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Thermodynamic Propertiesproperties ThermodynamicThermodynamic PropertiesProperties PropertyProperty TableTable -- from direct measurement 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? Good approximation for P-v-T behaviors of real gases at low densities (low pressure and high temperature). Air, nitrogen, oxygen, hydrogen, helium, argon, neon, carbon dioxide, …. ( < 1% error). PercentPercent errorerror forfor applyingapplying idealideal gasgas equationequation ofof statestate toto steamsteam CompressibilityCompressibility FactorFactor It accounts mainly for two things •• MolecularMolecular structurestructure •• IntermolecularIntermolecular attractiveattractive forcesforces Ideal Real Gas Gases Z > 1 or Z=1 Z < 1 What is it really doing? CompressibilityCompressibility FactorFactor The 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 The 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 ∂TP ∂v ∂TP 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 ∂Tv ∂u ∂Tv T Other Thermodynamic Properties: Specific Heat at Const. Pressure h P ∂h CP = > 0 ∂TP ∂h ∂TP T SpecificSpecific HeatsHeats forfor SomeSome GasesGases 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 h = u + Pv where Pv can be replaced by RT because Pv = RT. 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! 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 For ideal gases u, h, Cv, and Cp are functions of temperature alone. For 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 ∆u = u2 -u1 (table) ∆h = h2 -h1 (table) 2 2 ∆u = C v (T) dT ∆h = C p(T) dT ∫ 1 ∫ 1 ∆u = Cv,av ∆T ∆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). ∆h ≈ Cp,300k ∆T = (tab. A-7) 1,0416x(1773-373) = 1457kJ/kg 2 ∆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, 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 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 The path is described by: Pvn = C The ideal gas state equation is: Pv = RT 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 γ 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 In a reversible and adiabatic process P, T and v follows: P f (γ−1) 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.
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