上海交通大学：《材料热力学》教学资源_07电子教案（英文课件）lecture 3 second law I

Contents of Today S.J.T.U. Phase Transformation and Applications Continue /Review previous 2nd law Entropy Reversible process Carnot cycle etc. SJTU Thermodynamics of Materials Spring2006©X.J.Jin Lecture 3 Second law I

Phase Transformation and Applications S. J. T. U. SJTU Thermodynamics of Materials Spring 2006 © X. J. Jin Lecture 3 Second law I Contents of Today Continue / Review previous 2nd law Entropy Reversible process Carnot cycle etc

1.18 Enthalpies of Formation (1) S.J.T.U. Phase Transformation and Applications The enthalpy change of a material between states 1 and 2: △H=H2-H1 We could consider the mass itself to be a form of energy. Standard state A reference state:to assign a value of zero to enthalpy for certain materials in that reference state. Elements in their equilibrium states at 298 K and one atmosphere pressure. EX., Enthalpy of diatomic oxygen at 298 K and one atm:ZERO Enthalpy of monatomic oxygen at 298 K and one atm:NOT ZERO Enthalpy of graphite at 298 K and one atm:ZERO Enthalpy of diamond at 298 K and one atm:NOT ZERO SJTU Thermodynamics of Materials Spring2006©X.J.Jin Lecture 3 Second law I

Phase Transformation and Applications S. J. T. U. SJTU Thermodynamics of Materials Spring 2006 © X. J. Jin Lecture 3 Second law I 1.18 Enthalpies of Formation (1) The enthalpy change of a material between states 1 and 2: Δ = − HHH 12 We could consider the mass itself to be a form of energy. A reference state: to assign a value of zero to enthalpy for certain materials in that reference state. Elements in their equilibrium states at 298 K and one atmosphere pressure. Ex., Enthalpy of diatomic oxygen at 298 K and one atm: ZERO Enthalpy of monatomic oxygen at 298 K and one atm: NOT ZERO Enthalpy of graphite at 298 K and one atm: ZERO Enthalpy of diamond at 298 K and one atm: NOT ZERO Standard state

1.19 Enthalpy Change in Chemical Reactions(2) S.J.T.U. Phase Transformation and Applications In general,the enthalpy change for a chemical reaction at a specified temperature may be expressed as the sum of the enthalpies of formation of the products less the enthalpies of formation of the reactions at that temperature. 1.e. AHr=∑n,AHr-n,AHf.r products reac tan ts where np and n,are the stoichiometric coefficients of the reaction. Enthalpy of combusion the enthalpy change for methane when it is reacted with oxygen,or the heat of combustion of methane. Low heat of combustion:gas water;High heat of combustion:liquid water. SJTU Thermodynamics of Materials Spring 2006 ©X.J.Jin Lecture 3 Second law I

Phase Transformation and Applications S. J. T. U. SJTU Thermodynamics of Materials Spring 2006 © X. J. Jin Lecture 3 Second law I 1.19 Enthalpy Change in Chemical Reactions (2) ∑ ∑ Δ−Δ=Δ tsreac r Tf products T p HnH Tf Hn tan , , i. e., In general, the enthalpy change for a chemical reaction at a specified temperature may be expressed as the sum of the enthalpies of formation of the products less the enthalpies of formation of the reactions at that temperature. where np and nr are the stoichiometric coefficients of the reaction. Enthalpy of combusion : the enthalpy change for methane when it is reacted with oxygen, or the heat of combustion of methane. Low heat of combustion: gas water; High heat of combustion: liquid water

1.19 Enthalpy Change in Chemical Reactions (4) S.J.T.U. Phase Transformation and Applications T-T △HT T-T R aInjeledwaI (C,dT). F(CdT). T=298 △H298 T=298 R P (Reactants (Products) Enthalpy is a point function s∑(nC,dT+△H,=AHs+f∑(nC,dr) △H,=AHs+f[∑(nCr）nias-∑6nCp）)enH SJTU Thermodynamics of Materials Spring 2006 ©X.J.Jin Lecture 3 Second law I

Phase Transformation and Applications S. J. T. U. SJTU Thermodynamics of Materials Spring 2006 © X. J. Jin Lecture 3 Second law I 1.19 Enthalpy Change in Chemical Reactions (4) T=T R T=298 R T=T P T=298 P ΔH298 ΔHT ( ) ∫T P R dTC 298 ( ) ∫T P P dTC 298 Temperature (Reactants) (Products) Enthalpy is a point function ( ) ( ) ∫ ∑ +Δ=Δ+ ∫ ∑T T P products T P tsreac dTnC HH dTnC 298 298 298 tan HH [ () () nC nC ]dT T T ∫ ∑ P products ∑ P tsreac +Δ=Δ − 298 298 tan

1.20 Adiabatic Temperature Change in Chemical Reactions(1) S.J.T.U. Phase Transformation and Applications Adiabatic flame temperature Fuel gas 20% Carbon monoxide(CO) 30% Carbon dioxide(CO2) 50% Nitrogen(N2) Gas Moles in Moles out Exit gas composition(%) co 0.20 0 C02 0.30 0.50 36 02 0.10 0 N2 0.5+0.1×(79/21) 0.88 64 Total 1.38 100 SJTU Thermodynamics of Materials Spring2006©X.J.Jin Lecture 3 Second law I

Phase Transformation and Applications S. J. T. U. SJTU Thermodynamics of Materials Spring 2006 © X. J. Jin Lecture 3 Second law I 1.20 Adiabatic Temperature Change in Chemical Reactions (1) Adiabatic flame temperature Gas Moles in Moles out Exit gas composition (%) CO 0.20 0 CO2 0.30 0.50 36 O2 0.10 0 N2 0.5+0.1×(79/21) 0.88 64 Total 1.38 100 Fuel gas 20 % Carbon monoxide (CO) 30 % Carbon dioxide (CO2) 50 % Nitrogen (N2) 2 2 2 1 →+ COOCO

1.20 Adiabatic Temperature Change in Chemical Reactions(3) S.J.T.U. Phase Transformation and Applications ∑H。n。=∑H,n co+0,→c0 AHf.c0.298=-110,500J Cp.N,=34.3 J /(mol.K) △Hf.c0,.298=-393,500J Cp.co,=57.3 J/(mol.K) T=1260K AFT:adiabatic flame temperature The maximum temperature of these gases leaving the burner. Moisture,combustion reaction completion or not. Chemical equilibrium SJTU Thermodynamics of Materials Spring2006©X.J.Jimn Lecture 3 Second law I

Phase Transformation and Applications S. J. T. U. SJTU Thermodynamics of Materials Spring 2006 © X. J. Jin Lecture 3 Second law I 1.20 Adiabatic Temperature Change in Chemical Reactions (3) ∑ oo = ∑ iinHnH 2 2 21 →+ COOCO H J Δ COf 298,, = − 500,110 H J COf 298,, 500,393 2 Δ = − C ( KmolJ ) NP = /3.34 ⋅ 2 , C ( KmolJ ) COP = /3.57 ⋅ 2 , =1260 KT AFT: adiabatic flame temperature The maximum temperature of these gases leaving the burner. Moisture, combustion reaction completion or not. Chemical equilibrium ?

Review Key points S.J.T.U. Phase Transformation and Applications 1.Adiabatic process:Joule-Thomson expansion 2.Equations of state 3.Adiabatic compression or expansion 4.Enthalpies of formation 5.Enthalpy changes in chemical reactions 6.Adiabatic temperature change in chemical reactions SJTU Thermodynamics of Materials Spring2006©X.J.Jin Lecture 3 Second law I

Phase Transformation and Applications S. J. T. U. SJTU Thermodynamics of Materials Spring 2006 © X. J. Jin Lecture 3 Second law I Review / Key points 1. Adiabatic process: Joule-Thomson expansion 2. Equations of state 3. Adiabatic compression or expansion 4. Enthalpies of formation 5. Enthalpy changes in chemical reactions 6. Adiabatic temperature change in chemical reactions

1.17 Adiabatic Compression or Expansion (1) S.J.T.U. Phase Transformation and Applications Air arises rapidly (adiabatically)up a mountainside. At the higher altitude,the pressure is lower,and the temperature is also lower. Adiabatic Reversible:Presisting=Psystem Ideal gas System boundary +oW -dl -Pdv du nCvdT P system X Presisting PV=ART Pav +Vap nrdT 绝热压缩和膨胀 SJTU Thermodynamics of Materials Spring2006©X.J.Jin Lecture 3 Second law I

Phase Transformation and Applications S. J. T. U. SJTU Thermodynamics of Materials Spring 2006 © X. J. Jin Lecture 3 Second law I 1.17 Adiabatic Compression or Expansion (1) Air arises rapidly (adiabatically) up a mountainside. At the higher altitude, the pressure is lower, and the temperature is also lower. Presisting System boundary Psystem Adiabatic Reversible: Presisting=Psystem Ideal gas δ +δ = dUWQ =− = V dTnCdUPdV nRdTVdPPdV nRTPV =+ = 绝热压缩和膨胀

1.17 Adiabatic Compression or Expansion (4) S.J.T.U. Phase Transformation and Applications Helium,ideal gas Valve First gas, Insulated 50-liter,25 C,20 atm → Quench chamber,?10 atm (L,)im,-(H。)m。+Q+δW=dU=0 System boundary om, H,=H。 T,=T, H(T,P)=H(T) Quench chamber,?10 atm,Close system Define the system as the quantity of gas remaining in the tank when T the pressure reaches 10 atm Adiabatic expansion T=T(0.758)=298×0.758=226K 226K/-47C SJTU Thermodynamics of Materials Spring 2006 ©X.J.Jin Lecture 3 Second law I

Phase Transformation and Applications S. J. T. U. SJTU Thermodynamics of Materials Spring 2006 © X. J. Jin Lecture 3 Second law I 1.17 Adiabatic Compression or Expansion (4) Helium, ideal gas Insulated 50-liter, 25 °C, 20 atm Quench chamber, ?, 10 atm − )()( + + = dUWQmHmH = 0 i δ i o δ o δ δ Valve = HH oi ( ) 25 4.0 1 2 1 2 )5.0() 2010 == ()( = R R CR P PP TT = TT 12 = × = 226758.0298)758.0( K 226K / -47 °C System boundary δmi δmo First gas, ? Ti = To Quench chamber, ?, 10 atm， Close system H(T, P) = H(T) Define the system as the quantity of gas remaining in the tank when the pressure reaches 10 atm Adiabatic expansion

Questions and Answers S.J.T.U. Phase Transformation and Applications 0=dU CydT dU =CydT V1,T1 V2,T2 dU=CdT dU =0 Ideal gas V1,T2 20 atm 10 atm 20 atm 50L 50L ?L 10 atm 25℃ 25C 50L Adiabatic ? expansion SJTU Thermodynamics of Materials Spring2006©X.J.Jimn Lecture 3 Second law I

Phase Transformation and Applications S. J. T. U. SJTU Thermodynamics of Materials Spring 2006 © X. J. Jin Lecture 3 Second law I Questions and Answers V1,T1 V2,T2 V1,T2 dU = 0 Ideal gas ⎟ = 0 ⎠⎞ ⎜⎝⎛ ∂∂V T U = V dTCdU dTCdUQ = V dTCdU δ = = V 20 atm 50 L 25 °C 10 atm 50 L ? 10 atm 50 L ? 20 atm ? L 25 °C Adiabatic expansion