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

Similar documents
= T. (kj/k) (kj/k) 0 (kj/k) int rev. Chapter 6 SUMMARY

Previous lecture. Today lecture

Minimizing and maximizing compressor and turbine work respectively

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

Chapter 7. Entropy: A Measure of Disorder

Thermodynamics is the Science of Energy and Entropy

Thermodynamics: An Engineering Approach Seventh Edition Yunus A. Cengel, Michael A. Boles McGraw-Hill, Chapter 7 ENTROPY

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

ENTROPY. Chapter 7. Mehmet Kanoglu. Thermodynamics: An Engineering Approach, 6 th Edition. Yunus A. Cengel, Michael A. Boles.

Thermodynamics: An Engineering Approach Seventh Edition in SI Units Yunus A. Cengel, Michael A. Boles McGraw-Hill, 2011.

Lecture 44: Review Thermodynamics I

King Fahd University of Petroleum & Minerals

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

Entropy and the Second Law of Thermodynamics

ME Thermodynamics I

Lecture 35: Vapor power systems, Rankine cycle

CHAPTER 7 ENTROPY. Copyright Hany A. Al-Ansary and S. I. Abdel-Khalik (2014) 1

Teaching schedule *15 18

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)

ME 2322 Thermodynamics I PRE-LECTURE Lesson 23 Complete the items below Name:

Thermodynamics II. Week 9

Week 5. Energy Analysis of Closed Systems. GENESYS Laboratory

CLAUSIUS INEQUALITY. PROOF: In Classroom

Week 8. Steady Flow Engineering Devices. GENESYS Laboratory

Using the Entropy Rate Balance to Determine the Heat Transfer and Work in an Internally Reversible, Polytropic, Steady State Flow Process

Chapter 6. Using Entropy

Exergy and the Dead State

Thermodynamics ENGR360-MEP112 LECTURE 7

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

+ 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.

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

Spring_#8. Thermodynamics. Youngsuk Nam

5/6/ :41 PM. Chapter 6. Using Entropy. Dr. Mohammad Abuhaiba, PE

Entropy and the Second Law of Thermodynamics

Course: MECH-341 Thermodynamics II Semester: Fall 2006

= 1 T 4 T 1 T 3 T 2. W net V max V min. = (k 1) ln ( v 2. v min

ME 300 Thermodynamics II Spring 2015 Exam 3. Son Jain Lucht 8:30AM 11:30AM 2:30PM

Chemical Engineering Thermodynamics

Chapter 5: The First Law of Thermodynamics: Closed Systems

I. (20%) Answer the following True (T) or False (F). If false, explain why for full credit.

Chapter One Reviews of Thermodynamics Update on 2013/9/13

ME Thermodynamics I. Lecture Notes and Example Problems

ME 354 THERMODYNAMICS 2 MIDTERM EXAMINATION. Instructor: R. Culham. Name: Student ID Number: Instructions

7. Development of the 2nd Law

ENERGY TRANSFER BY WORK: Electrical Work: When N Coulombs of electrical charge move through a potential difference V

Introduction to Chemical Engineering Thermodynamics. Chapter 7. KFUPM Housam Binous CHE 303

first law of ThermodyNamics

Chapter Two. Basic Thermodynamics, Fluid Mechanics: Definitions of Efficiency. Laith Batarseh

Chapter 4. Energy Analysis of Closed Systems

Modeling and Analysis of Dynamic Systems

First Law of Thermodynamics

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

This follows from the Clausius inequality as a consequence of the second law of thermodynamics. Therefore. (for reversible process only) (22.

Chapter 7. Dr Ali Jawarneh. Department of Mechanical Engineering Hashemite University

Fundamentals of Thermodynamics Applied to Thermal Power Plants

Thermodynamics General

NOTE: Only CHANGE in internal energy matters

Chapter 1 Introduction and Basic Concepts

The First Law of Thermodynamics. By: Yidnekachew Messele

Dr Ali Jawarneh. Hashemite University

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

Consequences of Second Law of Thermodynamics. Entropy. Clausius Inequity

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

AME 436. Energy and Propulsion. Lecture 7 Unsteady-flow (reciprocating) engines 2: Using P-V and T-s diagrams

(1)5. Which of the following equations is always valid for a fixed mass system undergoing an irreversible or reversible process:

CHAPTER 8 ENTROPY. Blank

Consequences of Second Law of Thermodynamics. Entropy. Clausius Inequity

Isentropic Efficiency in Engineering Thermodynamics

The Second Law of Thermodynamics

More Thermodynamics. Specific Specific Heats of a Gas Equipartition of Energy Reversible and Irreversible Processes

Chapter 4. Entropy and the Second Law. Soong Ho Um Sungkyunkwan University Chemical Engineering

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

Unified Quiz: Thermodynamics

Chapter 5. Mass and Energy Analysis of Control Volumes

열과유체, 에너지와친해지기 KAIST 기계공학과정상권

CHEMICAL ENGINEERING THERMODYNAMICS. Andrew S. Rosen

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

Introduction to Turbomachinery

ECE309 INTRODUCTION TO THERMODYNAMICS & HEAT TRANSFER. 13 June 2007

Lecture 10: Heat Engines and Reversible Processes

Earlier Topics. Introduction to Cryogenic Engineering An introductory knowledge of Cryogenic Engineering.

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

ESO201A: Thermodynamics

Chapter 2 Carnot Principle

Chapter 5. Mass and Energy Analysis of Control Volumes. by Asst. Prof. Dr.Woranee Paengjuntuek and Asst. Prof. Dr.Worarattana Pattaraprakorn

ME 2322 Thermodynamics I PRE-LECTURE Lesson 10 Complete the items below Name:

Fundamentals of Thermodynamics. Chapter 8. Exergy

Objectives. Conservation of mass principle: Mass Equation The Bernoulli equation Conservation of energy principle: Energy equation

3. First Law of Thermodynamics and Energy Equation

Chapter 7: The Second Law of Thermodynamics

UNIT I Basic concepts and Work & Heat Transfer

Classical thermodynamics

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

University of Engineering & Technology Lahore. (KSK Campus)

Practice Examinations Chem 393 Fall 2005 Time 1 hr 15 min for each set.

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

Physical Biochemistry. Kwan Hee Lee, Ph.D. Handong Global University

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

Reversible Processes. Furthermore, there must be no friction (i.e. mechanical energy loss) or turbulence i.e. it must be infinitely slow.

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

Transcription:

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 can be changed by Heat transfer between system and surroundings (here modeled as isothermal) Mass transfer (only for open systems (CVs)) Irersibilities (for all actual systems) ds dt CV Qɺ ɺ ɺ = + misi mese + Sgen T ɺ

Entropy balance special forms dscv Qɺ = + mɺ isi mɺ ese + Sɺ gen dt T Qɺ + mɺ isi mɺ ese + Sɺ gen = 0 steady T Qɺ + m ɺ ( s i s e) + S ɺ gen = 0 SISO T closed system Q = = + s s s Sgen T Quasi (QE) process Recall moving boundary wor formula w = Pdv Valid for closed system undergoing quasi compression or expansion A quasi process involves small departures from All states through which quasi process passes can be considered states Initial state Near Near Near Final state QE process is ersible Reversible process must proceed through a series of states Otherwise there would a tendency for system to change spontaneously which is irersible Hence, a quasiequilbrium process is an internally ersible process

Wor for internally ersible process Consider st law in differential form on a per unit mass bass flowing through a CV for an internally ersible process δ q δ w = dh + de + dpe δ q = Tds δ q = dh vdp Tds = dh vdp δ w = vdp + de + dpe w = vdp e pe Special cases For incompressible fluid, we have: w = v( P P ) e pe For devices with no wor interaction: v( P P ) + e + pe = 0 Above is called Bernoulli s equation (fluid mechanics) When one can neglect inetic and potential energy changes we have: w vdp = Implications w What does this equation tell us? The larger the specific volume, the larger the ersible wor produced or consumed by steady flow devices Conclusion is valid for actual steady flow devices Hence, we desire specific volume to be as small as possible during compression processes and as large as possible during expansion processes This tell us to compress (pump) liquids and expand gases (minimum wor input and maximum wor output) It also tells us to cool during compression and heat during expansion = vdp 3

What is the difference between these two formulas? w = Pdv w One is for closed system undergoing QE (int..) boundary wor The other is for open system, steady flow, internally ersible process Don t confuse the two! Stop! Warning! = vdp Proof that steady flow devices deliver most wor and consume least wor when process is ersible δ q δ w = dh + de + dpe act act δ q δ w = dh + de + dpe δ q δ w = δ q δ w act act δ w δ w = δ q δ q δ q act act = Tds δ w δ wact δ qact = ds 0 T T δ w δ w w w act act Minimizing compressor wor w, in = vdp How? Approach ersible process by minimizing irersibilities due to friction, turbulence, non- QE compression Keep specific volume of gas as small as possible during compression For gases, volume is proportional to temperature So, eep temperature as low as possible e.g. cool as you compress 4

Wor comparison Isentropic compression ( ) / R( T T ) RT P wcomp, in = vdp = = Pv = const P Polytropic compression w ( n ) / n nr( T T ) nrt P comp, in = vdp = = n Pv = const n n P Isothermal compression P wcomp, in = vdp = RT ln Pv= const P On P-v diagram area to left of curve is ersible steady flow wor Adiabatic compression requires maximum wor Isothermal compression requires minimum wor Motivates use of cooling water jacets around compressor casing to approach isothermal Wor comparison Multistage compression with intercooling Maintaining near isothermal conditions using cooling jacets is not always practically possible Gas is compressed in stages and cooled between each stage as it passes through a heat exchanger called an intercooler Ideally intercooling taes place at constant pressure and gas is cooled to initial temperature upon entry to intercooler 5

P-v and T-s diagrams for two-stage steady flow compression process What is best intercooler pressure? To minimize compression wor during two-stage compression, the pressure ratio across each stage of the compressor should be the same e.g. wor across each stage is identical w = w + w comp, in comp comp I, in II, in minimize ( n ) / n ( n ) / n x x nrt P nrt P = + n P n P P P P = ( PP ) or = x / x P Px Turbine Compressor Pump Isentropic efficiencies Actual turbine wor wa h h ηt = = Isentropic turbine wor w h h s a s Isentropic compressor wor ws h h ηc = = Actual compressor wor w h h a s a Isentropic pump wor ws v( P P ) ηp = = Actual pump wor w h h a a 6

Illustration Turbine Compressor Isentropic efficiency of nozzles η = N Actual KE at nozzle exit Isentropic KE at nozzle exit V h h = V h h a a s s Note exit pressure is same for both actual and isentropic processes But exit state is different To be done in class Examples 7

Summary Ideal compressor or turbine operation under ersible conditions To minimize actual compression worinput eep specific volume small; cooling for gases Ideally isothermal; practically multistage comrpession with intercooling and equal pressure ratios To maximize actual expansion wor output eep specific volume large; heating for gases; multistage expansion with reheating 8