Power factor and VA vs Watts

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Smart $ said:
Ummm... forgot to mention your "P" is not an accurate representation of P_real. You reduced your eq. 8 by P=Vm*Im cos(0v-0i)/2. The waveform of P_real is Vm*cos(wt)*Im*cos(0v-0i)
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If my thinking is correct, your "P" is actually [scalar] P_avg.
 
Hameedulla-Ekhlas said:
My only concern, p = p + pcos2wt should be ploted by consin not by sine. Because

1- According to Electric Circuits 7th Edition Nilsson-Riedel 2004 book it says.

We are operating in the sinusoidal steady state, so we may choose any convenient reference for zero time. Engineers concerned with operating systems involving the transfer of large blocks of power have found that a zero time corresponding to the instant when the current is passing through a positive maximum is convenient.

2- According to Transforms and Applications Handbook, Third Edition by Alexander D. Poularikas.page-51


p = cos(0i-0v) = P/S

"is called the power factor. The power factor may be regarded as a normalized correlation coefficient of the voltage and current signals while sin(0i - 0v) = SQR(1 - sqr(p)) may be called the anticorrelation of coefficient.

Now I need your help and I might be wrong but I want to clear my self thanks
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You're using too many p's. It appeared you first used "p" to denote apparent power. Now you're using "p" to denote power factor. Major confusion ensues... ;)
 
Cold Fusion said:
I don't think so. That is why I asked, the first "p" is the function p(t).

cf
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There was only one "p" before you rewrote it. So is it the original "p" that is p(t), or the original "P"?

More than one "p", even if changing cases, is still too many p's. Reminds me of my mid-night crises... he's peeing too much :)
 
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Hameedulla-Ekhlas said:
...yes, ofcourse you can incase you have pf = 1.
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I would have thought that it is implicit that you were discussing pf =1 for your equation. If you were not, then I would have expected your equation to have another term

For pf ≠ 1

p(t) = VI cosθ (1 + cos2ωt) + VI sinθ ( cos2ωt = π/2)

where θ is the V, I displacement

I'm just trying to follow along with your reasoning.

cf
 
Smart $ said:
If my thinking is correct, your "P" is actually [scalar] P_avg.
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For Vector and Scalar:

In vector notation, the effective power is expressed as the dot product of the vector voltage and the vector current. Effective power is a scalar quantity not a vector quantity, not phasor quantity.


for p:

p = Vm*Im cos(0v - 0i)/2 + Vm*Im*cos(0v-0i)cos2wt/2 - Vm*Im*sin(0v-0i)sin2wt/2

small p represent the instantaneous power which contain both real and reactive power

for pure resistive 0i=0v we have

P = P+Pcos2wt

CAPs lock P represents the real power which is one part of the instantaneous power.


what Gar was saying, he is completely right but I wanted to know make it clear that the graph for (P +Pcos2wt) should be based on cosin not sine because.

1- It is a real power
2- the phase angle between the voltage and current is cosine.
3- The power factor which means = 1
 
Hameedulla-Ekhlas said:
For Vector and Scalar:

In vector notation, the effective power is expressed as the dot product of the vector voltage and the vector current. Effective power is a scalar quantity not a vector quantity, not phasor quantity.


for p:

p = Vm*Im cos(0v - 0i)/2 + Vm*Im*cos(0v-0i)cos2wt/2 - Vm*Im*sin(0v-0i)sin2wt/2

small p represent the instantaneous power which contain both real and reactive power

for pure resistive 0i=0v we have

P = P+Pcos2wt

CAPs lock P represents the real power which is one part of the instantaneous power.


what Gar was saying, he is completely right but I wanted to know make it clear that the graph for (P +Pcos2wt) should be based on cosin not sine because.

1- It is a real power
2- the phase angle between the voltage and current is cosine.
3- The power factor which means = 1
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Still confusing. If you are working under the convention power is scalar, a graph of such is a horizontal line when graphed (y-axis value is constant as, x-axis varies).

Now you are saying P = P + Pcos2wt when φ=0? (or pf=1). This equation can only be true if cos2wt=0. We know t=0 only at the start instant and w≠0 for an AC load. So the equation fails except for a DC condition.

I don't know why you are going through all this. If your original "p" is the vector dot product:
v?i = |v||i|cos(φ)​
and where φ=0?
v?i = |v||i|​
The relationship with P (as in scalar P, otherwise known as Pavg) is:
P=v?i/2​
Proof:
|v|=Vrms*√2
|i|=Irms*√2
P=Vrms*Irms​
Therefore
P=|v|/√2*|i|/√2
P=|v||i|/2​
And where pf<1
P=|v||i|cos(φ)/2​
 
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I suppose I should note that |v| and |i| above are peak, not rms values.

If using rms values:
P=|V||I|cos(φ)​
 
Smart $ said:
Now you are saying P = P + Pcos2wt when φ=0? (or pf=1). This equation can only be true if cos2wt=0 -----

[/INDENT]
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Yeah, I have told you that and I am only only talking about real power.

If the circuit between the terminal is purely resistive, the voltage and current are in phase, wich means that 0i = 0v

p = P + Pcos2wt -----1

The instantaneous power expressed in Eq.1 is referred to as the instanteous real power. Note that this designation implies that average power(P) is also referred to as real power.

The average power associated with sinusoidal signals is the average of the instantaneous power over one period or equation form

Pavg = (1/T) ∫v(t) i(t) dt


Since there is no reactive power flow 0i = 0v that is why I mentioned pf becomes 1.


I am not talking about scalar and vector which is already explained.
 
Hameedulla-Ekhlas said:
Yeah, I have told you that and I am only only talking about real power.

If the circuit between the terminal is purely resistive, the voltage and current are in phase, wich means that 0i = 0v

p = P + Pcos2wt -----1

The instantaneous power expressed in Eq.1 is referred to as the instanteous real power. Note that this designation implies that average power(P) is also referred to as real power.

The average power associated with sinusoidal signals is the average of the instantaneous power over one period or equation form

Pavg = (1/T) ∫v(t) i(t) dt


Since there is no reactive power flow 0i = 0v that is why I mentioned pf becomes 1.


I am not talking about scalar and vector which is already explained.
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Now you are back to using "p" and "P"... you're going to have to keep these correct if you keep using two different p's in the same equation.

Anyway, I gave you the means to verify... apparently you didn't see it ;)

If you are wanting to graph p(t) based on the value of Pavg, we can start verification with p(t) = (1+cos(2wt))/2 where φ=0? and v and i vectors are referenced to 0? and their moduli (magnitudes) are peak values.

We have:
p(t) = |v||i|(1+cos(2wt))/2
P = |v||i|/2​
Verify:
p = P + Pcos(2wt)​
Substituting:
|v||i|(1+cos(2wt))/2 ?=? |v||i|/2 + |v||i|cos(2wt)/2
|v||i|(1+cos(2wt))/2 = |v||i|(1+cos(2wt))/2
​
 
Smart $ said:
Now you are back to using "p" and "P"... you're going to have to keep these correct if you keep using two different p's in the same equation.

p = P + Pcos2wt.


[/INDENT]
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Smart$:

p = P + Pcos2wt.
yes, there are two kinds p, If i have not written any were in previous post, it would be by typing mistake.

small p to the right site of equation represents the intantanous power in which the reactive power is zero and has only real part

Caps lock P represents real part which is one part of instantanous power.

If you have any doubt please refer to

Electric Circuits 7th Edition Nilsson-Riedel 2004 Page-431 you will find all infromation.
 
Hameedulla-Ekhlas said:
Smart$:

p = P + Pcos2wt.
yes, there are two kinds p, If i have not written any were in previous post, it would be by typing mistake.

small p to the right site of equation represents the intantanous power in which the reactive power is zero and has only real part

Caps lock P represents real part which is one part of instantanous power.

If you have any doubt please refer to

Electric Circuits 7th Edition Nilsson-Riedel 2004 Page-431 you will find all infromation.
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Not about to buy that book just to have a common basis for discussion... and I can even get it pretty cheap.

I don't have any doubts... until you are inconsistent with your "peeing" :D
 
Smart $ said:
Not about to buy that book just to have a common basis for discussion... and I can even get it pretty cheap.

I don't have any doubts... until you are inconsistent with your "peeing" :D
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Smart$:

If you need it, i will send you the free link and you can download for free its pdf. You dont need it to buy it.
 
Smart $ said:
Not about to buy that book just to have a common basis for discussion... and I can even get it pretty cheap.

I don't have any doubts... until you are inconsistent with your "peeing" :D
Click to expand...

I would like thanks you, gan, gunn, CF and others who took part in this dicussion. I really enjoyed it and you may have learn at least one word from me but I have learned sentences. So, I would like to finish this discussion here and it was really useful.
 
Hameedulla-Ekhlas said:
by the way, which program are you using for graph.
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I use a program called Canvas.

It has a feature which will plot basic 2D Cartesian [y=f(x)] and Polar [r=f(x)]... but it has some bugs and other limitations which I have to compensate for. For example, it plots all Cartesian waveforms centered on the x axis, even if they are offset from it. And it always plots centered on the view, so if I want to overlay plots, my view has to be centered on where the plot is to be made. Fortunately it remembers the last plot or else that would be very annoying.

I also used to do some graphs in TurboCAD, but the newer versions don't support vbscript (IIRC), and I have not converted over to the scripting methods it currently does support.
 
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