Lagging Power Factor

[va_gator]

Given the power factor, we can find the power factor angle (@), with: @ = arccos(PF).

Then if we are given a current (I) and it is said to be lagging the voltage, we can write the current as I<-@ ('<' is the angle symbol, for polar representation). In some of the calculations in the sample exam, the '-' is applied to the angles for lagging power factors (I<-@), but not all the time. In other solutions, the '-' for the angles is omitted on lagging power factors. It appears on the motor problems, and I can’t seem to find a trend when to include and when to drop the negative for the angles for lagging PFs

This is one of the last things I’m struggling with, I may be overlooking a very simple concept, and would appreciate it if any of you can help me out. Thanks.

 

[Flyer_PE]

It all depends on if you are dealing with the angle of current with respect to voltage or the angle for apparent power.

Apparent power: S=VI*. A load with a lagging power factor will have a negative angle for current and a positive angle for apparent power.

 

[cableguy]

Lagging means the current lags the voltage. Therefore the angle is negative whenever you see the word lagging.

35 degrees lagging means -35 degrees.

 

[va_gator]

It all depends on if you are dealing with the angle of current with respect to voltage or the angle for apparent power.
Apparent power: S=VI*. A load with a lagging power factor will have a negative angle for current and a positive angle for apparent power.

That is probably the closest "trend" that I can use, when dealing with apparent power (S), even on a system with lagging PF, S is represented with a positive angle:

S<@

But when you are using the current, I to solve for example V=IR, if the PF is lagging, use I= I<-@.

But there are still situations when that rule is broken by the sample test solution (Problem 517 of the 2009 NCEES Sample). Where S and the negative angle representation for the PF is used to solve for Reactive Power.

 

[sam314159]

Y-System:

The power factor angle is the angle by which V_phase LEADS the corresponding I_line

Delta-System:

The power factor angle is the angle by which V_line LEADS the corresponding I_phase

Just in general though, the power factor angle is the angle by which voltage LEADS current.

Lagging Power Factor (Inductive Load)

I<-@ : Negative Angle (If you take your voltage as the zero reference, then I needs to be LAGGING or -@ behind the 0)

S<+@: Positive Angle (P+jQ) (A load taking in S<+@, or P+jQ, means it is taking in WATTS and VARS, both are same sign)

Z<+@: Positive Angle (R+jX) (X is inductive, X=jwL which is positive)

Leading Power Factor (Capacitive Load)

I<+@: Positive Angle (If you take your voltage as the zero reference, then I needs to be LEADING or +@ ahead of the 0)

S<-@ : Negative Angle (P-jQ) (A load taking in S<-@, or P-jQ, means it is taking in WATTS and putting out VARS, opposite signs)

Z<-@ : Negative Angle (R-jX) (X is capacitive, X=-j(1/wC) which is negative)

 

[Flyer_PE]

It all depends on if you are dealing with the angle of current with respect to voltage or the angle for apparent power.
Apparent power: S=VI*. A load with a lagging power factor will have a negative angle for current and a positive angle for apparent power.

That is probably the closest "trend" that I can use, when dealing with apparent power (S), even on a system with lagging PF, S is represented with a positive angle:

S<@

But when you are using the current, I to solve for example V=IR, if the PF is lagging, use I= I<-@.

But there are still situations when that rule is broken by the sample test solution (Problem 517 of the 2009 NCEES Sample). Where S and the negative angle representation for the PF is used to solve for Reactive Power.

They didn't break the rule in their solution to #517, they just didn't include all the details.

The synchronous motor has a leading power factor which gives a negative angle for apparent power (S).

What they didn't show for the induction motor is that I = 14.43/-53.13o. However, S=sqrt(3)VI* -> SInd Mot = sqrt3*480*14.43/+53.13o

 

[va_gator]

Thanks for the help, I somewhat understand it now, but more importantly, I can use the above as a guideline. Just a few minutes ago I found a thread with a related topic.

http://engineerboards.com/index.php?showtopic=11450

 

[Jonjo]

Thanks for the help, I somewhat understand it now, but more importantly, I can use the above as a guideline. Just a few minutes ago I found a thread with a related topic.
http://engineerboards.com/index.php?showtopic=11450

More over , if your problem is what to use? , find first is (+A°) Cap Circuit , or (- A°) Inductive Circuit , and than no problem because

Cos(+A°) = Cos(-A°) => , P = 1.7321 V.I Cos(+A°) = 1.7321 V.I Cos(-A°) is the Module of the Power assets (Watts)