Cyclopedia of applied electricity : a general reference work on direct-current generators and motors, storage batteries, electrochemistry, welding, electric wiring, meters, electric lighting, electric railways, power stations, switchboards, power transmission, alternating-current machinery, telegraphy, etc. . ion of the wheel advances the car —X feet ( is the ratio of the circumference of a circle to its diameter). One revolution of the motor armature then advances the car y^or feet. As the motor armature makes 588 in this case, the corresponding rate of car
Cyclopedia of applied electricity : a general reference work on direct-current generators and motors, storage batteries, electrochemistry, welding, electric wiring, meters, electric lighting, electric railways, power stations, switchboards, power transmission, alternating-current machinery, telegraphy, etc. . ion of the wheel advances the car —X feet ( is the ratio of the circumference of a circle to its diameter). One revolution of the motor armature then advances the car y^or feet. As the motor armature makes 588 in this case, the corresponding rate of car travel is speed= 1,252 feet per minute miles per hour= 1,252X ^^^7^77= ,280 (c) The mechanical power produced by each motor is the prod-uct of the speed of a point on the torque circle—that is, of a circleof one foot radius—and the torque. Remembering that 1 horse-power is 33,000 feet-pounds per minute, we have , ,_ _^•^•P- = 337)00 = ^^-^ (d) The electrical power input in watts is the product of voltsand amperes, or watts = 500X 80=40,000. As 746 watts equal 1 electrical horse-power, the input in this unit is , 40,000 (e) The efficiency is the ratio of output to input, in this case ofm. h. p. to e. h. p. This is usually referred to in per cent, so that 138. ELECTRIC RAILWAYS 43 TABLE IMotor Efficiencies for Different Tracic Grades Grade Trac-tiveEffort PERMOTOK Torque Current Speed Efficiency M. M. p. H. 1 per cent 2 per cent 3 per cent5 per cent 10 per cent 400 600 800 1200 2200 829737673588497 per per per per per cent the ratio must be multiplied by 100. Then efficiency= X100=85 per cent The student should, for practice, make the above calculationsfor other grades, say for 1 per cent, 2 per cent, 3 per cent, 10 percent. The answers for these grades are given in Table I. Thesame principle can be applied to leve
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