The Steam Engine Familiarly Explained and Illustrated / With an historical sketch of its invention and progressive improvement; its applications to navigation and railways; with plain axioms for railway speculators

by Dionysius Lardner

Sixth, Force is required to open and close the several valves, to pump up the water for condensation, and to overcome the friction of the various axes.

Seventh, Force is expended upon working the air-pump.

In engines which do not condense the steam, and which, therefore, work with steam of high-pressure, some of these sources of waste are absent, but others are of increased amount. If we suppose the total effective force of the water evaporated per hour in the boiler to be expressed by 1000, it is calculated that the waste in a high-pressure engine will be expressed by the number 392; or, in other words, taking the whole undiminished force obtained by evaporation as expressed by 10, very nearly 4 of these parts will be consumed in moving the engine, and the other 6 only will be available.

In a single-acting engine which condenses the steam, taking, as before, 1000 to express the total mechanical power of the water evaporated in the boiler, 402 will express the part of this consumed in moving the engine, and 598, therefore, will express the portion of the power practically available; or, taking round numbers, we shall have the same result as in the non-condensing engine, viz. the whole force of the water evaporated being expressed by 10, 4 will express the waste, and 6 the available part.

In a double-acting engine the available part of the power bears a somewhat greater proportion to the whole. Taking, as before, 1000 to express the whole force of the water evaporated, 368 will express the proportion of that force expended on the engine, and 632 the proportion which is available for work.

In general, then, taking round numbers, we may consider that the mechanical force of four tenths of the water evaporated in the boiler is intercepted by the engine, and the other six tenths are available as a moving force. In this calculation, however, the resistance produced in the condensing engine by the uncondensed steam is not taken into account: the amount of this force will depend upon the temperature at which the water is maintained in the condenser. If this water be kept at the temperature of 120°, the vapour arising from it will have a pressure expressed by three inches seven tenths of mercury; if we suppose the pressure of steam in the boiler to be measured by 37 inches of mercury, then the resistance from the uncondensed steam will amount to one tenth of the whole power of the boiler; this, added to the four tenths already accounted for, would show a waste amounting to half the whole power of the boiler, and consequently only half the water evaporated would be available as a moving power.

If the temperature of the condenser be kept down to 100°, then the pressure of uncondensed steam will be expressed by two inches of mercury, and the loss of power consequent upon it would amount to a proportionally less fraction of the whole power.

The following example will illustrate the method of estimating the effective power of an engine.

In a double-acting engine, in good working condition, the total power of steam in the boiler being expressed by 1000, the proportion intercepted by the engine, exclusive of the resistance of the uncondensed steam, will be 368, and the effective part 632. Now, suppose the pressure of steam in the boiler to be measured by a column of 35 inches of mercury; the thousandth part of this will be seven two hundredths of an inch of mercury, and 632 of these parts will express the effective portion of the power. By multiplying seven two hundredths by 632, we obtain 22 nearly. Now, suppose the temperature in the condenser is 1200, the pressure of steam corresponding to that temperature will be measured by 3-7/10 inches of mercury. Subtracting this from 22, there will remain 18-3/10 inches of mercury, as the effective moving force upon the piston; this will be equivalent to about 7 lbs. on each circular inch.

If the diameter of the piston then be 24 inches, its surface will consist of a number of circular inches expressed by the square of 24, or 24 × 24 = 576; and, as upon each of these circular inches there is an effective pressure of 7 lbs., we shall find the total pressure in pounds by multiplying 576 by 7, which gives 4032 lbs.