ber of strokes of the engine being 25 per minute, what diameter of pump will be necessary, supposing the pump's length of stroke to be two feet three inches? 105. What is any common rule through which to ascertain, by calculation, the proper capacity of a cold-water pump for a stationary condensing engine? RULE.-Multiply the square of the cylinder's diameter in inches by the length of the stroke in feet, and by the decimal 5, the product is the capacity of the pump in cylindrical inches. Or, if the capacity in cylindrical inches be multiplied by .00284, the product equal the capacity in imperial gallons. Ex.-Suppose an engine with a cylinder 60 inches in diameter, the length of stroke 5 feet, required the proper capacity for a cold-water pump. 602× 5.5=19800×.5=9900 cylin drical inches Capacity. And 9900 x .00284 = 28.116 impe rial gallons 106. How is the capacity of an air-pump determined for condensing engines? RULE.-Multiply the square of the cylinder's diameter in inches by the length of the piston's stroke, also in inches, and by 142, the product equal the capacity of air-pump in cylindrical inches. Ex.-Required the capacity of air-pump for an engine with a cylinder 45 inches in diameter, and a stroke of four feet; also, what will be the diameter when worked at two-thirds from the centre of the beam or side-levers, half the length of which equal six feet three inches. 452 × 48 97200 x 142 13802.4 cylindrical inches capacity. 75 x 2 - 50-inches 3 from centre of beam. Then 75:48::50:32 inches, the length of stroke of air-pump. Hence 13802.4 = √431.3 = 20.76 inches dia meter. 107. How is the proper weight ascertained whereby to make the rim of the fly-wheel of a rotary engine? RULE.-Multiply the diameter of the wheel in feet by the number of revolutions per minute, and by .0523; divide 2004 times the nominal horsepower of the engine by the square of the product, and the quotient is the weight in hundredweights. Ex.-What must be the weight of a fly-wheel rim for a thirty-horse engine making 19 revolutions per minute, and diameter of the wheel 22 feet. = 22 x 19418x.052321.8614 and 126 cwts. weight of the rim. 30 x 2004 21.86142 108. What is the rule by which to find the proper velocity for the governor of a steam-engine? The governor of a steam-engine in the usual form consists of two pendulums suspended from the upper extremity of a vertical axis; and, when made to revolve about that axis, assumes the coni cal form, the point of suspension being the apex of the cone, and the distance from centre to centre of the balls the plane of revolution; hence, when a certain expansion of the balls is desired, and a certain given length of pendulum fixed, so that the vertical height of the cone is known, multiply the square root of the height of the cone in inches by 0.31986, and 60 divided by the product equal the proper number of revolutions per minute. Ex.-Suppose the distance between the centre of suspension and plane of revolution equal 25 inches, required the number of revolutions per minute. ✔255 and 31986 x 51.5993, then 60+ 1.5993 = 38. 109. How are the various valves in marine engines generally distinguished? 110. At page 24, question 23, a rule is given by which to determine the nominal horse-power of condensing engines. How is the nominal horsepower of non-condensing engines estimated? RULE.-Multiply the square of the cylinder's diameter in inches by the pressure of the steam in pounds per square inch above the atmospheric pressure, and by the cube root of the stroke in feet, also by 00106, and the product is the nominal horse-power, when the velocity is 128 times the cube root of the stroke. Ex. Let the diameter of the cylinder of a high-pressure engine equal 16 inches, the length of stroke 3 feet, and the pressure of steam 45 pounds per square inch, as indicated by the safety |