Part 18

ωdv − pφ(dv)/(v), or (vω − pφ)(dv)/(v),

for the aggregate amount of mechanical effect gained during the cycle of operations. It only remains for us to express this result in terms of dq and τ, on which the given thermal agency depends. For this purpose we remark that φ and ω are alterations of volume and pressure which take place along with a change of temperature τ, and hence, by the laws of compressibility and expansion, we may establish a relation between them in the following manner:

Let p{0} be the pressure of the mass of air when reduced to the temperature zero, and confined in a volume v{0}; then, whatever be v{0}, the product p{0}v{0} will, by the law of compressibility, remain constant; and, if the temperature be elevated from 0 to t + τ, and the gas be allowed to expand freely without any change of pressure, its volume will be increased in the ratio of 1 to 1 + E(t + τ), where E is very nearly equal to .00366 (the Centigrade scale of the air-thermometer being referred to), whatever be the gas employed, according to the researches of Regnault and of Magnus on the expansion of gases by heat. If, now, the volume be altered arbitrarily with the temperature continually at t_ + τ, the product of the pressure and volume will remain constant; and therefore we have

pv = p{0}v{0}{1 + E(t + τ)}.

Similarly,

(p − ω)(v + φ) = p{0}v{0}{1 + Et}.

Hence, by subtraction, we have

vω − pφ + ωφ = p{0}v{0}Eτ,

or, neglecting the product ωφ,

vω − pφ = p{0}v{0}Eτ.

Hence the preceding expression for mechanical effect, gained in the cycle of operations, becomes

p{0}v{0}. Eτ . dv/v.

Or, as we may otherwise express it,

(Ep{0}v{0})/(vdq/dv). dq. τ.

Hence, if we denote by M the mechanical effect due to H units of heat descending through the same interval τ, which might be obtained by repeating the cycle of operations described above, (H)/(dq) times, we have

M = (Ep{0}v{0})/(vdq/dv). Hτ. (3)

27. If the amplitudes of the operations had been finite, so as to give rise to an absorption of H units of heat during the first operation, and a lowering of temperature from S to T during the second, the amount of work obtained would have been found to be expressed by means of a double definite integral thus:

M = ∫{0}^{H} dq ∫{T}^{S} dt. (Ep{0}v{0})/(vdq/dv), ⎫ or ⎬. (4) M = Ep{0}v{0} ∫{0}^{H} ∫{T}^{S} (1)/(v) (dv)/(dq). dtdq; ⎭

this second form being sometimes more convenient.

28. The preceding investigations, being founded on the approximate laws of compressibility and expansion (known as the law of Mariotte and Boyle, and the law of Dalton and Gay-Lussac), would require some slight modifications to adapt them to cases in which the gaseous medium employed is such as to present sensible deviations from those laws. Regnault’s very accurate experiments show that the deviations are insensible, or very nearly so, for the ordinary gases at ordinary pressures; although they may be considerable for a medium, such as sulphurous acid, or carbonic acid under high pressure, which approaches the physical condition of a vapor at saturation; and therefore, in general, and especially in practical applications to real air-engines, it will be unnecessary to make any modification in the expressions. In cases where it may be necessary, there is no difficulty in making the modifications, when the requisite data are supplied by experiment.

29. Either the steam-engine or the air-engine, according to the arrangements described above, gives all the mechanical effect that can possibly be obtained from the thermal agency employed. For it is clear that in either case the operations may be performed in the reverse order, with every thermal and mechanical effect reversed. Thus, in the steam-engine, we may commence by placing the cylinder on the impermeable stand, allow the piston to rise, performing work, to the position E{3}F{3}; we may then place it on the body B, and allow it to rise, performing work, till it reaches E{2}F{2} after that the cylinder may be placed again on the impermeable stand, and the piston may be pushed down to E{1}F{1}; and, lastly, the cylinder being removed to the body A, the piston may be pushed down to its primitive position. In this inverse cycle of operations a certain amount of work has been spent, precisely equal, as we readily see, to the amount of mechanical effect gained in the direct cycle described above; and heat has been abstracted from B, and deposited in the body A, at a higher temperature, to an amount precisely equal to that which in the direct style was let down from A to B. Hence it is impossible to have an engine which will derive more mechanical effect from the same thermal agency than is obtained by the arrangement described above; since, if there could be such an engine, it might be employed to perform, as a part of its whole work, the inverse cycle of operations, upon an engine of the kind we have considered, and thus to continually restore the heat from B to A, which has descended from A to B for working itself; so that we should have a complex engine, giving a residual amount of mechanical effect without any thermal agency, or alteration of materials, which is an impossibility in nature. The same reasoning is applicable to the air-engine; and we conclude, generally, that any two engines, constructed on the principles laid down above, whether steam-engines with different liquids, an air-engine and a steam-engine, or two air-engines with different gases, must derive the same amount of mechanical effect from the same thermal agency.

30. Hence, by comparing the amounts of mechanical effect obtained by the steam-engine and the air-engine from the letting down of the H units of heat from A at the temperature (t + τ) to B at t, according to the expressions (2) and (3), we have

M = (1 − σ)(dp)/(kdt). Hτ = (Ep{0}v{0})/(vdq/dv). Hτ. (5)

If we denote the coefficient of Ητ in these equal expressions by μ, which maybe called “Carnot’s coefficient,” we have

μ = (1 − σ)(dp)/(kdt) = (Ep{0}v{0})/(vdq/dv), (6)

and we deduce the following very remarkable conclusions:

(1) For the saturated vapors of all different liquids, at the same temperature, the value of (1 − σ)(dp/kdt) must be the same.

(2) For any different gaseous masses, at the same temperature, the value of Ep{0}v{0}/(vdq/dv) must be the same.