Page:A Treatise on the Steam Engine (1847).djvu/101

88 Slide Valve. respective cranks C V and C D; or, in other words, we shall consider these rods to be always parallel to the lines C P and C p’. This, of course, can never be exactly true; but in almost every case the error thus introduced is small in amount, or at least of such a kind as to produce no error in the results It may be satisfactory to the reader that we should make a few remarks, to show how it is that the errors thus introduced into our investigations are of no practical importance. First, with regard to the shortness of the connecting-rod, the error introduced by it arises from the circumstance that the crank is not horizontal at half stroke of the piston. Thus suppose C O in the figure to be the crank, and O P the connecting rod. Suppose P’ and P’’ to be the extremities of the stroke of the point P. P’ D and P’’ E will both be equal to P O, and therefore, if P is the position of the cross head at half stroke, and the shorter the connecting-rod is in proportion to the crank, the more will the crank at half stroke deviate from the horizontal position. The motion of the valve, however, not being affected by the shortness of the connecting rod, will, so far as it is concerned, work in relation to the crank, exactly as if the connecting rod were infinitely long. That is, when the crank is horizontal, the valve will be in the position properly corresponding to half-stroke. This, it will readily be perceived, is of no consequence to the working of the engine. The grand point is to get the valve to work correctly at the critical moment when the piston approaches either end of the stroke, and is changing the direction of its motion. The derangement produced by a short connecting rod is of little practical importance from the same cause. The valve being placed correctly when the piston is at the end of the stroke, the only effect of the short valve connecting-rod, will be to make the valve travel a little farther one way than it does the other, or, what is the same thing, the valve will be a little more open at half-stroke, when the piston is going one way, than when it is going the other. of any practical importance.

When the crank is in any position C D’, the distance s will evidently be equal to D U; that is, to r - r cos x, and this will be the value of s in any part of the revolution; for, after the crank passes the horizontal line C D’’, the cosines of s become negative. Thus, when the crank is in the position C D’’’, s will be equal to D U’’’, or to r - r cos x, the same as before.

We have, therefore,

$$$$[equation]

or $$$$[equation]

Again, when the crank is in any position C D’, the eccentric will also have moved over an arc V V’, equal to D D’ or to x. So that the distance s’ will be equal to k f or to r’ + V’m.

That is, [equation]

But $$$$[equation]

Now the arc V’ V k may be considered as made up of three parts, viz. k V or 90°, V V’ or x, and lastly, Q V, the length of which is determined by the quantity of cover (c) and the lead (l); the arc V Q being always such that its sine multiplied by r’ may be equal c + l. We shall therefore have,

$$$$[equation]

For simplicity in the notation let us put

$$$$[equation]

Then, $$$$[equation]

And $$$$[equation]

But $$$$[equation]

.·. $$$$[equation]

Or $$$$[equation]

Or $$$$[equation]

Now the arc x is evidently equal to the arc whose versed sine to the radius r is equal to s, that is,

$$$$[equation]

$$$$[equation]

Substituting, therefore, in the last two equations, the values of x and d, we shall have

$$s'=r' \left(1- \cos \left( 90^\circ + \sin^{-1} {c+l\over r} + \cos^{-1} {r-s \over r} \right) \right) $$ (A)

Or $$$$[equation]

Again, from formula (b) we have

$$$$[equation]

$$$$[equation]

Hence, $$$$[equation]

$$$$[equation]

Substituting this value of x in formula (a) we have,

$$$$[equation]

Or $$$$[equation]

Substituting for d its value, we get,

$$s=r \left( 1-\cos \left( \cos^{-1} {r'-s' \over r'} - 90^\circ - \sin^{-1} {c+l \over r'} \right) \right) $$ (B)

Or $$$$[equation]

Either of the formulas (A) will readily enable us to find, by the aid of a table of natural sines, the value of s’, or the position of the valve for any value of s, and conversely either of the formulas (B) will give us the value of s, corresponding to any value of s’.

We may now proceed to investigate a formula that will give the amount of the cover on the steam side that is requisite to cause the steam to be cut off at any required part of the stroke.

The steam, when tbe piston is going down, will be cut off when the edge of the valve E comes to the edge of the port F during the upward motion of the valve. Now this will happen when the eccentric is in some position C V’’’, such that the sine of the arc V’’’ Q’ is equal to the cover on the steam side. But when the crank has made half a revolution, or reached its lowest position, the eccentric will also have made half a revolution, and will be in the position C V’’’’. It thus appears that, when the steam is cut off, the crank has described an arc of 180 degrees, minus the arc V’’’ V’’’’. Supposing, therefore, that C D’’’ represents the position of the crank when the steam is cut off, we will have the arc D D’’ D’’’ = arc V V’ V’’’. But the distance of the piston from the top of the cylinder, when the steam is cut off (call this distance s’’) is equal to the radius of the crank (r), plus the cosine C U’’’ of the arc D’’’ D’’’’. We have, therefore,

$$$$[equations]

Again, this arc V’’’ V’’’’ is composed of two arcs Q’ V’’’ and Q’ V’’’’. Sine Q’ V’’’ to radius r’ we have already seen is equal to the required cover on the steam side, that is, sin Q’ V’’’ = c/r’, and since Q’ V’’’’ = Q V, we have

$$$$[equation]

The lead is generally very small, and therefore these two arcs, Q’ V’’’’ and Q’ V’’’, are are very nearly equal: so that it will be sufficiently accurate for practical purposes if we make c, the required cover on the steam side, equal to

r' sin ½ arc V’’’ V’’’’ - ½ l. That is, c = sin. ½ arc V’’’ V’’’’ - ½ l (g)

By a well-known trigonometrical formula,

$$$$[equation]

Substituting in this the value of V’’’ V’’’’ from formula (f), we get

$$$$[equation]

Hence (g), $$c=r' \sqrt{\left(\frac {2r-s''}{2r}\right)}-\tfrac{1}{2} l$$ (C)

The cover that is necessary on the steam side to cause the steam to be cut off at any proposed part of the stroke, may be aacertained from this formula, by substituting for s’’, in the second member, the proposed length of stroke to be made before the steam is cut off.

The three formulas A, B, and C, which we have now investigated, will enable the engineer to solve any questions that are likely to arise respecting the working of the slide-valve, and will frequently obviate the necessity of the very common practice of making wooden models of the valve to see how it will work, or how much cover it ought to have. The formulas are easily worked, and require in general but little calculation. The only auxiliary required is a table of natural sines, which is to be found in almost every collection of trigonometrical tables.

As an example of the application of formula (C)

inches

Let the Stroke of the engine = 60

Or r = 30

The stroke of the valve = 10

Or r’ = 5

Lead = ¼

s’’ = 45

It is required to find how much cover the valve should have on the steam side to make the steam be cut off, when the piston is 45 inches from the beginning of the stroke. We have (C)

$$c = 5 \sqrt{\left(\tfrac{60 - 45}{120}\right)} - \tfrac{1}{8}$$ = 2⅜ inches, the amount of cover required to produce the given amount of expansion.