In the absence of any direct method of determining the general relation between the pressure and volume of common steam, empirical formulæ expressing it have been proposed by different mathematicians.

The late Professor Navier proposed the following:- Let S express the volume of steam into which an unit of volume of water is converted under the pressure P, this pressure being expressed in kilogrammes per square mètre. Then the relation between S and P will be

where a 1000, b = 0·09, and m = 0·0000484.

This formula, however, does not agree with experiment at pressures less than an atmosphere. M. de Pambour, therefore, proposes the following changes in the values of its co-efficients :-Let P express the pressure in pounds per square foot; and let

a = 10000

b=0·4227 m = = 0.00258, and the formula will be accurate for all pressures. For pressures above two atmospheres the following values give more accuracy to the calculation:

a = 10000 b=1.421 m=0.0023.

In these investigations I shall adopt the following modified formula. The symbols S and P retaining their signification, we shall have

These values of a and b will be sufficiently accurate for practical purposes for all pressures, and may be used in reference to lowpressure engines of every form, as well as for high-pressure engines which work expansively.

When the pressure is not less than 30 pounds per square inch, the following values of a and b will be more accurate :—

On the Expansive Action of Steam.

The investigation of the effect of the expansion of steam which has been given in the text, is intended to convey to those who are not conversant with the principles and language of analysis, some notion of the nature of that mechanical effect to which the advantages attending the expansive principle are due. We shall now, however, explain these effects more accurately.

The dynamical effect produced by any mechanical agent is expressed by the product of the resistance overcome and the space through which that resistance is moved.

Let P

the pressure of steam expressed in pounds per square foot.

=

S the number of cubic feet of steam of that pressure produced by the evaporation of a cubic foot of water. E the mechanical effect produced by the evaporation of a cubic foot of water expressed in pounds raised one foot. Then we shall have E = PS; and if W be a volume of water evaporated under the pressure P, the mechanical effect produced by it will be WPS.

By (10.) we have

Hence, for the mechanical effect of a cubic foot of water evaporated under the pressure P we have

Let a cubic foot of water be evaporated under the pressure P', and let it produce a volume of steam S' of that pressure. Let this steam afterwards be allowed to expand to the increased volume S and the diminished pressure P; and let it be required to determine the mechanical effect produced during the expansion of the steam from the volume S' to the volume S.

Let E'

E"

E

the mechanical effect produced by the evaporation of the water under the pressure P' without expansion.

the mechanical effect produced during the expansion of the steam.

the mechanical effect which would be produced by the evaporation under the pressure P without expansion. E the total mechanical effect produced by the evaporation under the pressure P' and subsequent expansion.

Thus we have

EE+E".

Let s be any volume of the steam during the process of expansion, p the corresponding pressure, and e" the mechanical effect produced by the expansion of the steam. We have then by (10.)

Hence it appears that the mechanical effect of a cubic foot of water evaporated under the pressure P may be increased by the quantity a log.

S
S

if it be first evaporated under the greater pres

sure P', and subsequently expanded to the lesser pressure P.

The logarithms in these formulæ are hyberbolic.

To apply these principles to the actual case of a double acting steam engine,

Let L

the stroke of the piston in feet.

A = the area of the piston in square feet.

n = the number of strokes of the piston per minute.

...2n AL the number of cubic feet of space through which

=

the piston moves per minute.

Let cLA = the clearage, or the space between the steam valve and the piston at each end of the stroke.

Let V

The volume of steam admitted through the steam valve
at each stroke of the engine will be 2n AL (1 + c).
the mean speed of the piston in feet per minute,

• 2nL = V.

The volume of steam admitted to the cylinder per minute will therefore be VA (1 + c), the part of it employed in working the piston being VA.

Let W

=

the water in cubic feet admitted per minute in the form of steam through the steam valve.

S = the number of cubic feet of steam produced by a cubic

foot of water.

L L