Kelvin Planck Statement of the Second Law. Clausius Statement of the Second Law

Similar documents
NEWCOMEN ATMOSPHERIC ENGINE

SIMPLE RANKINE CYCLE. 3 expander. boiler. pump. condenser 1 W Q. cycle cycle. net. shaft

(b) The heat transfer can be determined from an energy balance on the system

Lecture 10: Carnot theorem

Thermodynamics Lecture Series

300 kpa 77 C. (d) If we neglect kinetic energy in the calculation of energy transport by mass

Chapters 19 & 20 Heat and the First Law of Thermodynamics

the first derivative with respect to time is obtained by carefully applying the chain rule ( surf init ) T Tinit

Chapter 9 Practical cycles

General Physics I. New Lecture 27: Carnot Cycle, The 2nd Law, Entropy and Information. Prof. WAN, Xin

Lecture 13. Heat Engines. Thermodynamic processes and entropy Thermodynamic cycles Extracting work from heat

Lecture 13 Heat Engines

Notes on the function gsw_enthalpy_first_derivatives_ct_exact(sa,ct,p)

The average velocity of water in the tube and the Reynolds number are Hot R-134a

8-4 P 2. = 12 kw. AIR T = const. Therefore, Q &

Chapter 20: Exercises: 3, 7, 11, 22, 28, 34 EOC: 40, 43, 46, 58

Physics 207 Lecture 23

6-5. H 2 O 200 kpa 200 C Q. Entropy Changes of Pure Substances

Then the amount of water that flows through the pipe during a differential time interval dt is (1) 4

CHEE 221: Chemical Processes and Systems

The Second Law of Thermodynamics. (Second Law of Thermodynamics)

Compressor 1. Evaporator. Condenser. Expansion valve. CHE 323, October 8, Chemical Engineering Thermodynamics. Tutorial problem 5.

1 = The rate at which the entropy of the high temperature reservoir changes, according to the definition of the entropy, is

Carnot Factor of a Vapour Power Cycle with Regenerative Extraction

Thermodynamics Lecture Series

First Name Last Name CIRCLE YOUR LECTURE BELOW: Div. 1 10:30 am Div. 2 2:30 pm Div. 3 4:30 pm Prof. Gore Prof. Udupa Prof. Chen

Thermodynamics [ENGR 251] [Lyes KADEM 2007]

The isentropic exponent in plasmas

Basic thermodynamics. heat to the high temperature reservoir.

At 1atm = 101,325Pa, one mole of gas at 20 0 C = 293K has volume V = 2.40 x10-2 m 3 = 24 litres

Numerical Differentiation


SOLUTION: Consider the system to be the refrigerator (shown in the following schematic), which operates over a cycle in normal operation.

Department of Mechanical Engineering ME 322 Mechanical Engineering Thermodynamics. Calculation of Entropy Changes. Lecture 19

Physics 2A (Fall 2012) Chapters 11:Using Energy and 12: Thermal Properties of Matter

Actual exergy intake to perform the same task

EF 152 Exam #3, Fall, 2012 Page 1 of 6

A short note on Reitlinger thermodynamic cycles

Spring_#7. Thermodynamics. Youngsuk Nam.

NODIA AND COMPANY. GATE SOLVED PAPER Mechanical Engineering Thermodynamics. Copyright By NODIA & COMPANY

Quotes. Review - First Law. Review - First Law. Review - First Law. Review - First Law. Thermodynamics Lecture Series

PHYSICAL PROCESSES IN ANISOTROPIC THERMOELEMENT AND THEIR FEATURES

Reminder: Exam 3 Friday, July 6. The Compton Effect. General Physics (PHY 2140) Lecture questions. Show your work for credit.

Chapter 2 Solutions. 2.1 (D) Pressure and temperature are dependent during phase change and independent when in a single phase.

FINITE TIME THERMODYNAMIC MODELING AND ANALYSIS FOR AN IRREVERSIBLE ATKINSON CYCLE. By Yanlin GE, Lingen CHEN, and Fengrui SUN

Engineering Services Examination - UPSC MECHANICAL ENGINEERING. Topic-wise Conventional Papers I & II (Last 30 Years Questions)

Physics 231 Lecture 35

Chapter 1 Fundamentals

BME-A PREVIOUS YEAR QUESTIONS

V. The Second Law of Thermodynamics. V. The Second Law of Thermodynamics

1 Power is transferred through a machine as shown. power input P I machine. power output P O. power loss P L. What is the efficiency of the machine?

Analysis: The speed of the proton is much less than light speed, so we can use the

PV/T = k or PV = kt Describe the difference between an ideal gas and a real gas.

Department of Mechanical Engineering ME 322 Mechanical Engineering Thermodynamics. Introduction to 2 nd Law and Entropy.

Example problems. Chapter 2: Temperature, heat, and the 1 st law of Thermodynamic. Homework: 2, 3, 4, 5, 6, 10, 15, 19, 21 (pages )

Work and Energy. Introduction. Work. PHY energy - J. Hedberg

(British) (SI) British Metric L T [V] = L T. [a] = 2 [F] = F = 2 T

F = U TS G = H TS. Availability, Irreversibility S K Mondal s Chapter 5. max. actual. univ. Ns =

Classical Approach to 2 nd Law for CM

Lecture 38: Vapor-compression refrigeration systems

Combining functions: algebraic methods

Developing Transfer Functions from Heat & Material Balances

Chapter-6: Entropy. 1 Clausius Inequality. 2 Entropy - A Property

02. Equilibrium Thermodynamics II: Engines

Chapter 5. The Second Law of Thermodynamics (continued)

CHAPTER 8 ENTROPY. Blank

1. Which one of the following expressions is not equal to all the others? 1 C. 1 D. 25x. 2. Simplify this expression as much as possible.

Chapter 4 Optimal Design

The Foundations of Chemistry 1

4. Energy balances Partly based on Chapter 4 of the De Nevers textbook.

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

Where F1 is the force and dl1 is the infinitesimal displacement, but F1 = p1a1

Chapter 3 Thermoelectric Coolers

2.8 The Derivative as a Function

3 Minority carrier profiles (the hyperbolic functions) Consider a

Exergy and the Dead State

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

Fluids and Buoyancy. 1. What will happen to the scale reading as the mass is lowered?

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

Lecture XVII. Abstract We introduce the concept of directional derivative of a scalar function and discuss its relation with the gradient operator.

MAT 145. Type of Calculator Used TI-89 Titanium 100 points Score 100 possible points

Unsteady State Simulation of Vapor Compression Heat Pump Systems With Modular Analysis (This paper will not be presented.)

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

Maths for Computer Graphics

5.1 We will begin this section with the definition of a rational expression. We

f self = 1/T self (b) With revolution, rotaton period T rot in second and the frequency Ω rot are T yr T yr + T day T rot = T self > f self

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

Chemistry 420/523 Chemical Thermodynamics (Spring ) Examination 1

Solutions to Homework #10

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

SECOND LAW OF THERMODYNAMICS

Phy 231 Sp 02 Homework #6 Page 1 of 4

SECOND LAW OF THERMODYNAMICS

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

ME 201 Thermodynamics

JJMIE Jordan Journal of Mechanical and Industrial Engineering

EVAPORATION. Robert evaporator. Balance equations Material balance (total) Component balance. Heat balance

kg m kg kg m =1 slope

3. Semiconductor heterostructures and nanostructures

Unit code: H/ QCF level: 5 Credit value: 15 OUTCOME 1 - THERMODYNAMIC SYSTEMS TUTORIAL 2

Transcription:

Kelv Planck Statement of te Second aw It is imossible to construct an enge wic, oeratg a cycle, will roduce no oter effect tan te extraction of eat from a sgle reservoir and te erformance of an equivalent amount of work. Clausius Statement of te Second aw It is imossible to ave a system oeratg a cycle wic transfers eat from a cooler to a otter body witout work beg done on te system by te surroundgs Reversible eat Enge Reversible Refrigerator

Actual eat Enge

Actual Refrigeration Mace

Carnot Power Cycle Reversible constant temerature rocess Reversible adiabatic, rocess exansion, rocess, rocess eat transfer, Efficiency Desired Effect Required Inut cycle ork

Carnot Refrigeration Cycle Reversible constant temerature eat transfer, rocess, rocess Reversible adiabatic exansion, rocess rocess Coefficien t of Performance CP CP refrigerator eat um Desired Effect Required Inut ork out ork

NECMEN AMSPERIC ENGINE Ford Museum,Detroit, 760, strokes/m

NECMEN AMSPERIC ENGINE Ford Museum, 760 strokes/m atmoseric ressure F 7 stroke 5 sia steam 8 50 F water

NECMEN AMSPERIC ENGINE atmoseric ressure Ford Museum, 760 strokes/m 5 sia steam 8 F 7 stroke 50 F water A πr F A F d Power Power π 6 (.7 sia.8 sia) 9 lbs ( 7/) ft ft lb 5,68 stroke 765,000 ft lb/m,000 /ft lb/m strokes m 6 9 lbs 5,68 ft lb/ stroke 5 sia ( F) ft lb 765,000 m P or 7.5 kw 50 F (.8 sia) v m V v 6 /stroke strokes ft / strokes ft /6.6 m lb. lb/m

NECMEN AMSPERIC ENGINE atmoseric ressure Ford Museum, 760 strokes/m 5 sia steam F A F d m 8 F A πr π 6 (.7 sia.8 sia) 9 lbs ( 7/) ft ft lb Power 5,68 stroke 765,000 ft lb/m Power,000 /ft lb/m 7 stroke strokes m 50 F water 6 9 lbs 5,68 ft lb/ stroke ft lb 765,000 m P or 7.5 kw V 6 /stroke strokes ft strokes /6.6. lb/m v ft / m lb g 5 sia ( F) v05.5 BU/lb 80 F (.507 sia) l9.08 v @5, sia 50.5 BU/lbm @ 80 l m F 9.08 BU/lbm,759. BU/lbm 765,000 ftlb/m 778 ftlb/bu ( 9.008 BU/lb). lb/m 50.5 BU/lb 9. BU/m CYCE CYCE CARN l 9.BU/m,759. BU/lbm 6.% 80 0.% 60 +

Carnot Prciles. No enge oeratg between two eat reservoirs, eac avg a fixed temerature, can be more efficient tan a reversible enge oeratg between te same reservoirs. actual Carnot. All reversible enges oeratg between two eat reservoirs, eac avg its own fixed temerature, ave te same efficiency.. e efficiency of any reversible enge oeratg between two reservoirs is deendent of te nature of te workg fluid and deends only on te temerature of te reservoirs.. An absolute temerature scale can be defed a manner deendent of te termometric material.

FIGURE 5-7 Proof of te first Carnot rcile. Coyrigt e McGraw-ill Comanies, Inc. Permission required for reroduction or dislay. 5-5

function(, ) from enge scematics by identity, function(, ) f(, substitutg, ) f(, ) f(, ) f(, ) f(, ) ermodynamic emerature Scale f(, ) tis equation can be satisfied only if, l and l l l A reversibleenge (or a real enge corrected to reversibe)can be used to measure temerature difference. Second aw eat Enge ermodynamic emerature Scale

SECND A and ut and s s.66 i i.6 i substituti ng.66 Scale Range.66 Scale Range for, is meaasured, - absolute temeratures. emerature scales can be setu for any arbitrarily sleected scale 0 ot and Scale Range of degrees between ice and steam. ScaleRange, en a reversible enge (or a real enge correctable to reversible ) is run between ice and steam temeratures wit a constant eat s s i out i i i s For ice as Scale 0 00 i 7.5.66 Celsius 0 7.5 K For : : Celsius 00 80 Scale Scale less 80 i 9.68 K.66 Fareneigt 0 9.68 - Fareneigt 0 Scale Range Fareneig t Scale Range tan ice as Scale 0 59.68 K R

Carnot Cycle Performance ork CP ork CP ork - CP are, and Carnot efficiency e scale, temerature termodynamic absolute te Usg out CYCE REVERSIBE CARN EA PUMP CYCE REVERSIBE CARN REFRIGERA R CYCE REVERSIBE CARN ENGINE

alf te work of an enge oeratg between 800 C and 0 C is used to ower a refrigeration mace absorbg eat at C and rejectg 6,000 kj/r at C ow muc eat is sulied to te enge? CP eat um eat um eat enge out CP ea ea,556 out out tum l tum kj/r out eat um 6,000 kj/r.8 89..75 l 89. kj/kg 780 K.75 800 + 7.5 7.5 + 0 eat enge.8 800 C eat um eat enge eat um C 0 C C 6,000 kj/kg 5-6

.00 kg steam executes te followg cycle. e absolute ig temerature is twice te absolute low temerature and te net work outut is 5 kj. eat is rejected durg a ase cange from a vaor to a liquid. at is te rejection temerature? l out l l l 5 kj.5 l 5 kj 50 kj.5 out m 5 kj out 5 kj 7. kj/kg m.00 kg @ fg 7. fg 5 8.6. 7. 0 0.5 5 ( 7. - 8.6) ( 0.5-7.) 5. C fg.00 kg steam S

00 K 00 K wo Carnot enges oerate series at te same efficiency. e ig temerature enge receives eat at 00 K and te low temerature enge rejects eat at 00. at is te temerature between te enges? 00 00 l ( 00 ) 00 ( 00 00) 88.5 00 00 00 00 00 00 00 K.5 5-

Sce cycle l oneof l 0 may be deendent of l te caracteristics at, of a termodynamocroerty. l In First aw, cycle ( d ) 0 lead to te defitionof termodynamic roerty E + energy as a

Ideal Gas Carnot Cycle roerty reversible cycle like a beaves tis d 0 P ln R ln R n d Note R ln P R ln R ln P P or P P constant v for P R ln R ln P R ln R ln l net n n n n n net rocess, rocess Reversible adiabatic exansion,,rocess rocess Reversible constant temerature eat transfer,

An engeer roosed an attemted to imrove te efficiency of a ower cycle by transferrg eat from te available ig temerature source to am alternate iger temerature source usg a eat um. at do you tk of tis suggestion?, enge CP eat[um enge enge eatum Enge enge eat um, eat Pum eat um were enge enge eat um eat um tere is no net work ga wit reversible maces and tere would be a net loss wit real maces.,