Thermodynamics EAS 204 Spring 2004Class Month Day Chapter Topic Reading Due

1 January 12 M Introduction2 14 W Chapter 1 Concepts Chapter 1

19 M MLK Holiday no class3 21 W Chapter 2 Properties Chapter 2 PS14 26 M Chapter 2 Properties PS 25 28 W Chapter 2 Properties6 February 2 M Chapter 3 Heat&Work PS 37 4 W Chapter 3 Heat&Work8 9 M Chapter 4 First Law Chapter 4 PS 49 11 W Chapter 4 First Law

10 16 M Chapter 4 First Law PS 511 18 W Chapter 4 First Law12 23 M Review Chapter 5 PS 613 25 W Exam 1-414 March 1 M Chapter 5 Second Law Chapter 5 PS 715 3 W Chapter 5 Second Law16 8 M Chapter 5 Second Law17 10 W Chapter 6 Entropy Chapter 6

15 M Spring Recess no class PS 817 W Spring Recess no class

18 22 M Chapter 6 Entropy PS 919 24 W Review20 29 M Exam 5-6 PS 1021 31 W Chapter 8 Gas Power Chapter 822 April 5 M Chapter 8 Gas Power PS 1123 7 W Chapter 9 Vapor Power Chapter 924 12 M Chapter 9 Vapor Power PS 1225 14 W Chapter 9 Vapor Power26 19 M Refrigeration Cycles Chapter 10 PS 1327 21 W Refrigeration Cycles28 26 M Review PS 14

TBA FINAL

Course Summary Chapter 5 Second LawStatement and CorollariesHeat EnginesReversible engines and refrigeratorsCarnot Cycle

Chapter 1 ConceptsThermodynamic system, properties, state

point, process, cycle, heat and work.Thermodynamic problem solving technique. Chapter 6 Entropy

Second Law and heat enginesThe Entropy propertyIsentropic processEntropy change calculation

Chapter 8 Gas Power CyclesBrayton (gas turbine) cycleOtto (spark ignition engine) cycleDiesel cycle

Chapter 9 Vapor Power CyclesRankine (steam power) reheat, superheat

and regeneration cycles

Chapter 10 Refrigeration CyclesVapor Compression CycleHeat PumpsReversed Brayton Cycle

Chapter 2 Fluid PropertiesReal gases - steam, air, refrigerants tablesIdeal gasesEquations of state - EES CD

Chapter 3 Heat and WorkWork in non-flow, steady flow

and unsteady flow systems.Adiabatic Process

Chapter 4 First LawFirst Law for processes and cyclesHeat and work in closed non-flow, open

flow and unsteady flow systems.

THERMODYNAMICS OVERVIEWThermodynamics

Thermodynamics is the study of the relationship between allforms of Energy beginning historically with the relationshipbetween Heat and Work.

Laws of Thermodynamics ( fundamental observations)Mass Balance (should be a law)

Mass can not be created or destroyed and is conserved.

First LawHeat and work are equivalentProperty energy is definedEnergy can change form, can not be destroyed, and is conserved

Second LawHeat can not be converted completely to work.Property entropy defined.Ideal and actual heat engine efficiency.

THERMODYNAMICS IN DESIGN

Thermodynamic Analysis – Process Analysis

Thermodynamic analysis is the first step in energy system design.

Through Thermodynamic Analysis the required mass flows, volume flows, temperatures and pressuresare established. Performance is determined.

ENERGY SYSTEMS

Food Processing PlantsRocket EnginesAir Conditioning SystemsRefrigeration SystemsHeating SystemsGas Turbine Engines

Gasoline EnginesDiesel EnginesSteam Power PlantsChemical PlantsCompression SystemsGas Liquefaction

Thermodynamics Concepts

Thermodynamic SystemPropertiesState Point ProcessCycleHeatWorkEnergy

Thermodynamic

HEAT

Control Volume

3 Types of SYSTEMSClosedOpenUnsteady

WORK

MASS

ConceptsSystemPropertiesState PointProcessCycle

CLOSED THERMODYNAMIC SYSTEMNON FLOW SYSTEM

A MASS OF MATERIAL. Heat and Work can

cross the system boundaries. Mass can not.

Examples:A closed tank.A piston cylinder. A balloon

WorkHeat

OPEN THERMODYNAMIC SYSTEMSTEADY FLOW THERMODYNAMIC SYSTEM

A FIXED REGION IN SPACEMass, heat and work can

cross the system boundaries.Examples:

turbine, compressors, boilers, heat exchangers

ConceptsSystemPropertiesState PointProcessCycle

Mass in Mass out

HeatWork

ConceptsSystemPropertiesState PointProcessCycle

Work

UNSTEADY FLOW THERMODYNAMIC SYSTEM

A VARIABLE MASSMass, heat and work cancross the system boundaries

Examples:A filling tank.An emptying tankPiston-Cylinder Filling

Heat

Mass

ConceptsSystemPropertiesState PointProcessCycle

Identify the thermodynamic system, 1) open,steady flow thermodynamic system2) closed,non-flow thermodynamic system3) unsteady flow thermodynamic system

in the following problems.

2-48 closed system, mass of water in the piston cylinder.2-123 closed system, mass of hydrogen in both tanks3-50 open system, region in space occupied by the nozzle3-74 closed system, mass of ball4-11 closed system, mass of steam in the radiator4-84 open system, region in space occupied by the turbine4-155 unsteady system, mass initially in the tank5-84 open system, region in space occupied by the heat engine6-100 open system, region in space occupied by the compressor6-132 open system, region in space occupied by the mixing chamber

Thermodynamic Properties ConceptsSystemPropertiesState PointProcessCycle

BTU kJ,FBTU/lbm C,/kgm kG/jg, olume,constant vat heat specifcc

dTcdu BTU/lb kJ/kg,

vmassV ft,m

lbm/ft,kg/m/lbft/kg,m

pressureAmbient pressure Gagepressure Absolute lb/in bar, s,atmosphere kPa,

459.69FR273.15CK

32.C1.8FRK, absolute, C,F,

oo3v

v

33

33

33

2

oo

oo

oo

oooo

−−

=−

×=−

−

−

+=−

+=

+=

+×=

−

ergyInternalEn U

gyternalEnerSpecificIn u

VolumeV Density ρ

lumeSpecificVo v

Pressure p

eTemperatur

Thermodynamic Properties ConceptsSystemPropertiesState PointProcessCycle

smS KkJ/

FBTU/lbm C,kJ/kghmH

BTU kJ,

FBTU/lbm C,kJ/kg pressure,constant at heat specificc

dTcdh pvuh

BTU/lbm kJ/kg,

o

oo

oop

p

×=−

−

×=−

=+=

−

Entropy SEntropy Specific s

Enthalpy H

Enthalpy Specific h

PRESSURE

gageatmabs PPP +=

Force Units - force = mass x acceleration

2

2

2

m/sec9.807 ofon acceleratian at N 9.807 weighsmass kg l exerts.it force nalgravitatio

themeasuringby indirectly measured is Mass

m/sec 9.807 kg 1N 9.807m/sec 1 kg 1N 1•=

•=

2fm

2f

m

2mf

2cmf

ec32.174ft/s ofon acceleratian at lb 1 weighslb 1

ft/sec 1slug llb 1

slugs32.174

1 lb 1

ft/sec 32.174lblb 1

ft/sec glblb 1

•=

=

•=

•=

Energy Units - force x distance, mass and temperature change

C

C

oo

oo

raised15 C15at water kg 4.1816kJ 1calories 4.1868J 1

15 raised C15at water g 1Calorie1m1NJ 1

=

=−

•=

f

oom

ff

ftlb 7781BTUF1 raised F60at water lb 1BTU

1ft1lb1ftlb

=−

•=

Power Units - energy per time

1kJ/sec1kw1J/sec1watt

−−

.7457HP1kw/sec550ftlb1HP f

==

THERMODYNAMIC STATE POINT

Properties are measured.

Equations and models are fitted to the data resulting in :

Tables of Property ValuesEquations of StateComputer property modules

Two properties define thestate point of a single phasefluid.

One property defines the state point of a multiphasefluid.

abscissa property

ordinate property

(T,p,v,u,h,s)

ConceptsSystemPropertiesState PointProcessCycle

THERMODYNAMIC PROCESSA thermodynamic process is an interaction between a thermodynamic system and

its surroundings which results in a change in the state point of the system

ConceptsSystemPropertiesState PointProcessCycle

Reversible ProcessA process is reversible if the state points of all affectedthermodynamic systems, including the external systemor surroundings, are returned to their original state point values.Examples:

- movement of a frictionless pendulum- transfer of work to potential energy without loss- movement of a frictionless spring

Irreversible ProcessA process which can not be reversing bringing all the affected thermodynamic properties back to their original values is irreversible.

Examples:

- applying brakes to a moving wheel

- mixing hot and cold water

- transfer of heat through a finite temperature difference

ConceptsSystemPropertiesState PointProcessCycle

THERMODYNAMIC CYCLE

A thermodynamic system undergoesa cycle when the system is subjected to a seriesof processes and all of the state point properties of the system are returned to their initial values.

p

v

∫∫

∑

∑

=

=

=

WQ

LawFirst

WW

QQ

cycleprocessnetcycle

cycleprocess

netcycle

δδ

initial point

1. Problem StatementCarbon dioxide is contained in a cylinder

with a piston. The carbon dioxide is compressedwith heat removal from T1,p1 to T2,p2. The gasis then heated from T2, p2 to T3, p3 at constant volume and then expanded without heat transfer to the original state point.

4. Property Diagramstate points - processes - cycle

T1,p1

T3,p3

T2,p2

p

Thermodynamic Problem Solving Technique

QW

QW

2. Schematicv

5. Property Determination

T2,p2

1 2 3 T pvuhs

3. Select Thermodynamic Systemopen - closed - control volume

a closed thermodynamic systemcomposed to the mass of carbondioxide in the cylinder

p

v

2CO

6. Laws of ThermodynamicsQ=? W=? E=? material flows=?