Investigation of Barilo Pneumatic Hammers

S. K. Konyukhov.

The invention of pneumatic hammers, which at the present time have such extensive and varied application in engineering, dates back to the seventies of the last century. It is difficult to decide to whom in fact belongs the fruitful idea of applying compressed air to such small operations as riveting, chasing, stamping, trimming, etc., since the Americans name the dentists Benister and Green as the first inventors of this kind of pneumatic tools, while the English name Macferlen Gray. One thing is nevertheless beyond doubt: the first practical application of these tools must unquestionably be ascribed to the practical Americans.

However varied in their construction pneumatic impact tools, and in particular hammers, may be—now numbering, judging by patents, literally in the hundreds—three essential parts may still be distinguished in them: 1) a distributing organ for regulating the admission and exhaust of compressed air into and out of the hammer; 2) a cylinder, usually open on one side for the passage of the shank of the working tool, such as, for example, a chisel, die, punch, etc.; 3) a piston.

The hammers may then very conveniently be divided into two groups: in one, the distribution of air is effected by the moving piston itself. This is the group of hammers without valves. In the other, a special valve is used for distributing the air, usually placed near the handle or in the handle of the tool itself. This is the group of hammers with a valve. In the present article a hammer without a valve is considered, bearing in engineering the mark $Q$ and $C$. Its arrangement is clearly seen from drawing 1 and reduces to the following: the cylinder $B$, piston $A$, handle $E$, ring $C$, nut $D$, pawl $F$, etc., are made of cast steel. In the piston the working, or striking, part has a somewhat larger dimension than the part on the opposite side, upon which the air presses. The lower part of the piston is solid, while the upper is hollow. In the latter, four windows $m$ and $n$ are arranged around the periphery, intended for the passage of air. Above the hollow part there is a short nipple for convenient removal of the piston when disassembling the hammer. The air-supply hose adjoins the nipple $X$. By pressing with the finger on the nose of the pawl, we cause it to incline in the direction of the arrow, whereby the end of the rectangular knee rests against the pin $I$; the latter presses on the small piston $K$, which bears against the spring $R$. Under the influence of this pressure the small piston $K$ moves downward and opens the channel $r$. Under these conditions the air gains access to the channel $Y$, and from there through the channel $a$ enters the chamber $H$. The principal positions of the piston $A$ are shown in drawings 4–8. Almost in the middle of the cylinder are arranged the openings $b—b_1—b$ and $c$ for the discharge of air. In addition, in the upper part of the piston there is a chamber $L$, which in particular positions of the piston is reduced to nothing.

Let the piston occupy the upper position. If at this time air is admitted into channel \(Y\), channel \(a\), and chamber \(H\), then tube \(C\) will press with its lower edge against the narrow rim of cylinder \(B\), while the piston will move downward. The air that was in chamber \(G\) will escape through openings \(b\) into the atmosphere. But as soon as openings \(b\) are closed by the thickened part of the piston, the air in \(G\) will begin to be compressed. The piston still continues to move downward, while air enters chamber \(G\) through openings \(m\) and \(n\). The blow follows after some time, Fig. 7. A view after the blow indicates what, besides the elasticity of the colliding bodies, contributes to it: the tendency of the air to expand. Then the piston comes into the position shown in Fig. 6. At this moment the admission of air ceases, and the period of expansion begins, continuing until the piston occupies the position shown in Fig. 8, i.e., when openings \(b—b\) open. Opening \(C\), located just at the rim against which ring \(C\) abuts, serves to discharge air from chamber \(L\). However, not all the air leaves this chamber; the remaining part, after the piston closes opening \(C\), is strongly compressed and in this state serves as an elastic cushion to protect the piston from striking the lower part of ring \(C\).

In the hammer one phenomenon is constantly observed: the smaller upper side of the piston is subjected to the constant pressure of air, while on the lower, larger side the air pressure acts successively in the following order: 1) when the piston moves from top to bottom, i.e., during the working stroke: \(a)\) exhaust; \(b)\) pure compression; \(c)\) compression and admission together; 2) for the return stroke from bottom to top: \(d)\) expansion and admission together; \(e)\) simple expansion; \(i)\) exhaust.

The relation between the forces acting on the piston may be expressed by the following equality

\[ \frac{\pi d^2}{4} P_r + \frac{\pi (D^2 - d^2)}{4} P_a + p - \frac{\pi D^2}{4} P_x = 0, \tag{1} \]

Here \(d\) denotes the smaller diameter of the piston.
\(D\) ” the larger diameter of the piston.
\(P_r\) ” the pressure of the compressed air entering the hammer. This pressure differs somewhat from the pressure feeding the reservoir, and the more technically perfect the whole installation, the smaller this difference.
\(P_x\)—the current variable compression of the air located in chamber \(G\).
\(P_a\)—pressure per unit of the piston.

From the written equation it follows that

\[ P_x = \frac{ \frac{\pi d^2}{4} P_r + \frac{\pi (D^2 - d^2)}{4} P_a + p }{ \pi D^2/4 }. \tag{2} \]

Since on the right-hand side all terms have a constant character, a change in \(P_x\) must always entail a disturbance of equi-

substance. Thus, if \(P_x\) exceeds the right-hand side, then the piston rises, and under the opposite relation it descends. Consequently, the motion of the piston is determined by the law of variation of \(P_x\). And so everything now reduces to determining the regular variation of \(P_x\). But, however simple such a formulation of the question may be, nevertheless a theoretical solution presents a very difficult problem. An approximate solution of the question that interests us may be found in Hugoniot’s article (Hugoniot, Journal de l’école politechnique 58, 1889, “Memoire sur la propagation du mouvement dans les corps et specialement dans les gaz parfaits”). In this article the following problem is posed: a gas is enclosed in a cylinder, one of whose bases is fixed, while the other is a piston of mass \(M\). The external pressure on the piston is not taken into account. The system is in equilibrium and is left to itself. It is required to determine the motion of the piston and of the gas. In its time this problem interested many geometers, but Lagrange solved it more successfully in a memoir published by Poisson after his death.

Approximately, the matter reduces to the following: to determine, for each very small displacement of the piston, corresponding to a very short interval of time, the velocity of the piston and the pressure \(P_x\) in the chamber \(G\). Without risking a practically large error, one may suppose that during \(0.001\) second both the pressure and the velocity remain constant. The piston is vertical in the upper position and is ready to strike a steel strip. For the sake of simplicity Baril neglects friction.

Let us consider the useful stroke of the piston from top to bottom. Here one may distinguish the three following phases.

1. Exhaust on the large side of the piston into chamber \(G\). The equation of motion will be as follows

\[ \frac{mV_1^2}{2}-\frac{mV_0^2}{2} = \left[ \frac{\pi d^2}{4}P_r + \frac{\pi(D^2-d^2)}{4}P_a + p - \frac{\pi D^2}{4}P_x \right]l \]

Here \(V_0=0\), since the motion is without initial velocity; \(V^1\) denotes the velocity of the piston after it has traversed the distance \(0.03\) m. In the hammer under investigation the dimensions of the piston diameters and areas were as follows

\[ d=25.8\ \text{mm.}\quad \frac{\pi d^2}{4}=5.23\ \text{cm}^2. \]

\[ D=34.2\ \text{mm.}\quad \frac{\pi D^2}{4}=9.51\ \text{cm}^2. \]

\[ \frac{\pi(D^2-d^2)}{4}=4.28\ \text{cm}^2. \]

\[ p=0.515\ \text{kg};\quad m=\frac{p}{g}=0.0524;\quad l=0.013\ \text{m}. \]

As for the pressure, \(P_r=5\ \text{kg}/\text{cm}^2;\quad P_a=1\ \text{kg}/\text{cm}^2.\)

\(P_x=1\) kgf/cm\(^2\). With regard to \(P_x\) it should be noted that, in taking it equal to 1 kgf/cm\(^2\), one is guided by the following consideration: during the period under observation the discharge of air from chamber \(G\) into the atmosphere predominates, and the sections \(b—b\) are so large that there will be no back pressure. Substituting the numerical data into the equation of motion, we obtain

\[ \frac{mV_1^2}{2}=0.279;\quad V_1=3.65\ \text{m per second}. \]

II phase. Compression. Pure compression lasts over a length of 9 millimeters. By the end of the period the pressure \(P_x\) becomes equal to

\[ P_x=\frac{73.22}{64.66}=1.13. \]

Here \(73.22\) cm\(^3\) represents the volume occupied by the air at the beginning of compression. This volume consists of three parts: 1) the air in chamber \(G\); 2) the air in the piston box; and 3) the air in the piston chamber or tube. Since the initial pressure \(P_x\) was equal to 1 kgf/cm\(^2\), and the final pressure to 1.13, the mean pressure will be

\[ P_x=\frac{1+1.13}{2}=1.065\ \text{kgf/cm}^2. \]

Next one may write

\[ \frac{mV_2^2}{2}-\frac{mV_1^2}{2} = \left[ \frac{\pi d^2}{4}P_r+ \frac{\pi(D^2-d^2)}{4}P_a+p- \right. \]

\[ \left. -\frac{\pi D^2}{4}P_x \right]l_2 = (30.945-951\cdot1.065)\cdot0.009=0.187. \]

But since

\[ \frac{mV_1^2}{2}=0.279, \]

then

\[ \frac{mV_2^2}{2}+0.279+0.187=0.466, \]

and

\[ V_2= \sqrt{\frac{0.466\times2}{0.0524}} = 4.22\ \text{m/sec}. \]

III phase. Compression is combined with admission. This phase extends over a section of 15 mm. According to Weisbach the air velocity is determined by the formula

\[ w= \sqrt{2g\frac{k}{k-1}}\, \sqrt{RT_0}\, \sqrt{1-\left(\frac{P_1}{P}\right)^{\frac{k-1}{k}}}. \]

For the case when \(p_1\le 0.52p_0\), this formula may be reduced to the simpler form

\[ w=\varphi\cdot 11.6\sqrt{T_0}. \]

In this simplified formula \(T_0\) denotes the absolute temperature, and \(\varphi\) is a coefficient characterizing the cross-section of the passages through which

air will flow through them. For our case it may be taken as equal to 0.41.

The cross-section of the channels \(n\) is equal to \(0.40\ \mathrm{cm}^2\). Consequently, in \(0.001\) sec. there will enter an amount of air, reduced to atmospheric pressure,

\[ Q=\frac{\omega}{1000}\cdot 0.40\cdot P_r . \]

Taking the temperature of the entering air to be \(17^\circ\mathrm{C}\), we obtain

\[ W=0.41\cdot 11.6\sqrt{290}=0.41\cdot 11.6\times 17 \simeq 81\ \mathrm{m./sec.} \]

And then

\[ Q=\frac{8100\times 0.40\times 5}{1000}=16.2\ \mathrm{cm}^3 . \]

Thus, the quantity of air contained in chamber \(G\) and measured at atmospheric pressure will be

\[ 64.66+1.13+16.2=89.27\ \mathrm{cm}^3 . \]

Assuming, as before, that during \(0.001\) sec. the piston moves with constant velocity, we easily find the distance traversed \(4.22:1000=0.422\ \mathrm{cm}\). The new volume of air will therefore be equal to

\[ 64.66-(9.51\times 0.422)=60.66\ \mathrm{cm}^3 . \]

Consequently,

\[ P_x=\frac{89.27}{60.66}=1.46\ \mathrm{kg/cm}^2, \]

and the mean compression

\[ \frac{1.46+1.13}{2}\simeq 1.30 . \]

Let us again write the equation of live forces

\[ \frac{mV_3^2}{2}-\frac{mV_2^2}{2}=\bigl[30.945-(9.45\times 1.30)\bigr]0.00422=0.078745 . \]

But, since

\[ \frac{mV_2^2}{2}=0.466,\quad \text{then}\quad \frac{mV_3^2}{2}=0.466+0.078745=0.544745 \]

\[ V_3=\sqrt{\frac{0.5447\times 2}{0.0524}}=4.55\ \mathrm{m./sec.} \]

In this way we shall find, for the following intervals of time \(b_0=c_0=0.001\) sec., the pressures \(P_x\) and the velocities \(V\)

\[ \begin{aligned} b_0&=0.001 & P_x&=1.87 & V_4&=4.84\\ c_0&=0.001 & P_x&=2.33 & V_5&=5.04\\ d_0&=0.001 & P_x&=2.50 & V_6&=5.10 \end{aligned} \]

The velocity \(V_6\) will be the velocity of the piston’s impact against the cutter. The kinetic energy corresponding to this instant is converted partly into useful work, and partly into imparting rebound to the piston. Thus this rebound absorbs part of the useful work. This is why, in air hammers intended for more or less heavy forgings, designers try to obtain an adherent blow.

Return Stroke.

In the return stroke it is also possible to distinguish three characteristic phases.

I. Expansion and admission together. The beginning of this phase is characterized by the pushing of the piston in the reverse direction, so that chamber \(G\) gradually increases. Air continuously flows into the rarefied space being formed. Under these conditions, in chamber \(G\) there will be a combination of expansion and admission. By means of experiments, Baril found that the piston, being positioned vertically with respect to the steel plate being worked and, striking the latter at a velocity of \(5.1\ \mathrm{m/sec.}\), rebounds at a velocity of \(0.82\ \mathrm{m/sec.}\)

During one thousandth of a second the piston will travel a distance of \(0.082\ \mathrm{cm.}\), and the volume of the air admitted will be \(16.10\ \mathrm{cm^3}\). In chamber \(G\), when the piston was in contact with the cutter, there was \(50.41\ \mathrm{cm^3}\) of air; and during the upward motion of the piston, over \(0.001\ \mathrm{sec.}\), there will enter \(9.52 \times 0.082 = 0.78\ \mathrm{cm^3}\).

Thus, when the piston rebounds during \(0.001\ \mathrm{sec}\), beneath the hammer there will be the volume

\[ 50.41 + 9.51 \times 0.082 = 51.19\ \mathrm{cm^3}. \]

If the former air is reduced to a volume of \(50.41\) at a pressure of \(2.5\ \mathrm{kg/cm^2}\), to atmospheric pressure, and to it is added the volume \(16.10\ \mathrm{cm^3}\), then we obtain the volume \(142.12\ \mathrm{cm^3}\). Since now this air occupies the volume \(51.19\ \mathrm{cm^3}\), the pressure \(P_x\) will be

\[ P_x = \frac{142.12}{51.19} = 2.78\ \mathrm{kg/cm^2}. \]

But at the end of the downward motion of the piston this pressure was equal to \(2.5\ \mathrm{kg/cm^2}\); consequently, the mean pressure will be

\[ \frac{2.78 + 2.50}{2} = 2.64\ \mathrm{kg/cm^2}. \]

To find the velocity at the end of \(0.001\ \mathrm{sec.}\) after the rebound, we write the equation of kinetic energies

\[ \frac{mV_1^2}{2} - \frac{mV_0^2}{2} = (9.51 \times 2.64 - 30.945)0.00082 = -0.0042. \]

Hence, substituting the values \(V_0 = 0.82\) and \(\frac{mV_0^2}{2} = 0.175\), we obtain

\[ V_1=\sqrt{\frac{(0.0175-0.0042)^2}{0.0524}}=0.71\ \text{m./sec.} \]

Table I gives the values for \(P_x\) and \(V\) at the end of the periods \(b—c—d—e\ldots\), equal to 0.001 seconds.

Table 1.

II phase. Expansion. The process of expansion takes place over a length of 9 mm. The pressure \(P\) at the end of expansion is 4.06 kg./cm². If we again write the equation of live forces, then from it the velocity \(V_{15}\) at the end of expansion is easily found. Thus,

\[ \frac{mV_{15}^{2}}{2}-\frac{mV_{14}^{2}}{2} = \left[ 9.51\left(\frac{4.60+4.06}{2}\right)-30.945 \right]0.009 = 0.0922; \]

\[ \frac{mV_{14}^{2}}{2}=0.1432;\quad \frac{mV_{15}^{2}}{2}=0.1432+0.0922=0.2354;\quad V_{15}= \]

\[ =\sqrt{\frac{2\cdot0.2354}{0.0524}}=2.99\ \text{m./sec.} \]

III phase. Discharge. We shall trace this period in exactly the same way, successively by points, taking for the time intervals periods of 0.001 sec. The discharge velocity \(\omega\) will be determined in the same way as the velocity

the intake speed, but only here \(\varphi\) is taken equal to \(0.8\); then \(\omega_1 = 158\) m/sec. The beginning of this phase occurs when the piston is in the position shown in Fig. 5. Still assuming that during \(0.001\) sec. the piston moves with constant speed, we shall find how far the channels \(b—b\) open, and consequently we shall also know how much air escapes. This quantity of air will be equal to \(4.62\ \mathrm{cm}^3\). Since the former volume of air at pressure \(P = 4.06\ \mathrm{kg}/\mathrm{cm}^2\) was \(68.6\ \mathrm{cm}^3\), and the new volume is \(76.07\ \mathrm{cm}^3\), the new pressure \(P_x\) will be equal to

\[ P_x=\frac{4.06+68.6}{76.06}=3.66\ \mathrm{kg}/\mathrm{cm}^2. \]

and the mean pressure

\[ \frac{4.06+3.66}{2}=3.86\ \mathrm{kg}/\mathrm{cm}^2. \]

Again we use the equation of live forces to determine

\[ \frac{mV_{16}^{2}}{2}-\frac{mV_{15}^{2}}{2} = \left[(3.86\times 9.51)-30.945\right]0.00299 = 0.0175. \]

Whence we find that

\[ V_{16}=\sqrt{\frac{0.2529\times 2}{0.0524}}=3.10\ \mathrm{m}/\mathrm{sec}. \]

In this way we shall successively find the values of \(P_x\) and \(V\) for the periods \(o—p—q—z\). The results are entered in Table 2.

Table 2.

Only when the piston, moving toward the handle, uncovers the openings \(c\), Fig. 4, will the air present in a small quantity between the piston tube and the wall of the latter be compressed and impede the blow. Beginning with this moment, the equation of live forces will be valid.

\[ \frac{mV_{21}^{2}}{2}-\frac{mV_{20}^{2}}{2}+\frac{m_{1}U_{1}^{2}}{2}-\frac{m_{1}U_{0}^{2}}{2}=T. \]

Here \(m_{1}\) is the mass of the tube \(=0.221:9.81=0.0225\);

\[ U_{0} \]
is the initial velocity of the tube;

\[ \ldots\ U_{1} \]
is the velocity of the same tube at the end of \(0.001\) sec. Since \(U_{0}=0\), and \(U_{1}\) may be taken equal to \(V_{21}\), the equation of live forces will take the form

\[ \frac{mV_{21}^{2}}{2}-\frac{mV_{20}^{2}}{2}+\frac{m_{1}V_{21}^{2}}{2}=T. \]

The quantity \(T\) entering into the right-hand side of this equation is determined from the consideration that the surface receiving the impulse has diameter \(D_{1}\), corresponding to the largest diameter of the body of the cylinder toward the light. Thus, \(U_{21}=1.17\) m/sec. Comparing this value of the velocity with the velocity \(V_{20}\), we are convinced that here a noticeable fall occurs. This gives us the right, without great error, to determine that additional travel when the velocity becomes equal to \(0\). To solve the problem, we again write the equation of live forces

\[ \frac{(m+m_{1})V_{22}^{2}}{2}-\frac{(m+m_{1})V_{21}^{2}}{2}=T=Fx. \]

Here \(F\) represents the resultant of the forces applied to our moving system, and \(x\) is the sought additional travel. In this equation, in accordance with what has just been said, one must put \(V_{22}\) equal to zero, and then

\[ x=-\frac{(m+m_{1})}{2F}V_{21}^{2}. \]

However, the value of the resultant is negative, and therefore the quantity \(x\) takes the value

\[ x=0.001075. \]

Consequently, from the point of departure the piston will have traveled the distance

\[ 2.52+1.075=3.595\ \text{mm}. \]

The characteristics of the second stroke of the piston are given by Tables 3 and 4.

Table 3.—2nd working stroke.

Table 4.—2nd return stroke.

Graphically these results are represented in Figs. 9–11. Here the abscissa axis shows the distances traveled, and the ordinate axis the corresponding velocities.

Thus, in Fig. 9 the curves drawn denote:

$A B C D$ — the first forward stroke;
$E F G A$ — the return stroke;
$A^1 B^1 C^1 D^1$ — the second forward stroke;
$E_1 F_1 G_1 H_1$ — the return stroke.

Finally, the third forward stroke begins at point $H$.

By means of these first curves, the mean velocities are plotted in Fig. 9 as horizontal lines. Namely:

$N N^1$ — for $A B C D$, the first forward stroke;
$L L^1$ — for $E F G A^1$, “ ” backward;
$I I^1$ — for $A^1 B^1 C^1 D^1$, the second forward stroke;
$M M^1$ — for $E^1 F^1 G^1 H^1$, “ ” backward;
$K K^1$ — for “ the third forward stroke.

The form of these curves gives the right to draw the following conclusions: the greater the stroke of the piston, the greater also the velocity at the moment of impact, the weaker the return velocity, and the shorter the path.

In Fig. 9 the starting point $A$ is chosen arbitrarily, and for this stroke the minimum velocity $O D$ was obtained. As a consequence of this, for the first stroke in the reverse direction we obtain the maximum stroke length $O A^1$. In exactly the same way the maximum value of the velocity is obtained. All the velocity curves will lie between the curves $A B C D$ and $A^1 B^1 C^1 D^1$, precisely because the first curve corresponds to the minimum, and the second to the maximum of the velocities, as a consequence of the maximum stroke length.

Since the interval between these curves is rather large, it is necessary to find a third curve for which the mean velocity will be $K K^1$.

From the preceding it is perfectly clear that all the other curves of velocity of the working stroke will be situated between the second and third curves, etc. In the limit all the curves will merge into one, coinciding with the arithmetic mean passing between $I I^1$ and $K K^1$.

Quite analogously to this, the mean velocity for the return stroke will pass between $L L^1$ and $M M^1$.

We now pass to the drawing of the compression diagram. In Fig. 10, for the working stroke, the curves of the change of pressures on the large side of the piston are drawn. Namely:

$A V B C$ — for the forward stroke (the first);
$A V B F$ — for the second stroke;
$A V B D$ — for the third stroke;
$A V B E$ — the general mean pressure.

Thus, if at the end of the stroke the velocity is minimal, then the pressure becomes greatest, and conversely.

As for the counterpressure, which acts from right to left, it is constant and therefore is represented by the horizontal straight line $R S$. The magnitude of this counterpressure is determined by the following formula:

\[ \frac{\frac{\pi d^2}{4} P_r + \frac{\pi(D^2-d^2)}{4} P_a + p}{\pi D^2/4} \]

Here

\[ \frac{\pi d^2}{4} P_r \]

is the pressure on the small area of the piston;

\[ \frac{\pi(D^2-d^2)}{4} P_a \]

is the pressure on the annular area;

\(p\) is the weight of the piston in kg.

\[ \frac{\pi D^2}{4} \]

is the large area of the piston, to the unit of surface of which the back pressure is referred.

Combining the pressure and back-pressure curves yields a new curve representing the resultant of the pressures on the piston. The curve plotted in Fig. 10, \(GTHM\), represents the mean pressure curve. For the return stroke the compression curve, or rather a series of curves, is given in Fig. 11. Thus the values of \(P_x\) are seen from the curves

\[ \begin{aligned} &A_3\ A_2\ A_1\ A &&\text{for the first stroke;}\\ &C_3\ C_2\ C_1\ C &&\text{for the second stroke;}\\ &B_3\ B_2\ B_1\ B &&\text{mean value.} \end{aligned} \]

And here the back pressure is expressed by the straight line \(R.S.\)

Here attention should be paid to the negative resultant \(OD_3\) and \(OF_3\). Their presence indicates that before the beginning of the return stroke there occurs a rebound, which is a function of the impact velocity. But we have already seen that the highest impact velocity corresponds to the beginning of the second return stroke. In subsequent strokes the action of the air replaces this rebound. For the second stroke the mean compression is less than for the first stroke, and therefore the mean velocity for the second stroke will lie between the velocity of the second and that of the first stroke.

Determination of the number of blows per minute. To find the number of blows per minute, it is necessary to know separately the duration of the forward and return strokes. Since the mean stroke of the piston is \(40.5\) mm, and the mean velocity of the forward stroke is \(3.97\) m, the duration of the forward stroke in thousandths will be equal to

\[ 0.0405 \times 1000 : 3.98 = 10.15. \]

For the return stroke the mean velocity will already be smaller, namely \(2.16\) m, and the duration of the stroke increases

\[ 0.0405 \times 1000 : 2.16 = 18.75. \]

Consequently, a complete stroke will take \(10.15 + 18.75 = 28.90\) thousandths of a second, and therefore the number of blows will be equal to

\[ n = \frac{60.1000}{28.90} \simeq 2070. \]

Determination of air consumption. The most natural way of determining the air consumption consists in counting the amount of air entering during the separate periods. Thus, for the first phase, the working stroke, when intake and compression predominate in the hammer, in the separate periods there will enter

\[ \begin{array}{c@{\ \ .\ \ .\ \ .\ \ .\ \ .\ \ }r@{\ }l} a & 16{,}10 & \text{cm}^3,\\ b & 16{,}10 & \text{”}\\ c & 15{,}20 & \text{”} \end{array} \left\} \quad \text{Together }47{,}40\ \text{cm}^3. \right. \]

On the return stroke, when in the hammer intake is combined with compression, there will enter air

\[ \begin{array}{c@{\ \ .\ \ .\ \ .\ \ .\ \ .\ \ .\ \ .\ \ .\ \ .\ \ .\ \ .\ \ .\ \ }r@{\ }l} a & 16{,}10 & \text{cm}^3,\\ b & 15{,}75 & \text{”}\\ c & 15{,}70 & \text{”}\\ d & 15{,}45 & \text{”}\\ e & 14{,}80 & \text{”}\\ f & 13{,}70 & \text{”}\\ g & 13{,}50 & \text{”}\\ h & 12{,}88 & \text{”}\\ i & 11{,}90 & \text{”}\\ j & 11{,}27 & \text{”}\\ k & 10{,}31 & \text{”}\\ l & 10{,}10 & \text{”}\\ m & 9{,}35 & \text{”}\\ n & 4{,}85 & \text{”}\\ \hline \multicolumn{1}{c}{} & \multicolumn{2}{c}{\text{Total . . . }174{,}65\ \text{cm}^3,} \end{array} \]

and for the whole stroke there will enter \(47{,}40 + 174{,}65 = 222{,}05\ \text{cm}^3\) of air.

For the second complete stroke the air consumption will be \(52{,}75\ \text{cm}^3\) (on the working stroke) and \(170{,}72\ \text{cm}^3\) (on the return stroke), and in all \(223{,}47\ \text{cm}^3\), or on the average

\[ \frac{222{,}05 + 223{,}47}{2} = \frac{445{,}52}{2} = 222{,}76\ \text{cm}^3. \]

Per minute, at 2070 complete strokes, this will amount to \(222{,}76 \times 2070 = 461113\ \text{cm}^3\), or about 461 liters, measured at atmospheric pressure. Measurements made with the aid of a counter placed on the supply hose gave 500—530 liters.

Work developed by the hammer. To compute this work one may use the usual formula

\[ L = \frac{\pi D^2}{4}\cdot p \cdot c. \]

where \(p\) represents the pressure, \(c\)—the length of the stroke, and \(\frac{\pi D^2}{4}\) the surface subjected to the pressure \(p\). Substituting the numerical data into the preceding formula, we obtain

\[ \begin{aligned} 9.51 \times 1.94 \times 0.0406 &= 0.748 \text{ kg·m},\\ 9.51 \times 0.18 \times 0.0406 &= 0.0693 \quad " \end{aligned} \]

\[ \text{Total . . } 0.8173 \text{ kg·m}. \]

per one blow. But in the course of 1 sec. the hammer can make \(2070 : 60 = 34.5\) blows, and therefore its work will be equal to

\[ 0.8173 \times 34.5 : 75 = 0.376 \text{ horsepower}. \]

To determine the efficiency coefficient of the hammer, it is necessary to know the work that must be expended in polytropic compression (the polytrope in the present case is close to the adiabat) in order to bring one liter of air at atmospheric pressure to compression at 5 kg/cm\(^2\) abs. Approximately this work is equal to 19.4 kg·m. But we have determined the minute consumption of air as 461 liters. Consequently, theoretically, in 1 sec. this quantity of air can give the work

\[ \frac{19.4 \times 461}{60 \times 75} = 1.98 \text{ hp}, \]

and since we obtained only 0.376 hp, the efficiency coefficient is equal to

\[ \eta = 0.376 : 1.98 = 0.185. \]

In well-designed pneumatic hammers with valves this coefficient sometimes reaches 0.40.

It should also be noted that if the mean rebound velocity is approximately equal to 0.90 m, then in the useful work of the hammer a part of the kinetic energy is not utilized. This loss will be equal to

\[ T = \frac{mV^2}{2} = \frac{0.0524 \times 0.90^2}{2} = 0.0424 \text{ kg·m}. \]

Thus, only

\[ 0.748 - 0.0424 = 0.706 \text{ kg·m} \]

are utilized, or, in horsepower,

\[ \frac{0.706 \times 34.5}{75} = 0.325 \text{ hp}. \]

In this way, the efficiency coefficient is

\[ \eta = 0.325 : 1.98 = 0.164. \]

Fig. 9.

Velocity diagram

Labels visible in the diagram:

• Reduced velocity (forward motion)
• Reduced velocity (backward motion)
• I release
• II compression
• III release
• II expansion
• III release
• compression
• path
• \(x\)
• \(45\,\mathrm{mm}\)

Fig. 11.

Labels visible in the diagram:

• smaller side of the strip
• Average compression (large side of the strip)
• Average compression (resultant)
• I release and expansion
• II expansion
• III release
• \(x\)
• \(45\,\mathrm{mm}\)

Stress diagram
(reverse motion)

Stress diagram
forward motion

Labels visible in the diagram:

• smaller side of the strip
• Average compression (large side of the strip)
• Average compression (resultant)
• III compression and release
• II compression
• I release
• \(x\)
• \(45\,\mathrm{mm}\)

Fig. 10.

Handle details
Fig. 14.

Fig. 1.

section along \(x x\)
Fig. 2.

along \(y y\)
Fig. 3.

Fig. 4.

Fig. 5.

Fig. 6.

Fig. 7.

Fig. 8.

Various positions of piston \(A\).

Fig. 9.

Release and expansion

III compression and discharge

Fig. 12.

Fig. 13.

Pump handle.