Efficiency at Maximum Power in Weak Dissipation Regimes R. Kawai University of Alabama at Birmingham M. Esposito (Brussels) C. Van den Broeck (Hasselt) Delmenhorst, Germany (October 10-13, 2010)
Contents Introduction: Efficiency at maximum power General derivation of the efficiency at maximum power with a weak dissipation approximation A case study: Quantum dot heat engine Esposito, Kawai, Van den Broeck, Lindenberg, PRL 105, 150603 (2010) Esposito, Kawai, Van den Broeck, Lindenberg, PRE 81, 041106 (2010)
Carnot Cycle t T H Stop the piston when T sys =T H. T sys 3 4 Sadi Carnot 2 1 Stop the piston when T sys =T C. 1st Law: W Net Q H Q C =0 2nd law: = W Net Q H = 1 T C T H Quasi static (reversible) limit (Highest efficiency) P= W 0 (Lowest power)
Finite Time Thermodynamics What will happen if is finite? (beyond ) T H T sys Mismatch T H T C Mismatch 3 2 4 1 t / T C T sys Izumida and Okuda, EPL 83 (2008), 60003
Efficiency at maximum power Efficiency decreases as increases. But Power reaches its maximum at a certain. 1/ * maximum power: P * = W =Q * H H Q C * C * * H efficiency at maximum power: * =1 Q * C C Q H * H Maximize P with respect to? only assuming (t/ ) the form of (t) (t) and other system parameters
Curzon-Ahlborn efficiency T H Q H ' T H Q C ' T C W Efficiency at maximum power T C Novikov, J. Nucl. Energy II (1958), 125 Curzon-Ahlborn, Am. J. Phys 43 (1975), 22 T C T H A Observed Steam power plant (USA) 298 923 67.6% 43.2% 40%
How universal is the Curzon-Ahlbone efficiency? Many case studies: Brownian heat engine [Schmiedl&Seifert, EPL 81 (2008), 20003] *= 2 T T H C = 3T H T C 2 2 3 C 8 32 Linear non-equilibrium thermodynamics+strong coupling *= 2 Van den Broeck, PRL 95 (2005), 190602 Non-linear correction (with left-right symmetry)+strong coupling *= 2 2 C 8 o 3 Esposito et al., PRL 102 (2009), 130602
Weak dissipation limit S = S e S i Change in system entropy Irreversible entropy production Entropy flow: S e = Q T Quasi static limit : S e S rev, S i 0 Asymptotic expansion (Weak dissipation approximation)
General case: d dt P t = W T t P t m W T mn t =0 W T t P qs t =0, W T mn t P qs n t =W T nm t P qs m t Thermodynamics state H 1 H t / H Hot bath (T H, H ) H 0 adiabatic Cold bath (T C, H ) adiabatic 0 t / 1 Scaling time s= t W T t = W T [ t / ] d d s P s = W T s P s 1 Control parameter: Four corners are fixed such that the cycle is a Carnot cycle at the quasi static limit.
Asymptotic expansion W T s 1 s = d d s P qs s, W T s 2 s = d d s 1 s, m m 1 = m m 2 =0 Entropy flow S rev = k m P m qs 1 ln P m qs 1 k m P m qs 0 ln P m qs 0 Entropy production
Asymptotic expansion (Weak dissipation approximation) Heat Q H Q C e = T H S H e = T C S C T H S rev T C S rev T e H H H e T C Power P= Q H Q C H = T T S rev T e H H H C H H T e C Maximizing power with respect to H and
Efficiency at Maximum Power e e H e e H e e H = * = =0 * = =1 * = 2 2 2 4 3 C 8 = 2 2 2 C 8 3 C 16 = A 2 * 2 A In the middle of two bounds
T C T H A Observed Lower bound L Upper bound U 32% 47% 24% 31% 16% 19% Steam power plant (USA) 25 650 67.6% 43.2% 40% 34% 51%
I think I have already said so. T. Schmiedl and U. Seifert, Europhys. Lett. 81, 20003 (2008). B. Gaveau, M. Moreau, and L. S. Schulman, Phys. Rev. Lett. 105, 060601 (2010). L. Chen and Z. Yan, J. Chem. Phys. 90, 3740 (1989) S. Velasco et al., J. Phys. D 34, 1000 (2001).
isothermal Quantum Dot Carnot Engine Control parameter (protocol) t p 1 = 1 1 e 1 H /kt H 1 = e 2 / kt C 1 1 p 1 p 1 adiabatic 2 H T C 0 isothermal p 0 p 0 adiabatic 3 T H p 0 = 1 1 e 0 H / kt H 1 = e 3 /kt C 1 1 0 T Work done to the system W = 1 0 Heat flows out Q=
Steps to find the efficiency at maximum power Step 1: Optimize the protocol ( t or p t ) Step 2: Maximize power with respect to H and Step 3: Evaluate efficiency at maximum power Master equation ṗ t = 1 p t 2 [1 p t ] C 1 = e t / kt 1, 2= e t /kt 1 C ṗ t = C p t e t /kt 1 C=tunneling rate Work: W [ p t ]= 0 t p t dt Heat: Q[ p t ]= 0 t ṗ t dt
Optimizing protocol K measures the degree of dissipation. Carnot limit: K 0
Determination of K Exact Entropy flow
Solution at Weak Dissipation Limit K 1 For given, H, p 0, p 1 Optimum protocol i =arcsin 1 2 p i i=0,1 Entropy change S H = S rev 1 0 2 C H 1 H S C = S rev 1 0 2 C C 1 Maximum power P= Q H Q C H = T H T C S rev 1 0 2 [ H T H C H H T C C C ]
isothermal Further maximization w.r.t. H and 2 * 2 Lower bound: C H C C Upper bound: C H C C 0 1 p 1 p 1 adiabatic 2 H T C 0 isothermal p 0 p 0 adiabatic 3 T H
Conclusions 1)The efficiency at maximum power is derived without a specific model at the weak dissipation limit,. 2)The efficiency at maximum power is bounded between /2 and /(2- ) 3)Exact Curzon-Ahlborn efficiency is obtained when left-right symmetry holds and it lies between the lower and upper bounds. 4)Only maximization of power with respect to operation times is necessary to get the Curzon-Ahlborn efficiency 5) The method of asymptotic expansion (weak dissipation limit) is justified for general Markovian processes. 6) The present results are demonstrated using analytically solvable model based on a quantum dot Carnot engine.
Brownian heat engine h isotherm sudden adiabatic adiabatic sudden c isotherm Controle parameter: spring constant Thermodynamic state: density p(x) *= 2 T T h c = 2 C 3T h T c 2 8 3 C 32 Schmiedl & Seifert (2008)