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Home Losses, Efficiency, and Rating of DirectCurrent Machines Efficiency of DC Motors  
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Efficiency of DC MotorsAuthor: E.E. Kimberly The losses in a directcurrent motor are of the same type as those in a directcurrent generator and may be found in the same manner. For a given power output the shuntfield loss and the armatureiron loss of a directcurrent machine used as a motor are less than when the machine is used as a generator because of the lower flux required in the motor.
Example 121.  The efficiency of a directcurrent, selfexcited, flat compound generator of 500 kw, 250 volts is to be determined. A 15hp calibrated motor is available. Give instructions for obtaining the test data required for calculating the conventional efficiency at any load.
Solution.  Drive the generator at rated speed with the calibrated motor. Excite it separately (if separate dc supply is available), and plot a curve of noload terminal voltage vs. power from the driving motor. The generated voltage is proportional to the field flux. Therefore, the curve so drawn will show the values of fixed losses (windage, friction, and iron), as they vary with the generated voltage. Measure the armaturecircuit resistance (terminal to terminal), using 125, 100, 75, 50, and 25 per cent of the rated fuJlload current. Draw a curve of armaturecircuit resistance vs. load current. Measure the shuntfield current.
Example 122.  The noload losses of the generator in Example 121, including shuntfield circuit loss, amount to 15 kw. The armaturecircuit resistance at 31.6°C is as follows:
Calculate the efficiency of this generator at rated load at 75 C. Solution.  The fullload current is
At fullload current and 75 C, the armaturecircuit resistance is 0.0043 ohm. The generated voltage is then 250 + (0.0043 X 2000) = 258.6 volts Assume that, from the curve of generated voltage vs. armature power input, the power input at 258.6 volts is 15,000 watts; and assume that the stray load loss is 0.01 X 500,000 = 5000 watts. Then,
The foregoing solution includes numerous small losses, such as poleface losses and eddycurrent losses in the armature conductors, and also the increased shuntfield current necessary to obtain the rated voltage when the generator is loaded. When it is not practicable to measure the armaturecircuit resistance at the requisite percentage of rated current, it is recommended that the resistance of only the copper portion of the armature circuit be used in calculating the I_{2}aRa loss. To this loss then must be added 2I0, which is the brushcontact loss discussed on page 144.


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