1Chapter 3: Evaluating PropertiesProperty Relations in Engineering ThermodynamicsLava flowing into the Pacific Ocean in Hawaii. Photo courtesy of Mike Benson.Objectives1. How to define the state of a system.2. How many properties are needed to define the state of a system?3. What happens when phase changes occur, or when the system is composed of more than one phase?4. How to retrieve values for the properties ?2Definitions• Simple system: a system is referred to as simple system if there is only one way the system energy can be significantly altered by work as the system undergoes quasiequilibriumprocesses.• Simple compressible system: simple system where the energy change by work is associated with volume change.The State principleThe number of independent properties is one plus the number of relevant work interactions.3The State PrincipleTwo independent, intensive, thermodynamic properties are required to fix the state of a simple compressible system.For example:P and vT and ux and hIntensive thermodynamic properties:Less used:g - Gibbs free energy a - Helmholz free energyv – specific volumeT – absolute temperatureP –absolute pressures –specific entropyx – quality(steam only)u – specific internal energyh – specific enthalpyP-v-T RelationNote the location of the following:- Single phase regions- Two phase regions- Saturation states- Triple line- Critical Point4T-v diagramDiagram courtesy of Jerry M. Seitzman, 2001.Phase ChangesConstant Pressure phase change of water:a) Subcooled or Compressed Liquidb) Two-phase liquid-vapor mixture (steam)c) Superheated vapor5Vapor and Liquid TablesFor Water:– Superheated vapor properties in Tables A-4, A-4E– Compressed liquid properties in Tables A-5, A-5EThermodynamic Tables6Linear Interpolation:Between values in the tablesLLHLHLTT vvTTvv−−=−−Subscripts:L – Value in table at lower endH – Value in table at upper endNone – value of interestQualityFor use in Tables A-2 and A-3• For Saturated Mixture (Liquid-Vapor) Region– Quality; x; an intensive property– x gives fraction that is vapor (gas)– (1-x) gives Moisture Contentgfgmmmx+≡0 ≤ x ≤ 1; x = 0 → Saturated Liquid (subscript ‘f’) x = 1 → Saturated Vapor (subscript ‘g’)‘fg’→ ‘g’-’f’7Example 1• Determine the phase of H2O ata. p=5 bar, T=151.9 oCb. P=5 bar, T=200 oCc. p=1.32 bar, T=100 oCd. p=0.221 bar, T=100 oCe. T=200 oC , v=0.1 m3/KGf. T=200 oC, v=0.2 m3/KGExample 2• 1 Kg of H2O goes from state 1 to states 2, 3, and 4.1. State 1: T1=100 oC, P1=3 bar2. State 2: T2=200 oC, P2=3 bar 3. State 3: P3=1 bar, v3=0.716 m3/Kg4. State 4: P4=1 bar, T4=100 oC• Show the transitions on the p-v and T-v diagrams.8Example: Problem 3.61Quality RelationsLET b = ANY INTENSIVE PROPERTY– (b = v, u, h, s, etc.)(1 )ffgf fgffgfg g fgfbb bbxbb bbb xbbbbbxb xb−−==−=+⋅=−=⋅ + − ⋅9EnthalpyHUPVhuPv=+⋅=+⋅Enthalpy is a property constructed due to the frequent occurrence of the above combination of properties.vvucT∂⎞=⎟∂⎠pphcT∂⎞=⎟∂⎠Specific Heats (Heat Capacities)pvckc=10Approximations for liquids and solids• Using Saturated Liquid Data (‘Compressed Liquid Rule’)• Using ‘Incompressible Substance Model’ffffvvhhuuss≈≈≈≈21 2121 21 2121 21()()()()pvcccuucTThhcTT vPPhhcTT==−= −−= − + −−≈ −Example: Problem 3.6911Generalized Gas Models • Pressure and temperature of a gas determines its state• Other gas properties, such as speicificvolume and internal energy, vary as a function of pressure and temperature• All gases exhibit similar behavior (ideal gas model) at high temperature and low pressureUniversal Gas ConstantTvpRp 0→= limUniversal gas constant:Where nVv =n: number of moles=m/M⎪⎩⎪⎨⎧=R lbf/lbmole-ft 1545R Btu/lbmole 9861K KJ/mole 3148ooo..R12Compressibility Factor• The dimensionless ratio is called the compressibility factor• Z tends to 1 as pressure approaches zero• This is because at low pressures the molecular forces become negligible• Under this condition, for a fixed number of molecules (say 1 Mole), pressure increases proportionally to temperature.TRvpZ =Compressibility ChartcRppp =cRTTT =13Properties for Ideal GasesPvnRTPvRTPVmRT===The Ideal Gas Model:()() ()uuThhT uT RT===+()()vpducTdTdhcTdT==Tables A-22(E) and A-23 (E)When specific heats are assumed constant Table A-20(E):21 2121 21()()vpuucTThhcTT−=−−= −Requirements:1ccZPPTT≈RRM=Example: Problem 3.9614Example: Problem 3.97Polytropic Process of an Ideal GasFor a closed system:2112constantPVPVnnPV =⎛⎞=⎜⎟⎝⎠Expansion/Compression (Moving Boundary) Work (Ideal Gas OR liquid):222 111221111,( 1)1ln ,( 1)PV PVPdV nnVPdV PV nV−⋅= ≠−⎛⎞⋅= =⎜⎟⎝⎠∫∫Ideal Gases ONLY:(1)/ (1)22 111 222112211(),( 1)1lnnn nTP VTP VmR T TPdV nnVPdV mRTV−−⎛⎞ ⎛⎞==⎜⎟ ⎜⎟⎝⎠ ⎝⎠−⋅=≠−⎛⎞⋅=⎜⎟⎝⎠∫∫15ExampleShow that for an ideal gas undergoing a polytropicadiabatic process n=k.Summary• State Principle: Number of independent properties of a system=1+Number of “relevant” work interactions.• For a pure substance 2 independent properties (P,T,v,u,h) fix the state.16How to find properties & phase• Compressed Liquid– Compressed liquid tables(only water A-5)–v,u≈vf,ufat the same temperature– Incompressible substances model-cp,cv,ρ (table A-19)• Tow-phase liquid-vapor mixture– P,T not independent– X-quality is needed– Liquid and vapor components have saturated properties; use saturation table• Superheated vapor– Vapor tables– Ideal gas model• Saturation & superheated tables are given for water, R22, R-134, Ammonia, and