89 Slide Valve.
As an example of the application of formula (C), we may apply it to the
same case.
We have found that the cover should be 2⅜ inches to make the steam be
cut off at three-quarters of the stroke. By formula (C) we may now verify
our calculation, and — supposing the valve to be made with a cover of 2⅜
inches — we can see whether or not it will cut the steam off at the proper
part of the stroke.
Supposing the piston to be moving downards, when the steam is cut off, the valve is below its middle position by the length of the cover 2⅜ inches, and when at its middle position, it is half the length of its stroke (5 inches) below its highest position ; hence s’ = 7⅜, or s’ = 7.375. The values of the other letters will be the same as before. We will have, therefore,
arccos((r′-s′)/r′) = arccos(-0.475) = 118½° or 241½°
Both these arcs have their cosines equal to -.475 ; so that we may take
either of them.
Also, arcsin((c + l)/r’) = arcsin(0.525) = 32°
and we have (C)
s = r*(1 - cos(118.5 - 90 - 32)) = r*(1 - cos 3½°)
Or, s = r*(1 - cos(241.5 - 90 - 32)) = r*(1 - cos 119½°)
according as we take the first or second value of arccos((r′-s′)/r′)
Now, cos. 3½° = .9981
And, cos. 119½° = -.4924
Hence, s = 30(1 - .9981)
Or, s = 30(1 + .4924)
.·. s = -0.57, or 44.76.
The second value of s is evidently the one required for our present purpose, and shows that the steam will be cut off at 44.76 inches from the
beginning of the stroke, which agrees very nearly with our former calculation, which made a cover of 2⅜ inches cut the steam off at 45 inches from
the beginning of the stroke.
With regard to the above calculation it may be remarked, that there are
an infinite number of arcs which have their cosines equal to -.475 ; but if
we were to take more of these arcs it would be found, that they would
ultimately all give one or other of the results we have already obtained, so
that it is unnecessary to take more than the first two of them.
The reason that we get two values for s, is this. For each position of
tbe valve there are two corresponding positions of the piston; the one
being its corresponding position when going downwards, and the other
when going upwards. The second value we got for s, therefore, shows us
that when the piston is moving upwards, the valve will begin to open the
port, c, when the piston is .057 inch from the top of its stroke. By increasing the cover on the steam side of the valve, any amount of expansion
may be obtained, and we might thus obviate the necessity of using expansion valves, were it not that increasing the cover on the steam side beyond
moderate limits deranges the working of the exhausting ports. A valve
with much cover on the steam side must always shut the exhausting port
considerably before the piston has reached the end of the stroke, while it
at the same time opens a passage to the condenser for the steam that it
acting expansively, and thereby entirely removes the propelling power
before the piston has completed its stroke. In the valve we have taken as
an example, the operation of this kind of valve in producing expansion is
carried too far, or at least to the utmost allowable limits. As another
example of the use of formula (C) we shall apply it to the same valve to
ascertain to what extent the objections we have just described apply in this
instance. Let us suppose that the cover on the exhausting side (c’) is ½
inch, we would then have the exhaustion below the piston cut off when the
valve is 5½ inches below its highest position.
In other words we will have s’ = 5½
Then, arccos((r′ - s′)/r′) = arccos(-0.1) = 96° or 264°
And as before, arcsin((c + l)/r’) = 32°
Hence s = 30(1 - cos. 26°) or 30(1 - cos. 142°)
.·.s = 3.03 or s = 53.64
We thus see that the exhaustion below the piston would be cut off when the piston is still about 6½ inches from the bottom of its stroke. The other value of s shows that, when the piston is going upwards, the exhausting passage for the steam below it would be opened when the piston was still 3·03 inches from the top of the stroke.
In what precedes we have spoken exclusively of the common short slide-valve, such as is represented in the figure; but the long D valve works on exactly the same principle; only in it the exhausting sides of the ports are usually reversed. The steam generally enters the cylinder from the insides of the ports, that is, from K and M, and exhausts at F and G.
The formula (C), however, will apply to the long slide without alteration, and the other two formulae (A and B) require only one sign to be changed.
For the long D valve these two formulae will stand thus:—
s’ = r*(1 - cos(90° - arcsin((c + l)/r’) + arccos((r - s)/r)) (A)
or s’ = r*versine(90° - arcsin((c+l)/r’) + versine(s/r)
s = r*(1 - cos(arccos((r’ - s’)/r’) - 90° + arcsin((c + l)/r) (B)
or s = r*versine(arcversine(s’/r’) - 90° + arcsin((c + l)/r))
The principal results of the foregoing observations may be expressed in the four following practical rules, applicable alike to short slide and long D valves.
RULE I.—To to find out how much cover must be given on the steam side in order to cut the steam off at any given part of the stroke.
From the length of the stroke of the piston, subtract the length of that part of the stoke that is to be made before the steam is cut off. Divide the remainder by the length of the stroke of the piston, and extract the square root of the quotient. Multiply the square root thus found by half the length of the stroke of the valve, and from the product take half the lead, and the remainder will be the cover required.
RULE II.—To find at what part of the stroke any given amount of cover on the steam side will cut off the steam.
Add the lap on the steam side to the lead; divide the sum by half the length of stroke of the valve. In a table of natural sines find the arc whose sine is equal to the quotient thus obtained. To this arc add 90°, and from the sum of these two arcs subtract the arc whose cosine is equal to the lap on the steam side divided by half the stroke of the valve. Find the cosine of the remaining arc, add 1 to it, and multiply the sum by half the stroke of the piston, and the product is the length of that part of the stroke that will be made by the
piston before the steam is cut off.
RULE III.—To find out how much before the end of the stroke, the exhaustion of the steam in front of the piston will be cut off.
To the lap on the steam side add the lead, and divide the sum by half the length of the stroke of the valve. Find the arc whose sine is equal to the quotient, and add 90° to it. Divide the lap on the exhausting side by half the stroke of the valve, and find the arc whose cosine is equal to the quotient. Subtract this arc from the one last obtained, and find the cosine of the remainder. Subtract this cosine from 2, and multiply the remainder by half the stroke of the piston. The product is the distance of the piston from the end of
its stroke when the exhaustion is cut off.
RULE IV.—To find out how far the piston is from the end of its stroke, when the steam that is propelling it by expansion is allowed to escape to the condenser.
To the lap on the steam side add the lead; divide the sum by half the stroke of the valve and find the arc whose sine is equal to the quotient. Find the arc whose cosine is equal to the lap on the exhausting side, divided by half the stroke of the valve. Add these two arcs together, and subtract 90°. Find the cosine of the residue, subtract it from 1, and multiply the remainder by half the stroke of the piston. The product is the distance of the piston from
the end of its stroke, when the steam that is propelling it is allowed to escape to the condenser. In using these rules all the dimensions are to be taken in inches, and the answers will be found in inches also.
From an examination of the formulae we have given on this subject, it will be perceived (supposing that there is no lead) that the part of the stroke where the steam is cut off, is determined by the proportion which the cover on the steam side bears to the length of the stroke of the valve: so that in all cases where the cover bears the same proportion to the length of the stroke of the valve, the steam will be cut off at the same part of the stroke of the piston.
In the first line, accordingly, of Table I., will be found eight different parts of the stroke of the piston designated; and directly below each, in the second line, is given the quantity of cover requisite to cause the steam to be cut off at that particular part of the stroke. The different sizes of the cover are given in the second line, in decimal parts of the length of the stroke of the valve; so that, to get the quantity of cover corresponding to any of the given degrees of expansion, it is only necessary to take the decimal in the second line, which stands under the fraction in the first, that marks that degree of expansion, and multiply that decimal by the length you intend to make the stroke of the valve. Thus suppose you have an engine in which you wish to have the steam cut off when the piston is a quarter of the length of its stroke from the end of it, look in the first line of the table, and you will find in the third column from the left, ¼. Directly under that, in the second line, you have the decimal, .250. Suppose that you think 18 inches will be a convenient length for the stroke of the valve, multiply the decimal .250 by 18, which gives 4½. Hence we learn, that with an 18-inch
