Compressed air; theory and computations . described above. Othermeans that could be used to move the switch are: The differencein pressures in the intake and discharge pipes. This differencegradually increases until the switch is thrown and would be util-ized when the tanks are deeply submerged. Another meanswould be to switch after an assigned number of revolutions ofthe compressor—the number being that necessary to run a cycleas described above. The switch is usually made in the form of a piston would be inappropriate in this volume. Example 36.—Design a return-air pump to deli
Compressed air; theory and computations . described above. Othermeans that could be used to move the switch are: The differencein pressures in the intake and discharge pipes. This differencegradually increases until the switch is thrown and would be util-ized when the tanks are deeply submerged. Another meanswould be to switch after an assigned number of revolutions ofthe compressor—the number being that necessary to run a cycleas described above. The switch is usually made in the form of a piston would be inappropriate in this volume. Example 36.—Design a return-air pump to deliver 200 cu. minute under a head of 100 ft.; the tanks to be placed atwater level (as in Fig. 13a) and air pipes to be 500 ft. long. Solution.—The ratio po -5- pn is the same as the ratio of thewater heads (taking amtospheric pressure as a head of ft.).Then this ratio is -h = 4. Assume for first trial thatthe tanks have a volume of 400 cu. ft. each, and that effective SPECIAL APPLICATIONS OF COMPRESSED AIR 73. Fig. 13a. 74 COMPRESSED AIR intake capacity of the compressor is cu. ft. per stroke. Thenthe number of strokes required in a cycle is 400 log 4 n = TT + 1 am g i Ann = 266 + 37° = 636- log — log 400 636 strokes = 318 revolutions to deliver 400 cu. ft., or 159revolutions to deliver 200 cu. ft. per minute. This speed is excessive, but before we make another trial wewill see what size air pipe will be necessary in order to prescribethe correct size of tanks. 159 X 2 X = 477 cu. ft. per minute intake to the com-pressor. This is the maximum rate of passage of air througheither air pipe and occurs once in a cycle, and that just at theend when the pressure in the air pipe is about that of the atmos-phere. It is at this time that friction in the air pipe is greatestand we may allow a drop (/) of 5 lb. in order to economize in sizeof both air pipes and tanks. Then by formula (27d) X 500 X (^Y\ or by Plate III we see that a 3^-
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