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)

2016 Workshop on Special Topics in Statistical Physics & Complex Systems (SPnCS) @ Chosun Univ., Gwangju, 22 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.

quasi-static


=

|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

hW i = (E1 E2)P1 and hW 0i = (E1 E2)P2where P1 (P2) is the population in E1 (E2) at R1 (R2), respectively

independent of t1

and t2

(no P1

and P2

dependency)

and ! Carnot

= 1 T2

/T1

when ' q

total entropy change S = Q1T1

+

Q2T2

hQ1i = (P1 P2)T1 ln[(1 q)/q], hQ2i = (P1 P2)T2 ln[(1 )/]

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]

0

= (P1 P2)E1 = (P1 P2)E2(from the Schnakenberg entropy formula, or equivalently, the 1st law:

hEi = hQi hW i for each half of the cycle)


eciency =hWnetihQ1i

=hW i hW 0i

hQ1i= 1 T2

T1

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

P1(t1 !1, t2 !1) = q and P2(t1 !1, t2 !1) = , as expected

meaningful only for q > , or hWneti > 0P1 = (1 q)P1 + q(1 P1)P2 = (1 )P2 + (1 P2)

! P1 = q A1et01 , P2 = A2et

02

(0 t01 t1 and 0 t02 t2)P1(t

01 = 0, t2) = P2(t1, t2) and P2(t1, t

02 = 0) = P1(t1, t2)

! A1 = q P2, A2 = P1 and let t01 = t1, t02 = t2

! P1 =q(1 et1) + (1 et2)et1

1 e(t1+t2)

P2 =(1 et2) + q(1 et1)et2

1 e(t1+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 )/]}

(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


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)


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

! T2q(1 q)

T1(1 )= 1

at (q, ) (global optimum)

3

equation system for given q and values,

dP1

dt

= (1 q)P1 + q(1 P1) ,dP2

dt

= (1 )P2 + (1 P2) .(7)

With the circular boundary condition P1(t01 = 0, t02 = t2) =

P2(t01 = t1, t02 = t2) and P2(t

01 = t1, t

02 = 0) = P1(t

01 = t1, t

02 =

t2) where the primed variables describe intermediate time, thesolution at t01 = t1 and t

02 = t2 is

P1 =q(1 et1 ) + (1 et2 )et1

1 e(t1+t2) ,

P2 =(1 et2 ) + q(1 et1 )et2

1 e(t1+t2) ,(8)

and limt1,t2!1 P1 = q and limt1,t2!1 P2 = as expected.

Let t1 = t2 = /2 (each reservoir contacts with the systemduring the equal time), then

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

1 e"T1 ln

1 q

q

! T2 ln

1

!#,

(9)so the monotonically increasing factor (1 e/2)2/(1 e)for the time scale is decoupled from the rest of the formulaand only plays the role of overall factor. The amount of network is increased up to

lim!1hWneti = (q )

"T1 ln

1 q

q

! T2 ln

1

!#. (10)

The average power output hPi, which is the amount of network per unit time, is given by

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

(1 e)"T1 ln

1 q

q

! T2 ln

1

!#.

(11)In terms of , the maximum power is achieved forlim!0hPi = hWneti/4 and the power is monotonically de-creased as is increased. Therefore, from now on, we discard

the time dependence altogether and focus on other parame-ters, i.e., denoting hWneti = hPi without considering the over-all factor involving for notational convenience. Numerically,we obtain the net work and eciency for (q, ) combination,as shown in Fig. 2. In Sec. III B, we derive the condition forthe eciency at the maximum power output.

B. Eciency at the maximum power output

1. The condition for the maximum power output

For a given T2/T1 value, the maximum power output con-dition for the two-variable function is

@hPi@q

q=q,=

=@hPi@

q=q,=

= 0 , (12)

which leads to

1 T2T1

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

q

q

(1 q) ln[(1 q)/q] , (13a)

and

1 T2T1

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

(T2/T1)(q )(1 ) ln[(1 q)/q] , (13b)

from Eq. (11). By eliminating the left-hand side of Eqs. (13a)and (13b), we obtain the following simple relation

T2q(1 q)

T1(1 ) = 1 , (14a)

or

=12

2666641

r1 4T2

T1q

(1 q)377775 , (14b)

because 0 < < q < 1/2. By substituting as a function ofq

in Eq. (14b) to Eq. (13a) or Eq. (13b), we get the optimumcondition f (T2/T1, q) = 0, which is explicitly

f (T2/T1, q) = ln

1 qq

! T2

T1ln

26666666666666664

1 +r

1 4T2T1

q

(1 q)

1 r

1 4T2T1

q

(1 q)

37777777777777775

q

12+

12

r1 4T2

T1q

(1 q)q

(1 q)

= ln

1 qq

! (1

C

) ln

266666666664

1 +q

(1 2q)2 + 4C

q

(1 q)1

q(1 2q)2 + 4

C

q

(1 q)

377777777775

q

12+

12

q(1 2q)2 + 4

C

q

(1 q)q

(1 q) = 0 .

(15)

Furthermore, the condition in Eq. (15) leads to the followingform of op from Eq. (6),

op =q

12+

12

r1 4T2

T1q

(1 q)q

(1 q) ln[(1 q)/q] (16a)

or in terms of C

= 1 T2/T1,

op =q

12+

12

q(1 2q)2 + 4

C

q

(1 q)q

(1 q) ln[(1 q)/q] . (16b)

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)

quadratic form

FIG. 3. Numerically found q and values satisfying Eq. (15), 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 quadratic form indicates Eq. (19).

We verify that the (q, ) is indeed the maximum point byusing the relations of second derivatives

0BBBBB@@2hPi@q2

q=q,=

1CCCCCA

0BBBBB@@2hPi@2

q=q,=

1CCCCCA

0BBBBB@@2hPi@q@

q=q,=

1CCCCCA

2

=

T

21(1 )q

(1 q) (2q q )2 4T 21 2(1 )2

2(1 )2q2(1 q)2

= T1(q )2

(1 )q3(1 q)3 > 0 ,(17)

where we use the relation in Eq. (14a), and

@2hPi@q2

q=q,=

= T1[q + (1 2q)]q

2(1 q)2 < 0 ,

@2hPi@2

q=q,=

= T2[q + (1 2q)]2(1 )2 < 0 .

(18)

Therefore, the procedure to calculate the eciency forgiven T2/T1 at the maximum power output seems straight-forward now. First, find the q value satisfying Eq. (15) andsubstitute the q value to Eq. (16a). The only (but very crucial)problem is that f (T2/T1, q) = 0 is a transcendental equationwhose closed-form solution is unattainable.

2. Asymptotic behaviors for extreme cases

The upper bound for q is given by the condition T2/T1 = 0,satisfying ln[(1 q)/q] = 1/(1 q) and q(T2/T1 = 0) '0.217 811 705 719 800 found numerically and (T2/T1 =0) = 0 exactly from Eq. (14b). T2/T1 = 1 always satisfiesf (T2/T1, q) = 0 regardless of q values, but the T2/T1 ' 1asymptotic case yields the lower bound for q(T2/T1 ' 1) =(T2/T1 ' 1) ' 0.083 221 720 199 517 7 found numerically.

schematically . . .

q

= q

no net work

q(T2/T1 1) 0.0832217201995177= (T2/T1 1)

q(T2/T1 = 0) 0.217811705719800(T2/T1 = 0) = 0

as T2/T1 (C) is decreased (increased),respectively

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

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1

op

c = 1 T2 / T1

at (q*, *)CA = 11c

c/(2c)c/2

FIG. 5. The eciency at the maximum power op as the functionof the Carnot eciency

C

in Eq. (16b) using numerically found op-timal q values, along with various asymptotic cases: the Curzon-Ahlborn eciency CA in Eq. (21), the upper bound C/(2 C) andthe lower bound

C

/2 in Ref. [19].

Figure 3 shows the numerical solution (q, ) as a functionof

C

= 1 T2/T1, where the asymptotic behaviors derivedabove hold when T2/T1 ' 0 and T2/T1 ' 1. It seems thatq

is monotonically decreased and is monotonically in-creased, as T2/T1 is increased, i.e., qmin = q

(T2/T1 ' 1),q

max = q

(T2/T1 = 0), min = 0, and max =

(T2/T1 ' 1).Figure 4 illustrates the situation on the (q, ) plane. We numer-ically obtain the quadratic function q(

C

) = qmin + aC + b2C

by fitting it to the numerical solution near C

= 0, in the range

C

2 [0, 0.05] in this case, which isq

(C

) ' qmin + 0.0457524C + 0.02767322C . (19)To proceed further, we have to take into account that q

is also a function of C

. Unfortunately, it is not possible toobtain the closed form solution for q(

C

), so we rely on the

! T2q(1 q)

T1(1 )= 1


! T2q(1 q)

T1(1 )= 1

! T2q(1 q)

T1(1 )= 1


) = 12

1

r1 4T2

T1q(1 q)

!

substituting (q) to

1 T2T1

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

=

T2T1

q

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

Finding the global maximum theoretically

1 T2T1

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

=

q

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

! f(T2

/T1

, q) = 0 ! q(T1

, T2

) ! (T1

, T2

) ! (hWnet

imax

, op

)

! f(T2

/T1

, q) = 0 ! q(T1

, T2

) ! (T1

, T2

) ! (hWnet

imax

, op

)

f(T2/T1, q) = ln

1 q

q

T2

T1ln

0

BB@1 +

r1 4T2

T1q(1 q)

1r

1 4T2T1

q(1 q)

1

CCAq 1

2+

1

2

r1 4T2

T1q(1 q)

q(1 q)

numerically found q(C) and (C)

0

0.05

0.1

0.15

0.2

0 0.2 0.4 0.6 0.8 1

q*(c0) = *(c0)

q*(c=1)

optim

al tr

ansi

tion

rate

s

c

q**

c0 and 1 asymptotes

schematically . . .

q

= q

net power < 0

schematically . . .

q

= q

net power < 0

C from 0 to 1(T2/T1 from 1 to 0)

( ), T1 = 1, T2 = 1/100

0.1 0.2 0.3 0.4 0.5q

0.1

0.2

0.3

0.4

0.5

0

0.05

0.1

0.15

0.2

0.25

0.3

( ), T1 = 1, T2 = 1/10

0.1 0.2 0.3 0.4 0.5q

0.1

0.2

0.3

0.4

0.5

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2

( ), T1 = 1, T2 = 1/2

0.1 0.2 0.3 0.4 0.5q

0.1

0.2

0.3

0.4

0.5

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04

( ), T1 = 1, T2 = 9/10

0.1 0.2 0.3 0.4 0.5q

0.1

0.2

0.3

0.4

0.5

0

0.0002

0.0004

0.0006

0.0008

0.001

0.0012

P (T2/T1 = 9/10)

P (T2/T1 = 1/2)

P (T2/T1 = 1/10)

P (T2/T1 = 1/100)

q

q

q

q


q(C ! 0) = (C ! 0) ' 0.083 221 720 199 517 7 q0

the root of

2

1 2q = ln1 q

q

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

the root of

1

1 q = ln1 q

q

) = 12

1

r1 4T2

T1q(1 q)

!

) op

= 1 T2T1

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

=

q 12+

1

2

r1 4T2

T1

q(1 q)

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

=q 1

2+

1

2

p(1 2q)2 + 4Cq(1 q)

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

= ln

1 q

q

(1 C) ln

1 +

p(1 2q)2 + 4Cq(1 q)

1p

(1 2q)2 + 4Cq(1 q)

!q 1

2+

1

2

p(1 2q)2 + 4Cq(1 q)

q(1 q)

f(T2/T1, q) = ln

1 q

q

T2

T1ln

0

BB@1 +

r1 4T2

T1q(1 q)

1r

1 4T2T1

q(1 q)

1

CCAq 1

2+

1

2

r1 4T2

T1q(1 q)

q(1 q)

! f(T2

/T1

, q) = 0 ! q(T1

, T2

) ! (T1

, T2

) ! (hWnet

imax

, op

)

the series expansion at C ! 0

0

0.05

0.1

0.15

0.2

0 0.2 0.4 0.6 0.8 1

q*(c0) = *(c0)

q*(c=1)

optim

al tr

ansi

tion

rate

s

c

q**

c0 and 1 asymptotes

the numerically found functional form of

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)

quadratic form near c=0


C


C




0BBBBB@@2hPi@q2

q=q,=

1CCCCCA

0BBBBB@@2hPi@2

q=q,=

1CCCCCA

0BBBBB@@2hPi@q@

q=q,=

1CCCCCA

2

=

T

21(1 )q

(1 q) (2q q )2 4T 21 2(1 )2

2(1 )2q2(1 q)2

= T1(q )2

(1 )q3(1 q)3 > 0 ,(18)


@2hPi@q2

q=q,=

= T1[q + (1 2q)]q

2(1 q)2 < 0 ,

@2hPi@2

q=q,=

= T2[q + (1 2q)]2(1 )2 < 0 .

(19)


2. Asymptotic behaviors based on series expansion

If we treat q as constant for a moment (which is clearly notthe case, of course, as q is also a function of T2/T1 to satisfy

schematically . . .

q

= q

no net work

q(T2/T1 1) 0.0832217201995177= (T2/T1 1)

q(T2/T1 = 0) 0.217811705719800(T2/T1 = 0) = 0



f = 0), the function f has the expansion form

f (T2/T1, q) ="ln

1 q

q

! 1

1 q#+

(1 ln

T2

T1

!+ ln

q

(1 q))

T2

T1

!

+ O2666664

T2

T1

!23777775

(20a)

when T2/T1 ' 0, and

f (T2/T1, q) ="ln

1 q

q

! 2

1 2q#

1 T2T1

!+

12(1 2q)

1 T2

T1

!2

+ O2666664

1 T2

T1

!33777775

(20b)

when T2/T1 ' 1. The upper bound for q is given by the con-dition T2/T1 = 0, satisfying ln[(1 q)/q] = 1/(1 q) andq

(T2/T1 = 0) ' 0.217 811 705 719 800 found numericallyand (T2/T1 = 0) = 0 exactly from Eq. (15b). T2/T1 = 1always satisfies f (T2/T1, q) = 0 regardless of q values, butthe T2/T1 ' 1 asymptotic case yields the lower bound forq

(T2/T1 ' 1) = (T2/T1 ' 1) ' 0.083 221 720 199 517 7found numerically. Figure 3 shows the numerical solution(q, ) as a function of

C

= 1 T2/T1, where the asymp-totic behaviors derived above hold when T2/T1 ' 0 andT2/T1 ' 1. It seems that q is monotonically decreasedand is monotonically increased, as T2/T1 is increased, i.e.,q

min = q

(T2/T1 ' 1), qmax = q(T2/T1 = 0), min = 0, andmax = (T2/T1 ' 1). For later, we numerically obtain thefunctional form of q(

C

) up to the quadratic term by fitting aquadratic function near

C

= 0, in the range C

2 [0, 0.05] inthis case, which is

q

(C

) ' qmin + 0.0457524C + 0.02767322C . (21)

0

0.05

0.1

0.15

0.2

0 0.2 0.4 0.6 0.8 1

q*(c0) = *(c0)

q*(c=1)

optim

al tr

ansi

tion

rate

s

c

q**

c0 and 1 asymptotes


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)



C


C




0BBBBB@@2hPi@q2

q=q,=

1CCCCCA

0BBBBB@@2hPi@2

q=q,=

1CCCCCA

0BBBBB@@2hPi@q@

q=q,=

1CCCCCA

2

=

T

21(1 )q

(1 q) (2q q )2 4T 21 2(1 )2

2(1 )2q2(1 q)2

= T1(q )2

(1 )q3(1 q)3 > 0 ,(18)


@2hPi@q2

q=q,=

= T1[q + (1 2q)]q

2(1 q)2 < 0 ,

@2hPi@2

q=q,=

= T2[q + (1 2q)]2(1 )2 < 0 .

(19)




schematically . . .

q

= q

no net work

q(T2/T1 1) 0.0832217201995177= (T2/T1 1)

q(T2/T1 = 0) 0.217811705719800(T2/T1 = 0) = 0




f (T2/T1, q) ="ln

1 q

q

! 1

1 q#+

(1 ln

T2

T1

!+ ln

q

(1 q))

T2

T1

!

+ O2666664

T2

T1

!23777775

(20a)

when T2/T1 ' 0, and

f (T2/T1, q) ="ln

1 q

q

! 2

1 2q#

1 T2T1

!+

12(1 2q)

1 T2

T1

!2

+ O2666664

1 T2

T1

!33777775

(20b)




C


min = q


C


C

= 0, in the range C


q

(C

) ' qmin + 0.0457524C + 0.02767322C . (21)f(T2/T1 = 0, q

) = ln

1 q

q

1

1 q = 0

! q ' 0.217 811 705 719 800

0

0.05

0.1

0.15

0.2

0 0.2 0.4 0.6 0.8 1

q*(c0) = *(c0)

q*(c=1)

optim

al tr

ansi

tion

rate

s

c

q**

c0 and 1 asymptotes


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)



C


C




0BBBBB@@2hPi@q2

q=q,=

1CCCCCA

0BBBBB@@2hPi@2

q=q,=

1CCCCCA

0BBBBB@@2hPi@q@

q=q,=

1CCCCCA

2

=

T

21(1 )q

(1 q) (2q q )2 4T 21 2(1 )2

2(1 )2q2(1 q)2

= T1(q )2

(1 )q3(1 q)3 > 0 ,(18)


@2hPi@q2

q=q,=

= T1[q + (1 2q)]q

2(1 q)2 < 0 ,

@2hPi@2

q=q,=

= T2[q + (1 2q)]2(1 )2 < 0 .

(19)




schematically . . .

q

= q

no net work

q(T2/T1 1) 0.0832217201995177= (T2/T1 1)

q(T2/T1 = 0) 0.217811705719800(T2/T1 = 0) = 0




f (T2/T1, q) ="ln

1 q

q

! 1

1 q#+

(1 ln

T2

T1

!+ ln

q

(1 q))

T2

T1

!

+ O2666664

T2

T1

!23777775

(20a)

when T2/T1 ' 0, and

f (T2/T1, q) ="ln

1 q

q

! 2

1 2q#

1 T2T1

!+

12(1 2q)

1 T2

T1

!2

+ O2666664

1 T2

T1

!33777775

(20b)




C


min = q


C


C

= 0, in the range C


q

(C

) ' qmin + 0.0457524C + 0.02767322C . (21)f(T2/T1 = 0, q

) = ln

1 q

q

1

1 q = 0

! q ' 0.217 811 705 719 800

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


= 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


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)

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)

= ln

1 q

q

(1 C) ln

1 +

p(1 2q)2 + 4Cq(1 q)

1p

(1 2q)2 + 4Cq(1 q)

!q 1

2+

1

2

p(1 2q)2 + 4Cq(1 q)

q(1 q)0 = f(C , q)

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)

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)

asymptotic form


C


C


= 1) = 0(the horizontal axis). The asymptotic form on the right side indicatesEq. (31).

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





C


C


C


C


C


C

, as

q

= q0 + a1C + a22C

+ O 3C

. (22)

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1

op

c = 1 T2 / T1

at (q*, *)CA = 11c

c/(2c)c/2

1+a[ln(1c)b](1c)

FIG. 5. The eciency at the maximum power op as the function ofthe Carnot eciency

C

in Eq. (19) using numerically found optimalq

values, along with various asymptotic cases: the Curzon-Ahlborneciency CA in Eq. (28), the upper bound C/(2C) and the lowerbound

C

/2 in Ref. [19], and the function in Eq. (32) for C

0.65.


C

again, we obtain

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

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

2(1 a0)a0(1 2a0)32

C

+O 3C

= 0 .

(23)


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 =a0(1 a0)2(1 2a0) . (25)


C


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)


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

+ O 3C

.

(26)

Using Eqs. (24) and (25), Eq. (26) becomes simply

op =12

C

+182

C

+ O 3C

, (27)

which has exactly the same coecients up to the quadraticterm to those of the Curzon-Ahlborn eciency [35] definedas

CA = 1 p

T2/T1 = 1 p

1 C

, (28)

with the expansion form

CA =12

C

+182

C

+1

163

C

+5

1284

C

+ O(5C

) , (29)

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)

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]+(1

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)


0

0.05

0.1

0.15

0.2

0 0.2 0.4 0.6 0.8 1

q*(c0) = *(c0)

q*(c=1)

optim

al tr

ansi

tion

rate

s

c

q**

c0 and 1 asymptotes

= ln

1 q

q

(1 C) ln

1 +

p(1 2q)2 + 4Cq(1 q)

1p

(1 2q)2 + 4Cq(1 q)

!q 1

2+

1

2

p(1 2q)2 + 4Cq(1 q)

q(1 q)0 = f(C , q)

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)

asymptotic form


C


C



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





C


C


C


C


C


C

, as

q

= q0 + a1C + a22C

+ O 3C

. (22)

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1

op

c = 1 T2 / T1

at (q*, *)CA = 11c

c/(2c)c/2

1+a[ln(1c)b](1c)


C



C


0.65.


C

again, we obtain

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

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

2(1 a0)a0(1 2a0)32

C

+O 3C

= 0 .

(23)


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 =a0(1 a0)2(1 2a0) . (25)


C


0

q(C ! 0) = q0 = (C ! 0) ' 0.083 221 720 199 517 7

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)

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)

asymptotic form


C


C



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





C


C


C


C


C


C

, as

q

= q0 + a1C + a22C

+ O 3C

. (22)

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1

op

c = 1 T2 / T1

at (q*, *)CA = 11c

c/(2c)c/2

1+a[ln(1c)b](1c)


C



C


0.65.


C

again, we obtain

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

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

2(1 a0)a0(1 2a0)32

C

+O 3C

= 0 .

(23)


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 =a0(1 a0)2(1 2a0) . (25)


C


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)


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

+ O 3C

.

(26)

Using Eqs. (24) and (25), Eq. (26) becomes simply

op =12

C

+182

C

+ O 3C

, (27)

which has exactly the same coecients up to the quadraticterm to those of the Curzon-Ahlborn eciency [35] definedas

CA = 1 p

T2/T1 = 1 p

1 C

, (28)

with the expansion form

CA =12

C

+182

C

+1

163

C

+5

1284

C

+ O(5C

) , (29)

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)


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)


0

0.05

0.1

0.15

0.2

0 0.2 0.4 0.6 0.8 1

q*(c0) = *(c0)

q*(c=1)

optim

al tr

ansi

tion

rate

s

c

q**

c0 and 1 asymptotes

= ln

1 q

q

(1 C) ln

1 +

p(1 2q)2 + 4Cq(1 q)

1p

(1 2q)2 + 4Cq(1 q)

!q 1

2+

1

2

p(1 2q)2 + 4Cq(1 q)

q(1 q)0 = f(C , q)

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)

asymptotic form


C


C



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





C


C


C


C


C


C

, as

q

= q0 + a1C + a22C

+ O 3C

. (22)

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1

op

c = 1 T2 / T1

at (q*, *)CA = 11c

c/(2c)c/2

1+a[ln(1c)b](1c)


C



C


0.65.


C

again, we obtain

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

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

2(1 a0)a0(1 2a0)32

C

+O 3C

= 0 .

(23)


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 =a0(1 a0)2(1 2a0) . (25)


C


0 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)

asymptotic form


C


C


= 1) = 0(the horizon

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

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