Leading Power Factor - Voltage drop

[wfan]

Hi, I am hoping you can help me understand this concept.

I've always known that generators operating a leading power factor lowers voltage drop and a lagging power factor increases voltage rise (as seen here: http://books.google.ca/books?id=nIc... generator at a leading power factor"&f=false ), but never understood why.

How does the source/sink of VARs affect voltage rise/drop?

Thanks

 

[ScottyUK]

I don't fully agree with what that book says, except in the specific case of a lightly loaded line or cable which has a dominant capacitance which tends to cause voltage rise at the remote end. You might want to Google the Ferranti Effect for an explanation. A leading generator usually pulls the local system voltage down slightly, much as a lagging motor does.

Think about how a generator would behave it it were islanded, and compare that with how it behaves when tied to a large system. In an islanded system raising the field increases terminal voltage. On a large system raising the field increases the reactive export of the machine (the generator becomes more lagging). Unless the system is very stiff compared to the generator then a lagging generator usually causes a local rise in system voltage.


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[wfan]

Thanks, that makes sense. Do you mind if I ask a question a bit off-topic. In the case of capacitors that also raise voltage, is this accomplished by simply reducing the net reactance, leading to a voltage rise due to less voltage drop?

 

[electricpete]

Capacitors add vars.

As a rough rule, the voltage profile is determined by reactive power.... reactive power flowing through series inductance of transmission lines and transformers creates voltage drop. Wherever you add vars usually tends to increase voltage at that location.

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[wfan]

Thanks. But I guess my question is aimed at a more fundamental level. What is it about the "supply" of VARs that creates the voltage rise?

 

[electricpete]

I may be going off a little on a tangent, but this is the way it makes sense to be.

The load flow equations are the first equation here:

http://en.wikipedia.org/wiki/Power_flow_study

As it turns out, the off-diagonal elements of J and J^-1 are very close to zero for typical series transmission elements which are primarily inductive.

That is the basis for "decoupling" the real and reactive power equations during "fast decoupled load flow" method and also very useful for quick back-of-the-envelope qualitative analysis as we are discussing.

When real and reactive power are decoupled, real power flow through a series inductive element is dependent upon (or controls, depending on your viewpoint) the voltage PHASE angle difference accross that element (and is independent of the voltage magnitude difference accross that element). More importantly, reactive power flow through a series inductive element is dependent upon (or controls, depending on your viewpoint) the voltage MAGNITUDE difference accross that element (and is independent of voltage phase angle difference).

So again the last sentence is the important one. Vars control voltage magnitude. If there is a net vars flowing out of a node where caps are added, then that node will be higher voltage than whatever node those vars flow to. By mapping out the var flow in the system you can get a pretty good idea of the voltage distribution.

Sorry, if it's not what you're looking for.

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[electricpete]

To clarify, the equation I was referring to was not the very first equation on the page... it was the first equation under "Newton-Raphson solution method"

The off-diagonal elements which are close to zero are:
dP/d|V| and dQ/d(theta)

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[trchambe]

I don't think it's that the addition of capacitors causes a voltage rise. It does, but that's not the best way to say it. Rather, it causes a reduction in voltage drop. So I believe you were on track.

Due to electromechanical devices and miles of conductor, a power system is of course inductive. Adding capacitance reactance negates some of the inductive reactice, reducing the total impedance. A reduction of impedance will of course cause the voltage to rise.

 

[davidbeach]

No, capacitors cause voltage rise. An open circuited capacitive line will have a higher voltage at the open end than at the sending end.

 

[waross]

There are a couple of factors.
1> Power factor correction reduces line current and line voltage drop.
2> An open circuited capacitive line or an unloaded line with capacitors connected at the open end will flow current towards the source. As a result the voltage drop will be seen as a voltage rise.
This is a gross simplification to illustrate the effects.
In real life the various voltage drops are at different phase angles and the calculations are a little more complex than simple addition or subtraction of voltage drops.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[ScottyUK]

trchambe,

You are mistaken - perhaps this will help:

http://tkne.net/downloads/power/transmissionlines1/transmission lines/_Ferranti Effect.pdf

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[trchambe]

Thank you, I was not aware of that. I had alwas assumed it was an issue related to power factor.

I'm not certain I fully understand it, but will run through those calculations to see if it makes more sense.

I apologize for my error.

 

[ScottyUK]

No need for any apology: the whole purpose of this site is for us help each other and learn from each other!


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If we learn from our mistakes I'm getting a great education!

 

[trchambe]

I looked over the sheet that ScottyUK provided, but I'm having some difficulty in grasping the concept.

My thought is that the current of the resonant harmonic (primarily) charges the capacitance, and all voltage losses are through the conductor resistance. As the current discharges, the voltage at the end of the line is highest, as it is now the souce and hasn't yet dropped across the line resistance back to the generator.

Am I on the right track or do I have a bit more to think about?

 

[electricpete]

I think it all applies to fundamental... nothing to do with resonance.

The tricky part to understand is why is a long transmission line modeled as series inductor with shunt capacitance on each end (pi circuit). I don't have any handy explanation for that model.... but once you accept that model it's easy to see why the voltage at the open end is higher (the vars injected by the cap have only one direction to flow....toward the system end and this has to create higher voltage at the open end than the system end.



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