Mechanical Engineer D. V. SLEPTSII

CALCULATION AND INVESTIGATION

OF CAM VALVE STEAM DISTRIBUTION

IN THE DÖRFEL SYSTEM

Tomsk
1929

Engineer-Mechanic D. V. Spetstsi

CALCULATION AND INVESTIGATION

OF CAM VALVE STEAM DISTRIBUTION

OF THE DÖRFEL SYSTEM

Steam distribution and regulation of a steam engine are its soul. The steam-distribution system determines not only the external appearance of the engine, but also its dimensions, influencing the choice of the number of revolutions and, thereby, the cost of the engine.

The steam-distribution system is of still greater importance from the operational standpoint. Simplicity of design, a small number of parts, little wear, ease of assembly, disassembly, and setting of the steam-distribution periods, tightness and precision of the steam-controlling organs (steam consumption), controllability, noiseless operation, etc.—these are a whole series of requirements that must be imposed on every steam distribution, and which none of the existing steam-distribution systems fully satisfies.

In examining certain systems of steam-distribution mechanisms, one sometimes asks oneself whether, in designing the steam distribution, the designer did not set himself the aim of breaking the record for the number of articulated joints; thus, in the disengaging valve steam distribution of the Sulzer system (1900), which is considered the most perfect of its steam distributions, we count 18 joints. Matters are not much better in certain forced valve steam distributions, as, for example, in the forced steam distribution of the Kollmann system, in which we find 11 articulated joints.

In modern steam engines the principal types of steam-distribution mechanisms are slide-valve and valve steam distributions; among slide-valve steam distributions should be included flat slide valves, piston slide valves, and the so-called piston valves, which are sometimes combined with valve steam-distribution mechanisms.

Dwelling on valve steam-distribution mechanisms, as the most widespread at present in medium- and high-power engines, it should be noted that the forced cam steam distributions of the systems of Lentz, Proell-Schwebe, Proell, Dörfel, Stumpf, and Stein stand out for their particular simplicity. These valve steam distributions, especially the Lentz steam distribution, are so simple and easy in valve lift that they make it possible to increase the number of revolutions of the engine considerably. In vertical stationary steam engines with Lentz steam distribution, the number of revolutions reaches 200 and more.

Among cam steam distributions, the springless steam distribution of the Dörfel system occupies a special place.

The absence of a closing spring simplifies this steam distribution, despite the fact that it has 7 hinged joints, whereas the Lentz system has only 4 hinges. The absence of a spring promises simplification of the operations for regulating accuracy and noiselessness of operation. This external advantage arouses a certain interest in the steam distribution.

In the literature, the steam distributions of Lentz, Prell-Schwabe, and Stumpf have been most fully developed. Therefore, in pausing here to attempt to bring out the features of the design and operation of Dörfel’s steam distribution, using data from the literature, we can compare this steam distribution with others similar to it.

In order to identify the features of the steam distribution, the author considered it possible to examine a specific numerical example, without omitting, for the sake of completeness of the example, certain generally known details.

To determine the magnitude of valve lift at normal cut-off, to construct the curve of valve lift as a function of time, and to construct the regulator’s center curves, the author used the simplest and most obvious methods; moreover, the method of constructing the center curve—an arc of a circle—is, so far as is known, proposed here for the first time.

The example under consideration is connected with a calculation of the principal dimensions of the engine carried out by the author, given in the booklet Determination of the Principal Dimensions of Double-Expansion Steam Engines by Particular Examples, by I. N. Butakov, Prof., and D. V. Spetstsi, engineer-mechanic.

In the example, Dörfel’s steam distribution was worked out for the HP cylinder of a tandem engine (example No. 2 of the above-mentioned booklet). For this purpose it was necessary to use the following principal design data:

\(Nef = 190\) h.p.; \(n = 120\) revolutions per minute;
\(p_1 = 13\ \text{kg/cm}^2\) abs.; \(t_1 = 325^\circ C\); \(p_e = 0.2\ \text{kg/cm}^2\) abs.

HP cylinder: cut-off 17.5%;
compression 20%;
lead at admission 1%;
lead at exhaust 5%;
clearance volume 8%;
piston stroke \(S = 0.67\) m;
cylinder diameter \(D_{\mathrm{в}} = 35.2\) cm; \(O_{\mathrm{л}} = 950\ \text{cm}^2\); mean piston speed \(C_m = 2.68\) m/sec.

The cylinder type was taken from Die Dampfmaschine von Gutermuth (Zweiter Band. I Teil. 1928, pp. 58, 254. Erst “ ” p. 560).

The dimensions of the cylinder walls were determined by the formula

\[ \delta = \frac{D_{\mathrm{в}}}{40} + 1.5\ \text{cm.} = 2.38\ \text{cm.} \]

The adopted dimension is \(\delta = 2.5\) cm; in this case the circumferential stress is equal to

\[ \kappa_1 = \frac{D_{\mathrm{в}}\cdot(p_1 - 1)}{2\cdot\delta} = \frac{35.2\cdot 12}{2\cdot 2.5} = 85\ \text{kg/cm}^2 \]

(Gutermuth. I Band. 28, p. 342, \(\kappa_1 = 60\text{—}80\ \text{kg/cm}^2\))

We take the thickness of the walls of the valve box as equal to

\[ \delta_1 = 0.8\,\delta = 0.8 \cdot 2.5 = 2\ \text{cm} \]

and check it as a pipe subjected to internal pressure, by the formula

\[ r_a = r_i \cdot \sqrt{\frac{K_z + 0.4(p_1 - p_a)}{K_z - 1.3(p_1 - p_a)}} = 10.5 \cdot \sqrt{\frac{200 + 0.4 \cdot 12}{200 - 1.3 \cdot 12}} = 11.1\ \text{cm}, \]

that is, \(\delta_1' = 11.1 - 10.5 = 0.6\) cm,
where \(r_i = 10.5\) cm—the internal diameter of the valve box;
\(K_z = 200\) kg/cm² for cast iron; \(p_1 - p_a\) kg/cm²—the steam overpressure.

The design of the steam distribution for admission and exhaust for the cover-side cavity of the HP cylinder is shown in drawing I.

The size of the admission valve is determined from the condition of a sufficiently free passage of steam at all possible cutoffs in the cylinder. From the condition of continuity of steam flow we have

\[ d_c = \sqrt{\frac{4}{\Pi}\cdot \frac{O_p \cdot C_m}{W \cdot k}}. \]

Notation: \(d_c\)—clear diameter of the upper valve seat, in cm; \(O_p\)—effective piston area, in cm²; \(C_m\)—mean piston speed, in m/sec; \(W\)—velocity coefficient for steam, in m/sec; \(k\)—coefficient of contraction of the opening of the valve seat by the valve body, dimensionless.

The choice of the velocity coefficient \(W\) is made on the basis of experimental data on the mean value of this coefficient. It is difficult to find uniform indications in the literature on this question; therefore we shall use the table given below, taken from Crassmann (S.495).

Values of \(W\), in m/sec

Selection of value \(W\)	Admission	Exhaust to atmosphere	Exhaust to condenser
1. Single-cylinder engine:
Saturated steam,
cylinder and cover are not heated	35	25	25
cylinder and cover are heated	38	30	30
Superheated steam	45	35	35
2. Compound engines:		Into receiver
a) HP cylinder, saturated steam, without wall heating	35	28	—
HP cylinder, saturated steam, with wall heating	38	30	—
HP cylinder, superheated steam	45	35	—
b) LP cylinder, without wall heating	45	—	35
LP cylinder, with good heating	50	—	40

The coefficient \(k\) is selected depending on the diameter of the valve:

\[ k = 0.6\text{–}0.8, \]

smaller values of \(k\) for small valves,
larger—for large valves,

\[ d_c'=\sqrt{\frac{4}{\Pi}\cdot\frac{950\cdot 2.68}{45\cdot 0.6}}=11.2\ \text{cm}. \]

The adopted size of the inlet valve is \(d_c = 11.5\ \text{cm}\).
The required free area for the passage of steam is

\[ f=\frac{O\cdot C_m}{W}=\frac{950\cdot 2.68}{45}=56.6\ \text{cm}^2. \]

The dimensions and construction of the valve are given in Drawing 2.

Drawing 2

We check the actual value of the coefficient \(k\) and of the velocity coefficient. For the selected \(d_c = 11.5\ \text{cm}\) and the construction in Drawing 2, the free area for the passage of steam is obtained as

\[ f_e=\frac{\Pi}{4}\cdot d_c^2-\frac{\Pi}{4}\left[d_x^2-(d_x-2s)^2\right]-z\cdot a\cdot 0.4\cdot d_x-\frac{\Pi\cdot d_{\mathrm{в}}}{4}, \]

where \(z\) is the number, \(a\) and \(0.4\,d_x\) are the dimensions of the ribs connecting the hub and the valve pipe*); \(s\) is the thickness of the valve pipe.

*) The valve ribs are arranged tangentially to the circumference of the valve hub.

The valve dimension \(d_x\) is not known and must be determined from the condition of equality of the areas for the passage of steam in directions A and B, i.e.,

\[ \frac{\Pi}{4}\left(d_c^2-d_x^2\right) = \frac{\Pi}{4}\left(d_x-2s\right)^2 - z\cdot a\cdot 0.4\cdot d_x - \frac{\Pi}{4}d_B^2; \]

whence

\[ d_x=s+0.127\cdot z\cdot a+ \]

\[ +\sqrt{(s+0.127\cdot z\cdot a)^2+\frac{d_c^2}{2}-2.55\cdot s^2+\frac{d_B^2}{2}}. \]

Substituting the valve dimensions in cm and the number of ribs \(z=4\), we obtain

\[ d_x=89\ \text{mm}, \]

therefore,

\[ f_e=\frac{\Pi}{4}\left(11.5^2-8.9^2+7.9^2-4.8^2\right)-4\cdot0.5\cdot0.4\cdot8.9 =65.4\ \text{cm}^2. \]

\[ \text{Coefficient } k'=\frac{f_e}{\frac{\Pi\cdot d_c}{4}} =\frac{65.4}{104}=0.63, \]

i.e., somewhat greater than the selected \(k=0.6\).

The actual velocity coefficient is

\[ W_e=\frac{O\cdot C_m}{f_e} = \frac{950\cdot2.68}{65.4} = 39\ \text{m/sec}. \]

For small machines operating with superheated steam, reducing \(W\) to \(39\ \text{m/sec}\) will not cause a noticeable increase in the cost of the valve, but will improve the appearance of the indicator diagram.

The height of the valve is determined by its greatest lift.

\[ h=\frac{O_n\cdot C_m}{W_e\cdot2\cdot\Pi\cdot d_c} \]

\[ h=\frac{950\cdot2.68}{39\cdot2\cdot\Pi\cdot11.5} =0.9^{*)}\ \text{cm}. \]

The valve is drawn in two positions: on the seat (solid line) and in the position of greatest lift (dotted line); then the adequacy of the openings exposed for the passage of steam is checked.

The annular opening with diameter \(d_y\) must pass one half of the steam jet:

\[ \frac{1}{2}f_e \le e\cdot\Pi\cdot d_y\cdot\beta, \]

where \(\beta\) is the coefficient of contraction of the seat opening by the ribs.

Taking \(\beta=0.8\), we obtain:

\[ e\cdot\Pi\cdot d_y\cdot\beta = 1.3\cdot3.14\cdot12\cdot0.8 = 41\ \text{cm}^2 \]

\[ *)\ \text{For cam steam distributions, when designing the valve height, the valve lift must be taken into account with a margin up to }2h.\ \text{For steam distributions of other types, it is necessary to take into account the unproductively large valve lifts obtained at large fillings.} \]

The size of the valve tube is taken as 5 mm (material: cast iron); the valve ribs are of the same thickness as the tube.

Usually the thickness of the valve tube is taken within the range 5–7 mm, the thickness of the ribs 5–10 mm, and the number of ribs \(z = 3–5\).

The diameter of the valve spindle at its weakest point is determined by the formula (Heat Engineering I, 28th year, p. 560)

\[ d_s' = \sqrt{d_c} + (1–5)\ \text{mm} \]

\[ d_s' = \sqrt{115} + 1.3\ \text{mm} = 12\ \text{mm}\quad \text{(at the thread).} \]

The outside diameter of the spindle is

\[ d_s = 15\ \text{mm} \]

The number of labyrinths in the labyrinth packing for the HPC may be determined by the formula

\[ Z^{*}) = \frac{3 \cdot 10^4 \cdot (p_1 - p_a) \cdot d^4}{O_p^2 \cdot v_1}, \]

where: \(p_1\)—steam pressure in the valve chest, in kg/cm²; \(p_a\)—atmospheric pressure, in kg/cm²; \(d\)—diameter of the valve spindle, in cm; \(O_p\)—effective piston area, in cm²; \(v_1\)—specific volume of steam, in m³/kg;

\[ Z = \frac{30000 \cdot 12 \cdot 1.5^4}{950^2 \cdot 0.156} = 13. \]

If one labyrinth is taken per 1 cm of spindle length, then the length of the labyrinth packing will be 13 cm.

The holes in the seat bushing are selected constructively, so that each can pass the entire jet of steam. The admission opening E is calculated for an area of 65.4 cm² in the smallest section of the admission branch pipe.

The area of the opening G—G (drawing 2) is selected from the condition of free passage of 0.5–0.75 of the steam jet, that is,

\[ \text{area } G = (0.5–0.75) \cdot 65.4. \]

In the drawing, \(G = 40\ \text{cm}^2\) was obtained.

The remaining dimensions up to the axis of the hinge that sets in motion the spindle of the admission and exhaust valve are determined from design considerations. Thus the distances from the cylinder axis to the axes of the hinges of the admission and exhaust valves were determined.

The distance \(L\) from the cylinder axis to the axis of the distribution shaft was determined as the sum of: \(1/2\) the length of the crank pin, the thickness of the crank arm at the point of its connection with the shaft, the length of the main-bearing liner, the greatest radius of the gear meshing with the distribution shaft, and a certain allowance (the gear of the distribution shaft is located closer to the main bearing than the gear of the main shaft).

\[ L = 950\ \text{mm}. \]

\[ \text{*)} \]

The empirical formula proposed by me is derived from the expression \(G_{\text{sec}} = aF\)

\[ \sqrt{\frac{2 \cdot g(p_1 - p_n)}{z \cdot v_1}} \]

under the assumption of a certain, very small, steam consumption for leakage through the labyrinth packing. Naturally, for low pressures a minimum number of labyrinths must be established.

The rotation of the distributing shaft will take place in the direction toward the cylinder*) (in drawing 1, counterclockwise).

Steam distribution for the admission valve.

The spindle joint, the transmitting lever with rollers, and the exhaust cam were selected constructively; in drawing 3 they are shown at the moment preceding admission.

When choosing the dimensions in the plane of drawing 3, the basic point of departure was the maximum and normal valve lift, as well as the minimum dimensions of the shafts and rollers.

The maximum valve lift was determined earlier:

\[ h_{\max} = 9\ \text{mm}. \]

The normal valve lift (that is, the lift at normal filling) is determined by the same formula as the maximum, but only taking into account the mean piston speed not over the whole stroke, but over the filling time under normal operation of the engine:

\[ h_{\text{n}} = \frac{O_{\text{n}} \cdot C_{\text{n}}} {2 \cdot \Pi \cdot d_{\text{c}} \cdot W_{\text{e}}} \]

*) If the gear wheel of the main shaft is placed near the main bearing, then the overhang \(L\) will be greater.

The mean piston speed during the time of normal admission is

\[ C_{\mathrm{н}}^*)=R(1-\cos\varphi_{\mathrm{н}}):\frac{60\cdot\varphi_{\mathrm{н}}^\circ}{n\cdot360^\circ} =\frac{R(1-\cos\varphi_{\mathrm{н}})\cdot 6\cdot n}{\varphi^\circ}, \]

where \(R(1-\cos\varphi_{\mathrm{н}})\) is the path traveled by the piston from the dead point during the time of normal admission;

\[ \frac{60\cdot\varphi_{\mathrm{н}}^\circ}{n\cdot360^\circ} \]

is the time of rotation of the crank from the dead position to cutoff (through the angle \(\varphi_{\mathrm{н}}^\circ\), Drawing 4).

Taking into account the strong crumpling of the steam at the moment of cutoff and the influence of the finite length of the eccentric rod\({}^{**}\), we increase the theoretical admission (\(E=17.5\%\)) in the HPC to 19%. With such a cutoff, the angle of rotation \(\varphi_{\mathrm{н}}^\circ\) of the crank from the extreme position to cutoff will be \(\varphi_{\mathrm{н}}^\circ=47^\circ 40'\) (see Drawing 4) for the cover-end cavity of the HPC; then

\[ C_{\mathrm{н}}=\frac{R(1-\cos\varphi_{\mathrm{н}})\cdot 6\cdot n}{\varphi_{\mathrm{н}}^\circ} =\frac{0.67(1-0.67)\cdot 6\cdot 120}{2\cdot 47.67} =1.66\ \text{m/sec}. \]

The required valve lift at 19% admission will be equal to:

\[ h_{\mathrm{н}}=\frac{O\cdot C_{\mathrm{н}}}{W_e\cdot 2\cdot \Pi\cdot d_c} =\frac{950\cdot 1.66}{39\cdot 2\cdot \Pi\cdot 11.5} =0.56\ \text{cm}\simeq 6\ \text{mm}. \]

We place the supporting axis of rotation \(B\) of the transmitting lever

\[ \frac{h_{\mathrm{н}}}{2}=\frac{6}{2}=3\ \text{mm} \]

above center \(A\)—the hinged joint of the lever with the spindle—in order to avoid large displacements of the pin block in the jaws of the lever; moreover, we choose this elevation of point \(B\) according to normal admission, because at this admission the machine will chiefly operate.

We position roller \(D\) so that the curve \(DF\) of its displacements (a circle from point \(B\), Drawing 3), in its working part, is a continuation of the straight line \(ED\) and, as far as possible, coincides with it.

Rotating cam \(F\) (its axis of rotation is also denoted by \(E\)) counterclockwise so that roller \(D\) runs onto the projection

\({}^*)\) In determining \(C_{\mathrm{н}}\) in this way, the finite length of the connecting rod is not taken into account, which, in essence, plays no role for the present calculation.

\({}^{**}\) Compare p. 9, and also Drawing 10; in the latter it is clear why the finite length of the eccentric rod (the arc near the straight line \(PP\)) reduces the period of valve opening.

cam \(G\), we shall thereby move roller \(D\) away from the support of cam \(E\), while the center of roller \(D\) will move along the arc \(DE\). Together with roller \(D\), about point \(B\) the arm \(A\) and roller \(C\) will also rotate, while the end \(A\) of the arm and the center of roller \(C\) will move along arcs of radii \(BA\) and \(BC\). The magnitudes of the displacements of points \(A\) and \(C\) will be greater than the displacement of point \(D\), owing to the fact that arms \(AB\) and \(BC\) are greater than arm \(BD\). The transmission ratio to the spindle hinge \(A\) is

\[ P_A=\frac{AB}{BD}=\frac{75}{60}=1.25; \]

the transmission ratio to roller \(C\) is

\[ P_c=\frac{BC}{BD}=\frac{86}{60}=1.44 \]

Thus, a valve lift \(h_{\max}=9\) mm corresponds to a displacement of the center of roller \(D\) by \(h_{D\max}=9:1.25\) mm; a valve lift \(h_n=6\) mm corresponds to a displacement of the center of roller \(D\) by \(h_{Dn}=6:1.25=4.8\) mm.

We choose the diameter of roller \(D\) as 30 mm; of roller \(C\), 25 mm (since the latter works on the larger arm).

Since roller \(D\) governs the lifting of the valve, and roller \(C\) the seating of the valve, the profiles for the run-on and run-off of these rollers over cam \(E\) must be drawn conjugately.

The construction of the lifting profile for roller \(D\) on cam \(E\) is begun in the same way as is done for spring valve gears.

The transition curve \(GH\) is first drawn from two centers \(J\) and \(K\). Radius \(DJ\) is chosen as 5 mm (usually 5–8 mm*), and center \(J\) is located on the extension of the straight line \(ED\). Radius \(KG\) is equal to radius \(JD\), and center \(K\) is selected by conjugating the circle of radius \(JH\) with the circle of radius \(EG\). The curve of displacement of the center of roller \(D\), \(DL\), is drawn by arcs from centers \(J\) and \(K\).

The displacement of the center of roller \(D\), corresponding to the valve lift on the scale of the transmission ratio 1.25, when cam \(E\) is rotated through some angle \(\alpha\), will be found as the segment of radius \(EN\) between the circle \(DM\) and the curve \(DL\), which for profile \(HG\) is the lift curve of the center of roller \(D\) (and, on the corresponding scale, the valve-lift curve).

The displacement of the center of the roller \(en=4.8\) mm corresponds exactly to the valve lift at normal admission; to obtain such a valve lift, the angle of rotation of cam \(E\) must be equal to \(\alpha\), as indicated in drawing 3.

We choose the position of hinge \(O\)—the point of attachment of the eccentric rod to the cam. Point \(O\), under the action of the eccentric and the eccentric rod, moves along an arc of a circle of radius \(EO\), described from center \(E\) (drawing 3). The center of hinge \(O\) in drawing 3 corresponds to the moment of admission lead, if the cam is rotating counterclockwise, and to the moment of cut-off when the cam is rotating clockwise.

* Here and below all dimensions are given full size.

In view of the considerable length of the eccentric rod in comparison with the radii of the eccentrics, this length, in order to simplify the constructions, may be regarded as infinite. The direction of the eccentric rod \(OP\) with respect to cam \(E\) is determined from drawing 1 by connecting, with a straight line, the center \(O\) of the hinge with the center of the distributing shaft.

When the cam is turned through an angle \(\alpha\), the rod \(OP\) will take the position \(O'P'\), while the end of the eccentric rod \(O\), along the direction of the rod, will be displaced by the amount of the segment \(O'Q = 4.6\) mm (taken from the drawing).

Here attention should be paid to the fact that the sharper (or blunter) the angle \(EOP\), the smaller the eccentric radius required, obviously, to produce the specified valve lift. Therefore it is recommended*) that the position of point \(O\) be chosen so that, in the extreme positions at the maximum eccentric, the angle \(EOP\) be not less than \(35^\circ\) and not greater than \(145^\circ\).

Drawing 4

From the valve diagram (drawing 4) we determine the angle of rotation of the crank during the admission of steam (from admission lead—point 1—to the moment of cut-off—point II).

\[ \gamma_{\mathrm{n}} = 62.5^\circ . \]

The radius of the eccentric at normal filling is determined from the drawing:

\[ O'Q = 4.6\ \text{mm} = r_{\mathrm{n}} \cdot \left(1 - \cos \frac{62.5^\circ}{2}\right) \]

\[ r_{\mathrm{n}} = \frac{4.6}{0.15} = 30.6\ \text{mm}. \]

Further constructions depend on the choice of regulator.

) G. Dübel, Steam Engines and Steam Turbines*. Leningrad, 1925.

We shall examine two cases: 1) the center curve is a straight line (the Lentz regulator) and 2) the center curve is an arc of a circle (the Prell regulator).

In drawing 4 we construct the Müller–Reuleaux slide-valve diagram. From the center \(O_1\) we draw a circle of radius \(R_0 = 50\) mm, whose horizontal diameter will represent, to scale, the piston stroke.

Taking into account the finite length of the inlet-valve rod, we choose the lead of admission into the cover-end cavity of the high-pressure cylinder as \(2\%\) (instead of the previously specified \(1\%\)); for the same reasons the cut-off has previously been increased to \(19\%\) (instead of \(17.5\%\)). The finite length of the eccentric rod in this case reduces the period of admission, because the eccentric acts in opening the valve away from itself (the eccentric rod is compressed). With respect to the exhaust eccentric, which (according to drawing 1) opens the valve toward itself (the eccentric rod is stretched), the opposite phenomenon is obtained, that is, the period of exhaust, owing to the finite length of the rod, is somewhat increased. Since a slight increase in exhaust lead and a reduction of compression are not harmful to us, we leave the previously specified quantities: \(V_a = 5\%\) and \(C = 19\%\).

The quantities \(V_e = 2\%\), \(E = 19\%\), \(V_a = 5\%\), and \(C = 19\%\) are laid off along the piston stroke—\(M_\lambda M_\rho\) in drawing 4.

From the pole \(O_b\), located from \(O_1\) to the right by \(1/2 \psi R_0\) (the construction is made for the cover-end cavity of the high-pressure cylinder), we describe a circle of radius \(r_n = 30.6\) mm, and also an auxiliary circle \(R_b = 50\) mm. We divide the admission angle \(I\,O_b\,II\) by the line \(O_b m_{\mathrm{nor}}\) in half, and through point 5—the intersection of the straight line \(O_b m_{\mathrm{nor}}\) with the circle of radius \(r_n\)—we draw the perpendicular \(7\!-\!8\) to the position of the crank \(O_b I\).

The straight line \(7\!-\!8\) (the center curve) is the geometric locus of the vertices of the variable eccentric (the line of displacement of the eccentric center by the regulator), which in fact effects the steam distribution.

For starting the engine from rest under load, the cut-off must be at least \(50\%\)*). The position of the crank at the moment of cut-off at \(50\%\) cut-off will be \(O_b II_{\max}\), and the bisector \(O_b m_{\max}\) of the admission angle \(I\,O_b\,II_{\max}\) will intersect the center straight line \(7\!-\!8\) at point 7, whereby the eccentric radius is determined as \(r_{\max} = 40\) mm, at which a cut-off of \(50\%\) will be effected.

Zero cut-off (there will be no admission at all) is reached at \(r_0 = O_b 8 = 26\) mm. Thus the regulator must be so designed that it can vary the magnitude of the eccentricity within the limits from 26 to 40 mm, while the center curve must be a straight line.

In drawing 4, for the three cut-offs \(E_n\), \(E_{\mathrm{per}}\), and \(E_{\max}\), we have obtained three different bisectors—the directions of the eccentric rod—and their angles of advance \(\delta_n\), \(\delta_{\mathrm{per}}\), and \(\delta_{\max}\). In drawing 3a the mean direction of the rod \(m_\rho m_\lambda\) (the continuation of \(OP\) or \(O'P'\)) remains constant**), and since the lines of the extreme positions of the piston and eccentricity at

*) In choosing the cut-offs \(E_{\mathrm{per}}\) and \(E_{\max}\) the finite length of the rod was not taken into account.

**) In reality this is far from so, since the eccentric rod changes its direction not only owing to its finite length, but also owing to the change in the position of point \(O\) (drawing 4). The latter circumstance has no effect on the moments of inlet lead and cut-off.

at different eccentricities they are located at different angles; then (in drawing 3a), for the three fillings indicated above, the lines of the extreme positions of the piston \(M_{\mathrm{nor}}O_B\), \(M_{\mathrm{per}}O_B\), and \(M_{\max}O_B\) are plotted. The center line \(7\text{--}8\) will be situated perpendicular to the line of the extreme eccentric positions at a distance \(r_o = 26\) mm from the mean line.

In drawing 3a five circles are plotted: \(r_o\), \(r_n\), \(r_{\mathrm{per}}\), \(r_{\max}\), and the auxiliary \(R_B = 50\) mm.

The center straight line and the points 5, 6, 7, and 8 on it determine the position of the end of the rod at the moment of admission advance (the beginning of valve opening), when the other end of the rod will be at point \(O\) (Fig. 3).

The length of the rod \(e = 962\) mm is determined as the distance between points \(O\) and 8 (drawings 3 and 3a) according to drawing 1.

With this, in essence, the construction of the slide-valve diagram is completed.

Let us return to the construction of the profile of cam \(E\). The extreme positions (at the greatest eccentric radius \(r_{\max} = 40\) mm) of the hinge \(O\) of cam \(E\) (drawing 3) are marked at the points \(O_{m\lambda}\) and \(O_{mp}\), while the cam itself at the greatest opening of the valve is shown by a dashed line, and at the greatest deviation to the opposite side—by a dash-dotted line.

Having obtained the angle of deflection \(\alpha_{\max}\) of cam \(E\) at the greatest eccentric radius, we extend the straight line \(JK\) to its intersection at point \(K'\) with the straight line \(EL'\), and now describe the upper transition part of the profile \(HG'\) from the center \(K'\). By this we increase the valve lift at maximum filling as compared with the calculated value, but at the same time we observe the condition that, even at maximum filling, roller \(D\) should not pass beyond the outer outline of the cam. The lift curve of the roller will now be not \(DeL\), as we obtained previously, but \(DeL'\), and the point \(e\) will change its position only very slightly.

The position of roller \(C\) and of its center at the moment of admission advance is determined in the following manner.

Cam \(E\) is set in the admission-advance position, and the center of roller \(D\) is connected with the center \(E\) of the cam. After the profile curve \(HG'\) has been drawn, and the section \(G_m\) has been marked by a dashed arc from center \(E\), we turn the cam to the extreme position \(O_{m\lambda}\) (dash-dotted line). The length of arm \(BC\) is chosen so that in this position of the cam roller \(C\) does not run onto the profile \(HG'\). Having determined arm \(BC\) and selected the diameter of roller \(C\), we find the position of the center of the latter at the moment of admission advance on the arc \(RS\), so that the circumference of the roller is tangent to the dashed circle \(G_m\), when cam \(E\) is again set at the moment of admission advance.

The center \(C\) of the supporting roller, when the transmission lever is turned about \(B\), moves along the arc \(RS\).

When working roller \(D\) begins to run onto the projection \(HG'\) of cam \(E\), the supporting roller \(C\) must approach the center of cam \(E\), passing onto the profile \(mp\). When cam \(E\) is turned through the angle \(\alpha\) to the left, the center of working roller \(D\) will move away from center \(E\) by \(en = 4.8\) mm; the center \(C\) of the supporting roller will approach the center \(E\) of the cam by

\[ yz = en \cdot \frac{86}{60} = 4.8 \cdot \frac{86}{60} = 6.88\ \text{mm}. \]

Assuming further that the cam is fixed, while roller C rolls over the cam, we find that the center of roller C, when the cam is turned through the angle \(\alpha\), would fall at point \(y\) if this center approached center E along a straight line; but since center C moves along the arc RS, as a result it falls not at \(y\), but at \(x\). The segment of the arc \(yx\) is equal to the segment of the arc of the circle drawn from center E through point \(y\) (and \(x\)), between the straight line CE and the arc RS.

Turning the cam through various angles up to the full lift of the valve and finding the corresponding displacements of center C, as the displacement \(yz\) was found, we obtain the curve Cyt; and by shifting the points of this curve by the corresponding displacements of the center of roller C along the arc RS from the straight line CE, we draw the curve Cxt—the displacement of center C.

The curve \(mp\) of the cam profile for the supporting roller C is drawn as the envelope of a series of circles of the diameter of the supporting roller C, the centers of which lie on the curve Cxt.

The envelope curve \(mp\) drawn by us forms an angle at point \(m\) with the circle Gm. The portion of the cam profile at point \(m\) obtained by us is not suitable for us, since at the instant of admission lead roller C will not bear against the cam, and at the instant of cutoff, which will be delayed, roller D will not bear against the cam. To draw a suitable profile at point \(m\), we proceed as follows. At point \(m\) we erect a perpendicular to the envelope curve \(pm\), and on this perpendicular, at point \(u\), we select the center of a circle passing through point \(m\) and tangent to the circumference of roller C.

Thus the portion \(mq\) of the profile of cam E is drawn from center \(u\) by the arc of a circle, and the portion of the curve CV—the lowering and lifting of the center of roller C within the limits of the angle CuV—is also described by an arc of a circle (of radius Cu). The continuation of the circle VC to \(V'\) shows that after the instant of cutoff (which occurs when the center of the roller is at C), when the cam is turned to the right the center of the roller continues to move away from cam E, that is, lever ABC turns, lowering the spindle after cutoff has already occurred (if the valve is without lap, then after it has seated), while roller D approaches the center of cam E.

The roller passes onto the outer surface of the cam at point \(q\). Points \(q\) and \(V'\) lie on the straight line Eu. The outer outline of the cam is drawn by the arc of the circle \(qG'\) of radius \(E_q\); in doing so, the center of this arc is chosen so that it merges smoothly at point \(G'\) with the left-hand profile \(G'H\).

When roller C rolls along the arc \(qG'\) to its extreme left position, the spindle must lower somewhat further, loading the safety spring (roller C will rise), and the total lowering of the spindle from the instant of cutoff to the extreme left position of roller C (the cam at \(O_{\mathrm{pr}}\)) must amount to no more than 3 mm.

If the lifting of the roller along the arc \(qG'\) does not occur, this is due to the fact that the position of the center C of the roller on the arc RS at the instant of admission lead was taken too high; it must be taken somewhat lower.

Having thus finally established the right-hand profile of cam E, we proceed to correcting the left-hand profile. By the method described earlier for obtaining the curve Cxt, but in reverse order, along the arc of the circle \(VCV'\) we obtain the modified curve \(ZDV''\) of the lift of roller D, and the portion of the profile \(Z'HW\) is obtained as the envelope of circles.

of roller diameter \(D\), the centers of which lie on the curve \(ZDV''\). The section of the profile \(WW'\) is described by an arc from center \(E'\), with such a calculation that toward \(W'\) roller \(D\) gradually moves away from the cam. The construction of the remaining outlines of the cam is clear from the drawing.

The curve \(V''DZeL'\) of the roller lift shows that the detachment of the valve is already produced by the moving spindle, that is, with impact. This circumstance is a defect of the given steam distribution, and it cannot be avoided, but can only be mitigated by a successful selection of profiles.

The transmitting lever \(ABDC\), in its part between the rollers and the cam, must have such a recess as not to touch the projections of the cam when the rollers approach the center of the latter.

Valve-lift curves.

We construct the valve-lift curves for three cutoffs \(E_{\mathrm{n}}, E_{\mathrm{per}}\), and \(E_{\max}\).

In drawings 5, 6, and 7 the valve-lift curves are constructed on the developed arcs

\[ \frac{2\Pi \cdot r_{\mathrm{n}} \cdot \gamma_{\mathrm{n}}^{\circ}}{360^{\circ}},\quad \frac{2\Pi \cdot r_{\mathrm{per}} \cdot \gamma_{\mathrm{per}}^{\circ}}{360^{\circ}} \quad \text{and} \quad \frac{2\Pi \cdot r_{\max} \cdot \gamma_{\max}^{\circ}}{360^{\circ}}, \]

taken from drawing 3a. Each of these arcs is divided into parts so that the time of rotation of the crank through a small angle \(d\gamma\) amounts to \(dt \cong 0.01\) seconds.

The number of divisions, consequently, is determined as follows:

\[ Z_{\mathrm{n}}= \frac{60\cdot \gamma_{\mathrm{n}}^{\circ}}{n\cdot 360\cdot 0.01} = \frac{62.5^{\circ}}{120\cdot 6\cdot 0.01} = 8.7 \sim 8. \]

\[ Z_{\mathrm{p}}= \frac{\gamma_{\mathrm{per}}^{\circ}}{6\cdot n\cdot 0.01} = \frac{74^{\circ}}{6\cdot 120\cdot 0.01} = 10.3 \sim 10. \]

\[ Z_{\mathrm{m}}= \frac{\gamma_{\max}^{\circ}}{6\cdot n\cdot 0.01} = \frac{99.4^{\circ}}{6\cdot 120\cdot 0.01} = 13.8 \sim 14. \]

Rounding to the nearest even numbers, we obtain the time of rotation of the crank per one division:

\[ dt_{\mathrm{n}}= \frac{60\cdot \gamma_{\mathrm{n}}^{\circ}}{n\cdot 360\cdot Z_{\mathrm{n}}} = \frac{62.5^{\circ}}{6\cdot 120\cdot 8} = 0.01085\ \text{sec} \]

\[ dt_{\mathrm{p}}= \frac{74}{6\cdot 120\cdot 10} = 0.0103\ \text{sec} \]

\[ dt_{\mathrm{m}}= \frac{99.4}{6\cdot 120\cdot 14} = 0.0099\ \text{sec} \]

For constructing the valve-lift curves in drawings 5, 6, and 7, the scales of abscissas adopted are respectively:

\[ d\gamma_{\mathrm{n}} \cong dt_{\mathrm{n}} = 0.01085\ \text{sec} = 6.3\ \text{mm}, \]

\[ d\gamma_{\mathrm{p}} \cong dt_{\mathrm{p}} = 0.0103\ \text{sec} = 6.3\ \text{mm}, \]

\[ d\gamma_{\mathrm{m}} \cong dt_{\mathrm{m}} = 0.0099\ \text{sec} = 6.3\ \text{mm}. \]

The construction was carried out in the following manner.

For normal cutoff, at the moment of admission lead, the center of the eccentric will be at point 5 (drawing 3a, zero division), and the end of the eccentric rod connected to cam \(E\) will be at point \(O\).

The lift of the valve for such a position of the eccentric and the cam is equal to zero; this value is plotted in drawing 5 above point O. When the center of the eccentric, rotating in the direction of the arrow, reaches the 4th division (on the circumference of radius \(O_E 5 = r_n\)), the hinge O will fall at \(O'\), and the cam will turn through the angle \(\alpha\). With the new position of the cam, the center of roller D will move away from center E by the amount \(en = 4.8\) mm, and the valve will rise by

\[ h_n = 4.8 \cdot 1.25 = 6 \text{ mm}, \]

where

\[ 1.25 = \frac{\mathrm{AB}}{\mathrm{BD}} \]

is the transmission ratio of the lever.

The valve lift \(h_n\) in drawing 5 is plotted above division 4 at the scale \(1\ \text{mm} = 2.52\ \text{mm}\).

Drawing 5.

For the remaining positions of the eccentric, the distances of the division points are found along the perpendicular from the center line \(7\,7''\); by the same amount the center O (drawing 3) also moves away from the line OQ, parallel to the center line \(7\,7''\); for example, O will shift to \(O_{mp}\) when the center of the eccentric from point 7, at maximum cut-off (drawing 3a), falls into division 7 on the same circumference.

Drawing 6.

In drawing 8 the valve-lift curves are drawn as a function of the position of the piston along its stroke. The construction is shown for maximum cut-off. In drawing 4 the circumference of radius \(r_{\max}\) is divided into 14 parts, as in drawings 3a and 7. The division points are projected radially onto the circumference of diameter \(M_\lambda M_\rho\) at the divisions marked with a sign. Dropping the points marked with the sign vertically onto the diameter \(M_\lambda M_\rho\), we obtain the corresponding positions of the piston along its stroke, marked with a double sign. Above these piston positions, transferred to drawing 8, we lay off (in accordance with the numerical designations) the corresponding valve lifts taken from drawing 7. We proceed in exactly the same way for the other cut-offs.

Curve of steam throttling.

The piston speed at a given moment is determined as

\[ c=\frac{dx}{dt}= \frac{d\left[R\cdot(1-\cos\Theta)\pm 0.5\cdot\mu\cdot R\cdot\sin\Theta\right]}{dt}, \]

hence

\[ c=v\cdot(\sin\Theta\pm 0.5\cdot\mu\cdot\sin2\Theta), \]

where

\[ v=\frac{2\Pi R_n}{60} \]

is the speed of the crank pin. Substituting the value of \(v\), we obtain

\[ c=\frac{\Pi R_n}{30}(\sin\Theta\pm 0.5\cdot\mu\cdot\sin2\Theta). \]

Fig. 7.

If it is assumed that steam crushing begins at the moment when its speed reaches \(W=80\ \text{m/sec}\), then the valve lift corresponding to this speed is determined by

\[ h_{80}=\frac{O_n\cdot c}{2\cdot\Pi\cdot d_c\cdot W}. \]

Assigning different values to \(\Theta\), we obtain for \(c\) the following table \(\left(\text{for } \mu=\frac{R}{L}=\frac{1}{5}\right)\).

\(\Theta^\circ=\)	15	30	45	60	75	90	105	120	135	150	165	180
\(\sin\Theta\pm0.5\cdot\mu\cdot\sin2\Theta\)	0.31	0.59	0.81	0.95	1.02	1	0.92	0.78	0.61	0.42	0.21	0
\(c=\dfrac{\Pi R_n}{30}(\ldots)\)	1.3	2.47	3.4	4.0	4.27	4.2	3.85	3.28	2.56	1.75	0.88	0
\(h=\dfrac{O_n\cdot c}{2\cdot\Pi\cdot d_c\cdot W}\)	0.21	0.40	0.56	0.65	0.70	0.69	0.63	0.54	0.42	0.29	0.14	0

The braking curve, also on the same scale (1 mm = 2.52 mm), as well as the valve-lift curves, is constructed in Drawing 8, so as not to clutter Drawing 4.

The portion of the valve-lift curve lying within the compression curve, measured by piston travel in fractions of cutoff, determines the degree of perfection of the steam distribution. In fractions of cutoff this portion must under no circumstances exceed 0.5. At the selected velocity of 80 m/sec the valve-lift curve at normal cutoff does not satisfy this condition; but since we took \(W = 80\) m/sec with a margin, and could have taken \(W = 100\) m/sec as the limiting value for superheated steam (a small portion of this curve is plotted in Drawing 8), we consider that this condition too has been fulfilled.

Dynamics of the Steam-Distribution Mechanism

The diagrams of valve lift as a function of time, \(h = f(t)\), make it possible easily, at any given instant, to determine the valve velocity

\[ v = \frac{dh}{dt} \]

and the valve acceleration

\[ p = \frac{dv}{dt} = \frac{d^2 h}{dt^2}. \]

In Drawings 5, 6, and 7 the derivatives have been found graphically; for the given cutoff all three curves—lift, velocity, and acceleration—are plotted on one drawing.

The method of constructing the spindle velocity curve is as follows. As our element of time we previously selected one division \(dt\). In order not to draw tangents to the valve-lift curve at every point (Drawing 5), we select on the time axis \(0\)—\(8\) the pole \(p_1\), from which along the axis \(0\)—\(8\) we lay off the element of time \(dt\) (in Drawing 5 this element, for constructing the velocity curve, is enlarged 2.5 times; hence, of course, \(dh\) is increased by the same factor).

Parallel to the tangent to the valve-lift curve at point \(2h\), we draw a straight line through the pole \(p_1\) to its intersection at point \(b\) with the vertical \(ab\).

The segment of the vertical \(ab\) represents the increment of travel over the element \(dt\) (both are enlarged 2½ times). Dropping point \(b\) horizontally to the vertical through the second division (and through point \(2h\)), we find point \(2v\) of the velocity curve. The scale of the velocity curve is determined as follows.

At point \(2v\) the velocity is

\[ v = \frac{dh}{dt} = \frac{2.5 \cdot dh}{2.5 \cdot dt} = \frac{2.5 \cdot 0.00165}{2.5 \cdot 0.0108} = 0.153\ \text{m/sec}, \]

since in Drawing 5 the segment \(2.5 \cdot dh = 10.4\) mm \(= 2.5 \cdot 4.15\); and, introducing the scale of the valve-lift curve and converting mm into meters, we obtain

\[ 2.5\,dh = 2.5 \cdot \frac{4.15}{2.52 \cdot 1000} = 2.5 \cdot 0.00165. \]

Thus, the scale for velocities (Fig. 5) is: 1 mm of ordinate

\[ = \frac{0.153}{10.4} = 0.0147\ \text{m/sec}. \]

To construct the acceleration curve, the pole \(q_1\) is placed on the right-hand side of the curves in Fig. 5. The polar distance is taken equal to the time element on the scale \(dt = 0.0108\) sec. \(= 6.3\) mm.

Parallel to the tangent to the velocity curve, we draw a straight line through the pole \(q_1\), and the segment \(dc = 6.4\) mm will be taken as the ordinate of the acceleration curve at the moment of valve separation. For this moment,

\[ p = \frac{dv}{dt} = \frac{6.4 \cdot 0.0147}{0.0108} = 8.75\ \text{m/sec}^2. \]

Scale of the acceleration curve in Fig. 5:

\[ 1\ \text{mm of ordinate} = \frac{8.75}{6.4} = 1.37\ \text{m/sec}. \]

Signs: positive quantities are located above the horizontal \(0—8\), negative ones below the horizontal. Thus, with the always positive lift of the valve, its velocity in the first half is positive, and in the second half negative; the acceleration at the beginning is positive, in the middle negative (deceleration up to the middle and negative acceleration beyond the middle), and at the end positive (deceleration of the negative velocity).

The method of constructing the remaining points of the curves, as well as the velocity and acceleration curves in Figs. 6 and 7, is the same as that described, and the construction is indicated in the drawings.

Scales:

1 mm of ordinate on the drawing gives

Drawing	Valve lift, mm	Valve velocity, m/sec	Valve acceleration, m/sec²
5	0.4	0.0147	1.37
6	0.4	0.0192	1.85
7	0.4	0.02	2.00

The valve accelerations found for the three cutoffs make it possible to determine the inertial pressures of the masses of the valve and of the moving parts connected with it. In spring valve gears this is done chiefly for the calculation of the springs; here, however, the determination of these forces will allow us to refine the calculation of the cam mechanism.

Considering Figs. 5, 6, and 7, we see that the most unfavorable case of acceleration action (and, consequently, of inertial pressures) will occur at a cutoff of 50%. The calculation of the machine is carried out on the assumption of guaranteeing a 25% overload of its normal power. Thus, even under a 25% overload, the machine must still maintain the calculated number of revolutions, and the time element \(dt\) is determined precisely for this number of revolutions.

A further increase of overload (above 25%) entails a reduction in the number of revolutions. Filling of 50% is achieved only when starting from rest, that is, at a small number of revolutions.

From what has been said it follows that the time element $dt$, even at 25% overload, retains its value; with a further increase in filling, however, the time element begins to increase, that is, in fact, owing to the reduction in the number of revolutions, the power of the engine no longer increases. Therefore the calculation of the valve gear should be carried out for the greatest load at which the engine still maintains its normal number of revolutions, and such a load is operation with 25% overload, that is, the case of drawing 6.

The weight of the valve, spindle, nuts, connecting hinge with its roller, and guide piston is 3.35 kg.

The weight of the part of the transmitting lever situated on the right side of the point of rotation B is 0.954 kg, and the distance of the center of gravity of this part from the point of rotation B is 5.4 cm.

The weight of the left part of the transmitting lever is 1.15 kg; the distance of the center of gravity of this part from the point of rotation B is 3.5 cm.

The reduced weight of the right part of the transmitting lever is:

\[ \frac{0.954 \cdot 5.4}{7.5} = 0.687\ \text{kg}. \]

The same for the left part:

\[ \frac{1.15 \cdot 3.5}{7.5} = 0.538\ \text{kg}. \]

The reduced weight of the masses acting along the valve axis, consequently, is:

\[ 3.35 + 0.687 = 4.575 \cong 4.6\ \text{kg}. \]

The mass reduced to the valve axis is

\[ M_k = \frac{P}{g} = \frac{4.6}{9.81} = 0.47\ \frac{\text{kg}(\text{sec}^2)}{\text{m}}. \]

The mass thus determined now makes it possible to establish the magnitudes of the forces acting along the valve axis and, consequently, to carry out the calculation of all parts.

Forces acting along the valve axis.

A) At the moment of valve breakaway:

1) Impact.

2) Steam pressure on the unbalanced part of the valve (see drawing 2)

\[ \frac{\Pi}{4} \cdot (d_h^2 - d_b^2) \cdot (p_1 - p_c) = \frac{3.14}{4}\,(12.1^2 - 10.8^2)\,(13 - 7.5)^*) = 130\ \text{kg}. \]

3) Weight of the valve and the moving parts connected with it

\[ 3.35\ \text{kg}. \]

4) Weight of the unbalanced part of the lever

\[ 0.149\ \text{kg}. \]

*) $p_c = 7.5\ \text{kg/cm}^2$ at normal load. At small loads and zero filling, the value of $p_c$ falls to $3\ \text{kg/cm}^2$.

5) Friction in the labyrinth and guides

\[ F = 3.35 - 0.149 = 3.2 \text{ kgf maximum }*). \]

6) The inertial pressure of the moving masses is equal to: the acceleration, with opposite sign, of the moving masses, multiplied by the mass of the moving parts.

7) Steam pressure on the valve spindle

\[ P_{\text{sh}}=\frac{\Pi \cdot d_{\text{sh}}^{2}}{4}\cdot(p_1-1) =\frac{\Pi \cdot 2.25}{4}\cdot 12=21.2\text{ kg}. \]

B) After separation:

The same forces, but without pressure on the unbalanced part of the valve (impact only at the moment the spindle head comes into contact with the valve at the instant of admission lead).

The construction of the diagram of acting forces for the case of overload has been carried out in Drawing 9 on the force scale: 1 mm of the drawing = 1.54 kg. The curve of variation of the inertial pressures in this drawing is plotted from the acceleration curve in Drawing 6, after multiplying the ordinates of the latter by

\[ M_k=0.47\,\frac{\text{kg}(\text{sec}^2)}{\text{m}}, \]

and is laid off on the force scale.

Drawing 9.

The ordinates of the left half of the shaded diagram (Drawing 9) give the resultant forces on the valve spindle during the lift of the latter (after separation of the valve from the seat); the ordinates of the right half give the resultant forces during lowering of the valve. The construction was made as follows: above the zero-force line, for valve lift, \(P_{\text{sh}}\) was laid off—the steam pressure on the spindle, acting in this case always upward; from this pressure were subtracted the weight of the valve and the friction in the labyrinth and guides, since these forces act downward during lifting; from the resultant \(r_{\text{p}} r_{\text{p}}\) of these forces were laid off, taking into account action upward or downward, the inertial pressures obtained from the diagram of Drawing 6.

For lowering of the valve the whole difference from lifting consists in the fact that weight and friction, as acting in opposite directions, cancel each other.

From the diagram of acting forces it is seen that all the resultant forces acting on the spindle are directed upward both during lifting and during lowering of the valve. This diagram emphasizes the significance of the right-hand profile, especially for high \(P_{\text{sh}}\).

*) For labyrinth packing one may take the friction force as equal to the weight of the valve with the moving parts.

At the moment the valve is torn from its seat, an additional impact occurs between the stationary valve and the moving spindle. The duration of the impact may be taken as

\[ \frac{1}{100}\,dt = 0.01 \cdot 0.0103 = 0.0001 \text{ sec.} \]

Taking part in the impact are, on the one hand, the mass of the valve

\[ M_1=\frac{0.992}{9.81}, \]

and, on the other, the spindle with the reduced mass of the lever

\[ M_2=\frac{3.68}{9.81}. \]

The loss of energy in the impact, considering it inelastic, is determined from:

\[ E=\frac{1}{2}\cdot \frac{M_1\cdot M_2}{M_1+M_2}\cdot (V_2-V_1)^2, \]

where \(V_1=0\) and \(V_2=0.115\) m/sec (according to drawing 6),

\[ E=\frac{1}{2}\cdot \frac{0.922\cdot 3.68}{9.81\cdot (0.922+3.68)}\cdot 0.115^2 \approx 0.0005 \text{ kg m}, \]

whence

\[ \frac{E}{0.0001}=\frac{0.0005}{0.0001}=P\cdot v=P\cdot 0.115=5\,\frac{\text{kg m}}{\text{sec}}, \]

Consequently, the force acting at the moment of impact may be taken as approximately equal to

\[ P=\frac{5}{0.115}=43.5 \text{ kgf.} \]

On the unbalanced part of the valve the pressure amounts to 130 kgf, and when the receiver pressure is reduced (under underload) this pressure may reach the figure of 325 kgf; consequently, the force on the spindle is equal to \(325+43.0=368\) kgf.

Through block \(A\) (drawing 3) this force is transmitted in the amount

\[ 368\cdot \frac{75}{58}=475 \text{ kgf} \]

to the roller and shaft \(D\), and to shaft \(E\).

The direction of the tearing-off force \(P\) and its arms are found from drawing 3, as the normal to curve \(HG^1\) from roller \(D\) to cam \(E\). This pressure creates a moment that must be overcome by the eccentric rod when opening the valve, working in longitudinal bending.

The deviation of the force transmitted by roller \(D\) from the center of roller \(E\) is 27 mm, and, consequently, the following force will be transmitted to the eccentric rod:

\[ P=\frac{475\cdot 27}{58}=221 \text{ kgf.} \]

After the valve is torn off, the forces acting on the valve spindle change their sign; moreover, until the valve closes, they act upward, pressing roller \(C\) against the profile \(mp\) of cam \(E\).

The greatest upward-directed force on the spindle, according to the diagram, is equal to (drawing 9)

\[ 26.5\ \text{kg}, \]

and on roller E at point m there is transmitted

\[ P_m = 26.5 \cdot \frac{75}{54} = 36\ \text{kg}. \]

(since in the present case the reaction is not directed along the tangent to the circumference RS, but along the normal to mp at the point of contact, near m.

The eccentric rod is stretched by the force

\[ P_{\mathrm{e}} = 36 \cdot \frac{60}{58} \simeq 36\ \text{kg}. \]

At the moment when the force \(P_{\mathrm{e}} = 36\ \text{kg}\) acts, the eccentric block and the eccentric rod are at an angle of \(15^\circ\) to one another.

Normal to the block, the eccentric will be pressed by the force

\[ Q_{\mathrm{n}} = 36 \cdot \sin 15^\circ = 36 \cdot 0.259 = 9.4\ \text{kg}. \]

Parallel to the block, the eccentric will be shifted by the force

\[ Q_{\mathrm{p}} = 36 \cdot \cos 15^\circ - Q_{\mathrm{n}} \cdot \mu_{\mathrm{pr}} = 36 \cdot 0.966 - 9.4 \cdot 0.06 = 33\ \text{kg}. \]

This force will act on the governor finger, tending to increase the radius of the eccentric, and consequently the admission, that is, to move the cut-off point away. This force is the greatest force acting over a full revolution of the distributing shaft, since during the lifting of the valve from its seat the eccentric rod is almost perpendicular to the block, and therefore gives no component along the block.

We calculate the eccentric rod for longitudinal bending by Euler’s formula:

\[ 8P = \Pi^2 \cdot \frac{E \cdot J}{l^2} \qquad d = \sqrt[4]{\frac{8 \cdot P \cdot l^2 \cdot 64}{\Pi^3 E}}. \]

\[ d = \sqrt[4]{\frac{8 \cdot 221 \cdot 96^2 \cdot 64}{31.2200000}} = 2.1\ \text{cm}. \qquad \text{Adopted } d = 2\ \text{cm}. \]

The greatest force acting on the roller makes it possible to determine the length of the latter, using Hertz’s formula*)

\[ P = c \cdot l \cdot d, \]

where \(c\) is an empirical constant; \(l\) is the length of the roller, in cm; \(d\) is the diameter of the roller, in cm. We choose \(c = 60\), then

\[ 475 = 60 \cdot l \cdot 3,\quad \text{and } l = 2.5\ \text{cm}. \]

Smaller dimensions were chosen because the case of zero admission was taken, that is, the most unfavorable one.

Design and installation of the eccentrics.

The diameter of the steam-distributing shaft is taken as

\[ d = 60\ \text{mm}. \]

The dimensions of the block were chosen so that, at the maximum radius of the eccentric, the axis of the eccentric rod would pass from the edge of the block no closer than one third of its length.

*) See Hütte, vol. I.

The smallest dimension of the cam at the opening must not be less than 1 cm. The length of the cutout in the eccentric for the cam is determined by the magnitude of the eccentric’s travel along the cam when the radius of the eccentric changes from the minimum to the maximum and the regulator finger \(P\) passes into its extreme positions.

The size of the cutout determines the diameter of the eccentric so that, at the most weakened place, the net section of the metal of the eccentric is not less than 1.5 cm.

The procedure is the same both for drawing and for installation and marking out. The piston is set to the moment of admission lead into the cover-end cavity (for our case, and in drawing such a setting of the piston is understood). The cam is set to the moment at which the valve begins to lift (drawing 1) (the beginning of the roller’s running onto the curve \(HG\), drawing 3). Point \(O\) of the cam is connected with the center of the distributing shaft.

At a distance \(r_{\min}\) from the center of the distributing shaft, toward*) the cam, a perpendicular is drawn to the direction of the eccentric rod \(OT\). This perpendicular is the locus of the vertices of the eccentric radii—the centerline. With radius \(r_n\) from the center of the distributing shaft we describe a circle, which intersects the centerline at two points: the first point \(V_e\)—the beginning of the small arc in the direction of rotation—corresponds to the moment of admission lead; the second, \(II\), to cutoff.

The length of the eccentric rod is determined as the shortest distance between point \(O\) (at the moment of running on) and the centerline.

Position of the exhaust eccentric.

The dimension of the exhaust eccentric is taken as 32 mm.

Point \(B\)) of the exhaust cam is connected with the center of the distributing shaft, which gives the direction \(m_b m_b\) of the eccentric rod of the exhaust valve (leader on drawing I). About the mean line \(m_b m_b\) (leader, drawing I) we lay off the exhaust angle \(III\,O_b\)*) taken from drawing 4.

The obtained points \(III\) and \(IV\) correspond to the positions of the center of the exhaust eccentric at the moments of exhaust lead (\(III\)) and the beginning of compression (\(IV\)).

The length of the eccentric rod of the exhaust eccentric is determined as the distance from points \(III\) or \(IV\) to point \(B\) of the cam. The position of the exhaust eccentric at the moment chosen by us, when the piston stands at admission lead and the admission eccentric has correspondingly already been positioned, is determined if, from position \(O_b IV\), the exhaust eccentric is turned in the direction of rotation through the angle \(IV\,O_b\,I\) (compression period from drawing 4) to point \(V_e\) (leader, drawing I). The position of the exhaust cam shown in drawing I is obtained by making an intersection with the length of the eccentric rod from the position of the exhaust eccentric at point \(V_e\) (leader of drawing I).

*) Since the lifting of the valve occurs when the eccentric rod moves from the distributing shaft toward the cam.

**) Point \(B\)—the position of the hinged connection of the eccentric rod with the cam at the moment of either the beginning or the end of exhaust.

***) In the direction of motion of the eccentric rod when the exhaust valve opens.

The profile of the exhaust cam is drawn in the same manner as that of the admission cam, with the same condition: that, when the valve is closed, the latter is pressed against its seat by the safety spring. The position of the exhaust eccentric is fixed on the distributing shaft by a key.

Case in which the central curve is an arc of a circle

(Pröll regulator).

For this steam distribution, the magnitude of the eccentric radius at normal cut-off remains the same as determined above, that is \(r_{\mathrm{н}} = 30.6\) mm.

It is best to select the central curve according to Drawing 10, in which the radii of the eccentrics are shown to scale. The construction of Drawing 10 was carried out as follows.

Through the center of the distributing shaft \(O_{\mathrm{б}}\) (Drawing 10) we draw the direction of the eccentric rod \(TT\), perpendicular to which, at a distance \(r_{\mathrm{н}} - 4.6 = 30.6 - 4.6 = 26\) mm from the center of the distributing shaft toward the cam, we draw the straight line \(PP\). This straight line replaces an arc of a circle whose radius is equal to the length of the eccentric rod; the length of this rod was determined above. Here, as above, to simplify the constructions we take the length of the eccentric rod as infinite. The arc drawn beside the straight line \(PP\), with a radius equal to the length of the eccentric rod, shows what difference is obtained thereby.

At the moment of admission lead, the eccentric rod will take the position \(T'T'\), and only in the case where we place the center of the main (wedged) eccentric on the straight line \(T'T'\), the force acting along the rod at the moment of admission lead will not be transmitted to the regulator finger, since this force will be entirely absorbed by the key of the main eccentric. Thus, the distance between the positions of the eccentric rods \(TT\) and \(T'T'\) determines ...

the smallest possible radius of the main eccentric; the final choice of the size of this eccentric depends on the choice of the size of the movable eccentric.

Considering a series of center curves \(CC\) of various radii (with the center on the straight line \(T'T'\)), we come to the conclusion that, from the standpoint of design (small eccentrics), the least friction losses, and also the smallest (reverse) action on the governor at the moment of admission lead and cutoff, the radius of the movable eccentric should be chosen as small as possible; but, on the other hand, too great a reduction in the radius of the shifting eccentric would lead to a great reduction, and even to the complete elimination, of the admission lead at cutoffs differing from the normal one, that is, at cutoffs smaller and larger than the normal. Large radii of the shifting eccentric give a much smaller change in the admission lead, but increase the reverse action on the governor and increase the work of the governor (the required working capacity), since the angle of rotation of the resultant eccentric changes insignificantly.

Taking the above circumstances into account, the choice of the size of the shifting eccentric may be made by the following graphical method. In drawing 10, the position of the crank \(O_{\mathrm{B}} I\) corresponds to the moment of admission lead at normal cutoff, and the line \(O_{\mathrm{B}}M\) to the line of the extreme positions of the piston when the eccentrics are set for normal cutoff. These positions have been transferred from drawing 11. We impose the condition that, at the greatest cutoff, the admission lead be not less than \(0.5\%\), and that the cutoff be \(50\%\).

On the diagram of drawing 11 this gives the crank positions \(O_{\mathrm{B}} I'\) and \(O_{\mathrm{B}} II''\). Joining the points \(I''\) and \(II''\) by a straight line and drawing through the point \(O_{\mathrm{B}}\) a perpendicular to the straight line \(I''II''\), we obtain the direction of the rod \(O_{\mathrm{B}}T''\) and the angle \(\delta_{\max}\) under the conditions specified by us.

We lay off the angle \(\delta_{\max}\), in the direction of rotation, from the perpendicular to the line \(TT\) in drawing 10, and obtain \(O_{\mathrm{B}}M''\)—the line of the extreme piston positions for setting the eccentrics to the maximum cutoff.

If now, against the direction of rotation in drawing 10, from the straight line \(O_{\mathrm{B}}M''\), we lay off the angle \(M''O_{\mathrm{B}}I''_{\max}\) (taken from drawing 11, the angle \(M_{\lambda}O_{\mathrm{B}}I'\)), then at the intersection of the line \(O_{\mathrm{B}}I''_{\max}\) with the straight line \(PP\) we obtain the point \(V''_{e}\)—the moment of admission lead—and the resultant eccentric radius \(O_{\mathrm{B}}V''_{e}\), which will produce a cutoff of \(50\%\) with an admission lead of \(0.5\%\). If the admission lead remained unchanged for us at the maximum cutoff, as at the normal one, then with respect to the line of the extreme piston positions \(O_{\mathrm{B}}M''\) it should begin at the crank position \(O_{\mathrm{B}}I'_{\mathrm{norm}}\). Setting off, with radius \(O_{\mathrm{B}}V''_{e}\), on the latter straight line, the point \(a\), we obtain the position of the end of the resultant eccentric radius at point \(a\) at the moment when, in essence, it should be on the straight line \(PP\), so that the admission lead would remain unchanged (about \(2\%\)). Thus we have found the second point of the center circle (the first is \(V_e\)), whose center must lie on the straight line \(T'T'\). At the intersection of the perpendicular to the middle

By the straight line \(V_e a\), with the position of the rod \(T'T\), we find the point \(J\), which determines the dimensions \(r_o\)—the radius of the main eccentric—and \(r_p\)—the radius of the reversing eccentric.

The further construction of the diagrams is clear from drawings 10 and 11. The setting of the exhaust eccentric is carried out by the method described above, as in the case of the central straight line.

It is also evident that for maximum admission we are dealing with a smoother course of the valve-lift curve with a central curve than with a central straight line. For normal admission, the valve-lift curves are identical for both central curves. The crushing curve for maximum admission with a central curve will run lower than with a central straight line (under the same assumption of the normal number of revolutions).

There will be no reverse action on the governor with a central curve only at the moment of preadmission for normal admission; at other admissions and at the moment of preadmission this action will take place. At the moment of cut-off, for all admissions, this action will be in the direction of increasing the admission.

The two examples of central curves for a flat governor that have been analyzed may be regarded as sufficient for characterizing the given steam distribution. Consideration of other types of central curves will add nothing new, since, in essence, the difference from other steam distributions of a similar type will lie only in the direction of the action of the forces along the eccentric rod. It should only be noted that in this steam distribution the forces acting along the eccentric rod are smaller than in other steam distributions.

The principal advantage of the steam distribution under consideration is the absence of a spring; but this absence is also its great shortcoming, since instead of the spring there is added a roller, an extra profile, the complexity of profiling and manufacturing the cam, and, finally, the very design of the steam distribution already admits the possibility of impact. This last circumstance must contribute to accelerated derangement of the steam distribution and to its noisy operation.

Nevertheless, despite the shortcomings indicated, this steam distribution, by virtue of its simplicity, may be considered one of the best cam valve steam distributions after the Lentz steam distribution.

In preparing the article the author used the following literature.

I. N. Butakov, Prof., and D. V. Spetsshi, mech. eng., Calculation of the Basic Dimensions of Compound Steam Engines by Particular Examples, Tomsk, 1927.

Dubbel, Prof., Steam Engines and Steam Turbines. Leningrad, 1925.

A. Pohlhausen, Dipl.-Ing., Die Kolbendampfmaschinen, Leipzig, 1925.

Gutermuth und Watzinger, Die Dampfmaschinen, 1928.

C. Leist, Prof., Die Steuerungen der Dampfmaschinen, Berlin, 1905.

R. Grassmann, Anleitung zur Berechnung einer Dampfmaschine, Berlin, 1924.

Section along \(AB\)

Fig. 1

Section along \(EF\)

Section along \(CD\)

Center line of the eccentric rod.