Electrical Engineering is a free introductory textbook to the basics of electrical engineering. See the editorial for more information....  # Efficiency of D-C Motors

Author: E.E. Kimberly

The losses in a direct-current motor are of the same type as those in a direct-current generator and may be found in the same manner. For a given power output the shunt-field loss and the armature-iron loss of a direct-current 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 12-1. - The efficiency of a direct-current, self-excited, flat compound generator of 500 kw, 250 volts is to be determined. A 15-hp 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 d-c supply is available), and plot a curve of no-load 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 armature-circuit resistance (terminal to terminal), using 125, 100, 75, 50, and 25 per cent of the rated fuJl-load current. Draw a curve of armature-circuit resistance vs. load current. Measure the shunt-field current.

Example 12-2. - The no-load losses of the generator in Example 12-1, including shunt-field circuit loss, amount to 15 kw. The armature-circuit resistance at 31.6°C is as follows:

 % Full-Load Current Resistance[Ohms] 125 0.0035 100 0.0037 75 0.0039 50 0.0042 25 0.0046

Calculate the efficiency of this generator at rated load at 75 C.

Solution. - The full-load current is At full-load current and 75 C, the armature-circuit 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 pole-face losses and eddy-current losses in the armature conductors, and also the increased shunt-field current necessary to obtain the rated voltage when the generator is loaded.

When it is not practicable to measure the armature-circuit 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 I2aRa loss. To this loss then must be added 2I0, which is the brush-contact loss discussed on page 144.

Last Update: 2010-10-06