First Law Second La · T3,p3 T2,p2 p Thermodynamic Problem Solving Technique W Q Q W 2. Schematic v 5. Property Determination T2,p2 1 2 3 T p v u h s 3. Select Thermodynamic System

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

First Law - Energy Balance lead the analysis

Thermodynamic SystemControl Volume

HEATWORK

MASS

3 Types of SYSTEMSClosedOpenUnsteady

Heat added to a system and Work doneby a system are positive (+) by convention.

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


Mass in Mass out

HeatWork


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


The mass of water is a closed thermodynamic system. Heat(+) is added to the closed thermodynamic system from the surroundings system. Work (+) is done by the closed thermodynamic system on the surroundings system.

The boiler, turbine, condenser and pump are open thermodynamic systems Heat (+) is transferred into the open boiler system from the surroundings system. Work (+) is done by the open turbine system on the surrounding system. Heat (-) is rejected from the condenser open system to the surrounding system. Work(-) is done on the open pump open system by the surroundings system.


CONCEPTSSystemPropertiesState PointProcessCycle

Thermodynamic PropertiesTemperaturePressure VolumeInternal EnergyEnthalpyEntropy

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,

33

33

33

2

oo

oo

oo

oooo

×=−

−

−

+=−

+=

+=

+×=

−

VolumeV Density ρ

lumeSpecificVo v

Pressure p

eTemperatur


Thermodynamic Properties

BTU kJ,FBTU/lbm C,/kgm kG/jg,

olume,constant vat heat specifcc dTcdu

BTU/lb kJ/kg,

oo3v

v

−

−=

−

ergyInternalEn U

gyternalEnerSpecificIn u

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)


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


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

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

vinitial point

dependent.path is that ederiviativ a al,differentiinexact an is δ

WQ

LawFirst

WW

QQ

cycleprocessnetcycle

cycleprocess

netcycle

∫∫

∑

∑

δ=δ

=

=



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=?

Documents

First Law Second La · T3,p3 T2,p2 p Thermodynamic Problem Solving Technique W Q Q W 2. Schematic v 5. Property Determination T2,p2 1 2 3 T p v u h s 3. Select Thermodynamic System