b) Wat is die buite-lug temperatuur by die

Size: px
Start display at page:

Download "b) Wat is die buite-lug temperatuur by die"

Transcription

1 Voorlttrs n Van MTX 0 KLASTOETS 6A / CLASSTES 6A 5 mn. (5 punt/marks) *Sal hrwrk tot 0 punt vr klaslys dolnds / Wll b workd back to 0 marks for class lst purposs Voorlttrs n Van MTX 0 KLASTOETS 6B / CLASSTES 6B 5 mn. (5 punt/marks) *Sal hrwrk tot 0 punt vr klaslys dolnds / Wll b workd back to 0 marks for class lst purposs as n httpomp wat htt ut d hus onttrk. In d wntr ontvang d kondnsator huslug tn m /s n vrht dt van T = 0ºC tot T = 0ºC trwyl d lug dur d kondnsator vlo. D vrdampr ontvang butlug tn T = 5ºC, tn kg/s. D komprssor word aangdryf dur n kw motor. a) Brkn d d β ' van d httpomp utlaat van d vrdampr? (T ) htt kan ut d hus onttrk word n d somr? In th wntr th condnsr rcvs hous-ar at m /s and hats t from T = 0ºC to T = 0ºC as th ar flows through th condnsr. Th vaporator rcvs xtrnal ar at T = 5ºC, at kg/s. Th comprssor s powrd by a kw motor. a) Dtrmn β ' of th hat pump. th xt of th vaporator? (T ) hat can b xtractd from th hous n th summr? as n httpomp wat htt ut d hus onttrk. In d wntr ontvang d kondnsator huslug tn 0.9 m /s n vrht dt van T = 0ºC tot T = 5ºC trwyl d lug dur d kondnsator vlo. D vrdampr ontvang butlug tn T = 7ºC, tn. kg/s. D komprssor word aangdryf dur n kw motor. a) Brkn d β ' van d httpomp utlaat van d vrdampr? (T ) htt kan ut d hus onttrk word n d somr? In th wntr th condnsr rcvs hous-ar at 0.9 m /s and hats t from T = 0ºC to T = 5ºC as th ar flows through th condnsr. Th vaporator rcvs xtrnal ar at T = 7ºC, at. kg/s. Th comprssor s powrd by a kw motor. a) Dtrmn β ' of th hat pump. th xt of th vaporator? (T ) hat can b xtractd from th hous n th summr? P = 00 kpa R = 87 J/kgK C p0 = 00 J/kgK T =0ºC P = 00 kpa R = 87 J/kgK C p0 = 00 J/kgK T =5ºC T = 5ºC T =0ºC T = 7ºC T =0ºC

2 Voorlttrs n Van MTX 0 KLASTOETS 6C / CLASSTES 6C 5 mn. (5 punt/marks) *Sal hrwrk tot 0 punt vr klaslys dolnds / Wll b workd back to 0 marks for class lst purposs Voorlttrs n Van MTX 0 KLASTOETS 6D / CLASSTES 6D 5 mn. (5 punt/marks) *Sal hrwrk tot 0 punt vr klaslys dolnds / Wll b workd back to 0 marks for class lst purposs as n httpomp wat htt ut d hus onttrk. In d somr ontvang d vrdampr hus-lug tn m /s n vrkol dt van T = 5ºC tot T = 8ºC trwyl d lug dur d vrdampr vlo. D kondnsator ontvang butlug tn T = 5ºC, tn kg/s. D komprssor word aangdryf dur n kw motor. a) Brkn d β van d stlsl. utlaat van d kondnsator? (T ) htt kan aan d hus glwr word n d wntr? In th summr th vaporator rcvs housar at m /s and cools t from T = 5ºC to T = 8ºC as th ar flows through th vaporator. Th condnsr rcvs xtrnal ar at T = 5ºC, at kg/s. Th comprssor s powrd by a kw motor. a) Dtrmn β of th systm. th xt of th condnsr? (T ) hat can b dlvrd to th hous n th wntr? as n httpomp wat htt ut d hus onttrk. In d somr ontvang d vrdampr hus-lug tn 0.5 m /s n vrkol dt van T = ºC tot T = 8ºC trwyl d lug dur d vrdampr vlo. D kondnsator ontvang butlug tn T = 5ºC, tn. kg/s. D komprssor word aangdryf dur n.8 kw motor. a) Brkn d β van d stlsl utlaat van d kondnsator? (T ) htt kan aan d hus glwr word n d wntr? In th summr th vaporator rcvs housar at 0.5 m /s and cools t from T = ºC to T = 8ºC as th ar flows through th vaporator. Th condnsr rcvs xtrnal ar at T = 5ºC, at. kg/s. Th comprssor s powrd by a.8 kw motor. a) Dtrmn β of th systm. th xt of th condnsr? (T ) hat can b dlvrd to th hous n th wntr? P = 00 kpa R = 87 J/kgK C p0 = 00 J/kgK T =8ºC P = 00 kpa R = 87 J/kgK C p0 = 00 J/kgK T =8ºC T =5ºC T =ºC T =5ºC T =5ºC

3 Utrksl ut Formul Blad / Extract from Formula Sht: φ = φ f + xφ fg Pv = ZRT H = U + PV δ W = PdV PV n = const. u h Cv = C p = T v T p W Q η = β = L / H Q H W TdS = du + PdV TdS = dh VdP mv E = U + + mgz dm = m& m& = 0 dt δ Q = δw mv mv Q + U + + mgz = U + + mgz + W V de V Q & + m& ( h + + gz ) = + m& ( h + + gz + W& dt )

4

5

6

7

φ = φ n Kompressor neem 0.5 m 3 /s CO2 in teen 100 kpa en 26.85ºC. Die kompressor is adiabaties en ontvang 10 kw drywing.

φ = φ n Kompressor neem 0.5 m 3 /s CO2 in teen 100 kpa en 26.85ºC. Die kompressor is adiabaties en ontvang 10 kw drywing. v v Voorlttrs n Van Intals and Surnam Studnt nommr Studnt numbr Datum / Dat MX 0 KLASOES 5A / CLASSES 5A 5 mn. (5 unt/marks) *Sal hrwrk tot 0 unt vr klaslys dolnds * Wll b workd back to 0 marks for class

More information

Entropy Equation for a Control Volume

Entropy Equation for a Control Volume Fudamtals of Thrmodyamcs Chaptr 7 Etropy Equato for a Cotrol Volum Prof. Syoug Jog Thrmodyamcs I MEE2022-02 Thrmal Egrg Lab. 2 Q ds Srr T Q S2 S1 1 Q S S2 S1 Srr T t t T t S S s m 1 2 t S S s m tt S S

More information

1. (10) Calorically perfect ideal air at 300 K, 100 kpa, 1000 m/s, is brought to rest isentropically. Determine its final temperature.

1. (10) Calorically perfect ideal air at 300 K, 100 kpa, 1000 m/s, is brought to rest isentropically. Determine its final temperature. AME 5053 Intermediate Thermodynamics Examination Prof J M Powers 30 September 0 0 Calorically perfect ideal air at 300 K, 00 kpa, 000 m/s, is brought to rest isentropically Determine its final temperature

More information

Voorletters en Van Initials and Surname Studente nommer Student number Datum / Date

Voorletters en Van Initials and Surname Studente nommer Student number Datum / Date Voorletters en Van Initials and Surname Studente nommer Student number Datum / Date MTX221 2011 KLASTOETS 3E / CLASSTES 3E 25 min. (14 punte/marks) *Sal herwerk tot 10 punte vir klaslys doeleindes * Will

More information

c Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME 6101 (2017) 2 / 25 T145 = Q + W cv + i h 2 = h (V2 1 V 2 2)

c Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME 6101 (2017) 2 / 25 T145 = Q + W cv + i h 2 = h (V2 1 V 2 2) Thermodynamic Processes & Isentropic Efficiency Dr. Md. Zahurul Haq Professor Department of Mechanical Engineering Bangladesh University of Engineering & Technology (BUET Dhaka-1000, Bangladesh zahurul@me.buet.ac.bd

More information

Chapter 3 First Law of Thermodynamics and Energy Equation

Chapter 3 First Law of Thermodynamics and Energy Equation Fundamentals of Thermodynamics Chapter 3 First Law of Thermodynamics and Energy Equation Prof. Siyoung Jeong Thermodynamics I MEE0-0 Spring 04 Thermal Engineering Lab. 3. The energy equation Thermal Engineering

More information

Consequences of Second Law of Thermodynamics. Entropy. Clausius Inequity

Consequences of Second Law of Thermodynamics. Entropy. Clausius Inequity onsequences of Second Law of hermodynamics Dr. Md. Zahurul Haq Professor Department of Mechanical Engineering Bangladesh University of Engineering & echnology BUE Dhaka-000, Bangladesh zahurul@me.buet.ac.bd

More information

Thermodynamics is the Science of Energy and Entropy

Thermodynamics is the Science of Energy and Entropy Definition of Thermodynamics: Thermodynamics is the Science of Energy and Entropy - Some definitions. - The zeroth law. - Properties of pure substances. - Ideal gas law. - Entropy and the second law. Some

More information

T098. c Dr. Md. Zahurul Haq (BUET) First Law of Thermodynamics ME 201 (2012) 2 / 26

T098. c Dr. Md. Zahurul Haq (BUET) First Law of Thermodynamics ME 201 (2012) 2 / 26 Conservation of Energy for a Closed System Dr. Md. Zahurul Haq Professor Department of Mechanical Engineering Bangladesh University of Engineering & Technology (BUET Dhaka-, Bangladesh zahurul@me.buet.ac.bd

More information

AE1104 Physics 1. List of equations. Made by: E. Bruins Slot

AE1104 Physics 1. List of equations. Made by: E. Bruins Slot i AE04 Physics List of equations Made by: E. Bruins Slot Chapter Introduction and basic concepts Newton s second law Weight F = M a (N) W = m g J = N m (N) Density Specific volume ρ = m V m 3 v = V m =

More information

Some Useful Formulae

Some Useful Formulae ME - hrmodynamcs I Som Usful Formula Control Mass Contnuty Equaton m constant Frst Law Comprsson-xpanson wor U U m V V mg Z Z Q W For polytropc procs, PV n c, Scond Law W W PdV P V P V n n P V ln V V n

More information

Chapter 4. Energy Analysis of Closed Systems

Chapter 4. Energy Analysis of Closed Systems Chapter 4 Energy Analysis of Closed Systems The first law of thermodynamics is an expression of the conservation of energy principle. Energy can cross the boundaries of a closed system in the form of heat

More information

Chapter 7. Entropy: A Measure of Disorder

Chapter 7. Entropy: A Measure of Disorder Chapter 7 Entropy: A Measure of Disorder Entropy and the Clausius Inequality The second law of thermodynamics leads to the definition of a new property called entropy, a quantitative measure of microscopic

More information

Entropy and the Second Law of Thermodynamics

Entropy and the Second Law of Thermodynamics Entropy and the Second Law of hermodynamics Reading Problems 6-, 6-2, 6-7, 6-8, 6-6-8, 6-87, 7-7-0, 7-2, 7-3 7-39, 7-46, 7-6, 7-89, 7-, 7-22, 7-24, 7-30, 7-55, 7-58 Why do we need another law in thermodynamics?

More information

Dishwasher. Heater. Homework Solutions ME Thermodynamics I Spring HW-1 (25 points)

Dishwasher. Heater. Homework Solutions ME Thermodynamics I Spring HW-1 (25 points) HW-1 (25 points) (a) Given: 1 for writing given, find, EFD, etc., Schematic of a household piping system Find: Identify system and location on the system boundary where the system interacts with the environment

More information

King Fahd University of Petroleum & Minerals

King Fahd University of Petroleum & Minerals King Fahd University of Petroleum & Minerals Mechanical Engineering Thermodynamics ME 04 BY Dr. Haitham Bahaidarah My Office Office Hours: :00 0:00 am SMW 03:00 04:00 pm UT Location: Building Room # 5.4

More information

ME Thermodynamics I

ME Thermodynamics I HW-03 (25 points) i) Given: for writing Given, Find, Basic equations Rigid tank containing nitrogen gas in two sections initially separated by a membrane. Find: Initial density (kg/m3) of nitrogen gas

More information

Consequences of Second Law of Thermodynamics. Entropy. Clausius Inequity

Consequences of Second Law of Thermodynamics. Entropy. Clausius Inequity onsequences of Second Law of hermodynamics Dr. Md. Zahurul Haq Professor Department of Mechanical Engineering Bangladesh University of Engineering & echnology BUE Dhaka-000, Bangladesh zahurul@me.buet.ac.bd

More information

The First Law of Thermodynamics. By: Yidnekachew Messele

The First Law of Thermodynamics. By: Yidnekachew Messele The First Law of Thermodynamics By: Yidnekachew Messele It is the law that relates the various forms of energies for system of different types. It is simply the expression of the conservation of energy

More information

Eng Thermodynamics I conservation of mass; 2. conservation of energy (1st Law of Thermodynamics); and 3. the 2nd Law of Thermodynamics.

Eng Thermodynamics I conservation of mass; 2. conservation of energy (1st Law of Thermodynamics); and 3. the 2nd Law of Thermodynamics. Eng3901 - Thermodynamics I 1 1 Introduction 1.1 Thermodynamics Thermodynamics is the study of the relationships between heat transfer, work interactions, kinetic and potential energies, and the properties

More information

Section 2: Lecture 1 Integral Form of the Conservation Equations for Compressible Flow

Section 2: Lecture 1 Integral Form of the Conservation Equations for Compressible Flow Section 2: Lecture 1 Integral Form of the Conservation Equations for Compressible Flow Anderson: Chapter 2 pp. 41-54 1 Equation of State: Section 1 Review p = R g T " > R g = R u M w - R u = 8314.4126

More information

IV. First Law of Thermodynamics. Cooler. IV. First Law of Thermodynamics

IV. First Law of Thermodynamics. Cooler. IV. First Law of Thermodynamics D. Applcatons to stady flow dvcs. Hat xchangrs - xampl: Clkr coolr for cmnt kln Scondary ar 50 C, 57,000 lbm/h Clkr? C, 5 ton/h Coolr Clkr 400 C, 5 ton/h Scondary ar 0 C, 57,000 lbm/h a. Assumptons. changs

More information

Lecture 44: Review Thermodynamics I

Lecture 44: Review Thermodynamics I ME 00 Thermodynamics I Lecture 44: Review Thermodynamics I Yong Li Shanghai Jiao Tong University Institute of Refrigeration and Cryogenics 800 Dong Chuan Road Shanghai, 0040, P. R. China Email : liyo@sjtu.edu.cn

More information

c Dr. Md. Zahurul Haq (BUET) Entropy ME 203 (2017) 2 / 27 T037

c Dr. Md. Zahurul Haq (BUET) Entropy ME 203 (2017) 2 / 27 T037 onsequences of Second Law of hermodynamics Dr. Md. Zahurul Haq Professor Department of Mechanical Engineering Bangladesh University of Engineering & echnology BUE Dhaka-000, Bangladesh zahurul@me.buet.ac.bd

More information

Dr Ali Jawarneh. Hashemite University

Dr Ali Jawarneh. Hashemite University Dr Ali Jawarneh Department of Mechanical Engineering Hashemite University Examine the moving boundary work or P d work commonly encountered in reciprocating devices such as automotive engines and compressors.

More information

Engineering Thermodynamics. Chapter 6. Entropy: a measure of Disorder 6.1 Introduction

Engineering Thermodynamics. Chapter 6. Entropy: a measure of Disorder 6.1 Introduction Engineering hermodynamics AAi Chapter 6 Entropy: a measure of Disorder 6. Introduction he second law of thermodynamics leads to the definition of a new property called entropy, a quantitative measure of

More information

Chapter 5: The First Law of Thermodynamics: Closed Systems

Chapter 5: The First Law of Thermodynamics: Closed Systems Chapter 5: The First Law of Thermodynamics: Closed Systems The first law of thermodynamics can be simply stated as follows: during an interaction between a system and its surroundings, the amount of energy

More information

Chapter 7. Entropy. by Asst.Prof. Dr.Woranee Paengjuntuek and Asst. Prof. Dr.Worarattana Pattaraprakorn

Chapter 7. Entropy. by Asst.Prof. Dr.Woranee Paengjuntuek and Asst. Prof. Dr.Worarattana Pattaraprakorn Chapter 7 Entropy by Asst.Prof. Dr.Woranee Paengjuntuek and Asst. Prof. Dr.Worarattana Pattaraprakorn Reference: Cengel, Yunus A. and Michael A. Boles, Thermodynamics: An Engineering Approach, 5th ed.,

More information

MAE 110A. Homework 3: Solutions 10/20/2017

MAE 110A. Homework 3: Solutions 10/20/2017 MAE 110A Homework 3: Solutions 10/20/2017 3.10: For H 2O, determine the specified property at the indicated state. Locate the state on a sketch of the T-v diagram. Given a) T 140 C, v 0.5 m 3 kg b) p 30MPa,

More information

Thermodynamics II. Week 9

Thermodynamics II. Week 9 hermodynamics II Week 9 Example Oxygen gas in a piston cylinder at 300K, 00 kpa with volume o. m 3 is compressed in a reversible adiabatic process to a final temperature of 700K. Find the final pressure

More information

MAE 110A. Homework 6: Solutions 11/9/2017

MAE 110A. Homework 6: Solutions 11/9/2017 MAE 110A Hoework 6: Solutions 11/9/2017 H6.1: Two kg of H2O contained in a piston-cylinder assebly, initially at 1.0 bar and 140 C undergoes an internally ersible, isotheral copression to 25 bar. Given

More information

Thermodynamics Free E and Phase D. J.D. Price

Thermodynamics Free E and Phase D. J.D. Price Thermodynamics Free E and Phase D J.D. Price Force - the acceleration of matter (N, kg m/s 2 ) Pressure (P)( ) - a force applied over an area (N/m 2 ) Work (W) - force multiplied by distance (kg( m 2 /s

More information

UNIVERSITY OF SOUTHAMPTON VERY IMPORTANT NOTE. Section A answers MUST BE in a separate blue answer book. If any blue

UNIVERSITY OF SOUTHAMPTON VERY IMPORTANT NOTE. Section A answers MUST BE in a separate blue answer book. If any blue UNIVERSITY OF SOUTHAMPTON PHYS1013W1 SEMESTER 2 EXAMINATION 2011/12 ENERGY AND MATTER SOLUTIONS Duration: 120 MINS VERY IMPORTANT NOTE Section A answers MUST BE in a separate blue answer book. If any blue

More information

Chapter 3. Property Relations The essence of macroscopic thermodynamics Dependence of U, H, S, G, and F on T, P, V, etc.

Chapter 3. Property Relations The essence of macroscopic thermodynamics Dependence of U, H, S, G, and F on T, P, V, etc. Chapter 3 Property Relations The essence of macroscopic thermodynamics Dependence of U, H, S, G, and F on T, P, V, etc. Concepts Energy functions F and G Chemical potential, µ Partial Molar properties

More information

first law of ThermodyNamics

first law of ThermodyNamics first law of ThermodyNamics First law of thermodynamics - Principle of conservation of energy - Energy can be neither created nor destroyed Basic statement When any closed system is taken through a cycle,

More information

CHAPTER 8 ENTROPY. Blank

CHAPTER 8 ENTROPY. Blank CHAPER 8 ENROPY Blank SONNAG/BORGNAKKE SUDY PROBLEM 8-8. A heat engine efficiency from the inequality of Clausius Consider an actual heat engine with efficiency of η working between reservoirs at and L.

More information

First Law of Thermodynamics

First Law of Thermodynamics First Law of Thermodynamics During an interaction between a system and its surroundings, the amount of energy gained by the system must be exactly equal to the amount of energy lost by the surroundings.

More information

Chapter 18 Heat and the First Law of Thermodynamics

Chapter 18 Heat and the First Law of Thermodynamics Chapter 18 Heat and the First Law of Thermodynamics Heat is the transfer of energy due to the difference in temperature. The internal energy is the total energy of the object in its centerofmass reference

More information

MAE 11. Homework 8: Solutions 11/30/2018

MAE 11. Homework 8: Solutions 11/30/2018 MAE 11 Homework 8: Solutions 11/30/2018 MAE 11 Fall 2018 HW #8 Due: Friday, November 30 (beginning of class at 12:00p) Requirements:: Include T s diagram for all cycles. Also include p v diagrams for Ch

More information

Entropy balance special forms. Quasiequilibrium (QE) process. QE process is reversible. dt Tk = = +

Entropy balance special forms. Quasiequilibrium (QE) process. QE process is reversible. dt Tk = = + Entropy balance Outline Closed systems Open systems Reversible steady flow wor Minimizing compressor wor Isentropic efficiencies Examples Entropy balance Sin Sout + Sgen = Ssys Entropy balance Entropy

More information

+ m B1 = 1. u A1. u B1. - m B1 = V A. /v A = , u B1 + V B. = 5.5 kg => = V tot. Table B.1.

+ m B1 = 1. u A1. u B1. - m B1 = V A. /v A = , u B1 + V B. = 5.5 kg => = V tot. Table B.1. 5.6 A rigid tank is divided into two rooms by a membrane, both containing water, shown in Fig. P5.6. Room A is at 200 kpa, v = 0.5 m3/kg, VA = m3, and room B contains 3.5 kg at 0.5 MPa, 400 C. The membrane

More information

Chapter 6. Using Entropy

Chapter 6. Using Entropy Chapter 6 Using Entropy Learning Outcomes Demonstrate understanding of key concepts related to entropy and the second law... including entropy transfer, entropy production, and the increase in entropy

More information

Today lecture. 1. Entropy change in an isolated system 2. Exergy

Today lecture. 1. Entropy change in an isolated system 2. Exergy Today lecture 1. Entropy change in an isolated system. Exergy - What is exergy? - Reversible Work & Irreversibility - Second-Law Efficiency - Exergy change of a system For a fixed mass For a flow stream

More information

ENT 254: Applied Thermodynamics

ENT 254: Applied Thermodynamics ENT 54: Applied Thermodynamics Mr. Azizul bin Mohamad Mechanical Engineering Program School of Mechatronic Engineering Universiti Malaysia Perlis (UniMAP) azizul@unimap.edu.my 019-4747351 04-9798679 Chapter

More information

du = δq + δw = δq rev + δw rev = δq rev + 0

du = δq + δw = δq rev + δw rev = δq rev + 0 Chem 4501 Introduction to hermodynamics, 3 Credits Kinetics, and Statistical Mechanics Module Number 6 Active Learning Answers and Optional Problems/Solutions 1. McQuarrie and Simon, 6-6. Paraphrase: Compute

More information

TEST DATA ON COPPER MICRO-CHANNEL HEAT SINKS

TEST DATA ON COPPER MICRO-CHANNEL HEAT SINKS TEST DATA ON COPPER MICRO-CHANNEL HEAT SINKS By: Ralph Webb and Hasan Nasir State College, PA 16801 www.omega.com inemi Liquid Cooling Symposium San Diego, CA May 31, 2006 Objective Test results on low

More information

Lecture 4 Clausius Inequality

Lecture 4 Clausius Inequality Lecture 4 Clausius Inequality Entropy distinguishes between irreversible and reversible processes. irrev S > 0 rev In a spontaneous process, there should be a net increase in the entropy of the system

More information

THERMODYNAMICS, FLUID AND PLANT PROCESSES. The tutorials are drawn from other subjects so the solutions are identified by the appropriate tutorial.

THERMODYNAMICS, FLUID AND PLANT PROCESSES. The tutorials are drawn from other subjects so the solutions are identified by the appropriate tutorial. THERMODYNAMICS, FLUID AND PLANT PROCESSES The tutorials are drawn from other subjects so the solutions are identified by the appropriate tutorial. THERMODYNAMICS TUTORIAL 2 THERMODYNAMIC PRINCIPLES SAE

More information

Lecture 38: Vapor-compression refrigeration systems

Lecture 38: Vapor-compression refrigeration systems ME 200 Termodynamics I Lecture 38: Vapor-compression refrigeration systems Yong Li Sangai Jiao Tong University Institute of Refrigeration and Cryogenics 800 Dong Cuan Road Sangai, 200240, P. R. Cina Email

More information

ENERGY TRANSFER BETWEEN FLUID AND ROTOR. Dr. Ir. Harinaldi, M.Eng Mechanical Engineering Department Faculty of Engineering University of Indonesia

ENERGY TRANSFER BETWEEN FLUID AND ROTOR. Dr. Ir. Harinaldi, M.Eng Mechanical Engineering Department Faculty of Engineering University of Indonesia ENERGY TRANSFER BETWEEN FLUID AND ROTOR Dr. Ir. Harinaldi, M.Eng Mechanical Engineering Department Faculty of Engineering University of Indonesia Basic Laws and Equations Continuity Equation m m ρ mass

More information

Outline. Property diagrams involving entropy. Heat transfer for internally reversible process

Outline. Property diagrams involving entropy. Heat transfer for internally reversible process Outline roperty diagrams involving entropy What is entropy? T-ds relations Entropy change of substances ure substances (near wet dome) Solids and liquids Ideal gases roperty diagrams involving entropy

More information

8.21 The Physics of Energy Fall 2009

8.21 The Physics of Energy Fall 2009 MIT OpenCourseWare http://ocw.mit.edu 8.21 The Physics of Energy Fall 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. 8.21 Lecture 9 Heat Engines

More information

Thermodynamics Introduction and Basic Concepts

Thermodynamics Introduction and Basic Concepts Thermodynamics Introduction and Basic Concepts by Asst. Prof. Channarong Asavatesanupap Mechanical Engineering Department Faculty of Engineering Thammasat University 2 What is Thermodynamics? Thermodynamics

More information

Lecture 4 Clausius Inequality

Lecture 4 Clausius Inequality Lecture 4 Clausius Inequality We know: Heat flows from higher temperature to lower temperature. T A V A U A + U B = constant V A, V B constant S = S A + S B T B V B Diathermic The wall insulating, impermeable

More information

Classical thermodynamics

Classical thermodynamics Classical thermodynamics More about irreversibility chap. 6 Isentropic expansion of an ideal gas Sudden expansion of a gas into vacuum cf Kittel and Kroemer end of Cyclic engines cf Kittel and Kroemer

More information

ANALYSIS: The mass rate balance for the one-inlet, one-exit control volume at steady state is

ANALYSIS: The mass rate balance for the one-inlet, one-exit control volume at steady state is Problm 4.47 Fgur P4.47 provds stady stat opratng data for a pump drawng watr from a rsrvor and dlvrng t at a prssur of 3 bar to a storag tank prchd 5 m abov th rsrvor. Th powr nput to th pump s 0.5 kw.

More information

Chapter 1: Basic Definitions, Terminologies and Concepts

Chapter 1: Basic Definitions, Terminologies and Concepts Chapter : Basic Definitions, Terminologies and Concepts ---------------------------------------. UThermodynamics:U It is a basic science that deals with: -. Energy transformation from one form to another..

More information

Lecture 4. Differential Analysis of Fluid Flow Navier-Stockes equation

Lecture 4. Differential Analysis of Fluid Flow Navier-Stockes equation Lecture 4 Differential Analysis of Fluid Flow Navier-Stockes equation Newton second law and conservation of momentum & momentum-of-momentum A jet of fluid deflected by an object puts a force on the object.

More information

Relationship to Thermodynamics. Chapter One Section 1.3

Relationship to Thermodynamics. Chapter One Section 1.3 Relationship to Thermodynamics Chapter One Section 1.3 Alternative Formulations Alternative Formulations Time Basis: CONSERVATION OF ENERGY (FIRST LAW OF THERMODYNAMICS) An important tool in heat transfer

More information

Chapter 12 PROPERTY RELATIONS. Department of Mechanical Engineering

Chapter 12 PROPERTY RELATIONS. Department of Mechanical Engineering Chapter 12 THERMODYNAMIC PROPERTY RELATIONS Dr Ali Jawarneh Department of Mechanical Engineering Hashemite University it Objectives Develop fundamental relations between commonly encountered thermodynamic

More information

Applied Thermodynamics for Marine Systems Prof. P. K. Das Department of Mechanical Engineering Indian Institute of Technology, Kharagpur

Applied Thermodynamics for Marine Systems Prof. P. K. Das Department of Mechanical Engineering Indian Institute of Technology, Kharagpur Applied Thermodynamics for Marine Systems Prof. P. K. Das Department of Mechanical Engineering Indian Institute of Technology, Kharagpur Lecture - 8 Introduction to Vapour Power Cycle Today, we will continue

More information

1 mol ideal gas, PV=RT, show the entropy can be written as! S = C v. lnt + RlnV + cons tant

1 mol ideal gas, PV=RT, show the entropy can be written as! S = C v. lnt + RlnV + cons tant 1 mol ideal gas, PV=RT, show the entropy can be written as! S = C v lnt + RlnV + cons tant (1) p, V, T change Reversible isothermal process (const. T) TdS=du-!W"!S = # "Q r = Q r T T Q r = $W = # pdv =

More information

Week 2. Energy, Energy Transfer, And General Energy Analysis

Week 2. Energy, Energy Transfer, And General Energy Analysis Week 2. Energy, Energy Transfer, And General Energy Analysis Objectives 1. Introduce the concept of energy and define its various forms 2. Discuss the nature of internal energy 3. Define the concept of

More information

Week 5. Energy Analysis of Closed Systems. GENESYS Laboratory

Week 5. Energy Analysis of Closed Systems. GENESYS Laboratory Week 5. Energy Analysis of Closed Systems Objectives 1. Examine the moving boundary work or PdV work commonly encountered in reciprocating devices such as automotive engines and compressors 2. Identify

More information

1 st Law Analysis of Control Volume (open system) Chapter 6

1 st Law Analysis of Control Volume (open system) Chapter 6 1 st Law Analysis of Control Volume (open system) Chapter 6 In chapter 5, we did 1st law analysis for a control mass (closed system). In this chapter the analysis of the 1st law will be on a control volume

More information

UNIVESITY OF SWAZILAND FACl.JLTY OF SCIENCE AND ENGINEERING DEPARTMENT OF PHYSICS

UNIVESITY OF SWAZILAND FACl.JLTY OF SCIENCE AND ENGINEERING DEPARTMENT OF PHYSICS UNIVESITY OF SWAZILAND FACl.LTY OF SCIENCE AND ENGINEERING DEPARTMENT OF PHYSICS Main Examination 2016/2017. COURSE NAME: Thermodynamics/Thermofluids COURSE CODE: PHY242/EEE202 TIME ALLOWED: 3 hours ANSWER

More information

ME Thermodynamics I

ME Thermodynamics I Homework - Week 01 HW-01 (25 points) Given: 5 Schematic of the solar cell/solar panel Find: 5 Identify the system and the heat/work interactions associated with it. Show the direction of the interactions.

More information

Chemistry. Lecture 10 Maxwell Relations. NC State University

Chemistry. Lecture 10 Maxwell Relations. NC State University Chemistry Lecture 10 Maxwell Relations NC State University Thermodynamic state functions expressed in differential form We have seen that the internal energy is conserved and depends on mechanical (dw)

More information

KNOWN: Data are provided for a closed system undergoing a process involving work, heat transfer, change in elevation, and change in velocity.

KNOWN: Data are provided for a closed system undergoing a process involving work, heat transfer, change in elevation, and change in velocity. Problem 44 A closed system of mass of 10 kg undergoes a process during which there is energy transfer by work from the system of 0147 kj per kg, an elevation decrease of 50 m, and an increase in velocity

More information

Introduction CHAPTER Prime Movers. 1.2 Sources of Energy

Introduction CHAPTER Prime Movers. 1.2 Sources of Energy Introduction CHAPTER 1 1.1 Prime Movers Prime mover is a device which converts natural source of energy into mechanical work to drive machines for various applications. In olden days, man had to depend

More information

THERMODYNAMICS Lecture 5: Second Law of Thermodynamics

THERMODYNAMICS Lecture 5: Second Law of Thermodynamics HERMODYNAMICS Lecture 5: Second Law of hermodynamics Pierwsza strona Second Law of hermodynamics In the course of discussions on the First Law of hermodynamics we concluded that all kinds of energy are

More information

Thermodynamic Variables and Relations

Thermodynamic Variables and Relations MME 231: Lecture 10 Thermodynamic Variables and Relations A. K. M. B. Rashid Professor, Department of MME BUET, Dhaka Today s Topics Thermodynamic relations derived from the Laws of Thermodynamics Definitions

More information

Entropy and the Second Law of Thermodynamics

Entropy and the Second Law of Thermodynamics Entropy and the Second Law of Thermodynamics Reading Problems 7-1 7-3 7-88, 7-131, 7-135 7-6 7-10 8-24, 8-44, 8-46, 8-60, 8-73, 8-99, 8-128, 8-132, 8-1 8-10, 8-13 8-135, 8-148, 8-152, 8-166, 8-168, 8-189

More information

Lecture 3 Evaluation of Entropy

Lecture 3 Evaluation of Entropy Lecture 3 Evaluation of Entropy If we wish to designate S by a proper name we can say of it that it is the transformation content of the body, in the same way that we say of the quantity U that it is the

More information

Thermodynamics of phase transitions

Thermodynamics of phase transitions Thermodynamics of phase transitions Katarzyna Sznajd-Weron Department of Theoretical of Physics Wroc law University of Science and Technology, Poland March 12, 2017 Katarzyna Sznajd-Weron (WUST) Thermodynamics

More information

Ttions, and its relation to states of matter. This chapter covers the

Ttions, and its relation to states of matter. This chapter covers the Related Commercial Resources CHAPTER 1 THERMODYNAMICS AND REFRIGERATION CYCLES THERMODYNAMICS... 1.1 First Law of Thermodynamics... 1.2 Second Law of Thermodynamics... 1.2 Thermodynamic Analysis of Refrigeration

More information

KNOWN: Pressure, temperature, and velocity of steam entering a 1.6-cm-diameter pipe.

KNOWN: Pressure, temperature, and velocity of steam entering a 1.6-cm-diameter pipe. 4.3 Steam enters a.6-cm-diameter pipe at 80 bar and 600 o C with a velocity of 50 m/s. Determine the mass flow rate, in kg/s. KNOWN: Pressure, temperature, and velocity of steam entering a.6-cm-diameter

More information

MS212 Thermodynamics of Materials ( 소재열역학의이해 ) Lecture Note: Chapter 7

MS212 Thermodynamics of Materials ( 소재열역학의이해 ) Lecture Note: Chapter 7 2017 Spring Semester MS212 Thermodynamics of Materials ( 소재열역학의이해 ) Lecture Note: Chapter 7 Byungha Shin ( 신병하 ) Dept. of MSE, KAIST Largely based on lecture notes of Prof. Hyuck-Mo Lee and Prof. WooChul

More information

S.E. (Chemical Engineering) (Second Semester)EXAMINATION, 2012 THERMODYNAMICS-I (2008 PATTERN) Time : Three Hours Maximum Marks : 100

S.E. (Chemical Engineering) (Second Semester)EXAMINATION, 2012 THERMODYNAMICS-I (2008 PATTERN) Time : Three Hours Maximum Marks : 100 Total No. of Questions 12] [Total No. of Printed Pages 7 Seat No. [4162]-189 S.E. (Chemical Engineering) (Second Semester)EXAMINATION, 2012 THERMODYNAMICS-I (2008 PATTERN) Time : Three Hours Maximum Marks

More information

Conservation of Energy for a Closed System. First Law of Thermodynamics. First Law of Thermodynamics for a Change in State

Conservation of Energy for a Closed System. First Law of Thermodynamics. First Law of Thermodynamics for a Change in State Conservation of Energy for a Closed System First Law of Thermodynamics Dr. Md. Zahurul Haq rofessor Department of Mechanical Engineering Bangladesh University of Engineering & Technology BUET Dhaka-000,

More information

Outline. Example. Solution. Property evaluation examples Specific heat Internal energy, enthalpy, and specific heats of solids and liquids Examples

Outline. Example. Solution. Property evaluation examples Specific heat Internal energy, enthalpy, and specific heats of solids and liquids Examples Outline Property ealuation examples Specific heat Internal energy, enthalpy, and specific heats of solids and liquids s A piston-cylinder deice initially contains 0.5m of saturated water apor at 00kPa.

More information

MME 2010 METALLURGICAL THERMODYNAMICS II. Fundamentals of Thermodynamics for Systems of Constant Composition

MME 2010 METALLURGICAL THERMODYNAMICS II. Fundamentals of Thermodynamics for Systems of Constant Composition MME 2010 METALLURGICAL THERMODYNAMICS II Fundamentals of Thermodynamics for Systems of Constant Composition Thermodynamics addresses two types of problems: 1- Computation of energy difference between two

More information

QUIZZES RIEPJCPIγPJEJJJY

QUIZZES RIEPJCPIγPJEJJJY Che 3021 Thermodynamics I QUIZZES RIEPJCPIγPJEJJJY QUIZ 1. Find Molecular Weights: 1 1 CO 2 2 NaCl 3 Aspirin C 9 H 8 O 4 CO2 = NaCl = C9H8O4 = PIgPJC Quiz 1. Temperature conversion 1 Convert 94 o F, to

More information

1. (10) True or False: A material with an ideal thermal equation of state must have a constant c v.

1. (10) True or False: A material with an ideal thermal equation of state must have a constant c v. AME 54531 Intermediate hermodynamics Examination : Prof. J. M. Powers 7 November 018 1. 10) rue or False: A material with an ideal thermal equation of state must have a constant c v. False. Forsuchamaterialc

More information

Relationships between WORK, HEAT, and ENERGY. Consider a force, F, acting on a block sliding on a frictionless surface. x 2

Relationships between WORK, HEAT, and ENERGY. Consider a force, F, acting on a block sliding on a frictionless surface. x 2 Relationships between WORK, HEAT, and ENERGY Consider a force, F, acting on a block sliding on a frictionless surface x x M F x Frictionless surface M dv v dt M dv dt v F F F ; v mass velocity in x direction

More information

Physical Fundamentals of Global Change Processes

Physical Fundamentals of Global Change Processes University of Applied Sciences Eberswalde Master Study Program Global Change Management Manfred Stock Potsdam Institute for Climate Impact Research Module: Physical Fundamentals of Global Change Processes

More information

Monday, October 21, 13. Copyright 2009 Pearson Education, Inc.

Monday, October 21, 13. Copyright 2009 Pearson Education, Inc. Lecture 4 1st Law of Thermodynamics (sections 19-4 to 19-9) 19-4 Calorimetry 19-5 Latent Heat 19-6 The 1st Law of Thermodynamics 19-7 Gas: Calculating the Work 19-8 Molar Specific Heats 19-9 Adiabatic

More information

Chapter 5. Mass and Energy Analysis of Control Volumes

Chapter 5. Mass and Energy Analysis of Control Volumes Chapter 5 Mass and Energy Analysis of Control Volumes Conservation Principles for Control volumes The conservation of mass and the conservation of energy principles for open systems (or control volumes)

More information

CHAPTER 5 MASS AND ENERGY ANALYSIS OF CONTROL VOLUMES

CHAPTER 5 MASS AND ENERGY ANALYSIS OF CONTROL VOLUMES Thermodynamics: An Engineering Approach 8th Edition in SI Units Yunus A. Çengel, Michael A. Boles McGraw-Hill, 2015 CHAPTER 5 MASS AND ENERGY ANALYSIS OF CONTROL VOLUMES Lecture slides by Dr. Fawzi Elfghi

More information

AAE COMBUSTION AND THERMOCHEMISTRY

AAE COMBUSTION AND THERMOCHEMISTRY 5. COMBUSTIO AD THERMOCHEMISTRY Ch5 1 Overview Definition & mathematical determination of chemical equilibrium, Definition/determination of adiabatic flame temperature, Prediction of composition and temperature

More information

2. Describe the second law in terms of adiabatic and reversible processes.

2. Describe the second law in terms of adiabatic and reversible processes. Lecture #3 1 Lecture 3 Objectives: Students will be able to: 1. Describe the first law in terms of heat and work interactions.. Describe the second law in terms of adiabatic and reversible processes. 3.

More information

4.1 LAWS OF MECHANICS - Review

4.1 LAWS OF MECHANICS - Review 4.1 LAWS OF MECHANICS - Review Ch4 9 SYSTEM System: Moving Fluid Definitions: System is defined as an arbitrary quantity of mass of fixed identity. Surrounding is everything external to this system. Boundary

More information

Clausius Clapeyron Equation

Clausius Clapeyron Equation Course - BSc. Applied Physical Science (Computer Science) Year & Semester - Ist, IInd Subject - Physics Paper No. - 6 Paper Title - Thermal Physics Lecture No. 18 Clausius Clapeyron Equation Hello friends,

More information

ME Thermodynamics I = = = 98.3% 1

ME Thermodynamics I = = = 98.3% 1 HW-08 (25 points) i) : a) 1 Since ν f < ν < ν g we conclude the state is a Saturated Liquid-Vapor Mixture (SLVM) 1, from the saturation tables we obtain p 3.6154 bar. 1 Calculating the quality, x: x ν

More information

UNIVERSITY OF SOUTHAMPTON

UNIVERSITY OF SOUTHAMPTON UNIVERSITY OF SOUTHAMPTON PHYS1013W1 SEMESTER 2 EXAMINATION 2014-2015 ENERGY AND MATTER Duration: 120 MINS (2 hours) This paper contains 8 questions. Answers to Section A and Section B must be in separate

More information

- Apply closed system energy balances, observe sign convention for work and heat transfer.

- Apply closed system energy balances, observe sign convention for work and heat transfer. CHAPTER : ENERGY AND THE FIRST LAW OF THERMODYNAMICS Objectives: - In this chapter we discuss energy and develop equations for applying the principle of conservation of energy. Learning Outcomes: - Demonstrate

More information

Chapter 7: The Second Law of Thermodynamics

Chapter 7: The Second Law of Thermodynamics Chapter 7: he Second Law of hermodynamics he second law of thermodynamics asserts that processes occur in a certain direction and that the energy has quality as well as quantity he first law places no

More information

UNIVERSITY OF WATERLOO. ECE 309 Thermodynamics and Heat Transfer. Final Examination Spring 1997

UNIVERSITY OF WATERLOO. ECE 309 Thermodynamics and Heat Transfer. Final Examination Spring 1997 UNIVERSITY OF WATERLOO DEPARTMENT OF ELECTRICAL ENGINEERING ECE 309 Thermodynamics and Heat Transfer Final Examination Spring 1997 M.M. Yovanovich August 5, 1997 9:00 A.M.-12:00 Noon NOTE: 1. Open book

More information

Physics 213. Practice Final Exam Spring The next two questions pertain to the following situation:

Physics 213. Practice Final Exam Spring The next two questions pertain to the following situation: The next two questions pertain to the following situation: Consider the following two systems: A: three interacting harmonic oscillators with total energy 6ε. B: two interacting harmonic oscillators, with

More information

Lecture 3 Clausius Inequality

Lecture 3 Clausius Inequality Lecture 3 Clausius Inequality Rudolf Julius Emanuel Clausius 2 January 1822 24 August 1888 Defined Entropy Greek, en+tropein content transformative or transformation content The energy of the universe

More information