TESTING MOTOR AT LOW VOLTAGE AND SEVERAL

FREQUENCIES

MOTOR 20 HP, 230 VOLTS,1800RPM, TESTING

VOLTAGE 10VAC

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.

Voltage at 300Hz

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.

Current=f (frequency)I=f[f] y = 0.1889x-0.7179

R2 = 0.9965

0

0.1

0.2

0.3

0.4

0.5

0.6

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

frequency [kHz]

curr

ent

[Am

ps]

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.

Watts = f [frequency]W=f[f], Total Watts

y = 0.636x-0.6083

R2 = 0.9976

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Freqency [kHz]

Wat

ts

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).

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

comparison Total Watts to RI^2

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

frequency [kHz]

Wat

ts

TOTAL WATTS

RI 2

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.

Ratio r = f [f]

Ratio r=f[f]

0

10

20

30

40

50

60

70

80

90

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

frequency [kHz]

rati

o r

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).

INDUCTANCE AS A FUNCTION OF FREQUENCY

Inductance=f(f)

7

7.5

8

8.5

9

9.5

10

10.5

11

11.5

12

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

frequency [kHz]

Ind

uct

ance

[m

H]

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).

2

2

2

*2

1000][

I

P

I

V

fmHL

ac

ac

Phase Angle = f [frequency]phase angle (phi)

0

10

20

30

40

50

60

70

80

90

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

frequency [ kHz]

ph

i [d

egre

es]

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.