Lecture 10 _Real Equations of state _

# Lecture 10 _Real Equations of state _ - Lecture 10 Towards...

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Unformatted text preview: Lecture 10: Towards real Equations of state ChE 141 1 Process calculations Isothermal process change at constant T Isobaric process change at constant P Isochoric process change at constant V Adiabatic process change with Q=0 polytropic process change at constant PV δ Question: What is the change in energy/enthalpy and what is the work/heat done on the system to achieve this? 2 Isochoric process Δ U = Q + W Energy : Isochoric: Enthalpy : Work: W =- V II V I P d V Heat: P v II I Δ T = T II- T I = V R ( P II- P I ) Δ U = T II T I C V d T Δ H = Δ U + V Δ P = Q = Δ U 3 Adiabatic process Δ U = Q + W Energy: Adiabatic: Work: P v II I d U = d Q + d W d Q = d U = C V d T d W =- P d V =- RT d V V From the frst law: C V d T =- RT d V V I¡ we assume C V independent o¡ the temperature d T T =- R C V d V V or T II T I = V I V II R/C V ln T II T I =- R C V ln V II V I or 4 Adiabatic process (2) Energy: The path: Enthalpy : Work: Heat: P v II I T II T I = V I V II R/C V Δ U = C V ( T II- T I ) Δ H = ( C V + R )( T II- T I ) W = Δ U Q = TV R/C V = const Defne γ ≡ C P C V TV R/C V = TV C p- C V C V = TV γ- 1 = C 5 polytropic process Δ U = Q + W Path Work: P v II I d U = d Q + d W PV δ = C d W =- P d V =- C d V V δ W =- C V II V I d V V δ =- C 1- δ V 1- δ | V II V I etc. etc. etc. 6 Example 3.2 Air is compressed from an initial state of 1 bar and 25ºC to a Fnal state of 5 bar and 25ºC by three different mechanically reversible processes in a closed system:...
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Lecture 10 _Real Equations of state _ - Lecture 10 Towards...

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