. Journal of electricity . 800 = 2300 watts. Efficiency = 2300/2470 = 93%. ■ It is interesting to inquire what the voltage at the lamps will be when the motor stops. We can add the lamp resistance to that of the line, and find the cuiTent from the generator voltage (which we may assume remains constant at volts). Then the lamp resistance times the current gives the lamp voltage. Mechanical Power.—One horsepower (hp.) maybe defined mechanically as the power required to liftone pound at the rate of 550 ft. per second, or to lift550 lbs. at one ft. per second. Hence the numberof hp. requiie
. Journal of electricity . 800 = 2300 watts. Efficiency = 2300/2470 = 93%. ■ It is interesting to inquire what the voltage at the lamps will be when the motor stops. We can add the lamp resistance to that of the line, and find the cuiTent from the generator voltage (which we may assume remains constant at volts). Then the lamp resistance times the current gives the lamp voltage. Mechanical Power.—One horsepower (hp.) maybe defined mechanically as the power required to liftone pound at the rate of 550 ft. per second, or to lift550 lbs. at one ft. per second. Hence the numberof hp. requiied for any operation = pounds pull Xfeet per second -^ 550. Then, if an automobile beingdriven at 42 ft. per second meets a total of 340 lbs. ENERGY LOSSES — MECHANICAL POWERLost Energy.—No mechanical machine operateswithout a loss of energy; the output is always lessthan the input by the amount of energy wasted infriction and turned into heat. Similarly every elec-trical device permits some of the energy to change. -Diagram of circuit containing generator lamps and motor,illUStrating calculations in power and losses. into heat, and (unless heating is the object of thedevice) this constitutes an electrical loss. If it takes 3 volts to drive 12 amperes througha certain piece of wire, the power used ^ 3 X 12= 36 watts. Evidently the 3 volts = the voltagedrop in that piece of line, and hence we get the rule:Watt loss in any conductar at 100% power factor =amperes X volt drop in that conductor. This leadsto another rule: Watt loss in any conductor =amperes X amperes X ohms, since the volt drop =amperes X ohms. It is found that this last ruleholds true for all cases of alternating currents aswell as direct currents, though the proof is beyondthe scope of this paper. A circuit is shown in Fig. A, containing genera-tor, lamps and motor, with which calculations inpower and losses may be illustrated. Assume thatthe motor absorbs an input of kw. while takinga current of 14 amp
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