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
Letsbe any volume of the steam during the process of expansion,pthe corresponding pressure, ande″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 limitss= S′ ands= 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 quantityalog.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 bye.
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:—
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 theuseful 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 theDUTYof 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 quantityrcan 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 is21⁄2lbs. per square inch more than in the condenser, and that the pressure in the latter being usually11⁄2lb. 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 byrin 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 thatc= 0·05.
That part of the resistance which is proportional to the load, and on which the value ofmin 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 ofe′corresponding to all values ofe, frome= ·10 toe= ·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 with51⁄2feet 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 thatr= 6 × 144 = 864andm= 0·1. Since the engine is a condensing engine, we haveb= 164ande′= 3691399. By the formulæ (25.) and (26.) we have
and since by the data we have
the formula, by these substitutions, becomes
Sincee= 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 be151⁄3lbs.
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, andr= 864, we shall have
and by the column for condensing engines, in table, p. 519, we havee′= 6029916, wheree= 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 be501⁄3inches.
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 be161⁄10lbs.