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How Does PF Change in an overexcited Synchronous Motor
2

How Does PF Change in an overexcited Synchronous Motor

How Does PF Change in an overexcited Synchronous Motor

(OP)
I am trying to figure out why, and more precisely how, the power factor is affected by a synchronous motor. My model is an input voltage source in series with a resistance and a negative voltage source representing the back EMF. I also am choosing insignificant inductance such as a servo might present, also because I'm trying to keep it simple. I notice that as the rotor position (phase difference between stator and armature) advances from the 90 degree position (max torque) to the 0 degree position (no torque), that the summed voltage (resultant) lages the input waveform from 0 to 90 degrees. I sort of thought that the resultant voltage waveform would tell me something about the current waveform which I expected to lead the input wave - not lag it.

RE: How Does PF Change in an overexcited Synchronous Motor

Suggestion: Synchronous motor:
1. Overexcited
Current Ia drawn by the motor is leading motor terminal voltage
2. Underexcited
Current Ia drawn by the motor is lagging motor terminal voltage
Reference:
1. William D. Stevenson, Jr., Elements of Power System Analysis," Third Edition, McGraw-Hill, 1975, Figure 9.6 on page 218

RE: How Does PF Change in an overexcited Synchronous Motor

(OP)
There is no question that it happens in just this way. The question I'm wrestling with is the mechanics of why. In other words, the physics of the process. All would be well if the input waveform and the back EMF were in phase but we both know this is not (and shouldn't be) the case. Try applying the formula's for magnetic moment and induced EMF - they come out 180 degrees apart.
Also, I've checked my local library. Do know of a site where I might find fig. 9.6 ? Thanks for the pointer.

RE: How Does PF Change in an overexcited Synchronous Motor

(OP)
Thanks a lot fella's. I am now convinced that I have been drawing the phasor's correctly. My interest is in low inductance motors and I now see that the leading PF is caused by the high ratio of L to R in the winding which results in a nearly 90 lag between resultant voltage(V in and back EMF)and the current vector. In my case the angle from resultant voltage to the current vector might be less than 90 degrees lagging because of the high R and low L in the windings. I suspect this to situation is probably not capable of a leading PF depending on
arc tan X(sub)L/R          Sound right?

RE: How Does PF Change in an overexcited Synchronous Motor

Framing1,

Does you comment mean that you are no longer interested in additional responses?

RE: How Does PF Change in an overexcited Synchronous Motor

Suggestion: Reference:
M. G. Say "Alternating Current Machines," John Wiley & Sons, 1978, Section 10 Synchronous Machine: Theory and Performance on Page 366...
Focus on the Excitation Voltage, Operating Characteristics, Equivalent Circuits, and relationships
P1=-(V1/Xa) Et sin(d)..............Eq10.1
Q1=(V1/Xa) [Et cos(d) - V1]....Eq10.2

RE: How Does PF Change in an overexcited Synchronous Motor

(OP)
I am still interested in trcking this down. Now trying to find the book referenced last (Alternating Current Machines), can't seem to come up with a copy in any of the library networks here (MA). Also a little confused on the phasor diagram you referenced first - it shows V(sub)G which is the back EMF leading by 90 - 180 which places it in quadrant II instead of quadrant III where I would expect it since it lags the terminal voltage by 90 to 180.
It sounds like I need to find a copy of "Alternating Current Machines" but have run out of options at this juncture. Thanks for sticking with me on this one!

RE: How Does PF Change in an overexcited Synchronous Motor

Try the following book:

"Electric Machinery and Control"
   Author:     Irving L. Kosow,
   Publisher:  Prentice-Hall Inc., 1964
   LoCCCN:     64-22802.

If you have difficulty contact me:  EpsiconInc@aol.com

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