efficiency at the maximum power output for simple two-level heat engine

Efficiency at the maximum power output for simple two-level heat engine

Sang Hoon Lee School of Physics, Korea Institute for Advanced Study

http://newton.kias.re.kr/~lshlj82

in collaboration with Jaegon Um (CCSS, CTP and Department of Physics and Astronomy, SNU) and Hyunggyu Park (School of Physics & Quantum Universe Center, KIAS)

DDAP9: 9th Dynamics Days Asia Pacific @ Hong Kong, 15 December, 2016.

SHL, J. Um, and H. Park, e-print arXiv:1612.00518.

Carnot engine

source: http://web2.uwindsor.ca/courses/physics/high_schools/2013/SteamEngine/images/PVgraph.jpg

S. Carnot, Rflexions Sur La Puissance Motrice Du Feu Et Sur Les Machines Propres Dvelopper Cette Puissance (Bachelier Libraire, Paris, 1824).

the

Sadi Carnot (1796-1832)

http://web2.uwindsor.ca/courses/physics/high_schools/2013/SteamEngine/images/PVgraph.jpg

Carnot engine

source: http://web2.uwindsor.ca/courses/physics/high_schools/2013/SteamEngine/images/PVgraph.jpg

S. Carnot, Rflexions Sur La Puissance Motrice Du Feu Et Sur Les Machines Propres Dvelopper Cette Puissance (Bachelier Libraire, Paris, 1824).

the

Sadi Carnot (1796-1832)

the Carnot eciency C =Weng|Qh|

=

|Qh| |Qc||Qh|

= 1 TcTh

quasi-static(the 1st law of thermodynamics)

http://web2.uwindsor.ca/courses/physics/high_schools/2013/SteamEngine/images/PVgraph.jpg


=

|Qh| |Qc||Qh|

= 1 TcTh



=

|Qh| |Qc||Qh|

= 1 TcTh


0: cyclic process

the 2nd law of thermodynamics: Stot

= Seng

+Sres

= QhTh

+

QcTc 0

(per cycle)

! |Qc||Qh| Tc

Th! = 1 |Qc||Qh|

1 TcTh

= C

) C in general, and C is the theoretically maximum eciency.


=

|Qh| |Qc||Qh|

= 1 TcTh


0: cyclic process

the 2nd law of thermodynamics: Stot

= Seng

+Sres

= QhTh

+

QcTc 0

(per cycle)

! |Qc||Qh| Tc

Th! = 1 |Qc||Qh|

1 TcTh

= C

) C in general, and C is the theoretically maximum eciency.

Weng reaches the maximum value for given |Qh| in the Carnot engine,but the power P = Weng/ ! 0 where is the operating time !1

quasi-static

Th

Tc

hot reservoir

cold reservoir

Thw

Tcw

Endoreversible engine P. Chambadal, Les Centrales Nuclaires (Armand Colin, Paris, 1957). I. I. Novikov, Efficiency of an atomic power generating installation, At. Energy 3, 1269 (1957);

The efficiency of atomic power stations, J. Nucl. Energy 7, 125 (1958). F. L. Curzon and B. Ahlborn, Efficiency of a Carnot engine at maximum power output, Am. J. Phys. 43, 22 (1975).

endoreversibility

Th

Tc

hot reservoir

cold reservoir

Thw

Tcw

during t1

irreversible heat conduction

the input energy (linear heat conduction) Qh = t1(Th Thw)

the reversible engine

operated at Thw and Tcw!QhThw

=QcTcw

during t2


the heat rejected (linear heat conduction) Qc = t2(Tcw Tc)


The efficiency of atomic power stations, J. Nucl. Energy 7, 125 (1958). F. L. Curzon and B. Ahlborn, Efficiency of a Carnot engine at maximum power output, Am. J. Phys. 43, 22 (1975).

endoreversibility

the (Chambadal-Novikov-)Curzon-Ahlborn eciency CA = 1r

TcTh

Th

Tc

hot reservoir

cold reservoir

Thw

Tcw

during t1


the input energy (linear heat conduction) Qh = t1(Th Thw)

the reversible engine

operated at Thw and Tcw!QhThw

=QcTcw

during t2


the heat rejected (linear heat conduction) Qc = t2(Tcw Tc)

maximizing power P =Qh Qct1 + t2

with respect to t1 and t2


The efficiency of atomic power stations, J. Nucl. Energy 7, 125 (1958). F. L. Curzon and B. Ahlborn, Efficiency of a Carnot engine at maximum power output, Am. J. Phys. 43, 22 (1975).3/31/16, 12:03Endoreversible thermodynamics - Wikipedia, the free encyclopedia

Page 2 of 3https://en.wikipedia.org/wiki/Endoreversible_thermodynamics

Power Plant (C) (C) (Carnot) (Endoreversible) (Observed)West Thurrock (UK) coal-fired power

plant 25 565 0.64 0.40 0.36

CANDU (Canada) nuclear power plant 25 300 0.48 0.28 0.30Larderello (Italy) geothermal power

plant 80 250 0.33 0.178 0.16

As shown, the endoreversible efficiency much more closely models the observed data. However, such anengine violates Carnot's principle which states that work can be done any time there is a difference intemperature. The fact that the hot and cold reservoirs are not at the same temperature as the working fluidthey are in contact with means that work can and is done at the hot and cold reservoirs. The result istantamount to coupling the high and low temperature parts of the cycle, so that the cycle collapses.[7] In theCarnot cycle there is strict necessity that the working fluid be at the same temperatures as the heat reservoirsthey are in contact with and that they are separated by adiabatic transformations which prevent thermalcontact. The efficiency was first derived by William Thomson [8] in his study of an unevenly heated body inwhich the adiabatic partitions between bodies at different temperatures are removed and maximum work isperformed. It is well known that the final temperature is the geometric mean temperature so that

the efficiency is the Carnot efficiency for an engine working between and .

Due to occasional confusion about the origins of the above equation, it is sometimes named theChambadal-Novikov-Curzon-Ahlborn efficiency.

See alsoHeat engine

An introduction to endoreversible thermodynamics is given in the thesis by Katharina Wagner.[4] It is alsointroduced by Hoffman et al.[9][10] A thorough discussion of the concept, together with many applications inengineering, is given in the book by Hans Ulrich Fuchs.[11]

References1. I. I. Novikov. The Efficiency of Atomic Power Stations. Journal Nuclear Energy II, 7:125128, 1958. translated from

Atomnaya Energiya, 3 (1957), 409.2. Chambadal P (1957) Les centrales nuclaires. Armand Colin, Paris, France, 4 1-583. F.L. Curzon and B. Ahlborn, American Journal of Physics, vol. 43, pp. 2224 (1975)4. M.Sc. Katharina Wagner, A graphic based interface to Endoreversible Thermodynamics, TU Chemnitz, Fakultt fr

Naturwissenschaften, Masterarbeit (in English). http://archiv.tu-chemnitz.de/pub/2008/0123/index.html5. A Bejan, J. Appl. Phys., vol. 79, pp. 11911218, 1 Feb. 1996 http://dx.doi.org/10.1016/S0035-3159(96)80059-66. Callen, Herbert B. (1985). Thermodynamics and an Introduction to Thermostatistics (2nd ed. ed.). John Wiley &

Sons, Inc.. ISBN 0-471-86256-8.7. B. H. Lavenda, Am. J. Phys., vol. 75, pp. 169-175 (2007)8. W. Thomson, Phil. Mag. (Feb. 1853)


TcTh


TcTh

Q. Is this a universal formula for power-maximizing efficiency, or does endoreversibility guarantee it?


TcTh

Q. Is this a universal formula for power-maximizing efficiency, or does endoreversibility guarantee it?

A. No. The linear heat conduction is essential.

Q = (Th Tc)

We introduce a different type of engine with non-(CN)CA optimal efficiency.

L. Chen and Z. Yan, J. Chem. Phys. 90, 3740 (1988): F. Angulo-Brown and R. Pez-Hernndez, J. Appl. Phys. 74, 2216 (1993):

(Dulong-Petit law of cooling)Q = (Th Tc)n

Q = (Tnh Tnc )

our simple two-level heat engine model

R1

R2

relaxation with

relaxation with

Q1

Q2

E1

E2

T1

T2

q

t1

t2

during t1

during t2

W = E1 E2W 0 = E1 E2

q/(1 q) = exp(E1/T1)

/(1 ) = exp(E2/T2)

0

E1

q(setting the Boltzmann constant

kB 1 for notational convenience)


R1

R2

relaxation with

relaxation with

Q1

Q2

E1

E2

T1

T2

q

t1

t2

during t1

during t2

W = E1 E2W 0 = E1 E2

q/(1 q) = exp(E1/T1)

/(1 ) = exp(E2/T2)

0

E2

(setting the Boltzmann constant



R1

R2

relaxation with

relaxation with

Q1

Q2

E1

E2

T1

T2

q

t1

t2

during t1

during t2

W = E1 E2W 0 = E1 E2

q/(1 q) = exp(E1/T1)

/(1 ) = exp(E2/T2)

0

q

E1

(setting the Boltzmann constant



R1

E1T1

q

R2

E2

T2

0

q

0

during duringstochastic Markov processes

1 2

|P1i(t1 = 0) |P1i(t1 = 1) = |P2i(t2 = 0) |P2i(t2 = 2)

|P2i(t2 = 2) = |P1i(t1 = 0)

W = E1 E2

W 0 = E1 E2

Q1 Q2

hWnetiT1

sh =T2T1

sc +hWnetiT1

sh =E1T1

scG1 (sc)

sc

sh

Let t1 = t2 = /2, then

in terms of , the maximum power is achieved for ! 0, asPower ! hWneti/4 and the power is monotonically decreased as is increased.

hWneti = (q )(1 e/2)2

1 e

{T1 ln[(1 q)/q] T2 ln[(1 )/]}

time: decoupled overall factor

hWneti( !1) = (q ) {T1 ln[(1 q)/q] T2 ln[(1 )/]}

Power hP i = hWneti

=

q

(1 e/2)2

1 e

{T1 ln[(1 q)/q] T2 ln[(1 )/]}

(still decoupled even when t1 6= t2)

Let t1 = t2 = /2, then

in terms of , the maximum power is achieved for ! 0, asPower ! hWneti/4 and the power is monotonically decreased as is increased.

hWneti = (q )(1 e/2)2

1 e

{T1 ln[(1 q)/q] T2 ln[(1 )/]}

time: decoupled overall factor

hWneti( !1) = (q ) {T1 ln[(1 q)/q] T2 ln[(1 )/]}

Power hP i = hWneti

=

q

(1 e/2)2

1 e

{T1 ln[(1 q)/q] T2 ln[(1 )/]}

our goal: to find (q, ) = (q, ) maximizing hP i@hP i@q

q=q,=

=@hP i@

q=q,=

= 0

(still decoupled even when t1 6= t2)

hWneti = hW i hW 0i = (P1 P2)(E1 E2)= (P1 P2){T1 ln[(1 q)/q] T2 ln[(1 )/]}

eciency =hWnetihQ1i

=hW i hW 0i

hQ1i= 1 T2

T1

ln[(1 )/]ln[(1 q)/q]

substitute (q, ) = (q, ) here,then

op

(q, ) is the eciencyat the maximum power output

4

0

0.05

0.1

0.15

0.2

0 0.2 0.4 0.6 0.8 1

q* a

nd

*

c = 1 T2 / T1

q**

q*(c0) = *(c0)q*(c=1)

c1 asymptote

FIG. 3. Numerically found q and values satisfying Eq. (18), as afunction of

C

= 1T2/T1, along with the q(C ! 0) = (C ! 0)and q(

C

= 1) values presented in Sec. III B 2. (C

= 1) = 0 (thehorizontal axis). The

C

! 1 asymptote indicates Eq. (34).schematically . . .

q

= q

no net work

as C is increased

q(C 0) = (C 0) 0.083 221 720 199 517 7

q(C = 1) 0.217 811 705 719 800(C = 1) = 0

FIG. 4. Illustration of the optimal transition rates (q, ) for the max-imum power output as the T2/T1 value varies.

2. Asymptotic behaviors obtained from series expansion

The upper bound for q is given by the condition C

= 1,satisfying ln[(1 q)/q] = 1/(1 q) and q(

C

= 1) '0.217 811 705 719 800 found numerically and (

C

= 1) = 0exactly from Eq. (16b).

C

= 0 always satisfies Eq. (18) re-gardless of q values, so finding the optimal q is meaningless(in fact, when

C

= 0, the operating regime for the engineis shrunk to the line q = and there cannot be any positivework). Therefore, let us examine the case

C

' 0 using theseries expansion of q with respect to

C

, as

q

= q0 + a1C + a22C

+ a33C

+ O 4C

. (22)

Substituting Eq. (22) into Eq. (18) and expanding the left-handside with respect to

C

again, we obtain

2 (1 2q0) ln[(1 q0)/q0]2q0 1 C

+q0(1 q0) 2a1(1 2q0)

2(1 q0)q0(1 2q0)32

C

+ c3(q0, a1, a2)3C

+ O 4C

= 0 ,

(23)

where c3(q0, a1, a2) = [10q60 + 3a21 6q0(a21 + a2) 6q50(5 +

6a1+8a2)12q30(1+6a1+16a21+9a2)+q20(1+18a1+132a21+42a2)+q40(31+90a1+96a

21+120a2)]/[6(12q0)5(1q0)2q20].

Letting the linear coecient to be zero yields

21 2q0 = ln

1 q0

q0

!, (24)

from which the lower bound for q(C

! 0) = q0 =(

C

! 0) ' 0.083 221 720 199 517 7 found numerically[lim

C

!0 U(C , q) = 1 2q, thus (C ! 0) = q(C ! 0)by Eq. (16b)]. Figure 3 shows the numerical solution (q, )as a function of

C

, where the asymptotic behaviors derivedabove hold when

C

' 0 and C

' 1. It seems that q ismonotonically increased and is monotonically decreased,as

C

is increased, i.e., qmin = q(

C

! 0), qmax = q(C = 1),min = 0, and

max =

(C

! 0). Figure 4 illustrates the situ-ation on the (q, ) plane. The linear coecient a1 in Eq. (22)can be written in terms of q0 when we let the coecient of thequadratic term in Eq. (23) to be zero, as

a1 =q0(1 q0)2(1 2q0) . (25)

Similarly, the coecient a2 in Eq. (22) can also be written interms of q0 alone, by letting c3(q0, a1, a2) = 0 in Eq. (23) andusing the relations in Eqs. (24) and (25), as

a2 =7q0(1 q0)24(1 2q0) . (26)

With the relations of coecients in hand, we find theasymptotic behavior of op in Eq. (19) by expanding it withrespect to

C

after substituting q as the series expansion of

C

in Eq. (22). Then,

op =1

(1 2q0) ln[(1 q0)/q0]C

+

a1q03q20+2q30

+[q20+2a1q0(1+4a1)] ln[(1q0)/q0]

(12q0)3

ln2[(1 q0)/q0]2

C

+ d3(q0, a1, a2)3C

+ O 4C

,

(27)

where d3(q0, a1, a2) = {2(1 2q0)2a1[q20 + 2a1 q0(1 +4a1)] ln[(1q0)/q0]+2[2q40+a14a212a2+4q0(4a21+3a2)+4q30(1+a1+4a2)2q20(1+3a1+8a21+12a2)] ln2[(1q0)/q0]+(12q0)4{2a21+[(12q0)a212(1q0)q0a2] ln[(1q0)/q0]}}/[(1q

20)

2q

20]. Using Eqs. (24), (25) and (26), Eq. (27) becomes sim-

ply

op =12

C

+182

C

+7 24q0 + 24q20

96(1 2q0)2 3C

+ O 4C

. (28)

No closed-form solution, but we get the series expansion at

cf) CA = 1p

1 C = 12C +

1

82C +

1

163C +

5

1284C +O(5C)

* C = 1

T2T1

C ! 0

4

0

0.05

0.1

0.15

0.2

0 0.2 0.4 0.6 0.8 1

q* a

nd

*

c = 1 T2 / T1

q**

q*(c0) = *(c0)q*(c=1)

c1 asymptote


C


C



C


q

= q

no net work

as C is increased

q(C 0) = (C 0) 0.083 221 720 199 517 7

q(C = 1) 0.217 811 705 719 800(C = 1) = 0





C


C


C


C


C


C

, as

q

= q0 + a1C + a22C

+ a33C

+ O 4C

. (22)


C

again, we obtain

2 (1 2q0) ln[(1 q0)/q0]2q0 1 C

+q0(1 q0) 2a1(1 2q0)

2(1 q0)q0(1 2q0)32

C

+ c3(q0, a1, a2)3C

+ O 4C

= 0 ,

(23)


6a1+8a2)12q30(1+6a1+16a21+9a2)+q20(1+18a1+132a21+42a2)+q40(31+90a1+96a

21+120a2)]/[6(12q0)5(1q0)2q20].


21 2q0 = ln

1 q0

q0

!, (24)


! 0) = q0 =(

C


C


C


C

' 0 and C


C


C

! 0), qmax = q(C = 1),min = 0, and

max =

(C


a1 =q0(1 q0)2(1 2q0) . (25)


a2 =7q0(1 q0)24(1 2q0) . (26)


C


C

in Eq. (22). Then,

op =1

(1 2q0) ln[(1 q0)/q0]C

+

a1q03q20+2q30

+[q20+2a1q0(1+4a1)] ln[(1q0)/q0]

(12q0)3

ln2[(1 q0)/q0]2

C

+ d3(q0, a1, a2)3C

+ O 4C

,

(27)


20)

2q


ply

op =12

C

+182

C

+7 24q0 + 24q20

96(1 2q0)2 3C

+ O 4C

. (28)

different!' 0.077 492

= 0.0625

strong coupling between the thermodynamics fluxes: the heat flux is directly proportional to the work-generating flux ref) C. Van den Broeck, PRL 95, 190602 (2005).

strong coupling + symmetry between the reservoirs (left-right symmetry) ref) M. Esposito, K. Lindenberg, and C. Van den Broeck, PRL 102, 130602 (2009).

The deviation from CA

for op

enters from the third order.

No closed-form solution, but we get the series expansion at

cf) CA = 1p

1 C = 12C +

1

82C +

1

163C +

5

1284C +O(5C)

* C = 1

T2T1

C ! 0

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1

op

c

at (q*, *)CA = 11c

c/(2c)c/2

c1 asymptote

0.88

0.92

0.96

1

0.97 0.98 0.99 1

op

c

: very similar to up to a certain point, but clearly different!

4

0

0.05

0.1

0.15

0.2

0 0.2 0.4 0.6 0.8 1

q* a

nd *

c = 1 T2 / T1

q**

q*(c0) = *(c0)q*(c=1)

c1 asymptote


C


C



C


q

= q

no net work

as C is increased

q(C 0) = (C 0) 0.083 221 720 199 517 7

q(C = 1) 0.217 811 705 719 800(C = 1) = 0





C


C


C


C


C


C

, as

q

= q0 + a1C + a22C

+ a33C

+ O 4C

. (22)


C

again, we obtain

2 (1 2q0) ln[(1 q0)/q0]2q0 1 C

+q0(1 q0) 2a1(1 2q0)

2(1 q0)q0(1 2q0)32

C

+ c3(q0, a1, a2)3C

+ O 4C

= 0 ,

(23)


6a1+8a2)12q30(1+6a1+16a21+9a2)+q20(1+18a1+132a21+42a2)+q40(31+90a1+96a

21+120a2)]/[6(12q0)5(1q0)2q20].


21 2q0 = ln

1 q0

q0

!, (24)


! 0) = q0 =(

C


C


C


C

' 0 and C


C


C

! 0), qmax = q(C = 1),min = 0, and

max =

(C


a1 =q0(1 q0)2(1 2q0) . (25)


a2 =7q0(1 q0)24(1 2q0) . (26)


C


C

in Eq. (22). Then,

op =1

(1 2q0) ln[(1 q0)/q0]C

+

a1q03q20+2q30

+[q20+2a1q0(1+4a1)] ln[(1q0)/q0]

(12q0)3

ln2[(1 q0)/q0]2

C

+ d3(q0, a1, a2)3C

+ O 4C

,

(27)


20)

2q


ply

op =12

C

+182

C

+7 24q0 + 24q20

96(1 2q0)2 3C

+ O 4C

. (28)cf) CA = 1

p1 C = 1

2C +

1

82C +

1

163C +

5

1284C +O(5C)

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1

op

c

at (q*, *)CA = 11c

c/(2c)c/2

c1 asymptote

0.88

0.92

0.96

1

0.97 0.98 0.99 1

op

c

M. Esposito et al., PRL 105, 150603 (2010)s upper and lower bounds, respectively

deviation

very similar

: very similar to up to a certain point, but clearly different!

4

0

0.05

0.1

0.15

0.2

0 0.2 0.4 0.6 0.8 1

q* a

nd *

c = 1 T2 / T1

q**

q*(c0) = *(c0)q*(c=1)

c1 asymptote


C


C



C


q

= q

no net work

as C is increased

q(C 0) = (C 0) 0.083 221 720 199 517 7

q(C = 1) 0.217 811 705 719 800(C = 1) = 0





C


C


C


C


C


C

, as

q

= q0 + a1C + a22C

+ a33C

+ O 4C

. (22)


C

again, we obtain

2 (1 2q0) ln[(1 q0)/q0]2q0 1 C

+q0(1 q0) 2a1(1 2q0)

2(1 q0)q0(1 2q0)32

C

+ c3(q0, a1, a2)3C

+ O 4C

= 0 ,

(23)


6a1+8a2)12q30(1+6a1+16a21+9a2)+q20(1+18a1+132a21+42a2)+q40(31+90a1+96a

21+120a2)]/[6(12q0)5(1q0)2q20].


21 2q0 = ln

1 q0

q0

!, (24)


! 0) = q0 =(

C


C


C


C

' 0 and C


C


C

! 0), qmax = q(C = 1),min = 0, and

max =

(C


a1 =q0(1 q0)2(1 2q0) . (25)


a2 =7q0(1 q0)24(1 2q0) . (26)


C


C

in Eq. (22). Then,

op =1

(1 2q0) ln[(1 q0)/q0]C

+

a1q03q20+2q30

+[q20+2a1q0(1+4a1)] ln[(1q0)/q0]

(12q0)3

ln2[(1 q0)/q0]2

C

+ d3(q0, a1, a2)3C

+ O 4C

,

(27)


20)

2q


ply

op =12

C

+182

C

+7 24q0 + 24q20

96(1 2q0)2 3C

+ O 4C

. (28)cf) CA = 1

p1 C = 1

2C +

1

82C +

1

163C +

5

1284C +O(5C)

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1

op

c

at (q*, *)CA = 11c

c/(2c)c/2

c1 asymptote

0.88

0.92

0.96

1

0.97 0.98 0.99 1

op

c

M. Esposito et al., PRL 105, 150603 (2010)s upper and lower bounds, respectively

deviation

very similar

: very similar to up to a certain point, but clearly different!Curzon-Ahlborn regime log correction regime

4

0

0.05

0.1

0.15

0.2

0 0.2 0.4 0.6 0.8 1

q* a

nd *

c = 1 T2 / T1

q**

q*(c0) = *(c0)q*(c=1)

c1 asymptote


C


C



C


q

= q

no net work

as C is increased

q(C 0) = (C 0) 0.083 221 720 199 517 7

q(C = 1) 0.217 811 705 719 800(C = 1) = 0





C


C


C


C


C


C

, as

q

= q0 + a1C + a22C

+ a33C

+ O 4C

. (22)


C

again, we obtain

2 (1 2q0) ln[(1 q0)/q0]2q0 1 C

+q0(1 q0) 2a1(1 2q0)

2(1 q0)q0(1 2q0)32

C

+ c3(q0, a1, a2)3C

+ O 4C

= 0 ,

(23)


6a1+8a2)12q30(1+6a1+16a21+9a2)+q20(1+18a1+132a21+42a2)+q40(31+90a1+96a

21+120a2)]/[6(12q0)5(1q0)2q20].


21 2q0 = ln

1 q0

q0

!, (24)


! 0) = q0 =(

C


C


C


C

' 0 and C


C


C

! 0), qmax = q(C = 1),min = 0, and

max =

(C


a1 =q0(1 q0)2(1 2q0) . (25)


a2 =7q0(1 q0)24(1 2q0) . (26)


C


C

in Eq. (22). Then,

op =1

(1 2q0) ln[(1 q0)/q0]C

+

a1q03q20+2q30

+[q20+2a1q0(1+4a1)] ln[(1q0)/q0]

(12q0)3

ln2[(1 q0)/q0]2

C

+ d3(q0, a1, a2)3C

+ O 4C

,

(27)


20)

2q


ply

op =12

C

+182

C

+7 24q0 + 24q20

96(1 2q0)2 3C

+ O 4C

. (28)cf) CA = 1

p1 C = 1

2C +

1

82C +

1

163C +

5

1284C +O(5C)

the entropy production relation in our model

6

R1

E1T1

q

R2

E2

T2

0

q

0


1 2

|P1(t1 = 0) |P1(t1 = 1) = |P2(t2 = 0) |P2(t2 = 2)

|P2(t2 = 2) = |P1(t1 = 0)

W = E1 E2

W = E1 E2

Q1 Q2

WnetT1

sh =T2T1

sc +Wnet

T1

sh =E1T1

scG1 (sc)

sc

sh

FIG. 6. The entropy relation between sh

and sc

, given by Eq. (34) andthe linear relation in Eq. (36) representing the first law of thermody-namics. The maximum value of hWneti (the intercept of the linearrelation on the vertical axis times T1) is achieved when the line be-comes the tangential one of the curve, as illustrated here.

based on the relations in Eqs. (30) and (31) [the same proce-dure as the one leading to (25)]. As shown in Fig. 5, however,the asymptotic form only holds in a rather limited range of

C

very close to unity, indicating the necessity to taking higherorder terms into account for more accurate asymptotic behav-ior.

C. The entropy production relation

In Ref. [9], it is argued that the necessary and sucientcondition for the Curzon-Ahlborn eciency at the maximumpower is that the entropy productions at the hot and cold reser-voirs (denoted by s

h

and sc

, respectively) should be relatedby a specific functional form, namely, s

h

= F (sc

) whereF (x) = x/(1 + x) with the system-specific constant .

The entropy production in our model is given by

s

h

=hQ1iT1=

(1 e/2)21 e (q )

E1

T1,

s

c

=hQ2iT2=

(1 e/2)21 e (q )

E2

T2,

(33)

Given T1 and T2 and putting E1 as a constant, we obtain theentropy relation given by

s

h

=E1

T1

s

c

G1 (sc

), (34)

where G1 is the inverse function of G defined from the rela-tion in Eq. (33), s

c

= G(E2/T2). Note that sh is an increasingfunction with respect to s

c

while dsh

/dsc

is a decreasing one,so s

h

(sc

) is an increasing and concave function of sc

as il-lustrated in Fig. 6. Therefore, the unique [guaranteed by theconcavity of s

h

(sc

)] optimal entropy production denoted by

s

c

, which makes the power reach its maximum value, is deter-mined by

ds

h

ds

c

s

c

=sc

=E1

T1

q + G1 sc

=T2

T1, (35)

where T2/T1 comes from the thermodynamic first law,

s

h

=T2

T1s

c

+hWneti

T1. (36)

Here is not the same with the value in previous section yet:Since s

c

or obtained from Eq. (35) is a function of E1 or q,after optimizing q, q and agree with the previous ones.

BecauseG1(sc

) is not a linear function of sc

in our case, weconclude that op , CA in general. However, we reveal that itis possible to find the regime where the entropy production ofour model approximately follows the functional form F (x),which indeed corresponds to the op ' CA regime, as weshow in the following section.

1. The linear regime

First, we take the regime where T = T1 T2 T1. Thenfrom Eq. (35), one can see the solution E = E1 E2 E1or s

c

1, which allows the small sc

expansion of G1 up tothe linear order as

G1(sc

) ' G1(0) + dG1

ds

c

s

c

=0s

c

=E1

T1+ (E1/T1)sc , (37)

where the constant (E1/T1) is given by

= 2

T1

E1

!2 "1 + cosh

E1

T1

!#1 e

(1 e/2)2 . (38)

Inserting Eq. (37) to the entropy relation in Eq. (34), we findthat the entropy production for hot and cold reservoirs actuallyfollows the functional form F (x) = x/(1 + x) for

C

! 0,which explains the Curzon-Ahlborn-like behavior for

C

! 0when it is actually the case that T T1. We have alreadychecked that op ' CA as C ! 0 due to the same linear andquadratic coecients from the series expansion in Sec. III B 2,but the entropy production relation suggests that there couldbe a deeper relation between our model and engines with theoptimal eciency of CA than the reasons for the linear andquadratic coecients, namely, the strong coupling betweenthe thermodynamics fluxes and the reservoir symmetry, re-spectively. In fact, the implication of the reservoir symmetryin the expression F 1(x) = F (x) holds only for

C

! 0.We emphasize that the behavior of eciency op ' CA in

C

! 0 is independent of the value q. However we can opti-mize hWnetiwith respect to q as following: The optimal condi-tion to maximize hWneti for a given T2/T1 value is equivalentto minimize , because

hWneti = T1

0BBBB@1

rT2

T1

1CCCCA

2

/ 1/ , (39)

sh: the entropy reduction of the hot reservoirsc: the entropy production of the cold reservoir

the entropy production relation in our model

6

R1

E1T1

q

R2

E2

T2

0

q

0


1 2

|P1(t1 = 0) |P1(t1 = 1) = |P2(t2 = 0) |P2(t2 = 2)

|P2(t2 = 2) = |P1(t1 = 0)

W = E1 E2

W = E1 E2

Q1 Q2

WnetT1

sh =T2T1

sc +Wnet

T1

sh =E1T1

scG1 (sc)

sc

sh


and sc



C




h

and sc


h

= F (sc



s

h

=hQ1iT1=

(1 e/2)21 e (q )

E1

T1,

s

c

=hQ2iT2=

(1 e/2)21 e (q )

E2

T2,

(33)


s

h

=E1

T1

s

c

G1 (sc

), (34)


c


c

while dsh

/dsc


h

(sc



h

(sc


s

c


ds

h

ds

c

s

c

=sc

=E1

T1

q + G1 sc

=T2

T1, (35)


s

h

=T2

T1s

c

+hWneti

T1. (36)


c


BecauseG1(sc





c



G1(sc

) ' G1(0) + dG1

ds

c

s

c

=0s

c

=E1

T1+ (E1/T1)sc , (37)


= 2

T1

E1

!2 "1 + cosh

E1

T1

!#1 e

(1 e/2)2 . (38)


C


C


C


C


hWneti = T1

0BBBB@1

rT2

T1

1CCCCA

2

/ 1/ , (39)

sh: the entropy reduction of the hot reservoirsc: the entropy production of the cold reservoir

6

R1

E1T1

q

R2

E2

T2

0

q

0


1 2

|P1(t1 = 0) |P1(t1 = 1) = |P2(t2 = 0) |P2(t2 = 2)

|P2(t2 = 2) = |P1(t1 = 0)

W = E1 E2

W = E1 E2

Q1 Q2

WnetT1

sh =T2T1

sc +Wnet

T1

sh =E1T1

scG1 (sc)

sc

sh


and sc



C




h

and sc


h

= F (sc



s

h

=hQ1iT1=

(1 e/2)21 e (q )

E1

T1,

s

c

=hQ2iT2=

(1 e/2)21 e (q )

E2

T2,

(33)


s

h

=E1

T1

s

c

G1 (sc

), (34)


c


c

while dsh

/dsc


h

(sc



h

(sc


s

c


ds

h

ds

c

s

c

=sc

=E1

T1

q + G1 sc

=T2

T1, (35)


s

h

=T2

T1s

c

+hWneti

T1. (36)


c


BecauseG1(sc





c



G1(sc

) ' G1(0) + dG1

ds

c

s

c

=0s

c

=E1

T1+ (E1/T1)sc , (37)


= 2

T1

E1

!2 "1 + cosh

E1

T1

!#1 e

(1 e/2)2 . (38)


C


C


C


C


hWneti = T1

0BBBB@1

rT2

T1

1CCCCA

2

/ 1/ , (39)

6

R1

E1T1

q

R2

E2

T2

0

q

0


1 2

|P1(t1 = 0) |P1(t1 = 1) = |P2(t2 = 0) |P2(t2 = 2)

|P2(t2 = 2) = |P1(t1 = 0)

W = E1 E2

W = E1 E2

Q1 Q2

WnetT1

sh =T2T1

sc +Wnet

T1

sh =E1T1

scG1 (sc)

sc

sh


and sc



C




h

and sc


h

= F (sc



s

h

=hQ1iT1=

(1 e/2)21 e (q )

E1

T1,

s

c

=hQ2iT2=

(1 e/2)21 e (q )

E2

T2,

(33)


s

h

=E1

T1

s

c

G1 (sc

), (34)


c


c

while dsh

/dsc


h

(sc



h

(sc


s

c


ds

h

ds

c

s

c

=sc

=E1

T1

q + G1 sc

=T2

T1, (35)


s

h

=T2

T1s

c

+hWneti

T1. (36)


c


BecauseG1(sc





c



G1(sc

) ' G1(0) + dG1

ds

c

s

c

=0s

c

=E1

T1+ (E1/T1)sc , (37)


= 2

T1

E1

!2 "1 + cosh

E1

T1

!#1 e

(1 e/2)2 . (38)


C


C


C


C


hWneti = T1

0BBBB@1

rT2

T1

1CCCCA

2

/ 1/ , (39)

where

6

R1

E1T1

q

R2

E2

T2

0

q

0


1 2

|P1(t1 = 0) |P1(t1 = 1) = |P2(t2 = 0) |P2(t2 = 2)

|P2(t2 = 2) = |P1(t1 = 0)

W = E1 E2

W = E1 E2

Q1 Q2

WnetT1

sh =T2T1

sc +Wnet

T1

sh =E1T1

scG1 (sc)

sc

sh


and sc



C




h

and sc


h

= F (sc



s

h

=hQ1iT1=

(1 e/2)21 e (q )

E1

T1,

s

c

=hQ2iT2=

(1 e/2)21 e (q )

E2

T2,

(33)


s

h

=E1

T1

s

c

G1 (sc

), (34)


c


c

while dsh

/dsc


h

(sc



h

(sc


s

c


ds

h

ds

c

s

c

=sc

=E1

T1

q + G1 sc

=T2

T1, (35)


s

h

=T2

T1s

c

+hWneti

T1. (36)


c


BecauseG1(sc





c



G1(sc

) ' G1(0) + dG1

ds

c

s

c

=0s

c

=E1

T1+ (E1/T1)sc , (37)


= 2

T1

E1

!2 "1 + cosh

E1

T1

!#1 e

(1 e/2)2 . (38)


C


C


C


C


hWneti = T1

0BBBB@1

rT2

T1

1CCCCA

2

/ 1/ , (39)

the 1st law of thermodynamics:6

R1

E1T1

q

R2

E2

T2

0

q

0


1 2

|P1(t1 = 0) |P1(t1 = 1) = |P2(t2 = 0) |P2(t2 = 2)

|P2(t2 = 2) = |P1(t1 = 0)

W = E1 E2

W = E1 E2

Q1 Q2

WnetT1

sh =T2T1

sc +Wnet

T1

sh =E1T1

scG1 (sc)

sc

sh


and sc



C




h

and sc


h

= F (sc



s

h

=hQ1iT1=

(1 e/2)21 e (q )

E1

T1,

s

c

=hQ2iT2=

(1 e/2)21 e (q )

E2

T2,

(33)


s

h

=E1

T1

s

c

G1 (sc

), (34)


c


c

while dsh

/dsc


h

(sc



h

(sc


s

c


ds

h

ds

c

s

c

=sc

=E1

T1

q + G1 sc

=T2

T1, (35)


s

h

=T2

T1s

c

+hWneti

T1. (36)


c


BecauseG1(sc





c



G1(sc

) ' G1(0) + dG1

ds

c

s

c

=0s

c

=E1

T1+ (E1/T1)sc , (37)


= 2

T1

E1

!2 "1 + cosh

E1

T1

!#1 e

(1 e/2)2 . (38)


C


C


C


C


hWneti = T1

0BBBB@1

rT2

T1

1CCCCA

2

/ 1/ , (39)

the maximum hWneti condition

6

R1

E1T1

q

R2

E2

T2

0

q

0


1 2

|P1(t1 = 0) |P1(t1 = 1) = |P2(t2 = 0) |P2(t2 = 2)

|P2(t2 = 2) = |P1(t1 = 0)

W = E1 E2

W = E1 E2

Q1 Q2

WnetT1

sh =T2T1

sc +Wnet

T1

sh =E1T1

scG1 (sc)

sc

sh


and sc



C




h

and sc


h

= F (sc



s

h

=hQ1iT1=

(1 e/2)21 e (q )

E1

T1,

s

c

=hQ2iT2=

(1 e/2)21 e (q )

E2

T2,

(33)


s

h

=E1

T1

s

c

G1 (sc

), (34)


c


c

while dsh

/dsc


h

(sc



h

(sc


s

c


ds

h

ds

c

s

c

=sc

=E1

T1

(1 )q + G1 s

c

(1 ) =

T2

T1, (35)


s

h

=T2

T1s

c

+hWneti

T1. (36)


c


BecauseG1(sc





c



G1(sc

) ' G1(0) + dG1

ds

c

s

c

=0s

c

=E1

T1+ (E1/T1)sc , (37)


= 2

T1

E1

!2 "1 + cosh

E1

T1

!#1 e

(1 e/2)2 . (38)


C


C


C


C


hWneti = T1

0BBBB@1

rT2

T1

1CCCCA

2

/ 1/ , (39)

concave

6

R1

E1T1

q

R2

E2

T2

0

q

0


1 2

|P1(t1 = 0) |P1(t1 = 1) = |P2(t2 = 0) |P2(t2 = 2)

|P2(t2 = 2) = |P1(t1 = 0)

W = E1 E2

W = E1 E2

Q1 Q2

WnetT1

sh =T2T1

sc +Wnet

T1

sh =E1T1

scG1 (sc)

sc

sh


and sc



C




h

and sc


h

= F (sc



s

h

=hQ1iT1=

(1 e/2)21 e (q )

E1

T1,

s

c

=hQ2iT2=

(1 e/2)21 e (q )

E2

T2,

(33)


s

h

=E1

T1

s

c

G1 (sc

), (34)


c


c

while dsh

/dsc


h

(sc



h

(sc


s

c


ds

h

ds

c

s

c

=sc

=E1

T1

q + G1 sc

=T2

T1, (35)


s

h

=T2

T1s

c

+hWneti

T1. (36)


c


BecauseG1(sc





c



G1(sc

) ' G1(0) + dG1

ds

c

s

c

=0s

c

=E1

T1+ (E1/T1)sc , (37)


= 2

T1

E1

!2 "1 + cosh

E1

T1

!#1 e

(1 e/2)2 . (38)


C


C


C


C


hWneti = T1

0BBBB@1

rT2

T1

1CCCCA

2

/ 1/ , (39)

the maximum hWneti condition

6

R1

E1T1

q

R2

E2

T2

0

q

0


1 2

|P1(t1 = 0) |P1(t1 = 1) = |P2(t2 = 0) |P2(t2 = 2)

|P2(t2 = 2) = |P1(t1 = 0)

W = E1 E2

W = E1 E2

Q1 Q2

WnetT1

sh =T2T1

sc +Wnet

T1

sh =E1T1

scG1 (sc)

sc

sh


and sc



C




h

and sc


h

= F (sc



s

h

=hQ1iT1=

(1 e/2)21 e (q )

E1

T1,

s

c

=hQ2iT2=

(1 e/2)21 e (q )

E2

T2,

(33)


s

h

=E1

T1

s

c

G1 (sc

), (34)


c


c

while dsh

/dsc


h

(sc



h

(sc


s

c


ds

h

ds

c

s

c

=sc

=E1

T1

(1 )q + G1 s

c

(1 ) =

T2

T1, (35)


s

h

=T2

T1s

c

+hWneti

T1. (36)


c


BecauseG1(sc





c



G1(sc

) ' G1(0) + dG1

ds

c

s

c

=0s

c

=E1

T1+ (E1/T1)sc , (37)


= 2

T1

E1

!2 "1 + cosh

E1

T1

!#1 e

(1 e/2)2 . (38)


C


C


C


C


hWneti = T1

0BBBB@1

rT2

T1

1CCCCA

2

/ 1/ , (39)

the linear regime

6

R1

E1T1

q

R2

E2

T2

0

q

0


1 2

|P1(t1 = 0) |P1(t1 = 1) = |P2(t2 = 0) |P2(t2 = 2)

|P2(t2 = 2) = |P1(t1 = 0)

W = E1 E2

W = E1 E2

Q1 Q2

WnetT1

sh =T2T1

sc +Wnet

T1

sh =E1T1

scG1 (sc)

sc

sh


and sc



C




h

and sc


h

= F (sc



s

h

=hQ1iT1=

(1 e/2)21 e (q )

E1

T1,

s

c

=hQ2iT2=

(1 e/2)21 e (q )

E2

T2,

(33)


s

h

=E1

T1

s

c

G1 (sc

), (34)


c


c

while dsh

/dsc


h

(sc



h

(sc


s

c


ds

h

ds

c

s

c

=sc

=E1

T1

q + G1 sc

=T2

T1, (35)


s

h

=T2

T1s

c

+hWneti

T1. (36)


c


BecauseG1(sc





c



G1(sc

) ' G1(0) + dG1

ds

c

s

c

=0s

c

=E1

T1+ (E1/T1)sc , (37)


= 2

T1

E1

!2 "1 + cosh

E1

T1

!#1 e

(1 e/2)2 . (38)


C


C


C


C


hWneti = T1

0BBBB@1

rT2

T1

1CCCCA

2

/ 1/ , (39)

6

R1

E1T1

q

R2

E2

T2

0

q

0


1 2

|P1(t1 = 0) |P1(t1 = 1) = |P2(t2 = 0) |P2(t2 = 2)

|P2(t2 = 2) = |P1(t1 = 0)

W = E1 E2

W = E1 E2

Q1 Q2

WnetT1

sh =T2T1

sc +Wnet

T1

sh =E1T1

scG1 (sc)

sc

sh


and sc



C




h

and sc


h

= F (sc



s

h

=hQ1iT1=

(1 e/2)21 e (q )

E1

T1,

s

c

=hQ2iT2=

(1 e/2)21 e (q )

E2

T2,

(33)


s

h

=E1

T1

s

c

G1 (sc

), (34)


c


c

while dsh

/dsc


h

(sc



h

(sc


s

c


ds

h

ds

c

s

c

=sc

=E1

T1

q + G1 sc

=T2

T1, (35)


s

h

=T2

T1s

c

+hWneti

T1. (36)


c


BecauseG1(sc





c



G1(sc

) ' G1(0) + dG1

ds

c

s

c

=0s

c

=E1

T1+ (E1/T1)sc , (37)


= 2

T1

E1

!2 "1 + cosh

E1

T1

!#1 e

(1 e/2)2 . (38)


C


C


C


C


hWneti = T1

0BBBB@1

rT2

T1

1CCCCA

2

/ 1/ , (39)

6

R1

E1T1

q

R2

E2

T2

0

q

0


1 2

|P1(t1 = 0) |P1(t1 = 1) = |P2(t2 = 0) |P2(t2 = 2)

|P2(t2 =

efficiency at the maximum power output for simple two-level heat engine

Science