# Open shaft current?

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### #1 d-mark

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Posted 28 January 2003 - 06:44 PM

I am assuming that open shaft or magnetizing current refers to current drawn by motor to rotate rotor shaft with no-load applied. Although no real work is being done, power is still being expended to spin the shaft and the weight of the rotor, correct?

Aren't we trying to determine the portion of the current that creates excess magnetic flux in the coils that does no work? And is still present when the motor is under fully loaded condition?

;p;

### #2 marke

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Posted 28 January 2003 - 06:56 PM

Hello d-mark
Welcome to the forum
Yes, the open shaft current is the current drawn when the motor is operating at full voltage but with nothing coupled to the shaft. Under these conditions, the motor is doing very little work, just overcomming friction and fan losses. The current is predominantly magnetising current. This can be quite different from leaving the machery coupled to the motor shaft and doing no work. An unloaded machine can still present a high loading on the motor. For example, an unloaded screw compressor will typically require 30 - 40% torque to keep spinning.
The magnetising current is essentially constant and independent of load.
Best regards,

### #3 d-mark

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Posted 28 January 2003 - 08:15 PM

Thanks for the amazingly fast reply. In our private email discussion about the Motor A Example from the paper you provided the following response:

Quote:

"Motor A is a 200 KW 6 pole motor with a magnetizing current of 124A. From tables, the correction applied would be 37KVAR. From the calculations, this would require a correction of 68.7 KVAR"

These examples will be at 400 Volt AC 50Hz but the same theory applies at other voltages and frequencies.
Line current would be sqrt ( (200000 / 1.73 / 400 / 0.92 )^2 + 124^2) = 337.46 Amps
Full load pf = 0.929 ignoring the leakage reactance which would reduce the full load power factor a bit.

Adding the power factor correction would lift the full load power factor to 0.99 ignoring the leakage reactance."

End quote

If I calculate the resultant magnetizing current after PF correction to 16 kvar (85-69), I get 23 amps. A reduction of around 100 amps! However the overall line current is reduced from 337 to about 314 amps. This 23 amp reduction at 400 volts rms would equate to a savings of ~9200 watts drawn or a 4.6% reduction in the rated 200Kw demand of the motor.

### #4 marke

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Posted 29 January 2003 - 09:06 AM

Hello d-mark.
Adding power factor correction will reduce the current drawn from the supply, but will not alter the current flowing into the motor. It will not alter the KW drawn by the motor either. The only saving is in the supply. The reduction in current will reduce the I^2 R losses in the cables etc supplying the current from the gnerator. This will not impact on the KWHr meter readings.
There will be an improvement in the KVA drawn by the motor, and if you pay a charge for KVA as well as KWHr, then you will see a saving.

Best regards,

### #5 d-mark

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Posted 29 January 2003 - 05:38 PM

QUOTE
Originally posted by marke
The only saving is in the supply. The reduction in current will reduce the I^2 R losses in the cables etc supplying the current from the gnerator. This will not impact on the KWHr meter readings.

Thanks Mark

I certainly concur that there will be no savings in electricity consumption used by the motor and should not have stated a percentage of the motors demand as being saved through PF correction.

However, I believe there are significant kw/kwh savings that are achieved at the service entry meter due to the reduction in losses in the distribution system you referred to.

I have a small 1/4hp motor fitted with a properly sized capacitor that can be switched on and off. This is connected to the 120 volt wall socket via a simple meter that measures watts drawn.

When the assembly is connected close to the wall socket there is no measureable difference in the watts drawn with the capacitor switch open or closed. Although the amperage drawn decreases from 4.5 amps to 1.0 amps from the wall. The motor of course receives the full 4.5 amps as with the capacitor off.

When 200 feet of extension cord is inserted between the assembly and the wall socket the difference in wattage drawn is as much as 17%. The more cable, connectors, etc added to simulate a distribution system the more the savings in watts drawn.

I have been led to believe that this effect is real and applicable to large scale motorized plants. The savings due to reduction of PF penalties are one benefit, but even more are the savings in demand (kw) charges and consumption (kwh) costs.

By reducing current flowing in the distribution system and associated losses aren't we also reducing A/C resistance in the system thereby allowing less power to do more work?

I would very much appreciate your professional opinion on this matter as I have heard there is wide-spread disagreement among EEs as to the benefit of PFC on energy savings.

### #6 marke

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Posted 29 January 2003 - 06:23 PM

Hello d-mark

Yes, there will certainly be a reduction in the I^2 R losses between the meter and the motor, but these should be small. If they are significant, then one must question the size of the cabling used. You can easily get an indication by comparing the voltage drop on the cable between the motor running and not running. If the voltage drop along the cable is significant when the motor is running, then there will be a saving. If the voltage drop can not be measured, there will be no saving. A 5% voltage drop (relative to the supply voltage) suggests that you could save a percentage of 5% by adding power factor correction. I would not expect to see actual savings in excess of 5% unless you have very small, undersized cables, or very long cable runs supplying your motors.
Best regards,

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