# Help explain Lagging/Leading PF



## applepieordie

Hi,

I've always used this rule of thumb:


lagging generator (over-excited): exporting Watt and exporting VAR

leading generator (under-excited): exporting Watt and importing VAR

lagging motor (under-excited): importing Watt and importing VAR

leading motor (over-excited): importing Watt and exporting VAR

so along this line of thinking, lagging generator θ is positive, leading generator θ is negative, lagging motor θ is negative, leading motor θ is positive. where θ is the impedance phase angle in the "power triangle."

Is this correct?

I'm getting confused with the phase difference between voltage and current (see attached photos from graffeo book). in his lagging generator explanation, the θ is negative. is the θ between voltage and current different than the θ in the power triangle? I thought these were the same. Or is the rule of thumb i've been using wrong from the start?

can someone please shed some light here for me. thank you.

View attachment 9841


View attachment 9842


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## BigWheel

It might help to think of "leading" vs. "lagging" in terms of current and voltage instead of positive and negative. Power factor is generally used in calculations as an absolute value, so the sign of the PF is irrelevant.

A leading power factor means the current is "leading" the voltage (i.e., current crosses the zero-axis first) while a lagging power factor means that the current is "lagging" behind the voltage (i.e., current crosses the zero axis second).


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## rg1

applepieordie said:


> Hi,
> 
> I've always used this rule of thumb:
> 
> 
> lagging generator (over-excited): exporting Watt and exporting VAR
> 
> leading generator (under-excited): exporting Watt and importing VAR
> 
> lagging motor (under-excited): importing Watt and importing VAR
> 
> leading motor (over-excited): importing Watt and exporting VAR
> 
> so along this line of thinking, lagging generator θ is positive, leading generator θ is negative, lagging motor θ is negative, leading motor θ is positive. where θ is the impedance phase angle in the "power triangle."
> 
> Is this correct?
> 
> I'm getting confused with the phase difference between voltage and current (see attached photos from graffeo book). in his lagging generator explanation, the θ is negative. is the θ between voltage and current different than the θ in the power triangle? I thought these were the same. Or is the rule of thumb i've been using wrong from the start?
> 
> can someone please shed some light here for me. thank you.
> 
> View attachment 9841
> 
> 
> View attachment 9842


Your thumb rule is correct. PF angle is same as power triangle when we have one voltage across the load. These two concepts and the note below should not be mixed till one is thorough in the all the concepts. So it will be a good idea to understand them separately and thoroughly.

Note -This concept is only for the machines connected to Grid (Vt is constant) else for stand alone cases the scenario will be different.

Let me give a try for Gen?Motor? connected to grid. In generator and motor case the pf angle theta, if not specified otherwise, is the angle between terminal voltage, repeat terminal voltage, repeat terminal voltage and the current. There is no change in the notation of this angle (negative when behind the voltage and positive when ahead)  leading implies ahead of Vt and lagging means behind Vt. Now build your phasor diagram from here; E=Vt+IX  for generator and  E=Vt-IX for motor. If E obtained like this is more than Vt, the machine (whether G or M) is over excited and if less it is under excited. 

Let us come to whether VARs are consumed or supplied.  For this, one has to compare reactive power with real power and decide. Say a generator delivering a power to a lagging pf load. By Vt x I* method, we see that both P and jQ are positive ,so both, real and reactive powers are supplied by Generator and consumed by load and from phasor we get Gen is over excited. This makes sense as the excitation current in the generator is responsible for generation of reactive power and turbine input for generation of active power.

Similarly for leading pf, Vt x I* will give us reactive power as negative and real power positive means load is consuming and hence Gen is supplying the real power and load is supplying and Gen is consuming the reactive power. From phasor Gen is under excited. This also makes sense when we say Gen is under excited the E will be less than Vt, at the same time the flux required to produce power will be less (due to less excitation), so some one from out side has to strengthen the Gen flux else terminal voltage will drop which is not possible because the thing is connected to grid.

For motor, as it is a load, the explanation is a bit simple. When P and Q are both positive, it is consuming both, real and reactive power and again from phaser we will see that it is under excited and I is lagging. When the motor is over excited, it will have I as leading resulting into negative real power implying it is consuming real power and supplying reactive power.

One has to go on thinking about it till one gets it, because the stuff involves mixing of many concepts. If it has not helped, you can frame your question  pin pointed at one point (instead of understanding the whole lot of concepts in one go) so that we can discuss that part in detail.


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## rg1

applepieordie said:


> Hi,
> 
> I've always used this rule of thumb:
> 
> 
> lagging generator (over-excited): exporting Watt and exporting VAR
> 
> leading generator (under-excited): exporting Watt and importing VAR
> 
> lagging motor (under-excited): importing Watt and importing VAR
> 
> leading motor (over-excited): importing Watt and exporting VAR
> 
> The above is correct
> 
> so along this line of thinking, lagging generator θ is positive, leading generator θ is negative, lagging motor θ is negative, leading motor θ is positive. where θ is the impedance phase angle in the "power triangle."
> 
> The pf angle has only one notation and the reference in all these cases if not specified otherwise is Terminal voltage of the machine ( G or M)
> 
> Is this correct?
> 
> I'm getting confused with the phase difference between voltage and current (see attached photos from graffeo book). in his lagging generator explanation, the θ is negative. is the θ between voltage and current different than the θ in the power triangle? I thought these were the same. Or is the rule of thumb i've been using wrong from the start?
> 
> Draw voltage phasors yourself without seeing any book, by taking reference as terminal voltage, you may reach the same conclusion with better understanding. There are only three voltages, E (internal generated Voltage), Vt ( terminal Voltage) and IX drop in armature which always leads the current through it by 90 degrees. For example for lagging case of Gen
> 
> 1. Draw Vt on X axis
> 
> 2. Draw current I, lagging Vt by Theta angle or say at -theta degrees
> 
> 3 Draw IX, leading I by 90 degrees (Reason is, in an inductor Voltage drop across it, always leads its current)
> 
> 4 Add Vt and IX you get E which is more than Vt.
> 
> Draw 6 Phasors - 3 for generator and 3 for motor ( unity, lag and lead each), you may get a very good understanding. If it helps, good, else we can discuss it further.
> 
> can someone please shed some light here for me. thank you.
> 
> View attachment 9841
> 
> 
> View attachment 9842


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## TNPE

The diagrams from Graffeo are merely graphically illustrating, in a counterclockwise and clockwise direction,  depending on the PF, what is meant by leading/lagging PF.  

There are a few things to keep up with, and you can think of it or remember it in many ways, but PF is with respect to CURRENT (I) leading or lagging VOLTAGE (V).  ELI the ICE man is a moniker most commonly used to determine leading/lagging PF (ELI, E before I in an inductive circuit (hence, I lags); ICE, I before E in a capacitive circuit (hence, I leads)).

Thus, PF=cos(theta), where theta=Vangle - Iangle, or as you stated, the impedance angle.  A lagging PF entails a given circuit consuming VARS (motors, XFMRS etc.), conversely, a  leading PF entails VARS being generated (think capacitors).  In terms of the power/impedance triangle, a lagging PF has a positive angle (measured between the positive x-axis and moving counterclockwise), whereas, a leading PF has a negative angle (measured between the positive x-axis and moving clockwise).  

Glad to expand further if needed!


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## Zach Stone P.E.

applepieordie said:


> Hi,
> 
> I've always used this rule of thumb:
> 
> 
> lagging generator (over-excited): exporting Watt and exporting VAR
> 
> leading generator (under-excited): exporting Watt and importing VAR
> 
> lagging motor (under-excited): importing Watt and importing VAR
> 
> leading motor (over-excited): importing Watt and exporting VAR
> 
> so along this line of thinking, lagging generator θ is positive, leading generator θ is negative, lagging motor θ is negative, leading motor θ is positive. where θ is the impedance phase angle in the "power triangle."
> 
> Is this correct?
> 
> I'm getting confused with the phase difference between voltage and current (see attached photos from graffeo book). in his lagging generator explanation, the θ is negative. is the θ between voltage and current different than the θ in the power triangle? I thought these were the same. Or is the rule of thumb i've been using wrong from the start?
> 
> can someone please shed some light here for me. thank you.
> 
> View attachment 9841
> 
> 
> View attachment 9842


Hi @applepieordie

Take a look at this quick sketch I made, I think it might help you:




Remember that when a power quantity such as P or Q are positive, it is being absorbed (like a load).

When P or Q negative positive, it is being supplied (like a source).

For example, a leading synchronous load (capacitive) will supply reactive power (Q, Vars) while consuming real power, P (watts).  

We start at the origin and draw a positive horizontal component for P to the right, and then draw our negative vertical component for Q downward. 

Make sense?

For more information there is a great article and cheat sheet you can print out on leading and lagging on our website here:

Electrical PE Review - Leading and Lagging


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## applepieordie

Electrical PE Review said:


> Hi @applepieordie
> 
> Take a look at this quick sketch I made, I think it might help you:
> 
> View attachment 9855
> 
> 
> Remember that when a power quantity such as P or Q are positive, it is being absorbed (like a load).
> 
> When P or Q negative positive, it is being supplied (like a source).
> 
> For example, a leading synchronous load (capacitive) will supply reactive power (Q, Vars) while consuming real power, P (watts).
> 
> We start at the origin and draw a positive horizontal component for P to the right, and then draw our negative vertical component for Q downward.
> 
> Make sense?
> 
> For more information there is a great article and cheat sheet you can print out on leading and lagging on our website here:
> 
> Electrical PE Review - Leading and Lagging


Yes, this makes sense...though I'm more used treating positive P or Q as generating/outputting power rather than as loads because I work for a utility company. But the concept is the same but reversed.

My confusion was addressed when I took a look at the cheat sheet on your website, particularly rule #2 “The phase current angle and apparent power angle will *always* be opposite in polarity (when one is negative, the other is positive and vice versa)”

I was confused why the theta in Graffeo's solution was negative while I thought theta should be positive as I mentioned in the rule of thumb I follow.  It is because the theta in the solution is the phase current angle while I was thinking of the apparent power angle. S = 3(Vp)(Ip*) ... I finally know why the complex conjugate is there lol.

Thank you for the great resource. And thank you to everyone for the replies. I now have a more profound understanding of leading and lagging.


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## TNPE

@applepieordie

Just for clarity and standardized convention, S=VI* is not apparent power, rather, it is complex power (also P+jQ).  Yes, I suppose it could be defined as such, but from academic, theoretical doctrine, there is a difference in "definition."  Apparent power is defined as VI, regardless of phase relationship between V and I (think DC circuits).  The conjugate stems directly from the definition I gave of theta in my previous post.  Lagging current yields a positive angle on the power/impedance triangle (e.g. this vector lands in the first quadrant of a Cartesian coordinate system..positive-x and positive-y..unless if unity, then it's on the x-axis and on the y-axis if entirely reactive).  Likewise, a leading current yields a negative angle which lands in the fourth quadrant of a Cartesian coordinate system (positive-x and negative-y).  Some of these topics are highly misleading when investigated superficially.  I think anyone who has completed an EE curriculum in power can attest to how cumbersome these topics can be, at least until the dots are connected and the fundamental concepts are broken down to minuscule components and built up.  Glad you're understanding is progressing.  Keep the questions coming.  Though I have passed the PE, this board helps keep me fresh on topics/concepts, and I enjoy helping out where I can.


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## Zach Stone P.E.

applepieordie said:


> Yes, this makes sense...though I'm more used treating positive P or Q as generating/outputting power rather than as loads because I work for a utility company. But the concept is the same but reversed.
> 
> My confusion was addressed when I took a look at the cheat sheet on your website, particularly rule #2 “The phase current angle and apparent power angle will *always* be opposite in polarity (when one is negative, the other is positive and vice versa)”
> 
> I was confused why the theta in Graffeo's solution was negative while I thought theta should be positive as I mentioned in the rule of thumb I follow.  It is because the theta in the solution is the phase current angle while I was thinking of the apparent power angle. S = 3(Vp)(Ip*) ... I finally know why the complex conjugate is there lol.
> 
> Thank you for the great resource. And thank you to everyone for the replies. I now have a more profound understanding of leading and lagging.




Glad it helped. 

Feel free to print the article and include it in your study binder.


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## applepieordie

TNPE said:


> @applepieordie
> 
> Just for clarity and standardized convention, S=VI* is not apparent power, rather, it is complex power (also P+jQ).  Yes, I suppose it could be defined as such, but from academic, theoretical doctrine, there is a difference in "definition."  Apparent power is defined as VI, regardless of phase relationship between V and I (think DC circuits).  The conjugate stems directly from the definition I gave of theta in my previous post.  Lagging current yields a positive angle on the power/impedance triangle (e.g. this vector lands in the first quadrant of a Cartesian coordinate system..positive-x and positive-y..unless if unity, then it's on the x-axis and on the y-axis if entirely reactive).  Likewise, a leading current yields a negative angle which lands in the fourth quadrant of a Cartesian coordinate system (positive-x and negative-y).  Some of these topics are highly misleading when investigated superficially.  I think anyone who has completed an EE curriculum in power can attest to how cumbersome these topics can be, at least until the dots are connected and the fundamental concepts are broken down to minuscule components and built up.  Glad you're understanding is progressing.  Keep the questions coming.  Though I have passed the PE, this board helps keep me fresh on topics/concepts, and I enjoy helping out where I can.


Thanks for the clarification. I didn't take any power courses in college and have been learning power through books and online material so I really appreciate the effort other forum members put in to answering my questions and explaining the fundamentals.


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## Zach Stone P.E.

applepieordie said:


> Thanks for the clarification. I didn't take any power courses in college and have been learning power through books and online material so I really appreciate the effort other forum members put in to answering my questions and explaining the fundamentals.


No problem.

This forum was a great resource when I was studying for the PE exam so I enjoy paying it forward and helping the community.


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