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Energy-Efficient Use of Entropy

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Energy-Efficient Use of Entropy Energy-efficient use of Entropy Problem: to minimize energy E spent to lift a weight using a stretched rubber band. Lift by hand, no help Lift by hand at constant Lift by warming the from the rubber band T, rubber band helps rubber band E = W '= ? EΓ=const = Q = ? E0 = W = Mgh T =const Spontaneous heat flow h h h min (EΓ=const , ET =const ) = ? Notes Example of a heat engine: isothermal rubber band T1 T1 Heat source Heat source Rubber band Rubber T =T T2=T 1 Rubber band Rubber 2 1 h = L 2 –L1 M Applying force to the weight so that the rubber band contracts from L1 to L2 isothermally (L − L ) L2 Γ = cBT 0 cBT 2 2 WRWS = − ∫ ΓdL = ()()()L1 − Lo − L2 − Lo L0 2L L1 0 Notes Example of a heat engine: isothermal rubber band U = cL 0T ∆U = 0 ∆U = ∆QRHS + ∆WRWS ∆QRHS = −∆WRWS 2 T1 cB (L − L0 ) Sband = cL 0 ln ()U − + const 2 L0 Rubber band Rubber Heat source ∆W ∆Q ∆S = RWS ∆S = RHS T =T band RHS 2 1 T T ∆ = ∆ + ∆ ∆ = 0 Reversible heat engine Stotal Sband SRHS Stotal Notes Example of a heat engine: isobaric rubber band System: rubber band T1 T2 Heat source band Rubber Heat source Rubber band Rubber T2>T 1 T2>T 1 M Free-hanging weight (constant tension Γ) Notes Example of a heat engine: isobaric rubber band U = cL 0T ∆U = cL 0 (T2 −T1 ) T 2 (L − L ) Γ = cBT 0 = Mg L0 Rubber band Rubber Heat source T2>T 1 MgL 1 1 0 h = L2 − L1 = − Free-hanging weight cB T2 T1 (constant tension Γ) Notes Example of a heat engine: isobaric rubber band ∆U = cL 0 (T2 −T1 ) 2 2 M g L0 (T2 −T1 ) Work done by the rubber band ∆WRWS = Mgh = > 0 cBT 1T2 Heat transferred from the rubber band M 2 g 2 ∆ = −∆ − ∆ = − − − = − − + < QRHS U WRWS cL 0 (T2 T1) Mgh cL 0 (T2 T1)1 2 0 c BT 1T2 Heat transferred to the rubber band EΓ=const = Q = ∆QRHS = cL 0 (T2 −T1) + Mgh > Mgh Notes Example of a heat engine: isobaric rubber band 2 cB (L − L0 ) S = cL 0 ln ()U − + const 2 L0 Change of entropy of the band: T cB 2 2 2 ∆S = cL 0 ln − ()()()L2 − L0 − L1 − L0 = T1 2L0 2 T cB MgL (T −T )(T +T ) = 2 + 0 2 1 2 1 cL 0 ln 2 2 T1 2L0 cB T1 T2 2 T cB MgL T 2 ∆ = ln 2 + 0 2 −1 S cL 0 2 2 T1 2L0T2 cB T1 Notes Example of a heat engine: isobaric rubber band Change of entropy of the heat source: ∆Q T −T M 2 g 2 ∆ = RHS = − 2 1 1+ SRHS cL 0 2 T2 T2 c BT 1T2 Total entropy change of the system: ∆Stotal = ∆SRHS + ∆S 2 (T −T ) M 2 g 2 T cB MgL T 2 ∆ = − 2 1 1+ + ln 2 + 0 2 −1 Stotal cL 0 2 cL 0 2 2 T2 c BT 1T2 T1 2L0T2 cB T1 2 T T M 2 g 2 L T ∆ = ln 2 − 1− 1 + 0 2 −1 Stotal cL 0 cL 0 2 T1 T2 2cBT 2 T1 Notes Example of a heat engine: isobaric rubber band 2 T T M 2 g 2 L T ∆ = ln 2 − 1− 1 + 0 2 −1 Stotal cL 0 cL 0 2 T1 T2 2cBT 2 T1 2 M 2 g 2 L T T T 0 2 −1 ≥ 0 and ln 2 − 1− 1 ≥ 0 ∆Stotal ≥ 0 because 2 cL 0 cL 0 2cBT 2 T1 T1 T2 Plot @Log @xD+1ê x- 1,8x,0.4,4<,PlotRange -> All D 0.6 1 ∆ > 0 0.5 Stotal The latter is equivalent to: ln ()x + −1 0.4 x If T ≠ T 1 0.3 2 1 ln ()x + ≥1 0.2 x Isobaric rubber band 0.1 engine is irreversible! Which is a true statement 0.5 1.5 2 2.5 3 3.5 4 Ü Graphics Ü Notes Harvesting Low Entropy to do work: rubber band Case 1 – low entropy wasted Spontaneous adiabatic T 1 contraction T1 Stretched rubber band rubber rubber band rubber Unstretched ∆Q = ∆W = ∆U = 0 Notes Harvesting Low Entropy to do work: rubber band ∆Q = ∆W = ∆U = 0 Initial entropy final entropy − 2 cB (L L0 ) S = cL ln (U )+ const Si = cL 0 ln ()U − + const f 0 2 L0 cB (L − L )2 ∆S = ∆S = S − S = 0 > 0 No work done, total total f i entropy increased 2 L0 Notes Summary Lift by hand, no help Lift by hand at constant T, Lift by warming the from the rubber band rubber band helps rubber band = < E = Q = Mgh + ∆U E0 = W = Mgh ET =const W Mgh Γ=const Spontaneous heat flow h h h Emin = ET =const Least energy is spent in the reversible process to lift the weight ∆S > 0 ∆Stotal > 0 ∆Stotal = 0 total Notes The Maximum Work Theorem We model the environment of the system as two objects: - reversible work source (sink) RWS - reversible heat source (sink) RHS Heat engine model ∆U = U B –UA -δQ RHS -δW System A →B RWS Heat sink Work sink TRHS T Rigid walls, so no work can Adiabatic walls, so no heat can be be done on the heat sink ! transferred to the work sink! ( dS RWS =0 ) Notes The Maximum Work Theorem Maximum work theorem: for a given ∆U, -δW is maximum for a reversible process System goes from state A to state B. Which process delivers maximum work AB to the reversible work source? The change of internal energy ∆U is the same for all A →B processes ∆U is distributed between work done on RWS, ∆W, and heat flow to the RHS, ∆Q Therefore, maximum work delivered to RWS implies minimum heat flow to RHS. Notes The Maximum Work Theorem Maximum work theorem: for a given ∆U, -δW is maximum for a reversible process AB The change of entropy of the working body ∆S is the same for all A →B processes The change of entropy of the RHS is proportional to ∆Q, therefore maximum work corresponds to the minimum entropy change of the RHS and (since ∆S is independent on the process A →B) of the entire composite system (engine). The absolute minimum entropy increase is zero and achieved in a reversible process . Therefore, maximum work done by a heat engine is achieved in a reversible process. Notes The Maximum Work Theorem Maximum work theorem: for a given ∆U, -δW is maximum for a reversible process Proof: δQRHS dS total = dS + dS RHS = dS + ≥ 0 δQRHS ≥ −TRHS dS TRHS dU = −δWRWS −δQRHS dU ≤ −δWRWS +TRHS dS Functions of states A and B δWRWS ≤ TRHS dS − dU only, not of the process A→B Notes The Maximum Work Theorem δWRWS ≤ TRHS dS − dU The work is maximum when £ → = max δWRWS = TRHS dS − dU The origin of the § sign is traced back to dS total ¥0. Therefore, for maximum work we have dS total =0 , hence reversible process maximizes work . Notes 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.
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