+ Post New Thread
Results 1 to 9 of 9
 10th January 2006, 06:20 #1
 Join Date
 Nov 2005
 Posts
 440
 Helped
 4 / 4
 Points
 3,853
 Level
 14
negative power factor
Could someone show me the relationship, if any, withing the following terms:
Lagging/Leading Power Factor (PF)
Lagging/Leading Reactive Power (Q)
Positive/Negative Reactive Power (Q)
Can I say that Lagging PF = Lagging Q = Positive Q?
Thanks.
 10th January 2006, 07:12 #2
 Join Date
 Aug 2005
 Location
 iran(ahwaz)
 Posts
 191
 Helped
 23 / 23
 Points
 3,066
 Level
 13
negative reactive power
hi
we have a traveling power (reactive) that it is saved in magnetic and electric fields .for a half of priod it saved in mag. field (L)and for next half priod is saved in elec. field (C). for a inductanc load (L) ,current is lagging from voltage .so we say a inductive load has lagging reactive power.for a Cap. load we say it has leading reactive power.
Negative Reactive Power (Q) used for generation of Reactive Power and positve for consumption.
http://Can I say that Lagging PF = L... = Positive Q?
Do you mean PF=Q ??
 10th January 2006, 07:12
 10th January 2006, 08:30 #3
 Join Date
 Nov 2005
 Posts
 440
 Helped
 4 / 4
 Points
 3,853
 Level
 14
leading pf
Originally Posted by smxx
I'm confused. Any problem with the following statements:
For a load of capacitor VAr is produced and considered negative or lagging as in Figure 3.10.
A capacitive load on the other hand will be represented by RjX as shown in Figure 3.12 and hence produces lagging VArs.
Or, pls check my another thread here:
http://www.edaboard.com/viewtopic.php?t=148522
Please give me your advices.
 10th January 2006, 08:30
 11th January 2006, 03:12 #4
 Join Date
 Nov 2004
 Location
 USA
 Posts
 211
 Helped
 44 / 44
 Points
 4,144
 Level
 15
negative vars
hi
Here is the understanding I have:
Lagging power factor is where the current lags the voltage (current peak occurs after voltage peak). This occurs if the load is inductive. In this case, the load consumes VArs. The question is now, is this positive VArs or negative VArs. Regardless, you must supply VArs. Most people call this positive VArs or positive Q.
Therefore:
Lagging power factor <=> load consumes VArs <=> positive Q <=> inductive load
(I'm using <=> to denote an implication and an inverse implication. Basically, this is an equality)
Leading power factor is where the current leads the voltage (current peak occurs before voltage peak). This occurs if the load is capacitive. In this case, the load supplies VArs. Most people call this negative VArs or negative Q.
Therefore:
Leading power factor <=> load supplies VArs <=> negative Q <=> capacitive load
Remember, that leading and lagging is always measured using the voltage (or real power, which is in phase of the voltage). Therefore, negative Q is lagging the voltage and positive Q is leading the voltage. Using the convention, you should rewrite your implication to be:
Lagging PF = Leading Q = Positive Q
Also,
Leading PF = Lagging Q = Negative Q
 11th January 2006, 03:12
 11th January 2006, 22:11 #5
 Join Date
 Nov 2004
 Location
 USA
 Posts
 211
 Helped
 44 / 44
 Points
 4,144
 Level
 15
lagging vars
Actually, thinking about this some more, when
Lagging PF then the VArs are lagging the P. Therefore, your statement is correct, but your text appears to be using the convention I mentioned in the post above. The book is backwards and bizarre, it confused me too. Sorry, but
Lagging PF = Lagging Q = Positive Q.
Everything else I said above is correct. I never really think in terms of leading of lagging Q, but rather leading and lagging current and the consuming or supplying of VArs.
 12th January 2006, 06:37 #6
 Join Date
 Nov 2005
 Posts
 440
 Helped
 4 / 4
 Points
 3,853
 Level
 14
pf lagging
I agree with u. The author of the text used kind of convention that confuses reader. Thanks for your explanation.
Originally Posted by jonw0224
 12th January 2006, 07:01 #7
 Join Date
 Aug 2005
 Location
 iran(ahwaz)
 Posts
 191
 Helped
 23 / 23
 Points
 3,066
 Level
 13
pf leading lagging
hi all
PF is a number between 1 and 1
and reactive power Q may have any value.
I don't understand this
Lagging PF = Lagging Q = Positive Q.
pls tell me.
for fully sine wave we have
V=Vm cos(Θ) where Θ=ωt
I=Im cos(Θ+Φ)
S=V*I=Vm *Im cos(Θ)* cos(Θ+Φ)=½ Vm *Im [cos(2Θ+Φ)+cos(Φ)]
S=[Vrms*Irms *cos(Φ)]+[Vrms*Irms*cos(2Θ+Φ)]=P+Q
Q=Vrms*Irms*cos(2Θ+Φ)
so freq of reactive power is 2Θ
Q=Vrms*Irms*cos(2Θ+Φ)=Vrms*Irms*[cos(2Θ)cos(Φ)sin(2Θ)sin(Φ)]
for a period mean values for term cos(2Θ)cos(Φ) is 0 and for sin(2Θ)sin(Φ) is sin(Φ)
so the mean value for Q is :Qm=Vrms*Irms*sin(Φ)
this value is the Amplitude of a sine wave only.
for dynamic PFC we must design a filter to measuering real time reactive power .
thanks all
 12th January 2006, 17:20 #8
 Join Date
 Nov 2005
 Posts
 440
 Helped
 4 / 4
 Points
 3,853
 Level
 14
pf leading
Hello,
I thought cos(θ) = cos(θ) ≡ "positive value". So, PF should be a number between 0 ~ 1. Pls correct me if I am wrong.
Sorry for my post that confusing u. Actually I tried to ask as follows:
Lagging PF ≡ Lagging Q ≡ Positive Q
... meaning if a device gives "lagging PF", can we say that the Q associated to the device is "lagging Q" and "positive Q"?
Thanks.
Originally Posted by smxx
 12th January 2006, 18:13 #9
 Join Date
 Nov 2004
 Location
 USA
 Posts
 211
 Helped
 44 / 44
 Points
 4,144
 Level
 15
positive vars
Yes, pf is never negative. But you designate a "leading" power factor or a "lagging" power factor.
Maybe some people use a negative sign to denote leading?
We've been using equal signs for a "reversable implication". Basically, we've been saying A = B to mean that
A implies B and B implies A
It's a logical equality not a numeric equality. Sorry for the confusion.
+ Post New Thread
Please login
LinkBacks (?)

08 Lagging Pf at Askives
Refback This thread14th December 2013, 15:56