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Synchronous Generator Power Factor

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Chin2

Electrical
Aug 22, 2010
8
Hello everyone,

I am a new member to this forum and would like to ask a little help from you guys. I am designing a Permanent Magnet Synchronous Generator (PMSG) for a Wind turbine application which is of direct drive (DD) type. Being a DD type PMSG the frequency of the output is very low and hence I had to employ converter Inverter pair to improve the frequency. Now, how do I calculate the power factor (PF) of the PMSG? As the PMSG is not directly connected to the load, I believe that the PF does not depend on the load nature. Can someone help me by providing any equations or any research papers and any other material that can help me finding out the PF of the PMSG.

Thank you very much.
 
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The power factor is determined by the load, that being the rectifier. Power factor at the generator would not necessarily be influenced by the load on the inverter, although it may.

Alan
“The engineer's first problem in any design situation is to discover what the problem really is.” Unk.
 
Thank you for your reply Alehman. The point you made is correct in my opinion. The load to the Generator in my case is the Rectifier and depending on its characteristics the power factor of the Generator will result. Can you suggest me how to proceed to get the power factor value of the generator i.e., any suggestions for the equations, research papers or books. I tried looking in the internet but could not really find anything helpful. Thank you very much once again.
 
I'm not sure I completely know the setup, but if the load is a rectifier, then the current is not sinusoidal, power factor will have two components: displacement power factor and distortion power factor.

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(2B)+(2B)' ?
 
Thank you Electricpete, kindly let me know if you find anything related to this issue. I appreciate your time and help.
 
First thing you need to to is to calculate synchronous reactance (I assume you plan to use surface magnets so you have equal reactances in both d- and q-axis). Then you apply synchronous machine load angle equation to get the load angle. From this you will get load angle (angle between terminal voltage and back-emf voltage) for given power and terminal voltage.

Then you simply use some trigonometry to get the components for voltage and currents (in d- and q-axis). You need also reactances here (X=LI). When you have the components, you can calculate the angle between voltage and current vectors and then you have power factor. It is a bit tricky, and needs understanding of vector diagram presentation of electric machines.

In general, power factor with PMSM is not very important, usually it is more practical to get minimum current. Power factor of PMSM can be set to unity for example, by adjusting the terminal voltage from converter, but usually operation at unity power factor is not practical as it results in higher current and higher (copper) losses. By choosing lower power factor is often more feasible as it results in lower current.

I hope this helps
 
Thank you jpts. I think I will try the first option you suggested with the phasor diagrams. If I think its tricky and if it gives me issues can I go with some assumed power factor number close to unity. May be something around 0.95 lagging. If yes, is there any research paper that can be used as a reference for that value. Thanks once again.
 
Power Factor = Distortion Power Factor * Displacement Power Factor
where
Power Factor = real power / (Vrms * Irms) [true rms]
Displacement power factor = cos of angle between voltage and fundamental current
Distortion Power Factor = rms of current fundamental / rms of current total
assumes sinusoidal voltage

If you have a full wave bridge rectifier feeding a resistive load, then the currrent in the half-cycle w*t =-Pi/2 to +Pi/2 (which is all we need since the interval Pi/2 to to 3*Pi/2 is half-wave symmetric) looks like:
I(t) = 0 for –Pi/2 < w*t < –Pi/6
I(t) = Imax * sin(w*t) for –Pi/6 < w*t < Pi/6
I(t) = 0 for Pi/6 < w*t < Pi/2

You could calculate the rms and fundamental of the waveform above to get an idea of the distortion power factor if you were feeding a resistive load.

For what purpose will this power factor number be used?

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(2B)+(2B)' ?
 
Correction in bold:
I(t) = 0 for –Pi/2 < w*t < –Pi/6
I(t) = Imax * cos(w*t) for –Pi/6 < w*t < Pi/6
I(t) = 0 for Pi/6 < w*t < Pi/2


Rather than doing the algebra I solved it numerically as shown in attached (tab data1... ignore the rest).
Results were:
I1rms = 0.431335137 *Imax
Itotrms = 0.552267595 *Imax
Distortion PF = I1rms/Itotrms = 0.781025613


=====================================
(2B)+(2B)' ?
 
 http://files.engineering.com/getfile.aspx?folder=8f1b5e29-ebec-4901-95c7-12433dfb1954&file=MyFFT4.xls
Electricpete, The generator used here is for a Wind power applications. The power factor will be used to find out the apparent power of the generator and also to know the behavior of the generator at various wind speeds.
 
OK, I was addressing p.f. of PMG assuming it was feeding a rectifier load. Apparently the wrong thing.

=====================================
(2B)+(2B)' ?
 
I will explain the question in little more clear. I have the Active power of the generator and the voltage at its terminals. I could not decide the amount of output current since the load here in my case is a rectifier-inverter pair and also could not find the power factor of the machine. I know the field current of the generator and the d,q axes reactances too. I think now the question is a bit clear for you guys to help me out. Thank you so much :)
 
Still not very clear. I didn't know a PMSG used a field...

A PMSG feeding a rectifier is similar to line power feeding the input of a common 6-pulse VFD. Maybe you can look at those examples and extract some useful info.

I'm not sure what use the apparent power is beyond sizing the conductors and rectifier. In that case, assuming something like the power factor is -0.7 should suffice.
 
My bad. The field current I mentioned is the equivalent field current from the permanent magnet. I am actually planning to assume a pf of around 0.9 - 0.95 lagging if at all I do not get any info about it. Thanks once again
 
Attached you can find some steps that might help you. In general, if you have surface magnets you should end up with quite high power factor (0.9...0.95) due to low inductances. By the way, what is the approximate power rating of your generator?

 
-0.9 to 0.95???

How exactly is the below no "information about it"????

electricpete said:
Distortion PF = I1rms/Itotrms = 0.781025613

 
Power factor of a stand alone synch generator will be defined by its load. The voltage will vary with the excitation in this case.

When a synchronous gen is paralleled with another stiff source such as a grid, power factor of the gen will vary with the excitation. The voltage (and frequency) will be dictated by the stronger source. I am sure there are books on the subject.

Rafiq Bulsara
 
Be careful, folks! A lot of you are misreading the OP and subsequent comments.

This is a permanent-magnet generator, so no excitation current is required. It will have much lower inductance than an induction generator, so its displacement power factor will be much higher (0.9 to 0.95 would not surprise me). Distortion power factor could be much lower, as Pete has shown, depending on how the rectification to DC is done. But the rectification definitely "isolates" it from the grid.

Chin2, how are you planning to do the rectification? A simple passive diode bridge is inexpensive, but leads to large distortions (as per Pete's estimations) with the CD bus "gulping" current at the peaks. It may also limit your performance so much as to make the system impractical, because the speed must be high enough to create a back EMF higher than needed to create the grid voltage.

An active front end is more expensive, but virtually eliminates both displacement and distortion power factor concerns, and permits generation over a far larger range of operation. All of the projects of this type I know of use such an active front end.

Curt Wilson
Delta Tau Data Systems
 
I guess I shouldn't assume that "rectifier" means a diode rectifier...

If the "rectifier" is an active solution then a good power factor might be the result. Even sinusoidal current at close to unity power factor is possible.

At this point, the origional poster hasn't really given enough details to know what he wants and doesn't seem to know what the load will be. Still, the power factor will be a result of the applied load in this application.
 
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