20.000 13.333 10.000 -8:000 51.333 = sum of the odd pressures. 2 102.666 twice the sum of the odd pressures. 281.020 46.666 3)430.352 143.451 work done by expansion. 80.000 work done before expansion. 223.451 36.000 187.451 whole work done upon each square inch. whole effective work. Pambour, who has made a great number of experiments, estimates the resistances as follows The pressure of the atmosphere about 15 lbs. per square inch; also the resistance arising from the various parts of the engine, at a mean, the estimate is about 1 lb. to the square inch for the unloaded engine, and an additional friction of 14, or of the effective pressure or useful load, for overcoming the friction of the loaded engine. Supposing the pressure of the steam is 60 lbs. per square inch; the resistances are 15 lbs. per square inch from the atmosphere, 1 lb. for the friction of the unloaded engine; then 15+1=16, and this taken from 60, leaves 44; therefore the load of the load is equal to 44 lbs., or 44 x 7 of the load 44; therefore the load= 8 = 38, or 38.5. The load is, therefore, the effective pressure on the piston. For high-pressure engines we have the Load+load + 16 whole pressure of the steam. Hence the following rule: To find the load when friction is taken into account, viz., THE USEFUL LOAD. From the pressure of the steam in the cylinder, subtract 16; multiply the remainder by 7, and divide this product by 8, and the quotient is the useful load. Example.-Given the pressure of steam in the cylinder 50 lbs. per square inch, to find the load. Then, by the rule, 50 16 34; then 34 x 7 8 =29.75. For the condensing or low-pressure engine, instead of the resistance of the atmosphere, we must use the resistance of the vapour in the condenser, which is generally estimated at about 4 lbs. to the square inch: in this case we have the Load+load + 1 + 4 = whole pressure of steam. Hence we have the following rule: From the mean pressure of the steam subtract 5; multiply the remainder by 7; this product divided by 8 gives the useful load. On the Evaporating Power of the Boiler. The evaporating power of the boiler is of the greatest importance, and as it is the source of all work in the steam engine, many ingenious contrivances have been made to increase this evaporating power; these various contrivances will hereafter be explained. The quantity of work done depends on the quantity of water evaporated, the temperature, and the pressure at which the steam is generated; we shall also hereafter give formulæ to find the relation between the volume of steam and the pressure; but the application of the following experimental table will give results sufficiently correct for all practical purposes. It shows the volume which a cubic foot of water has in the form of steam at the different pressures, as well as the corresponding temperatures. EXAMPLES. 1. Given the area of the piston of a high-pressure engine 200 square inches, the length of the stroke 4 feet, the evaporation of the boiler half a cubic foot of water per minute, the pressure of steam in the cylinder 60 lbs. per square inch find the useful load and the horse power. To find the Useful Load. By the rule (page 10), 60 38.5 lbs. useful load. 1644; then 44 x 8 To find the Volume of Steam evaporated per Minute. Multiply the evaporation of the boiler in cubic feet per minute by the volume in the table corresponding to the given pressure, and the product is the volume of steam evaporated per minute. Here the pressure is 60 lbs., the corresponding volume in the table is 470. The evaporating power of the boiler being half a cubic foot per minute, we have 470 x 235 for the number of cubic feet evaporated per minute. The number of cubic feet discharged per stroke is equal to the area of the piston in feet, multiplied by the length of the stroke in feet = 200 ×4= 800 = 5.55. The whole volume discharged per minute is equal to the number of strokes per minute, multiplied by the volume discharged at one stroke; and the volume discharged per minute must be equal to the volume evaporated per minute. Number of strokes per minute × 5·55, whole discharge in one minute = 235 lbs. evaporated also in one minute. Hence the number of strokes = 235 5.55 = 42. But the useful work done in one stroke is 38.5 x 200 x 4 = 30800, therefore the useful work per minute is 38.5 × 200 × 4 × 42, or 30800 × 42; therefore the horse power is 30800 × 42 = 39. Example 2.-In a condensing engine the area of the cylinder is 1440 square inches; the length of the stroke, including clearance, is 5 feet; the steam is cut off at 1 foot; the clearance is of a foot; the pressure of steam is 30 lbs.; the elasticity of the vapour in the condenser is 4 lbs.; the effective evaporation of the boiler is 2 of a cubic foot per minute, and the resistances as usual: required the useful load, and the useful horse power.Tate's Mechanics. The space through which the piston moves before the Hyp. log. of 6.33 = 1·8453002. 1.8453002 × 22.5 41.5192545 work done by expan sion. Hence the whole work in one stroke is 22.541.5192545 64-0192545. |