Generalized golden mean and the efficiency of thermal machines
GGeneralized golden mean and the efficiency of thermal machines
Ramandeep S. Johal ∗ Department of Physical Sciences,Indian Institute of Science Education and Research Mohali,Sector 81, S.A.S. Nagar,Manauli PO 140306, Punjab, India
Abstract
We investigate generic heat engines and refrigerators operating between two heat reservoirs, forthe condition when their efficiencies are equal to each other. It is shown that the correspondingvalue of efficiency is given as the inverse of the generalized golden mean, φ p , where the parameter p depends on the degrees of irreversibility of both engine and refrigerator. The reversible case( p = 1) yields the efficiency in terms of the standard golden mean. We also extend the analysis toa three-heat-resrervoir setup. a r X i v : . [ phy s i c s . c l a ss - ph ] J un . INTRODUCTION Although the golden mean (golden ratio) has engaged artists, mathematicians andphilosophers since antiquity , it has been appreciated more recently that it is not aunique number as far as many of its algebraic and geometric properties are concerned . Infact, one of the simplest generalizations of the golden mean may be defined by the positivesolution, φ p , of the following quadratic equation: x − px − , (1)where p is a given positive number. The number φ p is given by: φ p = (cid:112) p + 4 + p , (2)also referred as the p th order extreme mean (POEM) . More specifically, when p is a positiveinteger n , it is addressed as the n th order extreme mean (NOEM) or a member of the familyof metallic means. For example, p = 1 gives the golden mean φ = ( √ / p = 2 yieldsthe silver mean, φ = √ F n = pF n − + F n − , tend tothis number: lim n →∞ F n +1 F n = φ p . (3)It also relates the lengths of diagonals of a regular odd n -gon ( n ≥ φ p , see Ref. .The generalized golden mean ( φ p ) appears as the optimal solution determining the shapeof Newton’s frustum that faces the least resistance while moving through a rare medium .The metallic means family has been related to quasiperiodic dynamics . Another simplephysical example is a semi-infinite resistor network as shown in Fig. 1a, whose equivalentresistance ( r (cid:48) ) between points A and B satisfies the quadratic euquation: ( r (cid:48) ) − rr (cid:48) − r (cid:48) = φ r . In this article, we describe an occurrence of the generalized goldenmean in the context of thermodynamics by relating this number to the instance of equalefficiencies of a heat engine and refrigerator.2 r r1 / r1 / rr1 / r r'ABABa ) b ) FIG. 1. a) An semi-infinite resistor network where each resistance in series is r while a resistancein parallel is 1 /r . b) The effective resistance between points A and B is r (cid:48) = φ r . II. TWO-HEAT-RESERVOIR SETUP
A heat engine and a refrigerator constitute basic mechanisms of a thermal machine op-erating between two heat reservoirs at unequal temperatures, say, T h and T c ( T c < T h ) .An irreversible machine always leads to a net entropy increase of the universe. On the otherhand, the ideal limit of a reversible operation implies no net entropy change of the universe.Further, irreversibility leads to a decrease in performance of the machine, and so the effi-ciency of a thermal machine is limited by its maximal value, known as the Carnot efficiency,obtained in the reversible case. 3 eat engine : We first consider an irreversible heat engine working in a cycle, and supposethat W amount of work is extracted per cycle when Q h amount of heat is absorbed fromthe hot reservoir. The total entropy generated per cycle is:∆ tot S = − Q h T h + Q c T c > . (4)where Q c = Q h − W , is the heat rejected to the cold reservoir. All quantities defined aboveare positive. From Eq. (4), we can express the work output of an irreversible cycle as: W = (1 − θ ) Q h − T c ∆ tot S, (5)where θ = T c /T h . Let ∆ rev S = Q h /T h be the entropy transferred from the hot reservoir tothe working medium, in the reversible case. Then, the efficiency, E = W/Q h , can be writtenas E = 1 − (cid:18) tot S ∆ rev S (cid:19) θ. (6)Here, we define the “irreversibility” parameter, z = 1 + ∆ tot S/ ∆ rev S >
1, so that theefficiency of the engine can be expressed as E = 1 − zθ . The reversible case corresponds to∆ tot S = 0, or z = 1, yielding Carnot efficiency equal to 1 − θ . In other words, 0 ≤ E ≤ − θ implies 1 ≤ z ≤ /θ . Refrigerator : Now, let us consider the machine in the refrigerator mode. Suppose, aninput work W is required to transport Q c amount of heat against the temperature gradient.Let Q h = Q c + W , be the amount of heat deposited in the hot reservoir. The efficiency ofa refrigerator is defined as R = Q c / W . The total entropy generated per cycle is:∆ tot S = Q h T h − Q c T c > . (7)Let ∆ rev S be the amount of heat transferred reversibly, from the cold reservoir to theworking medium. Then we have Q c = T c ∆ rev S , and so, we can write W = 1 − θθ Q c + T h ∆ tot S . (8)The efficiency is then given by: R = θ (cid:18) tot S ∆ rev S − θ (cid:19) − . (9)Analogous to the case of engine, we define the parameter z (cid:48) = 1 + ∆ tot S / ∆ rev S >
1, andexpress the efficiency of the refrigerator as : R = θ ( z (cid:48) − θ ) − . In the reversible case, z (cid:48) = 1,and so R = θ (1 − θ ) − . Note that, unlike the parameter z , z (cid:48) is not bounded from above.4n the above, we have identified the expressions for efficiencies as functions of the ratioof temperatures as well as the irreversibility parameter z or z (cid:48) . These expressions refer to any irreversible thermal cycle between the two reservoirs, with given values of θ, z and z (cid:48) .We may now ask, for what ratio of temperatures, a heat engine and a refrigerator with thegiven values of their respective irreversibility parameters, have the same efficiency, and whatis its value?Thereby, setting E = R , and solving for 0 < θ <
1, we obtain: θ = zz (cid:48) + 2 − (cid:112) ( zz (cid:48) ) + 42 z . (10)Then, the efficiency, E = 1 − zθ , at the above condition is given by: E = R = (cid:112) ( zz (cid:48) ) + 4 − zz (cid:48) . (11)Interestingly, the above expression for the efficiency depends only on the product of theirreversibility parameters. Thus, we may define zz (cid:48) ≡ p , and rewrite Eq. (11) as follows E = (cid:112) p + 4 − p φ p , (12)where E is expressed in terms of φ p from Eq. (2). Note that, in the above p (cid:62)
1, where p = 1implies the reversible case, with z = z (cid:48) = 1, yielding E = R = ( √ − / /φ , whichwas earlier noted in Ref. . Correspondingly, the ratio of temperatures in the reversible caseis required to be: θ = (3 − √ / − φ . III. THREE-HEAT-RESERVOIR SETUP
Next, we show the occurrence of the generalized golden mean in a slightly different setting.Let us consider three heat reservoirs with temperatures ordered as T l < T c < T h . Assumethat we can operate an engine between the reservoirs at T h and T l , and a refrigerator betweenthe reservoirs at T c and T l (see Fig. 2). Further, let these be ideal or reversible machines,so that the respective thermal efficiencies are given as: E hl = T h − T l T h , R cl = T l T c − T l . (13)Now, for given values of T h and T c , we look for the temperature T l such that these two5 E T h T c T l FIG. 2. A three heat-reservoir setup, in which a heat engine runs between the hottest ( T h ) and thecoldest ( T l ) reservoirs, while a refrigerator runs between the coldest and an intermediate reservoir( T c ). For given values of T h and T c , we look for T l at which efficiencies of the engine and therefrigerator are equal. efficiencies become equal. Setting E hl = R cl , we obtain a quadratic equation for T l : T l − (2 T h + T c ) T l + T h T l = 0 , (14)whose solutions are: T l = T h (cid:104) θ + 2 ± √ θ + 4 (cid:105) , (15)6here θ = T c /T h . Only the negative root above yields an allowed value of the efficiency,given by: E hl = R cl = √ θ + 4 − θ φ θ . (16)Note that the generalized golden mean appears above within a reversible setup. Secondly,due to 0 < θ <
1, it covers a range of parameters complementary to the expression (12),where p (cid:62) T h and T l , specified by the ir-reversibility parameter z . Likewise, let the corresponding parameter for the refrigeratoroperating between T c and T l be z (cid:48) . Then, following Section II, the efficiencies of thesemachines can be written as: ¯ E hl = 1 − z T l T h ; ¯ R cl = T l /T c z (cid:48) − T l /T c . (17)Now setting the condtion ¯ E hl = ¯ R cl , we can solve for T l : T l = T h z (cid:104) pθ + 2 ± (cid:112) ( pθ ) + 4 (cid:105) , (18)where p = zz (cid:48) . Consequently, the allowed solution for efficiency is given by:¯ E hl = (cid:112) ( pθ ) + 4 − pθ φ pθ . (19)For p = 1, we obtain the reversible case discussed above (Eq. (16)). Moreover, letting θ →
1, which may be realized by taking T c → T h , we revert to the case of two reservoirs (at T h and T l ), as discussed in Section II. IV. CONCLUSIONS
In the above, we have observed that the generalized golden mean determines the efficien-cies of a heat engine and a refrigerator when the latter are set equal to each other. Whenboth the machines are reversible, the efficiency is related to the standard golden mean. Toextend to irreversible domain, we have first recast the efficiency of a generic heat engine anda refrigerator in terms of an irreversibility parameter and the ratio of the reservoir tempera-tures. Then, the efficiency depends only on the product of the two irreversibility parameters.7he importance of this step can be appreciated by noting that only for the reversible case,we have R = Q c /W h along with E = W/Q h , where Q c = Q h − W , i.e. the same amountsof heat and work appear for a reversible heat engine as well as for a refrigerator. So, in thiscase, we can express the condition E = R , as follows: WQ h = Q c W . (20)From this, we obtain the equation: E = E − −
1, whose solution is E = R = 1 /φ .However, the efficiencies in the irreversible case are E = W/Q h and R = Q c / W h , whichdon’t seem useful for applying the E = R condition. Thereby, the forms (6) and (9) havebeen employed. Further, we have also extended this condition to three-reservoir scenariowhich also includes the limiting case of two-reservoir setup. It will be interesting to identifyphysical situations leading to the equality of the efficiencies of engine and refrigerator underthe given conditions. Here, we have discussed one possible setup using three heat reservoirs.The interested reader can identify other engine-refrigerator pairs in the three-reservoir setupwhich lead to equal efficiencies, expressed in terms of suitable generalized golden means. ∗ [email protected] H. S. M. Coxeter,
Introduction to Geometry , 2nd ed. (Wiley, New York, 1969). C. S. Ogilvy,
Excursions in Geometry (Dover, New York, 1990) pp. 122-125. G. Markowsky, The College Mathematics Journal , 2 (1992),https://doi.org/10.1080/07468342.1992.11973428. M. Livio,
The Golden Ratio: The Story of Phi, The World’s Most Astonishing Number (Broad-way Books, New York, 2002). C. Falbo, The College Mathematics Journal , 123 (2005),https://doi.org/10.1080/07468342.2005.11922119. D. H. Fowler, Fibonacci Quarterly , 146 (1982). J. Cruz-Sampedro and M. Tetlalmatzi-Montiel, The College Mathematics Journal , 145(2010), https://doi.org/10.4169/074683410X480258. V. W. D. Spinadel, Chaos, Solitons & Fractals , 1631 (1997). T. P. Srinivasan, American Journal of Physics , 461 (1992), https://doi.org/10.1119/1.16849. M. Kasperski and W. K(cid:32)lobus, European Journal of Physics , 015008 (2013). M. Zemansky and R. Dittman,
Heat and Thermodynamics: An Intermediate Textbook , Inter-national Series in Pure and Applied Physics (McGraw-Hill, 1997). V. V. Popkov and E. V. Shipitsyn, Physics-Uspekhi , 1155 (2000)., 1155 (2000).