1. TESTING MOTOR AT LOW VOLTAGE AND SEVERAL FREQUENCIES MOTOR 20 HP, 230 VOLTS,1800RPM, TESTING VOLTAGE 10VAC

2. CONDITIONS: The voltage source was a sine wave inverter. The voltage was kept sinusoidal by variation of the capacitance between 40  F to 300  F. The current was monitored on 1 Ohm resistor in series with the winding. Both voltage and current were fed to the two channel analyzer. The watts were calculated from the collected data in Excel. Following slide is a sample of the voltage wave form on the terminals of the tested motor. You can notice that the distortion was very low.

3. Voltage at 300Hz

4. TEST RESULTS Following are several graphs showing the impact of the frequency while the voltage is kept constant at 10 Vac. Note that some trend-lines are shown with analytical equations (blue thin lines). The analytical equations can be compared to what would be expected under simplified conditions.

5. Current=f (frequency)

6. Comment: Though the test was also performed at 60 Hz, it is impossible to make a reasonable graph, because the current is so much higher (looks like a hockey stick). The current at 60 Hz was 4.77 Amps One would expect decreasing the current with the power of –1. Compare to the power of the trend-line: -0.7179. It means the current is dropping with frequency at much slower rate.

7. Watts = f [frequency]

8. Comments: In case of a linear circuit, the curve should be dropping with the power of –2. However it is not the case. The watts drop very slowly with the power of –0.6083. The effect of iron at high frequencies is really profound. The watts at 60Hz: 15.65 W (and 10 V ac).

9. COMPARISON OF THE TOTAL WATTS TO THE WATTS LOST IN THE RESISTANCE OF THE STATOR WINDING

10. Comments: The watts in the copper of the winding were calculated as: P=RI^2. The resistance R from line to line was 0.2064 Ohm. For comparison, the total losses at 60 Hz were 15.649 W. The loss in the stator winding only was 4.7 W. (60 Hz) Notice how much closer those numbers are compared to higher frequencies. It is obvious, that those two curves have to meet at one point at zero frequency. For the ease of the comparison a ratio was created from those two sets of watt readings. The ratio r = (W-RI^2)/RI^2. W are the total watts.

11. Ratio r = f [f]

12. Comments: Note that the ratio for e.g. 800 Hz is about 70. It means that the losses other than those in the stator winding were 70 times higher. For the comparison, the ratio for 60 Hz was only 2.33. This curve shows clearly, how much different the low voltage is from the conditions at normal operation. The next slide will show the inductance, once calculated from the McKinnon-Smolleck formula and once from the correct formula (REAL).

13. INDUCTANCE AS A FUNCTION OF FREQUENCY

14. Comments: The inductance of the circuit is far away form being a constant. The formula for calculating the inductance: The formula takes into consideration all losses P in the circuit. The inductance at 60 Hz was only 5.284 mH (10 VAC). This inductance drops further to 4.45 mH at 50 V (60 Hz).

15. Phase Angle = f [frequency]

16. Comments: It is remarkable,that the phase angle is virtually constant with frequency. It would be interesting to perform the same test on the motor with open rotor slots.