movement or change of position caused by a force. A constant force F, (such as a push

Win Wout

+ -

Qin Qout

_ +




or a pull or otherwise known as contact forces), acting on a box and changes the position of the box from an initial position s1 to a final position s2, will do work on

the box an amount of W = F(s2 – s1) = Fs. If the push

or pull is changing in strength throughout the movement of the box, then the work done for each of the different forces which moves the box a distance,

ds, will be dsFW . The total amount of work

done by all the different forces in moving the box the

whole distance is

2

1

2

1

FdsWW . What we have

done is actually finding the sum of all the little amount of work done by the different forces. If in the integration above, the force is constant throughout the movement of the box, then

sFssFdsFFdsW )( 12

2

1

2

1

, which

is the same as the one above.

17. Why is this mechanical work important in thermodynamics? The figure below depicts an expansion process.

Processes in thermodynamics include expansion and compression for a gas in a piston-cylinder device. In an expansion (compression), the gas (piston) exerts pressure on the piston (gas) and causes the piston to move up (down) during an expansion (compression). Since the pressure acts on the surface of the piston and pressure is related to force by the relation,

A

F

Area

Forceessure Pr , then the force, F, in the

work relation can be replaced by PAF . During an expansion (compression), the piston moves to a new position. Since there is a force acting to change the position of the piston, then work had been done by the gas (piston) on the piston (gas). Hence for a constant

force, the work done is dsPAdsFW . The

quantity Ads is recognized as the volume change of

the gas, dV = Ads. Notice that the boundary of the

system has moved. Hence the work done, known as

the boundary work done, is PdVW and the total

work done in expanding (compressing) a gas is

kg

kJ Pd or kJ; ,PdVWW bb ,

2

1

2

1

2

1

2

1

If in the process, the pressure is constant (not changing at all throughout the expansion), then,

kJ ,VPVVPdVPPdVWb )( 12

2

1

2

1

For specific work done, then

kg

kJ ,PPdPPdb )( 12

2

1

2

1

18. Can the boundary work done be determined graphically?

Try plotting graphs of pressure versus volume, P-V, or

pressure versus specific volume, P-. Try doing this for gases of pure substances such as water undergoing isothermal expansion or compression with temperatures above the critical temperature or even those below the critical temperature. What will the area under the graph represent? Hint: Check the units.

19. Can we have some examples to Try doing examples 3-7, 3-8, and 3-9 in the text.

Refer to the textbook for the diagram.

ds




apply what we had obtained about the boundary work?

20. What other forms of mechanical work are relevant?

Gravitational and accelerational work. Gravitational work is the work done against the gravitational pull of

the earth. So, when you move an object vertically away from a position z1 to a position z2, the work done

is

2

1

1

2

1

2

dzmgmgdzFdzWg

1000

kJ ,PEzzmgWg )( 12

or kg

kJ pe,

zzgg

1000

)( 12

Notice that mg(z2 –z1) is the change in potential energy of an object when raised (lowered) from a position z1 to a position z2. Accelerational work needs to be done to accelerate an object to a certain velocity. The force

involved is just accelerational force,dt

vdmmaF

and so the work done is

2

1

2

1

1

2

1

2

vdvmdtvdt

vdmmadsFdsWa

Then 1000

kJ KE

vvmvdvmWa ,)

2(

21

22

2

1

Or kg

kJ ke

vva ,)

2000(

21

22

Notice that the last expression represents the change in kinetic energy of the object during the accelerational process. Note also that the units should be in Joule. In order to change to kilojoule, then you multiply by 1000 then divide by 1000. The 1000 on the numerator are named kilo while the 1000 on the denominator should remain.

kJ 1000

1J

1000

1000 J 1

Refer to the text for further explaination and diagrams.

Thermodynamics: First Law for Closed System Notes for Peer Leaders: Lecturer: Dr. JJ. FSG, UiTM Page 1 of 3

First Law of Thermodynamics

Statement of the First Law Energy Balance – Math expressions of the first law of thermodynamics

Energy must be conserved in any thermodynamics process. The total energy before and after a system undergoes a process must be the same. Any energy that crosses the system’s boundaries (interaction or dynamic energies) can

change the system’s energy. This change (Esys) must have the same magnitude as the cause of the change (Ein – Eout). So the system’s total energy must be balanced like you balance your bank account (money deposited minus money withdrawn must be equal to the amount of money left in the account). The system’s energy will not change if there is no energy interaction.

Energy entering system – Energy leaving the system = Change in the system’s energy

Ein – Eout = Esys = E2 – E1

Esys = 0, if Ein = Eout = 0. None of the system’s properties change since there is no agent to change it.

For any system, the energy interactions are the heat transfer, Q, the work done, W, and the energy of the mass, Emass, flowing in and out of the system. Note that we do not use the

symbol delta () for the interaction energy difference. The symbol is reserved for energy change within a system.

(Qin – Qout) + (W in – Wout) + (Emass,in – Emass,out) = Esys, kJ - total.

or (qin – qout) + (in - out) + (mass,in – mass,out) = esys, kJ/kg – unit-mass basis.

The left side of this mathematical relationship is the energy interaction or simply the agents (cause) of change The right side of the relationship is the change itself, the systems’ energy change after a process has taken place where E1 and E2 are the system’s total energy before and after a process respectively. . The total energy of a system is the sum of the internal (U), kinetic (KE) and potential energies (PE).

Ein – Eout = Esys = E2 – E1 : System’s final energy – system’s initial energy, where E1 = U1 + KE1 + PE1, & E2 = U2 + KE2 + PE2, kJ or e1 = u1 + ke1 + pe1 & e2 = u2 + ke2 + pe2, kJ/kg.

Then Esys = E2 – E1 = (U2 – U1) + (KE2 – KE1) + (PE2 – PE1) = U + KE + PE, kJ or

esys = e2 – e1 = (u2 – u1) + (ke2 – ke1) + (pe2 – pe1) = u +ke + pe, kJ/kg

For a closed or control mass system such as a rigid tank, there is no mass flowing in or flowing out of the system. Hence only heat transfer and work done is responsible for

energy changes within a system. The symbol mass is used to

represent the specific total energy for the moving mass.

Emass,in = 0 & Emass,out = 0, kJ or mass,in = 0 & mass,out= 0, kJ/kg. Hence,

(Qin – Qout) + (W in – Wout) + 0 = Esys, kJ - total

or (qin – qout) + (in - out) + 0) = esys, kJ/kg – unit-mass basis

Esys = E2 – E1 Ein

Eout Esys = E2 – E1

Ein



Combining both the causes of change (the energy interactions), and the changes itself (the system’s total energy change), for a closed system, the energy balance is written as shown on the right.

(Qin – Qout) + (W in – Wout) + 0 =U + KE + PE, kJ

or (qin – qout) + (in - out) + 0 = u +ke + pe, kJ/kg

For historical purposes, the energy interactions are written as Q and W only to represent the net heat received and the net work output respectively. But we will not use historical representation to avoid confusion.

Q – W = Qnet,in – W net,out = (Qin – Qout) - (Wout – W in). = (Qin – Qout) + (W in – Wout).

or q – w = qnet,in - net,out = (qin – qout) - (out - in) = (qin – qout) + (in - out)

Q – W = Qnet,in – Wnet,out = U + KE + PE, kJ

or q - = qnet,in – net,out = u +ke + pe, kJ/kg

In thermodynamics, the first law energy balance reduces to simple forms when special processes are considered. In most cases, closed systems (no mass flow) are assumed to be stationary. Hence, the kinetic energies and the potential energies do not change: ke1 = ke2 & pe1 = pe2, kJ/kg.

Qnet,in – Wnet,out = (Qin – Qout) + (W in – Wout) = U + 0 + 0, kJ

or qnet,in – net,out = (qin – qout) + (in - out) = u +0 + 0, kJ/kg

For stationary closed systems that undergo an isobaric process (constant pressure, i.e., P1 = P2), then the work done can be written as the sum of the boundary work done, Wb, (due to boundary movement when the system expands or contracts) and work done by other sources, Wother, such as the electrical work, [Welectrical = (electrical potential)*(electric

current)*(time that the current was passed) = it] and the shaft (paddle-wheel) work which are both W in. The boundary work can be either W in or Wout and is related to the pressure and volume as shown in the exercise in the adjacent column.

Win – Wout = Wother,in - Wb

21112212

2

1

2

1

b

21112212

2

1

2

1

b

PP where ;kg

kJ ,PP)(PdPPd

or PP where ; kJ ,VPVP)VV(PdVPPdVW

So the first law energy balance for a stationary closed system

undergoing an isobaric process is Q + Wother,in - Wb = U. Bringing the boundary work to the right, then the energy balance is as shown on the right.

Q + Wother,in = U + Wb = (U2 –U1) + (P2V2 – P1V1) = (U2 + P2V2) – (U1 + P1V1) = H2 – H1 = H where H is the enthalpy, H = U + PV.

Hence the energy balance is Q + Wother,in = H, kJ or q + other,in = h +0 + 0, kJ/kg

The diagrams below show a rigid tank (volume is a constant) stationary closed system interacting with heat and also electrical work with the states before and after interaction shown by the different properties. The energy balance for the process is shown on the right.

Since isochoric process (constant volume), then the boundary work done is zero. Hence Wout = 0

(Qin – 0) + (W in – 0) =U + 0 + 0, kJ or better written as Qin + Welec,in = U2 – U1.

(qin – 0) + (in - 0) = u + 0 + 0, kJ/kg or qin + elec,in = u2 – u1.



In writing the energy balance for a stationary-closed system, always consider special cases involved since it will reduce the mathematical expressions and the time involved in determining missing properties of a system. An isobaric process maintains a constant pressure throughout the process. An isochoric process is applied to a rigid tank, hence no boundary work can be done (Wb,out = 0). An adiabatic process is a process where the system is insulated to avoid heat transfer into and out of the system. A cyclic process is where the system’s final state is returned to its initial state at the end of the process.

Isobaric process: P1 = P2. So, Q + Wother,in = H, kJ or q + other,in = h +0 + 0, kJ/kg.

Isochoric process: V1 = V2. So, Qin + Wother,in = U = U2 – U1 or qin + other,in = u = u2 – u1.

Adiabatic process: Qin = Qout = 0. So, W in – Wout = U or in - out = u

Cyclic process: E2 = E1. So, Qin – Qout + W in – Wout = E = E2 – E1 = 0. Determining values when solving the energy balance. When dealing with water, use property table A4 to A6. When using refrigerant-134a, refer to property table A-11 to A-13. When the system is gases which obey the ideal gas condition, then apply the ideal has equation of state and refer to table A-1 and a minimum of A-17.

Closed System – Control

Mass P1, T1 , V1

kg

kJ ,pekeue or

kJ ,PEKEU1

E

1111

111

m1, 1 Initial State State 1

Qin

+

_

Welec,in

Closed System – Control

Mass P2, T2 , v2 = v1

kg

kJ ,pekeue or

kJ ,PEKEUE

2222

2222

m2 = m1; 2 = 1 Final State

State 2

formative self-assessment instrument Pefasai

Thermodynamics CMT 251/408

Self-Evaluation Before and After Facilitating Session. Topic: First Law Control Mass 3-2

Thermodynamics CMT 251/408 Facilitators's Name: ___________________ Your Name: _________________________

Topic: First Law Control Mass 3-2 Lecturer: Dr. JJ. Sem: Nov01. Page 2 of 10 Sem: Nov01. Topic: 3-2; Lecturer: Dr. JJ. Page 2 of 10

Please use the following codes: 1 - Strongly Disagree: 5 - Strongly Agree

Please circle your selection. Before peer teaching After peer teaching

values, then I can approximate it to be between the two values or I can perform the interpolation technique to obtain the exact value.

6. I am able to define the term quality in the wet-mix phase as the ratio of the vapor mass to the system’s total mass or

mathematically represented as Quality, x = (mass of vapor / mass of system) = gf

gg

mm

m

m

m

. The more vapor in

the wet mix phase, the higher the quality. The highest value is one (saturated vapor) if all the liquid has vaporized, hence mf = 0 and mg = 1, thus making x = 1. The lowest value is zero (saturated liquid) since at this state, mg = 0, (mf = 1) hence x = 0. Values of x along the saturated liquid line and saturated vapor line on the property diagrams are 0 and 1 respectively. In the wet-mix phase, 0 < x < 1, and quality is not defined in any other phases.are 0 and 1 respectively. In the wet-mix phase, 0 < x < 1, and quality is not defined in any other phases.

1 2 3 4 5 1 2 3 4 5

7. I can derive the average value of y in the wet-mix phase by doing the following. Consider a tank of total volume V containing a mixture of liquid (mass is mf) and vapor (mass is mg) of total mass, m, I can write V = Vf + Vg. Replacing

the total volume by the specific volume, mavg = mff + mgg. Dividing both sides by the total mass, m, then

g

g

f

fggff

avgm

m

m

m

m

mm

. Realize that m = mf + mg, hence by replacing mf by m – mg, and recalling that x =

m

mg, then gf

g

g

g

f

g

avg xm

m

m

m

m

m

m

mm

Canceling the masses and replacing the mass ratios with

the quality, then, fgffgfgffgfavg xxxxxx 1 The expression fgx is often

written in place of fgx . In the wet-mix phase, it is understood that the property y is an average value, hence

fgffgf xyyyyxyy and this expression is ONLY VALID while the system is boiling. At the saturated liquid

and saturated vapor line, y = yf and yg respectively since x = 0 and x = 1 respectively. Note that the term yfg means yg – yf.

1 2 3 4 5 1 2 3 4 5

8. I am able to determine the phase of a system if I know the state of the system. For example if the temperature of

refrigerant-134a in a tank is 0C, and the quality of the system is 0.2000, then the phase is wet-mix since the quality is 0 < x < 1. Then the system’s pressure must be the saturation (boiling) pressure since the system is in a wet-mix

phase (boiling). The other properties are determined by using the expression y = yf + xyfg. The volume, = f+xfg =

0.014552 m3/kg (use values from table A-11), will be used as one of the property to determine the final state of the

system since a rigid tank means no change in volume (and hence the specific volume since the mass remains

constant for closed systems). Thus if the system’s temperature is increased to 20C, then the phase for the final state

(T = 20C, = 0.014552 m3/kg) is determined by comparing = 0.014552 m

3/kg to the saturated liquid volume, f ,

and the saturated vapor volume, g, at T = 20C. Since f < < g, at T = 20C (read Table A-11), then the final

state is still wet-mix phase. Hence the pressure is the saturation (boiling) pressure at 20C (P=Psat@20C). Before

1 2 3 4 5 1 2 3 4 5







other properties such as u and h can be determined using the expression y = yf + xyfg, the quality for this final phase

must be calculated. Note that f and g which are used must be the values read at T = 20C. It can be shown that the

quality is 3926.0

20@

Cfg

fx

. It can be shown that if the initial state had a quality of 0.8, then the final

state when the temperature is increased to, 20C will be superheated vapor since > g. Hence use table A-13 to

determine u and h.

9. I am aware that a system’s property changes due to interaction energies or dynamic energies that cross the imaginary boundary of the system. When a system’s property has changed, we say a process has taken place. The forms of dynamic energy include heat, work and energy of mass transfer. Mass transfer is only involved when an open system (control volume) such as cold water entering a chamber (the system) and hot water exiting the chamber is discussed.

1 2 3 4 5 1 2 3 4 5

10. I am able to identify heat (short for heat transfer) as a form of thermal energy in motion due to a temperature difference. In other words thermal energy that crosses a system’s boundary because the temperature inside the system and the temperature of the surrounding is not the same. The name heat is only used when the thermal energy is crossing the system’s boundary. If there is absolutely no temperature difference or when the temperature difference

is very small or sometimes called infinitely small, T = 0, between the system and the surrounding, then the system and surrounding is said to be in thermal equilibrium. An infinitesimal heat transfer is achieved over very long time period and the heat transfer process is referred to as an isothermal process.

1 2 3 4 5 1 2 3 4 5

11. I realize that the term heat is usually represented by the symbol, Q, in units of kilojoule (kJ). Q, designates the total amount of heat received or lost by a system. Often times, the symbol, q, is used to represent specific heat or the mount of heat received or lost for each unit mass ( 1 kg) of the system. Hence the unit for q (q = Q/m) is kilojoule per kilogram (kJ/kg). For example, if 5 kg of water gains 10 kJ of heat then the specific heat is q = 10 kJ / 5 kg = 2 kJ/kg. Note that if there is no heat transfer to or from a system then q = 0 kJ/kg.

1 2 3 4 5 1 2 3 4 5

12. I realize that in other cases, the symbol,

Q , is used to represent the rate of heat transfer which is the amount of heat

received or lost by a system for each unit time (1 second). Hence, the unit for

Q (

Q =Q/t) is kilojoule per second

(kJ/s) or kilowatt (kW). Note also that kilo means 1000. Hence, 1 kJ = 1000 J and 1 kW = 1 kJ/s. For example if 10

kJ of heat is gained by water over a period of 10 seconds, then the rate of heat transfer into the water is

Q = 10 kJ /

10 s = 1 kJ/s = 1 kW.

1 2 3 4 5 1 2 3 4 5

13. I am aware that the usual convention used for heat transfer is positive value or usually Qin for heat received or gain by a system and negative value or Qout for heat lost by a system. Generally, the symbol Q represented in textbooks refers to the net (total) heat transfer Q = Qnet,in = Qin – Qout. As a note, heat can be transferred by either conduction

(direct contact), or convection (movement) or radiation (no medium required).

1 2 3 4 5 1 2 3 4 5







14. I realize that if the energy interaction is not heat, then it is either work done or mass transfer. In a closed system (control mass) where no mass enters or leaves the system, then the interaction must be heat transfer or work done or both. As a reminder, changes in a system’s property such as T, P, U, V or H are usually caused by energy interactions that cross the system’s boundary. Hence, Q and W are not states of a system, but rather the cause for the states of a system to change. It is ridiculous to write Q1 or Q2 to represent initial and final heat transfer. So, DO

NOT WRITE Q or W. The symbol delta, , is reserved to mean a change between final and initial values for a

system’s property. The expression u = u2 – u1 represents the internal energy change in a process.

1 2 3 4 5 1 2 3 4 5

15. I realize that work done is usually represented by the symbol, W, in units of kilojoule (kJ). W, designates the total

amount of work done by or on a system. Often times, the symbol, , is used to represent specific work done by or on

the system for each unit mass ( 1 kg) of the system. Hence the unit for ( = W/m) is kilojoule per kilogram

(kJ/kg). For example, if 10 kJ of work is done on a 5 kg system, then the specific work is = 10 kJ / 5 kg = 2 kJ/kg.

1 2 3 4 5 1 2 3 4 5

16. I realize that in other cases, the symbol,

W , is used to represent the rate of work done (power) which is the amount of

work done by or on a system for each unit time (1 second). Hence, the unit for

W (

W = W/t) is kilojoule per second

(kJ/s) or kilowatt (kW). Note also that kilo means 1000. Hence, 1 kJ = 1000 J and 1 kW = 1 kJ/s. For example if 10

kJ of work is done on a system over a period of 10 seconds, then the input power is

W = 10 kJ / 10 s = 1 kJ/s = 1

kW.

1 2 3 4 5 1 2 3 4 5

17. I realize that the convention used for work done by a system (output power) is positive or Wout and the work done on a system (input power) is negative or Win.

1 2 3 4 5 1 2 3 4 5

18. I am aware that when I place an electrical resistor into water and pass a current, i, through the resistor, then electrical work, Win,e is done on the system. The amount of work can be determined by multiplying the voltage source, , the current, i, and the amount of time or clock reading, t, that the current is allowed to pass through the resistor. Hence,

the electrical work done is Win,e = it. Generally, the symbol W represented in textbooks refers to the net work done

W = Wnet,out = Wout – Win (

W =

W net,out =

W out –

W in.).

1 2 3 4 5 1 2 3 4 5

19. I know that another common form of work done on a system, Win,pw, is the work done by a paddle-wheel or a fan which is rotating inside a system such as a fan rotating inside a room.

1 2 3 4 5 1 2 3 4 5

20. I realize that another important form of work done is the mechanical work done which is due to contact forces such as pushing and pulling. Pushing an object with a force, F, and causing the object to move from position, s1 to a final

position, s2, which is a total distance, s, will yield a total work done of, FsssFW 12 . If the force changes in

strength all the time between the initial position and the final position, then the work done, W , by each force over the

distance, ds, is FdsW . Hence, the total work done to move the object from its initial position to its final position

must be the sum of all the incremental work done, n

nk

k

WWWWW

....211

. For infinitesimally small (very

1 2 3 4 5 1 2 3 4 5







small) ds, the total work done is, 2

1

2

1

FdsWW . If the force is constant (both the strength and direction) from

the initial to the final position, then FsssFdsFFdsW 12

2

1

2

1

, which is the same as the one obtained before.

21. I realize that the mechanical work done is important in thermodynamics since there is always expansion (Wout) and compression (Win) of systems contained in a piston-cylinder device. When a system expands due to heat that it received, a force, F, pushes up on the surface of the piston of area, A, and causing the piston to rise, hence the word

expand. Since the force exerted on the piston is related to pressure, P, defined as, A

F

Area

Force essure Pr , hence F =

PA. Then the work done to push the piston a distance, ds, is PAdsFdsW . Notice that the product, Ads, is just the

volume change, dV = Ads. Hence the total work done by the system in changing the position of the piston from state

1 to state 2 during the expansion must be 2

1

2

1

2

1

PdVFdsWW . The specific work done is easily obtained by

dividing both sides by the mass, m, which then yields 2

1

2

1

Pd . Notice that in the case where the pressure is

constant throughout the expansion process, then 12

2

1

PdP or for the total work done,

12

2

1

VVPdVPW . A sketch of a P - graph will reveal that, P(2- 1), is the area under the graph and hence

the specific work done. In cases where the pressure is changing during the expansion (such as an isothermal

process), the specific form of the pressure as a function of the volume is necessary before the integration can be solved. Note that this type of work where it involves movement of a system’s boundary (piston-cylinder device) is called the boundary work done, Wb.

1 2 3 4 5 1 2 3 4 5

22. I realize that an isothermal process is a process where the temperature of the system remains the same throughout the expansion or compression process (T1 = T2). In this case, PV = constant = k. Hence the pressure is

inversely proportional to the volume, V

k

volume

tcons essure

tanPr . The boundary work done in this case is then,

1

212

2

1

2

1

2

1

2

1

2

1

lnlnln1

V

VkVVkdV

VkdV

V

kPdVFdsWWb . The fact that PV = constant means that P1V1 =

1 2 3 4 5 1 2 3 4 5







P2V2 = P3V3 = k. Hence, the boundary work done can also be written as

2

111

1

211

1

2 lnlnlnP

PVP

V

VVP

V

VkWb . At

this juncture, I must make sure that I am able to obtain the last expression by deriving it using the first principle which

is 2

1

2

1

2

1

PdVFdsWWb and the knowledge of PV = constant. In cases where the system contains real gases such

as air which can be treated as an ideal gas, then the equation of state, PV = NRuT, or PV = mRT can be used. Then, the constant, k, can be replaced by mR or NRu.

23. I realize that the first law of thermodynamics states that energy is always conserved in any processes. In other words, total energy entering or leaving a system must be the same as the total energy change of the system. For example, if the amount of energy received by 1 kg of compressed liquid water at room temperature and pressure is 100 kJ, then the total energy of the water must increase by 100 kJ. On the other hand, if the system loses energy by 100 kJ, then the system’s total energy must drop by 100 kJ also.

1 2 3 4 5 1 2 3 4 5

24. I can write the energy balance for the first law as

systema by

gainEnergy -

systema by

lostEnergy =

systemthe within

changeenergy Total. Mathematically, I

can represent it by writing, Ein - Eout = Esys ,in units of kilojoule OR

sysoutin EEE in units of kilowatt or

alternatively, ein - eout = esys in units of kilojoule per kilogram. Note that the energy change within the system, means

the difference between the system’s initial energy and final energy, Esys = Efinal – Einitial = E2 – E1.

1 2 3 4 5 1 2 3 4 5

25. I can identify the mechanism for energy transfer as the interaction or dynamic energies (energies crossing the boundaries of a system) which are heat, Q, work done, W, and the energy of mass transfer for an open (control volume) system, Emass.

1 2 3 4 5 1 2 3 4 5

26. I can rewrite the energy gain by the system by replacing it with the interaction energies,

systema by

gainEnergy or Ein is the sum

of the heat received, Qin, the work done on the system, Win, and the energy of the mass flowing into the system,

Emass,in. On the other hand, the energy lost by the system can be written the same way, systema by

lostEnergy or Eout is the sum

of the heat lost, Qout, the work done by the system, Wout, and the energy of the mass flowing leaving the system, Emass,out. Mathematically, the expression is, Ein = Qin + Win + Emass,in and Eout = Qout + Wout + Emass,out. Notice that the energy interactions Q, W and Emass are the cause for the energy change within a system. If there aren’t any energy crossing into and out of a system’s boundary that is, Ein = Eout = 0, then the system’s total energy remains

unchanged, Esys = 0 and hence E2 = E1.

1 2 3 4 5 1 2 3 4 5







27. I am very familiar with the symbols )( , , );( , , ,

kg

kJ

m

Ee

m

Qq

m

WkW

t

EE

t

WW

t

QQ

sys

sys

. 1 2 3 4 5 1 2 3 4 5

28. I am aware that the energies of a system is the sum of its thermal energy associated with its temperature, U = mcT,

(where c is the system’s specific heat) its kinetic energy associated with its speed, 2000

2

mKE and its potential

energy associated with its vertical position, 1000

mgyPE . Hence the energy change within the system is

PEKEUPEPEKEKEUUEsys 121212 . Note that TmcU ,

2000

21

22

m

KE and

10001000

12 mghyymgPE

. Note that in most cases the specific internal energy, u, can be obtained by reading

the property table and that the expression Tcu is only valid in the compressed liquid or superheated vapor phases

only.

1 2 3 4 5 1 2 3 4 5

29. I realize that I can write all the expressions for the first law in units of kilojoule, kilowatts or kilojoule per kilogram.

Hence the energy balance can be expressed in the following ways: systema by

gainEnergy -

systema by

lostEnergy =

systemthe within

changeenergy Total.

Or mathematically, sysoutin EEE in kJ, or sysoutin EEE

, in kW, or sysoutin eee in kJ/kg. Alternatively, it

can written as PEKEUEEWWQQ outmassinmassoutinoutin ,, in kJ or

PEKEUEEWWQQ outmassinmassoutinoutin ,, in kW or

pekeuqq outmassinmassoutinoutin ,, , in kJ/kg.

1 2 3 4 5 1 2 3 4 5

30. I am aware that a stationary system means: KE = 0 and PE = 0. So, Esys = U in kJ,

UE sys in kW and

esys = u in kJ/kg,

1 2 3 4 5 1 2 3 4 5







31. I know that a closed system means Emass,in = 0 and Emass,out = 0 or 0,,

outmassinmass EE or 0,, outmassinmass . 1 2 3 4 5 1 2 3 4 5

32. Hence, for a stationary closed system, the first law of thermodynamics energy balance can be written in any of the following ways:

UWWQQ outinoutin or

UWWQQ outinoutin or uqq outinoutin . Care must be

taken in using the symbol (pronounced as delta), to mean the change between the final value and the initial values

within the system. Hence, we DO NOT WRITE Q = Qin – Qout. This is because Q, or W are energy interactions (CAUSES OF CHANGE) and do not have any values at the initial or final state. ONLY PROPERTIES SUCH AS E, U, H, KE, PE, AND T HAVE INITIAL AND FINAL VALUES DURING A PROCESS (CHANGE OF STATE).

1 2 3 4 5 1 2 3 4 5

33. I know that for closed system undergoing an isochoric (constant volume) process, the boundary work done, Wb = 0, &

since the total work done is bothersoutin WWWW , hence UWQQ inoutin . Note that

epwinothers WWWW where Wpw is the work done by a paddle-wheel (like a fan) and We is the electrical work done

by a resistor or a heating element. Note also that the electrical work done is the product of the current through the resistor, , the potential difference across the resistor, , and the amount of time the current is allowed to pass, t.

Hence the electrical work done by a resistor is We = i t/1000 in kW. Another important fact to realize is that Wout is Wb for an expansion process in a piston-cylinder device.

1 2 3 4 5 1 2 3 4 5

34. I realize that for closed system undergoing an isobaric (constant pressure) process, the volume will change and hence I need to consider the boundary work done by the system. Recall that the boundary work done is obtained by summing over the infinitesimal work done to move a piston form its initial to its final position. Mathematically, the

boundary work done is, 112212 VPVPVVPdVPPdVWb where the constant pressure P = P1 = P2.

Again, since the total work done, for an expansion process is, bpweoutin WWWWW , hence the energy

balance can be written as 111222121122bpweoutin VPUVPUUUVPVPUWWWQQ . Recall

that in chapter 2 we had define the enthalpy of a system as H = U + PV. Hence, we can write

1212111222pweoutin hhmHHVPUVPUWWQQ . Writing the energies in terms of the enthalpies

of the system is economical since values for the enthalpy can be found in property tables and I have reduced my working time to analyze the energy balance by a significant amount. So, in units of kilojoule per kilogram, the energy

balance for a stationary-closed system undergoing an isobaric process is 12pweoutin hhqq . CAUTION:

This expression is ONLY VALID for isobaric (P2 = P1) processes. At times, the expression others is encountered

in the textbook in place of pwe

1 2 3 4 5 1 2 3 4 5







35. I know that for a stationary-closed system undergoing a cyclic (starts and end at the same state) process, the total

energy for the final and initial states are the same, hence Esys = U = 0. Due to this, the energy balance can be

written as 0 outinoutin WWQQ . This means that inoutoutin WWQQ . The physical interpretation of this is

very significant since many engineering devices operate in a cycle. This expression clearly indicates that if the left side of the expression is positive then the right side must also be positive and vice-versa. In other words, for a system such as water in as a turbine to produce work and hence electricity, a net amount of heat must be transferred to the system. On the other hand, in order to have a net amount of heat transferred out of a system such as an air conditioner, a net amount of work must be done on the system.

1 2 3 4 5 1 2 3 4 5

36. I am aware that for a stationary-closed system undergoing an adiabatic (no heat transfer, Q = 0) process, the energy

balance can be written as .UWW outin Hence, for a system in a piston-cylinder-device to expand (spontaneous

expansion) which means that work is done by the system (Win = 0), U must be negative. In other words, expansion will make the thermal energy of the system to drop since the energy had to be used to do work in the expansion process. As a result, a drop in temperature will be recorded. On the other hand, if the piston is pushed down (Wout = 0), work is being done on the system resulting in an increase in the system’s thermal energy. As a result, an increase in temperature will be observed.

1 2 3 4 5 1 2 3 4 5

37. I realize that if the system involves gases such as air or nitrogen, I need a way to determine the change in internal

energy u and enthalpy h respectively, since tables providing values for these are very limited. Hence I need to

know about the energy storage ability for each gas. I am also aware that the internal energy for gases has been experimentally demonstrated to depend only on the temperature, u = u(T). In other words, the internal energy at any state is specified when the temperature is known and will only change when there is a physical temperature change.

1 2 3 4 5 1 2 3 4 5

38. I can define specific heat as the amount of heat required to change the temperature of a unit mass of any substance

by 1C or 1K. The higher the specific heat, the more heat is required to change its temperature by one degree. Using

the first law for a stationary closed system I can define the specific heat at constant volume to be

T

uC and the

specific heat at constant pressure to be p

pT

hC

. Using the definition of enthalpy (h = u+ P) and equation of

state for ideal gases (P = RT), I am able to show that the enthalpy for gases depend only on the gas‘s temperature,

just like the internal energy. Since both the internal energy and the enthalpy depends only on temperature, then the

differential form of the energies are dT)T(Cdu and dT)T(Cdh p . The energy change in a process is obtained

by integrating the functions above and will yield 2

1

dT)T(Cu and 2

1

p dT)T(Ch respectively. The specific heat

1 2 3 4 5 1 2 3 4 5







is a function of temperature over large temperature variations but not when the temperature variation in a process is only a few hundred. Hence obtaining the internal and enthalpy change for ideal gases can be obtained by estimation

methods and some available table for C and Cp for certain gases. Hence, TCu avg, and the enthalpy change is

TCh avg,p . The average specific heat can be estimated by taking its value at the average temperature between

the initial and final temperature (C,avg = C read at 2

TT 12 from the property table such as table A-2(b)).

39. I am able to show that by differentiating the enthalpy function, h = u + P, I can relate the specific heats Cp = C +R in

units of kJ/kgK. In addition, another ideal-gas property is commonly cited in the literature is the specific heat ratio,

C

Ck

p . For mono-atomic gases such as helium, its value is 1.667 and is constant for any temperature. For diatomic

gases such as carbon dioxide, 4.1k at room temperature.

1 2 3 4 5 1 2 3 4 5

40. I am aware that for incompressible substances such as solids and liquids, the specific heat is just C = Cp = C for

small temperature intervals (a few hundred degrees). Then the internal energy change can be approximated to be

12avgavg TTCTCu . I realize that the volume change in solids and liquids is negligible. Hence the estimation

for u and h can made by again differentiating the enthalpy, h = u+ P which yields the term PPuh .

For solids, the term P and P (zero for constant volume) is not significant and hence TCuh avg . For

liquids undergoing constant pressure process, TCuh avg and for liquids undergoing constant temperature

process, Ph .

1 2 3 4 5 1 2 3 4 5

AS 115/4 DIC AS 225/2 BSAC

Self Assesment for First Law Control Mass. Blank spaces = No Valid Digital Submission till 4 pm Wednesday

Semester May 02 - Oct 02

Fill in the cells for each of the statement in your self assessment. Be honest. I just need to know your confidence level.

Save your work and attach it with your email. Due Wednesday by noon.

AfterPeer Discussion

Bil. KP ITM NAMA Nama Komersil 1 2 3 4 5 6 7 8 9 10 11 12 13 14

1 2001131685 Ahmad Lutfi b. Md. Yusoff Lutfi

2 2001131900 Ainul Wahida bt Kandak Ainul

3 2001357940 Ariffah Adilah bt Dzulkhefli Ifah

4 2000409496 Azlan Abu Bakar Lan

5 2000635809 Badrul Hisham b. Abdullah Badrul

6 2001132157 Bibi Normi Md Noor Bibi

7 2001176617 Faranurhazrin Zainal Fara

8 2001360016 Hairul Adini b. Idros Adini

9 2001357958 Harith Fitry B. Mohd Azali Muke

10 2001131294 Hazlelina binte Mohd Tahir Lin

11 2001357972 Khairienna Ahtar bt Khairuddin Rina

12 2001357997 Masitah bt Abdul Hadi Mas

13 2000409583 Mohd Faisal Othman Faisal

14 2001426486 Mohd Hainur Hapis b. Moh Esa Apis

15 2001131303 Mohd Helmi b. Othman Helmi

16 2001357965 Mohd Izwan b. Abdullah Wan

17 2001131932 Mohd Razif b. Aziz Razif

18 2000409688 Mohd Suhaimi bin Chik Mie

19 2001357933 Nor Hamizah bt Abd. Majid Hamizah

20 2001358014 Noraida bt Md Yusof Aida

21 2001176512 Norhafizah bt Abd. Razak Fiza

22 2001426639 Norzalilawaty Che Mod Nor

23 2001132196 Nur Huwaina binti Yunus Ina

24 2001343201 Nurhaizum Aimi bt Adnan Ajom

25 2001132205 Ramona bt Abdul Razak Mon

26 2000118724 Saiful Hizam b. Mohamed Shaer Zam

27 99004615 Subainah Lahimin Nana

1=Strongly Disagree 2=Disagree 3=Unsure/Undecided 4=Agree 5=Strongly Agree

Statement Number on the Self Assessment

Prepared by DR. JJ Thermodynamics

AS 115/4 DIC AS 225/2 BSAC

Self Assesment for First Law Control Mass. Blank spaces = No Valid Digital Submission till 4 pm Wednesday

Semester May 02 - Oct 02

15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

1=Strongly Disagree 2=Disagree 3=Unsure/Undecided 4=Agree 5=Strongly Agree

Statement Number on the Self Assessment

Prepared by DR. JJ Thermodynamics


Self-Evaluation Before and After Facilitating Session.Topic: Pure Substance – Part 2


Please use the following codes and please mark your selection. 1=- Strongly Disagree: 2=Disagree 3 = Unsure/Undecided 4=Agree 5 = Strongly Agree Pre Post

1. I have read the chapter before coming to the discussion session today. 4.3 4.6 2. I am able to state that a system is simply a space or mass that I choose to study. 4.3 4.6 3. I am aware that a process has taken place whenever there are changes in the system’s properties such as

changes in T, P, E, V or the mass, m. 4.1 4.6 4. I realize that intensive properties can easily be obtained by considering the system to have a mass of 1 kg. Hence,

to obtain an intensive property, I will just divide all of the size-dependent extensive properties by the mass of the system. Examples are the specific internal energy, u = U/m, the specific total energy, e = E/m, the specific volume, ν = V/m, the specific kinetic energy, ke = KE/m, the specific potential energy, pe = PE/m, and many more. Notice that the intensive properties are represented by using lower case alphabets and the term “specific” is used to mean “for a unit mass (mass of I kg) of the system”. 3.9 4.6

5. I am aware that the common system involved in thermodynamics are the pure substances; substances that have homogeneous chemical composition throughout. Examples include water, refrigerants used for cooling, nitrogen, helium, air in gaseous phase, a two-phase such as ice-water mixture and water-vapor mixture. Oil-water mixture is not a pure substance since the two substances do not mix homogeneously. 4.3 4.7

6. I realize that water can exist in the solid phase (ice), the liquid phase and the gaseous phase (vapor). 4.3 4.7 7. I realize that the solid-liquid mixture (ice-water mixture) and the liquid-gaseous mixture (water-vapor mixture)

respectively can coexist in real life as exhibited by iced-tea (ice-water mixture at its melting point) that I always drink or the water that is boiling (liquid-vapor mixture at its boiling point). 4.1 4.5

8. I can sketch a temperature versus specific volume graph (T - ν) for the phase change process of the ice that I had placed on the stove in my kitchen. Given a pressure of 101 kPa or atmospheric pressure, water begins to melt (ice changing to liquid) at 0°C and begins to boil (liquid begin to vaporize) at 100°C. The boiling temperature is known as the saturation (boiling) temperature at 101 kPa and is normally represented by T = Tsat@101kPa. 3.8 4.4

9. I know that the state at the start of boiling (phase change) is called the saturated liquid (liquid ready to vaporize) phase. The state at the end of boiling is called the saturated vapor (vapor ready to condense) phase and in between the two states where liquid and vapor coexist, it is called the saturated liquid-vapor mixture or simply wet-mix phase. The state where the temperature is below the boiling point is known as compressed liquid (liquid not ready to vaporize) phase while the state where the temperature is higher than the boiling point is referred to as superheated vapor phase (vapor not ready to condense). 4.0 4.6

10. I am aware that the specific volume at the start of boiling (liquid starts to vaporize) increases with an increase in pressure while the specific volume at the end of boiling (all the liquid turns into vapor) decreases with an increase in pressure. The subscript f and subscript g is used to designate properties at the saturated liquid state and the saturated vapor respectively. For example, νf and νg are the specific volume for the saturated liquid and saturated 4.0 4.4

Topic: Pure Substances-Part 2 Lecturer: Dr. JJ. Sem: Nov01. Page 1 of 6




Please use the following codes and please mark your selection.

1=- Strongly Disagree: 2=Disagree 3 = Unsure/Undecided 4=Agree 5 = Strongly Agree Pre Post

vapor respectively. Note that the phase change occur at the constant boiling (saturation) temperature. These values can be found in a property table such as given in Appendix A-5 in my Thermodynamics textbook.

11. I realize that as the pressure increases, the horizontal line connecting the saturated liquid and saturated vapor points gets shorter and eventually vanishes. The point where the line vanishes and the saturated liquid and saturated vapor point merges is called the critical point. This point has a pressure of Pcr = 22.09 MPa (1MPa = 1000 kPa), a temperature of Tcr = 374.14°C, and a specific volume of νcr = 0.003155 m3/kg. Beyond this point there is no wet-mix phase, only the superheated vapor phase. 3.9 4.4

12. I realize that by drawing a line to connect the saturated liquid points and a line to connect the saturated vapor points through the critical point, for pressure lines at and below the critical pressure, I will see a dome being formed. States to the left of the saturated liquid line but below the critical temperature is compressed liquid, those within the line (underneath the dome) is wet-mix and the phase to the right of the saturated vapor line and also above the critical temperature is the superheated vapor phase. The T - ν diagram containing the saturated liquid and saturated vapor lines and the pressure lines is known as a property diagram. 4.0 4.6

13. I know that values for the saturation temperatures, saturated liquid volume, νf and saturated vapor volume, νg for any pure substances at any given pressure can be obtained from tables such as table A1 through A13 in the Yunus & Cengel Thermodynamics textbook. 3.9 4.4

14. Using table A-5 (saturated water-pressure table), for example, I found that water at a pressure of P = 50 kPa starts to boil at 81.33°C (Tsat@50 kPa = 81.33°C). The volume for the saturated liquid state is νf = 0.001030 m3/kg and the volume when all the water has turned into vapor (saturated vapor phase) is νg = 3.240 m3/kg. Hence, if I had 1 kg of water at a temperature of 25°C, the water is in compressed liquid phase (since the given temperature, T < Tsat@50 kPa) which means that its volume can be estimated to be ν = νf@25°C (read the temperature table A-4, νf@25°C). It will start boiling if I can increase the temperature to the saturation temperature, 81.33°C. At this temperature, further addition of energy will force the water to begin vaporizing. The energy required to change 1 kg of water at its boiling point completely into vapor is called specific latent heat of vaporization. On the contrary, if the water temperature is given as 100°C, then the water is superheated vapor because this temperature is bigger that the saturation (boiling) temperature at 50 kPa (T > Tsat@50 kPa ). If this was the case, then the other properties for water at the given T and P, can be obtained by reading table A-6 (superheated water table). 3.3 4.1

15. I can sketch a pressure versus specific volume diagram (P - ν) for the phase change process of water at room

temperature T = 25°C and a room pressure of P = 101 kPa. If the temperature remains constant, water in a container will start to boil at the saturation (boiling) pressure, Psat@25°C. = 3.169 kPa. Since the room pressure P > Psat@25°C, (101 kPa > 3.169 kPa), hence at room temperature and pressure, water is in a compressed liquid phase. Hence, to vaporize it, the pressure needs to be reduced to the boiling or saturation pressure. When the pressure is reduced lower than 3.169 kPa (sat.pressure), the water will be in the superheated vapor phase. 3.3 4.2







16. The state at the end of boiling is called the saturated vapor (vapor ready to condense) phase and in between the

two states where liquid and vapor coexist, it is called the saturated liquid-vapor mixture or simply wet-mix phase. The state where the pressure is higher than the boiling point is known as compressed liquid (liquid not ready to vaporize) phase while the state where the pressure is lower than the boiling point is referred to as superheated vapor phase (vapor not ready to condense). 3.8 4.4

17. I am aware that the specific volume at the start of boiling (liquid starts to vaporize) increases with an increase in temperature while the specific volume at the end of boiling (all the liquid turns into vapor) decreases with an increase in temperature. 3.8 4.3

18. I realize that as the temperature increases, the horizontal line connecting the saturated liquid and saturated vapor gets shorter and eventually vanishes. The point where the line vanishes and the saturated liquid and saturated vapor point merges is called the critical point. This point has a pressure of Pcr = 22.09 MPa (1MPa = 1000 kPa), a temperature of Tcr = 374.14°C, and a specific volume of νcr = 0.003155 m3/kg. Beyond this point there is no wet-mix phase, only the superheated vapor phase. 4.0 4.4

19. I realize that by drawing a line to connect the saturated liquid points and a line to connect the saturated vapor points through the critical point, for temperature lines at and below the critical temperature, I will see a dome being formed. States to the left of the saturated liquid line but below the critical pressure is compressed liquid, those within the line (underneath the dome) is wet-mix and the phase to the right of the saturated vapor line and also above the critical pressure is the superheated vapor phase. The P - ν diagram containing the saturated liquid and saturated vapor lines and the pressure lines is known as a property diagram. 4.0 4.3

20. I know that values for the saturation pressures, saturated liquid volume, νf and saturated vapor volume, νg for any pure substances at any given pressure can be obtained from tables such as table A1 through A13 in the Yunus & Cengel Thermodynamics textbook. 3.9 4.5

21. Using table A-4 (saturated water-temperature table), for example, I found that water at a temperature of T = 50°C starts to boil at 12.349 kPa (Psat@50°C = 12.349 kPa). The volume for the saturated liquid state is νf = 0.001012 m3/kg and the volume when all the water has turned into vapor (saturated vapor phase) is νg = 12.03 m3/kg. Hence, if I had 1 kg of water at T = 50°C and a pressure of P = 50 kPa, the water is in compressed liquid phase (since the given pressure, P > Psat@50°C) which means that its volume can be estimated to be ν = νf@50°C (read the temperature table A-4, νf@50°C). It will start boiling if I can reduce the pressure to the saturation pressure, 12.349 kPa. At this pressure, water will vaporize. Further reduction in pressure will convert the wet-mix phase to superheated vapor. On the contrary, if the water pressure is given as 5 kPa and T = 50°C, then the water is superheated vapor because its pressure is smaller that the saturation pressure at 50°C (P < Psat@50°C). If this was the case, then the other properties for water at the given T and P, can be obtained by reading table A-6 (superheated water table). 3.5 4.3

22. I am aware that if I have determined the phase of a system as compressed liquid, than the specific properties such d h b i d b d h H I d f h bl

3.2 4.3







as ν, u, and h can be approximated to be νf@T, uf@T, and hf@T. Hence, I need to refer to the temperature table to obtain the properties or values of ν, u, and h.

23. I realize that if I have determined that the phase of my system as superheated vapor, than the specific properties such as ν, u and h are ν@T,P, u@T,P, and h@T,P. Hence, I need to refer to the superheated water or superheated refrigerant 134a table to obtain the properties ν, u and h. If the values cannot be read from the table directly but lies between two values, then I can approximate it to be between the two values or I can perform the interpolation technique to obtain the exact value. 3.2 4.2

24. I am able to define the term quality in the wet-mix phase as the ratio of the vapor mass to the system’s total mass

or mathematically represented as Quality, x = (mass of vapor / mass of system) = gf

gg

mmm +=

mm. The more

vapor in the wet mix phase, the higher the quality. The highest value is one (saturated vapor) if all the liquid has vaporized, hence mf = 0 and mg = 1, thus making x = 1. The lowest value is zero (saturated liquid) since at this state, mg = 0, (mf = 1) hence x = 0. Values of x along the saturated liquid line and saturated vapor line on the property diagrams are 0 and 1 respectively. In the wet-mix phase, 0 < x < 1, and quality is not defined in any other phases. 3.4 4.0

25. I realize that for a wet-mix phase, the temperature for any given pressure or the pressure for any given temperature, are the saturation temperature and saturation pressure respectively. I also realize that properties such as ν, u and h must have values lower that that of the saturated vapor value but higher than the saturated liquid value. If I use the symbol y to represent properties such as ν, u or h, then yf < y < yg in the wet mix phase. 3.5 4.3

26. I can derive the average value of y in the wet-mix phase by doing the following. Consider a tank of total volume V containing a mixture of liquid (mass is mf) and vapor (mass is mg) of total mass, m, I can write V = Vf + Vg. Replacing the total volume by the specific volume, mνavg = mfνf + mgνg. Dividing both sides by the total mass, m,

then gg

ffggff

avg mmmννν ==

mmmm νν+

+. Realize that m = mf + mg, hence by replacing mf by m – mg, and

recalling that x = m

mg , then gfg

gg

f xm

mmm

mm

νννν

−=+

gavg m

mmν

−= + Canceling the masses and replacing

the mass ratios with the quality, then, ( ) ( ) fgffgfgffgf xxxxxx νννννννννν =−+=+−=+−= 1avgν +

fgxν

The expression is often written in place of ( )fgx νν − . In the wet-mix phase, it is understood that the property y is an average value, hence ( ) fgff xyy +=gf yyxyy −+= and this expression is ONLY VALID while the system is boiling. At the saturated liquid and saturated vapor line, y = yf and yg respectively since x = 0 and x = 1 3.2 4.2







respectively. Note that the term yfg means yg – yf.

27. I am able to read the property table in the textbook such as table A-4 and A-5 to determine the saturation pressures, Psat@T, or temperatures, Tsat@P, the saturated liquid properties, yf, the saturated vapor properties, yg and the property, y, (superheated vapor in table A-6 or compressed liquid phase) for any pressure or temperature that I know. In addition, I can determine the phase of the system once I have 2 independent intensive properties (such as T, ν, or T, P) and to mark the phase of the system on a T - ν or P - ν diagram with respect to the saturation lines (the dome) that I have sketched. 3.5 4.4

28. I am able to determine the phase of a system if I know the state of the system. For example if the temperature of refrigerant-134a in a tank is 0°C, and the quality of the system is 0.2000, then the phase is wet-mix since the quality is 0 < x < 1. Then the system’s pressure must be the saturation (boiling) pressure since the system is in a wet-mix phase (boiling). The other properties are determined by using the expression y = yf + xyfg. The volume, ν = νf+xνfg = 0.014552 m3/kg (use values from table A-11), will be used as one of the property to determine the final state of the system since a rigid tank means no change in volume (and hence the specific volume since the mass remains constant for closed systems). Thus if the system’s temperature is increased to 20°C, then the phase for the final state (T = 20°C, ν = 0.014552 m3/kg) is determined by comparing ν = 0.014552 m3/kg to the saturated liquid volume, νf , and the saturated vapor volume, νg, at T = 20°C. Since νf < ν < νg, at T = 20°C (read Table A-11), then the final state is still wet-mix phase. Hence the pressure is the saturation (boiling) pressure at 20°C (P=Psat@20 C° ). Before other properties such as u and h can be determined using the expression y = yf + xyfg, the quality for this final phase must be calculated. Note that νf and νg which are used must be the values read at T =

20°C. It can be shown that the quality is 3926.020@

=

−

−=

Cfg

fxo

νν

νν . It can be shown that if the initial state had a

quality of 0.8, then the final state when the temperature is increased to, 20°C will be superheated vapor since ν > νg. Hence use table A-13 to determine u and h. 3.2 4.3

29. I know that when my system is not water and refrigerant-134a but is a gas such as air, nitrogen or carbon dioxide,

I need to determine if these real gases behave like ideal gases. 3.4 4.1 30. I am aware that the properties of ideal gases are related by the ideal gas equation of state. The need to know this

relation is significant since not all property tables such as A-4, A-5 and A-6 can be made available for all known pure substances. 3.8 4.3

31. I can write the ideal gas equation of state as P = RT where R is the ν gas constant in units of kJ/kg.K, and T is the absolute temperature for the gas measured in units of Kelvin. 3.9 4.3

32. I realize that I can also rewrite the ideal gas equation in terms of the total volume, V, and the mass of the gas, 3.5 4.1







m. Since the total volume is V = mν and by replacing the specific volume, ν = mV , the ideal gas equation is

mPV =

RT or PV = mRT. Note that the units are kilojoule on both sides where 1 kJ = 1 kPa.m3. The expression can also be written in term of the number of kilomoles of the gas. The mass of the gas, m =MN where M is the molar mass in units of kg/kmol and N is the number of kilomoles in units of kmol. Writing the gas constant, R in terms of the

universal gas constant Kkmol

.M

Ruo

3148==kJR , then the ideal gas equation is

MPV u

/=

TNRM/ which is PV =NRuT.

33. I know that in order for real gases to behave like ideal gases, the gas must have low density (low mass to volume ratio). In other words, the pressure must be much lower than the critical pressure (P << Pcr) and the temperature must be higher than the critical temperature (T >> Tcr). If these 2 conditions are met, then the real gases can be treated as ideal gases since its density is low and the gas molecules are very far apart from each other. Hence it obeys the ideal gas equation of state : PV = NRuT or PV = mRT or Pν = RT. Note that T is the absolute temperature measured in Kelvin. The typical values of R, Tcr and Pcr for some gases can be read from Table A-1 in the appendix of Cengel & Yunus Thermodynamics textbook. 3.3 4.0


AS 115/4 DIC, 225/2 - BSAC THERMODYNAMICS CMT 251408 SEMESTER Dec 2002 - Apr 2003

Bil. NAMA Nama Komersil Prog. Group1 Group21 Aidah Abdul jalil Aidah bsac 3 62 Ain Wahidah Badri Ain bsac 3 83 Dalilah Yusoff Dell bsac 4 64 Diana Indim Diana bsac 3 95 Elmy Haliliy Halili Sanggam bsac 4 96 Farhain Misrudin Princess bsac 5 77 Fatimah Md Zaki Fatimah bsac 1 98 Fatimah Sarah Mohd Hassan Ati bsac 3 99 Grace Lawrence Grace bsac 2 6

10 Haryanty Abd Karim Yanty bsac 5 911 Julieyana Hussain Julie bsac 3 812 Liza Salleh Liza bsac 4 713 Masdandy Salamat Dandy bsac 1 714 Mohd Khairul Ngah Yoe bsac 4 815 Mohd Zarizi Abd Ghani Zie bsac 2 816 Nizla Banu Balladin Nizla bsac 2 717 Noor Hafizoh Saidan Noor bsac 5 718 Noor Haida Mohd Kaus Haida bsac 1 719 Noor Wahida Mahasim Wahid bsac 2 820 Nor Ashikin Mustapha Shikin bsac 2 721 Nor Asmah Hashim Asmah bsac 5 822 Nor Hanisah Mohd Zaki Nisah bsac 2 823 Nor Hasimah Md Sudin Shimah bsac 1 824 Nor Rizan Osman @ Mohd Nor Rizan bsac 5 625 Norazatul Akmal Zakarian Azah bsac 4 926 Nurul Izza Husin Nurul bsac 4 627 Nuwaira Muahamadun Amin Wayer bsac 1 628 Ramlah Mohamad Ramlah bsac 1 629 Rasidah Bakar Shidah bsac 2 930 Razdiah Yurad Razdiah bsac 5 631 Salina Ni Lin bsac 4 732 Solhan Yahya An bsac 3 733 Abdul Karim Mohtey Karim dic4 5 834 Ahmad Firdaus Ali Firdaus dic4 3 835 Ahmad Shahir b Abd Muri Shahir dic4 5 936 Aimi Hana Binti Rosli Mie dic4 3 637 Farah Diyana Binti Rusli Farah dic4 4 738 Hilmie Zaifruez Bin Bakhari Hilmie dic4 4 939 Jamil Bin Mohamed Sapari Jimmy dic4 4 740 Jayson Anak Lindong Jayson dic4 2 941 Johari Bin Abu Kasim Jo dic4 5 642 Julaina Binti Baistaman Ju dic4 3 743 Kamaru Elamir Bin Ahmad Ameq dic4 1 844 Khairil Anuar Bin Jantan Nuar dic4 5 845 Mohamad Kamal Bin Rohman Rock dic4 2 7

NAME LIST FOR PEER FACILITATING. THERMODYNAMICS CMT 251/408. Dec 2002 - Apr

2003.

Please check course outline for dates that each group will be facilitators.

Sort by Name

LECTURER: DR. JJ 2/18/2003 6:10 PM


46 Mohamad Yusri Bin Mohamad Yusof Yuz dic4 5 747 Mohd Azli Bin Yusof Lik dic4 4 648 Mohd Dede Bin Jamaludin Borzack dic4 2 949 Mohd Faizal Bin Mohd Zaidin Pejal dic4 4 850 Mohd Farouk Afizal Bin Ismail Farouk dic4 4 651 Mohd Hadzrul Harith Bin Ibrahim Erol dic4 2 752 Mohd Rohaizad Bin Zakaria Ijat dic4 2 653 Muhammad Fadly Bin Ahmad Usul Pali dic4 2 854 Noor Hafidzah Binti Jamal Alang dic4 1 655 Nor Rahafza Binti Abdul Manap Angah dic4 2 756 Norfarhaniza Binti Nordin Farhan dic4 1 957 Norhafizah Binti Abdul Halim Pizah dic4 3 658 Norley Ermayani Binti Mokhtar Elly dic4 5 959 Nurbaizura Edayu Binti Mohd Annuar Ayu dic4 1 660 Nurfaradila Binti Haris Dila dic4 1 961 Roseliza Binti Ghazali Liza dic4 1 762 Rossuriati Binti Dol Hamid Ros dic4 5 863 Sally Shirlyna Binti Mejeni Sally dic4 1 864 Siti Asiah Binti Md Nawawi Echah dic4 4 965 Siti Khairunnisa Binti Zaki Nisa dic4 3 666 Siti Sharah Binti Mohd Nor Sha dic4 3 967 Wan Noor Syamimi Binti Md Noor Mimi dic4 5 968 Zurain Binti Kadir Joy dic4 3 8

14 17

Group1 Group25 7

Bil. NAMA Nama Komersil Prog. Group1 Group21 Nuwaira Muahamadun Amin Wayer bsac 1 62 Noor Haida Mohd Kaus Haida bsac 1 73 Fatimah Md Zaki Fatimah bsac 1 94 Ramlah Mohamad Ramlah bsac 1 65 Masdandy Salamat Dandy bsac 1 76 Nor Hasimah Md Sudin Shimah bsac 1 87 Sally Shirlyna Binti Mejeni Sally dic4 1 88 Kamaru Elamir Bin Ahmad Ameq dic4 1 89 Roseliza Binti Ghazali Liza dic4 1 7

10 Nurbaizura Edayu Binti Mohd Annuar Ayu dic4 1 611 Norfarhaniza Binti Nordin Farhan dic4 1 912 Noor Hafidzah Binti Jamal Alang dic4 1 613 Nurfaradila Binti Haris Dila dic4 1 914 Nizla Banu Balladin Nizla bsac 2 715 Noor Wahida Mahasim Wahid bsac 2 816 Mohd Zarizi Abd Ghani Zie bsac 2 817 Rasidah Bakar Shidah bsac 2 918 Nor Ashikin Mustapha Shikin bsac 2 719 Grace Lawrence Grace bsac 2 6


Sort by Group for First Half of the Semester



20 Nor Hanisah Mohd Zaki Nisah bsac 2 822 Muhammad Fadly Bin Ahmad Usul Pali dic4 2 823 Mohd Rohaizad Bin Zakaria Ijat dic4 2 624 Mohd Dede Bin Jamaludin Borzack dic4 2 925 Nor Rahafza Binti Abdul Manap Angah dic4 2 726 Jayson Anak Lindong Jayson dic4 2 927 Mohd Hadzrul Harith Bin Ibrahim Erol dic4 2 721 Mohamad Kamal Bin Rohman Rock dic4 2 728 Ain Wahidah Badri Ain bsac 3 829 Diana Indim Diana bsac 3 930 Aidah Abdul jalil Aidah bsac 3 631 Solhan Yahya An bsac 3 732 Julieyana Hussain Julie bsac 3 833 Fatimah Sarah Mohd Hassan Ati bsac 3 934 Julaina Binti Baistaman Ju dic4 3 735 Siti Sharah Binti Mohd Nor Sha dic4 3 936 Zurain Binti Kadir Joy dic4 3 837 Aimi Hana Binti Rosli Mie dic4 3 638 Siti Khairunnisa Binti Zaki Nisa dic4 3 639 Norhafizah Binti Abdul Halim Pizah dic4 3 640 Ahmad Firdaus Ali Firdaus dic4 3 841 Elmy Haliliy Halili Sanggam bsac 4 942 Nurul Izza Husin Nurul bsac 4 643 Liza Salleh Liza bsac 4 744 Mohd Khairul Ngah Yoe bsac 4 845 Norazatul Akmal Zakarian Azah bsac 4 946 Dalilah Yusoff Dell bsac 4 647 Salina Ni Lin bsac 4 748 Mohd Farouk Afizal Bin Ismail Farouk dic4 4 649 Siti Asiah Binti Md Nawawi Echah dic4 4 950 Farah Diyana Binti Rusli Farah dic4 4 751 Mohd Faizal Bin Mohd Zaidin Pejal dic4 4 852 Jamil Bin Mohamed Sapari Jimmy dic4 4 753 Hilmie Zaifruez Bin Bakhari Hilmie dic4 4 954 Mohd Azli Bin Yusof Lik dic4 4 655 Nor Rizan Osman @ Mohd Nor Rizan bsac 5 656 Farhain Misrudin Princess bsac 5 757 Haryanty Abd Karim Yanty bsac 5 958 Nor Asmah Hashim Asmah bsac 5 859 Razdiah Yurad Razdiah bsac 5 660 Noor Hafizoh Saidan Noor bsac 5 761 Rossuriati Binti Dol Hamid Ros dic4 5 862 Wan Noor Syamimi Binti Md Noor Mimi dic4 5 963 Abdul Karim Mohtey Karim dic4 5 864 Khairil Anuar Bin Jantan Nuar dic4 5 865 Johari Bin Abu Kasim Jo dic4 5 666 Mohamad Yusri Bin Mohamad Yusof Yuz dic4 5 767 Ahmad Shahir b Abd Muri Shahir dic4 5 968 Norley Ermayani Binti Mokhtar Elly dic4 5 9


AS 115/4 DIC THERMODYNAMICS CMT 251 SEMESTER June - Oct 2003

Bil. KP ITM NAMA Nama Komersil Prog. Group1 Group21 2001131685 Ahmad Lutfi b. Md. Yusoff Lutfi 3 92 2001131900 Ainul Wahida bt Kandak Ainul 4 93 2001357940 Ariffah Adilah bt Dzulkhefli Ifah 4 64 2000409496 Azlan Abu Bakar Lan 3 65 2000635809 Badrul Hisham b. Abdullah Badrul 4 96 2001132157 Bibi Normi Md Noor Bibi 3 87 2001176617 Faranurhazrin Zainal Fara 5 88 2001360016 Hairul Adini b. Idros Adini 5 99 2001357958 Harith Fitry B. Mohd Azali Muke 1 7

10 2001131294 Hazlelina binte Mohd Tahir Lin 2 911 2001357972 Khairienna Ahtar bt Khairuddin Rina 1 712 2001357997 Masitah bt Abdul Hadi Mas 5 613 2000409583 Mohd Faisal Othman Faisal 3 714 2001426486 Mohd Hainur Hapis b. Moh Esa Apis 3 815 2001131303 Mohd Helmi b. Othman Helmi 2 716 2001357965 Mohd Izwan b. Abdullah Wan 2 817 2001131932 Mohd Razif b. Aziz Razif 4 718 2000409688 Mohd Suhaimi bin Chik Mie 1 819 2001357933 Nor Hamizah bt Abd. Majid Hamizah 2 720 2001358014 Noraida bt Md Yusof Aida 1 621 2001176512 Norhafizah bt Abd. Razak Fiza 2 822 2001426639 Norzalilawaty Che Mod Nor 1 623 2001132196 Nur Huwaina binti Yunus Ina 5 624 2001343201 Nurhaizum Aimi bt Adnan Ajom 2 625 2001132205 Ramona bt Abdul Razak Mon 1 926 2000118724 Saiful Hizam b. Mohamed Shaer Zam 5 727 99004615 Subainah Lahimin Nana 4 8

NAME LIST FOR PEER FACILITATING. THERMODYNAMICS CMT 251. June - Oct. 2003.


Sort by Name


grades

AS 15/4, AS 25/3 Grades Distribution For Thermodynamics CMT 251/408 Semesters Jun99 - Jun00

Submitted by Dr. JJ 2/3/2003 2:11 PM

Grades for Thermodynamics CMT 251/408 forAS115/4 (DIC) / AS225/3 (BSAC) Jun00 - Jun99 (3 semesters)

0.0%

5.0%

10.0%

15.0%

20.0%

25.0%

30.0%

A+ A A- B+ B B- C+ C C- D+ D E F

Perc

enta

ge

Jun-00 (N=46)

Jan-00 (N=46)

Jun-99 (N=41)

A+ = 90 - 100 A = 80 - 89 A- = 75 -79B+ = 70 - 74 B = 65 - 69 B- = 60 - 64

C+ = 55 - 59 C = 50 - 54 C- = 47 - 49D+ = 44 - 46 D = 40 - 43 E = 30 - 39 F = < 29

AS 15/4, AS 25/3 Grades Distribution For Thermodynamics CMT 251/408 Semesters Jun00 - Nov01


Grades for Thermodynamics CMT 251/408 forAS115/4 (DIC) / AS225/3 (BSAC) Nov01 - Jun00 (4 semesters)

0%

5%

10%

15%

20%

25%

30%

35%


Perc

enta

ge

Jan01 (N=58)

Jun-00 (N=46)

Jun01 (N=68)

Nov-01 (N=75)

A+ = 90 - 100 A = 80 - 89 A- = 75 -79B+ = 70 - 74 B = 65 - 69 B- = 60 - 64

C+ = 55 - 59 C = 50 - 54 C- = 47 - 49D+ = 44 - 46 D = 40 - 43 E = 30 - 39 F = < 29

AS 15/4, AS 25/3 Grades Distribution For Thermodynamics CMT 251/408 Semesters Jun99 - Jun00


Grades for Thermodynamics CMT 251/408 forAS115/4 (DIC) / AS225/3 (BSAC) May02 - Jan01 (4 semesters)

0.0%

5.0%

10.0%

15.0%

20.0%

25.0%

30.0%

35.0%

40.0%


Perc

enta

ge

Jan-01 (N=58)

May-01 (N=68)

Nov-01 (N=75)

May-02 (N=58)

A+ = 90 - 100 A = 80 - 89 A- = 75 -79B+ = 70 - 74 B = 65 - 69 B- = 60 - 64

C+ = 55 - 59 C = 50 - 54 C- = 47 - 49D+ = 44 - 46 D = 40 - 43 E = 30 - 39 F = < 29

students ratings

Thermodynamics - CMT 251 (DIC) CMT 408 (BSAC) Evaluation for the Course Semester Dec 98 - May 99

Self Evaluation: Semester Dec 99 - May 2000

Course: Thermodynamics. Groups: CMT 251 (DIC 4) & CMT 408 (BSAC 00). N=41

Please use the following codes: 1-Srongly Disagree:: 5-Strongly AgreeMean Median Std. Dev. %mean min max

Thermodynamics is a very difficult class. 3.34 3.00 0.62 66.8% 2.72 3.96The concepts in thermodynamics are too complicated to learn. 3.37 3.00 0.80 67.3% 2.57 4.16The mathematical expression in thermodynamics are too complicated to learn. 3.37 4.00 0.80 67.3% 2.57 4.16The terminologies in thermodynamics are too difficult to learn. 3.34 3.00 0.69 66.8% 2.65 4.03The language used in the text is too difficult to comprehend. 3.17 3.00 0.95 63.4% 2.22 4.12The five-hour per week is not enough for good comprehension of the subject matter. 3.38 3.00 1.00 67.5% 2.37 4.38

The structure (format) for this class had force me to work harder than I would normally do. 3.68 4.00 0.76 73.7% 2.93 4.44The structure (format) for this class has helped me to focus my learning better. 3.68 4.00 0.72 73.7% 2.96 4.41I always do my reading and submit my reading assignment on time. 3.51 4.00 0.81 70.2% 2.70 4.32I always do the problem assignment and submit them on time. 2.71 3.00 0.78 54.1% 1.92 3.49I always do my preparation before my consultancy. 3.07 3.00 0.75 61.5% 2.32 3.83I always ask questions during consultancy, during peer-lecturering and during lecture hour. 3.05 3.00 0.89 61.0% 2.16 3.94I always discuss with my peers outside the classroom regarding the subject matter. 2.98 3.00 1.04 59.5% 1.94 4.01The reading assignment is a very good exercise for me in preparing myself for class. 3.78 4.00 0.85 75.6% 2.93 4.63The consultancy, peer-discussion is a very effective method in preparing myself before class. 3.76 4.00 0.83 75.1% 2.93 4.59The lectures after peer discussion helped made the learning easier. 3.95 4.00 0.77 79.0% 3.18 4.72I have worked very hard in this class to understand the materials but I still cannot understand. 3.34 3.00 0.82 66.8% 2.52 4.17

The learning format for this class is very good and should be employed in many science classses. 3.93 4.00 0.79 78.5% 3.14 4.71I feel that the class should only be held for three hours only and not five hours. 2.54 2.00 1.07 50.7% 1.46 3.61I feel that the consultancy, peer-discussion learning cycle format is a very efficient method. 3.68 4.00 0.69 73.7% 3.00 4.37I feel that the traditional lecturer talk in front and student listen is the best way of learning. 2.80 3.00 1.05 56.1% 1.75 3.86

The lecturer always arrived on time. 4.29 4.00 0.78 85.9% 3.51 5.08The lecturer is always very motivated and enthusiastic. 4.61 5.00 0.49 92.2% 4.12 5.10The lecturer is always very patient in answering our questions. 4.59 5.00 0.55 91.7% 4.04 5.13The lecturer always explained the materials clearly during consultancy and lecture. 4.54 5.00 0.67 90.7% 3.86 5.21The lecturer presents his lectures in very interesting manner and always joke around in between. 4.56 5.00 0.55 91.2% 4.01 5.11

Lecturer: Dr. JJ 20/05/2011 6:57 PM

Thermodynamics - CMT 251 (DIC) CMT 408 (BSAC) Evaluation for the Course Semester Dec 98 - May 99

Self Evaluation: Semester Dec 99 - May 2000

Course: Thermodynamics. Groups: CMT 251 (DIC 4) & CMT 408 (BSAC 00). N=41

Please use the following codes: 1-Srongly Disagree:: 5-Strongly AgreeMean Median Std. Dev. %mean min max

The lecturer is very confident and has a good mastery of the topic discussed. 4.78 5.00 0.42 95.6% 4.36 5.20The lecturer is very organized with the flow of the content presented. 4.76 5.00 0.43 95.1% 4.32 5.19The lecturer always do examples to help made the material more comprehensible. 4.49 5.00 0.60 89.8% 3.89 5.08The lecturer always test my understanding during lecture by asking us fundamental questions. 4.34 4.00 0.76 86.8% 3.58 5.10The lecturer encouraged me to be more open in asking questions and motivate me. 4.39 4.00 0.54 87.8% 3.85 4.93The lecturer has succeeded in imparting valuable knowledge about thermodynamics. 4.54 5.00 0.50 90.7% 4.03 5.04I look forward to be lectured by this lecturer again in the future. 4.68 5.00 0.47 93.7% 4.21 5.15

Please write your comments about this course, the learning format and lecturer.

All comments will be kept confidential.

Lecturer: Dr. JJ 20/05/2011 6:57 PM

Thermodynamics - CMT 251 (DIC) CMT 408 (BSAC)

End of Semester Course Evaluation

Semester Nov 00 - April 01

Thermodynamics. CMT 251 (DIC 4) & CMT 408 (BSAC). Nov00 - Apr01 Code: 1-Strongly Disagree: 5-Strongly Agree

Please use the following codes: 1-Strongly Disagree: 5-Strongly Agree N=56 N=41Jan-01 Jun-00 Dis-99 Jun-99 Dis-98

MY INITIAL KSA & SELF EFFORT1 My conceptual understanding in physics is weak . 3 3 na na na

2 I memorised most of the equations in physics and not really understanding them. 3 3 na na na

3 I found thermodynamics to be a very difficult course. 3 3 3 3 3

4 The concepts in thermodynamics are too difficult to learn. 3 3 3 4 4

5 There are too many mathematical expressions in thermodynamics for me to memorise. 3 3 na na na

6 The are too many terminologies in thermodynamics for me to comprehend. 3 3 na na na

7 The textbook helped me learn thermodynamics better than if there is no textbook. 4 3 na na na

8 The problems at the end of each chapter are very useful to increase comprehension. 4 4 na na na

9 The five-hour per week is more than enough for good comprehension of the subject matter. 4 3 3 3 3

MEDIAN 3 3 3 3 3

TLAs & SELF EFFORT

10 The structure (format) for this class had force me to work harder than I would normally do. 4 4 4 4 4

11 The structure (format) for this class has helped me to focus my learning better. 4 4 4 4 4

12 I always do my reading and submit my reading assignment on time. 4 5 4 3 3

13 I always do the problem assignment and submit them on time. 3 4 3 2 2

14 I always do my preparation before my consultation with the lecturer. 3 3 3 3 3

15 I always ask questions during consultancy, during peer-lecturing and during lecture hour. 3 3 3 3 3

16 I always discuss with my peers outside the classroom regarding the topic of the week. 3 3 3 3 3

17 The reading assignment is a very good exercise for me in preparing myself for class. 4 4 4 3 3

18 The consultancy, peer-discussion is a very effective method in preparing myself before class. 4 4 4 4 4

19 The lectures after peer discussion helped made the learning easier. 5 4 4 4 4

20 I have worked very hard in this class to understand the materials but I still cannot understand. 3 3 3 4 4

MEDIAN 4 4 4 3 3

LEARNING FORMAT

21 The learning format for this class should be employed in other science classses. 5 5 4 3 3

22 The class should be held for only three hours and not five hours. 1 3 2 2 2

23 I feel that the consultancy, peer-discussion learning cycle format is a waste of time. 1 1 4 4 4

24 I feel that the traditional lecturer talk in front and student listen is the best way of learning. 1 3 3 2 2

MEDIAN 1 3 4 3 3

DR JJ's EFFORT

25 The lecturer always arrived on time. 4 4 4 4 4

26 The lecturer is always very motivated and enthusiastic. 5 5 5 4 4

27 The lecturer is always very patient in answering our questions. 5 5 5 5 5

28 The lecturer always explained the materials clearly during consultancy and lecture. 5 5 5 5 5

29 The lecturer presents his lectures in very interesting manner and always joke around in between. 5 5 5 5 5

30 The lecturer is very confident and has a good mastery of the topic discussed. 5 5 5 5 5

31 The lecturer is very organized with the flow of the content presented. 5 5 5 4 4

32 The lecturer always do examples to help make the material more comprehensible. 5 5 5 4 4

33 The lecturer always test my understanding during lecture by asking fundamental questions. 5 4 5 4 4

34 The lecturer encouraged me to be more open in asking questions and motivate me. 5 5 4 4 4

MODE

Lecturer: Dr. JJ 21/05/2011 3:38 AM



Semester Nov 00 - April 01

Thermodynamics. CMT 251 (DIC 4) & CMT 408 (BSAC). Nov00 - Apr01 Code: 1-Strongly Disagree: 5-Strongly Agree

Please use the following codes: 1-Strongly Disagree: 5-Strongly Agree N=56 N=41Jan-01 Jun-00 Dis-99 Jun-99 Dis-98

MODE

35 The lecturer has succeeded in imparting valuable knowledge about thermodynamics. 5 5 5 4 4

36 I look forward to be lectured by this lecturer again in the future. 5 5 5 4 4

MEDIAN 5 5 5 4 4

Please write your comments about this course, the learning format and lecturer.

Make your suggestions on how learning of this course can be improved in the future.

All comments will be kept confidential. # of responses

The course is very interesting 4

Want to take the course again for fun 'cos its interesting 1

FC should be more well-versed in the subject matter 1

Thought the course was hard but turned out the opposite 1

Enjoy learning 'cos the lecturer is good 1

Thermo is subject that is easy to understand 1

Learning method used is very effective 2

The subject stresses more on conceptual understanding & not on formula memorisation 1

Learning method used is very effective & attracts students to like the subject & learn it well 1

A hard course 4

OK lecturer 4

OK but sometimes stressful format 1

Lecturer need to have more patience 1

Wish that the learning format be empolyed in future classes 3

Dr JJ is just great 1

Very happy to have taken the class 1

Not confident with the FC, so cannot optimize learning 1

Format is excellent 1


Thermodynamics CMT 251/408 AS15/4 & AS25/2 End of Semester Course- Evaluation

Page 1 of 4

Copyright © Dr. J.J. FSG, UiTM Lecturer: Dr. JJ, Course: Thermodynamics 5/21/2011 1:14 AM

Subject #: ______ Group:DIC / BSAC

Please use the following codes: 1 - Strongly Disagree (Sangat tidak setuju): 2. Disagree (Tidak Setuju) 3. Agree (Setuju) 4 - Strongly Agree (Sangat bersetuju) Please write your response (1 or 2 or 3 or 4) in the square to the right of the statement.

0. Example: I have trouble sleeping at night 4

1. My conceptual understanding in physics is weak. [Pemahaman konsep fizik saya adalah lemah]

2. I memorized most of the equations in physics and not really understanding them. [Saya hafal kebanyakan formula dalam fizik dan sebenarnya tidak memahami maksudnya.]

3. I found thermodynamics to be a very difficult course. [Saya merasakan yang subjek termodinamik adalah sangat sukar.]

4. The concepts in thermodynamics are too difficult to learn. [Konsep-konsep dalam termodinamik amat sukar untuk dipelajari.]

5. There are too many mathematical expressions in thermodynamics for me to memorise. [Terlalu banyak ungkapan matematik yang perlu dihafal dalam termodinamik]

6. There are too many terminologies in thermodynamics for me to comprehend. [Terlalu banyak terminologi yang perlu dihayati dalam termodinamik.]

7. The course outline/syllabi which was distributed at the beginning of the semester had helped me focus my learning and structure my time appropriately and should be employed in all classes. [Silibus serta garispanduan kursus yang diedarkan pada awal semester telah membantu saya untuk memfokus pembelajaran saya dan membantu saya mengatur masa saya dan saya rasa perlu diamalkan untuk semua kelas.]

8. My command of English comprehension is not good and I found difficulties understanding the textbook, lecture and the questions in tests and quizzes. [Pemahaman Bahasa Inggeris saya adalah lemah dan saya menghadapi masalah memahami buku teks, kuliah dan soalan-soalan dalam ujian dan quiz.]

9. The textbook helped me learn thermodynamics better than if there is no textbook. [Adanya buku teks dalam kursus ini memudahkan pembelajaran saya berbanding dengan jika tiada buku teks.]

10. The problems at the end of each chapter are very useful to increase comprehension. [Soalan-soalan pada penghujung setiap tajuk amat berguna untuk meningkatkan pemahaman.]

11. The structure (format) for this class had force me to work harder than I would normally do. [Kaedah pembelajaran untuk kursus ini telah memaksa saya berusaha lebih banyak daripada biasa.]

12. The structure (format) for this class has helped me to focus my learning better. [Kaedah pembelajaran kursus ini telah membantu saya memfokus pembelajaran saya dengan lebih baik.]

13. I always do my reading and submit my reading assignment on time. [Saya sentiasa membaca dan menghantar hasil pembacaan saya pada waktu yang telah ditetapkan.]

14. I always tryout the problem assignment (at the end of the chapter) and ask questions when I found difficulties.


Page 2 of 4




[Saya sentiasa mencuba soalan-soalan pada penghujung tajuk dan bertanya soalan jika menghadapi masalah.]

15. I always read the textbook before my consultation with the lecturer on Fridays and Sundays. Saya sentiasa membaca buku teks sebelum berjumpa dengan pensyarah untuk sesi perbincangan pada hari Jumaat & Ahad.

16. Before and after consultation periods with the lecturer, I always read and assess my understanding by using the self-assessment provided. [Saya sentiasa menilai tahap kefahaman saya dengan menggunakan penilaian kendiri sebelum dan selepas sesi perbincangan dengan pensyarah.]

17. When I am not a facilitator, I always assess my understanding by using the self-assessment provided, before and after peer discussion [Saya sentiasa menilai kefahaman saya menggunakan penilaian kendiri sebelum dan selepas perbincangan rakan sekursus.]

18. Sending my self-assessment to the lecturer by using email and file attachment is useful in making me learn the use of ICT. [Menghantar penilaian kendiri saya melalui internet dan kepilan fail adalah cara yang baik untuk membantu saya mempelajari penggunaan ICT.]

19. The self-assessment materials developed by the lecturer is an important tool and must be developed in other courses to help students identify their strength and weaknesses. [Bahan-bahan penilaian kendiri yang dihasilkan oleh pensyarah untuk kursus ini adalah alat pembelajaran yang penting dan perlu juga dihasilkan dalam kursus-kursus lain bagi membantu pelajar-pelajar mengenalpasti kelemahan dan kekuatan mereka.]

20. Obtaining information about this course is very easy because the course outline, the due dates, events dates and all the necessary documents are available on the lecturer’s website. [Mendapatkan maklumat mengenai kursus ini amatlah mudah kerana panduan perjalanan kursus termasuklah tarikh penghantaran kerja kursus, tarikh-tarikh penting serta dokumen yang berkaitan dengan kursus ini ada dalam halaman web pensyarah.]

21. I feel that utilizing the world wide web and using email must be utilized in all courses to facilitate learning. [Saya rasa penggunaan jaringan web dan penggunaan email perlu dimenafaatkan dalam semua kursus untuk memudahkan pembelajaran.]

22. I always ask questions during consultancy, during peer-lecturing and during lecture hour.[Saya sentiasa bertanya soalan sewaktu sessi perbincangan dengan pensyarah, sessi perbincangan dengan rakan sekursus dan sewaktu sessi kuliah.]

23. I always discuss with my peers outside the classroom regarding the topic of the week. Saya sentiasa berbincang di luar kelas dengan rakan-rakan sekursus mengenai tajuk untuk sesuatu minggu.

24. The reading assignment summary is a very good exercise for me in preparing myself for class. Ringkasan pembacaan yang perlu dihantar setiap minggu adalah berguna untuk meningkatkan persediaan saya.

25. The consultancy, peer-discussion is a very effective method in preparing myself before class. [Sesi perbincangan dengan pensyarah dan sesi perbincangan dengan rakan sekursus amat berguna untuk menyediakan saya sebelum kuliah.]

26. The lectures after peer discussion helped made the learning easier. [Sesi kuliah selepas sesi perbincangan dengan rakan sekursus telah membuatkan pembelajaran lebih mudah.]

27. I have worked very hard in this class to get good grade. [Saya telah bekerja keras didalam kursus ini untuk mendapatkan gred yang baik.]

28. I have worked very hard to ensure understanding of the subject matter. [Saya telah berusaha dengan gigih untuk memastikan pemahaman isi kandungan subjek.]


Page 3 of 4




29. The learning format for this class should be employed in other science classes. [Kaedah pengajaran & pembelajaran untuk kursus ini perlu digunakan untuk kursus-kursus yang lain.]

30.

The class should be held for only three hours in a week and we can still finish the syllabus. There is no need for Friday and Sunday sessions even though the Fri and Sun sessions are only done twice a semester. [Kursus ini perlu diadakan hanya 3 jam sahaja seminggu dan kita masih boleh memastikan keperluan silibus dipenuhi. Sesi hari Jumaat dan Ahad tidak perlu diadakan walaupun sesi tersebut diadakan hanya 2 kali sahaja dalam satu semester..]

31. I feel that the consultations with the lecturer in order to train facilitators are necessary .[Saya berpendapat bahawa perbincangan dengan pensyarah untuk melatih fasilitator adalah penting]

32 I feel that consultations with the lecturer should only be done on Fridays and not Fridays and Sundays. [Saya merasakan bahawa sesi perbincangan dengan pensyarah patut diadakan pada hari Jumaat sahaja dan bukannya pada hari Jumaat dan Ahad].

33. I feel that the consultancy, peer-discussion learning cycle format is a waste of time. [Saya merasakan bahawa format untuk kursus ini hanya membuang masa sahaja.]

34.

I feel that the traditional transmittalist method (lecturer talk in front and student listen and taking notes) is the best way of learning as compared to the peer-facilitating learning-cycle that is presently used for m this class. [Saya merasakan bahawa kaedah pengajaran biasa dimana pensyarah sentiasa sahaja memberi kuliah sementara pelajar pula asyik mendengar dan menyalin nota adalah kaedah pembelajaran yang paling baik sekali berbanding dengan kaedah “Peer-facilitating Learning-cycle” yang sekarang ini diamalkan untuk kelas ini.]

35. The lecturer always arrived on time.

36. The lecturer is always very motivated, enthusiastic and always willing to sacrifice his time for us.

37. The lecturer had worked very hard in preparing materials for us and helping us learn the subject matter.

38. The lecturer is always very patient in answering our questions during consultation and during lecture hour.

39. The lecturer always explained the materials clearly during consultancy and during lecture.

40. The lecture materials are organized and legible from where I sit in the classroom.

41. The lecturer’s voice can be heard clearly from where I sit in the classroom.

42. The lecturer presents his lectures in very interesting manner and always joke around in between.

43. The lecturer is very confident and has a good mastery of the topic discussed.

44. The lecturer is very organized with the flow of the content presented.

45. The lecturer will always show an example and make us do problems to help make the material more comprehensible.

46. The lecturer will always ask fundamental (basic) questions to test my understanding during consultation and lecture.

47. The lecturer will always encourage me to ask questions and will always motivate me to learn.


Page 4 of 4




48. The lecturer has succeeded in imparting valuable knowledge about thermodynamics.

49. I look forward to take a course taught by this lecturer again in the future.

Please circle the following. 50. How challenging do you find this course? Too easy A bit easy Just right A bit difficult Too difficult

51. How much do you think you have learned in this course compared with other courses you have taken this semester?

Almost nothing Little A fair amount Much A great deal l 52. How well has this course met your expectations? Not at all Not very well Adequately Well Very well

Comment on the course format. 53. Overall, what contributed most significantly to your learning in this course? In other words, what are the important features to retain the next time this

course is taught? 54. Overall, what made your learning in this course more difficult? In other words, what are the most important changes you would suggest for the next time

this course is taught? 55. Any comments about the lecturer?



Semester Dec 02 - April 03

Course Evaluation: Semester Nov 01 - March 02. N=67

Thermodynamics. Groups: CMT 251 (DIC 4) & CMT 408 (BSAC)

Thermodynamics. CMT 251 (DIC 4) & CMT 408 (BSAC). Dec 02 - April 03

Please use the following codes: 1-Strongly Disagree: 2=Diasagree 3=Agree 4-Strongly Agree

Median Mode %mean Nov-01 Jun-01 Jan-01 Jun-00 Dis-99 Jun-99 Dis-98

MY INITIAL KSA & SELF EFFORT1 My conceptual understanding in physics is weak . 3 3 67.5% 3 3 3 na na na

2 I memorised most of the equations in physics and not really understanding them. 3 3 66.0% 3 3 3 na na na

3 I found thermodynamics to be a very difficult course. 2 2 59.0% 3 2 3 3 3 3

4 The concepts in thermodynamics are too difficult to learn. 2 2 59.3% 3 2 3 3 4 4

5 There are too many mathematical expressions in thermodynamics for me to memorise. 3 3 66.4% 3 3 3 na na na

6 The are too many terminologies in thermodynamics for me to comprehend. 3 3 74.6% 4 3 3 na na na

7 The course outline/syllabi which was distributed at the beginning of the semester had helped me focus my

learning and structure my time appropriately and should be employed in all classes. 3 3 78.0% 4 na na na na na

8 My command of English comprehension is not good and I found that I have difficulties understanding the

textbook, lecture and the questions in tests and quizzes. 3 3 68.7% 3 na na na na na

9 The textbook helped me learn thermodynamics better than if there is no textbook. 4 4 86.9% 4 4 3 na na na

10 The problems at the end of each chapter are very useful to increase comprehension. 3 3 81.9% 5 3 4 na na na

MEDIAN 3 3 3 3 3 3 4 4

TLAs & SELF EFFORT 0.0%

11 The structure (format) for this class had forced me to work harder than I would normally do. 3 4 86.2% 4 4 4 4 4 4

12 The structure (format) for this class has helped me to focus my learning better. 3 3 69.8% 4 3 4 4 4 4

13 I always do my reading and submit my reading assignment on time. 3 3 69.4% 4 3 5 4 3 3

14

I always tryout the problem assignment (at the end of the chapter) and ask questions when I found

difficulties. 2 2 60.4% 3 2 4 3 2 2

15 I always read the textbook before my consultation with the lecturer on Fridays and Sundays. 3 3 73.1% 3 3 3 3 3 3

16

Before and after consultation periods with the lecturer, I always read and assess my understanding by

using the self-assessment provided. 3 3 68.7%

17

When I am not a facilitator, I always assess my understanding by using the self-assessment provided,

before and after peer discussion 3 3 67.8%

18

Sending my self-assessment to the lecturer by using email and file attachment is useful in making me

learn the use of ICT. 3 3 80.2%

19

The self-assessment materials developed by the lecturer is an important tool and must be developed in

other courses to help students identify their strength and weaknesses. 3 3 79.5%

20

Obtaining information about this course is very easy because the course outline, the due dates, events

dates and all the necessary documents are available on the lecturer’s website 4 4 87.3%

21 I feel that utilizing the world wide web and using email must be utilized in all courses to facilitate learning. 3 3 71.6%

MEDIAN 3 3 4 3 4 4 3 3

MY ACTIVE ENGAGEMENT EFFORT & MY EXPECTATION

22 I always ask questions during consultancy, during peer-lecturing and during lecture hour. 3 3 63.8% 3 3 3 3 3 3

23 I always discuss with my peers outside the classroom regarding the topic of the week. 2 2 63.1% 3 2 3 3 3 3

24 The reading assignment is a very good exercise for me in preparing myself for class. 3 3 75.0% 4 3 4 4 3 3

25 The consultancy, peer-discussion is a very effective method in preparing myself before class. 3 3 79.1% 4 3 4 4 4 4

26 The lectures after peer discussion helped made the learning easier. 3 3 75.4% 4 3 4 4 4 4

27 I have worked very hard in this class to get good grade 3 3 81.4% 3 3 3 3 4 4

28 I have worked very hard to ensure understanding of the subject matter 3 3 0.0%

MEDIAN 3 3 4 3 4 4 4 4

LEARNING FORMAT

29 The learning format for this class should be employed in other science classses. 3 2 65.7% 3 2 5 4 3 3

30 The class should be held for only three hours and not five hours. 3 3 68.3% 3 3 3 2 2 2

31 I feel that the consultations with the lecturer in order to train facilitators are necessary. 3 0.0%

32 I feel that the consultations with the lecturer should not be done on Friday afternoons and Sundays but the

consultations are necessary and should be done only once on Friday. 3 4 74.2% 3 na na na na na

33 I feel that the consultancy, peer-discussion learning cycle format is a waste of time. 2 2 48.5% 1 2 1 4 4 4

Code: 1-Strongly Disagree: 5-Strongly Agree

Mode




Semester Dec 02 - April 03

Course Evaluation: Semester Nov 01 - March 02. N=67

Thermodynamics. Groups: CMT 251 (DIC 4) & CMT 408 (BSAC)

Thermodynamics. CMT 251 (DIC 4) & CMT 408 (BSAC). Dec 02 - April 03

Please use the following codes: 1-Strongly Disagree: 2=Diasagree 3=Agree 4-Strongly Agree

Median Mode %mean Nov-01 Jun-01 Jan-01 Jun-00 Dis-99 Jun-99 Dis-98

Code: 1-Strongly Disagree: 5-Strongly Agree

Mode

34

I feel that the traditional transmittalist method (lecturer talk in front and student listen and taking notes) is

the best way of learning as compared to the peer-facilitating learning-cycle that is presently used for m this

class. 2 2 57.7% 3 2 3 3 2 2

MEDIAN 3 2 3 2 3 4 3 3

DR JJ's EFFORT

35 The lecturer always arrived on time. 3 3 77.6% 4 3 4 4 4 4

36 The lecturer is always very motivated, enthusiastic and always willing to sacrifice his time for us. 4 4 86.6% 5 4 5 5 4 4

37 The lecturer had worked very hard in preparing materials for us and helping us learn the subject matter. 3 3 85.8% 5 na na na na na

38 The lecturer is always very patient in answering our questions during consultation and during lecture hour. 3 3 78.0% 5 3 5 5 5 5

39 The lecturer always explained the materials clearly during consultancy and lecture. 3 3 75.0% 4 3 5 5 5 5

40 The lecture materials are organized and legible from where I sit in the classroom. 3 3 76.9% 5 na na na na na

41 The lecturer’s voice can be heard clearly from where I sit in the classroom. 4 4 89.9% 5 na na na na na

42 The lecturer presents his lectures in very interesting manner and always joke around in between. 3 3 81.0% 5 3 5 5 5 5

43 The lecturer is very confident and has a good mastery of the topic discussed. 4 4 88.1% 5 4 5 5 5 5

44 The lecturer is very organized with the flow of the content presented. 3 3 84.0% 5 3 5 5 4 4

45 The lecturer always do examples to help make the material more comprehensible. 3 3 71.3% 4 3 5 5 4 4

46 The lecturer always test my understanding during lecture by asking fundamental questions. 3 3 82.5% 5 3 4 5 4 4

47 The lecturer encouraged me to be more open in asking questions and motivate me. 3 3 81.3% 5 3 5 4 4 4

48 The lecturer has succeeded in imparting valuable knowledge about thermodynamics. 3 4 84.7% 4 4 5 5 4 4

49 I look forward to be lectured by this lecturer again in the future. 3 3 74.6% 4 3 5 5 4 4

MEDIAN 3 3 5 3 5 5 4 4

OVERALL CHALLENGES AND PERCEIVED ATTAINMENT

50

How challenging do you find this course? 1=Too easy 2= A bit easy 3=Just right 4=A bit difficult 5=Too

difficult 4 4 76.4% 4 na na na na na

51

How much do you think you have learned in this course compared with other courses you have taken this

year?

1=Almost nothing 2=Little 3=A fair amount 4=Much 5=A great deal 4 4 72.1% 4 na na na na na

52

How well has this course met your expectations? 1=Not at all 2=Not very well 3=Adequately Well

4=Very well 3 3 78.0% 3 na na na na na



Purpose:

Students evaluate their effort, the learning and teaching

methods and the lecturer’s performance. The analysis can be

used for assessing a lecturer’ s effort but most importantly

for self-improvement in terms of teaching styles and teaching

techniques for producing the most effective learning.

Method:

Students respond to statements in the absence of the

lecturer. Responses are anonymous. Responses are keyed in

and analyzed using both MS Excel and SPSS. For statements

1 through 49, the keys used are:

1=Strongly disagree 2=Disagree

3=Agree 4=Strongly agree

Semester Dec 2002 – April 2003

Course: Thermodynamics CMT 408

Program AS225/2

Lecturer: Assoc. Prof. Dr. Jaafar Jantan

Webpage:www.uitm.edu.my/faculties/fsg/drjj1.html. Voice: 019-355-1621 or 03-5544-4593 (Office) Page 1 of 19

Thermodynamics: Key: 1=S. Disagree, 2=Disagree, 3=Agree, 4=S. Agree,

End of Semester Course Evaluation-BSAC

Semester: Dec 02 – April 03 Lecturer: Dr. J.J., FSG, UiTM.

Email:[email protected] Q1A: My conceptual understanding in physics is weak.

Frequency Percent 2 13 40.6 3 14 43.8 4 5 15.6

Total 32 100.0

Q1_bsac

Q1

432

Freq

uenc

y

16

14

12

10

8

6

4

2

0

Q2A: I memorized most of the equations in physics and not really understanding them.

Frequency Percent 1 1 3.1 2 14 43.8 3 15 46.9 4 2 6.3

Total 32 100.0

Q2_bsac

Q2

4321

Freq

uenc

y

16

14

12

10

8

6

4

2

0

Q3A: I found thermodynamics to be a very difficult course.


Total 32 100.0

Q3_bsac

Q3

4321

Freq

uenc

y

30

20

10

0





Email:[email protected] Q4A: The concepts in thermodynamics are too difficult to learn.


Total 32 100.0

Q4_bsac

Q4

321

Freq

uenc

y

20

10

0

Q5A: There are too many mathematical expressions in thermodynamics for me to memorize.


Total 32 100.0

Q5_bsac

Q5

4321

Freq

uenc

y

16

14

12

10

8

6

4

2

0

Q6A: There are too many terminologies in thermodynamics for me to comprehend.


Total 31 100.0

Q6_bsac

Q6

4321

Freq

uenc

y

16

14

12

10

8

6

4

2

0





Email:[email protected] Q7A: The course outline which was distributed at the beginning of the semester had helped me focus my learning and structure my time appropriately and should be employed in all classes.


Total 32 100.0

Q7_bsac

Q7

432

Freq

uenc

y

20

10

0

Q8A: My command of English comprehension is not good and I found difficulties in understanding the textbook, lecture and the questions in tests and quizzes.


Total 32 100.0

Q8_bsac

Q8

4321

Freq

uenc

y

14

12

10

8

6

4

2

0

Q9A: The textbook helped me learn thermodynamics better than if there is no textbook.


Total 32 100.0

Q9_bsac

Q9

432

Freq

uenc

y

20

10

0





Email:[email protected] Q10A: The problems at the end of each chapter are very useful to increase comprehension.


Total 32 100.0

Q10_bsac

Q10

432

Freq

uenc

y

16

14

12

10

8

6

4

2

0

Q11A: The structure (format) for this class had force me to work harder than I would normally do.


Total 32 100.0

Q11_bsac

Q11

432

Freq

uenc

y

20

10

0

Q12A: The structure (format) for this class has helped me to focus my learning better.


Total 32 100.0

Q12_bsac

Q12

432

Freq

uenc

y

20

10

0





Email:[email protected] Q13A: I always do my reading and submit my reading assignment on time.


Total 32 100.0

Q13_bsac

Q13

432

Freq

uenc

y

20

10

0

Q14A: I always tryout the problem assignment (at the end of the chapter) and submit them on time.


Total 32 100.0

Q14_bsac

Q14

4321

Freq

uenc

y

20

10

0

Q15A: I always do my preparation before my consultation with the lecturer on Fridays and Sundays.


Total 32 100.0

Q15_bsac

Q15

5432

Freq

uenc

y

16

14

12

10

8

6

4

2

0





Email:[email protected] Q16A: Before and after consultation periods with the lecturer, I always read and assess my understanding by using the self-assessment provided.


Total 32 100.0

Q16_bsac

Q16

4321

Freq

uenc

y

16

14

12

10

8

6

4

2

0

Q17A: When I am not a facilitator, I always assess my understanding by using the self-assessment provided, before and after peer discussion.


Total 32 100.0

Q17_bsac

Q17

4321

Freq

uenc

y

30

20

10

0

Q18A: Sending my self-assessment to the lecturer by using email and file attachment is useful in making me learn the use of ICT.


Total 32 100.0

Q18_bsac

Q18

432

Freq

uenc

y

20

10

0





Email:[email protected] Q19A: The self-assessment materials developed by the lecturer is an important tool and must be developed in other courses to help students identify their strength and weaknesses.


Total 32 100.0

Q19_bsac

Q19

4321

Freq

uenc

y

16

14

12

10

8

6

4

2

0

Q20A: Obtaining information about this course is very easy because the course outline, the due dates, events dates and all the necessary documents are available on the lecturer’s website.


Total 32 100.0

Q20_bsac

Q20

432

Freq

uenc

y

20

10

0

Q21A: I feel that utilizing the world-wide-web and using email must be utilized in all courses to facilitate learning.


Total 32 100.0

Q21_bsac

Q21

432

Freq

uenc

y

20

10

0





Email:[email protected] Q22A: I always ask questions during consultancy, during peer-lecturing and during lecture hour.


Total 32 100.0

Q22_bsac

Q22

4321

Freq

uenc

y

16

14

12

10

8

6

4

2

0

Q23A: I always discuss with my peers outside the classroom regarding the topic of the week.


Total 32 100.0

Q23_bsac

Q23

4321

Freq

uenc

y

20

10

0

Q24A: The reading assignment is a very good exercise for me in preparing myself for class.


Total 32 100.0

Q24_bsac

Q24

432

Freq

uenc

y

30

20

10

0





Email:[email protected] Q25A: The consultancy, peer-discussion is a very effective method in preparing myself before class.


Total 32 100.0

Q25_bsac

Q25

432

Freq

uenc

y

30

20

10

0

Q26A: The lectures after peer discussion helped made the learning easier.


Total 32 100.0

Q26_bsac

Q26

4321

Freq

uenc

y

30

20

10

0

Q27A: I have worked very hard in this class to get good grade.

Frequency Percent 3 18 56.3 4 14 43.8

Total 32 100.0

Q27_bsac

Q27

43

Freq

uenc

y

20

10

0





Email:[email protected] Q28A: I have worked very hard to ensure understanding of the subject matter.


Total 32 100.0

Q28_bsac

Q28

432

Freq

uenc

y

20

10

0

Q29A: The learning format for this class should be employed in other science classes.


Total 32 100.0

Q29_bsac

Q29

4321

Freq

uenc

y

14

12

10

8

6

4

2

0

Q30A: The class should be held for only three hours and not five hours.


Total 32 100.0

Q30_bsac

Q30

4321

Freq

uenc

y

14

12

10

8

6

4

2

0





Email:[email protected] Q31A: I feel that the consultations with the lecturer in order to train facilitators are necessary.


Total 32 100.0

Q31_bsac

Q31

432

Freq

uenc

y

20

10

0

Q32A: I feel that the consultations with the lecturer should not be done on Friday afternoons and Sundays but the consultations are necessary and should be done only once on Friday.


Total 32 100.0

Q32_bsac

Q32

4321

Freq

uenc

y

16

14

12

10

8

6

4

2

0

Q33A: I feel that the consultancy, peer-discussion learning cycle format is a waste of time.


Total 31 100.0

Q33_bsac

Q33

321

Freq

uenc

y

30

20

10

0





Email:[email protected] Q34A: I feel that the traditional transmittalist method (lecturer talk in front and student listen and taking notes) is the best way of learning as compared to the peer-facilitating learning-cycle that is presently used for this class.


Total 31 100.0

Q34_bsac

Q34

321

Freq

uenc

y

20

10

0

Q35A: The lecturer always arrived on time.


Total 32 100.0

Q35_bsac

Q35

4321

Freq

uenc

y

14

12

10

8

6

4

2

0

Q36A: The lecturer is always very motivated, enthusiastic and always willing to sacrifice his time for us.


Total 32 100.0

Q36_bsac

Q36

432

Freq

uenc

y

20

10

0





Email:[email protected] Q37A: The lecturer had worked very hard in preparing materials for us and helping us learn the subject matter.


Total 32 100.0

Q37_bsac

Q37

432

Freq

uenc

y

20

10

0

Q38A: The lecturer is always very patient in answering our questions during consultation and during lecture hour.


Total 32 100.0

Q38_bsac

Q38

4321

Freq

uenc

y

20

10

0

Q39A: The lecturer always explained the materials clearly during consultancy and during lecture.


Total 32 100.0

Q39_bsac

Q39

432

Freq

uenc

y

14

12

10

8

6

4

2

0





Email:[email protected] Q40A: The lecture materials are organized and legible from where I sit in the classroom.


Total 32 100.0

Q40_bsac

Q40

4321

Freq

uenc

y

30

20

10

0

Q41A: The lecturer’s voice can be heard clearly from where I sit in the classroom.


Total 32 100.0

Q41_bsac

Q41

432

Freq

uenc

y

30

20

10

0

Q42A: The lecturer presents his lectures in very interesting manner and always joke around in between.


Total 32 100.0

Q42_bsac

Q42

4321

Freq

uenc

y

14

12

10

8

6

4

2

0





Email:[email protected] Q43A: The lecturer is very confident and has a good mastery of the topic discussed.


Total 32 100.0

Q43_bsac

Q43

432

Freq

uenc

y

20

10

0

Q44A: The lecturer is very organized with the flow of the content presented.


Total 32 100.0

Q44_bsac

Q44

432

Freq

uenc

y

30

20

10

0

Q45A: The lecturer will always show an example to help make the material more comprehensible.


Total 32 100.0

Q45_bsac

Q45

4321

Freq

uenc

y

16

14

12

10

8

6

4

2

0





Email:[email protected] Q46A: The lecturer will always ask fundamental (basic) questions to test my understanding during consultation and lecture.


Total 32 100.0

Q46_bsac

Q46

4321

Freq

uenc

y

16

14

12

10

8

6

4

2

0

Q47A: The lecturer will always encourage me to be more open in asking questions and motivate me.


Total 32 100.0

Q47_bsac

Q47

432

Freq

uenc

y

20

10

0

Q48A: The lecturer has succeeded in imparting valuable knowledge about thermodynamics.


Total 32 100.0

Q48_bsac

Q48

432

Freq

uenc

y

16

14

12

10

8

6

4

2

0





Email:[email protected] Q49A: I look forward to be lectured by this lecturer again in the future.


Total 31 100.0

Q49_bsac

Q49

4321

Freq

uenc

y

14

12

10

8

6

4

2

0


Thermodynamics End of Semester Course Evaluation-DIC


Email:[email protected] Q50A: How challenging do you find this course?

Key: 1=Too Easy, 2=A bit easy, 3=Just Right, 4=A bit difficult, 5=Too Difficult. Frequency Percent

3 9 28.1 4 20 62.5 5 3 9.4

Total 32 100.0

Q50_bsac

Q50

543

Freq

uenc

y

30

20

10

0

Q51A: How much do you think you have learned in this course compared with other courses you have taken this year? Key: 1=Almost nothing, 2= Little, 3=A fair amount, 4= Much, 5=A lot


Total 32 100.0

Q51_bsac

Q51

5432

Freq

uenc

y

30

20

10

0

Q52A: How well has this course met your expectations?

Key: 1= Not at all, 2=Not very well, 3=Adequately, 4=Well, 5=Very well Frequency Percent

2 5 15.6 3 16 50.0 4 10 31.3 5 1 3.1

Total 32 100.0

Q52_bsac

Q52

5432

Freq

uenc

y

20

10

0

End of Analysis: Thermodynamics CMT408: Dec 02 – Apr 03 Contact Assoc. Prof. DR. J.J. for more info on the evaluation & the analysis.



Purpose:

Students evaluate their effort, the learning and teaching

methods and the lecturer’s performance. The analysis can be

used for assessing a lecturer’ s effort but most importantly

for self-improvement in terms of teaching styles and teaching

techniques for producing the most effective learning.

Method:

Students respond to statements in the absence of the

lecturer. Responses are anonymous. Responses are keyed in

and analyzed using both MS Excel and SPSS. For statements

1 through 49, the keys used are:

1=Strongly disagree 2=Disagree

3=Agree 4=Strongly agree

Semester Dec 2002 – April 2003

Course: Thermodynamics CMT 251

Program AS115/4

Lecturer: Assoc. Prof. Dr. Jaafar Jantan



End of Semester Course Evaluation-DIC


Email:[email protected] Q1A: My conceptual understanding in physics is weak.


Total 36 100.0

Q1_DIC

Q1_DIC

432

Freq

uenc

y

20

10

0

Q2A: I memorized most of the equations in physics and not really understanding them.


Total 36 100.0

Q2_DIC

Q2_DIC

4321

Freq

uenc

y

30

20

10

0

Q3A: I found thermodynamics to be a very difficult course.


Total 36 100.0

Q3_DIC

Q3_DIC

4321

Freq

uenc

y

30

20

10

0





Email:[email protected] Q4A: The concepts in thermodynamics are too difficult to learn.


Total 36 100.0

Q4_DIC

Q4_DIC

4321

Freq

uenc

y

30

20

10

0

Q5A: There are too many mathematical expressions in thermodynamics for me to memorize.


Total 36 100.0

Q5_DIC

Q5_DIC

432

Freq

uenc

y

30

20

10

0

Q6A: There are too many terminologies in thermodynamics for me to comprehend.


Total 35 100.0

Q6_DIC

Q6_DIC

4321

Freq

uenc

y

30

20

10

0





Email:[email protected] Q7A: The course outline which was distributed at the beginning of the semester had helped me focus my learning and structure my time appropriately and should be employed in all classes.


Total 36 100.0

Q7_DIC

Q7_DIC

4321

Freq

uenc

y

20

10

0

Q8A: My command of English comprehension is not good and I found difficulties in understanding the textbook, lecture and the questions in tests and quizzes.


Total 36 100.0

Q8_DIC

Q8_DIC

4321

Freq

uenc

y

16

14

12

10

8

6

4

2

0

Q9A: The textbook helped me learn thermodynamics better than if there is no textbook.


Total 36 100.0

Q9_DIC

Q9_DIC

432

Freq

uenc

y

30

20

10

0





Email:[email protected] Q10A: The problems at the end of each chapter are very useful to increase comprehension.


Total 34 100.0

Q10_DIC

Q10_DIC

432

Freq

uenc

y

30

20

10

0

Q11A: The structure (format) for this class had force me to work harder than I would normally do.


Total 36 100.0

Q11_DIC

Q11_DIC

432

Freq

uenc

y

30

20

10

0

Q12A: The structure (format) for this class has helped me to focus my learning better.


Total 36 100.0

Q12_DIC

Q12_DIC

432

Freq

uenc

y

20

10

0





Email:[email protected] Q13A: I always do my reading and submit my reading assignment on time.


Total 36 100.0

Q13_DIC

Q13_DIC

432

Freq

uenc

y

20

10

0

Q14A: I always tryout the problem assignment (at the end of the chapter) and submit them on time.


Total 36 100.0

Q14_DIC

Q14_DIC

4321

Freq

uenc

y

20

10

0

Q15A: I always do my preparation before my consultation with the lecturer on Fridays and Sundays.


Total 34 100.0

Q15_DIC

Q15_DIC

432

Freq

uenc

y

20

10

0





Email:[email protected] Q16A: Before and after consultation periods with the lecturer, I always read and assess my understanding by using the self-assessment provided.


Total 36 100.0

Q16_DIC

Q16_DIC

432

Freq

uenc

y

30

20

10

0

Q17A: When I am not a facilitator, I always assess my understanding by using the self-assessment provided, before and after peer discussion.


Total 35 100.0

Q17_DIC

Q17_DIC

4321

Freq

uenc

y

30

20

10

0

Q18A: Sending my self-assessment to the lecturer by using email and file attachment is useful in making me learn the use of ICT.


Total 36 100.0

Q18_DIC

Q18_DIC

4321

Freq

uenc

y

20

10

0





Email:[email protected] Q19A: The self-assessment materials developed by the lecturer is an important tool and must be developed in other courses to help students identify their strength and weaknesses.


Total 36 100.0

Q19_DIC

Q19_DIC

432

Freq

uenc

y

30

20

10

0

Q20A: Obtaining information about this course is very easy because the course outline, the due dates, events dates and all the necessary documents are available on the lecturer’s website.


Total 36 100.0

Q20_DIC

Q20_DIC

432

Freq

uenc

y

30

20

10

0

Q21A: I feel that utilizing the world-wide-web and using email must be utilized in all courses to facilitate learning.


Total 36 100.0

Q21_DIC

Q21_DIC

4321

Freq

uenc

y

16

14

12

10

8

6

4

2

0





Email:[email protected] Q22A: I always ask questions during consultancy, during peer-lecturing and during lecture hour.


Total 36 100.0

Q22_DIC

Q22_DIC

4321

Freq

uenc

y

20

10

0

Q23A: I always discuss with my peers outside the classroom regarding the topic of the week.


Total 36 100.0

Q23_DIC

Q23_DIC

4321

Freq

uenc

y

20

10

0

Q24A: The reading assignment is a very good exercise for me in preparing myself for class.


Total 36 100.0

Q24_DIC

Q24_DIC

432

Freq

uenc

y

20

10

0





Email:[email protected] Q25A: The consultancy, peer-discussion is a very effective method in preparing myself before class.


Total 36 100.0

Q25_DIC

Q25_DIC

432

Freq

uenc

y

20

10

0

Q26A: The lectures after peer discussion helped made the learning easier.


Total 36 100.0

Q26_DIC

Q26_DIC

4321

Freq

uenc

y

20

10

0

Q27A: I have worked very hard in this class to get good grade.


Total 35 100.0

Q27_DIC

Q27_DIC

432

Freq

uenc

y

30

20

10

0





Email:[email protected] Q28A: I have worked very hard to ensure understanding of the subject matter.


Total 36 100.0

Q28_DIC

Q28_DIC

432

Freq

uenc

y

30

20

10

0

Q29A: The learning format for this class should be employed in other science classes.


Total 36 100.0

Q29_DIC

Q29_DIC

4321

Freq

uenc

y

20

10

0

Q30A: The class should be held for only three hours and not five hours.


Total 36 100.0

Q30_DIC

Q30_DIC

4321

Freq

uenc

y

16

14

12

10

8

6

4

2

0





Email:[email protected] Q31A: I feel that the consultations with the lecturer in order to train facilitators are necessary.


Total 36 100.0

Q31_DIC

Q31_DIC

432

Freq

uenc

y

20

10

0

Q32A: I feel that the consultations with the lecturer should not be done on Friday afternoons and Sundays but the consultations are necessary and should be done only once on Friday.


Total 35 100.0

Q32_DIC

Q32_DIC

4321

Freq

uenc

y

16

14

12

10

8

6

4

2

0

Q33A: I feel that the consultancy, peer-discussion learning cycle format is a waste of time.


Total 36 100.0

Q33_DIC

Q33_DIC

4321

Freq

uenc

y

30

20

10

0





Email:[email protected] Q34A: I feel that the traditional transmittalist method (lecturer talk in front and student listen and taking notes) is the best way of learning as compared to the peer-facilitating learning-cycle that is presently used for this class.


Total 35 100.0

Q34_DIC

Q34_DIC

4321

Freq

uenc

y

30

20

10

0

Q35A: The lecturer always arrived on time.


Total 36 100.0

Q35_DIC

Q35_DIC

432

Freq

uenc

y

20

10

0

Q36A: The lecturer is always very motivated, enthusiastic and always willing to sacrifice his time for us.


Total 36 100.0 Total 52 100.0

Q36_DIC

Q36_DIC

43

Freq

uenc

y

30

20

10

0





Email:[email protected] Q37A: The lecturer had worked very hard in preparing materials for us and helping us learn the subject matter.


Total 36 100.0

Q37_DIC

Q37_DIC

432

Freq

uenc

y

30

20

10

0

Q38A: The lecturer is always very patient in answering our questions during consultation and during lecture hour.


Total 36 100.0

Q38_DIC

Q38_DIC

432

Freq

uenc

y

30

20

10

0

Q39A: The lecturer always explained the materials clearly during consultancy and during lecture.


Total 36 100.0

Q39_DIC

Q39_DIC

432

Freq

uenc

y

20

10

0





Email:[email protected] Q40A: The lecture materials are organized and legible from where I sit in the classroom.


Total 36 100.0

Q40_DIC

Q40_DIC

432

Freq

uenc

y

20

10

0

Q41A: The lecturer’s voice can be heard clearly from where I sit in the classroom.


Total 36 100.0

Q41_DIC

Q41_DIC

432

Freq

uenc

y

30

20

10

0

Q42A: The lecturer presents his lectures in very interesting manner and always joke around in between.


Total 36 100.0

Q42_DIC

Q42_DIC

432

Freq

uenc

y

20

10

0





Email:[email protected] Q43A: The lecturer is very confident and has a good mastery of the topic discussed.


Total 36 100.0

Q43_DIC

Q43_DIC

43

Freq

uenc

y

30

20

10

0

Q44A: The lecturer is very organized with the flow of the content presented.


Total 36 100.0

Q44_DIC

Q44_DIC

432

Freq

uenc

y

20

10

0

Q45A: The lecturer will always show an example to help make the material more comprehensible.


Total 36 100.0

Q45_DIC

Q45_DIC

432

Freq

uenc

y

30

20

10

0





Email:[email protected] Q46A: The lecturer will always ask fundamental (basic) questions to test my understanding during consultation and lecture.


Total 36 100.0

Q46_DIC

Q46_DIC

432

Freq

uenc

y

20

10

0

Q47A: The lecturer will always encourage me to be more open in asking questions and motivate me.


Total 36 100.0

Q47_DIC

Q47_DIC

432

Freq

uenc

y

20

10

0

Q48A: The lecturer has succeeded in imparting valuable knowledge about thermodynamics.


Total 36 100.0

Q48_DIC

Q48_DIC

432

Freq

uenc

y

20

10

0





Email:[email protected] Q49A: I look forward to be lectured by this lecturer again in the future.


Total 33 100.0

Q49_DIC

Q49_DIC

4321

Freq

uenc

y

20

10

0


Thermodynamics End of Semester Course Evaluation-DIC


Email:[email protected] Q50A: How challenging do you find this course?

Key: 1=Too Easy, 2=A bit easy, 3=Just Right, 4=A bit difficult, 5=Too Difficult. Frequency Percent

2 1 2.8 3 9 25.0 4 21 58.3 5 5 13.9

Total 36 100.0

Q50_DIC

Q50_DIC

5432Fr

eque

ncy

30

20

10

0

Q51A: How much do you think you have learned in this course compared with other courses you have taken this year? Key: 1=Almost nothing, 2= Little, 3=A fair amount, 4= Much, 5=A lot


Total 35 100.0

Q51_DIC

Q51_DIC

5432

Freq

uenc

y

14

12

10

8

6

4

2

0

Q52A: How well has this course met your expectations?

Key:1= Not at all, 2=Not very well, 3=Adequately, 4=Well, 5=Very well Frequency Percent

1 1 2.9 2 9 25.7 3 14 40.0 4 11 31.4

Total 35 100.0

Q52_DIC

Q52_DIC

4321

Freq

uenc

y

16

14

12

10

8

6

4

2

0

End of Analysis: Thermodynamics CMT251: Dec 02 – Apr 03 Contact Assoc. Prof. DR. J.J. for more info on the evaluation & the analysis.


ALUMNI TESTIMONIALS

http://drjj.uitm.edu.my Anugerah Akademik Negara 2010-DrJJ

Page 1 of 9

Student Name: Naziatul Nazrah Md Rauf. Process Engineer, Winmore Engineering Sdn. Bhd

Program: Diploma in Industrial Chemistry. Semester 4.

Course Taken: Thermodynamics. Semester: 2002.

Email: [email protected] .

Tel : +603-91732180

Fax : +603-91736180

Cell: +6012-3735731

My Testimonial of Associate Professor Dr Jaafar Jantan’s Teaching. 15th January 2010

“Give a Man a Fish, Feed Him For a Day. Give Man a Fishing Pole, Feed Him For a Lifetime – Lao Tzu”. Dr

JJ can be described as a teacher who does not believe in spoon-fed his students to make them learn. He

believe in active learning where students actively engage with learning material and hence more likely

to recall / retain information. One of such active learning activities is think-pair-share activity, held after

the unit has been sufficiently covered already by Dr JJ, and it is conducted between 5 to 6 students. One

student will act as facilitator to facilitate discussion where the students are required to take a minute to

reflect what has been learned, then discuss it among themselves and share it with the rest of the class.

This activity certainly force students to be better prepared with sound instruction before expecting them

to discuss it on their own.

This one activity certainly helps me in building up my career as Chemical Engineer / Project Manager

where I have to conduct on-job training for my engineers as well as formal technical training for Client.

This activity alone teaches me on how important it is to think critically on the subject matter so that I

can be well prepared to discuss it to my engineers and Client during training session. As there is a

saying,“The mediocre teacher tells. The good teacher explains. The superior teacher demonstrates. The

great teacher inspires". He certainly inspires me to be the best that I can be and for that I am always

grateful for having him as my Thermodynamic Lecturer. God bless him.

mailto:[email protected]


Page 2 of 9

Name: Hanni Salehan. . Business Analyst, Managing Director’s Office Malaysia Airlines

Program: Diploma in Industrial Chemisrty

Course Taken: Thermodynamics. Semester: Between 2002

Email: Hanni Salehan [email protected]: Handphone #: +60177933797

A typical style of teaching and learning where the interaction is only one-way is somewhat dull and

boring. That is however the most common method used by most lecturers, as it is indeed the easiest.

Despite having the experience pursuing my undergraduate studies in the UK, I will always remember the

first moment when I started to feel passionate about learning, which was through Dr JJ’s unorthodox,

yet effective method in teaching Thermodynamics. His creative approach made the subject so easy and

it did get the best out of me as a student.

As a lecturer, Dr JJ stepped out of his comfort zone and introduced a new style of teaching-learning

where there was active engagement between lecture-students as well as among the students. Dr JJ

allowed everyone in class to participate in an active process and students were given the opportunity to

question, debate and challenge each other’s points which I absolutely loved. From my experience as a

student, I always had the problem in concentrating and listening to lectures, but in Dr JJ’s class I never

had that problem. Through active discussions, my brain was able to work better compared to merely

passively listening to a lecture. The class activities made all of us awake and concentrate better, we

shared thoughts and we had fun while learning and in my opinion, that was the crucial part. Most

lecturers forgot to understand this; learning is not just about facts but there are a lot more into it. I

personally learnt better by doing, and Dr JJ’s creative way of teaching helped me to absorb and react

better to lecture material and ultimately, learnt the subject not by memorizing but by understanding it.



Page 3 of 9

Student Name: Khashaf bin Amir. Process Engineer, Senior Project Engineer servicing for a global Oil &

Gas Pipeline company in Kuantan

Program: Diploma in Industrial Chemistry. Semester 4.

Course Taken: Thermodynamics. Semester: 2002.

Email: [email protected] Cell: +6019-2824292

My Testimonial of Associate Professor Dr Jaafar Jantan’s Teaching. 15th January 2010

Dr. Jaafar Jantan, mainly known as Dr. JJ was one of the commonly known lecturers during my student

days in UiTM. He taught me Thermodynamics back in 2002 which gave a significant positive impact

towards my learning process and interest.

Dr. JJ introduced and applied a different method of teaching which I found very creative, effective and

suitable considering student's level of understanding. Where majority of the students found the subject

quite tough, Dr. JJ made it attractive through mediums of internet, simplified presentations, group

discussions and most of all his interactive class engagements. It's very obvious that his approach

enhanced our communication and intellectual skills whilst increasing our affection towards learning. He

encouraged students to conduct mock teaching sessions in class which was interesting.

I duly acknowledge Dr.JJ's continuous effort to make lecturing a 2-way communication session. Honestly,

creating interest in students will ease the teaching process; and at the end of the day, 90% input is what

the students expect from the class. Dr. JJ has that ability and I hope his method of teaching could be

extended to other faculties as well. When students enjoy lectures, UiTM will produce quality graduates.

Lecturers are the roots to success.


summative assessment

DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO

This examination paper consists of 14 printed pages. CONFIDENTIAL

APPENDIX 2 SAMPLE OF FINAL EXAMINATION AND ANSWER SCHEME

CONFIDENTIAL

AS/APR 2003/CMT408

UNIVERSITI TEKNOLOGI MARA

FINAL EXAMINATION

COURSE : THERMODYNAMICS

COURSE CODE : CMT408

DATE :

TIME : 3 HOURS

FACULTY : Applied Sciences

SEMESTER : Dec 2002 – April 2003

PROGRAMME / CODE : Bachelor of Science (Hons.) in Applied Chemistry / AS225

Grading Scheme to the final paper (40%) Prepared By Dr. J.J., Faculty of Applied Sciences

UiTM, Shah Alam

DO NOT PRINT FOR DISTRIBUTION TO STUDENTS

General Rule for grading to preserve consistency:

� Grade by the question NOT by the script. Start on the 2nd question only after finishing all scripts for question 1.

� Credit is given as indicated but for those that are vague, use judgment. For example, if a student had calculated a wrong answer for a certain part and use that wrong value for solving other parts requiring the use of that number, then he is penalized only once; i.e. only the part where the wrong value was calculated. On the other parts, if the techniques are correct, he gets full credit.

� Words that I had underlined are the keywords that deserve credits. Again, use judgment. Please verify answers or values and inform me of any errors. Contact me via SMS or handphone: 019-355-1621 or email [email protected] or [email protected] . My website: http://www.uitm.edu.my/faculties/fsg/drjj1.html

� Take lots of breaks to secure fair assessment.

CONFIDENTIAL 2 AS/APR 2003/CMT408

CONFIDENTIAL

SECTION A (20 marks for each question. Total is 100 MARKS) 1. (a) Briefly, give a description that best fit each of the following:

i) isolated iii) saturated liquid-vapor mixture ii) state iv) entropy

ANSWER: (2 marks each)

(i) A system that prohibits both energy and mass to cross its boundary (ii) A set of properties that completely describes the condition of the system. (iii) A mixture phase that is made up of both liquid and vapor coexisting together. (iv) Entropy is a measure of disorder or chaos in a system High entropy indicates high

disorder. (8 marks)

b) i) Using the property table, read the saturation pressures and the saturated

volumes for water at temperatures of 50°C and 350°C, respectively. ii) Using the property table, read the saturation temperatures and the saturated

volumes for refrigerant-134a at pressures of 60 kPa and 1.2 MPa, respectively.

ANSWER: (6 marks for each section. Values without units receive no credit)

System T, °C Psat, kPa νf, m3/kg νg,m

3/kg

Water 50 12.349 0.001012 12.03 (i)

Water 350 16,513 0.001740 0.008813

System P, kPa Tsat, °C νf, m3/kg νg,m

3/kg

Ref-134a 60 -37.07 0.0007097 0.3100 (ii)

Ref-134a 1200 46.32 0.0008928 0.0166

(12 marks)


CONFIDENTIAL

2. a) i) Name the TWO forms of energy interaction that can cross the boundary of a thermodynamic closed system.

ANSWER: Heat and work

ii) Briefly, describe what each of the energy interaction means. ANSWER:

Heat is thermal energy in motion due to temperature difference across the system’s boundary. Work is the energy transfer associated with a force that acts through a distance

(2 + 6 = 8 marks)

b) i) In your own words, state the conservation of mass principle.

ANSWER: The conservation of mass principle states that the net mass transfer, min – mout, in or out of a system in a process must equal the net change in total mass of the system,

∆msys, during the process. ii) Write down the rate-from mass balance for any system undergoing any

process ANSWER:

sysoutin

mmm•••

∆=−∑∑ (without the summation symbol is also accepted)

iii) Write down the unit-mass basis energy that is transported by a flowing fluid.

ANSWER:

pekehPpekeu ++=+++= νθ

(3 + 3+ 6 = 12 marks) 3. a) Write down a unit-mass basis energy balance for a closed stationary system.

ANSWER:

000 ++∆=+−+− uqq outinoutin ωω (8 marks)

b) Write down a unit-mass basis energy balance for a steady-flow device undergoing any

kind of process ANSWER:

0=−+−+− outinoutinoutin qq θθωω where ϑ is flow energy. (7 marks)

c) Write down the unit-mass basis energy balance for a throttle and explain the

implication on the temperature and pressure. ANSWER:

inoutinout hh −=−=+ θθ00 So, inout hh = or

( ) ( )inout PuPu υυ +=+ . = constant

If Pν increases, then the internal energy must decrease resulting in a drop of the exit temperature.

If Pν decreases, then the internal energy must increase resulting in an increase of the exit temperature.

(1 mark)

(2 marks)

(1mark)

(1 mark)


CONFIDENTIAL

4. (a) Draw an energy-flow diagram for a refrigerator. Label all the components and the energy exchanges in units of kJ/kg (unit-mass basis) including the reservoirs and its temperatures.

(1 for fridge, 3 for energies in kJ/kg, 2 for reservoirs with labeled T: Total = 6 marks)

(b) State the Carnot Principles. ANSWER: Reversible heat engines in contact with the same hot and cold reservoirs will

have the same efficiencies while irreversible engines will have efficiencies lower than the reversible engines.

(4 marks)

(c) What is a Carnot engine? ANSWER: A carnot heat engine is a dream (ideal) heat engine where all the processes

are reversible.

(2 marks)

(d) Using the relation Input quiredRe

Output DesiredePerformanc =

, obtain the coefficient of

performance for a Carnot refrigerator.

ANSWER:

1

1

1

1

,

,

−

=

−

=

−

=

=

L

H

revin

outrevinout

in

revinnet

in

revR

T

T

q

qqq

qqCOP

ω

3 marks 1 mark 2 mark 2 marks = 8 marks Symbols for other units such as kJ and kJ/s, are also acceptable.

Fridge

High Temperature Reservoir, TH

Hout qq =

Low Temperature Reservoir, TL

inω

Lin qq =


CONFIDENTIAL

5. a) In your own words, state the Increase of Entropy Principle. ANSWER: The principle states that the entropy of an isolated system during a process always

increase or, in the limiting case of a reversible process, remains constant.

(5 marks)

b) Write down the mathematical expression to represent the entropy generated in an isolated system to reflect the Increase of Entropy Principle.

ANSWER: 0≥∆+∆=∆= surrsystotalgen SSSS

1 mark 1 mark 2 marks 2 marks = 6 marks (6 marks)

c) A piston-cylinder device contains 1 kg of water at 100 kPa and 70°C. Heat crosses

into the system until the system’s temperature increases to 150°C. Determine the initial and final phase, the initial and final entropies of water and hence obtain the total entropy change of water during this process.

ANSWER:

State 1: P1= 100 kPa

T1= 70°C

Tsat = 99.63°C Since T < Tsat, hence phase is comp. liq.

Then s1 = sf@70°C s1 = 0.9549 kJ/kg.K

1 mark 2 marks

1 mark

State 2: P1= 100 kPa

T1= 150°C

Tsat = 99.63°C Since T > Tsat, hence phase is superheated vapor.

Then s2 = s@100 kPa, 150°C s2 = 7.6134 kJ/kg.K

2 marks

1 mark

Entropy change of water during the process is

( )syssyssys ssmsmS 12 −=∆=∆

( )Kkg

kJkgS sys

•

−=∆ 9549.06134.71

or K

kJS sys 6585.6=∆

1 marks

1 mark

(9 marks)


CONFIDENTIAL

SECTION B (150 MARKS) 1. (a) Using the property table, determine the missing properties in the table below. [Note:

The symbols used are the following: P for pressure, T for temperature, and ν for the specific volume. When indicating the quality, use NA for the compressed liquid and the superheated vapor phase and use 0 < x < 1 for the wet mix phase. When

indicating properties such as ν, u, h or s in the wet mix phase, just write down an expression on how to get it. NO CALCULATIONS ARE REQUIRED.]

Substance P

kPa

Psat@T

kPa

T

°C

Tsat@P

°C

Quality

x

ν

m3/kg

Phase

& Reason

i) 2,000 - 212.42 212.42 1 0.09963

Sat.

Vapor

ν =νg

Water

ii) 50 31.19 70 81.33 NA

ν = ν f@-

70°C

=0.001023

Comp.

Liquid

T<Tsat or P>Psat

iii) 240 571.60 20 -5.37 NA 0.09339 Sup. Vapor

ν > νg

Ref-134a

iv) 217.04 217.04 -8 - 0.2 ν =νf+xνfg Wet mix

0 < x <1

(Note that 2 marks are credited for answers in the last column: 5+6+6+5 = 22 marks)


CONFIDENTIAL

(b) In the question that follows, you need to use the data in the table above along with the property table to sketch property diagrams with respect to the saturation lines to indicate the state of the system. In your diagram, draw and label the temperature or the pressure lines. Place an X mark on the graph to indicate the state. Insert values

for T, Tsat, P, Psat, ν, νf and νg.

(i) Use the space below to draw a T - ν diagram for the system in part (ii) of the table.

Accepted pairs of saturated values and must be marked

correctly on the graph in m3/kg are:

νf@70 °C = 0.001023and νg @70 °C = 5.042

νf@50 kPa = 0.001030 and νg @50 kPa =3.240

(2 for T’s, 2 for P’s, 3 for ν’s, & 1 for the dot. Total = 8 marks)

T, °C

81.3

70

ν=νf@70 °C = 0.001023

31.9 kPa

50 kPa

ν, m3/kg

5.042


CONFIDENTIAL

(ii) Use the space below to draw a P – ν diagram for the system in part (iii) of the table.

(8 marks)

Accepted pairs of saturated values and must be marked correctly on the graph in m

3/kg are:

νf@240 kPa = 0.00076182 νg @240 kPa = 0.0834

νf@20 °C = 0.0008157 νg @20 °C = 0.0358

(2 for T’s, 2 for P’s, 3 for ν’s, & 1 for the dot. Total = 8 marks)

571.6

240

νf =0.00076182 νg = 0.0834

-5.37 °C

ν, m3/kg

P, kPa

20 °C

ν = 0.09339


CONFIDENTIAL

(c) Steam at 5,000 kPa, 300 o

C in a piston-cylinder device is allowed to cool isobarically

until the temperature drops to 250 o

C. Use the space below to draw a P - ν diagram with respect to the saturation lines for this process. Indicate the initial and final states and the direction of the process. Draw and label the temperature lines and insert

values for T1, T2, Tsat, P1, P2, Psat, ν1, ν2, νf, and νg.

Accepted pairs of saturated values and must be marked correctly on the graph in m

3/kg are:

νf@250 °C = 0.001251 νg @250 °C = 0.05013

νf@5,000 kPa = 0.001286 νg @5,000 kPa = 0.3944

νf@300 °C = 0.001404 νg @300 °C = 0.02167

(3 for T’s, 3 for P’s, 4 for ν’s, & 1 for the pair of dots, 1 for direction. Total = 12 marks)

Total marks for the question according to parts:

22+16+12 = 50 marks

P, kPa

8,581

3,973

νf = 0.001404

250oC

300oC

ν, m3/kg

νg = 0.02167 ν1 = 0.04532

5,000

263.99oC

2

1

ν2 = νf@250 °C = 0.001251


CONFIDENTIAL

2. a) i) State all the characteristics of heat engines ANSWER: Receive heat from a high temperature source

Convert part of the heat into work Reject excess heat into a low temperature sink Operates in a cycle

(2+2+2+1 = 7 marks)

ii) Sketch an energy-flow diagram for a steam power plant. Label all the interaction energies and the reservoirs.

ANSWER:

(2 marks for reservoirs, 3 marks for energy interactions (symbols for any units can be accepted) and 1 mark for SPP = 6 marks)

iii) State the desired output and required input for a steam power plant.

ANSWER: Desired output is out,netω . Required input is qin

(2 marks)

iv) Write down the energy balance for the steam power plant.

ANSWER: 00qq outinoutin =+−+− ωω

1 mark

1 mark

1 mark

1 mark

1 mark

1 mark

= 6 marks

(6 marks)

v) Using parts (iii) and (iv), obtain the thermal efficiency for the steam power

plant.

ANSWER: in

out

in

outin

in

inout

in

out,net

q

q1

q

qq

qq−=

−

=

−

==

ωωω

η

2 marks 1 mark 1 mark = 4 marks

(4 marks)

SPP

High Temperature Reservoir, TH

Hin qq =

Low Temperature Reservoir, TL

out,netω

Lout qq =


CONFIDENTIAL

b) A heat pump with a COP of 2.4 is used to heat a house. When running, the heat pump consumes 8 kW of electric power. The house is losing heat to the outside at an

average of 40,000 kJ/hr and the temperature of the house is 3°C when the pump is

turned on. The final temperature to achieve is 22°C. Assume that the house is well-sealed (no air leaks) and take the entire mass within the house (air, furniture, etc.) to

be equivalent to 2,000 kg of air. (Use the specific heat capacity for air as Cν = 0.718 kJ/kg.K) i) Draw a schematic diagram to represent the situation described in the above

problem. ANSWER:

(1 for low T surrounding, 1 for HP, 1 for W in, 1 for house, 1 for change in T from 3 to 22, 3 for

Q’s: Total = 8 marks)

ii) Determine the rate of heat supplied by the pump to the house.

ANSWER:

in,net

HHP

W

QCOP

•

•

= . So, kW 2.19kW 8x4.2WxCOPQ in,netHPH ===

••

1 mark 1 mark 1 mark = 3 marks

(3 marks)

iii) Determine the total internal energy change in kJ, for the air in the house when

the temperature changes from 3°C to 22°C.

ANSWER

kJ284,27K)273327322(x)Kkg/kJ718.0(kgx2000)TT(mCU 12v =−−+•=−=∆ .

1 mark 1 mark 1 mark (3 marks)

Heat Pump COP = 2.4

House, 3°C to

22°C

Hout QQ••

=

Surrounding, 3°C

kW 8W in =

•

Lin QQ••

=

hr/kJ 000,40Q lost =

•


CONFIDENTIAL

iv) Write down the rate form energy balance for the air in the house. Include all the assumptions that you make.

ANSWER:

Energy Balance: sysoutin EEE•••

=− ∆ . For stationary closed system,

assume that the potential and kinetic energy change of the air are negligible. Taking the air inside the house as a system, then;

sysoutinoutin U0WWQQ•••••

=+−+− ∆ , then,

t/)TT(mC00QQ 12voutin ∆−=++−

••

where ∆t is the time taken

2 marks

5 marks

(Total 7 marks)

v) Using parts (iii), and (iv), determine the time taken (in minutes), to raise the

temperature of the air in the house from 3°C to 22°C. ANSWER:

The the time taken is

outin

12v

QQ

)TT(mCt

••

−

−

=∆ .

∆t = 27,284 kJ/ [19.2 kW –(40,000kJ/hr)*(1hr/3600s)]

∆t = 27,284/ [19.2 – 11.11]

So ∆t = 27,284/8.089 = 3,373.02 s = 56.21 mins.

3 marks

1 mark

(Total 4 marks)


CONFIDENTIAL

3. Consider a steam power plant that operates on a simple ideal Rankine cycle and has a net power output of 45 MW. Steam enters the high-pressure turbine at 7 MPa and 500

oC and is

cooled in the condenser at a pressure of 10 kPa by running cool water from a lake through the tubes of the condenser at a rate of 2,000 kg/s.

a) Using the property table, read the saturation temperature, specific volume ν1, specific enthalpy h1, and the specific entropy s1 for state 1 (pump inlet).

ANSWER

Sat liquid. T1 = Tsat@10kPa = 45.81 oC; ν1= νf@10kPa = 0.001010 m

3/kg.

h1= hf@10kPa = 191.83 kJ/kg, s1= sf@10kPa = 0.6493 kJ/kg K

2 marks 2 marks

(4 marks) b) Using the property table, read the saturation temperature for the pressure 7 MPa, the

specific enthalpy h3, and the specific entropy s3 for state 3 (turbine inlet).

ANSWER

P3 = 7 MPa, T3 = 500 oC.

Hence, Superheated vapor. So, h3= 3410.3 kJ/kg, s3= 6.7975 kJ/kg K (3 marks)

c) Sketch a T - s diagram with respect to the saturation lines for this cycle. Label all the

energy interactions, pressures, temperatures, entropies, the states, and the direction of the process.

ANSWER

(3 for T’s, 2 for P’s, 2 for s, 4 for energy interactions, 1 for the cycle. Total = 12 marks)

T, °°°°C

500

45.8

Tcrit

0.6493

1

qout = h4 – h1

2

ωin

qin = h3-h2

3

10 kPa

ωout,

7 MPa

4

6.7975 s, kJ/kg°K

285.9


CONFIDENTIAL

d) Obtain the quality and hence the enthalpy h4, of state 4 (condenser inlet).

ANSWER

Since state 3 to 4 is isentropic, then s4 = s3 = 6.6975 kJ/kg K.

Since for liquid –vapor mixture, y = yf+xyfg,

then s4 = (sf + xsfg)@10 kPa. Then the quality is

8197.05009.7

6493.06975.6

s

ssx

fg

f4=

−=

−

= . Then

kg/kJ12.2153)kg/kJ8.2392(*8197.0kg/kJ83.191xhhh fgf4 =+=+=

1 mark

1 mark 3 marks

4 marks

(9 marks)

e) Using ( )122 PPin −= νω where 12 νν ≅ for incompressible liquid, and applying the

energy balance at the pump, determine and calculate the enthalpy, h2, at state 2 (boiler inlet).

ANSWER

inletexitinoutin hh00 −=−=−+ ϑϑω

)PP(vhh 12112in −=−=ω .

So, )PP(vhh 12112 −+=

kPa)10000,7(*)kg/m001010.0(kg/kJ83.191h 32 −+=

Since the unit kJ is kPa.m3, then kg/kJ89.198h2 =

3 marks 1 mark

1 mark 1 mark 1 mark

(7 marks) f) Write down the energy balance at the boiler and at the condenser.

ANSWER

Boiler: inletexitinoutoutinoutin hhqq −=−=−+− ϑϑωω .

23in hh000q −=−+−

Condenser: inletexitinoutoutinoutin hhqq −=−=−+− ϑϑωω .

41out hh00q0 −=−+− . 14out hhq −=

4 mark

5 mark

(9 marks)

g) Using the results from all the parts above, determine the thermal efficiency of this power plant.

ANSWER

23

14

in

outin

in

out,net

hh

hh1

q

qq

q −

−−=

−

==

ω

η

%9.383893.09.1983.3410

8.1911.21531

hh

hh1

23

14==

−

−−=

−

−

−=η

5 marks

1 mark

(6 marks)

END OF QUESTION PAPER

PowerPoint dynamics lecture notes

8/22/2005

Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005 1

Thermodynamics Lecture Series

email: [email protected]://www3.uitm.edu.my/staff/drjj/

Applied Sciences Education Research Group (ASERG)

Faculty of Applied SciencesUniversiti Teknologi MARA

First Law First Law –– Control Control VolumeVolume

First Law First Law -- QuotesQuotes

“Education is not the piling on of learning, information, data, facts, skills, or abilities--that's training or instruction--but is rather a making visible what is hidden as a seed... To be educated, a person doesn't have to know much or be informed, but he or she does have to have been exposed vulnerably to the transformative events of an engaged human life... One of the greatest problems of our time is that many are schooled but few are educated.”

Author:Thomas MooreSource:he Education of the Heart by Thomas Moore

QuoteQuote

Introduction Introduction -- ObjectivesObjectives

1. State the conservation of mass principle.

2. State the meaning of steady-flow process and the implications on a system’s properties.

3. Write the unit-mass basis and unit-time basis (or rate-form basis) energy balance for a general steady-flow process.

4. Write the unit-time basis (or rate-form basis) mass balance for a general steady-flow process.

Objectives:Objectives:


4. State the assumptions for steady-flow devices such as nozzles, diffusers, turbines, compressors, throttle valves, heat exchangers and mixing chambers.

5. State the purpose for each of the steady-flow device noted above.

6. Write the unit-mass basis and unit-time basis (or rate-form basis) energy balance for each of the steady-flow devices noted above



7. Write the unit-time basis (or rate-form basis) mass balance for each of the steady-flow device.

8. Use the energy and mass balance to solve problems related to each of the steady-flow device.


8/22/2005 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005 6

First Law - GeneralFirst Law - General

How to relate changes to the cause

Dynamic Energies as causes (agents) of

change

SystemE1, P1, T1, V1

ToE2, P2, T2, V2

Properties will change indicating change of

state

Mass out

Mass inWinWout

QinQout

8/22/2005



First Law – Energy BalanceFirst Law – Energy Balance

Energy Balance

Amount of energy causing energy causing changechange must be equal to amount

of energy changeenergy change of system

Change of system’s energy

=Energy Leaving a system

-Energy

Entering a system


First Law – Energy BalanceFirst Law – Energy Balance

Energy Balance

Ein – Eout = ∆Esys, kJ orein – eout = ∆esys, kJ/kg or

Change of system’s energy

=Energy Leaving a system

-Energy

Entering a system

kW,EEE sysoutin•••

=− ∆


First Law – Energy BalanceFirst Law – Energy BalanceEnergy Balance –General system

kW ,PEKEU

EEWWQQ out,massin,massoutinoutin

•••

••••••

++=

−+−+−

∆∆∆

Qin – Qout+ Win – Wout+ Emass,in - Emass,out= ∆U+ ∆KE + ∆PE, kJ

qin – qout+ ωin – ωout + θin – θout

= ∆u+ ∆ke + ∆pe, kJ/kg


First Law – Stationary ClosedFirst Law – Stationary ClosedEnergy Balance –Stationary Closed

system

00 −+−+−••••

outinoutin WWQQ

Qin – Qout+ Win – Wout+ 0 – 0 = ∆U+ 0 + 0, kJ

qin – qout+ ωin – ωout + 0 – 0 = ∆u+ 0 + 0, kJ/kg

kW ,00 ++∆=•

U


Mass Balance

min – mout = ∆msys, kg or

First Law – Mass BalanceFirst Law – Mass BalanceChange of system’s

mass=

Mass Leaving a system

-Mass

Entering a system

s/kg,mmm sysoutin•••

=− ∆


First Law - Mass flow ratesFirst Law - Mass flow ratesMass Balance –Mass & Volume

Flow Rate

sm ,AmV

3

υνr

==••

The volume of the cylinder is

The volume flow rate is

sm ,

3

νtmA

ttV

∆=

∆=

∆l

The velocity of the mass is

s/m ,t∆

υ lr=

skg ,Am

νυr

=•

Then, mass flow rate is

AV l= 3m ,mν=

8/22/2005



First Law Energy balance CVFirst Law Energy balance CVMass & Energy

Balance–Steady-Flow: Single Stream

qin – qout+ ωin – ωout = θout – θin, kJ/kgEnergy balance:

kg/s ,mmm 21•••

==Mass balance:

= h2 – h1 + ke2 – ke1 + pe2 – pe1, kJ/kg

= ∆h + ∆ke + ∆pe, kJ/kg8/22/2005 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005 20

First Law – Energy Balance CVFirst Law – Energy Balance CV

initialfinal kekeke −=∆

initialfinal pepepe −=∆

Mass & Energy Balance–Steady-Flow:

Single StreamEnergy balance:

= ∆h + ∆ke + ∆pe, kJ/kgqin – qout+ ωin – ωout

kgkJ ,

2000

21

22 υυ

rr−

=

( )kgkJ ,

1000yyg 12 −

=

where

and


First Law of – Ideal GasFirst Law of – Ideal GasMass & Energy

Balance–Steady-Flow CV: Ideal Gases

Use Ideal Gas Equation of StateIdeal Gas Equation of State for real gases that behave like ideal gases. Criteria:

Where ν is the specific volume,m3/kg, R is gas constant, kJ/kg•K, T is absolute temperature in Kelvin

For known P and T, use to determine ν, and hence the mass flow rate.

Pgas << Pcrit and/or Tgas >> Tcrit. PPνν = RT= RT


First Law of – Ideal GasFirst Law of – Ideal GasMass & Energy

Balance–Steady-Flow CV: Ideal Gases

Use of property table of property table for real gases that behave like ideal gases.

Knowing T, read value for h and viceKnowing T, read value for h and vice--versa. versa. If T or h not found, do interpolationIf T or h not found, do interpolation


First Law - NozzlesFirst Law - NozzlesNozzles

outinoutin WWQQ••••

−+−

Energy balance

kg/s ,mmm 21•••

==Mass balance


−+−

( ) kW ,m inout ϑϑ −=•

( ) kW ,pekehm ∆∆∆ ++=•

PurposePurpose: Increase velocity: Increase velocity EffectEffect: Pressure drops: Pressure drops

0000 −+− ( ) kW ,0kehm ++=•

∆∆

InInState 1State 1

OutOutState 2State 2

AA22 << A<< A11



Energy balance:

kg/s , 2

22

1

11

νυ

νυ AA rr

=

PurposePurpose: Increase velocity: Increase velocity EffectEffect: Pressure drops: Pressure drops

0000 −+− ( ) kW ,0kehm ++=•

∆∆

kJ/kg ,keh0 ∆∆ += kJ/kg ,kekehh0 1212 −+−=

Mass balance:

kJ/kg ,2000

kekehh2

22

12112

υυrr

−=−=−

21 ••

= mm

8/22/2005




Energy balance:PurposePurpose: Increase velocity: Increase velocity EffectEffect: Pressure drops: Pressure drops

kJ/kg ,2000

kekehh2

22

12112

υυrr

−=−=−

AirAir:Use energy balance to find h:Use energy balance to find h22 and use table Aand use table A--17 (& interpolation technique) to determine T17 (& interpolation technique) to determine T22..

AirAir: Use table A: Use table A--17 to find h17 to find h11(& interpolation technique) (& interpolation technique) for a given Tfor a given T11..

InInState 1State 1


AA22 << A<< A118/22/2005 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005 26

First Law - DiffusersFirst Law - DiffusersDiffusers

Energy balance:PurposePurpose: Increase Pressure: Increase Pressure EffectEffect: Velocity drops: Velocity drops

kJ/kg ,2000

kekehh2

22

12112

υυrr

−=−=−

AirAir:Use energy balance to find h:Use energy balance to find h22 and use table Aand use table A--17 (& interpolation technique) to determine T17 (& interpolation technique) to determine T22..

AirAir: Use table A: Use table A--17 to 17 to find hfind h11 (& interpolation (& interpolation technique) for a given Ttechnique) for a given T11..

InInState 1State 1


AA22 >> A>> A11


First Law - TurbinesFirst Law - TurbinesTurbines

Energy balance:

PurposePurpose: Produce Work: Produce Work

Mass balance:


−+− ( ) kW ,pekehm ∆∆∆ ++=•

outW000•

−+− ( ) kW ,00hhm 12 ++−=•

= h2 – h1+ ke2 – ke1+ pe2 – pe1qin – qout+ ωin - ωout

EffectEffect: Pressure Drops: Pressure Drops

= h2 – h1, kJ/kg– ωout

In

Out= h1 – h2, kJ/kgωout

kg/s , 2

22

1

11

νυ

νυ AA rr

=21 ••

= mm


First Law - CompressorsFirst Law - Compressors

Energy balance:Mass balance:


−+− ( ) kW ,pekehm ∆∆∆ ++=•

Compressors

PurposePurpose: Increase Pressure: Increase Pressure

inW000•

++− ( ) kW ,00hhm 12 ++−=•

SacrificeSacrifice: Work supplied: Work supplied

= h2 – h1, kJ/kgωin

Out

In


kg/s , 2

22

1

11

νυ

νυ AA rr

=21 ••

= mm


First Law – Increase PressureFirst Law – Increase PressurePressure Increasing

Devices

CompressorsCompressors: Increase Pressure of : Increase Pressure of gas to high Pto high P

SacrificeSacrifice: Work supplied: Work supplied

FansFans: Increase Pressure of : Increase Pressure of gas slightlygas slightly to move airto move air

PumpsPumps: Increase Pressure of : Increase Pressure of liquid to high Pto high P


First Law - ThrottleFirst Law - ThrottleThrottle

8/22/2005



First Law - ThrottleFirst Law - Throttle

Energy balance:Mass balance:


−+− ( ) kW ,pekehm ∆∆∆ ++=•

Throttle

PurposePurpose: Reduce Pressure: Reduce Pressure

0000 −+−

kJ/kg ,hh 12 =

EffectEffect: Temp drops: Temp drops

isenthalpic processkJ/kg ,00hh 12 ++−=


In Out

kg/s , 2

22

1

11

νυ

νυ AA rr

=21 ••

= mm


First Law – Mixing ChamberFirst Law – Mixing Chamber

Energy balance:

kg/s ,mmm 321•••

=+Mass balance:

( )33321 )( pekehmm +++••

( ) ( ) kW ,11112222 pekehmpekehm +++++=••

= θ3 – θ1 - θ2, kJ/kgqnet,in – ωnet,out

Mixing Chamber

PurposePurpose: Mixing : Mixing

1

3

kW ),()( 311232 hhmhhm −=−•• 2

;03123 ≅−=− kekekeke 03123 ≅−=− pepepepe

Then,


First Law – Heat ExchangerFirst Law – Heat Exchanger

Heat Exchanger

Boundary has2 inlets and

2 exits


First Law – Heat ExchangerFirst Law – Heat Exchanger

Heat Exchanger

Boundary has1 inlet and

1 exit


First Law of ThermodynamicsFirst Law of ThermodynamicsHeat Exchanger

-no mixing-1 inlet and 1 exit

In, 1 Exit, 2

In, 3

Exit, 4



-no mixing-2 inlets and 2 exits

In, 1 Exit, 2

In, 3

Exit, 4

8/22/2005




1 2

4

3Case 1

1234 --0 ϑϑϑϑ +=

kg/s ,mm ,mm 4321••••

==8/22/2005 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005 38


1 2

4

3

Case 2

12in,net -0Q ϑϑ=+

kg/s ,mm 21••

=

kg/s ,mm 43••

=Qout


1 2

4

3

First Law of ThermodynamicsFirst Law of Thermodynamics

Energy balance: Case 1

kg/s ,mm ,mm 4321••••

==Mass balance:

( )44440000 pekehm ++=−+−•

( )3333 - pekehm ++•

Heat Exchanger

PurposePurpose: Remove or add heat : Remove or add heat

( )1212121 pepekekehhm −+−+−•

( ) ( ) kW ,11112222 pekehmpekehm ++−+++••

( )3434343 pepekekehhm −+−+−=•

0≅∆=∆ pekewhere8/22/2005 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005 40

1 2

4

3


Energy balance: Case 2kg/s ,mm ,mm 4321

••••==

Mass balance:

Heat Exchanger


kW ,hmhm00QQ 1122outin

••••

−=−+−

( ) kW ,hhm00Q 122out −=−+−••

( ) ( )1

43312

m

hhmhh •

•−

=−whereQout


1 2

4

3


Energy balance: Case 2kg/s ,mm ,mm 4321

••••==

Mass balance:

Heat Exchanger


kW ,hmhm00QQ 1122outin

••••−=−+−

( ) kW ,hhm000Q 122in −=−+−••

( ) ( )1

43312

m

hhmhh •

•−

=−where

Qin


40 studs. AC 5kW. Qperson=360 kJ/hr. 10 light bulbs, 100 W each. Qin = 15,000 kJ/hr. Need to maintain room at 21°C. How many AC require?

Example: Q7Example: Q7

40 guys x 360kJ/hr x1hr/3600s = 4 kW

10 x 100 W = 1kW

Qin=15,000 kJ/hr= 4.1 kW

Qincrease= 1+4.1+4=9.1 kWP=5kW

P=5kW

8/22/2005



Radiator V=20 L, phase superheated vapor 300 kPa, 250°C. Closed system. Find Qout when P=100 kPa, show on phase diagram.


Initial state: State 1V = 20 L x 10-3 m3 /L10-3 L = 1cm3 = (10-2)3 = 10-6

m3

P=300 kPa, 250°C, superheated vapor. u1 = 2728.7 kJ/kg

ν = 0.7964 m3/kg.

m=V/ ν =0.02m3/0.7964m3/kg

Final state: State 2

P=100 kPa, T?? Phase ??

ν= 0.7964 m3/kg.

νf<ν< νg, wet mix.

x = (ν- νf )/(νg - νf)=0.47

0 – qout = ∆u. Find quality x then use it to get u2=uf+xufg


Radiator V=20 L, phase superheated vapor 300 kPa, 250 °C. Closed system. Find Qout when P=100 kPa, show on phase diagram.


u1 = 2728.7 kJ/kg

m=V/ ν =0.02m3/0.7964m3/kg

u2=uf+xufg = 1399.049 kJ/kg

-qout = u2 – u1 = -1329.7 kJ/kg

Qout = m ∆u =

Final state: State 2

P=100 kPa, T?? Phase ??

ν= 0.7964 m3/kg.

νf<ν< νg, wet mix.

x = (ν- νf )/(νg - νf)=0.47

0 – qout = ∆u. Find quality x then use it to get u2=uf+xufg


“The illiterate of the 21st century will not be those who cannot read and write, but those who cannot learn, unlearn, and relearn.”

Author: Alvin TofflerSource: Lessons from the Art of Juggling; How to Achieve Your Full Potential in Business, Learning and Life by Michael Gelb and Tony Buzan

Concluding remarks Concluding remarks -- QuotesQuotes

knowing me knowing you

Students in Thermodynamics Class at Fac. of App. Sciences, UiTM. Nov01-Mar02

Students in Thermodynamics Class Fac. of App. Sciences, UiTM. Nov01-Mar02

4zack.jpg

afzan.jpg

aminah.jpg

azhan.jpg

azman.jpg

azrul.jpg

Page: 123456789101112131415161718192021

[ Prev ] [ Next ]

Back to Dr. J.J.'s Page

file:///G|/FSG/DRJJ/AAN-DRJJ-2011/THERMO/KNOWING/page_01.htm22/01/2011 5:33:20



azshairniz.jpg

azwani.jpg

azwani1.jpg

bakar.jpg

bakar2.jpg

faizatul farah.jpg

Page: 12 3456789101112131415161718192021

[ Prev ] [ Next ]





farah liyana.jpg

farah liyana2.jpg

fauzi.jpg

Hafizah.jpg

halim.jpg

halim2.jpg

Page: 123 456789101112131415161718192021

[ Prev ] [ Next ]





hanani.jpg

harlisa.jpg

harlisa2.jpg

hasan.jpg

herzy.jpg

kaidil.jpg

Page: 1234 56789101112131415161718192021

[ Prev ] [ Next ]





Lizman.jpg

maslia.jpg

mastura.jpg

mhalim.jpg

mjufry.jpg

mrafie.jpg

Page: 12345 6789101112131415161718192021

[ Prev ] [ Next ]





mshah.jpg

mshaifudin.jpg

nazeera.jpg

nizab.jpg

nooraslina.jpg

noorazril.jpg

Page: 123456 789101112131415161718192021

[ Prev ] [ Next ]





norafida.jpg

norahjelita.jpg

norasmah.jpg

norasmah2.jpg

norazaini.jpg

norazam.jpg

Page: 1234567 89101112131415161718192021

[ Prev ] [ Next ]





norazimah.jpg

norhafizah.jpg

norhasikin.jpg

norhayatie.jpg

norijah.jpg

normeliza.jpg

Page: 12345678 9101112131415161718192021

[ Prev ] [ Next ]





norshuhada.jpg

norsyuhada.jpg

norzakiah.jpg

norzatul.jpg

nur waheeda.jpg

nur waheeda2.jpg

Page: 123456789 101112131415161718192021

[ Prev ] [ Next ]





nuraziah.jpg

nurulain.jpg

nurzatul itri.jpg

nurzatul itri2.jpg

rabiatul.jpg

rabiatulpejam.jpg

Page: 12345678910 1112131415161718192021

[ Prev ] [ Next ]





rahida wati.jpg

rajamhairi.jpg

rashidah.jpg

roseni.jpg

roseyanti.jpg

roseyanti2.jpg

Page: 1234567891011 12131415161718192021

[ Prev ] [ Next ]





rozelawati.jpg

rulina.jpg

sabarina.jpg

salwana.jpg

sarida.jpg

shaffuddin.jpg

Page: 123456789101112 131415161718192021

[ Prev ] [ Next ]





shahir2.jpg

shaiful irwan.jpg

shaiful.jpg

siti hafidah.jpg

siti hamidah.jpg

siti khadijah.jpg

Page: 12345678910111213 1415161718192021

[ Prev ] [ Next ]





siti khadijah2.jpg

siti naziah.jpg

sitiaziah.jpg

sitiaziah2.jpg

sitinaziah.jpg

snoraishah.jpg

Page: 1234567891011121314 15161718192021

[ Prev ] [ Next ]





suhana.jpg

syaidatun.jpg

thermo2 001.jpg

thermo2 002.jpg

thermo2 003.jpg

thermo2 004.jpg

Page: 123456789101112131415 161718192021

[ Prev ] [ Next ]





thermo2 005.jpg

thermo2 006.jpg

thermo2 007.jpg

thermo2 008.jpg

thermo2 009.jpg

thermo2 010.jpg

Page: 12345678910111213141516 1718192021

[ Prev ] [ Next ]





thermo2 011.jpg

thermo2 012.jpg

thermo2 013.jpg

thermo2 014.jpg

thermo2 016.jpg

thermo2 017.jpg

Page: 1234567891011121314151617 18192021

[ Prev ] [ Next ]





thermo2 018.jpg

thermo2 019.jpg

thermo2 020.jpg

thermo2 021.jpg

thermo2 022.jpg

thermo2 023.jpg

Page: 123456789101112131415161718 192021

[ Prev ] [ Next ]





thermo2 024.jpg

thermo2 025.jpg

thermo2 026.jpg

thermo2 027.jpg

thermo2 028.jpg

thermo2 029.jpg

Page: 12345678910111213141516171819 2021

[ Prev ] [ Next ]





thermo2 030.jpg

thermo2 031.jpg

thermo2 032.jpg

thermo2 033.jpg

thermo2 061.jpg

zikri.jpg

Page: 1234567891011121314151617181920 21

[ Prev ] [ Next ]





zool fahmi.jpg

zool fahmi2.jpg

zuaidah.jpg

zuaidah2.jpg

zulkifli.jpg

Page: 123456789101112131415161718192021

[ Prev ] [ Next ]



eviDences foR classRoom innovations, leaRning gains ...€¦ · Rankine Cycle Gas Mixtures...

Documents

Transcript of eviDences foR classRoom innovations, leaRning gains ...€¦ · Rankine Cycle Gas Mixtures...