[Pg505]
TOCINX

On the Relation between the Temperature, Pressure, and Density of Common Steam.

The following formula, proposed by M. Biot, represents with great accuracy the relation between the temperature and pressure of common steam, throughout all that part of the thermometric scale to which experiments have been extended.

Let

The relation between the temperature t with reference to the centesimal thermometer, and the pressure p in millimÈtres of mercury at the temperature of melting ice, will then be expressed by the following formula:—

FormulÆ have, however, been proposed, which, though not applicable to the whole scale of temperatures, are more manageable in their practical application than the preceding.

For pressures less than an atmosphere, Southern proposed the following formula, where the pressure is intended to be expressed [Pg506] in pounds per square inch, and the temperature in reference to Fahrenheit's thermometer,—

The following formula was proposed by Tredgold, where p expresses the pressure in inches of mercury:—

This was afterwards modified by Mellet, and represents with sufficient accuracy experiments from 1 to 4 atmospheres. Let p represent pounds per square inch, and t the temperature by Fahrenheit's thermometer,—

M. de Pambour has proposed the following formula, also applicable through the same limits of the scale:—

MM. Dulong and Arago have proposed the following formula for all pressures between 4 and 50 atmospheres:—

It was about the year 1801, that Dalton, at Manchester, and Gay-Lussac, at Paris, instituted a series of experiments on gaseous bodies, which conducted them to the discovery of the law mentioned in art. (96.), p. 171. These philosophers found that all gases whatever, and all vapours raised from liquids by heat, as well as all mixtures of gases and vapours, are subject to the same quantity of expansion between the temperatures of melting ice and boiling water; and by experiments subsequently made by Dulong and Petit, this uniformity of expansion has been proved to extend to all temperatures which can come under practical inquiries.

Dalton found that 1000 cubic inches of air at the temperature of melting ice dilated to 1325 cubic inches if raised to the temperature of boiling water. According to Gay-Lussac, the increased volume was 1375 cubic inches. The latter determination has been subsequently found to be the more correct one.[40]

[Pg507]

It appears, therefore, that for an increase of temperature from 32° to 212°, amounting to 180°, the increase of volume is 375 parts in 1000; and since the expansion is uniform, the increase of volume for 1° will be found by dividing this by 180, which will give an increase of 208¹/₃ parts in 100,000 for each degree of the common thermometer.

To reduce the expression of this important and general law to mathematical language, let v be the volume of an elastic fluid at the temperature of melting ice, and let nv be the increase which that volume would receive by being raised one degree of temperature under the same pressure. Let V be its volume at the temperature T. Then we shall have

If V' be its volume at any other temperature T', and under the same pressure, we shall have, in like manner,

Hence we obtain

which expresses the relation between the volumes of the same gas or vapour under the same pressure and at any two temperatures. The co-efficient n, as explained in the text, has the same value for the same gas or vapour throughout the whole thermometric scale. But it is still more remarkable that this constant has the same value for all gases and vapours. It is a number, therefore, which must have some essential relation to the gaseous or elastic state of fluid matter, independent of the peculiar qualities of any particular gas or vapour.

The value of n, according to the experiments of Gay-Lussac, is 0·002083, or ¹/₄₈₀.

To reduce the law of Mariotte, explained in (97.) p. 171., to mathematical language, let V, V' be the volumes of the same gas or vapour under different pressures P, P', but at the same temperature. We shall then have

If it be required to determine the relation between the volumes of the same gas or vapour, under a change of both temperature and pressure, let V be the volume at the temperature T and under the pressure P, and let V' be the volume at the temperature T' and under the pressure P'. Let v be the volume at the temperature T and under the pressure P'.

By formula (7.) we have

[Pg508]

and by formula (6.) we have

Eliminating v, we shall obtain

or,

which is the general relation between the volumes, pressures, and temperatures of the same gas or vapour in two different states.

To apply this general formula to the case of the vapour of water, let T' = 212°. It is known by experiment that the corresponding value of P', expressed in pounds per square inch, is 14·706; and that V', expressed in cubic inches, the water evaporated being taken as a cubic inch, is 1700. If, then, we take 0·002083 as the value of n, we shall have by (8.),

If, by means of this formula (9.), and any of the formulÆ (1.), (2.), (3.), (4.), (5.), T were eliminated, we should obtain a formula between V and P, which would enable us to compute the enlargement of volume which water undergoes in passing into steam under any proposed pressure. But such a formula would not be suitable for practical computations. By the formulÆ (1.) to (5.), a table of pressures and corresponding temperatures may be computed; and these being known, the formula (9.) will be sufficient for the computation of the corresponding values of V, or the enlargement of volume which water undergoes in passing into steam.

In the following table, the temperatures corresponding to pressures from 1 to 240 lbs. per square inch are given by computation from the formulÆ (2.) to (5.), and the volumes of steam produced by an unit of volume of water as computed from the formula (9.).

The mechanical effect is obtained by multiplying the pressure in pounds by the expansion of a cubic inch of water in passing into steam expressed in feet, and is therefore the number of pounds which would be raised one foot by the evaporation of a cubic inch of water under the given pressure. [Pg509]

[Pg511]

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

where

a = 3875969 b = 164.

These values of a and b will be sufficiently accurate for practical purposes for all pressures, and may be used in reference to low-pressure 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:—

a = 4347826 b = 618.

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. [Pg512]

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

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

Thus we have

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 by integrating we obtain

[Pg513]

which, taken between the limits s = S' and s = S, becomes

But by (11.) we have

Or,

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

The logarithms in these formulÆ are hyperbolic.

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

Let

Let

Let

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

[Pg514]

Hence we shall have

Since by (10.) we have

By which the pressure of steam in the cylinder will be known, when the effective evaporation, the diameter of the cylinder, and speed of the piston, are given.

If it be required to express the mechanical effect produced per minute by the action of steam on the piston, it is only necessary to multiply the pressure on the surface of the piston by the space per minute through which the piston moves. This will give

which expresses the whole mechanical effect per minute in pounds raised one foot.

If the steam be worked expansively, let it be cut off after the piston has moved through a part of the stroke expressed by e.

The volume of steam of the undiminished pressure P' admitted per minute through the valve would then be

and the ratio of this volume to that of the water producing it being expressed by S', we should have

The final volume into which this steam is subsequently expanded being VA(1 + c), its ratio to that of the water will be

The pressure P', till the steam is cut off, will be

The mechanical effect E' produced per minute by the steam of full pressure will be

and the effect E per minute produced by the expansion of the steam will by (12.) be [Pg515]

Hence the total effect per minute will be

If the engine work without expansion, e=1;

as before; and the effect per minute gained by expansion will therefore be

which therefore represents the quantity of power gained by the expansive action, with a given evaporating power.

In these formulÆ the total effect of the steam is considered without reference to the nature of the resistances which it has to overcome.

These resistances may be enumerated as follows:—

1. The resistance produced by the load which the engine is required to move.
2. The resistance produced by the vapour which remains uncondensed if the engine be a condensing engine, or of the atmospheric pressure if the engine do not condense the steam.
3. The resistance of the engine and its machinery, consisting of the friction of the various moving parts, the resistances of the feed pump, the cold water pump, &c. A part of these resistances are of the same amount, whether the engine be loaded or not, and part are increased, in some proportion depending on the load.

When the engine is maintained in a state of uniform motion, the sum of all these resistances must always be equal to the whole effect produced by the steam on the piston. The power expended on the first alone is the useful effect.

Let

The total resistance, therefore, being R + mR + r, which, when the mean motion of the piston is uniform, must be equal to the mean pressure on the piston. The total mechanical effect [Pg516] must therefore be equal to the total resistance multiplied by the space through which that resistance is driven. Hence we shall have

For brevity, let

By solving this for VA, we obtain

This quantity RVA, being the product of the resistance RA, of the load reduced to the surface of the piston, multiplied by the space through which the piston is moved, will be equal to the load itself multiplied by the space through which it is moved. This being, in fact, the useful effect of the engine, let it be expressed by U, and we shall have

Or by (22.),

The value of the useful effect obtained from these formulÆ will be expressed in pounds, raised one foot per minute, W being the effective evaporation in cubic feet per minute, A the area of the piston in square feet, and V the space per minute through which it is moved, in feet.

Since a resistance amounting to 33,000 pounds moved through one foot per minute is called one-horse power, it is evident that the horse power H of the engine is nothing more than the useful effect per minute referred to a larger unit of weight or resistance; that is to 33,000 pounds instead of one pound. Hence we shall have

Since the useful effect expressed in (24.) and (25.) is that due to a number of cubic feet of water, expressed by W, we shall obtain the effect due to one cubic foot of water, by dividing U by W. If, therefore, U' be the effect produced by the effective evaporation of a cubic foot of water, we shall have [Pg517]

If the quantity of fuel consumed per minute be expressed by F, the effect produced by the unit of fuel, called the DUTY of the engine, will, for like reason, be

If the fuel be expressed in hundredweights of coal, then D will express the number of pounds' weight raised one foot by a hundredweight of coal.

By solving (24.) and (25.) for W, we obtain

By eliminating U, by (26.), we shall have

The evaporation necessary per horse power per minute will be found by putting H=1 in these formulÆ.[41]

It will be observed that the quantities A and V, the area of the cylinder and the speed of the piston, enter all these formulÆ as factors of the same product. Other things, therefore, being the same, the speed of the piston will be always inversely as the area of the cylinder. In fact, VA is the volume of steam per minute employed in working the piston, and if the piston be increased or diminished in magnitude, its speed must be inversely [Pg518] varied by the necessity of being still moved through the same number of cubic feet by the same volume of steam.

It has been already stated in the text, that no satisfactory experiments have yet been made, by which the numerical value of the quantity r can be exactly known. In engines of different magnitudes and powers, this resistance bears very different proportions to the whole power of the machine. In general, however, the larger and more powerful the engine, the less that proportion will be.

That part of this resistance which arises from the reaction of the uncondensed vapour on the piston is very variable, owing to the more or less perfect action of the condensing apparatus, the velocity of the piston, and the magnitude and form of the steam passages. M. de Pambour states, that, by experiments made with indicators, the mean amount of this resistance in the cylinder is 2¹/₂ lbs. per square inch more than in the condenser, and that the pressure in the latter being usually 1¹/₂ lb. per square inch, the mean amount of the pressure of the condensed vapour in the cylinder is about 4 lbs. per square inch. Engineers, however, generally consider this estimate to be above the truth in well-constructed engines, when in good working order.

In condensing low pressure engines of forty horse power and upwards, working with an average load, it is generally considered that the resistance produced by the friction of the machine and the force necessary to work the pumps may be taken at about 2 lbs. per square inch of piston surface.

Thus the whole resistance represented by r in the preceding formulÆ, as applied to the larger class of low pressure engines, may be considered as being under 6 lbs. per square inch, or 864 lbs. per square foot, of the piston. It is necessary, however, to repeat, that this estimate must be regarded as a very rough approximation; and as representing the mean value of a quantity subject to great variation, not only in one engine compared with another, but even in the same engine compared with itself at different times and in different states.

In the same class of engines, the magnitude of the clearage is generally about a twentieth part of the capacity of the cylinder, so that c=0·05.

That part of the resistance which is proportional to the load, and on which the value of m in the preceding formulÆ depends, is still more variable, and depends so much on the form, magnitude, and the arrangement of its parts, that no general rule can be given for its value. It must, in fact, be determined in every particular case.

In the practical application of the preceding formulÆ in condensing engines we shall have [Pg519]

In engines which work without condensation, and therefore with high pressure steam, we shall have

To facilitate computation, the values of e' corresponding to all values of e, from e=·10 to e=·90, are given in the following table:—

[Pg520]

In engines which work without expansion we have

For condensing engines without expansion, we shall then have

and for non-condensing engines,

As the diameters of the cylinders of engines are generally expressed in inches, the corresponding areas of the pistons expressed in square feet are given in the following table, so that the values of A may be readily found:—

[Pg521]

The practical application of the preceding formulÆ will be shown by the following examples.

EXAMPLES.

1. A 36-inch cylinder with 5¹/₂ feet stroke is supplied by a boiler evaporating effectively 60 cubic feet of water per hour, and the piston makes 20 strokes per minute without expansion;—what is the power of the engine and the pressure of steam in the cylinder?

Let it be assumed that r=6 × 144=864 and m=0·1. Since the engine is a condensing engine, we have b=164 and e'=3691399. By the formulÆ (25.) and (26.) we have

and since by the data we have

the formula, by these substitutions, becomes

Since e=1, the pressure P of steam in the cylinder, by (18.), is

Therefore

which being the pressure in pounds per square foot, the pressure per square inch will be 15¹/₃ lbs.

2. To find the effective evaporation necessary to produce a power of 80 horses with the same engine. Also, find the pressure of steam in the cylinder, the speed of the piston being the same.

By the formula (32.), with the above substitutions, we have

The evaporating power would therefore be only increased 22 per cent., while the working power of the engine would be increased nearly 40 per cent.

The pressure P in the cylinder will be given, by (18.), as before.

which is equivalent to 19 lbs. per square inch. [Pg522]

3. What must be the diameter of a cylinder to work with a power of a hundred horses, supplied by a boiler evaporating effectively 70 cubic feet of water per hour, the mean speed of the piston being 240 feet per minute, and the steam being cut off at half stroke? Also, what will be the full pressure of steam on the piston?

Taking, as in the former examples, m=0·1, b=164, and r=864, we shall have

and by the column for condensing engines, in table, p. 519, we have e'=6029916, where e=0·50. Making these substitutions in

we shall have

Whence we find

and by the table, p. 520, the corresponding diameter of the cylinder will be 50¹/₃ inches.

If P' be the full pressure of the steam, we shall have, by (18.),

Making in this the proper substitutions, we have

which being in pounds per square foot, the pressure per square inch will be 16¹/₁₀ lbs.

[40] M. de Pambour states that the increased volume is 1364 cubic inches.