Bounds on thermodynamic efficiency at maximum power (Curzon ...

Finite-time thermodynamicsHow much the efficiency deteriorates when the cycle isoperated in a finite time? This is the central question in the fieldof finite-time thermodynamics.Endoreversible thermodynamics: wiews a thermodynamicsystem as a collection of reversible subsystems which interactin an irreversible mannerMany applications: heat engines, optimal paths in chemicalreactions, in solar energy capture in semiconductor solar cells(or in the chlorophyll of plants), etc. Often, even in nature,found processes close to maximum power production (efficientpower production).3

Maximum powerAssume the time spent in the adiabatic strokes negligible withrespect to the times of the isothermal strokes (fast relaxationprocesses within the working fluid)Output power:P = W t= Q 1 + Q 2t= Ξ 1t 1 (T 1 − T 1i )+Ξ 2 t 2 (T 2 − T 2i )The internal Carnot engine has efficiencyη Ci =1− T 2iT 1i=1+ Q 2Q 1t 1 + t 25

Duration of isothermal process proportional to temperaturedifferenceQ 1t 1 =Ξ 1 (T 1 − T 1i ) , t Q 22 =Ξ 2 (T 2 − T 2i )Output power:By maximizing the power with respect to the internaltemperatures we obtain6

Efficiency at maximum powerThe efficiency at the maximum power of the CA engine is givenby the so-called Curzon-Ahlborn efficiency:η CA =1−T2T 1Quite remarkably, the CA efficiency is independent of thethermal conductances Ξ1 and Ξ2.[Curzon and Ahlborn, Am. J. Phys. 43, 22 (1975); Yvon (1955);Chambadal (1957), Novikov (1958)]7

CA efficiency is not a universal upper bound but reproduces theefficiency of real thermal plants quite well(Curzon and Ahlborn, 1975)[Esposito, Kawai, Lindenberg, Van denBroeck, PRL 105, 150603 (2010)]8

For a Carnot cycle in the infnite-time cycle, nondissipativelimit, the system entropy changes during isothermal strokes as∆S = Q 1T 1, −∆S = Q 2T 2In the low dissipation regime entropy production proportionalto 1/t1 and 1/t2, so that it vanishes in the limit of infinite-timecycleQ 1 = T 1∆S − Σ 1t 1+ ..., Q 2 = T 2 −∆S − Σ 2+ ...t 2From Clausius inequality BA δQTirr< BA δQTrev⇒ Σ 1 , Σ 2 > 09

Output powerP =Wt 1 + t 2= Q 1 + Q 2t 1 + t 2= (T 1 − T 2 )∆S − T 1 Σ 1 /t 1 − T 2 Σ 2 /t 2t 1 + t 2maximum when∂P= ∂P =0∂t 1 ∂t 2This leads tot 1 =2T 1 Σ 1(T 1 − T 2 )∆S1+T2 Σ 2T 1 Σ 1t 210

Efficiency at the maximum output power:η(P max )=1+η C1+T 2 Σ 2T 1 Σ 1 2 T 2 Σ 2T 1 Σ 1+T 2T 11 − Σ 2Σ 1[Esposito, Kawai, Lindenberg, Van den Broeck, PRL 105, 150603(2010); Schmiedl and Seifert, EPL 81, 20003 (2008)]Symmetric dissipation Σ 1 =Σ 2 :η(P max )=η CAΣ 2 /Σ 1 → 0: η(P max )= η C2 − η CΣ 2 /Σ 1 →∞:η(P max )= η C211

Linear response steady state heat to work conversionStochastic baths: idealgases at fixed temperatureand chemical potentialX 1 = β∆µX 2 = −∆β =∆T/T 2β =1/T∆µ = µ 1 − µ 2∆β = β 1 − β 2∆T = T 1 − T 2(we assume T 1 >T 2 ,µ 1

Restrictions from thermodynamicsLet us consider the time-symmetric caseOnsager relation:Positivity of entropy production:dSdt = J ρX 1 + J q X 2 ≥ 013

Onsager and transport coefficientsG =Jρ∆µ/e JqΞ=∆T ∆µ/eS =∆T∆T =0J ρ =0J ρ =0⇒G = e2T L ρρ⇒ Ξ= 1T 2 det LL ρρ⇒ S = − 1eTL ρqL ρρ14Note that the positivity of entropy production impliesthat the (isothermal) electric conductance G>0 and thethermal conductance Ξ>0

Seebeck and Peltier coefficientsS = ∆µ/e∆TJ ρ =0⇒ S = − 1eTL ρqL ρρΠ=JqeJ ρ∆T =0= L qρeL ρρ= L ρqeL ρρ= TSSeebeck and Peltier coefficients are trivially related byOnsager reciprocal relations (when time symmetry isnot broken)15

Steady state power generation efficiencyη = W Q 1=(T 1 >T 2 )(µ 1 0,J q > 0)η ≤ η C =1− T 2T 116

Thermoelectric figure of meritZT = L2 ρqdetL = GS2Ξ T 17The maximum of η over X1, for fixed X2 (i.e., over theapplied voltage ΔV for fixed temperature differenceΔT) is (for systems with time-reversal symmetry)

Efficiency at maximum powerWithin linear response, the output powerP = −TX 1 J ρ = −TX 1 (L ρρ X 1 + L ρq X 2 )is maximum whenP18

Maximum output powerP max = T 4L 2 ρqX2 2 = η CL ρρ 4L 2 ρqL ρρX 2Efficiency at maximum powerη(P max )= η C2ZTZT +2Both maximum efficiency and efficiency at maximumpower are monotonous growing functions of thethermoelectric figure of merit ZTη(P max ) → η C2when ZT →∞η CA =1−T1T 2=1− 1 − η C = η C2 + O(η2 C)19

η maxη(P max )20

And when time-reversal is broken?T 1 ,µ 1 BT 2 ,µ 2B applied magnetic field or anyparameter breaking time-reversibilitysuch as the Coriolis force, etc.21

Constraints from thermodynamicsPOSITIVITY OF THE ENTROPY PRODUCTION:ONSAGER-CASIMIR RELATIONS:22

Efficiency at maximum power depends on two parametersη(P max )= η C2xy2+y23

maximum η of η(P max )η(P max ) ≤ η = η Cx 2x 2 +124

The Curzon-Ahlborn limit can beovercome within linear response25