Carnot cycle
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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, pVο§ = 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 ππ΄ , ππ΄ , ππ΄ Isotherm 1 ⇒ ππ΄ = ππ΅ = πβ p ππ΅ , ππ΅ , ππ΅ Adiabat 2 Adiabat 1 ππ· , ππ· , ππ· Isotherm 2 ⇒ ππΆ = ππ· = ππ V ππΆ , ππΆ , ππΆ Efficiency of a Carnot engine ππ΄ , ππ΄ , πβ Isotherm 1 p ⇒ Δπβ = π πβ ln ππ΅ ππ΄ Adiabat 2 ππ΅ , ππ΅ , πβ ⇒ Δπ = 0 πΎ−1 πΎ−1 ⇒ ππ ππ· = πβ ππ΄ Adiabat 1 ⇒ Δπ = 0 πΎ−1 πΎ−1 ⇒ πβ ππ΅ = ππ ππΆ ππ· , ππ· , ππ Isotherm 2 ππ· ⇒ Δππ = π ππ ln ππΆ V ππΆ , ππΆ , ππ Efficiency of a Carnot engine ππ΅ Δπβ = π πβ ln ππ΄ πΎ−1 πβ ππ΅ πΎ−1 = ππ ππΆ ππ· Δππ = π ππ ln ππΆ πΎ−1 ππ ππ· (1) From isotherm 1 (2) From adiabat 1 (3) From isotherm 2 πΎ−1 = πβ ππ΄ (4) From adiabat 2 From the first law of thermodynamics: Δπ = Δπ + Δπ For the complete Carnot cycle Δπ = 0 since π is a state variable Efficiency of a Carnot engine ππ΅ Δπβ = π πβ ln ππ΄ πΎ−1 πβ ππ΅ πΎ−1 = ππ ππΆ ππ· Δππ = π ππ ln ππΆ πΎ−1 ππ ππ· (1) From isotherm 1 (2) From adiabat 1 (3) From isotherm 2 πΎ−1 = πβ ππ΄ (4) From adiabat 2 From the first law of thermodynamics: Δπ = Δπ + Δπ For the complete Carnot cycle Δπ = 0 since π is a state variable ⇒ Δπ = −Δπ Δπis the work done on the engine (system), let π be the work done by the engine ⇒ π = −Δπ = Δπ π π From (1) and (3): Δπ = Δπβ +Δππ = π πβ ln π΅ + π ππ ln π· π π π΄ πΆ Efficiency of a Carnot engine ππ΅ Δπβ = π πβ ln ππ΄ πΎ−1 πβ ππ΅ (1) From isotherm 1 πΎ−1 = ππ ππΆ (2) From adiabat 1 ππ· Δππ = π ππ ln ππΆ πΎ−1 ππ ππ· (3) From isotherm 2 πΎ−1 = πβ ππ΄ (4) From adiabat 2 π π From (1) and (3): Δπ = Δπβ +Δππ = π πβ ln π΅ + π ππ ln π· = π π π π΄ Efficiency is defined as: πΆ Output Input Output is the work done by the engine i.e. π and input is the heat absorbed by the engine i.e. Δπβ ⇒ π efficiency = π Δπβ +Δππ Δππ = =1+ Δπβ Δπβ Δπβ Efficiency of a Carnot engine ππ΅ Δπβ = π πβ ln ππ΄ πΎ−1 πβ ππ΅ (1) From isotherm 1 πΎ−1 = ππ ππΆ (2) From adiabat 1 ππ· Δππ = π ππ ln ππΆ πΎ−1 ππ ππ· (3) From isotherm 2 πΎ−1 = πβ ππ΄ ⇒ π efficiency = (4) From adiabat 2 π Δπβ +Δππ Δππ = =1+ Δπβ Δπβ Δπβ ππ· ππ· ln ππ ππΆ ππΆ ⇒π = 1+ =1− π πβ ln ππ΄ π πβ ln ππ΅ ππ΅ π΄ π ππ ln πβ ππΆ ⇒ = (2) ππ ππ΅ πΎ−1 and (4) πβ ππ· ⇒ = ππ ππ΄ (from (1) and (3)) πΎ−1 ππΆ ππ· ⇒ = ππ΅ ππ΄ ππ΄ ππ· ⇒ = ππ΅ ππΆ Efficiency of a Carnot engine ππ ⇒π =1− πβ 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 πβ πβ = Δπβ π = Δπβ +Δππ = πβ − ππ Carnot ππ = Δππ = −Δππ ππ Carnot engine: Schematic representation πβ πβ π = πβ − ππ Carnot ππ ππ Carnot engine is reversible πβ πβ π = πβ − ππ Carnot ππ ππ Carnot engine is reversible (refrigerator) πβ πβ π = πβ − ππ Carnot ππ ππ 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) πβ πβ′ πβ π ′ = πβ′ − ππ′ π = πβ − ππ R Carnot ππ′ ππ ππ Adjust the cycles so that π = π ′ Carnot engine is reversible (refrigerator) πβ πβ′ πβ π = πβ − ππ = πβ′ − ππ′ R Carnot ππ′ ππ ππ Carnot engine is reversible (refrigerator) πβ πβ′ πβ π π > πβ′ πβ π = πβ − ππ = πβ′ − ππ′ ⇒ πβ′ < πβ ⇒ πβ − πβ′ > 0 R Carnot If π′ > π then: Also: πβ − ππ = πβ′ − ππ′ ππ′ ππ ππ ⇒ πβ − πβ′ = ππ − ππ′ > 0 Is this possible? πβ πβ πβ − πβ′ πβ′ ⇒ πβ − πβ′ = ππ − ππ′ > 0 π = πβ − ππ = πβ′ − ππ′ R Carnot ππ ππ − ππ′ ππ ⇒ πβ > πβ′ πππ ππ > ππ′ ππ′ 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. ⇒ π′ β― π Carnot’s theorem reversible Of all heat engines working between two given temperatures, none is more efficient than a Carnot engine ππππ < ππππ£ ⇒ ππππ < ππππ£ For reversible engines πβ πβ′ πβ π = πβ − ππ = πβ′ − ππ′ R Carnot ππ′ ππ ππ ⇒ π′ β― π πππ π β― π′ ⇒ π = π′ 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 πCarnot 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 πβ πβ′ πβ ⇒ πβ −πβ′ = ππ > 0 π = πβ′ = πβ − ππ Kelvin violator Carnot ππ ππ ⇒ πβ −ππ = πβ′ Carnot engine and Claussius violator πβ πβ ππ π = πβ − ππ Claussius violator Carnot ππ ππ ππ
