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Reflections on the Motive Power of Heat · Sadi Carnot — chapter 17 of 39 · ~1,135 words · public domain

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M = H ∫{T}^{S}(1 - σ)((dp/dt)/k)dt_. (1)

for the total amount of mechanical effect gained by the operations described above.

21. If the interval of temperatures be extremely small,—so small that (1 − σ)(dp)/(dt/k) will not sensibly vary for values of t between T and S,—the preceding expression becomes simply

Μ = (1 - σ)(dp)/(dt)/(k). Η(S - Τ). (2)

This might, of course, have been obtained at once by supposing the breadth of the quadrilateral figure AA{1}A{2}A to be extremely small compared with its length, and then taking for its area, as an approximate value, the product of the breadth into the line AA{1}, or the line A{3}A{2}_, or any line of intermediate magnitude.

The expression (2) is rigorously correct for any interval S − T, if the mean value of (1 − σ)((dp/dt)/k) for that interval be employed as the coefficient of H(S − T).

CARNOT’S THEORY OF THE AIR-ENGINE.

22. In the ideal air-engine imagined by Carnot four operations performed upon a mass of air or gas enclosed in a closed vessel of variable volume constitute a complete cycle, at the end of which the medium is left in its primitive physical condition; the construction being the same as that which was described above for the steam-engine, a body A, permanently retained at the temperature S, and B at the temperature T; an impermeable stand K; and a cylinder and piston, which in this case contains a mass of air at the temperature S, instead of water in the liquid state, at the beginning and end of a cycle of operations. The four successive operations are conducted in the following manner:

(1) The cylinder is laid on the body A, so that the air in it is kept at the temperature S; and the piston is allowed to rise, performing work.

(2) The cylinder is placed on the impermeable stand K, so that its contents can neither gain nor lose heat, and the piston is allowed to rise farther, still performing work, till the temperature of the air sinks to T.

(3) The cylinder is placed on B, so that the air is retained at the temperature T, and the piston is pushed down till the air gives out to the body B as much heat as it had taken in from A, during the first operation.

(4) The cylinder is placed on K, so that no more heat can be taken in or given out, and the piston is pushed down to its primitive position.

23. At the end of the fourth operation the temperature must have reached its primitive value S, in virtue of CARNOT’S axiom.

24. Here, again, as in the former case, we observe that work is performed by the piston during the first two operations; and during the third and fourth work is spent upon it, but to a less amount, since the pressure is on the whole less during the third and fourth operations than during the first and second, on account of the temperature being lower. Thus, at the end of a complete cycle of operations, mechanical effect has been obtained; and the thermal agency from which it is drawn is the taking of a certain quantity of heat from A, and letting it down, through the medium of the engine, to the body B at a lower temperature.

25. To estimate the actual amount of effect thus obtained, it will be convenient to consider the alterations of volume of the mass of air in the several operations as extremely small. We may afterwards pass by the integral calculus, or, practically, by summation to determine the mechanical effect whatever be the amplitudes of the different motions of the piston.

26. Let dq be the quantity of heat absorbed during the first operation, which is evolved again during the third; and let dv be the corresponding augmentation of volume which takes place while the temperature remains constant, as it does during the first operation. The diminution of volume in the third operation must be also equal to dv, or only differ from it by an infinitely small quantity of the second order. During the second operation we may suppose the volume to be increased by an infinitely small quantity φ; which will occasion a diminution of pressure and a diminution of temperature, denoted respectively by ω and τ. During the fourth operation there will be a diminution of volume and an increase of pressure and temperature, which can only differ, by infinitely small quantities of the second order, from the changes in the other direction, which took place in the second operation, and they also may, therefore, be denoted by φ, ω, and τ, respectively. The alteration of pressure during the first and third operations may at once be determined by means of Mariotte’s law, since in them the temperature remains constant. Thus, if, at the commencement of the cycle, the volume and pressure be v and p, they will have become v + dv and pv/(v + dv) at the end of the first operation. Hence the diminution of pressure during the first operation is p − pv/(v + dv) or pdv/(v + dv) and therefore, if we neglect infinitely small quantities of the second order, we have pdv/v for the diminution of pressure during the first operation; which to the same degree of approximation, will be equal to the increase of pressure during the third. If t + τ and t be taken to denote the superior and inferior limits of temperature, we shall thus have for the volume, the temperature, and the pressure at the commencements of the four successive operations, and at the end of the cycle, the following values respectively:

(1) v, t + τ, p; (2) v + dv, t + τ, p(1 − (dv)/(v)); (3) v + dv + φ, t, p(1 − (dv)/(v)) − ω; (4) v + φ, t, p − ω; (5) v, t + τ, p.

Taking the mean of the pressures at the beginning and end of each operation, we find

(1) p(1 − ½(dv)/(v)),

(2) p(1 − (dv)/(v)) − ½ω,

(3) p(1 − ½(dv)/(v))) − ω,

(4) p − ½ω,

which, as we are neglecting infinitely small quantities of the second order, will be the expressions for the mean pressures during the four successive operations. Now, the mechanical effect gained or spent, during any of the operations, will be found by multiplying the mean pressure by the increase or diminution of volume which takes place; and we thus find

(1) p(1 − ½(dv)/(v))dv,

(2) {p(1 − (dv)/(v)) − ½ω}φ,

(3) {p(1 − ½(dv)/(v)) − ω}dv,

(4) (p − ½ω)φ.

for the amounts gained during the first and second, and spent during the third and fourth operations; and hence, by addition and subtraction, we find

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