ThermodynamicThermodynamic PropertiesProperties

Š PropertyProperty TableTable -- from direct measurement ŠŠ EquationEquation ofof StateState -- any equation that relates P,v, and T of a substance

jump ExerciseExercise 33--1212

A bucket containing 2 liters of R-12 is left outside in the atmosphere (0.1 MPa) a) What is the R-12 temperature assuming it is in the saturated state. b) the surrounding transfer heat at the rate of 1KW to the liquid. How long will take for all R-12 vaporize? See R-12 (diclorindifluormethane) on Table A-2 SolutionSolution -- pagepage 11 Part a) From table A-2, at the saturation pressure of 0.1 MPa one finds: • Tsaturation = - 30oC

3 • vliq = 0.000672 m /kg 3 • vvap = 0.159375 m /kg

• hlv = 165KJ/kg (vaporization heat) SolutionSolution -- pagepage 22 Part b)

The mass of R-12 is m = Volume/vL, m = 0.002/0.000672 = 2.98 kg The vaporization energy: Evap = vap energy * mass = 165*2.98 = 492 KJ Time = Heat/Power = 492 sec or 8.2 min GASGAS PROPERTIESPROPERTIES Ideal -Gas Equation of State M PV = nR T; n = u mol Universal gas constant is given on

Ru = 8.31434 kJ/kmol-K = 8.31434 kPa-m3/kmol-k = 0.0831434 bar-m3/kmol-K = 82.05 L-atm/kmol-K = 1.9858 Btu/lbmol-R = 1545.35 ft-lbf/lbmol-R = 10.73 psia-ft3/lbmol-R EExamplexample Determine the particular gas constant for air (28.97 kg/kmol) and hydrogen (2.016 kg/kmol). kJ 8.1417 kJ = Ru = kmol − K = 0.287 Rair kg M 28.97 kg − K kmol

kJ 8.1417 kJ = kmol − K = 4.124 Rhydrogen kg 2.016 kg − K kmol IdealIdeal GasGas “Law”“Law” isis aa simplesimple EquationEquation ofof StateState

PV = MRT Pv = RT

PV = NRuT P V P V 1 1 = 2 2 T1 T2 QuestionQuestion …...…... Under what conditions is it appropriate to apply the ideal gas equation of state?

Š Good approximation for P-v-T behaviors of real gases at low densities (low pressure and high temperature).

Š Air, nitrogen, oxygen, hydrogen, helium, argon, neon, carbon dioxide, …. ( < 1% error). PercentPercent errorerror forfor applyingapplying idealideal gasgas equationequation ofof statestate toto steamsteam CompressibilityCompressibility FactorFactor Š It accounts mainly for two things •• MolecularMolecular structurestructure •• IntermolecularIntermolecular attractiveattractive forcesforces

Ideal Real Gas Gases Z > 1 or Z=1 Z < 1

Š What is it really doing? CompressibilityCompressibility FactorFactor ŠThe deviation from ideal-gas behavior can be properly accounted for by using the compressibility factor Z, defined as

•Z represents the volume ratio or compressibility, •Ideal Gas: Z =1; •Real Gases: Z>1 or Z<1. jump PrinciplePrinciple ofof correspondingcorresponding statesstates

ŠThe compressibility factor Z is approximately the same for all gases at the same reduced temperature and reduced pressure.

Z = Z(PR,TR) for all gases ReducedReduced PressurePressure andand TemperatureTemperature P T PR ≡ ; TR ≡ Pcr Tcr where:

PR and TR are reduced values. Pcr and Tcr are critical properties. CompressibilityCompressibility factorfactor forfor tenten substancessubstances (applicable(applicable forfor allall gasesgases TableTable AA--3)3) OTHEROTHER

THERMODYNAMICTHERMODYNAMIC

PROPERTIESPROPERTIES Other Thermodynamic Properties: Isobaric (c. pressure) Coefficient

v P 1  ∂v  β =   > 0 v  ∂TP  ∂v     ∂TP ForFor idealideal gas,gas, ββ == 1/T1/T

T Other Thermodynamic Properties: Isothermal (c. temp) Coefficient

v 1  ∂v  κ = −   > 0 v  ∂P T  ∂v     ∂P  T T ForFor idealideal gas,gas, κκ == 1/P1/P

P Other Thermodynamic Properties: We can think of the volume as being a function of pressure and temperature, v = v(P,T). Hence infinitesimal differences in volume are expressed as infinitesimal differences in P and T, using κ and β coefficients

 ∂v   ∂v  dv =   dT +   dP ≡ βvdT − κvdP  ∂T P  ∂P T

For ideal gas: dv/v = dT/T- dP/P →VP/T = const! For substances other than ideal gas, it can be approximated by:

 v  Ln  = β()()T − T0 − κ P − P0  v0  If κ and β are nearly constant, Other Thermodynamic Properties:

u = u(T,v) -- InternalInternal EnergyEnergy

h = h(T,P)≡ u+ Pv -- EnthalpyEnthalpy

s = s(u,v) -- EntropyEntropy Other Thermodynamic Properties: Specific Heat at Const. Volume

u v  ∂u  Cv =   > 0  ∂Tv

 ∂u     ∂Tv

T Other Thermodynamic Properties: Specific Heat at Const. Pressure

h P  ∂h  CP =   > 0  ∂TP

 ∂h     ∂TP

T SpecificSpecific HeatsHeats forfor SomeSome GasesGases

Š Cp = Cp(T) a function of temperature IdealIdeal Gases:Gases: uu == u(T)u(T)

 ∂u   ∂u  0 du =   dT +   dv  ∂T  v  ∂v  T Therefore,  ∂u  du =   dT = C v (T )dT  ∂T  v The internal energy change is: T ∆=uu − u =∫ 2 C(T) dT 21 T1 v

As Cv changes with temperature it cannot be pulled out from the integral (for general cases). EnthalpyEnthalpy forfor anan IdealIdeal GasGas

Š h = u + Pv where Pv can be replaced by RT because Pv = RT.

Š Therefore, h = u + RT => since u is only a function of T, R is a constant, then h is also only a function of T!

Š so h = h(T) Similarly,Similarly, forfor aa changechange inin enthalpyenthalpy forfor idealideal gases:gases:

 ∂h  C p = C p (T ) &   ≡ 0  ∂P 

dh = C pdT, and

T2 ∆h = h2 − h1 = C p (T)dT ∫ T 1 Summary:Summary: IdealIdeal GasesGases

ŠFor ideal gases u, h, Cv, and Cp are functions of temperature alone.

ŠFor ideal gases, Cv and Cp are written in terms of ordinary differentials as

 du   dh  C v =   ; C p =    dT  ideal gas  dT  ideal gas ThreeThree WaysWays toto CalculateCalculate ∆∆uu andand ∆∆hh

Š ∆u = u2 -u1 (table) Š ∆h = h2 -h1 (table)

2 2 Š ∆u = C v (T) dT Š ∆h = C p(T) dT ∫ 1 ∫ 1

Š ∆u = Cv,av ∆T Š ∆h = Cp,av ∆T Problem 3.31 – N2 is heated from 373K to 1773K at constant pressure. Evaluate the specific entalpy change (KJ per kg).

Š ∆h ≈ Cp,300k ∆T = (tab. A-7) 1,0416x(1773-373) = 1457kJ/kg

2 Š ∆h = C (T) dT tab A-5, θ = t(K)/100 ∫ 1 p Cp = 39,060 - 512,79θ -1.5+1072,7θ -2-820,40θ -3 ∆h =1635.98 kJ/kg

There is an error of 11% for using Cp constant when the temperature span is of 1400K! CpCp xx CvCv RelationshipRelationship forfor anan IdealIdeal Gas,Gas,

Š h = u + Pv = u + RT dh du = + R dT dT  kJ  Cp = Cv + R   kg ⋅ K  RatioRatio ofof SpecificSpecific HeatsHeats γγ Cp Cp(T) γ ≡ = = γ(T) Cv Cv(T) ForFor mostmost gasesgases γγ isis almostalmost constantconstant withwith temperaturetemperature andand equalsequals toto 1.41.4

Cp R = 1 + = γ > 1 Cv Cv

R γR C = and C = v γ-1 p γ-1 ISOTHERMAL, POLITROPIC AND ADIABATIC PROCESSES FOR AN IDEAL GAS IsothermalIsothermal ProcessProcess

Š Ideal gas: PV = mRT = constant ForFor idealideal gas,gas, PVPV == mRTmRT WeWe substitutesubstitute intointo thethe integralintegral

2 2 mRT Wb = PdV = dV ∫∫ 1 1 V

Collecting terms and integrating yields:

2 dV V2  Wb = mRT = mRTAn  ∫ 1   V  V1  PolytropicPolytropic Process: Process: PVn = C; 11 ≤≤ nn ≤≤γ≤ γγ Isothermal …………… n = 1 Adiabatic (Q=0)………. n = γ = Cp/Cv Others: Constant pressure………n = 0 Constant volume ……… n = ∞ PV1 = c

PVγ = c

The politropic lines are always to the right of the isothermal lines. P,P, vv andand TT RelationshipRelationship inin aa PolytropicPolytropic ProcessProcess,, 11≤≤nn ≤≤11

Š The path is described by: Pvn = C

Š The ideal gas state equation is: Pv = RT

Š Combining these two expressions is possible to relate the initial to the final states

nn1− Pv  Tv 21==  21 Pv12  Tv 12 n1− TPn 22= TP 11 jump BoundaryBoundary workwork forfor aa gasgas whichwhich obeysobeys thethe polytropicpolytropic equationequation

2 2 dV Wb = ∫ 1 PdV = c∫ 1 Vn

1-n v2 1−n 1−n V  V2 − V1  = c   = c  para ()n ≠ 1  1-n   1 − n  v1  

Where C stands for the constant Pvn = C WeWe cancan furtherfurther simplifysimplify

n n The constant c = P1V1 = P2V2

n 1−n n 1−n 2V 2 (V )− 1V 1 (V 1 ) W = P 2 P b 1-n P V − PV = 2 2 1 1 , n ≠ 1 1− n SummarySummary forfor PolytropicPolytropic ProcessProcess

2 2 c Wb = PdV = dV ∫ 1 ∫ 1 V n P V − PV = 2 2 1 1 , n ≠ 1 1− n

V 2  = PVAn  , n = 1  V1  Ideal Gas Adiabatic Process and Reversible Work 㠊 Why Pv = C P f represents a process f f

where the volume is Q = T = 0 expanding or c o n st contracting without .. heat flux, Q = 0? i v Ideal Gas Adiabatic Process and Reversible Work (cont)

First Law: dQN − dWN = dUN =0 PdV MCvdT

Substituting  dV  C  dT   = − V   P = MRT/V  V  NR  T  ()1−γ −1

Integrating from (1−γ ) γ  T2   V2   P2   V1  (1) to (2)   =   ⇒   =    T1   V1   P1   V2 

Which are the polytropic relations seen before! ExerciseExercise 33--3030 Air is compressed reversibly and adiabatically from a pressure of 0.1 MPa and a temperature of 20oC to a pressure of 1.0 MPa. a) Find the air temperature after the compression b) What is the density ratio (after to before compression) c) How much work is done in compressing 2 kg of air? d) How much power is required to compress 2kg per second of air? SolutionSolution -- pagepage 11

Š In a reversible and adiabatic process P, T and v follows: P f (γ−1) f f  T2   V1    =   Q T V =  1   2  T = 0 c o γ n  P   V  st  2  =  1  ..  P1   V2  i

(γ−1) γ v  T   P   2  =  2   T1   P1  SolutionSolution -- pagepage 22

Part a) The temperature after compression is (γ−1) γ 0.4 1.4  P2   1  T2 = T1 ⋅   → T2 = 293⋅   = 566K (293oC)  P1   0.1

Part b) The density ratio is −1 γ 1 1.4  V   ρ   P   ρ   1   2  ≡  1  =  2  →  2  =   = 5.179  V1   ρ2   P1   ρ1   0.1 SolutionSolution -- pagepage 33 Part c) The reversible work: [(PV) − (PV) ] M ⋅ R(T − T ) W = − 2 1 ≡ 2 1 = REV ()γ − 1 ()γ − 1 2 ⋅ 287 ⋅ ()566 − 293 = = 391KJ Part d) 0.4 The power is: dW M R(T − T ) P = = − 2 1 = 391KW dt ()γ − 1 Exercícios – Capítulo 3 Propriedades das Substâncias Puras

Exercícios Propostos: 3.6 / 3.9 / 3.12 / 3.16 / 3.21 / 3.22 / 3.26 / 3.30 / 3.32 / 3.34

Team Play: 3.1 / 3.2 / 3.4