ON FLY-WHEELS. The effect of a fly-wheel to remove irregularity in the motion of an engine is proportional to its velocity, if the weight and diameter of the fly be given. Mr. Buchanan gives the following rule, from Mr. Murray, for finding the weights of fly-wheels : Multiply the number of horses' power of the engine by 2000, and divide this product by the square of the velocity of the circumference of the wheel per second; the quotient will give the weight in cwts. Ex. What must be the weight of a fly-wheel for an engine 20 horse power, 20 feet diameter, and making 20 revolutions in a minute? 20 feet x 3.1416 62.832 feet circumference; and 62-832 × 20 = 1256-64 feet the velocity per minute, Therefore, 1256-6460 20-944 feet, velocity per se Multiply 40 times the pressure on the piston in pounds by the radius of the crank in feet, and divide this product by the cube of the radius of the fly-wheel in feet, and by the number of its revolutions per minute; the result is the area of the rim of the fly in inches. Ex.-Suppose the pressure on the piston to be 6200 lbs. the radius of the crank 3 feet, and the number of revolutions per minute 20; required the area of the section of the rim of the fly of 10 feet radius. ON THE EXPANSIVE ENGINE. Mr. Watt found that by cutting off the steam before the engine had performed its whole length of stroke, a greater quantity of work could be done with the same quantity of steam. His theory of the expansive engine is Let A B C D (Fig. 11) represent a section of the cylinder, and E F the surface of its piston. Let us suppose that the steam is admitted while E F was in contact with A B, and that, as soon as it had pressed it down to the situation E F, the steam cock is shut. The steam will continue to press it down; and, as the steam expands, its pressure diminishes. We may express its pressure (exerted all the while the piston moves from the situation A B to the situation E F) by the line E F. If we suppose the elasticity of the steam proportional to its density, as is nearly the case with air, we may express the pressure on the piston, such as K L or D C, by Kl and Dc, the ordinates of a rectangular hyperbola F 1 c, of which A E and A B are the asymptotes, and A the centre. The accumulated pressure, during the motion of the piston from E F to DC, will be expressed by the area EFcDE; and the pressure, during the whole motion, by the area ABFCDA. Now, it is well known that the area E Fc D E is equal to ABFEX by the hyp. log. AD A E A D and the A B FE (1 + LAD) Thus, let the diameter of the piston be 24 inches, and the pressure of the atmosphere on a square inch be 14 lbs. ; the pressure on the piston is 6333 lbs. ; let the whole stroke be 6 feet; and let the steam be stopped when the piston has descended 18 inches, or 1.5 foot. The hyperbolic lo The length of ayinder and The for a yinder laving his promotion staser to condensation han my her, netosing tity of meam. The area of the tean jassages hoiat of the diameter if he winder equal to the area of the inter city of the piston in feet per minute, and divided by 4800. -Tredgold. The diameter of the air pump should be about two-thirds of the diameter of the cylinder, and half the length of stroke; and the larger the passages through the air bucket and the discharging flap are, the better. The quantity of water for injection should be about 23 times that required for steam, or about 26 cubic inches to each cubic foot of the contents of the stroke of the piston. Mr. Watt considered a wine pint, or 287 cubic inches, quite sufficient. There should be 62 times as much water in the boiler as is introduced at one feed.-Tredgold. Mr. Tredgold gives the following method of determining the number of horses' power : If the force of the steam in the boiler be de Then, besides the loss from uncondensed steam, there is a loss, ... First, by the force producing the motion of the ... ..·007 -016 -125 ... ⚫007 ... .063 ... •100 Fifth, by the force required to open and close the valves, raise the injection water, and the friction of the axis, ... ... ... Sixth, by the steam being cut off before the end of the stroke, ... ... ... Seventh, by the power necessary to work the nearly 400 .600 From the above it appears that there is a loss of about of the force of steam in the boiler, and therefore only 6 of that force is effective. Rule. Multiply the number of square inches in the area of the piston by the mean effective pressure on each square inch, and that product by the velocity in feet per minute ; the result will be the effective power in pounds raised one foot high per minute. To find the horses' power, divide this result by 33000. Some engineers consider that a horse can draw 200 lbs. at the rate of 220 feet per minute, over a pulley; and if this be taken as a standard, a horse will draw 44000 lbs. one foot high in a minute. The force of the steam in the boiler is generally 35 inches of mercury, the temperature of the uncondensed steam 120°, its force 3-7 inches. Hence, 35 x 6 = 21·0; and 21.03.7 17.3 inches, which is about 81⁄2 lbs. per square inch for the mean effective pressure. Ex. Given the diameter of the cylinder of a double acting steam engine 24 inches, the length of the stroke 5 feet, and the number of strokes per minute 22; find the horses' power, if the mean effective pressure on each square inch is 9 lbs. The velocity of the piston is 2 x 5 × 22 = 220 feet per minute. Then, 24 × 7854 × 9 × 220 = 895733 lbs. nearly, raised one foot high per minute. Note. We have increased the loss of power in working the air pump, as we are fully persuaded that Mr. Tredgold's rule makes the diameter of the air pump too small. Mr. Tredgold gives the following method of calculating the power of a high pressure engine : U |