Chapter 19: The Second Law of Thermodynamics

Similar documents
Thermal energy 2 U Q W. 23 April The First Law of Thermodynamics. Or, if we want to obtain external work: The trick of using steam

U Q W The First Law of Thermodynamics. Efficiency. Closed cycle steam power plant. First page of S. Carnot s paper. Sadi Carnot ( )

Thermodynamics. Question 1. Question 2. Question 3 3/10/2010. Practice Questions PV TR PV T R

The Stirling Engine: The Heat Engine

Chapter 4 The second law of thermodynamics

a) Read over steps (1)- (4) below and sketch the path of the cycle on a P V plot on the graph below. Label all appropriate points.

Lecture 6: Coding theory

CHAPTER 20: Second Law of Thermodynamics

Thermal & Kinetic Physics: Lecture Notes Kevin Donovan 4. HEAT ENGINES.

Momentum and Energy Review

Counting Paths Between Vertices. Isomorphism of Graphs. Isomorphism of Graphs. Isomorphism of Graphs. Isomorphism of Graphs. Isomorphism of Graphs

Factorising FACTORISING.

Now we must transform the original model so we can use the new parameters. = S max. Recruits

CS 491G Combinatorial Optimization Lecture Notes

Appendix A: HVAC Equipment Efficiency Tables

for all x in [a,b], then the area of the region bounded by the graphs of f and g and the vertical lines x = a and x = b is b [ ( ) ( )] A= f x g x dx

22: Union Find. CS 473u - Algorithms - Spring April 14, We want to maintain a collection of sets, under the operations of:

ERT 316: REACTION ENGINEERING CHAPTER 3 RATE LAWS & STOICHIOMETRY

Numbers and indices. 1.1 Fractions. GCSE C Example 1. Handy hint. Key point

Mid-Term Examination - Spring 2014 Mathematical Programming with Applications to Economics Total Score: 45; Time: 3 hours

8 THREE PHASE A.C. CIRCUITS

Surds and Indices. Surds and Indices. Curriculum Ready ACMNA: 233,

Chapter 4rth LIQUIDS AND SOLIDS MCQs

1 PYTHAGORAS THEOREM 1. Given a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Particle Physics. Michaelmas Term 2011 Prof Mark Thomson. Handout 3 : Interaction by Particle Exchange and QED. Recap

Logic, Set Theory and Computability [M. Coppenbarger]

Electronic Circuits I Revision after midterm

Solutions for HW9. Bipartite: put the red vertices in V 1 and the black in V 2. Not bipartite!

Vapour compression refrigeration system

18.06 Problem Set 4 Due Wednesday, Oct. 11, 2006 at 4:00 p.m. in 2-106

CARLETON UNIVERSITY. 1.0 Problems and Most Solutions, Sect B, 2005

NON-DETERMINISTIC FSA

AP CALCULUS Test #6: Unit #6 Basic Integration and Applications

The DOACROSS statement

CSE 332. Sorting. Data Abstractions. CSE 332: Data Abstractions. QuickSort Cutoff 1. Where We Are 2. Bounding The MAXIMUM Problem 4

1 This question is about mean bond enthalpies and their use in the calculation of enthalpy changes.

Appendix C Partial discharges. 1. Relationship Between Measured and Actual Discharge Quantities

03. Early Greeks & Aristotle

Data Structures LECTURE 10. Huffman coding. Example. Coding: problem definition

Probability. b a b. a b 32.

Chapter 8 Roots and Radicals

where the box contains a finite number of gates from the given collection. Examples of gates that are commonly used are the following: a b

CS 2204 DIGITAL LOGIC & STATE MACHINE DESIGN SPRING 2014

CIT 596 Theory of Computation 1. Graphs and Digraphs


2.4 Theoretical Foundations

Project 6: Minigoals Towards Simplifying and Rewriting Expressions

1.3 SCALARS AND VECTORS

I 3 2 = I I 4 = 2A

Finite State Automata and Determinisation

CS311 Computational Structures Regular Languages and Regular Grammars. Lecture 6

Algebra 2 Semester 1 Practice Final

Section 2.3. Matrix Inverses

First Law of Thermodynamics. Control Mass (Closed System) Conservation of Mass. Conservation of Energy

CS 573 Automata Theory and Formal Languages

6.5 Improper integrals

Let s divide up the interval [ ab, ] into n subintervals with the same length, so we have

For a, b, c, d positive if a b and. ac bd. Reciprocal relations for a and b positive. If a > b then a ab > b. then

A Study on the Properties of Rational Triangles

Introduction to Olympiad Inequalities

Name Class Date. Match each phrase with the correct term or terms. Terms may be used more than once.

Lesson 2.1 Inductive Reasoning

QUADRATIC EQUATION. Contents

Chapter 4 State-Space Planning

Equivalent fractions have the same value but they have different denominators. This means they have been divided into a different number of parts.

Linear Inequalities. Work Sheet 1

(a) A partition P of [a, b] is a finite subset of [a, b] containing a and b. If Q is another partition and P Q, then Q is a refinement of P.

VISIBLE AND INFRARED ABSORPTION SPECTRA OF COVERING MATERIALS FOR SOLAR COLLECTORS

Lesson 2.1 Inductive Reasoning

GM1 Consolidation Worksheet

Formula for Trapezoid estimate using Left and Right estimates: Trap( n) If the graph of f is decreasing on [a, b], then f ( x ) dx

SECTION A STUDENT MATERIAL. Part 1. What and Why.?

9.1 Day 1 Warm Up. Solve the equation = x x 2 = March 1, 2017 Geometry 9.1 The Pythagorean Theorem 1

Conservation Law. Chapter Goal. 6.2 Theory

AP Physics - Heat Engines

Eigenvectors and Eigenvalues

6.3.2 Spectroscopy. N Goalby chemrevise.org 1 NO 2 CH 3. CH 3 C a. NMR spectroscopy. Different types of NMR

= x x 2 = 25 2

University of Sioux Falls. MAT204/205 Calculus I/II

SIMPLE NONLINEAR GRAPHS

Chemical Equilibrium

Spacetime and the Quantum World Questions Fall 2010

Review of Gaussian Quadrature method

Probability The Language of Chance P(A) Mathletics Instant Workbooks. Copyright

Linear Algebra Introduction

Lecture 11 Binary Decision Diagrams (BDDs)

Solutions to Problem Set #1

Comparing the Pre-image and Image of a Dilation

p-adic Egyptian Fractions

Discrete Mathematics and Probability Theory Spring 2013 Anant Sahai Lecture 17

Algorithm Design and Analysis

Something found at a salad bar

AP Calculus BC Chapter 8: Integration Techniques, L Hopital s Rule and Improper Integrals

Part 4. Integration (with Proofs)

Applied. Grade 9 Assessment of Mathematics. Multiple-Choice Items. Winter 2005

Physics Lecture 14: MON 29 SEP

Polynomials. Polynomials. Curriculum Ready ACMNA:

50 AMC Lectures Problem Book 2 (36) Substitution Method

Discrete Structures Lecture 11

Math 32B Discussion Session Week 8 Notes February 28 and March 2, f(b) f(a) = f (t)dt (1)

Transcription:

hpter 9: he Seon Lw of hermoynmis Diretions of thermoynmi proesses Irreversile n reversile proesses hermoynmi proesses tht our in nture re ll irreversile proesses whih proee spontneously in one iretion ut not the other. In reversile proess the system must e ple of eing returne to its originl stte with no other hnge in the surrounings. A reversile proess proees slowly through equilirium sttes.

et engines et engine Any evie tht trnsforms het prtly into work or mehnil energy is lle het engine. A quntity of mtter insie the engine unergoes inflow n outflow of het, expnsion n ompression, n sometimes hnge of phse. his mtter insie the engine is lle the working sustne of the engine. he simplest kin of engine to nlyze is one in whih the working sustne unergoes yli proess. All het engines sor het from soure t reltively high temperture, perform some mehnil work, n isr or rejet some het t lower temperture. U U U 0 2 For yli proess

et engines Energy flow igrm for het engine > 0 < 0 emperture of hot reservoir et flow from hot reservoir per yle emperture of ol reservoir et flow from ol reservoir per yle wste! Net het sore per yle: + Net work one per yle: +

et engines (ont Energy flow igrm for het engine (ont Net het sore per yle: + Net work one per yle: + herml effiieny ε: + ε

et engines (ont Stem engines (externl omustion

Internl-omustion engines Gsoline engine ( het engine First mixture of ir n gsoline vpor flows into yliner through n open intke vlve while the piston esens, inresing the volume of the yliner from minimum of to mximum of r (r: ompression rtio 8-0. At the en of intke stroke, the intke vlve loses n the mixture is ompresse, pproximtely itilly, k to volume uring the ompression stroke. hen the mixture is ignite y the sprk plug, n the hete gs expns, pproximtely itilly, k to volume r, pushing on the piston-power stroke.

Internl-omustion engines (ont he Otto yle : : : : ompression stroke (iti ompression ignite fuel (heting t onstnt volume power stroke (iti expnsion Rejet het to environment (ooling t onstnt volume 0 ( : > n 0 ( : < n + + ε ( : γ γ r ( : γ γ r ( ( ( + γ γ γ γ γ γ ε r r r r r r γ γ γ ε r r r therml effiieny in Otto yle r8,γ.4->ε56%

Internl-omustion engines (ont he Otto yle

Internl-omustion engines (ont he Diesel yle : 0, U U n n ( (, : n p ( U, U n( p n ( (, : ε : net net 0, U U U 0, U n p + n n + + n ( ( ( n + (,, ( n ( ( 56% 5, γ ( : : : : 5 ompression stroke (iti ompression ignite fuel (heting t onstnt pressure power stroke (iti expnsion Rejet het to environment (ooling t onstnt volume

Internl-omustion engines (ont he Diesel yle

Refrigertor Refrigertors A refrigertor works s reverse het engine.

Refrigertors Energy flow of refrigertor A refrigertor tkes het from low temperture reservoir n gives it off to high temperture reservoir., From the first lw of thermoynmis + 0 + oeffiient of performne K ( rnot refrigertor

Air onitioner Refrigertors A ir onitioner works on extly the sme priniple s refrigertor. For ir onitioners, the quntities of gretest prtil importne re the rte of het removl (the het urrent from the region eing oole n the power input P/t. K t Pt P in Btu/h, P in wtts /P (energy effiieny rting EER in (Btu/h/. Note 3.43 Btu/h. EER~7-0.

he seon lw of thermoynmis he seon lw of thermoynmis It is impossile for ny system to unergo proess in whih it sors het from reservoir t single temperture n onverts the het ompletely into mehnil work, with the system ening in the sme stte in whih it egn (Kelvin-Plnk sttement he kineti n potentil (ue to intertions etween moleules energies ssoite with rnom motion onstitute the internl energy. hen oy sliing on surfe omes to rest s result of frition, the orgnize motion of the oy is onverte to rnom motion of moleules in the oy n in the surfe. Sine we n t ontrol the motions of iniviul moleules, we n t onvert this rnom motion OMPLELY k to orgnize motion. It is impossile for ny proess to hve s its sole result the trnsfer of het from ooler to hotter oy (lusius sttement.

he seon lw of thermoynmis (ont Proof : Kelvin-Plnk sttement lusius sttement impossile impossile ( A workless refrigertor violtes ( An impossile 00% effiient engine 2 n lw (K-P sttement. If it existe, violtes sttement. If it existe, it it oul e use to mke 00% oul e use to mke workless effiient engine, whih violte K-P refrigertor, whih violtes sttement. sttement. K-P sttement -sttement

he seon lw of thermoynmis (ont he seon lw of thermoynmis he onversion of work to het, s in frition or visous flui flow, n het flow from hot to ol ross finite temperture grient, re irreversile proesses. wo equivlent sttements of the 2 n lw stte tht these proesses n e only prtilly reverse Gses, for exmple, lwys seep spontneously through n opening from region of high pressure to region of low pressure. he 2 n lw is n expression of the inherent one-wy spet of this n mny other irreversile proesses. o hve the mximum effiieny of het engine, we must voi ll irreversile proesses. Answer: rnot yle In ny proess in whih the temperture of the working sustne of the engine is intermeite etween n, there must e NO het trnsfer etween the engine n either reservoir euse suh het trnsfer oul not e reversile. Any proess in whih the temperture of the working sustne hnges must e iti. As het flow through finite temperture rop is n irreversile, uring het trnsfer, there must e NO finite temperture ifferene.

he rnot yle he rnot yle reversile expnsion iti ompression iti expnsion reversile ompression

he rnot yle he rnot yle : iel gs pproximtion 0 0, : > > 0 0, : < < ( n nr l 0 ( ( ( n nr n nr > < l l n n / / / ( / (, / ( / ( ( γ γ γ γ γ γ l l isotherml expnsion isotherml expnsion Aiti proesses: ε therml effiieny of rnot engine

he rnot yle (ont he rnot refrigertor A reverse rnot engine rnot refrigertor K, / / K oeffiient of performne of rnot refrigertor hen the temperture ifferene is smll, K is muh lrger thn unity, in whih se lot of het n e pumpe from the lower to the higher temperture with only little expeniture of work.

he rnot yle (ont he rnot yle n the seon lw More effiient thn rnot engine impossile A rnot refrigertor Sine eh step in the rnot yle is reversile, the entire yle my e reverse. If you run the entire kwr, the engine eomes refrigertor.

he rnot yle (ont he rnot yle n the seon lw (ont he refrigertor oes work -, tkes in het from the ol reservoir, n expels her to the hot reservoir. he supereffiient het engine expels het, ut to o so, it tkes in greter mount of het +. Its work output is then +. he net effet of the two mhines together is to tke quntity of het n onvert it ompletely into work, whih violtes the 2 n lw. No engine n e more effiient thn rnot engine. All rnot engines operting etween the sme two tempertures hve the sme effiieny, irrespetive of the nture of the working sustne.

Entropy n isorer Entropy Entropy provies quntittive mesure of isorer. Define the infinitesiml entropy hnge S uring n infinitesiml reversile proess t solute temperture s: S If totl mount of het is e uring reversile isotherml proess t solute temperture, the totl entropy hnge is given y: 2 S S S igher temperture mens greter rnomness of motion Aing het uses sustntil frtionl inrese in moleulr motion n rnomness

Entropy (ont Entropy n isorer (ont Generliztion of efinition of entropy hnge is to inlue ANY reversile proess leing from one stte to nother, whether it is isotherml or not: Let us represent the proess s series of infinitesiml reversile steps. During step, n infinitesiml quntity of het is e to the system t solute temperture. hen the hnge of entropy for the entire proess is: S 2 entropy hnge in reversile proess he hnge in entropy oes not epen on the pth leing from the initil to the finl stte ut is the sme for ll possile proesses. Sine entropy is funtion only of the urrent stte of the system, we n ompute entropy hnge in irreversile proesses using the ove formul. ( we nee to invent pth onneting the given initil n finl stte tht onsist entirely of reversile, equilirium proesses n ompute the totl entropy hnge s for the tul pth.

I I Entropy (ont Entropy in yli proesses For ny reversile proess (e.g. rnot yle: + II II 0 0 II he entropy of system in given stte is inepenent of the pth tken to get there, n is thus stte vrile.

Entropy (ont Entropy in irreversile proesses onsier omine system with rnot engine n yli system. ( + R rev sys U het reservoir R R R R R R U R rnot yle R U, U 0 ( yli omine yli system rnot engine yli system rev sys

Entropy (ont Entropy in irreversile proesses onsier omine system with rnot engine n yli system. R U, U 0 ( yli het reservoir R R If > 0, yli evie exhnging het with single het reservoir n prouing n equivlent mount of work. - ioltion of K-P sttement of 2 n lw- 0 0 omine yli system -the 2n lw- he equlity is true for reversile proess. lusius inequlity rnot engine yli system R rev sys

Entropy (ont Entropy in irreversile proesses (ont Proesses: A n B reversile Proess : irreversile A reversile het trnsfer uses hnges in entropy of oth the system n the reservoir (t lest one neee exept for iti proess In n irreversile het trnsfer proess, there is finite soure of energy inste of n energy reservoir, n the temperture ifferene uring the het trnsfer proess is finite. A S B x S < 0 S + S A > + x x < 0 ( < 0 x A + B 0 sutrte (> for irreversile n for reversile proess A 0 + 0 AB A B B S For n isolte system 0, so S>0.

Entropy (ont Entropy n the seon lw hen ll systems tking prt in proess re inlue, the entropy either remins onstnt or inreses. OR No proess is possile in whih the totl entropy ereses, when ll systems tking prt in the proess re inlue.

Prolem Exerises A rnot engine tkes 2000 J of het from reservoir t 500 K, oes some work, n isrs some het to reservoir t 350 K. ow muh work oes it o, how muh het is isre, n wht is the effiieny? Solution he het isre y the engine is: 350K ( 2000J 400 J. 500K From the first lw, the work one y the engine is: + 2000J + ( 400J 600 J. he therml effiieny is: 350K ε 500K 0.30

Prolem 2 Exerises One kilogrm of wter t 0 o is hete to 00 o. ompute its hnge in entropy. Solution In prtie, the proess esrie woul e one irreversily, perhps y setting pn of wter on n eletri rnge whose ooking surfe is mintine t 00 o. But the entropy hnge of the wter epens only on the initil n finl sttes of the system, n is the sme whether the proess is reversile or not. ene we n imgine tht the temp. of the wter is inrese reversily in series of infinitesiml steps, in eh of whih the temp is rise y n infinitesiml mount. m S S 2 S 2 2 2 m mln.3 0 3 J / K.

Exmple: Refrigertor A refrigertor pumps het from the insie of the freezer (-5 to the room (25. ht is the mximum oeffiient of performne? L 5 o + 273 268K 25 o + 273 298K K iel 268K 298K 268K 8.9 i.e. 8.9 Joules of het woul e pumpe from the freezer for every Joule of work one y the ompressor. (typil K 3-5

Prolem 3 Exerises ht is the entropy hnge in free expnsion proess, when the volume is oule. Solution he work one y n moles of iel gs in n isotherml expnsion from to 2 is: nrln 2 / ( 2 nrl n nrln2. herefore the entropy hnge for n is: S nrln2 (mol[8.34j /( mol K]( ln2 5.76 J / K.

Prolem 4 Exerises A physis stuent immerses one en of opper ro in oiling wter t t 00 o n the other en in n ie-wter mixture t 0 o. he sies of the ro re insulte. After stey-stte onitions hve een hieve in the ro, 0.60 kg of ie melts in ertin time intervl. For this time intervl fin ( the entropy hnge of the oiling wter; ( the entropy hnge of the ie-wter mixture; ( the entropy hnge of the opper ro; ( the totl entropy hnge of the entire system. Solution ( S/-mL f /-(0.60 kg(334x0 3 J/kg/(373.5 K-43 J/K. ( S/mL f /(0.60 kg(334x0 3 J/kg/(273.5 K 96 J/K. ( From the time equilirium hs een rehe, there is no net het exhnge etween the ro n its surrounings, so the entropy hnge of the opper ro is zero. ( 96 J/K-43 J/K53 J/K.

Prolem 5 Exerises An experimentl power plnt t the Nturl Energy L genertes eletriity from the temperture grient of the oen. he surfe n eep-wter tempertures re 27 o n 6 o, respetively. ( ht is the mximum theoretil effiieny of this power plnt? ( If the power plnt is to proue 20 k of power, t wht rte must het e extrte from the wrm wter? At wht rte must het e sore y the ol wter? ( he ol wter tht enters the plnt leves it t temperture of 0 o. ht must e the flow rte of ol wter through the system? Solution 279.5K ( ε 7.0% 300.5K ( ( P out / ε 20 k / 0.070 3.0 M, 3.0 M 20 k (/ ε (20 k 2.8 M. m t / t 5 6 (2.8 0 (3600s / h [490J /( kg K](4K 5 6 0 kg / h 6 0 l / h.