Power Factor & VD

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Ken9876

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Does a very poor power factor have an effect on voltage drop? If so how much influence does it play and how would one calculate for voltage drop with a poor power factor.
 
Has anyone else noticed repeat topics, such as "Ghost voltages w/ high-Z meters" and this "voltage drop formula" are often duplicated when not seen on the front page, 27 lines apart this time.

We all know recent upgrades at mikeholt.com destroyed some key-word searches & links to older posts, but these duplicates are getting rediculous.

Shouldn't someone bring this broke-search website issue to the webmaster, or perhaps suggest a more intuitive way to show status of popular topics, no longer visible on the forum's front page?
 
Ken9876 said:
Does a very poor power factor have an effect on voltage drop? If so how much influence does it play and how would one calculate for voltage drop with a poor power factor.
Click to expand...

VD = IRCos(theta) + IXsin(theta)

I = amps
R = conductor resistance table 9
X = conductor reactance table 9
Cos(theta) = power factor
Caculate Sin(theta)
 
Ken9876 said:
Does a very poor power factor have an effect on voltage drop? If so how much influence does it play and how would one calculate for voltage drop with a poor power factor.
Click to expand...

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.
 
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.
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...
Rattus
Your calc I = P/(VxPF) Power = IVxPF is in watts. VD uses VA in the caculation not watts. VD = IRCos(theta) + IXsin(theta).
** Sample Caculations using R = 2 and X = 0.1 amps = 10 ******************************************************************** ************* Cos *** Sin *** R *** X *** amps *** VD
0.9 *** 0.017 **2 ***0.1 *** 10 *** 18.017
0.8 *** 0.600 **2 ***0.1 *** 10 ****16.6
0.7 *** 0.700 **2 ***0.1 *** 10 *** 14.7
0.6 *** 0.800 **2 ***0.1 ***10 *** 12.8
0.5 *** 0.870 **2 ***0.1 ***10 *** 10.87

I can not get the spacing correct in these type tables. HELP
 
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Huh?

Huh?

bob said:
Rattus
Your calc I = P/(VxPF) Power = IVxPR is in watts. VD uses VA in the caculation not watts. VD = IRCos(theta) + IXsin(theta).
**** Sample Caculations using R = 2 and X = 0.1 amps = 10 ************* Cos *** Sin *** R *** X *** amps *** VD
0.9 *** 0.017 **2 ***0.1 *** 10 *** 18.017
0.8 *** 0.600 **2 ***0.1 *** 10 ****16.6
0.7 *** 0.700 **2 ***0.1 *** 10 *** 14.7
0.6 *** 0.800 **2 ***0.1 ***10 *** 12.8
0.5 *** 0.870 **2 ***0.1 ***10 *** 10.87

I can not get the spacing correct in these type tables. HELP
Click to expand...

Of course it is in watts.

Preal = IxVxPF (watts)

I can't make any sense of your equation.
 
rattus said:
Of course it is in watts.
Preal = IxVxPF (watts)
I can't make any sense of your equation.
Click to expand...

Voltage drop cacuations use the load amps regardless of the power factor
to caculate VD. For an explanation of the equasion
VD = IRCos(theta) + IXsin(theta)
Check the IEEE Red Book "Electric Power Distribution". They explain it better that I can.
 
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Bob,

Yes your right, but I thought the discrepancy was over the VA versus the watts. That's why I showed the second equation demonstrating the substitution for "I", orignally posted by Rattus.
 
I take issue!

I take issue!

I finally doped out the equations, and there are two phase angles to consider. One (theta) determines the PF of the load; the other (phi) is the phase angle of the line impedance.

The equation should read,

VD = (Preal/(Vlnxcos(theta)))*(R*cos(phi) + X*sin(phi))
 
kingpb said:
Bob,

Yes your right, but I thought the discrepancy was over the VA versus the watts. That's why I showed the second equation demonstrating the substitution for "I", orignally posted by Rattus.
Click to expand...

I may have confused you. The current used in the equasion I posted is derived from I = VA/V = amps( ?√ where required). It is the actual current that the load requires. You are correct. The VA is not in the equasion. If you were to use watts to derive amps it would (VA x PF)/V which would give you a smaller value of amps and a smaller voltage drop.
Many people use this equasion cm = kx1.72xIxD/vdcm = kx1.72xIxD/vd
which give a close value for VD. However it does not account for PF which is what the OP was asking about..
 
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rattus said:
I finally doped out the equations, and there are two phase angles to consider. One (theta) determines the PF of the load; the other (phi) is the phase angle of the line impedance.

The equation should read,

VD = (Preal/(Vlnxcos(theta)))*(R*cos(phi) + X*sin(phi))
Click to expand...

(Preal/(Vlnxcos(theta))). I think this is = watts/(V x PF) = VA/V = I.
Which looks very much like VD = IRCos(theta) + IXsin(theta).
However theta is the PF angle not line impedance angle.
 
bob said:
(Preal/(Vlnxcos(theta))). I think this is = watts/(V x PF) = VA/V = I.
Which looks very much like VD = IRCos(theta) + IXsin(theta).
However theta is the PF angle not line impedance angle.
Click to expand...

Both angles must be included in the equation which is written in such a way that complex numbers are not required for its evaluation. You cannot assume that the line impedance angle and the PF angle are the same!

It is far simpler to write,

VD = Zline*Iline = Zline*Preal/(Vline*PF)

This simple equation answers the original question which in effect asks what is the effect of PF on voltage drop for constant real power?
 
This simple equation answers the original question...
Click to expand...

It does?

I get the impression, and I may be way off base, that the original question is regarding voltage drop on circuit conductors when loads have a low power factor. Wouldn't the circuit conductors be evaluated as components of the total load upon the source, then VD_load subtracted from V_source to determine VD_conductors?

Just off the top of my head, without performing any calculations to confirm, I'd say the votage drop across the circuit conductors due soley to the power factor of a load thereon would be negligible.
 
Yes it does answer the question.

Yes it does answer the question.

Smart $ said:
It does?

I get the impression, and I may be way off base, that the original question is regarding voltage drop on circuit conductors when loads have a low power factor. Wouldn't the circuit conductors be evaluated as components of the total load upon the source, then VD_load subtracted from V_source to determine VD_conductors?

Just off the top of my head, without performing any calculations to confirm, I'd say the votage drop across the circuit conductors due soley to the power factor of a load thereon would be negligible.
Click to expand...

The equation is a bit of an approximation, but there is no need for an exact solution here. The equation illustrates the fact that the load current, for constant real power, is inversely proportional to the power factor if we assume that Vline is constant.

And the effect is not neglible unless the power factor is quite high.
 
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bob said:
I can not get the spacing correct in these type tables. HELP
Click to expand...
Just wrap CODE tags around the selected text or table, or highlight and click the "#" button on the webpage editor.

Anyone care to splain how we differentiate circuit balancing with these formula, since voltage drop varies substantially between such circuits?
 
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ramsy said:
Anyone care to splain how we differentiate circuit balancing with these formula, since voltage drop varies substantially between such circuits?
Click to expand...

The current drawn by the load will not change based on power factor of the load since the rated current is based on KVA. (using KVA, or Watts and pf)

Since voltage does have an effect on current (as seen by calculation), a higher voltage drop will certainly change the current. But by limiting the voltage drop to the recommended 2or 3% would also require an increase in the size of conductor.

Increasing conductor size thusly reduces the effective impedance of the cable. The outcome would be to maintain nearly the same current draw and therefore, have negligible impact on balancing.
 
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.
 
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