Page:A Treatise on the Steam Engine (1847).djvu/100

87 therefore [equation] therefore [equation] We have stated in a former part of this work that a cubic foot of water evaporated per hour is equivalent to one horse power; therefore in this case N = 250 and t = 18.923 sq. in. As another example. Required the proper area of the safety-valve of a boiler suited to an engine of 500 horse power, when it is wished that the steam should never acquire an elastic force greater than 60 lbs. on the square inch above the atmosphere. In this case the whole elastic force of the steam is 75 lbs. ; and as 1 lb. corresponds in round numbers to 2 inches of mercury, it follows that f = 150. It will be necessary to calculate the temperature corresponding to this force. The operation, as performed by formula 8, page 41., is a4 follows : — [equations] Hence the required area -• -0653 x -20626 k 500 square inches. To show in what manner the proper opening of the safety-valve may be found when the temperature is giren, we shall calculate it for the following example: — Required the area of the safety-valve of a boiler suited for an engine of 500 horse power, when it is wished the steam should never acquire a greater temperature than 300°. In this example it will be necessary to calculate the elastic force corresponding to this temperature. The operation as performed by formula Q, [equations]

Hence the required area = 0653K '231 x50O-'0151 x 50O-7'S5 square inches. It will be perceived from these examples that the greater the elasticity and the higher the corresponding temperature the less is the area of the safety-valve. This is just as might have been expected, for then the steam can escape with increased velocity. We may repeat that the results we have arrived at are much less than those used in practice. For the sake of safety the orifices of the safety-valve are intentionally made much larger than what theory requires usually, 6 of a square inch per horse power.

We may here introduce some investigations touching the action of the slide valve, though we are not sure that they do not come prematurely.

SLIDE VALVE.

We shall first find a formula which — the dimension of the valve eccentric, &c., being given — will enable us to find the exact position of the valve, corresponding to any given part of the stroke of the piston; and then a formula to give conversely the position of the piston corresponding to any given position of the valve. We shall then find another formula to determine the cover and length of stroke of the valve that are requisite to cut the steam off at any given part of the stroke. These formulas we shall illustrate by examples, and subjoin a digest of the whole in three practical rules, with examples for the use of those who are not familiar with algebraic notation.

We shall suppose fig. 2. plate 4, to represent a steam engine cylinder, crank, and short slide-valve. A A is the cylinder, v v the valve, a a’ the steam ports, e the exhausting port. H S is the valve rod connected with the lever O S, which works on a shaft whose centre is at O. On the same shaft is another lever, O P, to which the eccentric connecting-rod is attached by a pin at P. C is the centre of the fly-wheel shaft, C D the length of the crank, and C V the radius of the eccentric. R D represents the connecting rod when the piston is at the top of the stroke. p p is the piston, and R p’ is the piston-rod, The valve, as shown in dotted lines, is at its middle position, or half stroke. When the piston is at the top of its stroke, however, the valve must be in the position shown by the sectional lines.

The space F E' which the valve overlaps the steam port, we call the cover on the steam side. J K, the overlap on the other side of the port, we call the cover on the exhausting side. It is usual to have the valve so adjusted that the steam port may be slightly open when the piston is at the top or bottom of its stroke. The space thus open, F E, we call the lead. In adjusting the length of the valve-rod P V, care must be taken that the length is such that when the eccentric is in the position C Q, at right angles to C P, the valve may be exactly in its middle position; that is, that the overlap F E' may be the same as the overlap G H at the other side. When the length of the valve-rod is thus determined, the proper position of the eccentric on the shaft may then be ascertained by placing the crank in the perpendicular position, as shown in the figures, and turning the eccentric round into a position C V, such that the valve will be pushed downwards so far as to cause the port a just to begin to be opened at F E. The eccentric may then be fixed firm to the shaft.

To facilitate the following investigations, we shall denote the various parts by different letters thus:—

r = radius of the crank.

r’ = radius of eccentric, or half the stroke of the valve.

s = distance of the piston at any moment from the top of its stroke.

s’ = distance of the valve at the corresponding time from top of its stroke.

x = arc travelled over by the crank (measuring from the perpendicular position C D) while the piston has moved over a space = s.

c = cover of the valve on the steam side.

c’ = cover of the valve on the exhausting side.

l = lead of the valve.

We are now prepared to investigate a formula to give the valve of s’ at any part of the stroke.

We shall, in what follows, suppose the valve-rod P V, and the connecting-rod, to be both indefinitely long, in comparison with the length of their