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v = 10:1:100; t = 100; r = 8.314; gamma = 1.67; p = r*t./v; k = (10^(gamma-1)).*r*t; pa = k./v.^gamma; plot(v,pa,v,p) 90 80 70 60 50 p Isotherm, pV = constant = NkBT 40 30 20 10 0 10 Adiabat, pV = constant 20 30 40 50 60 V 70 80 90 100 Carnot cycle v = 10:1:100; th = 100; tl = 50; r = 8.314; gamma = 1.67; p1 = r*th./v; k1 = (30^(gamma-1)).*r*th; pa1 = k1./v.^gamma; p2 = r*tl./v; k2 = (30^(gamma-1)).*r*tl; pa2 = k2./v.^gamma; plot(v,p1,v,pa1,v,p2,v,pa2) Carnot cycle Carnot cycle pA, VA, TA Isotherm 1 Th p Qh pB, VB, TB Adiabat 2 Adiabat 1 pD, VD, TD Isotherm 2 pC, VC, TC V Carnot cycle pA, VA, TA Isotherm 1 Th p Qh pB, VB, TB Adiabat 2 Adiabat 1 pD, VD, TD Isotherm 2 pC, VC, TC V Carnot cycle pA, VA, TA Isotherm 1 Tl p Qh pB, VB, TB Adiabat 2 Adiabat 1 pD, VD, TD Ql Isotherm 2 pC, VC, TC V Carnot cycle pA, VA, TA Isotherm 1 Tl p Qh pB, VB, TB Adiabat 2 Adiabat 1 pD, VD, TD Ql Isotherm 2 pC, VC, TC V Why such a strange engine? Will discuss in class Efficiency of a Carnot engine , , Isotherm 1 ⇒ = = ℎ p , , Adiabat 2 Adiabat 1 , , Isotherm 2 ⇒ = = V , , Efficiency of a Carnot engine , , ℎ Isotherm 1 p ⇒ Δℎ = ℎ ln Adiabat 2 , , ℎ ⇒ Δ = 0 −1 −1 ⇒ = ℎ Adiabat 1 ⇒ Δ = 0 −1 −1 ⇒ ℎ = , , Isotherm 2 ⇒ Δ = ln V , , Efficiency of a Carnot engine Δℎ = ℎ ln −1 ℎ −1 = Δ = ln −1 (1) From isotherm 1 (2) From adiabat 1 (3) From isotherm 2 −1 = ℎ (4) From adiabat 2 From the first law of thermodynamics: Δ = Δ + Δ For the complete Carnot cycle Δ = 0 since is a state variable Efficiency of a Carnot engine Δℎ = ℎ ln −1 ℎ −1 = Δ = ln −1 (1) From isotherm 1 (2) From adiabat 1 (3) From isotherm 2 −1 = ℎ (4) From adiabat 2 From the first law of thermodynamics: Δ = Δ + Δ For the complete Carnot cycle Δ = 0 since is a state variable ⇒ Δ = −Δ Δis the work done on the engine (system), let be the work done by the engine ⇒ = −Δ = Δ From (1) and (3): Δ = Δℎ +Δ = ℎ ln + ln Efficiency of a Carnot engine Δℎ = ℎ ln −1 ℎ (1) From isotherm 1 −1 = (2) From adiabat 1 Δ = ln −1 (3) From isotherm 2 −1 = ℎ (4) From adiabat 2 From (1) and (3): Δ = Δℎ +Δ = ℎ ln + ln = Efficiency is defined as: Output Input Output is the work done by the engine i.e. and input is the heat absorbed by the engine i.e. Δℎ ⇒ efficiency = Δℎ +Δ Δ = =1+ Δℎ Δℎ Δℎ Efficiency of a Carnot engine Δℎ = ℎ ln −1 ℎ (1) From isotherm 1 −1 = (2) From adiabat 1 Δ = ln −1 (3) From isotherm 2 −1 = ℎ ⇒ efficiency = (4) From adiabat 2 Δℎ +Δ Δ = =1+ Δℎ Δℎ Δℎ ln ⇒ = 1+ =1− ℎ ln ℎ ln ln ℎ ⇒ = (2) −1 and (4) ℎ ⇒ = (from (1) and (3)) −1 ⇒ = ⇒ = Efficiency of a Carnot engine ⇒ =1− ℎ Laws of thermodynamics 0. There is a game 1. You can never win 2. You cannot break even, either 3. You cannot quit the game Carnot engine: Schematic representation ℎ ℎ = Δℎ = Δℎ +Δ = ℎ − Carnot = Δ = −Δ Carnot engine: Schematic representation ℎ ℎ = ℎ − Carnot Carnot engine is reversible ℎ ℎ = ℎ − Carnot Carnot engine is reversible (refrigerator) ℎ ℎ = ℎ − Carnot Carnot’s theorem reversible Of all heat engines working between two given temperatures, none is more efficient than a Carnot engine Carnot engine is reversible (refrigerator) ℎ ℎ′ ℎ ′ = ℎ′ − ′ = ℎ − R Carnot ′ Adjust the cycles so that = ′ Carnot engine is reversible (refrigerator) ℎ ℎ′ ℎ = ℎ − = ℎ′ − ′ R Carnot ′ Carnot engine is reversible (refrigerator) ℎ ℎ′ ℎ > ℎ′ ℎ = ℎ − = ℎ′ − ′ ⇒ ℎ′ < ℎ ⇒ ℎ − ℎ′ > 0 R Carnot If ′ > then: Also: ℎ − = ℎ′ − ′ ′ ⇒ ℎ − ℎ′ = − ′ > 0 Is this possible? ℎ ℎ ℎ − ℎ′ ℎ′ ⇒ ℎ − ℎ′ = − ′ > 0 = ℎ − = ℎ′ − ′ R Carnot − ′ ⇒ ℎ > ℎ′ > ′ ′ The Second Law of Thermodynamics • Clausius’ statement: It is impossible to construct a device that operates in a cycle and whose sole effect is to transfer heat from a cooler body to a hotter body. ⇒ ′ ≯ Carnot’s theorem reversible Of all heat engines working between two given temperatures, none is more efficient than a Carnot engine < ⇒ < For reversible engines ℎ ℎ′ ℎ = ℎ − = ℎ′ − ′ R Carnot ′ ⇒ ′ ≯ ≯ ′ ⇒ = ′ Carnot’s theorem Of all heat engines working between two given temperatures, none is more efficient than a Carnot engine All reversible engines working between two temperatures have the same efficiency as Carnot The Second Law of Thermodynamics • Clausius’ statement: It is impossible to construct a device that operates in a cycle and whose sole effect is to transfer heat from a cooler body to a hotter body. • Kelvin-Planck statement: It is impossible to construct a device that operates in a cycle and produces no other effect than the performance of work and the exchange of heat from a single reservoir. Carnot refrigerator and Kelvin violator ℎ ℎ′ ℎ ⇒ ℎ −ℎ′ = > 0 = ℎ′ = ℎ − Kelvin violator Carnot ⇒ ℎ − = ℎ′ Carnot engine and Claussius violator ℎ ℎ = ℎ − Claussius violator Carnot