The strain to be resisted by the cross-sections of the rim may be formed by this Rule: Divide the sum of all the strains, ascertained as above, by the ratio of the circumference to the diameter of the wheel-that is, by 3*1416—and the quotient will be the amount of the strain that will have to be resisted by the combined strength of any two opposite cross-sections of the rim of the wheel. Taking for example the data from the previous example-Then 341000 lbs.÷31416 108,540lbs. the strains to be resisted by or in any two opposite cross-sections. A rim of this weight and diameter would probably have about 50 square inches in its cross-section area, or 100 square inches in the opposite areas. So the strain will be 1,085 lbs. on each square inch. Taking the strength of the rim at 10,000 lbs. per square inch, the factor of safety in this case will be a little over 9. Good cast-iron has a greater tensile strength than this, but in the case of heavy masses like fly-wheel rims, the metal in which has not been decarbonised at all by remelting, it is not safe to assume in any case a greater tensile strength than 10,000lbs. This wheel could be run, however, at 200 revolutions, and still on this basis would have a factor of safety of 21. The difficulty with large fly-wheels, however, is rarely or never on account of insufficient strength of their cross-section. The weakest part is generally the joints of the wheel, when made in segments. The segments are united by connections which are frequently weak relatively to the strength of the rim, and every wheel is as strong as its weakest place, and no stronger. In most cases two tons, or 4,480 pounds per square inch, is the greatest tensile stress that should be allowed on cast-iron. This rule may be explained as follows * :— A B H E In Fig. 63 let the circle A B C D represent the rim of a fly-wheel revolving about its centre E. At F the rim tends to move in the tangent F G, but it is compelled instead to follow the arc F B. It is deflected in the direction of the radius F E, and its resistance is necessarily in the opposite direction E F. The strain thus produced is, however, not in fact exerted on the line FE, because there is, by previous supposition, no metal there; but all these strains are transferred to or received in the rim itself. To understand how this is done, let the circle be bisected in any two directions at right angles with each other, as by the lines A C and B D. Now, every force, in whatever radial direction it is exerted, may be resolved into two forces at right angles with each other, and perpendicular to these two lines; and if the radius represents the amount of the force, then these two perpendiculars will represent the two rectangular components of it. Thus F L and F H are Porter in The Machinist. * Vide article on "Revolving Wheels," by Mr. C. T. D Fig. 63.-Diagram of Strains in a Fly-Wheel. thus rectangular components of the force F E, and represent the equivalent strains in these two directions. But F L is the sine of the angle A E F, and F H is equal to L E, the cosine of the same angle. And so, universally, the sines of all the angles in the circle represent the strains in the rim which are pependicular to the line A C, and the cosines represent those which are perpendicular to the line B D, and these together are equivalent to the sum of all the radial strains. The sum of each of these two components will then be less than the sum of all the centrifugal strains in the rim, as the average of all the sines, or of all the cosines, which is the same thing, is less than the radius. The average sine or cosine, 63662, has the same relation to the radius that the latter has to the quadrant of the circumference, or that I has to 15708. So, to get the sum of all the strains exerted in the two directions perpendicular to the line A C, the sum of all the radial strains must be divided by 15708, and those exerted in the two directions perpendicular to the line BD will be the same. But this gives the sum of the strains at these points of the rim in both directions, and of this one-half constitutes the resistance against which the other half is exerted. The bursting strain to be resisted at any two opposite points of the rim is therefore, as already given, the sum of all the radial strains divided by 3*1416. The fact is here assumed, without an explanation of the parallelogram of forces, that the sine and cosine of any angle represent the rectangular components of the force that is represented by the radius. The Energy of a Fly-Wheel may be found by this Rule :- Mass in pounds × (velocity in feet per second) 2 32 2 × 2 In which the number 32.2 is the intensity of gravity. Example.-Required the energy of a fly-wheel having a rim weighing 6,000 pounds, making 70 revolutions per minute supposing the momentum to be concentrated at the inside of the rim of the wheel, which is 10 feet diameter. Then 10 x 3'1416 31416 feet circumference × 70 revolutions per minute 2199 12÷60 seconds = 36.52 feet per second, 6000 lbs. x 36.522 the velocity of the mass, and = 102,488 foot - pounds, 33°2 × 2 HIGH-PRESSURE CONDENSING STEAM-ENGINES. Condenser. The object of a condenser is to remove the pressure of the atmosphere which opposes the advance of the piston in the cylinder, so that all the work performed by the steam may be brought to bear effectually upon the piston. The steam in a condenser is condensed instantaneously by the cooling medium. Jet-Condensers are usually employed for Stationary CondensingEngines. A jet condenser with horizontal air-pump and hot-well combined Fig. 64.-Sectional Elevation of Condenser and Air-Pump. in one casting is shown in sectional elevation in Fig. 64; in sectional plan in Fig. 65; and in sectional end-view in Fig. 66. The exhaust-steam from the engine enters the condensing-chamber where it is immediately con Fig. 65.-Sectional Plan of Condenser and Air-Pump. densed by a spray of water from a rose on the end of the injection-pipe as shown in Fig. 66. The condensed steam and condensing-water is pumped out of the condenser into the hot-well by the air-pump shown in Figs. 64 and 65, which is doubleacting. Part of the water from the hot-well is returned to the boiler as feed-water, and the remainder is discharged into a reservoir to be cooled. By condensing the exhaust-steam in this manner a partial vacuum is created behind the piston in the steam-cylinder and the back-pressure is diminished. The heat of the steam is imparted to the O Fig. 66.-Sectional End-View of Condenser and Air-Pump. water, and with an ample supply of injection-water, the vacuum would be practically perfect, if no air passed with the exhaust-steam to the condenser. Capacity of a Jet Condenser.-The capacity of the condenser should not be less than the capacity of the air-pump, but rather greater. A good proportion is to make the cubic contents of the condenser equal to threefourths that of the cylinder in communication with it. Diameter of injection-pipe = diameter of the cylinder divided by 7. Diameter of overflow-pipe from hot-well-diameter of cylinder x 3 Diameter of cold-water pump = diameter of the cylinder multiplied by *3, when its stroke equals the stroke of engine. = Condensing-water for Jet-condensers.-The quantity of injectionwater required per nominal horse-power of the engine, in cubic feet per minute is generally temperature of the steam in degrees Fahr. multiplied by 00304; approximately 5 gallons are required per nominal horsepower per minute. The quantity of injection-water required depends upon the temperature of the exhaust-steam and the temperature of the condensingOn an average, 2 gallons of injection-water are required per pound of steam consumed per hour by the engine. For instance, a steam-engine using 20 pounds of steam per indicated horse-power per hour requires 20 × 2=40 gallons of condensing-water per hour for each indicated horsepower developed by the engine. When the steam-consumption of an engine is not known, it may be assumed to be 23 pounds, and 23 pounds x 2 gallons = 46 gallons of injection-water should be provided per indicated horse-power per hour, which is probably the average maximum quantity of condensing-water used by stationary condensing steam-engines. The temperature of the hot-well should be 100° Fahr. Reservoirs for Cooling Condensation-Water flowing from the hotwell of condensing steam-engines should equal in capacity 130 gallons of water per indicated horse-power per hour. The area of the surface of the water should equal 75 square feet per indicated horse-power for an engine working 12 hours per day, or equal 150 square feet if working 24 hours, or day and night. Vacuum in the Condenser.-The vacuum is generally estimated by inches of mercury, each pound of vacuum representing 2 inches of mercury, therefore each inch of mercury represents a diminution of alb. in the back-pressure, and a gain of an equivalent amount of effective pressure on the piston of a steam-engine. The vacuum is never perfect in the condenser, owing to the presence of a small quantity of air and vapour, which is generally equal to a pressure of from 2 to 3 lbs. per square inch. This vapour-pressure forms a back-pressure which resists the movement of the low-pressure piston. If there were a vacuum of 24 inches in the condenser, the back-pressure would be reduced to the extent of 12 lbs., and the advance of the piston would be opposed by a back-pressure of 15-12 =3 lbs. per square inch, instead of 15 pounds per square inch, the pressure of the atmosphere. The vacuum depends upon the temperature of the condenser. A fall of vacuum in a cool condenser indicates defects in the air-pump; and in a hot condenser it indicates a defective supply of condensing-water. Imperfect Vacuum in a condenser may be caused by excessive clearance spaces in the air-pump, in which vapour will remain and expand, and other defects in the design of an air-pump: and also by contracted exhaust-passages in the cylinder. A bad vacuum may be caused by a deficient supply of cooling water, and by the presence in the condenser of air, drawn into the cylinder through imperfectly packed glands of pistonrods, valve-spindles and drain-cocks, or through defective joints, and fractures or blow-holes in castings. It may be caused by hot-water carried with the steam from the cylinder to the condenser as the result of boilerpriming. Surface-Condensers.-Steam is condensed in surface condensers by contact with the cold surfaces of a number of small metal-tubes, without its coming in contact with water. The tubes are cooled by water supplied by a circulating pump. A cooling surface of from 2 to 2 square feet is usually provided per indicated horse-power of the engine. The quantity of cooling water provided for each pound of steam to be condensed is usually from 40 to 50 lbs. Air-Pumps.—The function of an air-pump is to remove the air and condensation-water from the condenser. The power required to work an air-pump, when the plunger is connected to the tail-rod of the piston of the engine as shown in Fig. 69, is equivalent to an additional back-pressure on the piston of the steam-cylinder of from of a lb. to 1 lb. per square inch. The valves of the air-pump are either of metal, india-rubber, or vulcanised fibre. The bucket, plunger or piston, piston-rings, grids, and bolts are of gunmetal. The cubic contents, or the area × the length, of a double-acting air-pump should equal onetwelfth the cubic contents of the cylinder. The diameter of a double-acting air-pump, of the form shown in Figs. 64 and 65, when the length of stroke is equal to one-half that of the engine, may be diameter of cylinder multiplie. by 3. Fig. 67.-Vertical Single-Acting A vertical single-acting air-pump, shown in Fig. 67, is the most efficient form of air-pump. The clearance between the bucket and the foot-valve should be as small as practicable. The cubic contents of a single-acting air-pump should equal one-sixth the cubic contents of the cylinder. |