first law second la · t3,p3 t2,p2 p thermodynamic problem solving technique w q q w 2. schematic v...
TRANSCRIPT
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
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
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
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
CONCEPTSSystemPropertiesState PointProcessCycle
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)
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
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
cycleprocessnetcycle
cycleprocess
netcycle
∫∫
∑
∑
δ=δ
=
=
CONCEPTSSystemPropertiesState PointProcessCycle
ConceptsSystemPropertiesState PointProcessCycle
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=?