Comparative Steam Utilization Coefficient in Direct-Acting Steam Compressors: with 2 tables
Abstract Generated abstract
This paper develops a theoretical comparison of steam utilization in direct-acting steam compressors, motivated by conflicting experimental evaluations of railway brake compressors and by the need to separate system effects from construction and maintenance effects. Under simplifying assumptions of ideal steam admission, polytropic air compression, no clearance, leakage, resistance, friction, or inertia, it derives formulas for the relative amount of atmospheric air compressed per unit of steam reduced to atmospheric pressure across two-, three-, four-, and multi-cylinder arrangements, with and without steam expansion. The analysis identifies optimal cylinder volume ratios for selected systems and tabulates coefficients over a range of working pressures. It concludes that steam utilization declines as attainable air pressure rises, but that suitable cylinder proportions and steam expansion can raise the coefficient severalfold relative to a simple two-cylinder compressor.
Full Text
S. P. Gomellia
Comparative Coefficient
of Steam Utilization
in Direct-Acting Steam Compressors.
with two plates of drawings.
1–36
Tomsk
Steam Printing House of N. I. Orlova.
1911.
Comparative Steam Utilization Coefficient
in Direct-Acting Steam Compressors.
§ 1. Direct-acting steam compressors not only have extensive application on railways in train air brakes, but are gradually beginning to win a place for themselves in other branches of technology as well, wherever, for the storage of the energy of compressed air, a simple, compact, and inexpensive apparatus is required. Ordinarily, the question of the economical operation of a compressor plays a secondary role when the matter concerns the production of a small quantity of compressed air: in this case a low efficiency may be compensated for by simplicity of construction and by the absence of complexity in caring for the compressor. But if more serious work is required of the compressor, then the question of its efficiency can no longer be left without attention. Indeed, in the United States of North America the increase in the make-up of railway trains not only led to an increase in the dimensions of compressors for air brakes, but also brought to the fore the question of raising their coefficient of efficiency; for with that enormous quantity of steam which locomotives in some cases expend in actuating air brakes (from 5% to 10%), increasing the productivity of the air pump can yield a noticeable economy.
Several comparative tests have been carried out on compressors used for railway brakes, chiefly of the systems “Westinghouse,” “New York,” “Five-Lille,” and others; however, in view of the contradictory nature of the data from these experiments, it is difficult to draw from them any definite conclusion regarding the comparative merits of these systems. Usually, the compressors of the firm carrying out the test turn out to be the best, perhaps because the compressor of its own system being tested is in better condition than the compressor of another system being tested.
Generally speaking, the accuracy of operation and assembly, the condition of the valves, slide valves, packing glands, pistons—all this has a considerable influence on the operation of such a comparatively small machine as an ordinary direct-acting steam compressor. Thus, every separate investigation, even if it is repeated on a large number of compressors of the same system, will contain individual influences of the mechanism and sometimes may
...may give results that are much too inadequate for assessing the suitability of the given system.
Therefore a theoretical investigation of the operation of direct-acting compressors under conditions identical for all their designs and systems will not only provide a quite fair comparative evaluation, independent of the execution of the mechanism itself and other secondary conditions, for existing designs, but will also indicate the direction in which these compressors must be improved so that they may be used with the greatest economy of steam.
§ 2. In considering the various systems of direct-acting steam compressors, we shall adopt the following subdivision of them.
A) Compressors operating without expansion of steam:
1) two-cylinder;
2) three-cylinder;
3) four-cylinder:
a) of periodic action,
β) of simultaneous action.
B) Compressors with single expansion of steam:
1) two-cylinder;
2) four-cylinder, of periodic action and with final compression of the air:
a) by means of live steam;
β) by means of expanding steam;
3) four-cylinder, not of periodic action, and with final compression of the air:
a) by means of live steam;
β) by means of expanding steam.
C) Compressors with double and multiple expansion of steam:
1) three-cylinder,
2) four-cylinder, and
3) multi-cylinder.
In considering these compressor systems we shall adopt the following identical assumptions for all of them:
1) the pressure of the live steam on the piston is constant throughout its entire stroke;
2) the pressure of the steam during its expansion follows Mariotte’s law;
3) the pressure of compression of the air follows the polytropic curve \(pv^n=c\);
4) the clearance spaces for the steam and air cylinders, as well as air leakage, are equal to zero;
5) the resistance of steam and air in passing through openings and passages is equal to zero;
6) the friction of the mechanisms is equal to zero;
7) we neglect the inertia of the moving parts*).
We shall then adopt the following notation:
\(v\)—the volume of the cylinder into which live steam enters;
\(v_0\)—the volume of the cylinder in which the first expansion of the steam takes place;
\(v_{co}\)—the volume of the cylinder in which the secondary expansion of the steam takes place;
\(P\)—the pressure of the live steam;
\(P_0\)—the pressure of the steam at the end of its expansion;
\(v_1\)—the volume of the cylinder into which atmospheric air enters;
\(v_2\)—the volume of the cylinder in which the first compression of the air takes place;
\(v_3\)—the volume of the cylinder in which the secondary compression of the air takes place;
\(p_1\)—the pressure of the air at the end of the first compression;
\(p_2\)—the pressure of the air at the end of the secondary compression;
\(p\)—the final maximum compression of the air;
\(n\)—the constant of the polytropic curve; for compressors without artificial cooling, such as those considered by us, one may take \(n = 1.4\) (adiabatic curve).
All pressures are taken from absolute zero. The piston strokes are everywhere considered equal to one another.
The task of the present investigation is to determine the relative degree of steam utilization in the various systems of direct-acting steam compressors, i.e. the magnitudes of the ratios, for different systems of direct-acting compressors, between the volume of atmospheric air entering the compressor and the volume of steam expended in compressing this air, also reduced to atmospheric pressure. Since in the present case the steam is the source of the motive force, we may also call this coefficient the relative coefficient of useful action of the compressor, and take it as a measure of the comparative economy of one or another system of direct-acting steam compressors.
At the same time, as stated above, we exclude from the investigation all those circumstances of compressor operation which are rather difficult to evaluate without experimental data and which, in relation to the magnitude of the compressor’s work, may be regarded as small quantities of higher order. Consequently, the results following from such an investigation may be considered only as a first approximation; nevertheless, they are not only sufficient for clarifying this question, which has not yet been studied, but perhaps also the only—
*) The circumstances not adopted here, while very essential in determining the absolute coefficient of useful action, are of little significance in determining the comparative coefficient of steam utilization.
respectively possible, in view of the fact that the aforementioned quantities of higher order are for the most part variable factors, depending not on the system, but mainly on the design, and also on the execution and maintenance of the compressor mechanism.
§ 3. Let us first consider a two-cylinder simple compressor, in which the steam and air cylinders are not equal to one another (Fig. 1).
This system of direct-acting steam compressors is the most widespread. Let us denote the volume of the steam cylinder by \(v\) and the volume of the air cylinder by \(v_1\). At each stroke of the piston a volume of air \(v_1\) is drawn in, while the work of steam \(vP\) is expended; hence the steam utilization coefficient \(\eta_1\) will be equal to:
\[ \eta_1=\frac{1}{P}\frac{v_1}{v}\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots (1) \]
Consequently, the steam utilization coefficient in simple direct-acting compressors with two cylinders is inversely proportional to the magnitude of the live-steam pressure, inversely proportional to the volume of the steam cylinder, and directly proportional to the magnitude of the volume of the air cylinder.
The maximum compression of air \(p\), which can be obtained by means of the indicated compressor, will be:
\[ p=P\frac{v}{v_1}, \]
so that formula (1) may be represented in the following form:
\[ \eta_1=\frac{1}{p}\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots (2) \]
From formula (2) it is evident that the higher the limiting pressure that can be obtained by means of the given two-cylinder direct-acting steam compressor, the smaller its steam utilization coefficient, and conversely.
If both cylinders are equal to one another, then
\[ p=P \text{ and } \eta_1=\frac{1}{P}, \quad \text{or, what is the same: } \eta_1=\frac{1}{p}. \]
Thus, the steam utilization coefficient in simple direct-acting steam compressors is very low; thus, for exam-
meter; for a compressor with two equal cylinders, operating at 10 atm., the theoretical*) coefficient \(\eta=0.1\), while its practical value is still lower.
§ 4. In order to reduce the coefficient of steam consumption in direct-acting compressors, two-cylinder compressors soon gave way to three-cylinder and four-cylinder compressors with double compression of the air. Double compression in this type of compressor would have had a more substantial significance if the air, after the first compression, could be cooled; this, however, is not the case. Therefore, under the conditions considered by us, a gain in the coefficient of steam utilization can be obtained chiefly from the appropriate choice of the ratios of the cylinder volumes and of the distribution of the work of steam and air in them.
Let us note here that the coefficient of steam utilization for any system of direct-acting compressors, generally speaking, depends not on the pressure of the working steam \(P\), but on the greatest possible compression of the air \(p\). Indeed, if the greatest compression of the air is equal to the pressure of the fresh steam, then the coefficient of steam utilization will be:
\[ \eta=\frac{1}{p}\frac{v_1}{v},\quad \text{or}\quad \eta=\frac{1}{p}\cdot\frac{v_1}{v}, \]
where \(v_1\) is the volume of the air cylinder, and \(v\) is the volume of the steam cylinder into which the fresh steam enters.
If the same compression is attained at a steam pressure \(p\alpha\), then, obviously, the volume of each steam cylinder must be \(\frac{1}{\alpha}\) times larger, as a result of which:
\[ \eta=\frac{1}{p\alpha}\cdot\frac{v_1\alpha}{v}=\frac{1}{p}\frac{v_1}{v}\ . . . . . . . . . . . . . . . . . . . . . . . . . . . (3). \]
Consequently, the coefficient of steam utilization computed for the case when \(P=p\) will be the same also when \(P\ne p\), provided only that in both cases the greatest possible compressions of the air are equal to one another. Thus, in all subsequent derivations, for simplicity we shall assume that the greatest possible compression of the air is equal to the pressure of the fresh steam entering the compressor.
Let us now proceed to the consideration of various other systems of direct-acting steam compressors.
§ 5. A three-cylinder compressor consists of one steam cylinder and two air cylinders of unequal size, arranged—
*) I.e., calculated theoretically according to the formula (2) given above.
lying on one axis. The outside air enters the large air cylinder II, whence it is driven into the small cylinder III, where it is then finally compressed to the maximum pressure (see the diagram in Fig. 2).
For the compressor to be capable of operating, it is necessary that the pressure on the steam side should always be no less than the resistance from the compressed air, so that we may write:
\[ vP=(v_1-v_2)p_1+v_2p. \]
Since the maximum value \(P=p\), then
\[ vp=(v_1-v_2)p_1+v_2p; \]
whence
\[ p(v-v_2)=p_1(v_1-v_2); \]
but, on the other hand,
\[ p_1=\left(\frac{v_1}{v_2}\right)^n; \]
Consequently,
\[ p(v-v_2)=\left(\frac{v_1}{v_2}\right)^n(v_1-v_2). \]
By dividing both sides of the equation by \(v_1\) and after some transformations, we obtain:
\[ p\frac{v}{v_1}=\left(\frac{v_1}{v_2}\right)^n\left(1-\frac{v_2}{v_1}\right)+p\frac{v_2}{v_1}. \]
Consequently, the steam utilization coefficient \(\eta_2\) will be:
\[ \eta_2=\frac{1}{p}\cdot\frac{v_1}{v} =\frac{1}{ \left(\frac{v_1}{v_2}\right)^n\left(1-\frac{v_2}{v_1}\right)+p\frac{v_2}{v_1} } \ . . . . . . . . . . . . . . . \ (4). \]
Let us now determine what the value of \(\frac{v_1}{v_2}\) must be for \(\eta_2\) to be a maximum. It is not difficult to see that \(\eta_2\) will have a maximum when the denominator of expression (4) has a minimum, since the numerator is a constant. For this purpose, the first derivative with respect to
\[ \left(\frac{v_1}{v_2}\right) \quad \text{of} \quad \left[ \left(\frac{v_1}{v_2}\right)^n \left(1-\frac{v_1}{v_2}\right) +p\frac{v_1}{v_2} \right] \]
set equal to zero, whereby we obtain:
\[ n\left(\frac{v_1}{v_2}\right)^{n-1} \left[1-\left(\frac{v_1}{v_2}\right)^{-1}\right] + \left(\frac{v_1}{v_2}\right)^n \left(\frac{v_1}{v_2}\right)^{-2} - p\left(\frac{v_1}{v_2}\right)^{-2} =0 \ . . . . . . . . . . . \tag{5} \]
or
\[ n\left(\frac{v_1}{v_2}\right)^{n-1} -(n-1)\left(\frac{v_1}{v_2}\right)^{n-2} -p\left(\frac{v_1}{v_2}\right)^{-2} =0. \]
To make sure that this is a minimum, we take the second derivative with respect to \(\left(\frac{v_1}{v_2}\right)\):
\[ n(n-1)\left(\frac{v_1}{v_2}\right)^{n-2} -(n-1)(n-2)\left(\frac{v_1}{v_2}\right)^{n-3} +2p\left(\frac{v_1}{v_2}\right)^{-3}. \]
Since \(2>n\geq 1\), the second derivative is always greater than zero; consequently, we have obtained the required minimum.
Equation (5) may be represented in the following simpler form:
\[ n\left(\frac{v_1}{v_2}\right)^{n+1} -(n-1)\left(\frac{v_1}{v_2}\right)^n -p=0 \ . . . . . . . . . . . \tag{6} \]
If there were sufficient cooling of the air after its first compression, then one could take \(n=1\), and then our equation would take the form:
\[ \left(\frac{v_1}{v_2}\right)^2-p=0, \]
whence
\[ \frac{v_1}{v_2}=\sqrt{p}\ ^*) \]
and
\[ \eta_2=\frac{1}{2\sqrt{p}-1} = \frac{1}{p}\frac{p}{2\sqrt{p}-1} \ . . . . . . . . . . . \tag{7} \]
*) Negative roots and others having no real value are not given.
Since in direct-acting steam compressors there is no cooling, we must take \(n=1.4\); then we obtain the more complicated equation:
\[ 1.4\left(\frac{v_1}{v_2}\right)^{2.4} -0.4\left(\frac{v_1}{v_2}\right)^{1.4} -p=0. \]
To solve this equation, we multiply it by \(\left(\frac{v_1}{v_2}\right)^y\), where \(y\) is chosen so that \(2.4+y=(1.4+y)2\), i.e., so as to obtain a quadratic equation. Since \(y=-0.4\), we obtain:
\[ 1.4\left(\frac{v_1}{v_2}\right)^2 -0.4\left(\frac{v_1}{v_2}\right) -p\left(\frac{v_1}{v_2}\right)^{-0.4} =0 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots (8) \]
For the first approximation we take, as before, \(\frac{v_1}{v_2}=p^{0.5}\); consequently, our equation takes the form:
\[ 1.4\left(\frac{v_1}{v_2}\right)^2 -0.4\left(\frac{v_1}{v_2}\right)^{0.8} -p=0; \]
whence
\[ \frac{v_1}{v_2} = \frac{0.4+\sqrt{0.16+5.6p^{0.8}}}{2.8}, \]
which gives the following values for \(\frac{v_1}{v_2}\) at various pressures \(p\) (second approximation):
| \(p\). | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | atm. |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| \(\dfrac{v_1}{v_2}\) | 1.76 | 1.92 | 2.04 | 2.13 | 2.22 | 2.31 | 2.40 | 2.47 | 2.55 | 2.62 | 2.70 | 2.76 | 2.82 | 2.89 | 2.95 | 3.00 | — |
Substituting the obtained values of \(\frac{v_1}{v_2}\) into the last term of equation (8) and, solving it again, we obtain values of \(\frac{v_1}{v_2}\) closer to the truth, and so on. Thus, after the third approximation we obtain the following values, at which we may stop:
| \(p\). | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | atm. |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| \(\dfrac{v_1}{v_2}\) | 1.85 | 2.00 | 2.13 | 2.25 | 2.36 | 2.45 | 2.55 | 2.64 | 2.73 | 2.81 | 2.88 | 2.96 | 3.03 | 3.10 | 3.18 | 3.24 | — |
In this case, if the steam-utilization coefficient $\eta_2$ is represented in the form: $\eta_2 = \dfrac{1}{p}\dfrac{v_1}{v}$, then $\dfrac{v_1}{v}$ will have the following values:
| $p.$ | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | atm. |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| $\dfrac{v_1}{v}$ | 1.37 | 1.54 | 1.64 | 1.72 | 1.80 | 1.88 | 1.95 | 2.04 | 2.09 | 2.17 | 2.23 | 2.29 | 2.35 | 2.41 | 2.47 | 2.52 | $p\eta_2$ |
To illustrate the question more fully, let us present a theoretical diagram for one period of operation of the compressor (Fig. 3). Inside the rectangle on the left side is shown the combined diagram of the work of the steam and of the resistance from the compression of the air;*) from it one can see that at the beginning of the stroke there exists a considerable excess of steam pressure on the piston rod over the air pressure. It is clear that this excess goes toward increasing the speed of the pistons, as a result of which, together with the increase in speed, the resistance of the steam entering the cylinder also increases, as do the other resistances, and an inevitable fall of steam pressure takes place. At the end of the stroke of the pistons, as a result of the increase in resistance from useful work, the speed of the pistons decreases, and the steam pressure rises to normal. Thus, the excess work of the steam present at the beginning of the stroke is utilized the less, the greater the speed of the pistons, i.e., the smaller the mass of the reciprocating moving parts.
§ 6. Four-cylinder direct-acting steam compressors may be subdivided into compressors of periodic and non-periodic action. In the first case, when one pair of pistons is moving, the other pair is standing still, and vice versa. In non-periodic operation, both pairs of pistons work simultaneously. It is clear that, as a result of this, the capacity of compressors of the second kind (all other conditions being equal) is greater than that of compressors operating periodically.
Most often, four-cylinder compressors of periodic action are arranged so that both steam cylinders and the second air cylinder are equal to one another; the first air cylinder, into which outside air enters, has a volume twice as large as the volume of each steam cylinder (Fig. 4). Since in compressors of periodic action air is drawn into both air cylinders, the steam-utilization coefficient $\eta_3$ for the case under consideration will be:
\[ \eta_3 = \frac{1}{p}\,\frac{v_1 + v_2}{2v}; \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots (9) \]
*) Both in this and in the following diagrams, the work of the steam is marked by horizontal hatching, and the work of the air resistance by inclined or vertical hatching.
and since \(v_2 = v\) and \(v_1 = 2v_1\), we obtain:
\[ \eta_3 = \frac{3}{2}\frac{1}{p}. \ . . . . . . . . . . . . . . . . . . . . . . . \tag{9 bis} \]
Thus, with the usual ratios of cylinder volumes indicated above, this compressor is 50% more economical than a simple two-cylinder compressor.
The ratio \(\dfrac{v_1 + v_2}{2v}\) may also be made greater than \(\dfrac{3}{2}\), whereby it is possible to increase the steam-utilization coefficient of this compressor.
In fact, if the three cylinders have equal volumes:
\[ v_0 = v = v_2, \]
then the air pressure \(p_1\) in the second air cylinder, after the first compression of the air, cannot be greater than \(P\), but
\[ p_1 = \left(\frac{v_1 + v_2}{v_2}\right)^n = \left(1 + \frac{v_1}{v_2}\right)^n, \]
and since \(P = p\) and \(v_2 = v\), then
\[ p \ge \left(\frac{v_1}{v} + 1\right)^n . . . . . . . . . . . . . . . . . . . \tag{10} \]
On the other hand, the pressure on the piston rod in the first pair of cylinders from the compressed-air side must be no greater than the pressure from the steam, i.e.
\[ v_1 p_1 \le Pv, \]
which, on the basis of the relations given above, gives:
\[ \left(\frac{v_1}{v} + 1\right)^n \le p\frac{v}{v_1} . . . . . . . . . . . . . . . . . . \tag{11} \]
But since \(v_1 > v\) and \(p > 1\), condition (11) is entirely sufficient, because when condition (11) is fulfilled, condition (10) is also fulfilled at the same time.
Thus, the steam-utilization coefficient \(\eta'_3\) for the case under consideration is expressed as follows:
\[ \eta'_3 = \frac{1}{2}\frac{1}{p}\left(1 + \frac{v_1}{v}\right). . . . . . . . . . . . . . . . . . \tag{12} \]
where \(1+\dfrac{v_1}{v}\) is determined from the equation:
\[ \left(1+\frac{v_1}{v}\right)^{n+1} - \left(1+\frac{v_1}{v}\right)^n -p=0 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots (13) \]
The roots of this equation may be determined by Newton’s method. If we denote the left-hand side of this equation by \(f(x)\), where
\[ x=1+\frac{v_1}{v}, \]
then the approximate value of the root of this equation is expressed as follows:
\[ x=x_1-\frac{f(x_1)}{f'(x_1)}, \]
where \(x_1\) is some known approximate value of the root of equation (13).
We take, as the first approximate value of \(\dfrac{v_1}{v}\),
\[ x_1=\sqrt[n+1]{p}. \]
After several substitutions we obtain the following series of values of \(\dfrac{v_1}{v}\) for various pressures \(p\) from 5 to 20 atm.:
| \(p\) | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | Atm. |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| \(\dfrac{v_1}{v}\) | 1.37 | 1.56 | 1.67 | 1.80 | 1.92 | 2.04 | 2.13 | 2.27 | 2.33 | 2.42 | 2.50 | 2.58 | 2.67 | 2.75 | 2.84 | 2.91 |
Substituting the found quantities \(\dfrac{v_1}{v}\) into expression (12), we obtain the following series of values for \(\eta_3' p\) at various pressures \(p\):
| \(p\) | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | Atm. |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| \(\eta_3' p\) | 1.19 | 1.28 | 1.34 | 1.40 | 1.46 | 1.52 | 1.52 | 1.64 | 1.67 | 1.71 | 1.75 | 1.79 | 1.84 | 1.88 | 1.92 | 1.96 |
From the tables given it is evident that, for the usual ratios of cylinder volumes, when \(v_1=2v\), these compressors cannot—
…will function properly below 10 atm, since in this case the steam pressure cannot overcome the first compression of the air.
§ 7. The useful action of the compressor of the system under consideration will increase still more if the steam cylinders are not equal to one another. Let us denote the volume of the first steam cylinder by \(v'\), and that of the second by \(v\), and determine the relation between the volumes of the cylinders at which the coefficient of steam utilization will be greatest.
We write the equation expressing the condition of equality of the pressures on the pistons of the first pair of cylinders at the end of its stroke.
\[ PV'=\left(1+\frac{v}{v_1}\right)^n v_1, \]
or
\[ \frac{v'}{v}=\frac{1}{P}\,\frac{v_1}{v}\left(1+\frac{v_1}{v}\right)^n; \]
adding 1 to each side of the equation:
\[ \frac{v'+v}{v}=\frac{1}{P}\,\frac{v_1}{v}\left(1+\frac{v_1}{v}\right)^n+1; \]
multiplying each side of the equation by \(\dfrac{v}{v_1+v}\), we obtain:
\[ \frac{v'+v}{v_1+v} = \frac{1}{P}\,\frac{v}{v_1+v} \left[ \left(1+\frac{v_1}{v}\right)^n\frac{v_1}{v}+P \right] \]
The coefficient of steam utilization \(\eta''_3\) will be equal to:
\[ \eta''_3=\frac{1}{p}\,\frac{v_1+v}{v'+v}, \]
and since \(P=p\), then
\[ \eta''_3 = \frac{1}{p}\, \frac{ p\left(1+\frac{v_1}{v}\right) }{ \frac{v_1}{v}\left(1+\frac{v_1}{v}\right)^n+p } \ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots \tag{14} \]
Let us denote \(1+\dfrac{v_1}{v}\) by \(y\); then equality (14) may be represented in the following form:
\[ \eta''_3 = \frac{1}{p}\, \frac{p\,y}{(y-1)y^n+p}, \ldots\ldots\ldots\ldots\ldots\ldots\ldots \tag{15} \]
and \(\eta''_3\) will have a maximum when the expression
\[ \frac{p}{y} + y^{\,n-1}(y-1) \]
has its minimum.
We take from it the first derivative with respect to \(y\) and set it equal to 0:
\[ -\frac{p}{y^2} + (n-1)y^{\,n-2}(y-1)+y^{\,n-1} = \frac{-p+ny^{\,n+1}-(n-1)y^n}{y^2} =0; \]
Since the second derivative is always greater than zero, we do, indeed, obtain for \(y\) the value that turns \(\eta''_3\) into a maximum.
On considering the equation:
\[ ny^{\,n+1}-(n-1)y^n-p=0, \]
we see that it is identical with equation (6), and, consequently, for \(y\) at various values of \(p\) we obtain the same values as for \(\frac{v_1}{v_2}\) in the investigation of the three-cylinder compressor (§ 5).
Thus, we have the following values of \(\frac{v_1}{v}\) for the present case:
| \(p\) | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | atm. |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| \(\frac{v_1}{v}\) | 0.85 | 1.00 | 1.13 | 1.25 | 1.36 | 1.45 | 1.55 | 1.64 | 1.73 | 1.81 | 1.88 | 1.96 | 2.03 | 2.10 | 2.18 | 2.24 |
In exactly the same way, the steam utilization coefficient in the four-cylinder periodic-action compressor under consideration is identical with the steam utilization coefficient \(\eta_2\) in the three-cylinder compressor, if in both cases the most advantageous ratios of cylinder volumes are taken, since \(\eta_2\) and \(\eta''_3\) are identical with each other, provided only that
\[ y=\frac{v_1}{v_2}. \]
Thus, we obtain the following values for \(\eta''_3\) and \(\rho=\frac{v_1+v}{v'+v}\) at various pressures \(p\):
| \(p\) | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | atm. |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| \(\eta''_3 p\) | 1.37 | 1.54 | 1.64 | 1.72 | 1.80 | 1.88 | 1.95 | 2.04 | 2.09 | 2.17 | 2.23 | 2.29 | 2.35 | 2.41 | 2.47 | 2.52 |
As for the ratio \(\frac{v'}{v}\), it is determined from the equation:
\[ \frac{v'}{v} = \frac{1}{p} \left( 1 + \frac{v'}{v} \right)^n \frac{v_1}{v} \qquad \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots (16) \]
or, on the basis of the steam-utilization coefficient \(\eta''_3\), namely:
\[ \frac{v'}{v} = \frac{y-\eta''_3 p}{\eta''_3 p} \qquad \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots (17) \]
which gives for \(\frac{v'}{v}\) the following series of values for different magnitudes of \(p\):
| \(p\) | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | atm. |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| \(\frac{v'}{v}\) | 0.35 | 0.30 | 0.30 | 0.31 | 0.31 | 0.30 | 0.31 | 0.30 | 0.31 | 0.30 | 0.34 | 0.33 | 0.29 | 0.32 | 0.29 | 0.29 |
Thus, in the system of four-cylinder compressors considered by us, an increase in their useful action can be achieved if the volumes of the steam cylinders are changed accordingly, so that the steam cylinder of the second pair is approximately three times larger than the other steam cylinder.
The theoretical diagrams for this system of compressors are presented in Fig. 5.
§ 8. Much higher in its productivity is the system of four-cylinder compressors of non-periodic action, operating likewise with the full pressure of the live steam admitted into the cylinders. The principal difference of this system from the preceding one consists in the fact that the air compressed in the first pair of cylinders, on passing into the air cylinder of the second pair, assists the movement of the pistons in it (Fig. 6).
The condition for the possibility of the pistons moving to the end of their stroke in the first pair of cylinders is expressed by the following equation:
\[ P v' = \left( \frac{v_1}{v_2} \right)^n v_1 \qquad \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots (18) \]
The condition for the possibility of motion of the pistons in the second pair of cylinders will likewise be represented by the equation:
\[ P v=\left[p-\left(\frac{v_1}{v_2}\right)^n\right]v_2 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots (18\ \text{bis}) \]
To determine the coefficient of steam utilization we carry out a series of transformations, taking into account that \(P=p\).
\[ \frac{v'}{v_1}=\left(\frac{v_1}{v_2}\right)^n\frac{1}{p}; \]
\[ \frac{v}{v_2}=1-\frac{v'}{v_1}; \]
Multiplying both sides of the last equality by \(\frac{v_1}{v'}\) and adding 1 to each side, we obtain:
\[ \frac{v+v'}{v'}=\left(1-\frac{v'}{v_1}\right)\frac{v_2}{v'}+1. \]
Then we multiply each side by \(\frac{v'}{v_1}\), whereby we obtain:
\[ \frac{v'+v}{v_1}=\frac{v_2}{v_1}-\frac{v_2}{v_1}\cdot\frac{v'}{v_1}+\frac{v'}{v_1}; \]
since
\[ \frac{v'}{v_1}=\frac{1}{p}\left(\frac{v_1}{v_2}\right)^n, \]
then
\[ \frac{v'+v}{v_1}= \frac{v_2}{v_1} -\frac{1}{p}\left(\frac{v_1}{v_2}\right)^{n-1} +\frac{1}{p}\left(\frac{v_1}{v_2}\right)^n. \]
From this we derive the value of the coefficient of steam utilization \(\eta_4\) for the compressor under consideration:
\[ \eta_4= \frac{1}{p}\frac{v_1}{v'+v} = \frac{1}{p}\, \frac{p}{ p\frac{v_2}{v_1} +\left(\frac{v_1}{v_2}\right)^n \left(1-\frac{v_2}{v_1}\right) } \ldots \ldots \ldots \ldots \ldots \ldots (19) \]
It is not difficult to see that this is the very same formula as (4) and (15); therefore \(\eta_4=\eta_2=\eta_3''\), and its maximum value \(\eta_4'\) at \(n=1\) will likewise be
\[ \eta_4'=\eta_2'=\frac{1}{p}\frac{p}{2\sqrt{p}-1} \ldots \ldots \ldots \ldots \ldots \ldots \ldots (19\ \text{bis}) \]
Thus, if there are two compressors, of which one is three-cylinder and the other four-cylinder, of periodic or non-periodic action, and if for both the ratios \(\frac{v_1}{v_2}\) are equal, then the coefficients of utilization of steam in them are equal; consequently, the ratio of the sum of the volumes of the steam cylinders to the second, air volume is also the same in both.
The values of \(\frac{v_1}{v_2}\) have already been found by us (see § 5); let us try to determine the other relations between the cylinder volumes.
\[ \frac{v_2}{v+v_0} = \frac{v_1}{v+v_0} \left(\frac{v_1}{v_2}\right)^{-1} = \eta_4 p \left(\frac{v}{v_2}\right)^{-1} \qquad \ldots \ldots \ldots \ldots \ldots \ldots (20) \]
\[ \frac{v_0}{v+v_0} = \frac{v_1}{v+v_0}\, \frac{v_0}{v_1} = \eta_4 \left(\frac{v_1}{v_2}\right)^n \qquad \ldots \ldots \ldots \ldots \ldots \ldots (21) \]
\[ \frac{v}{v+v_0} = \frac{v_0}{v+v_0}\, \frac{v}{v_0} = \eta_4 \left(\frac{v_1}{v_2}\right)^n \frac{ p-\left(\frac{v_1}{v_2}\right)^n }{ \left(\frac{v_1}{v_2}\right)^{n+1} } \]
so that:
\[ \frac{v}{v+v_0} = \eta_4 \frac{ p-\left(\frac{v_1}{v_2}\right)^n }{ \frac{v_1}{v_2} } \qquad \ldots \ldots \ldots \ldots \ldots \ldots (22) \]
Taking \((v+v_0)\) as unity, we obtain the following values for \(v_0\), \(v\), \(v_2\), and \(v_1=\eta_4 p\) at pressures from \(p=5\) to \(p=20\) atmospheres:
| 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | Atm | |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| \(v_0\) | 0.65 | 0.67 | 0.67 | 0.67 | 0.67 | 0.66 | 0.66 | 0.67 | 0.66 | 0.66 | 0.65 | 0.65 | 0.65 | 0.65 | 0.65 | 0.65 | — |
| \(v\) | 0.35 | 0.33 | 0.33 | 0.33 | 0.33 | 0.34 | 0.34 | 0.33 | 0.34 | 0.34 | 0.35 | 0.35 | 0.35 | 0.35 | 0.35 | 0.35 | — |
| \(v_2\) | 0.74 | 0.77 | 0.77 | 0.76 | 0.76 | 0.76 | 0.76 | 0.77 | 0.76 | 0.77 | 0.78 | 0.78 | 0.78 | 0.78 | 0.78 | 0.78 | — |
| \(v_1\) | 1.37 | 1.54 | 1.64 | 1.72 | 1.80 | 1.88 | 1.95 | 2.04 | 2.09 | 2.17 | 2.23 | 2.29 | 2.35 | 2.41 | 2.47 | 2.52 | \(\eta_4 p\) |
The theoretical diagrams of the work of steam and air (Fig. 7) clarify in somewhat greater detail the circumstances of the operation of this system of compressors. Owing to the communication of the air cylinders with one another, the useful work of the first steam cylinder is dependent...
the operation of the second pair of cylinders. The first steam cylinder, through the air medium, makes up, if necessary, the deficiency of working force in the second pair of cylinders. Indeed, if the counterpressure from the second air cylinder is so great that the rod with the pistons cannot be moved by the steam pressure in the second cylinder, then the compression of the air will occur only in the first cylinder; the rod of the second pair remains at rest until in the first cylinder the pressure reaches the required value; in our case this value is \(p_1\). It is clear that by this amount the pressure in the second air cylinder can be increased over the entire stroke of the piston; but in fact this increase occurs only over the interval necessary for the motion of the second rod. In the case under consideration, in both the one and the other pair of cylinders there exists an excess of driving force over resistance at the beginning of the piston stroke, but this excess is evidently smaller than in the other systems we have examined.
§ 9. Despite the complete simplicity of the design of direct-acting steam compressors, for a long time they were used almost exclusively for railway air brakes. One of the chief reasons that hindered a wider distribution of these machines was the circumstance that, working without expansion of the steam, they were too uneconomical. In order to increase their useful effect, compressors began to be built with steam expansion and double compression of the air.
The simplest design of this kind is the two-cylinder steam compressor (Fig. 8). Its distinguishing feature is that each cylinder simultaneously serves both for steam and for air: steam enters beneath the piston of the small cylinder; then, when the piston reaches the extreme upper position, the two cylinders are connected, and by the difference of the steam pressures on the two pistons they are driven downward. The air entering beneath the lower side of the piston of the large cylinder is forced into the space above the piston of the small cylinder, and then, by the full pressure of the steam, is forced into the discharge pipe.
Let us consider the conditions under which the operation of this compressor is possible. The most dangerous position of the pistons will be at the greatest expansion of the steam; therefore for this position we write the equation on the basis of the equilibrium of the driving and resisting forces:
\[ v\left[\left(\frac{v_1}{v}\right)^n - P\frac{v}{v_1}\right] - \left[\left(\frac{v_1}{v}\right)^n - P\frac{v}{v_1}\right]v_1 = 0 \ldots\ldots\ldots (23) \]
Since \(v\) is not equal to \(v_1\), then
\[ \left(\frac{v_1}{v}\right)^n = P\frac{v}{v_1}; \]
and since \(P = p\), then
\[ \frac{v_1}{v} = \sqrt[n+1]{p}\ldots\ldots\ldots (24) \]
Thus, the steam-utilization coefficient \(\eta_5\) for a two-cylinder compressor with steam expansion is expressed as follows:
\[ \eta_5=\frac{1}{p}\frac{v_1}{v}=\frac{1}{p}\sqrt[n+1]{p}\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots (25) \]
The pressure \(P_0\) of the steam leaving the compressor depends on the relative volumes of the cylinders and is expressed as follows:
\[ P_0=P\frac{v}{v_1}=\frac{p}{\eta_5 p}=\frac{1}{\eta_5}\qquad (25\ \text{bis}). \]
Thus, for various pressures from 5 to 20 atmospheres, we obtain the following numbers for \(\dfrac{v_1}{v}=\eta_5 p\) and for \(P_0\):
| \(p.\) | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | atm. |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| \(\eta_5 p\) | 1.95 | 2.13 | 2.25 | 2.39 | 2.50 | 2.62 | 2.72 | 2.82 | 2.91 | 3.00 | 3.09 | 3.17 | 3.26 | 3.34 | 3.42 | 3.50 | \(\dfrac{v_1}{v}\) |
| \(P_0\) | 2.57 | 2.81 | 3.09 | 3.34 | 3.50 | 3.80 | 4.43 | 4.22 | 4.41 | 4.63 | 4.76 | 5.00 | 5.17 | 5.28 | 5.42 | 5.68 | atm. |
Thus, this system can provide a saving of from 95 to 250%, depending on the pressure to which the pump is intended to compress the air, in comparison with a simple two-cylinder compressor (§ 3). The diagram for this compressor is shown in Fig. 9.
§ 10. Let us turn to a four-cylinder direct-acting compressor with steam expansion, first considering the system in which the final compression of the air is performed by fresh steam, and each pair of cylinders works alternately (periodically) (Fig. 10). It is obvious that the lowest steam pressure in the second steam cylinder is equal to the pressure of the steam escaping into the atmosphere, \(P_0\); consequently,
\[ P_0=\frac{P\,v}{v+v_0}, \]
where \(v\) is the volume of the steam cylinder operating with fresh steam, and \(v_0\)—with expanded steam. The pressure of the air compressed in the first air cylinder will be
\[ p_1=\left(\frac{v_1+v}{v}\right)^n. \]
We form the equation expressing the condition of equilibrium of pressures on the rod of the second pair of cylinders at the end of steam expansion:
\[ P_0 v_0 = p_1 v_1, \]
or
\[ P \frac{v_0}{v_0+v} = \left(\frac{v_1+v}{v}\right)^n \frac{v_1}{v}; \]
but since
\[ \frac{P v_0}{v_0+v}=P-P_0, \]
then \(\dfrac{v_1}{v}\) is determined from the equation:
\[ P-P_0=\left(\frac{v_1}{v}+1\right)^n \frac{v_1}{v}. \]
If we denote \(1+\dfrac{v_1}{v}\) by \(x\), then we obtain an equation of the following form, analogous to equation (13):
\[ x^{n+1}-x^n-(P-P_0)=0 \ldots \ldots \ldots \ldots (26). \]
In this case the steam utilization coefficient \(\eta_6\) is expressed as follows:
\[ \eta_6=\frac{1}{p}\left(\frac{v_1}{v}+1\right)=\frac{1}{p}x \ldots \ldots \ldots \ldots (27). \]
If \(P_0=0\), then \(\eta_6\) would be, as is not difficult to see, twice as large as \(\eta'_3\), and \(\dfrac{v_0}{v}\) would have to be equal to infinity.
Let us determine the values of \(\dfrac{v_0}{v}\), \(\dfrac{v_1}{v}=\eta_6 p-1\) at an inlet steam pressure
\[ P_0=2 \text{ atm.} \]
To determine \(\dfrac{v_0}{v}\) we have the formula:
\[ \frac{v_0}{v}=\frac{P-P_0}{P_0}; \]
indeed:
\[ (v_0+v)P_0=Pv;\quad \frac{v_0}{v}+1=\frac{P}{P_0}; \]
therefore,
\[ \frac{v_0}{v}=\frac{P-P_0}{P_0}; \]
as for \(\dfrac{v_1}{v}\), it is easily determined once the roots are known
equation (26), which we solve by one of the methods of approximate determination of roots.
We give a table of values of \(\dfrac{v_0}{v}\), \(\dfrac{v_1}{v}\), and \(\eta_6 p\) for \(P_0 = 2\) atm.
| \(p\) | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | atm. |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| \(\dfrac{v_0}{v}\) | 1.5 | 2.0 | 2.5 | 3.0 | 3.5 | 4.0 | 4.5 | 5.0 | 5.5 | 6.0 | 6.5 | 7.0 | 7.5 | 8.0 | 8.5 | 9.0 | — |
| \(\dfrac{v_1}{v}\) | 1.11 | 1.28 | 1.37 | 1.56 | 1.67 | 1.80 | 1.92 | 2.04 | 2.13 | 2.27 | 2.33 | 2.42 | 2.50 | 2.58 | 2.67 | 2.75 | — |
| \(\eta_6 p\) | 2.11 | 2.28 | 2.37 | 2.56 | 2.67 | 2.80 | 2.92 | 3.04 | 3.13 | 3.27 | 3.33 | 3.42 | 3.50 | 3.58 | 3.67 | 3.75 | — |
Thus, when steam expansion is applied in four-cylinder compressors of the system described, their useful effect increases almost twofold in comparison with the useful effect of the same compressors, but without steam expansion.
In Fig. 11 the theoretical diagram of operation for the system considered is presented.
From a comparison of Figs. 9 and 11 it is evident that, during the initial compression of the air, in the latter system there is a greater excess of driving force over resistance than in the preceding one (§ 9). With respect to the secondary compression, however, both systems operate almost identically.
§ 11. With periodic action of a four-cylinder compressor, and with the final compression of the air by expanded steam, the air pressure at the end of compression in the first cylinder (Fig. 12) will be:
\[ p_1=\left(\frac{v_1+v_2}{v_2}\right)^n. \]
The condition for the possibility of a full stroke of the pistons in the first pair of cylinders is represented in the following form:
\[ Pv=\left(\frac{v_1+v_2}{v_2}\right)^n v_1,\quad \text{where } P=p. \]
The condition for the possibility of a full stroke of the pistons in the second pair of cylinders:
\[ p v_2=P_0 v_0, \]
but
\[ v_2=v\frac{P-P_0}{P}; \]
therefore,
\[ p=\left(1+\frac{v_1}{v}\frac{P}{P-P_0}\right)^n \frac{v_1}{v}. \ldots\ldots\ldots (28) \]
From this equation we can determine \(\frac{v_1}{v}\); for this purpose let us denote \(1 + \frac{v_1}{v}\frac{P}{P-P_0}\), equal to \(\frac{v_1+v_2}{v_2}\), by \(y\); we obtain a new equation:
\[ y^n(y-1)-\frac{P}{P-P_0}\,p=0 \qquad\ldots\ldots\ldots\ldots\ldots\ldots (29) \]
If \(P_0\) were equal to zero, then this equation would be identical with equations (13) and (26), and then in exactly the same way \(\frac{v_0}{v}\) would be equal to infinity.
Solving this equation by one of the methods for finding an approximate value of its roots, we also obtain the values of \(\frac{v_1}{v}\) from the equation:
\[ \frac{v_1}{v}=(y-1)\frac{P-P_0}{P}. \]
The coefficient of utilization of steam \(\eta_7\) for the system under consideration will be:
\[ \eta_7=\frac{1}{p}\frac{v_1+v_2}{v} =\frac{1}{p}\,y\,\frac{P-P_0}{P} \qquad\ldots\ldots\ldots\ldots\ldots\ldots (30) \]
It is not difficult to notice that, if \(P_0=0\), then \(\eta_7=\eta_6=2\eta_3\), and then, of course, the relation \(\frac{v_0}{v}=\infty\) would have to hold.
Let us note here that, as in the preceding case, when \(P_0\neq0\),
\[ \frac{v_0}{v}=\frac{p-P_0}{P_0}, \]
since \(P=p\); for the calculation of \(\frac{v_2}{v}\), however, we have:
\[ \frac{v_2}{v}=\frac{p-P_0}{p}. \]
As for the values of \(\frac{v_1}{v}\), we can still determine them by means of \(p\eta_7\), namely:
\[ \frac{v_1}{v}=p\eta_7-\frac{v_2}{v}. \]
Thus, we obtain the following table for various values of \(p\) from 5 to 20 atm.
| \(p.\) | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | atm. |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| \(\dfrac{v_0}{v}\) | 1.5 | 2.0 | 2.5 | 3.0 | 3.5 | 4.0 | 4.5 | 5.0 | 5.5 | 6.0 | 6.5 | 7.0 | 7.5 | 8.0 | 8.5 | 9.0 | — |
| \(\dfrac{v_2}{v}\) | 0.60 | 0.67 | 0.71 | 0.75 | 0.78 | 0.80 | 0.82 | 0.83 | 0.85 | 0.86 | 0.87 | 0.88 | 0.88 | 0.89 | 0.90 | 0.90 | — |
| \(\dfrac{v_1}{v}\) | 1.10 | 1.28 | 1.44 | 1.58 | 1.71 | 1.83 | 1.95 | 2.07 | 2.15 | 2.25 | 2.34 | 2.43 | 2.52 | 2.59 | 2.68 | 2.77 | — |
| \(\eta^0\) | 1.70 | 1.95 | 2.15 | 2.33 | 2.49 | 2.49 | 2.77 | 2.90 | 3.00 | 3.11 | 3.21 | 3.31 | 3.40 | 3.48 | 3.58 | 3.67 | — |
In Fig. 13 are shown the schemes of the theoretical diagrams for the compressor system under consideration.
§ 12. Let us now proceed to the consideration of a four-cylinder compressor operating simultaneously (not periodically) with both pairs of pistons, with expansion of the steam and final compression by live steam.
A compressor of this kind operates in the following manner (Fig. 14): the steam that has done work in the first cylinder passes into the second, so that the piston of the first cylinder is always subjected to the action of the difference between the pressures of the live and the expanding steam. As regards the second pair of cylinders, in them the moving force is composed of the pressure of the expanding steam and the pressure of the air compressed by the first piston.
Let us denote the lowest pressure of the expanding steam by \(P_0\), and the volume of the second cylinder by \(v_0\); then the moving force for the first pair of pistons at the end of their stroke will be equal to:
\[ (P - P_0)v, \]
i.e. the difference between the pressures of the live and the expanding steam; and the resistance of the compressed air will be:
\[ \left[p - \left(\frac{v_1}{v_2}\right)^n\right]v_2. \]
For operation to be possible in the first pair of cylinders of the compressor, it is necessary that
\[ (P - P_0)v \geq \left[p - \left(\frac{v_1}{v_2}\right)^n\right]v_2 \ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots (31) \]
On the other hand, for the possibility of motion of the pistons in the second pair of cylinders, it is necessary that
\[ P v=\left(\frac{v_1}{v_2}\right)^n v_1 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots (32) \]
If \(v_2=v\) and, as before, \(P=p\), then
\[ p=\left(\frac{v_1}{v}\right)^{n+1}, \]
whence
\[ \frac{v_1}{v}=\sqrt[n+1]{p}\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots (33). \]
At the same time it is also necessary, as is clear from (31), that
\[ P_0 \leq \left(\frac{v_1}{v}\right)^n \ldots \ldots \ldots \ldots \ldots \ldots (33\ \mathrm{bis}). \]
The steam-utilization coefficient \(\eta_8\) in this case will be equal to:
\[ \eta_8=\frac{1}{p}\,\frac{v_1}{v}=\frac{1}{p}\sqrt[n+1]{p}\ldots \ldots \ldots (34) \]
Thus, if the first pair of cylinders has equal volumes, then the steam-utilization coefficient \(\eta_8\) and the limiting pressure in the expansion of the steam \(P_0\) for the construction under consideration are the same as for a two-cylinder compressor with steam expansion (§ 9), i.e. \(\eta_8=\eta_5\).
§ 13. However, in a four-cylinder compressor of the system just considered, it is possible to bring the pressure of the exhaust steam down to the least possible limit \(P_0\), for which it is necessary to increase \(v_1\) and \(v_2\), so that \(v_2>v\). Then from equations (32) and (31) we obtain:
\[ p\,\frac{v}{v_2}=\left(\frac{v_1}{v_2}\right)^{n+1};\ldots \ldots \ldots \ldots \ldots (35) \]
\[ (p-P_0)v=\left[p-\left(\frac{v_1}{v_2}\right)^n\right]v_2;\ldots \ldots \ldots (36) \]
whence:
\[ (p-P_0)\frac{v}{v_2}-p+p^{\frac{n}{n+1}}\left(\frac{v}{v_2}\right)^{\frac{n}{n+1}}=0 \ldots \ldots \ldots (37) \]
In this case the steam utilization coefficient \(\eta'_s\) will be:
\[ \eta'_s = \frac{1}{p}\frac{v_1}{v} - \frac{1}{p}\frac{v_1}{v_2}\frac{v_2}{v} = \frac{1}{p} \left( p\frac{v}{v_2} \right)^{\frac{1}{n+1}} \left( \frac{v}{v_2} \right)^{-1}, \]
or
\[ \eta'_s = \frac{1}{p} \sqrt[n+1]{\,p\left(\frac{v_2}{v}\right)^n\,}. \tag{38} \]
Since \(v_2>v\), \(\eta'_s\) will always be greater than \(\eta_s\). The value of the quantity \(\left(\frac{v_2}{v}\right)^n\), entering into the expression for \(\eta'_s\), can be obtained from equation (37), which contains the unknown in the powers \(1\) and \(\frac{n}{n+1}\). If \(n=1\), then we would have the following equation:
\[ (p-P_0)\frac{v}{v_2} + p^{\frac{1}{2}} \left( \frac{v}{v_2} \right)^{\frac{1}{2}} - p = 0; \]
from which we would obtain:
\[ \frac{v_2}{v} = \left[ \frac{ -p^{\frac{1}{2}} + p^{\frac{1}{2}}\sqrt{\,1+4(p-P_0)\,} }{ 2(p-P_0) } \right]^{-\frac{1}{2}}, \]
which gives, for various values of \(p\) with \(P_0=2\) atmospheres, values close to unity; therefore let us take, as an approximate value of the root, \(\frac{v_2}{v}=1\) and compute from formula (35) the values of \(\frac{v_1}{v_2}\) at various pressures:
\[ \frac{v_1}{v_2} = \sqrt[n+1]{\frac{v}{v_1}\,p}. \]
If the values thus found are substituted into formula (36), we obtain the first approximate values for \(\frac{v_2}{v}\), which on average give the value \(1.28\); taking this value for all pressures, we again substitute it into formula (35), thereby obtaining the second series of approximate values for \(\frac{v_1}{v_2}\):
\[ \frac{v_1}{v_2} = \sqrt[n+1]{p\frac{1}{1.28}} = 0.9\sqrt[n+1]{p}. \]
Substituting the found values of \(\frac{v_1}{v_2}\) again into equation (36), we obtain the next approximate value of \(\frac{v_2}{v}\), equal on average
than 1.19. Substituting this quantity into equation (35), we obtain for \(\frac{v_1}{v_2}\) the value \(0.88 \sqrt[n-1]{p}\), and for \(\frac{v_2}{v_1}\) the average value 1.21. Repeating analogous operations once more, we obtain for \(\frac{v_2}{v}=1.20\). In order not to make further calculations, we may finally take \(\frac{v_2}{v}\) as equal to the average of the last two approximations, whereby we obtain the following values for
\[ \frac{v_2}{v},\quad \frac{v_1}{v_2},\quad \eta'_s p \quad \text{and} \quad \frac{v_0}{v}=\frac{P}{P_0} \]
at \(P_0=2\) atm.
| \(p\) | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | atm. |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| \(\frac{v_2}{v}\) | 1.14 | 1.16 | 1.19 | 1.20 | 1.20 | 1.20 | 1.20 | 1.20 | [[unclear: 1,?0]] | 1.20 | 1.21 | 1.21 | 1.21 | 1.20 | 1.20 | 1.20 | — |
| \(\frac{v_1}{v_2}\) | 1.86 | 1.98 | 2.11 | 2.23 | 2.38 | 2.47 | 2.54 | 2.61 | 2.70 | 2.78 | 2.82 | 2.90 | 2.98 | 3.09 | 3.16 | 3.23 | — |
| \(\frac{v_1}{v}\) | 2.12 | 2.33 | 2.50 | 2.65 | 2.84 | 2.97 | 3.05 | 3.14 | 3.25 | 3.34 | 3.51 | 3.61 | 3.70 | 3.72 | 3.80 | 3.89 | \(\eta'_s p\) |
| \(\frac{v_0}{v}\) | 2.5 | 3.0 | 3.5 | 4.0 | 4.5 | 5.0 | 5.5 | 6.0 | 6.5 | 7.0 | 7.5 | 8.0 | 8.5 | 9.0 | 9.5 | 10.0 | — |
Thus, by changing the cylinder volumes, a rather considerable economy is obtained (from 25% to 40%); moreover, in comparison with a simple two-cylinder compressor, the useful work increases by 2–3 and even almost 4 times.
The diagrams for the work of the steam and for the compression of the air have the form shown in Fig. 15. In the present case the second steam cylinder takes upon itself a considerable part of the work which the first cylinder would have had to perform; in this case the motion of the pistons has approximately the following character: from their extreme positions the pistons of the second pair of cylinders begin to move earlier, while in the first pair of cylinders the pistons stand still until the pressure difference \(P-P'\) (where \(P'\) is the pressure of the expanded steam) is in a condition to overcome the pressure of the compressed air on the first pair of pistons; moreover, subsequently it depends on the relation between the resistance and the driving force in each pair of cylinders: if in some pair of cylinders the former is greater than the latter, then the pistons of the other pair, having moved forward, supply the missing pressure with the aid of the compressed air. It can be proved,
that both steam cylinders can always deliver the necessary work for each of the air cylinders. Indeed, in the second pair of cylinders, owing to the fact that the steam pressure falls toward the end of the piston stroke, while the air pressure rises, and at the very end they become equal, all the work is performed only by the expanding steam. As for the driving force in the first pair of cylinders, it is composed of the pressures:
\[ (P - P') \frac{v}{v_2} \ \text{and}\ P' \frac{v_0}{v_1}. \]
When the pistons of the first pair of cylinders, owing to a deficiency of driving force, come to a stop, the compression of the air in both air cylinders increases (and together with it also the second term
\[ P' \frac{v_0}{v_1} \]
) until the total pressure on the rod, equal to
\[ \left[(P - P') \frac{v}{v_2} + P' \frac{v_0}{v_1}\right] v_2, \]
overcomes the pressure of the air being removed from the compressor, whose pressure is not higher than \(p\). From Figure 16 it is easy to see that
\[ (P - P') \frac{v}{v_2} + P' \frac{v_0}{v_1} > p, \]
provided that the condition exists that
\[ \frac{v_0}{v_1} > 1 \quad \text{and} \quad \frac{v}{v_2} < 1. \]
Indeed, from equations (35) and (36) we obtain:
\[ \frac{v}{v_2} = \frac{1 - \frac{v}{v_1}} {1 - \frac{P_0}{p}}; \]
therefore, since \(\frac{v}{v_2} < 1\), then
\[ \frac{P_0}{p} < \frac{v}{v_1}. \]
On the other hand, since
\[ \frac{v_0}{v_1} = \frac{v_0}{v} \frac{v}{v_1}, \]
we again obtain, as before:
\[ \frac{v_0}{v} \frac{v}{v_1} > 1 \quad \text{and, therefore,} \quad \frac{P_0}{p} < \frac{v}{v_1}. \]
Thus, the two inequalities
\[ \frac{v_0}{v} > 1 \quad \text{and} \quad \frac{v}{v_2} < 1 \]
are mutually ...
are equivalent, and therefore it is sufficient to adopt one of them. Taking the inequality \(\dfrac{v}{v_2}<1\) and comparing it with equation (37):
\[ p^{\frac{n}{n+1}}\left(\frac{v}{v_2}\right)^{\frac{n}{n+1}}+(p-P_0)\frac{v}{v_2}-p=0, \]
we observe that the first two terms, being always positive, increase together with \(\dfrac{v}{v_2}\), and since for \(\dfrac{v}{v_2}<1\) the equation becomes zero, then for \(\dfrac{v}{v_2}=1\) we obtain
\[ p^{\frac{n}{n+1}}+p-P_0-p>0, \]
whence
\[ P_0<p^{\frac{n}{n+1}}, \]
or
\[ P_0<\frac{p}{\sqrt[n+1]{p}}, \]
and in the limit:
\[ P_0=\frac{p}{\sqrt[n+1]{p}}=\frac{1}{\eta_s}. \]
This is the same value of \(P_0\) that we obtained for a two-cylinder compressor with steam expansion (§ 9).
Thus, \(P_0\) may have any values whatever, but not higher than the following limiting values:
| \(p\) | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | atm. |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| \(P_0\) | 2.57 | 2.81 | 3.09 | 3.34 | 3.50 | 3.80 | 4.03 | 4.22 | 4.41 | 4.62 | 4.76 | 5.00 | 5.17 | 5.28 | 5.42 | 5.68 |
If \(P_0=0\), then from equation (37) we could obtain the following equation with the new variable \(\dfrac{v_1}{v}\):
\[ \left(\frac{v_1}{v}\right)^{\frac{n+1}{n}}+\frac{v_1}{v}-p=0 \]
and, consequently, in this case the steam utilization coefficient
\[ (\eta_s')=\frac{1}{p}\frac{v_1}{v} \]
would be still higher; but then \(\dfrac{v_0}{v}\) would have to be equal to infinity.
§ 14. Let us now consider a system of four-cylinder compressors with steam expansion, likewise of non-periodic action, but in which the final compression of the air is effected by the expanding steam (Fig. 17). For the pistons of the first pair of cylinders we have the equation:
\[ \left(\frac{v_1}{v_2}\right)^n v_1 = (P-P_0)v, \]
whence
\[ \frac{v_1}{v_2} = (P-P_0)^{\frac{1}{n}} \left(\frac{v}{v_1}\right)^{\frac{1}{n}} \ldots\ldots\ldots\ldots\ldots\ldots\ldots \tag{39} \]
For the pistons of the second pair of cylinders we obtain:
\[ p v_2 - v_2\left(\frac{v_1}{v_2}\right)^n = P v, \]
whence
\[ p - \left(\frac{v_1}{v_2}\right)^n = P\frac{v}{v_2} \ldots\ldots\ldots\ldots\ldots\ldots \tag{39 bis} \]
Solving equations (39) and (39 bis) together and taking into account that \(P=p\), we obtain:
\[ p - \frac{v}{v_1}(p-P_0) - p(p-P_0)^{\frac{1}{n}} \left(\frac{v}{v_1}\right)^{\frac{n+1}{n}} =0; \]
whence
\[ p\left(\frac{v_1}{v}\right)^{\frac{n+1}{n}} - (p-P_0)\left(\frac{v_1}{v}\right)^{\frac{1}{n}} - p(p-P_0)^{\frac{1}{n}} =0 \ldots \tag{40} \]
In this case the steam utilization coefficient \(\eta_{\vartheta}\) is
\[ \eta_{\vartheta}=\frac{1}{p}\frac{v_1}{v} \]
If \(n=1\), then, for determining \(\frac{v_1}{v}\), we would have a quadratic equation, from which we would obtain:
\[ \frac{v_1}{v} = \frac{1}{2} \left\{ 1-\frac{P_0}{p} + \sqrt{ \left(1-\frac{P_0}{p}\right)^2 + 4p\left(1-\frac{P_0}{p}\right) } \right\} \ldots\ldots \tag{41} \]
If, at the same time, \(P_0 = 0\), then
\[ \frac{v_1}{v}=0.5\left(1+\sqrt{1+4p}\right); \]
in this case the steam utilization coefficient \(\eta'_9\) will be:
\[ \eta'_9=\frac{1}{p}\frac{v_1}{v} =\frac{1}{p}\left(\sqrt{p+\frac{1}{4}}+\frac{1}{2}\right). \]
Since \(n\) is not equal to unity, we can solve equation (40) by one of the approximate methods, for example, by Newton’s method. For this we take the first derivative of the left-hand side of equation (40), whereby we obtain:
\[ \frac{n+1}{n}\,p\left(\frac{v_1}{v}\right)^{\frac{1}{n}} -\frac{1}{n}(p-P_0)\left(\frac{v_1}{v}\right)^{\frac{1-n}{n}} =f'(x) \]
instead of \(\dfrac{v_1}{v}\) we substitute its approximate value:
\[ x_1=\sqrt{p+\frac{1}{4}}+\frac{1}{2}, \]
then the second approximate value of \(\dfrac{v_1}{v}\) will be
\[ x_2+\sqrt{p+\frac{1}{4}}+\frac{1}{2} -\frac{f(x_1)}{f'(x_1)} \]
The third approximate value will be:
\[ x_3=x_2-\frac{f(x_2)}{f'(x_2)} \]
and so on, until we obtain quantities differing very little from one another.
The ratios between the volumes of the cylinders are determined by the formulas:
\[ \frac{v_0}{v}=\frac{P}{P_0} \quad\text{and}\quad \frac{v}{v_2}=1-\left(1-\frac{P_0}{p}\right)\frac{v}{v_1} \]
We give a table of the values of \(\dfrac{v_0}{v}\), \(\dfrac{v_2}{v}\), and \(\dfrac{v_1}{v}=\eta_9 p\), for \(P_0=2\) atmospheres.
| \(p.\) | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | atm. |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| \(\dfrac{v_0}{v}\) | 2.5 | 3.0 | 3.5 | 4.5 | 5.0 | 5.0 | 5.5 | 6.0 | 6.5 | 7.0 | 7.5 | 8.0 | 8.5 | 9.0 | 9.5 | 10.0 | — |
| \(\dfrac{v_2}{v}\) | 1.52 | 1.47 | 1.47 | 1.45 | 1.43 | 1.41 | 1.39 | 1.39 | 1.39 | 1.37 | 1.35 | 1.35 | 1.33 | 1.33 | 1.33 | 1.31 | — |
| \(\dfrac{v_1}{v}\) | 1.82 | 2.06 | 2.25 | 2.60 | 2.60 | 2.73 | 2.86 | 2.99 | 3.09 | 3.18 | 3.28 | 3.37 | 3.46 | 3.53 | 3.64 | 3.70 | \(\eta_0\ p.\) |
As in the preceding case, for operation of the compressor of the system under consideration to be possible, it is necessary that \(\dfrac{v_0}{v_2} > 1\) and \(\dfrac{v}{v_1} < 1\).
In fact, we may represent \(\dfrac{v_0}{v}\) in the form of the following product:
\[ \frac{v_0}{v_2}=\frac{v_0}{v}\,\frac{v}{v_2}; \]
and since \(\dfrac{v_0}{v}=\dfrac{p}{P_0}\) and \(\dfrac{v}{v_2}=1-\left(1-\dfrac{P_0}{p}\right)\dfrac{v}{v_1}\), then
\[ \frac{v_0}{v_2}=\frac{p}{P_0}-\left(\frac{p}{P_0}-1\right)\frac{v}{v_1} \]
If \(\dfrac{v}{v_1}=1\), then also \(\dfrac{v_0}{v_2}=1\), but since \(\dfrac{v_0}{v_2}>1\), it follows that \(\dfrac{v}{v_1}<1\); thus, obviously, the two inequalities are mutually connected, one following from the other. Therefore it is sufficient to specify one of them: \(\dfrac{v}{v_1}<1\); comparing this condition with equation (40), as a result, when \(\dfrac{v}{v_1}=1\), we obtain:
\[ P_0^n=p^n(p-P_0). \]
The limiting values of \(P_0\) are determined from the equation:
\[ P_0^n+p^nP_0-p^{n+1}=0, \]
which gives the following maximum values of \(P_0\) for different values of \(p\):
| \(p\) | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | atm. |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| \(P_0\) | 4.2 | 5.2 | 6.2 | 7.0 | 7.9 | 8.8 | 9.6 | 10.6 | 11.6 | 12.6 | 13.6 | 14.6 | 15.5 | 16.5 | 17.5 | 18.5 |
Let us now consider the theoretical working diagram of the compressor of the above-described system (Fig. 18). For the first pair of cylinders, at the beginning of the stroke the driving force is insufficient; therefore the first pair of pistons does not move until a sufficient value \(P - P'\) is obtained. The second pair of pistons, however, will at this time be moving, for the driving force at the beginning considerably exceeds the resistance. After a certain interval of time the picture changes, and the first pair of cylinders in turn begins to assist the second, adding, through the air medium, to the pressure of the expanding steam the pressure of the compressed air as well.
§ 15. It remains to consider direct-acting steam compressors with double expansion of steam, in which the steam first works at full pressure and then expands successively in two cylinders. The simplest type of this kind is a three-cylinder compressor with double compression of air (Fig. 19). From the condition of equilibrium of the rod when one expanding pair is working, we obtain the following equation (downstroke):
\[ \left[\left(\frac{v_1}{v}\right)^n - P\frac{v}{v_2}\right]v - \left[\left(\frac{v_1}{v}\right)^n - P\frac{v}{v_2}\right]v_2 = 0 \qquad\ldots\ldots\ldots (43) \]
When the piston moves in the opposite direction (upward):
\[ \left[\left(\frac{v_1}{v_2}\right)^n - P\frac{v}{v_1}\right]v_1 - \left[\left(\frac{v_1}{v_2}\right)^n - P\frac{v}{v_1}\right]v_2 + v(P-p) = 0 . \qquad (44) \]
From these equations, for \(P=p\), we have:
\[ \left(\frac{v_1}{v}\right)^n = p\frac{v}{v_2} \quad \text{and} \quad \left(\frac{v_1}{v_2}\right)^n = p\frac{v}{v_1}. \]
But since:
\[ \frac{v_2}{v}=\frac{v_2}{v_1}\cdot\frac{v_1}{v}, \]
then, substituting into this expression the values \(\frac{v_2}{v_1}\) and \(\frac{v_1}{v}\), we find:
\[ \frac{v_2}{v} = p^{-\frac{1}{n}} \left(\frac{v}{v_1}\right)^{-\frac{1}{n}} \frac{v_1}{v}, \]
whence we obtain:
\[ \frac{v_2}{v}=p^{\frac{1}{n^2+n-1}} \]
For \(n=1.4\),
\[ \frac{v_2}{v}=p^{0.42}; \]
on the other hand:
\[ \left(\frac{v_1}{v}\right)^n = p\left(\frac{v_2}{v}\right)^{-1} = p^{\,1-\frac{1}{n^2+n-1}}; \]
therefore,
\[ \frac{v_1}{v} = p^{\frac{1}{n}-\frac{1}{n(n^2+n-1)}} = p^{\frac{n+1}{n^2+n-1}}; \]
Therefore the steam-utilization coefficient \(\eta_{10}\) will be:
\[ \eta_{10} = \frac{1}{p}\frac{v_1}{v} = \frac{1}{p}\,p^{\frac{n+1}{n^2+n-1}} \ldots\ldots\ldots\ldots\ldots\ldots (45) \]
for \(n=1\).
\[ \eta_{10}=\frac{1}{p}p^{\frac{2}{3}} \]
After substituting \(n=1.4\), we obtain:
\[ \eta_{10}=\frac{1}{p}p^{0.55} \ldots\ldots\ldots\ldots\ldots\ldots\ldots (46) \]
We give a table of values of \(\frac{v_2}{v_1}\), \(\frac{v_1}{v}=\eta_{10}p\), and \(P_0=\frac{p}{\eta_{10}p}\)
| \(p\) | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | atm. |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| \(\frac{v_2}{v}\) | 1.96 | 2.12 | 2.26 | 2.40 | 2.51 | 2.63 | 2.74 | 2.84 | 2.93 | 3.03 | 3.12 | 3.20 | 3.28 | 3.40 | 3.44 | 3.52 | |
| \(\frac{v_1}{v}\) | 2.42 | 2.68 | 2.91 | 3.14 | 3.35 | 3.55 | 3.74 | 3.92 | 4.11 | 4.27 | 4.44 | 4.60 | 4.75 | 4.90 | 5.05 | 5.19 | \(\eta_{10}p\) |
| \(P_0\) | 2.07 | 2.24 | 2.40 | 2.55 | 2.69 | 2.82 | 2.94 | 3.06 | 3.13 | 3.27 | 3.35 | 3.48 | 3.58 | 3.67 | 3.76 | 3.85 |
The theoretical diagrams of the compressor’s operation are presented in Fig. 20.
§ 16. If we connect two two-cylinder compressors, in which the cylinders simultaneously serve for air and for steam (§ 9), then in such a four-cylinder compressor we obtain a triple expansion of the steam; in this case the steam gradually expands from volume \(v\) to \(v_1\) and is then discharged; the air, on the contrary, is gradually compressed from the initial volume \(v_1\) to \(v\) (Fig. 21).
The condition for the compressor to be able to operate is expressed by three equations corresponding to the equilibrium positions of the pistons at the ends of the strokes:
\[ \left. \begin{aligned} \left(\frac{v_1}{v}\right)^n &= p\,\frac{v}{v_3}\\ \left(\frac{v_1}{v_3}\right)^n &= p\,\frac{v}{v_2}\\ \left(\frac{v_1}{v_2}\right)^n &= p\,\frac{v}{v_1} \end{aligned} \right\} \qquad .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ (47) \]
Solving these equations, we obtain:
\[ \left(\frac{v_1}{v}\right)^n = p\,\frac{v}{v_1}\,\frac{v_1}{v_3} = p\,\frac{v}{v_1}\,p^{\frac{1}{n}}\left(\frac{v}{v_2}\right)^{\frac{1}{n}} = \]
\[ = p\,\frac{v}{v_1}\, p^{\frac{1}{n}} \left(\frac{v}{v_1}\right)^{\frac{1}{n}} \left(\frac{v_1}{v_2}\right)^{\frac{1}{n}} = p\,\frac{v}{v_1}\, p^{\frac{1}{n}} \left(\frac{v}{v_1}\right)^{\frac{1}{n}} p^{\frac{1}{n^2}} \left(\frac{v}{v_1}\right)^{\frac{1}{n^2}}; \]
Thus, finally:
\[ \left(\frac{v_1}{v}\right)^n = p^{\frac{n^2+n+1}{n^2}} \left(\frac{v_1}{v}\right)^{\frac{n^2+n+1}{n^2}}, \]
whence
\[ \frac{v_1}{v} = p^{\frac{n^2+n+1}{n^3+n^2+n+1}} \]
Consequently, the steam-utilization coefficient of the compressor system under consideration, \(\eta_{11}\), will be:
\[ \eta_{11} = \frac{1}{p}\frac{v_1}{v} = \frac{1}{p}\, p^{\frac{n^2+n+1}{n^3+n^2+n+1}} \qquad .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ (48) \]
For \(n=1\) we obtain:
\[ \eta_{11}=\frac{1}{p}p^{0.75} \]
For \(n=1.4\) we obtain
\[ \eta_{11}=\frac{1}{p}p^{0.61} \qquad .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ (49) \]
Knowing the ratio \(\dfrac{v_1}{v}\), one can also find the relative volumes of the other cylinders, namely:
\[ \frac{v_3}{v} = \frac{p}{\left(\dfrac{v_1}{v}\right)^n} = p^{\frac{1}{n^3+n^2+n+1}} = p^{0.14}; \qquad \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots (50) \]
and then
\[ \frac{v_2}{v} = \frac{p}{\left(\dfrac{v_1}{v}\,\dfrac{v}{v_3}\right)^n} = p^{\frac{n+1}{n^3+n^2+n+1}} = p^{0.33}. \qquad \ldots \ldots \ldots (51) \]
Below we give a table of the values of \(\eta_{11}\), \(\dfrac{v_3}{v}\), \(\dfrac{v_2}{v}\), \(\dfrac{v_1}{v}=\eta_{11}p\), and \(P_0=\dfrac{p}{\eta_{11}}\) for various values of \(p\).
| 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | ||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| \(p\) | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | Atm. |
| \(\dfrac{v_3}{v}\) | 1.25 | 1.29 | 1.31 | 1.34 | 1.36 | 1.38 | 1.40 | 1.42 | 1.44 | 1.45 | 1.46 | 1.47 | 1.49 | 1.50 | 1.51 | 1.52 | |
| \(\dfrac{v_2}{v}\) | 1.72 | 1.82 | 1.92 | 2.00 | 2.08 | 2.16 | 2.23 | 2.29 | 2.35 | 2.41 | 2.47 | 2.52 | 2.57 | 2.62 | 2.67 | 2.72 | |
| \(\dfrac{v_1}{v}\) | 2.67 | 2.98 | 3.28 | 3.56 | 3.74 | 4.08 | 4.33 | 4.55 | 4.78 | 4.90 | 5.22 | 5.42 | 5.63 | 5.83 | 6.03 | 6.23 | \(\eta_{11}p\) |
| \(P_0\) | 1.87 | 2.01 | 2.13 | 2.20 | 2.41 | 2.45 | 2.54 | 2.64 | 2.72 | 2.85 | 2.87 | 2.95 | 3.02 | 3.09 | 3.15 | 3.21 | Atm. |
The diagrams of the pressures of the steam and compressed air, given in Fig. 22, are entirely analogous to the diagrams presented in Fig. 20.
§ 17. It should be noted that in the last system of compressors, with an increase in the number of steam cylinders, the coefficient of steam utilization \(\eta_k\) increases according to the law:
\[ \eta_k = p^{\frac{ \dfrac{n^{k-2}}{n^{k-1}} + \dfrac{n^{k-3}}{n^{k-2}} + \ldots + n^2+n+1 }{ n^{k-1}+n^{k-2}+\ldots+n^2+n+1 }} = \frac{\dfrac{1}{n}-1}{1-\dfrac{1}{n^k}}. \qquad \ldots (52) \]
If
\[ n=1,\ \text{then}\ \eta'_k=\frac{1}{p}\,p^{\frac{k-1}{k}}. \qquad \ldots \ldots \ldots \ldots \ldots (52) \]
where \(k\) is the number of all cylinders. It is not difficult to see that, as the number \(k\) increases, the coefficient of steam utilization \(\eta_\infty\) tends to
\(\dfrac{1}{p}\sqrt[n]{p}\). For \(p = 10\) atmospheres, we obtain the following values for \(\eta_k\) for various values of \(n\) and \(k\):
| \(n = 1.4\) | 2 | 3 | 4 | 6 | \(\infty\) |
|---|---|---|---|---|---|
| \(k\) | 2 | 3 | 4 | 6 | \(\infty\) |
| \(\eta_k\) | 0.26 | 0.35 | 0.41 | 0.45 | 0.52 |
| \(n = 1\) | 2 | 3 | 4 | 6 | \(\infty\) |
|---|---|---|---|---|---|
| \(k\) | 2 | 3 | 4 | 6 | \(\infty\) |
| \(\eta_k\) | 0.32 | 0.46 | 0.57 | 0.68 | 1.00 |
§ 18. We have considered only those systems of direct-acting steam compressors which either are already applied in engineering practice or, by their construction, are no more complicated than existing compressors. But the question is not exhausted by this, for it is possible to obtain a very large number of combinations of air and steam cylinders satisfying the necessary conditions for proper operation in compressing air. Passing directly from four-cylinder compressors to six-cylinder ones, it should be noted that the latter likewise can be divided into compressors of periodic and non-periodic action, and in them also the final compression may be performed by fresh steam or else during either the first or the secondary expansion of it. Various combinations may yield new systems. The analysis of these systems does not fall within the author’s task, since even four-cylinder direct-acting compressors have not yet become widely used in engineering practice. It may be sufficient merely to indicate that the investigation can be carried out in a manner completely analogous to the investigation of four-cylinder systems. A priori it may be said that the coefficient of steam utilization will not differ appreciably from the coefficient of steam utilization of a three-cylinder compressor with double expansion of steam.
§ 19. From all that has been said above, the following conclusions may be drawn:
1) The coefficient of steam utilization in direct-acting steam compressors decreases as the limiting air pressure which the compressor is capable of developing increases.
2) By an appropriate choice of cylinder dimensions and by applying steam expansion, the coefficient of steam utilization in direct-acting steam compressors can be increased several-fold in comparison with a simple two-cylinder direct-acting compressor.
3) The data presented in the present article may serve for a preliminary calculation of direct-acting compressors; a more detailed calculation may be made, using these data, graphically. In doing so, it will be necessary to take into account all those
circumstances which, in the present investigation, in order to obtain general results comparable with one another, were omitted.
In conclusion, we present a summary diagram for the relative steam-utilization coefficients of the various systems of direct-acting steam compressors considered by us (Fig. 23). On the abscissa axis in this diagram are plotted the pressures of the working steam \(p\), and on the ordinate axis—the quantities proportional to the products of the steam-utilization coefficients by the maximum air-compression pressure (equal to the pressure \(p\)); moreover, the relative coefficient for a simple two-cylinder compressor is denoted by the number 100, so that the comparative increase of the steam-utilization coefficient for the various other systems is expressed in percentages relative to the action of a simple two-cylinder direct-acting compressor.
To the article by S. P. Gomeliya: “Comparative Steam Utilization Coefficient in Direct-Acting Steam Compressors”
Fig. 1.
Fig. 2.
Fig. 3.
Fig. 4.
Fig. 5.
Fig. 6.
Fig. 7.
Fig. 8.
Fig. 9.
Fig. 10.
Fig. 11.
Fig. 12.
Table I.
Fig. 11.
Fig. 12.
Fig. 5.
Fig. 6.
Fig. 8.
Fig. 9.
For the article by S. P. Gomel, “Comparative Steam Utilization Coefficient in Direct-Acting Steam Compressors.”
Fig. 14.
Fig. 15.
Fig. 16.
Fig. 17.
Fig. 18.
Fig. 19.
Plate II.
Fig. 20a.
Visible labels and formulas in the figure:
- I
- II
- III
- \(l_1\)
- \(l_2\)
- \(l_3\)
- \(l\)
- \(v\)
- \(\vartheta_a=\left(\dfrac{H}{h}\right)\) [[unclear: remaining notation]]
- \(\vartheta_b=\left(\dfrac{H}{h}\right)\) [[unclear: remaining notation]]
Fig. 20b.
Visible labels and formulas in the figure:
- I
- II
- \(l_1\)
- \(l\)
- \(v\)
- \(\vartheta=\left(\dfrac{H}{h}\right)\) [[unclear: remaining notation]]
Fig. 20c.
Visible labels and formulas in the figure:
- I
- II
- III
- \(l_1\)
- \(l_2\)
- \(l_3\)
- \(l\)
- \(v\)
- \(\vartheta_a\)
- \(\vartheta_b\)
- \(\vartheta_c\)
Fig. 21.
Visible markings on the graph:
- Horizontal axis: \(5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20\) atm.
- Left vertical scale: \(0\), \(0{,}001\%\), \(0{,}002\%\), \(0{,}003\%\), \(0{,}004\%\), \(0{,}005\%\), \(0{,}006\%\)
- Right vertical scale: \(1\), \(3\), \(3'\), \(2\text{–}5'\text{–}4\), \(5\text{–}8\), \(6\), \(8'\), \(10\), \(11\)