. Electric railway journal . rating ofthis motor is amp., sothat we find this motor hassufficient capacity for thegiven service. The averagevoltage for each motor isfound from the curveOPRSTF to be 470. Thevoltage curve is constructedby drawing a line through R (270 volts at 3 seconds and S (540 volts at seconds)and then using the average line voltage of 540 forthe remainder of the time that power is on. The areaof this curve is squares or 10, volt-seconds,which divided by seconds, the time that power ison, gives 470 volts as the average that is applied to eachmotor d
. Electric railway journal . rating ofthis motor is amp., sothat we find this motor hassufficient capacity for thegiven service. The averagevoltage for each motor isfound from the curveOPRSTF to be 470. Thevoltage curve is constructedby drawing a line through R (270 volts at 3 seconds and S (540 volts at seconds)and then using the average line voltage of 540 forthe remainder of the time that power is on. The areaof this curve is squares or 10, volt-seconds,which divided by seconds, the time that power ison, gives 470 volts as the average that is applied to eachmotor during the run. Core Losses Are Determinedby Tests Fig. 3 shows graphs for the iron and core losses forthis motor. These losses depend on both the currentand the voltage. They follow no simple law as do thecopper losses, which are proportional to the square ofthe current. Owing to the great mass of metal in its frame, arailway motor has considerable heat storage the temperature of the windings is rising that 130. Accurate operating data are essential in choosing- motors as the average core loss plotting this curve, let us find the point on the core-loss curve at ten seconds after the start of the the curves showing current and volts per motorit is found that at the ten-second point there are 66amp. and 540 volts per motor. Then from the graphsshown in Fig. 3, we find the core loss corresponding to this current and voltage to be1020 watts. This value plottedat the ten-second point in will be the desired point inthe core-loss curve. The aver-age core loss for the run isthen obtained by dividing thearea under this curve in watt-seconds by the total time forthe run, including stop, or This gives 245 wattsThe power-input curve is con-structed from the car-current curve by multiplying thevarious current values by the average line voltage. Inthe table preceding, I have tabulated the various powervalues in watts corresponding to different curren
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