CONTENTS. Area of Ports-Lead-Point of Cut-off-Diagram of Simple

Valve-Percentages of Release and Compression-Application of the

Diagram-Divided Valves-Valve Diagram-Effect of Connecting - rod

Angularity and Equalisation of Steam Distribution-Ellipse Diagram-

Double-Ported Valve - Trick Valve - Thom's Trick Valve Valve

Diagram for Tandem Engines-Valve Spindles-Valve Pins-Balanced

Valves-Piston Valve.

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IN dealing with valve gears, the common slide valve suggests itself
as being a suitable starting point, for once the simple diagrams and
requirements relating to it are understood, the subject leads easily
and naturally to the consideration of more complex gears.

Area of Ports.-The first consideration when designing any
valve gear is the amount of steam and exhaust openings requisite
for the free passage of steam.
It has been found that the velocity

of steam entering the cylinder should not exceed 6,000 to 7,800

feet per minute; while that of the exhaust should not exceed

4,800 to 6,800 feet

per minute.

In deciding upon the values of V and V1, regard must be had to the length and shape of the ports, and to the intended pressure of steam. In long-stroke slide valve engines the lesser velocities should be taken; but in Corliss engines, where the ports are short and direct, the higher values of V and V1 may be taken without fear of wiredrawing. The frictional loss of steam in the ports is proportional to its density; hence it is necessary to consider the pressure when calculating the area of passages. An instance of the effect of this consideration may be found in a well-designed compound engine. In the high-pressure cylinder the ports may have such an area as will give a velocity of, say, 5,700 feet per minute to the steam and 5,200 to exhaust; but in the low-pressure cylinder it is not uncommon to find these velocities to be 7,200 and 6,500 respectively.

Having determined the area of the ports, the length must be fixed upon to find the width; or, if the width be settled, the length is easily found. It is to be noted that the steam port should be equal in area to the amount given by the formula for exhaust-that is, in slide valve engines-as it is evident that the exhaust must pass through the steam port. Allowance sho ld be made for any ribs which may be cast in the ports, as they contract the passages.

Lead. In fixing the amount of lead, the chief point requiring attention is the piston speed. Other things being equal, the lead increases as the piston speed. Large clearance engines require more lead than Corliss engines. A common practice is to allow in medium-stroke engines inch linear lead for every 100 feet of piston velocity. In vertical engines the lead on the top side of the cylinder is less than on the bottom, for two reasons. Firstly, because gravity is against the piston, piston-rod, crosshead, and connectingrod on the up-stroke; and secondly, because most of the wear in the joints of the link is in a downward direction. It is a common practice, when first setting vertical engines to work, to give on the top side half the lead given on the bottom end.

Point of Cut-off.-The percentage of cut-off with a common slide valve may be from to of the stroke. Greater expansion than the earlier cut-off given should be obtained by expansion valves; with a later cut-off than there is likely to be excessive back pressure. A cut-off at 75 per cent. is very convenient; the exhaust and compression then being quite satisfactory, and the travel of the valve not abnormal.

Having decided upon these points—namely, the port opening, the lead, and the percentage of cut-off-the design of the valve may be at once proceeded with.

Diagram of Simple Valve.-On a horizontal line A B (Fig. 1),

draw CD for the position of the crank when steam shall be cut-off.
From C, make CE equal to the lead, and draw EF perpendicular to
A B. Draw BG parallel to E F, making E B equal to the required
width of steam opening. Next, draw C H at right angles to CD,
cutting the line EF at J. Bisect HJF by line J K. On J K find
the centre of a circle which shall pass through C and touch the line
BG. Let L be the centre thus found. Then, in the diagram, LC or
LG equals the throw of the eccentric, or half the travel of the

Arekach and spoiling
of a good simple construction

in below and

proven.

how known? Not proven anywhere,

M

valve; LH is the amount of outside lap, and LC is the position of
the eccentric sheave relative to the position LM of the crank,
LM being parallel to AB. On LC describe the primary valve
circle CHL. This circle cuts the lap circle HK in H. Draw the
line LH R. This is the position of the crank when cut-off occurs.
The line L R should be parallel to CD; and the fulfilment of this
condition is a test for the accuracy of the diagram.

That this construction is correct is proved by the following
reasoning:-The line H C is drawn at right angles to DC; and in

the semicircle LHC, the angle CHL is a right angle (Euclid's Elements, Bk. iii., Prop. 31); LR is therefore at right angles to HC, and, consequently, parallel to DC.

Again, because the circle H K O is drawn so that the line H C is tangent, the line LR is at right angles to that tangent (Euclid, Bk. iii., Prop. 18), and therefore parallel to the given line of cut-off D C.

That the maximum port opening is equal to the given amount E B can readily be shown. The line E F is drawn at right angles to A B, and so is line BG. Therefore, E F and BG are parallel. The line EF is also tangent to the circle K HO, by construction; and, therefore, since the half travel of a slide valve is occupied by lap and port opening, the distance LG is equal to the port opening plus the lap. But it has been shown that V G is equal to the port Opening; therefore L V is the lap

The whole principle of the Zeuner diagram is directly founded on the 31st proposition of Euclid's third book; and the truth of all diagrams can be proved by reasoning similar to that given above.

Taking any position of the crank, as at LN, the condition of things is clearly shown by the diagram. The port will be open to steam an amount equal to O P, given by the distance between the lap circle and the valve circle on the line of the then position of the crank. Also, at this position, the valve is distant to the left of its central position an amount equal to LP. At crank position LO the valve is at its greatest distance to the left of its central position, the port opening being equal to QC; QC being equal to EB as arranged for. Again, at LR the valve has just closed the port; LH being its displacement to the left of its mid position. At LS the valve has arrived at its central position and travels to the right until LK is reached. It now returns and arrives at its central position at LT; the line ST being at right angles to LC. At LU the port is open ready for the commencement of the stroke.

In

Thus far only the action of the entering steam has been considered. In order to study the exhaust Fig. 2 is constructed. actual working this diagram would not be drawn separately, but it is to facilitate explanation and avoid confusion that two diagrams have been drawn in this instance.

Make LC, LH, and angle CLU, respectively, equal to LC, LH, and angle CLU in the first diagram. Produce C L to V, and on LV describe the secondary valve circle. Now, as the primary valve circle gave the port openings to steam for any crank position from T to S, so will the secondary valve circle give the port openings and valve positions for any point of the stroke from S to T, in the direction of the arrow.

Taking a case in which there is no inside lap, at crank position LS, the valve is in its central position, as previously explained; and it is evident that when it is in this position the valve is about to open for exhaust either one port or the other. LS, therefore, represents the position of the crank when release occurs. The port is being opened from S to V. Of course, the opening cannot