. Hawkins electrical guide. Questions, answers & illustrations; a progressive course of study for engineers, electricians, students and those desiring to acquire a working knowledge of electricity and its applications; a practical treatise. race is required, but for belts,rt be considerable distance between ; : i^:.-:~::i: Slow Speed Alternators.—By slow speed is here understoodrelat: v speed, such as the usual speeds of reciprocating engines. A slow speed alternator is one designed to run at aspeed slow enough that it may be direct connected to an engine. ALTERNATORS 1,173 Such alte
. Hawkins electrical guide. Questions, answers & illustrations; a progressive course of study for engineers, electricians, students and those desiring to acquire a working knowledge of electricity and its applications; a practical treatise. race is required, but for belts,rt be considerable distance between ; : i^:.-:~::i: Slow Speed Alternators.—By slow speed is here understoodrelat: v speed, such as the usual speeds of reciprocating engines. A slow speed alternator is one designed to run at aspeed slow enough that it may be direct connected to an engine. ALTERNATORS 1,173 Such alternators are of the revolving field type and a little con-sideration will show that they must have a multiplicity of fieldmagnets to attain the required frequency. In order that there be room for the magnets, the machineevidently must be of large size, especially for high FlG. 1,435.—Crocker-Wheeler 350 kva., slow speed alternator direct connected to a Corlissengine. In front is seen a belted exciter driven from a pulley on the main shaft betweenthe alternator and the large band wheel. The latter serves to give the additional fly wheeleffect needed for close speed regulation. EXAMPLE.—How many field magnets are required on a two phasealternator direct connected to an engine running 240 revolutions perminute, for a frequency of 60? An engine running 2-40 revolutions per minute will turn240 -:- 60 = 4 revolutions per second. A frequency of 60 requires 60 ■*■ 4 = 15 cycles per phase per revolution, or15 x 2 = 30 poles per phase. 1,174 HAWKIXS ELECTRICITY Hence for a two phase alternator the total number of poles required is 30 X 2 = 60. It is thus seen that a considerable length of spider rim is required to. the numerous poles, the exact size depending upon their dimen-sions and clearance. |W % IP 1 - ■PI Pig . —Three Crocker-Wheeler 75
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