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Carnot Cycle

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Carnot Cycle Summary of heat engines so far - A thermodynamic system in a process Reversible Heat Source connecting state A to state B ∆Q Th RHS T-dynamic - In this process, the system can do work and system emit/absorb heat ∆WRWS - What processes maximize work done by the system? Rev Work Source - We have proven that reversible processes (processes that do not change the total entropy of the system and its environment) maximize Process A →B work done by the system in the process A→B of the system Notes Cyclic Engine -To continuously generate work , a heat engine should go through a Heat source Heat sink cyclic process . ∆Q ∆Q Th h c Tc - Such an engine needs to have a heat source and a heat sink at different temperatures. ∆WRWS Working body - The net result of each cycle is to e. g. rubber band extract heat from the heat source and distribute it between the heat Rev Work Source transferred to the sink and useful work. Notes Carnot cycle As an example, we will consider a particular cycle consisting of two isotherms and two adiabats called the Carnot cycle applied to a rubber band engine. As we will see, Carnot cycle is reversible Here T and S are temperature and entropy of the working body Heat source Heat sink T AB ∆ ∆ T Th Qh Qc Tc h ∆WRWS Working body Tc e. g. rubber band D C Rev Work Source S SA SB Notes 1. Carnot Cycle: Isothermal Contraction (A) (B) T ∆ = 0 h U Th ∆Qsource LB LA Heat source Heat source T h Th T A B Force= Γ Th 2 2 D C cBT (L − L0 ) (L − L0 ) A→B h A B Tc ∆WRWS = ∆Qsource = Th∆S = − S 2 S S L0 L0 A B Notes 2. Carnot Cycle: Adiabatic Contraction (B) (C) Th Tc LC LB ∆Q = 0 ∆ B→C = −∆ WRWS U T A B Th Quasi-static adiabatic B→C D C ∆ = − = − Tc process WRWS U B U C cL 0 (Th Tc ) S SA SB Notes 3. Carnot Cycle: Isothermal Expansion (C) ∆U = 0 (D) Tc Tc ∆Qsink LC LD Heat sink Heat sink Tc Tc T A T B 2 2 h C→D cBT (L − L0 ) (L − L0 ) ∆W = ∆Q = T ∆S = c C − D D C RWS sink c T 2 c L0 L0 S SA SB Notes 4. Carnot Cycle: Adiabatic Expansion (D) (A) Th Tc LB L ∆Q = 0 A D→A ∆ = ∆ T WRWS U A B Th D C D→A Tc ∆WRWS = U D −U A = cL 0 (Tc −Th ) S SA SB Notes Carnot Cycle: Summary Total work done in one cycle: ∆WRWS = ∆Qsource + ∆Qsink + cL 0 (Th −Tc )+ cL 0 (Tc −Th ) = ∆Qsource + ∆Qsink where 2 2 2 2 cBT (L − L ) (L − L ) cBT (L − L ) (L − L ) ∆Q = h A 0 − B 0 ∆Q = c C 0 − D 0 source 2 sink 2 L0 L0 L0 L0 Carnot engine is reversible because each of the four parts of the cycle is reversible! Exercise: show that adiabatic expansion of a rubber band where all internal energy goes into work is a reversible process. Note that spontaneous adiabatic contraction (non-quasistatic) where internal energy does not change is irreversible! Notes Heat Engine Efficiency Efficiency can be defined for a cyclic engine Efficiency of a cyclic engine is defined as the ratio ∆WRWS of the net work done in a cycle to the net amount ε = ∆Q of heat extracted from the hot body in one cycle source For Carnot engine: 2 2 ∆W ∆Q T L − L − L − L T RWS 1 sink 1 c ( A 0 ) ( B 0 ) 1 c = + = − 2 2 = − ∆Qsource ∆Qsource Th ()()LD − L0 − LC − L0 Th Exercise: show that this is correct Note that since all reversible engines do the same work (maximum), they have the same efficiency (e. g. same as Carnot engine). Notes Heat Engine Efficiency T There is a simpler way to calculate Carnot AB engine efficiency using the (S,T) plot of the Th Carnot cycle. ∆W = ∆Q + ∆Q RWS source sink Tc D C Since ∆Q = T (S − S ) = T ∆S source h B A h S SA SB ∆Qsource = Tc (S A − SB ) = −Tc∆S ∆W (T −T )∆S T ε = RWS = h c =1− c We obtain ∆WRWS = (Th −Tc )∆S ∆Qsource Th∆S Th Notes Engine, refrigerator, heat pump Heat engine can run in direct and reverse cycles. If it runs in the reverse cycle, work is done to transfer heat from the cold to the hot system (refrigerator, heat pump). Engine Refrigerator or heat pump Heat source Heat sink Heat source Heat sink Th δQh δQc Tc Th δQh δQc Tc δWRWS δWRWS Rev Work Source Rev Work Source Notes Real versus Ideal Engines Completely reversible engines are impossible to realize due to friction and other loss mechanisms. Even if such engines were realizable, they would generate very little power since they have to employ very slow quasi-static processes in order to be reversible . The reason why reversible engines are slow is because they employ heat transfer between bodies of the same temperature. Since heat flow is proportional to the gradient of temperature (non-equilibrium thermodynamics result), the flow of heat in the reversible engine has to be infinitely slow. Real engines are always irreversible and thus have lower efficiency compared to the reversible engines. Notes Carnot Cycle: Isothermal Contraction – a very slow process (A) (B) Th Th ∆Qsource LB L A Heat source Heat source Th Th Force= Γ 2 2 cBT (L − L ) (L − L ) ∆W A→B = ∆Q = T ∆S = h A 0 − B 0 RWS source h 2 L0 L0 Notes Endoreversible Engine To transfer heat fast in real engines, we need to employ irreversible processes. An example of such an engine is endoreversible engine . For endoreversible Carnot engine, the isothermal processes are done at a temperatures Tw and Tt that are higher than the temperature of the heat sink Tc but lower than the temperature of the heat source Th. Since for both isothermal processes, the temperature difference between the heat source/sink and the working body of the engine is non-zero, the isothermal heat transfer processes are not infinitely slow. Notes Endoreversible Engine For endoreversible Carnot engine, the isothermal processes are done at a temperatures Tw and Tt that are higher than the temperature of the heat sink Tc but lower than the temperature of the heat source Th. T Rubber band is at T here Th w AB Tw Th > Tw > Tt > Tc D C Tt Rubber band is at Tt here Tc S A→B and C→D are irreversible now while SA SB B→C and D→A are still reversible Notes Endoreversible Engine Analysis Non equilibrium thermodynamics results for heat flow rate: ∂Q = σ ()T1 −T2 where σ is heat conductivity ∂t From this equation we can calculate time of the isothermal processes of the endoreversible engine ∆Qsource tsource = - time of isothermal process in contact with heat source σ h ()Th −Tw ∆Q t = sink - time of isothermal process in contact with heat sink sink σ − c ()Tt Tc Because adiabatic tcycle ≈ tsource + tsink processes can be very fast Notes Endoreversible Engine Analysis ∆ ∆ Qsource Qsink T tcycle ≈ + σ ()T −T σ ()T −T Th h h w c t c AB Tw Tw ∆Qsource = Tw∆S ∆Q = W D C source Tt Tw −Tt ∆Qsin k = Tt ∆S Tc Tt ∆Qsin k = W W = (Tw −Tt )∆S S S S Tw −Tt A B 1 1 T 1 1 T w t tcycle ≈ + W σ h Th −T w Tw −T t σ c Tt −T c Tw −T t Notes Maximum Endoreversible Engine Power 1 1 1 1 1 − W Tw Tt P = = + t cycle σ h Th −T w Tw −T t σ c Tt −T c Tw −T t To find maximum power delivered, we need to maximize P with respect to Tw and Tc ∂P = 0 Tw = c Th σ T + σ T ∂T where c = h h c c w σ + σ ∂P T = c T h c = 0 t c ∂Tt Exercise: show this Notes Maximum Endoreversible Engine Power Substituting Tw = c Th Tt = c Tc 1 into 1 1 1 1 − W Tw Tt P = = + t cycle σ h Th −T w Tw −T t σ c Tt −T c Tw −T t we obtain the maximum power delivered by endoreversible engine 2 W T − T P = = σ σ h c max h c t cycle σ h + σ c Notes Endoreversible Engine Efficiency T Endoreversible engine W T −T T ε = = w t h efficiency at max power: AB ∆Qsource Th Tw D C Substituting Tw = c Th Tt = c Tc Tt T σ T + σ T c h h c c S c = SA SB σ h + σ c Tc We obtain: ε =1− Not how this expression does not depend on Th the values of thermal conductivities! Notes.
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