TABLE Of Change Wheels for Screw-cutting; the leading screw being of inch pitch, or containing 2 threads in an inch. It has been found by experiment that the force requisite to overcome the friction of a locomotive engine and attendant machinery, the engine being without load, is equal to a pressure of about 1.15 lbs. effective on each square inch of the cylinder's cross sectional area, or equal to 1 lb. effective for the engine, and 0.15 lb. effective for the attendant machinery; this item, however, for practical purposes, is usually taken at 1 lbs. ; by effective pressure is meant a pressure over and above that of the atmosphere, or over and above 14.7 lbs. on each square inch of surface. If we let diameter of cylinder in inches, S= stroke of piston in feet, T= revolutions per minute, p = mean effective pressure in lbs. per square inch, as shown by the indicator, a=sXrX 2 = velocity of piston in feet per minute, h = 1 lbs. effective pressure on each square inch of cylinder's cross sectional area, requisite to overcome friction, then, To find the effective power or force of a steam engine. EXAMPLE. = effective force in horse-power. -What is the effective force of a steam engine, the diameter of the cylinder (d) being 36 inches, the stroke of the piston (s) 7 feet, the effective pressure (p) 30 lbs., and making 174 revolutions (r) per minute? 362 X (30—1.5) X 7 X 17.5 × 2 ÷ 42017 power. Ans. = 215.37 horse To find the nominal power of a low-pressure or condensing engine. ·To find the nominal power of a high-pressure, or non-condensing d3 XpX/s 940 = engine. nominal force in horse-power. Or, nominal force in horse-power, the piston moving at the ordinary speed of 128 times the cube root of the stroke. To find the pressure of the steam on each square inch of the boiler's surface. pressure in lbs. on safety valve, pressure in lbs. on each square inch of surface, w = sum of the weights of the lever, safety valve, and balancing weight, in lbs., s = length of lever in inches from its axis of motion to a point vertical to the centre of the valve, = length of lever in inches from its axis of motion to W, or to the point on the lever at which the spring balance is attached, a = area of safety valve in square inches. EXAMPLE. The length of the lever from its axis of motion to a point over the centre of the safety valve (s) is 3 inches, its length from the axis of motion to the point at which the weight or spring balance is attached (7) 24 inches, the weight, or pressure, as indicated by the spring balance (W) 40 lbs., the sum of the weights of the lever, valve, &c., (w) 6 lbs., and the area of the valve 64 square inches; required the pressure of the steam per square inch. To find the volume of steam compared with the volume of water. Let F elastic force of steam in pounds per square inch, = volume of steam compared with volume of water. V = EXAMPLE. - The elastic force of the steam is 40 lbs. to each square inch of the surface of its bulk; what space does it occupy, compared with the space it would occupy if it were condensed to water? 24250 40 60665671; that is, when the elastic force of the steam is 40 pounds to each square inch of the surface of its bulk, the volume of the steam, compared with its volume in water, is as 671 to 1. NOTE.-The preceding formulas are not strictly correct for all densities of steam, but give the mean of a general range. -= The volume of steam compared with the volume of water as 1, has been found, by experiment, when the elastic force is 14.7 lbs. = 1694; 20 lbs. 1280; 25 lbs. 1044; 30 lbs. 883; 35 lbs. = 767; 40 lbs. 679; 45 lbs. 610; 50 lbs. 554; 55 lbs. 508; 60 lbs. 470; 65 lbs. 437; 70 lbs. = 408; 75 lbs. Ibs. 325; 100 lbs. 295; 150 lbs. 205; 200 lbs. 158. 383; 80 lbs. =362; 90 To find the temperature of the steam, its elastic force being known. RULE. - Multiply the 6th root of the elastic force in inches of mercury by 177, and subtract 100 from the product; the remainder will be the temperature in degrees, Fahrenheit nearly. EXAMPLE. -The elastic force of the steam is 150 inches of mercury, (150 ÷ 2.04 = 734 pounds to each square inch of the surface of its bulk); what is the temperature? 150 2.305 × 177 = 408 · 100 308°. Ans. NOTE.The pressure of the steam in pounds per square inch X 2.04 pressure in inches of mercury. The temperature due to a pressure of 14.7 lbs. per square inch 2120; 20 lbs. = 2280; 25 lbs. 2410; 30 lbs. 252; 35 lbs. 2610; 40 lbs. = 2690; 45 lbs. = 2760; 50 lbs. 2830; 55 lbs. 2890; 60 lbs. 2960; 65 lbs. 3010; 70 lbs. lbs. 3110; 80 lbs. 3160; 85 lbs. 320°; 90 lbs. 3240; 100 lbs. lbs. 363°; 200 lbs. 3879. 1 === 3060; 75 332°; 150 To find the quantity of water required for steam per minute by an engine in motion. RULE.-Multiply the velocity of the piston in feet per minute, by the square of the cylinder's diameter in feet multiplied by 0.7854, and divide the product by the volume of steam compared with the volume of water, due to the pressure exerted. EXAMPLE. The diameter of the cylinder is 23 feet, the velocity of the piston 245 feet per minute, and the constant pressure exerted by the steam 60 pounds to the square inch; what quantity of water must be converted into steam per minute? 2.52.7854 × 245 ÷ 470 = 2.56 cubic feet. Ans. NOTE. For a single-acting engine half the quantity indicated by the above rule is required. To find the quantity of steam required to raise a given quantity of water of a given temperature to a required temperature. The sum of the latent and thermometrical or sensible heats of water in a state of vapor or steam is always the same, viz., 1212°. The latent heat, therefore, of vapor of water at 32° = 1212 32 = 140°. The latent heat of water not in a state of vapor = temperature of water to be raised, c = temperature to which the water is to be raised, d = volume of steam compared with the volume in water due to the temperature of the steam, w= quantity of water in cubic feet to be heated. Then quantity of water (say in cubic feet) that must be converted into steam having a temperature a, required to raise 1 cubic foot of water from b to c. (c-b) X d 1000+ ac = quantity of steam in cubic feet, at temperature a, required to raise 1 cubic foot of water from b to c. wx (c—b) 1000+ ac = quantity of water in cubic feet that must be converted into steam, and have a temperature at a, required to raise the given quantity of water from 6 to c. wx (c—b) X d quantity of steam in cubic feet at tempera = 1000+a-c ture a, required to raise the given number of cubic feet of water from b to c. EXAMPLE. -What quantity of water in steam at 212°, will raise 100 cubic feet of water from 60° to 200° ? 100 X 140 1000+212- 200 1694: = =13.83 cubic feet. Ans. Making 13.83 X 23428 cubic feet of steam at 212 degrees. To find the quantity of water of a given temperature required to reduce a given quantity of steam to a given temperature. Let Then α= temperature of steam to be reduced, C = temperature to which the steam is to be reduced, 1000+ ac c—b = number of cubic inches of the water required to reduce 1 cubic foot of the steam from a to c, nearly. |