Power Factor & VD

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rattus said:
Now, what do you say to my claim that we have two angles to consider?
First, we have the PF angle of the load itself. That is, arccos(PF)
Second, we have the phase angle of the line itself. That is, arctan(X/R)
I contend that you must use both angles in your formula. If you use only the PF, it falls out of the equation and therefore has no effect on VD.
Click to expand...

I say that you return to your text books and check again. This formula
"VD = IRCos(theta) + IXsin(theta)", is in every engineering text that I have
and it is used for VD caculations. The PF does not drop out of the equasion.
Why don't you take the time to check out the IEEE Red book and read their explanation of this caculation. I will say that in real applications that X is small
for short lengths of wire. You can get an accurate result using the actual given line current for the load, VA/V, x the wire resistance.
 
Rattus,

Your contention that there are two angles to contend with is correct, in a purely analytical sense. Although I cannot confirm that your formula you are proposing is correct, the fact that some miniscule amount of VD does occur based on the R and X power factor, is real.

However, the power factor between R and X is so small, for example, in the case of a #12AWG, the pf = 0.034, that the effect on the VD from Line end to Load end is inconsequential. The VD calculation is then reduced, or best approximated by using the effective Z as determined by using the load power factor.
 
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Don't have that book.

Don't have that book.

bob said:
VD = IRCos(theta) + IXsin(theta)

I = amps
R = conductor resistance table 9
X = conductor reactance table 9
Cos(theta) = power factor
Caculate Sin(theta)
Click to expand...

Bob,

I don't have a Red Book, and you admit you can't explain the equation. Either the book is wrong, or you are misinterpreting it. Why don't you provide an excerpt from it.

"theta" is the angle of the line impedance. The sum of the sine and cosine terms is the magnitude of the VD.

Your interpretation says that VD is proportional to PF, but common sense tells us that is wrong. Fact is that "I" is not a constant; it is inversely proportional to PF for constant real power. Then VD is inversely proportional to PF. Plain and simple.
 
rattus said:
Bob,

I don't have a Red Book, and you admit you can't explain the equation. Either the book is wrong, or you are misinterpreting it. Why don't you provide an excerpt from it.

"theta" is the angle of the line impedance. The sum of the sine and cosine terms is the magnitude of the VD.

Your interpretation says that VD is proportional to PF, but common sense tells us that is wrong. Fact is that "I" is not a constant; it is inversely proportional to PF for constant real power. Then VD is inversely proportional to PF. Plain and simple.
Click to expand...

Rattus
I tried to scan the pages in the IEEE book but the file was too big to upload.
Check this site. It is an example of the use of this method for caculating VD.

http://www.skm.com/support/kb/120000.pdf#search="ieee red book voltage drop formula".
I will try to find others.
 
rattus said:
Simply put, the current drawn by a load is inversely proprotional to the power factor:

I = P/(VxPF)

Where "P" is the real power, "V" is the applied voltage and "PF" is the power factor.

Then,

VD = IxZ = PxZ/(VxPF)

where "Z" is the impedance of the line.

For example, a PF of 0.8 would increase the voltage drop by a factor of 1.25.
Click to expand...

The current is not exactly inversely proportional to the PF. If the applied voltage were directly across the load, then you would be correct (but there would also be no voltage drop to calculate).

We have a constant voltage applied to one end of a transmission line, and a load on the other end of the line. As we vary PF of the load, it affects the voltage at the load. But you're equation I = P/(VxPF) would require a constant voltage for changing PF's.

I think that explains why you are getting a different equation.

Steve
 
rattus said:
King, this is far too complicated for the original question. Let's keep it simple.

Now, what do you say to my claim that we have two angles to consider?

First, we have the PF angle of the load itself. That is, arccos(PF)

Second, we have the phase angle of the line itself. That is, arctan(X/R)

I contend that you must use both angles in your formula. If you use only the PF, it falls out of the equation and therefore has no effect on VD.
Click to expand...

In most real world cases, the load impedence is much larger than the line impedence. So the equation Bob posted assumes the total PF is equal to the load PF. No need to include two different power factors.

So we ignore the PF of the line, but not the voltage drop across the line.

This is really a very simple equation. Remember the dot product?

Vd = I*Z = [I sin (theta) + j I cos (theta)]*[R+jX]

Remember which terms go to zero? From there you get two components that are 90 degrees out of phase. But instead of using A = sqrt(B^2 + C^2), we just approximate by using A = B+C. It's a conservative approximation, and pretty close when C<<B.

Steve
 
steve066 said:
In most real world cases, the load impedence is much larger than the line impedence. So the equation Bob posted assumes the total PF is equal to the load PF. No need to include two different power factors.

So we ignore the PF of the line, but not the voltage drop across the line.

This is really a very simple equation. Remember the dot product?

Vd = I*Z = [I sin (theta) + j I cos (theta)]*[R+jX]

Remember which terms go to zero? From there you get two components that are 90 degrees out of phase. But instead of using A = sqrt(B^2 + C^2), we just approximate by using A = B+C. It's a conservative approximation, and pretty close when C<<B.

Steve
Click to expand...

Steve, shouldn't it be Irms{cos(theta) + jsin(theta)}?

And, if we already know the current, PF has no effect. That is,

VD = I*(R + jX)

I don't get it.
 
kingpb said:
Rattus,

That site is talking about voltage regulation, not voltage drop, are those the same thing? :confused:





Just kidding :D :D :D

You guys take this stuff awful serious
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King,

Same, same! Voltage regulation and voltage drop are intertwined.

VR = VD/VS, no?
 
OK, I now see why we must consider PF of the load. One must consider phase differences when computing VD.

I made a mistake once! Thought I was wrong!
 
steve066 said:
From there you get two components that are 90 degrees out of phase. But instead of using A = sqrt(B^2 + C^2), we just approximate by using A = B+C. It's a conservative approximation, and pretty close when C<<B.

Steve
Click to expand...

Hmmm...the "edit" button seems to be gone. Now I have to quote myself to correct an error.

I forgot that the result of a dot product is a scalar, so the stuff quoted above is off on a wrong tangent.

Steve
 
rattus said:
OK already!

Bob's formula is an approximation which is valid only when the PF angle and impedance angle are approximately equal which makes it useless in my opinion.
Click to expand...
That not what Jim said.
"In most real world cases, the load impedence is much larger than the line impedence. So the equation Bob posted assumes the total PF is equal to the load PF. No need to include two different power factors."

Rattus
read the attached file. Fuzzy but best I could do.
 
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For real-world application, I derived the following formula:

View attachment 172

where:
  • VD is voltage drop (see note 1)
  • I is load current in amperes
  • L is length of one conductor in feet (source to load)
  • R is NEC Table 9 resistance for conductor in ohms per 1000 feet
  • X is NEC Table 9 reactance for conductor in ohms per 1000 feet
  • pf is the power factor of the load
Notes:
  1. Total source to load voltage drop is dependent upon circuit configuration. For three-phase loads, multiply VD by 1.732. For single phase loads, multiply VD by 2 (note such loads on a multiwire branch circuit may have a voltage drop less than 2 times VD when other loads on the circuit are conducting).

Would you guys be so kind as to check its validity?
 
Now I see it!

Now I see it!

It?s Like This:

At the start, I said that the formula made no sense to me. Now it does.

The load current phasor is,

I load = Irms[cos(theta) + jsin(theta)]

And, the line impedance is,

Zline = R + jX

The product of these two terms is the voltage drop.

VD = Irms{[Rcox(theta) ? Xsin(theta)]+ j[Rsin(theta) + Xcos(theta)]}

The real terms represent the components of VD in phase with the source voltage.

The imaginary terms are typically small and may be ignored if PF is high. In fact for a negative theta, they tend to cancel each other.

And, since theta is typically negative, sin(theta) is negative which makes this formula equivalent to the one in the IEEE Red Book.

Anyone care to double-check me on this?
 
Smart $ said:
For real-world application, I derived the following formula:

View attachment 172

where:
  • VD is voltage drop (see note 1)
  • I is load current in amperes
  • L is length of one conductor in feet (source to load)
  • R is NEC Table 9 resistance for conductor in ohms per 1000 feet
  • X is NEC Table 9 reactance for conductor in ohms per 1000 feet
  • pf is the power factor of the load
Notes:
  1. Total source to load voltage drop is dependent upon circuit configuration. For three-phase loads, multiply VD by 1.732. For single phase loads, multiply VD by 2 (note such loads on a multiwire branch circuit may have a voltage drop less than 2 times VD when other loads on the circuit are conducting).

Would you guys be so kind as to check its validity?
Click to expand...

Smart,

If you write this formula with sine and cosine, you have the approximation from the IEEE Red Book plus the length factor.
 
Smart $ said:
For real-world application, I derived the following formula:

View attachment 172

where:
  • VD is voltage drop (see note 1)
  • I is load current in amperes
  • L is length of one conductor in feet (source to load)
  • R is NEC Table 9 resistance for conductor in ohms per 1000 feet
  • X is NEC Table 9 reactance for conductor in ohms per 1000 feet
  • pf is the power factor of the load
Notes:
  1. Total source to load voltage drop is dependent upon circuit configuration. For three-phase loads, multiply VD by 1.732. For single phase loads, multiply VD by 2 (note such loads on a multiwire branch circuit may have a voltage drop less than 2 times VD when other loads on the circuit are conducting).

Would you guys be so kind as to check its validity?
Click to expand...
That is the same formula as VD = IRCos(theta) + IXsin(theta). You added
a divisor of 1000 to adjust the table values for R and X.
 
rattus said:
Smart,

If you write this formula with sine and cosine, you have the approximation from the IEEE Red Book plus the length factor.
Click to expand...
bob said:
That is the same formula as VD = IRCos(theta) + IXsin(theta). You added
a divisor of 1000 to adjust the table values for R and X.
Click to expand...
Sorry for the delayed response... working six tens at a power station outage. Coupled with commute and prep time leaves little play time :)

My formula was derived from the IEEE one... just that I put it in terms of pf and included the conductor length adjustment. After all, what good is a formula if you can't just plug in your numbers from the get-go.

However, I am a bit concerned with the replacement of sin(theta) with sqrt(1-pf^2) regarding sign of the result (i.e. positive or negative). With limited time I haven't had an oprortunity to evaluate the formula for potential problems, and I could not make out the vectorial diagram in the image Bob posted. Is there any chance you can post a more distinct (perhaps an enlarged....) image, Bob?
 
Smart $ said:
With limited time I haven't had an oprortunity to evaluate the formula for potential problems, and I could not make out the vectorial diagram in the image Bob posted. Is there any chance you can post a more distinct (perhaps an enlarged....) image, Bob?
Click to expand...


Smart
I tried but I am limited to about 39K for the file to upload. When I scanned a more detail drawing I was over the limit.
 
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bob said:
Smart
I tried but I am limited to about 39K for the file to upload. When I scanned a more detail drawing I was over the limit.
Click to expand...
Perhaps saving in GIF format will help. GIF has better compression than JPEG for images having fewer then 256 colors. The example below was extracted from your image, cropped to diagram, resolution doubled, colors reduced, and saved to GIF format...

[IMAGE REMOVED]

I can even make some of it out when enlarged, but the clarity of the original can only slightly be improved and not enough to make it legible.

If you still cannot manage to get it within the limit, email me the higher res' image and I'll see what I can do with it. I'll send you a PM with my email address...
 
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